the reverse triangle inequality in normed spacest€¦ · new zealand journal of mathematics volume...

NEW ZEALAND JOURNAL OF MATHEMATICS Volume 25 (1996), 181-193

THE REVERSE TRIANGLE INEQUALITY IN NORMEDSPACESt

M lTSU R U N A K A I* AND TOSH IM ASA T A D A *

(Received January 1995)

Abstract. The validity of the reverse triangle inequality in a normed space X is characterized by the finiteness of what we call the best constant cr(X) associated with X . The validity of the reverse triangle inequality in X , i.e. cr(X) < oo, is completely determined algebraically by d im X < oo. Certain practical estimates of cr(X) from above and from below for general normed space X are given. These are applied to our main purpose of this paper of describing the order of <j(lp(n)) as n —* oo for n-dimensional lp space (1 < p < oo). A typical result is that cr(lp(n)) « n 1 -1 / p as n —► oo for 2 < p < n.

1. Introduction

Consider a real normed space X equipped with the the norm || • || = || • ; X||. The triangle inequality, which is one of the defining properties of norms, may be formulated as follows: for any finite sequence {x/c}fceL> (D = {1 , . . . , h}) in X

We say that the reverse triangle inequality is valid in X if there exists a space constant C = C x with the following property: for,any finite sequence {xk}keD in X there exists a subset A c D depending on {xk}k&D such that

The smallest possible constant of such C ’s is said to be the best constant for the reverse triangle inequality in X and will be denoted by cr(X). For convenience we set cr(X) = oo if the reverse triangle inequality is invalid in X.

In this paper we first remark that the reverse triangle inequality is valid in X ,i.e. cr(X) < oo, if and only if X is finite dimensional, i.e. dim X < oo (Theorem 1). The main purpose of this paper is to determine the order of a(lp(n)) as n f oo, where lp(n) is the finite n dimensional real linear space with lp norm (1 < p < oo). For the purpose we first give certain general practical estimates of cr(X) from above and also from below for general normed spaces X (Theorem 2). These are used to determine the constants a(lp(n)) exactly for p = 1,2, and oo (Theorem 3), which

1991 AMS Mathematics Subject Classification: Primary 46B20; Secondary 46B45. t Dedicated to our old friend Professor Heppe O ’Malla on his sixtieth birthday.* The authors were supported in part by Grant-in-Aid for Scientific Research, No. 06640227, Japanese Ministry of Education, Science and Culture.

(1)

182 MITSURU NAKAI AND TOSHIMASA TADA

implies a complete result a{lp(n)) « n1 1//p for 2 < p < oo and also an estimate of the order of a(lp(n)) for 1 < p < 2 (Theorem 4).

2. Finite Dimensional Normed Spaces

It is convenient to associate two quantities with a finite sequence {x*;} = {xkjkeD (D = {1 , . . . ,/i}) as follows. The first is the total length T { x k} = J2ktD ll fcll {xfc} and the second is the maximal length M {x k } of {x*;} given by

M{xk\ = max L 3 a c d E i4

keA(3)

In terms of these two quantities the triangle inequality (1) is formulated as ini(T{xk}/M{xk}) = 1 where the infimum is taken with respect to {x*;} which runs over all finite sequences in X \{0}. In contrast the best constant cr(X) for the reverse triangle inequality (2) in X is given by

sup {T {xk}/M {xk}) = cr(X) (4)

where again the supremum is taken with respect to {xfc} which runs over all finite sequences in X \{0}. Thus the reverse triangle inequality is valid in X if and only if a(X) < oo. The property cr(X) < oo for X is completely characterized by the linear space structure of X as follows:

Theorem 1. The reverse triangle inequality is valid in a normed space X if and only if X is finite dimensional.

Proof. Since cr(X) = <r(X) for the completion X of X , we may assume that X is a Banach space. We first show that <x(X) = oo if dim X = oo. By the theorem of Dvoretzky and Rogers [1] (see also e.g. Marti [2, page 27] and Singer [5]), there exists a series J2ieN x i = {1 ,2 ,.. . }) in X which is not absolutely convergent, i.e. Y^ieN 11*11 = °°> but unconditionally convergent, i.e. J2 i eNxp(i) convergent in X for every permutation p of N . We can assume here that {x^} C X \{0}. The unconditional convergence of J2 ieN Xi is equivalent to its unordered convergence (cf. e.g. [2 , page 20]), i.e. setting x := YhieN x*> we ^ave x = lim7er whereT is the family of finite subsets of N , so that there exists a 70 € T such thatk - E i67 ' < 1 for every 7 G T with 7 D 70- Let K — 1 + ||x|| + 7o

which lies in (1 ,00). Then we see that

E Xf— E x i ~ x + x — Xi < K

ie 7 yiS7U7o y ie 70 \7

for every 7 € T. From this it follows that M {x ;}i< i< m < K for any m € N so that

E ||xj|| —>■ 00 (m —> 00),l< i< m

which implies that cr(X) = 00 as desired.

REVERSE TRIANGLE INEQUALITY 183

We next show that cr(X) < oo if dim X < oo (cf. Rudin [3, page 126], [4, page 118]). Contrary to the assertion assume that a ( X ) = oo. Then there are a sequence {xijiejv m and a strictly increasing sequence {^(A;)}*;^ inN with i/(l) = 1 such that

T { Xi} k > 2kM { Xi} k {k e N ) (5)

where {x j } fc := {xi : v(k) < i < u(k + 1)} (fc G JV). On replacing Xj by ('T {x i }k)~1xj (^(fc) < j < v(k + 1)), if necessary, we can assume that T { x i } k = 1 so that M { x i ] k < 2~k (k E N ). By virtue of the fact that

E in i = E TMk = E 1 = °°-ieN keN keN

the series J2ieN x i is not absolutely convergent in X . On the other hand, the series J2ieN Xi *s unconditionally convergent in X . To see this take an arbitrary permutation p on JV. Let e be any positive number. Fix a ko E N such that2-fco+i < £m There exists an m E N such that p(i) > u(ko) for any i > m. Then, for any y > m and s E N there exists at E N such that p(i) < u(t) (y (k + l )

< M { Xi} k < 2~ k o + 1 < £fco<fc<oo

for any y > m and s E N. This means that J2ieNxp{i) is convergent in X and thus J^ieN x i *s unconditionally convergent. This contradicts the Riemann theorem (c.f. e.g. [2, page 22]) which asserts the equivalence of absolute convergence of a series and unconditional convergence of a series in a finite dimensional Banach space X . □

3. U pper and Lower Estimates

We denote by X* the dual space of a normed space X. The value of £ E X* at x E X will be denoted by (x,£). We denote by Sx (S x *, resp.) the unit sphere in X (X *, resp.). For any A E R we write A+ = max(A, 0). We maintain the following

Theorem 2. For any finite nonzero Borel measure y on Sx* the following upper estimate holds:

\ (6)tsx .

for any finite nonzero Borel measure m on Sx the following lower estimate holds:- l

(7)a{X) > ( sup 77; r [ {x,£)+ dm{x) \ t e s x * m [bX ) J s x

P roof. First we prove (6). Choose an arbitrary sequence {xk }keD (.D = {1 , . . . ,h} ) in X \{0}. Since the function £ i—» YlkeD(x ki£)+ is weakly* continuous on S x • which is weakly* compact by the Banach-Alaoglu theorem,

there exists an 7? in S x* such that J2keD(x kiV)+ — m&x-$esx * ^2keD(x kiO+ so that Y^keD(x k^v)+ > J2keD(x ki£)+ f°r every £ in Sx*- Integrating both sides of the above over Sx* with respect to dfi(£) we obtain

^ i*k<o+ M «)•keD X ’ k e D JSx*

On setting A = {k e D : (xk, rj) > 0} we see that

M { x k} >

Hence we deduce

J 2 XkkeA

> (J2xk,v) = ^2(xk,v) = 2(xk,v)+= ^2ixk,vY\ ke A / keA keA ke D

M[Xk] - ) (\\xk\\ 1xk,^)+ d ^ ) I M

>

so that we conclude that

T { x k) / U { x k} <-1

Sx *On taking the supremum in the above with respect to {xfc} C X \ {0 } we obtain (6).

We proceed to the proof of (7). By Theorem 1, a ( X ) = 00 if dim X = 00, in which case (7) is trivially valid. We can thus assume that dim X < 00. Then Sx is strongly compact so that we can find for an arbitrary positive number e a finite family {E k}keD {D = {1 , . . . ,/i}) of Borel subsets Ek C Sx with the following properties:

(a) the diameter of Ek is less than e for every k e D ;(b) Ei D Ej = 0 (i ± j );(c) m(Ek) > 0 for every k e D;(d) m(Sx\ UkeD Ek) = 0 .

Choose and then fix an arbitrary point x k in Ek for each k e D and set yk = m(Ek)xk (k € D). We take a maximal subset A c D such that M {Vk} = || Efce^yjfell- Here the maximality of A means that if M {yk} = ||Efc€B|/fc|| and A C B C D, then A = B. By an application of e.g. Hahn-Banach theorem there exists an 77 in Sx * such that {Y^keAVk^) = II J2keAVk\\ (cf- e-S- Yosida [6 , page 108]). Hence we have M { y k} = J2keA{yk,v)-

We maintain that

A = { k e D : (yk,r]) > 0}.

Assume the existence of an i e A and yet (yi}rj) < 0. Then(8)

M{yk} = E (2/*’ ) < E (yk,v) <ke A &€A\{i}

E »*fce 4\{i}

< M { y k},

a contradiction. Conversely assume that (yi,r7) > 0 and yet i ^ A. Then

M { y k} = J2 (yk,v) < 2 -k£A fceAu{i}

X ! ykk&AU{i}

< M { y k},

which contradicts the maximality of A. We have thus established (8).In view of (8) we have the identity

M { y k} = ^2 (yk,v) = ^2 (yk,v)+ = £ < ^ > + = J2 {xk,r])+m(Ek).keA ke A k e D keD

For each k e D , because of (a), we have

\(xk,r])+ - (x,rj)+ 1 < \(xk,ri) - {x,rf)\ < Hr/H ||xfc - x|| < £

for every x e Ek and thus

(xk,rj)+ < (x,r})+ + £ ( x e E k).

On integrating both sides of the above over Ek with respect to dm(x) we have

(xk,r])+ m(Ek) < / (x,r])+dm(x) + £m(Ek).JEk

Thus we see by (b) and (d) that

M { y k} < 5 3 ( / (x ,rj)+dm(x) + £m(Ek) J = / (x,rj)+dm(x) + £m(Sx ) keD \JEk J Jsx

and therefore we have

M { y k} )+dm(x) + £m(Sx )- £ e S x * JSx

On the other hand, again by (d), we see that

T{yk] = $ 3 W l = 1 3 = 1 3 m(E*0 = ™(sx)-ke D k e D ke D

By (c) we see that { y k } k e D is a finite sequence in X \ {0} and a fortiori

a( X) > T {yk}/M{yk} > ( sup 1 [ (x,£)+dm(x) + .\££SX. rn{bx) Jsx j

Since £ > 0 is arbitrary, on letting £ j 0 in the above, (7) is deduced. □

4. Computations of Best Constants

We now apply Theorem 2 to compute a(lp(n)) (n e N ) for the space lp(n) (p = 1,2, oo). For the purpose we denote by Bn and Sn~l the unit ball and the unit sphere in the Euclidean n-space En. We denote by \Bn\ the Euclidean volume of Bn and by |5n-1| the Euclidean surface area of S'71-1 so that

\B2n\ = 7rn/n\, |B 2n~l \ = 22n7rn_1n!/(2n)!, IS "-1! = n\Bn\ (n e N ). (9)

Here we follow the usual convention |5°| = 2 and we also set \B°\ = 1 for convenience. We also need to consider the quantities Nn given by

where [ ] is the Gaussian symbol. For two sequences {am} meN and {&m} mejv of positive real numbers we write am >- bm or 6m -<: am if lim inf™ -^ am/6m > 0 and we also write am ~ bm if am y bm and am -< bm. We will write n-vectors x as x = (x1, . . . , x n). We have the following result.

Theorem 3. The best constant cr(/p(n)) (n £ N ) for three spaces lp(n) (p — 1,2, and oo) are given as follows:

Proof. Recall that /1(n)* = l°°(n) and l2 (n)* = l2 (n). Because of the finite dimensionality of /°°(n) we also have l°°(n)* = /1(n).

We first prove (13). Let £k = (<5fci,.. • G Sji(n) = Si°°(n)* where <5- is the Kronecker delta. Consider the Borel measure fi = X a<i<n(6£fc + ^-ek) on S'/i(n) where 6a is the Dirac measure at a. Then by (6) we have

(n e N ) y (10)

cr^^n)) = 1/Nn « y/n\

a(l2 (n)) = n\Bn\/\Bn~l \ w y/n;

(r(/°°(n)) = 2 n « n.

(12)(11)

(13)

n

xGSto0(n) 2n ' k=1

so that we obtain cr(Z°°(n)) < 2n.

Conversely, we take ek = (Ski, • • • , <5/cn) £ Consider the Borel measurem = Yh<k<n(^ek + <5-efc) on 5;oo(n). Then by (7) we have

so that we obtain a(l°°(n)) > 2n. We have thus established (13).Next we prove (12). It is easy to see directly that <j(/2(1)) = 2. By (9) and the

convention |jB°| = 1, we see that 1 • I#1! / !# 0! = 2. Thus (12) holds for n = 1. We thus assume hereafter in this proof of (12) that n > 2. Let ds be the Euclidean area element on S'"-1 . Since l2 (n) = E n and = S*2(n) = 5 n_1, by takingthe area element ds as dfi in (6) and by noting {x,£)+ = (£,x)+ , we obtain

For any £ 6 Sn 1 there exists an orthogonal transformation P of En = l2 (n) such that Pen = £ where en = (0 ,... ,0,1) G En. By the change of variable Py = x (y = (y1, . . . ,yn)), on noting P ~ 1 (Sn~1) = 5 n_1, ds(Py) = ds(y), (Py ,P en) = (y,en) = yn and ds(y) = (1 /yn)dyl . . . dyn~l on Sn_1 D {yn > 0} over 5 n_1, we deduce that f Sn-i (x, £)+ds(x) = |Bn-1|. This with (14) and (15) implies <r(l2 (n)) = \S " - 1! / ] # " " 1! = n\Bn\/\Bn~l \.

By using the Wallis formula limn_too(n!)2 22n/y/n(2n)\ = \Jts and (9), we see that limn_>00 cr(l2 (2 n — 1 ))/\ /n = lim ^oo a(l2 (2 n ))/y/n = 2 y/n so that we also conclude that a(l2 (n)) = n\Bn\/\Bn~1\ ~ y/n (n —> oo).

Finally we prove (11). The proof is not so simple as those of (12) and (13). Let D = {1 ,2 ,3 ,.. . , 2n} and {£& : k e D } be the set of 2n points Ek in Si<*>(„) = <Sfi(n). such that e%k (i = 1 ,.. . ,n) are either 1 or —1, where £k = (e*., ■ • • ,£%)• Consider the Borel measure = Y^keD 011 &i°°(n)• Then by (6) we have

<7(/°°(n)) 1 infV {S l°°(n)) J s l0O{n)

f (x,£)+dn(Z)

(16 )

J2 (x ,ek)+ -

Take e = (1 /n ,. . . , 1/n) G Sti(n) and set = {fc 6 D : (e,Ek) > 0}. We maintain that

i™n = = (17) Il(n) fceD fceo fceA

Observe that k e A is equivalent to X^i<i<n £fe — ® an< hence we see that J^keA £k is independent of i = 1 ,.. . , n. The common value is denoted by E:

= E (i = 1 ,. . . , n). (18)k eA

Since {£k : k G D } and also Sji(n) are symmetric with respect to the coordinate hyperplane considered in ^(n), we only have to show that

^ 2 ( x , e k)+ > ^ 2 ( e , e k)+ (19)keD keD

for every x G 5p(n) with x l > 0 (i = 1 ,... , n) to conclude the validity of (17). By (18) we infer as follows:

E<e’e*>+ = = E (E**) = Eei(Ee*)fceo fce.4 fceA \i=i / i=i \fceA /

= J V e = E ^ e 1 = E = e J 2 x { = ]^ x * E i=l i=l 1=1 i=l

= iy (E 4 ) = e (x>‘4)i=i VfceA / fceA \i=i /

= ]T (£ ,£ fc) < ^ < £ ,£ fc )+ < ]T (x ,£ fc) + keA ke A ke D

so that (19) and hence (17) has been shown. By (16) and (17) we conclude that

cr( 1(n))_1 > (20) keA

Recall that k e A is equivalent to that the number of components e\ = 1 is greater than or equal to that of elk = —1. Hence the number j of components e\ = — 1 for k e A is at most t := [n/2]. We denote by Aj the set of k e A such that the number of components e\ = -1 is j so that A = Uo<j<eAj (disjoint union) and the number of elements in Aj is (” ). To compute the right hand side of (20) we infer

E <«.«»> - sE(f>i) = e ( e (e 4))k£A keA \i=l / j=0 yfceAj \i=l / J

1 1 / / \ \ 1 [(^-t-l)/2]-l / v= ± z ( ( : M e <»-«>(:)■

Hence by (10) and (20) we conclude that cr(/1(n)) < 1/A^n.

We need to show that cr(/1(n)) > 1 /Nn to conclude the identity in (11). Using D = {1 ,2 ,3 ,... ,2n} above we consider the set {e k : fc G D } of 2n points ek in 5/i(n) such that e\ (i = 1, . . . , n) are either 1/n or —1/n , where ek = (e£,. . . , e£). Take the Borel measure m = YlkeD^ek on By (7) and ^(n)* = Z°°(n), wehave

a (Pin)) 1 S>+ = £ ( e*’ ?>+ -

Take e = (1 ,... , 1) G 5z«>(n) and set A = {fc € D : (ek,e) > 0}. We maintain that

“ a* £ ( e*.?>+ = £ ( efc>«')+ = £ (et.e ). (22)k€D k e D *€ A

Since {e* : fc G D } and Sj°o(n) axe symmetric with respect to the coordinate hyperplanes considered in /°°(n), we only have to show that

£ ( e fc,e>+ < £(e*,<r>+ (23)k e D keD

for every £ € Sjoo(n) with £* > 0 (i = 1 ,... ,n) to conclude the validity of (22). Put B = {fc € D : (ek,£) > 0} for an arbitrary fixed point £ G Si°°(n) with f* > 0 (« = 1 ,... , n). Before proving (23) we need to establish two auxiliary assertions. We first shows that

£ 4 > 0 (i = l , . . . ,n ) . (24)keB

Fix an arbitrary i = 1 ,... , n. If there is a fc G B such that ejj. = —1/n, then we consider an ek> for a fc' G D with the following properties: eJk, = 4 (j ¥> *) and ek' = 1/n. Since we have (ek>,£) = {ek,£) + 2£*/n > 0, we see that fc' G B. Therefore the number of fc G B with e\ = 1/n is not less than that of fc G B with e*. = —1/n for each fixed i = 1 ,.. . ,n. This assures the validity of (24). We next show that

E ( E 4 ) > E ( E 4 ) . (25)ke A \ i= l / keB \ i = l /

In fact, by the equivalence of fc G A and E i<i<n el — we ^ er that

After these two preperations above we proceed to the proof of (23). By (25), 0 < e < l ( i = l , . . . , n ) and (24), we deduce the following:

E<e*'e>+ = E<e*'e> = E (Xy*£i) = E(E4)k£D keA ke A \ i - 1 / ke A \i=l /

* E(E4) =E(E4) >E(E4)fkeB \ i = 1 / i= 1 \ k e B J i= 1 VfcGB /

= E(E4e) = E<e*-«> = E<e*-f>+ = E >«>+’keB \ i = l J k£B ke B keD

i.e. (23) and hence (22) has been proved. By (21) and (22) we conclude that

cr( 1(n) ) _1 < (26) keA

Recall that k € A is equivalent to X)i<i<n ek — ® which is then equivalent to that the number of components ek = l/n is at least that of ek = — 1/n. Here the number j of components ek = — l/n for k G A is at most I — [n/2], the number already considered above. We denote by Aj the set of k e A such that the number of components elk = —l/n is j so that A = Uo<j<eAj (disjoint union) and the number of elements in Aj is (n). Hence we see that

E<^> = E(E4) = e (e (E4keA ke A \i—1 / j —0 y k eA j \i=l /

- § ( Q H ' H a H § < - w C

Therefore by (10) and (26) we conclude that cr(/1(n)) > 1/-/Vn and the identity in(11) has been established.

The proof of (11) is complete if we show that Nn « 1/y/n. We can easily see that Nn -< Nn+i. Hence we only have to show that

N 2n = 2 ~ 2n( 2 n ) ~ 1 J 2 2 (n - j ) f 2U) ~ l / V n . j =o V 3 '

On changing j to k by n — j = k in the above expression of N2n we have

(2n)\y/n -tt A A:M = J - y / r ( 2n)Vn f r ,

2n n . P n ^ n n 2 o 2 n J .1 (nJn ^ (n!)222n 11 V n + ? fc=i K 7 j=i x

We use here the Wallis formula (2n)!i/n /(n !)222n | l/V^r (n | oo) to deducen k

n\fn N2n ~ IT (l - fc/(n + j ))- fc=i j = l

Hence we have to show that

k ' k

in order to assure N2n ~ 1/y/n. We divide the summation in the left side of (27) as the sum of an = and bn = Z)fc=[v^]+r T^en t ie ra t io n (27) is derivedif we can show that an ~ n and bn -< n. For the purpose we only have to show that cin + bn -< n and an y n.

Clearly an < = h/^Kh/ra] + l ) /2 -< n. Observe that

j = 1 x n + j j V 3 + n JJ y j r i n + J

n + 1 + k\ ( n + 1 xfc< exp —k log n + 1 J \ n + l + k

= 1 / 1 + — r r < 1n + 1 / / n + 1

2(n + l )2 < 4(n + l ) 5k3{k - 1) “ fc4

and therefore pTl

bn < 4(n + l ) 2 ^ 2 k~ 3 < 4(n + l ) 2 / x~3dx < 2(n + l ) 2 x [y/n] -<n. fc=[^]+i ■'Iv'"]

Thus we have seen that an + bn -< n.We turn to the proof of an >- n. We use the inequality log(l — x) > (log4)x

(0 < x < 1/2). If n > 4 and 1 < k < yfn, then kf(n + j ) < y/n/n < 1/2 for j = 1, . . . , k. Therefore

ft 0 - h) = exp ( § log 0 - 7

> exp ( - ( lo g 4) ~^Tj j - exp ( ~ ( loS4)fc loS

Since 1 < k < y/n in the summation of an, we obtain for n > 4 that

as required. The proof of (27) and thus that of Theorem 3 is herewith complete. □

5. Order of cr(Zp(n))

For simplicity we write || • ||p = || •; lp(n)\\ (1 n-1 / p||x||p for each fixed x in R n is increasing on [1, oo]. Similarly P ^ Ik I Ip f°r each fixed x in R n is decreasing on [1, oo]. Based on these two facts we can see that

n1//pcr(Z1(n)) < a(lp(n)) < n 1~1 pa( l 1 (n)) (1 < p < 2); (28)

n1/ 2- 1/p(J(;2(n)) < a(lp{n)) < n1/p_1/,2cr(Z2(n)) (1 < p < 2); (29)

a(lp(n)) < n1//2_1//pcr(/2(n)) (2 < p < oo); (30)

n-1//p<j(/°°(n)) < a(lp(n)) ( 2 < p < o o ) . (31)

Using these inequalities we prove the following result.

Theorem 4. The orders of the best constants cr(lp(n)) of lp(n) as n | oo are exactly given for 2 < p < oo as

a (lp(n)) « n1-1/p (32)

and are estimated for 1 < p < 2 as

n m a x (l /p -l /2 ,l - l /p ) ^ cr(lP(n)) ^ n m in (3 /2 -l/p ,l/p ) (3 3 )

Proof. Since a(l2 (n)) w y/n by (12), the above (30) implies that a{lp(n)) ^ n1_1/p (2 < p < oo). Similarly (13) and (31) yield n1_1/p -< a(lp(n)) (2 < p < oo). These two relations assure the validity of (32). By (11) and (28), we see that

ni /P- i /2 ^ a(lp(n)) ^ n3/2_1/p ( l < p < 2 ) . (34)

Similarly, from (13) and (29) it follows that

n i - i / p ^ a(lp(n)) n1//p (1 < p < 2). (35)

Combining (34) and (35), we deduce (33). □


References

1. A. Dvoretzky nd C.A. Rogers, Absolute and conditional convergence in normed linear spaces, Proc. Nat. Acad. Sci. U.S.A. 36 (1950), 192-197.

2. J.T. Marti, Introduction to the Theory of Bases, Springer, 1969.3. W. Rudin, Real and Complex Analysis, 2nd edition, McGraw-Hill, 1974.4. W. Rudin, Real and Complex Analysis, 3rd edition, McGraw-Hill, 1987.5. I. Singer, Bases in Banach Spaces I, Springer, 1970.6. K. Yosida, Functional Analysis, Springer, 1965.

Mitsuru NakaiNagoya Institute of TechnologyGokisoShowaNagoya [email protected]

[email protected]

Toshimasa TadaDaido Institute of TechnologyDaidoMinamiNagoya 457JAPAN

t ada@ asuke. daido-it .ac.jp

mailto:[email protected]

mailto:[email protected]

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