on the theory of fractional calculus in the pettis-function spaces · 2019. 7. 30. ·...

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Research Article On the Theory of Fractional Calculus in the Pettis-Function Spaces Hussein A. H. Salem Department of Mathematics and Computer Science, Faculty of Science, Alexandria University, Alexandria, Egypt Correspondence should be addressed to Hussein A. H. Salem; [email protected] Received 19 February 2018; Accepted 2 April 2018; Published 15 May 2018 Academic Editor: Gestur ´ Olafsson Copyright © 2018 Hussein A. H. Salem. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. roughout this paper, we outline some aspects of fractional calculus in Banach spaces. Some examples are demonstrated. In our investigations, the integrals and the derivatives are understood as Pettis integrals and the corresponding derivatives. Our results here extended all previous contributions in this context and therefore are new. To encompass the full scope of our paper, we show that a weakly continuous solution of a fractional order integral equation, which is modeled off some fractional order boundary value problem (where the derivatives are taken in the usual definition of the Caputo fractional weak derivative), may not solve the problem. 1. Introduction and Preliminaries e issue of fractional calculus for the functions that take values in Banach space where the integrals and the derivatives are understood as Pettis integrals and the corresponding derivatives has been studied for the first time by the authors of [1, 2]. Following the appearance of [1], there has been a sig- nificant interest in the study of this topic (see, e.g., [3–6]; see also [7–9]). is paper is devoted to presenting general results and examples for the existence of the fractional integral (and corresponding fractional differential) operators in arbitrary Banach space where it is endowed with its weak topology. In our investigations, we show that the well-known properties of the fractional calculus for functions taking values in finite dimensional spaces also hold in infinite dimensional spaces. Our results extend all previous contributions of the same type in the Bochner integrability setting and in the Pettis integrability one. For the readers convenience, here we present some notations and the main properties for the Pettis integrals. For further background, unexplained terminology and details pertaining to this paper can be found in Diestel et al. [10, 11] and Pettis [12]. roughout this paper, we consider the measure space (, Ω, ), where = [0,1], 0≤<<∞ denote a fixed interval of the real line, Ω denotes the Lebesgue -algebra L(), and stands for the Lebesgue measure. denotes a real Banach space with a norm ‖⋅‖ and is its dual. By we denotes the space when endowed with the weak topology generated by the continuous linear functionals on . We will let [, ] denote the Banach space of weakly continuous functions , with the topology of weak uniform convergence. And [, ] denotes the space of - valued Pettis integrable functions in the interval (see [10, 12] for the definition). Recall that (see, e.g., [10, 13–19]) the weakly measurable function : is said to be Dunford (or Gelfand) integrable on if and only if is Lebesgue integrable on for each . Definition 1. Let ∈ [1, ∞]. Define H () to be the class of all weakly measurable functions :→ having () for every . If =∞, the added condition l.u.b. ‖‖=1 (ess sup () )<∞ (1) must be satisfied by each H (). We also define the class H 0 () by H 0 () { ∈ [, ] : ∈ ()} . (2) Hindawi Journal of Function Spaces Volume 2018, Article ID 8746148, 13 pages https://doi.org/10.1155/2018/8746148

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Page 1: On the Theory of Fractional Calculus in the Pettis-Function Spaces · 2019. 7. 30. · JournalofFunctionSpaces Further,wedenethespace H (),˙≥0by H () fl ˛ ∈ ˚, ˜: isHolderainoforder

Research ArticleOn the Theory of Fractional Calculus inthe Pettis-Function Spaces

Hussein A H Salem

Department of Mathematics and Computer Science Faculty of Science Alexandria University Alexandria Egypt

Correspondence should be addressed to Hussein A H Salem hssdinaalex-sciedueg

Received 19 February 2018 Accepted 2 April 2018 Published 15 May 2018

Academic Editor Gestur Olafsson

Copyright copy 2018 Hussein A H SalemThis is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Throughout this paper we outline some aspects of fractional calculus in Banach spaces Some examples are demonstrated In ourinvestigations the integrals and the derivatives are understood as Pettis integrals and the corresponding derivatives Our resultshere extended all previous contributions in this context and therefore are new To encompass the full scope of our paper we showthat a weakly continuous solution of a fractional order integral equation which is modeled off some fractional order boundaryvalue problem (where the derivatives are taken in the usual definition of the Caputo fractional weak derivative) may not solve theproblem

1 Introduction and Preliminaries

The issue of fractional calculus for the functions that takevalues in Banach spacewhere the integrals and the derivativesare understood as Pettis integrals and the correspondingderivatives has been studied for the first time by the authorsof [1 2] Following the appearance of [1] there has been a sig-nificant interest in the study of this topic (see eg [3ndash6] seealso [7ndash9])This paper is devoted to presenting general resultsand examples for the existence of the fractional integral (andcorresponding fractional differential) operators in arbitraryBanach space where it is endowed with its weak topology Inour investigations we show that the well-known propertiesof the fractional calculus for functions taking values in finitedimensional spaces also hold in infinite dimensional spacesOur results extend all previous contributions of the sametype in the Bochner integrability setting and in the Pettisintegrability one

For the readers convenience here we present somenotations and the main properties for the Pettis integrals Forfurther background unexplained terminology and detailspertaining to this paper can be found in Diestel et al [10 11]and Pettis [12]

Throughout this paper we consider the measure space(119868 Ω 120583) where 119868 = [0 1] 0 le 119886 lt 119887 lt infin denote a fixed

interval of the real line Ω denotes the Lebesgue 120590-algebraL(119868) and 120583 stands for the Lebesgue measure 119864 denotes areal Banach space with a norm sdot and 119864lowast is its dual By119864120596 we denotes the space 119864 when endowed with the weaktopology generated by the continuous linear functionals on119864 We will let 119862[119868 119864120596] denote the Banach space of weaklycontinuous functions 119868 rarr 119864 with the topology of weakuniform convergence And 119875[119868 119864] denotes the space of 119864-valued Pettis integrable functions in the interval 119868 (see [10 12]for the definition) Recall that (see eg [10 13ndash19]) theweaklymeasurable function 119909 119868 rarr 119864 is said to be Dunford(or Gelfand) integrable on 119868 if and only if 120593119909 is Lebesgueintegrable on 119868 for each 120593 isin 119864lowastDefinition 1 Let 119901 isin [1infin] DefineH119901(119864) to be the class ofall weaklymeasurable functions119909 119868 rarr 119864 having120593119909 isin 119871119901(119868)for every 120593 isin 119864lowast

If 119901 = infin the added condition

lub120593=1 (ess sup119905isin119868

1003816100381610038161003816120593119909 (119905)1003816100381610038161003816) lt infin (1)

must be satisfied by each 119909 isin Hinfin(119864)We also define the classH

1199010 (119864) by

H1199010 (119864) fl 119909 isin 119875 [119868 119864] 120593119909 isin 119871119901 (119868) (2)

HindawiJournal of Function SpacesVolume 2018 Article ID 8746148 13 pageshttpsdoiorg10115520188746148

2 Journal of Function Spaces

Further we define the spaceH120582(119864) 120582 ge 0 byH120582 (119864) fl 119909 isin 119862 [119868 119864120596]

120593119909 is Holderain of order 120582 on 119868 for every 120593isin 119864lowast

(3)

We also defineH0(119864) fl 119862[119868 119864120596]In the following proposition we summarize some impor-

tant facts which are the main tool in carrying out ourinvestigations (see [10 12 16 17])

Proposition 2 Let 119901 119902 isin [1infin] be of conjugate exponents(that is 1119901 + 1119902 = 1 with the convention that 1infin = 0) If119909 119868 rarr 119864 is weakly measurable then

(1) if 119909(sdot) isin 119875[119868 119864] and 119906(sdot) isin 119871infin[119868R] then the follow-ing hold 119909(sdot)119906(sdot) isin 119875[119868 119864]

(2) 119909 isin H1199010 (119864) 119901 gt 1 if and only if 119909(sdot)119906(sdot) isin 119875[119868 119864] for

every 119906(sdot) isin 119871119902(119868)(3) if 119864 is reflexive (containing no isometric copy of 1198880)

the weakly (strongly) measurable function 119909 119868 rarr 119864is Pettis integrable on 119868 if and only if 119909 is Dunfordintegrable on 119868

(4) for any 119901 gt 1 we have 119909 isin H119901(119864) 119909 stronglymeasurable sube H

1199010 (119864) If 119864 is weakly sequentially

complete this is also true for 119901 = 1We remark that there is a bounded weakly measurable

function which is not Pettis integrable (see eg [19])A fundamental property of Pettis integral is contained in

the following

Proposition 3 (see [12] Corollary 251) If 119909 isin 119875[119868 119864] thenfor any bounded subset Ω of elements of 119864lowast the integrals

int119869

1003816100381610038161003816120593 (119909 (119904))1003816100381610038161003816 119889119904 120593 isin Ω (4)

are weakly equi-absolutely continuous

Theorem 4 (mean value theorem for Pettis integral) If thefunction 119909 119868 rarr 119864 is Pettis integrable on 119868 then

int119869119909 (119904) 119889119904 isin |119869| conv (119909 (119869)) (5)

where 119869 sub 119868 |119869| is the length of 119869 and conv(119909(119869)) is the closedconvex hull of 119909(119869)2 Fractional Integrals ofVector-Valued Functions

In this section we define and study the Riemann-Liouvillefractional integral operators and the corresponding fractionalderivatives in Banach spaces

Devoted by the definition of the Riemann-Liouvillefractional integral of real-valued function we introduce thefollowing

Definition 5 Let 119909 119868 rarr 119864 The Riemann-Liouvillefractional Pettis integral (shortly RFPI) of 119909 of order 120572 gt 0is defined by

I120572119909 (119905) fl 1Γ (120572) int

119905

0(119905 minus 119904)120572minus1 119909 (119904) 119889119904 119905 gt 0 (6)

In the preceding definition ldquoint rdquo stands for the Pettisintegral

When 119864 = R it is well known (see eg [20 21]) thatthe operator I120572 sends 119871119902[0 119887] 119887 isin (0infin) continuously to119871119901[0 119887] if 119901 isin [1infin] satisfy 119902 gt 1(120572 + (1119901))

This seems to be a good place to put the following

Example 6 Let 119864 be an infinite dimensional Banach spacethat fails cotype (see [22] and the references therein) Definethe strongly measurable function 119909 [0 1] rarr 119864 by

119909 (119905) fl infinsum119899=1

2119899sum119896=1

119888119899 120594119860119899119896 (119905)120583 (119860119899119896) 119890119899119896 120583 Lebesgue measure (7)

with similar notations as in ([13] Corollary 4) where wechoose 119888119899 = 2119870120595(23minus119899) 119870 gt 1 120595(119905) fl 119905120598 120598 = 09 and119860119899

119896 = [1198861198963119899 1198871198963119899 ] 119899 = 0 1 2 119896 = 1 2 3 2119899 (8)

to be the fat Cantor sets (that is 120583(119860119899119896) = 13119899 holds for every119896 isin 1 2 3 2119899)

As cited in ([13] Corollary 4) 119909 is Pettis integrablefunctions on [0 1] whose indefinite integral is nowhereweakly differentiable on [0 1] Here we will show that 119909 hasRFPI of all order 120572 ge 34 and

I120572119909 (119905)= infinsum

119899=1

2119899sum119896=1

119888119899 ( 119890119899119896120583 (119860119899119896) Γ (120572) int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1 119889119904)isin 119864

(9)

Arguing similarly as in ([13] page 368) we have in view ofsum2119899

119896=1 120576119899119896119906119899119896ℓ2119899infin = 1 that sum2119899

119896=1 |120593(119890119899119896)| le 2120593 holds for every120593 isin 119864lowastAlso for any 119905 isin [0 1] and fixed 119899 isin N we have for some1198990 lt 119899 that

2119899sum119896=1

[int119905

0(119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899

119896)119889119904]

2

= 2119899sum119896=1

32119899 [int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1 119889119904]2

= 32119899 21198990sum

119896=1

[int119860119899119896

(119905 minus 119904)120572minus1 119889119904]2

Journal of Function Spaces 3

+ 32119899 [int119905

11988621198990 +1

(119905 minus 119904)120572minus1 119889119904]2

le 3211989912057221198990sum119896=1

[(119905 minus 1198861198963119899 )120572 minus (119905 minus 1198871198963119899)

120572]2

+ 321198991205722 (119905 minus 11988621198990+13119899 )2120572

le 3211989912057221198990sum119896=1

[(119905 minus 1198861198963119899 ) minus (119905 minus 1198871198963119899)]2120572

+ 321198991205722 (11988721198990+12119899 minus 11988621198990+12119899 )2120572 le 32119899211989912057232119899120572= 21198991205722 32119899(1minus120572)

(10)

Therefore for any measurable 119869 sub [0 1] we arrive atint[0119905]cap119869

infinsum119899=1

100381610038161003816100381610038161003816100381610038161003816100381610038162119899sum119896=1

119888119899Γ (120572) (119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899119896)120593 (119890119899119896) 119889119904

10038161003816100381610038161003816100381610038161003816100381610038161003816le int119905

0

infinsum119899=1

100381610038161003816100381610038161003816100381610038161003816100381610038162119899sum119896=1

119888119899Γ (120572) (119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899119896)120593 (119890119899119896) 119889119904

10038161003816100381610038161003816100381610038161003816100381610038161003816le infinsum

119899=1

119888119899Γ (120572) int119905

0

100381610038161003816100381610038161003816100381610038161003816100381610038162119899sum119896=1

(119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899119896)120593 (119890119899119896) 119889119904

10038161003816100381610038161003816100381610038161003816100381610038161003816le infinsum

119899=1

119888119899Γ (120572)2119899sum119896=1

1003816100381610038161003816120593 (119890119899119896)1003816100381610038161003816 int119905

0(119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899

119896)119889119904

le infinsum119899=1

119888119899Γ (120572) (2119899sum119896=1

1003816100381610038161003816120593 (119890119899119896)10038161003816100381610038162)12

sdot ( 2119899sum119896=1

[int119905

0(119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899

119896)119889119904]

2)12

le 2 10038171003817100381710038171205931003817100381710038171003817sdot infinsum119899=1

radic21198993119899(1minus120572)119888119899120572Γ (120572) = 411987023120598 10038171003817100381710038171205931003817100381710038171003817infinsum119899=1

2119899(12minus120598)3119899(1minus120572)Γ (1 + 120572)

(11)

Obviously the latter series converges whenever 120572 ge 34120598 = 09 which allows us to interchange the integral andsummation below to see that

1Γ (120572) int[0119905]cap119869 120593 ((119905 minus 119904)120572minus1 119909 (119904))= 1Γ (120572) int[0119905]cap119869 (119905 minus 119904)120572minus1 120593 (119909 (119904))= int

[0119905]cap119869

infinsum119899=1

2119899sum119896=1

119888119899Γ (120572) (119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899119896)120593 (119890119899119896) 119889119904

= infinsum119899=1

2119899sum119896=1

int[0119905]cap119869

119888119899Γ (120572) (119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899119896)120593 (119890119899119896) 119889119904

= 120593[infinsum119899=1

2119899sum119896=1

119888119899 ( 119890119899119896120583 (119860119899119896) Γ (120572) int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1 119889119904)]

(12)

It remains to prove that

infinsum119899=1

2119899sum119896=1

119888119899 ( 119890119899119896120583 (119860119899119896) Γ (120572) int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1 119889119904) isin 119864 (13)

Evidently we have10038171003817100381710038171003817100381710038171003817100381710038171003817119898sum119899=119897

2119899sum119896=1

119890119899119896119888119899 (int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1120583 (119860119899119896) Γ (120572)119889119904)

10038171003817100381710038171003817100381710038171003817100381710038171003817119864=10038171003817100381710038171003817100381710038171003817100381710038171003817119898sum119899=119897

2119899sum119896=1

119888119899 (int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1120583 (119860119899119896) Γ (120572)119889119904)119879119899119906119899119896

10038171003817100381710038171003817100381710038171003817100381710038171003817119864le 2 119898sum

119899=119897

119888119899100381710038171003817100381710038171003817100381710038171003817100381710038172119899sum119896=1

(int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1120583 (119860119899119896) Γ (120572)119889119904) 119906119899119896

10038171003817100381710038171003817100381710038171003817100381710038171003817ℓ2119899infin= 2Γ (120572)

119898sum119899=119897

3119899119888119899 max1le119896le2119899

100381610038161003816100381610038161003816100381610038161003816int[0119905]cap119860119899119896 (119905 minus 119904)120572minus1 120594119860119899119896(119904) 119889119904100381610038161003816100381610038161003816100381610038161003816

le 2Γ (120572)119898sum119899=119897

3119899119888119899 max1le119896le2119899

100381610038161003816100381610038161003816100381610038161003816int1198871198963119899

1198861198963119899(119905 minus 119904)120572minus1 119889119904100381610038161003816100381610038161003816100381610038161003816

le 411987023120598Γ (1 + 120572)119898sum119899=119897

3119899(1minus120572)2120598119899

(14)

which approaches zero as 119897 119898 rarr infin as needed for (9)

Remark 7 Observe Example 6 We remark the following

(1) There is a reflexive Banach space for which theindefinite Pettis integral of the function 119909 defined by(7) is nowhere weakly differentiable on [0 1] (see [13]the remark below Corollary 4) Meanwhile 119909 has aRFPI of all order 120572 ge 34

(2) The function I120572119909(sdot) is weakly continuous on [0 1]this follows easily from the definition of the Pettisintegral In fact we have in view of (9) that

120593 (I120572119909 (119905)) = I120572120593 (119909 (119905))

= infinsum119899=1

2119899sum119896=1

119888119899120593 (119890119899119896)120583 (119860119899119896) Γ (120572) int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1 119889119904 (15)

holds for every 120593 isin 119864lowast Since100381610038161003816100381610038161003816100381610038161003816100381610038162119899sum119896=1

119888119899120593 (119890119899119896)120583 (119860119899119896) Γ (120572) int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1 11988911990410038161003816100381610038161003816100381610038161003816100381610038161003816

le 119888119899( 2119899sum119896=1

1003816100381610038161003816120593 (119890119899119896)10038161003816100381610038162)12

sdot ( 2119899sum119896=1

[int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1120583 (119860119899119896) Γ (120572)119889119904]

2)12

le 411987023120598 10038171003817100381710038171205931003817100381710038171003817 2119899(12minus120598)3119899(1minus120572)Γ (1 + 120572)

(16)

4 Journal of Function Spaces

then the infinite series of continuous functions in theleft hand side of (15) converges uniformly in [0 1]Hence the function 120593(I120572119909(sdot)) is continuous on [0 1](this yields the weak continuity ofI120572119909(sdot) on [0 1])

In the following lemma we gather together some simpleparticular sufficient conditions that ensure the existence ofthe Riemann-Liouville fractional integral of the functionsfromH119901(119864)Lemma 8 Let 119909 119868 rarr 119864 be weakly measurable function TheRFPI of the function 119909 of order 120572 gt 0 makes sense ae on 119868 ifat least one of the following cases holds

(a) 119909 isin H1199010 (119864) 119901 isin [1infin] and 120572 ge 1

(b) 119909 isin H1199010 (119864) 119901 gt max1 1120572

(c) 119909 is strongly measurable which lies in H119901(119864) where119901 isin [1infin] provided that 119864 is weakly complete orcontains no copy of 1198880

If 119864 is reflexive this is also true for any 119909 isin H119901(119864) 119901 isin [1infin]and 120572 gt 0

In all cases 120593(I120572119909) = I120572120593119909 holds for every 120593 isin 119864lowastIn the assertions ((a) and (b)) we find sufficient con-

ditions needed for the existence of I120572119909 in the situation inwhich no restriction is placed on 119864 In the third assertion theproperties of119864 allow us to characterize a function 119909 isin H119901(119864)for whichI120572119909 exists

Proof Firstly assertion (a) is direct consequence of Proposi-tion 2 (part (1)) since 119904 rarr (119905 minus 119904)120572minus1 isin 119871infin[0 119905] holds foralmost every 119905 isin 119868 whenever 120572 ge 1

Secondly to prove assertion (b) let 119901 gt max1 1120572 and119902 be the conjugate exponents to 119901 Since 119902(120572 minus 1) + 1 gt 0 wehave that 119904 rarr (119905 minus 119904)120572minus1 isin 119871119902[0 119905] holds for every 119905 isin 119868 Thusthe assertion (b) is now an easy consequence of Proposition 2(part (2))

Thirdly to prove (c) we let 120572 gt 0 119901 isin [1infin]and 119909 isin H119901(119864) be strongly measurable Since the strongmeasurability is preserved under multiplication operation onfunctions the product (119905 minus sdot)120572minus1119909(sdot) [0 119905] rarr 119864 is stronglymeasurable on [0 119905] for almost every 119905 isin 119868 In view of Youngrsquosinequality for every 120593 isin 119864lowast the real-valued function

119904 997891997888rarr 120593 ((119905 minus 119904)120572minus1 119909 (119904)) = (119905 minus 119904)120572minus1 120593119909 (119904) (17)

is Lebesgue integrable on [0 119905] for almost every 119905 isin 119868 So theassertion (c) follows immediately from Proposition 2 (parts(3 4))

Similarly when 119864 is reflexive the result follows from part(3) of Proposition 2However in all cases the function 119904 997891rarr (119905 minus 119904)120572minus1119909(119904) is

Pettis integrable on [0 119905] for almost every 119905 isin 119868 That is foralmost every 119905 isin 119868 there exists an element in 119864 denoted byI120572119909(119905) such that

120593 (I120572119909 (119905)) = int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904

= int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120593119909 (119904) 119889119904 = I120572120593119909 (119905)

(18)

holds for every 120593 isin 119864lowast This completes the proof

Remark 9 If 119909 isin H1199010 (119864) such that I120572119909(119905) does not exist for

some 119905 isin 119868 then it does not exist even when we ldquoenlargerdquo thespace 119864 into 119865 To see this let 119909 119868 rarr 119865 such that 119909(119868) sub 119864 IfI120572119909(119905) exists for some 119905 isin 119868 then (119905 minus sdot)120572minus1119909(sdot) isin 119875([0 119905] 119865)Since 119909 assumes only values in 119864 it follows by the mean valuetheorem for Pettis integral (Theorem 4) that the RFPI of 119909should lie in 119864

Before we come to a deep study of the mathematicalproperties of the RFPI operator let us take a look at thefollowing miscellaneous examples

Example 10 Let 120572 gt 0 Define the function 119909 [0 1] rarr119871infin[0 1] by 119909(119905) = 120594[0119905]This function is weakly measurable Pettis integrable on[0 1] and 120593119909 is a function of bounded variation (see eg

[18]) That is 119909 isin Hinfin0 (119871infin) Hence in view of Lemma 8 with119901 = infin the RFPI of 119909 exists on [0 1] Further calculations

(cf [4]) show that

I120572119909 (119905) (sdot) = (119905 minus sdot)120572Γ (1 + 120572)120594[0119905] (sdot) isin 119871infin [0 1] (19)

Example 11 Let120572 gt 0 Define the function119909 from the interval[0 1] into the Hilbert space ℓ2 as119909 119905 997891997888rarr 1(1 + 119905) 1(2 + 119905) 1(3 + 119905) 119905 isin [0 1] (20)

We note that

119909 (119905)2ℓ2 = sum119899

( 1119899 + 119905)2 le sum

119899

11198992 lt infin 119905 isin [0 1] (21)

Thus the function 119909 is well defined We claim that 119909 isDunford integrable on [0 1] Once our claim is establishedLemma 8 guarantees the existence ofI120572119909 on [0 1] It remainsto prove this claim and to calculate I120572119909

To see this let 120593 isin (ℓ2)lowast = ℓ2 According to theRiesz representation theorem on Hilbert spaces there existsa uniquely determined 120582 fl 120582119899 isin ℓ2 such that 120593119909(119905) =sum119899(120582119899(119899 + 119905)) A standard arguments using Beppo LevirsquosTheorem yields

int1

0

1003816100381610038161003816120593119909 (119905)1003816100381610038161003816 119889119905 le sum119899

int1

0

10038161003816100381610038161205821198991003816100381610038161003816(119899 + 119905)119889119905 = sum119899

10038161003816100381610038161205821198991003816100381610038161003816 ln(1 + 1119899)le sum

119899

10038161003816100381610038161205821198991003816100381610038161003816119899 le sum119899

10038161003816100381610038161205821198991003816100381610038161003816 lt infin(22)

So 119909 is Dunford integrable on [0 1] and hence Pettis inte-grable on [0 1] since ℓ2 is reflexive Consequently in view of

Journal of Function Spaces 5

Lemma 8 I120572119909 exists on [0 1] To calculate the RFPI of 119909 fix119905 isin [0 1] and let 120593 isin (ℓ2)lowast We have

int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904 = int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120593119909 (119904) 119889119904= I

120572 (infinsum119899=1

120582119899(119899 + 119905)) (23)

Since the series sum119899(120582119899(119899 + 119905)) is uniformly convergent on[0 1] it follows in view of the generalized linearity of thefractional integrals [23 Lemma 5] that (cf [21 Table 91])

int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904= infinsum

119899=1

120582119899119905120572119899Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 ) = 120593 (119892 (119905)) (24)

where 2F1 is the Gauss hypergeometric function evaluated at(1 1 120572 + 1 minus119905119899) and119892 (119905) fl 119905120572119899Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 ) (25)

Since

1003817100381710038171003817119892 (119905)10038171003817100381710038172ℓ2 = sum119899

( 119905120572Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 ))2

le 2F21 (1 1 120572 + 1 minus1)Γ2 (1 + 120572) sum

119899

11198992 lt infin119905 isin [0 1]

(26)

we see that 119892 isin ℓ2 Thus

I120572119909 (119905) = 119905120572119899Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 )

119905 isin [0 1] (27)

Example 12 Let 120572 isin [12 1) Define the countable-valuedfunction 119909 [0 1] rarr 1198880 by

119909 (119905) fl 119909119899 (119905) 119909119899 fl 119899120594(1(119899+1)1119899] (28)

This function is strongly measurable Pettis integrable func-tion on [0 1] (see eg [10 16]) We claim that 119909 isin H119875

0 (1198880)with 119901 isin (1120572 2] Once our claim is established Lemma 8guarantees the existence of I120572119909 120572 isin [12 1) on [0 1] Itremains to prove this claim by showing firstly that H119875(1198880)and to calculate I120572119909 To do this let 120593 isin 119888lowast0 Then therecorresponds to 120593 a unique 120582119899 isin ℓ1 such that 120593119909 = sum119899 120582119899119909119899By noting that

10038171003817100381710038171205821198991199091198991003817100381710038171003817119901 = (int1

0

1003816100381610038161003816120582119899119909119899 (119905)1003816100381610038161003816119901 119889119905)1119901

= 10038161003816100381610038161205821198991003816100381610038161003816 (int1119899

1(119899+1)119899119901119889119905)1119901 = 10038161003816100381610038161205821198991003816100381610038161003816 1198991119902(119899 + 1)1119901

le 10038161003816100381610038161205821198991003816100381610038161003816 11989911199021198991119901 le 100381610038161003816100381612058211989910038161003816100381610038161198992119901minus1 le 10038161003816100381610038161205821198991003816100381610038161003816

(29)

holds for any 119901 isin (1120572 2] a standard argument using LevirsquosTheorem (or Lebesgue dominated convergence theorem) andMinkowskirsquos inequality yields

[int1

0

1003816100381610038161003816120593119909 (119905)1003816100381610038161003816119901 119889119905]1119901 le [int1

0(infinsum119899=1

10038161003816100381610038161205821198991003816100381610038161003816 119909119899 (119905))119901 119889119905]1119901

= [[lim119896rarrinfin

int1

0( 119896sum119899=1

10038161003816100381610038161205821198991003816100381610038161003816 119909119899 (119905))119901

119889119905]]1119901

le [[lim119896rarrinfin

( 119896sum119899=1

10038171003817100381710038171205821198991199091198991003817100381710038171003817119901)119901]]1119901

le infinsum119899=1

10038171003817100381710038171205821198991199091198991003817100381710038171003817119901le infinsum

119899=1

10038161003816100381610038161205821198991003816100381610038161003816 lt infin

(30)

Thus 120593119909(sdot) isin 119871119901[0 1] 119901 isin (1120572 2] holds for every 120593 isin 119888lowast0(that is 119909 isin H119875(1198880)) Since 119901 gt 1120572 gt 1 it follows byProposition 2 in view of the strong measurability of 119909 that119909 isin H119875

0 (1198880) Owing to Lemma 8 we infer that the RFPI of119909 of any order 120572 isin [12 1) exists on the interval [0 1] Tocompute this integral fix 119905 isin 1198681198990 fl [1(1198990+1) 11198990] for some1198990 isin N the set of natural numbers and let 120593 isin 119888lowast0 Then

sum119899

int119905

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 12058211989910038161003816100381610038161003816 119909119899 (119904) 119889119904le 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816 int

119905

1(1198990+1)(119905 minus 119904)120572minus1 1198990119889119904

+ sum119899gt1198990

10038161003816100381610038161205821198991003816100381610038161003816 int1119899

1(119899+1)(119905 minus 119904)120572minus1 119899 119889119904

= 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572

+ sum119899gt1198990

119899 10038161003816100381610038161205821198991003816100381610038161003816120572 [(119905 minus 1119899 + 1)120572 minus (119905 minus 1119899)

120572]

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572

+ sum119899gt1198990

119899 10038161003816100381610038161205821198991003816100381610038161003816120572 [(119905 minus 1119899 + 1) minus (119905 minus 1119899)]120572

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572 + sum

119899gt1198990

119899 10038161003816100381610038161205821198991003816100381610038161003816120572 1119899120572 (119899 + 1)120572

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572 + sum

119899gt1198990

10038161003816100381610038161205821198991003816100381610038161003816120572 11198992120572minus1

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572 + sum

119899gt1198990

10038161003816100381610038161205821198991003816100381610038161003816120572 lt infin(because 120572 isin [12 1))

(31)

6 Journal of Function Spaces

Therefore by Beppo Levi Theorem it follows that

int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904 = int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120593 (119909 (119904)) 119889119904= int119905

0

(119905 minus 119904)120572minus1Γ (120572) sum119899

120582119899119899120594(1(119899+1)1119899] (119904) 119889119904= sum

119899

int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120582119899119899120594(1(119899+1)1119899] (119904) 119889119904= 11989901205821198990Γ (1 + 120572) (119905 minus 11198990 + 1)

120572

+ sum119899gt1198990

119899120582119899Γ (1 + 120572) [(119905 minus 1119899 + 1)120572 minus (119905 minus 1119899)

120572]

(32)

Consequently we conclude that

I120572119909 (119905) = 0 0⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

1198990minus1

1198990Γ (1 + 120572) (119905 minus 11198990 + 1)120572

(1198990 + 1)Γ (1 + 120572) [(119905 minus 11198990 + 2)

120572 minus (119905 minus 11198990 + 1)120572]

(33)

where the nonzero coordinate started from the 1198990th place Itcan be easily seen thatI120572119909(119905) isin 1198880 for any 119905 isin [0 1] Evidentlywe have that10038161003816100381610038161003816100381610038161003816119899 [(119905 minus

1119899 + 1)120572 minus (119905 minus 1119899)

120572]10038161003816100381610038161003816100381610038161003816le 119899 [(119905 minus 1119899 + 1) minus (119905 minus 1119899)]

120572 le 119899 [ 1119899 (119899 + 1)]120572

le 11198992120572minus1 997888rarr 0 as 119899 997888rarr infin1198990Γ (1 + 120572) (119905 minus 11198990 + 1)

120572

le 1198990Γ (1 + 120572) ( 11198990 minus 11198990 + 1)120572 le 11198992120572minus10 Γ (1 + 120572)

997888rarr 0 as 1198990 997888rarr infin

(34)

hold for any 119905 isin [0 1]This yieldsI1205721119909(119905) isin 1198880 for any 119905 isin [0 1]

(this is precisely what we would expect from Definition 5)

Look at part (c) of Lemma 8 with 120572 isin (0 1) and119864 = 1198880 Below we give an example showing that the stronglymeasurability hypothesis imposed on a function 119909 isin H1(119864)H1

0(119864) is not sufficient for the existence of I120572119909 even when 119909is Denjoy-Pettis integrable (cf [16])

Example 13 Let 120572 isin (0 1) Define the strongly measurablefunction 119909 [0 1] rarr 1198880 by119909 (119905) fl 119909119899 (119905)

119909119899 (119905) fl 119899 (119899 + 1) [120594119868119899 (119905) minus 120594119869119899 (119905)] (35)

where

119868119899 fl ( 1119899 + 1 119899 + 12119899 (119899 + 1)) 119869119899 fl ( 119899 + 12119899 (119899 + 1) 1119899)

(36)

It is immediate that (cf [16]) 119909 is a well-defined Denjoy-Pettis (but it is not Pettis) integrable on [0 1]

In what follows we will show that the RFPI of 119909 does notexist on a subinterval of positive measure on [0 1]

To see this we make use of Proposition 3 as followsdefine for each intervalI119899 = [1(119899 + 1) 1119899] the functionals120593119899 isin 119888lowast0 (required by Proposition 3) to be the correspondingto the element 120582119899 fl (0 0 0 1 0 0 ) isin ℓ1 wherethe nonzero coordinate is in the 119899th place Then 120593119899(119909(119905)) =119909119899(119905) and thus |120593119899(119909(119904))| = 119899(119899 + 1)[120594119868119899(119905) minus 120594119869119899(119905)] Clearlythe family 120593119899 runs through the unit ball of the dual of1198880 Evidently the isometric isomorphism between 119888lowast0 and ℓ1yields 120593119899119888lowast0 = 120582119899ℓ1 = 1

Now take 119905 isin [12 1] and define the continuous function119891 [12 1] rarr R+ by

119891 (119905) fl (119905 minus 1119899 + 1)120572 minus (119905 minus 1119899)

120572 119899 ge 2 119905 isin [12 1]

(37)

It can be easily seen that in view of 0 lt 120572 lt 1 1198911015840 is negativeon (0 1) By standard results from (classical) calculus itfollows that 119891 is strictly decreasing on [0 1] in particular119891(1) lt 119891(119905) for all 119905 isin [0 1) Thus for any 119899 ge 2 and any120593 isin 120593119899 we have

intI119899

100381610038161003816100381610038161003816100381610038161003816120593 ((119905 minus 119904)120572minus1Γ (120572) 119909 (119904))100381610038161003816100381610038161003816100381610038161003816 119889119904

= intI119899

(119905 minus 119904)120572minus1Γ (120572) 1003816100381610038161003816120593 (119909 (119904))1003816100381610038161003816 119889119904= int1119899

1(119899+1)

(119905 minus 119904)120572minus1Γ (120572) 119899 (119899 + 1) 119889119904= 119899 (119899 + 1)Γ (1 + 120572) [(119905 minus 1119899 + 1)

120572 minus (119905 minus 1119899)120572]

ge 1198992Γ (1 + 120572) [(1 minus 1119899 + 1)120572 minus (1 minus 1119899)

120572]

(38)

An explicit calculation using LrsquoHospitalrsquos rule two timesreveals

lim119899rarrinfin

(1 minus 1 (119899 + 1))120572 minus (1 minus 1119899)120572119899minus2 = minus1205722sdot lim119899rarrinfin

1198993 [( 119899119899 + 1)120572minus1 1

(119899 + 1)2 minus (1 minus 1119899)120572minus1 11198992 ]

= minus1205722 lim119899rarrinfin

119899 [( 119899119899 + 1)120572minus1 1198992

(119899 + 1)2 minus (1 minus 1119899)120572minus1]

Journal of Function Spaces 7

= minus1205722 lim119899rarrinfin

119899 [( 119899119899 + 1)120572+1 minus (1119899)

120572minus1] = minus1205722sdot lim119899rarrinfin

1198992 [(120572 + 1) ( 119899119899 + 1)120572 1(119899 + 1)2

minus (120572 minus 1) (1 minus 1119899)120572minus2 11198992 ] = minus1205722

sdot lim119899rarrinfin

[(120572 + 1) ( 119899119899 + 1)120572+2 minus (120572 minus 1) (1 minus 1119899)

120572minus2]= 120572

(39)

from which it follows that

intI119899

100381610038161003816100381610038161003816100381610038161003816120593 ((119905 minus 119904)120572minus1Γ (120572) 119909 (119904))100381610038161003816100381610038161003816100381610038161003816 119889119904997888rarr0 as 119899 997888rarr infin (40)

Therefore in view of Proposition 3 119904 997891rarr (119905 minus 119904)120572minus1119909(119904) notin119875([0 1] 1198880) Hence I120572119909 does not exist which is what wewished to show

The following theorem provides a useful characterizationof the space H119901(119864) for which the statements reveal howmuch the fractional integral I120572 is better than the function119909 isin H119901(119864) Indeed based on Lemma 8 using an inequalityof Young we can easily prove the following

Lemma 14 For any 120572 gt 0 the following holds(a) If 119864 is reflexive then for every 119901 isin [1infin] the operator

I120572 mapsH119901(119864) intoH1199010 (119864) In particular if 0 lt 120572 lt1 then I120572 H1(119864) rarr H1((1minus120572)minus120598)0 (119864) however small120598 gt 0 is

(b) If 119864 contains no copy of 1198880 then for every 119901 isin [1infin]the operator I120572 maps 119909 isin H

1199010 (119864) 119909 strongly

measurable into H1199010 (119864) In particular if 120572 ge 1 the

operator I120572 mapsH1199010 (119864) into itself

(c) If 119901 gt max1 1120572 the operator I120572 sends H1199010 (119864) to119862[119868 119864120596] (if we define I120572119909(0) fl 0) In particular if120572 isin (0 1) I120572 H119901

0 (119864) rarr H120572minus1119901(119864)Proof At the beginning let 120572 gt 0 and 119909 isin H119901(119864) with119901 isin [1infin] and define the real-valued function 119892 119868 rarr R

by 119892(119905) fl 119905120572minus1Γ(120572) If 119864 is reflexive it follows in view ofLemma 8 that J120572119909 exists ae in 119868 and 120593(I120572119909) = I120572120593119909for every 120593 isin 119864lowast Youngrsquos inequality guarantees that theconvolution product 119892 lowast 120593119909(sdot) lies in 119871119901[119868] Consequently120593(I120572119909) = I120572120593119909 = 119892 lowast 120593119909 isin 119871119901[119868] for every 120593 isin 119864lowastThe reflexivity of 119864 together with Proposition 2 yields I120572119909 isinH

1199010 (119864)In particular if 0 lt 120572 lt 1 it can be easily seen thatint

119868(119892(119905))119902119889119905 lt infin for every 119902 isin [1 1(1 minus 120572)) That is 119892 isin1198711(1minus120572)minus120598[119868] 120598 gt 0 By Youngrsquos inequality it follows that120593(I120572119909) = 119892 lowast 120593119909 isin 1198711(1minus120572)minus120598[119868] for every 120593 isin 119864lowast however

small 120598 gt 0 is Now the assertion (a) follows because of thereflexivity of 119864

Next in order to prove the assertion (b) let 119909 isin H1199010 (119864)

and note in view of Lemma 8 that I120572119909 exists on 119868 Define119910 119868 rarr 119864 by119910(119905) fl I120572119909(119905) 119909 isin H1199010 (119864) Since120593119910(sdot) isin 119871119901[119868]

for every120593 isin 119864lowast it follows that119910 isin H119901(119864) Moreover for any119886 119887 isin 119868 (119886 lt 119887) we haveint119887

119886120593119910 (119904) 119889119905 = int119887

119886I120593119909 (119905) 119889119905 = 1Γ (120572)

sdot int119887

119886int119905

0(119905 minus 119904)120572minus1 120593119909 (119904) 119889119904 119889119905

= 1Γ (120572) [int119887

119886int119887

119904(119905 minus 119904)120572minus1 120593119909 (119904) 119889119905 119889119904

+ int119886

0int119887

119886(119905 minus 119904)120572minus1 120593119909 (119904) 119889119905 119889119904]

= 1Γ (120572) [int119887

119886(119887 minus 119904)120572 120593119909 (119904) 119889119904

+ int119886

0[(119887 minus 119904)120572 minus (119886 minus 119904)120572] 120593119909 (119904) 119889119904]

= 1Γ (120572) [int119886

0(119887 minus 119904)120572 120593119909 (119904) 119889119904

+ int119887

119886(119887 minus 119904)120572 120593119909 (119904) 119889119904 minus int119886

0(119886 minus 119904)120572 120593119909 (119904) 119889119904]

= int119887

0

(119887 minus 119904)120572Γ (1 + 120572)120593119909 (119904) 119889119904 minus int119886

0

(119886 minus 119904)120572Γ (1 + 120572)120593119909 (119904) 119889119904= 120593 (119909[119886119887])

(41)

where

119909[119886119887] fl (119875) int119887

0

(119887 minus 119904)120572Γ (1 + 120572)119909 (119904) 119889119904minus (119875) int119886

0

(119886 minus 119904)120572Γ (1 + 120572)119909 (119904) 119889119904(42)

Since 119909 isin 119875[119868 119864] owing to Proposition 2 (part (1)) it followsthat (119887 minus sdot)120572119909(119904) and (119886 minus sdot)120572119909(119904) are Pettis integrable on 119868 andso 119909[119886119887] isin 119864

A combination of these results yields120593119910 isin 119871119901(119868) for every120593 isin 119864lowast and there exists an element 119909[119886119887] isin 119864 such that120601119909[119886119887] = int119887

119886120593119910(119905)119889119905 for every 119886 119887 isin 119868 Since 119864 contains no

copy of 1198880 it follows that (cf [17 Theorem 23]) 119910 isin 119875[119868 119864]Consequently 119910(sdot) isin H

1199010 (119864) This is the claim (b)

To prove the assertion (c) let 119901 gt max1 1120572 and 119909 isinH

1199010 (119864) By Lemma 8 we deduce that I120572119909 exists ae on 119868

Now let 119902 isin [1infin] be such that 1119901 + 1119902 = 1 Since119902(120572 minus 1) gt minus1 119892 fl (sdot)120572minus1Γ(120572) isin 119871119902[119868] Therefore as a directconsequence of Youngrsquo inequality it follows that

120593 (I120572119909 (sdot)) = I120572120593119909 (sdot) = 119892 lowast 120593119909 (sdot) isin 119862 [119868R] (43)

holds for every 120593 isin 119864lowast Now we claim that I120572119909(119905) rarr 0in 119864 as 119905 rarr 0 Once our claim is established the definition

8 Journal of Function Spaces

I120572119909(0) fl 0 guarantees thatI120572119909(sdot) isin 119862[119868 119864120596] which is whatwewished to show It remains to prove our claim without lossof generality assume that I120572119909(119905) = 0 Then there exists (asa consequence of the Hahn-Banach theorem) 120593 isin 119864lowast with120593 = 1 such that I120572119909(119905) = 120593(I120572119909(119905)) = I120572120593119909(119905) ByHolderrsquos inequality we obtain

1003816100381610038161003816120593 (I120572119909 (119905))1003816100381610038161003816le 1Γ (120572) (int

119905

0(119905 minus 119904)119902(120572minus1) 119889119904)1119902 (int119905

0

1003816100381610038161003816120593119909 (119904)1003816100381610038161003816119901 119889119904)1119901

le 1Γ (120572) ( 119905(120572minus1)119902+1(120572 minus 1) 119902 + 1)1119902 (int

119868

1003816100381610038161003816120593119909 (119904)1003816100381610038161003816119901 119889119904)1119901

= 119905120572minus1+1119902Γ (120572) [119902 (120572 minus 1) + 1]1119902

10038171003817100381710038171205931199091003817100381710038171003817119871119901

= 119905120572minus1119901 10038171003817100381710038171205931199091003817100381710038171003817119871119901Γ (120572) [119902 (120572 minus 1) + 1]1119902

(44)

This is equivalent with the following estimate

1003817100381710038171003817I120572119909 (119905)1003817100381710038171003817 le 119905120572minus1119901 10038171003817100381710038171205931199091003817100381710038171003817119871119901Γ (120572) [119902 (120572 minus 1) + 1]1119902 (45)

Owing to 120572 gt 1119901 we get I120572119909(119905) rarr 0 in 119864 as 119905 rarr 0 andconsequently in view ofI120572119909(0) fl 0I120572119909(sdot) isin 119862[119868 119864120596] Thisproves the first part of the assertion (c)

Finally let 120572 isin (0 1) and 119905 120591 isin 119868 Without loss ofgenerality assume 120591 lt 119905 Then for any 120593 isin 119864lowast we have byHolder inequality with 119901 gt 1120572 in view 119902(120572 minus 1) gt minus1 that

1003816100381610038161003816120593 (I120572119909 (119905) minusI120572119909 (120591))1003816100381610038161003816 Γ (120572)

= 10038161003816100381610038161003816100381610038161003816int119905

0(119905 minus 119904)120572minus1 120593119909 (119904) 119889119904 minus int120591

0(120591 minus 119904)120572minus1 120593119909 (119904) 11988911990410038161003816100381610038161003816100381610038161003816

le (int120591

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 minus (120591 minus 119904)120572minus110038161003816100381610038161003816 1003816100381610038161003816120593119909 (119904)1003816100381610038161003816 119889119904+ int119905

120591(119905 minus 119904)120572minus1 1003816100381610038161003816120593119909 (119904)1003816100381610038161003816 119889119904)

= [(int120591

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 minus (120591 minus 119904)120572minus110038161003816100381610038161003816119902 119889119904)1119902

+ (int119905

120591(119905 minus 119904)(120572minus1)119902 119889119904)1119902] 10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]

le [[(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

+ ((119905 minus 120591)(120572minus1)119902+1(120572 minus 1) 119902 + 1 )1119902]]10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]

= [(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

+ (119905 minus 120591)120572minus1119901[119902 (120572 minus 1) + 1]1119902]

10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868] (46)

By noting that (120572 minus 1)119902 isin (minus1 0) when 120572 isin (0 1) it can beeasily seen that

int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904le 1119902 (120572 minus 1) + 1 (119905 minus 120591)119902(120572minus1)+1

(47)

That is

(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

le (119905 minus 120591)120572minus1119901[119902 (120572 minus 1) + 1]1119902

(48)

A combination of these results yields1003816100381610038161003816120593 (I120572119909 (119905)) minus 120593 (I120572119909 (120591))1003816100381610038161003816= 1003816100381610038161003816120593 (I120572119909 (119905) minusI

120572119909 (120591))1003816100381610038161003816le10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]Γ (120572) 2 (119905 minus 120591)120572minus1119901

[119902 (120572 minus 1) + 1]1119902 (49)

ThusI120572119909 isin H120572minus1119901(119864) This completes the proof

Corollary 15 For any 120572 gt 0 I120572 119862[119868 119864120596] rarr 119862[119868 119864120596]Proof Let 119909 isin 119862[119868 119864120596] then 120593119909 isin 119871infin holds for every120593 isin 119864lowast By noting that the weak continuity implies a strongmeasurability ([24] page 73) it follows that 119909 is stronglymeasurable on 119868 Hence in view of Proposition 2 119909 isinHinfin

0 (119864) Consequently 119862[119868 119864120596] sube H1199010 (119864) for every 119901 isin[1infin] That isI120572 119862[119868 119864120596] rarr 119862[119868 119864120596] Hence the desired

result is obtained

Example 16 Let 120572 isin [12 1) Define the function 119909 [0 1] rarr1198880 by 119909(119905) = 119899120594(1(119899+1)1119899]We have (in view of Example 12)that

I120572119909 (119905) = 0 0 0 1198990 (119905 minus 1 (1198990 + 1))120572

Γ (1 + 120572) (1198990 + 1)Γ (1 + 120572) [(119905 minus 11198990 + 2)

120572 minus (119905 minus 11198990 + 1)120572]

(50)

To show that I120572119909 is norm continuous on [0 1] let 120591 119905 isin[0 1] With no loss of generality we may assume that 119905 120591 isin1198681198990 for some 1198990 isin N Since the nonzero terms of the sequencein (50) are nonincreasing then in view of

|119905 minus 120591| le ( 11198990 minus 11198990 + 1) = 11198990 (1198990 + 1) le 111989920 (51)

Journal of Function Spaces 9

we have1003817100381710038171003817I120572119909 (119905) minusI

1205721119909 (120591)10038171003817100381710038171198880

= 1198990Γ (1 + 120572)10038161003816100381610038161003816100381610038161003816(119905 minus

1119899 + 1)120572 minus (120591 minus 1119899 + 1)

12057210038161003816100381610038161003816100381610038161003816le 1198990Γ (1 + 120572)

1003816100381610038161003816100381610038161003816(119905 minus 1119899 + 1) minus (120591 minus 1119899 + 1)1003816100381610038161003816100381610038161003816120572

le 1198990Γ (1 + 120572) |119905 minus 120591|120572 le 1Γ (1 + 120572) |119905 minus 120591|120572radic|119905 minus 120591|= |119905 minus 120591|120572minus12Γ (1 + 120572)

(52)

Thus the RFPI of 119909 is norm continuous on [0 1] Preciselysince 0 lt 120572 minus 12 lt 120572 minus 1119901 lt 1 for any 119901 le 2 thenI120572119909 isin H120572minus12(119864) sub H120572minus1119901(119864) This is precisely what onewould expect from Lemma 14 (part (c))

Analogously an explicit calculation reveals that I120572119909 liesin 119862([0 1] 119871infin[0 1]) where 119909 [0 1] rarr 119871infin[0 1] defined byExample 10

We now consider additional mapping properties of theoperator I120572 Precisely we will show that the RFPI enjoysthe following commutative property which is folklore in case119864 = R However the proof is completely similar to that of [8]Lemma 35

Lemma 17 Let 120572 120573 gt 0 and 119901 gt max1 1120572 1120573 Then

I120572I120573119909 = I

120573I120572119909 = I

120572+120573119909 (53)

holds for every 119909 isin H1199010 (119864) If 119864 is reflexive this is also true for

every 119901 ge 1Proof Since 119901 gt max1 1120572 1120573 it follows that 119901 gtmax1 1(120572+120573) So by Lemmas 8 and 14 themappingsI120572119909I120573119909 andI120572+120573119909 belong toHinfin

0 (119864) for every 119909 isin H1199010 (119864) Now

repeating the same process as in ([8] Lemma 35) implies theclaim

When 119864 is reflexive the result follows as a direct applica-tion of Lemma 14

3 Fractional Derivatives ofVector-Valued Functions

After the notation of the fractional integrals of vector-valued functions the fractional derivatives become a naturalrequirement Before we come to the definitions and a detailedstudy of the mathematical properties of fractional differentialoperators we recall the following

Definition 18 Consider the vector-valued function 119909 119868 rarr119864(1) Let 120593119909 be differentiable on 119868 for every 120593 isin 119864lowast The

function 119909 is said to be weakly differentiable on 119868 if

there exists 119910 119868 rarr 119864 such that for every 120593 isin 119864lowast wehave

119889120593119909 (119905)119889119905 = 120593119910 (119905) for every 119905 isin 119868 (54)

The function 119910 is called the weak derivative of thefunction 119909

(2) Let 120593119909 be differentiable ae on 119868 for every 120593 isin 119864lowast(the null set may vary with 120593 isin 119864lowast) The function 119909is said to be pseudo differentiable on 119868 if there existsa function 119910 119868 rarr 119864 such that for every 120593 isin 119864lowast thereexists a null set119873(120593) sub 119868 such that

(120593119909 (119905))1015840 = 120593119910 (119905) for every 119905 isin 119868119873 (120593) (55)

In this case the function 119910 is called the pseudoderivative of 119909If 119909 is pseudo differentiable on 119868 and the null setinvariant for every 120593 isin 119864lowast then 119909 is ae weaklydifferentiable on 119868

Clearly if 119909 is ae weakly differentiable on 119868 then 120593(119909)is ae differentiable on 119868 The converse holds in a weaklysequentially complete space (see [25] Theorem 733)

For more details of the derivatives of vector-valuedfunctions we refer to [10 12 26]

The following results play a major role in our analysis

Proposition 19 (see [27] Theorem 51) The function 119910 119868 rarr119864 is an indefinite Pettis integrable if and only if 119910 is weaklyabsolutely continuous on 119868 and have a pseudo derivative on 119868In this case 119910 is an indefinite Pettis integral of any of its pseudoderivatives

Now we are in the position to define the fractional-typederivatives of vector-valued functions

Definition 20 Let 119909 119868 rarr 119864 For the positive integer119898 suchthat 120572 isin (119898 minus 1119898) 119898 isin N0 fl 0 1 2 we define theCaputo fractional-pseudo (weak) derivative ldquoshortly CFPD(CFWD)rdquo of 119909 of order 120572 by

119889120572119889119905120572 119909 (119905) fl I119898minus120572119863119898119909 (119905) (56)

where the sign ldquo119863rdquo denotes the pseudo (or weak) differentialoperator We use the notation 119889120572119901119889119905120572 and 119889120572120596119889119905120572 to charac-terize the Caputo fractional-pseudo derivatives and Caputofractional weak derivatives respectively

It is well known that although the weak derivative of aweakly differentiable function is uniquely determined thepseudo derivative of the pseudo differentiable function is notunique Also although any two pseudo derivatives 119910 119911 of afunction 119909 119868 rarr 119864 need not be ae equal (see [12 Example91] and [26 p 2]) the functions 119910 119911 are weakly equivalenton 119868 (that is 120593119910 = 120593119911 holds ae for every 120593 isin 119864lowast) The

10 Journal of Function Spaces

next lemma provides a useful characterization property of theCFPD Really it can be easily seen that the CFPD of a Caputofractional-pseudo differentiable function does not depend onthe choice of a pseudo derivative of the function

Lemma 21 Let 119909 119868 rarr 119864 be pseudo differentiable functionwhere pseudo derivatives lie in H

1199010 (119864) If (119889120572119901119889119905120572)119909 exists on119868 then the CFPD of 119909 depends on the choice of the pseudo

derivatives of 119909Proof Let 119910 119911 isin H

1199010 (119864) be two pseudo derivatives of the

pseudo differentiable function 119909 Since 119909 and 119910 are weaklyequivalent on 119868 then for every 120593 isin 119864lowast there exists a nullset N which depends on 120593 isin 119864lowast such that (119905 minus 119904)minus120572120593119910(119904) =(119905minus119904)minus120572120593119911(119904) for every 119904 isin [0 119905]N Consequently for almostevery 119905 isin 119868

120593 (I1minus120572119910 (119905)) = I1minus120572120593119910 (119905) = I

1minus120572120593119911 (119905)= 120593 (I1minus120572119911 (119905)) (57)

holds for every 120593 isin 119864lowast Thus

119889120572119901119889119905120572 119909 = I1minus120572119863119901119909 = I

1minus120572119910 = I1minus120572119911 on 119868 (58)

which is what we wished to show

We consider the following examples

Example 22 Define 119909 [0 1] rarr ℓ2 by119909 (119905) fl ln(1 + 119905119899) 119899 isin N (59)

We claim that the CFWD of 119909 exists on [0 1] To see this welet 120593 isin (ℓ2)lowast = ℓ2 Then there corresponds to 120593 a sequence120582119899 isin ℓ2 such that

120593119909 (119905) = sum119899

120582119899 ln(1 + 119905119899) (60)

Since the series in (60) converges uniformly on [0 1] theformal differentiation of 120593119909 yields

(120593119909 (119905))1015840 = sum119899

120582119899119899 + 119905 = 120593 ( 1119905 + 119899) (61)

It is not hard to justify the differentiation by noting that theseries in the right hand side of (61) is uniformly and absolutelyconvergent on [0 1] Consequently 120593119909 is differentiable on[0 1] for 120593 isin (ℓ2)lowast meaning that 119909 is weakly differentiable on[0 1] (because ℓ2 is weakly complete (cf [25] Theorem 733)and 119863120596119909(119905) = 1(119905 + 119899) Owing to Lemma 25 the CFWDof any order 120572 isin (0 1) of 119909 exists on [0 1] To calculate theCFWD of 119909 on [0 1] we observe in view of Example 11 that

119889120572120596119889119905120572 119909 (119905) = I1minus120572119863120596119909 (119905)

= 1199051minus120572119899Γ (2 minus 120572) 2F1 (1 1 2 minus 120572 minus119905119899 ) (62)

Example 23 Let119864 be an infinite dimensionBanach space thatfails cotype Define 119910 [0 1] rarr 119864 by 119910(119905) fl int119905

0119909(119904)119889119904

where 119909 [0 1] rarr 119864 given by formula (7) As cited in ([13]Corollary 4) 119910 is nowhere weakly differentiable on [0 1]Consequently the CFWD of any order 120572 isin [34 1) of 119910 losesits meaning on [0 1]Remark 24 As shown in Example 23 there is an infinitedimension Banach space 119864 and weakly absolutely continuousfunction 119910 119868 rarr 119864 which is nowhere weakly differentiable(hence the CFWD of 119910 does not exist) Also even when 119864 isseparable and 119910 is Lipschitz function the pseudo derivatives(hence the CFPD) of 119910 need not to exist [28]

However Definition 20 of the CFPD (CFWD) has thedisadvantage that it completely loses its meaning if thefunction119909 fails to be pseudo (weakly) differentiable Preciselythe CFPD (in particular the CFWD) of a function 119909 loses itsmeaning if 119909 is not weakly absolutely continuous

Thenext lemma gives sufficient conditions that ensure theexistence of the Caputo fractional derivatives of a function119909 isin H119901(119864)Lemma 25 Let 120572 isin (0 1) For the function 119909 119868 rarr 119864 thefollowing hold

(a) If 119909 has a pseudo derivative 119863119901119909 isin H1199010 (119864) where 119901 gt1(1 minus 120572) then the CFPD of 119909 of order 120572 exists on 119868

Moreover (119889120572119901119889119905120572)119909 isin 119862[119868 119864120596](b) If 119864 is weakly complete or contains no copy of 1198880 and if119909 has a weak derivative 119863120596119909 isin H119901(119864) where 119901 isin[1infin] then the CFWD of 119909 of order 120572 exists on 119868

Moreover (119889120572120596119889119905120572)119909 isin 119862[119868 119864120596]This holds for any 119909 isin H119901(119864) with 119901 isin [1infin] if 119864 is reflexive

In all cases 120593((119889120572119901119889119905120572)119909) = (119889120572119901119889119905120572)120593119909 (120593((119889120572120596119889119905120572)119909) =(119889120572120596119889119905120572)120593119909) holds for every 120593 isin 119864lowastProof Since the weak (pseudo) derivative of an ae weakly(pseudo) differentiable function is strongly (weakly) mea-surable [12 26 28] the proof is readily available in view ofthe definition of Caputo fractional derivatives and Lemma 8Moreover since 119863120596119909(119863119901119909) isin H

1199010 (119864) with 119901 gt 1(1 minus 120572)

it follows in view of Lemma 14 that I1minus120572119863120596119909(I1minus120572119863119901119909) isin119862[119868 119864120596] This completes the proof

Besides the Caputo fractional-pseudo (weak) derivativeswe define the Riemann-Liouville fractional-pseudo (weak)derivatives

Definition 26 Let 119909 119868 rarr 119864 For the positive integer 119898such that 120572 isin (119898 minus 1119898) 119898 isin N0 fl 0 1 2 we definethe Riemann-Liouville fractional-pseudo (weak) derivativeldquoshortly RFPD (RFWD)rdquo of 119909 of order 120572 by

D120572119909 (119905) fl 119863119898

I119898minus120572119909 (119905) (63)

where 119863 is defined as in Definition 20 We use the notationD120572

119901 (D120572120596) to characterize the Riemann-Liouville fractional-

pseudo (weak) derivatives

Journal of Function Spaces 11

Clearly in infinite dimension Banach spaces the weakabsolute continuity of I119898minus120572119909 is necessarily (but not suf-ficient) condition for the existence of RFPD (in particularRFWD) of 119909Lemma 27 Let 0 lt 120572 lt 1 For any 119909 isin H

1199010 (119864) with 119901 gt

max1120572 1(1 minus 120572) we haveD

120572119901I

120572119909 = 119909 ae (64)

If 119864 is reflexive this is also true for every 119901 ge 1Proof Our assumption 119909 isin H

1199010 (119864)with 119901 gt max1120572 1(1minus120572) yields in view of Lemma 17 that

D120572119901I

120572119909 = 119863119901I1minus120572

I120572119909 = 119863119901I

1119909 (65)

The claim now follows immediately since (cf Proposition 19)the indefinite integral of Pettis integrable function is weaklyabsolutely continuous and it is pseudo differentiable withrespect to the right endpoint of the integration interval andits pseudo derivative equals the integrand at that point

Remark 28 When we replace D120572119901 by D120572

120596 then Lemma 27is no longer necessarily true for arbitrary 119909 isin H1

0(119864) evenwhen 119864 is reflexive evidently in [13 remark below Corollary4] the existence of a reflexive Banach space 119864 and a strongmeasurable Pettis integrable function 119909 119868 rarr 119864 was provedsuch that 119909 has nowhere weakly differentiable integral In thiscase 119863120596I

1119909 lost its meaning This gives a reason to believethat (65) (hence (64)) withD120572

120596119909 could not happen

However we have the following result

Lemma 29 Let 0 lt 120572 lt 1 For any 119909 isin 119862[119868 119864120596] we haveD

120572120596I

120572119909 = 119909 on 119868 (66)

Proof Our assumption 119909 isin 119862[119868 119864120596] sub Hinfin0 (119864) (see the proof

of Corollary 15) yields in view of Lemma 17 that

D120572120596I

120572119909 = 119863120596I1minus120572

I120572119909 = 119863120596I

1119909 (67)

The claim now follows immediately since the indefiniteintegral of weakly contentious function is weakly absolutelycontinuous and it is weakly differentiable with respect tothe right endpoint of the integration interval and its weakderivative equals the integrand at that point

The following lemma is folklore in case 119864 = R but to seethat it also holds in the vector-valued case we provide a proof

Lemma 30 Let 0 lt 120572 lt 1 and 119901 gt max1120572 1(1minus120572) If thefunction 119909 119868 rarr 119864 is weakly absolutely continuous on 119868 andpasses a pseudo derivative inH

1199010 (119864) then119889120572119901119909119889119905120572 (119905) = D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (68)

In particular if 119909 passes a weak derivative in 119862[119868 119864120596] then119889120572120596119909119889119905120572 (119905) = D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (69)

If 119864 is reflexive this is also true for every 119901 ge 1

Proof We observe that under the assumption imposed on119863119901119909 together with Proposition 3 the weakly absolutelycontinuity of 119909 is equivalent to

119909 (119905) = 119909 (0) + int119905

0119863119901119909 (119904) 119889119904 (70)

Hence owing to Lemma 27 it follows that

119889120572119901119909119889119905120572 = I1minus120572119863119901119909 = (D120572

119901I120572)I1minus120572119863119901119909

= D120572119901 (I120572

I1minus120572)119863119901119909 = (119863119901I

1minus120572)I1119863119901119909= 119863119901I

1minus120572 (119909 (119905) minus 119909 (0))= D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0)

(71)

which is what we wished to showThe proof of (69) is very similar to that in (68) therefore

we omit the details

Remark 31 As we remark above the definition of the CFPDof a function 119909 loses completely its meaning if 119909 is notweakly absolutely continuous For this reason we are able touse Lemma 30 to define the Caputo fractional derivative ingeneral that is we put

119889120572119901119909119889119905120572 (119905) fl D120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) 0 lt 120572 lt 1 (72)

Similarly we define

119889120572120596119909119889119905120572 (119905) fl D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (73)

However Lemma 30 claims that for the weakly absolutelycontinuous functions having Pettis integrable pseudo [weak]derivatives definitions (72) and (73) of the Caputo fractional-pseudo [weak] derivatives coincide with Definition 20

4 An Application

Let 119891 119868 times 119864 rarr 119864 be given function Consider the boundaryvalue problem of the fractional type

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin [0 1] 120582 isin R

119910 (0) + 119887119910 (1) = ℎ(74)

with certain constants 119887 ℎ isin R 119887 = minus1 To obtain theintegral equation modeled off the problem (74) we let 119910 bea weakly continuous solution to problem (74) then formallywe have

I120572 119889120572120596119889119905120572119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (75)

12 Journal of Function Spaces

that is

I120572I1minus120572119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) 997904rArrI1119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (76)

This reads (cf [29])

119910 (119905) = 119910 (0) + 120582I120572119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin 119868 120582 isin R (77)

with some (presently unknown) quantity 119910(0)Now we solve (77) for 119910(0) by 119910(0) + 119887119910(1) = ℎ and it

follows that

(1 + 119887) 119910 (0) + 120582119887Γ (120572) int1

0(1 minus 119904)120572minus1 119891 (119904 119910 (119904)) 119889119904 = ℎ (78)

Thus

119910 (0) = ℎ1 + 119887 minus 1205821198871 + 119887119868120572119891 (1 119910 (1)) (79)

Assume that the function 119891 is weakly-weakly continuousfunction such that

(1) for any 119903 gt 0 there is a constant 119872 gt 0 such that119891(119905 119910) le 119872 for all 119905 isin 119868 and 119910 le 119903(2) there exists a nondecreasing continuous function 120595 [0infin) rarr [0infin) 120595(0) = 0 and 120595(119905) lt 119905 for all 119905 gt 0

such that 120573(119891(119879 times 119860)) le 120595(120573(119860)) for every boundedset119860 sub 119864 where 120573 stands for De Blasirsquos weakmeasureof noncompactness (see [30])

Occasionally if 119891 119868 times 119864 rarr 119864 is weakly-weakly continuousand 119864 is reflexive then the assumptions (1) and (2) areautomatically satisfied (see eg [31])

Theorem 32 (see [6] Theorem 33) Let 120572 isin (0 1) and |120582| ltΓ(1 + 120572) If 119891 119868 times 119864 rarr 119864 is weakly-weakly continuous andsatisfies the assumptions (1) and (2) then the integral equation(77) has a weakly continuous solution 119910 defined on [0 1]Proof We omit the proof since it is almost identical to thatin the proof in ([6] Theorem 33) with (small) necessarychanges

In the following example we assume that 119910 isin 119862[119868 119864120596]solves (77) and we will show that not only do we have that 119910no longer necessarily solves (74) (when the Caputo fractionalweak derivative is taken in the sense of Definition 20)but even worse it could happen that the problem (74) isldquomeaninglessrdquo on 119868Example 33 Let 120572 isin (0 14] Let 119864 be an indefinitedimensional reflexive Banach space that fails cotype Definethe weakly-weakly continuous function 119891 [0 1] times 119864 rarr 119864by 119891 fl I1minus120572119909 where 119909 119868 rarr 119864 is defined by formula (7)Obviously (cf Example 6 in view of Remark 7)119891 satisfies theassumptions of Theorem 32

Now consider the integral equation (77) with119891 = I1minus120572119909Namely we consider the integral equation

119910 (119905) = 119910 (0) + 120582I120572I1minus120572119909 (119905)

120572 isin (0 14] 119905 isin 119868 120582 isin R (80)

Obviously the solution 119910 is weakly continuous on [0 1] (thisis an immediate consequence of Remark 7 and Lemma 17)

Since 120593119909 isin 1198711[0 1] holds for every 120593 isin 119864lowast it follows bythe commutative property of the fractional integral operatorscalculus over the space of Lebesgue integrable functions (see[21] Section 23) that

120593 (I120572I1minus120572119909 (119905)) = I

120572120593 (I1minus120572119909 (119905))= I

120572I1minus120572120593 (119909 (119905)) = I

1120593 (119909 (119905))= 120593 (I1119909 (119905))

(81)

that is

120593 (I120572I1minus120572119909 (119905) minusI

1119909 (119905)) = 0 for every 120593 isin 119864lowast (82)

Hence I120572I1minus120572119909 = I1119909 Consequently problem (80) be-comes

119910 (119905) = 119910 (0) + 120582I1119909 (119905) 120572 isin (0 14] 119905 isin 119868 (83)

As showed in ([13] Corollary 4) the indefinite integral 119905 rarrint1199050119909(119904)119889119904 is nowhere weakly differentiable on [0 1] Thus 119910

is nowhere weakly differentiable on [0 1] hence the CFWDof 119910 (hence the boundary value problem) completely loses itsmeaning if the CFWD is taken in the sense of Definition 20This is what we wished to show

Now we are in the position to state and prove thefollowing existence theorem

Theorem 34 If 119891 119868 times 119864 rarr 119864 is a function such that allconditions fromTheorem 32 hold then problem (74) (where theCaputo fractionalweak derivative is taken in the sense (73)) hasa weak solution on [0 1]Proof Let 119910 isin 119862[119868 119864120596] be a solution to (77) Equip 119864 andI times E with weak topology and note that 119905 997891rarr (119905 119910(119905)) iscontinuous as a mapping from 119868 into 119868 times 119864 Since 119891 119868 times 119864 rarr119864 is weakly-weakly continuous on [0 1] then 119891(sdot 119910(sdot)) is acomposition of this mapping with 119891 and thus 119891(sdot 119910(sdot)) isweakly continuous on [0 1]

Now operating by the operatorD120572120596 on both sides of (77)

it follows by Lemma 29 that

D120572120596119910 = D

120572120596119910 (0) + 120582D120572

120596I120572119891 = 119905minus120572Γ (1 minus 120572)119910 (0) + 120582119891 (84)

Now insert definition (73) of the Caputo fractionalderivative and we get

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) (85)

Journal of Function Spaces 13

With some further efforts one can get the boundary condi-tion 119910(0) + 119887119910(1) = ℎ

Therefore 119910 satisfies problem (74) which is what wewished to show

Data Availability

No data were used to support this study

Conflicts of Interest

The author declares no conflicts of interest

References

[1] H A Salem and A M El-Sayed ldquoWeak solution for fractionalorder integral equations in reflexive Banach spacesrdquoMathemat-ica Slovaca vol 55 no 2 pp 169ndash181 2005

[2] H A Salem A M El-Sayed and O L Moustafa ldquoA note onthe fractional calculus in Banach spacesrdquo Studia ScientiarumMathematicarum Hungarica vol 42 no 2 pp 115ndash130 2005

[3] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoMulti-term fractional differential equations in a nonreflexiveBanach spacerdquo Advances in Difference Equations 2013302 18pages 2013

[4] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoFractional calculus and fractional differential equations innonreflexive Banach spacesrdquo Communications in NonlinearScience andNumerical Simulation vol 20 no 1 pp 59ndash73 2015

[5] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoNonlinear fractional differential equations in nonreflexiveBanach spaces and fractional calculusrdquo Advances in DifferenceEquations 2015112 18 pages 2015

[6] R P Agarwal V Lupulescu D OrsquoRegan and G RahmanldquoWeak solutions for fractional differential equations in nonre-flexive Banach spaces via Riemann-Pettis integralsrdquo Mathema-tische Nachrichten vol 289 no 4 pp 395ndash409 2016

[7] H A Salem ldquoOn the fractional order m-point boundaryvalue problem in reflexive Banach spaces and weak topologiesrdquoJournal of Computational andAppliedMathematics vol 224 no2 pp 565ndash572 2009

[8] H A Salem ldquoOn the fractional calculus in abstract spaces andtheir applications to the Dirichlet-type problem of fractionalorderrdquo Computers and Mathematics with Applications vol 59no 3 pp 1278ndash1293 2010

[9] HAH Salem andMCichon ldquoOn solutions of fractional orderboundary value problems with integral boundary conditions inBanach spacesrdquo journal of function spaces and applications vol2013 Article ID 428094 2013

[10] J Diestel and J J Uhl Jr Vector Measures vol 15 ofMathemat-ical Surveys American Mathematical Society Providence RIUSA 1977

[11] R H Martin and J J Uhl Jr Nonlinear Operators AndDifferential Equations in Banach Spaces John Wiley amp SonsNew York NY USA 1976

[12] B J Pettis ldquoOn integration in vector spacesrdquo Transactions of theAmericanMathematical Society vol 44 no 2 pp 277ndash304 1938

[13] S J Dilworth and M Girardi ldquoNowhere weak differentiabilityof the Pettis integralrdquo Quaestiones Mathematicae vol 18 no 4pp 365ndash380 1995

[14] G A Edgar ldquoMeasurability in a Banach spacerdquo Indiana Univer-sity Mathematics Journal vol 26 no 4 pp 663ndash677 1977

[15] G A Edgar ldquoMeasurability in a Banach space IIrdquo IndianaUniversityMathematics Journal vol 28 no 4 pp 559ndash579 1979

[16] J L Gamez and J Mendoza ldquoOn denjoy-dunford and denjoy-pettis integralsrdquo StudiaMathematica vol 130 no 2 pp 115ndash1331998

[17] R AGordon ldquoTheDenjoy extension of the Bochner Pettis andDUNford integralsrdquo Studia Mathematica vol 92 no 1 pp 73ndash91 1989

[18] R F Geitz ldquoPettis integrationrdquo Proceedings of the AmericanMathematical Society vol 82 no 1 pp 81ndash86 1981

[19] R F Geitz ldquoGeometry and the Pettis integralrdquo Transactions ofthe AmericanMathematical Society vol 269 no 2 pp 535ndash5481982

[20] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations New York NYUSA Elsevier 2006

[21] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Yverdon Switzerland 1993

[22] M Mendel and A Naor ldquoMetric cotyperdquo Annals of Mathemat-ics Second Series vol 168 no 1 pp 247ndash298 2008

[23] B Ross S G Samko and E R Love ldquoFunctions that have nofirst order derivative might have fractional derivatives of allorders less than onerdquo Real Analysis Exchange vol 20 no 1 pp140ndash157 199495

[24] E Hille and R S Phillips Functional analysis and semi-groupsAmerican Mathematical Society Colloquium Publications vol31 American Mathematical Society Providence RI USA 1957

[25] S Schwabik and Y Guoju Topics in Banach Space IntegrationWorld Scientific Singapore 2005

[26] DW Solomon ldquoOn differentiability of vector-valued functionsof a real variablerdquo Studia Mathematica vol 29 pp 1ndash4 1967

[27] K Naralenkov ldquoOn Denjoy type extensions of the Pettisintegralrdquo Czechoslovak Mathematical Journal vol 60(135) no3 pp 737ndash750 2010

[28] D W Solomon Denjoy integration in abstract spaces Memorisof the American Mathematical Society No 85 AmericanMathematical Society Providence RI 1969

[29] A R Mitchell and C Smith ldquoAn existence theorem for weaksolutions of differential equations in Banach spacesrdquo inNonlin-ear Equations in Abstract Spaces V Lakshmikantham Ed pp387ndash404 1978

[30] F S De Blasi ldquoOn a property of the unit sphere in a Banachspacerdquo Bulletin Mathematiques De La Societe Des SciencesMathematiques De Roumanie vol 21 no 3 pp 259ndash262 1977

[31] A Szep ldquoExistence theorem for weak solutions of ordinarydifferential equations in reflexive Banach spacesrdquo Studia Scien-tiarum Mathematicarum Hungarica vol 6 pp 197ndash203 1971

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Page 2: On the Theory of Fractional Calculus in the Pettis-Function Spaces · 2019. 7. 30. · JournalofFunctionSpaces Further,wedenethespace H (),˙≥0by H () fl ˛ ∈ ˚, ˜: isHolderainoforder

2 Journal of Function Spaces

Further we define the spaceH120582(119864) 120582 ge 0 byH120582 (119864) fl 119909 isin 119862 [119868 119864120596]

120593119909 is Holderain of order 120582 on 119868 for every 120593isin 119864lowast

(3)

We also defineH0(119864) fl 119862[119868 119864120596]In the following proposition we summarize some impor-

tant facts which are the main tool in carrying out ourinvestigations (see [10 12 16 17])

Proposition 2 Let 119901 119902 isin [1infin] be of conjugate exponents(that is 1119901 + 1119902 = 1 with the convention that 1infin = 0) If119909 119868 rarr 119864 is weakly measurable then

(1) if 119909(sdot) isin 119875[119868 119864] and 119906(sdot) isin 119871infin[119868R] then the follow-ing hold 119909(sdot)119906(sdot) isin 119875[119868 119864]

(2) 119909 isin H1199010 (119864) 119901 gt 1 if and only if 119909(sdot)119906(sdot) isin 119875[119868 119864] for

every 119906(sdot) isin 119871119902(119868)(3) if 119864 is reflexive (containing no isometric copy of 1198880)

the weakly (strongly) measurable function 119909 119868 rarr 119864is Pettis integrable on 119868 if and only if 119909 is Dunfordintegrable on 119868

(4) for any 119901 gt 1 we have 119909 isin H119901(119864) 119909 stronglymeasurable sube H

1199010 (119864) If 119864 is weakly sequentially

complete this is also true for 119901 = 1We remark that there is a bounded weakly measurable

function which is not Pettis integrable (see eg [19])A fundamental property of Pettis integral is contained in

the following

Proposition 3 (see [12] Corollary 251) If 119909 isin 119875[119868 119864] thenfor any bounded subset Ω of elements of 119864lowast the integrals

int119869

1003816100381610038161003816120593 (119909 (119904))1003816100381610038161003816 119889119904 120593 isin Ω (4)

are weakly equi-absolutely continuous

Theorem 4 (mean value theorem for Pettis integral) If thefunction 119909 119868 rarr 119864 is Pettis integrable on 119868 then

int119869119909 (119904) 119889119904 isin |119869| conv (119909 (119869)) (5)

where 119869 sub 119868 |119869| is the length of 119869 and conv(119909(119869)) is the closedconvex hull of 119909(119869)2 Fractional Integrals ofVector-Valued Functions

In this section we define and study the Riemann-Liouvillefractional integral operators and the corresponding fractionalderivatives in Banach spaces

Devoted by the definition of the Riemann-Liouvillefractional integral of real-valued function we introduce thefollowing

Definition 5 Let 119909 119868 rarr 119864 The Riemann-Liouvillefractional Pettis integral (shortly RFPI) of 119909 of order 120572 gt 0is defined by

I120572119909 (119905) fl 1Γ (120572) int

119905

0(119905 minus 119904)120572minus1 119909 (119904) 119889119904 119905 gt 0 (6)

In the preceding definition ldquoint rdquo stands for the Pettisintegral

When 119864 = R it is well known (see eg [20 21]) thatthe operator I120572 sends 119871119902[0 119887] 119887 isin (0infin) continuously to119871119901[0 119887] if 119901 isin [1infin] satisfy 119902 gt 1(120572 + (1119901))

This seems to be a good place to put the following

Example 6 Let 119864 be an infinite dimensional Banach spacethat fails cotype (see [22] and the references therein) Definethe strongly measurable function 119909 [0 1] rarr 119864 by

119909 (119905) fl infinsum119899=1

2119899sum119896=1

119888119899 120594119860119899119896 (119905)120583 (119860119899119896) 119890119899119896 120583 Lebesgue measure (7)

with similar notations as in ([13] Corollary 4) where wechoose 119888119899 = 2119870120595(23minus119899) 119870 gt 1 120595(119905) fl 119905120598 120598 = 09 and119860119899

119896 = [1198861198963119899 1198871198963119899 ] 119899 = 0 1 2 119896 = 1 2 3 2119899 (8)

to be the fat Cantor sets (that is 120583(119860119899119896) = 13119899 holds for every119896 isin 1 2 3 2119899)

As cited in ([13] Corollary 4) 119909 is Pettis integrablefunctions on [0 1] whose indefinite integral is nowhereweakly differentiable on [0 1] Here we will show that 119909 hasRFPI of all order 120572 ge 34 and

I120572119909 (119905)= infinsum

119899=1

2119899sum119896=1

119888119899 ( 119890119899119896120583 (119860119899119896) Γ (120572) int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1 119889119904)isin 119864

(9)

Arguing similarly as in ([13] page 368) we have in view ofsum2119899

119896=1 120576119899119896119906119899119896ℓ2119899infin = 1 that sum2119899

119896=1 |120593(119890119899119896)| le 2120593 holds for every120593 isin 119864lowastAlso for any 119905 isin [0 1] and fixed 119899 isin N we have for some1198990 lt 119899 that

2119899sum119896=1

[int119905

0(119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899

119896)119889119904]

2

= 2119899sum119896=1

32119899 [int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1 119889119904]2

= 32119899 21198990sum

119896=1

[int119860119899119896

(119905 minus 119904)120572minus1 119889119904]2

Journal of Function Spaces 3

+ 32119899 [int119905

11988621198990 +1

(119905 minus 119904)120572minus1 119889119904]2

le 3211989912057221198990sum119896=1

[(119905 minus 1198861198963119899 )120572 minus (119905 minus 1198871198963119899)

120572]2

+ 321198991205722 (119905 minus 11988621198990+13119899 )2120572

le 3211989912057221198990sum119896=1

[(119905 minus 1198861198963119899 ) minus (119905 minus 1198871198963119899)]2120572

+ 321198991205722 (11988721198990+12119899 minus 11988621198990+12119899 )2120572 le 32119899211989912057232119899120572= 21198991205722 32119899(1minus120572)

(10)

Therefore for any measurable 119869 sub [0 1] we arrive atint[0119905]cap119869

infinsum119899=1

100381610038161003816100381610038161003816100381610038161003816100381610038162119899sum119896=1

119888119899Γ (120572) (119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899119896)120593 (119890119899119896) 119889119904

10038161003816100381610038161003816100381610038161003816100381610038161003816le int119905

0

infinsum119899=1

100381610038161003816100381610038161003816100381610038161003816100381610038162119899sum119896=1

119888119899Γ (120572) (119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899119896)120593 (119890119899119896) 119889119904

10038161003816100381610038161003816100381610038161003816100381610038161003816le infinsum

119899=1

119888119899Γ (120572) int119905

0

100381610038161003816100381610038161003816100381610038161003816100381610038162119899sum119896=1

(119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899119896)120593 (119890119899119896) 119889119904

10038161003816100381610038161003816100381610038161003816100381610038161003816le infinsum

119899=1

119888119899Γ (120572)2119899sum119896=1

1003816100381610038161003816120593 (119890119899119896)1003816100381610038161003816 int119905

0(119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899

119896)119889119904

le infinsum119899=1

119888119899Γ (120572) (2119899sum119896=1

1003816100381610038161003816120593 (119890119899119896)10038161003816100381610038162)12

sdot ( 2119899sum119896=1

[int119905

0(119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899

119896)119889119904]

2)12

le 2 10038171003817100381710038171205931003817100381710038171003817sdot infinsum119899=1

radic21198993119899(1minus120572)119888119899120572Γ (120572) = 411987023120598 10038171003817100381710038171205931003817100381710038171003817infinsum119899=1

2119899(12minus120598)3119899(1minus120572)Γ (1 + 120572)

(11)

Obviously the latter series converges whenever 120572 ge 34120598 = 09 which allows us to interchange the integral andsummation below to see that

1Γ (120572) int[0119905]cap119869 120593 ((119905 minus 119904)120572minus1 119909 (119904))= 1Γ (120572) int[0119905]cap119869 (119905 minus 119904)120572minus1 120593 (119909 (119904))= int

[0119905]cap119869

infinsum119899=1

2119899sum119896=1

119888119899Γ (120572) (119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899119896)120593 (119890119899119896) 119889119904

= infinsum119899=1

2119899sum119896=1

int[0119905]cap119869

119888119899Γ (120572) (119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899119896)120593 (119890119899119896) 119889119904

= 120593[infinsum119899=1

2119899sum119896=1

119888119899 ( 119890119899119896120583 (119860119899119896) Γ (120572) int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1 119889119904)]

(12)

It remains to prove that

infinsum119899=1

2119899sum119896=1

119888119899 ( 119890119899119896120583 (119860119899119896) Γ (120572) int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1 119889119904) isin 119864 (13)

Evidently we have10038171003817100381710038171003817100381710038171003817100381710038171003817119898sum119899=119897

2119899sum119896=1

119890119899119896119888119899 (int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1120583 (119860119899119896) Γ (120572)119889119904)

10038171003817100381710038171003817100381710038171003817100381710038171003817119864=10038171003817100381710038171003817100381710038171003817100381710038171003817119898sum119899=119897

2119899sum119896=1

119888119899 (int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1120583 (119860119899119896) Γ (120572)119889119904)119879119899119906119899119896

10038171003817100381710038171003817100381710038171003817100381710038171003817119864le 2 119898sum

119899=119897

119888119899100381710038171003817100381710038171003817100381710038171003817100381710038172119899sum119896=1

(int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1120583 (119860119899119896) Γ (120572)119889119904) 119906119899119896

10038171003817100381710038171003817100381710038171003817100381710038171003817ℓ2119899infin= 2Γ (120572)

119898sum119899=119897

3119899119888119899 max1le119896le2119899

100381610038161003816100381610038161003816100381610038161003816int[0119905]cap119860119899119896 (119905 minus 119904)120572minus1 120594119860119899119896(119904) 119889119904100381610038161003816100381610038161003816100381610038161003816

le 2Γ (120572)119898sum119899=119897

3119899119888119899 max1le119896le2119899

100381610038161003816100381610038161003816100381610038161003816int1198871198963119899

1198861198963119899(119905 minus 119904)120572minus1 119889119904100381610038161003816100381610038161003816100381610038161003816

le 411987023120598Γ (1 + 120572)119898sum119899=119897

3119899(1minus120572)2120598119899

(14)

which approaches zero as 119897 119898 rarr infin as needed for (9)

Remark 7 Observe Example 6 We remark the following

(1) There is a reflexive Banach space for which theindefinite Pettis integral of the function 119909 defined by(7) is nowhere weakly differentiable on [0 1] (see [13]the remark below Corollary 4) Meanwhile 119909 has aRFPI of all order 120572 ge 34

(2) The function I120572119909(sdot) is weakly continuous on [0 1]this follows easily from the definition of the Pettisintegral In fact we have in view of (9) that

120593 (I120572119909 (119905)) = I120572120593 (119909 (119905))

= infinsum119899=1

2119899sum119896=1

119888119899120593 (119890119899119896)120583 (119860119899119896) Γ (120572) int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1 119889119904 (15)

holds for every 120593 isin 119864lowast Since100381610038161003816100381610038161003816100381610038161003816100381610038162119899sum119896=1

119888119899120593 (119890119899119896)120583 (119860119899119896) Γ (120572) int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1 11988911990410038161003816100381610038161003816100381610038161003816100381610038161003816

le 119888119899( 2119899sum119896=1

1003816100381610038161003816120593 (119890119899119896)10038161003816100381610038162)12

sdot ( 2119899sum119896=1

[int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1120583 (119860119899119896) Γ (120572)119889119904]

2)12

le 411987023120598 10038171003817100381710038171205931003817100381710038171003817 2119899(12minus120598)3119899(1minus120572)Γ (1 + 120572)

(16)

4 Journal of Function Spaces

then the infinite series of continuous functions in theleft hand side of (15) converges uniformly in [0 1]Hence the function 120593(I120572119909(sdot)) is continuous on [0 1](this yields the weak continuity ofI120572119909(sdot) on [0 1])

In the following lemma we gather together some simpleparticular sufficient conditions that ensure the existence ofthe Riemann-Liouville fractional integral of the functionsfromH119901(119864)Lemma 8 Let 119909 119868 rarr 119864 be weakly measurable function TheRFPI of the function 119909 of order 120572 gt 0 makes sense ae on 119868 ifat least one of the following cases holds

(a) 119909 isin H1199010 (119864) 119901 isin [1infin] and 120572 ge 1

(b) 119909 isin H1199010 (119864) 119901 gt max1 1120572

(c) 119909 is strongly measurable which lies in H119901(119864) where119901 isin [1infin] provided that 119864 is weakly complete orcontains no copy of 1198880

If 119864 is reflexive this is also true for any 119909 isin H119901(119864) 119901 isin [1infin]and 120572 gt 0

In all cases 120593(I120572119909) = I120572120593119909 holds for every 120593 isin 119864lowastIn the assertions ((a) and (b)) we find sufficient con-

ditions needed for the existence of I120572119909 in the situation inwhich no restriction is placed on 119864 In the third assertion theproperties of119864 allow us to characterize a function 119909 isin H119901(119864)for whichI120572119909 exists

Proof Firstly assertion (a) is direct consequence of Proposi-tion 2 (part (1)) since 119904 rarr (119905 minus 119904)120572minus1 isin 119871infin[0 119905] holds foralmost every 119905 isin 119868 whenever 120572 ge 1

Secondly to prove assertion (b) let 119901 gt max1 1120572 and119902 be the conjugate exponents to 119901 Since 119902(120572 minus 1) + 1 gt 0 wehave that 119904 rarr (119905 minus 119904)120572minus1 isin 119871119902[0 119905] holds for every 119905 isin 119868 Thusthe assertion (b) is now an easy consequence of Proposition 2(part (2))

Thirdly to prove (c) we let 120572 gt 0 119901 isin [1infin]and 119909 isin H119901(119864) be strongly measurable Since the strongmeasurability is preserved under multiplication operation onfunctions the product (119905 minus sdot)120572minus1119909(sdot) [0 119905] rarr 119864 is stronglymeasurable on [0 119905] for almost every 119905 isin 119868 In view of Youngrsquosinequality for every 120593 isin 119864lowast the real-valued function

119904 997891997888rarr 120593 ((119905 minus 119904)120572minus1 119909 (119904)) = (119905 minus 119904)120572minus1 120593119909 (119904) (17)

is Lebesgue integrable on [0 119905] for almost every 119905 isin 119868 So theassertion (c) follows immediately from Proposition 2 (parts(3 4))

Similarly when 119864 is reflexive the result follows from part(3) of Proposition 2However in all cases the function 119904 997891rarr (119905 minus 119904)120572minus1119909(119904) is

Pettis integrable on [0 119905] for almost every 119905 isin 119868 That is foralmost every 119905 isin 119868 there exists an element in 119864 denoted byI120572119909(119905) such that

120593 (I120572119909 (119905)) = int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904

= int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120593119909 (119904) 119889119904 = I120572120593119909 (119905)

(18)

holds for every 120593 isin 119864lowast This completes the proof

Remark 9 If 119909 isin H1199010 (119864) such that I120572119909(119905) does not exist for

some 119905 isin 119868 then it does not exist even when we ldquoenlargerdquo thespace 119864 into 119865 To see this let 119909 119868 rarr 119865 such that 119909(119868) sub 119864 IfI120572119909(119905) exists for some 119905 isin 119868 then (119905 minus sdot)120572minus1119909(sdot) isin 119875([0 119905] 119865)Since 119909 assumes only values in 119864 it follows by the mean valuetheorem for Pettis integral (Theorem 4) that the RFPI of 119909should lie in 119864

Before we come to a deep study of the mathematicalproperties of the RFPI operator let us take a look at thefollowing miscellaneous examples

Example 10 Let 120572 gt 0 Define the function 119909 [0 1] rarr119871infin[0 1] by 119909(119905) = 120594[0119905]This function is weakly measurable Pettis integrable on[0 1] and 120593119909 is a function of bounded variation (see eg

[18]) That is 119909 isin Hinfin0 (119871infin) Hence in view of Lemma 8 with119901 = infin the RFPI of 119909 exists on [0 1] Further calculations

(cf [4]) show that

I120572119909 (119905) (sdot) = (119905 minus sdot)120572Γ (1 + 120572)120594[0119905] (sdot) isin 119871infin [0 1] (19)

Example 11 Let120572 gt 0 Define the function119909 from the interval[0 1] into the Hilbert space ℓ2 as119909 119905 997891997888rarr 1(1 + 119905) 1(2 + 119905) 1(3 + 119905) 119905 isin [0 1] (20)

We note that

119909 (119905)2ℓ2 = sum119899

( 1119899 + 119905)2 le sum

119899

11198992 lt infin 119905 isin [0 1] (21)

Thus the function 119909 is well defined We claim that 119909 isDunford integrable on [0 1] Once our claim is establishedLemma 8 guarantees the existence ofI120572119909 on [0 1] It remainsto prove this claim and to calculate I120572119909

To see this let 120593 isin (ℓ2)lowast = ℓ2 According to theRiesz representation theorem on Hilbert spaces there existsa uniquely determined 120582 fl 120582119899 isin ℓ2 such that 120593119909(119905) =sum119899(120582119899(119899 + 119905)) A standard arguments using Beppo LevirsquosTheorem yields

int1

0

1003816100381610038161003816120593119909 (119905)1003816100381610038161003816 119889119905 le sum119899

int1

0

10038161003816100381610038161205821198991003816100381610038161003816(119899 + 119905)119889119905 = sum119899

10038161003816100381610038161205821198991003816100381610038161003816 ln(1 + 1119899)le sum

119899

10038161003816100381610038161205821198991003816100381610038161003816119899 le sum119899

10038161003816100381610038161205821198991003816100381610038161003816 lt infin(22)

So 119909 is Dunford integrable on [0 1] and hence Pettis inte-grable on [0 1] since ℓ2 is reflexive Consequently in view of

Journal of Function Spaces 5

Lemma 8 I120572119909 exists on [0 1] To calculate the RFPI of 119909 fix119905 isin [0 1] and let 120593 isin (ℓ2)lowast We have

int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904 = int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120593119909 (119904) 119889119904= I

120572 (infinsum119899=1

120582119899(119899 + 119905)) (23)

Since the series sum119899(120582119899(119899 + 119905)) is uniformly convergent on[0 1] it follows in view of the generalized linearity of thefractional integrals [23 Lemma 5] that (cf [21 Table 91])

int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904= infinsum

119899=1

120582119899119905120572119899Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 ) = 120593 (119892 (119905)) (24)

where 2F1 is the Gauss hypergeometric function evaluated at(1 1 120572 + 1 minus119905119899) and119892 (119905) fl 119905120572119899Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 ) (25)

Since

1003817100381710038171003817119892 (119905)10038171003817100381710038172ℓ2 = sum119899

( 119905120572Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 ))2

le 2F21 (1 1 120572 + 1 minus1)Γ2 (1 + 120572) sum

119899

11198992 lt infin119905 isin [0 1]

(26)

we see that 119892 isin ℓ2 Thus

I120572119909 (119905) = 119905120572119899Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 )

119905 isin [0 1] (27)

Example 12 Let 120572 isin [12 1) Define the countable-valuedfunction 119909 [0 1] rarr 1198880 by

119909 (119905) fl 119909119899 (119905) 119909119899 fl 119899120594(1(119899+1)1119899] (28)

This function is strongly measurable Pettis integrable func-tion on [0 1] (see eg [10 16]) We claim that 119909 isin H119875

0 (1198880)with 119901 isin (1120572 2] Once our claim is established Lemma 8guarantees the existence of I120572119909 120572 isin [12 1) on [0 1] Itremains to prove this claim by showing firstly that H119875(1198880)and to calculate I120572119909 To do this let 120593 isin 119888lowast0 Then therecorresponds to 120593 a unique 120582119899 isin ℓ1 such that 120593119909 = sum119899 120582119899119909119899By noting that

10038171003817100381710038171205821198991199091198991003817100381710038171003817119901 = (int1

0

1003816100381610038161003816120582119899119909119899 (119905)1003816100381610038161003816119901 119889119905)1119901

= 10038161003816100381610038161205821198991003816100381610038161003816 (int1119899

1(119899+1)119899119901119889119905)1119901 = 10038161003816100381610038161205821198991003816100381610038161003816 1198991119902(119899 + 1)1119901

le 10038161003816100381610038161205821198991003816100381610038161003816 11989911199021198991119901 le 100381610038161003816100381612058211989910038161003816100381610038161198992119901minus1 le 10038161003816100381610038161205821198991003816100381610038161003816

(29)

holds for any 119901 isin (1120572 2] a standard argument using LevirsquosTheorem (or Lebesgue dominated convergence theorem) andMinkowskirsquos inequality yields

[int1

0

1003816100381610038161003816120593119909 (119905)1003816100381610038161003816119901 119889119905]1119901 le [int1

0(infinsum119899=1

10038161003816100381610038161205821198991003816100381610038161003816 119909119899 (119905))119901 119889119905]1119901

= [[lim119896rarrinfin

int1

0( 119896sum119899=1

10038161003816100381610038161205821198991003816100381610038161003816 119909119899 (119905))119901

119889119905]]1119901

le [[lim119896rarrinfin

( 119896sum119899=1

10038171003817100381710038171205821198991199091198991003817100381710038171003817119901)119901]]1119901

le infinsum119899=1

10038171003817100381710038171205821198991199091198991003817100381710038171003817119901le infinsum

119899=1

10038161003816100381610038161205821198991003816100381610038161003816 lt infin

(30)

Thus 120593119909(sdot) isin 119871119901[0 1] 119901 isin (1120572 2] holds for every 120593 isin 119888lowast0(that is 119909 isin H119875(1198880)) Since 119901 gt 1120572 gt 1 it follows byProposition 2 in view of the strong measurability of 119909 that119909 isin H119875

0 (1198880) Owing to Lemma 8 we infer that the RFPI of119909 of any order 120572 isin [12 1) exists on the interval [0 1] Tocompute this integral fix 119905 isin 1198681198990 fl [1(1198990+1) 11198990] for some1198990 isin N the set of natural numbers and let 120593 isin 119888lowast0 Then

sum119899

int119905

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 12058211989910038161003816100381610038161003816 119909119899 (119904) 119889119904le 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816 int

119905

1(1198990+1)(119905 minus 119904)120572minus1 1198990119889119904

+ sum119899gt1198990

10038161003816100381610038161205821198991003816100381610038161003816 int1119899

1(119899+1)(119905 minus 119904)120572minus1 119899 119889119904

= 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572

+ sum119899gt1198990

119899 10038161003816100381610038161205821198991003816100381610038161003816120572 [(119905 minus 1119899 + 1)120572 minus (119905 minus 1119899)

120572]

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572

+ sum119899gt1198990

119899 10038161003816100381610038161205821198991003816100381610038161003816120572 [(119905 minus 1119899 + 1) minus (119905 minus 1119899)]120572

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572 + sum

119899gt1198990

119899 10038161003816100381610038161205821198991003816100381610038161003816120572 1119899120572 (119899 + 1)120572

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572 + sum

119899gt1198990

10038161003816100381610038161205821198991003816100381610038161003816120572 11198992120572minus1

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572 + sum

119899gt1198990

10038161003816100381610038161205821198991003816100381610038161003816120572 lt infin(because 120572 isin [12 1))

(31)

6 Journal of Function Spaces

Therefore by Beppo Levi Theorem it follows that

int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904 = int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120593 (119909 (119904)) 119889119904= int119905

0

(119905 minus 119904)120572minus1Γ (120572) sum119899

120582119899119899120594(1(119899+1)1119899] (119904) 119889119904= sum

119899

int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120582119899119899120594(1(119899+1)1119899] (119904) 119889119904= 11989901205821198990Γ (1 + 120572) (119905 minus 11198990 + 1)

120572

+ sum119899gt1198990

119899120582119899Γ (1 + 120572) [(119905 minus 1119899 + 1)120572 minus (119905 minus 1119899)

120572]

(32)

Consequently we conclude that

I120572119909 (119905) = 0 0⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

1198990minus1

1198990Γ (1 + 120572) (119905 minus 11198990 + 1)120572

(1198990 + 1)Γ (1 + 120572) [(119905 minus 11198990 + 2)

120572 minus (119905 minus 11198990 + 1)120572]

(33)

where the nonzero coordinate started from the 1198990th place Itcan be easily seen thatI120572119909(119905) isin 1198880 for any 119905 isin [0 1] Evidentlywe have that10038161003816100381610038161003816100381610038161003816119899 [(119905 minus

1119899 + 1)120572 minus (119905 minus 1119899)

120572]10038161003816100381610038161003816100381610038161003816le 119899 [(119905 minus 1119899 + 1) minus (119905 minus 1119899)]

120572 le 119899 [ 1119899 (119899 + 1)]120572

le 11198992120572minus1 997888rarr 0 as 119899 997888rarr infin1198990Γ (1 + 120572) (119905 minus 11198990 + 1)

120572

le 1198990Γ (1 + 120572) ( 11198990 minus 11198990 + 1)120572 le 11198992120572minus10 Γ (1 + 120572)

997888rarr 0 as 1198990 997888rarr infin

(34)

hold for any 119905 isin [0 1]This yieldsI1205721119909(119905) isin 1198880 for any 119905 isin [0 1]

(this is precisely what we would expect from Definition 5)

Look at part (c) of Lemma 8 with 120572 isin (0 1) and119864 = 1198880 Below we give an example showing that the stronglymeasurability hypothesis imposed on a function 119909 isin H1(119864)H1

0(119864) is not sufficient for the existence of I120572119909 even when 119909is Denjoy-Pettis integrable (cf [16])

Example 13 Let 120572 isin (0 1) Define the strongly measurablefunction 119909 [0 1] rarr 1198880 by119909 (119905) fl 119909119899 (119905)

119909119899 (119905) fl 119899 (119899 + 1) [120594119868119899 (119905) minus 120594119869119899 (119905)] (35)

where

119868119899 fl ( 1119899 + 1 119899 + 12119899 (119899 + 1)) 119869119899 fl ( 119899 + 12119899 (119899 + 1) 1119899)

(36)

It is immediate that (cf [16]) 119909 is a well-defined Denjoy-Pettis (but it is not Pettis) integrable on [0 1]

In what follows we will show that the RFPI of 119909 does notexist on a subinterval of positive measure on [0 1]

To see this we make use of Proposition 3 as followsdefine for each intervalI119899 = [1(119899 + 1) 1119899] the functionals120593119899 isin 119888lowast0 (required by Proposition 3) to be the correspondingto the element 120582119899 fl (0 0 0 1 0 0 ) isin ℓ1 wherethe nonzero coordinate is in the 119899th place Then 120593119899(119909(119905)) =119909119899(119905) and thus |120593119899(119909(119904))| = 119899(119899 + 1)[120594119868119899(119905) minus 120594119869119899(119905)] Clearlythe family 120593119899 runs through the unit ball of the dual of1198880 Evidently the isometric isomorphism between 119888lowast0 and ℓ1yields 120593119899119888lowast0 = 120582119899ℓ1 = 1

Now take 119905 isin [12 1] and define the continuous function119891 [12 1] rarr R+ by

119891 (119905) fl (119905 minus 1119899 + 1)120572 minus (119905 minus 1119899)

120572 119899 ge 2 119905 isin [12 1]

(37)

It can be easily seen that in view of 0 lt 120572 lt 1 1198911015840 is negativeon (0 1) By standard results from (classical) calculus itfollows that 119891 is strictly decreasing on [0 1] in particular119891(1) lt 119891(119905) for all 119905 isin [0 1) Thus for any 119899 ge 2 and any120593 isin 120593119899 we have

intI119899

100381610038161003816100381610038161003816100381610038161003816120593 ((119905 minus 119904)120572minus1Γ (120572) 119909 (119904))100381610038161003816100381610038161003816100381610038161003816 119889119904

= intI119899

(119905 minus 119904)120572minus1Γ (120572) 1003816100381610038161003816120593 (119909 (119904))1003816100381610038161003816 119889119904= int1119899

1(119899+1)

(119905 minus 119904)120572minus1Γ (120572) 119899 (119899 + 1) 119889119904= 119899 (119899 + 1)Γ (1 + 120572) [(119905 minus 1119899 + 1)

120572 minus (119905 minus 1119899)120572]

ge 1198992Γ (1 + 120572) [(1 minus 1119899 + 1)120572 minus (1 minus 1119899)

120572]

(38)

An explicit calculation using LrsquoHospitalrsquos rule two timesreveals

lim119899rarrinfin

(1 minus 1 (119899 + 1))120572 minus (1 minus 1119899)120572119899minus2 = minus1205722sdot lim119899rarrinfin

1198993 [( 119899119899 + 1)120572minus1 1

(119899 + 1)2 minus (1 minus 1119899)120572minus1 11198992 ]

= minus1205722 lim119899rarrinfin

119899 [( 119899119899 + 1)120572minus1 1198992

(119899 + 1)2 minus (1 minus 1119899)120572minus1]

Journal of Function Spaces 7

= minus1205722 lim119899rarrinfin

119899 [( 119899119899 + 1)120572+1 minus (1119899)

120572minus1] = minus1205722sdot lim119899rarrinfin

1198992 [(120572 + 1) ( 119899119899 + 1)120572 1(119899 + 1)2

minus (120572 minus 1) (1 minus 1119899)120572minus2 11198992 ] = minus1205722

sdot lim119899rarrinfin

[(120572 + 1) ( 119899119899 + 1)120572+2 minus (120572 minus 1) (1 minus 1119899)

120572minus2]= 120572

(39)

from which it follows that

intI119899

100381610038161003816100381610038161003816100381610038161003816120593 ((119905 minus 119904)120572minus1Γ (120572) 119909 (119904))100381610038161003816100381610038161003816100381610038161003816 119889119904997888rarr0 as 119899 997888rarr infin (40)

Therefore in view of Proposition 3 119904 997891rarr (119905 minus 119904)120572minus1119909(119904) notin119875([0 1] 1198880) Hence I120572119909 does not exist which is what wewished to show

The following theorem provides a useful characterizationof the space H119901(119864) for which the statements reveal howmuch the fractional integral I120572 is better than the function119909 isin H119901(119864) Indeed based on Lemma 8 using an inequalityof Young we can easily prove the following

Lemma 14 For any 120572 gt 0 the following holds(a) If 119864 is reflexive then for every 119901 isin [1infin] the operator

I120572 mapsH119901(119864) intoH1199010 (119864) In particular if 0 lt 120572 lt1 then I120572 H1(119864) rarr H1((1minus120572)minus120598)0 (119864) however small120598 gt 0 is

(b) If 119864 contains no copy of 1198880 then for every 119901 isin [1infin]the operator I120572 maps 119909 isin H

1199010 (119864) 119909 strongly

measurable into H1199010 (119864) In particular if 120572 ge 1 the

operator I120572 mapsH1199010 (119864) into itself

(c) If 119901 gt max1 1120572 the operator I120572 sends H1199010 (119864) to119862[119868 119864120596] (if we define I120572119909(0) fl 0) In particular if120572 isin (0 1) I120572 H119901

0 (119864) rarr H120572minus1119901(119864)Proof At the beginning let 120572 gt 0 and 119909 isin H119901(119864) with119901 isin [1infin] and define the real-valued function 119892 119868 rarr R

by 119892(119905) fl 119905120572minus1Γ(120572) If 119864 is reflexive it follows in view ofLemma 8 that J120572119909 exists ae in 119868 and 120593(I120572119909) = I120572120593119909for every 120593 isin 119864lowast Youngrsquos inequality guarantees that theconvolution product 119892 lowast 120593119909(sdot) lies in 119871119901[119868] Consequently120593(I120572119909) = I120572120593119909 = 119892 lowast 120593119909 isin 119871119901[119868] for every 120593 isin 119864lowastThe reflexivity of 119864 together with Proposition 2 yields I120572119909 isinH

1199010 (119864)In particular if 0 lt 120572 lt 1 it can be easily seen thatint

119868(119892(119905))119902119889119905 lt infin for every 119902 isin [1 1(1 minus 120572)) That is 119892 isin1198711(1minus120572)minus120598[119868] 120598 gt 0 By Youngrsquos inequality it follows that120593(I120572119909) = 119892 lowast 120593119909 isin 1198711(1minus120572)minus120598[119868] for every 120593 isin 119864lowast however

small 120598 gt 0 is Now the assertion (a) follows because of thereflexivity of 119864

Next in order to prove the assertion (b) let 119909 isin H1199010 (119864)

and note in view of Lemma 8 that I120572119909 exists on 119868 Define119910 119868 rarr 119864 by119910(119905) fl I120572119909(119905) 119909 isin H1199010 (119864) Since120593119910(sdot) isin 119871119901[119868]

for every120593 isin 119864lowast it follows that119910 isin H119901(119864) Moreover for any119886 119887 isin 119868 (119886 lt 119887) we haveint119887

119886120593119910 (119904) 119889119905 = int119887

119886I120593119909 (119905) 119889119905 = 1Γ (120572)

sdot int119887

119886int119905

0(119905 minus 119904)120572minus1 120593119909 (119904) 119889119904 119889119905

= 1Γ (120572) [int119887

119886int119887

119904(119905 minus 119904)120572minus1 120593119909 (119904) 119889119905 119889119904

+ int119886

0int119887

119886(119905 minus 119904)120572minus1 120593119909 (119904) 119889119905 119889119904]

= 1Γ (120572) [int119887

119886(119887 minus 119904)120572 120593119909 (119904) 119889119904

+ int119886

0[(119887 minus 119904)120572 minus (119886 minus 119904)120572] 120593119909 (119904) 119889119904]

= 1Γ (120572) [int119886

0(119887 minus 119904)120572 120593119909 (119904) 119889119904

+ int119887

119886(119887 minus 119904)120572 120593119909 (119904) 119889119904 minus int119886

0(119886 minus 119904)120572 120593119909 (119904) 119889119904]

= int119887

0

(119887 minus 119904)120572Γ (1 + 120572)120593119909 (119904) 119889119904 minus int119886

0

(119886 minus 119904)120572Γ (1 + 120572)120593119909 (119904) 119889119904= 120593 (119909[119886119887])

(41)

where

119909[119886119887] fl (119875) int119887

0

(119887 minus 119904)120572Γ (1 + 120572)119909 (119904) 119889119904minus (119875) int119886

0

(119886 minus 119904)120572Γ (1 + 120572)119909 (119904) 119889119904(42)

Since 119909 isin 119875[119868 119864] owing to Proposition 2 (part (1)) it followsthat (119887 minus sdot)120572119909(119904) and (119886 minus sdot)120572119909(119904) are Pettis integrable on 119868 andso 119909[119886119887] isin 119864

A combination of these results yields120593119910 isin 119871119901(119868) for every120593 isin 119864lowast and there exists an element 119909[119886119887] isin 119864 such that120601119909[119886119887] = int119887

119886120593119910(119905)119889119905 for every 119886 119887 isin 119868 Since 119864 contains no

copy of 1198880 it follows that (cf [17 Theorem 23]) 119910 isin 119875[119868 119864]Consequently 119910(sdot) isin H

1199010 (119864) This is the claim (b)

To prove the assertion (c) let 119901 gt max1 1120572 and 119909 isinH

1199010 (119864) By Lemma 8 we deduce that I120572119909 exists ae on 119868

Now let 119902 isin [1infin] be such that 1119901 + 1119902 = 1 Since119902(120572 minus 1) gt minus1 119892 fl (sdot)120572minus1Γ(120572) isin 119871119902[119868] Therefore as a directconsequence of Youngrsquo inequality it follows that

120593 (I120572119909 (sdot)) = I120572120593119909 (sdot) = 119892 lowast 120593119909 (sdot) isin 119862 [119868R] (43)

holds for every 120593 isin 119864lowast Now we claim that I120572119909(119905) rarr 0in 119864 as 119905 rarr 0 Once our claim is established the definition

8 Journal of Function Spaces

I120572119909(0) fl 0 guarantees thatI120572119909(sdot) isin 119862[119868 119864120596] which is whatwewished to show It remains to prove our claim without lossof generality assume that I120572119909(119905) = 0 Then there exists (asa consequence of the Hahn-Banach theorem) 120593 isin 119864lowast with120593 = 1 such that I120572119909(119905) = 120593(I120572119909(119905)) = I120572120593119909(119905) ByHolderrsquos inequality we obtain

1003816100381610038161003816120593 (I120572119909 (119905))1003816100381610038161003816le 1Γ (120572) (int

119905

0(119905 minus 119904)119902(120572minus1) 119889119904)1119902 (int119905

0

1003816100381610038161003816120593119909 (119904)1003816100381610038161003816119901 119889119904)1119901

le 1Γ (120572) ( 119905(120572minus1)119902+1(120572 minus 1) 119902 + 1)1119902 (int

119868

1003816100381610038161003816120593119909 (119904)1003816100381610038161003816119901 119889119904)1119901

= 119905120572minus1+1119902Γ (120572) [119902 (120572 minus 1) + 1]1119902

10038171003817100381710038171205931199091003817100381710038171003817119871119901

= 119905120572minus1119901 10038171003817100381710038171205931199091003817100381710038171003817119871119901Γ (120572) [119902 (120572 minus 1) + 1]1119902

(44)

This is equivalent with the following estimate

1003817100381710038171003817I120572119909 (119905)1003817100381710038171003817 le 119905120572minus1119901 10038171003817100381710038171205931199091003817100381710038171003817119871119901Γ (120572) [119902 (120572 minus 1) + 1]1119902 (45)

Owing to 120572 gt 1119901 we get I120572119909(119905) rarr 0 in 119864 as 119905 rarr 0 andconsequently in view ofI120572119909(0) fl 0I120572119909(sdot) isin 119862[119868 119864120596] Thisproves the first part of the assertion (c)

Finally let 120572 isin (0 1) and 119905 120591 isin 119868 Without loss ofgenerality assume 120591 lt 119905 Then for any 120593 isin 119864lowast we have byHolder inequality with 119901 gt 1120572 in view 119902(120572 minus 1) gt minus1 that

1003816100381610038161003816120593 (I120572119909 (119905) minusI120572119909 (120591))1003816100381610038161003816 Γ (120572)

= 10038161003816100381610038161003816100381610038161003816int119905

0(119905 minus 119904)120572minus1 120593119909 (119904) 119889119904 minus int120591

0(120591 minus 119904)120572minus1 120593119909 (119904) 11988911990410038161003816100381610038161003816100381610038161003816

le (int120591

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 minus (120591 minus 119904)120572minus110038161003816100381610038161003816 1003816100381610038161003816120593119909 (119904)1003816100381610038161003816 119889119904+ int119905

120591(119905 minus 119904)120572minus1 1003816100381610038161003816120593119909 (119904)1003816100381610038161003816 119889119904)

= [(int120591

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 minus (120591 minus 119904)120572minus110038161003816100381610038161003816119902 119889119904)1119902

+ (int119905

120591(119905 minus 119904)(120572minus1)119902 119889119904)1119902] 10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]

le [[(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

+ ((119905 minus 120591)(120572minus1)119902+1(120572 minus 1) 119902 + 1 )1119902]]10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]

= [(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

+ (119905 minus 120591)120572minus1119901[119902 (120572 minus 1) + 1]1119902]

10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868] (46)

By noting that (120572 minus 1)119902 isin (minus1 0) when 120572 isin (0 1) it can beeasily seen that

int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904le 1119902 (120572 minus 1) + 1 (119905 minus 120591)119902(120572minus1)+1

(47)

That is

(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

le (119905 minus 120591)120572minus1119901[119902 (120572 minus 1) + 1]1119902

(48)

A combination of these results yields1003816100381610038161003816120593 (I120572119909 (119905)) minus 120593 (I120572119909 (120591))1003816100381610038161003816= 1003816100381610038161003816120593 (I120572119909 (119905) minusI

120572119909 (120591))1003816100381610038161003816le10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]Γ (120572) 2 (119905 minus 120591)120572minus1119901

[119902 (120572 minus 1) + 1]1119902 (49)

ThusI120572119909 isin H120572minus1119901(119864) This completes the proof

Corollary 15 For any 120572 gt 0 I120572 119862[119868 119864120596] rarr 119862[119868 119864120596]Proof Let 119909 isin 119862[119868 119864120596] then 120593119909 isin 119871infin holds for every120593 isin 119864lowast By noting that the weak continuity implies a strongmeasurability ([24] page 73) it follows that 119909 is stronglymeasurable on 119868 Hence in view of Proposition 2 119909 isinHinfin

0 (119864) Consequently 119862[119868 119864120596] sube H1199010 (119864) for every 119901 isin[1infin] That isI120572 119862[119868 119864120596] rarr 119862[119868 119864120596] Hence the desired

result is obtained

Example 16 Let 120572 isin [12 1) Define the function 119909 [0 1] rarr1198880 by 119909(119905) = 119899120594(1(119899+1)1119899]We have (in view of Example 12)that

I120572119909 (119905) = 0 0 0 1198990 (119905 minus 1 (1198990 + 1))120572

Γ (1 + 120572) (1198990 + 1)Γ (1 + 120572) [(119905 minus 11198990 + 2)

120572 minus (119905 minus 11198990 + 1)120572]

(50)

To show that I120572119909 is norm continuous on [0 1] let 120591 119905 isin[0 1] With no loss of generality we may assume that 119905 120591 isin1198681198990 for some 1198990 isin N Since the nonzero terms of the sequencein (50) are nonincreasing then in view of

|119905 minus 120591| le ( 11198990 minus 11198990 + 1) = 11198990 (1198990 + 1) le 111989920 (51)

Journal of Function Spaces 9

we have1003817100381710038171003817I120572119909 (119905) minusI

1205721119909 (120591)10038171003817100381710038171198880

= 1198990Γ (1 + 120572)10038161003816100381610038161003816100381610038161003816(119905 minus

1119899 + 1)120572 minus (120591 minus 1119899 + 1)

12057210038161003816100381610038161003816100381610038161003816le 1198990Γ (1 + 120572)

1003816100381610038161003816100381610038161003816(119905 minus 1119899 + 1) minus (120591 minus 1119899 + 1)1003816100381610038161003816100381610038161003816120572

le 1198990Γ (1 + 120572) |119905 minus 120591|120572 le 1Γ (1 + 120572) |119905 minus 120591|120572radic|119905 minus 120591|= |119905 minus 120591|120572minus12Γ (1 + 120572)

(52)

Thus the RFPI of 119909 is norm continuous on [0 1] Preciselysince 0 lt 120572 minus 12 lt 120572 minus 1119901 lt 1 for any 119901 le 2 thenI120572119909 isin H120572minus12(119864) sub H120572minus1119901(119864) This is precisely what onewould expect from Lemma 14 (part (c))

Analogously an explicit calculation reveals that I120572119909 liesin 119862([0 1] 119871infin[0 1]) where 119909 [0 1] rarr 119871infin[0 1] defined byExample 10

We now consider additional mapping properties of theoperator I120572 Precisely we will show that the RFPI enjoysthe following commutative property which is folklore in case119864 = R However the proof is completely similar to that of [8]Lemma 35

Lemma 17 Let 120572 120573 gt 0 and 119901 gt max1 1120572 1120573 Then

I120572I120573119909 = I

120573I120572119909 = I

120572+120573119909 (53)

holds for every 119909 isin H1199010 (119864) If 119864 is reflexive this is also true for

every 119901 ge 1Proof Since 119901 gt max1 1120572 1120573 it follows that 119901 gtmax1 1(120572+120573) So by Lemmas 8 and 14 themappingsI120572119909I120573119909 andI120572+120573119909 belong toHinfin

0 (119864) for every 119909 isin H1199010 (119864) Now

repeating the same process as in ([8] Lemma 35) implies theclaim

When 119864 is reflexive the result follows as a direct applica-tion of Lemma 14

3 Fractional Derivatives ofVector-Valued Functions

After the notation of the fractional integrals of vector-valued functions the fractional derivatives become a naturalrequirement Before we come to the definitions and a detailedstudy of the mathematical properties of fractional differentialoperators we recall the following

Definition 18 Consider the vector-valued function 119909 119868 rarr119864(1) Let 120593119909 be differentiable on 119868 for every 120593 isin 119864lowast The

function 119909 is said to be weakly differentiable on 119868 if

there exists 119910 119868 rarr 119864 such that for every 120593 isin 119864lowast wehave

119889120593119909 (119905)119889119905 = 120593119910 (119905) for every 119905 isin 119868 (54)

The function 119910 is called the weak derivative of thefunction 119909

(2) Let 120593119909 be differentiable ae on 119868 for every 120593 isin 119864lowast(the null set may vary with 120593 isin 119864lowast) The function 119909is said to be pseudo differentiable on 119868 if there existsa function 119910 119868 rarr 119864 such that for every 120593 isin 119864lowast thereexists a null set119873(120593) sub 119868 such that

(120593119909 (119905))1015840 = 120593119910 (119905) for every 119905 isin 119868119873 (120593) (55)

In this case the function 119910 is called the pseudoderivative of 119909If 119909 is pseudo differentiable on 119868 and the null setinvariant for every 120593 isin 119864lowast then 119909 is ae weaklydifferentiable on 119868

Clearly if 119909 is ae weakly differentiable on 119868 then 120593(119909)is ae differentiable on 119868 The converse holds in a weaklysequentially complete space (see [25] Theorem 733)

For more details of the derivatives of vector-valuedfunctions we refer to [10 12 26]

The following results play a major role in our analysis

Proposition 19 (see [27] Theorem 51) The function 119910 119868 rarr119864 is an indefinite Pettis integrable if and only if 119910 is weaklyabsolutely continuous on 119868 and have a pseudo derivative on 119868In this case 119910 is an indefinite Pettis integral of any of its pseudoderivatives

Now we are in the position to define the fractional-typederivatives of vector-valued functions

Definition 20 Let 119909 119868 rarr 119864 For the positive integer119898 suchthat 120572 isin (119898 minus 1119898) 119898 isin N0 fl 0 1 2 we define theCaputo fractional-pseudo (weak) derivative ldquoshortly CFPD(CFWD)rdquo of 119909 of order 120572 by

119889120572119889119905120572 119909 (119905) fl I119898minus120572119863119898119909 (119905) (56)

where the sign ldquo119863rdquo denotes the pseudo (or weak) differentialoperator We use the notation 119889120572119901119889119905120572 and 119889120572120596119889119905120572 to charac-terize the Caputo fractional-pseudo derivatives and Caputofractional weak derivatives respectively

It is well known that although the weak derivative of aweakly differentiable function is uniquely determined thepseudo derivative of the pseudo differentiable function is notunique Also although any two pseudo derivatives 119910 119911 of afunction 119909 119868 rarr 119864 need not be ae equal (see [12 Example91] and [26 p 2]) the functions 119910 119911 are weakly equivalenton 119868 (that is 120593119910 = 120593119911 holds ae for every 120593 isin 119864lowast) The

10 Journal of Function Spaces

next lemma provides a useful characterization property of theCFPD Really it can be easily seen that the CFPD of a Caputofractional-pseudo differentiable function does not depend onthe choice of a pseudo derivative of the function

Lemma 21 Let 119909 119868 rarr 119864 be pseudo differentiable functionwhere pseudo derivatives lie in H

1199010 (119864) If (119889120572119901119889119905120572)119909 exists on119868 then the CFPD of 119909 depends on the choice of the pseudo

derivatives of 119909Proof Let 119910 119911 isin H

1199010 (119864) be two pseudo derivatives of the

pseudo differentiable function 119909 Since 119909 and 119910 are weaklyequivalent on 119868 then for every 120593 isin 119864lowast there exists a nullset N which depends on 120593 isin 119864lowast such that (119905 minus 119904)minus120572120593119910(119904) =(119905minus119904)minus120572120593119911(119904) for every 119904 isin [0 119905]N Consequently for almostevery 119905 isin 119868

120593 (I1minus120572119910 (119905)) = I1minus120572120593119910 (119905) = I

1minus120572120593119911 (119905)= 120593 (I1minus120572119911 (119905)) (57)

holds for every 120593 isin 119864lowast Thus

119889120572119901119889119905120572 119909 = I1minus120572119863119901119909 = I

1minus120572119910 = I1minus120572119911 on 119868 (58)

which is what we wished to show

We consider the following examples

Example 22 Define 119909 [0 1] rarr ℓ2 by119909 (119905) fl ln(1 + 119905119899) 119899 isin N (59)

We claim that the CFWD of 119909 exists on [0 1] To see this welet 120593 isin (ℓ2)lowast = ℓ2 Then there corresponds to 120593 a sequence120582119899 isin ℓ2 such that

120593119909 (119905) = sum119899

120582119899 ln(1 + 119905119899) (60)

Since the series in (60) converges uniformly on [0 1] theformal differentiation of 120593119909 yields

(120593119909 (119905))1015840 = sum119899

120582119899119899 + 119905 = 120593 ( 1119905 + 119899) (61)

It is not hard to justify the differentiation by noting that theseries in the right hand side of (61) is uniformly and absolutelyconvergent on [0 1] Consequently 120593119909 is differentiable on[0 1] for 120593 isin (ℓ2)lowast meaning that 119909 is weakly differentiable on[0 1] (because ℓ2 is weakly complete (cf [25] Theorem 733)and 119863120596119909(119905) = 1(119905 + 119899) Owing to Lemma 25 the CFWDof any order 120572 isin (0 1) of 119909 exists on [0 1] To calculate theCFWD of 119909 on [0 1] we observe in view of Example 11 that

119889120572120596119889119905120572 119909 (119905) = I1minus120572119863120596119909 (119905)

= 1199051minus120572119899Γ (2 minus 120572) 2F1 (1 1 2 minus 120572 minus119905119899 ) (62)

Example 23 Let119864 be an infinite dimensionBanach space thatfails cotype Define 119910 [0 1] rarr 119864 by 119910(119905) fl int119905

0119909(119904)119889119904

where 119909 [0 1] rarr 119864 given by formula (7) As cited in ([13]Corollary 4) 119910 is nowhere weakly differentiable on [0 1]Consequently the CFWD of any order 120572 isin [34 1) of 119910 losesits meaning on [0 1]Remark 24 As shown in Example 23 there is an infinitedimension Banach space 119864 and weakly absolutely continuousfunction 119910 119868 rarr 119864 which is nowhere weakly differentiable(hence the CFWD of 119910 does not exist) Also even when 119864 isseparable and 119910 is Lipschitz function the pseudo derivatives(hence the CFPD) of 119910 need not to exist [28]

However Definition 20 of the CFPD (CFWD) has thedisadvantage that it completely loses its meaning if thefunction119909 fails to be pseudo (weakly) differentiable Preciselythe CFPD (in particular the CFWD) of a function 119909 loses itsmeaning if 119909 is not weakly absolutely continuous

Thenext lemma gives sufficient conditions that ensure theexistence of the Caputo fractional derivatives of a function119909 isin H119901(119864)Lemma 25 Let 120572 isin (0 1) For the function 119909 119868 rarr 119864 thefollowing hold

(a) If 119909 has a pseudo derivative 119863119901119909 isin H1199010 (119864) where 119901 gt1(1 minus 120572) then the CFPD of 119909 of order 120572 exists on 119868

Moreover (119889120572119901119889119905120572)119909 isin 119862[119868 119864120596](b) If 119864 is weakly complete or contains no copy of 1198880 and if119909 has a weak derivative 119863120596119909 isin H119901(119864) where 119901 isin[1infin] then the CFWD of 119909 of order 120572 exists on 119868

Moreover (119889120572120596119889119905120572)119909 isin 119862[119868 119864120596]This holds for any 119909 isin H119901(119864) with 119901 isin [1infin] if 119864 is reflexive

In all cases 120593((119889120572119901119889119905120572)119909) = (119889120572119901119889119905120572)120593119909 (120593((119889120572120596119889119905120572)119909) =(119889120572120596119889119905120572)120593119909) holds for every 120593 isin 119864lowastProof Since the weak (pseudo) derivative of an ae weakly(pseudo) differentiable function is strongly (weakly) mea-surable [12 26 28] the proof is readily available in view ofthe definition of Caputo fractional derivatives and Lemma 8Moreover since 119863120596119909(119863119901119909) isin H

1199010 (119864) with 119901 gt 1(1 minus 120572)

it follows in view of Lemma 14 that I1minus120572119863120596119909(I1minus120572119863119901119909) isin119862[119868 119864120596] This completes the proof

Besides the Caputo fractional-pseudo (weak) derivativeswe define the Riemann-Liouville fractional-pseudo (weak)derivatives

Definition 26 Let 119909 119868 rarr 119864 For the positive integer 119898such that 120572 isin (119898 minus 1119898) 119898 isin N0 fl 0 1 2 we definethe Riemann-Liouville fractional-pseudo (weak) derivativeldquoshortly RFPD (RFWD)rdquo of 119909 of order 120572 by

D120572119909 (119905) fl 119863119898

I119898minus120572119909 (119905) (63)

where 119863 is defined as in Definition 20 We use the notationD120572

119901 (D120572120596) to characterize the Riemann-Liouville fractional-

pseudo (weak) derivatives

Journal of Function Spaces 11

Clearly in infinite dimension Banach spaces the weakabsolute continuity of I119898minus120572119909 is necessarily (but not suf-ficient) condition for the existence of RFPD (in particularRFWD) of 119909Lemma 27 Let 0 lt 120572 lt 1 For any 119909 isin H

1199010 (119864) with 119901 gt

max1120572 1(1 minus 120572) we haveD

120572119901I

120572119909 = 119909 ae (64)

If 119864 is reflexive this is also true for every 119901 ge 1Proof Our assumption 119909 isin H

1199010 (119864)with 119901 gt max1120572 1(1minus120572) yields in view of Lemma 17 that

D120572119901I

120572119909 = 119863119901I1minus120572

I120572119909 = 119863119901I

1119909 (65)

The claim now follows immediately since (cf Proposition 19)the indefinite integral of Pettis integrable function is weaklyabsolutely continuous and it is pseudo differentiable withrespect to the right endpoint of the integration interval andits pseudo derivative equals the integrand at that point

Remark 28 When we replace D120572119901 by D120572

120596 then Lemma 27is no longer necessarily true for arbitrary 119909 isin H1

0(119864) evenwhen 119864 is reflexive evidently in [13 remark below Corollary4] the existence of a reflexive Banach space 119864 and a strongmeasurable Pettis integrable function 119909 119868 rarr 119864 was provedsuch that 119909 has nowhere weakly differentiable integral In thiscase 119863120596I

1119909 lost its meaning This gives a reason to believethat (65) (hence (64)) withD120572

120596119909 could not happen

However we have the following result

Lemma 29 Let 0 lt 120572 lt 1 For any 119909 isin 119862[119868 119864120596] we haveD

120572120596I

120572119909 = 119909 on 119868 (66)

Proof Our assumption 119909 isin 119862[119868 119864120596] sub Hinfin0 (119864) (see the proof

of Corollary 15) yields in view of Lemma 17 that

D120572120596I

120572119909 = 119863120596I1minus120572

I120572119909 = 119863120596I

1119909 (67)

The claim now follows immediately since the indefiniteintegral of weakly contentious function is weakly absolutelycontinuous and it is weakly differentiable with respect tothe right endpoint of the integration interval and its weakderivative equals the integrand at that point

The following lemma is folklore in case 119864 = R but to seethat it also holds in the vector-valued case we provide a proof

Lemma 30 Let 0 lt 120572 lt 1 and 119901 gt max1120572 1(1minus120572) If thefunction 119909 119868 rarr 119864 is weakly absolutely continuous on 119868 andpasses a pseudo derivative inH

1199010 (119864) then119889120572119901119909119889119905120572 (119905) = D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (68)

In particular if 119909 passes a weak derivative in 119862[119868 119864120596] then119889120572120596119909119889119905120572 (119905) = D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (69)

If 119864 is reflexive this is also true for every 119901 ge 1

Proof We observe that under the assumption imposed on119863119901119909 together with Proposition 3 the weakly absolutelycontinuity of 119909 is equivalent to

119909 (119905) = 119909 (0) + int119905

0119863119901119909 (119904) 119889119904 (70)

Hence owing to Lemma 27 it follows that

119889120572119901119909119889119905120572 = I1minus120572119863119901119909 = (D120572

119901I120572)I1minus120572119863119901119909

= D120572119901 (I120572

I1minus120572)119863119901119909 = (119863119901I

1minus120572)I1119863119901119909= 119863119901I

1minus120572 (119909 (119905) minus 119909 (0))= D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0)

(71)

which is what we wished to showThe proof of (69) is very similar to that in (68) therefore

we omit the details

Remark 31 As we remark above the definition of the CFPDof a function 119909 loses completely its meaning if 119909 is notweakly absolutely continuous For this reason we are able touse Lemma 30 to define the Caputo fractional derivative ingeneral that is we put

119889120572119901119909119889119905120572 (119905) fl D120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) 0 lt 120572 lt 1 (72)

Similarly we define

119889120572120596119909119889119905120572 (119905) fl D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (73)

However Lemma 30 claims that for the weakly absolutelycontinuous functions having Pettis integrable pseudo [weak]derivatives definitions (72) and (73) of the Caputo fractional-pseudo [weak] derivatives coincide with Definition 20

4 An Application

Let 119891 119868 times 119864 rarr 119864 be given function Consider the boundaryvalue problem of the fractional type

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin [0 1] 120582 isin R

119910 (0) + 119887119910 (1) = ℎ(74)

with certain constants 119887 ℎ isin R 119887 = minus1 To obtain theintegral equation modeled off the problem (74) we let 119910 bea weakly continuous solution to problem (74) then formallywe have

I120572 119889120572120596119889119905120572119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (75)

12 Journal of Function Spaces

that is

I120572I1minus120572119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) 997904rArrI1119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (76)

This reads (cf [29])

119910 (119905) = 119910 (0) + 120582I120572119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin 119868 120582 isin R (77)

with some (presently unknown) quantity 119910(0)Now we solve (77) for 119910(0) by 119910(0) + 119887119910(1) = ℎ and it

follows that

(1 + 119887) 119910 (0) + 120582119887Γ (120572) int1

0(1 minus 119904)120572minus1 119891 (119904 119910 (119904)) 119889119904 = ℎ (78)

Thus

119910 (0) = ℎ1 + 119887 minus 1205821198871 + 119887119868120572119891 (1 119910 (1)) (79)

Assume that the function 119891 is weakly-weakly continuousfunction such that

(1) for any 119903 gt 0 there is a constant 119872 gt 0 such that119891(119905 119910) le 119872 for all 119905 isin 119868 and 119910 le 119903(2) there exists a nondecreasing continuous function 120595 [0infin) rarr [0infin) 120595(0) = 0 and 120595(119905) lt 119905 for all 119905 gt 0

such that 120573(119891(119879 times 119860)) le 120595(120573(119860)) for every boundedset119860 sub 119864 where 120573 stands for De Blasirsquos weakmeasureof noncompactness (see [30])

Occasionally if 119891 119868 times 119864 rarr 119864 is weakly-weakly continuousand 119864 is reflexive then the assumptions (1) and (2) areautomatically satisfied (see eg [31])

Theorem 32 (see [6] Theorem 33) Let 120572 isin (0 1) and |120582| ltΓ(1 + 120572) If 119891 119868 times 119864 rarr 119864 is weakly-weakly continuous andsatisfies the assumptions (1) and (2) then the integral equation(77) has a weakly continuous solution 119910 defined on [0 1]Proof We omit the proof since it is almost identical to thatin the proof in ([6] Theorem 33) with (small) necessarychanges

In the following example we assume that 119910 isin 119862[119868 119864120596]solves (77) and we will show that not only do we have that 119910no longer necessarily solves (74) (when the Caputo fractionalweak derivative is taken in the sense of Definition 20)but even worse it could happen that the problem (74) isldquomeaninglessrdquo on 119868Example 33 Let 120572 isin (0 14] Let 119864 be an indefinitedimensional reflexive Banach space that fails cotype Definethe weakly-weakly continuous function 119891 [0 1] times 119864 rarr 119864by 119891 fl I1minus120572119909 where 119909 119868 rarr 119864 is defined by formula (7)Obviously (cf Example 6 in view of Remark 7)119891 satisfies theassumptions of Theorem 32

Now consider the integral equation (77) with119891 = I1minus120572119909Namely we consider the integral equation

119910 (119905) = 119910 (0) + 120582I120572I1minus120572119909 (119905)

120572 isin (0 14] 119905 isin 119868 120582 isin R (80)

Obviously the solution 119910 is weakly continuous on [0 1] (thisis an immediate consequence of Remark 7 and Lemma 17)

Since 120593119909 isin 1198711[0 1] holds for every 120593 isin 119864lowast it follows bythe commutative property of the fractional integral operatorscalculus over the space of Lebesgue integrable functions (see[21] Section 23) that

120593 (I120572I1minus120572119909 (119905)) = I

120572120593 (I1minus120572119909 (119905))= I

120572I1minus120572120593 (119909 (119905)) = I

1120593 (119909 (119905))= 120593 (I1119909 (119905))

(81)

that is

120593 (I120572I1minus120572119909 (119905) minusI

1119909 (119905)) = 0 for every 120593 isin 119864lowast (82)

Hence I120572I1minus120572119909 = I1119909 Consequently problem (80) be-comes

119910 (119905) = 119910 (0) + 120582I1119909 (119905) 120572 isin (0 14] 119905 isin 119868 (83)

As showed in ([13] Corollary 4) the indefinite integral 119905 rarrint1199050119909(119904)119889119904 is nowhere weakly differentiable on [0 1] Thus 119910

is nowhere weakly differentiable on [0 1] hence the CFWDof 119910 (hence the boundary value problem) completely loses itsmeaning if the CFWD is taken in the sense of Definition 20This is what we wished to show

Now we are in the position to state and prove thefollowing existence theorem

Theorem 34 If 119891 119868 times 119864 rarr 119864 is a function such that allconditions fromTheorem 32 hold then problem (74) (where theCaputo fractionalweak derivative is taken in the sense (73)) hasa weak solution on [0 1]Proof Let 119910 isin 119862[119868 119864120596] be a solution to (77) Equip 119864 andI times E with weak topology and note that 119905 997891rarr (119905 119910(119905)) iscontinuous as a mapping from 119868 into 119868 times 119864 Since 119891 119868 times 119864 rarr119864 is weakly-weakly continuous on [0 1] then 119891(sdot 119910(sdot)) is acomposition of this mapping with 119891 and thus 119891(sdot 119910(sdot)) isweakly continuous on [0 1]

Now operating by the operatorD120572120596 on both sides of (77)

it follows by Lemma 29 that

D120572120596119910 = D

120572120596119910 (0) + 120582D120572

120596I120572119891 = 119905minus120572Γ (1 minus 120572)119910 (0) + 120582119891 (84)

Now insert definition (73) of the Caputo fractionalderivative and we get

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) (85)

Journal of Function Spaces 13

With some further efforts one can get the boundary condi-tion 119910(0) + 119887119910(1) = ℎ

Therefore 119910 satisfies problem (74) which is what wewished to show

Data Availability

No data were used to support this study

Conflicts of Interest

The author declares no conflicts of interest

References

[1] H A Salem and A M El-Sayed ldquoWeak solution for fractionalorder integral equations in reflexive Banach spacesrdquoMathemat-ica Slovaca vol 55 no 2 pp 169ndash181 2005

[2] H A Salem A M El-Sayed and O L Moustafa ldquoA note onthe fractional calculus in Banach spacesrdquo Studia ScientiarumMathematicarum Hungarica vol 42 no 2 pp 115ndash130 2005

[3] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoMulti-term fractional differential equations in a nonreflexiveBanach spacerdquo Advances in Difference Equations 2013302 18pages 2013

[4] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoFractional calculus and fractional differential equations innonreflexive Banach spacesrdquo Communications in NonlinearScience andNumerical Simulation vol 20 no 1 pp 59ndash73 2015

[5] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoNonlinear fractional differential equations in nonreflexiveBanach spaces and fractional calculusrdquo Advances in DifferenceEquations 2015112 18 pages 2015

[6] R P Agarwal V Lupulescu D OrsquoRegan and G RahmanldquoWeak solutions for fractional differential equations in nonre-flexive Banach spaces via Riemann-Pettis integralsrdquo Mathema-tische Nachrichten vol 289 no 4 pp 395ndash409 2016

[7] H A Salem ldquoOn the fractional order m-point boundaryvalue problem in reflexive Banach spaces and weak topologiesrdquoJournal of Computational andAppliedMathematics vol 224 no2 pp 565ndash572 2009

[8] H A Salem ldquoOn the fractional calculus in abstract spaces andtheir applications to the Dirichlet-type problem of fractionalorderrdquo Computers and Mathematics with Applications vol 59no 3 pp 1278ndash1293 2010

[9] HAH Salem andMCichon ldquoOn solutions of fractional orderboundary value problems with integral boundary conditions inBanach spacesrdquo journal of function spaces and applications vol2013 Article ID 428094 2013

[10] J Diestel and J J Uhl Jr Vector Measures vol 15 ofMathemat-ical Surveys American Mathematical Society Providence RIUSA 1977

[11] R H Martin and J J Uhl Jr Nonlinear Operators AndDifferential Equations in Banach Spaces John Wiley amp SonsNew York NY USA 1976

[12] B J Pettis ldquoOn integration in vector spacesrdquo Transactions of theAmericanMathematical Society vol 44 no 2 pp 277ndash304 1938

[13] S J Dilworth and M Girardi ldquoNowhere weak differentiabilityof the Pettis integralrdquo Quaestiones Mathematicae vol 18 no 4pp 365ndash380 1995

[14] G A Edgar ldquoMeasurability in a Banach spacerdquo Indiana Univer-sity Mathematics Journal vol 26 no 4 pp 663ndash677 1977

[15] G A Edgar ldquoMeasurability in a Banach space IIrdquo IndianaUniversityMathematics Journal vol 28 no 4 pp 559ndash579 1979

[16] J L Gamez and J Mendoza ldquoOn denjoy-dunford and denjoy-pettis integralsrdquo StudiaMathematica vol 130 no 2 pp 115ndash1331998

[17] R AGordon ldquoTheDenjoy extension of the Bochner Pettis andDUNford integralsrdquo Studia Mathematica vol 92 no 1 pp 73ndash91 1989

[18] R F Geitz ldquoPettis integrationrdquo Proceedings of the AmericanMathematical Society vol 82 no 1 pp 81ndash86 1981

[19] R F Geitz ldquoGeometry and the Pettis integralrdquo Transactions ofthe AmericanMathematical Society vol 269 no 2 pp 535ndash5481982

[20] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations New York NYUSA Elsevier 2006

[21] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Yverdon Switzerland 1993

[22] M Mendel and A Naor ldquoMetric cotyperdquo Annals of Mathemat-ics Second Series vol 168 no 1 pp 247ndash298 2008

[23] B Ross S G Samko and E R Love ldquoFunctions that have nofirst order derivative might have fractional derivatives of allorders less than onerdquo Real Analysis Exchange vol 20 no 1 pp140ndash157 199495

[24] E Hille and R S Phillips Functional analysis and semi-groupsAmerican Mathematical Society Colloquium Publications vol31 American Mathematical Society Providence RI USA 1957

[25] S Schwabik and Y Guoju Topics in Banach Space IntegrationWorld Scientific Singapore 2005

[26] DW Solomon ldquoOn differentiability of vector-valued functionsof a real variablerdquo Studia Mathematica vol 29 pp 1ndash4 1967

[27] K Naralenkov ldquoOn Denjoy type extensions of the Pettisintegralrdquo Czechoslovak Mathematical Journal vol 60(135) no3 pp 737ndash750 2010

[28] D W Solomon Denjoy integration in abstract spaces Memorisof the American Mathematical Society No 85 AmericanMathematical Society Providence RI 1969

[29] A R Mitchell and C Smith ldquoAn existence theorem for weaksolutions of differential equations in Banach spacesrdquo inNonlin-ear Equations in Abstract Spaces V Lakshmikantham Ed pp387ndash404 1978

[30] F S De Blasi ldquoOn a property of the unit sphere in a Banachspacerdquo Bulletin Mathematiques De La Societe Des SciencesMathematiques De Roumanie vol 21 no 3 pp 259ndash262 1977

[31] A Szep ldquoExistence theorem for weak solutions of ordinarydifferential equations in reflexive Banach spacesrdquo Studia Scien-tiarum Mathematicarum Hungarica vol 6 pp 197ndash203 1971

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Page 3: On the Theory of Fractional Calculus in the Pettis-Function Spaces · 2019. 7. 30. · JournalofFunctionSpaces Further,wedenethespace H (),˙≥0by H () fl ˛ ∈ ˚, ˜: isHolderainoforder

Journal of Function Spaces 3

+ 32119899 [int119905

11988621198990 +1

(119905 minus 119904)120572minus1 119889119904]2

le 3211989912057221198990sum119896=1

[(119905 minus 1198861198963119899 )120572 minus (119905 minus 1198871198963119899)

120572]2

+ 321198991205722 (119905 minus 11988621198990+13119899 )2120572

le 3211989912057221198990sum119896=1

[(119905 minus 1198861198963119899 ) minus (119905 minus 1198871198963119899)]2120572

+ 321198991205722 (11988721198990+12119899 minus 11988621198990+12119899 )2120572 le 32119899211989912057232119899120572= 21198991205722 32119899(1minus120572)

(10)

Therefore for any measurable 119869 sub [0 1] we arrive atint[0119905]cap119869

infinsum119899=1

100381610038161003816100381610038161003816100381610038161003816100381610038162119899sum119896=1

119888119899Γ (120572) (119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899119896)120593 (119890119899119896) 119889119904

10038161003816100381610038161003816100381610038161003816100381610038161003816le int119905

0

infinsum119899=1

100381610038161003816100381610038161003816100381610038161003816100381610038162119899sum119896=1

119888119899Γ (120572) (119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899119896)120593 (119890119899119896) 119889119904

10038161003816100381610038161003816100381610038161003816100381610038161003816le infinsum

119899=1

119888119899Γ (120572) int119905

0

100381610038161003816100381610038161003816100381610038161003816100381610038162119899sum119896=1

(119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899119896)120593 (119890119899119896) 119889119904

10038161003816100381610038161003816100381610038161003816100381610038161003816le infinsum

119899=1

119888119899Γ (120572)2119899sum119896=1

1003816100381610038161003816120593 (119890119899119896)1003816100381610038161003816 int119905

0(119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899

119896)119889119904

le infinsum119899=1

119888119899Γ (120572) (2119899sum119896=1

1003816100381610038161003816120593 (119890119899119896)10038161003816100381610038162)12

sdot ( 2119899sum119896=1

[int119905

0(119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899

119896)119889119904]

2)12

le 2 10038171003817100381710038171205931003817100381710038171003817sdot infinsum119899=1

radic21198993119899(1minus120572)119888119899120572Γ (120572) = 411987023120598 10038171003817100381710038171205931003817100381710038171003817infinsum119899=1

2119899(12minus120598)3119899(1minus120572)Γ (1 + 120572)

(11)

Obviously the latter series converges whenever 120572 ge 34120598 = 09 which allows us to interchange the integral andsummation below to see that

1Γ (120572) int[0119905]cap119869 120593 ((119905 minus 119904)120572minus1 119909 (119904))= 1Γ (120572) int[0119905]cap119869 (119905 minus 119904)120572minus1 120593 (119909 (119904))= int

[0119905]cap119869

infinsum119899=1

2119899sum119896=1

119888119899Γ (120572) (119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899119896)120593 (119890119899119896) 119889119904

= infinsum119899=1

2119899sum119896=1

int[0119905]cap119869

119888119899Γ (120572) (119905 minus 119904)120572minus1 120594119860119899119896 (119904)120583 (119860119899119896)120593 (119890119899119896) 119889119904

= 120593[infinsum119899=1

2119899sum119896=1

119888119899 ( 119890119899119896120583 (119860119899119896) Γ (120572) int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1 119889119904)]

(12)

It remains to prove that

infinsum119899=1

2119899sum119896=1

119888119899 ( 119890119899119896120583 (119860119899119896) Γ (120572) int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1 119889119904) isin 119864 (13)

Evidently we have10038171003817100381710038171003817100381710038171003817100381710038171003817119898sum119899=119897

2119899sum119896=1

119890119899119896119888119899 (int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1120583 (119860119899119896) Γ (120572)119889119904)

10038171003817100381710038171003817100381710038171003817100381710038171003817119864=10038171003817100381710038171003817100381710038171003817100381710038171003817119898sum119899=119897

2119899sum119896=1

119888119899 (int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1120583 (119860119899119896) Γ (120572)119889119904)119879119899119906119899119896

10038171003817100381710038171003817100381710038171003817100381710038171003817119864le 2 119898sum

119899=119897

119888119899100381710038171003817100381710038171003817100381710038171003817100381710038172119899sum119896=1

(int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1120583 (119860119899119896) Γ (120572)119889119904) 119906119899119896

10038171003817100381710038171003817100381710038171003817100381710038171003817ℓ2119899infin= 2Γ (120572)

119898sum119899=119897

3119899119888119899 max1le119896le2119899

100381610038161003816100381610038161003816100381610038161003816int[0119905]cap119860119899119896 (119905 minus 119904)120572minus1 120594119860119899119896(119904) 119889119904100381610038161003816100381610038161003816100381610038161003816

le 2Γ (120572)119898sum119899=119897

3119899119888119899 max1le119896le2119899

100381610038161003816100381610038161003816100381610038161003816int1198871198963119899

1198861198963119899(119905 minus 119904)120572minus1 119889119904100381610038161003816100381610038161003816100381610038161003816

le 411987023120598Γ (1 + 120572)119898sum119899=119897

3119899(1minus120572)2120598119899

(14)

which approaches zero as 119897 119898 rarr infin as needed for (9)

Remark 7 Observe Example 6 We remark the following

(1) There is a reflexive Banach space for which theindefinite Pettis integral of the function 119909 defined by(7) is nowhere weakly differentiable on [0 1] (see [13]the remark below Corollary 4) Meanwhile 119909 has aRFPI of all order 120572 ge 34

(2) The function I120572119909(sdot) is weakly continuous on [0 1]this follows easily from the definition of the Pettisintegral In fact we have in view of (9) that

120593 (I120572119909 (119905)) = I120572120593 (119909 (119905))

= infinsum119899=1

2119899sum119896=1

119888119899120593 (119890119899119896)120583 (119860119899119896) Γ (120572) int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1 119889119904 (15)

holds for every 120593 isin 119864lowast Since100381610038161003816100381610038161003816100381610038161003816100381610038162119899sum119896=1

119888119899120593 (119890119899119896)120583 (119860119899119896) Γ (120572) int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1 11988911990410038161003816100381610038161003816100381610038161003816100381610038161003816

le 119888119899( 2119899sum119896=1

1003816100381610038161003816120593 (119890119899119896)10038161003816100381610038162)12

sdot ( 2119899sum119896=1

[int[0119905]cap119860119899

119896

(119905 minus 119904)120572minus1120583 (119860119899119896) Γ (120572)119889119904]

2)12

le 411987023120598 10038171003817100381710038171205931003817100381710038171003817 2119899(12minus120598)3119899(1minus120572)Γ (1 + 120572)

(16)

4 Journal of Function Spaces

then the infinite series of continuous functions in theleft hand side of (15) converges uniformly in [0 1]Hence the function 120593(I120572119909(sdot)) is continuous on [0 1](this yields the weak continuity ofI120572119909(sdot) on [0 1])

In the following lemma we gather together some simpleparticular sufficient conditions that ensure the existence ofthe Riemann-Liouville fractional integral of the functionsfromH119901(119864)Lemma 8 Let 119909 119868 rarr 119864 be weakly measurable function TheRFPI of the function 119909 of order 120572 gt 0 makes sense ae on 119868 ifat least one of the following cases holds

(a) 119909 isin H1199010 (119864) 119901 isin [1infin] and 120572 ge 1

(b) 119909 isin H1199010 (119864) 119901 gt max1 1120572

(c) 119909 is strongly measurable which lies in H119901(119864) where119901 isin [1infin] provided that 119864 is weakly complete orcontains no copy of 1198880

If 119864 is reflexive this is also true for any 119909 isin H119901(119864) 119901 isin [1infin]and 120572 gt 0

In all cases 120593(I120572119909) = I120572120593119909 holds for every 120593 isin 119864lowastIn the assertions ((a) and (b)) we find sufficient con-

ditions needed for the existence of I120572119909 in the situation inwhich no restriction is placed on 119864 In the third assertion theproperties of119864 allow us to characterize a function 119909 isin H119901(119864)for whichI120572119909 exists

Proof Firstly assertion (a) is direct consequence of Proposi-tion 2 (part (1)) since 119904 rarr (119905 minus 119904)120572minus1 isin 119871infin[0 119905] holds foralmost every 119905 isin 119868 whenever 120572 ge 1

Secondly to prove assertion (b) let 119901 gt max1 1120572 and119902 be the conjugate exponents to 119901 Since 119902(120572 minus 1) + 1 gt 0 wehave that 119904 rarr (119905 minus 119904)120572minus1 isin 119871119902[0 119905] holds for every 119905 isin 119868 Thusthe assertion (b) is now an easy consequence of Proposition 2(part (2))

Thirdly to prove (c) we let 120572 gt 0 119901 isin [1infin]and 119909 isin H119901(119864) be strongly measurable Since the strongmeasurability is preserved under multiplication operation onfunctions the product (119905 minus sdot)120572minus1119909(sdot) [0 119905] rarr 119864 is stronglymeasurable on [0 119905] for almost every 119905 isin 119868 In view of Youngrsquosinequality for every 120593 isin 119864lowast the real-valued function

119904 997891997888rarr 120593 ((119905 minus 119904)120572minus1 119909 (119904)) = (119905 minus 119904)120572minus1 120593119909 (119904) (17)

is Lebesgue integrable on [0 119905] for almost every 119905 isin 119868 So theassertion (c) follows immediately from Proposition 2 (parts(3 4))

Similarly when 119864 is reflexive the result follows from part(3) of Proposition 2However in all cases the function 119904 997891rarr (119905 minus 119904)120572minus1119909(119904) is

Pettis integrable on [0 119905] for almost every 119905 isin 119868 That is foralmost every 119905 isin 119868 there exists an element in 119864 denoted byI120572119909(119905) such that

120593 (I120572119909 (119905)) = int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904

= int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120593119909 (119904) 119889119904 = I120572120593119909 (119905)

(18)

holds for every 120593 isin 119864lowast This completes the proof

Remark 9 If 119909 isin H1199010 (119864) such that I120572119909(119905) does not exist for

some 119905 isin 119868 then it does not exist even when we ldquoenlargerdquo thespace 119864 into 119865 To see this let 119909 119868 rarr 119865 such that 119909(119868) sub 119864 IfI120572119909(119905) exists for some 119905 isin 119868 then (119905 minus sdot)120572minus1119909(sdot) isin 119875([0 119905] 119865)Since 119909 assumes only values in 119864 it follows by the mean valuetheorem for Pettis integral (Theorem 4) that the RFPI of 119909should lie in 119864

Before we come to a deep study of the mathematicalproperties of the RFPI operator let us take a look at thefollowing miscellaneous examples

Example 10 Let 120572 gt 0 Define the function 119909 [0 1] rarr119871infin[0 1] by 119909(119905) = 120594[0119905]This function is weakly measurable Pettis integrable on[0 1] and 120593119909 is a function of bounded variation (see eg

[18]) That is 119909 isin Hinfin0 (119871infin) Hence in view of Lemma 8 with119901 = infin the RFPI of 119909 exists on [0 1] Further calculations

(cf [4]) show that

I120572119909 (119905) (sdot) = (119905 minus sdot)120572Γ (1 + 120572)120594[0119905] (sdot) isin 119871infin [0 1] (19)

Example 11 Let120572 gt 0 Define the function119909 from the interval[0 1] into the Hilbert space ℓ2 as119909 119905 997891997888rarr 1(1 + 119905) 1(2 + 119905) 1(3 + 119905) 119905 isin [0 1] (20)

We note that

119909 (119905)2ℓ2 = sum119899

( 1119899 + 119905)2 le sum

119899

11198992 lt infin 119905 isin [0 1] (21)

Thus the function 119909 is well defined We claim that 119909 isDunford integrable on [0 1] Once our claim is establishedLemma 8 guarantees the existence ofI120572119909 on [0 1] It remainsto prove this claim and to calculate I120572119909

To see this let 120593 isin (ℓ2)lowast = ℓ2 According to theRiesz representation theorem on Hilbert spaces there existsa uniquely determined 120582 fl 120582119899 isin ℓ2 such that 120593119909(119905) =sum119899(120582119899(119899 + 119905)) A standard arguments using Beppo LevirsquosTheorem yields

int1

0

1003816100381610038161003816120593119909 (119905)1003816100381610038161003816 119889119905 le sum119899

int1

0

10038161003816100381610038161205821198991003816100381610038161003816(119899 + 119905)119889119905 = sum119899

10038161003816100381610038161205821198991003816100381610038161003816 ln(1 + 1119899)le sum

119899

10038161003816100381610038161205821198991003816100381610038161003816119899 le sum119899

10038161003816100381610038161205821198991003816100381610038161003816 lt infin(22)

So 119909 is Dunford integrable on [0 1] and hence Pettis inte-grable on [0 1] since ℓ2 is reflexive Consequently in view of

Journal of Function Spaces 5

Lemma 8 I120572119909 exists on [0 1] To calculate the RFPI of 119909 fix119905 isin [0 1] and let 120593 isin (ℓ2)lowast We have

int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904 = int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120593119909 (119904) 119889119904= I

120572 (infinsum119899=1

120582119899(119899 + 119905)) (23)

Since the series sum119899(120582119899(119899 + 119905)) is uniformly convergent on[0 1] it follows in view of the generalized linearity of thefractional integrals [23 Lemma 5] that (cf [21 Table 91])

int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904= infinsum

119899=1

120582119899119905120572119899Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 ) = 120593 (119892 (119905)) (24)

where 2F1 is the Gauss hypergeometric function evaluated at(1 1 120572 + 1 minus119905119899) and119892 (119905) fl 119905120572119899Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 ) (25)

Since

1003817100381710038171003817119892 (119905)10038171003817100381710038172ℓ2 = sum119899

( 119905120572Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 ))2

le 2F21 (1 1 120572 + 1 minus1)Γ2 (1 + 120572) sum

119899

11198992 lt infin119905 isin [0 1]

(26)

we see that 119892 isin ℓ2 Thus

I120572119909 (119905) = 119905120572119899Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 )

119905 isin [0 1] (27)

Example 12 Let 120572 isin [12 1) Define the countable-valuedfunction 119909 [0 1] rarr 1198880 by

119909 (119905) fl 119909119899 (119905) 119909119899 fl 119899120594(1(119899+1)1119899] (28)

This function is strongly measurable Pettis integrable func-tion on [0 1] (see eg [10 16]) We claim that 119909 isin H119875

0 (1198880)with 119901 isin (1120572 2] Once our claim is established Lemma 8guarantees the existence of I120572119909 120572 isin [12 1) on [0 1] Itremains to prove this claim by showing firstly that H119875(1198880)and to calculate I120572119909 To do this let 120593 isin 119888lowast0 Then therecorresponds to 120593 a unique 120582119899 isin ℓ1 such that 120593119909 = sum119899 120582119899119909119899By noting that

10038171003817100381710038171205821198991199091198991003817100381710038171003817119901 = (int1

0

1003816100381610038161003816120582119899119909119899 (119905)1003816100381610038161003816119901 119889119905)1119901

= 10038161003816100381610038161205821198991003816100381610038161003816 (int1119899

1(119899+1)119899119901119889119905)1119901 = 10038161003816100381610038161205821198991003816100381610038161003816 1198991119902(119899 + 1)1119901

le 10038161003816100381610038161205821198991003816100381610038161003816 11989911199021198991119901 le 100381610038161003816100381612058211989910038161003816100381610038161198992119901minus1 le 10038161003816100381610038161205821198991003816100381610038161003816

(29)

holds for any 119901 isin (1120572 2] a standard argument using LevirsquosTheorem (or Lebesgue dominated convergence theorem) andMinkowskirsquos inequality yields

[int1

0

1003816100381610038161003816120593119909 (119905)1003816100381610038161003816119901 119889119905]1119901 le [int1

0(infinsum119899=1

10038161003816100381610038161205821198991003816100381610038161003816 119909119899 (119905))119901 119889119905]1119901

= [[lim119896rarrinfin

int1

0( 119896sum119899=1

10038161003816100381610038161205821198991003816100381610038161003816 119909119899 (119905))119901

119889119905]]1119901

le [[lim119896rarrinfin

( 119896sum119899=1

10038171003817100381710038171205821198991199091198991003817100381710038171003817119901)119901]]1119901

le infinsum119899=1

10038171003817100381710038171205821198991199091198991003817100381710038171003817119901le infinsum

119899=1

10038161003816100381610038161205821198991003816100381610038161003816 lt infin

(30)

Thus 120593119909(sdot) isin 119871119901[0 1] 119901 isin (1120572 2] holds for every 120593 isin 119888lowast0(that is 119909 isin H119875(1198880)) Since 119901 gt 1120572 gt 1 it follows byProposition 2 in view of the strong measurability of 119909 that119909 isin H119875

0 (1198880) Owing to Lemma 8 we infer that the RFPI of119909 of any order 120572 isin [12 1) exists on the interval [0 1] Tocompute this integral fix 119905 isin 1198681198990 fl [1(1198990+1) 11198990] for some1198990 isin N the set of natural numbers and let 120593 isin 119888lowast0 Then

sum119899

int119905

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 12058211989910038161003816100381610038161003816 119909119899 (119904) 119889119904le 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816 int

119905

1(1198990+1)(119905 minus 119904)120572minus1 1198990119889119904

+ sum119899gt1198990

10038161003816100381610038161205821198991003816100381610038161003816 int1119899

1(119899+1)(119905 minus 119904)120572minus1 119899 119889119904

= 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572

+ sum119899gt1198990

119899 10038161003816100381610038161205821198991003816100381610038161003816120572 [(119905 minus 1119899 + 1)120572 minus (119905 minus 1119899)

120572]

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572

+ sum119899gt1198990

119899 10038161003816100381610038161205821198991003816100381610038161003816120572 [(119905 minus 1119899 + 1) minus (119905 minus 1119899)]120572

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572 + sum

119899gt1198990

119899 10038161003816100381610038161205821198991003816100381610038161003816120572 1119899120572 (119899 + 1)120572

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572 + sum

119899gt1198990

10038161003816100381610038161205821198991003816100381610038161003816120572 11198992120572minus1

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572 + sum

119899gt1198990

10038161003816100381610038161205821198991003816100381610038161003816120572 lt infin(because 120572 isin [12 1))

(31)

6 Journal of Function Spaces

Therefore by Beppo Levi Theorem it follows that

int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904 = int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120593 (119909 (119904)) 119889119904= int119905

0

(119905 minus 119904)120572minus1Γ (120572) sum119899

120582119899119899120594(1(119899+1)1119899] (119904) 119889119904= sum

119899

int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120582119899119899120594(1(119899+1)1119899] (119904) 119889119904= 11989901205821198990Γ (1 + 120572) (119905 minus 11198990 + 1)

120572

+ sum119899gt1198990

119899120582119899Γ (1 + 120572) [(119905 minus 1119899 + 1)120572 minus (119905 minus 1119899)

120572]

(32)

Consequently we conclude that

I120572119909 (119905) = 0 0⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

1198990minus1

1198990Γ (1 + 120572) (119905 minus 11198990 + 1)120572

(1198990 + 1)Γ (1 + 120572) [(119905 minus 11198990 + 2)

120572 minus (119905 minus 11198990 + 1)120572]

(33)

where the nonzero coordinate started from the 1198990th place Itcan be easily seen thatI120572119909(119905) isin 1198880 for any 119905 isin [0 1] Evidentlywe have that10038161003816100381610038161003816100381610038161003816119899 [(119905 minus

1119899 + 1)120572 minus (119905 minus 1119899)

120572]10038161003816100381610038161003816100381610038161003816le 119899 [(119905 minus 1119899 + 1) minus (119905 minus 1119899)]

120572 le 119899 [ 1119899 (119899 + 1)]120572

le 11198992120572minus1 997888rarr 0 as 119899 997888rarr infin1198990Γ (1 + 120572) (119905 minus 11198990 + 1)

120572

le 1198990Γ (1 + 120572) ( 11198990 minus 11198990 + 1)120572 le 11198992120572minus10 Γ (1 + 120572)

997888rarr 0 as 1198990 997888rarr infin

(34)

hold for any 119905 isin [0 1]This yieldsI1205721119909(119905) isin 1198880 for any 119905 isin [0 1]

(this is precisely what we would expect from Definition 5)

Look at part (c) of Lemma 8 with 120572 isin (0 1) and119864 = 1198880 Below we give an example showing that the stronglymeasurability hypothesis imposed on a function 119909 isin H1(119864)H1

0(119864) is not sufficient for the existence of I120572119909 even when 119909is Denjoy-Pettis integrable (cf [16])

Example 13 Let 120572 isin (0 1) Define the strongly measurablefunction 119909 [0 1] rarr 1198880 by119909 (119905) fl 119909119899 (119905)

119909119899 (119905) fl 119899 (119899 + 1) [120594119868119899 (119905) minus 120594119869119899 (119905)] (35)

where

119868119899 fl ( 1119899 + 1 119899 + 12119899 (119899 + 1)) 119869119899 fl ( 119899 + 12119899 (119899 + 1) 1119899)

(36)

It is immediate that (cf [16]) 119909 is a well-defined Denjoy-Pettis (but it is not Pettis) integrable on [0 1]

In what follows we will show that the RFPI of 119909 does notexist on a subinterval of positive measure on [0 1]

To see this we make use of Proposition 3 as followsdefine for each intervalI119899 = [1(119899 + 1) 1119899] the functionals120593119899 isin 119888lowast0 (required by Proposition 3) to be the correspondingto the element 120582119899 fl (0 0 0 1 0 0 ) isin ℓ1 wherethe nonzero coordinate is in the 119899th place Then 120593119899(119909(119905)) =119909119899(119905) and thus |120593119899(119909(119904))| = 119899(119899 + 1)[120594119868119899(119905) minus 120594119869119899(119905)] Clearlythe family 120593119899 runs through the unit ball of the dual of1198880 Evidently the isometric isomorphism between 119888lowast0 and ℓ1yields 120593119899119888lowast0 = 120582119899ℓ1 = 1

Now take 119905 isin [12 1] and define the continuous function119891 [12 1] rarr R+ by

119891 (119905) fl (119905 minus 1119899 + 1)120572 minus (119905 minus 1119899)

120572 119899 ge 2 119905 isin [12 1]

(37)

It can be easily seen that in view of 0 lt 120572 lt 1 1198911015840 is negativeon (0 1) By standard results from (classical) calculus itfollows that 119891 is strictly decreasing on [0 1] in particular119891(1) lt 119891(119905) for all 119905 isin [0 1) Thus for any 119899 ge 2 and any120593 isin 120593119899 we have

intI119899

100381610038161003816100381610038161003816100381610038161003816120593 ((119905 minus 119904)120572minus1Γ (120572) 119909 (119904))100381610038161003816100381610038161003816100381610038161003816 119889119904

= intI119899

(119905 minus 119904)120572minus1Γ (120572) 1003816100381610038161003816120593 (119909 (119904))1003816100381610038161003816 119889119904= int1119899

1(119899+1)

(119905 minus 119904)120572minus1Γ (120572) 119899 (119899 + 1) 119889119904= 119899 (119899 + 1)Γ (1 + 120572) [(119905 minus 1119899 + 1)

120572 minus (119905 minus 1119899)120572]

ge 1198992Γ (1 + 120572) [(1 minus 1119899 + 1)120572 minus (1 minus 1119899)

120572]

(38)

An explicit calculation using LrsquoHospitalrsquos rule two timesreveals

lim119899rarrinfin

(1 minus 1 (119899 + 1))120572 minus (1 minus 1119899)120572119899minus2 = minus1205722sdot lim119899rarrinfin

1198993 [( 119899119899 + 1)120572minus1 1

(119899 + 1)2 minus (1 minus 1119899)120572minus1 11198992 ]

= minus1205722 lim119899rarrinfin

119899 [( 119899119899 + 1)120572minus1 1198992

(119899 + 1)2 minus (1 minus 1119899)120572minus1]

Journal of Function Spaces 7

= minus1205722 lim119899rarrinfin

119899 [( 119899119899 + 1)120572+1 minus (1119899)

120572minus1] = minus1205722sdot lim119899rarrinfin

1198992 [(120572 + 1) ( 119899119899 + 1)120572 1(119899 + 1)2

minus (120572 minus 1) (1 minus 1119899)120572minus2 11198992 ] = minus1205722

sdot lim119899rarrinfin

[(120572 + 1) ( 119899119899 + 1)120572+2 minus (120572 minus 1) (1 minus 1119899)

120572minus2]= 120572

(39)

from which it follows that

intI119899

100381610038161003816100381610038161003816100381610038161003816120593 ((119905 minus 119904)120572minus1Γ (120572) 119909 (119904))100381610038161003816100381610038161003816100381610038161003816 119889119904997888rarr0 as 119899 997888rarr infin (40)

Therefore in view of Proposition 3 119904 997891rarr (119905 minus 119904)120572minus1119909(119904) notin119875([0 1] 1198880) Hence I120572119909 does not exist which is what wewished to show

The following theorem provides a useful characterizationof the space H119901(119864) for which the statements reveal howmuch the fractional integral I120572 is better than the function119909 isin H119901(119864) Indeed based on Lemma 8 using an inequalityof Young we can easily prove the following

Lemma 14 For any 120572 gt 0 the following holds(a) If 119864 is reflexive then for every 119901 isin [1infin] the operator

I120572 mapsH119901(119864) intoH1199010 (119864) In particular if 0 lt 120572 lt1 then I120572 H1(119864) rarr H1((1minus120572)minus120598)0 (119864) however small120598 gt 0 is

(b) If 119864 contains no copy of 1198880 then for every 119901 isin [1infin]the operator I120572 maps 119909 isin H

1199010 (119864) 119909 strongly

measurable into H1199010 (119864) In particular if 120572 ge 1 the

operator I120572 mapsH1199010 (119864) into itself

(c) If 119901 gt max1 1120572 the operator I120572 sends H1199010 (119864) to119862[119868 119864120596] (if we define I120572119909(0) fl 0) In particular if120572 isin (0 1) I120572 H119901

0 (119864) rarr H120572minus1119901(119864)Proof At the beginning let 120572 gt 0 and 119909 isin H119901(119864) with119901 isin [1infin] and define the real-valued function 119892 119868 rarr R

by 119892(119905) fl 119905120572minus1Γ(120572) If 119864 is reflexive it follows in view ofLemma 8 that J120572119909 exists ae in 119868 and 120593(I120572119909) = I120572120593119909for every 120593 isin 119864lowast Youngrsquos inequality guarantees that theconvolution product 119892 lowast 120593119909(sdot) lies in 119871119901[119868] Consequently120593(I120572119909) = I120572120593119909 = 119892 lowast 120593119909 isin 119871119901[119868] for every 120593 isin 119864lowastThe reflexivity of 119864 together with Proposition 2 yields I120572119909 isinH

1199010 (119864)In particular if 0 lt 120572 lt 1 it can be easily seen thatint

119868(119892(119905))119902119889119905 lt infin for every 119902 isin [1 1(1 minus 120572)) That is 119892 isin1198711(1minus120572)minus120598[119868] 120598 gt 0 By Youngrsquos inequality it follows that120593(I120572119909) = 119892 lowast 120593119909 isin 1198711(1minus120572)minus120598[119868] for every 120593 isin 119864lowast however

small 120598 gt 0 is Now the assertion (a) follows because of thereflexivity of 119864

Next in order to prove the assertion (b) let 119909 isin H1199010 (119864)

and note in view of Lemma 8 that I120572119909 exists on 119868 Define119910 119868 rarr 119864 by119910(119905) fl I120572119909(119905) 119909 isin H1199010 (119864) Since120593119910(sdot) isin 119871119901[119868]

for every120593 isin 119864lowast it follows that119910 isin H119901(119864) Moreover for any119886 119887 isin 119868 (119886 lt 119887) we haveint119887

119886120593119910 (119904) 119889119905 = int119887

119886I120593119909 (119905) 119889119905 = 1Γ (120572)

sdot int119887

119886int119905

0(119905 minus 119904)120572minus1 120593119909 (119904) 119889119904 119889119905

= 1Γ (120572) [int119887

119886int119887

119904(119905 minus 119904)120572minus1 120593119909 (119904) 119889119905 119889119904

+ int119886

0int119887

119886(119905 minus 119904)120572minus1 120593119909 (119904) 119889119905 119889119904]

= 1Γ (120572) [int119887

119886(119887 minus 119904)120572 120593119909 (119904) 119889119904

+ int119886

0[(119887 minus 119904)120572 minus (119886 minus 119904)120572] 120593119909 (119904) 119889119904]

= 1Γ (120572) [int119886

0(119887 minus 119904)120572 120593119909 (119904) 119889119904

+ int119887

119886(119887 minus 119904)120572 120593119909 (119904) 119889119904 minus int119886

0(119886 minus 119904)120572 120593119909 (119904) 119889119904]

= int119887

0

(119887 minus 119904)120572Γ (1 + 120572)120593119909 (119904) 119889119904 minus int119886

0

(119886 minus 119904)120572Γ (1 + 120572)120593119909 (119904) 119889119904= 120593 (119909[119886119887])

(41)

where

119909[119886119887] fl (119875) int119887

0

(119887 minus 119904)120572Γ (1 + 120572)119909 (119904) 119889119904minus (119875) int119886

0

(119886 minus 119904)120572Γ (1 + 120572)119909 (119904) 119889119904(42)

Since 119909 isin 119875[119868 119864] owing to Proposition 2 (part (1)) it followsthat (119887 minus sdot)120572119909(119904) and (119886 minus sdot)120572119909(119904) are Pettis integrable on 119868 andso 119909[119886119887] isin 119864

A combination of these results yields120593119910 isin 119871119901(119868) for every120593 isin 119864lowast and there exists an element 119909[119886119887] isin 119864 such that120601119909[119886119887] = int119887

119886120593119910(119905)119889119905 for every 119886 119887 isin 119868 Since 119864 contains no

copy of 1198880 it follows that (cf [17 Theorem 23]) 119910 isin 119875[119868 119864]Consequently 119910(sdot) isin H

1199010 (119864) This is the claim (b)

To prove the assertion (c) let 119901 gt max1 1120572 and 119909 isinH

1199010 (119864) By Lemma 8 we deduce that I120572119909 exists ae on 119868

Now let 119902 isin [1infin] be such that 1119901 + 1119902 = 1 Since119902(120572 minus 1) gt minus1 119892 fl (sdot)120572minus1Γ(120572) isin 119871119902[119868] Therefore as a directconsequence of Youngrsquo inequality it follows that

120593 (I120572119909 (sdot)) = I120572120593119909 (sdot) = 119892 lowast 120593119909 (sdot) isin 119862 [119868R] (43)

holds for every 120593 isin 119864lowast Now we claim that I120572119909(119905) rarr 0in 119864 as 119905 rarr 0 Once our claim is established the definition

8 Journal of Function Spaces

I120572119909(0) fl 0 guarantees thatI120572119909(sdot) isin 119862[119868 119864120596] which is whatwewished to show It remains to prove our claim without lossof generality assume that I120572119909(119905) = 0 Then there exists (asa consequence of the Hahn-Banach theorem) 120593 isin 119864lowast with120593 = 1 such that I120572119909(119905) = 120593(I120572119909(119905)) = I120572120593119909(119905) ByHolderrsquos inequality we obtain

1003816100381610038161003816120593 (I120572119909 (119905))1003816100381610038161003816le 1Γ (120572) (int

119905

0(119905 minus 119904)119902(120572minus1) 119889119904)1119902 (int119905

0

1003816100381610038161003816120593119909 (119904)1003816100381610038161003816119901 119889119904)1119901

le 1Γ (120572) ( 119905(120572minus1)119902+1(120572 minus 1) 119902 + 1)1119902 (int

119868

1003816100381610038161003816120593119909 (119904)1003816100381610038161003816119901 119889119904)1119901

= 119905120572minus1+1119902Γ (120572) [119902 (120572 minus 1) + 1]1119902

10038171003817100381710038171205931199091003817100381710038171003817119871119901

= 119905120572minus1119901 10038171003817100381710038171205931199091003817100381710038171003817119871119901Γ (120572) [119902 (120572 minus 1) + 1]1119902

(44)

This is equivalent with the following estimate

1003817100381710038171003817I120572119909 (119905)1003817100381710038171003817 le 119905120572minus1119901 10038171003817100381710038171205931199091003817100381710038171003817119871119901Γ (120572) [119902 (120572 minus 1) + 1]1119902 (45)

Owing to 120572 gt 1119901 we get I120572119909(119905) rarr 0 in 119864 as 119905 rarr 0 andconsequently in view ofI120572119909(0) fl 0I120572119909(sdot) isin 119862[119868 119864120596] Thisproves the first part of the assertion (c)

Finally let 120572 isin (0 1) and 119905 120591 isin 119868 Without loss ofgenerality assume 120591 lt 119905 Then for any 120593 isin 119864lowast we have byHolder inequality with 119901 gt 1120572 in view 119902(120572 minus 1) gt minus1 that

1003816100381610038161003816120593 (I120572119909 (119905) minusI120572119909 (120591))1003816100381610038161003816 Γ (120572)

= 10038161003816100381610038161003816100381610038161003816int119905

0(119905 minus 119904)120572minus1 120593119909 (119904) 119889119904 minus int120591

0(120591 minus 119904)120572minus1 120593119909 (119904) 11988911990410038161003816100381610038161003816100381610038161003816

le (int120591

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 minus (120591 minus 119904)120572minus110038161003816100381610038161003816 1003816100381610038161003816120593119909 (119904)1003816100381610038161003816 119889119904+ int119905

120591(119905 minus 119904)120572minus1 1003816100381610038161003816120593119909 (119904)1003816100381610038161003816 119889119904)

= [(int120591

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 minus (120591 minus 119904)120572minus110038161003816100381610038161003816119902 119889119904)1119902

+ (int119905

120591(119905 minus 119904)(120572minus1)119902 119889119904)1119902] 10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]

le [[(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

+ ((119905 minus 120591)(120572minus1)119902+1(120572 minus 1) 119902 + 1 )1119902]]10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]

= [(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

+ (119905 minus 120591)120572minus1119901[119902 (120572 minus 1) + 1]1119902]

10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868] (46)

By noting that (120572 minus 1)119902 isin (minus1 0) when 120572 isin (0 1) it can beeasily seen that

int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904le 1119902 (120572 minus 1) + 1 (119905 minus 120591)119902(120572minus1)+1

(47)

That is

(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

le (119905 minus 120591)120572minus1119901[119902 (120572 minus 1) + 1]1119902

(48)

A combination of these results yields1003816100381610038161003816120593 (I120572119909 (119905)) minus 120593 (I120572119909 (120591))1003816100381610038161003816= 1003816100381610038161003816120593 (I120572119909 (119905) minusI

120572119909 (120591))1003816100381610038161003816le10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]Γ (120572) 2 (119905 minus 120591)120572minus1119901

[119902 (120572 minus 1) + 1]1119902 (49)

ThusI120572119909 isin H120572minus1119901(119864) This completes the proof

Corollary 15 For any 120572 gt 0 I120572 119862[119868 119864120596] rarr 119862[119868 119864120596]Proof Let 119909 isin 119862[119868 119864120596] then 120593119909 isin 119871infin holds for every120593 isin 119864lowast By noting that the weak continuity implies a strongmeasurability ([24] page 73) it follows that 119909 is stronglymeasurable on 119868 Hence in view of Proposition 2 119909 isinHinfin

0 (119864) Consequently 119862[119868 119864120596] sube H1199010 (119864) for every 119901 isin[1infin] That isI120572 119862[119868 119864120596] rarr 119862[119868 119864120596] Hence the desired

result is obtained

Example 16 Let 120572 isin [12 1) Define the function 119909 [0 1] rarr1198880 by 119909(119905) = 119899120594(1(119899+1)1119899]We have (in view of Example 12)that

I120572119909 (119905) = 0 0 0 1198990 (119905 minus 1 (1198990 + 1))120572

Γ (1 + 120572) (1198990 + 1)Γ (1 + 120572) [(119905 minus 11198990 + 2)

120572 minus (119905 minus 11198990 + 1)120572]

(50)

To show that I120572119909 is norm continuous on [0 1] let 120591 119905 isin[0 1] With no loss of generality we may assume that 119905 120591 isin1198681198990 for some 1198990 isin N Since the nonzero terms of the sequencein (50) are nonincreasing then in view of

|119905 minus 120591| le ( 11198990 minus 11198990 + 1) = 11198990 (1198990 + 1) le 111989920 (51)

Journal of Function Spaces 9

we have1003817100381710038171003817I120572119909 (119905) minusI

1205721119909 (120591)10038171003817100381710038171198880

= 1198990Γ (1 + 120572)10038161003816100381610038161003816100381610038161003816(119905 minus

1119899 + 1)120572 minus (120591 minus 1119899 + 1)

12057210038161003816100381610038161003816100381610038161003816le 1198990Γ (1 + 120572)

1003816100381610038161003816100381610038161003816(119905 minus 1119899 + 1) minus (120591 minus 1119899 + 1)1003816100381610038161003816100381610038161003816120572

le 1198990Γ (1 + 120572) |119905 minus 120591|120572 le 1Γ (1 + 120572) |119905 minus 120591|120572radic|119905 minus 120591|= |119905 minus 120591|120572minus12Γ (1 + 120572)

(52)

Thus the RFPI of 119909 is norm continuous on [0 1] Preciselysince 0 lt 120572 minus 12 lt 120572 minus 1119901 lt 1 for any 119901 le 2 thenI120572119909 isin H120572minus12(119864) sub H120572minus1119901(119864) This is precisely what onewould expect from Lemma 14 (part (c))

Analogously an explicit calculation reveals that I120572119909 liesin 119862([0 1] 119871infin[0 1]) where 119909 [0 1] rarr 119871infin[0 1] defined byExample 10

We now consider additional mapping properties of theoperator I120572 Precisely we will show that the RFPI enjoysthe following commutative property which is folklore in case119864 = R However the proof is completely similar to that of [8]Lemma 35

Lemma 17 Let 120572 120573 gt 0 and 119901 gt max1 1120572 1120573 Then

I120572I120573119909 = I

120573I120572119909 = I

120572+120573119909 (53)

holds for every 119909 isin H1199010 (119864) If 119864 is reflexive this is also true for

every 119901 ge 1Proof Since 119901 gt max1 1120572 1120573 it follows that 119901 gtmax1 1(120572+120573) So by Lemmas 8 and 14 themappingsI120572119909I120573119909 andI120572+120573119909 belong toHinfin

0 (119864) for every 119909 isin H1199010 (119864) Now

repeating the same process as in ([8] Lemma 35) implies theclaim

When 119864 is reflexive the result follows as a direct applica-tion of Lemma 14

3 Fractional Derivatives ofVector-Valued Functions

After the notation of the fractional integrals of vector-valued functions the fractional derivatives become a naturalrequirement Before we come to the definitions and a detailedstudy of the mathematical properties of fractional differentialoperators we recall the following

Definition 18 Consider the vector-valued function 119909 119868 rarr119864(1) Let 120593119909 be differentiable on 119868 for every 120593 isin 119864lowast The

function 119909 is said to be weakly differentiable on 119868 if

there exists 119910 119868 rarr 119864 such that for every 120593 isin 119864lowast wehave

119889120593119909 (119905)119889119905 = 120593119910 (119905) for every 119905 isin 119868 (54)

The function 119910 is called the weak derivative of thefunction 119909

(2) Let 120593119909 be differentiable ae on 119868 for every 120593 isin 119864lowast(the null set may vary with 120593 isin 119864lowast) The function 119909is said to be pseudo differentiable on 119868 if there existsa function 119910 119868 rarr 119864 such that for every 120593 isin 119864lowast thereexists a null set119873(120593) sub 119868 such that

(120593119909 (119905))1015840 = 120593119910 (119905) for every 119905 isin 119868119873 (120593) (55)

In this case the function 119910 is called the pseudoderivative of 119909If 119909 is pseudo differentiable on 119868 and the null setinvariant for every 120593 isin 119864lowast then 119909 is ae weaklydifferentiable on 119868

Clearly if 119909 is ae weakly differentiable on 119868 then 120593(119909)is ae differentiable on 119868 The converse holds in a weaklysequentially complete space (see [25] Theorem 733)

For more details of the derivatives of vector-valuedfunctions we refer to [10 12 26]

The following results play a major role in our analysis

Proposition 19 (see [27] Theorem 51) The function 119910 119868 rarr119864 is an indefinite Pettis integrable if and only if 119910 is weaklyabsolutely continuous on 119868 and have a pseudo derivative on 119868In this case 119910 is an indefinite Pettis integral of any of its pseudoderivatives

Now we are in the position to define the fractional-typederivatives of vector-valued functions

Definition 20 Let 119909 119868 rarr 119864 For the positive integer119898 suchthat 120572 isin (119898 minus 1119898) 119898 isin N0 fl 0 1 2 we define theCaputo fractional-pseudo (weak) derivative ldquoshortly CFPD(CFWD)rdquo of 119909 of order 120572 by

119889120572119889119905120572 119909 (119905) fl I119898minus120572119863119898119909 (119905) (56)

where the sign ldquo119863rdquo denotes the pseudo (or weak) differentialoperator We use the notation 119889120572119901119889119905120572 and 119889120572120596119889119905120572 to charac-terize the Caputo fractional-pseudo derivatives and Caputofractional weak derivatives respectively

It is well known that although the weak derivative of aweakly differentiable function is uniquely determined thepseudo derivative of the pseudo differentiable function is notunique Also although any two pseudo derivatives 119910 119911 of afunction 119909 119868 rarr 119864 need not be ae equal (see [12 Example91] and [26 p 2]) the functions 119910 119911 are weakly equivalenton 119868 (that is 120593119910 = 120593119911 holds ae for every 120593 isin 119864lowast) The

10 Journal of Function Spaces

next lemma provides a useful characterization property of theCFPD Really it can be easily seen that the CFPD of a Caputofractional-pseudo differentiable function does not depend onthe choice of a pseudo derivative of the function

Lemma 21 Let 119909 119868 rarr 119864 be pseudo differentiable functionwhere pseudo derivatives lie in H

1199010 (119864) If (119889120572119901119889119905120572)119909 exists on119868 then the CFPD of 119909 depends on the choice of the pseudo

derivatives of 119909Proof Let 119910 119911 isin H

1199010 (119864) be two pseudo derivatives of the

pseudo differentiable function 119909 Since 119909 and 119910 are weaklyequivalent on 119868 then for every 120593 isin 119864lowast there exists a nullset N which depends on 120593 isin 119864lowast such that (119905 minus 119904)minus120572120593119910(119904) =(119905minus119904)minus120572120593119911(119904) for every 119904 isin [0 119905]N Consequently for almostevery 119905 isin 119868

120593 (I1minus120572119910 (119905)) = I1minus120572120593119910 (119905) = I

1minus120572120593119911 (119905)= 120593 (I1minus120572119911 (119905)) (57)

holds for every 120593 isin 119864lowast Thus

119889120572119901119889119905120572 119909 = I1minus120572119863119901119909 = I

1minus120572119910 = I1minus120572119911 on 119868 (58)

which is what we wished to show

We consider the following examples

Example 22 Define 119909 [0 1] rarr ℓ2 by119909 (119905) fl ln(1 + 119905119899) 119899 isin N (59)

We claim that the CFWD of 119909 exists on [0 1] To see this welet 120593 isin (ℓ2)lowast = ℓ2 Then there corresponds to 120593 a sequence120582119899 isin ℓ2 such that

120593119909 (119905) = sum119899

120582119899 ln(1 + 119905119899) (60)

Since the series in (60) converges uniformly on [0 1] theformal differentiation of 120593119909 yields

(120593119909 (119905))1015840 = sum119899

120582119899119899 + 119905 = 120593 ( 1119905 + 119899) (61)

It is not hard to justify the differentiation by noting that theseries in the right hand side of (61) is uniformly and absolutelyconvergent on [0 1] Consequently 120593119909 is differentiable on[0 1] for 120593 isin (ℓ2)lowast meaning that 119909 is weakly differentiable on[0 1] (because ℓ2 is weakly complete (cf [25] Theorem 733)and 119863120596119909(119905) = 1(119905 + 119899) Owing to Lemma 25 the CFWDof any order 120572 isin (0 1) of 119909 exists on [0 1] To calculate theCFWD of 119909 on [0 1] we observe in view of Example 11 that

119889120572120596119889119905120572 119909 (119905) = I1minus120572119863120596119909 (119905)

= 1199051minus120572119899Γ (2 minus 120572) 2F1 (1 1 2 minus 120572 minus119905119899 ) (62)

Example 23 Let119864 be an infinite dimensionBanach space thatfails cotype Define 119910 [0 1] rarr 119864 by 119910(119905) fl int119905

0119909(119904)119889119904

where 119909 [0 1] rarr 119864 given by formula (7) As cited in ([13]Corollary 4) 119910 is nowhere weakly differentiable on [0 1]Consequently the CFWD of any order 120572 isin [34 1) of 119910 losesits meaning on [0 1]Remark 24 As shown in Example 23 there is an infinitedimension Banach space 119864 and weakly absolutely continuousfunction 119910 119868 rarr 119864 which is nowhere weakly differentiable(hence the CFWD of 119910 does not exist) Also even when 119864 isseparable and 119910 is Lipschitz function the pseudo derivatives(hence the CFPD) of 119910 need not to exist [28]

However Definition 20 of the CFPD (CFWD) has thedisadvantage that it completely loses its meaning if thefunction119909 fails to be pseudo (weakly) differentiable Preciselythe CFPD (in particular the CFWD) of a function 119909 loses itsmeaning if 119909 is not weakly absolutely continuous

Thenext lemma gives sufficient conditions that ensure theexistence of the Caputo fractional derivatives of a function119909 isin H119901(119864)Lemma 25 Let 120572 isin (0 1) For the function 119909 119868 rarr 119864 thefollowing hold

(a) If 119909 has a pseudo derivative 119863119901119909 isin H1199010 (119864) where 119901 gt1(1 minus 120572) then the CFPD of 119909 of order 120572 exists on 119868

Moreover (119889120572119901119889119905120572)119909 isin 119862[119868 119864120596](b) If 119864 is weakly complete or contains no copy of 1198880 and if119909 has a weak derivative 119863120596119909 isin H119901(119864) where 119901 isin[1infin] then the CFWD of 119909 of order 120572 exists on 119868

Moreover (119889120572120596119889119905120572)119909 isin 119862[119868 119864120596]This holds for any 119909 isin H119901(119864) with 119901 isin [1infin] if 119864 is reflexive

In all cases 120593((119889120572119901119889119905120572)119909) = (119889120572119901119889119905120572)120593119909 (120593((119889120572120596119889119905120572)119909) =(119889120572120596119889119905120572)120593119909) holds for every 120593 isin 119864lowastProof Since the weak (pseudo) derivative of an ae weakly(pseudo) differentiable function is strongly (weakly) mea-surable [12 26 28] the proof is readily available in view ofthe definition of Caputo fractional derivatives and Lemma 8Moreover since 119863120596119909(119863119901119909) isin H

1199010 (119864) with 119901 gt 1(1 minus 120572)

it follows in view of Lemma 14 that I1minus120572119863120596119909(I1minus120572119863119901119909) isin119862[119868 119864120596] This completes the proof

Besides the Caputo fractional-pseudo (weak) derivativeswe define the Riemann-Liouville fractional-pseudo (weak)derivatives

Definition 26 Let 119909 119868 rarr 119864 For the positive integer 119898such that 120572 isin (119898 minus 1119898) 119898 isin N0 fl 0 1 2 we definethe Riemann-Liouville fractional-pseudo (weak) derivativeldquoshortly RFPD (RFWD)rdquo of 119909 of order 120572 by

D120572119909 (119905) fl 119863119898

I119898minus120572119909 (119905) (63)

where 119863 is defined as in Definition 20 We use the notationD120572

119901 (D120572120596) to characterize the Riemann-Liouville fractional-

pseudo (weak) derivatives

Journal of Function Spaces 11

Clearly in infinite dimension Banach spaces the weakabsolute continuity of I119898minus120572119909 is necessarily (but not suf-ficient) condition for the existence of RFPD (in particularRFWD) of 119909Lemma 27 Let 0 lt 120572 lt 1 For any 119909 isin H

1199010 (119864) with 119901 gt

max1120572 1(1 minus 120572) we haveD

120572119901I

120572119909 = 119909 ae (64)

If 119864 is reflexive this is also true for every 119901 ge 1Proof Our assumption 119909 isin H

1199010 (119864)with 119901 gt max1120572 1(1minus120572) yields in view of Lemma 17 that

D120572119901I

120572119909 = 119863119901I1minus120572

I120572119909 = 119863119901I

1119909 (65)

The claim now follows immediately since (cf Proposition 19)the indefinite integral of Pettis integrable function is weaklyabsolutely continuous and it is pseudo differentiable withrespect to the right endpoint of the integration interval andits pseudo derivative equals the integrand at that point

Remark 28 When we replace D120572119901 by D120572

120596 then Lemma 27is no longer necessarily true for arbitrary 119909 isin H1

0(119864) evenwhen 119864 is reflexive evidently in [13 remark below Corollary4] the existence of a reflexive Banach space 119864 and a strongmeasurable Pettis integrable function 119909 119868 rarr 119864 was provedsuch that 119909 has nowhere weakly differentiable integral In thiscase 119863120596I

1119909 lost its meaning This gives a reason to believethat (65) (hence (64)) withD120572

120596119909 could not happen

However we have the following result

Lemma 29 Let 0 lt 120572 lt 1 For any 119909 isin 119862[119868 119864120596] we haveD

120572120596I

120572119909 = 119909 on 119868 (66)

Proof Our assumption 119909 isin 119862[119868 119864120596] sub Hinfin0 (119864) (see the proof

of Corollary 15) yields in view of Lemma 17 that

D120572120596I

120572119909 = 119863120596I1minus120572

I120572119909 = 119863120596I

1119909 (67)

The claim now follows immediately since the indefiniteintegral of weakly contentious function is weakly absolutelycontinuous and it is weakly differentiable with respect tothe right endpoint of the integration interval and its weakderivative equals the integrand at that point

The following lemma is folklore in case 119864 = R but to seethat it also holds in the vector-valued case we provide a proof

Lemma 30 Let 0 lt 120572 lt 1 and 119901 gt max1120572 1(1minus120572) If thefunction 119909 119868 rarr 119864 is weakly absolutely continuous on 119868 andpasses a pseudo derivative inH

1199010 (119864) then119889120572119901119909119889119905120572 (119905) = D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (68)

In particular if 119909 passes a weak derivative in 119862[119868 119864120596] then119889120572120596119909119889119905120572 (119905) = D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (69)

If 119864 is reflexive this is also true for every 119901 ge 1

Proof We observe that under the assumption imposed on119863119901119909 together with Proposition 3 the weakly absolutelycontinuity of 119909 is equivalent to

119909 (119905) = 119909 (0) + int119905

0119863119901119909 (119904) 119889119904 (70)

Hence owing to Lemma 27 it follows that

119889120572119901119909119889119905120572 = I1minus120572119863119901119909 = (D120572

119901I120572)I1minus120572119863119901119909

= D120572119901 (I120572

I1minus120572)119863119901119909 = (119863119901I

1minus120572)I1119863119901119909= 119863119901I

1minus120572 (119909 (119905) minus 119909 (0))= D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0)

(71)

which is what we wished to showThe proof of (69) is very similar to that in (68) therefore

we omit the details

Remark 31 As we remark above the definition of the CFPDof a function 119909 loses completely its meaning if 119909 is notweakly absolutely continuous For this reason we are able touse Lemma 30 to define the Caputo fractional derivative ingeneral that is we put

119889120572119901119909119889119905120572 (119905) fl D120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) 0 lt 120572 lt 1 (72)

Similarly we define

119889120572120596119909119889119905120572 (119905) fl D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (73)

However Lemma 30 claims that for the weakly absolutelycontinuous functions having Pettis integrable pseudo [weak]derivatives definitions (72) and (73) of the Caputo fractional-pseudo [weak] derivatives coincide with Definition 20

4 An Application

Let 119891 119868 times 119864 rarr 119864 be given function Consider the boundaryvalue problem of the fractional type

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin [0 1] 120582 isin R

119910 (0) + 119887119910 (1) = ℎ(74)

with certain constants 119887 ℎ isin R 119887 = minus1 To obtain theintegral equation modeled off the problem (74) we let 119910 bea weakly continuous solution to problem (74) then formallywe have

I120572 119889120572120596119889119905120572119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (75)

12 Journal of Function Spaces

that is

I120572I1minus120572119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) 997904rArrI1119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (76)

This reads (cf [29])

119910 (119905) = 119910 (0) + 120582I120572119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin 119868 120582 isin R (77)

with some (presently unknown) quantity 119910(0)Now we solve (77) for 119910(0) by 119910(0) + 119887119910(1) = ℎ and it

follows that

(1 + 119887) 119910 (0) + 120582119887Γ (120572) int1

0(1 minus 119904)120572minus1 119891 (119904 119910 (119904)) 119889119904 = ℎ (78)

Thus

119910 (0) = ℎ1 + 119887 minus 1205821198871 + 119887119868120572119891 (1 119910 (1)) (79)

Assume that the function 119891 is weakly-weakly continuousfunction such that

(1) for any 119903 gt 0 there is a constant 119872 gt 0 such that119891(119905 119910) le 119872 for all 119905 isin 119868 and 119910 le 119903(2) there exists a nondecreasing continuous function 120595 [0infin) rarr [0infin) 120595(0) = 0 and 120595(119905) lt 119905 for all 119905 gt 0

such that 120573(119891(119879 times 119860)) le 120595(120573(119860)) for every boundedset119860 sub 119864 where 120573 stands for De Blasirsquos weakmeasureof noncompactness (see [30])

Occasionally if 119891 119868 times 119864 rarr 119864 is weakly-weakly continuousand 119864 is reflexive then the assumptions (1) and (2) areautomatically satisfied (see eg [31])

Theorem 32 (see [6] Theorem 33) Let 120572 isin (0 1) and |120582| ltΓ(1 + 120572) If 119891 119868 times 119864 rarr 119864 is weakly-weakly continuous andsatisfies the assumptions (1) and (2) then the integral equation(77) has a weakly continuous solution 119910 defined on [0 1]Proof We omit the proof since it is almost identical to thatin the proof in ([6] Theorem 33) with (small) necessarychanges

In the following example we assume that 119910 isin 119862[119868 119864120596]solves (77) and we will show that not only do we have that 119910no longer necessarily solves (74) (when the Caputo fractionalweak derivative is taken in the sense of Definition 20)but even worse it could happen that the problem (74) isldquomeaninglessrdquo on 119868Example 33 Let 120572 isin (0 14] Let 119864 be an indefinitedimensional reflexive Banach space that fails cotype Definethe weakly-weakly continuous function 119891 [0 1] times 119864 rarr 119864by 119891 fl I1minus120572119909 where 119909 119868 rarr 119864 is defined by formula (7)Obviously (cf Example 6 in view of Remark 7)119891 satisfies theassumptions of Theorem 32

Now consider the integral equation (77) with119891 = I1minus120572119909Namely we consider the integral equation

119910 (119905) = 119910 (0) + 120582I120572I1minus120572119909 (119905)

120572 isin (0 14] 119905 isin 119868 120582 isin R (80)

Obviously the solution 119910 is weakly continuous on [0 1] (thisis an immediate consequence of Remark 7 and Lemma 17)

Since 120593119909 isin 1198711[0 1] holds for every 120593 isin 119864lowast it follows bythe commutative property of the fractional integral operatorscalculus over the space of Lebesgue integrable functions (see[21] Section 23) that

120593 (I120572I1minus120572119909 (119905)) = I

120572120593 (I1minus120572119909 (119905))= I

120572I1minus120572120593 (119909 (119905)) = I

1120593 (119909 (119905))= 120593 (I1119909 (119905))

(81)

that is

120593 (I120572I1minus120572119909 (119905) minusI

1119909 (119905)) = 0 for every 120593 isin 119864lowast (82)

Hence I120572I1minus120572119909 = I1119909 Consequently problem (80) be-comes

119910 (119905) = 119910 (0) + 120582I1119909 (119905) 120572 isin (0 14] 119905 isin 119868 (83)

As showed in ([13] Corollary 4) the indefinite integral 119905 rarrint1199050119909(119904)119889119904 is nowhere weakly differentiable on [0 1] Thus 119910

is nowhere weakly differentiable on [0 1] hence the CFWDof 119910 (hence the boundary value problem) completely loses itsmeaning if the CFWD is taken in the sense of Definition 20This is what we wished to show

Now we are in the position to state and prove thefollowing existence theorem

Theorem 34 If 119891 119868 times 119864 rarr 119864 is a function such that allconditions fromTheorem 32 hold then problem (74) (where theCaputo fractionalweak derivative is taken in the sense (73)) hasa weak solution on [0 1]Proof Let 119910 isin 119862[119868 119864120596] be a solution to (77) Equip 119864 andI times E with weak topology and note that 119905 997891rarr (119905 119910(119905)) iscontinuous as a mapping from 119868 into 119868 times 119864 Since 119891 119868 times 119864 rarr119864 is weakly-weakly continuous on [0 1] then 119891(sdot 119910(sdot)) is acomposition of this mapping with 119891 and thus 119891(sdot 119910(sdot)) isweakly continuous on [0 1]

Now operating by the operatorD120572120596 on both sides of (77)

it follows by Lemma 29 that

D120572120596119910 = D

120572120596119910 (0) + 120582D120572

120596I120572119891 = 119905minus120572Γ (1 minus 120572)119910 (0) + 120582119891 (84)

Now insert definition (73) of the Caputo fractionalderivative and we get

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) (85)

Journal of Function Spaces 13

With some further efforts one can get the boundary condi-tion 119910(0) + 119887119910(1) = ℎ

Therefore 119910 satisfies problem (74) which is what wewished to show

Data Availability

No data were used to support this study

Conflicts of Interest

The author declares no conflicts of interest

References

[1] H A Salem and A M El-Sayed ldquoWeak solution for fractionalorder integral equations in reflexive Banach spacesrdquoMathemat-ica Slovaca vol 55 no 2 pp 169ndash181 2005

[2] H A Salem A M El-Sayed and O L Moustafa ldquoA note onthe fractional calculus in Banach spacesrdquo Studia ScientiarumMathematicarum Hungarica vol 42 no 2 pp 115ndash130 2005

[3] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoMulti-term fractional differential equations in a nonreflexiveBanach spacerdquo Advances in Difference Equations 2013302 18pages 2013

[4] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoFractional calculus and fractional differential equations innonreflexive Banach spacesrdquo Communications in NonlinearScience andNumerical Simulation vol 20 no 1 pp 59ndash73 2015

[5] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoNonlinear fractional differential equations in nonreflexiveBanach spaces and fractional calculusrdquo Advances in DifferenceEquations 2015112 18 pages 2015

[6] R P Agarwal V Lupulescu D OrsquoRegan and G RahmanldquoWeak solutions for fractional differential equations in nonre-flexive Banach spaces via Riemann-Pettis integralsrdquo Mathema-tische Nachrichten vol 289 no 4 pp 395ndash409 2016

[7] H A Salem ldquoOn the fractional order m-point boundaryvalue problem in reflexive Banach spaces and weak topologiesrdquoJournal of Computational andAppliedMathematics vol 224 no2 pp 565ndash572 2009

[8] H A Salem ldquoOn the fractional calculus in abstract spaces andtheir applications to the Dirichlet-type problem of fractionalorderrdquo Computers and Mathematics with Applications vol 59no 3 pp 1278ndash1293 2010

[9] HAH Salem andMCichon ldquoOn solutions of fractional orderboundary value problems with integral boundary conditions inBanach spacesrdquo journal of function spaces and applications vol2013 Article ID 428094 2013

[10] J Diestel and J J Uhl Jr Vector Measures vol 15 ofMathemat-ical Surveys American Mathematical Society Providence RIUSA 1977

[11] R H Martin and J J Uhl Jr Nonlinear Operators AndDifferential Equations in Banach Spaces John Wiley amp SonsNew York NY USA 1976

[12] B J Pettis ldquoOn integration in vector spacesrdquo Transactions of theAmericanMathematical Society vol 44 no 2 pp 277ndash304 1938

[13] S J Dilworth and M Girardi ldquoNowhere weak differentiabilityof the Pettis integralrdquo Quaestiones Mathematicae vol 18 no 4pp 365ndash380 1995

[14] G A Edgar ldquoMeasurability in a Banach spacerdquo Indiana Univer-sity Mathematics Journal vol 26 no 4 pp 663ndash677 1977

[15] G A Edgar ldquoMeasurability in a Banach space IIrdquo IndianaUniversityMathematics Journal vol 28 no 4 pp 559ndash579 1979

[16] J L Gamez and J Mendoza ldquoOn denjoy-dunford and denjoy-pettis integralsrdquo StudiaMathematica vol 130 no 2 pp 115ndash1331998

[17] R AGordon ldquoTheDenjoy extension of the Bochner Pettis andDUNford integralsrdquo Studia Mathematica vol 92 no 1 pp 73ndash91 1989

[18] R F Geitz ldquoPettis integrationrdquo Proceedings of the AmericanMathematical Society vol 82 no 1 pp 81ndash86 1981

[19] R F Geitz ldquoGeometry and the Pettis integralrdquo Transactions ofthe AmericanMathematical Society vol 269 no 2 pp 535ndash5481982

[20] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations New York NYUSA Elsevier 2006

[21] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Yverdon Switzerland 1993

[22] M Mendel and A Naor ldquoMetric cotyperdquo Annals of Mathemat-ics Second Series vol 168 no 1 pp 247ndash298 2008

[23] B Ross S G Samko and E R Love ldquoFunctions that have nofirst order derivative might have fractional derivatives of allorders less than onerdquo Real Analysis Exchange vol 20 no 1 pp140ndash157 199495

[24] E Hille and R S Phillips Functional analysis and semi-groupsAmerican Mathematical Society Colloquium Publications vol31 American Mathematical Society Providence RI USA 1957

[25] S Schwabik and Y Guoju Topics in Banach Space IntegrationWorld Scientific Singapore 2005

[26] DW Solomon ldquoOn differentiability of vector-valued functionsof a real variablerdquo Studia Mathematica vol 29 pp 1ndash4 1967

[27] K Naralenkov ldquoOn Denjoy type extensions of the Pettisintegralrdquo Czechoslovak Mathematical Journal vol 60(135) no3 pp 737ndash750 2010

[28] D W Solomon Denjoy integration in abstract spaces Memorisof the American Mathematical Society No 85 AmericanMathematical Society Providence RI 1969

[29] A R Mitchell and C Smith ldquoAn existence theorem for weaksolutions of differential equations in Banach spacesrdquo inNonlin-ear Equations in Abstract Spaces V Lakshmikantham Ed pp387ndash404 1978

[30] F S De Blasi ldquoOn a property of the unit sphere in a Banachspacerdquo Bulletin Mathematiques De La Societe Des SciencesMathematiques De Roumanie vol 21 no 3 pp 259ndash262 1977

[31] A Szep ldquoExistence theorem for weak solutions of ordinarydifferential equations in reflexive Banach spacesrdquo Studia Scien-tiarum Mathematicarum Hungarica vol 6 pp 197ndash203 1971

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Page 4: On the Theory of Fractional Calculus in the Pettis-Function Spaces · 2019. 7. 30. · JournalofFunctionSpaces Further,wedenethespace H (),˙≥0by H () fl ˛ ∈ ˚, ˜: isHolderainoforder

4 Journal of Function Spaces

then the infinite series of continuous functions in theleft hand side of (15) converges uniformly in [0 1]Hence the function 120593(I120572119909(sdot)) is continuous on [0 1](this yields the weak continuity ofI120572119909(sdot) on [0 1])

In the following lemma we gather together some simpleparticular sufficient conditions that ensure the existence ofthe Riemann-Liouville fractional integral of the functionsfromH119901(119864)Lemma 8 Let 119909 119868 rarr 119864 be weakly measurable function TheRFPI of the function 119909 of order 120572 gt 0 makes sense ae on 119868 ifat least one of the following cases holds

(a) 119909 isin H1199010 (119864) 119901 isin [1infin] and 120572 ge 1

(b) 119909 isin H1199010 (119864) 119901 gt max1 1120572

(c) 119909 is strongly measurable which lies in H119901(119864) where119901 isin [1infin] provided that 119864 is weakly complete orcontains no copy of 1198880

If 119864 is reflexive this is also true for any 119909 isin H119901(119864) 119901 isin [1infin]and 120572 gt 0

In all cases 120593(I120572119909) = I120572120593119909 holds for every 120593 isin 119864lowastIn the assertions ((a) and (b)) we find sufficient con-

ditions needed for the existence of I120572119909 in the situation inwhich no restriction is placed on 119864 In the third assertion theproperties of119864 allow us to characterize a function 119909 isin H119901(119864)for whichI120572119909 exists

Proof Firstly assertion (a) is direct consequence of Proposi-tion 2 (part (1)) since 119904 rarr (119905 minus 119904)120572minus1 isin 119871infin[0 119905] holds foralmost every 119905 isin 119868 whenever 120572 ge 1

Secondly to prove assertion (b) let 119901 gt max1 1120572 and119902 be the conjugate exponents to 119901 Since 119902(120572 minus 1) + 1 gt 0 wehave that 119904 rarr (119905 minus 119904)120572minus1 isin 119871119902[0 119905] holds for every 119905 isin 119868 Thusthe assertion (b) is now an easy consequence of Proposition 2(part (2))

Thirdly to prove (c) we let 120572 gt 0 119901 isin [1infin]and 119909 isin H119901(119864) be strongly measurable Since the strongmeasurability is preserved under multiplication operation onfunctions the product (119905 minus sdot)120572minus1119909(sdot) [0 119905] rarr 119864 is stronglymeasurable on [0 119905] for almost every 119905 isin 119868 In view of Youngrsquosinequality for every 120593 isin 119864lowast the real-valued function

119904 997891997888rarr 120593 ((119905 minus 119904)120572minus1 119909 (119904)) = (119905 minus 119904)120572minus1 120593119909 (119904) (17)

is Lebesgue integrable on [0 119905] for almost every 119905 isin 119868 So theassertion (c) follows immediately from Proposition 2 (parts(3 4))

Similarly when 119864 is reflexive the result follows from part(3) of Proposition 2However in all cases the function 119904 997891rarr (119905 minus 119904)120572minus1119909(119904) is

Pettis integrable on [0 119905] for almost every 119905 isin 119868 That is foralmost every 119905 isin 119868 there exists an element in 119864 denoted byI120572119909(119905) such that

120593 (I120572119909 (119905)) = int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904

= int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120593119909 (119904) 119889119904 = I120572120593119909 (119905)

(18)

holds for every 120593 isin 119864lowast This completes the proof

Remark 9 If 119909 isin H1199010 (119864) such that I120572119909(119905) does not exist for

some 119905 isin 119868 then it does not exist even when we ldquoenlargerdquo thespace 119864 into 119865 To see this let 119909 119868 rarr 119865 such that 119909(119868) sub 119864 IfI120572119909(119905) exists for some 119905 isin 119868 then (119905 minus sdot)120572minus1119909(sdot) isin 119875([0 119905] 119865)Since 119909 assumes only values in 119864 it follows by the mean valuetheorem for Pettis integral (Theorem 4) that the RFPI of 119909should lie in 119864

Before we come to a deep study of the mathematicalproperties of the RFPI operator let us take a look at thefollowing miscellaneous examples

Example 10 Let 120572 gt 0 Define the function 119909 [0 1] rarr119871infin[0 1] by 119909(119905) = 120594[0119905]This function is weakly measurable Pettis integrable on[0 1] and 120593119909 is a function of bounded variation (see eg

[18]) That is 119909 isin Hinfin0 (119871infin) Hence in view of Lemma 8 with119901 = infin the RFPI of 119909 exists on [0 1] Further calculations

(cf [4]) show that

I120572119909 (119905) (sdot) = (119905 minus sdot)120572Γ (1 + 120572)120594[0119905] (sdot) isin 119871infin [0 1] (19)

Example 11 Let120572 gt 0 Define the function119909 from the interval[0 1] into the Hilbert space ℓ2 as119909 119905 997891997888rarr 1(1 + 119905) 1(2 + 119905) 1(3 + 119905) 119905 isin [0 1] (20)

We note that

119909 (119905)2ℓ2 = sum119899

( 1119899 + 119905)2 le sum

119899

11198992 lt infin 119905 isin [0 1] (21)

Thus the function 119909 is well defined We claim that 119909 isDunford integrable on [0 1] Once our claim is establishedLemma 8 guarantees the existence ofI120572119909 on [0 1] It remainsto prove this claim and to calculate I120572119909

To see this let 120593 isin (ℓ2)lowast = ℓ2 According to theRiesz representation theorem on Hilbert spaces there existsa uniquely determined 120582 fl 120582119899 isin ℓ2 such that 120593119909(119905) =sum119899(120582119899(119899 + 119905)) A standard arguments using Beppo LevirsquosTheorem yields

int1

0

1003816100381610038161003816120593119909 (119905)1003816100381610038161003816 119889119905 le sum119899

int1

0

10038161003816100381610038161205821198991003816100381610038161003816(119899 + 119905)119889119905 = sum119899

10038161003816100381610038161205821198991003816100381610038161003816 ln(1 + 1119899)le sum

119899

10038161003816100381610038161205821198991003816100381610038161003816119899 le sum119899

10038161003816100381610038161205821198991003816100381610038161003816 lt infin(22)

So 119909 is Dunford integrable on [0 1] and hence Pettis inte-grable on [0 1] since ℓ2 is reflexive Consequently in view of

Journal of Function Spaces 5

Lemma 8 I120572119909 exists on [0 1] To calculate the RFPI of 119909 fix119905 isin [0 1] and let 120593 isin (ℓ2)lowast We have

int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904 = int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120593119909 (119904) 119889119904= I

120572 (infinsum119899=1

120582119899(119899 + 119905)) (23)

Since the series sum119899(120582119899(119899 + 119905)) is uniformly convergent on[0 1] it follows in view of the generalized linearity of thefractional integrals [23 Lemma 5] that (cf [21 Table 91])

int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904= infinsum

119899=1

120582119899119905120572119899Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 ) = 120593 (119892 (119905)) (24)

where 2F1 is the Gauss hypergeometric function evaluated at(1 1 120572 + 1 minus119905119899) and119892 (119905) fl 119905120572119899Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 ) (25)

Since

1003817100381710038171003817119892 (119905)10038171003817100381710038172ℓ2 = sum119899

( 119905120572Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 ))2

le 2F21 (1 1 120572 + 1 minus1)Γ2 (1 + 120572) sum

119899

11198992 lt infin119905 isin [0 1]

(26)

we see that 119892 isin ℓ2 Thus

I120572119909 (119905) = 119905120572119899Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 )

119905 isin [0 1] (27)

Example 12 Let 120572 isin [12 1) Define the countable-valuedfunction 119909 [0 1] rarr 1198880 by

119909 (119905) fl 119909119899 (119905) 119909119899 fl 119899120594(1(119899+1)1119899] (28)

This function is strongly measurable Pettis integrable func-tion on [0 1] (see eg [10 16]) We claim that 119909 isin H119875

0 (1198880)with 119901 isin (1120572 2] Once our claim is established Lemma 8guarantees the existence of I120572119909 120572 isin [12 1) on [0 1] Itremains to prove this claim by showing firstly that H119875(1198880)and to calculate I120572119909 To do this let 120593 isin 119888lowast0 Then therecorresponds to 120593 a unique 120582119899 isin ℓ1 such that 120593119909 = sum119899 120582119899119909119899By noting that

10038171003817100381710038171205821198991199091198991003817100381710038171003817119901 = (int1

0

1003816100381610038161003816120582119899119909119899 (119905)1003816100381610038161003816119901 119889119905)1119901

= 10038161003816100381610038161205821198991003816100381610038161003816 (int1119899

1(119899+1)119899119901119889119905)1119901 = 10038161003816100381610038161205821198991003816100381610038161003816 1198991119902(119899 + 1)1119901

le 10038161003816100381610038161205821198991003816100381610038161003816 11989911199021198991119901 le 100381610038161003816100381612058211989910038161003816100381610038161198992119901minus1 le 10038161003816100381610038161205821198991003816100381610038161003816

(29)

holds for any 119901 isin (1120572 2] a standard argument using LevirsquosTheorem (or Lebesgue dominated convergence theorem) andMinkowskirsquos inequality yields

[int1

0

1003816100381610038161003816120593119909 (119905)1003816100381610038161003816119901 119889119905]1119901 le [int1

0(infinsum119899=1

10038161003816100381610038161205821198991003816100381610038161003816 119909119899 (119905))119901 119889119905]1119901

= [[lim119896rarrinfin

int1

0( 119896sum119899=1

10038161003816100381610038161205821198991003816100381610038161003816 119909119899 (119905))119901

119889119905]]1119901

le [[lim119896rarrinfin

( 119896sum119899=1

10038171003817100381710038171205821198991199091198991003817100381710038171003817119901)119901]]1119901

le infinsum119899=1

10038171003817100381710038171205821198991199091198991003817100381710038171003817119901le infinsum

119899=1

10038161003816100381610038161205821198991003816100381610038161003816 lt infin

(30)

Thus 120593119909(sdot) isin 119871119901[0 1] 119901 isin (1120572 2] holds for every 120593 isin 119888lowast0(that is 119909 isin H119875(1198880)) Since 119901 gt 1120572 gt 1 it follows byProposition 2 in view of the strong measurability of 119909 that119909 isin H119875

0 (1198880) Owing to Lemma 8 we infer that the RFPI of119909 of any order 120572 isin [12 1) exists on the interval [0 1] Tocompute this integral fix 119905 isin 1198681198990 fl [1(1198990+1) 11198990] for some1198990 isin N the set of natural numbers and let 120593 isin 119888lowast0 Then

sum119899

int119905

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 12058211989910038161003816100381610038161003816 119909119899 (119904) 119889119904le 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816 int

119905

1(1198990+1)(119905 minus 119904)120572minus1 1198990119889119904

+ sum119899gt1198990

10038161003816100381610038161205821198991003816100381610038161003816 int1119899

1(119899+1)(119905 minus 119904)120572minus1 119899 119889119904

= 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572

+ sum119899gt1198990

119899 10038161003816100381610038161205821198991003816100381610038161003816120572 [(119905 minus 1119899 + 1)120572 minus (119905 minus 1119899)

120572]

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572

+ sum119899gt1198990

119899 10038161003816100381610038161205821198991003816100381610038161003816120572 [(119905 minus 1119899 + 1) minus (119905 minus 1119899)]120572

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572 + sum

119899gt1198990

119899 10038161003816100381610038161205821198991003816100381610038161003816120572 1119899120572 (119899 + 1)120572

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572 + sum

119899gt1198990

10038161003816100381610038161205821198991003816100381610038161003816120572 11198992120572minus1

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572 + sum

119899gt1198990

10038161003816100381610038161205821198991003816100381610038161003816120572 lt infin(because 120572 isin [12 1))

(31)

6 Journal of Function Spaces

Therefore by Beppo Levi Theorem it follows that

int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904 = int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120593 (119909 (119904)) 119889119904= int119905

0

(119905 minus 119904)120572minus1Γ (120572) sum119899

120582119899119899120594(1(119899+1)1119899] (119904) 119889119904= sum

119899

int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120582119899119899120594(1(119899+1)1119899] (119904) 119889119904= 11989901205821198990Γ (1 + 120572) (119905 minus 11198990 + 1)

120572

+ sum119899gt1198990

119899120582119899Γ (1 + 120572) [(119905 minus 1119899 + 1)120572 minus (119905 minus 1119899)

120572]

(32)

Consequently we conclude that

I120572119909 (119905) = 0 0⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

1198990minus1

1198990Γ (1 + 120572) (119905 minus 11198990 + 1)120572

(1198990 + 1)Γ (1 + 120572) [(119905 minus 11198990 + 2)

120572 minus (119905 minus 11198990 + 1)120572]

(33)

where the nonzero coordinate started from the 1198990th place Itcan be easily seen thatI120572119909(119905) isin 1198880 for any 119905 isin [0 1] Evidentlywe have that10038161003816100381610038161003816100381610038161003816119899 [(119905 minus

1119899 + 1)120572 minus (119905 minus 1119899)

120572]10038161003816100381610038161003816100381610038161003816le 119899 [(119905 minus 1119899 + 1) minus (119905 minus 1119899)]

120572 le 119899 [ 1119899 (119899 + 1)]120572

le 11198992120572minus1 997888rarr 0 as 119899 997888rarr infin1198990Γ (1 + 120572) (119905 minus 11198990 + 1)

120572

le 1198990Γ (1 + 120572) ( 11198990 minus 11198990 + 1)120572 le 11198992120572minus10 Γ (1 + 120572)

997888rarr 0 as 1198990 997888rarr infin

(34)

hold for any 119905 isin [0 1]This yieldsI1205721119909(119905) isin 1198880 for any 119905 isin [0 1]

(this is precisely what we would expect from Definition 5)

Look at part (c) of Lemma 8 with 120572 isin (0 1) and119864 = 1198880 Below we give an example showing that the stronglymeasurability hypothesis imposed on a function 119909 isin H1(119864)H1

0(119864) is not sufficient for the existence of I120572119909 even when 119909is Denjoy-Pettis integrable (cf [16])

Example 13 Let 120572 isin (0 1) Define the strongly measurablefunction 119909 [0 1] rarr 1198880 by119909 (119905) fl 119909119899 (119905)

119909119899 (119905) fl 119899 (119899 + 1) [120594119868119899 (119905) minus 120594119869119899 (119905)] (35)

where

119868119899 fl ( 1119899 + 1 119899 + 12119899 (119899 + 1)) 119869119899 fl ( 119899 + 12119899 (119899 + 1) 1119899)

(36)

It is immediate that (cf [16]) 119909 is a well-defined Denjoy-Pettis (but it is not Pettis) integrable on [0 1]

In what follows we will show that the RFPI of 119909 does notexist on a subinterval of positive measure on [0 1]

To see this we make use of Proposition 3 as followsdefine for each intervalI119899 = [1(119899 + 1) 1119899] the functionals120593119899 isin 119888lowast0 (required by Proposition 3) to be the correspondingto the element 120582119899 fl (0 0 0 1 0 0 ) isin ℓ1 wherethe nonzero coordinate is in the 119899th place Then 120593119899(119909(119905)) =119909119899(119905) and thus |120593119899(119909(119904))| = 119899(119899 + 1)[120594119868119899(119905) minus 120594119869119899(119905)] Clearlythe family 120593119899 runs through the unit ball of the dual of1198880 Evidently the isometric isomorphism between 119888lowast0 and ℓ1yields 120593119899119888lowast0 = 120582119899ℓ1 = 1

Now take 119905 isin [12 1] and define the continuous function119891 [12 1] rarr R+ by

119891 (119905) fl (119905 minus 1119899 + 1)120572 minus (119905 minus 1119899)

120572 119899 ge 2 119905 isin [12 1]

(37)

It can be easily seen that in view of 0 lt 120572 lt 1 1198911015840 is negativeon (0 1) By standard results from (classical) calculus itfollows that 119891 is strictly decreasing on [0 1] in particular119891(1) lt 119891(119905) for all 119905 isin [0 1) Thus for any 119899 ge 2 and any120593 isin 120593119899 we have

intI119899

100381610038161003816100381610038161003816100381610038161003816120593 ((119905 minus 119904)120572minus1Γ (120572) 119909 (119904))100381610038161003816100381610038161003816100381610038161003816 119889119904

= intI119899

(119905 minus 119904)120572minus1Γ (120572) 1003816100381610038161003816120593 (119909 (119904))1003816100381610038161003816 119889119904= int1119899

1(119899+1)

(119905 minus 119904)120572minus1Γ (120572) 119899 (119899 + 1) 119889119904= 119899 (119899 + 1)Γ (1 + 120572) [(119905 minus 1119899 + 1)

120572 minus (119905 minus 1119899)120572]

ge 1198992Γ (1 + 120572) [(1 minus 1119899 + 1)120572 minus (1 minus 1119899)

120572]

(38)

An explicit calculation using LrsquoHospitalrsquos rule two timesreveals

lim119899rarrinfin

(1 minus 1 (119899 + 1))120572 minus (1 minus 1119899)120572119899minus2 = minus1205722sdot lim119899rarrinfin

1198993 [( 119899119899 + 1)120572minus1 1

(119899 + 1)2 minus (1 minus 1119899)120572minus1 11198992 ]

= minus1205722 lim119899rarrinfin

119899 [( 119899119899 + 1)120572minus1 1198992

(119899 + 1)2 minus (1 minus 1119899)120572minus1]

Journal of Function Spaces 7

= minus1205722 lim119899rarrinfin

119899 [( 119899119899 + 1)120572+1 minus (1119899)

120572minus1] = minus1205722sdot lim119899rarrinfin

1198992 [(120572 + 1) ( 119899119899 + 1)120572 1(119899 + 1)2

minus (120572 minus 1) (1 minus 1119899)120572minus2 11198992 ] = minus1205722

sdot lim119899rarrinfin

[(120572 + 1) ( 119899119899 + 1)120572+2 minus (120572 minus 1) (1 minus 1119899)

120572minus2]= 120572

(39)

from which it follows that

intI119899

100381610038161003816100381610038161003816100381610038161003816120593 ((119905 minus 119904)120572minus1Γ (120572) 119909 (119904))100381610038161003816100381610038161003816100381610038161003816 119889119904997888rarr0 as 119899 997888rarr infin (40)

Therefore in view of Proposition 3 119904 997891rarr (119905 minus 119904)120572minus1119909(119904) notin119875([0 1] 1198880) Hence I120572119909 does not exist which is what wewished to show

The following theorem provides a useful characterizationof the space H119901(119864) for which the statements reveal howmuch the fractional integral I120572 is better than the function119909 isin H119901(119864) Indeed based on Lemma 8 using an inequalityof Young we can easily prove the following

Lemma 14 For any 120572 gt 0 the following holds(a) If 119864 is reflexive then for every 119901 isin [1infin] the operator

I120572 mapsH119901(119864) intoH1199010 (119864) In particular if 0 lt 120572 lt1 then I120572 H1(119864) rarr H1((1minus120572)minus120598)0 (119864) however small120598 gt 0 is

(b) If 119864 contains no copy of 1198880 then for every 119901 isin [1infin]the operator I120572 maps 119909 isin H

1199010 (119864) 119909 strongly

measurable into H1199010 (119864) In particular if 120572 ge 1 the

operator I120572 mapsH1199010 (119864) into itself

(c) If 119901 gt max1 1120572 the operator I120572 sends H1199010 (119864) to119862[119868 119864120596] (if we define I120572119909(0) fl 0) In particular if120572 isin (0 1) I120572 H119901

0 (119864) rarr H120572minus1119901(119864)Proof At the beginning let 120572 gt 0 and 119909 isin H119901(119864) with119901 isin [1infin] and define the real-valued function 119892 119868 rarr R

by 119892(119905) fl 119905120572minus1Γ(120572) If 119864 is reflexive it follows in view ofLemma 8 that J120572119909 exists ae in 119868 and 120593(I120572119909) = I120572120593119909for every 120593 isin 119864lowast Youngrsquos inequality guarantees that theconvolution product 119892 lowast 120593119909(sdot) lies in 119871119901[119868] Consequently120593(I120572119909) = I120572120593119909 = 119892 lowast 120593119909 isin 119871119901[119868] for every 120593 isin 119864lowastThe reflexivity of 119864 together with Proposition 2 yields I120572119909 isinH

1199010 (119864)In particular if 0 lt 120572 lt 1 it can be easily seen thatint

119868(119892(119905))119902119889119905 lt infin for every 119902 isin [1 1(1 minus 120572)) That is 119892 isin1198711(1minus120572)minus120598[119868] 120598 gt 0 By Youngrsquos inequality it follows that120593(I120572119909) = 119892 lowast 120593119909 isin 1198711(1minus120572)minus120598[119868] for every 120593 isin 119864lowast however

small 120598 gt 0 is Now the assertion (a) follows because of thereflexivity of 119864

Next in order to prove the assertion (b) let 119909 isin H1199010 (119864)

and note in view of Lemma 8 that I120572119909 exists on 119868 Define119910 119868 rarr 119864 by119910(119905) fl I120572119909(119905) 119909 isin H1199010 (119864) Since120593119910(sdot) isin 119871119901[119868]

for every120593 isin 119864lowast it follows that119910 isin H119901(119864) Moreover for any119886 119887 isin 119868 (119886 lt 119887) we haveint119887

119886120593119910 (119904) 119889119905 = int119887

119886I120593119909 (119905) 119889119905 = 1Γ (120572)

sdot int119887

119886int119905

0(119905 minus 119904)120572minus1 120593119909 (119904) 119889119904 119889119905

= 1Γ (120572) [int119887

119886int119887

119904(119905 minus 119904)120572minus1 120593119909 (119904) 119889119905 119889119904

+ int119886

0int119887

119886(119905 minus 119904)120572minus1 120593119909 (119904) 119889119905 119889119904]

= 1Γ (120572) [int119887

119886(119887 minus 119904)120572 120593119909 (119904) 119889119904

+ int119886

0[(119887 minus 119904)120572 minus (119886 minus 119904)120572] 120593119909 (119904) 119889119904]

= 1Γ (120572) [int119886

0(119887 minus 119904)120572 120593119909 (119904) 119889119904

+ int119887

119886(119887 minus 119904)120572 120593119909 (119904) 119889119904 minus int119886

0(119886 minus 119904)120572 120593119909 (119904) 119889119904]

= int119887

0

(119887 minus 119904)120572Γ (1 + 120572)120593119909 (119904) 119889119904 minus int119886

0

(119886 minus 119904)120572Γ (1 + 120572)120593119909 (119904) 119889119904= 120593 (119909[119886119887])

(41)

where

119909[119886119887] fl (119875) int119887

0

(119887 minus 119904)120572Γ (1 + 120572)119909 (119904) 119889119904minus (119875) int119886

0

(119886 minus 119904)120572Γ (1 + 120572)119909 (119904) 119889119904(42)

Since 119909 isin 119875[119868 119864] owing to Proposition 2 (part (1)) it followsthat (119887 minus sdot)120572119909(119904) and (119886 minus sdot)120572119909(119904) are Pettis integrable on 119868 andso 119909[119886119887] isin 119864

A combination of these results yields120593119910 isin 119871119901(119868) for every120593 isin 119864lowast and there exists an element 119909[119886119887] isin 119864 such that120601119909[119886119887] = int119887

119886120593119910(119905)119889119905 for every 119886 119887 isin 119868 Since 119864 contains no

copy of 1198880 it follows that (cf [17 Theorem 23]) 119910 isin 119875[119868 119864]Consequently 119910(sdot) isin H

1199010 (119864) This is the claim (b)

To prove the assertion (c) let 119901 gt max1 1120572 and 119909 isinH

1199010 (119864) By Lemma 8 we deduce that I120572119909 exists ae on 119868

Now let 119902 isin [1infin] be such that 1119901 + 1119902 = 1 Since119902(120572 minus 1) gt minus1 119892 fl (sdot)120572minus1Γ(120572) isin 119871119902[119868] Therefore as a directconsequence of Youngrsquo inequality it follows that

120593 (I120572119909 (sdot)) = I120572120593119909 (sdot) = 119892 lowast 120593119909 (sdot) isin 119862 [119868R] (43)

holds for every 120593 isin 119864lowast Now we claim that I120572119909(119905) rarr 0in 119864 as 119905 rarr 0 Once our claim is established the definition

8 Journal of Function Spaces

I120572119909(0) fl 0 guarantees thatI120572119909(sdot) isin 119862[119868 119864120596] which is whatwewished to show It remains to prove our claim without lossof generality assume that I120572119909(119905) = 0 Then there exists (asa consequence of the Hahn-Banach theorem) 120593 isin 119864lowast with120593 = 1 such that I120572119909(119905) = 120593(I120572119909(119905)) = I120572120593119909(119905) ByHolderrsquos inequality we obtain

1003816100381610038161003816120593 (I120572119909 (119905))1003816100381610038161003816le 1Γ (120572) (int

119905

0(119905 minus 119904)119902(120572minus1) 119889119904)1119902 (int119905

0

1003816100381610038161003816120593119909 (119904)1003816100381610038161003816119901 119889119904)1119901

le 1Γ (120572) ( 119905(120572minus1)119902+1(120572 minus 1) 119902 + 1)1119902 (int

119868

1003816100381610038161003816120593119909 (119904)1003816100381610038161003816119901 119889119904)1119901

= 119905120572minus1+1119902Γ (120572) [119902 (120572 minus 1) + 1]1119902

10038171003817100381710038171205931199091003817100381710038171003817119871119901

= 119905120572minus1119901 10038171003817100381710038171205931199091003817100381710038171003817119871119901Γ (120572) [119902 (120572 minus 1) + 1]1119902

(44)

This is equivalent with the following estimate

1003817100381710038171003817I120572119909 (119905)1003817100381710038171003817 le 119905120572minus1119901 10038171003817100381710038171205931199091003817100381710038171003817119871119901Γ (120572) [119902 (120572 minus 1) + 1]1119902 (45)

Owing to 120572 gt 1119901 we get I120572119909(119905) rarr 0 in 119864 as 119905 rarr 0 andconsequently in view ofI120572119909(0) fl 0I120572119909(sdot) isin 119862[119868 119864120596] Thisproves the first part of the assertion (c)

Finally let 120572 isin (0 1) and 119905 120591 isin 119868 Without loss ofgenerality assume 120591 lt 119905 Then for any 120593 isin 119864lowast we have byHolder inequality with 119901 gt 1120572 in view 119902(120572 minus 1) gt minus1 that

1003816100381610038161003816120593 (I120572119909 (119905) minusI120572119909 (120591))1003816100381610038161003816 Γ (120572)

= 10038161003816100381610038161003816100381610038161003816int119905

0(119905 minus 119904)120572minus1 120593119909 (119904) 119889119904 minus int120591

0(120591 minus 119904)120572minus1 120593119909 (119904) 11988911990410038161003816100381610038161003816100381610038161003816

le (int120591

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 minus (120591 minus 119904)120572minus110038161003816100381610038161003816 1003816100381610038161003816120593119909 (119904)1003816100381610038161003816 119889119904+ int119905

120591(119905 minus 119904)120572minus1 1003816100381610038161003816120593119909 (119904)1003816100381610038161003816 119889119904)

= [(int120591

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 minus (120591 minus 119904)120572minus110038161003816100381610038161003816119902 119889119904)1119902

+ (int119905

120591(119905 minus 119904)(120572minus1)119902 119889119904)1119902] 10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]

le [[(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

+ ((119905 minus 120591)(120572minus1)119902+1(120572 minus 1) 119902 + 1 )1119902]]10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]

= [(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

+ (119905 minus 120591)120572minus1119901[119902 (120572 minus 1) + 1]1119902]

10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868] (46)

By noting that (120572 minus 1)119902 isin (minus1 0) when 120572 isin (0 1) it can beeasily seen that

int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904le 1119902 (120572 minus 1) + 1 (119905 minus 120591)119902(120572minus1)+1

(47)

That is

(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

le (119905 minus 120591)120572minus1119901[119902 (120572 minus 1) + 1]1119902

(48)

A combination of these results yields1003816100381610038161003816120593 (I120572119909 (119905)) minus 120593 (I120572119909 (120591))1003816100381610038161003816= 1003816100381610038161003816120593 (I120572119909 (119905) minusI

120572119909 (120591))1003816100381610038161003816le10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]Γ (120572) 2 (119905 minus 120591)120572minus1119901

[119902 (120572 minus 1) + 1]1119902 (49)

ThusI120572119909 isin H120572minus1119901(119864) This completes the proof

Corollary 15 For any 120572 gt 0 I120572 119862[119868 119864120596] rarr 119862[119868 119864120596]Proof Let 119909 isin 119862[119868 119864120596] then 120593119909 isin 119871infin holds for every120593 isin 119864lowast By noting that the weak continuity implies a strongmeasurability ([24] page 73) it follows that 119909 is stronglymeasurable on 119868 Hence in view of Proposition 2 119909 isinHinfin

0 (119864) Consequently 119862[119868 119864120596] sube H1199010 (119864) for every 119901 isin[1infin] That isI120572 119862[119868 119864120596] rarr 119862[119868 119864120596] Hence the desired

result is obtained

Example 16 Let 120572 isin [12 1) Define the function 119909 [0 1] rarr1198880 by 119909(119905) = 119899120594(1(119899+1)1119899]We have (in view of Example 12)that

I120572119909 (119905) = 0 0 0 1198990 (119905 minus 1 (1198990 + 1))120572

Γ (1 + 120572) (1198990 + 1)Γ (1 + 120572) [(119905 minus 11198990 + 2)

120572 minus (119905 minus 11198990 + 1)120572]

(50)

To show that I120572119909 is norm continuous on [0 1] let 120591 119905 isin[0 1] With no loss of generality we may assume that 119905 120591 isin1198681198990 for some 1198990 isin N Since the nonzero terms of the sequencein (50) are nonincreasing then in view of

|119905 minus 120591| le ( 11198990 minus 11198990 + 1) = 11198990 (1198990 + 1) le 111989920 (51)

Journal of Function Spaces 9

we have1003817100381710038171003817I120572119909 (119905) minusI

1205721119909 (120591)10038171003817100381710038171198880

= 1198990Γ (1 + 120572)10038161003816100381610038161003816100381610038161003816(119905 minus

1119899 + 1)120572 minus (120591 minus 1119899 + 1)

12057210038161003816100381610038161003816100381610038161003816le 1198990Γ (1 + 120572)

1003816100381610038161003816100381610038161003816(119905 minus 1119899 + 1) minus (120591 minus 1119899 + 1)1003816100381610038161003816100381610038161003816120572

le 1198990Γ (1 + 120572) |119905 minus 120591|120572 le 1Γ (1 + 120572) |119905 minus 120591|120572radic|119905 minus 120591|= |119905 minus 120591|120572minus12Γ (1 + 120572)

(52)

Thus the RFPI of 119909 is norm continuous on [0 1] Preciselysince 0 lt 120572 minus 12 lt 120572 minus 1119901 lt 1 for any 119901 le 2 thenI120572119909 isin H120572minus12(119864) sub H120572minus1119901(119864) This is precisely what onewould expect from Lemma 14 (part (c))

Analogously an explicit calculation reveals that I120572119909 liesin 119862([0 1] 119871infin[0 1]) where 119909 [0 1] rarr 119871infin[0 1] defined byExample 10

We now consider additional mapping properties of theoperator I120572 Precisely we will show that the RFPI enjoysthe following commutative property which is folklore in case119864 = R However the proof is completely similar to that of [8]Lemma 35

Lemma 17 Let 120572 120573 gt 0 and 119901 gt max1 1120572 1120573 Then

I120572I120573119909 = I

120573I120572119909 = I

120572+120573119909 (53)

holds for every 119909 isin H1199010 (119864) If 119864 is reflexive this is also true for

every 119901 ge 1Proof Since 119901 gt max1 1120572 1120573 it follows that 119901 gtmax1 1(120572+120573) So by Lemmas 8 and 14 themappingsI120572119909I120573119909 andI120572+120573119909 belong toHinfin

0 (119864) for every 119909 isin H1199010 (119864) Now

repeating the same process as in ([8] Lemma 35) implies theclaim

When 119864 is reflexive the result follows as a direct applica-tion of Lemma 14

3 Fractional Derivatives ofVector-Valued Functions

After the notation of the fractional integrals of vector-valued functions the fractional derivatives become a naturalrequirement Before we come to the definitions and a detailedstudy of the mathematical properties of fractional differentialoperators we recall the following

Definition 18 Consider the vector-valued function 119909 119868 rarr119864(1) Let 120593119909 be differentiable on 119868 for every 120593 isin 119864lowast The

function 119909 is said to be weakly differentiable on 119868 if

there exists 119910 119868 rarr 119864 such that for every 120593 isin 119864lowast wehave

119889120593119909 (119905)119889119905 = 120593119910 (119905) for every 119905 isin 119868 (54)

The function 119910 is called the weak derivative of thefunction 119909

(2) Let 120593119909 be differentiable ae on 119868 for every 120593 isin 119864lowast(the null set may vary with 120593 isin 119864lowast) The function 119909is said to be pseudo differentiable on 119868 if there existsa function 119910 119868 rarr 119864 such that for every 120593 isin 119864lowast thereexists a null set119873(120593) sub 119868 such that

(120593119909 (119905))1015840 = 120593119910 (119905) for every 119905 isin 119868119873 (120593) (55)

In this case the function 119910 is called the pseudoderivative of 119909If 119909 is pseudo differentiable on 119868 and the null setinvariant for every 120593 isin 119864lowast then 119909 is ae weaklydifferentiable on 119868

Clearly if 119909 is ae weakly differentiable on 119868 then 120593(119909)is ae differentiable on 119868 The converse holds in a weaklysequentially complete space (see [25] Theorem 733)

For more details of the derivatives of vector-valuedfunctions we refer to [10 12 26]

The following results play a major role in our analysis

Proposition 19 (see [27] Theorem 51) The function 119910 119868 rarr119864 is an indefinite Pettis integrable if and only if 119910 is weaklyabsolutely continuous on 119868 and have a pseudo derivative on 119868In this case 119910 is an indefinite Pettis integral of any of its pseudoderivatives

Now we are in the position to define the fractional-typederivatives of vector-valued functions

Definition 20 Let 119909 119868 rarr 119864 For the positive integer119898 suchthat 120572 isin (119898 minus 1119898) 119898 isin N0 fl 0 1 2 we define theCaputo fractional-pseudo (weak) derivative ldquoshortly CFPD(CFWD)rdquo of 119909 of order 120572 by

119889120572119889119905120572 119909 (119905) fl I119898minus120572119863119898119909 (119905) (56)

where the sign ldquo119863rdquo denotes the pseudo (or weak) differentialoperator We use the notation 119889120572119901119889119905120572 and 119889120572120596119889119905120572 to charac-terize the Caputo fractional-pseudo derivatives and Caputofractional weak derivatives respectively

It is well known that although the weak derivative of aweakly differentiable function is uniquely determined thepseudo derivative of the pseudo differentiable function is notunique Also although any two pseudo derivatives 119910 119911 of afunction 119909 119868 rarr 119864 need not be ae equal (see [12 Example91] and [26 p 2]) the functions 119910 119911 are weakly equivalenton 119868 (that is 120593119910 = 120593119911 holds ae for every 120593 isin 119864lowast) The

10 Journal of Function Spaces

next lemma provides a useful characterization property of theCFPD Really it can be easily seen that the CFPD of a Caputofractional-pseudo differentiable function does not depend onthe choice of a pseudo derivative of the function

Lemma 21 Let 119909 119868 rarr 119864 be pseudo differentiable functionwhere pseudo derivatives lie in H

1199010 (119864) If (119889120572119901119889119905120572)119909 exists on119868 then the CFPD of 119909 depends on the choice of the pseudo

derivatives of 119909Proof Let 119910 119911 isin H

1199010 (119864) be two pseudo derivatives of the

pseudo differentiable function 119909 Since 119909 and 119910 are weaklyequivalent on 119868 then for every 120593 isin 119864lowast there exists a nullset N which depends on 120593 isin 119864lowast such that (119905 minus 119904)minus120572120593119910(119904) =(119905minus119904)minus120572120593119911(119904) for every 119904 isin [0 119905]N Consequently for almostevery 119905 isin 119868

120593 (I1minus120572119910 (119905)) = I1minus120572120593119910 (119905) = I

1minus120572120593119911 (119905)= 120593 (I1minus120572119911 (119905)) (57)

holds for every 120593 isin 119864lowast Thus

119889120572119901119889119905120572 119909 = I1minus120572119863119901119909 = I

1minus120572119910 = I1minus120572119911 on 119868 (58)

which is what we wished to show

We consider the following examples

Example 22 Define 119909 [0 1] rarr ℓ2 by119909 (119905) fl ln(1 + 119905119899) 119899 isin N (59)

We claim that the CFWD of 119909 exists on [0 1] To see this welet 120593 isin (ℓ2)lowast = ℓ2 Then there corresponds to 120593 a sequence120582119899 isin ℓ2 such that

120593119909 (119905) = sum119899

120582119899 ln(1 + 119905119899) (60)

Since the series in (60) converges uniformly on [0 1] theformal differentiation of 120593119909 yields

(120593119909 (119905))1015840 = sum119899

120582119899119899 + 119905 = 120593 ( 1119905 + 119899) (61)

It is not hard to justify the differentiation by noting that theseries in the right hand side of (61) is uniformly and absolutelyconvergent on [0 1] Consequently 120593119909 is differentiable on[0 1] for 120593 isin (ℓ2)lowast meaning that 119909 is weakly differentiable on[0 1] (because ℓ2 is weakly complete (cf [25] Theorem 733)and 119863120596119909(119905) = 1(119905 + 119899) Owing to Lemma 25 the CFWDof any order 120572 isin (0 1) of 119909 exists on [0 1] To calculate theCFWD of 119909 on [0 1] we observe in view of Example 11 that

119889120572120596119889119905120572 119909 (119905) = I1minus120572119863120596119909 (119905)

= 1199051minus120572119899Γ (2 minus 120572) 2F1 (1 1 2 minus 120572 minus119905119899 ) (62)

Example 23 Let119864 be an infinite dimensionBanach space thatfails cotype Define 119910 [0 1] rarr 119864 by 119910(119905) fl int119905

0119909(119904)119889119904

where 119909 [0 1] rarr 119864 given by formula (7) As cited in ([13]Corollary 4) 119910 is nowhere weakly differentiable on [0 1]Consequently the CFWD of any order 120572 isin [34 1) of 119910 losesits meaning on [0 1]Remark 24 As shown in Example 23 there is an infinitedimension Banach space 119864 and weakly absolutely continuousfunction 119910 119868 rarr 119864 which is nowhere weakly differentiable(hence the CFWD of 119910 does not exist) Also even when 119864 isseparable and 119910 is Lipschitz function the pseudo derivatives(hence the CFPD) of 119910 need not to exist [28]

However Definition 20 of the CFPD (CFWD) has thedisadvantage that it completely loses its meaning if thefunction119909 fails to be pseudo (weakly) differentiable Preciselythe CFPD (in particular the CFWD) of a function 119909 loses itsmeaning if 119909 is not weakly absolutely continuous

Thenext lemma gives sufficient conditions that ensure theexistence of the Caputo fractional derivatives of a function119909 isin H119901(119864)Lemma 25 Let 120572 isin (0 1) For the function 119909 119868 rarr 119864 thefollowing hold

(a) If 119909 has a pseudo derivative 119863119901119909 isin H1199010 (119864) where 119901 gt1(1 minus 120572) then the CFPD of 119909 of order 120572 exists on 119868

Moreover (119889120572119901119889119905120572)119909 isin 119862[119868 119864120596](b) If 119864 is weakly complete or contains no copy of 1198880 and if119909 has a weak derivative 119863120596119909 isin H119901(119864) where 119901 isin[1infin] then the CFWD of 119909 of order 120572 exists on 119868

Moreover (119889120572120596119889119905120572)119909 isin 119862[119868 119864120596]This holds for any 119909 isin H119901(119864) with 119901 isin [1infin] if 119864 is reflexive

In all cases 120593((119889120572119901119889119905120572)119909) = (119889120572119901119889119905120572)120593119909 (120593((119889120572120596119889119905120572)119909) =(119889120572120596119889119905120572)120593119909) holds for every 120593 isin 119864lowastProof Since the weak (pseudo) derivative of an ae weakly(pseudo) differentiable function is strongly (weakly) mea-surable [12 26 28] the proof is readily available in view ofthe definition of Caputo fractional derivatives and Lemma 8Moreover since 119863120596119909(119863119901119909) isin H

1199010 (119864) with 119901 gt 1(1 minus 120572)

it follows in view of Lemma 14 that I1minus120572119863120596119909(I1minus120572119863119901119909) isin119862[119868 119864120596] This completes the proof

Besides the Caputo fractional-pseudo (weak) derivativeswe define the Riemann-Liouville fractional-pseudo (weak)derivatives

Definition 26 Let 119909 119868 rarr 119864 For the positive integer 119898such that 120572 isin (119898 minus 1119898) 119898 isin N0 fl 0 1 2 we definethe Riemann-Liouville fractional-pseudo (weak) derivativeldquoshortly RFPD (RFWD)rdquo of 119909 of order 120572 by

D120572119909 (119905) fl 119863119898

I119898minus120572119909 (119905) (63)

where 119863 is defined as in Definition 20 We use the notationD120572

119901 (D120572120596) to characterize the Riemann-Liouville fractional-

pseudo (weak) derivatives

Journal of Function Spaces 11

Clearly in infinite dimension Banach spaces the weakabsolute continuity of I119898minus120572119909 is necessarily (but not suf-ficient) condition for the existence of RFPD (in particularRFWD) of 119909Lemma 27 Let 0 lt 120572 lt 1 For any 119909 isin H

1199010 (119864) with 119901 gt

max1120572 1(1 minus 120572) we haveD

120572119901I

120572119909 = 119909 ae (64)

If 119864 is reflexive this is also true for every 119901 ge 1Proof Our assumption 119909 isin H

1199010 (119864)with 119901 gt max1120572 1(1minus120572) yields in view of Lemma 17 that

D120572119901I

120572119909 = 119863119901I1minus120572

I120572119909 = 119863119901I

1119909 (65)

The claim now follows immediately since (cf Proposition 19)the indefinite integral of Pettis integrable function is weaklyabsolutely continuous and it is pseudo differentiable withrespect to the right endpoint of the integration interval andits pseudo derivative equals the integrand at that point

Remark 28 When we replace D120572119901 by D120572

120596 then Lemma 27is no longer necessarily true for arbitrary 119909 isin H1

0(119864) evenwhen 119864 is reflexive evidently in [13 remark below Corollary4] the existence of a reflexive Banach space 119864 and a strongmeasurable Pettis integrable function 119909 119868 rarr 119864 was provedsuch that 119909 has nowhere weakly differentiable integral In thiscase 119863120596I

1119909 lost its meaning This gives a reason to believethat (65) (hence (64)) withD120572

120596119909 could not happen

However we have the following result

Lemma 29 Let 0 lt 120572 lt 1 For any 119909 isin 119862[119868 119864120596] we haveD

120572120596I

120572119909 = 119909 on 119868 (66)

Proof Our assumption 119909 isin 119862[119868 119864120596] sub Hinfin0 (119864) (see the proof

of Corollary 15) yields in view of Lemma 17 that

D120572120596I

120572119909 = 119863120596I1minus120572

I120572119909 = 119863120596I

1119909 (67)

The claim now follows immediately since the indefiniteintegral of weakly contentious function is weakly absolutelycontinuous and it is weakly differentiable with respect tothe right endpoint of the integration interval and its weakderivative equals the integrand at that point

The following lemma is folklore in case 119864 = R but to seethat it also holds in the vector-valued case we provide a proof

Lemma 30 Let 0 lt 120572 lt 1 and 119901 gt max1120572 1(1minus120572) If thefunction 119909 119868 rarr 119864 is weakly absolutely continuous on 119868 andpasses a pseudo derivative inH

1199010 (119864) then119889120572119901119909119889119905120572 (119905) = D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (68)

In particular if 119909 passes a weak derivative in 119862[119868 119864120596] then119889120572120596119909119889119905120572 (119905) = D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (69)

If 119864 is reflexive this is also true for every 119901 ge 1

Proof We observe that under the assumption imposed on119863119901119909 together with Proposition 3 the weakly absolutelycontinuity of 119909 is equivalent to

119909 (119905) = 119909 (0) + int119905

0119863119901119909 (119904) 119889119904 (70)

Hence owing to Lemma 27 it follows that

119889120572119901119909119889119905120572 = I1minus120572119863119901119909 = (D120572

119901I120572)I1minus120572119863119901119909

= D120572119901 (I120572

I1minus120572)119863119901119909 = (119863119901I

1minus120572)I1119863119901119909= 119863119901I

1minus120572 (119909 (119905) minus 119909 (0))= D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0)

(71)

which is what we wished to showThe proof of (69) is very similar to that in (68) therefore

we omit the details

Remark 31 As we remark above the definition of the CFPDof a function 119909 loses completely its meaning if 119909 is notweakly absolutely continuous For this reason we are able touse Lemma 30 to define the Caputo fractional derivative ingeneral that is we put

119889120572119901119909119889119905120572 (119905) fl D120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) 0 lt 120572 lt 1 (72)

Similarly we define

119889120572120596119909119889119905120572 (119905) fl D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (73)

However Lemma 30 claims that for the weakly absolutelycontinuous functions having Pettis integrable pseudo [weak]derivatives definitions (72) and (73) of the Caputo fractional-pseudo [weak] derivatives coincide with Definition 20

4 An Application

Let 119891 119868 times 119864 rarr 119864 be given function Consider the boundaryvalue problem of the fractional type

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin [0 1] 120582 isin R

119910 (0) + 119887119910 (1) = ℎ(74)

with certain constants 119887 ℎ isin R 119887 = minus1 To obtain theintegral equation modeled off the problem (74) we let 119910 bea weakly continuous solution to problem (74) then formallywe have

I120572 119889120572120596119889119905120572119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (75)

12 Journal of Function Spaces

that is

I120572I1minus120572119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) 997904rArrI1119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (76)

This reads (cf [29])

119910 (119905) = 119910 (0) + 120582I120572119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin 119868 120582 isin R (77)

with some (presently unknown) quantity 119910(0)Now we solve (77) for 119910(0) by 119910(0) + 119887119910(1) = ℎ and it

follows that

(1 + 119887) 119910 (0) + 120582119887Γ (120572) int1

0(1 minus 119904)120572minus1 119891 (119904 119910 (119904)) 119889119904 = ℎ (78)

Thus

119910 (0) = ℎ1 + 119887 minus 1205821198871 + 119887119868120572119891 (1 119910 (1)) (79)

Assume that the function 119891 is weakly-weakly continuousfunction such that

(1) for any 119903 gt 0 there is a constant 119872 gt 0 such that119891(119905 119910) le 119872 for all 119905 isin 119868 and 119910 le 119903(2) there exists a nondecreasing continuous function 120595 [0infin) rarr [0infin) 120595(0) = 0 and 120595(119905) lt 119905 for all 119905 gt 0

such that 120573(119891(119879 times 119860)) le 120595(120573(119860)) for every boundedset119860 sub 119864 where 120573 stands for De Blasirsquos weakmeasureof noncompactness (see [30])

Occasionally if 119891 119868 times 119864 rarr 119864 is weakly-weakly continuousand 119864 is reflexive then the assumptions (1) and (2) areautomatically satisfied (see eg [31])

Theorem 32 (see [6] Theorem 33) Let 120572 isin (0 1) and |120582| ltΓ(1 + 120572) If 119891 119868 times 119864 rarr 119864 is weakly-weakly continuous andsatisfies the assumptions (1) and (2) then the integral equation(77) has a weakly continuous solution 119910 defined on [0 1]Proof We omit the proof since it is almost identical to thatin the proof in ([6] Theorem 33) with (small) necessarychanges

In the following example we assume that 119910 isin 119862[119868 119864120596]solves (77) and we will show that not only do we have that 119910no longer necessarily solves (74) (when the Caputo fractionalweak derivative is taken in the sense of Definition 20)but even worse it could happen that the problem (74) isldquomeaninglessrdquo on 119868Example 33 Let 120572 isin (0 14] Let 119864 be an indefinitedimensional reflexive Banach space that fails cotype Definethe weakly-weakly continuous function 119891 [0 1] times 119864 rarr 119864by 119891 fl I1minus120572119909 where 119909 119868 rarr 119864 is defined by formula (7)Obviously (cf Example 6 in view of Remark 7)119891 satisfies theassumptions of Theorem 32

Now consider the integral equation (77) with119891 = I1minus120572119909Namely we consider the integral equation

119910 (119905) = 119910 (0) + 120582I120572I1minus120572119909 (119905)

120572 isin (0 14] 119905 isin 119868 120582 isin R (80)

Obviously the solution 119910 is weakly continuous on [0 1] (thisis an immediate consequence of Remark 7 and Lemma 17)

Since 120593119909 isin 1198711[0 1] holds for every 120593 isin 119864lowast it follows bythe commutative property of the fractional integral operatorscalculus over the space of Lebesgue integrable functions (see[21] Section 23) that

120593 (I120572I1minus120572119909 (119905)) = I

120572120593 (I1minus120572119909 (119905))= I

120572I1minus120572120593 (119909 (119905)) = I

1120593 (119909 (119905))= 120593 (I1119909 (119905))

(81)

that is

120593 (I120572I1minus120572119909 (119905) minusI

1119909 (119905)) = 0 for every 120593 isin 119864lowast (82)

Hence I120572I1minus120572119909 = I1119909 Consequently problem (80) be-comes

119910 (119905) = 119910 (0) + 120582I1119909 (119905) 120572 isin (0 14] 119905 isin 119868 (83)

As showed in ([13] Corollary 4) the indefinite integral 119905 rarrint1199050119909(119904)119889119904 is nowhere weakly differentiable on [0 1] Thus 119910

is nowhere weakly differentiable on [0 1] hence the CFWDof 119910 (hence the boundary value problem) completely loses itsmeaning if the CFWD is taken in the sense of Definition 20This is what we wished to show

Now we are in the position to state and prove thefollowing existence theorem

Theorem 34 If 119891 119868 times 119864 rarr 119864 is a function such that allconditions fromTheorem 32 hold then problem (74) (where theCaputo fractionalweak derivative is taken in the sense (73)) hasa weak solution on [0 1]Proof Let 119910 isin 119862[119868 119864120596] be a solution to (77) Equip 119864 andI times E with weak topology and note that 119905 997891rarr (119905 119910(119905)) iscontinuous as a mapping from 119868 into 119868 times 119864 Since 119891 119868 times 119864 rarr119864 is weakly-weakly continuous on [0 1] then 119891(sdot 119910(sdot)) is acomposition of this mapping with 119891 and thus 119891(sdot 119910(sdot)) isweakly continuous on [0 1]

Now operating by the operatorD120572120596 on both sides of (77)

it follows by Lemma 29 that

D120572120596119910 = D

120572120596119910 (0) + 120582D120572

120596I120572119891 = 119905minus120572Γ (1 minus 120572)119910 (0) + 120582119891 (84)

Now insert definition (73) of the Caputo fractionalderivative and we get

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) (85)

Journal of Function Spaces 13

With some further efforts one can get the boundary condi-tion 119910(0) + 119887119910(1) = ℎ

Therefore 119910 satisfies problem (74) which is what wewished to show

Data Availability

No data were used to support this study

Conflicts of Interest

The author declares no conflicts of interest

References

[1] H A Salem and A M El-Sayed ldquoWeak solution for fractionalorder integral equations in reflexive Banach spacesrdquoMathemat-ica Slovaca vol 55 no 2 pp 169ndash181 2005

[2] H A Salem A M El-Sayed and O L Moustafa ldquoA note onthe fractional calculus in Banach spacesrdquo Studia ScientiarumMathematicarum Hungarica vol 42 no 2 pp 115ndash130 2005

[3] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoMulti-term fractional differential equations in a nonreflexiveBanach spacerdquo Advances in Difference Equations 2013302 18pages 2013

[4] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoFractional calculus and fractional differential equations innonreflexive Banach spacesrdquo Communications in NonlinearScience andNumerical Simulation vol 20 no 1 pp 59ndash73 2015

[5] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoNonlinear fractional differential equations in nonreflexiveBanach spaces and fractional calculusrdquo Advances in DifferenceEquations 2015112 18 pages 2015

[6] R P Agarwal V Lupulescu D OrsquoRegan and G RahmanldquoWeak solutions for fractional differential equations in nonre-flexive Banach spaces via Riemann-Pettis integralsrdquo Mathema-tische Nachrichten vol 289 no 4 pp 395ndash409 2016

[7] H A Salem ldquoOn the fractional order m-point boundaryvalue problem in reflexive Banach spaces and weak topologiesrdquoJournal of Computational andAppliedMathematics vol 224 no2 pp 565ndash572 2009

[8] H A Salem ldquoOn the fractional calculus in abstract spaces andtheir applications to the Dirichlet-type problem of fractionalorderrdquo Computers and Mathematics with Applications vol 59no 3 pp 1278ndash1293 2010

[9] HAH Salem andMCichon ldquoOn solutions of fractional orderboundary value problems with integral boundary conditions inBanach spacesrdquo journal of function spaces and applications vol2013 Article ID 428094 2013

[10] J Diestel and J J Uhl Jr Vector Measures vol 15 ofMathemat-ical Surveys American Mathematical Society Providence RIUSA 1977

[11] R H Martin and J J Uhl Jr Nonlinear Operators AndDifferential Equations in Banach Spaces John Wiley amp SonsNew York NY USA 1976

[12] B J Pettis ldquoOn integration in vector spacesrdquo Transactions of theAmericanMathematical Society vol 44 no 2 pp 277ndash304 1938

[13] S J Dilworth and M Girardi ldquoNowhere weak differentiabilityof the Pettis integralrdquo Quaestiones Mathematicae vol 18 no 4pp 365ndash380 1995

[14] G A Edgar ldquoMeasurability in a Banach spacerdquo Indiana Univer-sity Mathematics Journal vol 26 no 4 pp 663ndash677 1977

[15] G A Edgar ldquoMeasurability in a Banach space IIrdquo IndianaUniversityMathematics Journal vol 28 no 4 pp 559ndash579 1979

[16] J L Gamez and J Mendoza ldquoOn denjoy-dunford and denjoy-pettis integralsrdquo StudiaMathematica vol 130 no 2 pp 115ndash1331998

[17] R AGordon ldquoTheDenjoy extension of the Bochner Pettis andDUNford integralsrdquo Studia Mathematica vol 92 no 1 pp 73ndash91 1989

[18] R F Geitz ldquoPettis integrationrdquo Proceedings of the AmericanMathematical Society vol 82 no 1 pp 81ndash86 1981

[19] R F Geitz ldquoGeometry and the Pettis integralrdquo Transactions ofthe AmericanMathematical Society vol 269 no 2 pp 535ndash5481982

[20] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations New York NYUSA Elsevier 2006

[21] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Yverdon Switzerland 1993

[22] M Mendel and A Naor ldquoMetric cotyperdquo Annals of Mathemat-ics Second Series vol 168 no 1 pp 247ndash298 2008

[23] B Ross S G Samko and E R Love ldquoFunctions that have nofirst order derivative might have fractional derivatives of allorders less than onerdquo Real Analysis Exchange vol 20 no 1 pp140ndash157 199495

[24] E Hille and R S Phillips Functional analysis and semi-groupsAmerican Mathematical Society Colloquium Publications vol31 American Mathematical Society Providence RI USA 1957

[25] S Schwabik and Y Guoju Topics in Banach Space IntegrationWorld Scientific Singapore 2005

[26] DW Solomon ldquoOn differentiability of vector-valued functionsof a real variablerdquo Studia Mathematica vol 29 pp 1ndash4 1967

[27] K Naralenkov ldquoOn Denjoy type extensions of the Pettisintegralrdquo Czechoslovak Mathematical Journal vol 60(135) no3 pp 737ndash750 2010

[28] D W Solomon Denjoy integration in abstract spaces Memorisof the American Mathematical Society No 85 AmericanMathematical Society Providence RI 1969

[29] A R Mitchell and C Smith ldquoAn existence theorem for weaksolutions of differential equations in Banach spacesrdquo inNonlin-ear Equations in Abstract Spaces V Lakshmikantham Ed pp387ndash404 1978

[30] F S De Blasi ldquoOn a property of the unit sphere in a Banachspacerdquo Bulletin Mathematiques De La Societe Des SciencesMathematiques De Roumanie vol 21 no 3 pp 259ndash262 1977

[31] A Szep ldquoExistence theorem for weak solutions of ordinarydifferential equations in reflexive Banach spacesrdquo Studia Scien-tiarum Mathematicarum Hungarica vol 6 pp 197ndash203 1971

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Page 5: On the Theory of Fractional Calculus in the Pettis-Function Spaces · 2019. 7. 30. · JournalofFunctionSpaces Further,wedenethespace H (),˙≥0by H () fl ˛ ∈ ˚, ˜: isHolderainoforder

Journal of Function Spaces 5

Lemma 8 I120572119909 exists on [0 1] To calculate the RFPI of 119909 fix119905 isin [0 1] and let 120593 isin (ℓ2)lowast We have

int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904 = int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120593119909 (119904) 119889119904= I

120572 (infinsum119899=1

120582119899(119899 + 119905)) (23)

Since the series sum119899(120582119899(119899 + 119905)) is uniformly convergent on[0 1] it follows in view of the generalized linearity of thefractional integrals [23 Lemma 5] that (cf [21 Table 91])

int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904= infinsum

119899=1

120582119899119905120572119899Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 ) = 120593 (119892 (119905)) (24)

where 2F1 is the Gauss hypergeometric function evaluated at(1 1 120572 + 1 minus119905119899) and119892 (119905) fl 119905120572119899Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 ) (25)

Since

1003817100381710038171003817119892 (119905)10038171003817100381710038172ℓ2 = sum119899

( 119905120572Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 ))2

le 2F21 (1 1 120572 + 1 minus1)Γ2 (1 + 120572) sum

119899

11198992 lt infin119905 isin [0 1]

(26)

we see that 119892 isin ℓ2 Thus

I120572119909 (119905) = 119905120572119899Γ (1 + 120572) 2F1 (1 1 120572 + 1 minus119905119899 )

119905 isin [0 1] (27)

Example 12 Let 120572 isin [12 1) Define the countable-valuedfunction 119909 [0 1] rarr 1198880 by

119909 (119905) fl 119909119899 (119905) 119909119899 fl 119899120594(1(119899+1)1119899] (28)

This function is strongly measurable Pettis integrable func-tion on [0 1] (see eg [10 16]) We claim that 119909 isin H119875

0 (1198880)with 119901 isin (1120572 2] Once our claim is established Lemma 8guarantees the existence of I120572119909 120572 isin [12 1) on [0 1] Itremains to prove this claim by showing firstly that H119875(1198880)and to calculate I120572119909 To do this let 120593 isin 119888lowast0 Then therecorresponds to 120593 a unique 120582119899 isin ℓ1 such that 120593119909 = sum119899 120582119899119909119899By noting that

10038171003817100381710038171205821198991199091198991003817100381710038171003817119901 = (int1

0

1003816100381610038161003816120582119899119909119899 (119905)1003816100381610038161003816119901 119889119905)1119901

= 10038161003816100381610038161205821198991003816100381610038161003816 (int1119899

1(119899+1)119899119901119889119905)1119901 = 10038161003816100381610038161205821198991003816100381610038161003816 1198991119902(119899 + 1)1119901

le 10038161003816100381610038161205821198991003816100381610038161003816 11989911199021198991119901 le 100381610038161003816100381612058211989910038161003816100381610038161198992119901minus1 le 10038161003816100381610038161205821198991003816100381610038161003816

(29)

holds for any 119901 isin (1120572 2] a standard argument using LevirsquosTheorem (or Lebesgue dominated convergence theorem) andMinkowskirsquos inequality yields

[int1

0

1003816100381610038161003816120593119909 (119905)1003816100381610038161003816119901 119889119905]1119901 le [int1

0(infinsum119899=1

10038161003816100381610038161205821198991003816100381610038161003816 119909119899 (119905))119901 119889119905]1119901

= [[lim119896rarrinfin

int1

0( 119896sum119899=1

10038161003816100381610038161205821198991003816100381610038161003816 119909119899 (119905))119901

119889119905]]1119901

le [[lim119896rarrinfin

( 119896sum119899=1

10038171003817100381710038171205821198991199091198991003817100381710038171003817119901)119901]]1119901

le infinsum119899=1

10038171003817100381710038171205821198991199091198991003817100381710038171003817119901le infinsum

119899=1

10038161003816100381610038161205821198991003816100381610038161003816 lt infin

(30)

Thus 120593119909(sdot) isin 119871119901[0 1] 119901 isin (1120572 2] holds for every 120593 isin 119888lowast0(that is 119909 isin H119875(1198880)) Since 119901 gt 1120572 gt 1 it follows byProposition 2 in view of the strong measurability of 119909 that119909 isin H119875

0 (1198880) Owing to Lemma 8 we infer that the RFPI of119909 of any order 120572 isin [12 1) exists on the interval [0 1] Tocompute this integral fix 119905 isin 1198681198990 fl [1(1198990+1) 11198990] for some1198990 isin N the set of natural numbers and let 120593 isin 119888lowast0 Then

sum119899

int119905

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 12058211989910038161003816100381610038161003816 119909119899 (119904) 119889119904le 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816 int

119905

1(1198990+1)(119905 minus 119904)120572minus1 1198990119889119904

+ sum119899gt1198990

10038161003816100381610038161205821198991003816100381610038161003816 int1119899

1(119899+1)(119905 minus 119904)120572minus1 119899 119889119904

= 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572

+ sum119899gt1198990

119899 10038161003816100381610038161205821198991003816100381610038161003816120572 [(119905 minus 1119899 + 1)120572 minus (119905 minus 1119899)

120572]

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572

+ sum119899gt1198990

119899 10038161003816100381610038161205821198991003816100381610038161003816120572 [(119905 minus 1119899 + 1) minus (119905 minus 1119899)]120572

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572 + sum

119899gt1198990

119899 10038161003816100381610038161205821198991003816100381610038161003816120572 1119899120572 (119899 + 1)120572

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572 + sum

119899gt1198990

10038161003816100381610038161205821198991003816100381610038161003816120572 11198992120572minus1

le 1198990 100381610038161003816100381610038161205821198990 10038161003816100381610038161003816120572 (119905 minus 11198990 + 1)120572 + sum

119899gt1198990

10038161003816100381610038161205821198991003816100381610038161003816120572 lt infin(because 120572 isin [12 1))

(31)

6 Journal of Function Spaces

Therefore by Beppo Levi Theorem it follows that

int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904 = int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120593 (119909 (119904)) 119889119904= int119905

0

(119905 minus 119904)120572minus1Γ (120572) sum119899

120582119899119899120594(1(119899+1)1119899] (119904) 119889119904= sum

119899

int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120582119899119899120594(1(119899+1)1119899] (119904) 119889119904= 11989901205821198990Γ (1 + 120572) (119905 minus 11198990 + 1)

120572

+ sum119899gt1198990

119899120582119899Γ (1 + 120572) [(119905 minus 1119899 + 1)120572 minus (119905 minus 1119899)

120572]

(32)

Consequently we conclude that

I120572119909 (119905) = 0 0⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

1198990minus1

1198990Γ (1 + 120572) (119905 minus 11198990 + 1)120572

(1198990 + 1)Γ (1 + 120572) [(119905 minus 11198990 + 2)

120572 minus (119905 minus 11198990 + 1)120572]

(33)

where the nonzero coordinate started from the 1198990th place Itcan be easily seen thatI120572119909(119905) isin 1198880 for any 119905 isin [0 1] Evidentlywe have that10038161003816100381610038161003816100381610038161003816119899 [(119905 minus

1119899 + 1)120572 minus (119905 minus 1119899)

120572]10038161003816100381610038161003816100381610038161003816le 119899 [(119905 minus 1119899 + 1) minus (119905 minus 1119899)]

120572 le 119899 [ 1119899 (119899 + 1)]120572

le 11198992120572minus1 997888rarr 0 as 119899 997888rarr infin1198990Γ (1 + 120572) (119905 minus 11198990 + 1)

120572

le 1198990Γ (1 + 120572) ( 11198990 minus 11198990 + 1)120572 le 11198992120572minus10 Γ (1 + 120572)

997888rarr 0 as 1198990 997888rarr infin

(34)

hold for any 119905 isin [0 1]This yieldsI1205721119909(119905) isin 1198880 for any 119905 isin [0 1]

(this is precisely what we would expect from Definition 5)

Look at part (c) of Lemma 8 with 120572 isin (0 1) and119864 = 1198880 Below we give an example showing that the stronglymeasurability hypothesis imposed on a function 119909 isin H1(119864)H1

0(119864) is not sufficient for the existence of I120572119909 even when 119909is Denjoy-Pettis integrable (cf [16])

Example 13 Let 120572 isin (0 1) Define the strongly measurablefunction 119909 [0 1] rarr 1198880 by119909 (119905) fl 119909119899 (119905)

119909119899 (119905) fl 119899 (119899 + 1) [120594119868119899 (119905) minus 120594119869119899 (119905)] (35)

where

119868119899 fl ( 1119899 + 1 119899 + 12119899 (119899 + 1)) 119869119899 fl ( 119899 + 12119899 (119899 + 1) 1119899)

(36)

It is immediate that (cf [16]) 119909 is a well-defined Denjoy-Pettis (but it is not Pettis) integrable on [0 1]

In what follows we will show that the RFPI of 119909 does notexist on a subinterval of positive measure on [0 1]

To see this we make use of Proposition 3 as followsdefine for each intervalI119899 = [1(119899 + 1) 1119899] the functionals120593119899 isin 119888lowast0 (required by Proposition 3) to be the correspondingto the element 120582119899 fl (0 0 0 1 0 0 ) isin ℓ1 wherethe nonzero coordinate is in the 119899th place Then 120593119899(119909(119905)) =119909119899(119905) and thus |120593119899(119909(119904))| = 119899(119899 + 1)[120594119868119899(119905) minus 120594119869119899(119905)] Clearlythe family 120593119899 runs through the unit ball of the dual of1198880 Evidently the isometric isomorphism between 119888lowast0 and ℓ1yields 120593119899119888lowast0 = 120582119899ℓ1 = 1

Now take 119905 isin [12 1] and define the continuous function119891 [12 1] rarr R+ by

119891 (119905) fl (119905 minus 1119899 + 1)120572 minus (119905 minus 1119899)

120572 119899 ge 2 119905 isin [12 1]

(37)

It can be easily seen that in view of 0 lt 120572 lt 1 1198911015840 is negativeon (0 1) By standard results from (classical) calculus itfollows that 119891 is strictly decreasing on [0 1] in particular119891(1) lt 119891(119905) for all 119905 isin [0 1) Thus for any 119899 ge 2 and any120593 isin 120593119899 we have

intI119899

100381610038161003816100381610038161003816100381610038161003816120593 ((119905 minus 119904)120572minus1Γ (120572) 119909 (119904))100381610038161003816100381610038161003816100381610038161003816 119889119904

= intI119899

(119905 minus 119904)120572minus1Γ (120572) 1003816100381610038161003816120593 (119909 (119904))1003816100381610038161003816 119889119904= int1119899

1(119899+1)

(119905 minus 119904)120572minus1Γ (120572) 119899 (119899 + 1) 119889119904= 119899 (119899 + 1)Γ (1 + 120572) [(119905 minus 1119899 + 1)

120572 minus (119905 minus 1119899)120572]

ge 1198992Γ (1 + 120572) [(1 minus 1119899 + 1)120572 minus (1 minus 1119899)

120572]

(38)

An explicit calculation using LrsquoHospitalrsquos rule two timesreveals

lim119899rarrinfin

(1 minus 1 (119899 + 1))120572 minus (1 minus 1119899)120572119899minus2 = minus1205722sdot lim119899rarrinfin

1198993 [( 119899119899 + 1)120572minus1 1

(119899 + 1)2 minus (1 minus 1119899)120572minus1 11198992 ]

= minus1205722 lim119899rarrinfin

119899 [( 119899119899 + 1)120572minus1 1198992

(119899 + 1)2 minus (1 minus 1119899)120572minus1]

Journal of Function Spaces 7

= minus1205722 lim119899rarrinfin

119899 [( 119899119899 + 1)120572+1 minus (1119899)

120572minus1] = minus1205722sdot lim119899rarrinfin

1198992 [(120572 + 1) ( 119899119899 + 1)120572 1(119899 + 1)2

minus (120572 minus 1) (1 minus 1119899)120572minus2 11198992 ] = minus1205722

sdot lim119899rarrinfin

[(120572 + 1) ( 119899119899 + 1)120572+2 minus (120572 minus 1) (1 minus 1119899)

120572minus2]= 120572

(39)

from which it follows that

intI119899

100381610038161003816100381610038161003816100381610038161003816120593 ((119905 minus 119904)120572minus1Γ (120572) 119909 (119904))100381610038161003816100381610038161003816100381610038161003816 119889119904997888rarr0 as 119899 997888rarr infin (40)

Therefore in view of Proposition 3 119904 997891rarr (119905 minus 119904)120572minus1119909(119904) notin119875([0 1] 1198880) Hence I120572119909 does not exist which is what wewished to show

The following theorem provides a useful characterizationof the space H119901(119864) for which the statements reveal howmuch the fractional integral I120572 is better than the function119909 isin H119901(119864) Indeed based on Lemma 8 using an inequalityof Young we can easily prove the following

Lemma 14 For any 120572 gt 0 the following holds(a) If 119864 is reflexive then for every 119901 isin [1infin] the operator

I120572 mapsH119901(119864) intoH1199010 (119864) In particular if 0 lt 120572 lt1 then I120572 H1(119864) rarr H1((1minus120572)minus120598)0 (119864) however small120598 gt 0 is

(b) If 119864 contains no copy of 1198880 then for every 119901 isin [1infin]the operator I120572 maps 119909 isin H

1199010 (119864) 119909 strongly

measurable into H1199010 (119864) In particular if 120572 ge 1 the

operator I120572 mapsH1199010 (119864) into itself

(c) If 119901 gt max1 1120572 the operator I120572 sends H1199010 (119864) to119862[119868 119864120596] (if we define I120572119909(0) fl 0) In particular if120572 isin (0 1) I120572 H119901

0 (119864) rarr H120572minus1119901(119864)Proof At the beginning let 120572 gt 0 and 119909 isin H119901(119864) with119901 isin [1infin] and define the real-valued function 119892 119868 rarr R

by 119892(119905) fl 119905120572minus1Γ(120572) If 119864 is reflexive it follows in view ofLemma 8 that J120572119909 exists ae in 119868 and 120593(I120572119909) = I120572120593119909for every 120593 isin 119864lowast Youngrsquos inequality guarantees that theconvolution product 119892 lowast 120593119909(sdot) lies in 119871119901[119868] Consequently120593(I120572119909) = I120572120593119909 = 119892 lowast 120593119909 isin 119871119901[119868] for every 120593 isin 119864lowastThe reflexivity of 119864 together with Proposition 2 yields I120572119909 isinH

1199010 (119864)In particular if 0 lt 120572 lt 1 it can be easily seen thatint

119868(119892(119905))119902119889119905 lt infin for every 119902 isin [1 1(1 minus 120572)) That is 119892 isin1198711(1minus120572)minus120598[119868] 120598 gt 0 By Youngrsquos inequality it follows that120593(I120572119909) = 119892 lowast 120593119909 isin 1198711(1minus120572)minus120598[119868] for every 120593 isin 119864lowast however

small 120598 gt 0 is Now the assertion (a) follows because of thereflexivity of 119864

Next in order to prove the assertion (b) let 119909 isin H1199010 (119864)

and note in view of Lemma 8 that I120572119909 exists on 119868 Define119910 119868 rarr 119864 by119910(119905) fl I120572119909(119905) 119909 isin H1199010 (119864) Since120593119910(sdot) isin 119871119901[119868]

for every120593 isin 119864lowast it follows that119910 isin H119901(119864) Moreover for any119886 119887 isin 119868 (119886 lt 119887) we haveint119887

119886120593119910 (119904) 119889119905 = int119887

119886I120593119909 (119905) 119889119905 = 1Γ (120572)

sdot int119887

119886int119905

0(119905 minus 119904)120572minus1 120593119909 (119904) 119889119904 119889119905

= 1Γ (120572) [int119887

119886int119887

119904(119905 minus 119904)120572minus1 120593119909 (119904) 119889119905 119889119904

+ int119886

0int119887

119886(119905 minus 119904)120572minus1 120593119909 (119904) 119889119905 119889119904]

= 1Γ (120572) [int119887

119886(119887 minus 119904)120572 120593119909 (119904) 119889119904

+ int119886

0[(119887 minus 119904)120572 minus (119886 minus 119904)120572] 120593119909 (119904) 119889119904]

= 1Γ (120572) [int119886

0(119887 minus 119904)120572 120593119909 (119904) 119889119904

+ int119887

119886(119887 minus 119904)120572 120593119909 (119904) 119889119904 minus int119886

0(119886 minus 119904)120572 120593119909 (119904) 119889119904]

= int119887

0

(119887 minus 119904)120572Γ (1 + 120572)120593119909 (119904) 119889119904 minus int119886

0

(119886 minus 119904)120572Γ (1 + 120572)120593119909 (119904) 119889119904= 120593 (119909[119886119887])

(41)

where

119909[119886119887] fl (119875) int119887

0

(119887 minus 119904)120572Γ (1 + 120572)119909 (119904) 119889119904minus (119875) int119886

0

(119886 minus 119904)120572Γ (1 + 120572)119909 (119904) 119889119904(42)

Since 119909 isin 119875[119868 119864] owing to Proposition 2 (part (1)) it followsthat (119887 minus sdot)120572119909(119904) and (119886 minus sdot)120572119909(119904) are Pettis integrable on 119868 andso 119909[119886119887] isin 119864

A combination of these results yields120593119910 isin 119871119901(119868) for every120593 isin 119864lowast and there exists an element 119909[119886119887] isin 119864 such that120601119909[119886119887] = int119887

119886120593119910(119905)119889119905 for every 119886 119887 isin 119868 Since 119864 contains no

copy of 1198880 it follows that (cf [17 Theorem 23]) 119910 isin 119875[119868 119864]Consequently 119910(sdot) isin H

1199010 (119864) This is the claim (b)

To prove the assertion (c) let 119901 gt max1 1120572 and 119909 isinH

1199010 (119864) By Lemma 8 we deduce that I120572119909 exists ae on 119868

Now let 119902 isin [1infin] be such that 1119901 + 1119902 = 1 Since119902(120572 minus 1) gt minus1 119892 fl (sdot)120572minus1Γ(120572) isin 119871119902[119868] Therefore as a directconsequence of Youngrsquo inequality it follows that

120593 (I120572119909 (sdot)) = I120572120593119909 (sdot) = 119892 lowast 120593119909 (sdot) isin 119862 [119868R] (43)

holds for every 120593 isin 119864lowast Now we claim that I120572119909(119905) rarr 0in 119864 as 119905 rarr 0 Once our claim is established the definition

8 Journal of Function Spaces

I120572119909(0) fl 0 guarantees thatI120572119909(sdot) isin 119862[119868 119864120596] which is whatwewished to show It remains to prove our claim without lossof generality assume that I120572119909(119905) = 0 Then there exists (asa consequence of the Hahn-Banach theorem) 120593 isin 119864lowast with120593 = 1 such that I120572119909(119905) = 120593(I120572119909(119905)) = I120572120593119909(119905) ByHolderrsquos inequality we obtain

1003816100381610038161003816120593 (I120572119909 (119905))1003816100381610038161003816le 1Γ (120572) (int

119905

0(119905 minus 119904)119902(120572minus1) 119889119904)1119902 (int119905

0

1003816100381610038161003816120593119909 (119904)1003816100381610038161003816119901 119889119904)1119901

le 1Γ (120572) ( 119905(120572minus1)119902+1(120572 minus 1) 119902 + 1)1119902 (int

119868

1003816100381610038161003816120593119909 (119904)1003816100381610038161003816119901 119889119904)1119901

= 119905120572minus1+1119902Γ (120572) [119902 (120572 minus 1) + 1]1119902

10038171003817100381710038171205931199091003817100381710038171003817119871119901

= 119905120572minus1119901 10038171003817100381710038171205931199091003817100381710038171003817119871119901Γ (120572) [119902 (120572 minus 1) + 1]1119902

(44)

This is equivalent with the following estimate

1003817100381710038171003817I120572119909 (119905)1003817100381710038171003817 le 119905120572minus1119901 10038171003817100381710038171205931199091003817100381710038171003817119871119901Γ (120572) [119902 (120572 minus 1) + 1]1119902 (45)

Owing to 120572 gt 1119901 we get I120572119909(119905) rarr 0 in 119864 as 119905 rarr 0 andconsequently in view ofI120572119909(0) fl 0I120572119909(sdot) isin 119862[119868 119864120596] Thisproves the first part of the assertion (c)

Finally let 120572 isin (0 1) and 119905 120591 isin 119868 Without loss ofgenerality assume 120591 lt 119905 Then for any 120593 isin 119864lowast we have byHolder inequality with 119901 gt 1120572 in view 119902(120572 minus 1) gt minus1 that

1003816100381610038161003816120593 (I120572119909 (119905) minusI120572119909 (120591))1003816100381610038161003816 Γ (120572)

= 10038161003816100381610038161003816100381610038161003816int119905

0(119905 minus 119904)120572minus1 120593119909 (119904) 119889119904 minus int120591

0(120591 minus 119904)120572minus1 120593119909 (119904) 11988911990410038161003816100381610038161003816100381610038161003816

le (int120591

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 minus (120591 minus 119904)120572minus110038161003816100381610038161003816 1003816100381610038161003816120593119909 (119904)1003816100381610038161003816 119889119904+ int119905

120591(119905 minus 119904)120572minus1 1003816100381610038161003816120593119909 (119904)1003816100381610038161003816 119889119904)

= [(int120591

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 minus (120591 minus 119904)120572minus110038161003816100381610038161003816119902 119889119904)1119902

+ (int119905

120591(119905 minus 119904)(120572minus1)119902 119889119904)1119902] 10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]

le [[(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

+ ((119905 minus 120591)(120572minus1)119902+1(120572 minus 1) 119902 + 1 )1119902]]10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]

= [(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

+ (119905 minus 120591)120572minus1119901[119902 (120572 minus 1) + 1]1119902]

10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868] (46)

By noting that (120572 minus 1)119902 isin (minus1 0) when 120572 isin (0 1) it can beeasily seen that

int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904le 1119902 (120572 minus 1) + 1 (119905 minus 120591)119902(120572minus1)+1

(47)

That is

(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

le (119905 minus 120591)120572minus1119901[119902 (120572 minus 1) + 1]1119902

(48)

A combination of these results yields1003816100381610038161003816120593 (I120572119909 (119905)) minus 120593 (I120572119909 (120591))1003816100381610038161003816= 1003816100381610038161003816120593 (I120572119909 (119905) minusI

120572119909 (120591))1003816100381610038161003816le10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]Γ (120572) 2 (119905 minus 120591)120572minus1119901

[119902 (120572 minus 1) + 1]1119902 (49)

ThusI120572119909 isin H120572minus1119901(119864) This completes the proof

Corollary 15 For any 120572 gt 0 I120572 119862[119868 119864120596] rarr 119862[119868 119864120596]Proof Let 119909 isin 119862[119868 119864120596] then 120593119909 isin 119871infin holds for every120593 isin 119864lowast By noting that the weak continuity implies a strongmeasurability ([24] page 73) it follows that 119909 is stronglymeasurable on 119868 Hence in view of Proposition 2 119909 isinHinfin

0 (119864) Consequently 119862[119868 119864120596] sube H1199010 (119864) for every 119901 isin[1infin] That isI120572 119862[119868 119864120596] rarr 119862[119868 119864120596] Hence the desired

result is obtained

Example 16 Let 120572 isin [12 1) Define the function 119909 [0 1] rarr1198880 by 119909(119905) = 119899120594(1(119899+1)1119899]We have (in view of Example 12)that

I120572119909 (119905) = 0 0 0 1198990 (119905 minus 1 (1198990 + 1))120572

Γ (1 + 120572) (1198990 + 1)Γ (1 + 120572) [(119905 minus 11198990 + 2)

120572 minus (119905 minus 11198990 + 1)120572]

(50)

To show that I120572119909 is norm continuous on [0 1] let 120591 119905 isin[0 1] With no loss of generality we may assume that 119905 120591 isin1198681198990 for some 1198990 isin N Since the nonzero terms of the sequencein (50) are nonincreasing then in view of

|119905 minus 120591| le ( 11198990 minus 11198990 + 1) = 11198990 (1198990 + 1) le 111989920 (51)

Journal of Function Spaces 9

we have1003817100381710038171003817I120572119909 (119905) minusI

1205721119909 (120591)10038171003817100381710038171198880

= 1198990Γ (1 + 120572)10038161003816100381610038161003816100381610038161003816(119905 minus

1119899 + 1)120572 minus (120591 minus 1119899 + 1)

12057210038161003816100381610038161003816100381610038161003816le 1198990Γ (1 + 120572)

1003816100381610038161003816100381610038161003816(119905 minus 1119899 + 1) minus (120591 minus 1119899 + 1)1003816100381610038161003816100381610038161003816120572

le 1198990Γ (1 + 120572) |119905 minus 120591|120572 le 1Γ (1 + 120572) |119905 minus 120591|120572radic|119905 minus 120591|= |119905 minus 120591|120572minus12Γ (1 + 120572)

(52)

Thus the RFPI of 119909 is norm continuous on [0 1] Preciselysince 0 lt 120572 minus 12 lt 120572 minus 1119901 lt 1 for any 119901 le 2 thenI120572119909 isin H120572minus12(119864) sub H120572minus1119901(119864) This is precisely what onewould expect from Lemma 14 (part (c))

Analogously an explicit calculation reveals that I120572119909 liesin 119862([0 1] 119871infin[0 1]) where 119909 [0 1] rarr 119871infin[0 1] defined byExample 10

We now consider additional mapping properties of theoperator I120572 Precisely we will show that the RFPI enjoysthe following commutative property which is folklore in case119864 = R However the proof is completely similar to that of [8]Lemma 35

Lemma 17 Let 120572 120573 gt 0 and 119901 gt max1 1120572 1120573 Then

I120572I120573119909 = I

120573I120572119909 = I

120572+120573119909 (53)

holds for every 119909 isin H1199010 (119864) If 119864 is reflexive this is also true for

every 119901 ge 1Proof Since 119901 gt max1 1120572 1120573 it follows that 119901 gtmax1 1(120572+120573) So by Lemmas 8 and 14 themappingsI120572119909I120573119909 andI120572+120573119909 belong toHinfin

0 (119864) for every 119909 isin H1199010 (119864) Now

repeating the same process as in ([8] Lemma 35) implies theclaim

When 119864 is reflexive the result follows as a direct applica-tion of Lemma 14

3 Fractional Derivatives ofVector-Valued Functions

After the notation of the fractional integrals of vector-valued functions the fractional derivatives become a naturalrequirement Before we come to the definitions and a detailedstudy of the mathematical properties of fractional differentialoperators we recall the following

Definition 18 Consider the vector-valued function 119909 119868 rarr119864(1) Let 120593119909 be differentiable on 119868 for every 120593 isin 119864lowast The

function 119909 is said to be weakly differentiable on 119868 if

there exists 119910 119868 rarr 119864 such that for every 120593 isin 119864lowast wehave

119889120593119909 (119905)119889119905 = 120593119910 (119905) for every 119905 isin 119868 (54)

The function 119910 is called the weak derivative of thefunction 119909

(2) Let 120593119909 be differentiable ae on 119868 for every 120593 isin 119864lowast(the null set may vary with 120593 isin 119864lowast) The function 119909is said to be pseudo differentiable on 119868 if there existsa function 119910 119868 rarr 119864 such that for every 120593 isin 119864lowast thereexists a null set119873(120593) sub 119868 such that

(120593119909 (119905))1015840 = 120593119910 (119905) for every 119905 isin 119868119873 (120593) (55)

In this case the function 119910 is called the pseudoderivative of 119909If 119909 is pseudo differentiable on 119868 and the null setinvariant for every 120593 isin 119864lowast then 119909 is ae weaklydifferentiable on 119868

Clearly if 119909 is ae weakly differentiable on 119868 then 120593(119909)is ae differentiable on 119868 The converse holds in a weaklysequentially complete space (see [25] Theorem 733)

For more details of the derivatives of vector-valuedfunctions we refer to [10 12 26]

The following results play a major role in our analysis

Proposition 19 (see [27] Theorem 51) The function 119910 119868 rarr119864 is an indefinite Pettis integrable if and only if 119910 is weaklyabsolutely continuous on 119868 and have a pseudo derivative on 119868In this case 119910 is an indefinite Pettis integral of any of its pseudoderivatives

Now we are in the position to define the fractional-typederivatives of vector-valued functions

Definition 20 Let 119909 119868 rarr 119864 For the positive integer119898 suchthat 120572 isin (119898 minus 1119898) 119898 isin N0 fl 0 1 2 we define theCaputo fractional-pseudo (weak) derivative ldquoshortly CFPD(CFWD)rdquo of 119909 of order 120572 by

119889120572119889119905120572 119909 (119905) fl I119898minus120572119863119898119909 (119905) (56)

where the sign ldquo119863rdquo denotes the pseudo (or weak) differentialoperator We use the notation 119889120572119901119889119905120572 and 119889120572120596119889119905120572 to charac-terize the Caputo fractional-pseudo derivatives and Caputofractional weak derivatives respectively

It is well known that although the weak derivative of aweakly differentiable function is uniquely determined thepseudo derivative of the pseudo differentiable function is notunique Also although any two pseudo derivatives 119910 119911 of afunction 119909 119868 rarr 119864 need not be ae equal (see [12 Example91] and [26 p 2]) the functions 119910 119911 are weakly equivalenton 119868 (that is 120593119910 = 120593119911 holds ae for every 120593 isin 119864lowast) The

10 Journal of Function Spaces

next lemma provides a useful characterization property of theCFPD Really it can be easily seen that the CFPD of a Caputofractional-pseudo differentiable function does not depend onthe choice of a pseudo derivative of the function

Lemma 21 Let 119909 119868 rarr 119864 be pseudo differentiable functionwhere pseudo derivatives lie in H

1199010 (119864) If (119889120572119901119889119905120572)119909 exists on119868 then the CFPD of 119909 depends on the choice of the pseudo

derivatives of 119909Proof Let 119910 119911 isin H

1199010 (119864) be two pseudo derivatives of the

pseudo differentiable function 119909 Since 119909 and 119910 are weaklyequivalent on 119868 then for every 120593 isin 119864lowast there exists a nullset N which depends on 120593 isin 119864lowast such that (119905 minus 119904)minus120572120593119910(119904) =(119905minus119904)minus120572120593119911(119904) for every 119904 isin [0 119905]N Consequently for almostevery 119905 isin 119868

120593 (I1minus120572119910 (119905)) = I1minus120572120593119910 (119905) = I

1minus120572120593119911 (119905)= 120593 (I1minus120572119911 (119905)) (57)

holds for every 120593 isin 119864lowast Thus

119889120572119901119889119905120572 119909 = I1minus120572119863119901119909 = I

1minus120572119910 = I1minus120572119911 on 119868 (58)

which is what we wished to show

We consider the following examples

Example 22 Define 119909 [0 1] rarr ℓ2 by119909 (119905) fl ln(1 + 119905119899) 119899 isin N (59)

We claim that the CFWD of 119909 exists on [0 1] To see this welet 120593 isin (ℓ2)lowast = ℓ2 Then there corresponds to 120593 a sequence120582119899 isin ℓ2 such that

120593119909 (119905) = sum119899

120582119899 ln(1 + 119905119899) (60)

Since the series in (60) converges uniformly on [0 1] theformal differentiation of 120593119909 yields

(120593119909 (119905))1015840 = sum119899

120582119899119899 + 119905 = 120593 ( 1119905 + 119899) (61)

It is not hard to justify the differentiation by noting that theseries in the right hand side of (61) is uniformly and absolutelyconvergent on [0 1] Consequently 120593119909 is differentiable on[0 1] for 120593 isin (ℓ2)lowast meaning that 119909 is weakly differentiable on[0 1] (because ℓ2 is weakly complete (cf [25] Theorem 733)and 119863120596119909(119905) = 1(119905 + 119899) Owing to Lemma 25 the CFWDof any order 120572 isin (0 1) of 119909 exists on [0 1] To calculate theCFWD of 119909 on [0 1] we observe in view of Example 11 that

119889120572120596119889119905120572 119909 (119905) = I1minus120572119863120596119909 (119905)

= 1199051minus120572119899Γ (2 minus 120572) 2F1 (1 1 2 minus 120572 minus119905119899 ) (62)

Example 23 Let119864 be an infinite dimensionBanach space thatfails cotype Define 119910 [0 1] rarr 119864 by 119910(119905) fl int119905

0119909(119904)119889119904

where 119909 [0 1] rarr 119864 given by formula (7) As cited in ([13]Corollary 4) 119910 is nowhere weakly differentiable on [0 1]Consequently the CFWD of any order 120572 isin [34 1) of 119910 losesits meaning on [0 1]Remark 24 As shown in Example 23 there is an infinitedimension Banach space 119864 and weakly absolutely continuousfunction 119910 119868 rarr 119864 which is nowhere weakly differentiable(hence the CFWD of 119910 does not exist) Also even when 119864 isseparable and 119910 is Lipschitz function the pseudo derivatives(hence the CFPD) of 119910 need not to exist [28]

However Definition 20 of the CFPD (CFWD) has thedisadvantage that it completely loses its meaning if thefunction119909 fails to be pseudo (weakly) differentiable Preciselythe CFPD (in particular the CFWD) of a function 119909 loses itsmeaning if 119909 is not weakly absolutely continuous

Thenext lemma gives sufficient conditions that ensure theexistence of the Caputo fractional derivatives of a function119909 isin H119901(119864)Lemma 25 Let 120572 isin (0 1) For the function 119909 119868 rarr 119864 thefollowing hold

(a) If 119909 has a pseudo derivative 119863119901119909 isin H1199010 (119864) where 119901 gt1(1 minus 120572) then the CFPD of 119909 of order 120572 exists on 119868

Moreover (119889120572119901119889119905120572)119909 isin 119862[119868 119864120596](b) If 119864 is weakly complete or contains no copy of 1198880 and if119909 has a weak derivative 119863120596119909 isin H119901(119864) where 119901 isin[1infin] then the CFWD of 119909 of order 120572 exists on 119868

Moreover (119889120572120596119889119905120572)119909 isin 119862[119868 119864120596]This holds for any 119909 isin H119901(119864) with 119901 isin [1infin] if 119864 is reflexive

In all cases 120593((119889120572119901119889119905120572)119909) = (119889120572119901119889119905120572)120593119909 (120593((119889120572120596119889119905120572)119909) =(119889120572120596119889119905120572)120593119909) holds for every 120593 isin 119864lowastProof Since the weak (pseudo) derivative of an ae weakly(pseudo) differentiable function is strongly (weakly) mea-surable [12 26 28] the proof is readily available in view ofthe definition of Caputo fractional derivatives and Lemma 8Moreover since 119863120596119909(119863119901119909) isin H

1199010 (119864) with 119901 gt 1(1 minus 120572)

it follows in view of Lemma 14 that I1minus120572119863120596119909(I1minus120572119863119901119909) isin119862[119868 119864120596] This completes the proof

Besides the Caputo fractional-pseudo (weak) derivativeswe define the Riemann-Liouville fractional-pseudo (weak)derivatives

Definition 26 Let 119909 119868 rarr 119864 For the positive integer 119898such that 120572 isin (119898 minus 1119898) 119898 isin N0 fl 0 1 2 we definethe Riemann-Liouville fractional-pseudo (weak) derivativeldquoshortly RFPD (RFWD)rdquo of 119909 of order 120572 by

D120572119909 (119905) fl 119863119898

I119898minus120572119909 (119905) (63)

where 119863 is defined as in Definition 20 We use the notationD120572

119901 (D120572120596) to characterize the Riemann-Liouville fractional-

pseudo (weak) derivatives

Journal of Function Spaces 11

Clearly in infinite dimension Banach spaces the weakabsolute continuity of I119898minus120572119909 is necessarily (but not suf-ficient) condition for the existence of RFPD (in particularRFWD) of 119909Lemma 27 Let 0 lt 120572 lt 1 For any 119909 isin H

1199010 (119864) with 119901 gt

max1120572 1(1 minus 120572) we haveD

120572119901I

120572119909 = 119909 ae (64)

If 119864 is reflexive this is also true for every 119901 ge 1Proof Our assumption 119909 isin H

1199010 (119864)with 119901 gt max1120572 1(1minus120572) yields in view of Lemma 17 that

D120572119901I

120572119909 = 119863119901I1minus120572

I120572119909 = 119863119901I

1119909 (65)

The claim now follows immediately since (cf Proposition 19)the indefinite integral of Pettis integrable function is weaklyabsolutely continuous and it is pseudo differentiable withrespect to the right endpoint of the integration interval andits pseudo derivative equals the integrand at that point

Remark 28 When we replace D120572119901 by D120572

120596 then Lemma 27is no longer necessarily true for arbitrary 119909 isin H1

0(119864) evenwhen 119864 is reflexive evidently in [13 remark below Corollary4] the existence of a reflexive Banach space 119864 and a strongmeasurable Pettis integrable function 119909 119868 rarr 119864 was provedsuch that 119909 has nowhere weakly differentiable integral In thiscase 119863120596I

1119909 lost its meaning This gives a reason to believethat (65) (hence (64)) withD120572

120596119909 could not happen

However we have the following result

Lemma 29 Let 0 lt 120572 lt 1 For any 119909 isin 119862[119868 119864120596] we haveD

120572120596I

120572119909 = 119909 on 119868 (66)

Proof Our assumption 119909 isin 119862[119868 119864120596] sub Hinfin0 (119864) (see the proof

of Corollary 15) yields in view of Lemma 17 that

D120572120596I

120572119909 = 119863120596I1minus120572

I120572119909 = 119863120596I

1119909 (67)

The claim now follows immediately since the indefiniteintegral of weakly contentious function is weakly absolutelycontinuous and it is weakly differentiable with respect tothe right endpoint of the integration interval and its weakderivative equals the integrand at that point

The following lemma is folklore in case 119864 = R but to seethat it also holds in the vector-valued case we provide a proof

Lemma 30 Let 0 lt 120572 lt 1 and 119901 gt max1120572 1(1minus120572) If thefunction 119909 119868 rarr 119864 is weakly absolutely continuous on 119868 andpasses a pseudo derivative inH

1199010 (119864) then119889120572119901119909119889119905120572 (119905) = D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (68)

In particular if 119909 passes a weak derivative in 119862[119868 119864120596] then119889120572120596119909119889119905120572 (119905) = D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (69)

If 119864 is reflexive this is also true for every 119901 ge 1

Proof We observe that under the assumption imposed on119863119901119909 together with Proposition 3 the weakly absolutelycontinuity of 119909 is equivalent to

119909 (119905) = 119909 (0) + int119905

0119863119901119909 (119904) 119889119904 (70)

Hence owing to Lemma 27 it follows that

119889120572119901119909119889119905120572 = I1minus120572119863119901119909 = (D120572

119901I120572)I1minus120572119863119901119909

= D120572119901 (I120572

I1minus120572)119863119901119909 = (119863119901I

1minus120572)I1119863119901119909= 119863119901I

1minus120572 (119909 (119905) minus 119909 (0))= D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0)

(71)

which is what we wished to showThe proof of (69) is very similar to that in (68) therefore

we omit the details

Remark 31 As we remark above the definition of the CFPDof a function 119909 loses completely its meaning if 119909 is notweakly absolutely continuous For this reason we are able touse Lemma 30 to define the Caputo fractional derivative ingeneral that is we put

119889120572119901119909119889119905120572 (119905) fl D120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) 0 lt 120572 lt 1 (72)

Similarly we define

119889120572120596119909119889119905120572 (119905) fl D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (73)

However Lemma 30 claims that for the weakly absolutelycontinuous functions having Pettis integrable pseudo [weak]derivatives definitions (72) and (73) of the Caputo fractional-pseudo [weak] derivatives coincide with Definition 20

4 An Application

Let 119891 119868 times 119864 rarr 119864 be given function Consider the boundaryvalue problem of the fractional type

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin [0 1] 120582 isin R

119910 (0) + 119887119910 (1) = ℎ(74)

with certain constants 119887 ℎ isin R 119887 = minus1 To obtain theintegral equation modeled off the problem (74) we let 119910 bea weakly continuous solution to problem (74) then formallywe have

I120572 119889120572120596119889119905120572119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (75)

12 Journal of Function Spaces

that is

I120572I1minus120572119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) 997904rArrI1119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (76)

This reads (cf [29])

119910 (119905) = 119910 (0) + 120582I120572119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin 119868 120582 isin R (77)

with some (presently unknown) quantity 119910(0)Now we solve (77) for 119910(0) by 119910(0) + 119887119910(1) = ℎ and it

follows that

(1 + 119887) 119910 (0) + 120582119887Γ (120572) int1

0(1 minus 119904)120572minus1 119891 (119904 119910 (119904)) 119889119904 = ℎ (78)

Thus

119910 (0) = ℎ1 + 119887 minus 1205821198871 + 119887119868120572119891 (1 119910 (1)) (79)

Assume that the function 119891 is weakly-weakly continuousfunction such that

(1) for any 119903 gt 0 there is a constant 119872 gt 0 such that119891(119905 119910) le 119872 for all 119905 isin 119868 and 119910 le 119903(2) there exists a nondecreasing continuous function 120595 [0infin) rarr [0infin) 120595(0) = 0 and 120595(119905) lt 119905 for all 119905 gt 0

such that 120573(119891(119879 times 119860)) le 120595(120573(119860)) for every boundedset119860 sub 119864 where 120573 stands for De Blasirsquos weakmeasureof noncompactness (see [30])

Occasionally if 119891 119868 times 119864 rarr 119864 is weakly-weakly continuousand 119864 is reflexive then the assumptions (1) and (2) areautomatically satisfied (see eg [31])

Theorem 32 (see [6] Theorem 33) Let 120572 isin (0 1) and |120582| ltΓ(1 + 120572) If 119891 119868 times 119864 rarr 119864 is weakly-weakly continuous andsatisfies the assumptions (1) and (2) then the integral equation(77) has a weakly continuous solution 119910 defined on [0 1]Proof We omit the proof since it is almost identical to thatin the proof in ([6] Theorem 33) with (small) necessarychanges

In the following example we assume that 119910 isin 119862[119868 119864120596]solves (77) and we will show that not only do we have that 119910no longer necessarily solves (74) (when the Caputo fractionalweak derivative is taken in the sense of Definition 20)but even worse it could happen that the problem (74) isldquomeaninglessrdquo on 119868Example 33 Let 120572 isin (0 14] Let 119864 be an indefinitedimensional reflexive Banach space that fails cotype Definethe weakly-weakly continuous function 119891 [0 1] times 119864 rarr 119864by 119891 fl I1minus120572119909 where 119909 119868 rarr 119864 is defined by formula (7)Obviously (cf Example 6 in view of Remark 7)119891 satisfies theassumptions of Theorem 32

Now consider the integral equation (77) with119891 = I1minus120572119909Namely we consider the integral equation

119910 (119905) = 119910 (0) + 120582I120572I1minus120572119909 (119905)

120572 isin (0 14] 119905 isin 119868 120582 isin R (80)

Obviously the solution 119910 is weakly continuous on [0 1] (thisis an immediate consequence of Remark 7 and Lemma 17)

Since 120593119909 isin 1198711[0 1] holds for every 120593 isin 119864lowast it follows bythe commutative property of the fractional integral operatorscalculus over the space of Lebesgue integrable functions (see[21] Section 23) that

120593 (I120572I1minus120572119909 (119905)) = I

120572120593 (I1minus120572119909 (119905))= I

120572I1minus120572120593 (119909 (119905)) = I

1120593 (119909 (119905))= 120593 (I1119909 (119905))

(81)

that is

120593 (I120572I1minus120572119909 (119905) minusI

1119909 (119905)) = 0 for every 120593 isin 119864lowast (82)

Hence I120572I1minus120572119909 = I1119909 Consequently problem (80) be-comes

119910 (119905) = 119910 (0) + 120582I1119909 (119905) 120572 isin (0 14] 119905 isin 119868 (83)

As showed in ([13] Corollary 4) the indefinite integral 119905 rarrint1199050119909(119904)119889119904 is nowhere weakly differentiable on [0 1] Thus 119910

is nowhere weakly differentiable on [0 1] hence the CFWDof 119910 (hence the boundary value problem) completely loses itsmeaning if the CFWD is taken in the sense of Definition 20This is what we wished to show

Now we are in the position to state and prove thefollowing existence theorem

Theorem 34 If 119891 119868 times 119864 rarr 119864 is a function such that allconditions fromTheorem 32 hold then problem (74) (where theCaputo fractionalweak derivative is taken in the sense (73)) hasa weak solution on [0 1]Proof Let 119910 isin 119862[119868 119864120596] be a solution to (77) Equip 119864 andI times E with weak topology and note that 119905 997891rarr (119905 119910(119905)) iscontinuous as a mapping from 119868 into 119868 times 119864 Since 119891 119868 times 119864 rarr119864 is weakly-weakly continuous on [0 1] then 119891(sdot 119910(sdot)) is acomposition of this mapping with 119891 and thus 119891(sdot 119910(sdot)) isweakly continuous on [0 1]

Now operating by the operatorD120572120596 on both sides of (77)

it follows by Lemma 29 that

D120572120596119910 = D

120572120596119910 (0) + 120582D120572

120596I120572119891 = 119905minus120572Γ (1 minus 120572)119910 (0) + 120582119891 (84)

Now insert definition (73) of the Caputo fractionalderivative and we get

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) (85)

Journal of Function Spaces 13

With some further efforts one can get the boundary condi-tion 119910(0) + 119887119910(1) = ℎ

Therefore 119910 satisfies problem (74) which is what wewished to show

Data Availability

No data were used to support this study

Conflicts of Interest

The author declares no conflicts of interest

References

[1] H A Salem and A M El-Sayed ldquoWeak solution for fractionalorder integral equations in reflexive Banach spacesrdquoMathemat-ica Slovaca vol 55 no 2 pp 169ndash181 2005

[2] H A Salem A M El-Sayed and O L Moustafa ldquoA note onthe fractional calculus in Banach spacesrdquo Studia ScientiarumMathematicarum Hungarica vol 42 no 2 pp 115ndash130 2005

[3] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoMulti-term fractional differential equations in a nonreflexiveBanach spacerdquo Advances in Difference Equations 2013302 18pages 2013

[4] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoFractional calculus and fractional differential equations innonreflexive Banach spacesrdquo Communications in NonlinearScience andNumerical Simulation vol 20 no 1 pp 59ndash73 2015

[5] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoNonlinear fractional differential equations in nonreflexiveBanach spaces and fractional calculusrdquo Advances in DifferenceEquations 2015112 18 pages 2015

[6] R P Agarwal V Lupulescu D OrsquoRegan and G RahmanldquoWeak solutions for fractional differential equations in nonre-flexive Banach spaces via Riemann-Pettis integralsrdquo Mathema-tische Nachrichten vol 289 no 4 pp 395ndash409 2016

[7] H A Salem ldquoOn the fractional order m-point boundaryvalue problem in reflexive Banach spaces and weak topologiesrdquoJournal of Computational andAppliedMathematics vol 224 no2 pp 565ndash572 2009

[8] H A Salem ldquoOn the fractional calculus in abstract spaces andtheir applications to the Dirichlet-type problem of fractionalorderrdquo Computers and Mathematics with Applications vol 59no 3 pp 1278ndash1293 2010

[9] HAH Salem andMCichon ldquoOn solutions of fractional orderboundary value problems with integral boundary conditions inBanach spacesrdquo journal of function spaces and applications vol2013 Article ID 428094 2013

[10] J Diestel and J J Uhl Jr Vector Measures vol 15 ofMathemat-ical Surveys American Mathematical Society Providence RIUSA 1977

[11] R H Martin and J J Uhl Jr Nonlinear Operators AndDifferential Equations in Banach Spaces John Wiley amp SonsNew York NY USA 1976

[12] B J Pettis ldquoOn integration in vector spacesrdquo Transactions of theAmericanMathematical Society vol 44 no 2 pp 277ndash304 1938

[13] S J Dilworth and M Girardi ldquoNowhere weak differentiabilityof the Pettis integralrdquo Quaestiones Mathematicae vol 18 no 4pp 365ndash380 1995

[14] G A Edgar ldquoMeasurability in a Banach spacerdquo Indiana Univer-sity Mathematics Journal vol 26 no 4 pp 663ndash677 1977

[15] G A Edgar ldquoMeasurability in a Banach space IIrdquo IndianaUniversityMathematics Journal vol 28 no 4 pp 559ndash579 1979

[16] J L Gamez and J Mendoza ldquoOn denjoy-dunford and denjoy-pettis integralsrdquo StudiaMathematica vol 130 no 2 pp 115ndash1331998

[17] R AGordon ldquoTheDenjoy extension of the Bochner Pettis andDUNford integralsrdquo Studia Mathematica vol 92 no 1 pp 73ndash91 1989

[18] R F Geitz ldquoPettis integrationrdquo Proceedings of the AmericanMathematical Society vol 82 no 1 pp 81ndash86 1981

[19] R F Geitz ldquoGeometry and the Pettis integralrdquo Transactions ofthe AmericanMathematical Society vol 269 no 2 pp 535ndash5481982

[20] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations New York NYUSA Elsevier 2006

[21] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Yverdon Switzerland 1993

[22] M Mendel and A Naor ldquoMetric cotyperdquo Annals of Mathemat-ics Second Series vol 168 no 1 pp 247ndash298 2008

[23] B Ross S G Samko and E R Love ldquoFunctions that have nofirst order derivative might have fractional derivatives of allorders less than onerdquo Real Analysis Exchange vol 20 no 1 pp140ndash157 199495

[24] E Hille and R S Phillips Functional analysis and semi-groupsAmerican Mathematical Society Colloquium Publications vol31 American Mathematical Society Providence RI USA 1957

[25] S Schwabik and Y Guoju Topics in Banach Space IntegrationWorld Scientific Singapore 2005

[26] DW Solomon ldquoOn differentiability of vector-valued functionsof a real variablerdquo Studia Mathematica vol 29 pp 1ndash4 1967

[27] K Naralenkov ldquoOn Denjoy type extensions of the Pettisintegralrdquo Czechoslovak Mathematical Journal vol 60(135) no3 pp 737ndash750 2010

[28] D W Solomon Denjoy integration in abstract spaces Memorisof the American Mathematical Society No 85 AmericanMathematical Society Providence RI 1969

[29] A R Mitchell and C Smith ldquoAn existence theorem for weaksolutions of differential equations in Banach spacesrdquo inNonlin-ear Equations in Abstract Spaces V Lakshmikantham Ed pp387ndash404 1978

[30] F S De Blasi ldquoOn a property of the unit sphere in a Banachspacerdquo Bulletin Mathematiques De La Societe Des SciencesMathematiques De Roumanie vol 21 no 3 pp 259ndash262 1977

[31] A Szep ldquoExistence theorem for weak solutions of ordinarydifferential equations in reflexive Banach spacesrdquo Studia Scien-tiarum Mathematicarum Hungarica vol 6 pp 197ndash203 1971

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Page 6: On the Theory of Fractional Calculus in the Pettis-Function Spaces · 2019. 7. 30. · JournalofFunctionSpaces Further,wedenethespace H (),˙≥0by H () fl ˛ ∈ ˚, ˜: isHolderainoforder

6 Journal of Function Spaces

Therefore by Beppo Levi Theorem it follows that

int119905

0120593((119905 minus 119904)120572minus1Γ (120572) 119909 (119904)) 119889119904 = int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120593 (119909 (119904)) 119889119904= int119905

0

(119905 minus 119904)120572minus1Γ (120572) sum119899

120582119899119899120594(1(119899+1)1119899] (119904) 119889119904= sum

119899

int119905

0

(119905 minus 119904)120572minus1Γ (120572) 120582119899119899120594(1(119899+1)1119899] (119904) 119889119904= 11989901205821198990Γ (1 + 120572) (119905 minus 11198990 + 1)

120572

+ sum119899gt1198990

119899120582119899Γ (1 + 120572) [(119905 minus 1119899 + 1)120572 minus (119905 minus 1119899)

120572]

(32)

Consequently we conclude that

I120572119909 (119905) = 0 0⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

1198990minus1

1198990Γ (1 + 120572) (119905 minus 11198990 + 1)120572

(1198990 + 1)Γ (1 + 120572) [(119905 minus 11198990 + 2)

120572 minus (119905 minus 11198990 + 1)120572]

(33)

where the nonzero coordinate started from the 1198990th place Itcan be easily seen thatI120572119909(119905) isin 1198880 for any 119905 isin [0 1] Evidentlywe have that10038161003816100381610038161003816100381610038161003816119899 [(119905 minus

1119899 + 1)120572 minus (119905 minus 1119899)

120572]10038161003816100381610038161003816100381610038161003816le 119899 [(119905 minus 1119899 + 1) minus (119905 minus 1119899)]

120572 le 119899 [ 1119899 (119899 + 1)]120572

le 11198992120572minus1 997888rarr 0 as 119899 997888rarr infin1198990Γ (1 + 120572) (119905 minus 11198990 + 1)

120572

le 1198990Γ (1 + 120572) ( 11198990 minus 11198990 + 1)120572 le 11198992120572minus10 Γ (1 + 120572)

997888rarr 0 as 1198990 997888rarr infin

(34)

hold for any 119905 isin [0 1]This yieldsI1205721119909(119905) isin 1198880 for any 119905 isin [0 1]

(this is precisely what we would expect from Definition 5)

Look at part (c) of Lemma 8 with 120572 isin (0 1) and119864 = 1198880 Below we give an example showing that the stronglymeasurability hypothesis imposed on a function 119909 isin H1(119864)H1

0(119864) is not sufficient for the existence of I120572119909 even when 119909is Denjoy-Pettis integrable (cf [16])

Example 13 Let 120572 isin (0 1) Define the strongly measurablefunction 119909 [0 1] rarr 1198880 by119909 (119905) fl 119909119899 (119905)

119909119899 (119905) fl 119899 (119899 + 1) [120594119868119899 (119905) minus 120594119869119899 (119905)] (35)

where

119868119899 fl ( 1119899 + 1 119899 + 12119899 (119899 + 1)) 119869119899 fl ( 119899 + 12119899 (119899 + 1) 1119899)

(36)

It is immediate that (cf [16]) 119909 is a well-defined Denjoy-Pettis (but it is not Pettis) integrable on [0 1]

In what follows we will show that the RFPI of 119909 does notexist on a subinterval of positive measure on [0 1]

To see this we make use of Proposition 3 as followsdefine for each intervalI119899 = [1(119899 + 1) 1119899] the functionals120593119899 isin 119888lowast0 (required by Proposition 3) to be the correspondingto the element 120582119899 fl (0 0 0 1 0 0 ) isin ℓ1 wherethe nonzero coordinate is in the 119899th place Then 120593119899(119909(119905)) =119909119899(119905) and thus |120593119899(119909(119904))| = 119899(119899 + 1)[120594119868119899(119905) minus 120594119869119899(119905)] Clearlythe family 120593119899 runs through the unit ball of the dual of1198880 Evidently the isometric isomorphism between 119888lowast0 and ℓ1yields 120593119899119888lowast0 = 120582119899ℓ1 = 1

Now take 119905 isin [12 1] and define the continuous function119891 [12 1] rarr R+ by

119891 (119905) fl (119905 minus 1119899 + 1)120572 minus (119905 minus 1119899)

120572 119899 ge 2 119905 isin [12 1]

(37)

It can be easily seen that in view of 0 lt 120572 lt 1 1198911015840 is negativeon (0 1) By standard results from (classical) calculus itfollows that 119891 is strictly decreasing on [0 1] in particular119891(1) lt 119891(119905) for all 119905 isin [0 1) Thus for any 119899 ge 2 and any120593 isin 120593119899 we have

intI119899

100381610038161003816100381610038161003816100381610038161003816120593 ((119905 minus 119904)120572minus1Γ (120572) 119909 (119904))100381610038161003816100381610038161003816100381610038161003816 119889119904

= intI119899

(119905 minus 119904)120572minus1Γ (120572) 1003816100381610038161003816120593 (119909 (119904))1003816100381610038161003816 119889119904= int1119899

1(119899+1)

(119905 minus 119904)120572minus1Γ (120572) 119899 (119899 + 1) 119889119904= 119899 (119899 + 1)Γ (1 + 120572) [(119905 minus 1119899 + 1)

120572 minus (119905 minus 1119899)120572]

ge 1198992Γ (1 + 120572) [(1 minus 1119899 + 1)120572 minus (1 minus 1119899)

120572]

(38)

An explicit calculation using LrsquoHospitalrsquos rule two timesreveals

lim119899rarrinfin

(1 minus 1 (119899 + 1))120572 minus (1 minus 1119899)120572119899minus2 = minus1205722sdot lim119899rarrinfin

1198993 [( 119899119899 + 1)120572minus1 1

(119899 + 1)2 minus (1 minus 1119899)120572minus1 11198992 ]

= minus1205722 lim119899rarrinfin

119899 [( 119899119899 + 1)120572minus1 1198992

(119899 + 1)2 minus (1 minus 1119899)120572minus1]

Journal of Function Spaces 7

= minus1205722 lim119899rarrinfin

119899 [( 119899119899 + 1)120572+1 minus (1119899)

120572minus1] = minus1205722sdot lim119899rarrinfin

1198992 [(120572 + 1) ( 119899119899 + 1)120572 1(119899 + 1)2

minus (120572 minus 1) (1 minus 1119899)120572minus2 11198992 ] = minus1205722

sdot lim119899rarrinfin

[(120572 + 1) ( 119899119899 + 1)120572+2 minus (120572 minus 1) (1 minus 1119899)

120572minus2]= 120572

(39)

from which it follows that

intI119899

100381610038161003816100381610038161003816100381610038161003816120593 ((119905 minus 119904)120572minus1Γ (120572) 119909 (119904))100381610038161003816100381610038161003816100381610038161003816 119889119904997888rarr0 as 119899 997888rarr infin (40)

Therefore in view of Proposition 3 119904 997891rarr (119905 minus 119904)120572minus1119909(119904) notin119875([0 1] 1198880) Hence I120572119909 does not exist which is what wewished to show

The following theorem provides a useful characterizationof the space H119901(119864) for which the statements reveal howmuch the fractional integral I120572 is better than the function119909 isin H119901(119864) Indeed based on Lemma 8 using an inequalityof Young we can easily prove the following

Lemma 14 For any 120572 gt 0 the following holds(a) If 119864 is reflexive then for every 119901 isin [1infin] the operator

I120572 mapsH119901(119864) intoH1199010 (119864) In particular if 0 lt 120572 lt1 then I120572 H1(119864) rarr H1((1minus120572)minus120598)0 (119864) however small120598 gt 0 is

(b) If 119864 contains no copy of 1198880 then for every 119901 isin [1infin]the operator I120572 maps 119909 isin H

1199010 (119864) 119909 strongly

measurable into H1199010 (119864) In particular if 120572 ge 1 the

operator I120572 mapsH1199010 (119864) into itself

(c) If 119901 gt max1 1120572 the operator I120572 sends H1199010 (119864) to119862[119868 119864120596] (if we define I120572119909(0) fl 0) In particular if120572 isin (0 1) I120572 H119901

0 (119864) rarr H120572minus1119901(119864)Proof At the beginning let 120572 gt 0 and 119909 isin H119901(119864) with119901 isin [1infin] and define the real-valued function 119892 119868 rarr R

by 119892(119905) fl 119905120572minus1Γ(120572) If 119864 is reflexive it follows in view ofLemma 8 that J120572119909 exists ae in 119868 and 120593(I120572119909) = I120572120593119909for every 120593 isin 119864lowast Youngrsquos inequality guarantees that theconvolution product 119892 lowast 120593119909(sdot) lies in 119871119901[119868] Consequently120593(I120572119909) = I120572120593119909 = 119892 lowast 120593119909 isin 119871119901[119868] for every 120593 isin 119864lowastThe reflexivity of 119864 together with Proposition 2 yields I120572119909 isinH

1199010 (119864)In particular if 0 lt 120572 lt 1 it can be easily seen thatint

119868(119892(119905))119902119889119905 lt infin for every 119902 isin [1 1(1 minus 120572)) That is 119892 isin1198711(1minus120572)minus120598[119868] 120598 gt 0 By Youngrsquos inequality it follows that120593(I120572119909) = 119892 lowast 120593119909 isin 1198711(1minus120572)minus120598[119868] for every 120593 isin 119864lowast however

small 120598 gt 0 is Now the assertion (a) follows because of thereflexivity of 119864

Next in order to prove the assertion (b) let 119909 isin H1199010 (119864)

and note in view of Lemma 8 that I120572119909 exists on 119868 Define119910 119868 rarr 119864 by119910(119905) fl I120572119909(119905) 119909 isin H1199010 (119864) Since120593119910(sdot) isin 119871119901[119868]

for every120593 isin 119864lowast it follows that119910 isin H119901(119864) Moreover for any119886 119887 isin 119868 (119886 lt 119887) we haveint119887

119886120593119910 (119904) 119889119905 = int119887

119886I120593119909 (119905) 119889119905 = 1Γ (120572)

sdot int119887

119886int119905

0(119905 minus 119904)120572minus1 120593119909 (119904) 119889119904 119889119905

= 1Γ (120572) [int119887

119886int119887

119904(119905 minus 119904)120572minus1 120593119909 (119904) 119889119905 119889119904

+ int119886

0int119887

119886(119905 minus 119904)120572minus1 120593119909 (119904) 119889119905 119889119904]

= 1Γ (120572) [int119887

119886(119887 minus 119904)120572 120593119909 (119904) 119889119904

+ int119886

0[(119887 minus 119904)120572 minus (119886 minus 119904)120572] 120593119909 (119904) 119889119904]

= 1Γ (120572) [int119886

0(119887 minus 119904)120572 120593119909 (119904) 119889119904

+ int119887

119886(119887 minus 119904)120572 120593119909 (119904) 119889119904 minus int119886

0(119886 minus 119904)120572 120593119909 (119904) 119889119904]

= int119887

0

(119887 minus 119904)120572Γ (1 + 120572)120593119909 (119904) 119889119904 minus int119886

0

(119886 minus 119904)120572Γ (1 + 120572)120593119909 (119904) 119889119904= 120593 (119909[119886119887])

(41)

where

119909[119886119887] fl (119875) int119887

0

(119887 minus 119904)120572Γ (1 + 120572)119909 (119904) 119889119904minus (119875) int119886

0

(119886 minus 119904)120572Γ (1 + 120572)119909 (119904) 119889119904(42)

Since 119909 isin 119875[119868 119864] owing to Proposition 2 (part (1)) it followsthat (119887 minus sdot)120572119909(119904) and (119886 minus sdot)120572119909(119904) are Pettis integrable on 119868 andso 119909[119886119887] isin 119864

A combination of these results yields120593119910 isin 119871119901(119868) for every120593 isin 119864lowast and there exists an element 119909[119886119887] isin 119864 such that120601119909[119886119887] = int119887

119886120593119910(119905)119889119905 for every 119886 119887 isin 119868 Since 119864 contains no

copy of 1198880 it follows that (cf [17 Theorem 23]) 119910 isin 119875[119868 119864]Consequently 119910(sdot) isin H

1199010 (119864) This is the claim (b)

To prove the assertion (c) let 119901 gt max1 1120572 and 119909 isinH

1199010 (119864) By Lemma 8 we deduce that I120572119909 exists ae on 119868

Now let 119902 isin [1infin] be such that 1119901 + 1119902 = 1 Since119902(120572 minus 1) gt minus1 119892 fl (sdot)120572minus1Γ(120572) isin 119871119902[119868] Therefore as a directconsequence of Youngrsquo inequality it follows that

120593 (I120572119909 (sdot)) = I120572120593119909 (sdot) = 119892 lowast 120593119909 (sdot) isin 119862 [119868R] (43)

holds for every 120593 isin 119864lowast Now we claim that I120572119909(119905) rarr 0in 119864 as 119905 rarr 0 Once our claim is established the definition

8 Journal of Function Spaces

I120572119909(0) fl 0 guarantees thatI120572119909(sdot) isin 119862[119868 119864120596] which is whatwewished to show It remains to prove our claim without lossof generality assume that I120572119909(119905) = 0 Then there exists (asa consequence of the Hahn-Banach theorem) 120593 isin 119864lowast with120593 = 1 such that I120572119909(119905) = 120593(I120572119909(119905)) = I120572120593119909(119905) ByHolderrsquos inequality we obtain

1003816100381610038161003816120593 (I120572119909 (119905))1003816100381610038161003816le 1Γ (120572) (int

119905

0(119905 minus 119904)119902(120572minus1) 119889119904)1119902 (int119905

0

1003816100381610038161003816120593119909 (119904)1003816100381610038161003816119901 119889119904)1119901

le 1Γ (120572) ( 119905(120572minus1)119902+1(120572 minus 1) 119902 + 1)1119902 (int

119868

1003816100381610038161003816120593119909 (119904)1003816100381610038161003816119901 119889119904)1119901

= 119905120572minus1+1119902Γ (120572) [119902 (120572 minus 1) + 1]1119902

10038171003817100381710038171205931199091003817100381710038171003817119871119901

= 119905120572minus1119901 10038171003817100381710038171205931199091003817100381710038171003817119871119901Γ (120572) [119902 (120572 minus 1) + 1]1119902

(44)

This is equivalent with the following estimate

1003817100381710038171003817I120572119909 (119905)1003817100381710038171003817 le 119905120572minus1119901 10038171003817100381710038171205931199091003817100381710038171003817119871119901Γ (120572) [119902 (120572 minus 1) + 1]1119902 (45)

Owing to 120572 gt 1119901 we get I120572119909(119905) rarr 0 in 119864 as 119905 rarr 0 andconsequently in view ofI120572119909(0) fl 0I120572119909(sdot) isin 119862[119868 119864120596] Thisproves the first part of the assertion (c)

Finally let 120572 isin (0 1) and 119905 120591 isin 119868 Without loss ofgenerality assume 120591 lt 119905 Then for any 120593 isin 119864lowast we have byHolder inequality with 119901 gt 1120572 in view 119902(120572 minus 1) gt minus1 that

1003816100381610038161003816120593 (I120572119909 (119905) minusI120572119909 (120591))1003816100381610038161003816 Γ (120572)

= 10038161003816100381610038161003816100381610038161003816int119905

0(119905 minus 119904)120572minus1 120593119909 (119904) 119889119904 minus int120591

0(120591 minus 119904)120572minus1 120593119909 (119904) 11988911990410038161003816100381610038161003816100381610038161003816

le (int120591

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 minus (120591 minus 119904)120572minus110038161003816100381610038161003816 1003816100381610038161003816120593119909 (119904)1003816100381610038161003816 119889119904+ int119905

120591(119905 minus 119904)120572minus1 1003816100381610038161003816120593119909 (119904)1003816100381610038161003816 119889119904)

= [(int120591

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 minus (120591 minus 119904)120572minus110038161003816100381610038161003816119902 119889119904)1119902

+ (int119905

120591(119905 minus 119904)(120572minus1)119902 119889119904)1119902] 10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]

le [[(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

+ ((119905 minus 120591)(120572minus1)119902+1(120572 minus 1) 119902 + 1 )1119902]]10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]

= [(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

+ (119905 minus 120591)120572minus1119901[119902 (120572 minus 1) + 1]1119902]

10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868] (46)

By noting that (120572 minus 1)119902 isin (minus1 0) when 120572 isin (0 1) it can beeasily seen that

int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904le 1119902 (120572 minus 1) + 1 (119905 minus 120591)119902(120572minus1)+1

(47)

That is

(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

le (119905 minus 120591)120572minus1119901[119902 (120572 minus 1) + 1]1119902

(48)

A combination of these results yields1003816100381610038161003816120593 (I120572119909 (119905)) minus 120593 (I120572119909 (120591))1003816100381610038161003816= 1003816100381610038161003816120593 (I120572119909 (119905) minusI

120572119909 (120591))1003816100381610038161003816le10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]Γ (120572) 2 (119905 minus 120591)120572minus1119901

[119902 (120572 minus 1) + 1]1119902 (49)

ThusI120572119909 isin H120572minus1119901(119864) This completes the proof

Corollary 15 For any 120572 gt 0 I120572 119862[119868 119864120596] rarr 119862[119868 119864120596]Proof Let 119909 isin 119862[119868 119864120596] then 120593119909 isin 119871infin holds for every120593 isin 119864lowast By noting that the weak continuity implies a strongmeasurability ([24] page 73) it follows that 119909 is stronglymeasurable on 119868 Hence in view of Proposition 2 119909 isinHinfin

0 (119864) Consequently 119862[119868 119864120596] sube H1199010 (119864) for every 119901 isin[1infin] That isI120572 119862[119868 119864120596] rarr 119862[119868 119864120596] Hence the desired

result is obtained

Example 16 Let 120572 isin [12 1) Define the function 119909 [0 1] rarr1198880 by 119909(119905) = 119899120594(1(119899+1)1119899]We have (in view of Example 12)that

I120572119909 (119905) = 0 0 0 1198990 (119905 minus 1 (1198990 + 1))120572

Γ (1 + 120572) (1198990 + 1)Γ (1 + 120572) [(119905 minus 11198990 + 2)

120572 minus (119905 minus 11198990 + 1)120572]

(50)

To show that I120572119909 is norm continuous on [0 1] let 120591 119905 isin[0 1] With no loss of generality we may assume that 119905 120591 isin1198681198990 for some 1198990 isin N Since the nonzero terms of the sequencein (50) are nonincreasing then in view of

|119905 minus 120591| le ( 11198990 minus 11198990 + 1) = 11198990 (1198990 + 1) le 111989920 (51)

Journal of Function Spaces 9

we have1003817100381710038171003817I120572119909 (119905) minusI

1205721119909 (120591)10038171003817100381710038171198880

= 1198990Γ (1 + 120572)10038161003816100381610038161003816100381610038161003816(119905 minus

1119899 + 1)120572 minus (120591 minus 1119899 + 1)

12057210038161003816100381610038161003816100381610038161003816le 1198990Γ (1 + 120572)

1003816100381610038161003816100381610038161003816(119905 minus 1119899 + 1) minus (120591 minus 1119899 + 1)1003816100381610038161003816100381610038161003816120572

le 1198990Γ (1 + 120572) |119905 minus 120591|120572 le 1Γ (1 + 120572) |119905 minus 120591|120572radic|119905 minus 120591|= |119905 minus 120591|120572minus12Γ (1 + 120572)

(52)

Thus the RFPI of 119909 is norm continuous on [0 1] Preciselysince 0 lt 120572 minus 12 lt 120572 minus 1119901 lt 1 for any 119901 le 2 thenI120572119909 isin H120572minus12(119864) sub H120572minus1119901(119864) This is precisely what onewould expect from Lemma 14 (part (c))

Analogously an explicit calculation reveals that I120572119909 liesin 119862([0 1] 119871infin[0 1]) where 119909 [0 1] rarr 119871infin[0 1] defined byExample 10

We now consider additional mapping properties of theoperator I120572 Precisely we will show that the RFPI enjoysthe following commutative property which is folklore in case119864 = R However the proof is completely similar to that of [8]Lemma 35

Lemma 17 Let 120572 120573 gt 0 and 119901 gt max1 1120572 1120573 Then

I120572I120573119909 = I

120573I120572119909 = I

120572+120573119909 (53)

holds for every 119909 isin H1199010 (119864) If 119864 is reflexive this is also true for

every 119901 ge 1Proof Since 119901 gt max1 1120572 1120573 it follows that 119901 gtmax1 1(120572+120573) So by Lemmas 8 and 14 themappingsI120572119909I120573119909 andI120572+120573119909 belong toHinfin

0 (119864) for every 119909 isin H1199010 (119864) Now

repeating the same process as in ([8] Lemma 35) implies theclaim

When 119864 is reflexive the result follows as a direct applica-tion of Lemma 14

3 Fractional Derivatives ofVector-Valued Functions

After the notation of the fractional integrals of vector-valued functions the fractional derivatives become a naturalrequirement Before we come to the definitions and a detailedstudy of the mathematical properties of fractional differentialoperators we recall the following

Definition 18 Consider the vector-valued function 119909 119868 rarr119864(1) Let 120593119909 be differentiable on 119868 for every 120593 isin 119864lowast The

function 119909 is said to be weakly differentiable on 119868 if

there exists 119910 119868 rarr 119864 such that for every 120593 isin 119864lowast wehave

119889120593119909 (119905)119889119905 = 120593119910 (119905) for every 119905 isin 119868 (54)

The function 119910 is called the weak derivative of thefunction 119909

(2) Let 120593119909 be differentiable ae on 119868 for every 120593 isin 119864lowast(the null set may vary with 120593 isin 119864lowast) The function 119909is said to be pseudo differentiable on 119868 if there existsa function 119910 119868 rarr 119864 such that for every 120593 isin 119864lowast thereexists a null set119873(120593) sub 119868 such that

(120593119909 (119905))1015840 = 120593119910 (119905) for every 119905 isin 119868119873 (120593) (55)

In this case the function 119910 is called the pseudoderivative of 119909If 119909 is pseudo differentiable on 119868 and the null setinvariant for every 120593 isin 119864lowast then 119909 is ae weaklydifferentiable on 119868

Clearly if 119909 is ae weakly differentiable on 119868 then 120593(119909)is ae differentiable on 119868 The converse holds in a weaklysequentially complete space (see [25] Theorem 733)

For more details of the derivatives of vector-valuedfunctions we refer to [10 12 26]

The following results play a major role in our analysis

Proposition 19 (see [27] Theorem 51) The function 119910 119868 rarr119864 is an indefinite Pettis integrable if and only if 119910 is weaklyabsolutely continuous on 119868 and have a pseudo derivative on 119868In this case 119910 is an indefinite Pettis integral of any of its pseudoderivatives

Now we are in the position to define the fractional-typederivatives of vector-valued functions

Definition 20 Let 119909 119868 rarr 119864 For the positive integer119898 suchthat 120572 isin (119898 minus 1119898) 119898 isin N0 fl 0 1 2 we define theCaputo fractional-pseudo (weak) derivative ldquoshortly CFPD(CFWD)rdquo of 119909 of order 120572 by

119889120572119889119905120572 119909 (119905) fl I119898minus120572119863119898119909 (119905) (56)

where the sign ldquo119863rdquo denotes the pseudo (or weak) differentialoperator We use the notation 119889120572119901119889119905120572 and 119889120572120596119889119905120572 to charac-terize the Caputo fractional-pseudo derivatives and Caputofractional weak derivatives respectively

It is well known that although the weak derivative of aweakly differentiable function is uniquely determined thepseudo derivative of the pseudo differentiable function is notunique Also although any two pseudo derivatives 119910 119911 of afunction 119909 119868 rarr 119864 need not be ae equal (see [12 Example91] and [26 p 2]) the functions 119910 119911 are weakly equivalenton 119868 (that is 120593119910 = 120593119911 holds ae for every 120593 isin 119864lowast) The

10 Journal of Function Spaces

next lemma provides a useful characterization property of theCFPD Really it can be easily seen that the CFPD of a Caputofractional-pseudo differentiable function does not depend onthe choice of a pseudo derivative of the function

Lemma 21 Let 119909 119868 rarr 119864 be pseudo differentiable functionwhere pseudo derivatives lie in H

1199010 (119864) If (119889120572119901119889119905120572)119909 exists on119868 then the CFPD of 119909 depends on the choice of the pseudo

derivatives of 119909Proof Let 119910 119911 isin H

1199010 (119864) be two pseudo derivatives of the

pseudo differentiable function 119909 Since 119909 and 119910 are weaklyequivalent on 119868 then for every 120593 isin 119864lowast there exists a nullset N which depends on 120593 isin 119864lowast such that (119905 minus 119904)minus120572120593119910(119904) =(119905minus119904)minus120572120593119911(119904) for every 119904 isin [0 119905]N Consequently for almostevery 119905 isin 119868

120593 (I1minus120572119910 (119905)) = I1minus120572120593119910 (119905) = I

1minus120572120593119911 (119905)= 120593 (I1minus120572119911 (119905)) (57)

holds for every 120593 isin 119864lowast Thus

119889120572119901119889119905120572 119909 = I1minus120572119863119901119909 = I

1minus120572119910 = I1minus120572119911 on 119868 (58)

which is what we wished to show

We consider the following examples

Example 22 Define 119909 [0 1] rarr ℓ2 by119909 (119905) fl ln(1 + 119905119899) 119899 isin N (59)

We claim that the CFWD of 119909 exists on [0 1] To see this welet 120593 isin (ℓ2)lowast = ℓ2 Then there corresponds to 120593 a sequence120582119899 isin ℓ2 such that

120593119909 (119905) = sum119899

120582119899 ln(1 + 119905119899) (60)

Since the series in (60) converges uniformly on [0 1] theformal differentiation of 120593119909 yields

(120593119909 (119905))1015840 = sum119899

120582119899119899 + 119905 = 120593 ( 1119905 + 119899) (61)

It is not hard to justify the differentiation by noting that theseries in the right hand side of (61) is uniformly and absolutelyconvergent on [0 1] Consequently 120593119909 is differentiable on[0 1] for 120593 isin (ℓ2)lowast meaning that 119909 is weakly differentiable on[0 1] (because ℓ2 is weakly complete (cf [25] Theorem 733)and 119863120596119909(119905) = 1(119905 + 119899) Owing to Lemma 25 the CFWDof any order 120572 isin (0 1) of 119909 exists on [0 1] To calculate theCFWD of 119909 on [0 1] we observe in view of Example 11 that

119889120572120596119889119905120572 119909 (119905) = I1minus120572119863120596119909 (119905)

= 1199051minus120572119899Γ (2 minus 120572) 2F1 (1 1 2 minus 120572 minus119905119899 ) (62)

Example 23 Let119864 be an infinite dimensionBanach space thatfails cotype Define 119910 [0 1] rarr 119864 by 119910(119905) fl int119905

0119909(119904)119889119904

where 119909 [0 1] rarr 119864 given by formula (7) As cited in ([13]Corollary 4) 119910 is nowhere weakly differentiable on [0 1]Consequently the CFWD of any order 120572 isin [34 1) of 119910 losesits meaning on [0 1]Remark 24 As shown in Example 23 there is an infinitedimension Banach space 119864 and weakly absolutely continuousfunction 119910 119868 rarr 119864 which is nowhere weakly differentiable(hence the CFWD of 119910 does not exist) Also even when 119864 isseparable and 119910 is Lipschitz function the pseudo derivatives(hence the CFPD) of 119910 need not to exist [28]

However Definition 20 of the CFPD (CFWD) has thedisadvantage that it completely loses its meaning if thefunction119909 fails to be pseudo (weakly) differentiable Preciselythe CFPD (in particular the CFWD) of a function 119909 loses itsmeaning if 119909 is not weakly absolutely continuous

Thenext lemma gives sufficient conditions that ensure theexistence of the Caputo fractional derivatives of a function119909 isin H119901(119864)Lemma 25 Let 120572 isin (0 1) For the function 119909 119868 rarr 119864 thefollowing hold

(a) If 119909 has a pseudo derivative 119863119901119909 isin H1199010 (119864) where 119901 gt1(1 minus 120572) then the CFPD of 119909 of order 120572 exists on 119868

Moreover (119889120572119901119889119905120572)119909 isin 119862[119868 119864120596](b) If 119864 is weakly complete or contains no copy of 1198880 and if119909 has a weak derivative 119863120596119909 isin H119901(119864) where 119901 isin[1infin] then the CFWD of 119909 of order 120572 exists on 119868

Moreover (119889120572120596119889119905120572)119909 isin 119862[119868 119864120596]This holds for any 119909 isin H119901(119864) with 119901 isin [1infin] if 119864 is reflexive

In all cases 120593((119889120572119901119889119905120572)119909) = (119889120572119901119889119905120572)120593119909 (120593((119889120572120596119889119905120572)119909) =(119889120572120596119889119905120572)120593119909) holds for every 120593 isin 119864lowastProof Since the weak (pseudo) derivative of an ae weakly(pseudo) differentiable function is strongly (weakly) mea-surable [12 26 28] the proof is readily available in view ofthe definition of Caputo fractional derivatives and Lemma 8Moreover since 119863120596119909(119863119901119909) isin H

1199010 (119864) with 119901 gt 1(1 minus 120572)

it follows in view of Lemma 14 that I1minus120572119863120596119909(I1minus120572119863119901119909) isin119862[119868 119864120596] This completes the proof

Besides the Caputo fractional-pseudo (weak) derivativeswe define the Riemann-Liouville fractional-pseudo (weak)derivatives

Definition 26 Let 119909 119868 rarr 119864 For the positive integer 119898such that 120572 isin (119898 minus 1119898) 119898 isin N0 fl 0 1 2 we definethe Riemann-Liouville fractional-pseudo (weak) derivativeldquoshortly RFPD (RFWD)rdquo of 119909 of order 120572 by

D120572119909 (119905) fl 119863119898

I119898minus120572119909 (119905) (63)

where 119863 is defined as in Definition 20 We use the notationD120572

119901 (D120572120596) to characterize the Riemann-Liouville fractional-

pseudo (weak) derivatives

Journal of Function Spaces 11

Clearly in infinite dimension Banach spaces the weakabsolute continuity of I119898minus120572119909 is necessarily (but not suf-ficient) condition for the existence of RFPD (in particularRFWD) of 119909Lemma 27 Let 0 lt 120572 lt 1 For any 119909 isin H

1199010 (119864) with 119901 gt

max1120572 1(1 minus 120572) we haveD

120572119901I

120572119909 = 119909 ae (64)

If 119864 is reflexive this is also true for every 119901 ge 1Proof Our assumption 119909 isin H

1199010 (119864)with 119901 gt max1120572 1(1minus120572) yields in view of Lemma 17 that

D120572119901I

120572119909 = 119863119901I1minus120572

I120572119909 = 119863119901I

1119909 (65)

The claim now follows immediately since (cf Proposition 19)the indefinite integral of Pettis integrable function is weaklyabsolutely continuous and it is pseudo differentiable withrespect to the right endpoint of the integration interval andits pseudo derivative equals the integrand at that point

Remark 28 When we replace D120572119901 by D120572

120596 then Lemma 27is no longer necessarily true for arbitrary 119909 isin H1

0(119864) evenwhen 119864 is reflexive evidently in [13 remark below Corollary4] the existence of a reflexive Banach space 119864 and a strongmeasurable Pettis integrable function 119909 119868 rarr 119864 was provedsuch that 119909 has nowhere weakly differentiable integral In thiscase 119863120596I

1119909 lost its meaning This gives a reason to believethat (65) (hence (64)) withD120572

120596119909 could not happen

However we have the following result

Lemma 29 Let 0 lt 120572 lt 1 For any 119909 isin 119862[119868 119864120596] we haveD

120572120596I

120572119909 = 119909 on 119868 (66)

Proof Our assumption 119909 isin 119862[119868 119864120596] sub Hinfin0 (119864) (see the proof

of Corollary 15) yields in view of Lemma 17 that

D120572120596I

120572119909 = 119863120596I1minus120572

I120572119909 = 119863120596I

1119909 (67)

The claim now follows immediately since the indefiniteintegral of weakly contentious function is weakly absolutelycontinuous and it is weakly differentiable with respect tothe right endpoint of the integration interval and its weakderivative equals the integrand at that point

The following lemma is folklore in case 119864 = R but to seethat it also holds in the vector-valued case we provide a proof

Lemma 30 Let 0 lt 120572 lt 1 and 119901 gt max1120572 1(1minus120572) If thefunction 119909 119868 rarr 119864 is weakly absolutely continuous on 119868 andpasses a pseudo derivative inH

1199010 (119864) then119889120572119901119909119889119905120572 (119905) = D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (68)

In particular if 119909 passes a weak derivative in 119862[119868 119864120596] then119889120572120596119909119889119905120572 (119905) = D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (69)

If 119864 is reflexive this is also true for every 119901 ge 1

Proof We observe that under the assumption imposed on119863119901119909 together with Proposition 3 the weakly absolutelycontinuity of 119909 is equivalent to

119909 (119905) = 119909 (0) + int119905

0119863119901119909 (119904) 119889119904 (70)

Hence owing to Lemma 27 it follows that

119889120572119901119909119889119905120572 = I1minus120572119863119901119909 = (D120572

119901I120572)I1minus120572119863119901119909

= D120572119901 (I120572

I1minus120572)119863119901119909 = (119863119901I

1minus120572)I1119863119901119909= 119863119901I

1minus120572 (119909 (119905) minus 119909 (0))= D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0)

(71)

which is what we wished to showThe proof of (69) is very similar to that in (68) therefore

we omit the details

Remark 31 As we remark above the definition of the CFPDof a function 119909 loses completely its meaning if 119909 is notweakly absolutely continuous For this reason we are able touse Lemma 30 to define the Caputo fractional derivative ingeneral that is we put

119889120572119901119909119889119905120572 (119905) fl D120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) 0 lt 120572 lt 1 (72)

Similarly we define

119889120572120596119909119889119905120572 (119905) fl D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (73)

However Lemma 30 claims that for the weakly absolutelycontinuous functions having Pettis integrable pseudo [weak]derivatives definitions (72) and (73) of the Caputo fractional-pseudo [weak] derivatives coincide with Definition 20

4 An Application

Let 119891 119868 times 119864 rarr 119864 be given function Consider the boundaryvalue problem of the fractional type

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin [0 1] 120582 isin R

119910 (0) + 119887119910 (1) = ℎ(74)

with certain constants 119887 ℎ isin R 119887 = minus1 To obtain theintegral equation modeled off the problem (74) we let 119910 bea weakly continuous solution to problem (74) then formallywe have

I120572 119889120572120596119889119905120572119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (75)

12 Journal of Function Spaces

that is

I120572I1minus120572119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) 997904rArrI1119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (76)

This reads (cf [29])

119910 (119905) = 119910 (0) + 120582I120572119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin 119868 120582 isin R (77)

with some (presently unknown) quantity 119910(0)Now we solve (77) for 119910(0) by 119910(0) + 119887119910(1) = ℎ and it

follows that

(1 + 119887) 119910 (0) + 120582119887Γ (120572) int1

0(1 minus 119904)120572minus1 119891 (119904 119910 (119904)) 119889119904 = ℎ (78)

Thus

119910 (0) = ℎ1 + 119887 minus 1205821198871 + 119887119868120572119891 (1 119910 (1)) (79)

Assume that the function 119891 is weakly-weakly continuousfunction such that

(1) for any 119903 gt 0 there is a constant 119872 gt 0 such that119891(119905 119910) le 119872 for all 119905 isin 119868 and 119910 le 119903(2) there exists a nondecreasing continuous function 120595 [0infin) rarr [0infin) 120595(0) = 0 and 120595(119905) lt 119905 for all 119905 gt 0

such that 120573(119891(119879 times 119860)) le 120595(120573(119860)) for every boundedset119860 sub 119864 where 120573 stands for De Blasirsquos weakmeasureof noncompactness (see [30])

Occasionally if 119891 119868 times 119864 rarr 119864 is weakly-weakly continuousand 119864 is reflexive then the assumptions (1) and (2) areautomatically satisfied (see eg [31])

Theorem 32 (see [6] Theorem 33) Let 120572 isin (0 1) and |120582| ltΓ(1 + 120572) If 119891 119868 times 119864 rarr 119864 is weakly-weakly continuous andsatisfies the assumptions (1) and (2) then the integral equation(77) has a weakly continuous solution 119910 defined on [0 1]Proof We omit the proof since it is almost identical to thatin the proof in ([6] Theorem 33) with (small) necessarychanges

In the following example we assume that 119910 isin 119862[119868 119864120596]solves (77) and we will show that not only do we have that 119910no longer necessarily solves (74) (when the Caputo fractionalweak derivative is taken in the sense of Definition 20)but even worse it could happen that the problem (74) isldquomeaninglessrdquo on 119868Example 33 Let 120572 isin (0 14] Let 119864 be an indefinitedimensional reflexive Banach space that fails cotype Definethe weakly-weakly continuous function 119891 [0 1] times 119864 rarr 119864by 119891 fl I1minus120572119909 where 119909 119868 rarr 119864 is defined by formula (7)Obviously (cf Example 6 in view of Remark 7)119891 satisfies theassumptions of Theorem 32

Now consider the integral equation (77) with119891 = I1minus120572119909Namely we consider the integral equation

119910 (119905) = 119910 (0) + 120582I120572I1minus120572119909 (119905)

120572 isin (0 14] 119905 isin 119868 120582 isin R (80)

Obviously the solution 119910 is weakly continuous on [0 1] (thisis an immediate consequence of Remark 7 and Lemma 17)

Since 120593119909 isin 1198711[0 1] holds for every 120593 isin 119864lowast it follows bythe commutative property of the fractional integral operatorscalculus over the space of Lebesgue integrable functions (see[21] Section 23) that

120593 (I120572I1minus120572119909 (119905)) = I

120572120593 (I1minus120572119909 (119905))= I

120572I1minus120572120593 (119909 (119905)) = I

1120593 (119909 (119905))= 120593 (I1119909 (119905))

(81)

that is

120593 (I120572I1minus120572119909 (119905) minusI

1119909 (119905)) = 0 for every 120593 isin 119864lowast (82)

Hence I120572I1minus120572119909 = I1119909 Consequently problem (80) be-comes

119910 (119905) = 119910 (0) + 120582I1119909 (119905) 120572 isin (0 14] 119905 isin 119868 (83)

As showed in ([13] Corollary 4) the indefinite integral 119905 rarrint1199050119909(119904)119889119904 is nowhere weakly differentiable on [0 1] Thus 119910

is nowhere weakly differentiable on [0 1] hence the CFWDof 119910 (hence the boundary value problem) completely loses itsmeaning if the CFWD is taken in the sense of Definition 20This is what we wished to show

Now we are in the position to state and prove thefollowing existence theorem

Theorem 34 If 119891 119868 times 119864 rarr 119864 is a function such that allconditions fromTheorem 32 hold then problem (74) (where theCaputo fractionalweak derivative is taken in the sense (73)) hasa weak solution on [0 1]Proof Let 119910 isin 119862[119868 119864120596] be a solution to (77) Equip 119864 andI times E with weak topology and note that 119905 997891rarr (119905 119910(119905)) iscontinuous as a mapping from 119868 into 119868 times 119864 Since 119891 119868 times 119864 rarr119864 is weakly-weakly continuous on [0 1] then 119891(sdot 119910(sdot)) is acomposition of this mapping with 119891 and thus 119891(sdot 119910(sdot)) isweakly continuous on [0 1]

Now operating by the operatorD120572120596 on both sides of (77)

it follows by Lemma 29 that

D120572120596119910 = D

120572120596119910 (0) + 120582D120572

120596I120572119891 = 119905minus120572Γ (1 minus 120572)119910 (0) + 120582119891 (84)

Now insert definition (73) of the Caputo fractionalderivative and we get

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) (85)

Journal of Function Spaces 13

With some further efforts one can get the boundary condi-tion 119910(0) + 119887119910(1) = ℎ

Therefore 119910 satisfies problem (74) which is what wewished to show

Data Availability

No data were used to support this study

Conflicts of Interest

The author declares no conflicts of interest

References

[1] H A Salem and A M El-Sayed ldquoWeak solution for fractionalorder integral equations in reflexive Banach spacesrdquoMathemat-ica Slovaca vol 55 no 2 pp 169ndash181 2005

[2] H A Salem A M El-Sayed and O L Moustafa ldquoA note onthe fractional calculus in Banach spacesrdquo Studia ScientiarumMathematicarum Hungarica vol 42 no 2 pp 115ndash130 2005

[3] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoMulti-term fractional differential equations in a nonreflexiveBanach spacerdquo Advances in Difference Equations 2013302 18pages 2013

[4] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoFractional calculus and fractional differential equations innonreflexive Banach spacesrdquo Communications in NonlinearScience andNumerical Simulation vol 20 no 1 pp 59ndash73 2015

[5] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoNonlinear fractional differential equations in nonreflexiveBanach spaces and fractional calculusrdquo Advances in DifferenceEquations 2015112 18 pages 2015

[6] R P Agarwal V Lupulescu D OrsquoRegan and G RahmanldquoWeak solutions for fractional differential equations in nonre-flexive Banach spaces via Riemann-Pettis integralsrdquo Mathema-tische Nachrichten vol 289 no 4 pp 395ndash409 2016

[7] H A Salem ldquoOn the fractional order m-point boundaryvalue problem in reflexive Banach spaces and weak topologiesrdquoJournal of Computational andAppliedMathematics vol 224 no2 pp 565ndash572 2009

[8] H A Salem ldquoOn the fractional calculus in abstract spaces andtheir applications to the Dirichlet-type problem of fractionalorderrdquo Computers and Mathematics with Applications vol 59no 3 pp 1278ndash1293 2010

[9] HAH Salem andMCichon ldquoOn solutions of fractional orderboundary value problems with integral boundary conditions inBanach spacesrdquo journal of function spaces and applications vol2013 Article ID 428094 2013

[10] J Diestel and J J Uhl Jr Vector Measures vol 15 ofMathemat-ical Surveys American Mathematical Society Providence RIUSA 1977

[11] R H Martin and J J Uhl Jr Nonlinear Operators AndDifferential Equations in Banach Spaces John Wiley amp SonsNew York NY USA 1976

[12] B J Pettis ldquoOn integration in vector spacesrdquo Transactions of theAmericanMathematical Society vol 44 no 2 pp 277ndash304 1938

[13] S J Dilworth and M Girardi ldquoNowhere weak differentiabilityof the Pettis integralrdquo Quaestiones Mathematicae vol 18 no 4pp 365ndash380 1995

[14] G A Edgar ldquoMeasurability in a Banach spacerdquo Indiana Univer-sity Mathematics Journal vol 26 no 4 pp 663ndash677 1977

[15] G A Edgar ldquoMeasurability in a Banach space IIrdquo IndianaUniversityMathematics Journal vol 28 no 4 pp 559ndash579 1979

[16] J L Gamez and J Mendoza ldquoOn denjoy-dunford and denjoy-pettis integralsrdquo StudiaMathematica vol 130 no 2 pp 115ndash1331998

[17] R AGordon ldquoTheDenjoy extension of the Bochner Pettis andDUNford integralsrdquo Studia Mathematica vol 92 no 1 pp 73ndash91 1989

[18] R F Geitz ldquoPettis integrationrdquo Proceedings of the AmericanMathematical Society vol 82 no 1 pp 81ndash86 1981

[19] R F Geitz ldquoGeometry and the Pettis integralrdquo Transactions ofthe AmericanMathematical Society vol 269 no 2 pp 535ndash5481982

[20] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations New York NYUSA Elsevier 2006

[21] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Yverdon Switzerland 1993

[22] M Mendel and A Naor ldquoMetric cotyperdquo Annals of Mathemat-ics Second Series vol 168 no 1 pp 247ndash298 2008

[23] B Ross S G Samko and E R Love ldquoFunctions that have nofirst order derivative might have fractional derivatives of allorders less than onerdquo Real Analysis Exchange vol 20 no 1 pp140ndash157 199495

[24] E Hille and R S Phillips Functional analysis and semi-groupsAmerican Mathematical Society Colloquium Publications vol31 American Mathematical Society Providence RI USA 1957

[25] S Schwabik and Y Guoju Topics in Banach Space IntegrationWorld Scientific Singapore 2005

[26] DW Solomon ldquoOn differentiability of vector-valued functionsof a real variablerdquo Studia Mathematica vol 29 pp 1ndash4 1967

[27] K Naralenkov ldquoOn Denjoy type extensions of the Pettisintegralrdquo Czechoslovak Mathematical Journal vol 60(135) no3 pp 737ndash750 2010

[28] D W Solomon Denjoy integration in abstract spaces Memorisof the American Mathematical Society No 85 AmericanMathematical Society Providence RI 1969

[29] A R Mitchell and C Smith ldquoAn existence theorem for weaksolutions of differential equations in Banach spacesrdquo inNonlin-ear Equations in Abstract Spaces V Lakshmikantham Ed pp387ndash404 1978

[30] F S De Blasi ldquoOn a property of the unit sphere in a Banachspacerdquo Bulletin Mathematiques De La Societe Des SciencesMathematiques De Roumanie vol 21 no 3 pp 259ndash262 1977

[31] A Szep ldquoExistence theorem for weak solutions of ordinarydifferential equations in reflexive Banach spacesrdquo Studia Scien-tiarum Mathematicarum Hungarica vol 6 pp 197ndash203 1971

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Page 7: On the Theory of Fractional Calculus in the Pettis-Function Spaces · 2019. 7. 30. · JournalofFunctionSpaces Further,wedenethespace H (),˙≥0by H () fl ˛ ∈ ˚, ˜: isHolderainoforder

Journal of Function Spaces 7

= minus1205722 lim119899rarrinfin

119899 [( 119899119899 + 1)120572+1 minus (1119899)

120572minus1] = minus1205722sdot lim119899rarrinfin

1198992 [(120572 + 1) ( 119899119899 + 1)120572 1(119899 + 1)2

minus (120572 minus 1) (1 minus 1119899)120572minus2 11198992 ] = minus1205722

sdot lim119899rarrinfin

[(120572 + 1) ( 119899119899 + 1)120572+2 minus (120572 minus 1) (1 minus 1119899)

120572minus2]= 120572

(39)

from which it follows that

intI119899

100381610038161003816100381610038161003816100381610038161003816120593 ((119905 minus 119904)120572minus1Γ (120572) 119909 (119904))100381610038161003816100381610038161003816100381610038161003816 119889119904997888rarr0 as 119899 997888rarr infin (40)

Therefore in view of Proposition 3 119904 997891rarr (119905 minus 119904)120572minus1119909(119904) notin119875([0 1] 1198880) Hence I120572119909 does not exist which is what wewished to show

The following theorem provides a useful characterizationof the space H119901(119864) for which the statements reveal howmuch the fractional integral I120572 is better than the function119909 isin H119901(119864) Indeed based on Lemma 8 using an inequalityof Young we can easily prove the following

Lemma 14 For any 120572 gt 0 the following holds(a) If 119864 is reflexive then for every 119901 isin [1infin] the operator

I120572 mapsH119901(119864) intoH1199010 (119864) In particular if 0 lt 120572 lt1 then I120572 H1(119864) rarr H1((1minus120572)minus120598)0 (119864) however small120598 gt 0 is

(b) If 119864 contains no copy of 1198880 then for every 119901 isin [1infin]the operator I120572 maps 119909 isin H

1199010 (119864) 119909 strongly

measurable into H1199010 (119864) In particular if 120572 ge 1 the

operator I120572 mapsH1199010 (119864) into itself

(c) If 119901 gt max1 1120572 the operator I120572 sends H1199010 (119864) to119862[119868 119864120596] (if we define I120572119909(0) fl 0) In particular if120572 isin (0 1) I120572 H119901

0 (119864) rarr H120572minus1119901(119864)Proof At the beginning let 120572 gt 0 and 119909 isin H119901(119864) with119901 isin [1infin] and define the real-valued function 119892 119868 rarr R

by 119892(119905) fl 119905120572minus1Γ(120572) If 119864 is reflexive it follows in view ofLemma 8 that J120572119909 exists ae in 119868 and 120593(I120572119909) = I120572120593119909for every 120593 isin 119864lowast Youngrsquos inequality guarantees that theconvolution product 119892 lowast 120593119909(sdot) lies in 119871119901[119868] Consequently120593(I120572119909) = I120572120593119909 = 119892 lowast 120593119909 isin 119871119901[119868] for every 120593 isin 119864lowastThe reflexivity of 119864 together with Proposition 2 yields I120572119909 isinH

1199010 (119864)In particular if 0 lt 120572 lt 1 it can be easily seen thatint

119868(119892(119905))119902119889119905 lt infin for every 119902 isin [1 1(1 minus 120572)) That is 119892 isin1198711(1minus120572)minus120598[119868] 120598 gt 0 By Youngrsquos inequality it follows that120593(I120572119909) = 119892 lowast 120593119909 isin 1198711(1minus120572)minus120598[119868] for every 120593 isin 119864lowast however

small 120598 gt 0 is Now the assertion (a) follows because of thereflexivity of 119864

Next in order to prove the assertion (b) let 119909 isin H1199010 (119864)

and note in view of Lemma 8 that I120572119909 exists on 119868 Define119910 119868 rarr 119864 by119910(119905) fl I120572119909(119905) 119909 isin H1199010 (119864) Since120593119910(sdot) isin 119871119901[119868]

for every120593 isin 119864lowast it follows that119910 isin H119901(119864) Moreover for any119886 119887 isin 119868 (119886 lt 119887) we haveint119887

119886120593119910 (119904) 119889119905 = int119887

119886I120593119909 (119905) 119889119905 = 1Γ (120572)

sdot int119887

119886int119905

0(119905 minus 119904)120572minus1 120593119909 (119904) 119889119904 119889119905

= 1Γ (120572) [int119887

119886int119887

119904(119905 minus 119904)120572minus1 120593119909 (119904) 119889119905 119889119904

+ int119886

0int119887

119886(119905 minus 119904)120572minus1 120593119909 (119904) 119889119905 119889119904]

= 1Γ (120572) [int119887

119886(119887 minus 119904)120572 120593119909 (119904) 119889119904

+ int119886

0[(119887 minus 119904)120572 minus (119886 minus 119904)120572] 120593119909 (119904) 119889119904]

= 1Γ (120572) [int119886

0(119887 minus 119904)120572 120593119909 (119904) 119889119904

+ int119887

119886(119887 minus 119904)120572 120593119909 (119904) 119889119904 minus int119886

0(119886 minus 119904)120572 120593119909 (119904) 119889119904]

= int119887

0

(119887 minus 119904)120572Γ (1 + 120572)120593119909 (119904) 119889119904 minus int119886

0

(119886 minus 119904)120572Γ (1 + 120572)120593119909 (119904) 119889119904= 120593 (119909[119886119887])

(41)

where

119909[119886119887] fl (119875) int119887

0

(119887 minus 119904)120572Γ (1 + 120572)119909 (119904) 119889119904minus (119875) int119886

0

(119886 minus 119904)120572Γ (1 + 120572)119909 (119904) 119889119904(42)

Since 119909 isin 119875[119868 119864] owing to Proposition 2 (part (1)) it followsthat (119887 minus sdot)120572119909(119904) and (119886 minus sdot)120572119909(119904) are Pettis integrable on 119868 andso 119909[119886119887] isin 119864

A combination of these results yields120593119910 isin 119871119901(119868) for every120593 isin 119864lowast and there exists an element 119909[119886119887] isin 119864 such that120601119909[119886119887] = int119887

119886120593119910(119905)119889119905 for every 119886 119887 isin 119868 Since 119864 contains no

copy of 1198880 it follows that (cf [17 Theorem 23]) 119910 isin 119875[119868 119864]Consequently 119910(sdot) isin H

1199010 (119864) This is the claim (b)

To prove the assertion (c) let 119901 gt max1 1120572 and 119909 isinH

1199010 (119864) By Lemma 8 we deduce that I120572119909 exists ae on 119868

Now let 119902 isin [1infin] be such that 1119901 + 1119902 = 1 Since119902(120572 minus 1) gt minus1 119892 fl (sdot)120572minus1Γ(120572) isin 119871119902[119868] Therefore as a directconsequence of Youngrsquo inequality it follows that

120593 (I120572119909 (sdot)) = I120572120593119909 (sdot) = 119892 lowast 120593119909 (sdot) isin 119862 [119868R] (43)

holds for every 120593 isin 119864lowast Now we claim that I120572119909(119905) rarr 0in 119864 as 119905 rarr 0 Once our claim is established the definition

8 Journal of Function Spaces

I120572119909(0) fl 0 guarantees thatI120572119909(sdot) isin 119862[119868 119864120596] which is whatwewished to show It remains to prove our claim without lossof generality assume that I120572119909(119905) = 0 Then there exists (asa consequence of the Hahn-Banach theorem) 120593 isin 119864lowast with120593 = 1 such that I120572119909(119905) = 120593(I120572119909(119905)) = I120572120593119909(119905) ByHolderrsquos inequality we obtain

1003816100381610038161003816120593 (I120572119909 (119905))1003816100381610038161003816le 1Γ (120572) (int

119905

0(119905 minus 119904)119902(120572minus1) 119889119904)1119902 (int119905

0

1003816100381610038161003816120593119909 (119904)1003816100381610038161003816119901 119889119904)1119901

le 1Γ (120572) ( 119905(120572minus1)119902+1(120572 minus 1) 119902 + 1)1119902 (int

119868

1003816100381610038161003816120593119909 (119904)1003816100381610038161003816119901 119889119904)1119901

= 119905120572minus1+1119902Γ (120572) [119902 (120572 minus 1) + 1]1119902

10038171003817100381710038171205931199091003817100381710038171003817119871119901

= 119905120572minus1119901 10038171003817100381710038171205931199091003817100381710038171003817119871119901Γ (120572) [119902 (120572 minus 1) + 1]1119902

(44)

This is equivalent with the following estimate

1003817100381710038171003817I120572119909 (119905)1003817100381710038171003817 le 119905120572minus1119901 10038171003817100381710038171205931199091003817100381710038171003817119871119901Γ (120572) [119902 (120572 minus 1) + 1]1119902 (45)

Owing to 120572 gt 1119901 we get I120572119909(119905) rarr 0 in 119864 as 119905 rarr 0 andconsequently in view ofI120572119909(0) fl 0I120572119909(sdot) isin 119862[119868 119864120596] Thisproves the first part of the assertion (c)

Finally let 120572 isin (0 1) and 119905 120591 isin 119868 Without loss ofgenerality assume 120591 lt 119905 Then for any 120593 isin 119864lowast we have byHolder inequality with 119901 gt 1120572 in view 119902(120572 minus 1) gt minus1 that

1003816100381610038161003816120593 (I120572119909 (119905) minusI120572119909 (120591))1003816100381610038161003816 Γ (120572)

= 10038161003816100381610038161003816100381610038161003816int119905

0(119905 minus 119904)120572minus1 120593119909 (119904) 119889119904 minus int120591

0(120591 minus 119904)120572minus1 120593119909 (119904) 11988911990410038161003816100381610038161003816100381610038161003816

le (int120591

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 minus (120591 minus 119904)120572minus110038161003816100381610038161003816 1003816100381610038161003816120593119909 (119904)1003816100381610038161003816 119889119904+ int119905

120591(119905 minus 119904)120572minus1 1003816100381610038161003816120593119909 (119904)1003816100381610038161003816 119889119904)

= [(int120591

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 minus (120591 minus 119904)120572minus110038161003816100381610038161003816119902 119889119904)1119902

+ (int119905

120591(119905 minus 119904)(120572minus1)119902 119889119904)1119902] 10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]

le [[(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

+ ((119905 minus 120591)(120572minus1)119902+1(120572 minus 1) 119902 + 1 )1119902]]10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]

= [(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

+ (119905 minus 120591)120572minus1119901[119902 (120572 minus 1) + 1]1119902]

10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868] (46)

By noting that (120572 minus 1)119902 isin (minus1 0) when 120572 isin (0 1) it can beeasily seen that

int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904le 1119902 (120572 minus 1) + 1 (119905 minus 120591)119902(120572minus1)+1

(47)

That is

(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

le (119905 minus 120591)120572minus1119901[119902 (120572 minus 1) + 1]1119902

(48)

A combination of these results yields1003816100381610038161003816120593 (I120572119909 (119905)) minus 120593 (I120572119909 (120591))1003816100381610038161003816= 1003816100381610038161003816120593 (I120572119909 (119905) minusI

120572119909 (120591))1003816100381610038161003816le10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]Γ (120572) 2 (119905 minus 120591)120572minus1119901

[119902 (120572 minus 1) + 1]1119902 (49)

ThusI120572119909 isin H120572minus1119901(119864) This completes the proof

Corollary 15 For any 120572 gt 0 I120572 119862[119868 119864120596] rarr 119862[119868 119864120596]Proof Let 119909 isin 119862[119868 119864120596] then 120593119909 isin 119871infin holds for every120593 isin 119864lowast By noting that the weak continuity implies a strongmeasurability ([24] page 73) it follows that 119909 is stronglymeasurable on 119868 Hence in view of Proposition 2 119909 isinHinfin

0 (119864) Consequently 119862[119868 119864120596] sube H1199010 (119864) for every 119901 isin[1infin] That isI120572 119862[119868 119864120596] rarr 119862[119868 119864120596] Hence the desired

result is obtained

Example 16 Let 120572 isin [12 1) Define the function 119909 [0 1] rarr1198880 by 119909(119905) = 119899120594(1(119899+1)1119899]We have (in view of Example 12)that

I120572119909 (119905) = 0 0 0 1198990 (119905 minus 1 (1198990 + 1))120572

Γ (1 + 120572) (1198990 + 1)Γ (1 + 120572) [(119905 minus 11198990 + 2)

120572 minus (119905 minus 11198990 + 1)120572]

(50)

To show that I120572119909 is norm continuous on [0 1] let 120591 119905 isin[0 1] With no loss of generality we may assume that 119905 120591 isin1198681198990 for some 1198990 isin N Since the nonzero terms of the sequencein (50) are nonincreasing then in view of

|119905 minus 120591| le ( 11198990 minus 11198990 + 1) = 11198990 (1198990 + 1) le 111989920 (51)

Journal of Function Spaces 9

we have1003817100381710038171003817I120572119909 (119905) minusI

1205721119909 (120591)10038171003817100381710038171198880

= 1198990Γ (1 + 120572)10038161003816100381610038161003816100381610038161003816(119905 minus

1119899 + 1)120572 minus (120591 minus 1119899 + 1)

12057210038161003816100381610038161003816100381610038161003816le 1198990Γ (1 + 120572)

1003816100381610038161003816100381610038161003816(119905 minus 1119899 + 1) minus (120591 minus 1119899 + 1)1003816100381610038161003816100381610038161003816120572

le 1198990Γ (1 + 120572) |119905 minus 120591|120572 le 1Γ (1 + 120572) |119905 minus 120591|120572radic|119905 minus 120591|= |119905 minus 120591|120572minus12Γ (1 + 120572)

(52)

Thus the RFPI of 119909 is norm continuous on [0 1] Preciselysince 0 lt 120572 minus 12 lt 120572 minus 1119901 lt 1 for any 119901 le 2 thenI120572119909 isin H120572minus12(119864) sub H120572minus1119901(119864) This is precisely what onewould expect from Lemma 14 (part (c))

Analogously an explicit calculation reveals that I120572119909 liesin 119862([0 1] 119871infin[0 1]) where 119909 [0 1] rarr 119871infin[0 1] defined byExample 10

We now consider additional mapping properties of theoperator I120572 Precisely we will show that the RFPI enjoysthe following commutative property which is folklore in case119864 = R However the proof is completely similar to that of [8]Lemma 35

Lemma 17 Let 120572 120573 gt 0 and 119901 gt max1 1120572 1120573 Then

I120572I120573119909 = I

120573I120572119909 = I

120572+120573119909 (53)

holds for every 119909 isin H1199010 (119864) If 119864 is reflexive this is also true for

every 119901 ge 1Proof Since 119901 gt max1 1120572 1120573 it follows that 119901 gtmax1 1(120572+120573) So by Lemmas 8 and 14 themappingsI120572119909I120573119909 andI120572+120573119909 belong toHinfin

0 (119864) for every 119909 isin H1199010 (119864) Now

repeating the same process as in ([8] Lemma 35) implies theclaim

When 119864 is reflexive the result follows as a direct applica-tion of Lemma 14

3 Fractional Derivatives ofVector-Valued Functions

After the notation of the fractional integrals of vector-valued functions the fractional derivatives become a naturalrequirement Before we come to the definitions and a detailedstudy of the mathematical properties of fractional differentialoperators we recall the following

Definition 18 Consider the vector-valued function 119909 119868 rarr119864(1) Let 120593119909 be differentiable on 119868 for every 120593 isin 119864lowast The

function 119909 is said to be weakly differentiable on 119868 if

there exists 119910 119868 rarr 119864 such that for every 120593 isin 119864lowast wehave

119889120593119909 (119905)119889119905 = 120593119910 (119905) for every 119905 isin 119868 (54)

The function 119910 is called the weak derivative of thefunction 119909

(2) Let 120593119909 be differentiable ae on 119868 for every 120593 isin 119864lowast(the null set may vary with 120593 isin 119864lowast) The function 119909is said to be pseudo differentiable on 119868 if there existsa function 119910 119868 rarr 119864 such that for every 120593 isin 119864lowast thereexists a null set119873(120593) sub 119868 such that

(120593119909 (119905))1015840 = 120593119910 (119905) for every 119905 isin 119868119873 (120593) (55)

In this case the function 119910 is called the pseudoderivative of 119909If 119909 is pseudo differentiable on 119868 and the null setinvariant for every 120593 isin 119864lowast then 119909 is ae weaklydifferentiable on 119868

Clearly if 119909 is ae weakly differentiable on 119868 then 120593(119909)is ae differentiable on 119868 The converse holds in a weaklysequentially complete space (see [25] Theorem 733)

For more details of the derivatives of vector-valuedfunctions we refer to [10 12 26]

The following results play a major role in our analysis

Proposition 19 (see [27] Theorem 51) The function 119910 119868 rarr119864 is an indefinite Pettis integrable if and only if 119910 is weaklyabsolutely continuous on 119868 and have a pseudo derivative on 119868In this case 119910 is an indefinite Pettis integral of any of its pseudoderivatives

Now we are in the position to define the fractional-typederivatives of vector-valued functions

Definition 20 Let 119909 119868 rarr 119864 For the positive integer119898 suchthat 120572 isin (119898 minus 1119898) 119898 isin N0 fl 0 1 2 we define theCaputo fractional-pseudo (weak) derivative ldquoshortly CFPD(CFWD)rdquo of 119909 of order 120572 by

119889120572119889119905120572 119909 (119905) fl I119898minus120572119863119898119909 (119905) (56)

where the sign ldquo119863rdquo denotes the pseudo (or weak) differentialoperator We use the notation 119889120572119901119889119905120572 and 119889120572120596119889119905120572 to charac-terize the Caputo fractional-pseudo derivatives and Caputofractional weak derivatives respectively

It is well known that although the weak derivative of aweakly differentiable function is uniquely determined thepseudo derivative of the pseudo differentiable function is notunique Also although any two pseudo derivatives 119910 119911 of afunction 119909 119868 rarr 119864 need not be ae equal (see [12 Example91] and [26 p 2]) the functions 119910 119911 are weakly equivalenton 119868 (that is 120593119910 = 120593119911 holds ae for every 120593 isin 119864lowast) The

10 Journal of Function Spaces

next lemma provides a useful characterization property of theCFPD Really it can be easily seen that the CFPD of a Caputofractional-pseudo differentiable function does not depend onthe choice of a pseudo derivative of the function

Lemma 21 Let 119909 119868 rarr 119864 be pseudo differentiable functionwhere pseudo derivatives lie in H

1199010 (119864) If (119889120572119901119889119905120572)119909 exists on119868 then the CFPD of 119909 depends on the choice of the pseudo

derivatives of 119909Proof Let 119910 119911 isin H

1199010 (119864) be two pseudo derivatives of the

pseudo differentiable function 119909 Since 119909 and 119910 are weaklyequivalent on 119868 then for every 120593 isin 119864lowast there exists a nullset N which depends on 120593 isin 119864lowast such that (119905 minus 119904)minus120572120593119910(119904) =(119905minus119904)minus120572120593119911(119904) for every 119904 isin [0 119905]N Consequently for almostevery 119905 isin 119868

120593 (I1minus120572119910 (119905)) = I1minus120572120593119910 (119905) = I

1minus120572120593119911 (119905)= 120593 (I1minus120572119911 (119905)) (57)

holds for every 120593 isin 119864lowast Thus

119889120572119901119889119905120572 119909 = I1minus120572119863119901119909 = I

1minus120572119910 = I1minus120572119911 on 119868 (58)

which is what we wished to show

We consider the following examples

Example 22 Define 119909 [0 1] rarr ℓ2 by119909 (119905) fl ln(1 + 119905119899) 119899 isin N (59)

We claim that the CFWD of 119909 exists on [0 1] To see this welet 120593 isin (ℓ2)lowast = ℓ2 Then there corresponds to 120593 a sequence120582119899 isin ℓ2 such that

120593119909 (119905) = sum119899

120582119899 ln(1 + 119905119899) (60)

Since the series in (60) converges uniformly on [0 1] theformal differentiation of 120593119909 yields

(120593119909 (119905))1015840 = sum119899

120582119899119899 + 119905 = 120593 ( 1119905 + 119899) (61)

It is not hard to justify the differentiation by noting that theseries in the right hand side of (61) is uniformly and absolutelyconvergent on [0 1] Consequently 120593119909 is differentiable on[0 1] for 120593 isin (ℓ2)lowast meaning that 119909 is weakly differentiable on[0 1] (because ℓ2 is weakly complete (cf [25] Theorem 733)and 119863120596119909(119905) = 1(119905 + 119899) Owing to Lemma 25 the CFWDof any order 120572 isin (0 1) of 119909 exists on [0 1] To calculate theCFWD of 119909 on [0 1] we observe in view of Example 11 that

119889120572120596119889119905120572 119909 (119905) = I1minus120572119863120596119909 (119905)

= 1199051minus120572119899Γ (2 minus 120572) 2F1 (1 1 2 minus 120572 minus119905119899 ) (62)

Example 23 Let119864 be an infinite dimensionBanach space thatfails cotype Define 119910 [0 1] rarr 119864 by 119910(119905) fl int119905

0119909(119904)119889119904

where 119909 [0 1] rarr 119864 given by formula (7) As cited in ([13]Corollary 4) 119910 is nowhere weakly differentiable on [0 1]Consequently the CFWD of any order 120572 isin [34 1) of 119910 losesits meaning on [0 1]Remark 24 As shown in Example 23 there is an infinitedimension Banach space 119864 and weakly absolutely continuousfunction 119910 119868 rarr 119864 which is nowhere weakly differentiable(hence the CFWD of 119910 does not exist) Also even when 119864 isseparable and 119910 is Lipschitz function the pseudo derivatives(hence the CFPD) of 119910 need not to exist [28]

However Definition 20 of the CFPD (CFWD) has thedisadvantage that it completely loses its meaning if thefunction119909 fails to be pseudo (weakly) differentiable Preciselythe CFPD (in particular the CFWD) of a function 119909 loses itsmeaning if 119909 is not weakly absolutely continuous

Thenext lemma gives sufficient conditions that ensure theexistence of the Caputo fractional derivatives of a function119909 isin H119901(119864)Lemma 25 Let 120572 isin (0 1) For the function 119909 119868 rarr 119864 thefollowing hold

(a) If 119909 has a pseudo derivative 119863119901119909 isin H1199010 (119864) where 119901 gt1(1 minus 120572) then the CFPD of 119909 of order 120572 exists on 119868

Moreover (119889120572119901119889119905120572)119909 isin 119862[119868 119864120596](b) If 119864 is weakly complete or contains no copy of 1198880 and if119909 has a weak derivative 119863120596119909 isin H119901(119864) where 119901 isin[1infin] then the CFWD of 119909 of order 120572 exists on 119868

Moreover (119889120572120596119889119905120572)119909 isin 119862[119868 119864120596]This holds for any 119909 isin H119901(119864) with 119901 isin [1infin] if 119864 is reflexive

In all cases 120593((119889120572119901119889119905120572)119909) = (119889120572119901119889119905120572)120593119909 (120593((119889120572120596119889119905120572)119909) =(119889120572120596119889119905120572)120593119909) holds for every 120593 isin 119864lowastProof Since the weak (pseudo) derivative of an ae weakly(pseudo) differentiable function is strongly (weakly) mea-surable [12 26 28] the proof is readily available in view ofthe definition of Caputo fractional derivatives and Lemma 8Moreover since 119863120596119909(119863119901119909) isin H

1199010 (119864) with 119901 gt 1(1 minus 120572)

it follows in view of Lemma 14 that I1minus120572119863120596119909(I1minus120572119863119901119909) isin119862[119868 119864120596] This completes the proof

Besides the Caputo fractional-pseudo (weak) derivativeswe define the Riemann-Liouville fractional-pseudo (weak)derivatives

Definition 26 Let 119909 119868 rarr 119864 For the positive integer 119898such that 120572 isin (119898 minus 1119898) 119898 isin N0 fl 0 1 2 we definethe Riemann-Liouville fractional-pseudo (weak) derivativeldquoshortly RFPD (RFWD)rdquo of 119909 of order 120572 by

D120572119909 (119905) fl 119863119898

I119898minus120572119909 (119905) (63)

where 119863 is defined as in Definition 20 We use the notationD120572

119901 (D120572120596) to characterize the Riemann-Liouville fractional-

pseudo (weak) derivatives

Journal of Function Spaces 11

Clearly in infinite dimension Banach spaces the weakabsolute continuity of I119898minus120572119909 is necessarily (but not suf-ficient) condition for the existence of RFPD (in particularRFWD) of 119909Lemma 27 Let 0 lt 120572 lt 1 For any 119909 isin H

1199010 (119864) with 119901 gt

max1120572 1(1 minus 120572) we haveD

120572119901I

120572119909 = 119909 ae (64)

If 119864 is reflexive this is also true for every 119901 ge 1Proof Our assumption 119909 isin H

1199010 (119864)with 119901 gt max1120572 1(1minus120572) yields in view of Lemma 17 that

D120572119901I

120572119909 = 119863119901I1minus120572

I120572119909 = 119863119901I

1119909 (65)

The claim now follows immediately since (cf Proposition 19)the indefinite integral of Pettis integrable function is weaklyabsolutely continuous and it is pseudo differentiable withrespect to the right endpoint of the integration interval andits pseudo derivative equals the integrand at that point

Remark 28 When we replace D120572119901 by D120572

120596 then Lemma 27is no longer necessarily true for arbitrary 119909 isin H1

0(119864) evenwhen 119864 is reflexive evidently in [13 remark below Corollary4] the existence of a reflexive Banach space 119864 and a strongmeasurable Pettis integrable function 119909 119868 rarr 119864 was provedsuch that 119909 has nowhere weakly differentiable integral In thiscase 119863120596I

1119909 lost its meaning This gives a reason to believethat (65) (hence (64)) withD120572

120596119909 could not happen

However we have the following result

Lemma 29 Let 0 lt 120572 lt 1 For any 119909 isin 119862[119868 119864120596] we haveD

120572120596I

120572119909 = 119909 on 119868 (66)

Proof Our assumption 119909 isin 119862[119868 119864120596] sub Hinfin0 (119864) (see the proof

of Corollary 15) yields in view of Lemma 17 that

D120572120596I

120572119909 = 119863120596I1minus120572

I120572119909 = 119863120596I

1119909 (67)

The claim now follows immediately since the indefiniteintegral of weakly contentious function is weakly absolutelycontinuous and it is weakly differentiable with respect tothe right endpoint of the integration interval and its weakderivative equals the integrand at that point

The following lemma is folklore in case 119864 = R but to seethat it also holds in the vector-valued case we provide a proof

Lemma 30 Let 0 lt 120572 lt 1 and 119901 gt max1120572 1(1minus120572) If thefunction 119909 119868 rarr 119864 is weakly absolutely continuous on 119868 andpasses a pseudo derivative inH

1199010 (119864) then119889120572119901119909119889119905120572 (119905) = D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (68)

In particular if 119909 passes a weak derivative in 119862[119868 119864120596] then119889120572120596119909119889119905120572 (119905) = D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (69)

If 119864 is reflexive this is also true for every 119901 ge 1

Proof We observe that under the assumption imposed on119863119901119909 together with Proposition 3 the weakly absolutelycontinuity of 119909 is equivalent to

119909 (119905) = 119909 (0) + int119905

0119863119901119909 (119904) 119889119904 (70)

Hence owing to Lemma 27 it follows that

119889120572119901119909119889119905120572 = I1minus120572119863119901119909 = (D120572

119901I120572)I1minus120572119863119901119909

= D120572119901 (I120572

I1minus120572)119863119901119909 = (119863119901I

1minus120572)I1119863119901119909= 119863119901I

1minus120572 (119909 (119905) minus 119909 (0))= D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0)

(71)

which is what we wished to showThe proof of (69) is very similar to that in (68) therefore

we omit the details

Remark 31 As we remark above the definition of the CFPDof a function 119909 loses completely its meaning if 119909 is notweakly absolutely continuous For this reason we are able touse Lemma 30 to define the Caputo fractional derivative ingeneral that is we put

119889120572119901119909119889119905120572 (119905) fl D120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) 0 lt 120572 lt 1 (72)

Similarly we define

119889120572120596119909119889119905120572 (119905) fl D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (73)

However Lemma 30 claims that for the weakly absolutelycontinuous functions having Pettis integrable pseudo [weak]derivatives definitions (72) and (73) of the Caputo fractional-pseudo [weak] derivatives coincide with Definition 20

4 An Application

Let 119891 119868 times 119864 rarr 119864 be given function Consider the boundaryvalue problem of the fractional type

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin [0 1] 120582 isin R

119910 (0) + 119887119910 (1) = ℎ(74)

with certain constants 119887 ℎ isin R 119887 = minus1 To obtain theintegral equation modeled off the problem (74) we let 119910 bea weakly continuous solution to problem (74) then formallywe have

I120572 119889120572120596119889119905120572119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (75)

12 Journal of Function Spaces

that is

I120572I1minus120572119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) 997904rArrI1119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (76)

This reads (cf [29])

119910 (119905) = 119910 (0) + 120582I120572119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin 119868 120582 isin R (77)

with some (presently unknown) quantity 119910(0)Now we solve (77) for 119910(0) by 119910(0) + 119887119910(1) = ℎ and it

follows that

(1 + 119887) 119910 (0) + 120582119887Γ (120572) int1

0(1 minus 119904)120572minus1 119891 (119904 119910 (119904)) 119889119904 = ℎ (78)

Thus

119910 (0) = ℎ1 + 119887 minus 1205821198871 + 119887119868120572119891 (1 119910 (1)) (79)

Assume that the function 119891 is weakly-weakly continuousfunction such that

(1) for any 119903 gt 0 there is a constant 119872 gt 0 such that119891(119905 119910) le 119872 for all 119905 isin 119868 and 119910 le 119903(2) there exists a nondecreasing continuous function 120595 [0infin) rarr [0infin) 120595(0) = 0 and 120595(119905) lt 119905 for all 119905 gt 0

such that 120573(119891(119879 times 119860)) le 120595(120573(119860)) for every boundedset119860 sub 119864 where 120573 stands for De Blasirsquos weakmeasureof noncompactness (see [30])

Occasionally if 119891 119868 times 119864 rarr 119864 is weakly-weakly continuousand 119864 is reflexive then the assumptions (1) and (2) areautomatically satisfied (see eg [31])

Theorem 32 (see [6] Theorem 33) Let 120572 isin (0 1) and |120582| ltΓ(1 + 120572) If 119891 119868 times 119864 rarr 119864 is weakly-weakly continuous andsatisfies the assumptions (1) and (2) then the integral equation(77) has a weakly continuous solution 119910 defined on [0 1]Proof We omit the proof since it is almost identical to thatin the proof in ([6] Theorem 33) with (small) necessarychanges

In the following example we assume that 119910 isin 119862[119868 119864120596]solves (77) and we will show that not only do we have that 119910no longer necessarily solves (74) (when the Caputo fractionalweak derivative is taken in the sense of Definition 20)but even worse it could happen that the problem (74) isldquomeaninglessrdquo on 119868Example 33 Let 120572 isin (0 14] Let 119864 be an indefinitedimensional reflexive Banach space that fails cotype Definethe weakly-weakly continuous function 119891 [0 1] times 119864 rarr 119864by 119891 fl I1minus120572119909 where 119909 119868 rarr 119864 is defined by formula (7)Obviously (cf Example 6 in view of Remark 7)119891 satisfies theassumptions of Theorem 32

Now consider the integral equation (77) with119891 = I1minus120572119909Namely we consider the integral equation

119910 (119905) = 119910 (0) + 120582I120572I1minus120572119909 (119905)

120572 isin (0 14] 119905 isin 119868 120582 isin R (80)

Obviously the solution 119910 is weakly continuous on [0 1] (thisis an immediate consequence of Remark 7 and Lemma 17)

Since 120593119909 isin 1198711[0 1] holds for every 120593 isin 119864lowast it follows bythe commutative property of the fractional integral operatorscalculus over the space of Lebesgue integrable functions (see[21] Section 23) that

120593 (I120572I1minus120572119909 (119905)) = I

120572120593 (I1minus120572119909 (119905))= I

120572I1minus120572120593 (119909 (119905)) = I

1120593 (119909 (119905))= 120593 (I1119909 (119905))

(81)

that is

120593 (I120572I1minus120572119909 (119905) minusI

1119909 (119905)) = 0 for every 120593 isin 119864lowast (82)

Hence I120572I1minus120572119909 = I1119909 Consequently problem (80) be-comes

119910 (119905) = 119910 (0) + 120582I1119909 (119905) 120572 isin (0 14] 119905 isin 119868 (83)

As showed in ([13] Corollary 4) the indefinite integral 119905 rarrint1199050119909(119904)119889119904 is nowhere weakly differentiable on [0 1] Thus 119910

is nowhere weakly differentiable on [0 1] hence the CFWDof 119910 (hence the boundary value problem) completely loses itsmeaning if the CFWD is taken in the sense of Definition 20This is what we wished to show

Now we are in the position to state and prove thefollowing existence theorem

Theorem 34 If 119891 119868 times 119864 rarr 119864 is a function such that allconditions fromTheorem 32 hold then problem (74) (where theCaputo fractionalweak derivative is taken in the sense (73)) hasa weak solution on [0 1]Proof Let 119910 isin 119862[119868 119864120596] be a solution to (77) Equip 119864 andI times E with weak topology and note that 119905 997891rarr (119905 119910(119905)) iscontinuous as a mapping from 119868 into 119868 times 119864 Since 119891 119868 times 119864 rarr119864 is weakly-weakly continuous on [0 1] then 119891(sdot 119910(sdot)) is acomposition of this mapping with 119891 and thus 119891(sdot 119910(sdot)) isweakly continuous on [0 1]

Now operating by the operatorD120572120596 on both sides of (77)

it follows by Lemma 29 that

D120572120596119910 = D

120572120596119910 (0) + 120582D120572

120596I120572119891 = 119905minus120572Γ (1 minus 120572)119910 (0) + 120582119891 (84)

Now insert definition (73) of the Caputo fractionalderivative and we get

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) (85)

Journal of Function Spaces 13

With some further efforts one can get the boundary condi-tion 119910(0) + 119887119910(1) = ℎ

Therefore 119910 satisfies problem (74) which is what wewished to show

Data Availability

No data were used to support this study

Conflicts of Interest

The author declares no conflicts of interest

References

[1] H A Salem and A M El-Sayed ldquoWeak solution for fractionalorder integral equations in reflexive Banach spacesrdquoMathemat-ica Slovaca vol 55 no 2 pp 169ndash181 2005

[2] H A Salem A M El-Sayed and O L Moustafa ldquoA note onthe fractional calculus in Banach spacesrdquo Studia ScientiarumMathematicarum Hungarica vol 42 no 2 pp 115ndash130 2005

[3] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoMulti-term fractional differential equations in a nonreflexiveBanach spacerdquo Advances in Difference Equations 2013302 18pages 2013

[4] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoFractional calculus and fractional differential equations innonreflexive Banach spacesrdquo Communications in NonlinearScience andNumerical Simulation vol 20 no 1 pp 59ndash73 2015

[5] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoNonlinear fractional differential equations in nonreflexiveBanach spaces and fractional calculusrdquo Advances in DifferenceEquations 2015112 18 pages 2015

[6] R P Agarwal V Lupulescu D OrsquoRegan and G RahmanldquoWeak solutions for fractional differential equations in nonre-flexive Banach spaces via Riemann-Pettis integralsrdquo Mathema-tische Nachrichten vol 289 no 4 pp 395ndash409 2016

[7] H A Salem ldquoOn the fractional order m-point boundaryvalue problem in reflexive Banach spaces and weak topologiesrdquoJournal of Computational andAppliedMathematics vol 224 no2 pp 565ndash572 2009

[8] H A Salem ldquoOn the fractional calculus in abstract spaces andtheir applications to the Dirichlet-type problem of fractionalorderrdquo Computers and Mathematics with Applications vol 59no 3 pp 1278ndash1293 2010

[9] HAH Salem andMCichon ldquoOn solutions of fractional orderboundary value problems with integral boundary conditions inBanach spacesrdquo journal of function spaces and applications vol2013 Article ID 428094 2013

[10] J Diestel and J J Uhl Jr Vector Measures vol 15 ofMathemat-ical Surveys American Mathematical Society Providence RIUSA 1977

[11] R H Martin and J J Uhl Jr Nonlinear Operators AndDifferential Equations in Banach Spaces John Wiley amp SonsNew York NY USA 1976

[12] B J Pettis ldquoOn integration in vector spacesrdquo Transactions of theAmericanMathematical Society vol 44 no 2 pp 277ndash304 1938

[13] S J Dilworth and M Girardi ldquoNowhere weak differentiabilityof the Pettis integralrdquo Quaestiones Mathematicae vol 18 no 4pp 365ndash380 1995

[14] G A Edgar ldquoMeasurability in a Banach spacerdquo Indiana Univer-sity Mathematics Journal vol 26 no 4 pp 663ndash677 1977

[15] G A Edgar ldquoMeasurability in a Banach space IIrdquo IndianaUniversityMathematics Journal vol 28 no 4 pp 559ndash579 1979

[16] J L Gamez and J Mendoza ldquoOn denjoy-dunford and denjoy-pettis integralsrdquo StudiaMathematica vol 130 no 2 pp 115ndash1331998

[17] R AGordon ldquoTheDenjoy extension of the Bochner Pettis andDUNford integralsrdquo Studia Mathematica vol 92 no 1 pp 73ndash91 1989

[18] R F Geitz ldquoPettis integrationrdquo Proceedings of the AmericanMathematical Society vol 82 no 1 pp 81ndash86 1981

[19] R F Geitz ldquoGeometry and the Pettis integralrdquo Transactions ofthe AmericanMathematical Society vol 269 no 2 pp 535ndash5481982

[20] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations New York NYUSA Elsevier 2006

[21] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Yverdon Switzerland 1993

[22] M Mendel and A Naor ldquoMetric cotyperdquo Annals of Mathemat-ics Second Series vol 168 no 1 pp 247ndash298 2008

[23] B Ross S G Samko and E R Love ldquoFunctions that have nofirst order derivative might have fractional derivatives of allorders less than onerdquo Real Analysis Exchange vol 20 no 1 pp140ndash157 199495

[24] E Hille and R S Phillips Functional analysis and semi-groupsAmerican Mathematical Society Colloquium Publications vol31 American Mathematical Society Providence RI USA 1957

[25] S Schwabik and Y Guoju Topics in Banach Space IntegrationWorld Scientific Singapore 2005

[26] DW Solomon ldquoOn differentiability of vector-valued functionsof a real variablerdquo Studia Mathematica vol 29 pp 1ndash4 1967

[27] K Naralenkov ldquoOn Denjoy type extensions of the Pettisintegralrdquo Czechoslovak Mathematical Journal vol 60(135) no3 pp 737ndash750 2010

[28] D W Solomon Denjoy integration in abstract spaces Memorisof the American Mathematical Society No 85 AmericanMathematical Society Providence RI 1969

[29] A R Mitchell and C Smith ldquoAn existence theorem for weaksolutions of differential equations in Banach spacesrdquo inNonlin-ear Equations in Abstract Spaces V Lakshmikantham Ed pp387ndash404 1978

[30] F S De Blasi ldquoOn a property of the unit sphere in a Banachspacerdquo Bulletin Mathematiques De La Societe Des SciencesMathematiques De Roumanie vol 21 no 3 pp 259ndash262 1977

[31] A Szep ldquoExistence theorem for weak solutions of ordinarydifferential equations in reflexive Banach spacesrdquo Studia Scien-tiarum Mathematicarum Hungarica vol 6 pp 197ndash203 1971

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Page 8: On the Theory of Fractional Calculus in the Pettis-Function Spaces · 2019. 7. 30. · JournalofFunctionSpaces Further,wedenethespace H (),˙≥0by H () fl ˛ ∈ ˚, ˜: isHolderainoforder

8 Journal of Function Spaces

I120572119909(0) fl 0 guarantees thatI120572119909(sdot) isin 119862[119868 119864120596] which is whatwewished to show It remains to prove our claim without lossof generality assume that I120572119909(119905) = 0 Then there exists (asa consequence of the Hahn-Banach theorem) 120593 isin 119864lowast with120593 = 1 such that I120572119909(119905) = 120593(I120572119909(119905)) = I120572120593119909(119905) ByHolderrsquos inequality we obtain

1003816100381610038161003816120593 (I120572119909 (119905))1003816100381610038161003816le 1Γ (120572) (int

119905

0(119905 minus 119904)119902(120572minus1) 119889119904)1119902 (int119905

0

1003816100381610038161003816120593119909 (119904)1003816100381610038161003816119901 119889119904)1119901

le 1Γ (120572) ( 119905(120572minus1)119902+1(120572 minus 1) 119902 + 1)1119902 (int

119868

1003816100381610038161003816120593119909 (119904)1003816100381610038161003816119901 119889119904)1119901

= 119905120572minus1+1119902Γ (120572) [119902 (120572 minus 1) + 1]1119902

10038171003817100381710038171205931199091003817100381710038171003817119871119901

= 119905120572minus1119901 10038171003817100381710038171205931199091003817100381710038171003817119871119901Γ (120572) [119902 (120572 minus 1) + 1]1119902

(44)

This is equivalent with the following estimate

1003817100381710038171003817I120572119909 (119905)1003817100381710038171003817 le 119905120572minus1119901 10038171003817100381710038171205931199091003817100381710038171003817119871119901Γ (120572) [119902 (120572 minus 1) + 1]1119902 (45)

Owing to 120572 gt 1119901 we get I120572119909(119905) rarr 0 in 119864 as 119905 rarr 0 andconsequently in view ofI120572119909(0) fl 0I120572119909(sdot) isin 119862[119868 119864120596] Thisproves the first part of the assertion (c)

Finally let 120572 isin (0 1) and 119905 120591 isin 119868 Without loss ofgenerality assume 120591 lt 119905 Then for any 120593 isin 119864lowast we have byHolder inequality with 119901 gt 1120572 in view 119902(120572 minus 1) gt minus1 that

1003816100381610038161003816120593 (I120572119909 (119905) minusI120572119909 (120591))1003816100381610038161003816 Γ (120572)

= 10038161003816100381610038161003816100381610038161003816int119905

0(119905 minus 119904)120572minus1 120593119909 (119904) 119889119904 minus int120591

0(120591 minus 119904)120572minus1 120593119909 (119904) 11988911990410038161003816100381610038161003816100381610038161003816

le (int120591

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 minus (120591 minus 119904)120572minus110038161003816100381610038161003816 1003816100381610038161003816120593119909 (119904)1003816100381610038161003816 119889119904+ int119905

120591(119905 minus 119904)120572minus1 1003816100381610038161003816120593119909 (119904)1003816100381610038161003816 119889119904)

= [(int120591

0

10038161003816100381610038161003816(119905 minus 119904)120572minus1 minus (120591 minus 119904)120572minus110038161003816100381610038161003816119902 119889119904)1119902

+ (int119905

120591(119905 minus 119904)(120572minus1)119902 119889119904)1119902] 10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]

le [[(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

+ ((119905 minus 120591)(120572minus1)119902+1(120572 minus 1) 119902 + 1 )1119902]]10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]

= [(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

+ (119905 minus 120591)120572minus1119901[119902 (120572 minus 1) + 1]1119902]

10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868] (46)

By noting that (120572 minus 1)119902 isin (minus1 0) when 120572 isin (0 1) it can beeasily seen that

int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904le 1119902 (120572 minus 1) + 1 (119905 minus 120591)119902(120572minus1)+1

(47)

That is

(int120591

0

10038161003816100381610038161003816(119905 minus 119904)119902(120572minus1) minus (120591 minus 119904)119902(120572minus1)10038161003816100381610038161003816 119889119904)1119902

le (119905 minus 120591)120572minus1119901[119902 (120572 minus 1) + 1]1119902

(48)

A combination of these results yields1003816100381610038161003816120593 (I120572119909 (119905)) minus 120593 (I120572119909 (120591))1003816100381610038161003816= 1003816100381610038161003816120593 (I120572119909 (119905) minusI

120572119909 (120591))1003816100381610038161003816le10038171003817100381710038171205931199091003817100381710038171003817119871119901[119868]Γ (120572) 2 (119905 minus 120591)120572minus1119901

[119902 (120572 minus 1) + 1]1119902 (49)

ThusI120572119909 isin H120572minus1119901(119864) This completes the proof

Corollary 15 For any 120572 gt 0 I120572 119862[119868 119864120596] rarr 119862[119868 119864120596]Proof Let 119909 isin 119862[119868 119864120596] then 120593119909 isin 119871infin holds for every120593 isin 119864lowast By noting that the weak continuity implies a strongmeasurability ([24] page 73) it follows that 119909 is stronglymeasurable on 119868 Hence in view of Proposition 2 119909 isinHinfin

0 (119864) Consequently 119862[119868 119864120596] sube H1199010 (119864) for every 119901 isin[1infin] That isI120572 119862[119868 119864120596] rarr 119862[119868 119864120596] Hence the desired

result is obtained

Example 16 Let 120572 isin [12 1) Define the function 119909 [0 1] rarr1198880 by 119909(119905) = 119899120594(1(119899+1)1119899]We have (in view of Example 12)that

I120572119909 (119905) = 0 0 0 1198990 (119905 minus 1 (1198990 + 1))120572

Γ (1 + 120572) (1198990 + 1)Γ (1 + 120572) [(119905 minus 11198990 + 2)

120572 minus (119905 minus 11198990 + 1)120572]

(50)

To show that I120572119909 is norm continuous on [0 1] let 120591 119905 isin[0 1] With no loss of generality we may assume that 119905 120591 isin1198681198990 for some 1198990 isin N Since the nonzero terms of the sequencein (50) are nonincreasing then in view of

|119905 minus 120591| le ( 11198990 minus 11198990 + 1) = 11198990 (1198990 + 1) le 111989920 (51)

Journal of Function Spaces 9

we have1003817100381710038171003817I120572119909 (119905) minusI

1205721119909 (120591)10038171003817100381710038171198880

= 1198990Γ (1 + 120572)10038161003816100381610038161003816100381610038161003816(119905 minus

1119899 + 1)120572 minus (120591 minus 1119899 + 1)

12057210038161003816100381610038161003816100381610038161003816le 1198990Γ (1 + 120572)

1003816100381610038161003816100381610038161003816(119905 minus 1119899 + 1) minus (120591 minus 1119899 + 1)1003816100381610038161003816100381610038161003816120572

le 1198990Γ (1 + 120572) |119905 minus 120591|120572 le 1Γ (1 + 120572) |119905 minus 120591|120572radic|119905 minus 120591|= |119905 minus 120591|120572minus12Γ (1 + 120572)

(52)

Thus the RFPI of 119909 is norm continuous on [0 1] Preciselysince 0 lt 120572 minus 12 lt 120572 minus 1119901 lt 1 for any 119901 le 2 thenI120572119909 isin H120572minus12(119864) sub H120572minus1119901(119864) This is precisely what onewould expect from Lemma 14 (part (c))

Analogously an explicit calculation reveals that I120572119909 liesin 119862([0 1] 119871infin[0 1]) where 119909 [0 1] rarr 119871infin[0 1] defined byExample 10

We now consider additional mapping properties of theoperator I120572 Precisely we will show that the RFPI enjoysthe following commutative property which is folklore in case119864 = R However the proof is completely similar to that of [8]Lemma 35

Lemma 17 Let 120572 120573 gt 0 and 119901 gt max1 1120572 1120573 Then

I120572I120573119909 = I

120573I120572119909 = I

120572+120573119909 (53)

holds for every 119909 isin H1199010 (119864) If 119864 is reflexive this is also true for

every 119901 ge 1Proof Since 119901 gt max1 1120572 1120573 it follows that 119901 gtmax1 1(120572+120573) So by Lemmas 8 and 14 themappingsI120572119909I120573119909 andI120572+120573119909 belong toHinfin

0 (119864) for every 119909 isin H1199010 (119864) Now

repeating the same process as in ([8] Lemma 35) implies theclaim

When 119864 is reflexive the result follows as a direct applica-tion of Lemma 14

3 Fractional Derivatives ofVector-Valued Functions

After the notation of the fractional integrals of vector-valued functions the fractional derivatives become a naturalrequirement Before we come to the definitions and a detailedstudy of the mathematical properties of fractional differentialoperators we recall the following

Definition 18 Consider the vector-valued function 119909 119868 rarr119864(1) Let 120593119909 be differentiable on 119868 for every 120593 isin 119864lowast The

function 119909 is said to be weakly differentiable on 119868 if

there exists 119910 119868 rarr 119864 such that for every 120593 isin 119864lowast wehave

119889120593119909 (119905)119889119905 = 120593119910 (119905) for every 119905 isin 119868 (54)

The function 119910 is called the weak derivative of thefunction 119909

(2) Let 120593119909 be differentiable ae on 119868 for every 120593 isin 119864lowast(the null set may vary with 120593 isin 119864lowast) The function 119909is said to be pseudo differentiable on 119868 if there existsa function 119910 119868 rarr 119864 such that for every 120593 isin 119864lowast thereexists a null set119873(120593) sub 119868 such that

(120593119909 (119905))1015840 = 120593119910 (119905) for every 119905 isin 119868119873 (120593) (55)

In this case the function 119910 is called the pseudoderivative of 119909If 119909 is pseudo differentiable on 119868 and the null setinvariant for every 120593 isin 119864lowast then 119909 is ae weaklydifferentiable on 119868

Clearly if 119909 is ae weakly differentiable on 119868 then 120593(119909)is ae differentiable on 119868 The converse holds in a weaklysequentially complete space (see [25] Theorem 733)

For more details of the derivatives of vector-valuedfunctions we refer to [10 12 26]

The following results play a major role in our analysis

Proposition 19 (see [27] Theorem 51) The function 119910 119868 rarr119864 is an indefinite Pettis integrable if and only if 119910 is weaklyabsolutely continuous on 119868 and have a pseudo derivative on 119868In this case 119910 is an indefinite Pettis integral of any of its pseudoderivatives

Now we are in the position to define the fractional-typederivatives of vector-valued functions

Definition 20 Let 119909 119868 rarr 119864 For the positive integer119898 suchthat 120572 isin (119898 minus 1119898) 119898 isin N0 fl 0 1 2 we define theCaputo fractional-pseudo (weak) derivative ldquoshortly CFPD(CFWD)rdquo of 119909 of order 120572 by

119889120572119889119905120572 119909 (119905) fl I119898minus120572119863119898119909 (119905) (56)

where the sign ldquo119863rdquo denotes the pseudo (or weak) differentialoperator We use the notation 119889120572119901119889119905120572 and 119889120572120596119889119905120572 to charac-terize the Caputo fractional-pseudo derivatives and Caputofractional weak derivatives respectively

It is well known that although the weak derivative of aweakly differentiable function is uniquely determined thepseudo derivative of the pseudo differentiable function is notunique Also although any two pseudo derivatives 119910 119911 of afunction 119909 119868 rarr 119864 need not be ae equal (see [12 Example91] and [26 p 2]) the functions 119910 119911 are weakly equivalenton 119868 (that is 120593119910 = 120593119911 holds ae for every 120593 isin 119864lowast) The

10 Journal of Function Spaces

next lemma provides a useful characterization property of theCFPD Really it can be easily seen that the CFPD of a Caputofractional-pseudo differentiable function does not depend onthe choice of a pseudo derivative of the function

Lemma 21 Let 119909 119868 rarr 119864 be pseudo differentiable functionwhere pseudo derivatives lie in H

1199010 (119864) If (119889120572119901119889119905120572)119909 exists on119868 then the CFPD of 119909 depends on the choice of the pseudo

derivatives of 119909Proof Let 119910 119911 isin H

1199010 (119864) be two pseudo derivatives of the

pseudo differentiable function 119909 Since 119909 and 119910 are weaklyequivalent on 119868 then for every 120593 isin 119864lowast there exists a nullset N which depends on 120593 isin 119864lowast such that (119905 minus 119904)minus120572120593119910(119904) =(119905minus119904)minus120572120593119911(119904) for every 119904 isin [0 119905]N Consequently for almostevery 119905 isin 119868

120593 (I1minus120572119910 (119905)) = I1minus120572120593119910 (119905) = I

1minus120572120593119911 (119905)= 120593 (I1minus120572119911 (119905)) (57)

holds for every 120593 isin 119864lowast Thus

119889120572119901119889119905120572 119909 = I1minus120572119863119901119909 = I

1minus120572119910 = I1minus120572119911 on 119868 (58)

which is what we wished to show

We consider the following examples

Example 22 Define 119909 [0 1] rarr ℓ2 by119909 (119905) fl ln(1 + 119905119899) 119899 isin N (59)

We claim that the CFWD of 119909 exists on [0 1] To see this welet 120593 isin (ℓ2)lowast = ℓ2 Then there corresponds to 120593 a sequence120582119899 isin ℓ2 such that

120593119909 (119905) = sum119899

120582119899 ln(1 + 119905119899) (60)

Since the series in (60) converges uniformly on [0 1] theformal differentiation of 120593119909 yields

(120593119909 (119905))1015840 = sum119899

120582119899119899 + 119905 = 120593 ( 1119905 + 119899) (61)

It is not hard to justify the differentiation by noting that theseries in the right hand side of (61) is uniformly and absolutelyconvergent on [0 1] Consequently 120593119909 is differentiable on[0 1] for 120593 isin (ℓ2)lowast meaning that 119909 is weakly differentiable on[0 1] (because ℓ2 is weakly complete (cf [25] Theorem 733)and 119863120596119909(119905) = 1(119905 + 119899) Owing to Lemma 25 the CFWDof any order 120572 isin (0 1) of 119909 exists on [0 1] To calculate theCFWD of 119909 on [0 1] we observe in view of Example 11 that

119889120572120596119889119905120572 119909 (119905) = I1minus120572119863120596119909 (119905)

= 1199051minus120572119899Γ (2 minus 120572) 2F1 (1 1 2 minus 120572 minus119905119899 ) (62)

Example 23 Let119864 be an infinite dimensionBanach space thatfails cotype Define 119910 [0 1] rarr 119864 by 119910(119905) fl int119905

0119909(119904)119889119904

where 119909 [0 1] rarr 119864 given by formula (7) As cited in ([13]Corollary 4) 119910 is nowhere weakly differentiable on [0 1]Consequently the CFWD of any order 120572 isin [34 1) of 119910 losesits meaning on [0 1]Remark 24 As shown in Example 23 there is an infinitedimension Banach space 119864 and weakly absolutely continuousfunction 119910 119868 rarr 119864 which is nowhere weakly differentiable(hence the CFWD of 119910 does not exist) Also even when 119864 isseparable and 119910 is Lipschitz function the pseudo derivatives(hence the CFPD) of 119910 need not to exist [28]

However Definition 20 of the CFPD (CFWD) has thedisadvantage that it completely loses its meaning if thefunction119909 fails to be pseudo (weakly) differentiable Preciselythe CFPD (in particular the CFWD) of a function 119909 loses itsmeaning if 119909 is not weakly absolutely continuous

Thenext lemma gives sufficient conditions that ensure theexistence of the Caputo fractional derivatives of a function119909 isin H119901(119864)Lemma 25 Let 120572 isin (0 1) For the function 119909 119868 rarr 119864 thefollowing hold

(a) If 119909 has a pseudo derivative 119863119901119909 isin H1199010 (119864) where 119901 gt1(1 minus 120572) then the CFPD of 119909 of order 120572 exists on 119868

Moreover (119889120572119901119889119905120572)119909 isin 119862[119868 119864120596](b) If 119864 is weakly complete or contains no copy of 1198880 and if119909 has a weak derivative 119863120596119909 isin H119901(119864) where 119901 isin[1infin] then the CFWD of 119909 of order 120572 exists on 119868

Moreover (119889120572120596119889119905120572)119909 isin 119862[119868 119864120596]This holds for any 119909 isin H119901(119864) with 119901 isin [1infin] if 119864 is reflexive

In all cases 120593((119889120572119901119889119905120572)119909) = (119889120572119901119889119905120572)120593119909 (120593((119889120572120596119889119905120572)119909) =(119889120572120596119889119905120572)120593119909) holds for every 120593 isin 119864lowastProof Since the weak (pseudo) derivative of an ae weakly(pseudo) differentiable function is strongly (weakly) mea-surable [12 26 28] the proof is readily available in view ofthe definition of Caputo fractional derivatives and Lemma 8Moreover since 119863120596119909(119863119901119909) isin H

1199010 (119864) with 119901 gt 1(1 minus 120572)

it follows in view of Lemma 14 that I1minus120572119863120596119909(I1minus120572119863119901119909) isin119862[119868 119864120596] This completes the proof

Besides the Caputo fractional-pseudo (weak) derivativeswe define the Riemann-Liouville fractional-pseudo (weak)derivatives

Definition 26 Let 119909 119868 rarr 119864 For the positive integer 119898such that 120572 isin (119898 minus 1119898) 119898 isin N0 fl 0 1 2 we definethe Riemann-Liouville fractional-pseudo (weak) derivativeldquoshortly RFPD (RFWD)rdquo of 119909 of order 120572 by

D120572119909 (119905) fl 119863119898

I119898minus120572119909 (119905) (63)

where 119863 is defined as in Definition 20 We use the notationD120572

119901 (D120572120596) to characterize the Riemann-Liouville fractional-

pseudo (weak) derivatives

Journal of Function Spaces 11

Clearly in infinite dimension Banach spaces the weakabsolute continuity of I119898minus120572119909 is necessarily (but not suf-ficient) condition for the existence of RFPD (in particularRFWD) of 119909Lemma 27 Let 0 lt 120572 lt 1 For any 119909 isin H

1199010 (119864) with 119901 gt

max1120572 1(1 minus 120572) we haveD

120572119901I

120572119909 = 119909 ae (64)

If 119864 is reflexive this is also true for every 119901 ge 1Proof Our assumption 119909 isin H

1199010 (119864)with 119901 gt max1120572 1(1minus120572) yields in view of Lemma 17 that

D120572119901I

120572119909 = 119863119901I1minus120572

I120572119909 = 119863119901I

1119909 (65)

The claim now follows immediately since (cf Proposition 19)the indefinite integral of Pettis integrable function is weaklyabsolutely continuous and it is pseudo differentiable withrespect to the right endpoint of the integration interval andits pseudo derivative equals the integrand at that point

Remark 28 When we replace D120572119901 by D120572

120596 then Lemma 27is no longer necessarily true for arbitrary 119909 isin H1

0(119864) evenwhen 119864 is reflexive evidently in [13 remark below Corollary4] the existence of a reflexive Banach space 119864 and a strongmeasurable Pettis integrable function 119909 119868 rarr 119864 was provedsuch that 119909 has nowhere weakly differentiable integral In thiscase 119863120596I

1119909 lost its meaning This gives a reason to believethat (65) (hence (64)) withD120572

120596119909 could not happen

However we have the following result

Lemma 29 Let 0 lt 120572 lt 1 For any 119909 isin 119862[119868 119864120596] we haveD

120572120596I

120572119909 = 119909 on 119868 (66)

Proof Our assumption 119909 isin 119862[119868 119864120596] sub Hinfin0 (119864) (see the proof

of Corollary 15) yields in view of Lemma 17 that

D120572120596I

120572119909 = 119863120596I1minus120572

I120572119909 = 119863120596I

1119909 (67)

The claim now follows immediately since the indefiniteintegral of weakly contentious function is weakly absolutelycontinuous and it is weakly differentiable with respect tothe right endpoint of the integration interval and its weakderivative equals the integrand at that point

The following lemma is folklore in case 119864 = R but to seethat it also holds in the vector-valued case we provide a proof

Lemma 30 Let 0 lt 120572 lt 1 and 119901 gt max1120572 1(1minus120572) If thefunction 119909 119868 rarr 119864 is weakly absolutely continuous on 119868 andpasses a pseudo derivative inH

1199010 (119864) then119889120572119901119909119889119905120572 (119905) = D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (68)

In particular if 119909 passes a weak derivative in 119862[119868 119864120596] then119889120572120596119909119889119905120572 (119905) = D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (69)

If 119864 is reflexive this is also true for every 119901 ge 1

Proof We observe that under the assumption imposed on119863119901119909 together with Proposition 3 the weakly absolutelycontinuity of 119909 is equivalent to

119909 (119905) = 119909 (0) + int119905

0119863119901119909 (119904) 119889119904 (70)

Hence owing to Lemma 27 it follows that

119889120572119901119909119889119905120572 = I1minus120572119863119901119909 = (D120572

119901I120572)I1minus120572119863119901119909

= D120572119901 (I120572

I1minus120572)119863119901119909 = (119863119901I

1minus120572)I1119863119901119909= 119863119901I

1minus120572 (119909 (119905) minus 119909 (0))= D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0)

(71)

which is what we wished to showThe proof of (69) is very similar to that in (68) therefore

we omit the details

Remark 31 As we remark above the definition of the CFPDof a function 119909 loses completely its meaning if 119909 is notweakly absolutely continuous For this reason we are able touse Lemma 30 to define the Caputo fractional derivative ingeneral that is we put

119889120572119901119909119889119905120572 (119905) fl D120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) 0 lt 120572 lt 1 (72)

Similarly we define

119889120572120596119909119889119905120572 (119905) fl D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (73)

However Lemma 30 claims that for the weakly absolutelycontinuous functions having Pettis integrable pseudo [weak]derivatives definitions (72) and (73) of the Caputo fractional-pseudo [weak] derivatives coincide with Definition 20

4 An Application

Let 119891 119868 times 119864 rarr 119864 be given function Consider the boundaryvalue problem of the fractional type

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin [0 1] 120582 isin R

119910 (0) + 119887119910 (1) = ℎ(74)

with certain constants 119887 ℎ isin R 119887 = minus1 To obtain theintegral equation modeled off the problem (74) we let 119910 bea weakly continuous solution to problem (74) then formallywe have

I120572 119889120572120596119889119905120572119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (75)

12 Journal of Function Spaces

that is

I120572I1minus120572119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) 997904rArrI1119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (76)

This reads (cf [29])

119910 (119905) = 119910 (0) + 120582I120572119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin 119868 120582 isin R (77)

with some (presently unknown) quantity 119910(0)Now we solve (77) for 119910(0) by 119910(0) + 119887119910(1) = ℎ and it

follows that

(1 + 119887) 119910 (0) + 120582119887Γ (120572) int1

0(1 minus 119904)120572minus1 119891 (119904 119910 (119904)) 119889119904 = ℎ (78)

Thus

119910 (0) = ℎ1 + 119887 minus 1205821198871 + 119887119868120572119891 (1 119910 (1)) (79)

Assume that the function 119891 is weakly-weakly continuousfunction such that

(1) for any 119903 gt 0 there is a constant 119872 gt 0 such that119891(119905 119910) le 119872 for all 119905 isin 119868 and 119910 le 119903(2) there exists a nondecreasing continuous function 120595 [0infin) rarr [0infin) 120595(0) = 0 and 120595(119905) lt 119905 for all 119905 gt 0

such that 120573(119891(119879 times 119860)) le 120595(120573(119860)) for every boundedset119860 sub 119864 where 120573 stands for De Blasirsquos weakmeasureof noncompactness (see [30])

Occasionally if 119891 119868 times 119864 rarr 119864 is weakly-weakly continuousand 119864 is reflexive then the assumptions (1) and (2) areautomatically satisfied (see eg [31])

Theorem 32 (see [6] Theorem 33) Let 120572 isin (0 1) and |120582| ltΓ(1 + 120572) If 119891 119868 times 119864 rarr 119864 is weakly-weakly continuous andsatisfies the assumptions (1) and (2) then the integral equation(77) has a weakly continuous solution 119910 defined on [0 1]Proof We omit the proof since it is almost identical to thatin the proof in ([6] Theorem 33) with (small) necessarychanges

In the following example we assume that 119910 isin 119862[119868 119864120596]solves (77) and we will show that not only do we have that 119910no longer necessarily solves (74) (when the Caputo fractionalweak derivative is taken in the sense of Definition 20)but even worse it could happen that the problem (74) isldquomeaninglessrdquo on 119868Example 33 Let 120572 isin (0 14] Let 119864 be an indefinitedimensional reflexive Banach space that fails cotype Definethe weakly-weakly continuous function 119891 [0 1] times 119864 rarr 119864by 119891 fl I1minus120572119909 where 119909 119868 rarr 119864 is defined by formula (7)Obviously (cf Example 6 in view of Remark 7)119891 satisfies theassumptions of Theorem 32

Now consider the integral equation (77) with119891 = I1minus120572119909Namely we consider the integral equation

119910 (119905) = 119910 (0) + 120582I120572I1minus120572119909 (119905)

120572 isin (0 14] 119905 isin 119868 120582 isin R (80)

Obviously the solution 119910 is weakly continuous on [0 1] (thisis an immediate consequence of Remark 7 and Lemma 17)

Since 120593119909 isin 1198711[0 1] holds for every 120593 isin 119864lowast it follows bythe commutative property of the fractional integral operatorscalculus over the space of Lebesgue integrable functions (see[21] Section 23) that

120593 (I120572I1minus120572119909 (119905)) = I

120572120593 (I1minus120572119909 (119905))= I

120572I1minus120572120593 (119909 (119905)) = I

1120593 (119909 (119905))= 120593 (I1119909 (119905))

(81)

that is

120593 (I120572I1minus120572119909 (119905) minusI

1119909 (119905)) = 0 for every 120593 isin 119864lowast (82)

Hence I120572I1minus120572119909 = I1119909 Consequently problem (80) be-comes

119910 (119905) = 119910 (0) + 120582I1119909 (119905) 120572 isin (0 14] 119905 isin 119868 (83)

As showed in ([13] Corollary 4) the indefinite integral 119905 rarrint1199050119909(119904)119889119904 is nowhere weakly differentiable on [0 1] Thus 119910

is nowhere weakly differentiable on [0 1] hence the CFWDof 119910 (hence the boundary value problem) completely loses itsmeaning if the CFWD is taken in the sense of Definition 20This is what we wished to show

Now we are in the position to state and prove thefollowing existence theorem

Theorem 34 If 119891 119868 times 119864 rarr 119864 is a function such that allconditions fromTheorem 32 hold then problem (74) (where theCaputo fractionalweak derivative is taken in the sense (73)) hasa weak solution on [0 1]Proof Let 119910 isin 119862[119868 119864120596] be a solution to (77) Equip 119864 andI times E with weak topology and note that 119905 997891rarr (119905 119910(119905)) iscontinuous as a mapping from 119868 into 119868 times 119864 Since 119891 119868 times 119864 rarr119864 is weakly-weakly continuous on [0 1] then 119891(sdot 119910(sdot)) is acomposition of this mapping with 119891 and thus 119891(sdot 119910(sdot)) isweakly continuous on [0 1]

Now operating by the operatorD120572120596 on both sides of (77)

it follows by Lemma 29 that

D120572120596119910 = D

120572120596119910 (0) + 120582D120572

120596I120572119891 = 119905minus120572Γ (1 minus 120572)119910 (0) + 120582119891 (84)

Now insert definition (73) of the Caputo fractionalderivative and we get

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) (85)

Journal of Function Spaces 13

With some further efforts one can get the boundary condi-tion 119910(0) + 119887119910(1) = ℎ

Therefore 119910 satisfies problem (74) which is what wewished to show

Data Availability

No data were used to support this study

Conflicts of Interest

The author declares no conflicts of interest

References

[1] H A Salem and A M El-Sayed ldquoWeak solution for fractionalorder integral equations in reflexive Banach spacesrdquoMathemat-ica Slovaca vol 55 no 2 pp 169ndash181 2005

[2] H A Salem A M El-Sayed and O L Moustafa ldquoA note onthe fractional calculus in Banach spacesrdquo Studia ScientiarumMathematicarum Hungarica vol 42 no 2 pp 115ndash130 2005

[3] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoMulti-term fractional differential equations in a nonreflexiveBanach spacerdquo Advances in Difference Equations 2013302 18pages 2013

[4] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoFractional calculus and fractional differential equations innonreflexive Banach spacesrdquo Communications in NonlinearScience andNumerical Simulation vol 20 no 1 pp 59ndash73 2015

[5] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoNonlinear fractional differential equations in nonreflexiveBanach spaces and fractional calculusrdquo Advances in DifferenceEquations 2015112 18 pages 2015

[6] R P Agarwal V Lupulescu D OrsquoRegan and G RahmanldquoWeak solutions for fractional differential equations in nonre-flexive Banach spaces via Riemann-Pettis integralsrdquo Mathema-tische Nachrichten vol 289 no 4 pp 395ndash409 2016

[7] H A Salem ldquoOn the fractional order m-point boundaryvalue problem in reflexive Banach spaces and weak topologiesrdquoJournal of Computational andAppliedMathematics vol 224 no2 pp 565ndash572 2009

[8] H A Salem ldquoOn the fractional calculus in abstract spaces andtheir applications to the Dirichlet-type problem of fractionalorderrdquo Computers and Mathematics with Applications vol 59no 3 pp 1278ndash1293 2010

[9] HAH Salem andMCichon ldquoOn solutions of fractional orderboundary value problems with integral boundary conditions inBanach spacesrdquo journal of function spaces and applications vol2013 Article ID 428094 2013

[10] J Diestel and J J Uhl Jr Vector Measures vol 15 ofMathemat-ical Surveys American Mathematical Society Providence RIUSA 1977

[11] R H Martin and J J Uhl Jr Nonlinear Operators AndDifferential Equations in Banach Spaces John Wiley amp SonsNew York NY USA 1976

[12] B J Pettis ldquoOn integration in vector spacesrdquo Transactions of theAmericanMathematical Society vol 44 no 2 pp 277ndash304 1938

[13] S J Dilworth and M Girardi ldquoNowhere weak differentiabilityof the Pettis integralrdquo Quaestiones Mathematicae vol 18 no 4pp 365ndash380 1995

[14] G A Edgar ldquoMeasurability in a Banach spacerdquo Indiana Univer-sity Mathematics Journal vol 26 no 4 pp 663ndash677 1977

[15] G A Edgar ldquoMeasurability in a Banach space IIrdquo IndianaUniversityMathematics Journal vol 28 no 4 pp 559ndash579 1979

[16] J L Gamez and J Mendoza ldquoOn denjoy-dunford and denjoy-pettis integralsrdquo StudiaMathematica vol 130 no 2 pp 115ndash1331998

[17] R AGordon ldquoTheDenjoy extension of the Bochner Pettis andDUNford integralsrdquo Studia Mathematica vol 92 no 1 pp 73ndash91 1989

[18] R F Geitz ldquoPettis integrationrdquo Proceedings of the AmericanMathematical Society vol 82 no 1 pp 81ndash86 1981

[19] R F Geitz ldquoGeometry and the Pettis integralrdquo Transactions ofthe AmericanMathematical Society vol 269 no 2 pp 535ndash5481982

[20] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations New York NYUSA Elsevier 2006

[21] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Yverdon Switzerland 1993

[22] M Mendel and A Naor ldquoMetric cotyperdquo Annals of Mathemat-ics Second Series vol 168 no 1 pp 247ndash298 2008

[23] B Ross S G Samko and E R Love ldquoFunctions that have nofirst order derivative might have fractional derivatives of allorders less than onerdquo Real Analysis Exchange vol 20 no 1 pp140ndash157 199495

[24] E Hille and R S Phillips Functional analysis and semi-groupsAmerican Mathematical Society Colloquium Publications vol31 American Mathematical Society Providence RI USA 1957

[25] S Schwabik and Y Guoju Topics in Banach Space IntegrationWorld Scientific Singapore 2005

[26] DW Solomon ldquoOn differentiability of vector-valued functionsof a real variablerdquo Studia Mathematica vol 29 pp 1ndash4 1967

[27] K Naralenkov ldquoOn Denjoy type extensions of the Pettisintegralrdquo Czechoslovak Mathematical Journal vol 60(135) no3 pp 737ndash750 2010

[28] D W Solomon Denjoy integration in abstract spaces Memorisof the American Mathematical Society No 85 AmericanMathematical Society Providence RI 1969

[29] A R Mitchell and C Smith ldquoAn existence theorem for weaksolutions of differential equations in Banach spacesrdquo inNonlin-ear Equations in Abstract Spaces V Lakshmikantham Ed pp387ndash404 1978

[30] F S De Blasi ldquoOn a property of the unit sphere in a Banachspacerdquo Bulletin Mathematiques De La Societe Des SciencesMathematiques De Roumanie vol 21 no 3 pp 259ndash262 1977

[31] A Szep ldquoExistence theorem for weak solutions of ordinarydifferential equations in reflexive Banach spacesrdquo Studia Scien-tiarum Mathematicarum Hungarica vol 6 pp 197ndash203 1971

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Page 9: On the Theory of Fractional Calculus in the Pettis-Function Spaces · 2019. 7. 30. · JournalofFunctionSpaces Further,wedenethespace H (),˙≥0by H () fl ˛ ∈ ˚, ˜: isHolderainoforder

Journal of Function Spaces 9

we have1003817100381710038171003817I120572119909 (119905) minusI

1205721119909 (120591)10038171003817100381710038171198880

= 1198990Γ (1 + 120572)10038161003816100381610038161003816100381610038161003816(119905 minus

1119899 + 1)120572 minus (120591 minus 1119899 + 1)

12057210038161003816100381610038161003816100381610038161003816le 1198990Γ (1 + 120572)

1003816100381610038161003816100381610038161003816(119905 minus 1119899 + 1) minus (120591 minus 1119899 + 1)1003816100381610038161003816100381610038161003816120572

le 1198990Γ (1 + 120572) |119905 minus 120591|120572 le 1Γ (1 + 120572) |119905 minus 120591|120572radic|119905 minus 120591|= |119905 minus 120591|120572minus12Γ (1 + 120572)

(52)

Thus the RFPI of 119909 is norm continuous on [0 1] Preciselysince 0 lt 120572 minus 12 lt 120572 minus 1119901 lt 1 for any 119901 le 2 thenI120572119909 isin H120572minus12(119864) sub H120572minus1119901(119864) This is precisely what onewould expect from Lemma 14 (part (c))

Analogously an explicit calculation reveals that I120572119909 liesin 119862([0 1] 119871infin[0 1]) where 119909 [0 1] rarr 119871infin[0 1] defined byExample 10

We now consider additional mapping properties of theoperator I120572 Precisely we will show that the RFPI enjoysthe following commutative property which is folklore in case119864 = R However the proof is completely similar to that of [8]Lemma 35

Lemma 17 Let 120572 120573 gt 0 and 119901 gt max1 1120572 1120573 Then

I120572I120573119909 = I

120573I120572119909 = I

120572+120573119909 (53)

holds for every 119909 isin H1199010 (119864) If 119864 is reflexive this is also true for

every 119901 ge 1Proof Since 119901 gt max1 1120572 1120573 it follows that 119901 gtmax1 1(120572+120573) So by Lemmas 8 and 14 themappingsI120572119909I120573119909 andI120572+120573119909 belong toHinfin

0 (119864) for every 119909 isin H1199010 (119864) Now

repeating the same process as in ([8] Lemma 35) implies theclaim

When 119864 is reflexive the result follows as a direct applica-tion of Lemma 14

3 Fractional Derivatives ofVector-Valued Functions

After the notation of the fractional integrals of vector-valued functions the fractional derivatives become a naturalrequirement Before we come to the definitions and a detailedstudy of the mathematical properties of fractional differentialoperators we recall the following

Definition 18 Consider the vector-valued function 119909 119868 rarr119864(1) Let 120593119909 be differentiable on 119868 for every 120593 isin 119864lowast The

function 119909 is said to be weakly differentiable on 119868 if

there exists 119910 119868 rarr 119864 such that for every 120593 isin 119864lowast wehave

119889120593119909 (119905)119889119905 = 120593119910 (119905) for every 119905 isin 119868 (54)

The function 119910 is called the weak derivative of thefunction 119909

(2) Let 120593119909 be differentiable ae on 119868 for every 120593 isin 119864lowast(the null set may vary with 120593 isin 119864lowast) The function 119909is said to be pseudo differentiable on 119868 if there existsa function 119910 119868 rarr 119864 such that for every 120593 isin 119864lowast thereexists a null set119873(120593) sub 119868 such that

(120593119909 (119905))1015840 = 120593119910 (119905) for every 119905 isin 119868119873 (120593) (55)

In this case the function 119910 is called the pseudoderivative of 119909If 119909 is pseudo differentiable on 119868 and the null setinvariant for every 120593 isin 119864lowast then 119909 is ae weaklydifferentiable on 119868

Clearly if 119909 is ae weakly differentiable on 119868 then 120593(119909)is ae differentiable on 119868 The converse holds in a weaklysequentially complete space (see [25] Theorem 733)

For more details of the derivatives of vector-valuedfunctions we refer to [10 12 26]

The following results play a major role in our analysis

Proposition 19 (see [27] Theorem 51) The function 119910 119868 rarr119864 is an indefinite Pettis integrable if and only if 119910 is weaklyabsolutely continuous on 119868 and have a pseudo derivative on 119868In this case 119910 is an indefinite Pettis integral of any of its pseudoderivatives

Now we are in the position to define the fractional-typederivatives of vector-valued functions

Definition 20 Let 119909 119868 rarr 119864 For the positive integer119898 suchthat 120572 isin (119898 minus 1119898) 119898 isin N0 fl 0 1 2 we define theCaputo fractional-pseudo (weak) derivative ldquoshortly CFPD(CFWD)rdquo of 119909 of order 120572 by

119889120572119889119905120572 119909 (119905) fl I119898minus120572119863119898119909 (119905) (56)

where the sign ldquo119863rdquo denotes the pseudo (or weak) differentialoperator We use the notation 119889120572119901119889119905120572 and 119889120572120596119889119905120572 to charac-terize the Caputo fractional-pseudo derivatives and Caputofractional weak derivatives respectively

It is well known that although the weak derivative of aweakly differentiable function is uniquely determined thepseudo derivative of the pseudo differentiable function is notunique Also although any two pseudo derivatives 119910 119911 of afunction 119909 119868 rarr 119864 need not be ae equal (see [12 Example91] and [26 p 2]) the functions 119910 119911 are weakly equivalenton 119868 (that is 120593119910 = 120593119911 holds ae for every 120593 isin 119864lowast) The

10 Journal of Function Spaces

next lemma provides a useful characterization property of theCFPD Really it can be easily seen that the CFPD of a Caputofractional-pseudo differentiable function does not depend onthe choice of a pseudo derivative of the function

Lemma 21 Let 119909 119868 rarr 119864 be pseudo differentiable functionwhere pseudo derivatives lie in H

1199010 (119864) If (119889120572119901119889119905120572)119909 exists on119868 then the CFPD of 119909 depends on the choice of the pseudo

derivatives of 119909Proof Let 119910 119911 isin H

1199010 (119864) be two pseudo derivatives of the

pseudo differentiable function 119909 Since 119909 and 119910 are weaklyequivalent on 119868 then for every 120593 isin 119864lowast there exists a nullset N which depends on 120593 isin 119864lowast such that (119905 minus 119904)minus120572120593119910(119904) =(119905minus119904)minus120572120593119911(119904) for every 119904 isin [0 119905]N Consequently for almostevery 119905 isin 119868

120593 (I1minus120572119910 (119905)) = I1minus120572120593119910 (119905) = I

1minus120572120593119911 (119905)= 120593 (I1minus120572119911 (119905)) (57)

holds for every 120593 isin 119864lowast Thus

119889120572119901119889119905120572 119909 = I1minus120572119863119901119909 = I

1minus120572119910 = I1minus120572119911 on 119868 (58)

which is what we wished to show

We consider the following examples

Example 22 Define 119909 [0 1] rarr ℓ2 by119909 (119905) fl ln(1 + 119905119899) 119899 isin N (59)

We claim that the CFWD of 119909 exists on [0 1] To see this welet 120593 isin (ℓ2)lowast = ℓ2 Then there corresponds to 120593 a sequence120582119899 isin ℓ2 such that

120593119909 (119905) = sum119899

120582119899 ln(1 + 119905119899) (60)

Since the series in (60) converges uniformly on [0 1] theformal differentiation of 120593119909 yields

(120593119909 (119905))1015840 = sum119899

120582119899119899 + 119905 = 120593 ( 1119905 + 119899) (61)

It is not hard to justify the differentiation by noting that theseries in the right hand side of (61) is uniformly and absolutelyconvergent on [0 1] Consequently 120593119909 is differentiable on[0 1] for 120593 isin (ℓ2)lowast meaning that 119909 is weakly differentiable on[0 1] (because ℓ2 is weakly complete (cf [25] Theorem 733)and 119863120596119909(119905) = 1(119905 + 119899) Owing to Lemma 25 the CFWDof any order 120572 isin (0 1) of 119909 exists on [0 1] To calculate theCFWD of 119909 on [0 1] we observe in view of Example 11 that

119889120572120596119889119905120572 119909 (119905) = I1minus120572119863120596119909 (119905)

= 1199051minus120572119899Γ (2 minus 120572) 2F1 (1 1 2 minus 120572 minus119905119899 ) (62)

Example 23 Let119864 be an infinite dimensionBanach space thatfails cotype Define 119910 [0 1] rarr 119864 by 119910(119905) fl int119905

0119909(119904)119889119904

where 119909 [0 1] rarr 119864 given by formula (7) As cited in ([13]Corollary 4) 119910 is nowhere weakly differentiable on [0 1]Consequently the CFWD of any order 120572 isin [34 1) of 119910 losesits meaning on [0 1]Remark 24 As shown in Example 23 there is an infinitedimension Banach space 119864 and weakly absolutely continuousfunction 119910 119868 rarr 119864 which is nowhere weakly differentiable(hence the CFWD of 119910 does not exist) Also even when 119864 isseparable and 119910 is Lipschitz function the pseudo derivatives(hence the CFPD) of 119910 need not to exist [28]

However Definition 20 of the CFPD (CFWD) has thedisadvantage that it completely loses its meaning if thefunction119909 fails to be pseudo (weakly) differentiable Preciselythe CFPD (in particular the CFWD) of a function 119909 loses itsmeaning if 119909 is not weakly absolutely continuous

Thenext lemma gives sufficient conditions that ensure theexistence of the Caputo fractional derivatives of a function119909 isin H119901(119864)Lemma 25 Let 120572 isin (0 1) For the function 119909 119868 rarr 119864 thefollowing hold

(a) If 119909 has a pseudo derivative 119863119901119909 isin H1199010 (119864) where 119901 gt1(1 minus 120572) then the CFPD of 119909 of order 120572 exists on 119868

Moreover (119889120572119901119889119905120572)119909 isin 119862[119868 119864120596](b) If 119864 is weakly complete or contains no copy of 1198880 and if119909 has a weak derivative 119863120596119909 isin H119901(119864) where 119901 isin[1infin] then the CFWD of 119909 of order 120572 exists on 119868

Moreover (119889120572120596119889119905120572)119909 isin 119862[119868 119864120596]This holds for any 119909 isin H119901(119864) with 119901 isin [1infin] if 119864 is reflexive

In all cases 120593((119889120572119901119889119905120572)119909) = (119889120572119901119889119905120572)120593119909 (120593((119889120572120596119889119905120572)119909) =(119889120572120596119889119905120572)120593119909) holds for every 120593 isin 119864lowastProof Since the weak (pseudo) derivative of an ae weakly(pseudo) differentiable function is strongly (weakly) mea-surable [12 26 28] the proof is readily available in view ofthe definition of Caputo fractional derivatives and Lemma 8Moreover since 119863120596119909(119863119901119909) isin H

1199010 (119864) with 119901 gt 1(1 minus 120572)

it follows in view of Lemma 14 that I1minus120572119863120596119909(I1minus120572119863119901119909) isin119862[119868 119864120596] This completes the proof

Besides the Caputo fractional-pseudo (weak) derivativeswe define the Riemann-Liouville fractional-pseudo (weak)derivatives

Definition 26 Let 119909 119868 rarr 119864 For the positive integer 119898such that 120572 isin (119898 minus 1119898) 119898 isin N0 fl 0 1 2 we definethe Riemann-Liouville fractional-pseudo (weak) derivativeldquoshortly RFPD (RFWD)rdquo of 119909 of order 120572 by

D120572119909 (119905) fl 119863119898

I119898minus120572119909 (119905) (63)

where 119863 is defined as in Definition 20 We use the notationD120572

119901 (D120572120596) to characterize the Riemann-Liouville fractional-

pseudo (weak) derivatives

Journal of Function Spaces 11

Clearly in infinite dimension Banach spaces the weakabsolute continuity of I119898minus120572119909 is necessarily (but not suf-ficient) condition for the existence of RFPD (in particularRFWD) of 119909Lemma 27 Let 0 lt 120572 lt 1 For any 119909 isin H

1199010 (119864) with 119901 gt

max1120572 1(1 minus 120572) we haveD

120572119901I

120572119909 = 119909 ae (64)

If 119864 is reflexive this is also true for every 119901 ge 1Proof Our assumption 119909 isin H

1199010 (119864)with 119901 gt max1120572 1(1minus120572) yields in view of Lemma 17 that

D120572119901I

120572119909 = 119863119901I1minus120572

I120572119909 = 119863119901I

1119909 (65)

The claim now follows immediately since (cf Proposition 19)the indefinite integral of Pettis integrable function is weaklyabsolutely continuous and it is pseudo differentiable withrespect to the right endpoint of the integration interval andits pseudo derivative equals the integrand at that point

Remark 28 When we replace D120572119901 by D120572

120596 then Lemma 27is no longer necessarily true for arbitrary 119909 isin H1

0(119864) evenwhen 119864 is reflexive evidently in [13 remark below Corollary4] the existence of a reflexive Banach space 119864 and a strongmeasurable Pettis integrable function 119909 119868 rarr 119864 was provedsuch that 119909 has nowhere weakly differentiable integral In thiscase 119863120596I

1119909 lost its meaning This gives a reason to believethat (65) (hence (64)) withD120572

120596119909 could not happen

However we have the following result

Lemma 29 Let 0 lt 120572 lt 1 For any 119909 isin 119862[119868 119864120596] we haveD

120572120596I

120572119909 = 119909 on 119868 (66)

Proof Our assumption 119909 isin 119862[119868 119864120596] sub Hinfin0 (119864) (see the proof

of Corollary 15) yields in view of Lemma 17 that

D120572120596I

120572119909 = 119863120596I1minus120572

I120572119909 = 119863120596I

1119909 (67)

The claim now follows immediately since the indefiniteintegral of weakly contentious function is weakly absolutelycontinuous and it is weakly differentiable with respect tothe right endpoint of the integration interval and its weakderivative equals the integrand at that point

The following lemma is folklore in case 119864 = R but to seethat it also holds in the vector-valued case we provide a proof

Lemma 30 Let 0 lt 120572 lt 1 and 119901 gt max1120572 1(1minus120572) If thefunction 119909 119868 rarr 119864 is weakly absolutely continuous on 119868 andpasses a pseudo derivative inH

1199010 (119864) then119889120572119901119909119889119905120572 (119905) = D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (68)

In particular if 119909 passes a weak derivative in 119862[119868 119864120596] then119889120572120596119909119889119905120572 (119905) = D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (69)

If 119864 is reflexive this is also true for every 119901 ge 1

Proof We observe that under the assumption imposed on119863119901119909 together with Proposition 3 the weakly absolutelycontinuity of 119909 is equivalent to

119909 (119905) = 119909 (0) + int119905

0119863119901119909 (119904) 119889119904 (70)

Hence owing to Lemma 27 it follows that

119889120572119901119909119889119905120572 = I1minus120572119863119901119909 = (D120572

119901I120572)I1minus120572119863119901119909

= D120572119901 (I120572

I1minus120572)119863119901119909 = (119863119901I

1minus120572)I1119863119901119909= 119863119901I

1minus120572 (119909 (119905) minus 119909 (0))= D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0)

(71)

which is what we wished to showThe proof of (69) is very similar to that in (68) therefore

we omit the details

Remark 31 As we remark above the definition of the CFPDof a function 119909 loses completely its meaning if 119909 is notweakly absolutely continuous For this reason we are able touse Lemma 30 to define the Caputo fractional derivative ingeneral that is we put

119889120572119901119909119889119905120572 (119905) fl D120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) 0 lt 120572 lt 1 (72)

Similarly we define

119889120572120596119909119889119905120572 (119905) fl D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (73)

However Lemma 30 claims that for the weakly absolutelycontinuous functions having Pettis integrable pseudo [weak]derivatives definitions (72) and (73) of the Caputo fractional-pseudo [weak] derivatives coincide with Definition 20

4 An Application

Let 119891 119868 times 119864 rarr 119864 be given function Consider the boundaryvalue problem of the fractional type

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin [0 1] 120582 isin R

119910 (0) + 119887119910 (1) = ℎ(74)

with certain constants 119887 ℎ isin R 119887 = minus1 To obtain theintegral equation modeled off the problem (74) we let 119910 bea weakly continuous solution to problem (74) then formallywe have

I120572 119889120572120596119889119905120572119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (75)

12 Journal of Function Spaces

that is

I120572I1minus120572119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) 997904rArrI1119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (76)

This reads (cf [29])

119910 (119905) = 119910 (0) + 120582I120572119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin 119868 120582 isin R (77)

with some (presently unknown) quantity 119910(0)Now we solve (77) for 119910(0) by 119910(0) + 119887119910(1) = ℎ and it

follows that

(1 + 119887) 119910 (0) + 120582119887Γ (120572) int1

0(1 minus 119904)120572minus1 119891 (119904 119910 (119904)) 119889119904 = ℎ (78)

Thus

119910 (0) = ℎ1 + 119887 minus 1205821198871 + 119887119868120572119891 (1 119910 (1)) (79)

Assume that the function 119891 is weakly-weakly continuousfunction such that

(1) for any 119903 gt 0 there is a constant 119872 gt 0 such that119891(119905 119910) le 119872 for all 119905 isin 119868 and 119910 le 119903(2) there exists a nondecreasing continuous function 120595 [0infin) rarr [0infin) 120595(0) = 0 and 120595(119905) lt 119905 for all 119905 gt 0

such that 120573(119891(119879 times 119860)) le 120595(120573(119860)) for every boundedset119860 sub 119864 where 120573 stands for De Blasirsquos weakmeasureof noncompactness (see [30])

Occasionally if 119891 119868 times 119864 rarr 119864 is weakly-weakly continuousand 119864 is reflexive then the assumptions (1) and (2) areautomatically satisfied (see eg [31])

Theorem 32 (see [6] Theorem 33) Let 120572 isin (0 1) and |120582| ltΓ(1 + 120572) If 119891 119868 times 119864 rarr 119864 is weakly-weakly continuous andsatisfies the assumptions (1) and (2) then the integral equation(77) has a weakly continuous solution 119910 defined on [0 1]Proof We omit the proof since it is almost identical to thatin the proof in ([6] Theorem 33) with (small) necessarychanges

In the following example we assume that 119910 isin 119862[119868 119864120596]solves (77) and we will show that not only do we have that 119910no longer necessarily solves (74) (when the Caputo fractionalweak derivative is taken in the sense of Definition 20)but even worse it could happen that the problem (74) isldquomeaninglessrdquo on 119868Example 33 Let 120572 isin (0 14] Let 119864 be an indefinitedimensional reflexive Banach space that fails cotype Definethe weakly-weakly continuous function 119891 [0 1] times 119864 rarr 119864by 119891 fl I1minus120572119909 where 119909 119868 rarr 119864 is defined by formula (7)Obviously (cf Example 6 in view of Remark 7)119891 satisfies theassumptions of Theorem 32

Now consider the integral equation (77) with119891 = I1minus120572119909Namely we consider the integral equation

119910 (119905) = 119910 (0) + 120582I120572I1minus120572119909 (119905)

120572 isin (0 14] 119905 isin 119868 120582 isin R (80)

Obviously the solution 119910 is weakly continuous on [0 1] (thisis an immediate consequence of Remark 7 and Lemma 17)

Since 120593119909 isin 1198711[0 1] holds for every 120593 isin 119864lowast it follows bythe commutative property of the fractional integral operatorscalculus over the space of Lebesgue integrable functions (see[21] Section 23) that

120593 (I120572I1minus120572119909 (119905)) = I

120572120593 (I1minus120572119909 (119905))= I

120572I1minus120572120593 (119909 (119905)) = I

1120593 (119909 (119905))= 120593 (I1119909 (119905))

(81)

that is

120593 (I120572I1minus120572119909 (119905) minusI

1119909 (119905)) = 0 for every 120593 isin 119864lowast (82)

Hence I120572I1minus120572119909 = I1119909 Consequently problem (80) be-comes

119910 (119905) = 119910 (0) + 120582I1119909 (119905) 120572 isin (0 14] 119905 isin 119868 (83)

As showed in ([13] Corollary 4) the indefinite integral 119905 rarrint1199050119909(119904)119889119904 is nowhere weakly differentiable on [0 1] Thus 119910

is nowhere weakly differentiable on [0 1] hence the CFWDof 119910 (hence the boundary value problem) completely loses itsmeaning if the CFWD is taken in the sense of Definition 20This is what we wished to show

Now we are in the position to state and prove thefollowing existence theorem

Theorem 34 If 119891 119868 times 119864 rarr 119864 is a function such that allconditions fromTheorem 32 hold then problem (74) (where theCaputo fractionalweak derivative is taken in the sense (73)) hasa weak solution on [0 1]Proof Let 119910 isin 119862[119868 119864120596] be a solution to (77) Equip 119864 andI times E with weak topology and note that 119905 997891rarr (119905 119910(119905)) iscontinuous as a mapping from 119868 into 119868 times 119864 Since 119891 119868 times 119864 rarr119864 is weakly-weakly continuous on [0 1] then 119891(sdot 119910(sdot)) is acomposition of this mapping with 119891 and thus 119891(sdot 119910(sdot)) isweakly continuous on [0 1]

Now operating by the operatorD120572120596 on both sides of (77)

it follows by Lemma 29 that

D120572120596119910 = D

120572120596119910 (0) + 120582D120572

120596I120572119891 = 119905minus120572Γ (1 minus 120572)119910 (0) + 120582119891 (84)

Now insert definition (73) of the Caputo fractionalderivative and we get

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) (85)

Journal of Function Spaces 13

With some further efforts one can get the boundary condi-tion 119910(0) + 119887119910(1) = ℎ

Therefore 119910 satisfies problem (74) which is what wewished to show

Data Availability

No data were used to support this study

Conflicts of Interest

The author declares no conflicts of interest

References

[1] H A Salem and A M El-Sayed ldquoWeak solution for fractionalorder integral equations in reflexive Banach spacesrdquoMathemat-ica Slovaca vol 55 no 2 pp 169ndash181 2005

[2] H A Salem A M El-Sayed and O L Moustafa ldquoA note onthe fractional calculus in Banach spacesrdquo Studia ScientiarumMathematicarum Hungarica vol 42 no 2 pp 115ndash130 2005

[3] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoMulti-term fractional differential equations in a nonreflexiveBanach spacerdquo Advances in Difference Equations 2013302 18pages 2013

[4] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoFractional calculus and fractional differential equations innonreflexive Banach spacesrdquo Communications in NonlinearScience andNumerical Simulation vol 20 no 1 pp 59ndash73 2015

[5] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoNonlinear fractional differential equations in nonreflexiveBanach spaces and fractional calculusrdquo Advances in DifferenceEquations 2015112 18 pages 2015

[6] R P Agarwal V Lupulescu D OrsquoRegan and G RahmanldquoWeak solutions for fractional differential equations in nonre-flexive Banach spaces via Riemann-Pettis integralsrdquo Mathema-tische Nachrichten vol 289 no 4 pp 395ndash409 2016

[7] H A Salem ldquoOn the fractional order m-point boundaryvalue problem in reflexive Banach spaces and weak topologiesrdquoJournal of Computational andAppliedMathematics vol 224 no2 pp 565ndash572 2009

[8] H A Salem ldquoOn the fractional calculus in abstract spaces andtheir applications to the Dirichlet-type problem of fractionalorderrdquo Computers and Mathematics with Applications vol 59no 3 pp 1278ndash1293 2010

[9] HAH Salem andMCichon ldquoOn solutions of fractional orderboundary value problems with integral boundary conditions inBanach spacesrdquo journal of function spaces and applications vol2013 Article ID 428094 2013

[10] J Diestel and J J Uhl Jr Vector Measures vol 15 ofMathemat-ical Surveys American Mathematical Society Providence RIUSA 1977

[11] R H Martin and J J Uhl Jr Nonlinear Operators AndDifferential Equations in Banach Spaces John Wiley amp SonsNew York NY USA 1976

[12] B J Pettis ldquoOn integration in vector spacesrdquo Transactions of theAmericanMathematical Society vol 44 no 2 pp 277ndash304 1938

[13] S J Dilworth and M Girardi ldquoNowhere weak differentiabilityof the Pettis integralrdquo Quaestiones Mathematicae vol 18 no 4pp 365ndash380 1995

[14] G A Edgar ldquoMeasurability in a Banach spacerdquo Indiana Univer-sity Mathematics Journal vol 26 no 4 pp 663ndash677 1977

[15] G A Edgar ldquoMeasurability in a Banach space IIrdquo IndianaUniversityMathematics Journal vol 28 no 4 pp 559ndash579 1979

[16] J L Gamez and J Mendoza ldquoOn denjoy-dunford and denjoy-pettis integralsrdquo StudiaMathematica vol 130 no 2 pp 115ndash1331998

[17] R AGordon ldquoTheDenjoy extension of the Bochner Pettis andDUNford integralsrdquo Studia Mathematica vol 92 no 1 pp 73ndash91 1989

[18] R F Geitz ldquoPettis integrationrdquo Proceedings of the AmericanMathematical Society vol 82 no 1 pp 81ndash86 1981

[19] R F Geitz ldquoGeometry and the Pettis integralrdquo Transactions ofthe AmericanMathematical Society vol 269 no 2 pp 535ndash5481982

[20] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations New York NYUSA Elsevier 2006

[21] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Yverdon Switzerland 1993

[22] M Mendel and A Naor ldquoMetric cotyperdquo Annals of Mathemat-ics Second Series vol 168 no 1 pp 247ndash298 2008

[23] B Ross S G Samko and E R Love ldquoFunctions that have nofirst order derivative might have fractional derivatives of allorders less than onerdquo Real Analysis Exchange vol 20 no 1 pp140ndash157 199495

[24] E Hille and R S Phillips Functional analysis and semi-groupsAmerican Mathematical Society Colloquium Publications vol31 American Mathematical Society Providence RI USA 1957

[25] S Schwabik and Y Guoju Topics in Banach Space IntegrationWorld Scientific Singapore 2005

[26] DW Solomon ldquoOn differentiability of vector-valued functionsof a real variablerdquo Studia Mathematica vol 29 pp 1ndash4 1967

[27] K Naralenkov ldquoOn Denjoy type extensions of the Pettisintegralrdquo Czechoslovak Mathematical Journal vol 60(135) no3 pp 737ndash750 2010

[28] D W Solomon Denjoy integration in abstract spaces Memorisof the American Mathematical Society No 85 AmericanMathematical Society Providence RI 1969

[29] A R Mitchell and C Smith ldquoAn existence theorem for weaksolutions of differential equations in Banach spacesrdquo inNonlin-ear Equations in Abstract Spaces V Lakshmikantham Ed pp387ndash404 1978

[30] F S De Blasi ldquoOn a property of the unit sphere in a Banachspacerdquo Bulletin Mathematiques De La Societe Des SciencesMathematiques De Roumanie vol 21 no 3 pp 259ndash262 1977

[31] A Szep ldquoExistence theorem for weak solutions of ordinarydifferential equations in reflexive Banach spacesrdquo Studia Scien-tiarum Mathematicarum Hungarica vol 6 pp 197ndash203 1971

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Page 10: On the Theory of Fractional Calculus in the Pettis-Function Spaces · 2019. 7. 30. · JournalofFunctionSpaces Further,wedenethespace H (),˙≥0by H () fl ˛ ∈ ˚, ˜: isHolderainoforder

10 Journal of Function Spaces

next lemma provides a useful characterization property of theCFPD Really it can be easily seen that the CFPD of a Caputofractional-pseudo differentiable function does not depend onthe choice of a pseudo derivative of the function

Lemma 21 Let 119909 119868 rarr 119864 be pseudo differentiable functionwhere pseudo derivatives lie in H

1199010 (119864) If (119889120572119901119889119905120572)119909 exists on119868 then the CFPD of 119909 depends on the choice of the pseudo

derivatives of 119909Proof Let 119910 119911 isin H

1199010 (119864) be two pseudo derivatives of the

pseudo differentiable function 119909 Since 119909 and 119910 are weaklyequivalent on 119868 then for every 120593 isin 119864lowast there exists a nullset N which depends on 120593 isin 119864lowast such that (119905 minus 119904)minus120572120593119910(119904) =(119905minus119904)minus120572120593119911(119904) for every 119904 isin [0 119905]N Consequently for almostevery 119905 isin 119868

120593 (I1minus120572119910 (119905)) = I1minus120572120593119910 (119905) = I

1minus120572120593119911 (119905)= 120593 (I1minus120572119911 (119905)) (57)

holds for every 120593 isin 119864lowast Thus

119889120572119901119889119905120572 119909 = I1minus120572119863119901119909 = I

1minus120572119910 = I1minus120572119911 on 119868 (58)

which is what we wished to show

We consider the following examples

Example 22 Define 119909 [0 1] rarr ℓ2 by119909 (119905) fl ln(1 + 119905119899) 119899 isin N (59)

We claim that the CFWD of 119909 exists on [0 1] To see this welet 120593 isin (ℓ2)lowast = ℓ2 Then there corresponds to 120593 a sequence120582119899 isin ℓ2 such that

120593119909 (119905) = sum119899

120582119899 ln(1 + 119905119899) (60)

Since the series in (60) converges uniformly on [0 1] theformal differentiation of 120593119909 yields

(120593119909 (119905))1015840 = sum119899

120582119899119899 + 119905 = 120593 ( 1119905 + 119899) (61)

It is not hard to justify the differentiation by noting that theseries in the right hand side of (61) is uniformly and absolutelyconvergent on [0 1] Consequently 120593119909 is differentiable on[0 1] for 120593 isin (ℓ2)lowast meaning that 119909 is weakly differentiable on[0 1] (because ℓ2 is weakly complete (cf [25] Theorem 733)and 119863120596119909(119905) = 1(119905 + 119899) Owing to Lemma 25 the CFWDof any order 120572 isin (0 1) of 119909 exists on [0 1] To calculate theCFWD of 119909 on [0 1] we observe in view of Example 11 that

119889120572120596119889119905120572 119909 (119905) = I1minus120572119863120596119909 (119905)

= 1199051minus120572119899Γ (2 minus 120572) 2F1 (1 1 2 minus 120572 minus119905119899 ) (62)

Example 23 Let119864 be an infinite dimensionBanach space thatfails cotype Define 119910 [0 1] rarr 119864 by 119910(119905) fl int119905

0119909(119904)119889119904

where 119909 [0 1] rarr 119864 given by formula (7) As cited in ([13]Corollary 4) 119910 is nowhere weakly differentiable on [0 1]Consequently the CFWD of any order 120572 isin [34 1) of 119910 losesits meaning on [0 1]Remark 24 As shown in Example 23 there is an infinitedimension Banach space 119864 and weakly absolutely continuousfunction 119910 119868 rarr 119864 which is nowhere weakly differentiable(hence the CFWD of 119910 does not exist) Also even when 119864 isseparable and 119910 is Lipschitz function the pseudo derivatives(hence the CFPD) of 119910 need not to exist [28]

However Definition 20 of the CFPD (CFWD) has thedisadvantage that it completely loses its meaning if thefunction119909 fails to be pseudo (weakly) differentiable Preciselythe CFPD (in particular the CFWD) of a function 119909 loses itsmeaning if 119909 is not weakly absolutely continuous

Thenext lemma gives sufficient conditions that ensure theexistence of the Caputo fractional derivatives of a function119909 isin H119901(119864)Lemma 25 Let 120572 isin (0 1) For the function 119909 119868 rarr 119864 thefollowing hold

(a) If 119909 has a pseudo derivative 119863119901119909 isin H1199010 (119864) where 119901 gt1(1 minus 120572) then the CFPD of 119909 of order 120572 exists on 119868

Moreover (119889120572119901119889119905120572)119909 isin 119862[119868 119864120596](b) If 119864 is weakly complete or contains no copy of 1198880 and if119909 has a weak derivative 119863120596119909 isin H119901(119864) where 119901 isin[1infin] then the CFWD of 119909 of order 120572 exists on 119868

Moreover (119889120572120596119889119905120572)119909 isin 119862[119868 119864120596]This holds for any 119909 isin H119901(119864) with 119901 isin [1infin] if 119864 is reflexive

In all cases 120593((119889120572119901119889119905120572)119909) = (119889120572119901119889119905120572)120593119909 (120593((119889120572120596119889119905120572)119909) =(119889120572120596119889119905120572)120593119909) holds for every 120593 isin 119864lowastProof Since the weak (pseudo) derivative of an ae weakly(pseudo) differentiable function is strongly (weakly) mea-surable [12 26 28] the proof is readily available in view ofthe definition of Caputo fractional derivatives and Lemma 8Moreover since 119863120596119909(119863119901119909) isin H

1199010 (119864) with 119901 gt 1(1 minus 120572)

it follows in view of Lemma 14 that I1minus120572119863120596119909(I1minus120572119863119901119909) isin119862[119868 119864120596] This completes the proof

Besides the Caputo fractional-pseudo (weak) derivativeswe define the Riemann-Liouville fractional-pseudo (weak)derivatives

Definition 26 Let 119909 119868 rarr 119864 For the positive integer 119898such that 120572 isin (119898 minus 1119898) 119898 isin N0 fl 0 1 2 we definethe Riemann-Liouville fractional-pseudo (weak) derivativeldquoshortly RFPD (RFWD)rdquo of 119909 of order 120572 by

D120572119909 (119905) fl 119863119898

I119898minus120572119909 (119905) (63)

where 119863 is defined as in Definition 20 We use the notationD120572

119901 (D120572120596) to characterize the Riemann-Liouville fractional-

pseudo (weak) derivatives

Journal of Function Spaces 11

Clearly in infinite dimension Banach spaces the weakabsolute continuity of I119898minus120572119909 is necessarily (but not suf-ficient) condition for the existence of RFPD (in particularRFWD) of 119909Lemma 27 Let 0 lt 120572 lt 1 For any 119909 isin H

1199010 (119864) with 119901 gt

max1120572 1(1 minus 120572) we haveD

120572119901I

120572119909 = 119909 ae (64)

If 119864 is reflexive this is also true for every 119901 ge 1Proof Our assumption 119909 isin H

1199010 (119864)with 119901 gt max1120572 1(1minus120572) yields in view of Lemma 17 that

D120572119901I

120572119909 = 119863119901I1minus120572

I120572119909 = 119863119901I

1119909 (65)

The claim now follows immediately since (cf Proposition 19)the indefinite integral of Pettis integrable function is weaklyabsolutely continuous and it is pseudo differentiable withrespect to the right endpoint of the integration interval andits pseudo derivative equals the integrand at that point

Remark 28 When we replace D120572119901 by D120572

120596 then Lemma 27is no longer necessarily true for arbitrary 119909 isin H1

0(119864) evenwhen 119864 is reflexive evidently in [13 remark below Corollary4] the existence of a reflexive Banach space 119864 and a strongmeasurable Pettis integrable function 119909 119868 rarr 119864 was provedsuch that 119909 has nowhere weakly differentiable integral In thiscase 119863120596I

1119909 lost its meaning This gives a reason to believethat (65) (hence (64)) withD120572

120596119909 could not happen

However we have the following result

Lemma 29 Let 0 lt 120572 lt 1 For any 119909 isin 119862[119868 119864120596] we haveD

120572120596I

120572119909 = 119909 on 119868 (66)

Proof Our assumption 119909 isin 119862[119868 119864120596] sub Hinfin0 (119864) (see the proof

of Corollary 15) yields in view of Lemma 17 that

D120572120596I

120572119909 = 119863120596I1minus120572

I120572119909 = 119863120596I

1119909 (67)

The claim now follows immediately since the indefiniteintegral of weakly contentious function is weakly absolutelycontinuous and it is weakly differentiable with respect tothe right endpoint of the integration interval and its weakderivative equals the integrand at that point

The following lemma is folklore in case 119864 = R but to seethat it also holds in the vector-valued case we provide a proof

Lemma 30 Let 0 lt 120572 lt 1 and 119901 gt max1120572 1(1minus120572) If thefunction 119909 119868 rarr 119864 is weakly absolutely continuous on 119868 andpasses a pseudo derivative inH

1199010 (119864) then119889120572119901119909119889119905120572 (119905) = D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (68)

In particular if 119909 passes a weak derivative in 119862[119868 119864120596] then119889120572120596119909119889119905120572 (119905) = D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (69)

If 119864 is reflexive this is also true for every 119901 ge 1

Proof We observe that under the assumption imposed on119863119901119909 together with Proposition 3 the weakly absolutelycontinuity of 119909 is equivalent to

119909 (119905) = 119909 (0) + int119905

0119863119901119909 (119904) 119889119904 (70)

Hence owing to Lemma 27 it follows that

119889120572119901119909119889119905120572 = I1minus120572119863119901119909 = (D120572

119901I120572)I1minus120572119863119901119909

= D120572119901 (I120572

I1minus120572)119863119901119909 = (119863119901I

1minus120572)I1119863119901119909= 119863119901I

1minus120572 (119909 (119905) minus 119909 (0))= D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0)

(71)

which is what we wished to showThe proof of (69) is very similar to that in (68) therefore

we omit the details

Remark 31 As we remark above the definition of the CFPDof a function 119909 loses completely its meaning if 119909 is notweakly absolutely continuous For this reason we are able touse Lemma 30 to define the Caputo fractional derivative ingeneral that is we put

119889120572119901119909119889119905120572 (119905) fl D120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) 0 lt 120572 lt 1 (72)

Similarly we define

119889120572120596119909119889119905120572 (119905) fl D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (73)

However Lemma 30 claims that for the weakly absolutelycontinuous functions having Pettis integrable pseudo [weak]derivatives definitions (72) and (73) of the Caputo fractional-pseudo [weak] derivatives coincide with Definition 20

4 An Application

Let 119891 119868 times 119864 rarr 119864 be given function Consider the boundaryvalue problem of the fractional type

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin [0 1] 120582 isin R

119910 (0) + 119887119910 (1) = ℎ(74)

with certain constants 119887 ℎ isin R 119887 = minus1 To obtain theintegral equation modeled off the problem (74) we let 119910 bea weakly continuous solution to problem (74) then formallywe have

I120572 119889120572120596119889119905120572119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (75)

12 Journal of Function Spaces

that is

I120572I1minus120572119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) 997904rArrI1119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (76)

This reads (cf [29])

119910 (119905) = 119910 (0) + 120582I120572119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin 119868 120582 isin R (77)

with some (presently unknown) quantity 119910(0)Now we solve (77) for 119910(0) by 119910(0) + 119887119910(1) = ℎ and it

follows that

(1 + 119887) 119910 (0) + 120582119887Γ (120572) int1

0(1 minus 119904)120572minus1 119891 (119904 119910 (119904)) 119889119904 = ℎ (78)

Thus

119910 (0) = ℎ1 + 119887 minus 1205821198871 + 119887119868120572119891 (1 119910 (1)) (79)

Assume that the function 119891 is weakly-weakly continuousfunction such that

(1) for any 119903 gt 0 there is a constant 119872 gt 0 such that119891(119905 119910) le 119872 for all 119905 isin 119868 and 119910 le 119903(2) there exists a nondecreasing continuous function 120595 [0infin) rarr [0infin) 120595(0) = 0 and 120595(119905) lt 119905 for all 119905 gt 0

such that 120573(119891(119879 times 119860)) le 120595(120573(119860)) for every boundedset119860 sub 119864 where 120573 stands for De Blasirsquos weakmeasureof noncompactness (see [30])

Occasionally if 119891 119868 times 119864 rarr 119864 is weakly-weakly continuousand 119864 is reflexive then the assumptions (1) and (2) areautomatically satisfied (see eg [31])

Theorem 32 (see [6] Theorem 33) Let 120572 isin (0 1) and |120582| ltΓ(1 + 120572) If 119891 119868 times 119864 rarr 119864 is weakly-weakly continuous andsatisfies the assumptions (1) and (2) then the integral equation(77) has a weakly continuous solution 119910 defined on [0 1]Proof We omit the proof since it is almost identical to thatin the proof in ([6] Theorem 33) with (small) necessarychanges

In the following example we assume that 119910 isin 119862[119868 119864120596]solves (77) and we will show that not only do we have that 119910no longer necessarily solves (74) (when the Caputo fractionalweak derivative is taken in the sense of Definition 20)but even worse it could happen that the problem (74) isldquomeaninglessrdquo on 119868Example 33 Let 120572 isin (0 14] Let 119864 be an indefinitedimensional reflexive Banach space that fails cotype Definethe weakly-weakly continuous function 119891 [0 1] times 119864 rarr 119864by 119891 fl I1minus120572119909 where 119909 119868 rarr 119864 is defined by formula (7)Obviously (cf Example 6 in view of Remark 7)119891 satisfies theassumptions of Theorem 32

Now consider the integral equation (77) with119891 = I1minus120572119909Namely we consider the integral equation

119910 (119905) = 119910 (0) + 120582I120572I1minus120572119909 (119905)

120572 isin (0 14] 119905 isin 119868 120582 isin R (80)

Obviously the solution 119910 is weakly continuous on [0 1] (thisis an immediate consequence of Remark 7 and Lemma 17)

Since 120593119909 isin 1198711[0 1] holds for every 120593 isin 119864lowast it follows bythe commutative property of the fractional integral operatorscalculus over the space of Lebesgue integrable functions (see[21] Section 23) that

120593 (I120572I1minus120572119909 (119905)) = I

120572120593 (I1minus120572119909 (119905))= I

120572I1minus120572120593 (119909 (119905)) = I

1120593 (119909 (119905))= 120593 (I1119909 (119905))

(81)

that is

120593 (I120572I1minus120572119909 (119905) minusI

1119909 (119905)) = 0 for every 120593 isin 119864lowast (82)

Hence I120572I1minus120572119909 = I1119909 Consequently problem (80) be-comes

119910 (119905) = 119910 (0) + 120582I1119909 (119905) 120572 isin (0 14] 119905 isin 119868 (83)

As showed in ([13] Corollary 4) the indefinite integral 119905 rarrint1199050119909(119904)119889119904 is nowhere weakly differentiable on [0 1] Thus 119910

is nowhere weakly differentiable on [0 1] hence the CFWDof 119910 (hence the boundary value problem) completely loses itsmeaning if the CFWD is taken in the sense of Definition 20This is what we wished to show

Now we are in the position to state and prove thefollowing existence theorem

Theorem 34 If 119891 119868 times 119864 rarr 119864 is a function such that allconditions fromTheorem 32 hold then problem (74) (where theCaputo fractionalweak derivative is taken in the sense (73)) hasa weak solution on [0 1]Proof Let 119910 isin 119862[119868 119864120596] be a solution to (77) Equip 119864 andI times E with weak topology and note that 119905 997891rarr (119905 119910(119905)) iscontinuous as a mapping from 119868 into 119868 times 119864 Since 119891 119868 times 119864 rarr119864 is weakly-weakly continuous on [0 1] then 119891(sdot 119910(sdot)) is acomposition of this mapping with 119891 and thus 119891(sdot 119910(sdot)) isweakly continuous on [0 1]

Now operating by the operatorD120572120596 on both sides of (77)

it follows by Lemma 29 that

D120572120596119910 = D

120572120596119910 (0) + 120582D120572

120596I120572119891 = 119905minus120572Γ (1 minus 120572)119910 (0) + 120582119891 (84)

Now insert definition (73) of the Caputo fractionalderivative and we get

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) (85)

Journal of Function Spaces 13

With some further efforts one can get the boundary condi-tion 119910(0) + 119887119910(1) = ℎ

Therefore 119910 satisfies problem (74) which is what wewished to show

Data Availability

No data were used to support this study

Conflicts of Interest

The author declares no conflicts of interest

References

[1] H A Salem and A M El-Sayed ldquoWeak solution for fractionalorder integral equations in reflexive Banach spacesrdquoMathemat-ica Slovaca vol 55 no 2 pp 169ndash181 2005

[2] H A Salem A M El-Sayed and O L Moustafa ldquoA note onthe fractional calculus in Banach spacesrdquo Studia ScientiarumMathematicarum Hungarica vol 42 no 2 pp 115ndash130 2005

[3] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoMulti-term fractional differential equations in a nonreflexiveBanach spacerdquo Advances in Difference Equations 2013302 18pages 2013

[4] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoFractional calculus and fractional differential equations innonreflexive Banach spacesrdquo Communications in NonlinearScience andNumerical Simulation vol 20 no 1 pp 59ndash73 2015

[5] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoNonlinear fractional differential equations in nonreflexiveBanach spaces and fractional calculusrdquo Advances in DifferenceEquations 2015112 18 pages 2015

[6] R P Agarwal V Lupulescu D OrsquoRegan and G RahmanldquoWeak solutions for fractional differential equations in nonre-flexive Banach spaces via Riemann-Pettis integralsrdquo Mathema-tische Nachrichten vol 289 no 4 pp 395ndash409 2016

[7] H A Salem ldquoOn the fractional order m-point boundaryvalue problem in reflexive Banach spaces and weak topologiesrdquoJournal of Computational andAppliedMathematics vol 224 no2 pp 565ndash572 2009

[8] H A Salem ldquoOn the fractional calculus in abstract spaces andtheir applications to the Dirichlet-type problem of fractionalorderrdquo Computers and Mathematics with Applications vol 59no 3 pp 1278ndash1293 2010

[9] HAH Salem andMCichon ldquoOn solutions of fractional orderboundary value problems with integral boundary conditions inBanach spacesrdquo journal of function spaces and applications vol2013 Article ID 428094 2013

[10] J Diestel and J J Uhl Jr Vector Measures vol 15 ofMathemat-ical Surveys American Mathematical Society Providence RIUSA 1977

[11] R H Martin and J J Uhl Jr Nonlinear Operators AndDifferential Equations in Banach Spaces John Wiley amp SonsNew York NY USA 1976

[12] B J Pettis ldquoOn integration in vector spacesrdquo Transactions of theAmericanMathematical Society vol 44 no 2 pp 277ndash304 1938

[13] S J Dilworth and M Girardi ldquoNowhere weak differentiabilityof the Pettis integralrdquo Quaestiones Mathematicae vol 18 no 4pp 365ndash380 1995

[14] G A Edgar ldquoMeasurability in a Banach spacerdquo Indiana Univer-sity Mathematics Journal vol 26 no 4 pp 663ndash677 1977

[15] G A Edgar ldquoMeasurability in a Banach space IIrdquo IndianaUniversityMathematics Journal vol 28 no 4 pp 559ndash579 1979

[16] J L Gamez and J Mendoza ldquoOn denjoy-dunford and denjoy-pettis integralsrdquo StudiaMathematica vol 130 no 2 pp 115ndash1331998

[17] R AGordon ldquoTheDenjoy extension of the Bochner Pettis andDUNford integralsrdquo Studia Mathematica vol 92 no 1 pp 73ndash91 1989

[18] R F Geitz ldquoPettis integrationrdquo Proceedings of the AmericanMathematical Society vol 82 no 1 pp 81ndash86 1981

[19] R F Geitz ldquoGeometry and the Pettis integralrdquo Transactions ofthe AmericanMathematical Society vol 269 no 2 pp 535ndash5481982

[20] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations New York NYUSA Elsevier 2006

[21] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Yverdon Switzerland 1993

[22] M Mendel and A Naor ldquoMetric cotyperdquo Annals of Mathemat-ics Second Series vol 168 no 1 pp 247ndash298 2008

[23] B Ross S G Samko and E R Love ldquoFunctions that have nofirst order derivative might have fractional derivatives of allorders less than onerdquo Real Analysis Exchange vol 20 no 1 pp140ndash157 199495

[24] E Hille and R S Phillips Functional analysis and semi-groupsAmerican Mathematical Society Colloquium Publications vol31 American Mathematical Society Providence RI USA 1957

[25] S Schwabik and Y Guoju Topics in Banach Space IntegrationWorld Scientific Singapore 2005

[26] DW Solomon ldquoOn differentiability of vector-valued functionsof a real variablerdquo Studia Mathematica vol 29 pp 1ndash4 1967

[27] K Naralenkov ldquoOn Denjoy type extensions of the Pettisintegralrdquo Czechoslovak Mathematical Journal vol 60(135) no3 pp 737ndash750 2010

[28] D W Solomon Denjoy integration in abstract spaces Memorisof the American Mathematical Society No 85 AmericanMathematical Society Providence RI 1969

[29] A R Mitchell and C Smith ldquoAn existence theorem for weaksolutions of differential equations in Banach spacesrdquo inNonlin-ear Equations in Abstract Spaces V Lakshmikantham Ed pp387ndash404 1978

[30] F S De Blasi ldquoOn a property of the unit sphere in a Banachspacerdquo Bulletin Mathematiques De La Societe Des SciencesMathematiques De Roumanie vol 21 no 3 pp 259ndash262 1977

[31] A Szep ldquoExistence theorem for weak solutions of ordinarydifferential equations in reflexive Banach spacesrdquo Studia Scien-tiarum Mathematicarum Hungarica vol 6 pp 197ndash203 1971

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Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

Page 11: On the Theory of Fractional Calculus in the Pettis-Function Spaces · 2019. 7. 30. · JournalofFunctionSpaces Further,wedenethespace H (),˙≥0by H () fl ˛ ∈ ˚, ˜: isHolderainoforder

Journal of Function Spaces 11

Clearly in infinite dimension Banach spaces the weakabsolute continuity of I119898minus120572119909 is necessarily (but not suf-ficient) condition for the existence of RFPD (in particularRFWD) of 119909Lemma 27 Let 0 lt 120572 lt 1 For any 119909 isin H

1199010 (119864) with 119901 gt

max1120572 1(1 minus 120572) we haveD

120572119901I

120572119909 = 119909 ae (64)

If 119864 is reflexive this is also true for every 119901 ge 1Proof Our assumption 119909 isin H

1199010 (119864)with 119901 gt max1120572 1(1minus120572) yields in view of Lemma 17 that

D120572119901I

120572119909 = 119863119901I1minus120572

I120572119909 = 119863119901I

1119909 (65)

The claim now follows immediately since (cf Proposition 19)the indefinite integral of Pettis integrable function is weaklyabsolutely continuous and it is pseudo differentiable withrespect to the right endpoint of the integration interval andits pseudo derivative equals the integrand at that point

Remark 28 When we replace D120572119901 by D120572

120596 then Lemma 27is no longer necessarily true for arbitrary 119909 isin H1

0(119864) evenwhen 119864 is reflexive evidently in [13 remark below Corollary4] the existence of a reflexive Banach space 119864 and a strongmeasurable Pettis integrable function 119909 119868 rarr 119864 was provedsuch that 119909 has nowhere weakly differentiable integral In thiscase 119863120596I

1119909 lost its meaning This gives a reason to believethat (65) (hence (64)) withD120572

120596119909 could not happen

However we have the following result

Lemma 29 Let 0 lt 120572 lt 1 For any 119909 isin 119862[119868 119864120596] we haveD

120572120596I

120572119909 = 119909 on 119868 (66)

Proof Our assumption 119909 isin 119862[119868 119864120596] sub Hinfin0 (119864) (see the proof

of Corollary 15) yields in view of Lemma 17 that

D120572120596I

120572119909 = 119863120596I1minus120572

I120572119909 = 119863120596I

1119909 (67)

The claim now follows immediately since the indefiniteintegral of weakly contentious function is weakly absolutelycontinuous and it is weakly differentiable with respect tothe right endpoint of the integration interval and its weakderivative equals the integrand at that point

The following lemma is folklore in case 119864 = R but to seethat it also holds in the vector-valued case we provide a proof

Lemma 30 Let 0 lt 120572 lt 1 and 119901 gt max1120572 1(1minus120572) If thefunction 119909 119868 rarr 119864 is weakly absolutely continuous on 119868 andpasses a pseudo derivative inH

1199010 (119864) then119889120572119901119909119889119905120572 (119905) = D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (68)

In particular if 119909 passes a weak derivative in 119862[119868 119864120596] then119889120572120596119909119889119905120572 (119905) = D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (69)

If 119864 is reflexive this is also true for every 119901 ge 1

Proof We observe that under the assumption imposed on119863119901119909 together with Proposition 3 the weakly absolutelycontinuity of 119909 is equivalent to

119909 (119905) = 119909 (0) + int119905

0119863119901119909 (119904) 119889119904 (70)

Hence owing to Lemma 27 it follows that

119889120572119901119909119889119905120572 = I1minus120572119863119901119909 = (D120572

119901I120572)I1minus120572119863119901119909

= D120572119901 (I120572

I1minus120572)119863119901119909 = (119863119901I

1minus120572)I1119863119901119909= 119863119901I

1minus120572 (119909 (119905) minus 119909 (0))= D

120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0)

(71)

which is what we wished to showThe proof of (69) is very similar to that in (68) therefore

we omit the details

Remark 31 As we remark above the definition of the CFPDof a function 119909 loses completely its meaning if 119909 is notweakly absolutely continuous For this reason we are able touse Lemma 30 to define the Caputo fractional derivative ingeneral that is we put

119889120572119901119909119889119905120572 (119905) fl D120572119901119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) 0 lt 120572 lt 1 (72)

Similarly we define

119889120572120596119909119889119905120572 (119905) fl D120572120596119909 (119905) minus 119905120572Γ (1 + 120572)119909 (0) (73)

However Lemma 30 claims that for the weakly absolutelycontinuous functions having Pettis integrable pseudo [weak]derivatives definitions (72) and (73) of the Caputo fractional-pseudo [weak] derivatives coincide with Definition 20

4 An Application

Let 119891 119868 times 119864 rarr 119864 be given function Consider the boundaryvalue problem of the fractional type

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin [0 1] 120582 isin R

119910 (0) + 119887119910 (1) = ℎ(74)

with certain constants 119887 ℎ isin R 119887 = minus1 To obtain theintegral equation modeled off the problem (74) we let 119910 bea weakly continuous solution to problem (74) then formallywe have

I120572 119889120572120596119889119905120572119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (75)

12 Journal of Function Spaces

that is

I120572I1minus120572119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) 997904rArrI1119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (76)

This reads (cf [29])

119910 (119905) = 119910 (0) + 120582I120572119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin 119868 120582 isin R (77)

with some (presently unknown) quantity 119910(0)Now we solve (77) for 119910(0) by 119910(0) + 119887119910(1) = ℎ and it

follows that

(1 + 119887) 119910 (0) + 120582119887Γ (120572) int1

0(1 minus 119904)120572minus1 119891 (119904 119910 (119904)) 119889119904 = ℎ (78)

Thus

119910 (0) = ℎ1 + 119887 minus 1205821198871 + 119887119868120572119891 (1 119910 (1)) (79)

Assume that the function 119891 is weakly-weakly continuousfunction such that

(1) for any 119903 gt 0 there is a constant 119872 gt 0 such that119891(119905 119910) le 119872 for all 119905 isin 119868 and 119910 le 119903(2) there exists a nondecreasing continuous function 120595 [0infin) rarr [0infin) 120595(0) = 0 and 120595(119905) lt 119905 for all 119905 gt 0

such that 120573(119891(119879 times 119860)) le 120595(120573(119860)) for every boundedset119860 sub 119864 where 120573 stands for De Blasirsquos weakmeasureof noncompactness (see [30])

Occasionally if 119891 119868 times 119864 rarr 119864 is weakly-weakly continuousand 119864 is reflexive then the assumptions (1) and (2) areautomatically satisfied (see eg [31])

Theorem 32 (see [6] Theorem 33) Let 120572 isin (0 1) and |120582| ltΓ(1 + 120572) If 119891 119868 times 119864 rarr 119864 is weakly-weakly continuous andsatisfies the assumptions (1) and (2) then the integral equation(77) has a weakly continuous solution 119910 defined on [0 1]Proof We omit the proof since it is almost identical to thatin the proof in ([6] Theorem 33) with (small) necessarychanges

In the following example we assume that 119910 isin 119862[119868 119864120596]solves (77) and we will show that not only do we have that 119910no longer necessarily solves (74) (when the Caputo fractionalweak derivative is taken in the sense of Definition 20)but even worse it could happen that the problem (74) isldquomeaninglessrdquo on 119868Example 33 Let 120572 isin (0 14] Let 119864 be an indefinitedimensional reflexive Banach space that fails cotype Definethe weakly-weakly continuous function 119891 [0 1] times 119864 rarr 119864by 119891 fl I1minus120572119909 where 119909 119868 rarr 119864 is defined by formula (7)Obviously (cf Example 6 in view of Remark 7)119891 satisfies theassumptions of Theorem 32

Now consider the integral equation (77) with119891 = I1minus120572119909Namely we consider the integral equation

119910 (119905) = 119910 (0) + 120582I120572I1minus120572119909 (119905)

120572 isin (0 14] 119905 isin 119868 120582 isin R (80)

Obviously the solution 119910 is weakly continuous on [0 1] (thisis an immediate consequence of Remark 7 and Lemma 17)

Since 120593119909 isin 1198711[0 1] holds for every 120593 isin 119864lowast it follows bythe commutative property of the fractional integral operatorscalculus over the space of Lebesgue integrable functions (see[21] Section 23) that

120593 (I120572I1minus120572119909 (119905)) = I

120572120593 (I1minus120572119909 (119905))= I

120572I1minus120572120593 (119909 (119905)) = I

1120593 (119909 (119905))= 120593 (I1119909 (119905))

(81)

that is

120593 (I120572I1minus120572119909 (119905) minusI

1119909 (119905)) = 0 for every 120593 isin 119864lowast (82)

Hence I120572I1minus120572119909 = I1119909 Consequently problem (80) be-comes

119910 (119905) = 119910 (0) + 120582I1119909 (119905) 120572 isin (0 14] 119905 isin 119868 (83)

As showed in ([13] Corollary 4) the indefinite integral 119905 rarrint1199050119909(119904)119889119904 is nowhere weakly differentiable on [0 1] Thus 119910

is nowhere weakly differentiable on [0 1] hence the CFWDof 119910 (hence the boundary value problem) completely loses itsmeaning if the CFWD is taken in the sense of Definition 20This is what we wished to show

Now we are in the position to state and prove thefollowing existence theorem

Theorem 34 If 119891 119868 times 119864 rarr 119864 is a function such that allconditions fromTheorem 32 hold then problem (74) (where theCaputo fractionalweak derivative is taken in the sense (73)) hasa weak solution on [0 1]Proof Let 119910 isin 119862[119868 119864120596] be a solution to (77) Equip 119864 andI times E with weak topology and note that 119905 997891rarr (119905 119910(119905)) iscontinuous as a mapping from 119868 into 119868 times 119864 Since 119891 119868 times 119864 rarr119864 is weakly-weakly continuous on [0 1] then 119891(sdot 119910(sdot)) is acomposition of this mapping with 119891 and thus 119891(sdot 119910(sdot)) isweakly continuous on [0 1]

Now operating by the operatorD120572120596 on both sides of (77)

it follows by Lemma 29 that

D120572120596119910 = D

120572120596119910 (0) + 120582D120572

120596I120572119891 = 119905minus120572Γ (1 minus 120572)119910 (0) + 120582119891 (84)

Now insert definition (73) of the Caputo fractionalderivative and we get

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) (85)

Journal of Function Spaces 13

With some further efforts one can get the boundary condi-tion 119910(0) + 119887119910(1) = ℎ

Therefore 119910 satisfies problem (74) which is what wewished to show

Data Availability

No data were used to support this study

Conflicts of Interest

The author declares no conflicts of interest

References

[1] H A Salem and A M El-Sayed ldquoWeak solution for fractionalorder integral equations in reflexive Banach spacesrdquoMathemat-ica Slovaca vol 55 no 2 pp 169ndash181 2005

[2] H A Salem A M El-Sayed and O L Moustafa ldquoA note onthe fractional calculus in Banach spacesrdquo Studia ScientiarumMathematicarum Hungarica vol 42 no 2 pp 115ndash130 2005

[3] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoMulti-term fractional differential equations in a nonreflexiveBanach spacerdquo Advances in Difference Equations 2013302 18pages 2013

[4] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoFractional calculus and fractional differential equations innonreflexive Banach spacesrdquo Communications in NonlinearScience andNumerical Simulation vol 20 no 1 pp 59ndash73 2015

[5] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoNonlinear fractional differential equations in nonreflexiveBanach spaces and fractional calculusrdquo Advances in DifferenceEquations 2015112 18 pages 2015

[6] R P Agarwal V Lupulescu D OrsquoRegan and G RahmanldquoWeak solutions for fractional differential equations in nonre-flexive Banach spaces via Riemann-Pettis integralsrdquo Mathema-tische Nachrichten vol 289 no 4 pp 395ndash409 2016

[7] H A Salem ldquoOn the fractional order m-point boundaryvalue problem in reflexive Banach spaces and weak topologiesrdquoJournal of Computational andAppliedMathematics vol 224 no2 pp 565ndash572 2009

[8] H A Salem ldquoOn the fractional calculus in abstract spaces andtheir applications to the Dirichlet-type problem of fractionalorderrdquo Computers and Mathematics with Applications vol 59no 3 pp 1278ndash1293 2010

[9] HAH Salem andMCichon ldquoOn solutions of fractional orderboundary value problems with integral boundary conditions inBanach spacesrdquo journal of function spaces and applications vol2013 Article ID 428094 2013

[10] J Diestel and J J Uhl Jr Vector Measures vol 15 ofMathemat-ical Surveys American Mathematical Society Providence RIUSA 1977

[11] R H Martin and J J Uhl Jr Nonlinear Operators AndDifferential Equations in Banach Spaces John Wiley amp SonsNew York NY USA 1976

[12] B J Pettis ldquoOn integration in vector spacesrdquo Transactions of theAmericanMathematical Society vol 44 no 2 pp 277ndash304 1938

[13] S J Dilworth and M Girardi ldquoNowhere weak differentiabilityof the Pettis integralrdquo Quaestiones Mathematicae vol 18 no 4pp 365ndash380 1995

[14] G A Edgar ldquoMeasurability in a Banach spacerdquo Indiana Univer-sity Mathematics Journal vol 26 no 4 pp 663ndash677 1977

[15] G A Edgar ldquoMeasurability in a Banach space IIrdquo IndianaUniversityMathematics Journal vol 28 no 4 pp 559ndash579 1979

[16] J L Gamez and J Mendoza ldquoOn denjoy-dunford and denjoy-pettis integralsrdquo StudiaMathematica vol 130 no 2 pp 115ndash1331998

[17] R AGordon ldquoTheDenjoy extension of the Bochner Pettis andDUNford integralsrdquo Studia Mathematica vol 92 no 1 pp 73ndash91 1989

[18] R F Geitz ldquoPettis integrationrdquo Proceedings of the AmericanMathematical Society vol 82 no 1 pp 81ndash86 1981

[19] R F Geitz ldquoGeometry and the Pettis integralrdquo Transactions ofthe AmericanMathematical Society vol 269 no 2 pp 535ndash5481982

[20] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations New York NYUSA Elsevier 2006

[21] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Yverdon Switzerland 1993

[22] M Mendel and A Naor ldquoMetric cotyperdquo Annals of Mathemat-ics Second Series vol 168 no 1 pp 247ndash298 2008

[23] B Ross S G Samko and E R Love ldquoFunctions that have nofirst order derivative might have fractional derivatives of allorders less than onerdquo Real Analysis Exchange vol 20 no 1 pp140ndash157 199495

[24] E Hille and R S Phillips Functional analysis and semi-groupsAmerican Mathematical Society Colloquium Publications vol31 American Mathematical Society Providence RI USA 1957

[25] S Schwabik and Y Guoju Topics in Banach Space IntegrationWorld Scientific Singapore 2005

[26] DW Solomon ldquoOn differentiability of vector-valued functionsof a real variablerdquo Studia Mathematica vol 29 pp 1ndash4 1967

[27] K Naralenkov ldquoOn Denjoy type extensions of the Pettisintegralrdquo Czechoslovak Mathematical Journal vol 60(135) no3 pp 737ndash750 2010

[28] D W Solomon Denjoy integration in abstract spaces Memorisof the American Mathematical Society No 85 AmericanMathematical Society Providence RI 1969

[29] A R Mitchell and C Smith ldquoAn existence theorem for weaksolutions of differential equations in Banach spacesrdquo inNonlin-ear Equations in Abstract Spaces V Lakshmikantham Ed pp387ndash404 1978

[30] F S De Blasi ldquoOn a property of the unit sphere in a Banachspacerdquo Bulletin Mathematiques De La Societe Des SciencesMathematiques De Roumanie vol 21 no 3 pp 259ndash262 1977

[31] A Szep ldquoExistence theorem for weak solutions of ordinarydifferential equations in reflexive Banach spacesrdquo Studia Scien-tiarum Mathematicarum Hungarica vol 6 pp 197ndash203 1971

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

Page 12: On the Theory of Fractional Calculus in the Pettis-Function Spaces · 2019. 7. 30. · JournalofFunctionSpaces Further,wedenethespace H (),˙≥0by H () fl ˛ ∈ ˚, ˜: isHolderainoforder

12 Journal of Function Spaces

that is

I120572I1minus120572119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) 997904rArrI1119863120596119910 (119905) = 120582I120572119891 (119905 119910 (119905)) (76)

This reads (cf [29])

119910 (119905) = 119910 (0) + 120582I120572119891 (119905 119910 (119905)) 120572 isin (0 1) 119905 isin 119868 120582 isin R (77)

with some (presently unknown) quantity 119910(0)Now we solve (77) for 119910(0) by 119910(0) + 119887119910(1) = ℎ and it

follows that

(1 + 119887) 119910 (0) + 120582119887Γ (120572) int1

0(1 minus 119904)120572minus1 119891 (119904 119910 (119904)) 119889119904 = ℎ (78)

Thus

119910 (0) = ℎ1 + 119887 minus 1205821198871 + 119887119868120572119891 (1 119910 (1)) (79)

Assume that the function 119891 is weakly-weakly continuousfunction such that

(1) for any 119903 gt 0 there is a constant 119872 gt 0 such that119891(119905 119910) le 119872 for all 119905 isin 119868 and 119910 le 119903(2) there exists a nondecreasing continuous function 120595 [0infin) rarr [0infin) 120595(0) = 0 and 120595(119905) lt 119905 for all 119905 gt 0

such that 120573(119891(119879 times 119860)) le 120595(120573(119860)) for every boundedset119860 sub 119864 where 120573 stands for De Blasirsquos weakmeasureof noncompactness (see [30])

Occasionally if 119891 119868 times 119864 rarr 119864 is weakly-weakly continuousand 119864 is reflexive then the assumptions (1) and (2) areautomatically satisfied (see eg [31])

Theorem 32 (see [6] Theorem 33) Let 120572 isin (0 1) and |120582| ltΓ(1 + 120572) If 119891 119868 times 119864 rarr 119864 is weakly-weakly continuous andsatisfies the assumptions (1) and (2) then the integral equation(77) has a weakly continuous solution 119910 defined on [0 1]Proof We omit the proof since it is almost identical to thatin the proof in ([6] Theorem 33) with (small) necessarychanges

In the following example we assume that 119910 isin 119862[119868 119864120596]solves (77) and we will show that not only do we have that 119910no longer necessarily solves (74) (when the Caputo fractionalweak derivative is taken in the sense of Definition 20)but even worse it could happen that the problem (74) isldquomeaninglessrdquo on 119868Example 33 Let 120572 isin (0 14] Let 119864 be an indefinitedimensional reflexive Banach space that fails cotype Definethe weakly-weakly continuous function 119891 [0 1] times 119864 rarr 119864by 119891 fl I1minus120572119909 where 119909 119868 rarr 119864 is defined by formula (7)Obviously (cf Example 6 in view of Remark 7)119891 satisfies theassumptions of Theorem 32

Now consider the integral equation (77) with119891 = I1minus120572119909Namely we consider the integral equation

119910 (119905) = 119910 (0) + 120582I120572I1minus120572119909 (119905)

120572 isin (0 14] 119905 isin 119868 120582 isin R (80)

Obviously the solution 119910 is weakly continuous on [0 1] (thisis an immediate consequence of Remark 7 and Lemma 17)

Since 120593119909 isin 1198711[0 1] holds for every 120593 isin 119864lowast it follows bythe commutative property of the fractional integral operatorscalculus over the space of Lebesgue integrable functions (see[21] Section 23) that

120593 (I120572I1minus120572119909 (119905)) = I

120572120593 (I1minus120572119909 (119905))= I

120572I1minus120572120593 (119909 (119905)) = I

1120593 (119909 (119905))= 120593 (I1119909 (119905))

(81)

that is

120593 (I120572I1minus120572119909 (119905) minusI

1119909 (119905)) = 0 for every 120593 isin 119864lowast (82)

Hence I120572I1minus120572119909 = I1119909 Consequently problem (80) be-comes

119910 (119905) = 119910 (0) + 120582I1119909 (119905) 120572 isin (0 14] 119905 isin 119868 (83)

As showed in ([13] Corollary 4) the indefinite integral 119905 rarrint1199050119909(119904)119889119904 is nowhere weakly differentiable on [0 1] Thus 119910

is nowhere weakly differentiable on [0 1] hence the CFWDof 119910 (hence the boundary value problem) completely loses itsmeaning if the CFWD is taken in the sense of Definition 20This is what we wished to show

Now we are in the position to state and prove thefollowing existence theorem

Theorem 34 If 119891 119868 times 119864 rarr 119864 is a function such that allconditions fromTheorem 32 hold then problem (74) (where theCaputo fractionalweak derivative is taken in the sense (73)) hasa weak solution on [0 1]Proof Let 119910 isin 119862[119868 119864120596] be a solution to (77) Equip 119864 andI times E with weak topology and note that 119905 997891rarr (119905 119910(119905)) iscontinuous as a mapping from 119868 into 119868 times 119864 Since 119891 119868 times 119864 rarr119864 is weakly-weakly continuous on [0 1] then 119891(sdot 119910(sdot)) is acomposition of this mapping with 119891 and thus 119891(sdot 119910(sdot)) isweakly continuous on [0 1]

Now operating by the operatorD120572120596 on both sides of (77)

it follows by Lemma 29 that

D120572120596119910 = D

120572120596119910 (0) + 120582D120572

120596I120572119891 = 119905minus120572Γ (1 minus 120572)119910 (0) + 120582119891 (84)

Now insert definition (73) of the Caputo fractionalderivative and we get

119889120572120596119889119905120572119910 (119905) = 120582119891 (119905 119910 (119905)) (85)

Journal of Function Spaces 13

With some further efforts one can get the boundary condi-tion 119910(0) + 119887119910(1) = ℎ

Therefore 119910 satisfies problem (74) which is what wewished to show

Data Availability

No data were used to support this study

Conflicts of Interest

The author declares no conflicts of interest

References

[1] H A Salem and A M El-Sayed ldquoWeak solution for fractionalorder integral equations in reflexive Banach spacesrdquoMathemat-ica Slovaca vol 55 no 2 pp 169ndash181 2005

[2] H A Salem A M El-Sayed and O L Moustafa ldquoA note onthe fractional calculus in Banach spacesrdquo Studia ScientiarumMathematicarum Hungarica vol 42 no 2 pp 115ndash130 2005

[3] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoMulti-term fractional differential equations in a nonreflexiveBanach spacerdquo Advances in Difference Equations 2013302 18pages 2013

[4] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoFractional calculus and fractional differential equations innonreflexive Banach spacesrdquo Communications in NonlinearScience andNumerical Simulation vol 20 no 1 pp 59ndash73 2015

[5] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoNonlinear fractional differential equations in nonreflexiveBanach spaces and fractional calculusrdquo Advances in DifferenceEquations 2015112 18 pages 2015

[6] R P Agarwal V Lupulescu D OrsquoRegan and G RahmanldquoWeak solutions for fractional differential equations in nonre-flexive Banach spaces via Riemann-Pettis integralsrdquo Mathema-tische Nachrichten vol 289 no 4 pp 395ndash409 2016

[7] H A Salem ldquoOn the fractional order m-point boundaryvalue problem in reflexive Banach spaces and weak topologiesrdquoJournal of Computational andAppliedMathematics vol 224 no2 pp 565ndash572 2009

[8] H A Salem ldquoOn the fractional calculus in abstract spaces andtheir applications to the Dirichlet-type problem of fractionalorderrdquo Computers and Mathematics with Applications vol 59no 3 pp 1278ndash1293 2010

[9] HAH Salem andMCichon ldquoOn solutions of fractional orderboundary value problems with integral boundary conditions inBanach spacesrdquo journal of function spaces and applications vol2013 Article ID 428094 2013

[10] J Diestel and J J Uhl Jr Vector Measures vol 15 ofMathemat-ical Surveys American Mathematical Society Providence RIUSA 1977

[11] R H Martin and J J Uhl Jr Nonlinear Operators AndDifferential Equations in Banach Spaces John Wiley amp SonsNew York NY USA 1976

[12] B J Pettis ldquoOn integration in vector spacesrdquo Transactions of theAmericanMathematical Society vol 44 no 2 pp 277ndash304 1938

[13] S J Dilworth and M Girardi ldquoNowhere weak differentiabilityof the Pettis integralrdquo Quaestiones Mathematicae vol 18 no 4pp 365ndash380 1995

[14] G A Edgar ldquoMeasurability in a Banach spacerdquo Indiana Univer-sity Mathematics Journal vol 26 no 4 pp 663ndash677 1977

[15] G A Edgar ldquoMeasurability in a Banach space IIrdquo IndianaUniversityMathematics Journal vol 28 no 4 pp 559ndash579 1979

[16] J L Gamez and J Mendoza ldquoOn denjoy-dunford and denjoy-pettis integralsrdquo StudiaMathematica vol 130 no 2 pp 115ndash1331998

[17] R AGordon ldquoTheDenjoy extension of the Bochner Pettis andDUNford integralsrdquo Studia Mathematica vol 92 no 1 pp 73ndash91 1989

[18] R F Geitz ldquoPettis integrationrdquo Proceedings of the AmericanMathematical Society vol 82 no 1 pp 81ndash86 1981

[19] R F Geitz ldquoGeometry and the Pettis integralrdquo Transactions ofthe AmericanMathematical Society vol 269 no 2 pp 535ndash5481982

[20] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations New York NYUSA Elsevier 2006

[21] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Yverdon Switzerland 1993

[22] M Mendel and A Naor ldquoMetric cotyperdquo Annals of Mathemat-ics Second Series vol 168 no 1 pp 247ndash298 2008

[23] B Ross S G Samko and E R Love ldquoFunctions that have nofirst order derivative might have fractional derivatives of allorders less than onerdquo Real Analysis Exchange vol 20 no 1 pp140ndash157 199495

[24] E Hille and R S Phillips Functional analysis and semi-groupsAmerican Mathematical Society Colloquium Publications vol31 American Mathematical Society Providence RI USA 1957

[25] S Schwabik and Y Guoju Topics in Banach Space IntegrationWorld Scientific Singapore 2005

[26] DW Solomon ldquoOn differentiability of vector-valued functionsof a real variablerdquo Studia Mathematica vol 29 pp 1ndash4 1967

[27] K Naralenkov ldquoOn Denjoy type extensions of the Pettisintegralrdquo Czechoslovak Mathematical Journal vol 60(135) no3 pp 737ndash750 2010

[28] D W Solomon Denjoy integration in abstract spaces Memorisof the American Mathematical Society No 85 AmericanMathematical Society Providence RI 1969

[29] A R Mitchell and C Smith ldquoAn existence theorem for weaksolutions of differential equations in Banach spacesrdquo inNonlin-ear Equations in Abstract Spaces V Lakshmikantham Ed pp387ndash404 1978

[30] F S De Blasi ldquoOn a property of the unit sphere in a Banachspacerdquo Bulletin Mathematiques De La Societe Des SciencesMathematiques De Roumanie vol 21 no 3 pp 259ndash262 1977

[31] A Szep ldquoExistence theorem for weak solutions of ordinarydifferential equations in reflexive Banach spacesrdquo Studia Scien-tiarum Mathematicarum Hungarica vol 6 pp 197ndash203 1971

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

Page 13: On the Theory of Fractional Calculus in the Pettis-Function Spaces · 2019. 7. 30. · JournalofFunctionSpaces Further,wedenethespace H (),˙≥0by H () fl ˛ ∈ ˚, ˜: isHolderainoforder

Journal of Function Spaces 13

With some further efforts one can get the boundary condi-tion 119910(0) + 119887119910(1) = ℎ

Therefore 119910 satisfies problem (74) which is what wewished to show

Data Availability

No data were used to support this study

Conflicts of Interest

The author declares no conflicts of interest

References

[1] H A Salem and A M El-Sayed ldquoWeak solution for fractionalorder integral equations in reflexive Banach spacesrdquoMathemat-ica Slovaca vol 55 no 2 pp 169ndash181 2005

[2] H A Salem A M El-Sayed and O L Moustafa ldquoA note onthe fractional calculus in Banach spacesrdquo Studia ScientiarumMathematicarum Hungarica vol 42 no 2 pp 115ndash130 2005

[3] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoMulti-term fractional differential equations in a nonreflexiveBanach spacerdquo Advances in Difference Equations 2013302 18pages 2013

[4] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoFractional calculus and fractional differential equations innonreflexive Banach spacesrdquo Communications in NonlinearScience andNumerical Simulation vol 20 no 1 pp 59ndash73 2015

[5] R P Agarwal V Lupulescu D OrsquoRegan and G ur RahmanldquoNonlinear fractional differential equations in nonreflexiveBanach spaces and fractional calculusrdquo Advances in DifferenceEquations 2015112 18 pages 2015

[6] R P Agarwal V Lupulescu D OrsquoRegan and G RahmanldquoWeak solutions for fractional differential equations in nonre-flexive Banach spaces via Riemann-Pettis integralsrdquo Mathema-tische Nachrichten vol 289 no 4 pp 395ndash409 2016

[7] H A Salem ldquoOn the fractional order m-point boundaryvalue problem in reflexive Banach spaces and weak topologiesrdquoJournal of Computational andAppliedMathematics vol 224 no2 pp 565ndash572 2009

[8] H A Salem ldquoOn the fractional calculus in abstract spaces andtheir applications to the Dirichlet-type problem of fractionalorderrdquo Computers and Mathematics with Applications vol 59no 3 pp 1278ndash1293 2010

[9] HAH Salem andMCichon ldquoOn solutions of fractional orderboundary value problems with integral boundary conditions inBanach spacesrdquo journal of function spaces and applications vol2013 Article ID 428094 2013

[10] J Diestel and J J Uhl Jr Vector Measures vol 15 ofMathemat-ical Surveys American Mathematical Society Providence RIUSA 1977

[11] R H Martin and J J Uhl Jr Nonlinear Operators AndDifferential Equations in Banach Spaces John Wiley amp SonsNew York NY USA 1976

[12] B J Pettis ldquoOn integration in vector spacesrdquo Transactions of theAmericanMathematical Society vol 44 no 2 pp 277ndash304 1938

[13] S J Dilworth and M Girardi ldquoNowhere weak differentiabilityof the Pettis integralrdquo Quaestiones Mathematicae vol 18 no 4pp 365ndash380 1995

[14] G A Edgar ldquoMeasurability in a Banach spacerdquo Indiana Univer-sity Mathematics Journal vol 26 no 4 pp 663ndash677 1977

[15] G A Edgar ldquoMeasurability in a Banach space IIrdquo IndianaUniversityMathematics Journal vol 28 no 4 pp 559ndash579 1979

[16] J L Gamez and J Mendoza ldquoOn denjoy-dunford and denjoy-pettis integralsrdquo StudiaMathematica vol 130 no 2 pp 115ndash1331998

[17] R AGordon ldquoTheDenjoy extension of the Bochner Pettis andDUNford integralsrdquo Studia Mathematica vol 92 no 1 pp 73ndash91 1989

[18] R F Geitz ldquoPettis integrationrdquo Proceedings of the AmericanMathematical Society vol 82 no 1 pp 81ndash86 1981

[19] R F Geitz ldquoGeometry and the Pettis integralrdquo Transactions ofthe AmericanMathematical Society vol 269 no 2 pp 535ndash5481982

[20] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations New York NYUSA Elsevier 2006

[21] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Yverdon Switzerland 1993

[22] M Mendel and A Naor ldquoMetric cotyperdquo Annals of Mathemat-ics Second Series vol 168 no 1 pp 247ndash298 2008

[23] B Ross S G Samko and E R Love ldquoFunctions that have nofirst order derivative might have fractional derivatives of allorders less than onerdquo Real Analysis Exchange vol 20 no 1 pp140ndash157 199495

[24] E Hille and R S Phillips Functional analysis and semi-groupsAmerican Mathematical Society Colloquium Publications vol31 American Mathematical Society Providence RI USA 1957

[25] S Schwabik and Y Guoju Topics in Banach Space IntegrationWorld Scientific Singapore 2005

[26] DW Solomon ldquoOn differentiability of vector-valued functionsof a real variablerdquo Studia Mathematica vol 29 pp 1ndash4 1967

[27] K Naralenkov ldquoOn Denjoy type extensions of the Pettisintegralrdquo Czechoslovak Mathematical Journal vol 60(135) no3 pp 737ndash750 2010

[28] D W Solomon Denjoy integration in abstract spaces Memorisof the American Mathematical Society No 85 AmericanMathematical Society Providence RI 1969

[29] A R Mitchell and C Smith ldquoAn existence theorem for weaksolutions of differential equations in Banach spacesrdquo inNonlin-ear Equations in Abstract Spaces V Lakshmikantham Ed pp387ndash404 1978

[30] F S De Blasi ldquoOn a property of the unit sphere in a Banachspacerdquo Bulletin Mathematiques De La Societe Des SciencesMathematiques De Roumanie vol 21 no 3 pp 259ndash262 1977

[31] A Szep ldquoExistence theorem for weak solutions of ordinarydifferential equations in reflexive Banach spacesrdquo Studia Scien-tiarum Mathematicarum Hungarica vol 6 pp 197ndash203 1971

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

Page 14: On the Theory of Fractional Calculus in the Pettis-Function Spaces · 2019. 7. 30. · JournalofFunctionSpaces Further,wedenethespace H (),˙≥0by H () fl ˛ ∈ ˚, ˜: isHolderainoforder

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom