reﬁnements of s-orlicz convex functions in normed linear ... · also, we deﬁne the class of...

Int. J. Contemp. Math. Sciences, Vol. 3, 2008, no. 32, 1569 - 1594

Refinements of s-Orlicz Convex Functions

in Normed Linear Spaces

Mohammad Alomari 1 and Maslina Darus 2

School of Mathematical SciencesFaculty of Science and TechnologyUniversiti Kebangsaan MalaysiaBangi 43600 Selangor, Malaysia

Abstract

In this paper a generalization of s–convexity 0 < s ≤ 1 in bothsense are established. It is proved among others, that s–convexity inthe second sense is essentially stronger than the s–convexity in first,original sense, whenever 0 < s < 1. Some of properties of s–convexfunction in both sense are considered and various examples are given.

Keywords: s–Convex function in the first sense, Jensen’s inequality, Lin-ear spaces, s–Orlicz convex function, s–Orlicz convex set

1 Introduction

In [9], Orlicz introduced two definitions of s–convexity of real valued functions.A function f : R+ → R, where R+ = [0,∞), is said to be s–convex in the firstsense if

f (αx + βy) ≤ αsf (x) + βsf (y) (1)

for all x, y ∈ [0,∞), α, β ≥ 0 with αs + βs = 1 and for some fixed s ∈ (0, 1].We denote this class of functions by K1

s .

Also, a function f : R+ → R, where R+ = [0,∞), is said to be s–convexin the second sense if

f (αx + βy) ≤ αsf (x) + βsf (y) (2)

1First author: [email protected] author: [email protected]

1570 M. Alomari and M. Darus

for all x, y ∈ [0,∞), α, β ≥ 0 with α + β = 1 and for some fixed s ∈ (0, 1]. Wedenote this class of functions by K2

s .

This definition of s–convexity, for so called ϕ–functions, was introduced byOrlicz in [9] and was used in the theory of Orlicz spaces (see [7], [8], [10]). Afunction f : R+ → R+ is said to be ϕ–function if f (0) = 0 and f is non–decreasing and continuous. Its easily to check that the both s–convexity meanjust the convexity when s = 1.

In [5], Hudzik and Maligranda established some results about s–convexfunction in the first sense. The property that f is in the first sense then f isnon–decreasing on (0,∞) is given as follows:

Theorem 1.1 Let 0 < s ≤ 1. If f ∈ K1s , then f is non–decreasing on

(0,∞) and limx→0+

f (x) ≤ f (0).

The above result does not mean hold in general in the case of convex functions.that is, where s = 1, as the convex function f : R+ → R need not be non–decreasing nor non–negative. If 0 < s < 1, then the function f ∈ K1

s isnon–decreasing in (0,∞) but not necessarily on [0,∞) (see [2]).

In [5] H. Hudzik and L. Maligranda introduced some results on s–convexfunctions in the first sense, as follows:

Theorem 1.2 Let 0 < s < 1 and let p : [0,∞) → [0,∞) be a non–decreasing function. Then the function f defined for x ∈ [0,∞) by

f (x) = xs/(1−s)p (x) ∈ K1

s . (3)

Also, we define the class of s–Orlicz convex functions defined on s–Orlicz setsin a linear space.

Definition 1.3 Let V be a linear space and s ∈ (0,∞). The set M ⊆ Vwill be called s–Orlicz convex in L if the following condition is true:

If x, y ∈ M and α, β ≥ 0 with αs + βs = 1. Then (αx + βy) ∈ M .

The main goal in this paper is to pave the way for us to talk about the applica-tions of Orlicz convexity in functional analysis. Indeed, this paper is devotedto extend the circle of convexity in the Orlicz spaces; which in turn, going tobe used in the future to give another vision to some applications in real andfunctional analysis.

s-Orlicz convex functions 1571

2 Preliminary Results

Firstly, we introduce the definitions of E– and E–s–Orlicz convex set M in anormed linear space V as follows:

Definition 2.1 Let V, W be two finite dimensional normed linear spacesover R. Let E : M ⊆ V → W be a continuous linear operator. A set M ⊆ Vis said to be E–Orlicz convex iff for each x, y ∈ M , α, β ≥ 0 with α + β = 1.We have {αE (x) + βE (y)} ∈ M .

Definition 2.2 Let V, W be two finite dimensional normed linear spaceover R and s ∈ (0,∞). Let E : M ⊆ V → W be a continuous linear op-erator. The set M ⊆ V will be called E–s–Orlicz convex in V over R if thefollowing conditions is true:

If x, y ∈ M and α, β ≥ 0 with αs + βs = 1. Then (αE(x) + βE(y)) ∈ M .

For instance, if every M ⊆ V is E–s–Orlicz convex set, then V will be calledE–s–Orlicz convex linear space, and if M is a linear subspace of V then Mwill be called E–s–Orlicz convex linear subspace.

Theorem 2.3 If M ⊆ V is an E–s–Orlicz convex subset, then E (M) ⊆ Mis an E–s–Orlicz convex.

Proof. Since M is an E–s–Orlicz convex, then for any x, y ∈ M and α, β ≥ 0such that αs + βs = 1 with s ∈ (0,∞) we have,

{αE (x) + βE (y)} ∈ M.

Thus, if α = 1 then E (y) ∈ M and since y ∈ M is an arbitrary elementtherefore E (M) ⊆ M is an E–s–Orlicz convex.

Theorem 2.4 Let E (M) be an E–s–Orlicz convex and E (M) ⊆ M . ThenM is an E–s–Orlicz convex.

Proof. Assume that x, y ∈ M , since E (M) is an E–s–Orlicz convex thenE (x) and E (y) ∈ E (M) and for each α, β ≥ 0 such that αs + βs = 1 withs ∈ (0,∞), we have

{αE (x) + βE (y)} ∈ E (M) ⊆ M. (4)

Hence, M is an E–s–Orlicz convex.

Theorem 2.5 Let M1, M2 ⊆ V be two E–s–Orlicz convex subsets. Then,M1 ∩ M2 is an E–s–Orlicz convex subspace.


Proof. Let x, y ∈ M1 ∩ M2. Since M1, M2 ⊂ V are two E–s–Orlicz con-vex subspaces, then {αE (x) + βE (y)} ∈ M1 and {αE (x) + βE (y)} ∈ M2.Hence, {αE (x) + βE (y)} ∈ M1∩M2, which shows that M1∩M2 is E–s–Orliczconvex subspace.

Theorem 2.6 Let M ⊆ V be an E1– and E2–s–Orlicz convex. Then M isan E1 ◦ E2– and E2 ◦ E1–s–Orlicz convex.

Proof. Assume that x, y ∈ M , we show that

{α (E1 ◦ E2) (x) + β (E1 ◦ E2) (y)} ∈ M.

That is,

{αE1 (E2 (x)) + βE1 (E2 (y))} ∈ M.

Set z1 = E1 (x) and z2 = E2 (y). Since by Theorem 2.3 and for all x, y ∈ M ,we have E1 (x) , E2 (y) ∈ M , which gives

{αE1 (z1) + βE2 (z2)} ∈ M.

Hence, E1 ◦E2–s–Orlicz convex. Similarly, one can show that E2 ◦E1–s–Orliczconvex.

Theorem 2.7 Let M1, M2 ⊂ V be two E–s–Orlicz convex subspaces. ThenM1 + M2 is an E–s–Orlicz convex subspace.

Proof. Suppose that E : V → W is a continuous linear operator. Let(p + q) , (x + y) ∈ M1 + M2, where p, x ∈ M1 and q, y ∈ M2. If α, β ≥ 0such that αs + βs = 1 with s ∈ (0,∞). Then we have,

αE (p + q) + βE (x + y) = α {E (p) + E (q)} + β [E (x) + E (y)]

= {αE (p) + βE (x)} + {αE (q) + βE (y)}∈ M1 + M2

Thus, M1 + M2 is an E–s–Orlicz convex subspace.

Remark 2.8 There exists E–convex set in normed linear space which isnot E–s–Orlicz for some s ∈ (0,∞) \ {1}.

Example 2.9 Let E : R2 → R2 be a continuous linear operator, definedsuch as, E (x, y) = (x, y). consider M =

{(x, y) ∈ R2 : x2 + y2 ≤ 1

}. We will

show that M is not E–2–Orlicz convex.


Let x = (x1, y1), y = (x2, y2) with x2i + y2

i = 1, (i = 1, 2) and x1x2, y1y2 >0. Consider α, β with α2 + β2 = 1. Then, αE (x) + βE (y) = αx + βy =(αx1 + βx2, αy1 + βy2). Therefore, we have

(αx1 + βx2)2 + (αy1 + βy2)

2 = α2(x2

1 + x22

)+ β2

(y2

1 + y22

)+ 2αβ (x1x2 + y1y2)

= α2 + β2 + 2αβ (x1x2 + y1y2)

= 1 + 2αβ (x1x2 + y1y2) > 1.

Thus, x, y ∈ M but αx + βy /∈ M , which shows that M is E–Orlicz but notE–2–Orlicz convex.

Remark 2.10 There exists E–s–Orlicz convex set in normed linear spacewhich is not E–Orlicz for some s ∈ (0,∞) \ {1}.

Example 2.11 Let E : R2 → R2 be a continuous linear operator, defined

such as, E (x, y) = (x, y). Consider M ={(x, y) ∈ R2 :

√|x| +

√|y| ≤ 1

}.

We will show that M is not E–Orlicz convex. Let x = (x1, y1), y = (x2, y2)

with x12i +y

12i = 1, (i = 1, 2) which imply α

12

√|x1|+α

12

√|y1| ≤ α

12 and β

12

√|x2|+

β12

√|y2| ≤ β

12 , then

α12

√|x1| + α

12

√|y1| + β

12

√|x2| + β

12

√|y2| ≤ α

12 + β

12 = 1.

But we know the inequalities

α12

√|x1| + β

12

√|x2| ≥

√α |x1| + β |x2| ≥

√|αx1 + βx2|

andα

12

√|y1| + β

12

√|y2| ≥

√α |y1| + β |y2| ≥

√|αy1 + βy2|

which give by addition that√|αx1 + βx2| +

√|αy1 + βy2| ≤ 1.

Therefore, we deduce that αx + βy ∈ M , i.e., M is E–12–Orlicz convex in R2.

Also, it is clear that M is not E–convex.

Corollary 2.12 Any subset M of an E–s–Orlicz convex linear space V withthe properties:

(i) For every x, y ∈ M , x + y ∈ M .

(ii) For every x ∈ M and α ≥ 0, αx ∈ M .

is E–s–Orlicz convex linear subspace for every s ∈ (0,∞).


The following theorem generalize a standard result about s–Orlicz convexsets.

Theorem 2.13 The linear space V is E–s–Orlicz convex iff every subsetM of V is E–s–Orlicz convex.

Proof. (⇐) is done by Definition 2.2 .

(⇒) Suppose on the contrary there is a subset M0 of V which is not E–s–Orlicz convex. Let x, y ∈ M0 and α, β ≥ 0 with αs + βs = 1. Let E : M0 ⊆V → W be a linear operator and set z = αx+βy ∈ M0, then E(z) = αE(x)+βE(y) ∈ W . Now, let x1, x1, ..., xr be a basis of M0 and d1, d2, ..., dr ∈ R, then

we can write z as z =r∑

i=1dixi . Thus, E(z) =

r∑i=1

diE(xi) ∈ E(M0). Hence,

E−1(E(z)) =r∑

i=1diE

−1(E(xi)) ∈ E−1(E(M0)), and so, z =r∑

i=1dixi ∈ M0 which

contradicts our assumption.

Corollary 2.14 The normed linear space V is E–1–Orlicz convex or simplyE–convex iff every subset M of V is E–Orlicz convex.

Proof. The proof is straight forward and we will omit the details.

Theorem 2.15 Let V, W be two normed linear spaces, s ∈ (0,∞), andE : V → W be a continuous linear operator. For a given subset M ⊆ V , thefollowing statements are equivalent:

(i) M is E–s–Orlicz convex;

(ii) For every x1, x2, ..., xn ∈ M and α1, α2, ..., αn ≥ 0 withn∑

i=1αs

i = 1 we

have thatn∑

i=1αiE(xi) ∈ M .

Proof. (ii) ⇒ (i). By Corollary 2.12 and Theorem 2.13 done.

(i) ⇒ (ii). The proof carries by induction. For n = 2, the arguments followsby Definition 2.2. . Suppose the statement holds for all 2 ≤ k ≤ n − 1. Let

x1, x2, ..., xn ∈ M and α1, α2, ..., αn ≥ 0 withn∑

i=1αs

i = 1, then by Theorem 2.3

E(M) is E–s–Orlicz, and if α1 = α2 = ... = αn−1 = 0, then αn = 1 and thusn∑

i=1αixi ∈ M . Since E(M) is E–s–Orlicz, and E is linear then

E

(n∑

i=1

αixi

)=

n∑i=1

αiE(xi) (5)


Now, suppose thatn−1∑i=1

αsixi > 0, and consider βj = αj(

n−1∑i=1

αsi

)1/s, then

n−1∑j=1

βj = 1,

j = 1, 2, ..., n − 1. Put y =n−1∑j=1

βjxj , and therefore E (y) =n−1∑j=1

βjE (xj). By

the induction hypothesis we have that y ∈ M and by Theorem 2.3 E (y) ∈E (M) ⊆ M . Now, by (5) one can get the following:

n∑i=1

αiE(xi) =

n−1∑j=1

αjE (xj)

(n−1∑i=1

αsi

)1/s

(n−1∑i=1

αsi

)1/s

+ αnE(xn)

=

(n−1∑i=1

αsi

)1/s

E(y) + αnE(xn) .

As(

n−1∑i=1

αsi

)+αs

n =n∑

i=1αi = 1, and E (xi) ∈ E (M) ⊆ M , for all i (i = 1, 2, ..., n − 1)

it follows that(

n−1∑i=1

αsi

)1/s

E(y) + αnE (xn) ∈ E (M) ⊆ M and the theorem is

thus proved.

Corollary 2.16 Let V, W be two normed linear spaces and E : V → Wbe a continuous linear operator. For a given subset M ⊆ V , the followingstatements are equivalent:

(i) M is E–1–Orlicz convex;

(ii) For every x1, x2, ..., xn ∈ M and α1, α2, ..., αn ≥ 0 withn∑

i=1αi = 1 we

have thatn∑

i=1αiE(xi) ∈ M .


Remark 2.17 Definition 2.1 and Theorem 2.3 in [4] is a special case ofDefinition 2.2 and Theorem 2.15 respectively, when s = 1 and consider theoperator E to be the identity operator.

Next, we generalize the concept of a convex hull to E–s–Orlicz convex hulland E–Orlicz convex hull, respectively; as follows:

Theorem 2.18 Let V, W be two normed linear spaces and E : V → W bea linear mapping.

cosE (M) =

{n∑

i=1

αiE (xi) : αi ≥ 0,n∑

i=1

αsi = 1, xi ∈ M, n ≥ 2

}.


Then, cosE (M) is an E–s–Orlicz convex set and will be called the E–s–Orlicz

convex hull of the set M .

Proof. Let x, y ∈ cosE (M), then x, y can be represented by

x =n∑

i=1αiE (xi), with αi ≥ 0, xi ∈ M and

n∑i=1

αsi = 1, n ≥ 2 ,

and

y =m∑

j=1βjE (yj), with βj ≥ 0, yj ∈ M and

m∑j=1

βsj = 1, m ≥ 2 .

Consider α, β ≥ 0 with α + β = 1. Then

αx + βy = αn∑

i=1αiE (xi) + β

m∑j=1

βjE (yj) =n+m∑k=1

λkE (zk),

where, λi = ααi ; (i = 1, 2, ..., n), λi = ββj ; (j = 1, 2, ..., m).

and

zi = xi ; (i = 1, 2, ..., n), zj = yj−n ; (j = n + 1, n + 2, ..., n + m) .

We have

n+m∑k=1

λsk = αs

n∑i=1

αsi + βs

m∑j=1

βsj = αs + βs = 1

which shows that αx + βy ∈ cosE (M) and the statement is proved.

Now, by Theorem 2.18 it’s easy to define the E–Orlicz convex hull, asfollows:

Corollary 2.19 In Theorem 2.18 , if s = 1, then

co1E (M) =

{n∑

i=1

αiE (xi) : αi ≥ 0,n∑

i=1

αi = 1, xi ∈ M, n ≥ 2

}

is an E–1–Orlicz convex set and will be called the E–Orlicz convex hull of theset M .


3 E– and E–s–Orlicz Mapping

Let V be a real normed linear space, s fixed positive number, and M ⊆ V anE–s–Orlicz convex set.


Definition 3.1 Let V, W be two normed linear spaces. A mapping f : M →R is said to be E–s–Orlicz convex on an E–s–Orlicz convex set M ⊆ V iffthere is a continuous linear map E : M → W such that for each x, y ∈ M andα, β ≥ 0 so that αs + βs = 1 and s ∈ (0,∞) one has

f (αE (x) + βE (y)) ≤ αsf (E (x)) + βsf (E (y)) . (6)

On the other hand, if

f (αE (x) + βE (y)) ≥ αsf (E (x)) + βsf (E (y)) . (7)

Then f is called E–s–Orlicz concave on M .

Definition 3.2 In Definition 3.1. set s = 1, then for all x, y ∈ M andα, β ≥ 0 so that α + β = 1, and

f (αE (x) + βE (y)) ≤ αf (E (x)) + βf (E (y)) . (8)

Then, f is said to be E–Orlicz convex mapping on M .


f (αE (x) + βE (y)) ≥ αf (E (x)) + βf (E (y)) . (9)

f is called E–Orlicz concave on M .

Theorem 3.3 Let E : M → W be a continuous linear operator on E–s–Orlicz convex subset M of linear space V , and f : M → R be E–s–Orliczconvex mapping on M . If for each x, y ∈ M , α, β ≥ 0 so that αs + βs = 1 ands ∈ (0,∞). Then the set

fxE (η) = {x ∈ M : f (E (x)) ≤ η} ,

is nonempty E–s–Orlicz convex subset of M .

Proof. Let x, y ∈ fxE (η) and α, β ≥ 0 so that αs + βs = 1. Then

f (E (x)) ≤ η and f (E (y)) ≤ η,

which imply that

αsf (E (x)) ≤ αsη and βsf (E (y)) ≤ βsη

and thus

f (αE(x) + βE(y)) ≤ αsf (E (x)) + βsf (E (y)) ≤ (αs + βs) η = η.

which show that fxE (η) is an E–s–Orlicz convex subset of M .


Definition 3.4 A mapping f : M → R is said to be a semi–E–s–Orliczconvex on a set M ⊆ V iff there is a continuous linear operator E : M → Wsuch that M is E–s–Orlicz convex set and for each x, y ∈ M and α, β ≥ 0 sothat αs + βs = 1 and s ∈ (0,∞) one has

f (αE (x) + βE (y)) ≤ αsf (x) + βsf (y) . (10)


f (αE (x) + βE (y)) ≥ αsf (x) + βsf (y) . (11)

Then f is called semi–E–s–Orlicz concave on M .

Note that, In Definition 3.4 if s = 1, then for each x, y ∈ M and α, β ≥ 0so that α + β = 1, and

f (αE (x) + βE (y)) ≤ αf (x) + βf (y) . (12)

Then, f is said to be a semi–E–Orlicz mapping on M . On the other hand, if

f (αE (x) + βE (y)) ≥ αf (x) + βf (y) . (13)

Then f is called semi–E–Orlicz concave on M .

Theorem 3.5 If a function f : M → R is semi–E–s–Orlicz convex on anE–s–Orlicz convex set M ⊆ V then f (E (x)) ≤ f (x) for each x ∈ M .

Proof. Since f is semi–E–s–convex on an E–convex set M ⊂ V then forany x, y ∈ M , α, β ≥ 0 so that αs + βs = 1 and s ∈ (0,∞), we have{αE (x) + βE (y)} ∈ M and

f (αE (x) + βE (y)) ≤ αsf (x) + βsf (y)

Thus, for α = 1, f (E (x)) ≤ f (x) for each x ∈ M .

Corollary 3.6 Let E : M → W , be a continuous linear operator on E–s–Orlicz convex subset M of a normed linear space V , and f : M → R besemi–E–s–Orlicz convex mapping on M . If for each x, y ∈ M , α, β ≥ 0 sothat αs + βs = 1 and s ∈ (0,∞). Then the set

Bx (η) = {x ∈ M : f (x) ≤ η} ,

is E–s–Orlicz convex subset of M .

Proof. By Theorem 3.5 f (E (x)) ≤ f (x) for each x ∈ M and by Theorem3.3 we get the result.


Remark 3.7 An E–s–convex function on E–s–convex set need not be semi–E–s–convex function.

Example 3.8 Let E : R2 → R2 be defined as E(x, y) = (1 + x, y) andlet f : R2 → R be defined as f(x, y) = x2 + y2, f is an E–2–Orlicz convex

function on a set M ={(x, y) ∈ R2 : x2 + y2 ≥ 1

}. Take (x, y) = (1, 2) there-

fore f (E (1, 2)) = f (2, 2) = 8 > 5 = f (2, 1), then by Theorem 3.5 f is not asemi–E–2–Orlicz convex function.

However, let E : R2 → R2 be defined as E (x, y) =(

1x, 1

1+y

)with x2 + y2 ≥ 1,

and consider f as above on the same set M . Take (x, y) =(√

2, 0), then,

f(E(√

2, 0))

= f(

1√2, 1)

= 32

< 2 = f(√

2, 0).

Theorem 3.9 Let V , W be two normed linear space and M be E–s–Orliczsubset in V . Let E : M → W be a continuous linear operator on M and f :M → R be a mapping defined on M . The following statements are equivalent:

(i) f is E–s–Orlicz convex on M .

(ii) For every αi ≥ 0 and s ∈ (0,∞) such thatn∑

i=1αs

i = 1, one has the

inequality

f

(n∑

i=1

αiE (xi)

)≤

n∑i=1

αsif (E (xi)) (14)

Proof. (ii) ⇒ (i). This is obvious and we are done.

(i) ⇒ (ii). Since M be E–s–Orlicz convex of V and For every αi ≥ 0 such

thatn∑

i=1αs

i = 1, thenn∑

i=1αiE (xi) ∈ M . The proof carries by induction over

n ≥ 2. If n = 2, then the inequality holds by Definition 3.1 .

Now, for 2 ≤ k ≤ n − 1, let x1, x2, ..., xn ∈ M and α1, α2, ..., αn ≥ 0 withn∑

i=1αs

i = 1, therefore E(x1), E(x2), ..., E(xn) ∈ E(M) ⊆ M . If α1 = ... =

αn−1 = 0 then αn = 1 and the inequality (14) is hold.

Suppose thatn−1∑i=1

αsi ≥ 0 and put

βj =αj(

n−1∑i=1

αsi

)1/s, (1 ≤ j ≤ n − 1)


Thenn−1∑j=1

βsj = 1 and it’s easy to see that

n−1∑j=1

βjE(xj) ∈ M . By using the

induction hypothesis we also can state

f

⎛⎝n−1∑

j=1

βjE(xj)

⎞⎠ ≤

n−1∑i=1

βjf (E (xj))

Now, we observe that

f

(n∑

i=1

αiE (xi)

)≤ f

⎛⎜⎜⎜⎜⎝(

n−1∑i=1

αsi

)1/sn−1∑i=1

αiE (xi)

(n−1∑i=1

αsi

)1/s+ αnE (xn)

⎞⎟⎟⎟⎟⎠

≤(

n−1∑i=1

αsi

)f

⎛⎜⎜⎜⎜⎝

n−1∑i=1

αiE (xi)

(n−1∑i=1

αsi

)1/s

⎞⎟⎟⎟⎟⎠+ αs

nf (E (xn)) (15)

we note that inequality (15) was obtained by Definition 3.1 in(

n−1∑i=1

αsi

)1/s

and

αn such as

((n−1∑i=1

αsi

)1/s)s

+ αsn = 1.

On the other hand,

f

⎛⎜⎜⎜⎜⎝

n−1∑i=1

αiE (xi)

(n−1∑i=1

αsi

)1/s

⎞⎟⎟⎟⎟⎠ ≤ f

⎛⎝n−1∑

j=1

βjE (xj)

⎞⎠

≤n−1∑j=1

βjf (E (xj))

=

n−1∑i=1

αif (E (xi))

n−1∑i=1

αsi

.

Therefore, by using the inequality (15) we get,

f

(n∑

i=1

αiE (xi)

)≤

(n−1∑i=1

αsi

) n−1∑i=1

αsi f (E (xi))

n−1∑i=1

αsi

+ αnf (E (xn))

=n∑

i=1

αsi f (E (xi))


and the theorem is proved.

Corollary 3.10 In Theorem 3.9 put s = 1. Then, the following statementsare equivalent:

(i) f is E–Orlicz convex on M .

(ii) For every αi ≥ 0 and s ∈ (0,∞) such thatn∑

i=1αi = 1, one has the

inequality

f

(n∑

i=1

αiE (xi)

)≤

n∑i=1

αif (E (xi)). (16)

Corollary 3.11 In Theorem 3.9, if the condition on the function f whichis E–s–Orlicz convex function replaced by semi–E–s–Orlicz convex functionthe same result will hold with a bit of changes in the inequality (14), such as

f

(n∑

i=1

αixi

)≤

n∑i=1

αsif (xi) (17)

Proof. It’s easy to see that by using Theorem 3.5 that is, f (E (x)) ≤ f (x)for each x ∈ M .

One can see that the result in Corollary 3.11 is somehow similar to theresult of Theorem 3.3 in [4], yet with different function f .

In [4], Dragomir and Fitzpatrick established the following functional :

Θ (I, p, f, x) =

(∑i∈I

psi

)f

⎛⎜⎜⎜⎜⎜⎝

∑i∈I

pixi(∑i∈I

psi

)1/s

⎞⎟⎟⎟⎟⎟⎠

where pi > 0, i ∈ I, f is a s–Orlicz convex mapping on the s–Orlicz convexset M ⊆ V , xi ∈ M , and I is from Pf (N), where Pf (N) denotes the finitesubsets of the natural numbers set N.

Now, we will construct a new functional ΘE from the above functional Θby modifying the function f to be semi–E–s–Orlicz convex as follows :

Suppose that

ΘE (I, p, f, E (x)) =

(∑i∈I

psi

)f

⎛⎜⎜⎜⎜⎜⎝∑i∈I

piE (xi)(∑i∈I

psi

)1/s

⎞⎟⎟⎟⎟⎟⎠ ,


with the same conditions on the functional above, then we state the following:

Theorem 3.12 Let f : M ⊆ V → R be an E–s–Orlicz convex mapping onthe E–s–Orlicz convex set M , pi > 0 (i ∈ N) and xi ∈ M . Then

(1) for all I, J ∈ Pf (N) with I ∩ J = ∅, one has the inequality

ΘE (I ∪ J, p, f, E (x)) ≥ ΘE (I, p, f, E (x)) + ΘE (J, p, f, E (x)) ≥ 0 (18)

that is, the mapping ΘE is superadditive on the first variable, and

(2) for all I, J ∈ Pf (N) with ∅ �= J ⊆ I, one has the inequality

ΘE (I, p, f, E (x)) ≥ ΘE (J, p, f, E (x)) ≥ 0 (19)

that is, the mapping ΘE is monotonic non–decreasing in the first variable onPf (N).

Proof. (1) Let I, J ∈ Pf (N) with I ∩ J = ∅. Then we have

ΘE (I ∪ J, p, f, E (x)) =∑i∈I

psif (E(xi)) +

∑j∈J

psjf (E(xj))

−⎛⎝ ∑

k∈I∪J

psk

⎞⎠ f

⎛⎜⎜⎜⎜⎜⎝∑i∈I

piE(xi) +∑j∈J

pjE(xj)

( ∑k∈I∪J

psk

)1/s

⎞⎟⎟⎟⎟⎟⎠

=∑i∈I

psif (E(xi)) +

∑j∈J

psjf (E(xj))

−⎛⎝ ∑

k∈I∪J

psk

⎞⎠ f

⎡⎢⎢⎢⎢⎢⎣

⎛⎜⎜⎜⎜⎝∑i∈I

psiE(xi)( ∑

k∈I∪Jps

k

)⎞⎟⎟⎟⎟⎠

1/s ∑i∈I

piE(xi)(∑i∈I

psi

)1/s

+

⎛⎜⎜⎜⎜⎝∑j∈J

psjE(xj)( ∑

k∈I∪Jps

k

)⎞⎟⎟⎟⎟⎠

1/s ∑j∈J

pjE(xj)

(∑j∈J

psj

)1/s

⎤⎥⎥⎥⎥⎥⎦

≥ ∑i∈I

psif (E (xi)) +

∑j∈J

psjf (E (xj))

−⎛⎝ ∑

k∈I∪J

psk

⎞⎠⎡⎢⎢⎢⎢⎢⎣

⎛⎜⎜⎜⎜⎝∑i∈I

psiE (xi)( ∑

k∈I∪Jps

k

)⎞⎟⎟⎟⎟⎠

1/s

f

⎛⎜⎜⎜⎜⎜⎝∑i∈I

piE (xi)(∑i∈I

psi

)1/s

⎞⎟⎟⎟⎟⎟⎠


+

⎛⎜⎜⎜⎜⎝∑j∈J

psjE (xj)( ∑

k∈I∪Jps

k

)⎞⎟⎟⎟⎟⎠

1/s

f

⎛⎜⎜⎜⎜⎜⎝∑j∈J

pjE (xj)

(∑j∈J

psj

)1/s

⎞⎟⎟⎟⎟⎟⎠

⎤⎥⎥⎥⎥⎥⎦

=∑i∈I

psif (E (xi)) −

∑i∈I

psif

⎛⎜⎜⎜⎜⎜⎝∑i∈I

piE (xi)(∑i∈I

psi

)1/s

⎞⎟⎟⎟⎟⎟⎠

+∑j∈J

psjf (E (xj)) −

∑j∈J

psjf

⎛⎜⎜⎜⎜⎜⎝∑j∈J

pjE (xj)

(∑j∈J

psj

)1/s

⎞⎟⎟⎟⎟⎟⎠

= ΘE (I, p, f, E (x)) + ΘE (J, p, f, E (x))

and thus inequality (18) is proved.

(2) Suppose that J ⊂ I with J �= ∅ and J �= I. Then we know

ΘE (I, p, f, E (x)) = ΘE (I ∪ (I\J) , p, f, E (x))

≥ ΘE (J, p, f, E (x)) + ΘE (I\J, p, f, E (x)) .

Therefore,

ΘE (I, p, f, E (x)) − ΘE (J, p, f, E (x)) ≥ ΘE (I\J, p, f, E (x)) ≥ 0

and the inequality (19) is proved.

Corollary 3.13 In Theorem 3.12, consider f to be a semi–E–s–Orlicz con-vex mapping, then we have:


ΘE (I ∪ J, p, f, x) ≥ ΘE (I, p, f, x) + ΘE (J, p, f, x) ≥ 0



ΘE (I, p, f, x) ≥ ΘE (J, p, f, x) ≥ 0



Proof. Just we note that since f is semi–E–s–Orlicz convex mapping, thenby Theorem 3.5 we get f (E (x)) ≤ f (x) for each x ∈ M , and the proof isstraight forward.

Corollary 3.14 In Corollary 3.13, since f is a semi–E–s–Orlicz convexmapping, then we have:


ΘE (I ∪ J, p, f, x) ≥ ΘE (I ∪ J, p, f, E (x)) ≥ 0



ΘE (I, p, f, x) ≥ ΘE (J, p, f, E (x)) ≥ 0


Corollary 3.15 In Theorem 3.12, let E : M → E (M) be the identityoperator E (x) = x. Then


Θx (I ∪ J, p, f, x) ≥ Θx (I, p, f, x) + Θx (J, p, f, x) ≥ 0

that is, the mapping Θx is superadditive on the first variable, and


Θx (I, p, f, x) ≥ Θx (J, p, f, x) ≥ 0

that is, the mapping Θx is monotonic non–decreasing in the first variable onPf (N).

Proof. In Theorem 3.12, set E : M → W be the identity operator E (x) = xand straight forward.

4 Generalized Inequalities For Double Sums

We will start with the following lemma which will be used in the next section.


Theorem 4.1 Let f : M ⊂ V → R be an E–s–Orlicz convex map on the

E–s–Orlicz set M and αsi ≥ 0 so that

n∑i=1

αsi = 1. Let xij ∈ V , 1 ≤ i, j ≤ n.

Then one has the inequalities

n∑i=1

n∑j=1

αsiα

sjf (E (xij)) ≥ max {A, B} ≥ min {A, B} ≥ f

⎛⎝ n∑

i=1

n∑j=1

αsi α

sjE (xij)

⎞⎠

where;

A =n∑

i=1

αsif

⎛⎝ n∑

j=1

αjE (xij)

⎞⎠ , B =

n∑j=1

αsjf

(n∑

i=1

αiE (xij)

).

Proof. Fix i ∈ {1, 2, ..., n}. Thus, by Theorem 3.9 we can state

f

⎛⎝ n∑

j=1

αiE (xij)

⎞⎠ ≤

n∑j=1

αsjf (E (xij))

by multiplying the both side by αsi ≥ 0, we have

αsi f

⎛⎝ n∑

j=1

αiE (xij)

⎞⎠ ≤

n∑j=1

αsi α

sjf (E (xij))

which gives, by addition that

f

⎛⎝ n∑

i=1

n∑j=1

αiαjE (xij)

⎞⎠ ≤

n∑i=1

αsi f

⎛⎝ n∑

j=1

αiE (xij)

⎞⎠ ≤

n∑i=1

n∑j=1

αsiα

sjf (E (xij)).

Thus,

f

⎛⎝ n∑

i=1

n∑j=1

αiαjE (xij)

⎞⎠ ≤ A ≤

n∑i=1

n∑j=1

αsiα

sjf (E (xij)).

The second part is proved similarly.

Corollary 4.2 Let f : M ⊂ V → R be a semi–E–s–Orlicz convex map on

the E–s–Orlicz set M and αsi ≥ 0 so that

n∑i=1

αsi = 1. Let xij ∈ V , 1 ≤ i, j ≤ n.

Then one has the inequalities

n∑i=1

n∑j=1

αsi α

sjf (xij) ≥ max {A, B} ≥ min {A, B} ≥ f

⎛⎝ n∑

i=1

n∑j=1

αsi α

sjxij

⎞⎠

where;

A =n∑

i=1

αsi f

⎛⎝ n∑

j=1

αjxij

⎞⎠ , B =

n∑j=1

αsjf

(n∑

i=1

αixij

).



Theorem 4.3 Let f : M ⊂ V → R be an E–s–Orlicz convex map on

the E–s–Orlicz convex set M and αsi ≥ 0 such that

n∑i=1

αsi = 1. Then for all

α, β ≥ 0 with αs + βs = 1 one has the inequality

f

⎛⎝(α + β)

21s−1

n∑i=1

αi

n∑i=1

αiE(xi)

⎞⎠ ≤

n∑i=1

n∑j=1

αsiα

sjf

⎛⎝(α + β)E (xi + xj)

21s

⎞⎠

≤n∑

i=1

n∑j=1

αsiα

sjf (E (αxi + βxj))

≤n∑

i=1

αsi f (E (xi)) (20)

Proof. Firstly, we note that since f is E–s–Orlicz convex map, and for allx, y ∈ M one has the inequalities:

f (αE(x) + βE(y)) ≤ αsf (E(x)) + βsf (E(y))

and

f (αE(y) + βE(x)) ≤ αsf (E(y)) + βsf (E(x))

which gives by addition

f (αE(x) + βE(y)) + f (αE(y) + βE(x))

≤ (αs + βs) (f (E(x)) + f (E(y)))

= f (E(x)) + f (E(y)) . (21)

Moreover, if in the definition of s–Orlicz convexity of f , we take α = β = 2−1s ,

we have αs + βs = 12

+ 12

= 1 then the inequality

f

⎛⎝E(x + y)

21s

⎞⎠ = f

⎛⎝E (x) + E (y)

21s

⎞⎠

≤ f (E(x)) + f (E(y))

2(22)

holds for all x, y ∈ M . Thus, by inequality (21) we conclude that

f

⎛⎝(α + β) (E(x + y))

21s

⎞⎠ = f

⎛⎝(α + β) (E (x) + E (y))

21s

⎞⎠

≤ f (αE(x) + βE(y)) + f (αE(y) + βE(x))

2.


Now, by (22), we can state

f (E (αxi + βxj)) + f (E (βxi + αxj)) ≤ f (E (xi)) + f (E (xj))

for all i, j ∈ {1, 2, ..., n}.

If we multiply this inequality by αsi α

sj ≥ 0 and sum over i, j from 1 to n,

we have

2n∑

i=1

n∑j=1

αsi α


=n∑

i=1

n∑j=1

αsiα

sj [f (E (αxi + βxj)) + f (E (βxi + αxj))]

≤ 2n∑

i=1

αsi

n∑j=1

αsi f (E (xi))

Thus,

n∑i=1

n∑j=1

αsiα

sjf (E (αxi + βxj)) ≤

n∑i=1

αsif (E (xi))

and the first inequality in (20) is proved. To prove the second inequality in(20) we observe that

f

⎛⎝(α + β)E (xi + xj)

21s

⎞⎠ = f

⎛⎝(α + β) (E (xi) + E (xj))

21s

⎞⎠

≤ f (αE (xi) + βE (xj)) + f (αE (xj) + βE (xi))

2

for all i, j ∈ {1, 2, ..., n}. If we multiply this inequality by αsi α

sj ≥ 0 and sum

over i, j from 1 to n, we get

n∑i=1

n∑j=1

αsiα

sjf

⎛⎝(α + β)E (xi + xj)

21s

⎞⎠ ≤

n∑i=1

n∑j=1

αsiα


which gives the second inequality in (20).

Now, set

xij =(α + β) (xi + xj)

21s

,

and since E is linear we have

E (xij) =(α + β) E (xi + xj)

21s

.


Thus, by using inequality (6), that is the definition of E–s–Orlicz convexity off for xij we get

f

⎛⎝α + β

21s−1

n∑i=1

αi

n∑i=1

αiE (xi)

⎞⎠

= f

⎛⎝α + β

21s

n∑i=1

n∑j=1

αiαjE (xi + xj)

⎞⎠

= f

⎛⎝ n∑

i=1

n∑j=1

αiαj(α + β) E (xi + xj)

21s

⎞⎠

≤n∑

i=1

n∑j=1

αsi α

sjf

⎛⎝(α + β)E (xi + xj)

21s

⎞⎠

and the theorem has been proved.

Corollary 4.4 Let f : M ⊂ V → R be a semi–E–s–Orlicz convex map on

the E–s–Orlicz convex set M and αsi ≥ 0 such that

n∑i=1

αsi = 1. Then for all

α, β ≥ 0 with αs + βs = 1 one has the inequality

f

⎛⎝(α + β)

21s−1

n∑i=1

αi

n∑i=1

αixi

⎞⎠ ≤

n∑i=1

n∑j=1

αsiα

sjf

⎛⎝(α + β) xi + xj

21s

⎞⎠

≤n∑

i=1

n∑j=1

αsiα

sjf (αxi + βxj)

≤n∑

i=1

αsif (xi) (23)

Proof. Since f is semi–E–s–Orlicz convex map then f (E (x)) ≤ f (x), andthe details goes likewise the proof of Theorem 4.3 with semi–E–s–Orlicz convexmapping f .

Corollary 4.5 Suppose that f is E–s–Orlicz convex with the above assump-tions for M , αi one has the inequalities:

f

⎛⎜⎜⎝

n∑i=1

αi

n∑i=1

αiE (xi)

22s−2

⎞⎟⎟⎠ ≤

n∑i=1

n∑j=1

αsiα

sjf

⎛⎝E (xi) + E (xj)

22s−1

⎞⎠

≤n∑

i=1

n∑j=1

αsiα

sjf

⎛⎝E (xi) + E (xj)

21s

⎞⎠

≤n∑

i=1

αsif (E (xi)) .


Remark 4.6 If in Theorem 4.3 we assume that f is an E–Orlicz convexand α = t, β = 1 − t with t ∈ [0, 1], we have the inequalities:

f

(n∑

i=1

αiE (xi)

)≤

n∑i=1

n∑j=1

αiαjf

(E (xi) + E (xj)

2

)

≤n∑

i=1

n∑j=1

αiαjf (tE (xi) + (1 − t) E (xj))

≤n∑

i=1

αif (E (xi))

where, αi ≥ 0 andn∑

i=1αi = 1.

In [4], Dragomir and Fitzpatrick, established a variant result of Jensen’s in-equality, in the following we generalize this result for E–s–Orlicz convexity.

Theorem 4.7 Suppose that f , M , αi, xi, and α, β are as in Theorem 3.3.. Then one has the inequalities:

f

[(α + β

n∑i=1

αi

)αiE (xi)

]≤

n∑i=1

αsi f

⎛⎝αE (xi) + β

n∑j=1

αjE (xj)

⎞⎠

≤n∑

i=1

αsi f (E (xi)) (24)

Proof. By the E–s–Orlicz convexity of f one has

f

⎛⎝αE (xi) + β

n∑j=1

αjE (xj)

⎞⎠ ≤ αsf (E (xi)) + βsf

⎛⎝ n∑

j=1

αjE (xj)

⎞⎠

for all i ∈ {1, ..., n}. If we multiply with αsi ≥ 0 and sum over i we get

n∑i=1

αsi f

⎛⎝αE (xi) + β

n∑j=1

αjE (xj)

⎞⎠

≤ αsn∑

i=1

αsi f (E (xi)) + βsf

⎛⎝ n∑

j=1

αjE (xj)

⎞⎠ . (25)

By Jensen’s inequality (6) we have

αsn∑

i=1

αsif (E (xi)) + βsf

⎛⎝ n∑

j=1

αjE (xj)

⎞⎠

≤ αsn∑

i=1

αsif (E (xi)) + βs

n∑i=1

αsi f (E (xi))

≤n∑

i=1

αsif (E (xi))


and then the first inequality in (24) is proved.

The second inequality follows by Jensen’s result (6). Indeed one has

f

[(α + β

n∑i=1

αi

)αiE (xi)

]= f

⎛⎝ n∑

i=1

αi

⎛⎝αE (xi) + β

n∑j=1

αjE (xj)

⎞⎠⎞⎠

≤n∑

i=1

αif

⎛⎝αE (xi) + β

n∑j=1

αjE (xj)

⎞⎠

and the proof is established.

Corollary 4.8 Suppose that f is E–s–Orlicz convex with the above assump-tions for M , αi, and xi one has the inequalities:

f

⎛⎜⎜⎜⎝

1 +n∑

j=1αj

21s

n∑j=1

αjE (xj)

⎞⎟⎟⎟⎠ ≤

n∑i=1

αsif

⎛⎜⎜⎜⎝

E (xi) +n∑

j=1αjE (xj)

21s

⎞⎟⎟⎟⎠

≤n∑

i=1

αsif (E (xi)) .

Remark 4.9 If in Theorem 4.7 we assume that f is an E–Orlicz convexand α = t, β = 1 − t with t ∈ [0, 1], we have the inequalities:

f

(n∑

i=1

αiE (xi)

)≤

n∑i=1

αif

⎛⎝tE (xi) + (1 − t)

n∑j=1

αjE (xj)

⎞⎠

≤n∑

i=1

αif (E (xi))


i=1αi = 1.

Finally, we prove the following inequalities:

Theorem 4.10 Let f , M , xi, αi, and α, β as above. Then one has theinequalities:

2f

⎡⎢⎢⎣

(α + β)(1 +

n∑i=1

αi

)

21s

n∑i=1

αiE (xi)

⎤⎥⎥⎦

≤ 2n∑

i=1

αsif

⎡⎣α + β

21s

⎛⎝E (xi) +

n∑j=1

αjE (xj)

⎞⎠⎤⎦


≤n∑

i=1

αsi

⎡⎣f⎛⎝αE (xi) + β

n∑j=1

αjE (xj)

⎞⎠+ f

⎛⎝βE (xi) + α

n∑j=1

αjE (xj)

⎞⎠⎤⎦

≤n∑

i=1

n∑j=1

αsiα

sj

⎡⎢⎢⎣f⎛⎜⎜⎝α

E (xi)n∑

i=1αi

+ βE (xj)

⎞⎟⎟⎠+ f

⎛⎜⎜⎝β

E (xi)n∑

i=1αi

+ αE (xj)

⎞⎟⎟⎠⎤⎥⎥⎦

≤n∑

i=1

αsi f

⎛⎜⎜⎝E (xi)

n∑i=1

αi

+ f (xi)

⎞⎟⎟⎠ . (26)

Proof. By the E–s–Orlicz convexity of f we can state

f

⎛⎜⎜⎝α

E (xi)n∑

i=1αi

+ βE (xj)

⎞⎟⎟⎠ ≤ αsf

⎛⎜⎜⎝E (xi)

n∑i=1

αi

⎞⎟⎟⎠+ βsf (E (xj))

and

f

⎛⎜⎜⎝β

E (xi)n∑

i=1αi

+ αE (xj)

⎞⎟⎟⎠ ≤ βsf

⎛⎜⎜⎝E (xi)

n∑i=1

αi

⎞⎟⎟⎠+ αsf (E (xj))

for all i ∈ {1, ..., n}. Adding these inequalities we obtain

f

⎛⎜⎜⎝α

E (xi)n∑

i=1αi

+ βE (xj)

⎞⎟⎟⎠+ f

⎛⎜⎜⎝β

E (xi)n∑

i=1αi

+ αE (xj)

⎞⎟⎟⎠

≤ f

⎛⎜⎜⎝E (xi)

n∑i=1

αi

⎞⎟⎟⎠+ f (E (xj))

for all i ∈ {1, ..., n}. Multiplying this inequality by αsj ≥ 0 and summing over

j from 1 to n we get

n∑j=1

αsj

⎡⎢⎢⎣f⎛⎜⎜⎝α

E (xi)n∑

i=1αi

+ βE (xj)

⎞⎟⎟⎠+ f

⎛⎜⎜⎝β

E (xi)n∑

i=1αi

+ αE (xj)

⎞⎟⎟⎠⎤⎥⎥⎦

≤ f

⎛⎜⎜⎝E (xi)

n∑i=1

αi

⎞⎟⎟⎠+

n∑j=1

αsjf (E (xj)) .


By Jensen’s inequality (6) we also have

f

⎛⎝αE (xi) + β

n∑j=1

αjE (xj)

⎞⎠+ f

⎛⎝βE (xi) + α

n∑j=1

αjE (xj)

⎞⎠

≤n∑

j=1

αsjf

⎛⎜⎜⎝α

E (xi)n∑

i=1αi

+ βE (xj)

⎞⎟⎟⎠+

n∑j=1

αsjf

⎛⎜⎜⎝β

E (xi)n∑

i=1αi

+ αE (xj)

⎞⎟⎟⎠

and by the above inequalities we can state

f

⎛⎝αE (xi) + β

n∑j=1

αjE (xj)

⎞⎠+ f

⎛⎝βE (xi) + α

n∑j=1

αjE (xj)

⎞⎠

≤n∑

j=1

αsj

⎡⎢⎢⎣f⎛⎜⎜⎝α

E (xi)n∑

i=1αi

+ βE (xj)

⎞⎟⎟⎠+ f

⎛⎜⎜⎝β

E (xi)n∑

i=1αi

+ αE (xj)

⎞⎟⎟⎠⎤⎥⎥⎦

≤ f

⎛⎜⎜⎝E (xi)

n∑i=1

αi

⎞⎟⎟⎠+

n∑j=1

αsjf (E (xj))

for all i ∈ {1, ..., n}.

Now, if we multiply this inequality by αsi ≥ 0 and summing over i, we get

the last two inequalities in (26).

To prove the remaining part of inequality (26), we shall use the fact that

f

⎛⎝a + b

21s

⎞⎠ ≤ f (a) + f (b)

2, ∀ a, b ∈ M

which gives us

2f

⎡⎣α + β

21s

⎛⎝E (xi) +

n∑j=1

αjE (xj)

⎞⎠⎤⎦

≤ f

⎛⎝αE (xi) + β

n∑j=1

αjE (xj)

⎞⎠+ f

⎛⎝βE (xi) + α

n∑j=1

αjE (xj)

⎞⎠

by multiplying this inequality by αsi ≥ 0 and sum over i we deduce the third

inequality in (26). The last inequality is oblivious by Jensen’s inequality

f

⎡⎢⎢⎣

(α + β)(1 +

n∑i=1

αi

)

21s

n∑i=1

αiE (xi)

⎤⎥⎥⎦


= f

⎛⎝ n∑

i=1

αi

⎡⎣α + β

2

⎛⎝E (xi) +

n∑j=1

αjE (xj)

⎞⎠⎤⎦⎞⎠

≤n∑

i=1

αsif

⎛⎝⎡⎣α + β

21s

⎛⎝E (xi) +

n∑j=1

αjE (xj)

⎞⎠⎤⎦⎞⎠.

and the theorem is proved.

Remark 4.11 If in Theorem 4.10 we assume that f is an E–Orlicz convexand α = t, β = 1 − t with t ∈ [0, 1], we get a refinement of Jensen’s classicalinequality as follows:

f

(n∑

i=1

αiE (xi)

)≤

n∑i=1

αif

⎛⎜⎜⎜⎝

E (xi) +n∑

j=1αjE (xj)

2

⎞⎟⎟⎟⎠

≤ 1

2

n∑i=1

αi

⎡⎣f⎛⎝tE (xi) + (1 − t)

n∑j=1

αjE (xj)

⎞⎠

+f

⎛⎝(1 − t)E (xi) + t

n∑j=1

αjE (xj)

⎞⎠⎤⎦

≤n∑

i=1

n∑j=1

αiαjf (tE (xi) + (1 − t) E (xj))

≤n∑

i=1

αif (E (xi))


i=1αi = 1.

Remark 4.12 The results in the previous two sections are hold for E–Orlicz, semi–E–Orlicz, semi–E–s–Orlicz convex functions.

ACKNOWLEDGEMENTS. The work here is supported by the Grant:UKM–GUP–TMK–07–02–107.

References

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[2] M. Alomari and M. Darus, The Hadamard’s inequality for s–convex func-tion of 2–variables On The co–ordinates, Int. Journal of Math. Analysis,2 (13) (2008), 629-638.

[3] M. Alomari and M. Darus, The Hadamard’s inequality for s–convex func-tion, Int. Journal of Math. Analysis, 2 (13) (2008), 639-646.

[4] S. S. Dragomir and S. Fitzpatrick, s–Orlicz convex functions in linearspaces and Jensen’s discrete inequality, J. of Math. Ana. Appl., 210(1997), 419-439.

[5] H. Hudzik and L. Maligranda, Some remarks on s–convex functions, Ae-qua. Math., Univ. of Waterloo, 48 (1994) 100–111.

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[9] S. Rolewicz, Metric Linear Spaces, 2nd ed., PW N, Warsaw, 1984.

Received: May 5, 2008

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