R E S E A R C H
Open Access
New convex functions in linear spaces and
Jensen’s discrete inequality
Xiusu Chen
**Correspondence:
xousuchen10@sina.com School of Mathematics and Statistics, Chongqing Key laboratory of Electronic Commerce & Supply Chain System, Chongqing Technology and Business University, Chongqing, 400067, P.R. China
Abstract
In the present paper we introduce a new class ofs-convex functions defined on a convex subset of a real linear space, establish some inequalities of Jensen’s type for this class of functions. Our results in special cases yield some of the recent results on classic convex functions.
Keywords: s-Orlicz convex functions;s-convex functions; Jensen’s inequalities
1 Introduction
The research on convexity and generalized convexity is one of the important subjects in mathematical programming, numerous generalizations of convex functions have been proved useful for developing suitable optimization problems (see [–]).s-convex func-tions defined on a space of real numbers was introduced by Orlicz in [] and was used in the theory of Orlicz spaces.s-Orlicz convex sets ands-Orlicz convex mappings in lin-ear spaces were introduced by Dragomir in []. Some properties of inequalities of Jensen’s type for this class of mappings were discussed.
Definition .[] LetXbe a linear space ands∈(,∞). The setK⊂Xis calleds-Orlicz
convex inXif the following condition is true:
x,y∈Xandα,β≥ withαs+βs= imply αx+βy∈X.
Remark Ifs= , then, by the above definition, we recapture the concept of convex sets in linear spaces.
Definition .[] LetXbe a linear space ands∈(,∞). LetK⊂Xbe ans-Orlicz convex set. The mappingf :K→Ris calleds-Orlicz convex onKif for allx,y∈Kandα,β≥ withαs+βs= , one has the inequality
f(αx+βy)≤αsf(x) +βsf(y). (.)
Note that fors= we recapture the class of convex functions.
In this paper we introduce another class ofs-convex functions defined on convex sets in a linear space. Some discrete inequalities of Jensen’s type are also obtained.
2 The relations amongs-convex functions,s-Orlicz convex functions and convex functions in linear spaces
LetXbe a linear space ands∈(,∞) be a fixed positive number, letK⊂Xbe a convex subset. It is natural to consider the following class of functions.
Definition . The mappingf:K→Ris calleds-convex onKif
f(αx+βy)≤αsf(x) +βsf(y) (.)
for allx,y∈Kandα,β≥ withα+β= .
Remark . Ifs= , then, by the above definition, we recapture the concept of convex functions in linear spaces.
Remark . s-convex functions defined on a convex set in linear spaces are different from convex functions.
() There exists-convex mappings in linear spaces which are not convex for some
s∈(,∞)|{}(see the following Example ).
() When <s≤, every non-negative convex function defined on a convex set in a linear space is also ans-convex function. Whens≥, every non-positive convex function defined on a convex set in a linear space is also ans-convex function.
Example Let X be a normed linear space, letK=Xand <s≤, definef(x) =xsfor allx∈K. For everyx,y∈Kandα,β≥ withα+β= , whenαx= , eitherα= or
x= , therefore,αx= , and
f(αx+βy) =f(βy) =βys=βsys=βsf(y);
whenαx = , from the monotonous increasing of the functiong(t) = +ts– ( +t)sin
the interval [,∞), we haveg(t) = +ts– ( +t)s≥g() = ,∀t> , then
f(αx+βy) =αx+βys≤αx+βys=αsxs
+βy
αx s
≤αsxs
+
βy
αx
s
=αsxs+βsys=αsf(x) +βsf(y),
hencef is ans-convex function onX; however,f is not a convex function onXwith <
s< .
Remark . There exists-Orlicz convex functions in linear spaces which are nots-convex functions for somes∈(,∞)|{}.
Example <s≤,K=R. Consider the setK={(x,y)∈R:|x|s+|y|s≤} ⊂R. Let
X= (x,y),X= (x,y)∈K, and letα,β≥ withαs+βs= . Then
which imply
|αx+βx|s≤
α|x|+β|x|
s
=
α|x|
+β|x|
α|x|
s
=α|x|
s
+β|x|
α|x|
s
≤α|x|
s
+
β|x|
α|x|
s
=α|x|
s
+β|x|
s
≤αs|x|s+βs|x|s
and
|αy+βy|s≤αs|y|s+βs|y|s.
As
αX+βX= (αx+βx,αy+βy) and
|αx+βx|s+|αy+βy|s≤αs|x|s+βs|x|s+αs|y|s+βs|y|s =αs|x|s+|y|s
+βs|x|s+|y|s
=αs+βs= ,
we deduce thatαX+βX∈K,i.e.,Kiss-Orlicz convex inR.
Definef(u) = (x+y)s,u= (x,y)∈K, we will show thatfis ans-Orlicz convex function,
f is not ans-convex function, forKis not a convex subset when <s< . Sinceu= (, )∈
K,v= (, )∈K, but u+v= (,) /∈Kfor ()s+ ()s=s> with <s< .
3 Properties ofs-convex functions in linear spaces
We consider the following properties ofs-convex functions in linear spaces.
Theorem . Let K⊂X be a convex set,fi:K→R be s-convex functions,i= , , . . . ,n.
Then
() h(x) =max≤i≤nfi(x)is ans-convex function onK.
() af(x) +bf(x)is ans-convex function onKfor alla,b≥.
The proof is omitted.
Proposition . Let f :K→R be an s-convex function on K andγ ∈R so that Kγ(f) =
{x∈K:f(x)≤γ}is nonempty.Then Kγ(f)is a convex subset of K when either one of the
following two conditions is satisfied:
() γ ≥ands≥, () γ ≤and >s> .
Proof Letx,x∈Kγ(f) andα,β≥ so thatα+β= . Thenf(x)≤γ andf(x)≤γ which imply that
f(αx+βx)≤αsf(x) +βsf(x)≤αsγ +βsγ=
αs+βsγ.
Thenf(αx+βx)≤(αs+βs)γ ≤γ when either one of () and () is satisfied, which shows
Theorem . Let X be a linear space,let K⊂X be a convex set,mapping f :K→R,the
following statements are equivalent:
() f is ans-convex function onK.
() For everyx,x, . . . ,xn∈Kand non-negative real numberα,α, . . . ,αnwith
n
i=αi= ,we have that
f
n
i=
αixi
≤ n
i=
αisf(xi). (.)
Proof ()⇒(). This is obvious.
()⇒(). We will prove by induction overn∈N,n≥. For n= , the inequality is obvious by Definition .. Suppose that the above inequality is valid for alln– . For natural numbern, letx,x, . . . ,xn∈Kandα,α, . . . ,αn≥ with
n
i=αi= .
If there is someαi= , then delete the numberαiand for the remainingn– number,
inequality (.) is obvious by using the inductive hypothesis. Now supposeαi= ,∀i= , . . . ,n, letβ =
n–
i= αi,β=αn,y=β
n–
i= αixi, y=xn,
sinceni=–αiβ
= . Using the inductive hypothesis, we have
f(y) =f n–
i=
αi
β
xi
≤ n–
i=
αi
β
s
f(xi) =
βs
n–
i=
αisf(xi).
Then
f
n
i=
αixi
=f(βy+βy)≤βsf(y) +βsf(xi)
≤ n–
i=
αisf(xi) +αnsf(xn) =
n
i=
αsif(xi),
and the theorem is proved.
The following corollaries are different formulations of the above inequalities of Jensen’s type.
Corollary . Let f :K→R be an s-convex function,α,α, . . . ,αnbe non-negative real numbers,Pn=ni=αi> .For every x,x, . . . ,xn∈K,we have that
f
Pn n
i=
αixi
≤
Ps n
n
i=
αsif(xi).
Corollary . Let f :K→R be an s-convex function,x,x, . . . ,xn∈K,then we have
f
n n
i=
xi
≤
ns n
i=
Corollary . Let f :K→R be an s-convex function,x,x, . . . ,xn∈K.For every
non-negative real numberα,α, . . . ,αn,when Qn=
n
i=α
s
i > ,then we have
f Qn n i= α s i xi ≤ Qs n n i=
αif(xi).
We consider the following function:
θ(I,P,f,x) :=
i∈I
αisf(xi) –
i∈I
αi s f i∈Iαi
i∈I
αixi
,
whereαi> ,i∈I,f is an s-convex function on the convex set K,xi∈K,i∈I,I∈P(N), where P(N)denotes the finite subsets of the natural number set N.
Theorem . Let f :K→R be an s-convex function,αi> ,xi∈K,i∈N,then
() ∀I,J∈P(N),I∩J=,we have that
θ(I∪J,P,f,x)≥θ(I,P,f,x) +θ(J,P,f,x)≥; (.)
() ∀I,J∈P(N),=J⊂I,we have
θ(I,P,f,x)≥θ(J,P,f,x)≥, (.) that is,the mappingθ(I,P,f,x)is monotonic non-decreasing in the first variable onP(N).
Proof () Let∀I,J∈P(N),I∩J=. Then we have
θ(I∪J,P,f,x) =
i∈I
αsif(xi) +
i∈J
αisf(xi) –
i∈I∪J
αi s
f
i∈Iαixi+
i∈Jαixi
i∈I∪Jαi
=
i∈I
αsif(xi) +
i∈J
αisf(xi)
–
i∈I∪J
αi s
f i∈Iαi i∈I∪Jαi
i∈Iαixi
i∈Iαi
+
i∈Jαi
i∈I∪Jαi
i∈Jαixi
i∈Jαi
≥
i∈I
αsif(xi) –
i∈I
αi s
f
i∈Iαixi
i∈Iαi
+
i∈J
αsif(xi) –
i∈J
αi s
f
i∈Jαixi
i∈Jαi
=θ(I,P,f,x) +θ(J,P,f,x) and inequality (.) is proved.
() Suppose thatI,J∈P(N), with=J⊂IandJ=I. With the above assumptions, consider the sequence
whence we get
θ(I,P,f,x) –θ(J,P,f,x)≥θ(I|J,P,f,x)≥
and inequality (.) is proved.
With the above assumptions, consider the sequence
θn:= n
i=
αisf(xi) –
n i= αi s f n
i=αi n
i=
αixi
, n> .
Corollary . With the above assumptions,
θn≥θn–≥ · · · ≥θ≥
i.e.,the sequenceθnis non-decreasing and one has the inequality
θn≥ max
≤i≤j≤n
αisf(xi) +αjsf(xj) – (αi+αj)sf
αixi+αjxj
αi+αj
≥.
Theorem . Let f :K⊂X→R be an s-convex function,andαi≥so that
n
i=αi= . Let xij∈X, ≤i,j≤n.Then one has the inequalities
n
i=
αisαsjf(xij)≥max{A,B} ≥min{A,B} ≥f
n
i=
αiαjxij
, (.)
where A:=ni=αisf(nj=αjxij),B:=nj=αsjf(ni=αixij).
Theorem . Let f:K⊂X→R be an s-convex function,andαi≥so that
n
i=αi= . Then,for everyα,β≥,withα+β= ,we have
()
n
j=
αjs
αs+βs
n
i=
αisf(xi)≥
n
i,j=
αsiαjsf(αxi+βxj),
()
n
i,j=
αisαjsf(αxi+βxj)≥s–
n
i,j=
αisαsjf
xi+xj
, () n
i,j=
αisαjsf
xi+xj
≥f n i= αixi ,
() αs+βs
n
i=
αif(xi)≥ n
i,j=
αiαjf(αxi+βxj).
Proof By thes-convexity off, we can state
f(αxi+βxj)≤αsf(xi) +βsf(xj)
⇒ n
i,j=
αisαjsf(αxi+βxj)≤αs
n
i,j=
αisαsjf(xi) +βs
n
i,j=
=αs
n
j=
αsj
n
i=
αsif(xi) +βs
n
i=
αis
n
j=
αjsf(xj)
=
n
j=
αsj
αs+βs
n
i=
αisf(xi).
We get the first inequality () in Theorem .. By the following inequality
f
xi+xj
=f
αxi+βxj+αxj+βxi
≤
s
f(αxi+βxj) +f(αxj+βxi)
,
we have
n
i,j=
αisαsjf
xi+xj
≤ s n
i,j=
αisαsjf(αxi+βxj) +
n
i,j=
αisαsjf(αxj+βxi)
= s–
n
i,j=
αisαjsf(αxi+βxj)
.
We get the second inequality () in Theorem .. Letxij=xi+xj, by the inequality in Theorem ., we have
n
i,j=
αisαsjf
xi+xj
≥f
n
i,j=
αiαj xi+xj
=f n i= αixi ,
i.e., inequality () is obtained.
n
i,j=
αiαjf(αxi+βxj)≤ n
i,j=
αiαjαsf(xi) + n
i,j=
αiαjβsf(xj)
=αs+βs
n
i=
αif(xi)
i.e., inequality () is proved.
Competing interests
The author declares that they have no competing interests.
Acknowledgements
The author has greatly benefited from the referee’s report. So I wish to express our gratitude to the reviewer and the associate editor SS Dragomir for their valuable suggestions which improved the content and presentation of the paper. The work was supported by the project of Chongqing Municipal Education Commission (No. KJ090732) and special fund project of Chongqing Key laboratory of Electronic Commerce & Supply Chain System (CTBU).
Received: 20 March 2013 Accepted: 30 August 2013 Published:07 Nov 2013 References
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