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\begin{document}
\title[Weighted Hermite-Hadamard and Simpson inequalities...]{Weighted
Hermite-Hadamard and Simpson type inequalities for double integrals}
\address{Department of Mathematics, \ Faculty of Science and Arts, D\"{u}zce
University, D\"{u}zce-TURKEY}
\author{H\"{u}seyin BUDAK}
\email{hsyn.budak@gmail.com}
\author{Fatma Ertu\u{g}ral}
\address{Department of Mathematics, Faculty of Science and Arts, D\"{u}zce
University, D\"{u}zce, Turkey}
\email{fatmaertugral14@gmail.com}
\author{Mehmet Zeki SARIKAYA}
\address{Department of Mathematics, \ Faculty of Science and Arts, D\"{u}zce
University, D\"{u}zce-TURKEY}
\email{sarikayamz@gmail.com}
\keywords{\textbf{\thanks{\textbf{2010 Mathematics Subject Classification.}
26D07, 26D10, 26D15, 26B15, 26B25.} }Hermite-Hadamard-Fejer inequality,
Simpson inequality, co-ordinated convex, integral inequalities.}

\begin{abstract}
In this paper, we first obtain two weighted identities for twice partially
differentiable mappings. Moreover, utilizing these equalities, we establish
the weighted Herrmite-Hadamard type inequalities and weighted Simpson type
inequalities for co-ordinated convex functions in a rectangle from the plane
$%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{2}$, respectivelly. The results given in this paper provide
generalizations of some result established in earlier works.
\end{abstract}

\maketitle

\section{Introduction}

The Hermite-Hadamard inequality discovered by C. Hermite and J. Hadamard
see, e.g., \cite{Dragomir1}, \cite[p.137]{Pecaric}) is one of the most well
established inequalities in the theory of convex functions with a
geometrical interpretation and many applications. 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
\limits_{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.
Hermite-Hadamard inequality for convex functions has received renewed
attention in recent years and a remarkable variety of refinements and
generalizations have been studied.

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}g(x)dx\leq \int
\limits_{a}^{b}f(x)g(x)dx\leq \frac{f(a)+f(b)}{2}\int \limits_{a}^{b}g(x)dx
\label{H1}
\end{equation}%
holds, where $g:[a,b]\rightarrow \mathbb{R}$ is nonnegative, integrable, and
symmetric about $x=\frac{a+b}{2}$ (i.e. $g(x)=g(a+b-x)$)$.$
\end{theorem}

Onthe other hand, the following inequality is well known in the literature
as Simpson's inequality.

\begin{theorem}
Let $\ f:\left[ a,b\right] \rightarrow
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ be a four times continuously differentiable mapping on $(a,b)$ and $%
\left
\Vert f^{(4)}\right \Vert _{\infty }=\sup \left \vert
f^{(4)}(x)\right \vert <\infty .$Then, the following inequality holds:
\end{theorem}

\begin{equation*}
\left \vert \frac{1}{3}\left[ \frac{f(a)+f(b)}{2}+2f\left( \frac{a+b}{2}%
\right) \right] -\frac{1}{b-a}\int_{a}^{b}f\left( x\right) dx\right \vert
\leq \frac{1}{2880}\left \Vert f^{(4)}\right \Vert _{\infty }(b-a)^{4}.
\end{equation*}

For recent refinements, counterparts, generalizations and new Simpson's type
inequalities, see (\cite{alomari3}, \cite{dragomir5}-\cite{du}, \cite%
{hussain}, \cite{kav}, \cite{liu},\  \cite{pecaric2}, \cite{qasiar}, \cite%
{sarikaya1}-\cite{sarikaya4}, \cite{set1}, \cite{set2}, \cite{tseng2}, \cite%
{ujevic}, \cite{yang3}).

A formal defination for co-ordinated convex function may be stated as
follows:

\begin{definition}
A function $f:\Delta \rightarrow
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ is called co-ordinated convex on $\Delta ,$ for all $(x,u),(y,v)\in \Delta
$ and $t,s\in \lbrack 0,1]$, if it satifies the following inequality:%
\begin{eqnarray}
&&f(tx+(1-t)\text{ }y,su+(1-s)\text{ }v)  \label{T4} \\
&&  \notag \\
&\leq &ts\text{ }f(x,u)+t(1-s)f(x,v)+s(1-t)f(y,u)+(1-t)(1-s)f(y,v).  \notag
\end{eqnarray}
\end{definition}

The mapping $f$ is a co-ordinated concave on $\Delta $ if the inequality (%
\ref{T4}) holds in reversed direction for all $t,s\in \lbrack 0,1]$ and $%
(x,u),(y,v)\in \Delta $.

In \cite{dragomir}, Dragomir proved the following inequalities which is
Hermite-Hadamard type inequalities for co-ordinated convex functions on the
rectangle from the plane $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{2}.$

\begin{theorem}
Suppose that $f:\Delta \rightarrow
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ is co-ordinated convex, then we have the following inequalities:%
\begin{eqnarray}
f\left( \frac{a+b}{2},\frac{c+d}{2}\right) &\leq &\frac{1}{2}\left[ \frac{1}{%
b-a}\int \limits_{a}^{b}f\left( x,\frac{c+d}{2}\right) dx+\frac{1}{d-c}\int
\limits_{c}^{d}f\left( \frac{a+b}{2},y\right) dy\right]  \notag \\
&\leq &\frac{1}{(b-a)(d-c)}\int \limits_{a}^{b}\int \limits_{c}^{d}f(x,y)dydx
\label{E2} \\
&\leq &\frac{1}{4}\left[ \frac{1}{b-a}\int \limits_{a}^{b}f(x,c)dx+\frac{1}{%
b-a}\int \limits_{a}^{b}f(x,d)dx\right.  \notag \\
&+&\left. \frac{1}{d-c}\int \limits_{c}^{d}f(a,y)dy+\frac{1}{d-c}\int
\limits_{c}^{d}f(b,y)dy\right]  \notag \\
&&  \notag \\
&\leq &\frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4}.  \notag
\end{eqnarray}%
The above inequalities are sharp. The inequalities in (\ref{E2}) hold in
reverse direction if the mapping $f$ is a co-ordinated concave mapping.
\end{theorem}

Over the years, many papers are dedicated on the generalizations and new
versions of the inequalities (\ref{E2}) using the different type convex
functions. For the other Hermite-Hadamard type inequalities for co-ordinated
convex functions, please refer to (\cite{alomari}, \cite{alomari2}, \cite%
{bakula} ,\cite{chen}, \cite{ozdemir}, \cite{sarikaya}, \cite{set}, \cite%
{ozdemir}, \cite{wang}, \cite{xi}, \cite{yil})

In \cite{ozdemir2}, \"{O}zdemir et al. gave the following identity and using
the this identity, the authors established some Simpson type inequalities
for double integrals:

\begin{lemma}
\label{l1} $f:\Delta :=\left[ a,b\right] \times \left[ c,d\right]
\rightarrow
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ be a twice partially differentiable mapping on $\Delta ^{\circ }$. If $%
\dfrac{\partial ^{2}f}{\partial t\partial s}\in L(\Delta ),$ then we have
the following equality%
\begin{eqnarray*}
&&\frac{f\left( a,\frac{c+d}{2}\right) +f\left( b,\frac{c+d}{2}\right)
+4f\left( \frac{a+b}{2},\frac{c+d}{2}\right) +f\left( \frac{a+b}{2},c\right)
+f\left( \frac{a+b}{2},d\right) }{9} \\
&&+\frac{f\left( a,c\right) +f\left( b,c\right) +f\left( a,d\right) +f\left(
b,d\right) }{36} \\
&&-\frac{1}{6\left( b-a\right) }\int_{a}^{b}\left[ f\left( x,c\right)
+4f\left( x,\frac{c+d}{2}\right) +f\left( x,d\right) \right] dx \\
&&-\frac{1}{6\left( d-c\right) }\int_{c}^{d}\left[ f\left( a,y\right)
+4f\left( \frac{a+b}{2},y\right) +f\left( b,y\right) \right] dy \\
&&+\frac{1}{\left( b-a\right) \left( d-c\right) }\int_{a}^{b}\int_{c}^{d}f%
\left( x,y\right) dydx \\
&=&\left( b-a\right) \left( d-c\right) \int_{0}^{1}\int_{0}^{1}q\left(
t,s\right) \frac{\partial ^{2}f}{\partial t\partial s}\left( at+\left(
1-t\right) b,cs+\left( 1-s\right) d\right) dsdt
\end{eqnarray*}%
which the mapping $q:\left[ 0,1\right] \times \left[ 0,1\right] \rightarrow
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ is defined by%
\begin{equation*}
q\left( t,s\right) =\left \{
\begin{array}{ccc}
\left( t-\frac{1}{6}\right) \left( s-\frac{1}{6}\right) & 0\leq t\leq \frac{1%
}{2}, & 0\leq s\leq \frac{1}{2} \\
&  &  \\
\left( t-\frac{1}{6}\right) \left( s-\frac{5}{6}\right) & 0\leq t\leq \frac{1%
}{2}, & \frac{1}{2}\leq s\leq 1 \\
&  &  \\
\left( t-\frac{5}{6}\right) \left( s-\frac{1}{6}\right) & \frac{1}{2}\leq
t\leq 1, & 0\leq s\leq \frac{1}{2} \\
&  &  \\
\left( t-\frac{5}{6}\right) \left( s-\frac{5}{6}\right) & \frac{1}{2}\leq
t\leq 1, & \frac{1}{2}\leq s\leq 1.%
\end{array}%
\right.
\end{equation*}
\end{lemma}

Budak and Sar\i kaya proved the following Hermite-Hadamard-Fej\'{e}r
inequalities for double integrals in \cite{budak}:

\begin{theorem}
Let $p:\Delta :=\left[ a,b\right] \times \left[ c,d\right] \rightarrow
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ be a positive, integrable and symmetric about $\frac{a+b}{2}$ and $\frac{%
c+d}{2}.$ Let $f:\Delta \rightarrow
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ be a co-ordinated convex on $\Delta $, then we have the following
Hermite-Hadamard-Fejer type inequality%
\begin{eqnarray*}
&&f\left( \frac{a+b}{2},\frac{c+d}{2}\right) \int \limits_{a}^{b}\int
\limits_{c}^{d}p(x,y)dydx \\
&\leq &\frac{1}{2}\int \limits_{a}^{b}\int \limits_{c}^{d}\left[ f\left( x,%
\frac{c+d}{2}\right) +f\left( \frac{a+b}{2},y\right) \right] p(x,y)dydx \\
&& \\
&\leq &\int \limits_{a}^{b}\int \limits_{c}^{d}f(x,y)p(x,y)dydx \\
&& \\
&\leq &\frac{1}{4}\int \limits_{a}^{b}\int \limits_{c}^{d}\left[
f(x,c)+f(x,d)+f(a,y)+f(b,y)\right] p(x,y)dydx \\
&& \\
&\leq &\frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4}\int \limits_{a}^{b}\int
\limits_{c}^{d}p(x,y)dydx.
\end{eqnarray*}
\end{theorem}

Moreover, Farid et al. established a weighted version of the inequalities (%
\ref{E2}) in \cite{farid}. Please see (\cite{lat}-\cite{latif3}, \cite{xi2})
for other papers focused on Hermite-Hadamard-Fej\'{e}r inequalities for
co-ordinated convex functions.

The aim of this paper is to establish some weighed generalizations of
Hermite-Hadamard and Simpson type integral inequalities. The results
presented in this paper provide extensions of those given in \cite{ozdemir2}
and \cite{sarikaya}.

\section{Weighted Hermite-Hadamard type Inequalities}

In this section, we first prove a weighted identity for twice partially
differentiable mapping. Then, using this identity, we established a weighted
Hermite-Hadamard type inequality for co-ordinated convex mapping.

\begin{lemma}
\label{l2} Let $p:\Delta :=\left[ a,b\right] \times \left[ c,d\right]
\rightarrow
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ be a positive and integrable function on $\Delta $ and let $f:\Delta
\rightarrow
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ be a twice partially differentiable mapping on $\Delta ^{\circ }$. If $%
\dfrac{\partial ^{2}f}{\partial t\partial s}\in L(\Delta ),$ then we have
the following equality%
\begin{eqnarray}
&&\frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4}\int \limits_{a}^{b}\int
\limits_{c}^{d}p(x,y)dydx  \label{e20} \\
&&  \notag \\
&&-\frac{1}{4}\int \limits_{a}^{b}\int \limits_{c}^{d}\left[
f(x,c)+f(x,d)+f(a,y)+f(b,y)\right] p(x,y)dydx+\int \limits_{a}^{b}\int
\limits_{c}^{d}f(x,y)p(x,y)dydx  \notag \\
&&  \notag \\
&=&\frac{\left( b-a\right) \left( d-c\right) }{4}\int \limits_{0}^{1}\int
\limits_{0}^{1}\left[ \int \limits_{U_{1}(t)}^{U_{2}(t)}\int
\limits_{V_{1}(s)}^{V_{2}(s)}p(u,v)dudv\right] \dfrac{\partial ^{2}f}{%
\partial t\partial s}\left( U_{2}(t),V_{2}(s)\right) dsdt  \notag
\end{eqnarray}%
where $U_{1}(t)=(1-t)a+tb,$ $U_{2}(t)=ta+(1-t)b,$ $V_{1}(s)=(1-s)c+sd$ and $%
V_{2}(s)=sc+(1-s)d.$
\end{lemma}

\begin{proof}
Integrating the by parts we have,

\begin{eqnarray}
&&\int \limits_{0}^{1}\int \limits_{0}^{1}\left[ \int%
\limits_{U_{1}(t)}^{U_{2}(t)}\int \limits_{V_{1}(s)}^{V_{2}(s)}p(u,v)dudv%
\right] \dfrac{\partial ^{2}f}{\partial t\partial s}\left(
U_{2}(t),V_{2}(s)\right) dsdt  \label{e21} \\
&&  \notag \\
&=&\int \limits_{0}^{1}\left \{ \left. \frac{1}{a-b}\left[
\int \limits_{U_{1}(t)}^{U_{2}(t)}\int
\limits_{V_{1}(s)}^{V_{2}(s)}p(u,v)dudv\right] \dfrac{\partial f}{\partial s}%
\left( U_{2}(t),V_{2}(s)\right) \right \vert _{0}^{1}\right.  \notag \\
&&\left. -\int \limits_{0}^{1}\left[ \int%
\limits_{V_{1}(s)}^{V_{2}(s)}p(U_{2}(t),v)dv+\int%
\limits_{V_{1}(s)}^{V_{2}(s)}p(U_{1}(t),v)dv\right] \dfrac{\partial f}{%
\partial s}\left( U_{2}(t),V_{2}(s)\right) dt\right \} ds  \notag \\
&=&\int \limits_{0}^{1}\left \{ \frac{1}{b-a}\left[ \int
\limits_{a}^{b}\int \limits_{V_{1}(s)}^{V_{2}(s)}p(u,v)dudv\right] \dfrac{%
\partial f}{\partial s}\left( a,V_{2}(s)\right) +\frac{1}{b-a}\left[ \int
\limits_{a}^{b}\int \limits_{V_{1}(s)}^{V_{2}(s)}p(u,v)dudv\right] \dfrac{%
\partial f}{\partial s}\left( b,V_{2}(s)\right) \right.  \notag \\
&&\left. -\int \limits_{0}^{1}\left[ \int%
\limits_{V_{1}(s)}^{V_{2}(s)}p(U_{2}(t),v)dv\right] \dfrac{\partial f}{%
\partial s}\left( U_{2}(t),V_{2}(s)\right) dt-\int \limits_{0}^{1}\left[
\int \limits_{V_{1}(s)}^{V_{2}(s)}p(U_{1}(t),v)dv\right] \dfrac{\partial f}{%
\partial s}\left( U_{2}(t),V_{2}(s)\right) dt\right \} ds  \notag \\
&=&\frac{1}{b-a}\left \{ \int \limits_{0}^{1}\left[ \int
\limits_{a}^{b}\int \limits_{V_{1}(s)}^{V_{2}(s)}p(u,v)dudv\right] \dfrac{%
\partial f}{\partial s}\left( a,V_{2}(s)\right) ds+\int \limits_{0}^{1}%
\left[ \int \limits_{a}^{b}\int \limits_{V_{1}(s)}^{V_{2}(s)}p(u,v)dudv%
\right] \dfrac{\partial f}{\partial s}\left( b,V_{2}(s)\right) ds\right \}
\notag \\
&&-\int \limits_{0}^{1}\int \limits_{0}^{1}\left[ \int%
\limits_{V_{1}(s)}^{V_{2}(s)}p(U_{2}(t),v)dv\right] \dfrac{\partial f}{%
\partial s}\left( U_{2}(t),V_{2}(s)\right) dtds-\int \limits_{0}^{1}\int
\limits_{0}^{1}\left[ \int \limits_{V_{1}(s)}^{V_{2}(s)}p(U_{1}(t),v)dv%
\right] \dfrac{\partial f}{\partial s}\left( U_{2}(t),V_{2}(s)\right) dtds
\notag \\
&=&\frac{1}{b-a}\left \{ I_{1}+I_{2}\right \} -I_{3}-I_{4}.  \notag
\end{eqnarray}%
Using again the intergration by parts, we obtain%
\begin{eqnarray}
I_{1} &=&\int \limits_{0}^{1}\left[ \int
\limits_{a}^{b}\int \limits_{V_{1}(s)}^{V_{2}(s)}p(u,v)dudv\right] \dfrac{%
\partial f}{\partial s}\left( a,V_{2}(s)\right) ds  \label{e22} \\
&=&\frac{1}{c-d}\left. \left[ \int
\limits_{a}^{b}\int \limits_{V_{1}(s)}^{V_{2}(s)}p(u,v)dudv\right] f\left(
a,V_{2}(s)\right) \right \vert _{0}^{1}  \notag \\
&&-\int \limits_{0}^{1}\left[ \int
\limits_{a}^{b}p(u,V_{2}(s))du+\int \limits_{a}^{b}p(u,V_{1}(s))du\right]
f\left( a,V_{2}(s)\right) ds  \notag \\
&=&\frac{1}{d-c}\left( \int \limits_{a}^{b}\int
\limits_{c}^{d}p(u,v)dudv\right) \left[ f\left( a,c\right) +f(a,d)\right]
\notag \\
&&-\int \limits_{0}^{1}\int \limits_{a}^{b}p(u,V_{2}(s))f\left(
a,V_{2}(s)\right) duds-\int \limits_{0}^{1}\int
\limits_{a}^{b}p(u,V_{1}(s))f\left( a,V_{2}(s)\right) duds  \notag \\
&=&\frac{1}{d-c}\left( \int \limits_{a}^{b}\int
\limits_{c}^{d}p(u,v)dudv\right) \left[ f\left( a,c\right) +f(a,d)\right] -%
\frac{2}{d-c}\int \limits_{a}^{b}\int \limits_{c}^{d}p(u,v)f(a,v)dudv,
\notag
\end{eqnarray}%
and similarly,%
\begin{eqnarray}
I_{2} &=&\int \limits_{0}^{1}\left[ \int
\limits_{a}^{b}\int \limits_{V_{1}(s)}^{V_{2}(s)}p(u,v)dudv\right] \dfrac{%
\partial f}{\partial s}\left( b,V_{2}(s)\right) ds  \label{e23} \\
&=&\left. \frac{1}{c-d}\left[ \int
\limits_{a}^{b}\int \limits_{V_{1}(s)}^{V_{2}(s)}p(u,v)dudv\right] f\left(
b,V_{2}(s)\right) \right \vert _{0}^{1}  \notag \\
&&-\int \limits_{0}^{1}\left[ \int
\limits_{a}^{b}p(u,V_{2}(s))dudv+\int \limits_{a}^{b}p(u,V_{1}(s))dudv\right]
f\left( b,V_{2}(s)\right) ds  \notag \\
&=&\frac{1}{d-c}\left( \int \limits_{a}^{b}\int
\limits_{c}^{d}p(u,v)dudv\right) \left[ f\left( b,c\right) +f(b,d)\right] -%
\frac{2}{d-c}\int \limits_{a}^{b}\int \limits_{c}^{d}p(u,v)f(b,v)dudv.
\notag
\end{eqnarray}%
Onthe other hand, we get
\begin{eqnarray}
I_{3} &=&\int \limits_{0}^{1}\int \limits_{0}^{1}\left[ \int%
\limits_{V_{1}(s)}^{V_{2}(s)}p(U_{2}(t),v)dv\right] \dfrac{\partial f}{%
\partial s}\left( U_{2}(t),V_{2}(s)\right) dtds  \label{e24} \\
&=&\int \limits_{0}^{1}\left \{ \left. \frac{1}{c-d}\left[
\int \limits_{V_{1}(s)}^{V_{2}(s)}p(U_{2}(t),v)dv\right] f\left(
U_{2}(t),V_{2}(s)\right) \right \vert _{0}^{1}\right.  \notag \\
&&\left. -\int \limits_{0}^{1}\left[
p(U_{2}(t),V_{2}(s))+p(U_{2}(t),V_{1}(s))\right] f\left(
U_{2}(t),V_{2}(s)\right) ds\right \} dt  \notag
\end{eqnarray}%
\begin{eqnarray*}
&=&\int \limits_{0}^{1}\left \{ \frac{1}{d-c}\left( \int
\limits_{c}^{d}p(U_{2}(t),v)dv\right) \left[ f\left( U_{2}(t),c\right)
+f\left( U_{2}(t),d\right) \right] \right. \\
&&\left. -\int \limits_{0}^{1}p(U_{2}(t),V_{2}(s))f\left(
U_{2}(t),V_{2}(s)\right) ds-\int \limits_{0}^{1}p(U_{2}(t),V_{1}(s))f\left(
U_{2}(t),V_{2}(s)\right) ds\right \} \\
&=&\frac{1}{\left( b-a\right) \left( d-c\right) }\int
\limits_{a}^{b}\int \limits_{c}^{d}p(u,v)\left[ f\left( u,c\right) +f\left(
u,d\right) \right] dvdu \\
&&-\frac{2}{\left( b-a\right) \left( d-c\right) }\int
\limits_{a}^{b}\int \limits_{c}^{d}p(u,v)f\left( u,v\right) dvdu
\end{eqnarray*}%
and similarly,
\begin{eqnarray}
I_{4} &=&\int \limits_{0}^{1}\int \limits_{0}^{1}\left[ \int%
\limits_{V_{1}(s)}^{V_{2}(s)}p(U_{1}(t),v)dv\right] \dfrac{\partial f}{%
\partial s}\left( U_{2}(t),V_{2}(s)\right) dtds  \label{e25} \\
&=&\frac{1}{\left( b-a\right) \left( d-c\right) }\int
\limits_{a}^{b}\int \limits_{c}^{d}p(u,v)\left[ f\left( u,c\right) +f\left(
u,d\right) \right] dvdu  \notag \\
&&-\frac{2}{\left( b-a\right) \left( d-c\right) }\int
\limits_{a}^{b}\int \limits_{c}^{d}p(u,v)f\left( u,v\right) dvdu.  \notag
\end{eqnarray}%
If we substitute the equalities (\ref{e22})-(\ref{e25}) in (\ref{e21}), we
obtain%
\begin{eqnarray}
&&\int \limits_{0}^{1}\int \limits_{0}^{1}\left[ \int%
\limits_{U_{1}(t)}^{U_{2}(t)}\int \limits_{V_{1}(s)}^{V_{2}(s)}p(u,v)dudv%
\right] \dfrac{\partial ^{2}f}{\partial t\partial s}\left(
U_{2}(t),V_{2}(s)\right) dsdt  \label{e26} \\
&=&\frac{1}{\left( b-a\right) \left( d-c\right) }\left( \int
\limits_{a}^{b}\int \limits_{c}^{d}p(u,v)dudv\right) \left[ f\left(
a,c\right) +f(a,d)\right] -\frac{2}{\left( b-a\right) \left( d-c\right) }%
\int \limits_{a}^{b}\int \limits_{c}^{d}p(u,v)f(a,v)dudv  \notag \\
&&+\frac{1}{\left( b-a\right) \left( d-c\right) }\left( \int
\limits_{a}^{b}\int \limits_{c}^{d}p(u,v)dudv\right) \left[ f\left(
b,c\right) +f(b,d)\right] -\frac{2}{\left( b-a\right) \left( d-c\right) }%
\int \limits_{a}^{b}\int \limits_{c}^{d}p(u,v)f(b,v)dudv  \notag \\
&&-\frac{1}{\left( b-a\right) \left( d-c\right) }\int
\limits_{a}^{b}\int \limits_{c}^{d}p(u,v)\left[ f\left( u,c\right) +f\left(
u,d\right) \right] dvdu+\frac{2}{\left( b-a\right) \left( d-c\right) }\int
\limits_{a}^{b}\int \limits_{c}^{d}p(u,v)f\left( u,v\right) dvdu  \notag \\
&&-\frac{1}{\left( b-a\right) \left( d-c\right) }\int
\limits_{a}^{b}\int \limits_{c}^{d}p(u,v)\left[ f\left( u,c\right) +f\left(
u,d\right) \right] dvdu+\frac{2}{\left( b-a\right) \left( d-c\right) }\int
\limits_{a}^{b}\int \limits_{c}^{d}p(u,v)f\left( u,v\right) dvdu  \notag \\
&=&\frac{1}{\left( b-a\right) \left( d-c\right) }\left( \int
\limits_{a}^{b}\int \limits_{c}^{d}p(u,v)dudv\right) \left[ f\left(
a,c\right) +f(a,d)+f\left( b,c\right) +f(b,d)\right]  \notag \\
&&-\frac{2}{\left( b-a\right) \left( d-c\right) }\int
\limits_{a}^{b}\int \limits_{c}^{d}p(u,v)\left[ ff\left( u,c\right) +f\left(
u,d\right) +(a,v)+f(b,v)\right] dvdu  \notag \\
&&+\frac{4}{\left( b-a\right) \left( d-c\right) }\int
\limits_{a}^{b}\int \limits_{c}^{d}p(u,v)f\left( u,v\right) dvdu.  \notag
\end{eqnarray}%
If we multiply the equality (\ref{e26}) by $\frac{\left( b-a\right) \left(
d-c\right) }{4},$ then we obtain the desired result (\ref{e20}).
\end{proof}

\begin{remark}
If we choose $p(x,y)=1$ in Lemma \ref{l2}, then the Lemma \ref{l2} reduces
to Lemma 1 in \cite{sarikaya}.
\end{remark}

\begin{theorem}
Let $p:\Delta :=\left[ a,b\right] \times \left[ c,d\right] \rightarrow
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ be a positive and integrable function on $\Delta $ and let $f:\Delta
\rightarrow
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ be a twice partially differentiable mapping on $\Delta ^{\circ }$. If $%
\left \vert \dfrac{\partial ^{2}f}{\partial t\partial s}\right \vert ^{q},$ $%
q>1,$ is a co-ordinated convex function on $\Delta ,$ then we have the
following inequality%
\begin{eqnarray}
&&\left \vert \frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4}\int
\limits_{a}^{b}\int \limits_{c}^{d}p(x,y)dydx\right.  \label{e27} \\
&&  \notag \\
&&\left. -\frac{1}{4}\int \limits_{a}^{b}\int \limits_{c}^{d}\left[
f(x,c)+f(x,d)+f(a,y)+f(b,y)\right] p(x,y)dydx+\int \limits_{a}^{b}\int
\limits_{c}^{d}f(x,y)p(x,y)dydx\right \vert  \notag \\
&&  \notag \\
&=&\frac{\left( b-a\right) \left( d-c\right) }{4}\left( \int
\limits_{0}^{1}\int \limits_{0}^{1}\left( H(t,s)\right) ^{p}dsdt\right) ^{%
\frac{1}{p}}  \notag \\
&&\times \left[ \frac{\left \vert \tfrac{\partial ^{2}f}{\partial t\partial s%
}\left( a,c\right) \right \vert ^{q}+\left \vert \tfrac{\partial ^{2}f}{%
\partial t\partial s}\left( a,d\right) \right \vert ^{q}+\left \vert \tfrac{%
\partial ^{2}f}{\partial t\partial s}\left( b,c\right) \right \vert
^{q}+\left \vert \tfrac{\partial ^{2}f}{\partial t\partial s}\left(
b,d\right) \right \vert ^{q}}{4}\right] ^{\frac{1}{q}}  \notag
\end{eqnarray}%
where $\frac{1}{p}+\frac{1}{q}=1$ and $H(t,s)$ defined by%
\begin{equation*}
H(t,s)=\left \vert \int \limits_{U_{1}(t)}^{U_{2}(t)}\int
\limits_{V_{1}(s)}^{V_{2}(s)}p(u,v)dudv\right \vert
\end{equation*}%
with $U_{1}(t)=(1-t)a+tb,$ $U_{2}(t)=ta+(1-t)b,$ $V_{1}(s)=(1-s)c+sd$ and $%
V_{2}(s)=sc+(1-s)d.$
\end{theorem}

\begin{proof}
Taking the modulus in Lemma \ref{l2} and by using the well-known H\"{o}%
lder's inequality, we have%
\begin{eqnarray}
&&\left \vert \frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4}\int
\limits_{a}^{b}\int \limits_{c}^{d}p(x,y)dydx\right.  \label{e29} \\
&&  \notag \\
&&\left. -\frac{1}{4}\int \limits_{a}^{b}\int \limits_{c}^{d}\left[
f(x,c)+f(x,d)+f(a,y)+f(b,y)\right] p(x,y)dydx+\int \limits_{a}^{b}\int
\limits_{c}^{d}f(x,y)p(x,y)dydx\right \vert  \notag \\
&&  \notag \\
&=&\frac{\left( b-a\right) \left( d-c\right) }{4}\left \vert \int
\limits_{0}^{1}\int \limits_{0}^{1}\left[ \int%
\limits_{U_{1}(t)}^{U_{2}(t)}\int \limits_{V_{1}(s)}^{V_{2}(s)}p(u,v)dudv%
\right] \dfrac{\partial ^{2}f}{\partial t\partial s}\left(
U_{2}(t),V_{2}(s)\right) dsdt\right \vert  \notag \\
&&  \notag \\
&\leq &\frac{\left( b-a\right) \left( d-c\right) }{4}\int
\limits_{0}^{1}\int \limits_{0}^{1}\left \vert \int
\limits_{U_{1}(t)}^{U_{2}(t)}\int
\limits_{V_{1}(s)}^{V_{2}(s)}p(u,v)dudv\right \vert \left \vert \dfrac{%
\partial ^{2}f}{\partial t\partial s}\left( U_{2}(t),V_{2}(s)\right) \right
\vert dsdt  \notag \\
&&  \notag \\
&\leq &\frac{\left( b-a\right) \left( d-c\right) }{4}\left( \int
\limits_{0}^{1}\int \limits_{0}^{1}\left \vert \int
\limits_{U_{1}(t)}^{U_{2}(t)}\int
\limits_{V_{1}(s)}^{V_{2}(s)}p(u,v)dudv\right \vert ^{p}dsdt\right) ^{\frac{1%
}{p}}\left( \int \limits_{0}^{1}\int \limits_{0}^{1}\left \vert \dfrac{%
\partial ^{2}f}{\partial t\partial s}\left( U_{2}(t),V_{2}(s)\right) \right
\vert ^{q}dsdt\right) ^{\frac{1}{q}}.  \notag
\end{eqnarray}%
Since $\left \vert \dfrac{\partial ^{2}f}{\partial t\partial s}\right \vert
^{q},$ $q>1,$ is a co-ordinated convex function on $\Delta ,$ we obtain%
\begin{eqnarray}
&&\int \limits_{0}^{1}\int \limits_{0}^{1}\left \vert \dfrac{\partial ^{2}f%
}{\partial t\partial s}\left( U_{2}(t),V_{2}(s)\right) \right \vert ^{q}dsdt
\label{e28} \\
&=&\int \limits_{0}^{1}\int \limits_{0}^{1}\left \vert \dfrac{\partial
^{2}f}{\partial t\partial s}\left( ta+(1-t)b,sa+(1-s)d\right) \right \vert
^{q}dsdt  \notag \\
&&  \notag \\
&\leq &\int \limits_{0}^{1}\int \limits_{0}^{1}\left[ ts\left \vert \tfrac{%
\partial ^{2}f}{\partial t\partial s}\left( a,c\right) \right \vert
^{q}+t(1-s)\left \vert \tfrac{\partial ^{2}f}{\partial t\partial s}\left(
a,d\right) \right \vert ^{q}\right.  \notag \\
&&  \notag \\
&&\left. +(1-t)s\left \vert \tfrac{\partial ^{2}f}{\partial t\partial s}%
\left( b,c\right) \right \vert ^{q}+(1-t)(1-s)\left \vert \tfrac{\partial
^{2}f}{\partial t\partial s}\left( b,d\right) \right \vert ^{q}\right] dsdt
\notag \\
&&  \notag \\
&=&\frac{\left \vert \tfrac{\partial ^{2}f}{\partial t\partial s}\left(
a,c\right) \right \vert ^{q}+\left \vert \tfrac{\partial ^{2}f}{\partial
t\partial s}\left( a,d\right) \right \vert ^{q}+\left \vert \tfrac{\partial
^{2}f}{\partial t\partial s}\left( b,c\right) \right \vert ^{q}+\left \vert
\tfrac{\partial ^{2}f}{\partial t\partial s}\left( b,d\right) \right \vert
^{q}}{4}.  \notag
\end{eqnarray}%
If we substitute the inequality (\ref{e28}) in (\ref{e29}), the we establish
desired result.
\end{proof}

\begin{remark}
If we choose $p(x,y)=1,$ then we obtain the following inequality%
\begin{eqnarray*}
&&\left \vert \frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4}\right. \\
&& \\
&&-\frac{1}{4}\left[ \frac{1}{b-a}\int \limits_{a}^{b}f(x,c)dx+\frac{1}{b-a}%
\int \limits_{a}^{b}f(x,d)dx\frac{1}{d-c}\int \limits_{c}^{d}f(a,y)dy+\frac{1%
}{d-c}\int \limits_{c}^{d}f(b,y)dy\right] \\
&& \\
&&\left. +\frac{1}{\left( b-a\right) \left( d-c\right) }\int
\limits_{a}^{b}\int \limits_{c}^{d}f(x,y)dydx\right \vert \\
&& \\
&\leq &\frac{\left( b-a\right) \left( d-c\right) }{4(p+1)^{\frac{2}{p}}}%
\left[ \frac{\left \vert \tfrac{\partial ^{2}f}{\partial t\partial s}\left(
a,c\right) \right \vert ^{q}+\left \vert \tfrac{\partial ^{2}f}{\partial
t\partial s}\left( a,d\right) \right \vert ^{q}+\left \vert \tfrac{\partial
^{2}f}{\partial t\partial s}\left( b,c\right) \right \vert ^{q}+\left \vert
\tfrac{\partial ^{2}f}{\partial t\partial s}\left( b,d\right) \right \vert
^{q}}{4}\right] ^{\frac{1}{q}}
\end{eqnarray*}%
which was proved by Sarikaya et al. in \cite{sarikaya}.
\end{remark}

\section{Weighted Simpson Type Inequalities}

In this section, we first define the following mapping%
\begin{eqnarray*}
&&\Theta (a,b;f,p) \\
&=&f\left( \frac{a+b}{2},\frac{c+d}{2}\right) \int_{\frac{1}{6}}^{\frac{5}{6}%
}\int_{\frac{1}{6}}^{\frac{5}{6}}p\left( u,v\right) dvdu \\
&&+f\left( a,\frac{c+d}{2}\right) \int_{\frac{5}{6}}^{1}\int_{\frac{1}{6}}^{%
\frac{5}{6}}p\left( u,v\right) dvdu+f\left( b,\frac{c+d}{2}\right) \int_{0}^{%
\frac{1}{6}}\int_{\frac{1}{6}}^{\frac{5}{6}}p\left( u,v\right) dvdu \\
&&+f\left( \frac{a+b}{2},c\right) \int_{\frac{1}{6}}^{\frac{5}{6}}\int_{%
\frac{5}{6}}^{1}p\left( u,v\right) dvdu+f\left( \frac{a+b}{2},d\right) \int_{%
\frac{1}{6}}^{\frac{5}{6}}\int_{0}^{\frac{1}{6}}p\left( u,v\right) dvdu \\
&&+f\left( a,c\right) \int_{\frac{5}{6}}^{1}\int_{\frac{5}{6}}^{1}p\left(
u,v\right) dvdu+f\left( b,c\right) \int_{0}^{\frac{1}{6}}\int_{\frac{5}{6}%
}^{1}p\left( u,v\right) dvdu \\
&&+f\left( a,d\right) \int_{\frac{5}{6}}^{1}\int_{0}^{\frac{1}{6}}p\left(
u,v\right) dvdu+f\left( b,d\right) \int_{0}^{\frac{1}{6}}\int_{0}^{\frac{1}{6%
}}p\left( u,v\right) dvdu \\
&&-\frac{1}{b-a}\int_{a}^{b}\left( \int_{\frac{5}{6}}^{1}p\left( \frac{b-x}{%
b-a},v\right) dv\right) f\left( x,c\right) dx-\frac{1}{b-a}%
\int_{a}^{b}\left( \int_{\frac{1}{2}}^{\frac{5}{6}}p\left( \frac{b-x}{b-a}%
,v\right) dv\right) f\left( x,\frac{c+d}{2}\right) dx \\
&&-\frac{1}{b-a}\int_{a}^{b}\left( \int_{0}^{\frac{1}{6}}p\left( \frac{b-x}{%
b-a},v\right) dv\right) f\left( x,d\right) dx-\frac{1}{d-c}%
\int_{c}^{d}\left( \int_{\frac{5}{6}}^{1}p\left( u,\frac{d-y}{d-c}\right)
du\right) f\left( a,y\right) dy \\
&&-\frac{1}{d-c}\int_{c}^{d}\left( \int_{\frac{1}{2}}^{\frac{5}{6}}p\left( u,%
\frac{d-y}{d-c}\right) du\right) f\left( \frac{a+b}{2},y\right) dy-\frac{1}{%
d-c}\int_{c}^{d}\left( \int_{0}^{\frac{1}{6}}p\left( u,\frac{d-y}{d-c}%
\right) du\right) f\left( b,y\right) dy \\
&&+\frac{1}{\left( b-a\right) \left( d-c\right) }\int_{a}^{b}\int_{c}^{d}p%
\left( \frac{b-x}{b-a},\frac{d-y}{d-c}\right) f\left( x,y\right) dydx.
\end{eqnarray*}%
Now, we give the following equality.

\begin{lemma}
\label{l3} Let the mappings $p,$ $U_{2}$ and $V_{2}$ be as in Lemma \ref{l2}
and let $f:\Delta :=\left[ a,b\right] \times \left[ c,d\right] \rightarrow
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ be a twice partially differentiable mapping on $\Delta ^{\circ }$. If $%
\dfrac{\partial ^{2}f}{\partial t\partial s}\in L(\Delta ),$ then we have
the following equality%
\begin{equation*}
\Theta (a,b;f,p)=\left( b-a\right) \left( d-c\right)
\int_{0}^{1}\int_{0}^{1}w\left( t,s\right) \frac{\partial ^{2}f}{\partial
t\partial s}\left( U_{2}(t),V_{2}(s)\right) dsdt
\end{equation*}%
where the mapping $w:\left[ 0,1\right] \times \left[ 0,1\right] \rightarrow
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ is defined by%
\begin{equation*}
w\left( t,s\right) =\left \{
\begin{array}{ccc}
\int_{\frac{1}{6}}^{t}\int_{\frac{1}{6}}^{s}p\left( u,v\right) dudv &
0\leq t\leq \frac{1}{2}, & 0\leq s\leq \frac{1}{2} \\
\int_{\frac{1}{6}}^{t}\int_{\frac{5}{6}}^{s}p\left( u,v\right) dudv &
0\leq t\leq \frac{1}{2}, & \frac{1}{2}\leq s\leq 1 \\
\int_{\frac{5}{6}}^{t}\int_{\frac{1}{6}}^{s}p\left( u,v\right) dudv &
\frac{1}{2}\leq t\leq 1, & 0\leq s\leq \frac{1}{2} \\
\int_{\frac{5}{6}}^{t}\int_{\frac{5}{6}}^{s}p\left( u,v\right) dudv &
\frac{1}{2}\leq t\leq 1, & \frac{1}{2}\leq s\leq 1.%
\end{array}%
\right.
\end{equation*}
\end{lemma}

\begin{proof}
From the definition of the mapping $w$, we have%
\begin{eqnarray*}
&&\left( b-a\right) \left( d-c\right) \int_{0}^{1}\int_{0}^{1}w\left(
t,s\right) \frac{\partial ^{2}f}{\partial t\partial s}\left(
U_{2}(t),V_{2}(s)\right) dsdt \\
&=&\left( b-a\right) \left( d-c\right) \int_{0}^{\frac{1}{2}}\int_{0}^{\frac{%
1}{2}}\left( \int_{\frac{1}{6}}^{t}\int_{\frac{1}{6}}^{s}p\left( u,v\right)
dvdu\right) \frac{\partial ^{2}f}{\partial t\partial s}\left(
U_{2}(t),V_{2}(s)\right) dtds \\
&&+\left( b-a\right) \left( d-c\right) \int_{0}^{\frac{1}{2}}\int_{\frac{1}{2%
}}^{1}\left( \int_{\frac{1}{6}}^{t}\int_{\frac{5}{6}}^{s}p\left( u,v\right)
dudv\right) \frac{\partial ^{2}f}{\partial t\partial s}\left(
U_{2}(t),V_{2}(s)\right) dtds \\
&&+\left( b-a\right) \left( d-c\right) \int_{\frac{1}{2}}^{1}\int_{0}^{\frac{%
1}{2}}\left( \int_{\frac{5}{6}}^{t}\int_{\frac{1}{6}}^{s}p\left( u,v\right)
dudv\right) \frac{\partial ^{2}f}{\partial t\partial s}\left(
U_{2}(t),V_{2}(s)\right) dtds \\
&&+\left( b-a\right) \left( d-c\right) \int_{\frac{1}{2}}^{1}\int_{\frac{1}{2%
}}^{1}\left( \int_{\frac{5}{6}}^{t}\int_{\frac{5}{6}}^{s}p\left( u,v\right)
dudv\right) \frac{\partial ^{2}f}{\partial t\partial s}\left(
U_{2}(t),V_{2}(s)\right) dtds \\
&=&\left( b-a\right) \left( d-c\right) \left[ J_{1}+J_{2}+J_{3}+J_{4}\right]
.
\end{eqnarray*}%
Integration by parts, we obtain%
\begin{eqnarray*}
J_{1} &=&\int_{0}^{\frac{1}{2}}\int_{0}^{\frac{1}{2}}\left( \int_{\frac{1}{6}%
}^{t}\int_{\frac{1}{6}}^{s}p\left( u,v\right) dvdu\right) \frac{\partial
^{2}f}{\partial t\partial s}\left( U_{2}(t),V_{2}(s)\right) dtds \\
&=&\frac{1}{c-d}\int_{0}^{\frac{1}{2}}\left( \int_{\frac{1}{6}}^{t}\int_{%
\frac{1}{6}}^{\frac{1}{2}}p\left( u,v\right) dvdu\right) \frac{\partial f}{%
\partial t}\left( U_{2}(t),\frac{c+d}{2}\right) dt \\
&&-\frac{1}{c-d}\int_{0}^{\frac{1}{2}}\left( \int_{\frac{1}{6}}^{t}\int_{%
\frac{1}{6}}^{0}p\left( u,v\right) dvdu\right) \frac{\partial f}{\partial t}%
\left( U_{2}(t),d\right) dt \\
&&\left. -\frac{1}{c-d}\int_{0}^{\frac{1}{2}}\int_{0}^{\frac{1}{2}}\left(
\int_{\frac{1}{6}}^{t}p\left( s,v\right) dv\right) \frac{\partial f}{%
\partial t}\left( U_{2}(t),V_{2}(s)\right) dtds\right]  \\
&=&\frac{1}{c-d}\left[ \frac{1}{a-b}\left( \int_{\frac{1}{6}}^{\frac{1}{2}%
}\int_{\frac{1}{6}}^{\frac{1}{2}}p\left( u,v\right) dvdu\right) f\left(
\frac{a+b}{2},\frac{c+d}{2}\right) \right.  \\
&&\left. -\frac{1}{a-b}\left( \int_{\frac{1}{6}}^{0}\int_{\frac{1}{6}}^{%
\frac{1}{2}}p\left( u,v\right) dvdu\right) f\left( b,\frac{c+d}{2}\right)
\right.  \\
&&\left. -\frac{1}{a-b}\int_{0}^{\frac{1}{2}}\left( \int_{\frac{1}{6}}^{%
\frac{1}{2}}p\left( t,v\right) dv\right) f\left( U_{2}(t),\frac{c+d}{2}%
\right) dt\right]  \\
&&-\frac{1}{c-d}\left[ \frac{1}{a-b}\left( \int_{\frac{1}{6}}^{\frac{1}{2}%
}\int_{\frac{1}{6}}^{0}p\left( u,v\right) dvdu\right) f\left( \frac{a+b}{2}%
,d\right) \right.  \\
&&\left. -\frac{1}{a-b}\left( \int_{\frac{1}{6}}^{0}\int_{\frac{1}{6}%
}^{0}p\left( u,v\right) dvdu\right) f\left( b,d\right) \right.  \\
&&\left. -\frac{1}{a-b}\int_{0}^{\frac{1}{2}}\left( \int_{\frac{1}{6}%
}^{0}p\left( t,v\right) dv\right) f\left( U_{2}(t),d\right) dt\right]  \\
&&-\left[ \frac{1}{c-d}\int_{0}^{\frac{1}{2}}\frac{1}{a-b}\left( \int_{\frac{%
1}{6}}^{\frac{1}{2}}p\left( s,v\right) dv\right) f\left( \frac{a+b}{2}%
,V_{2}(s)\right) \right.  \\
&&\left. -\frac{1}{a-b}\left( \int_{\frac{1}{6}}^{0}\int_{\frac{1}{6}%
}^{0}p\left( u,v\right) dvdu\right) f\left( b,d\right) \right.  \\
&&\left. -\frac{1}{a-b}\int_{0}^{\frac{1}{2}}p\left( s,t\right) f\left(
U_{2}(t),V_{2}(s)\right) dt\right] ds
\end{eqnarray*}%
\begin{eqnarray*}
&=&\frac{1}{\left( b-a\right) \left( d-c\right) }\left[ f\left( \frac{a+b}{2}%
,\frac{c+d}{2}\right) \int_{\frac{1}{6}}^{\frac{1}{2}}\int_{\frac{1}{6}%
}^{1}p\left( u,v\right) dvdu+f\left( b,\frac{c+d}{2}\right) \int_{0}^{\frac{1%
}{6}}\int_{\frac{5}{6}}^{1}p\left( u,v\right) dvdu\right.  \\
&&-\int_{0}^{\frac{1}{2}}\left( \int_{\frac{5}{6}}^{\frac{1}{2}}p\left(
t,v\right) dv\right) f\left( U_{2}(t),\frac{c+d}{2}\right) dt+f\left( \frac{%
a+b}{2},d\right) \int_{\frac{1}{6}}^{\frac{1}{2}}\int_{0}^{\frac{1}{6}%
}p\left( u,v\right) dudv \\
&&+f\left( b,d\right) \int_{0}^{\frac{1}{6}}\int_{\frac{1}{2}}^{\frac{1}{6}%
}p\left( u,v\right) dudv-\int_{0}^{\frac{1}{2}}\left( \int_{0}^{\frac{1}{6}%
}p\left( t,v\right) dv\right) f\left( U_{2}(t),d\right) dt \\
&&-\int_{0}^{\frac{1}{2}}\left( \int_{\frac{1}{6}}^{\frac{1}{2}}p\left(
s,v\right) dv\right) f\left( \frac{a+b}{2},V_{2}(s)\right) ds-\int_{0}^{%
\frac{1}{2}}\left( \int_{0}^{\frac{1}{6}}p\left( s,v\right) dv\right)
f\left( b,V_{2}(s)\right) ds \\
&&\left. +\int_{0}^{\frac{1}{2}}\int_{0}^{\frac{1}{2}}p\left( s,t\right)
f\left( U_{2}(t),V_{2}(s)\right) dtds\right] .
\end{eqnarray*}%
Similarly, we get%
\begin{eqnarray*}
J_{2} &=&\int_{0}^{\frac{1}{2}}\int_{\frac{1}{2}}^{1}\left( \int_{\frac{1}{6}%
}^{t}\int_{\frac{5}{6}}^{s}p\left( u,v\right) dudv\right) \frac{\partial
^{2}f}{\partial t\partial s}\left( U_{2}(t),V_{2}(s)\right) dtds \\
&=&\frac{1}{\left( b-a\right) \left( d-c\right) }\left[ f\left( \frac{a+b}{2}%
,c\right) \int_{\frac{1}{6}}^{\frac{1}{2}}\int_{\frac{5}{6}}^{1}p\left(
u,v\right) dudv+f\left( b,c\right) \int_{0}^{\frac{1}{6}}\int_{\frac{5}{6}%
}^{1}p\left( u,v\right) dudv\right.  \\
&&-\int_{0}^{\frac{1}{2}}\left( \int_{\frac{5}{6}}^{1}p\left( t,v\right)
dv\right) f\left( at+\left( 1-t\right) b,c\right) dt+f\left( \frac{a+b}{2},%
\frac{c+d}{2}\right) \int_{\frac{1}{6}}^{\frac{1}{2}}\int_{\frac{1}{2}}^{%
\frac{5}{6}}p\left( u,v\right) dudv \\
&&+f\left( b,\frac{c+d}{2}\right) \int_{0}^{\frac{1}{6}}\int_{\frac{1}{2}}^{%
\frac{5}{6}}p\left( u,v\right) dudv-\int_{0}^{\frac{1}{2}}\left( \int_{\frac{%
1}{2}}^{\frac{5}{6}}p\left( t,v\right) dv\right) f\left( U_{2}(t),\frac{c+d}{%
2}\right) dt \\
&&-\int_{\frac{1}{2}}^{1}\left( \int_{\frac{1}{6}}^{\frac{1}{2}}p\left(
u,s\right) ds\right) f\left( \frac{a+b}{2},V_{2}(s)\right) ds-\int_{\frac{1}{%
2}}^{1}\left( \int_{0}^{\frac{1}{6}}p\left( u,s\right) ds\right) f\left(
b,V_{2}(s)\right) ds \\
&&\left. +\int_{\frac{1}{2}}^{1}\int_{0}^{\frac{1}{2}}p\left( t,s\right)
f\left( U_{2}(t),V_{2}(s)\right) dtds\right] ,
\end{eqnarray*}%
\begin{eqnarray*}
J_{3} &=&\int_{\frac{1}{2}}^{1}\int_{0}^{\frac{1}{2}}\left( \int_{\frac{5}{6}%
}^{t}\int_{\frac{1}{6}}^{s}p\left( u,v\right) dudv\right) \frac{\partial
^{2}f}{\partial t\partial s}\left( U_{2}(t),V_{2}(s)\right) dtds \\
&=&\frac{1}{\left( b-a\right) \left( d-c\right) }f\left[ \left( a,\frac{c+d}{%
2}\right) \int_{\frac{5}{6}}^{1}\int_{\frac{1}{6}}^{\frac{1}{2}}p\left(
u,v\right) dvdu+f\left( \frac{a+b}{2},\frac{c+d}{2}\right) \int_{\frac{1}{2}%
}^{\frac{5}{6}}\int_{\frac{1}{6}}^{\frac{1}{2}}p\left( u,v\right)
dvdu\right.  \\
&&-\int_{\frac{1}{2}}^{1}\left( \int_{\frac{1}{6}}^{\frac{1}{2}}p\left(
t,v\right) dv\right) f\left( U_{2}(t),\frac{c+d}{2}\right) dt+f\left(
a,d\right) \int_{\frac{5}{6}}^{1}\int_{0}^{\frac{1}{6}}p\left( u,v\right)
dvdu \\
&&+f\left( \frac{a+b}{2},d\right) \int_{\frac{1}{2}}^{\frac{5}{6}}\int_{0}^{%
\frac{1}{6}}p\left( u,v\right) dvdu-\int_{\frac{1}{2}}^{1}\left( \int_{0}^{%
\frac{1}{6}}p\left( t,v\right) dv\right) f\left( U_{2}(t),d\right) dt \\
&&-\int_{0}^{\frac{1}{2}}\left( \int_{\frac{5}{6}}^{1}p\left( u,s\right)
du\right) f\left( a,V_{2}(s)\right) ds-\int_{0}^{\frac{1}{2}}\left( \int_{%
\frac{1}{2}}^{\frac{5}{6}}p\left( u,s\right) du\right) f\left( \frac{a+b}{2}%
,V_{2}(s)\right) ds \\
&&\left. +\int_{0}^{\frac{1}{2}}\int_{\frac{1}{2}}^{1}p\left( t,s\right)
f\left( U_{2}(t),V_{2}(s)\right) dtds\right] ,
\end{eqnarray*}%
and%
\begin{eqnarray*}
J_{4} &=&\int_{\frac{1}{2}}^{1}\int_{\frac{1}{2}}^{1}\left( \int_{\frac{5}{6}%
}^{t}\int_{\frac{5}{6}}^{s}p\left( u,v\right) dudv\right) \frac{\partial
^{2}f}{\partial t\partial s}\left( U_{2}(t),V_{2}(s)\right) dtds \\
&=&\frac{1}{\left( b-a\right) \left( d-c\right) }\left[ f\left( a,c\right)
\int_{\frac{5}{6}}^{\frac{1}{2}}\int_{\frac{5}{6}}^{1}p\left( u,v\right)
dvdu+f\left( \frac{a+b}{2},c\right) \int_{\frac{1}{2}}^{\frac{5}{6}}\int_{%
\frac{5}{6}}^{1}p\left( u,v\right) dvdu\right.  \\
&&-\int_{\frac{1}{2}}^{1}\left( \int_{\frac{5}{6}}^{1}p\left( t,v\right)
dv\right) f\left( U_{2}(t),c\right) dt+f\left( a,\frac{c+d}{2}\right) \int_{%
\frac{5}{6}}^{1}\int_{\frac{1}{2}}^{\frac{5}{6}}p\left( u,v\right) dvdu \\
&&+f\left( \frac{a+b}{2},\frac{c+d}{2}\right) \int_{\frac{1}{2}}^{\frac{5}{6}%
}\int_{\frac{1}{2}}^{\frac{5}{6}}p\left( u,v\right) dvdu-\int_{\frac{1}{2}%
}^{1}\left( \int_{\frac{1}{2}}^{\frac{5}{6}}p\left( t,v\right) dv\right)
f\left( U_{2}(t),\frac{c+d}{2}\right) dt \\
&&-\int_{\frac{1}{2}}^{1}\left( \int_{\frac{5}{6}}^{1}p\left( u,s\right)
du\right) f\left( a,V_{2}(s)\right) ds-\int_{\frac{1}{2}}^{1}\left( \int_{%
\frac{1}{2}}^{\frac{5}{6}}p\left( u,s\right) du\right) f\left( \frac{a+b}{2}%
,V_{2}(s)\right) ds \\
&&\left. +\int_{\frac{1}{2}}^{1}\int_{\frac{1}{2}}^{1}p\left( t,s\right)
f\left( U_{2}(t),V_{2}(s)\right) dtds\right] .
\end{eqnarray*}%
Using the change of variable, we obtain%
\begin{eqnarray*}
&&\left( b-a\right) \left( d-c\right) \left[ J_{1}+J_{2}+J_{3}+J_{4}\right]
\\
&=&f\left( \frac{a+b}{2},\frac{c+d}{2}\right) \int_{\frac{1}{6}}^{\frac{5}{6}%
}\int_{\frac{1}{6}}^{\frac{5}{6}}p\left( u,v\right) dvdu \\
&&+f\left( a,\frac{c+d}{2}\right) \int_{\frac{5}{6}}^{1}\int_{\frac{1}{6}}^{%
\frac{5}{6}}p\left( u,v\right) dvdu+f\left( b,\frac{c+d}{2}\right) \int_{0}^{%
\frac{1}{6}}\int_{\frac{1}{6}}^{\frac{5}{6}}p\left( u,v\right) dvdu \\
&&+f\left( \frac{a+b}{2},c\right) \int_{\frac{1}{6}}^{\frac{5}{6}}\int_{%
\frac{5}{6}}^{1}p\left( u,v\right) dvdu+f\left( \frac{a+b}{2},d\right) \int_{%
\frac{1}{6}}^{\frac{5}{6}}\int_{0}^{\frac{1}{6}}p\left( u,v\right) dvdu \\
&&+f\left( a,c\right) \int_{\frac{5}{6}}^{1}\int_{\frac{5}{6}}^{1}p\left(
u,v\right) dvdu+f\left( b,c\right) \int_{0}^{\frac{1}{6}}\int_{\frac{5}{6}%
}^{1}p\left( u,v\right) dvdu \\
&&+f\left( a,d\right) \int_{\frac{5}{6}}^{1}\int_{0}^{\frac{1}{6}}p\left(
u,v\right) dvdu+f\left( b,d\right) \int_{0}^{\frac{1}{6}}\int_{0}^{\frac{1}{6%
}}p\left( u,v\right) dvdu \\
&&-\frac{1}{b-a}\int_{a}^{b}\left( \int_{\frac{5}{6}}^{1}p\left( \frac{b-x}{%
b-a},v\right) dv\right) f\left( x,c\right) dx-\frac{1}{b-a}%
\int_{a}^{b}\left( \int_{\frac{1}{2}}^{\frac{5}{6}}p\left( \frac{b-x}{b-a}%
,v\right) dv\right) f\left( x,\frac{c+d}{2}\right) dx \\
&&-\frac{1}{b-a}\int_{a}^{b}\left( \int_{0}^{\frac{1}{6}}p\left( \frac{b-x}{%
b-a},v\right) dv\right) f\left( x,d\right) dx-\frac{1}{d-c}%
\int_{c}^{d}\left( \int_{\frac{5}{6}}^{1}p\left( u,\frac{d-y}{d-c}\right)
du\right) f\left( a,y\right) dy \\
&&-\frac{1}{d-c}\int_{c}^{d}\left( \int_{\frac{1}{2}}^{\frac{5}{6}}p\left( u,%
\frac{d-y}{d-c}\right) du\right) f\left( \frac{a+b}{2},y\right) dy-\frac{1}{%
d-c}\int_{c}^{d}\left( \int_{0}^{\frac{1}{6}}p\left( u,\frac{d-y}{d-c}%
\right) du\right) f\left( b,y\right) dy \\
&&+\frac{1}{\left( b-a\right) \left( d-c\right) }\int_{a}^{b}\int_{c}^{d}p%
\left( \frac{b-x}{b-a},\frac{d-y}{d-c}\right) f\left( x,y\right) dydx
\end{eqnarray*}%
which completes the proof.
\end{proof}

\begin{remark}
If we choose $p(x,y)=1$ in Lemma \ref{l3}, then Lemma \ref{l3} reduses to
Lemma \ref{l1} which is proved by \cite{ozdemir2}.
\end{remark}

\begin{theorem}
\label{t2} Let the mappings $p,$ $U_{2}$ and $V_{2}$ be as in Lemma \ref{l2}%
. If $\left \vert \dfrac{\partial ^{2}f}{\partial t\partial s}\right \vert
^{q},$ $q>1,$ is a co-ordinated convex function on $\Delta ,$ then we have
the following inequality%
\begin{eqnarray*}
\left \vert \Theta (a,b;f,p)\right \vert &\leq &\left( b-a\right) \left(
d-c\right) \left( \int_{0}^{1}\int_{0}^{1}\left \vert w\left( t,s\right)
\right \vert ^{p}dsdt\right) ^{\frac{1}{p}} \\
&&\times \left( \frac{\left \vert \tfrac{\partial ^{2}f}{\partial t\partial s%
}\left( a,c\right) \right \vert ^{q}+\left \vert \tfrac{\partial ^{2}f}{%
\partial t\partial s}\left( a,d\right) \right \vert ^{q}+\left \vert \tfrac{%
\partial ^{2}f}{\partial t\partial s}\left( b,c\right) \right \vert
^{q}+\left \vert \tfrac{\partial ^{2}f}{\partial t\partial s}\left(
b,d\right) \right \vert ^{q}}{4}\right) ^{\frac{1}{q}}
\end{eqnarray*}%
where $\frac{1}{p}+\frac{1}{q}=1$.
\end{theorem}

\begin{proof}
Taking the modulus in Lemma \ref{l3} and using the H\"{o}lders inequality we
have%
\begin{eqnarray*}
\left \vert \Theta (a,b;f,p)\right \vert &\leq &\left( b-a\right) \left(
d-c\right) \int_{0}^{1}\int_{0}^{1}\left \vert w\left( t,s\right) \right
\vert \left \vert \frac{\partial ^{2}f}{\partial t\partial s}\left(
U_{2}(t),V_{2}(s)\right) \right \vert dsdt \\
&& \\
&\leq &\left( b-a\right) \left( d-c\right) \left(
\int_{0}^{1}\int_{0}^{1}\left \vert w\left( t,s\right) \right \vert
^{p}dsdt\right) ^{\frac{1}{p}}\left( \int_{0}^{1}\int_{0}^{1}\left \vert
\frac{\partial ^{2}f}{\partial t\partial s}\left( U_{2}(t),V_{2}(s)\right)
\right \vert ^{q}dsdt\right) ^{\frac{1}{q}}.
\end{eqnarray*}%
Since $\left \vert \dfrac{\partial ^{2}f}{\partial t\partial s}\right \vert
^{q},$ $q>1,$ is a co-ordinated convex function on $\Delta ,$ using the
inequality (\ref{e28}), we obtain%
\begin{eqnarray*}
\left \vert \Theta (a,b;f,p)\right \vert &\leq &\left( b-a\right) \left(
d-c\right) \left( \int_{0}^{1}\int_{0}^{1}\left \vert w\left( t,s\right)
\right \vert ^{p}dsdt\right) ^{\frac{1}{p}} \\
&&\times \left( \frac{\left \vert \tfrac{\partial ^{2}f}{\partial t\partial s%
}\left( a,c\right) \right \vert ^{q}+\left \vert \tfrac{\partial ^{2}f}{%
\partial t\partial s}\left( a,d\right) \right \vert ^{q}+\left \vert \tfrac{%
\partial ^{2}f}{\partial t\partial s}\left( b,c\right) \right \vert
^{q}+\left \vert \tfrac{\partial ^{2}f}{\partial t\partial s}\left(
b,d\right) \right \vert ^{q}}{4}\right) ^{\frac{1}{q}}.
\end{eqnarray*}%
This completes the proof.
\end{proof}

\begin{corollary}
If we choose $p(x,y)=1$ in Theorem \ref{t2}, then we have the following
inequality%
\begin{eqnarray*}
&&\left \vert \frac{f\left( a,\frac{c+d}{2}\right) +f\left( b,\frac{c+d}{2}%
\right) +4f\left( \frac{a+b}{2},\frac{c+d}{2}\right) +f\left( \frac{a+b}{2}%
,c\right) +f\left( \frac{a+b}{2},d\right) }{9}\right. \\
&& \\
&&+\frac{f\left( a,c\right) +f\left( b,c\right) +f\left( a,d\right) +f\left(
b,d\right) }{36}-\frac{1}{6\left( b-a\right) }\int_{a}^{b}\left[ f\left(
x,c\right) +4f\left( x,\frac{c+d}{2}\right) +f\left( x,d\right) \right] dx \\
&& \\
&&\left. -\frac{1}{6\left( d-c\right) }\int_{c}^{d}\left[ f\left( a,y\right)
+4f\left( \frac{a+b}{2},y\right) +f\left( b,y\right) \right] dy+\frac{1}{%
\left( b-a\right) \left( d-c\right) }\int_{a}^{b}\int_{c}^{d}f\left(
x,y\right) dydx\right \vert \\
&& \\
&\leq &\frac{\left( b-a\right) \left( d-c\right) }{36}\left[ 2^{p+1}+1\right]
^{\frac{2}{p}}\left( \frac{1}{9\left( p+1\right) ^{2}}\right) ^{\frac{1}{p}}
\\
&&\times \left( \frac{\left \vert \tfrac{\partial ^{2}f}{\partial t\partial s%
}\left( a,c\right) \right \vert ^{q}+\left \vert \tfrac{\partial ^{2}f}{%
\partial t\partial s}\left( a,d\right) \right \vert ^{q}+\left \vert \tfrac{%
\partial ^{2}f}{\partial t\partial s}\left( b,c\right) \right \vert
^{q}+\left \vert \tfrac{\partial ^{2}f}{\partial t\partial s}\left(
b,d\right) \right \vert ^{q}}{4}\right) ^{\frac{1}{q}}.
\end{eqnarray*}
\end{corollary}

\begin{theorem}
\label{t3} Let the mappings $p,$ $U_{2}$ and $V_{2}$ be as in Lemma \ref{l2}%
. If $\dfrac{\partial ^{2}f}{\partial t\partial s}$ is bounded on $\Delta ,$
i.e.%
\begin{equation*}
\left \Vert \dfrac{\partial ^{2}f}{\partial t\partial s}\right \Vert
_{\infty }=\sup_{\left( x,y\right) \in \Delta }\left \vert \dfrac{\partial
^{2}f}{\partial t\partial s}(x,y)\right \vert <\infty ,
\end{equation*}%
then we have the following inequality%
\begin{equation*}
\left \vert \Theta (a,b;f,p)\right \vert \leq \left( b-a\right) \left(
d-c\right) \left \Vert \dfrac{\partial ^{2}f}{\partial t\partial s}\right
\Vert _{\infty }\int_{0}^{1}\int_{0}^{1}\left \vert w\left( t,s\right)
\right \vert dsdt.
\end{equation*}
\end{theorem}

\begin{proof}
From Lemma \ref{l3}, we have%
\begin{equation*}
\left \vert \Theta (a,b;f,p)\right \vert \leq \left( b-a\right) \left(
d-c\right) \int_{0}^{1}\int_{0}^{1}\left \vert w\left( t,s\right) \right \vert
\left \vert \frac{\partial ^{2}f}{\partial t\partial s}\left(
U_{2}(t),V_{2}(s)\right) \right \vert dsdt.
\end{equation*}%
Since $\dfrac{\partial ^{2}f}{\partial t\partial s}$ is bounded on $\Delta ,$
we obtain%
\begin{equation*}
\left \vert \Theta (a,b;f,p)\right \vert \leq \left( b-a\right) \left(
d-c\right) \left \Vert \dfrac{\partial ^{2}f}{\partial t\partial s}%
\right \Vert _{\infty }\int_{0}^{1}\int_{0}^{1}\left \vert w\left( t,s\right)
\right \vert dsdt
\end{equation*}%
which completes the proof.
\end{proof}

\begin{remark}
If we choose $p(x,y)=1$ in Theorem \ref{t3}, then we have the following
inequality%
\begin{eqnarray*}
&&\left \vert \frac{f\left( a,\frac{c+d}{2}\right) +f\left( b,\frac{c+d}{2}%
\right) +4f\left( \frac{a+b}{2},\frac{c+d}{2}\right) +f\left( \frac{a+b}{2}%
,c\right) +f\left( \frac{a+b}{2},d\right) }{9}\right. \\
&& \\
&&+\frac{f\left( a,c\right) +f\left( b,c\right) +f\left( a,d\right) +f\left(
b,d\right) }{36}-\frac{1}{6\left( b-a\right) }\int_{a}^{b}\left[ f\left(
x,c\right) +4f\left( x,\frac{c+d}{2}\right) +f\left( x,d\right) \right] dx \\
&& \\
&&\left. -\frac{1}{6\left( d-c\right) }\int_{c}^{d}\left[ f\left( a,y\right)
+4f\left( \frac{a+b}{2},y\right) +f\left( b,y\right) \right] dy+\frac{1}{%
\left( b-a\right) \left( d-c\right) }\int_{a}^{b}\int_{c}^{d}f\left(
x,y\right) dydx\right \vert \\
&& \\
&\leq &\frac{25\left( b-a\right) \left( d-c\right) }{1296}\left \Vert \dfrac{%
\partial ^{2}f}{\partial t\partial s}\right \Vert _{\infty }
\end{eqnarray*}%
which was proved by \"{O}zdemir et al. in \cite{ozdemir2}.
\end{remark}

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