Refinements of s-Orlicz Convex Functions in Normed Linear Spaces 1

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 Sciences Faculty of Science and Technology Universiti Kebangsaan Malaysia Bangi 43600 Selangor, Malaysia Abstract In this paper a generalization of s–convexity 0 < s ≤ 1 in both sense are established. It is proved among others, that s–convexity in the second sense is essentially stronger than the s–convexity in first, original sense, whenever 0 < s < 1. Some of properties of s–convex function in both sense are considered and various examples are given.

Keywords: s–Convex function in the first sense, Jensen’s inequality, Linear 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 first sense if f (αx + βy) ≤ αs f (x) + β s f (y)

(1)

for all x, y ∈ [0, ∞), α, β ≥ 0 with αs + β s = 1 and for some fixed s ∈ (0, 1]. We denote this class of functions by Ks1 . Also, a function f : R+ → R, where R+ = [0, ∞), is said to be s–convex in the second sense if f (αx + βy) ≤ αs f (x) + β s f (y) 1 2

First author: [email protected] Corresponding author: [email protected]

(2)

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]. We denote this class of functions by Ks2 . This definition of s–convexity, for so called ϕ–functions, was introduced by Orlicz in [9] and was used in the theory of Orlicz spaces (see [7], [8], [10]). A function 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 mean just the convexity when s = 1. In [5], Hudzik and Maligranda established some results about s–convex function in the first sense. The property that f is in the first sense then f is non–decreasing on (0, ∞) is given as follows: Theorem 1.1 Let 0 < s ≤ 1. If f ∈ Ks1 , then f is non–decreasing on (0, ∞) and lim+ f (x) ≤ f (0). x→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 ∈ Ks1 is non–decreasing in (0, ∞) but not necessarily on [0, ∞) (see [2]). In [5] H. Hudzik and L. Maligranda introduced some results on s–convex functions 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 s

f (x) = x /(1−s) p (x) ∈ Ks1 .

(3)

Also, we define the class of s–Orlicz convex functions defined on s–Orlicz sets in a linear space. Definition 1.3 Let V be a linear space and s ∈ (0, ∞). The set M ⊆ V will be called s–Orlicz convex in L if the following condition is true: If x, y ∈ M and α, β ≥ 0 with αs + β s = 1. Then (αx + βy) ∈ M. The main goal in this paper is to pave the way for us to talk about the applications of Orlicz convexity in functional analysis. Indeed, this paper is devoted to extend the circle of convexity in the Orlicz spaces; which in turn, going to be used in the future to give another vision to some applications in real and functional analysis.

s-Orlicz convex functions

2

1571

Preliminary Results

Firstly, we introduce the definitions of E– and E–s–Orlicz convex set M in a normed linear space V as follows: Definition 2.1 Let V, W be two finite dimensional normed linear spaces over R. Let E : M ⊆ V → W be a continuous linear operator. A set M ⊆ V is 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 space over R and s ∈ (0, ∞). Let E : M ⊆ V → W be a continuous linear operator. The set M ⊆ V will be called E–s–Orlicz convex in V over R if the following conditions is true: If x, y ∈ M and α, β ≥ 0 with αs + β s = 1. Then (αE(x) + βE(y)) ∈ M. For instance, if every M ⊆ V is E–s–Orlicz convex set, then V will be called E–s–Orlicz convex linear space, and if M is a linear subspace of V then M will be called E–s–Orlicz convex linear subspace. Theorem 2.3 If M ⊆ V is an E–s–Orlicz convex subset, then E (M) ⊆ M is an E–s–Orlicz convex. Proof. Since M is an E–s–Orlicz convex, then for any x, y ∈ M and α, β ≥ 0 such that αs + β s = 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 element therefore 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 . Then M is an E–s–Orlicz convex. Proof. Assume that x, y ∈ M, since E (M) is an E–s–Orlicz convex then E (x) and E (y) ∈ E (M) and for each α, β ≥ 0 such that αs + β s = 1 with s ∈ (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.

1572

M. Alomari and M. Darus

Proof. Let x, y ∈ M1 ∩ M2 . Since M1 , M2 ⊂ V are two E–s–Orlicz convex 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–Orlicz convex subspace. Theorem 2.6 Let M ⊆ V be an E1 – and E2 –s–Orlicz convex. Then M is an 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–Orlicz convex. Theorem 2.7 Let M1 , M2 ⊂ V be two E–s–Orlicz convex subspaces. Then M1 + 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 ≥ 0 such that αs + β s = 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 is not E–s–Orlicz for some s ∈ (0, ∞) \ {1}. Example 2.9 Let E : R2 → R2 be acontinuous linear operator,  defined such as, E (x, y) = (x, y). consider M = (x, y) ∈ R2 : x2 + y 2 ≤ 1 . We will show that M is not E–2–Orlicz convex.

s-Orlicz convex functions

1573

Let x = (x1 , y1 ), y = (x2 , y2) with x2i + yi2 = 1, (i = 1, 2) and x1 x2 , y1 y2 > 0. Consider α, β with α2 + β 2 = 1. Then, αE (x) + βE (y) = αx + βy = (αx1 + βx2 , αy1 + βy2). Therefore, we have 







(αx1 + βx2 )2 + (αy1 + βy2)2 = α2 x21 + x22 + β 2 y12 + y22 + 2αβ (x1 x2 + y1 y2 ) = α2 + β 2 + 2αβ (x1 x2 + y1 y2 ) = 1 + 2αβ (x1 x2 + y1 y2 ) > 1. Thus, x, y ∈ M but αx + βy ∈ / M, which shows that M is E–Orlicz but not E–2–Orlicz convex. Remark 2.10 There exists E–s–Orlicz convex set in normed linear space which is not E–Orlicz for some s ∈ (0, ∞) \ {1}. linearoperator, Example 2.11 Let E : R2 → R2 be a continuous  defined   2 such as, E (x, y) = (x, y). Consider M = (x, y) ∈ R : |x| + |y| ≤ 1 . We will show that M is not E–Orlicz convex. Let x = (x1 , y1), y = (x2 , y2) 1

1

1



1



1

1



with xi2 +yi2 = 1, (i = 1, 2) which imply α 2 |x1 |+α 2 |y1 | ≤ α 2 and β 2 |x2 |+ 1



1

β 2 |y2 | ≤ β 2 , then 1



1



1



1



1

1

α 2 |x1 | + α 2 |y1 | + β 2 |x2 | + β 2 |y2 | ≤ α 2 + β 2 = 1. But we know the inequalities 1



1









α 2 |x1 | + β 2 |x2 | ≥ and

1



1

α 2 |y1 | + β 2 |y2 | ≥

α |x1 | + β |x2 | ≥ α |y1 | + β |y2 | ≥



|αx1 + βx2 |



|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 with the 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, ∞).

1574

M. Alomari and M. Darus

The following theorem generalize a standard result about s–Orlicz convex sets. Theorem 2.13 The linear space V is E–s–Orlicz convex iff every subset M 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 + β s = 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 = E −1 (E(z)) =

r  i=1

r 

i=1

di xi . Thus, E(z) =

r 

i=1

di E(xi ) ∈ E(M0 ). Hence,

di E −1 (E(xi )) ∈ E −1 (E(M0 )), and so, z =

contradicts our assumption.

r  i=1

di xi ∈ M0 which

Corollary 2.14 The normed linear space V is E–1–Orlicz convex or simply E–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, ∞), and E : V → W be a continuous linear operator. For a given subset M ⊆ V , the following statements are equivalent: (i) M is E–s–Orlicz convex; (ii) For every x1 , x2 , ..., xn ∈ M and α1 , α2 , ..., αn ≥ 0 with have that

n  i=1

n  i=1

αis = 1 we

αi E(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 follows by Definition 2.2. . Suppose the statement holds for all 2 ≤ k ≤ n − 1. Let n  x1 , x2 , ..., xn ∈ M and α1 , α2 , ..., αn ≥ 0 with αis = 1, then by Theorem 2.3 i=1

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

i=1

αi xi ∈ M. Since E(M) is E–s–Orlicz, and E is linear then E

 n i=1



αi xi =

n i=1

αi E(xi )

(5)

s-Orlicz convex functions

Now, suppose that

1575

n−1 

n−1 

αis xi > 0, and consider βj =  αj 1/ , then βj = 1, s n−1 i=1 j=1 

j = 1, 2, ..., n − 1. Put y =

n−1  j=1

i=1

αsi

βj xj , and therefore E (y) =

n−1  j=1

βj E (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

n−1 

αi E(xi ) =

j=1

αj E (xj ) n−1

n−1 1/s  s α i

i=1

= n−1  s As α +αs i=1

i

n

it follows that

n−1 i=1

n 

=

i=1

αis

thus proved.

1/s

+ αn E(xn )

1/s

E(y) + αn E(xn ) .

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

n−1 1/s  s α E(y) + α i=1

i=1

αis

nE

i

(xn ) ∈ E (M) ⊆ M and the theorem is

Corollary 2.16 Let V, W be two normed linear spaces and E : V → W be a continuous linear operator. For a given subset M ⊆ V , the following statements are equivalent: (i) M is E–1–Orlicz convex; (ii) For every x1 , x2 , ..., xn ∈ M and α1 , α2 , ..., αn ≥ 0 with have that

n  i=1

n  i=1

αi = 1 we

αi E(xi ) ∈ M.

Proof. The proof is straight forward and we will omit the details. Remark 2.17 Definition 2.1 and Theorem 2.3 in [4] is a special case of Definition 2.2 and Theorem 2.15 respectively, when s = 1 and consider the operator E to be the identity operator. Next, we generalize the concept of a convex hull to E–s–Orlicz convex hull and E–Orlicz convex hull, respectively; as follows: Theorem 2.18 Let V, W be two normed linear spaces and E : V → W be a linear mapping. cosE (M) =

n i=1

αi E (xi ) : αi ≥ 0,

n i=1



αis = 1, xi ∈ M, n ≥ 2

.

1576

M. Alomari and M. Darus

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= and y=

n  i=1 m  j=1

αi E (xi ), with αi ≥ 0, xi ∈ M and

βj E (yj ), with βj ≥ 0, yj ∈ M and

n  i=1

αis = 1, n ≥ 2 ,

m  j=1

βjs = 1, m ≥ 2 .

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

n  i=1

αi E (xi ) + β

m  j=1

βj E (yj ) =

where, λi = ααi ; (i = 1, 2, ..., n),

n+m  k=1

λk E (zk ),

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 s

m s

i=1

j=1

αi + β s

βj = αs + β s = 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, as follows: Corollary 2.19 In Theorem 2.18 , if s = 1, then co1E

(M) =

n i=1

αi E (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 the set M. Proof. The proof is straight forward and we will omit the details.

3

E– and E–s–Orlicz Mapping

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

s-Orlicz convex functions

1577

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 iff there is a continuous linear map E : M → W such that for each x, y ∈ M and α, β ≥ 0 so that αs + β s = 1 and s ∈ (0, ∞) one has f (αE (x) + βE (y)) ≤ αs f (E (x)) + β s f (E (y)) .

(6)

On the other hand, if f (αE (x) + βE (y)) ≥ αs f (E (x)) + β s f (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 and α, 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. On the other hand, if 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–Orlicz convex mapping on M. If for each x, y ∈ M, α, β ≥ 0 so that αs + β s = 1 and s ∈ (0, ∞). Then the set fEx (η) = {x ∈ M : f (E (x)) ≤ η} , is nonempty E–s–Orlicz convex subset of M. Proof. Let x, y ∈ fEx (η) and α, β ≥ 0 so that αs + β s = 1. Then f (E (x)) ≤ η

and

f (E (y)) ≤ η,

αs f (E (x)) ≤ αs η

and

β s f (E (y)) ≤ β s η

which imply that

and thus f (αE(x) + βE(y)) ≤ αs f (E (x)) + β s f (E (y)) ≤ (αs + β s ) η = η. which show that fEx (η) is an E–s–Orlicz convex subset of M.

1578

M. Alomari and M. Darus

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

(10)

On the other hand, if f (αE (x) + βE (y)) ≥ αs f (x) + β s f (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 ≥ 0 so 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 an E–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 for any x, y ∈ M, α, β ≥ 0 so that αs + β s = 1 and s ∈ (0, ∞), we have {αE (x) + βE (y)} ∈ M and f (αE (x) + βE (y)) ≤ αs f (x) + β s f (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 be semi–E–s–Orlicz convex mapping on M. If for each x, y ∈ M, α, β ≥ 0 so that αs + β s = 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 Theorem 3.3 we get the result.

s-Orlicz convex functions

1579

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) and as f (x, y) = x2 + y 2 , f is an E–2–Orlicz convex let f : R2 → R be defined  function on a set M = (x, y) ∈ R2 : x2 + y 2 ≥ 1 . Take (x, y) = (1, 2) therefore f (E (1, 2)) = f (2, 2) = 8 > 5 = f (2, 1), then by Theorem 3.5 f is not a semi–E–2–Orlicz convex function. However, let E : R2 → R2 be defined as E (x, y) =



1 , 1 x 1+y



with x2 + y 2 ≥ 1, √  and consider f as above on the same set M. Take (x, y) = 2, 0 , then,  √    √  2, 0 = f √12 , 1 = 32 < 2 = f 2, 0 . f E Theorem 3.9 Let V , W be two normed linear space and M be E–s–Orlicz subset 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 that inequality  n

f

i=1



αi E (xi ) ≤

n i=1

n  i=1

αis = 1, one has the

αis f (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 that

n 

i=1

αis = 1, then

n 

i=1

αi E (xi ) ∈ M. The proof carries by induction over

n ≥ 2. If n = 2, then the inequality holds by Definition 3.1 . n  i=1

Now, for 2 ≤ k ≤ n − 1, let x1 , x2 , ..., xn ∈ M and α1 , α2 , ..., αn ≥ 0 with αis = 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 that

n−1  i=1

αis ≥ 0 and put

αj βj =  , n−1  s 1/s αi i=1

(1 ≤ j ≤ n − 1)

1580

M. Alomari and M. Darus

Then

n−1  j=1

n−1 

βjs = 1 and it’s easy to see that

j=1

βj E(xj ) ∈ M. By using the

induction hypothesis we also can state ⎛

f⎝

n−1 j=1

⎞

βj E(xj )⎠ ≤

n−1

βj f (E (xj ))

i=1

Now, we observe that ⎛

f

 n i=1



≤ f

αi E (xi )

n−1

1/s  α E (x ) ⎜n−1 i i ⎜ s i=1 ⎜ αi n−1 1/s ⎜  s ⎝ i=1

⎛

≤

n−1 i=1



αis

f

αi

i=1

⎞ ⎟ ⎟ + αn E (xn )⎟ ⎟ ⎠

⎞

n−1 

⎜ α E (xi ) ⎟ ⎜ i=1 i ⎟ s ⎜ ⎟ ⎜ n−1 1/s ⎟ + αn f ⎝  s ⎠ i=1

(E (xn ))

αi

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

 s n−1  s 1/s α i=1

i

(15)

n−1 1/s  s α i=1

i

+ αns = 1.

On the other hand, ⎛

f

⎞

n−1 

⎜ α E (xi ) ⎟ ⎜ i=1 i ⎟ ⎜ ⎟ ⎜ n−1 1/s ⎟ ⎝  s ⎠ i=1

⎛

⎞

≤ f⎝

βj E (xj )⎠

n−1 j=1

αi

≤

n−1 j=1 n−1 

=

i=1

βj f (E (xj )) αi f (E (xi )) n−1  i=1

αis

.

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

f

 n i=1



αi E (xi )

≤

n−1 i=1

=

n i=1

αis

n−1

 αs f i=1

i

(E (xi ))

n−1  i=1

αis f (E (xi ))

αis

+ αn f (E (xn ))

and

s-Orlicz convex functions

1581

and the theorem is proved. Corollary 3.10 In Theorem 3.9 put s = 1. Then, the following statements are equivalent: (i) f is E–Orlicz convex on M. (ii) For every αi ≥ 0 and s ∈ (0, ∞) such that inequality f

 n i=1



αi E (xi ) ≤

n i=1

n  i=1

αi = 1, one has the

αi f (E (xi )).

(16)

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

 n i=1



αi xi ≤

n i=1

αis f (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 the result 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

⎞



⎜ pi xi ⎟ ⎜ ⎟ ⎜ i∈I ⎟ ⎜

1/s ⎟ ⎜ ⎟ ⎝  s ⎠ i∈I

pi

where pi > 0, i ∈ I, f is a s–Orlicz convex mapping on the s–Orlicz convex set M ⊆ V , xi ∈ M, and I is from Pf (N), where Pf (N) denotes the finite subsets of the natural numbers set N. Now, we will construct a new functional ΘE from the above functional Θ by 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

⎞

 ⎜ pi E (xi ) ⎟ ⎜ ⎟ ⎜ i∈I ⎟ , ⎜

1/s ⎟ ⎜ ⎟ ⎝  s ⎠ i∈I

pi

1582

M. Alomari and M. Darus

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 on the 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 on Pf (N). Proof. (1) Let I, J ∈ Pf (N) with I ∩ J = ∅. Then we have ΘE (I ∪ J, p, f, E (x)) =

i∈I

⎛

− ⎝

psi f (E(xi )) + ⎛



k∈I∪J

=

i∈I

⎛

− ⎝

⎞

psk ⎠ f



k∈I∪J

psk ⎠ f

j∈J

psj f (E(xj ))



⎡⎛

j∈J

k∈I∪J

+

≥

⎞1/s   ⎢⎜ psi E(xi ) ⎟ pi E(xi ) ⎢⎜ ⎟ ⎢⎜  i∈I i∈I

⎟ ⎢⎜ 

1/s ⎟ ⎢⎝   s s ⎠ ⎣ ⎞1/s

s

p E(xj ) ⎟ ⎜ ⎜ j∈J j ⎟ ⎜

⎟ ⎜ ⎟  ⎝ s ⎠ i∈I

−



k∈I∪J

pk

psj f (E(xj ))

k∈I∪J

⎛

⎞



⎜ pi E(xi ) + pj E(xj ) ⎟ ⎜ ⎟ j∈J ⎜ i∈I ⎟ ⎜ ⎟ 

1/s ⎜ ⎟  ⎝ ⎠ s

psi f (E(xi )) + ⎞



 j∈J



pk



j∈J

i∈I

⎤

pi

pj E(xj ) ⎥ ⎥

j∈J

psi f (E (xi )) +

pk

⎥

1/s ⎥ ⎥ ⎦ s

pj

psj f (E (xj ))

⎡⎛ ⎞1/s ⎛ ⎞⎢  s pi E (xi ) ⎟ ⎢⎜ ⎟ ⎢⎜ i∈I ⎝ 

⎟ psk ⎠ ⎢⎜ ⎟ f ⎜ ⎢⎝  k∈I∪J s ⎠ ⎣ k∈I∪J

pk

⎛

⎞

 ⎜ pi E (xi ) ⎟ ⎜ ⎟ ⎜ i∈I ⎟ ⎜

1/s ⎟ ⎜ ⎟ ⎝  s ⎠ i∈I

pi

s-Orlicz convex functions

1583 ⎛

+



psj E

⎜ ⎜ j∈J ⎜ ⎜  ⎝

k∈I∪J

=

s i∈I

pi f

⎛

⎞1/s

(xj ) ⎟

⎟

⎟ ⎟ ⎠ s p k

f

⎞⎤



⎜ ⎟⎥ ⎜ j∈J pj E (xj ) ⎟⎥ ⎜ ⎟⎥ ⎥ ⎜

1/s ⎟ ⎜ ⎟⎥  ⎝ ⎠⎦ s j∈J

⎛

⎞



⎜ pi E (xi ) ⎟ ⎟ s ⎜ ⎜ i∈I ⎟ (E (xi )) − pi f ⎜

1/s ⎟ ⎜ ⎟ i∈I ⎝  s ⎠ ⎛

+

pj

i∈I

pi

⎞



⎜ pj E (xj ) ⎟ ⎜ ⎟ j∈J ⎜ ⎟ ⎟ psj f (E (xj )) − psj f ⎜ 

1/s ⎟ ⎜ j∈J j∈J ⎝  s ⎠



j∈J

pj

= 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 convex mapping, then we have: (1) for all I, J ∈ Pf (N) with I ∩ J = ∅, one has the inequality ΘE (I ∪ J, p, f, x) ≥ ΘE (I, p, f, x) + ΘE (J, p, f, x) ≥ 0 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, x) ≥ ΘE (J, p, f, x) ≥ 0 that is, the mapping ΘE is monotonic non–decreasing in the first variable on Pf (N).

1584

M. Alomari and M. Darus

Proof. Just we note that since f is semi–E–s–Orlicz convex mapping, then by Theorem 3.5 we get f (E (x)) ≤ f (x) for each x ∈ M, and the proof is straight forward. Corollary 3.14 In Corollary 3.13, since f is a semi–E–s–Orlicz convex mapping, then we have: (1) for all I, J ∈ Pf (N) with I ∩ J = ∅, one has the inequality ΘE (I ∪ J, p, f, x) ≥ ΘE (I ∪ J, p, f, E (x)) ≥ 0 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, x) ≥ ΘE (J, p, f, E (x)) ≥ 0 that is, the mapping ΘE is monotonic non–decreasing in the first variable on Pf (N). Corollary 3.15 In Theorem 3.12, let E : M → E (M) be the identity operator E (x) = x. Then (1) for all I, J ∈ Pf (N) with I ∩ J = ∅, one has the inequality 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 (2) for all I, J ∈ Pf (N) with ∅ = J ⊆ I, one has the inequality Θx (I, p, f, x) ≥ Θx (J, p, f, x) ≥ 0 that is, the mapping Θx is monotonic non–decreasing in the first variable on Pf (N). Proof. In Theorem 3.12, set E : M → W be the identity operator E (x) = x and straight forward.

4

Generalized Inequalities For Double Sums

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

s-Orlicz convex functions

1585

Theorem 4.1 Let f : M ⊂ V → R be an E–s–Orlicz convex map on the n  E–s–Orlicz set M and αis ≥ 0 so that αis = 1. Let xij ∈ V , 1 ≤ i, j ≤ n. i=1

Then one has the inequalities n n i=1 j=1

αis αjs f

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

where; A=

⎛

n i=1

⎛

αis f

⎝

n j=1

⎞

αj E (xij )⎠ ,

B=

n j=1

αjs f

 n i=1

n n i=1 j=1

⎞

αis αjs E

(xij )⎠



αi E (xij ) .

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

f⎝

n j=1

⎞

αi E (xij )⎠ ≤

n j=1

αjs f (E (xij ))

by multiplying the both side by αis ≥ 0, we have ⎛

αis f ⎝

n j=1

⎞

αi E (xij )⎠ ≤

which gives, by addition that f

⎛ ⎞ n n ⎝ αi αj E (xij )⎠

≤

i=1 j=1

Thus,

⎛

f

⎝

n n i=1 j=1

n i=1

⎛

αis f ⎝

n j=1

n j=1

⎞

αi E (xij )⎠ ≤

⎞

αi αj E (xij )⎠

αis αjs f (E (xij ))

≤A≤

n n i=1 j=1

n n i=1 j=1

αis αjs f (E (xij )).

αis αjs f (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 n  the E–s–Orlicz set M and αis ≥ 0 so that αis = 1. Let xij ∈ V , 1 ≤ i, j ≤ n. i=1

Then one has the inequalities n n i=1 j=1

⎛

αis αjs f (xij ) ≥ max {A, B} ≥ min {A, B} ≥ f ⎝

where; A=

n i=1

⎛

αis f ⎝

n

j=1

⎞

αj xij ⎠ ,

B=

n j=1

αjs f

n n i=1 j=1

 n i=1

⎞

αis αjs xij ⎠



αi xij .

1586

M. Alomari and M. Darus

Proof. The proof is straight forward and we will omit the details. Theorem 4.3 Let f : M ⊂ V → R be an E–s–Orlicz convex map on n  the E–s–Orlicz convex set M and αis ≥ 0 such that αis = 1. Then for all i=1

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

f

n (α + β) ⎝ 1 2 s −1

i=1

αi

n i=1

⎞

αi E(xi )⎠ ≤ ≤

n n i=1 j=1 n n i=1 j=1

≤

n i=1

⎛

αis αjs f

⎞

(α + β) E (xi + xj ) ⎠ ⎝ 1

2s

αis αjs f (E (αxi + βxj ))

αis f (E (xi ))

(20)

Proof. Firstly, we note that since f is E–s–Orlicz convex map, and for all x, y ∈ M one has the inequalities: f (αE(x) + βE(y)) ≤ αs f (E(x)) + β s f (E(y)) and f (αE(y) + βE(x)) ≤ αs f (E(y)) + β s f (E(x)) which gives by addition f (αE(x) + βE(y)) + f (αE(y) + βE(x)) ≤ (αs + β s ) (f (E(x)) + f (E(y))) = f (E(x)) + f (E(y)) .

(21) 1

Moreover, if in the definition of s–Orlicz convexity of f , we take α = β = 2− s , we have αs + β s = 12 + 12 = 1 then the inequality ⎛

f

⎞

E(x + y) ⎠ ⎝ 1

2s

⎛

= f

⎞

E (x) + E (y) ⎠ ⎝ 1

2s 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)) ⎠ ⎝ 1

2s

⎛

= f

⎞

(α + β) (E (x) + E (y)) ⎠ ⎝ 1

2s f (αE(x) + βE(y)) + f (αE(y) + βE(x)) . ≤ 2

s-Orlicz convex functions

1587

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 αis αjs ≥ 0 and sum over i, j from 1 to n, we have 2

n n i=1 j=1

=

n n i=1 j=1 n

≤ 2

i=1

αis αjs f (E (αxi + βxj ))

αis αjs [f (E (αxi + βxj )) + f (E (βxi + αxj ))]

αis

n s j=1

αi f (E (xi ))

Thus, n n i=1 j=1

αis αjs f (E (αxi + βxj )) ≤

n

αis f (E (xi ))

i=1

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

f

⎞

(α + β) E (xi + xj ) ⎠ ⎝ 1 2s

⎛

= f

⎞

(α + β) (E (xi ) + E (xj )) ⎠ ⎝ 1

2s f (αE (xi ) + βE (xj )) + f (αE (xj ) + βE (xi )) ≤ 2

for all i, j ∈ {1, 2, ..., n}. If we multiply this inequality by αis αjs ≥ 0 and sum over i, j from 1 to n, we get n n i=1 j=1

⎛

αis αjs f

⎞

(α + β) E (xi + xj ) ⎠ ⎝ 1 2s

≤

n n i=1 j=1

αis αjs f (E (αxi + βxj ))

which gives the second inequality in (20). Now, set xij =

(α + β) (xi + xj ) 1

,

2s and since E is linear we have E (xij ) =

(α + β) E (xi + xj ) 1

2s

.

1588

M. Alomari and M. Darus

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

f

n α+β ⎝

⎛

= f

αi αi E 1 2 s −1 i=1 i=1

n n α+β ⎝

⎛

= f⎝

1 2s

i=1 j=1

n n i=1 j=1

n n

≤

⎞

n

i=1 j=1

αi αj

(xi )⎠ ⎞

αi αj E (xi + xj )⎠ ⎞

(α + β) E (xi + xj ) ⎠ 1

2s

⎛

αis αjs f ⎝

⎞

(α + β) E (xi + xj ) ⎠ 1

2s

and the theorem has been proved. Corollary 4.4 Let f : M ⊂ V → R be a semi–E–s–Orlicz convex map on n  the E–s–Orlicz convex set M and αis ≥ 0 such that αis = 1. Then for all i=1

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

f

n (α + β) ⎝ 1 2 s −1

i=1

αi

n i=1

⎞

αi xi ⎠ ≤ ≤ ≤

n n i=1 j=1 n n

⎛

αis αjs f

⎞

(α + β) xi + xj ⎠ ⎝ 1

2s

αis αjs f (αxi + βxj )

i=1 j=1 n αis f i=1

(xi )

(23)

Proof. Since f is semi–E–s–Orlicz convex map then f (E (x)) ≤ f (x), and the details goes likewise the proof of Theorem 4.3 with semi–E–s–Orlicz convex mapping f . Corollary 4.5 Suppose that f is E–s–Orlicz convex with the above assumptions for M, αi one has the inequalities: ⎛ n

f

⎜ i=1 ⎜ ⎝

αi

n 

⎞

αi E (xi ) ⎟

i=1 2 2 s −2

⎟ ⎠

≤

n n i=1 j=1

≤ ≤

⎛

αis αjs f

⎞

E (xi ) + E (xj ) ⎠ ⎝ 2

2 s −1

⎛ ⎞ n n E (x ) + E (x ) i j ⎠ αs αs f ⎝ i=1 j=1 n αis f i=1

i

j

(E (xi )) .

1

2s

s-Orlicz convex functions

1589

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

 n i=1



≤

αi E (xi )

≤ ≤

n n i=1 j=1 n n i=1 j=1 n i=1

where, αi ≥ 0 and

n  i=1



αi αj f

E (xi ) + E (xj ) 2



αi αj f (tE (xi ) + (1 − t) E (xj ))

αi f (E (xi ))

αi = 1.

In [4], Dragomir and Fitzpatrick, established a variant result of Jensen’s inequality, 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 αi E (xi )

n i=1

≤

n i=1

⎛

αis f ⎝αE (xi ) + β

f ⎝αE (xi ) + β

n j=1

j=1

⎞

αj E (xj )⎠

αis f (E (xi ))

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

n

⎞

(24)

⎛

αj E (xj )⎠ ≤ αs f (E (xi )) + β s f ⎝

n

j=1

⎞

αj E (xj )⎠

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

n

αis f

i=1

≤ αs

n i=1

⎝αE (xi ) + β

n j=1

αs ≤ αs

i=1 n i=1

≤

n i=1

αj E (xj )⎠ ⎛

αis f (E (xi )) + β s f ⎝

By Jensen’s inequality (6) we have n

⎞

n j=1

⎛

αis f (E (xi )) + β s f ⎝

n

j=1

αis f (E (xi )) + β s

αis f (E (xi ))

n i=1

⎞

αj E (xj )⎠ . ⎞

αj E (xj )⎠

αis f (E (xi ))

(25)

1590

M. Alomari and M. Darus

and then the first inequality in (24) is proved. The second inequality follows by Jensen’s result (6). Indeed one has 

α+β

f

n i=1



⎛



= f⎝

αi αi E (xi )

n i=1

≤

n i=1

⎛

αi ⎝αE (xi ) + β ⎛

αi f ⎝αE (xi ) + β

n j=1

n j=1

⎞⎞

αj E (xj )⎠⎠ ⎞

αj E (xj )⎠

and the proof is established. Corollary 4.8 Suppose that f is E–s–Orlicz convex with the above assumptions for M, αi , and xi one has the inequalities: ⎛

f

⎞

n 

αj n ⎜1 + ⎟ j=1 ⎜ ⎜ ⎟ αj E (xj )⎟ 1 ⎝ ⎠

⎛

≤

j=1

2s

n i=1

≤

αis f

⎞

n 

αj E (xj ) ⎟ ⎜ E (xi ) + j=1 ⎜ ⎟ ⎜ ⎟ 1 ⎝ ⎠ 2s

n s i=1

αi f (E (xi )) .

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

 n i=1



≤

αi E (xi )

≤

n i=1 n i=1

where, αi ≥ 0 and

n  i=1

⎛

αi f ⎝tE (xi ) + (1 − t)

n j=1

⎞

αj E (xj )⎠

αi f (E (xi ))

αi = 1.

Finally, we prove the following inequalities: Theorem 4.10 Let f , M, xi , αi , and α, β as above. Then one has the inequalities: 

⎡

(α + β) 1 + ⎢

2f ⎢ ⎣

n

⎡

1 2s

n 



⎤

αi n i=1 i=1

⎛

⎥

αi E (xi )⎥ ⎦ ⎞⎤

n α+β ≤ 2 αis f ⎣ 1 ⎝E (xi ) + αj E (xj )⎠⎦ i=1 j=1 s 2

s-Orlicz convex functions

≤

n i=1

≤

⎡ ⎛

αis

⎣f ⎝αE (xi ) + β ⎡ ⎛

n j=1

⎞

⎛

αj E (xj )⎠ + f ⎝βE (xi ) + α ⎞

⎛

n j=1

⎞⎤

αj E (xj )⎠⎦ ⎞⎤

n n

⎢ ⎜ E (xi ) ⎟ ⎜ E (xi ) ⎟⎥ ⎜ ⎜ ⎥ αis αjs ⎢ + βE (xj )⎟ + αE (xj )⎟ n n ⎣f ⎝α  ⎠ + f ⎝β  ⎠⎦ i=1 j=1 αi αi i=1

⎛

≤

1591

i=1

⎞

n

⎜ E (xi ) ⎟ αis f ⎜ + f (xi )⎟ n ⎝  ⎠ . i=1 αi

(26)

i=1

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

⎞

⎛

⎞

⎜ E (xi ) ⎟ ⎜ ⎟ s ⎜ E (xi ) ⎟ s f⎜ + βE (xj )⎟ n n ⎝α  ⎠≤α f⎝  ⎠ + β f (E (xj )) αi αi i=1

i=1

and ⎛

⎞

⎛

⎞

⎜ E (xi ) ⎟ ⎜ ⎟ s ⎜ E (xi ) ⎟ s + αE (xj )⎟ f⎜ n n ⎝β  ⎠≤β f⎝  ⎠ + α f (E (xj )) αi αi i=1

i=1

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

⎞

⎛

⎞

⎜ E (xi ) ⎟ ⎜ E (xi ) ⎟ ⎟ + f ⎜β ⎟ α + βE (x ) + αE (x ) f⎜ j j n n ⎝  ⎠ ⎝  ⎠ αi αi ⎛

i=1

i=1

⎞

⎜ E (xi ) ⎟ ⎟ + f (E (xj )) ≤ f⎜ n ⎝  ⎠ αi i=1

for all i ∈ {1, ..., n}. Multiplying this inequality by αjs ≥ 0 and summing over j from 1 to n we get ⎡ ⎛

⎞

⎛

⎞⎤

n ⎢ ⎜ E (xi ) ⎜ αjs ⎢ n ⎣f ⎝α 

j=1

⎛

⎞

i=1

⎟ ⎜ E (xi ) ⎟⎥ ⎜ ⎥ + βE (xj )⎟ + αE (xj )⎟ n ⎠ + f ⎝β  ⎠⎦ αi αi

n ⎜ E (xi ) ⎟ ⎟+ ≤ f⎜ αjs f (E (xj )) . n ⎝  ⎠ j=1 αi i=1

i=1

1592

M. Alomari and M. Darus

By Jensen’s inequality (6) we also have ⎛

f ⎝αE (xi ) + β ⎛

≤

⎞

n

⎛

αj E (xj )⎠ + f ⎝βE (xi ) + α

j=1

⎞

⎛

j=1

⎞

αj E (xj )⎠ ⎞

n

n ⎜ E (xi ) ⎟ ⎜ E (xi ) ⎟ ⎟ αjs f ⎜ α + βE (x ) + αjs f ⎜ + αE (xj )⎟ j ⎠ n n ⎝  ⎝β  ⎠ j=1 j=1 αi αi i=1

i=1

and by the above inequalities we can state ⎛

f ⎝αE (xi ) + β ⎡ ⎛

≤

n

⎞

n

⎛

αj E (xj )⎠ + f ⎝βE (xi ) + α

j=1

⎞

⎛

n j=1

⎞

αj E (xj )⎠ ⎞⎤

n

⎢ ⎜ E (xi ) ⎟ ⎜ E (xi ) ⎟⎥ ⎜ ⎜ ⎥ αjs ⎢ + βE (xj )⎟ + αE (xj )⎟ n n ⎣ f ⎝α  ⎠ + f ⎝β  ⎠⎦ j=1 αi αi ⎛

⎞

i=1

i=1

n ⎜ E (xi ) ⎟ ⎟+ ≤ f⎜ αjs f (E (xj )) n ⎝  ⎠ j=1 αi i=1

for all i ∈ {1, ..., n}. Now, if we multiply this inequality by αis ≥ 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 which gives us

⎡

⎞

a + b⎠ ⎝ 1 2s

≤

f (a) + f (b) , 2

⎛

∀ a, b ∈ M ⎞⎤

n α+β 2f ⎣ 1 ⎝E (xi ) + αj E (xj )⎠⎦ j=1 2s ⎛ ⎞ ⎛

≤ f ⎝αE (xi ) + β

n

j=1

αj E (xj )⎠ + f ⎝βE (xi ) + α

n j=1

⎞

αj E (xj )⎠

by multiplying this inequality by αis ≥ 0 and sum over i we deduce the third inequality in (26). The last inequality is oblivious by Jensen’s inequality ⎡



(α + β) 1 + ⎢

f⎢ ⎣

1 2s

n 



⎤

αi n i=1 i=1

⎥

αi E (xi )⎥ ⎦

s-Orlicz convex functions

1593

⎛ ⎡ ⎛ ⎞⎤⎞ n n α + β ⎝ ⎝E (xi ) + αi ⎣ αj E (xj )⎠⎦⎠

= f

i=1

≤

n i=1

αis f

⎛⎡

2

⎛

j=1

⎞⎤⎞

n α+β ⎝ ⎝⎣ E (xi ) + αj E (xj )⎠⎦⎠. 1 j=1 2s

and the theorem is proved. Remark 4.11 If in Theorem 4.10 we assume that f is an E–Orlicz convex and α = t, β = 1 − t with t ∈ [0, 1], we get a refinement of Jensen’s classical inequality as follows:

f

 n i=1

⎛



αi E (xi )

≤

n i=1

⎞

n 

αj E (xj ) ⎟ ⎜ E (xi ) + j=1 ⎜ ⎟ αi f ⎜ ⎟ ⎝ ⎠ 2 ⎡ ⎛

⎞

n n 1 ≤ αi ⎣f ⎝tE (xi ) + (1 − t) αj E (xj )⎠ 2 i=1 j=1

⎛

+f ⎝(1 − t) E (xi ) + t ≤ ≤

n n i=1 j=1 n i=1

where, αi ≥ 0 and

n  i=1

n j=1

⎞⎤

αj E (xj )⎠⎦

αi αj f (tE (xi ) + (1 − t) E (xj ))

αi f (E (xi ))

αi = 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 [1] M. Alomari and M. Darus, A mapping connected with Hadamard–type inequalities in 4–variables, Int. Journal of Math. Analysis, 2 (13) (2008), 601-628.

1594

M. Alomari and M. Darus

[2] M. Alomari and M. Darus, The Hadamard’s inequality for s–convex function 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 function, Int. Journal of Math. Analysis, 2 (13) (2008), 639-646. [4] S. S. Dragomir and S. Fitzpatrick, s–Orlicz convex functions in linear spaces 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, Aequa. Math., Univ. of Waterloo, 48 (1994) 100–111. [6] W. Matuszewska and W. Orlicz, A note on the theorey of s–normed spaces of ϕ–integrable functions, studia Math., 21 (1981), 107–115. [7] J. Musielak, Orlicz soaces and modular spaces, Lecture Notes in Mathematics, Vol. 1034, Springer–Verlag, New York / Berlin, 1983. [8] W. Orlicz, A note on modular spaces, I, Bull. Acad. Polon. Sci. Math. Astronom. Phys., 9 (1961), 157–162. [9] S. Rolewicz, Metric Linear Spaces, 2nd ed., PW N, Warsaw, 1984. Received: May 5, 2008