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%
\title{On Fej\'{e}r type inequalities for products convex and $s$-convex
functions}
\author{H\"{u}seyin Budak}
\address{ D\"{u}zce University, \\ Faculty of Science and Arts\\
Department of Mathematics,\\
D\"{u}zce,Turkey}
\email{hsyn.budak@gmail.com}
%
\author{Yonca Bak\i \c{s}}
\address{ D\"{u}zce University, \\ Faculty of Science and Arts\\
Department of Mathematics,\\
D\"{u}zce,Turkey}
\email{yonca.bakis93@hotmail.com}
%

\subjclass{26D07, 26D10, 26D15, 26A33}
\keywords{Fej\'{e}r type inequalities, convex function,
integral inequalities}
\begin{abstract}
In this paper, we first obtain some new Fej\'{e}r type inequalities for
products of convex and $s$-convex mappings. Then, some Fej\'{e}r type
inequalities for products of two $s$-convex function are established.
\end{abstract}
\maketitle

\section{Introduction}

The inequalities discovered by C. Hermite and J. Hadamard for convex
functions are considerable significant in the literature (see, e.g.,\cite%
{Dragomir1}, \cite[p.137]{Pecaric}). These inequalities state that if $%
f:I\rightarrow \mathbb{R}$ is a convex function on the interval $I$ of real
numbers and $a,b\in I$ with $a<b$, then 
\begin{equation}
f\left( \frac{a+b}{2}\right) \leq \frac{1}{b-a}\int_{a}^{b}f(x)dx\leq \frac{%
f\left( a\right) +f\left( b\right) }{2}.  \label{E1}
\end{equation}

Both inequalities hold in the reversed direction if $f$ is concave. We note
that Hermite-Hadamard inequality may be regarded as a refinement of the
concept of convexity and it follows easily from Jensen's inequality. Over
the years, many studies have focused on to establish generalization of the
inequality (\ref{E1}) and to obtain new bounds for left hand side and right
hand side of the inequality (\ref{E1}).

The overall structure of the paper takes the form of five sections including
introduction. The remainder of this work is organized as follows: we first
give some Hermite-Hadamard and Fej\'{e}r type inequalities.. Moreover, we
give some Hermite-Hadamard type inequalities for products two convex
functions. In Section 2 and Section 3, we obtain some integral inequalities
of Hermite-Hadamard-Fej\'{e}r type for products convex and $s$-convex
functions and for products two $s$-convex functions. We give also some
special cases of these inequalities. Finally, conclusions and future
directions of research are discussed in Section 4.

The weighted version of the inequalities (\ref{E1}), so-called
Hermite-Hadamard-Fej\'{e}r inequalities, was given by Fejer in \cite{Fejer}
as follow:

\begin{theorem}
$f:[a,b]\rightarrow \mathbb{R}$, be a convex function, then the inequality%
\begin{equation}
f\left( \frac{a+b}{2}\right) \int \limits_{a}^{b}w(x)dx\leq \int
\limits_{a}^{b}f(x)w(x)dx\leq \frac{f(a)+f(b)}{2}\int \limits_{a}^{b}w(x)dx
\label{H1}
\end{equation}%
holds, where $w:[a,b]\rightarrow \mathbb{R}$ is non-negative, integrable,
and symmetric about $x=\frac{a+b}{2}$ (i.e. $w(x)=w(a+b-x)$)$.$
\end{theorem}

In \cite{pac}, Pachpatte established the Hermite-Hadamard type inequalities
for products of two convex functions.

\begin{theorem}
\label{t} Let $f$ and $g$ be real-valued, non-negative and convex functions
on $\left[ a,b\right] .$ Then we have%
\begin{equation}
\frac{1}{b-a}\int \limits_{a}^{b}f(x)g(x)dx\leq \frac{1}{3}M(a,b)+\frac{1}{6}%
N(a,b),  \label{1}
\end{equation}%
and 
\begin{equation}
2f\left( \frac{a+b}{2}\right) g\left( \frac{a+b}{2}\right) \leq \frac{1}{b-a}%
\int \limits_{a}^{b}f(x)g(x)dx+\frac{1}{6}M(a,b)+\frac{1}{3}N(a,b)  \label{2}
\end{equation}%
where $M(a,b)=f(a)g(a)+f(b)g(b)$ and $N(a,b)=f(a)g(b)+f(b)g(a).$
\end{theorem}

In recent years, the generalized versions of inequalities (\ref{1}) and (\ref%
{2}) for several convexity have been proved. For some of them please refer
to (\cite{chen}-\cite{chen3}, \cite{hue}, \cite{set}, \cite{yin}). Kirmaci
et al. gave the proved inequalities (\ref{1}) and (\ref{2}) for products of
convex and $s$-convex functions in \cite{kir}. On the other hand, Budak\cite%
{budak2} proved the weighted versions of the inequalities (\ref{1}) and (\ref%
{2}) which generalize the several obtained inequalities. Moreover in \cite%
{lat}, Latif and Alomari proved some inequalities for product of two
co-ordinated convex function. Furthermore in \cite{ozd} and \cite{ozd2},
Ozdemir et al. gave some generalizations of results given by Latif and
Alomari using the product of two coordinated $s$-convex mappings and
product of two coordinated $h$-convex mappings, respectively. In \cite%
{budak}, Budak and Sar\i kaya proved Hermite-Hadamard type inequalities for
products of two co-ordinated convex mappings via fractional integrals.

\section{Fej\'{e}r type inequalities for products convex and $s$-convex functions}

In this section, we present some Fej\'{e}r type inequalities for products
convex and $s$-convex functions.

\begin{theorem}
\label{t1} Suppose that $w:I\rightarrow \mathbb{R}$ is non-negative,
integrable, and symmetric about $x=\frac{a+b}{2}$ (i.e. $w(x)=w(a+b-x)$). If 
$f:I\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ is a real-valued, non-negative and convex functions on $I$ and if $%
g:I\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ is a $s$-convex on $I$ for some fixed $s\in (0,1],$ then for any $a,b\in I$%
, we have%
\begin{eqnarray}
 \int \limits_{a}^{b}f(x)g(x)w(x)dx &\leq & \frac{M(a,b)}{\left( b-a\right) ^{s+1}%
}\int \limits_{a}^{b}\left( b-x\right) ^{s+1}w(x)dx  \label{b1}\\
&& +\frac{N(a,b)}{\left(
b-a\right) ^{s+1}}\int \limits_{a}^{b}\left( b-x\right) \left( x-a\right)
^{s}w(x)dx  \notag
\end{eqnarray}
where 
\begin{equation*}
M(a,b)=f(a)g(a)+f(b)g(b)\text{ and }N(a,b)=f(a)g(b)+f(b)g(a).
\end{equation*}
\end{theorem}

\begin{proof}
Since $f$ is convex and $g$ is $s$-convex functions on $\left[ a,b\right] ,$
then we have%
\begin{equation}
f\left( ta+\left( 1-t\right) b\right) \leq tf(a)+(1-t)f(b)  \label{b2}
\end{equation}%
and%
\begin{equation}
g\left( ta+\left( 1-t\right) b\right) \leq t^{s}g(a)+\left( 1-t\right)
^{s}g(b).  \label{b3}
\end{equation}%
By adding the inequalities (\ref{b2}) and (\ref{b3}), we get%
\begin{eqnarray}
&&f\left( ta+\left( 1-t\right) b\right) g\left( ta+\left( 1-t\right) b\right)
\label{b4} \\
&&  \notag \\
&\leq &t^{s+1}f(a)g(a)+\left( 1-t\right) ^{s+1}f(b)g(b) \notag\\
&&  \notag \\
&+&t\left( 1-t\right)
^{s}f(a)g(b)+t^{s}\left( 1-t\right) f(b)g(a).  \notag
\end{eqnarray}%
Multiplying both sides of (\ref{b4}) by $w\left( ta+\left( 1-t\right)
b\right) ,$ then integrating the resulting inequality with respect to $t$
from $0$ to $1,$ we obtain%
\begin{eqnarray}
&&\int \limits_{0}^{1}f\left( ta+\left( 1-t\right) b\right) g\left(
ta+\left( 1-t\right) b\right) w\left( ta+\left( 1-t\right) b\right) dt
\label{b5} \\
&\leq &f(a)g(a)\int \limits_{0}^{1}t^{s+1}w\left( ta+\left( 1-t\right)
b\right) dt \notag\\
&&+ f(b)g(b)\int \limits_{0}^{1}\left( 1-t\right) ^{s+1}w\left(
ta+\left( 1-t\right) b\right) dt  \notag \\
&&+f(a)g(b)\int \limits_{0}^{1}t\left( 1-t\right) ^{s}w\left( ta+\left(
1-t\right) b\right) dt \notag\\
&&+f(b)g(a)\int \limits_{0}^{1}t^{s}\left( 1-t\right)
w\left( ta+\left( 1-t\right) b\right) dt.  \notag
\end{eqnarray}%
By change of variable $x=ta+\left( 1-t\right) b$ with $dx=-(b-a)dt,$ we get%
\begin{eqnarray}
&&\int \limits_{0}^{1}f\left( ta+\left( 1-t\right) b\right) g\left( ta+\left(
1-t\right) b\right) w\left( ta+\left( 1-t\right) b\right) dt\label{b6}\\
&&=\frac{1}{b-a}%
\int \limits_{a}^{b}f(x)g(x)w(x)dx. \notag 
\end{eqnarray}%
Moreover, it is easily observe that%
\begin{equation}
\int \limits_{0}^{1}t^{s+1}w\left( ta+\left( 1-t\right) b\right) dt=\frac{1}{%
\left( b-a\right) ^{s+2}}\int \limits_{a}^{b}\left( b-x\right) ^{s+1}w(x)dx
\label{b7}
\end{equation}%
and since $w$ is symmetric about $\frac{a+b}{2},$ we have%
\begin{eqnarray}
&&\int \limits_{0}^{1}\left( 1-t\right) ^{s+1}w\left( ta+\left( 1-t\right)
b\right) dt \notag \\
 &&=\frac{1}{\left( b-a\right) ^{s+2}}\int \limits_{a}^{b}\left(
x-a\right) ^{s+1}w(x)dx  \label{b8} \\
&&=\frac{1}{\left( b-a\right) ^{s+2}}\int \limits_{a}^{b}\left( b-u\right)
^{s+1}w(a+b-u)du.  \notag \\
&&=\frac{1}{\left( b-a\right) ^{s+2}}\int \limits_{a}^{b}\left( b-u\right)
^{s+1}w(u)du.  \notag
\end{eqnarray}%
We also have%
\begin{eqnarray}
&&\int \limits_{0}^{1}t\left( 1-t\right) ^{s}w\left( ta+\left( 1-t\right)
b\right) dt\label{b9}\\ 
&&=\frac{1}{\left( b-a\right) ^{s+2}}\int \limits_{a}^{b}\left(
b-x\right) \left( x-a\right) ^{s}w(x)dx  \notag
\end{eqnarray}%
and%
\begin{eqnarray}
&&\int \limits_{0}^{1}t^{s}\left( 1-t\right) w\left( ta+\left( 1-t\right)
b\right) dt  \label{b10} \\
 &&=\frac{1}{\left( b-a\right) ^{s+2}}\int \limits_{a}^{b}\left(
b-x\right) ^{s}\left( x-a\right) w(x)dx \notag \\
&&=\frac{1}{\left( b-a\right) ^{s+2}}\int \limits_{a}^{b}\left( b-u\right)
\left( u-a\right) ^{s}w(a+b-u)du  \notag \\
&&=\frac{1}{\left( b-a\right) ^{s+2}}\int \limits_{a}^{b}\left( b-u\right)
\left( u-a\right) ^{s}w(u)du.  \notag
\end{eqnarray}%
By substituting the equalities (\ref{b6})-(\ref{b10}) in (\ref{b5}), then we
have the following inequality%
\begin{eqnarray}
&&\frac{1}{b-a}\int \limits_{a}^{b}f(x)g(x)w(x)dx \label{b11} \\ &&\leq \frac{\left[
f(a)g(a)+f(b)g(b)\right] }{\left( b-a\right) ^{s+2}}\int
\limits_{a}^{b}\left( b-x\right) ^{s+1}w(x)dx  \notag \\
&&+\frac{f(a)g(b)+f(b)g(a)}{\left( b-a\right) ^{s+2}}\int
\limits_{a}^{b}\left( b-x\right) \left( x-a\right) ^{s}w(x)dx.  \notag
\end{eqnarray}%
If we multiply both sides of (\ref{b11}) by $\left( b-a\right) ,$ then we
obtain the desired result.
\end{proof}

\begin{remark}
If we choose $w(x)=1$ for all $x\in \left[ a,b\right] $ in Theorem \ref{t1},
then we have the following inequality%
\begin{equation*}
\frac{1}{b-a}\int \limits_{a}^{b}f(x)g(x)dx\leq \frac{1}{s+2}M(a,b)+\frac{1}{%
\left( s+1\right) \left( s+2\right) }N(a,b)
\end{equation*}%
which is proved by K\i rmac\i \ et al. in \cite{kir}.
\end{remark}

\begin{remark}
If we choose $s=1$ in Theorem \ref{t1}, then we have the following inequality%
\begin{eqnarray*}
\int \limits_{a}^{b}f(x)g(x)w(x)dx &\leq & \frac{M(a,b)}{\left( b-a\right) ^{2}}%
\int \limits_{a}^{b}\left( b-x\right) ^{2}w(x)dx\\
&&+\frac{N(a,b)}{\left(
b-a\right) ^{2}}\int \limits_{a}^{b}\left( b-x\right) \left( x-a\right)
w(x)dx
\end{eqnarray*}%
which is proved by Budak in \cite{budak2}.
\end{remark}

\begin{remark}
If we choose $f(x)=1$ for all $x\in \left[ a,b\right] $ in Theorem \ref{t1},
then we have the following inequality%
\begin{equation*}
\int \limits_{a}^{b}g(x)w(x)dx\leq \frac{g(a)+g(b)}{2\left( b-a\right) ^{s}}%
\int \limits_{a}^{b}\left[ \left( b-x\right) ^{s}+\left( x-a\right) ^{s}%
\right] w(x)dx
\end{equation*}%
which is proved by Sar\i kata et al. in \cite[for $h(t)=t^{s}$]{sarikaya}.
\end{remark}

\begin{proof}
From\ the inequality (\ref{b1}) for $f(x)=1$ for all $x\in \left[ a,b\right]
,$ we have%
\begin{eqnarray}
&& \int \limits_{a}^{b}g(x)w(x)dx \notag \\
&\leq &\frac{g(a)+g(b)}{\left( b-a\right)
^{s+1}}\int \limits_{a}^{b}\left( b-x\right) ^{s+1}w(x)dx  \label{b12} \\
&&+\frac{g(a)+g(b)}{\left( b-a\right) ^{s+1}}\int \limits_{a}^{b}\left(
b-x\right) \left( x-a\right) ^{s}w(x)dx  \notag \\
&=&\frac{g(a)+g(b)}{\left( b-a\right) ^{s+1}}\left[ \int
\limits_{a}^{b}\left( b-x\right) ^{s+1}w(x)dx+\int \limits_{a}^{b}\left(
b-x\right) \left( x-a\right) ^{s}w(x)dx\right] .  \notag
\end{eqnarray}%
Since $w$ is symmetric about $\frac{a+b}{2},$ we have%
\begin{equation*}
\int \limits_{a}^{b}\left( b-x\right) ^{s+1}w(x)dx=\int
\limits_{a}^{b}\left( x-a\right) ^{s+1}w(x)dx.
\end{equation*}%
Using this equality in (\ref{b12}), we get%
\begin{eqnarray*}
&&\int \limits_{a}^{b}g(x)w(x)dx \\
&\leq &\frac{g(a)+g(b)}{\left( b-a\right)
^{s+1}}\left[ \int \limits_{a}^{b}\left( x-a\right) ^{s+1}w(x)dx+\int
\limits_{a}^{b}\left( b-x\right) \left( x-a\right) ^{s}w(x)dx\right] \\
&& \\
&=&\frac{g(a)+g(b)}{\left( b-a\right) ^{s}}\int \limits_{a}^{b}\left(
x-a\right) ^{s}w(x)dx \\
&& \\
&=&\frac{g(a)+g(b)}{2\left( b-a\right) ^{s}}\int \limits_{a}^{b}\left[
\left( x-a\right) ^{s}+\left( b-x\right) ^{s}\right] w(x)dx
\end{eqnarray*}%
which completes the proof.
\end{proof}

\begin{theorem}
\label{t2} Suppose that conditions of Theorem \ref{t1} hold, then we have
the following inequality%
\begin{eqnarray}
&&2^{s}f\left( \frac{a+b}{2}\right) g\left( \frac{a+b}{2}\right) \int
\limits_{a}^{b}w\left( x\right) dx  \label{b14}\\
&&  \notag \\
&\leq &\int \limits_{a}^{b}f(x)g(x)w(x)dx+\frac{M(a,b)}{\left( b-a\right)
^{s+1}}\int \limits_{a}^{b}\left( x-a\right) ^{s}\left( b-x\right) w(x)dx\notag \\
&&+\frac{N(a,b)}{\left( b-a\right) ^{s+1}}\int \limits_{a}^{b}\left( b-x\right)
^{s+1}w(x)dx.  \notag
\end{eqnarray}%
where $M(a,b)$ and $N(a,b)$ are defined as in Theorem \ref{t1}.
\end{theorem}

\begin{proof}
For $t\in \left[ 0,1\right] $, we can write%
\begin{equation*}
\frac{a+b}{2}=\frac{(1-t)a+tb}{2}+\frac{ta+(1-t)b}{2}.
\end{equation*}%
Using the convexity of $f$ and $s$-convexity of $g,$ we have%
\begin{eqnarray*}
&&f\left( \frac{a+b}{2}\right) g\left( \frac{a+b}{2}\right) \\
&& \\
&=&f\left( \frac{(1-t)a+tb}{2}+\frac{ta+(1-t)b}{2}\right) g\left( \frac{%
(1-t)a+tb}{2}+\frac{ta+(1-t)b}{2}\right) \\
&& \\
&\leq &\frac{1}{2^{s+1}}\left[ f((1-t)a+tb)+f(ta+(1-t)b)\right] 
\\
&& \\
 && \times  \left[
g((1-t)a+tb)+g(ta+(1-t)b)\right] \\
&& \\
&=&\frac{1}{2^{s+1}}\left[ f((1-t)a+tb)g((1-t)a+tb)+f(ta+(1-t)b)g(ta+(1-t)b)%
\right] \\
&& \\
&&+\frac{1}{2^{s+1}}\left[ f((1-t)a+tb)g(ta+(1-t)b)+f(ta+(1-t)b)g((1-t)a+tb)%
\right] .
\end{eqnarray*}%
By using again the convexity of $f$ and $s$-convexity of $g$, we obtain%
\begin{eqnarray}
&&f\left( \frac{a+b}{2}\right) g\left( \frac{a+b}{2}\right)  \label{b15} \\
&\leq &\frac{1}{2^{s+1}}\left[
f((1-t)a+tb)g((1-t)a+tb)+f(ta+(1-t)b)g(ta+(1-t)b)\right]  \notag \\
&&+\frac{1}{2^{s+1}}\left[ t^{s}\left( 1-t\right) +t(1-t)^{s}\right] \left[
f(a)g(a)+f(b)g(b)\right]  \notag \\
&&+\frac{1}{2^{s+1}}\left[ t^{s+1}+\left( 1-t\right) ^{s+1}\right] \left[
f(a)g(b)+f(b)g(a)\right] .  \notag
\end{eqnarray}%
Multiplying both sides of (\ref{b15}) by $w\left( \left( 1-t\right)
a+tb\right) ,$ then integrating the resulting inequality with respect to $t$
from $0$ to $1,$ we obtain 
\begin{eqnarray}
&&f\left( \frac{a+b}{2}\right) g\left( \frac{a+b}{2}\right) \int
\limits_{0}^{1}w\left( \left( 1-t\right) a+tb\right) dt  \label{b16} \\
&\leq &\frac{1}{2^{s+1}}\int \limits_{0}^{1}\left[
f((1-t)a+tb)g((1-t)a+tb) \right. \notag \\
&&\notag \\
&& \left.+f(ta+(1-t)b)g(ta+(1-t)b)\right] w\left( \left(
1-t\right) a+tb\right) dt  \notag \\
&&\notag \\
&&+\frac{M(a,b)}{2^{s+1}}\int \limits_{0}^{1}\left[ t^{s}\left( 1-t\right)
+t(1-t)^{s}\right] w\left( \left( 1-t\right) a+tb\right) dt  \notag \\
&&+\frac{N(a,b)}{2^{s+1}}\int \limits_{0}^{1}\left[ t^{s+1}+\left(
1-t\right) ^{s+1}\right] w\left( \left( 1-t\right) a+tb\right) dt.  \notag
\end{eqnarray}%
Using the change of variable, we have 
\begin{equation}
\int \limits_{0}^{1}w\left( \left( 1-t\right) a+tb\right) dt=\frac{1}{b-a}%
\int \limits_{a}^{b}w\left( x\right) dx,  \label{b17}
\end{equation}%
\begin{eqnarray}
&&\int \limits_{0}^{1}f((1-t)a+tb)g((1-t)a+tb)w\left( \left( 1-t\right)
a+tb\right) dt  \label{b18} \\
&&+\int \limits_{0}^{1}f(ta+(1-t)b)g(ta+(1-t)b)w\left( \left( 1-t\right)
a+tb\right) dt  \notag \\
&=&\frac{1}{b-a}\int \limits_{a}^{b}f(x)g(x)w(x)dx+\frac{1}{b-a}\int
\limits_{a}^{b}f(x)g(x)w(a+b-x)dx  \notag \\
&=&\frac{2}{b-a}\int \limits_{a}^{b}f(x)g(x)w(x)dx,  \notag
\end{eqnarray}%
\begin{eqnarray}
&&\int \limits_{0}^{1}\left[ t^{s}\left( 1-t\right) +t(1-t)^{s}\right]
w\left( \left( 1-t\right) a+tb\right) dt  \label{b19} \\
&&  \notag \\
&=&\int \limits_{0}^{1}\left[ t^{s}\left( 1-t\right) w\left( \left(
1-t\right) a+tb\right) +t(1-t)^{s}w\left( \left( 1-t\right) a+tb\right) %
\right] dt  \notag 
\end{eqnarray}%

\begin{eqnarray}
&&=\frac{1}{\left( b-a\right) ^{s+2}}\int \limits_{a}^{b}\left( x-a\right)
^{s}\left( b-x\right) w(x)dx\notag \\
&&+\frac{1}{\left( b-a\right) ^{s+2}}\int
\limits_{a}^{b}\left( x-a\right) ^{s}\left( b-x\right) w(a+b-x)dx  \notag \\
&&  \notag \\
&=&\frac{2}{\left( b-a\right) ^{s+2}}\int \limits_{a}^{b}\left( x-a\right)
^{s}\left( b-x\right) w(x)dx  \notag
\end{eqnarray}%
and%
\begin{eqnarray}
&&\int \limits_{0}^{1}\left[ t^{s+1}+\left( 1-t\right) ^{s+1}\right] w\left(
\left( 1-t\right) a+tb\right) dt  \label{b20} \\
&&  \notag \\
&=&\int \limits_{0}^{1}\left[ t^{s+1}w\left( \left( 1-t\right) a+tb\right)
+\left( 1-t\right) ^{s+1}w\left( \left( 1-t\right) a+tb\right) \right] dt 
\notag \\
&&  \notag \\
&=&\frac{1}{\left( b-a\right) ^{s+2}}\int \limits_{a}^{b}\left( b-x\right)
^{s+1}w(a+b-x)dx\notag \\
&&+\frac{1}{\left( b-a\right) ^{s+2}}\int
\limits_{a}^{b}\left( b-x\right) ^{s+1}w(x)dx  \notag \\
&&  \notag \\
&=&\frac{2}{\left( b-a\right) ^{s+2}}\int \limits_{a}^{b}\left( b-x\right)
^{s+1}w(x)dx.  \notag
\end{eqnarray}%
If we substitute the equalities (\ref{b17})-(\ref{b20}) in (\ref{b16}), then
we have the following inequality 
\begin{eqnarray}
&&f\left( \frac{a+b}{2}\right) g\left( \frac{a+b}{2}\right) \frac{1}{b-a}%
\int \limits_{a}^{b}w\left( x\right) dx  \label{b21} \\
&&\leq \frac{1}{2^{s}\left( b-a\right) }\int \limits_{a}^{b}f(x)g(x)w(x)dx+\frac{M(a,b)}{2^{s}\left( b-a\right) ^{s+2}}\int \limits_{a}^{b}\left(
x-a\right) ^{s}\left( b-x\right) w(x)dx\notag \\
&&+\frac{N(a,b)}{2^{s}\left( b-a\right)
^{s+2}}\int \limits_{a}^{b}\left( b-x\right) ^{s+1}w(x)dx.  \notag
\end{eqnarray}%
By multiplying the both sides of (\ref{b21}) by $2^{s}(b-a)$ then we obtain
the desired result (\ref{b14}).
\end{proof}

\begin{remark}
If we choose $w(x)=1$ for all $x\in \left[ a,b\right] $ in Theorem \ref{t2},
then we have the following inequality%
\begin{equation}
2^{s}f\left( \frac{a+b}{2}\right) g\left( \frac{a+b}{2}\right) \leq \frac{1}{%
b-a}\int \limits_{a}^{b}f(x)g(x)dx+\frac{M(a,b)}{\left( s+1\right) \left(
s+2\right) }+\frac{N(a,b)}{s+2}  \notag
\end{equation}%
which is proved by K\i rmac\i \ et al. in \cite{kir}.
\end{remark}

\begin{remark}
If we choose $s=1$ in Theorem \ref{t1}, then we have the following inequality%
\begin{eqnarray}
&&2f\left( \frac{a+b}{2}\right) g\left( \frac{a+b}{2}\right) \int
\limits_{a}^{b}w\left( x\right) dx  \notag \\
&&  \notag \\
&&\leq \int \limits_{a}^{b}f(x)g(x)w(x)dx+\frac{M(a,b)}{\left( b-a\right)
^{2}}\int \limits_{a}^{b}\left( x-a\right) \left( b-x\right) w(x)dx\notag \\
&&+\frac{%
N(a,b)}{\left( b-a\right) ^{2}}\int \limits_{a}^{b}\left( b-x\right)
^{2}w(x)dx.  \notag
\end{eqnarray}%
which is proved by Budak in \cite{budak2}.
\end{remark}

\begin{corollary}
If we choose $f(x)=1$ for all $x\in \left[ a,b\right] $ in Theorem \ref{t2},
then we have the following the following Fej\'{e}r type inequality%
\begin{eqnarray*}
&&2^{s}g\left( \frac{a+b}{2}\right) \int \limits_{a}^{b}w\left( x\right)
dx\\
&&\leq \int \limits_{a}^{b}g(x)w(x)dx+\frac{g(a)+g(b)}{2\left( b-a\right)
^{s}}\int \limits_{a}^{b}\left[ \left( x-a\right) ^{s}+\left( b-x\right) %
\right] ^{s}w(x)dx.
\end{eqnarray*}
\end{corollary}

\begin{proof}
From inequality (\ref{b14}) for $f(x)=1$ for all $x\in \left[ a,b\right] ,$
we have 
\begin{eqnarray*}
&&2g\left( \frac{a+b}{2}\right) \int \limits_{a}^{b}w\left( x\right) dx \\
&\leq &\int \limits_{a}^{b}g(x)w(x)dx+\frac{g(a)+g(b)}{\left( b-a\right)
^{s+1}}\int \limits_{a}^{b}\left( x-a\right) ^{s}\left( b-x\right) w(x)dx\\
&&
+\frac{g(a)+g(b)}{\left( b-a\right) ^{s+1}}\int \limits_{a}^{b}\left(
b-x\right) ^{s+1}w(x)dx \\
&& \\
&=&\int \limits_{a}^{b}g(x)w(x)dx\\
&&+\frac{g(a)+g(b)}{\left( b-a\right) ^{s+1}}%
\left[ \int \limits_{a}^{b}\left( x-a\right) ^{s}\left( b-x\right)
w(x)dx+\int \limits_{a}^{b}\left( b-x\right) ^{s+1}w(x)dx\right] \\
&& \\
&=&\int \limits_{a}^{b}g(x)w(x)dx+\frac{g(a)+g(b)}{2\left( b-a\right) ^{s}}%
\int \limits_{a}^{b}\left[ \left( x-a\right) ^{s}+\left( b-x\right) ^{s}%
\right] w(x)dx.
\end{eqnarray*}%
This completes the proof.
\end{proof}

\section{Fej\'{e}r type inequalities for products two $s$-convex functions}

In this section, we present some Fej\'{e}r type inequalities for products
two $s$-convex functions which generalize the results in Section 2.

\begin{theorem}
\label{t11} Suppose that $w:I\rightarrow \mathbb{R}$ is non-negative,
integrable, and symmetric about $x=\frac{a+b}{2}$ (i.e. $w(x)=w(a+b-x)$). If 
$f:I\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ is $s_{1}$-convex functions on $I$ and if $g:I\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ is $s_{2}$-convex on $I$ for some fixed $s_{1},s_{2}\in (0,1],$ then for
any $a,b\in I$, we have%
\begin{eqnarray}
\int \limits_{a}^{b}f(x)g(x)w(x)dx &\leq &\frac{M(a,b)}{\left( b-a\right)
^{s_{1}+s_{2}}}\int \limits_{a}^{b}\left( b-x\right) ^{s_{1}+s_{2}}w(x)dx
\label{b22} \\
&&  \notag \\
&&+\frac{N(a,b)}{\left( b-a\right) ^{s_{1}+s_{2}}}\int \limits_{a}^{b}\left(
b-x\right) ^{s_{1}}\left( x-a\right) ^{s_{2}}w(x)dx.  \notag
\end{eqnarray}%
where $M(a,b)$ and $N(a,b)$ are defined as in Theorem \ref{t1}.
\end{theorem}

\begin{proof}
Since $f$ is $s_{1}$-convex and $g$ is $s_{2}$-convex functions on $\left[
a,b\right] ,$ then we have%
\begin{equation}
f\left( ta+\left( 1-t\right) b\right) \leq t^{s_{1}}f(a)+(1-t)^{s_{1}}f(b)
\label{b23}
\end{equation}%
and%
\begin{equation}
g\left( ta+\left( 1-t\right) b\right) \leq t^{s_{2}}g(a)+\left( 1-t\right)
^{s_{2}}g(b).  \label{b24}
\end{equation}%
By (\ref{b23}) and (\ref{b24}), we have%
\begin{eqnarray}
&&f\left( ta+\left( 1-t\right) b\right) g\left( ta+\left( 1-t\right) b\right)
\label{b25} \\
&&  \notag \\
&\leq &t^{s_{1}+s_{2}}f(a)g(a)+\left( 1-t\right)
^{s_{1}+s_{2}}f(b)g(b)\notag\\
&&  \notag \\
&&+t^{s_{1}}\left( 1-t\right)
^{s_{2}}f(a)g(b)+t^{s_{2}}\left( 1-t\right) ^{s_{1}}f(b)g(a).  \notag
\end{eqnarray}%
Multiplying both sides of (\ref{b25}) by $w\left( ta+\left( 1-t\right)
b\right) ,$ then integrating the resulting inequality with respect to $t$
from $0$ to $1,$ we obtain%
\begin{eqnarray}
&&\int \limits_{0}^{1}f\left( ta+\left( 1-t\right) b\right) g\left(
ta+\left( 1-t\right) b\right) w\left( ta+\left( 1-t\right) b\right) dt
\label{b26} \\
&&  \notag \\
&\leq &f(a)g(a)\int \limits_{0}^{1}t^{s_{1}+s_{2}}w\left( ta+\left(
1-t\right) b\right) dt\notag \\
&&  \notag \\
&&+f(b)g(b)\int \limits_{0}^{1}\left( 1-t\right)
^{s_{1}+s_{2}}w\left( ta+\left( 1-t\right) b\right) dt  \notag \\
&&  \notag \\
&&+f(a)g(b)\int \limits_{0}^{1}t^{s_{1}}\left( 1-t\right) ^{s_{2}}w\left(
ta+\left( 1-t\right) b\right) dt\notag \\
&&  \notag \\
&&+f(b)g(a)\int \limits_{0}^{1}t^{s_{2}}\left(
1-t\right) ^{s_{1}}w\left( ta+\left( 1-t\right) b\right) dt.  \notag
\end{eqnarray}%
By change of variable $x=ta+\left( 1-t\right) b,$ we get%
\begin{equation}
\int \limits_{0}^{1}t^{s_{1}+s_{2}}w\left( ta+\left( 1-t\right) b\right) dt=%
\frac{1}{\left( b-a\right) ^{s_{1}+s_{2}+1}}\int \limits_{a}^{b}\left(
b-x\right) ^{s_{1}+s_{2}}w(x)dx  \label{b27}
\end{equation}%
and since $w$ is symmetric about $\frac{a+b}{2},$ we have%
\begin{eqnarray}
&&\int \limits_{0}^{1}\left( 1-t\right) ^{s_{1}+s_{2}}w\left( ta+\left(
1-t\right) b\right) dt \label{b28} \\
 &=&\frac{1}{\left( b-a\right) ^{s_{1}+s_{2}+1}}\int
\limits_{a}^{b}\left( x-a\right) ^{s_{1}+s_{2}}w(x)dx  \notag \\
&=&\frac{1}{\left( b-a\right) ^{s_{1}+s_{2}+1}}\int \limits_{a}^{b}\left(
b-u\right) ^{s_{1}+s_{2}}w(a+b-u)du.  \notag \\
&=&\frac{1}{\left( b-a\right) ^{s_{1}+s_{2}+1}}\int \limits_{a}^{b}\left(
b-u\right) ^{s_{1}+s_{2}}w(u)du.  \notag
\end{eqnarray}%
We also have%
\begin{eqnarray}
&&\int \limits_{0}^{1}t^{s_{1}}\left( 1-t\right) ^{s_{2}}w\left( ta+\left(
1-t\right) b\right) dt\label{b29}\\
&&=\frac{1}{\left( b-a\right) ^{s_{1}+s_{2}+1}}\int
\limits_{a}^{b}\left( b-x\right) ^{s_{1}}\left( x-a\right) ^{s_{2}}w(x)dx
\notag 
\end{eqnarray}%
and%
\begin{eqnarray}
&&\int \limits_{0}^{1}t^{s_{2}}\left( 1-t\right) ^{s_{1}}w\left( ta+\left(
1-t\right) b\right) dt \label{b30}\\
 &=&\frac{1}{\left( b-a\right) ^{s_{1}+s_{2}+1}}\int
\limits_{a}^{b}\left( b-x\right) ^{s_{2}}\left( x-a\right) ^{s_{1}}w(x)dx
 \notag\\
&=&\frac{1}{\left( b-a\right) ^{s_{1}+s_{2}+1}}\int \limits_{a}^{b}\left(
b-u\right) ^{s_{1}}\left( u-a\right) ^{s_{2}}w(a+b-u)du  \notag \\
&=&\frac{1}{\left( b-a\right) ^{s_{1}+s_{2}+1}}\int \limits_{a}^{b}\left(
b-u\right) ^{s_{1}}\left( u-a\right) ^{s_{2}}w(u)du  \notag
\end{eqnarray}%
By substituting the equalities (\ref{b27})-(\ref{b30}) in (\ref{b26}), then
we have the following inequality%
\begin{eqnarray}
&&\frac{1}{b-a}\int \limits_{a}^{b}f(x)g(x)w(x)dx  \label{b31}\\
&&\leq \frac{%
f(a)g(a)+f(b)g(b)}{\left( b-a\right) ^{s_{1}+s_{2}+1}}\int
\limits_{a}^{b}\left( b-x\right) ^{s_{1}+s_{2}}w(x)dx  \notag \\
&&  \notag \\
&&+\frac{f(a)g(b)+f(b)g(a)}{\left( b-a\right) ^{s_{1}+s_{2}+1}}\int
\limits_{a}^{b}\left( b-x\right) ^{s_{1}}\left( x-a\right) ^{s_{2}}w(x)dx. 
\notag
\end{eqnarray}%
If we multiply both sides of (\ref{b31}) by $\left( b-a\right) ,$ then we
obtain the desired result.
\end{proof}

\begin{remark}
If we choose $w(x)=1$ for all $x\in \left[ a,b\right] $ in Theorem \ref{t11}%
, then we have the following inequality%
\begin{equation*}
\frac{1}{b-a}\int \limits_{a}^{b}f(x)g(x)dx\leq \frac{1}{s_{1}+s_{2}+1}%
M(a,b)+B(s_{1}+1,s_{2}+1)N(a,b)
\end{equation*}%
which is proved by K\i rmac\i \ et al. in \cite{kir}. Here $B(x,y)$ is the
Beta Euler function.
\end{remark}

\begin{remark}
If we choose $s_{1}=1$ and $s_{2}=s$ in Theorem \ref{t11}, then the
inequality (\ref{b22}) reduces to the inequality (\ref{b1}).
\end{remark}

\begin{corollary}
If we choose $s_{1}=s_{2}=s$ in Theorem \ref{t11}, then we have the
following inequality%
\begin{eqnarray*}
\int \limits_{a}^{b}f(x)g(x)w(x)dx &\leq &\frac{M(a,b)}{\left( b-a\right)
^{2s}}\int \limits_{a}^{b}\left( b-x\right) ^{2s}w(x)dx \\
&& \\
&&+\frac{N(a,b)}{\left( b-a\right) ^{2s}}\int \limits_{a}^{b}\left(
b-x\right) ^{s}\left( x-a\right) ^{s}w(x)dx.
\end{eqnarray*}
\end{corollary}

\begin{theorem}
\label{t22} Suppose that conditions of Theorem \ref{t11} hold, then we have
the following inequality%
\begin{eqnarray}
&&2^{s_{1}+s_{2}-1}f\left( \frac{a+b}{2}\right) g\left( \frac{a+b}{2}\right)
\int \limits_{a}^{b}w\left( x\right) dx  \label{b32} \\
&\leq &\int \limits_{a}^{b}f(x)g(x)w(x)dx +\frac{M(a,b)}{\left( b-a\right) ^{s_{1}+s_{2}}}\int \limits_{a}^{b}\left(
x-a\right) ^{s_{1}}\left( b-x\right) ^{s_{2}}w(x)dx \notag \\
&&+\frac{N(a,b)}{\left(
b-a\right) ^{s_{1}+s_{2}}}\int \limits_{a}^{b}\left( b-x\right)
^{s_{1}+s_{2}}w(x)dx.  \notag
\end{eqnarray}%
where $M(a,b)$ and $N(a,b)$ are defined as in Theorem \ref{t1}.
\end{theorem}

\begin{proof}
Using the $s_{1}$-convexity of $f$ and $s_{2}$-convexity of $g,$ we have%
\begin{eqnarray*}
&&f\left( \frac{a+b}{2}\right) g\left( \frac{a+b}{2}\right) \\
&& \\
&=&f\left( \frac{(1-t)a+tb}{2}+\frac{ta+(1-t)b}{2}\right) g\left( \frac{%
(1-t)a+tb}{2}+\frac{ta+(1-t)b}{2}\right) \\
&& \\
&\leq &\frac{1}{2^{s_{1}+s_{2}}}\left[ f((1-t)a+tb)+f(ta+(1-t)b)\right] \\
&&\\
&& \times \left[ g((1-t)a+tb)+g(ta+(1-t)b)\right] \\
&& \\
&=&\frac{1}{2^{s_{1}+s_{2}}}\left[
f((1-t)a+tb)g((1-t)a+tb)+f(ta+(1-t)b)g(ta+(1-t)b)\right] \\
&& \\
&&+\frac{1}{2^{s_{1}+s_{2}}}\left[
f((1-t)a+tb)g(ta+(1-t)b)+f(ta+(1-t)b)g((1-t)a+tb)\right] .
\end{eqnarray*}%
By using again the $s_{1}$-convexity of $f$ and $s_{2}$-convexity, we obtain%
\begin{eqnarray}
&&f\left( \frac{a+b}{2}\right) g\left( \frac{a+b}{2}\right)  \label{b33} \\
&&  \notag \\
&\leq &\frac{1}{2^{s_{1}+s_{2}}}\left[
f((1-t)a+tb)g((1-t)a+tb)+f(ta+(1-t)b)g(ta+(1-t)b)\right]  \notag \\
&&  \notag \\
&&+\frac{1}{2^{s_{1}+s_{2}}}\left[ t^{s_{1}}\left( 1-t\right)
^{s_{2}}+t^{s_{2}}(1-t)^{s_{1}}\right] \left[ f(a)g(a)+f(b)g(b)\right] 
\notag \\
&&  \notag \\
&&+\frac{1}{2^{s_{1}+s_{2}}}\left[ t^{s_{1}+s_{2}}+\left( 1-t\right)
^{s_{1}+s_{2}}\right] \left[ f(a)g(b)+f(b)g(a)\right] .  \notag
\end{eqnarray}%
Multiplying both sides of (\ref{b33}) by $w\left( \left( 1-t\right)
a+tb\right) ,$ then integrating the resulting inequality with respect to $t$
from $0$ to $1,$ we obtain 
\begin{eqnarray}
&&f\left( \frac{a+b}{2}\right) g\left( \frac{a+b}{2}\right) \int
\limits_{0}^{1}w\left( \left( 1-t\right) a+tb\right) dt  \label{b34} \\
&\leq &\frac{1}{2^{s_{1}+s_{2}}}\int \limits_{0}^{1}\left[
f((1-t)a+tb)g((1-t)a+tb)  \right.\notag \\
&&\notag \\
&&\left. +f(ta+(1-t)b)g(ta+(1-t)b)\right] w\left( \left(
1-t\right) a+tb\right) dt  \notag \\&&\notag \\
&&+\frac{M(a,b)}{2^{s_{1}+s_{2}}}\int \limits_{0}^{1}\left[ t^{s_{1}}\left(
1-t\right) ^{s_{2}}+t^{s_{2}}(1-t)^{s_{1}}\right] w\left( \left( 1-t\right)
a+tb\right) dt  \notag \\
&&+\frac{N(a,b)}{2^{s_{1}+s_{2}}}\int \limits_{0}^{1}\left[
t^{s_{1}+s_{2}}+\left( 1-t\right) ^{s_{1}+s_{2}}\right] w\left( \left(
1-t\right) a+tb\right) dt.  \notag
\end{eqnarray}%
Using the change of variable, we have 
\begin{eqnarray}
&&\int \limits_{0}^{1}\left[ t^{s_{1}}\left( 1-t\right)
^{s_{2}}+t^{s_{2}}(1-t)^{s_{1}}\right] w\left( \left( 1-t\right) a+tb\right)
dt  \label{b35} \\
&=&\int \limits_{0}^{1}\left[ t^{s_{1}}\left( 1-t\right) ^{s_{2}}w\left(
\left( 1-t\right) a+tb\right) +t^{s_{2}}(1-t)^{s_{1}}w\left( \left(
1-t\right) a+tb\right) \right] dt  \notag \\
&=&\frac{1}{\left( b-a\right) ^{s_{1}+s_{2}+1}}\int \limits_{a}^{b}\left(
x-a\right) ^{s_{1}}\left( b-x\right) ^{s_{2}}w(x)dx  \notag \\
&&+\frac{1}{\left( b-a\right) ^{s_{1}+s_{2}+1}}\int \limits_{a}^{b}\left(
x-a\right) ^{s_{1}}\left( b-x\right) ^{s_{2}}w(a+b-x)dx  \notag \\
&=&\frac{2}{\left( b-a\right) ^{s_{1}+s_{2}+1}}\int \limits_{a}^{b}\left(
x-a\right) ^{s_{1}}\left( b-x\right) ^{s_{2}}w(x)dx  \notag
\end{eqnarray}%
and%
\begin{eqnarray}
&&\int \limits_{0}^{1}\left[ t^{s_{1}+s_{2}}+\left( 1-t\right) ^{s_{1}+s_{2}}%
\right] w\left( \left( 1-t\right) a+tb\right) dt  \label{b36} \\
&=&\int \limits_{0}^{1}\left[ t^{s_{1}+s_{2}}w\left( \left( 1-t\right)
a+tb\right) +\left( 1-t\right) ^{s_{1}+s_{2}}w\left( \left( 1-t\right)
a+tb\right) \right] dt  \notag \\
&=&\frac{1}{\left( b-a\right) ^{s_{1}+s_{2}+1}}\int \limits_{a}^{b}\left(
b-x\right) ^{s_{1}+s_{2}}w(a+b-x)dx\notag \\
&&+\frac{1}{\left( b-a\right)
^{s_{1}+s_{2}+1}}\int \limits_{a}^{b}\left( b-x\right) ^{s_{1}+s_{2}}w(x)dx 
\notag \\
&=&\frac{2}{\left( b-a\right) ^{s_{1}+s_{2}+1}}\int \limits_{a}^{b}\left(
b-x\right) ^{s_{1}+s_{2}}w(x)dx.  \notag
\end{eqnarray}%
If we substitute the equalities (\ref{b17}), (\ref{b18}), (\ref{b35}) and (%
\ref{b36}) in (\ref{b34}), then we have the following inequality 
\begin{eqnarray}
&&f\left( \frac{a+b}{2}\right) g\left( \frac{a+b}{2}\right) \frac{1}{b-a}%
\int \limits_{a}^{b}w\left( x\right) dx  \label{b37} \\
&\leq &\frac{1}{2^{s_{1}+s_{2}-1}\left( b-a\right) }\int%
\limits_{a}^{b}f(x)g(x)w(x)dx  \notag \\
&&+\frac{M(a,b)}{2^{s_{1}+s_{2}-1}\left( b-a\right) ^{s_{1}+s_{2}+1}}\int
\limits_{a}^{b}\left( x-a\right) ^{s_{1}}\left( b-x\right) ^{s_{2}}w(x)dx 
\notag \\
&&+\frac{N(a,b)}{2^{s_{1}+s_{2}-1}\left( b-a\right) ^{s_{1}+s_{2}+1}}\int
\limits_{a}^{b}\left( b-x\right) ^{s_{1}+s_{2}}w(x)dx.  \notag
\end{eqnarray}%
By multiplying the both sides of (\ref{b37}) by $2^{s_{1}+s_{2}-1}\left(
b-a\right) $ then we obtain the desired result (\ref{b32}).
\end{proof}

\begin{corollary}
If we choose $w(x)=1$ for all $x\in \left[ a,b\right] $ in Theorem \ref{t22}%
, then we have the following inequality%
\begin{eqnarray*}
&&2^{s_{1}+s_{2}-1}f\left( \frac{a+b}{2}\right) g\left( \frac{a+b}{2}\right)
\\
&& \\
&\leq &\frac{1}{b-a}\int \limits_{a}^{b}f(x)g(x)dx+B(s_{1}+1,s_{2}+1)M(a,b)+%
\frac{1}{s_{1}+s_{2}+1}N(a,b).
\end{eqnarray*}
\end{corollary}

\begin{remark}
If we choose $s_{1}=1$ and $s_{2}=s$ in Theorem \ref{t22}, then the
inequality (\ref{b32}) reduces to the inequality (\ref{b14}).
\end{remark}

\begin{corollary}
If we choose $s_{1}=s_{2}=s$ in Theorem \ref{t22}, then we have the
following inequality%
\begin{eqnarray*}
&& \\
&&2^{2s-1}f\left( \frac{a+b}{2}\right) g\left( \frac{a+b}{2}\right) \int
\limits_{a}^{b}w\left( x\right) dx \\
&& \\
&\leq &\int \limits_{a}^{b}f(x)g(x)w(x)dx \\
&&+\frac{M(a,b)}{\left( b-a\right) ^{2s}}\int \limits_{a}^{b}\left(
x-a\right) ^{s}\left( b-x\right) ^{s}w(x)dx+\frac{N(a,b)}{\left( b-a\right)
^{2s}}\int \limits_{a}^{b}\left( b-x\right) ^{2s}w(x)dx.
\end{eqnarray*}
\end{corollary}

\section{Concluding Remarks}

In this paper, we present some Hermite-Hadamard-Fej\'{e}r type inequalities
for products convex and $s$-convex functions. For further investigations we
propose to consider the Fej\'{e}r type inequalities for products other type
convex functions or for fractional integral operators.

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\bibitem{budak2} Budak, H,   \textit{On Fej\'{e}r type inequalities for
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