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\begin{document}
\title[On generalization of midpoint and trapezoid type inequalities...]{On
generalization of midpoint and trapezoid type inequalities involving
fractional integrals}
\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}
\author{S\"{u}meyye S\"{O}NMEZO\u{G}LU}
\email{smyy.mat@gmail.com}
\keywords{\textbf{\thanks{\textbf{2010 Mathematics Subject Classification.}
26D15, 26B25, 26D10.} }Hermite-Hadamard inequality, midpoint inequality,
fractional integral operators, convex function.}

\begin{abstract}
In this paper, we first prove a lemma for twice differentiable functions .
Then we establish some inequalities for mapping whose second derivatives in
absolute value are convex via Riemann-Liouville fractional integrals. These
results generalize the midpoint and trapezoid inequalities involving
Riemann-Liouville fractional integrals given in earlier studies.
\end{abstract}

\maketitle

\section{{\protect\large {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{had}, \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 Hadamard's inequality may be regarded as a refinement of the concept of
convexity and it follows easily from Jensen's inequality. Hadamard's
inequality for convex functions has received renewed attention in recent
years and a remarkable variety of refinements and generalizations have been
found (see, for example, \cite{alomari}-\cite{cal}, \cite{dragomir}, \cite%
{kavurmaci}, \cite{kirmaci}, \cite{ozdemir2}, \cite{ozdemir3}, \cite{saglam}%
, \cite{sarikaya2}, \cite{sarikaya6}, \cite{set}, \cite{xi}, \cite{xi1}) and
the references cited therein.

In the following we will give some necessary definitions and mathematical
preliminaries of fractional calculus theory which are used further in this
paper. More details, one can consult (\cite{Gorenflo}, \cite{kilbas}, \cite%
{Miller}, \cite{Podlubni})

\begin{definition}
Let $f\in L_{1}[a,b].$ The Riemann-Liouville integrals $J_{a+}^{\alpha }f$
and $J_{b-}^{\alpha }f$ of order $\alpha >0$ with $a\geq 0$ are defined by 
\begin{equation*}
J_{a+}^{\alpha }f(x)=\frac{1}{\Gamma (\alpha )}\int_{a}^{x}\left( x-t\right)
^{\alpha -1}f(t)dt,\ \ x>a
\end{equation*}%
and%
\begin{equation*}
J_{b-}^{\alpha }f(x)=\frac{1}{\Gamma (\alpha )}\int_{x}^{b}\left( t-x\right)
^{\alpha -1}f(t)dt,\ \ x<b
\end{equation*}%
respectively. Here, $\Gamma (\alpha )$ is the Gamma function and $%
J_{a+}^{0}f(x)=J_{b-}^{0}f(x)=f(x).$
\end{definition}

It is remarkable that Sarikaya et al.\cite{sarikaya4} first give the
following interesting integral inequalities of Hermite-Hadamard type
involving Riemann-Liouville fractional integrals.

\begin{theorem}
Let $f:\left[ a,b\right] \rightarrow \mathbb{R}$ be a positive function with 
$0\leq a<b$ and $f\in L_{1}\left[ a,b\right] .$ If $f$ is a convex function
on $[a,b]$, then the following inequalities for fractional integrals hold:%
\begin{equation}
f\left( \frac{a+b}{2}\right) \leq \frac{\Gamma (\alpha +1)}{2\left(
b-a\right) ^{\alpha }}\left[ J_{a+}^{\alpha }f(b)+J_{b-}^{\alpha }f(a)\right]
\leq \frac{f\left( a\right) +f\left( b\right) }{2}  \label{S1}
\end{equation}%
with $\alpha >0.$
\end{theorem}

Sar\i kaya and Y\i ld\i r\i m also give the following Hermite-Hadamard type
inequality for the Riemann-Lioville fractional integrals in \cite{sarikaya1}.

\begin{theorem}
Let $f:\left[ a,b\right] \rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ be a positive function with $a<b$ and $f\in L_{1}\left[ a,b\right] .$ If $%
f $ is a convex function on $\left[ a,b\right] ,$ then the following
inequalities for fractional integrals hold:%
\begin{equation}
f\left( \frac{a+b}{2}\right) \leq \frac{2^{\alpha -1}\Gamma (\alpha +1)}{%
\left( b-a\right) ^{\alpha }}\left[ J_{\left( \frac{a+b}{2}\right)
^{+}}^{\alpha }f(b)+J_{\left( \frac{a+b}{2}\right) ^{-}}^{\alpha }f(a)\right]
\leq \frac{f(a)+f(b)}{2}.  \label{hh0}
\end{equation}
\end{theorem}

For the more information fractional calculus and related inequalities please
refer to (\cite{chen}, \cite{hussain}, \cite{iqbal}, \cite{noor}, \cite%
{ozdemir}, \cite{sarikaya3}, \cite{sarikaya5}, \cite{wang}, \cite{whang}, 
\cite{zhang})

\section{{\protect\large {Generalized Midpoint and Trapezoid Type
Inequalities}}}

In this section, we will first present a lemma for twice differentiable
functions . Then we establish some inequalities which generalize the
midpoint and trapezoid inequalities involving Riemann-Liouville fractional
integrals obtained in previous works.

\begin{lemma}
\label{l1} Let $I\subset 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ be an open interval, $a,b\in I$ with $a<b.$ If $f:I\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ is a twice differentiable mapping such that $f^{\prime \prime }$ is
integrable and $0\leq \lambda \leq 1,\ \alpha \geq 1,$ then we have

\begin{eqnarray}
&&\left[ \left( \lambda -\frac{\alpha +1}{2^{\alpha }}\right) f\left( \frac{%
a+b}{2}\right) -\lambda \left( \frac{f\left( a\right) +f\left( b\right) }{2}%
\right) +\frac{\Gamma \left( \alpha +2\right) }{2\left( b-a\right) ^{\alpha }%
}\left( J_{\left( \frac{a+b}{2}\right) ^{+}}^{\alpha }f\left( b\right)
+J_{\left( \frac{a+b}{2}\right) ^{-}}^{\alpha }f\left( a\right) \right) %
\right]   \label{2.1} \\
&&  \notag \\
&=&\frac{\left( b-a\right) ^{2}}{2}\int_{0}^{1}k\left( t\right) f^{\prime
\prime }\left( ta+\left( 1-t\right) b\right) dt  \notag
\end{eqnarray}%
where%
\begin{equation*}
k\left( t\right) =\left\{ 
\begin{array}{ll}
t\left( t^{\alpha }-\lambda \right)  & 0\leq t\leq \frac{1}{2} \\ 
&  \\ 
\left( 1-t\right) \left( \left( 1-t\right) ^{\alpha }-\lambda \right)  & 
\frac{1}{2}\leq t\leq 1.%
\end{array}%
\right. 
\end{equation*}
\end{lemma}

\begin{proof}
It suffices to note that

\begin{eqnarray}
I &=&\int_{0}^{1}k\left( t\right) f^{\prime \prime }\left( ta+\left(
1-t\right) b\right) dt  \label{2.2} \\
&&  \notag \\
&=&\int_{0}^{\frac{1}{2}}t\left( t^{\alpha }-\lambda \right) f^{\prime
\prime }\left( ta+\left( 1-t\right) b\right) dt+\int_{\frac{1}{2}}^{1}\left(
1-t\right) \left( \left( 1-t\right) ^{\alpha }-\lambda \right) f^{\prime
\prime }\left( ta+\left( 1-t\right) b\right) dt  \notag \\
&&  \notag \\
&=&I_{1}+I_{2}.  \notag
\end{eqnarray}%
Integrating by parts twice, we can state:%
\begin{eqnarray}
I_{1} &=&\int_{0}^{\frac{1}{2}}t\left( t^{\alpha }-\lambda \right) f^{\prime
\prime }\left( ta+\left( 1-t\right) b\right) dt  \label{2.3} \\
&&  \notag \\
&=&\left. t\left( t^{\alpha }-\lambda \right) \frac{f^{\prime }\left(
ta+\left( 1-t\right) b\right) }{a-b}\right\vert _{0}^{\frac{1}{2}}-\int_{0}^{%
\frac{1}{2}}\frac{f^{\prime }\left( ta+\left( 1-t\right) b\right) }{a-b}%
\left( \left( \alpha +1\right) t^{\alpha }-\lambda \right) dt  \notag \\
&&  \notag \\
&=&\frac{-1}{2\left( b-a\right) }\left( \frac{1}{2^{\alpha }}-\lambda
\right) f^{\prime }\left( \frac{a+b}{2}\right) -\frac{1}{\left( b-a\right)
^{2}}\left( \frac{\alpha +1}{2^{\alpha }}-\lambda \right) f\left( \frac{a+b}{%
2}\right)   \notag \\
&&  \notag \\
&&-\frac{\lambda }{\left( b-a\right) ^{2}}f\left( b\right) +\frac{\alpha
\left( \alpha +1\right) }{\left( b-a\right) ^{2}}\int_{0}^{\frac{1}{2}%
}f\left( ta+\left( 1-t\right) b\right) t^{\alpha -1}dt  \notag \\
&&  \notag \\
&&+\frac{\alpha \left( \alpha +1\right) }{\left( b-a\right) ^{2}}\frac{1}{%
\left( b-a\right) ^{\alpha }}\Gamma \left( \alpha \right) J_{\left( \frac{a+b%
}{2}\right) ^{+}}^{\alpha }f\left( b\right)   \notag
\end{eqnarray}%
and similarly, we get%
\begin{eqnarray}
I_{2} &=&\int_{\frac{1}{2}}^{1}\left( 1-t\right) \left( \left( 1-t\right)
^{\alpha }-\lambda \right) f^{\prime \prime }\left( ta+\left( 1-t\right)
b\right) dt  \label{2.4} \\
&&  \notag \\
&=&\left. \left( 1-t\right) \left( \left( 1-t\right) ^{\alpha }-\lambda
\right) \frac{f^{\prime }\left( ta+\left( 1-t\right) b\right) }{a-b}%
\right\vert _{\frac{1}{2}}^{1}  \notag \\
&&  \notag \\
&&+\int_{\frac{1}{2}}^{1}\frac{f^{\prime }\left( ta+\left( 1-t\right)
b\right) }{a-b}\left( \left( \alpha +1\right) \left( 1-t\right) ^{\alpha
}-\lambda \right)   \notag \\
&&  \notag \\
&=&\frac{1}{2\left( b-a\right) }\left( \frac{1}{2^{\alpha }}-\lambda \right)
f^{\prime }\left( \frac{a+b}{2}\right) -\lambda \frac{f\left( a\right) }{%
\left( b-a\right) ^{2}}  \notag \\
&&  \notag \\
&&-\frac{1}{\left( b-a\right) ^{2}}\left( \frac{\alpha +1}{2^{\alpha }}%
-\lambda \right) f\left( \frac{a+b}{2}\right) +\frac{\alpha \left( \alpha
+1\right) }{\left( b-a\right) ^{\alpha +2}}\Gamma \left( \alpha \right)
J_{\left( \frac{a+b}{2}\right) ^{-}}^{\alpha }f\left( a\right) .  \notag
\end{eqnarray}%
Using (\ref{2.3}) and (\ref{2.4}) in (\ref{2.2}), it follows that%
\begin{eqnarray*}
I &=&I_{1}+I_{2}=\frac{-2}{\left( b-a\right) ^{2}}\left( \frac{\alpha +1}{%
2^{\alpha }}-\lambda \right) f\left( \frac{a+b}{2}\right) -\frac{2\lambda }{%
\left( b-a\right) ^{2}}\left( \frac{f\left( a\right) +f\left( b\right) }{2}%
\right)  \\
&& \\
&&+\frac{\alpha \left( \alpha +1\right) }{\left( b-a\right) ^{\alpha +2}}%
\Gamma \left( \alpha \right) \left( J_{\left( \frac{a+b}{2}\right)
^{+}}^{\alpha }f\left( b\right) +J_{\left( \frac{a+b}{2}\right)
^{-}}^{\alpha }f\left( a\right) \right)  \\
&& \\
&=&\frac{2}{\left( b-a\right) ^{2}}\left[ \left( \lambda -\frac{\alpha +1}{%
2^{\alpha }}\right) f\left( \frac{a+b}{2}\right) -\lambda \left( \frac{%
f\left( a\right) +f\left( b\right) }{2}\right) \right]  \\
&& \\
&&+\frac{\Gamma \left( \alpha +2\right) }{\left( b-a\right) ^{\alpha +2}}%
\left( J_{\left( \frac{a+b}{2}\right) ^{+}}^{\alpha }f\left( b\right)
+J_{\left( \frac{a+b}{2}\right) ^{-}}^{\alpha }f\left( a\right) \right) .
\end{eqnarray*}%
Then by multipling the above equality with $\frac{\left( b-a\right) ^{2}}{2},
$ this completes the proof.
\end{proof}

\begin{theorem}
\label{t1} Let $I\subset 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ be an open intervial, $a,b\in I$ with $a<b$ and $f:I\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ be a twice differentiable mapping such that $f^{\prime \prime }$ is
integrable and \ $0\leq \lambda \leq 1,\ \alpha \geq 1.$ If $\left\vert
f^{\prime \prime }\right\vert $ is a convex on $\left[ a,b\right] ,$ then
the following inequalities hold:
\end{theorem}

\begin{eqnarray}
&&\left\vert \left( \lambda -\frac{\alpha +1}{2^{\alpha }}\right) f\left( 
\frac{a+b}{2}\right) -\lambda \left( \frac{f\left( a\right) +f\left(
b\right) }{2}\right) +\frac{\Gamma \left( \alpha +2\right) }{2\left(
b-a\right) ^{\alpha }}\left( J_{\left( \frac{a+b}{2}\right) ^{+}}^{\alpha
}f\left( b\right) +J_{\left( \frac{a+b}{2}\right) ^{-}}^{\alpha }f\left(
a\right) \right) \right\vert   \label{2.5} \\
&&  \notag \\
&\leq &\frac{\left( b-a\right) ^{2}}{2}\left\{ 
\begin{array}{ll}
\left( \dfrac{1}{2^{\alpha +2}\left( \alpha +2\right) }-\dfrac{\lambda }{8}+%
\dfrac{\alpha \lambda ^{1+\frac{2}{\alpha }}}{\alpha +2}\right) \left[
\left\vert f^{\prime \prime }\left( a\right) \right\vert +\left\vert
f^{\prime \prime }\left( b\right) \right\vert \right] , & 0\leq \lambda \leq 
\frac{1}{2} \\ 
&  \\ 
\left( \dfrac{\lambda }{8}-\dfrac{1}{2^{\alpha +2}\left( \alpha +2\right) }%
\right) \left[ \left\vert f^{\prime \prime }\left( a\right) \right\vert
+\left\vert f^{\prime \prime }\left( b\right) \right\vert \right] , & \frac{1%
}{2}\leq \lambda \leq 1.%
\end{array}%
\right.   \notag
\end{eqnarray}

\begin{proof}
From Lemma \ref{l1} and by defination of $k\left( t\right) $, we get

\begin{eqnarray}
&&\left\vert \left( \lambda -\frac{\alpha +1}{2^{\alpha }}\right) f\left( 
\frac{a+b}{2}\right) -\lambda \left( \frac{f\left( a\right) +f\left(
b\right) }{2}\right) +\frac{\Gamma \left( \alpha +2\right) }{2\left(
b-a\right) ^{\alpha }}\left( J_{\left( \frac{a+b}{2}\right) ^{+}}^{\alpha
}f\left( b\right) +J_{\left( \frac{a+b}{2}\right) ^{-}}^{\alpha }f\left(
a\right) \right) \right\vert   \label{2.6} \\
&&  \notag \\
&\leq &\frac{\left( b-a\right) ^{2}}{2}\int_{0}^{1}\left\vert k\left(
t\right) \right\vert \left\vert f^{\prime \prime }\left( ta+\left(
1-t\right) b\right) \right\vert dt  \notag \\
&&  \notag \\
&=&\frac{\left( b-a\right) ^{2}}{2}\left\{ \int_{0}^{\frac{1}{2}}\left\vert
t\left( t^{\alpha }-\lambda \right) \right\vert \left\vert f^{\prime \prime
}\left( ta+\left( 1-t\right) b\right) \right\vert dt\right.   \notag \\
&&  \notag \\
&&\left. +\int_{\frac{1}{2}}^{1}\left\vert \left( 1-t\right) \left( \left(
1-t\right) ^{\alpha }-\lambda \right) \right\vert \left\vert f^{\prime
\prime }\left( ta+\left( 1-t\right) b\right) \right\vert dt\right\}   \notag
\\
&&  \notag \\
&=&\frac{\left( b-a\right) ^{2}}{2}\left\{ J_{1}+J_{2}\right\} .  \notag
\end{eqnarray}%
We assume that $0\leq \lambda \leq \frac{1}{2},$ then using the convexity of 
$\left\vert f^{\prime \prime }\right\vert ,$we get%
\begin{eqnarray}
J_{1} &\leq &\int_{0}^{\frac{1}{2}}\left\vert t\left( t^{\alpha }-\lambda
\right) \right\vert \left[ t\left\vert f^{\prime \prime }\left( a\right)
\right\vert +\left( 1-t\right) \left\vert f^{\prime \prime }\left( b\right)
\right\vert \right] dt  \label{2.7} \\
&&  \notag \\
&=&\int_{0}^{\lambda ^{\frac{1}{\alpha }}}t\left( \lambda -t^{\alpha
}\right) \left[ t\left\vert f^{\prime \prime }\left( a\right) \right\vert
+\left( 1-t\right) \left\vert f^{\prime \prime }\left( b\right) \right\vert %
\right] dt  \notag \\
&&  \notag \\
&&+\int_{\lambda ^{\frac{1}{\alpha }}}^{\frac{1}{2}}t\left( t^{\alpha
}-\lambda \right) \left[ t\left\vert f^{\prime \prime }\left( a\right)
\right\vert +\left( 1-t\right) \left\vert f^{\prime \prime }\left( b\right)
\right\vert \right] dt  \notag \\
&&  \notag \\
&=&\left\vert f^{\prime \prime }\left( a\right) \right\vert \left[ \frac{%
2\alpha \lambda ^{1+\frac{3}{\alpha }}}{3\left( \alpha +3\right) }+\frac{1}{%
2^{\alpha +3}\left( \alpha +3\right) }-\frac{\lambda }{24}\right]   \notag \\
&&  \notag \\
&&+\left\vert f^{\prime \prime }\left( b\right) \right\vert \left[ \frac{%
\alpha \lambda ^{1+\frac{2}{\alpha }}}{\alpha +2}-\frac{2\alpha \lambda ^{1+%
\frac{3}{\alpha }}}{3\left( \alpha +3\right) }+\frac{\alpha +4}{2^{\alpha
+3}\left( \alpha +2\right) \left( \alpha +3\right) }-\frac{\lambda }{12}%
\right]   \notag
\end{eqnarray}%
and similarly, we have%
\begin{eqnarray}
J_{2} &\leq &\int_{\frac{1}{2}}^{1-\lambda ^{\frac{1}{\alpha }}}\left(
1-t\right) \left( \left( 1-t\right) ^{\alpha }-\lambda \right) \left[
t\left\vert f^{\prime \prime }\left( a\right) \right\vert +\left( 1-t\right)
\left\vert f^{\prime \prime }\left( b\right) \right\vert \right] dt
\label{2.8} \\
&&  \notag \\
&&+\int_{1-\lambda ^{\frac{1}{\alpha }}}^{1}\left( 1-t\right) \left( \lambda
-\left( 1-t\right) ^{\alpha }\right) \left[ t\left\vert f^{\prime \prime
}\left( a\right) \right\vert +\left( 1-t\right) \left\vert f^{\prime \prime
}\left( b\right) \right\vert \right] dt  \notag \\
&&  \notag \\
&=&\left\vert f^{\prime \prime }\left( a\right) \right\vert \left[ \frac{%
\alpha \lambda ^{1+\frac{2}{\alpha }}}{\alpha +2}-\frac{2\alpha \lambda ^{1+%
\frac{3}{\alpha }}}{3\left( \alpha +3\right) }+\frac{\alpha +4}{2^{\alpha
+3}\left( \alpha +2\right) \left( \alpha +3\right) }-\frac{\lambda }{12}%
\right]   \notag \\
&&  \notag \\
&&+\left\vert f^{\prime \prime }\left( b\right) \right\vert \left[ \frac{%
2\alpha \lambda ^{1+\frac{3}{\alpha }}}{3\left( \alpha +3\right) }+\frac{1}{%
2^{\alpha +3}\left( \alpha +3\right) }-\frac{\lambda }{24}\right] .  \notag
\end{eqnarray}%
Using (\ref{2.7}) and (\ref{2.8}) in (\ref{2.6}),we see thatthe first
inequality of (\ref{2.5}) holds. On the other hand, let $\frac{1}{2}\leq
\lambda \leq 1,$ then, using the convexity of $\left\vert f^{\prime \prime
}\right\vert $ and by simple computation we have%
\begin{eqnarray}
J_{1}^{\prime } &\leq &\int_{0}^{\frac{1}{2}}\left\vert t\left( t^{\alpha
}-\lambda \right) \right\vert \left[ t\left\vert f^{\prime \prime }\left(
a\right) \right\vert +\left( 1-t\right) \left\vert f^{\prime \prime }\left(
b\right) \right\vert \right] dt  \label{2.9} \\
&&  \notag \\
&=&\int_{0}^{\frac{1}{2}}t\left( \lambda -t^{\alpha }\right) \left[
t\left\vert f^{\prime \prime }\left( a\right) \right\vert +\left( 1-t\right)
\left\vert f^{\prime \prime }\left( b\right) \right\vert \right] dt  \notag
\\
&&  \notag \\
&=&\left( \frac{\lambda }{24}-\frac{1}{2^{\alpha +3}\left( \alpha +3\right) }%
\right) \left\vert f^{\prime \prime }\left( a\right) \right\vert +\left( 
\frac{\lambda }{12}-\frac{\alpha +4}{2^{\alpha +3}\left( \alpha +2\right)
\left( \alpha +3\right) }\right) \left\vert f^{\prime \prime }\left(
b\right) \right\vert   \notag
\end{eqnarray}%
and similarly%
\begin{eqnarray}
J_{2}^{\prime } &\leq &\int_{\frac{1}{2}}^{1}\left\vert \left( 1-t\right)
\left( \left( 1-t\right) ^{\alpha }-\lambda \right) \right\vert \left\vert
f^{\prime \prime }\left( ta+\left( 1-t\right) b\right) \right\vert dt
\label{2.10} \\
&&  \notag \\
&=&\int_{\frac{1}{2}}^{1}\left( 1-t\right) \left( \lambda -\left( 1-t\right)
^{\alpha }\right) \left[ t\left\vert f^{\prime \prime }\left( a\right)
\right\vert +\left( 1-t\right) \left\vert f^{\prime \prime }\left( b\right)
\right\vert \right] dt  \notag \\
&&  \notag \\
&=&\left( \frac{\lambda }{12}-\frac{\alpha +4}{2^{\alpha +3}\left( \alpha
+2\right) \left( \alpha +3\right) }\right) \left\vert f^{\prime \prime
}\left( a\right) \right\vert +\left( \frac{\lambda }{24}-\frac{1}{2^{\alpha
+3}\left( \alpha +3\right) }\right) \left\vert f^{\prime \prime }\left(
b\right) \right\vert .  \notag
\end{eqnarray}%
Thus if we (\ref{2.9}) and (\ref{2.10}) in (\ref{2.6}), we obtain the second
inequality of (\ref{2.5}). This completes the proof.
\end{proof}

\begin{corollary}
\label{c2} Under the assumptions of Theorem \ref{t1} with $\lambda =0$, then
we get the following inequality
\end{corollary}

\begin{eqnarray*}
&&\left\vert \frac{2^{\alpha -1}\Gamma \left( \alpha +1\right) }{\left(
b-a\right) ^{\alpha }}\left( J_{\left( \frac{a+b}{2}\right) ^{+}}^{\alpha
}f\left( b\right) +J_{\left( \frac{a+b}{2}\right) ^{-}}^{\alpha }f\left(
a\right) \right) -f\left( \frac{a+b}{2}\right) \right\vert  \\
&& \\
&\leq &\frac{\left( b-a\right) ^{2}}{\left( \alpha +1\right) \left( \alpha
+2\right) }\left[ \frac{\left\vert f^{\prime \prime }\left( a\right)
\right\vert +\left\vert f^{\prime \prime }\left( b\right) \right\vert }{8}%
\right] 
\end{eqnarray*}%
which is proved by Noor and Awan in \cite[Theorem 2 (for $s$=1)]{noor}.

\begin{remark}
If we take $\alpha =1$ in Corollary \ref{c2}, then we get the following
inequality%
\begin{equation*}
\left\vert \frac{1}{b-a}\dint\limits_{a}^{b}f(t)dt-f\left( \frac{a+b}{2}%
\right) \right\vert \leq \frac{\left( b-a\right) ^{2}}{24}\left[ \frac{%
\left\vert f^{\prime \prime }\left( a\right) \right\vert +\left\vert
f^{\prime \prime }\left( b\right) \right\vert }{2}\right]
\end{equation*}%
which is given by Sarikaya et al. in \cite{sarikaya6}.
\end{remark}

\begin{corollary}
\label{c1} Under the assumptions of Theorem \ref{t1} with $\lambda =\frac{%
\alpha +1}{2^{\alpha }}$, then we get the following inequality%
\begin{eqnarray*}
&&\left\vert \frac{f\left( a\right) +f\left( b\right) }{2}-\frac{2^{\alpha
-1}\Gamma \left( \alpha +1\right) }{\left( b-a\right) ^{\alpha }}\left(
J_{\left( \frac{a+b}{2}\right) ^{+}}^{\alpha }f\left( b\right) +J_{\left( 
\frac{a+b}{2}\right) ^{-}}^{\alpha }f\left( a\right) \right) \right\vert  \\
&& \\
&\leq &\frac{\left( b-a\right) ^{2}}{8\left( \alpha +1\right) \left( \alpha
+2\right) }\left( \alpha \left( \alpha +1\right) ^{1+\frac{2}{\alpha }}+1-%
\frac{\left( \alpha +1\right) \left( \alpha +2\right) }{2}\right) \left[
\left\vert f^{\prime \prime }\left( a\right) \right\vert +\left\vert
f^{\prime \prime }\left( b\right) \right\vert \right] 
\end{eqnarray*}%
for $\alpha \geq 3$ and 
\begin{eqnarray*}
&&\left\vert \frac{f\left( a\right) +f\left( b\right) }{2}-\frac{2^{\alpha
-1}\Gamma \left( \alpha +1\right) }{\left( b-a\right) ^{\alpha }}\left(
J_{\left( \frac{a+b}{2}\right) ^{+}}^{\alpha }f\left( b\right) +J_{\left( 
\frac{a+b}{2}\right) ^{-}}^{\alpha }f\left( a\right) \right) \right\vert  \\
&& \\
&\leq &\frac{\left( b-a\right) ^{2}}{8\left( \alpha +1\right) \left( \alpha
+2\right) }\left( \frac{\left( \alpha +1\right) \left( \alpha +2\right) }{2}%
-1\right) \left[ \left\vert f^{\prime \prime }\left( a\right) \right\vert
+\left\vert f^{\prime \prime }\left( b\right) \right\vert \right] 
\end{eqnarray*}%
for $1\leq \alpha \leq 3.$
\end{corollary}

\begin{remark}
If we take $\alpha =1$ in Corollary \ref{c1}, then we get the following
inequality%
\begin{equation*}
\left\vert \frac{f\left( a\right) +f\left( b\right) }{2}-\frac{1}{b-a}%
\dint\limits_{a}^{b}f(t)dt\right\vert \leq \frac{\left( b-a\right) ^{2}}{12}%
\left[ \frac{\left\vert f^{\prime \prime }\left( a\right) \right\vert
+\left\vert f^{\prime \prime }\left( b\right) \right\vert }{2}\right] 
\end{equation*}%
which is given by Sarikaya and Aktan in \cite{sarikaya2}.
\end{remark}

\begin{remark}
Under the assumptions of Theorem \ref{t1} with $\lambda =\frac{1}{3}$ and $%
\alpha =1,$ then we get the following inequality%
\begin{equation*}
\left\vert \frac{1}{6}\left[ f\left( a\right) +4f\left( \frac{a+b}{2}\right)
+f\left( b\right) \right] -\frac{1}{b-a}\dint\limits_{a}^{b}f(t)dt\right%
\vert \leq \frac{\left( b-a\right) ^{2}}{81}\left[ \frac{\left\vert
f^{\prime \prime }\left( a\right) \right\vert +\left\vert f^{\prime \prime
}\left( b\right) \right\vert }{2}\right] 
\end{equation*}%
which is given by Sarikaya and Aktan in \cite{sarikaya2}.
\end{remark}

\begin{remark}
Under the assumptions of Theorem \ref{t1} with $\lambda =\frac{1}{2}$ and $%
\alpha =1,$ then we get the following inequality%
\begin{equation*}
\left\vert \frac{1}{b-a}\dint\limits_{a}^{b}f(t)dt-\frac{1}{2}\left[ \frac{%
f\left( a\right) +f\left( b\right) }{2}+f\left( \frac{a+b}{2}\right) \right]
\right\vert \leq \frac{\left( b-a\right) ^{2}}{48}\left[ \frac{\left\vert
f^{\prime \prime }\left( a\right) \right\vert +\left\vert f^{\prime \prime
}\left( b\right) \right\vert }{2}\right] 
\end{equation*}%
which is given by Sarikaya and Aktan in \cite{sarikaya2}.
\end{remark}

\begin{theorem}
\label{t2} Let $I\subset 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ be an open intervial, $a,b\in I$ with $a<b$ and $f:I\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ be a twice differentiable mapping such that $f^{\prime \prime }$ is
integrable and $0\leq \lambda \leq 1,\ \alpha \geq 1.$ If $\left\vert
f^{\prime \prime }\right\vert ^{q}$ is a convex on $\left[ a,b\right] ,$ $%
q\geq 1$ then the following inequalities hold:%
\begin{eqnarray}
&&\left\vert \left[ \left( \lambda -\frac{\alpha +1}{2^{\alpha }}\right)
f\left( \frac{a+b}{2}\right) -\lambda \left( \frac{f\left( a\right) +f\left(
b\right) }{2}\right) +\frac{\Gamma \left( \alpha +2\right) }{2\left(
b-a\right) ^{\alpha }}\left( J_{\left( \frac{a+b}{2}\right) ^{+}}^{\alpha
}f\left( b\right) +J_{\left( \frac{a+b}{2}\right) ^{-}}^{\alpha }f\left(
a\right) \right) \right] \right\vert   \label{h1} \\
&&  \notag \\
&\leq &\frac{\left( b-a\right) ^{2}}{2}\left( \frac{\alpha \lambda ^{1+\frac{%
2}{\alpha }}}{\alpha +2}+\frac{1}{2^{\alpha +2}\left( \alpha +2\right) }-%
\frac{\lambda }{8}\right) ^{1-\frac{1}{q}}  \notag \\
&&  \notag \\
&&\times \left\{ \left[ C_{1}\left\vert f^{\prime \prime }\left( a\right)
\right\vert ^{q}+C_{2}\left\vert f^{\prime \prime }\left( b\right)
\right\vert ^{q}\right] ^{\frac{1}{q}}+\left[ C_{2}\left\vert f^{\prime
\prime }\left( a\right) \right\vert ^{q}+C_{1}\left\vert f^{\prime \prime
}\left( b\right) \right\vert ^{q}\right] ^{\frac{1}{q}}\right\}   \notag
\end{eqnarray}%
for $0\leq \lambda \leq \frac{1}{2}$ and%
\begin{eqnarray}
&&\left\vert \left( \lambda -\frac{\alpha +1}{2^{\alpha }}\right) f\left( 
\frac{a+b}{2}\right) -\lambda \left( \frac{f\left( a\right) +f\left(
b\right) }{2}\right) +\frac{\Gamma \left( \alpha +2\right) }{2\left(
b-a\right) ^{\alpha }}\left( J_{\left( \frac{a+b}{2}\right) ^{+}}^{\alpha
}f\left( b\right) +J_{\left( \frac{a+b}{2}\right) ^{-}}^{\alpha }f\left(
a\right) \right) \right\vert   \label{h2} \\
&&  \notag \\
&\leq &\frac{\left( b-a\right) ^{2}}{2}\left( \frac{\lambda }{8}-\frac{1}{%
2^{\alpha +2}\left( \alpha +2\right) }\right) ^{^{1-\frac{1}{q}}}  \notag \\
&&  \notag \\
&&\times \left\{ \left[ C_{3}\left\vert f^{\prime \prime }\left( a\right)
\right\vert ^{q}+C_{4}\left\vert f^{\prime \prime }\left( b\right)
\right\vert ^{q}\right] ^{^{\frac{1}{q}}}+\left[ C_{4}\left\vert f^{\prime
\prime }\left( a\right) \right\vert ^{q}+C_{3}\left\vert f^{\prime \prime
}\left( b\right) \right\vert ^{q}\right] ^{^{^{\frac{1}{q}}}}\right\} , 
\notag
\end{eqnarray}%
for $\frac{1}{2}\leq \lambda \leq 1$ where $\frac{1}{p}+\frac{1}{q}=1,$%
\begin{eqnarray*}
C_{1} &=&\left( \frac{2\alpha \lambda ^{1+\frac{3}{\alpha }}}{3\left( \alpha
+3\right) }+\frac{1}{2^{\alpha +3}\left( \alpha +3\right) }-\frac{\lambda }{%
24}\right)  \\
C_{2} &=&\left( \frac{\alpha \lambda ^{1+\frac{2}{\alpha }}}{\alpha +2}-%
\frac{2\alpha \lambda ^{1+\frac{3}{\alpha }}}{3\left( \alpha +3\right) }+%
\frac{\alpha +4}{2^{\alpha +3}\left( \alpha +2\right) \left( \alpha
+3\right) }-\frac{\lambda }{12}\right)  \\
C_{3} &=&\left( \frac{\lambda }{24}-\frac{1}{2^{\alpha +3}\left( \alpha
+3\right) }\right)  \\
C_{4} &=&\left( \frac{\lambda }{12}-\frac{\alpha +4}{2^{\alpha +3}\left(
\alpha +2\right) \left( \alpha +3\right) }\right) .
\end{eqnarray*}
\end{theorem}

\begin{proof}
Suppose that $q\geq 1.$ From Lemma \ref{l1} and using the well known power
mean inequality, we have%
\begin{eqnarray}
&&\left\vert \left( \lambda -\frac{\alpha +1}{2^{\alpha }}\right) f\left( 
\frac{a+b}{2}\right) -\lambda \left( \frac{f\left( a\right) +f\left(
b\right) }{2}\right) +\frac{\Gamma \left( \alpha +2\right) }{2\left(
b-a\right) ^{\alpha }}\left( J_{\left( \frac{a+b}{2}\right) ^{+}}^{\alpha
}f\left( b\right) +J_{\left( \frac{a+b}{2}\right) ^{-}}^{\alpha }f\left(
a\right) \right) \right\vert   \label{2.12} \\
&&  \notag \\
&\leq &\frac{\left( b-a\right) ^{2}}{2}\int_{0}^{1}\left\vert k\left(
t\right) \right\vert \left\vert f^{\prime \prime }\left( ta+\left(
1-t\right) b\right) \right\vert dt  \notag \\
&&  \notag \\
&\leq &\frac{\left( b-a\right) ^{2}}{2}\left\{ \int_{0}^{\frac{1}{2}%
}\left\vert t\left( t^{\alpha }-\lambda \right) \right\vert \left\vert
f^{\prime \prime }\left( ta+\left( 1-t\right) b\right) \right\vert dt\right. 
\notag \\
&&  \notag \\
&&\left. +\int_{\frac{1}{2}}^{1}\left\vert \left( 1-t\right) \left( \left(
1-t\right) ^{\alpha }-\lambda \right) \right\vert \left\vert f^{\prime
\prime }\left( ta+\left( 1-t\right) b\right) \right\vert dt\right\}   \notag
\\
&&  \notag \\
&=&\frac{\left( b-a\right) ^{2}}{2}\left\{ \left( \int_{0}^{\frac{1}{2}%
}\left\vert t\left( t^{\alpha }-\lambda \right) \right\vert dt\right) ^{1-%
\frac{1}{q}}\left( \int_{0}^{\frac{1}{2}}\left\vert t\left( t^{\alpha
}-\lambda \right) \right\vert \left\vert f^{\prime \prime }\left( ta+\left(
1-t\right) b\right) \right\vert ^{q}dt\right) ^{\frac{1}{q}}\right.   \notag
\\
&&  \notag \\
&&\left. +\left( \int_{\frac{1}{2}}^{1}\left\vert \left( 1-t\right) \left(
\left( 1-t\right) ^{\alpha }-\lambda \right) \right\vert dt\right) ^{^{1-%
\frac{1}{q}}}\left( \int_{\frac{1}{2}}^{1}\left\vert \left( 1-t\right)
\left( \left( 1-t\right) ^{\alpha }-\lambda \right) \right\vert \left\vert
f^{\prime \prime }\left( ta+\left( 1-t\right) b\right) \right\vert
^{q}dt\right) ^{^{\frac{1}{q}}}\right\}   \notag
\end{eqnarray}%
Let $0\leq \lambda \leq \frac{1}{2}.$ Then since $\left\vert f^{\prime
}\right\vert ^{q}$ is convex on $\left[ a,b\right] ,$we know that for $t\in %
\left[ 0,1\right] $%
\begin{equation*}
\left\vert f^{\prime }\left( ta+\left( 1-t\right) b\right) \right\vert
^{q}\leq t\left\vert f^{\prime }\left( a\right) \right\vert ^{q}+\left(
1-t\right) \left\vert f^{\prime }\left( b\right) \right\vert ^{q}
\end{equation*}%
hence, by simple computation%
\begin{eqnarray}
&&\int_{0}^{\frac{1}{2}}\left\vert t\left( t^{\alpha }-\lambda \right)
\right\vert \left\vert f^{\prime \prime }\left( ta+\left( 1-t\right)
b\right) \right\vert ^{q}dt  \label{2.13} \\
&&  \notag \\
&\leq &\int_{0}^{\lambda ^{\frac{1}{\alpha }}}t\left( \lambda -t^{\alpha
}\right) \left[ t\left\vert f^{\prime \prime }\left( a\right) \right\vert
^{q}+\left( 1-t\right) \left\vert f^{\prime \prime }\left( b\right)
\right\vert ^{q}\right] dt  \notag \\
&&  \notag \\
&&+\int_{\lambda ^{\frac{1}{\alpha }}}^{\frac{1}{2}}t\left( t^{\alpha
}-\lambda \right) \left[ t\left\vert f^{\prime \prime }\left( a\right)
\right\vert ^{q}+\left( 1-t\right) \left\vert f^{\prime \prime }\left(
b\right) \right\vert ^{q}\right] dt  \notag \\
&&  \notag \\
&=&\left\vert f^{\prime \prime }\left( a\right) \right\vert ^{q}\left[ \frac{%
2\alpha \lambda ^{1+\frac{3}{\alpha }}}{3\left( \alpha +3\right) }+\frac{1}{%
2^{\alpha +3}\left( \alpha +3\right) }-\frac{\lambda }{24}\right]   \notag \\
&&  \notag \\
&&+\left\vert f^{\prime \prime }\left( b\right) \right\vert ^{q}\left[ \frac{%
\alpha \lambda ^{1+\frac{2}{\alpha }}}{\alpha +2}-\frac{2\alpha \lambda ^{1+%
\frac{3}{\alpha }}}{3\left( \alpha +3\right) }+\frac{\alpha +4}{2^{\alpha
+3}\left( \alpha +2\right) \left( \alpha +3\right) }-\frac{\lambda }{12}%
\right] ,  \notag
\end{eqnarray}%
\begin{eqnarray}
&&\int_{\frac{1}{2}}^{1}\left\vert \left( 1-t\right) \left( \left(
1-t\right) ^{\alpha }-\lambda \right) \right\vert \left\vert f^{\prime
\prime }\left( ta+\left( 1-t\right) b\right) \right\vert ^{q}dt  \label{2.14}
\\
&&  \notag \\
&\leq &\int_{\frac{1}{2}}^{1-\lambda ^{\frac{1}{\alpha }}}\left( 1-t\right)
\left( \left( 1-t\right) ^{\alpha }-\lambda \right) \left[ t\left\vert
f^{\prime \prime }\left( a\right) \right\vert ^{q}+\left( 1-t\right)
\left\vert f^{\prime \prime }\left( b\right) \right\vert ^{q}\right] dt 
\notag \\
&&  \notag \\
&&+\int_{1-\lambda ^{\frac{1}{\alpha }}}^{1}\left( 1-t\right) \left( \lambda
-\left( 1-t\right) ^{\alpha }\right) \left[ t\left\vert f^{\prime \prime
}\left( a\right) \right\vert ^{q}+\left( 1-t\right) \left\vert f^{\prime
\prime }\left( b\right) \right\vert ^{q}\right] dt  \notag \\
&&  \notag \\
&=&\left\vert f^{\prime \prime }\left( a\right) \right\vert ^{q}\left[ \frac{%
\alpha \lambda ^{1+\frac{2}{\alpha }}}{\alpha +2}-\frac{2\alpha \lambda ^{1+%
\frac{3}{\alpha }}}{3\left( \alpha +3\right) }+\frac{\alpha +4}{2^{\alpha
+3}\left( \alpha +2\right) \left( \alpha +3\right) }-\frac{\lambda }{12}%
\right]   \notag \\
&&  \notag \\
&&+\left\vert f^{\prime \prime }\left( b\right) \right\vert ^{q}\left[ \frac{%
2\alpha \lambda ^{1+\frac{3}{\alpha }}}{3\left( \alpha +3\right) }+\frac{1}{%
2^{\alpha +3}\left( \alpha +3\right) }-\frac{\lambda }{24}\right] ,  \notag
\end{eqnarray}%
\begin{equation}
\int_{0}^{\frac{1}{2}}\left\vert t\left( t^{\alpha }-\lambda \right)
\right\vert dt=\int_{0}^{\lambda ^{\frac{1}{\alpha }}}t\left( \lambda
-t^{\alpha }\right) dt+\int_{\lambda ^{\frac{1}{\alpha }}}^{\frac{1}{2}%
}t\left( t^{\alpha }-\lambda \right) dt=\frac{\alpha \lambda ^{1+\frac{2}{%
\alpha }}}{\alpha +2}+\frac{1}{2^{\alpha +2}\left( \alpha +2\right) }-\frac{%
\lambda }{8}  \label{2.15}
\end{equation}%
and%
\begin{eqnarray}
&&\int_{\frac{1}{2}}^{1}\left\vert \left( 1-t\right) \left( \left(
1-t\right) ^{\alpha }-\lambda \right) \right\vert dt  \label{2.16} \\
&&  \notag \\
&=&\int_{\frac{1}{2}}^{1-\lambda ^{\frac{1}{\alpha }}}\left( 1-t\right)
\left( \left( 1-t\right) ^{\alpha }-\lambda \right) dt+\int_{1-\lambda ^{%
\frac{1}{\alpha }}}^{1}\left( 1-t\right) \left( \lambda -\left( 1-t\right)
^{\alpha }\right) dt  \notag \\
&&  \notag \\
&=&\frac{\alpha \lambda ^{1+\frac{2}{\alpha }}}{\alpha +2}+\frac{1}{%
2^{\alpha +2}\left( \alpha +2\right) }-\frac{\lambda }{8}.  \notag
\end{eqnarray}%
Substituting the equalities (\ref{2.13})-(\ref{2.16}) in (\ref{2.12}), the
we obtain the inequality (\ref{h1}). One can prove the inequality (\ref{h2})
similar to (\ref{h1}). It is omited to readers.
\end{proof}

\begin{remark}
Under the assumptions Theorem \ref{t2} with $\alpha =1$, then Theorem \ref%
{t2} reduces to Theorem 4 in \cite{sarikaya2}.
\end{remark}

\begin{remark}
Under the assumptions of Theorem \ref{t2}  with $\lambda =\frac{1}{3}$ and $%
\alpha =1,$ then we get the following inequality%
\begin{eqnarray*}
&&\left\vert \frac{1}{6}\left[ f\left( a\right) +4f\left( \frac{a+b}{2}%
\right) +f\left( b\right) \right] -\frac{1}{b-a}\dint\limits_{a}^{b}f(t)dt%
\right\vert  \\
&& \\
&\leq &\frac{\left( b-a\right) ^{2}}{162}\left[ \left( \frac{59\left\vert
f^{\prime \prime }\left( a\right) \right\vert ^{q}+133\left\vert f^{\prime
\prime }\left( b\right) \right\vert ^{q}}{2^{6}\times 3}\right) ^{\frac{1}{q}%
}+\left( \frac{133\left\vert f^{\prime \prime }\left( a\right) \right\vert
^{q}+59\left\vert f^{\prime \prime }\left( b\right) \right\vert ^{q}}{%
2^{6}\times 3}\right) ^{\frac{1}{q}}\right] 
\end{eqnarray*}%
which is given by Sarikaya and Aktan in \cite{sarikaya2}.
\end{remark}

\begin{corollary}
Under the assumptions Theorem \ref{t2} with $\lambda =0$, then we get the
following inequality%
\begin{eqnarray}
&&\left\vert \frac{2^{\alpha -1}\Gamma \left( \alpha +1\right) }{\left(
b-a\right) ^{\alpha }}\left( J_{\left( \frac{a+b}{2}\right) ^{+}}^{\alpha
}f\left( b\right) +J_{\left( \frac{a+b}{2}\right) ^{-}}^{\alpha }f\left(
a\right) \right) -f\left( \frac{a+b}{2}\right) \right\vert   \label{2.17} \\
&&  \notag \\
&\leq &\frac{\left( b-a\right) ^{2}2^{\alpha -1}}{\alpha +1}\left\{ \left( 
\frac{1}{2^{\alpha +2}\left( \alpha +2\right) }\right) ^{1-\frac{1}{q}%
}\right.   \notag \\
&&  \notag \\
&&\times \left[ \left\vert f^{\prime \prime }\left( a\right) \right\vert ^{q}%
\frac{1}{2^{\alpha +3}\left( \alpha +3\right) }+\left\vert f^{\prime \prime
}\left( b\right) \right\vert ^{q}\frac{\alpha +4}{2^{\alpha +3}\left( \alpha
+2\right) \left( \alpha +3\right) }\right] ^{\frac{1}{q}}  \notag \\
&&  \notag \\
&&+\left. \left[ \left\vert f^{\prime \prime }\left( a\right) \right\vert
^{q}\frac{\alpha +4}{2^{\alpha +3}\left( \alpha +2\right) \left( \alpha
+3\right) }+\left\vert f^{\prime \prime }\left( b\right) \right\vert ^{q}%
\frac{1}{2^{\alpha +3}\left( \alpha +3\right) }\right] ^{\frac{1}{q}%
}\right\} .  \notag
\end{eqnarray}
\end{corollary}

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