\def\papertitle{Results in Semi-$E$-Convex Functions}
\def\paperauthorA{Ayache Benhadid}
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\title{\papertitle}
\affiliation{\paperauthorA}	% This command replaces \name{The DAFx Crew} 
{Abbes Laghrour University, Faculty of Science and Technology\\ Departement of Mathematics and Informatics, Khenchela, Algeria.\\
	 e-mail: benhadidayache@yahoo.fr}

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\begin{abstract}
The concept of convexity and its various generalizations is important for quantitative and qualitative studies in operations research or applied mathematics. Recently, E-convex sets and E-convex functions were introduced by Youness \cite{Ref3}, and they have some important applications in various branches of mathematical sciences. Youness in \cite{Ref3} introduced a class of sets and functions which is called $E$-convex sets and $E$-convex functions by relaxing the definition of convex sets and convex functions. Xiusu Chen \cite{Ref2} introduced a new concept of semi $E$-convex functions and discussed its properties. According to Xiusu Chen \cite{Ref2}, if a function $f:M\rightarrow \mathbb{R}$ is semi-$E$-convex on an $E$-convex set $M\subset \mathbb{R}^{n}$ then, $f(E(x))\leq f(x)$ for each $x\in M$. In this article we have discussed the inverse of this proposition and present some results for convex functions.
\end{abstract}
\section{Introduction}
\indent
Youness in \cite{Ref3} introduced a class of sets and functions which is called
$E$-convex sets and $E$-convex functions by relaxing the definition of convex
sets and convex functions. Following this Xiusu Chen \cite{Ref2} introduced a new
class of semi-$E$-convex functions and applied these functions to non linear
programming problems see for instance \cite{Ref1,Ref5} . In this paper, we give weak condition for a lower
semi-continuous function on $\mathbb{R}^{n}$ to be a semi-$E$-convex function, we also present some
results for convex functions.
\section{Preliminaries}

Let $M$ be a nonempty subset of $%
\mathbb{R}^{n}$ and let $E$ : $\mathbb{R}^{n}\rightarrow \mathbb{R}^{n}$ be
a map. We recall:
\begin{definition}  \cite{Ref3} A set $M\subseteq
\mathbb{R}^{n}$ is said to be $E$-convex in $\mathbb{R}^{n}$ if
\[
tE(x)+(1-t)E(y)\in M,
\]
for each $x,y\in M$ and all $t\in \left[ 0,1\right] $.
\end{definition}
\begin{definition}\cite{Ref3} A function $f:M\rightarrow \mathbb{R}$ is said to be $E$-convex on $M$ if $M$
is $E$-convex and
%\begin{equation*}
\[
f\left( tE(x)+(1-t)E(y)\right) \leq tf\left( E\left( x\right) \right)
+(1-t)f\left( E\left( y\right) \right) ,
\]
%\end{equation*}
for each $x,y\in M$ and all $t\in \left[ 0,1\right] $.
\end{definition}
\begin{definition}\cite{Ref2}\label{ds}  A function $f:M\rightarrow \mathbb{R}$ is said to be semi-$E$-convex on $M$ if
$M$ is $E$-convex and
\[
f\left( tE(x)+(1-t)E(y)\right) \leq tf\left( x\right) +(1-t)f\left( y\right),
\]
for each $x,y\in M$ and all $t\in \left[ 0,1\right] $.
\end{definition}
\begin{definition}\cite{Ref2} We define a map $E\times I$ as follows:
\[ \left.
\begin{array}{rcl}
E\times I:\mathbb{R}^{n}\times \mathbb{R} & \rightarrow & \mathbb{R}^{n}\times \mathbb{R} \\
\left(x,t\right) & \rightarrow & (E\times I)(x,t)=\left( E(x),t\right).
\end{array}%
\right.\]
\end{definition}
This Proposition gives a characterization of a semi-$E$-convex function in
term of its $epi\left( f\right) $.
\begin{proposition}\label{p1} \cite{Ref2}   The function $ f:\mathbb{R}^{n}\rightarrow \mathbb{R}$ is semi-$E$-convex on $%
\mathbb{R}^{n}$ if and only if  its epigraph $epi\left( f\right) =\left\{ \left( x,\alpha
\right) \in \mathbb{R}^{n}\times \mathbb{R}:f\left( x\right) \leq \alpha
\right\} $ is $E\times I$-convex
on $\mathbb{R}^{n}\mathbb{\times R}$.
\end{proposition}

\begin{definition} \label{p2} \cite{Ref4} A function $f:\mathbb{R}^{n}\rightarrow \mathbb{R}$ is lower semi-continuous if and only if, for every real number $%
\alpha $, the set $\left\{ x\in\mathbb{R}^{n}:f\left( x\right) \leq \alpha \right\} $ is closed.
\end{definition}
In the following we introduce a Proposition about lower semi-continuous
functions which shall be used in the sequal. For details and for the missing proofs we refer G. Dal Maso  \cite{Ref4}.
\begin{proposition} \label{p3} \cite{Ref4} A function $f:\mathbb{R}^{n}\rightarrow \mathbb{R}$ is lower semi-continuous
if and only if its epigraph is closed.
\end{proposition}
\begin{definition} Let $(x,s),(y,t)\in \mathbb{R}^{n+1}$, with $x,y\in \mathbb{R}^{n}$ and $%
s,t\in\mathbb{R}$. The line segment $[(x,s),(y,t)]$ (with endpoints $(x,s)$ and $%
(y,t))$ is the segment
\[
\left\{ \alpha (x,s)+(1-\alpha )(y,t):0\leq \alpha \leq 1\right\} .
\]
If $(x,s)\neq (y,t)$, the interior $](x,s),(y,t)[$ of $[(x,s),(y,t)]$ is the
segment
\[
\left\{ \alpha (x,s)+(1-\alpha )(y,t):0<\alpha <1\right\} .
\]
In a similar way, we can define $[(x,s),(y,t))$ and $((x,s),(y,t)]$.

\end{definition}%
\section{Main results for semi-$E$-convex functions}
\begin{lemma}\label{l1}
Let $E:\mathbb{R}^{n}\rightarrow \mathbb{R}^{n}$ be a linear and idempotent
map. Consider \\$\left( \overline{x},u\right) \in \lbrack (E\left( x\right)
,s),(E\left( y\right) ,t)].$ Then%
\begin{equation*}
E\left( \overline{x}\right) =\overline{x}.
\end{equation*}
\end{lemma}
\begin{proof} Let $\left( \overline{x},u\right) \in \lbrack (E\left( x\right)
,s),(E\left( y\right) ,t)], $ then there exist $\alpha\in \left[ 0,1\right] $, such that \\$\left( \overline{x},u\right)=\alpha (E\left( x\right) ,s)+(1-\alpha )(E\left( y\right) ,t). $
 % \end{eqnarray* }
Using the fact that $E$ is linear and idempotent map, we have
\begin{eqnarray*}
(E\times I)\left( \overline{x},u\right) &=&\left( E\left( \alpha E\left(
x\right) +(1-\alpha )E\left( y\right) \right) ,\alpha s+(1-\alpha )t\right)
\\
&=&\left( \alpha E\left( x\right) +(1-\alpha )E\left( y\right) ,\alpha
s+(1-\alpha )t\right) \\
&=&\left( \overline{x},u\right).
\end{eqnarray*}
On the other hand
$(E\times I)\left( \overline{x},u\right) =\left( E\left( \overline{x}\right), u\right)$,
therfore $E\left( \overline{x}\right) =\overline{x}$.
\end{proof}

We shall make use the following three sets:\newline
\begin{align}
H_{Sci}&=\left\{ f:\mathbb{R}^{n}\rightarrow \mathbb{R},\mbox{ $f$ is lower
semi continuous}\right\}, \\
H_{L,I}&=\left\{ E:\mathbb{R}^{n}\rightarrow \mathbb{R}^{n},\,E \mbox{
is linear and idempotent}\right\} \\
\intertext{and for each $E\in H_{L,I}$ we define $H_{E}$ as follows:}
H_{E}&=\left\{ f\in H_{Sci},\text{ }f\left( E\left( x\right) \right)\leq
f\left( x\right) \mbox{ for all }x\in \mathbb{R}^{n}\right\}
\end{align}
\begin{theorem}\label{t1}
Let $E\in H_{L,I}$, and $f\in H_{E}$. Suppose that there
exists an $\alpha \in \left] 0,1\right[ $ such that for all $x,y\in \mathbb{R%
}^{n}$, $s,t\in \mathbb{R}$ such that $f\left( x\right) <s,$ $f\left(
y\right) <t,$ 
\[
f\left( \alpha E(x)+(1-\alpha )E(y)\right) <\alpha s+(1-\alpha )t.
\]
Then $f$ is semi-$E$-convex.
\end{theorem}
\begin{proof}  By Proposition (\ref{p1}), it is sufficient to show that $epi\left( f\right) $ is $%
E\times I$-convex as a subset of $\mathbb{R}^{n}\mathbb{\times R}$. By
contradiction, suppose that there exist $(x_{1},\alpha _{1}),(x_{2},\alpha
_{2})\in epi\left( f\right) $ (with $x_{1},x_{2}\in \mathbb{R}^{n}$ and $%
\alpha _{1},\alpha _{2}\in \mathbb{R}$) and $\alpha _{0}\in \left] 0,1\right[
$ such that, \\$\left( \alpha _{0}E\left( x_{1}\right) +\left( 1-\alpha
_{0}\right) E\left( x_{2}\right) ,\alpha _{0}\alpha _{1}+\left( 1-\alpha
_{0}\right) \alpha _{2}\right) \notin epi\left( f\right) .\\$ Let $%
x_{0}=\alpha _{0}E\left( x_{1}\right) +\left( 1-\alpha _{0}\right) E\left(
x_{2}\right) $ and $\lambda _{0}=\alpha _{0}\alpha _{1}+\left( 1-\alpha
_{0}\right) \alpha _{2}$, then $\left( x_{0},\lambda _{0}\right) \notin
epi\left( f\right) $. Using the fact that $f\in H_{ E }$, we
see that $(E\left( x_{1}\right) ,\alpha _{1}),(E\left( x_{2}\right) ,\alpha
_{2})\in epi\left( f\right) . $ Let%
%\begin{equation*}
\[
A=epi\left( f\right) \cap \left[ (E\left( x_{1}\right) ,\alpha _{1}),\left(
x_{0},\lambda _{0}\right) \right]
\]
%\end{equation*} %
 and %
%\begin{equation*}
\[
B=epi\left( f\right) \cap \left[ \left( x_{0},\lambda _{0}\right) ,(E\left(
x_{2}\right) ,\alpha _{2})\right].
\]
%\end{equation*}  %
Since $f\in H_{Sci}$, by Proposition (\ref{p3}), $epi\left( f\right) $ is a closed
subset of $\mathbb{R}^{n}\mathbb{\times R}$. Consequently, $A$ and $B$ are
bounded and closed subsets of $\mathbb{R}^{n}\mathbb{\times R}$.\\ Also we have $\left(
x_{0},\lambda _{0}\right) \notin A$ and $\left( x_{0},\lambda _{0}\right)
\notin B$. Thus there exist $Z_{A}=\left( x_{3},\alpha _{3}\right) \in A$
and $Z_{B}=\left( x_{4},\alpha _{4}\right) \in B$ such that,
\[\min\limits_{Z\in A}\left\Vert Z-\left( x_{0},\lambda _{0}\right)\right\Vert =\left\Vert Z_{A}-\left( x_{0},\lambda _{0}\right) \right\Vert \]
and \[\min\limits_{Z\in B}\left\Vert Z-\left( x_{0},\lambda _{0}\right)\right\Vert =\left\Vert Z_{B}-\left( x_{0},\lambda _{0}\right) \right\Vert. \]
Hence, we have
\begin{equation}
\left] Z_{A},Z_{B}\right[ \cap epi\left( f\right) =\emptyset. \label{ZaZb1}
\end{equation}
On the other hand, since $Z_{A}\in epi\left( f\right) $ and $Z_{B}\in
epi\left( f\right) $, we get\newline
$f\left( x_{3}\right) <\alpha _{3}+\varepsilon ,$ $f\left( x_{4}\right)
<\alpha _{4}+\varepsilon $ for each $\varepsilon >0.$\newline
Since $\alpha \left( \alpha _{3}+\varepsilon \right) +(1-\alpha )\left(
\alpha _{4}+\varepsilon \right) =\alpha \alpha _{3}+(1-\alpha )\alpha
_{4}+\varepsilon.$ By the hypothesis of the theorem, we obtian
%\begin{equation*}
\[
f\left( \alpha E(x_{3})+(1-\alpha )E(x_{4})\right) <\alpha \alpha
_{3}+(1-\alpha )\alpha _{4}+\varepsilon.
\]
%\end{equation*}%
Since $\varepsilon $ is an arbitrary positive real number, it follows that%
\begin{equation}
f\left( \alpha E(x_{3})+(1-\alpha )E(x_{4})\right) \leq \alpha \alpha
_{3}+(1-\alpha )\alpha _{4}.\label{q}
\end{equation}%
Since $Z_{A}\in A\subset \left[ \left( E\left( x_{1}\right) ,\alpha
_{1}\right) ,(E\left( x_{2}\right) ,\alpha _{2})\right] $ and \newline
$Z_{B}\in B\subset \left[ \left( E\left( x_{1}\right) ,\alpha _{1}\right)
,(E\left( x_{2}\right) ,\alpha _{2})\right] $. By Lemma  (\ref{l1}) we have $E\left(
x_{3}\right) =x_{3}$ and \\$E\left( x_{4}\right) =x_{4}.$ Using (\ref{q}) we get%
\[
\left( \alpha x_{3}+(1-\alpha )x_{4},\alpha \alpha _{3}+(1-\alpha )\right)
\alpha _{4})\in epi\left( f\right).
\] %
Therfore%
\[
\alpha Z_{A}+(1-\alpha )Z_{B}\in epi\left( f\right)
\] %
which contradicts (\ref{ZaZb1}). Thus, we conclude that $epi\left( f\right) $ is $%
E\times I$-convex.
\end{proof}
\begin{theorem}\label{t5}
Let $E:\mathbb{R}^{n}\rightarrow \mathbb{R}^{n}$ be a linear and idempotent
map, $f:\mathbb{R}^{n}\rightarrow \mathbb{R}$ be lower semi-continuous and $%
f\left( E\left( x\right) \right) \leq f\left( x\right)$ for all $x\in 
\mathbb{R}^{n}$. Then $f$ is semi-$E$-convex if and only if there exists an $%
\alpha \in \left] 0,1\right[ $ such that for all $x,y\in \mathbb{R}^{n}$
\[
f\left( \alpha E(x)+(1-\alpha )E(y)\right) \leq \alpha f\left( x\right)
+(1-\alpha )f\left( y\right).
\]
\end{theorem}
\begin{proof} Follows from Theorem (\ref{t1}) with $s=f\left( x\right) +\varepsilon $ and $t=f\left( y\right) +\varepsilon $ for each $\varepsilon >0,$ then taking $\varepsilon \rightarrow 0.$
\end{proof}
\begin{corollary}
Let $E:\mathbb{R}^{n}\rightarrow \mathbb{R}^{n}$ be a linear and idempotent
map, $f:\mathbb{R}^{n}\rightarrow \mathbb{R}$ be lower semi-continuous and $%
f\left( E\left( x\right) \right) \leq f\left( x\right) $ for all $x\in 
\mathbb{R}^{n}$. Then $f$ is semi-$E$-convex if and only if for all $x,y\in
\mathbb{R}^{n}$,%
\[
f\left( \frac{1}{2}\left( E(x)+E(y)\right) \right) \leq \frac{1}{2}\left[
f\left( x\right) +f\left( y\right) \right] .
\]
\end{corollary}

\begin{theorem}\label{t6}
Let $E:\mathbb{R}^{n}\rightarrow \mathbb{R}^{n}$ be a linear and idempotent
map, $f:\mathbb{R}^{n}\rightarrow \mathbb{R}$ be lower semi-continuous and $%
f\left( E\left( x\right) \right) \leq f\left( x\right) $ for all $x\in 
\mathbb{R}^{n}$. Then f is semi-$E$-convex
if and only if for all $x,y\in \mathbb{R}^{n}$, there exists an $\alpha \in %
\left] 0,1\right[ $ ($\alpha $ depends on $x,y$) such that%
\begin{equation}
f\left( \alpha E(x)+(1-\alpha )E(y)\right) \leq \alpha f\left( x\right)\label{ZaZb3}
+(1-\alpha )f\left( y\right).
\end{equation}
\end{theorem}


If we take $E=Id_{\mathbb{R}^{n},}$ we get $E\in H_{L,I}$, and $H_{E}=H_{Sci}.$ Then we find results about convex functions.

\begin{theorem}  Let  $f:\mathbb{R}^{n}\rightarrow \mathbb{R}$ be lower
semi-continuous.Then f is convex if and only if there exists an $%
\alpha \in \left] 0,1\right[ $ such that, for all $x,y\in \mathbb{R}^{n}$,%
\[
f\left( \alpha x+(1-\alpha )y\right) \leq \alpha f\left( x\right) +(1-\alpha
)f\left( y\right) .
\]%
\end{theorem}
\begin{theorem} Let  $f:\mathbb{R}^{n}\rightarrow \mathbb{R}$ be lower
semi-continuous. Then f is convex if and only if for all $x,y\in
\mathbb{R}^{n}$, there exists an $\alpha \in \left] 0,1\right[ $ ($\alpha $
depends on $x,y$) such that%
\[
f\left( \alpha x+(1-\alpha )y\right) \leq \alpha f\left( x\right) +(1-\alpha
)f\left( y\right).
\]
\end{theorem}
\begin{corollary}
 Let  $f:\mathbb{R}^{n}\rightarrow \mathbb{R}$ be lower
semi-continuous. Then $f$ is convex if and only if for all $x,y\in
\mathbb{R}^{n}$,%
\[
f\left( \frac{1}{2}\left( x+y\right) \right) \leq \frac{1}{2}\left[
f\left( x\right) +f\left( y\right) \right] .
\]
\end{corollary}

\begin{thebibliography}{}
\bibitem{Ref2} X. Chen, Some properties of semi-E-convex functions, Journal of mathematical Anaysis andApplication, 275, 251-262, (2002).
\bibitem{Ref3} E. A. Youness, E-convex sets, E-convex functions, and E-convex programming, Journal of Optimization Theory and Applications, 102, 439-450, (1999).
\bibitem{Ref4}G. Dal maso, An Introduction to ${\Gamma}$-convergence,10-11. BirKhauser, Boston, (1993).
\bibitem{Ref1} A. A. Megahed, H. G. Gomma, E. A. Youness, A. H. Banna, Optimality conditions of $E$-convex programing for an $E$-differentiable function, journal of Inequalities and Applications, 246, 1-11, (2013).
\bibitem{Ref5} M. A. Noor, M. U. Awan, K. I. Noor, On some inequalities for relative semi-convex functions, journal of Inequalities and Applications, 332, 1-16,(2013).
\end{thebibliography}
\end{document}
