View
7
Download
0
Category
Preview:
Citation preview
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: alomari@math.com2Corresponding author: maslina@pkrisc.cc.ukm.my
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.
1572 M. Alomari and M. Darus
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.
s-Orlicz convex functions 1573
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,∞).
1574 M. Alomari and M. Darus
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)
s-Orlicz convex functions 1575
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 .
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 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
}.
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 =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 .
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 anE–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 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 .
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–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 .
1578 M. Alomari and M. Darus
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)
On the other hand, if
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.
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) 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)
1580 M. Alomari and M. Darus
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))
s-Orlicz convex functions 1581
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
⎞⎟⎟⎟⎟⎟⎠ ,
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 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
⎞⎟⎟⎟⎟⎟⎠
s-Orlicz convex functions 1583
+
⎛⎜⎜⎜⎜⎝∑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:
(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 onPf (N).
1584 M. Alomari and M. Darus
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:
(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 onPf (N).
Corollary 3.15 In Theorem 3.12, let E : M → E (M) be the identityoperator 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 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.
s-Orlicz convex functions 1585
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
).
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
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.
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 αsi α
sj ≥ 0 and sum over i, j from 1 to n,
we have
2n∑
i=1
n∑j=1
αsi α
sjf (E (αxi + βxj))
=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
sjf (E (αxi + βxj))
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
.
1588 M. Alomari and M. Darus
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)) .
s-Orlicz convex functions 1589
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))
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
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))
where, αi ≥ 0 andn∑
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)
⎞⎠⎤⎦
s-Orlicz convex functions 1591
≤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)) .
1592 M. Alomari and M. Darus
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)
⎤⎥⎥⎦
s-Orlicz convex functions 1593
= 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))
where, αi ≥ 0 andn∑
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
[1] M. Alomari and M. Darus, A mapping connected with Hadamard–typeinequalities 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 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.
[6] W. Matuszewska and W. Orlicz, A note on the theorey of s–normed spacesof ϕ–integrable functions, studia Math., 21 (1981), 107–115.
[7] J. Musielak, Orlicz soaces and modular spaces, Lecture Notes in Mathe-matics, 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
Recommended