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\begin{document}
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\begin{center}{\footnotesize Khayyam J. Math. 1 (2015), no. 1, 1--35}\\\end{center}
\noindent\parbox{2.85cm}{\includegraphics*[keepaspectratio=true,scale=0.24]{KJM.jpg}}
\vspace{0.5cm}

\title[A Survey on Ostrowski Type Inequalities]{A Survey on Ostrowski Type
Inequalities Related to Pompeiu's Mean Value Theorem}

\author[S.S. Dragomir]{Silvestru S. Dragomir}


\address{Mathematics, College of Engineering \& Science\\
Victoria University, P.O. Box 14428\\
Melbourne City, MC 8001, Australia.
\newline
School of Computational \& Applied Mathematics, University of
the Witwatersrand, Private Bag 3, Johannesburg 2050, South Africa.}
\email{sever.dragomir@vu.edu.au}
\urladdr{http://rgmia.org/dragomir}
\dedicatory{{\rm Communicated by M.S. Moslehian}}
\subjclass[2010]{Primary 26D10;  Secondary 26D15}
\keywords{Ostrowski inequality, Pompeiu's mean inequality, integral
inequalities, special means.}

\date{Received: 18 June 2014;  Accepted: 18 July 2014.}


\begin{abstract}
In this paper we survey some recent results obtained by the author related
to Pompeiu's mean value theorem and inequality. Natural applications to
Ostrowski type inequalities that play an important role in Numerical
Analysis, Approximation Theory, Probability Theory \& Statistics,
Information Theory and other fields, are given as well.
\end{abstract}
\maketitle


\section{Introduction}
\noindent In 1946, Pompeiu \cite{opaDP} derived a variant of Lagrange's mean value
theorem, now known as \textit{Pompeiu's mean value theorem} (see also \cite[%
p. 83]{opaPKSTR}).

\begin{theorem}[Pompeiu, 1946 \protect\cite{opaDP}]
\label{l.op.1}For every real valued function $f$ differentiable on an
interval $\left[ a,b\right] $ not containing $0$ and for all pairs $%
x_{1}\neq x_{2}$ in $\left[ a,b\right] ,$ there exists a point $\xi $
between $x_{1}$ and $x_{2}$ such that
\begin{equation}
\frac{x_{1}f\left( x_{2}\right) -x_{2}f\left( x_{1}\right) }{x_{1}-x_{2}}%
=f\left( \xi \right) -\xi f^{\prime }\left( \xi \right) .  \label{e.op.1}
\end{equation}
\end{theorem}

Following \cite[p. 84 -- 85]{opaPKSTR}, we will mention here a geometrical
interpretation of Pompeiu's theorem.

The equation of the secant line joining the points $\left( x_{1},f\left(
x_{1}\right) \right) $ and $\left( x_{2},f\left( x_{2}\right) \right) $ is
given by
\begin{equation*}
y=f\left( x_{1}\right) +\frac{f\left( x_{2}\right) -f\left( x_{1}\right) }{%
x_{2}-x_{1}}\left( x-x_{1}\right) .
\end{equation*}%
This line intersects the $y-$axis at the point $\left( 0,y\right) ,$ where $%
y $ is
\begin{align*}
y& =f\left( x_{1}\right) +\frac{f\left( x_{2}\right) -f\left( x_{1}\right) }{%
x_{2}-x_{1}}\left( 0-x_{1}\right) \\
& =\frac{x_{1}f\left( x_{2}\right) -x_{2}f\left( x_{1}\right) }{x_{1}-x_{2}}.
\end{align*}%
The equation of the tangent line at the point $\left( \xi ,f\left( \xi
\right) \right) $ is
\begin{equation*}
y=\left( x-\xi \right) f^{\prime }\left( \xi \right) +f\left( \xi \right) .
\end{equation*}%
The tangent line intersects the $y-$axis at the point $\left( 0,y\right) ,$
where
\begin{equation*}
y=-\xi f^{\prime }\left( \xi \right) +f\left( \xi \right) .
\end{equation*}%
Hence, the geometric meaning of Pompeiu's mean value theorem is that \textit{%
the tangent of the point }$\left( \xi ,f\left( \xi \right) \right) $\textit{%
\ intersects on the }$y-$\textit{axis at the same point as the secant line
connecting the points }$\left( x_{1},f\left( x_{1}\right) \right) $\textit{\
and }$\left( x_{2},f\left( x_{2}\right) \right) .$

The following inequality is a simple consequence of \textit{Pompeiu's mean
value theorem.}

\begin{corollary}[Pompeiu's Inequality]
\label{c.opa.1}With the assumptions of Theorem \ref{l.op.1} and if $%
\left\Vert f-\ell f^{\prime }\right\Vert _{\infty }=\sup_{t\in \left(
a,b\right) }\left\vert f\left( t\right) -tf^{\prime }\left( t\right)
\right\vert <\infty $ where $\ell \left( t\right) =t,$ $t\in \left[ a,b%
\right] ,$ then
\begin{equation}
\left\vert tf\left( x\right) -xf\left( t\right) \right\vert \leq \left\Vert
f-\ell f^{\prime }\right\Vert _{\infty }\left\vert x-t\right\vert
\label{e.opa.2}
\end{equation}%
for any $t,x\in \left[ a,b\right] .$
\end{corollary}

The inequality (\ref{e.opa.2}) was obtained by the author in \cite{opaSSD},
see also \cite{opaSSD0}.

In 1938, A. Ostrowski \cite{ompAO} proved the following result in the
estimating the integral mean:

\begin{theorem}[Ostrowski, 1938 \protect\cite{ompAO}]
\label{t.op.3.0}Let $f:\left[ a,b\right] \rightarrow \mathbb{R}$ be
continuous on $\left[ a,b\right] $ and differentiable on $\left( a,b\right) $
with $\left\vert f^{\prime }\left( t\right) \right\vert \leq M<\infty $ for
all $t\in \left( a,b\right) .$ Then for any $x\in \left[ a,b\right] ,$ we
have the inequality%
\begin{equation}
\left\vert f\left( x\right) -\frac{1}{b-a}\int_{a}^{b}f\left( t\right)
dt\right\vert \leq \left[ \frac{1}{4}+\left( \frac{x-\frac{a+b}{2}}{b-a}%
\right) ^{2}\right] M\left( b-a\right) .  \label{e.op.3.0}
\end{equation}%
The constant $\frac{1}{4}$ is best possible in the sense that it cannot be
replaced by a smaller quantity.
\end{theorem}

In order to provide another approximation of the integral mean, by making
use of the Pompeiu's mean value theorem, the author proved the following
result:

\begin{theorem}[Dragomir, 2005 \protect\cite{opaSSD}]
\label{t.op.3.1}Let $f:\left[ a,b\right] \rightarrow \mathbb{R}$ be
continuous on $\left[ a,b\right] $ and differentiable on $\left( a,b\right) $
with $\left[ a,b\right] $ not containing $0.$ Then for any $x\in \left[ a,b%
\right] ,$ we have the inequality
\begin{align}
\left\vert \frac{a+b}{2}\cdot \frac{f\left( x\right) }{x}-\frac{1}{b-a}%
\int_{a}^{b}f\left( t\right) dt\right\vert & \leq \frac{b-a}{\left\vert
x\right\vert }\left[ \frac{1}{4}+\left( \frac{x-\frac{a+b}{2}}{b-a}\right)
^{2}\right]  \label{e.op.3.1} \\
& \times \left\Vert f-\ell f^{\prime }\right\Vert _{\infty },  \notag
\end{align}%
where $\ell \left( t\right) =t,$ $t\in \left[ a,b\right] .$

The constant $\frac{1}{4}$ is sharp in the sense that it cannot be replaced
by a smaller constant.
\end{theorem}

In \cite{ompECP}, E. C. Popa using a mean value theorem obtained a
generalization of (\ref{e.op.3.1}) as follows:

\begin{theorem}[Popa, 2007 \protect\cite{ompECP}]
\label{t.op.3.2}Let $f:\left[ a,b\right] \rightarrow \mathbb{R}$ be
continuous on $\left[ a,b\right] $ and differentiable on $\left( a,b\right)
. $ Assume that $\alpha \notin \left[ a,b\right] .$ Then for any $x\in \left[
a,b\right] ,$ we have the inequality
\begin{align}
& \left\vert \left( \frac{a+b}{2}-\alpha \right) f\left( x\right) +\frac{%
\alpha -x}{b-a}\int_{a}^{b}f\left( t\right) dt\right\vert  \label{e.op.3.2}
\\
& \leq \left[ \frac{1}{4}+\left( \frac{x-\frac{a+b}{2}}{b-a}\right) ^{2}%
\right] \left( b-a\right) \left\Vert f-\ell _{\alpha }f^{\prime }\right\Vert
_{\infty },  \notag
\end{align}%
where $\ell _{\alpha }\left( t\right) =t-\alpha ,$ $t\in \left[ a,b\right] .$
\end{theorem}

In \ \cite{ompJPSU}, J. Pe\v{c}ari\'{c} and S. Ungar have proved a general
estimate with the $p$-norm, $1\leq p\leq \infty $ which for $p=\infty $ give
Dragomir's result.

\begin{theorem}[Pe\v{c}ari\'{c} \& Ungar, 2006 \protect\cite{ompJPSU}]
\label{t.op.3.3}Let $f:\left[ a,b\right] \rightarrow \mathbb{R}$ be
continuous on $\left[ a,b\right] $ and differentiable on $\left( a,b\right) $
with $0<a<b.$ Then for $1\leq p,q\leq \infty $ with $\frac{1}{p}+\frac{1}{q}%
=1$ we have the inequality
\begin{equation}
\left\vert \frac{a+b}{2}\cdot \frac{f\left( x\right) }{x}-\frac{1}{b-a}%
\int_{a}^{b}f\left( t\right) dt\right\vert \leq PU\left( x,p\right)
\left\Vert f-\ell f^{\prime }\right\Vert _{p},  \label{e.op.3.3}
\end{equation}%
for $x\in \left[ a,b\right] ,$ where
\begin{eqnarray*}
PU\left( x,p\right) &:&=\left( b-a\right) ^{\frac{1}{p}-1}\left[ \left(
\frac{a^{2-q}-x^{2-q}}{\left( 1-2q\right) \left( 2-q\right) }+\frac{%
x^{2-q}-a^{1+q}x^{1-2q}}{\left( 1-2q\right) \left( 1+q\right) }\right)
^{1/q}\right. \\
&&\left. +\left( \frac{b^{2-q}-x^{2-q}}{\left( 1-2q\right) \left( 2-q\right)
}+\frac{x^{2-q}-b^{1+q}x^{1-2q}}{\left( 1-2q\right) \left( 1+q\right) }%
\right) ^{1/q}\right] .
\end{eqnarray*}%
In the cases $\left( p,q\right) =\left( 1,\infty \right) ,\left( \infty
,1\right) $ and $\left( 2,2\right) $ the quantity $PU\left( x,p\right) $ has
to be taken as the limit as $p\rightarrow 1,\infty $ and $2,$ respectively.
\end{theorem}

For other inequalities in terms of the $p$-norm of the quantity $f-\ell
_{\alpha }f^{\prime },$ where $\ell _{\alpha }\left( t\right) =t-\alpha ,$ $%
t\in \left[ a,b\right] $ and $\alpha \notin \left[ a,b\right] $ see \cite%
{ompAMAFS} and \cite{ompAMAABFS}.

In this paper we survey some recent results obtained by the author related
to Pompeiu's inequality presented above. Natural applications to Ostrowski
type inequalities that play an important role in Numerical Analysis,
Approximation Theory, Probability Theory \& Statistics, Information Theory
and other fields are given as well.
\section{Ostrowski Via a Generalized Pompeiu's Inequality}

\subsection{Pompeiu's Inequality for $p$-Norms}

We can generalize the above inequality (\ref{e.opa.2}) for the larger class
of functions that are absolutely continuous and complex valued as well as
for other norms of the difference $f-\ell f^{\prime }$ as follows:

\begin{lemma}[Dragomir, 2013 \protect\cite{opaSSD1}]
\label{l.opa.2}Let $f:\left[ a,b\right] \rightarrow \mathbb{C}$ be an
absolutely continuous function on the interval $\left[ a,b\right] $ with $%
b>a>$ $0.$ Then for any $t,x\in \left[ a,b\right] $ we have%
\begin{align}
& \left\vert tf\left( x\right) -xf\left( t\right) \right\vert
\label{e.opa.3} \\
& \leq \left\{
\begin{array}{ll}
\left\Vert f-\ell f^{\prime }\right\Vert _{\infty }\left\vert x-t\right\vert
& \text{if }f-\ell f^{\prime }\in L_{\infty }\left[ a,b\right] \\
&  \\
\frac{1}{2q-1}\left\Vert f-\ell f^{\prime }\right\Vert _{p}\left\vert \frac{%
x^{q}}{t^{q-1}}-\frac{t^{q}}{x^{q-1}}\right\vert ^{1/q} &
\begin{array}{l}
\text{if }f-\ell f^{\prime }\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f-\ell f^{\prime }\right\Vert _{1}\frac{\max \left\{ t,x\right\}
}{\min \left\{ t,x\right\} } &
\end{array}%
\right.  \notag
\end{align}%
or, equivalently%
\begin{align}
& \left\vert \frac{f\left( x\right) }{x}-\frac{f\left( t\right) }{t}%
\right\vert  \label{e.opa.3.a} \\
& \leq \left\{
\begin{array}{ll}
\left\Vert f-\ell f^{\prime }\right\Vert _{\infty }\left\vert \frac{1}{t}-%
\frac{1}{x}\right\vert & \text{if }f-\ell f^{\prime }\in L_{\infty }\left[
a,b\right] \\
&  \\
\frac{1}{2q-1}\left\Vert f-\ell f^{\prime }\right\Vert _{p}\left\vert \frac{1%
}{t^{2q-1}}-\frac{1}{x^{2q-1}}\right\vert ^{1/q} &
\begin{array}{c}
\text{if }f-\ell f^{\prime }\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f-\ell f^{\prime }\right\Vert _{1}\frac{1}{\min \left\{
t^{2},x^{2}\right\} } &
\end{array}%
\right.  \notag
\end{align}
\end{lemma}

\begin{proof}
If $f$ is absolutely continuous, then $f/\ell $ is absolutely continuous on
the interval $\left[ a,b\right] $ that does not containing $0$ and
\begin{equation*}
\int_{t}^{x}\left( \frac{f\left( s\right) }{s}\right) ^{\prime }ds=\frac{%
f\left( x\right) }{x}-\frac{f\left( t\right) }{t}
\end{equation*}%
for any $t,x\in \left[ a,b\right] $ with $x\neq t.$

Since%
\begin{equation*}
\int_{t}^{x}\left( \frac{f\left( s\right) }{s}\right) ^{\prime
}ds=\int_{t}^{x}\frac{f^{\prime }\left( s\right) s-f\left( s\right) }{s^{2}}%
ds
\end{equation*}%
then we get the following identity%
\begin{equation}
tf\left( x\right) -xf\left( t\right) =xt\int_{t}^{x}\frac{f^{\prime }\left(
s\right) s-f\left( s\right) }{s^{2}}ds  \label{e.opa.4}
\end{equation}%
for any $t,x\in \left[ a,b\right] .$

We notice that the equality (\ref{e.opa.4}) was proved for the smaller class
of differentiable function and in a different manner in \cite{ompJPSU}.

Taking the modulus in (\ref{e.opa.4}) we have%
\begin{eqnarray}
\left\vert tf\left( x\right) -xf\left( t\right) \right\vert &=&\left\vert
xt\int_{t}^{x}\frac{f^{\prime }\left( s\right) s-f\left( s\right) }{s^{2}}%
ds\right\vert  \label{e.opa.5} \\
&\leq &xt\left\vert \int_{t}^{x}\left\vert \frac{f^{\prime }\left( s\right)
s-f\left( s\right) }{s^{2}}\right\vert ds\right\vert :=I  \notag
\end{eqnarray}%
and utilizing H\"{o}lder's integral inequality we deduce%
\begin{eqnarray}
I &\leq &xt\left\{
\begin{array}{ll}
\sup_{s\in \left[ t,x\right] \left( \left[ x,t\right] \right) }\left\vert
f^{\prime }\left( s\right) s-f\left( s\right) \right\vert \left\vert
\int_{t}^{x}\frac{1}{s^{2}}ds\right\vert &  \\
&  \\
\left\vert \int_{t}^{x}\left\vert f^{\prime }\left( s\right) s-f\left(
s\right) \right\vert ^{p}ds\right\vert ^{1/p}\left\vert \int_{t}^{x}\frac{1}{%
s^{2q}}ds\right\vert ^{1/q} &
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1%
\end{array}
\\
&  \\
\left\vert \int_{t}^{x}\left\vert f^{\prime }\left( s\right) s-f\left(
s\right) \right\vert ds\right\vert \sup_{s\in \left[ t,x\right] \left( \left[
x,t\right] \right) }\left\{ \frac{1}{s^{2}}\right\} &
\end{array}%
\right.  \label{e.opa.6} \\
&\leq &\left\{
\begin{array}{ll}
\left\Vert f-\ell f^{\prime }\right\Vert _{\infty }\left\vert x-t\right\vert
&  \\
&  \\
\frac{1}{2q-1}\left\Vert f-\ell f^{\prime }\right\Vert _{p}\left\vert \frac{%
x^{q}}{t^{q-1}}-\frac{t^{q}}{x^{q-1}}\right\vert ^{1/q} &
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1%
\end{array}
\\
&  \\
\left\Vert f-\ell f^{\prime }\right\Vert _{1}\frac{\max \left\{ t,x\right\}
}{\min \left\{ t,x\right\} } &
\end{array}%
\right.  \notag
\end{eqnarray}%
and the inequality (\ref{e.opa.3.a}) is proved.
\end{proof}

\begin{remark}
\label{r.opa.2}The first inequality in (\ref{e.opa.3}) also holds in the
same form for $0>b>a.$
\end{remark}

\begin{remark}
\label{r.opa.3}If we take in (\ref{e.opa.3}) $x=A=A\left( a,b\right) :=\frac{%
a+b}{2}$ (the arithmetic mean) and $t=G=G\left( a,b\right) :=\sqrt{ab}$ (the
geometric mean) then we get the simple inequality for functions of means:
\begin{align}
& \left\vert Gf\left( A\right) -Af\left( G\right) \right\vert
\label{e.opa.6.a} \\
& \leq \left\{
\begin{array}{ll}
\left\Vert f-\ell f^{\prime }\right\Vert _{\infty }\left( A-G\right) & \text{%
if }f-\ell f^{\prime }\in L_{\infty }\left[ a,b\right] \\
&  \\
\frac{1}{2q-1}\left\Vert f-\ell f^{\prime }\right\Vert _{p}\frac{\left(
A^{2q-1}-G^{2q-1}\right) ^{1/q}}{A^{1/p}G^{1/p}} &
\begin{array}{l}
\text{if }f-\ell f^{\prime }\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f-\ell f^{\prime }\right\Vert _{1}\frac{A}{G} &
\end{array}%
\right.  \notag
\end{align}
\end{remark}

\subsection{Evaluating the Integral Mean}

The following new result holds.

\begin{theorem}[Dragomir, 2013 \protect\cite{opaSSD1}]
\label{t.opa.3.1}Let $f:\left[ a,b\right] \rightarrow \mathbb{C}$ be an
absolutely continuous function on the interval $\left[ a,b\right] $ with $%
b>a>$ $0.$ Then for any $x\in \left[ a,b\right] $ we have
\begin{align}
& \left\vert \frac{a+b}{2}\cdot \frac{f\left( x\right) }{x}-\frac{1}{b-a}%
\int_{a}^{b}f\left( t\right) dt\right\vert  \label{e.opa.3.1} \\
& \leq \left\{
\begin{array}{ll}
\frac{b-a}{x}\left[ \frac{1}{4}+\left( \frac{x-\frac{a+b}{2}}{b-a}\right)
^{2}\right] \left\Vert f-\ell f^{\prime }\right\Vert _{\infty } & \text{if }%
f-\ell f^{\prime }\in L_{\infty }\left[ a,b\right] \\
&  \\
\frac{1}{\left( 2q-1\right) x\left( b-a\right) ^{1/q}}\left\Vert f-\ell
f^{\prime }\right\Vert _{p}\left[ B_{q}\left( a,b;x\right) \right] ^{1/q} &
\begin{array}{l}
\text{if }f-\ell f^{\prime }\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1%
\end{array}%
\end{array}
\\
&  \\
\frac{1}{b-a}\left\Vert f-\ell f^{\prime }\right\Vert _{1}\left( \ln \frac{x%
}{a}+\frac{b^{2}-x^{2}}{2x^{2}}\right) , &
\end{array}%
\right. ,  \notag
\end{align}%
where
\begin{equation}
B_{q}\left( a,b;x\right) =\left\{
\begin{array}{ll}
\begin{array}{l}
\frac{x^{q}}{2-q}\left( 2x^{q-2}-a^{q-2}-b^{q-2}\right) \\
\\
+\frac{1}{x^{q-1}\left( q+1\right) }\left( b^{q+1}+a^{q+1}-2x^{q+1}\right) ,%
\end{array}
& q\neq 2 \\
&  \\
x^{2}\ln \frac{x^{2}}{ab}+\frac{b^{3}+a^{3}-2x^{3}}{3x}, & q=2%
\end{array}%
\right.  \label{e.opa.3.1.a}
\end{equation}
\end{theorem}

\begin{proof}
The first inequality can be proved in an identical way to the case of
differentiable functions from \cite{opaSSD} by utilizing the first
inequality in (\ref{e.opa.3}).

Utilising the second inequality in (\ref{e.opa.3}) we have%
\begin{align}
& \left\vert \frac{a+b}{2}\cdot f\left( x\right) -\frac{x}{b-a}%
\int_{a}^{b}f\left( t\right) dt\right\vert  \label{e.opa.3.2} \\
& \leq \frac{1}{b-a}\int_{a}^{b}\left\vert tf\left( x\right) -xf\left(
t\right) \right\vert dt  \notag \\
& \leq \frac{1}{\left( 2q-1\right) \left( b-a\right) }\left\Vert f-\ell
f^{\prime }\right\Vert _{p}\int_{a}^{b}\left\vert \frac{x^{q}}{t^{q-1}}-%
\frac{t^{q}}{x^{q-1}}\right\vert ^{1/q}dt  \notag
\end{align}%
Utilising H\"{o}lder's integral inequality we have%
\begin{equation}
\int_{a}^{b}\left\vert \frac{x^{q}}{t^{q-1}}-\frac{t^{q}}{x^{q-1}}%
\right\vert ^{1/q}dt\leq \left( b-a\right) ^{1/p}\left(
\int_{a}^{b}\left\vert \frac{x^{q}}{t^{q-1}}-\frac{t^{q}}{x^{q-1}}%
\right\vert dt\right) ^{1/q}.  \label{e.opa.3.3}
\end{equation}%
For $q\neq 2$ we have

\begin{align*}
& \int_{a}^{b}\left\vert \frac{x^{q}}{t^{q-1}}-\frac{t^{q}}{x^{q-1}}%
\right\vert dt \\
& =\frac{x^{q}}{2-q}\left( 2x^{q-2}-a^{q-2}-b^{q-2}\right) +\frac{1}{%
x^{q-1}\left( q+1\right) }\left( b^{q+1}+a^{q+1}-2x^{q+1}\right) \\
& =B_{q}\left( a,b;x\right) .
\end{align*}%
For $q=2$ we have
\begin{equation*}
\int_{a}^{b}\left\vert \frac{x^{2}}{t}-\frac{t^{2}}{x}\right\vert
dt=x^{2}\ln \frac{x^{2}}{ab}+\frac{1}{x}\frac{b^{3}+a^{3}-2x^{3}}{3}%
=B_{2}\left( a,b;x\right) .
\end{equation*}%
Utilizing (\ref{e.opa.3.2}) and (\ref{e.opa.3.3}) we get the second
inequality in (\ref{e.opa.3.1}).

Utilising the third inequality in (\ref{e.opa.3}) we have%
\begin{align}
\left\vert \frac{a+b}{2}\cdot f\left( x\right) -\frac{x}{b-a}%
\int_{a}^{b}f\left( t\right) dt\right\vert & \leq \frac{1}{b-a}%
\int_{a}^{b}\left\vert tf\left( x\right) -xf\left( t\right) \right\vert dt
\label{e.opa.3.4} \\
& \leq \frac{1}{b-a}\left\Vert f-\ell f^{\prime }\right\Vert _{1}\int_{a}^{b}%
\frac{\max \left\{ t,x\right\} }{\min \left\{ t,x\right\} }dt.  \notag
\end{align}%
Since%
\begin{equation*}
\int_{a}^{b}\frac{\max \left\{ t,x\right\} }{\min \left\{ t,x\right\} }%
dt=\int_{a}^{x}\frac{x}{t}dt+\int_{x}^{b}\frac{t}{x}dt=x\ln \frac{x}{a}+%
\frac{1}{x}\frac{b^{2}-x^{2}}{2},
\end{equation*}%
then by (\ref{e.opa.3.4}) we have%
\begin{align*}
\left\vert \frac{a+b}{2}\cdot f\left( x\right) -\frac{x}{b-a}%
\int_{a}^{b}f\left( t\right) dt\right\vert & \leq \frac{1}{b-a}%
\int_{a}^{b}\left\vert tf\left( x\right) -xf\left( t\right) \right\vert dt \\
& \leq \frac{1}{b-a}\left\Vert f-\ell f^{\prime }\right\Vert _{1}\left[ x\ln
\frac{x}{a}+\frac{1}{x}\frac{b^{2}-x^{2}}{2}\right] ,
\end{align*}%
and the last part of (\ref{e.opa.3.1}) is thus proved.
\end{proof}

\begin{remark}
\label{r.opa.4}If we take in (\ref{e.opa.3.1}) $x=A=A\left( a,b\right) :=%
\frac{a+b}{2}$, then we get
\begin{align}
& \left\vert f\left( A\right) -\frac{1}{b-a}\int_{a}^{b}f\left( t\right)
dt\right\vert  \label{e.opa.3.4.a} \\
& \leq \left\{
\begin{array}{ll}
\frac{b-a}{4A}\left\Vert f-\ell f^{\prime }\right\Vert _{\infty } & \text{if
}f-\ell f^{\prime }\in L_{\infty }\left[ a,b\right] \\
&  \\
\frac{1}{\left( 2q-1\right) A\left( b-a\right) ^{1/q}}\left\Vert f-\ell
f^{\prime }\right\Vert _{p}\left[ B_{q}\left( a,b;A\right) \right] ^{1/q} &
\begin{array}{l}
\text{if }f-\ell f^{\prime }\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1%
\end{array}%
\end{array}
\\
&  \\
\frac{1}{b-a}\left\Vert f-\ell f^{\prime }\right\Vert _{1}\left( \ln \frac{A%
}{a}+\frac{1}{2}\left( b-a\right) \frac{a+3b}{4}A\right) , &
\end{array}%
\right. ,  \notag
\end{align}%
where
\begin{equation*}
B_{q}\left( a,b;A\right) =\left\{
\begin{array}{ll}
\begin{array}{l}
\frac{2A^{q}}{2-q}\left( A^{q-2}-A\left( a^{q-2},b^{q-2}\right) \right) \\
\\
+\frac{2}{\left( q+1\right) A^{q-1}}\left( A\left( b^{q+1},a^{q+1}\right)
-A^{q+1}\right) ,%
\end{array}
& q\neq 2 \\
&  \\
2A^{2}\ln \frac{A}{G}+\frac{1}{2}\left( b-a\right) ^{2}, & q=2%
\end{array}%
\right.
\end{equation*}
\end{remark}

\subsection{A Related Result}

The following new result also holds.

\begin{theorem}[Dragomir, 2013 \protect\cite{opaSSD1}]
\label{t.opa.3.2}Let $f:\left[ a,b\right] \rightarrow \mathbb{C}$ be an
absolutely continuous function on the interval $\left[ a,b\right] $ with $%
b>a>$ $0.$ Then for any $x\in \left[ a,b\right] $ we have
\begin{align}
& \left\vert \frac{f\left( x\right) }{x}-\frac{1}{b-a}\int_{a}^{b}\frac{%
f\left( t\right) }{t}dt\right\vert  \label{e.opa.4.1} \\
& \leq \left\{
\begin{array}{ll}
\frac{2}{b-a}\left\Vert f-\ell f^{\prime }\right\Vert _{\infty }\left( \ln
\frac{x}{\sqrt{ab}}+\frac{\frac{a+b}{2}-x}{x}\right) & \text{if }f-\ell
f^{\prime }\in L_{\infty }\left[ a,b\right] \\
&  \\
\frac{1}{\left( 2q-1\right) \left( b-a\right) ^{1/q}}\left\Vert f-\ell
f^{\prime }\right\Vert _{p}\left( C_{q}\left( a,b;x\right) \right) ^{1/q} &
\begin{array}{l}
\text{if }f-\ell f^{\prime }\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1%
\end{array}%
\end{array}
\\
&  \\
\frac{1}{b-a}\left\Vert f-\ell f^{\prime }\right\Vert _{1}\frac{x^{2}+ab-2ax%
}{x^{2}a}, &
\end{array}%
\right. ,  \notag
\end{align}%
where
\begin{equation}
C_{q}\left( a,b;x\right) =\frac{1}{x^{2q-1}}\left( b+a-2x\right) +\frac{%
a^{2-2q}+b^{2-2q}-2x^{2-2q}}{2\left( q-1\right) },q>1.  \label{e.opa.4.2}
\end{equation}
\end{theorem}

\begin{proof}
From the first inequality in (\ref{e.opa.3.1.a}) we have%
\begin{align}
\left\vert \frac{f\left( x\right) }{x}-\frac{1}{b-a}\int_{a}^{b}\frac{%
f\left( t\right) }{t}dt\right\vert & \leq \frac{1}{b-a}\int_{a}^{b}\left%
\vert \frac{f\left( x\right) }{x}-\frac{f\left( t\right) }{t}\right\vert dt
\label{e.opa.4.3} \\
& \leq \left\Vert f-\ell f^{\prime }\right\Vert _{\infty }\frac{1}{b-a}%
\int_{a}^{b}\left\vert \frac{1}{t}-\frac{1}{x}\right\vert dt.  \notag
\end{align}%
Since%
\begin{align*}
\int_{a}^{b}\left\vert \frac{1}{x}-\frac{1}{t}\right\vert dt& =\left[
\int_{a}^{x}\left( \frac{1}{t}-\frac{1}{x}\right) dt+\int_{x}^{b}\left(
\frac{1}{x}-\frac{1}{t}\right) dt\right] \\
& =\left( \ln \frac{x^{2}}{ab}+\frac{a+b-2x}{x}\right)
\end{align*}%
for any $x\in \left[ a,b\right] ,$ then we deduce from (\ref{e.opa.4.3}) the
first inequality in (\ref{e.opa.4.1}).

From the second inequality in (\ref{e.opa.3.1.a}) we have
\begin{align}
& \left\vert \frac{f\left( x\right) }{x}-\frac{1}{b-a}\int_{a}^{b}\frac{%
f\left( t\right) }{t}dt\right\vert   \label{e.opa.4.4} \\
& \leq \frac{1}{b-a}\int_{a}^{b}\left\vert \frac{f\left( x\right) }{x}-\frac{%
f\left( t\right) }{t}\right\vert dt  \notag \\
& \leq \frac{1}{\left( 2q-1\right) \left( b-a\right) }\left\Vert f-\ell
f^{\prime }\right\Vert _{p}\int_{a}^{b}\left\vert \frac{1}{t^{2q-1}}-\frac{1%
}{x^{2q-1}}\right\vert ^{1/q}dt.  \notag
\end{align}%
Utilising H\"{o}lder's integral inequality we have%
\begin{equation}
\int_{a}^{b}\left\vert \frac{1}{t^{2q-1}}-\frac{1}{x^{2q-1}}\right\vert
^{1/q}dt\leq \left( b-a\right) ^{1/p}\left( \int_{a}^{b}\left\vert \frac{1}{%
t^{2q-1}}-\frac{1}{x^{2q-1}}\right\vert dt\right) ^{1/q}.  \label{e.opa.4.5}
\end{equation}%
Since%
\begin{align*}
& \int_{a}^{b}\left\vert \frac{1}{t^{2q-1}}-\frac{1}{x^{2q-1}}\right\vert dt
\\
& =\frac{1}{x^{2q-1}}\left( b+a-2x\right) +\frac{a^{2-2q}+b^{2-2q}-2x^{2-2q}%
}{2\left( q-1\right) }=C_{q}\left( a,b;x\right)
\end{align*}%
then by (\ref{e.opa.4.4}) and (\ref{e.opa.4.5}) we get%
\begin{align*}
& \left\vert \frac{f\left( x\right) }{x}-\frac{1}{b-a}\int_{a}^{b}\frac{%
f\left( t\right) }{t}dt\right\vert  \\
& \leq \frac{1}{\left( 2q-1\right) \left( b-a\right) }\left\Vert f-\ell
f^{\prime }\right\Vert _{p}\left( b-a\right) ^{1/p}\left( C_{q}\left(
a,b;x\right) \right) ^{1/q}
\end{align*}%
and the second inequality in (\ref{e.opa.4.1}) is proved.

From the third inequality in (\ref{e.opa.3.1.a}) we have%
\begin{align}
\left\vert \frac{f\left( x\right) }{x}-\frac{1}{b-a}\int_{a}^{b}\frac{%
f\left( t\right) }{t}dt\right\vert & \leq \frac{1}{b-a}\int_{a}^{b}\left%
\vert \frac{f\left( x\right) }{x}-\frac{f\left( t\right) }{t}\right\vert dt
\label{e.opa.3.6} \\
& \leq \frac{1}{b-a}\left\Vert f-\ell f^{\prime }\right\Vert _{1}\int_{a}^{b}%
\frac{1}{\min \left\{ t^{2},x^{2}\right\} }dt.  \notag
\end{align}%
Since%
\begin{equation*}
\int_{a}^{b}\frac{1}{\min \left\{ t^{2},x^{2}\right\} }dt=\int_{a}^{x}\frac{%
dt}{t^{2}}+\int_{x}^{b}\frac{dt}{x^{2}}=\frac{x^{2}+ab-2ax}{x^{2}a},
\end{equation*}%
then by (\ref{e.opa.3.6}) we deduce the last part of (\ref{e.opa.4.1}).
\end{proof}

\begin{remark}
\label{r.opa.5}If we take in (\ref{e.opa.4.1}) $x=A=A\left( a,b\right) :=%
\frac{a+b}{2}$, then we get
\begin{align}
& \left\vert \frac{f\left( A\right) }{A}-\frac{1}{b-a}\int_{a}^{b}\frac{%
f\left( t\right) }{t}dt\right\vert  \label{e.opa.3.7} \\
& \leq \left\{
\begin{array}{ll}
\frac{2}{b-a}\left\Vert f-\ell f^{\prime }\right\Vert _{\infty }\ln \left(
\frac{A}{G}\right) & \text{if }f-\ell f^{\prime }\in L_{\infty }\left[ a,b%
\right] \\
&  \\
\frac{1}{\left( 2q-1\right) \left( b-a\right) ^{1/q}}\left\Vert f-\ell
f^{\prime }\right\Vert _{p}\left( C_{q}\left( a,b;A\right) \right) ^{1/q} &
\begin{array}{l}
\text{if }f-\ell f^{\prime }\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1%
\end{array}%
\end{array}
\\
&  \\
\frac{1}{2}\left\Vert f-\ell f^{\prime }\right\Vert _{1}\frac{A+a}{A^{2}a},
&
\end{array}%
\right. ,  \notag
\end{align}%
where
\begin{equation*}
C_{q}\left( a,b;A\right) =\frac{A\left( a^{2-2q},b^{2-2q}\right) -A^{2-2q}}{%
q-1},q>1.
\end{equation*}
\end{remark}

\section{Ostrowski Via Power Pompeiu's Inequality}

\subsection{Power Pompeiu's Inequality}

We can generalize the above (\ref{e.opa.2}) inequality for the power
function as follows$.$

\begin{lemma}[Dragomir, 2013 \protect\cite{oppSSD1}]
\label{l.opp.2}Let $f:\left[ a,b\right] \rightarrow \mathbb{C}$ be an
absolutely continuous function on the interval $\left[ a,b\right] $ with $%
b>a>$ $0.$ If $r\in \mathbb{R}$, $r\neq 0,$ then for any $t,x\in \left[ a,b%
\right] $ we have%
\begin{align}
& \left\vert t^{r}f\left( x\right) -x^{r}f\left( t\right) \right\vert
\label{e.opp.3} \\
& \leq \left\{
\begin{array}{ll}
\frac{1}{\left\vert r\right\vert }\left\Vert f^{\prime }\ell -rf\right\Vert
_{\infty }\left\vert t^{r}-x^{r}\right\vert ,\text{if }f^{\prime }\ell
-rf\in L_{\infty }\left[ a,b\right] &  \\
&  \\
\begin{array}{l}
\left\Vert f^{\prime }\ell -rf\right\Vert _{p}\left\{
\begin{array}{l}
\frac{1}{\left\vert 1-q\left( r+1\right) \right\vert }\left\vert \frac{t^{r}%
}{x^{1-q\left( r+1\right) -r}}-\frac{x^{r}}{t^{1-q\left( r+1\right) -r}}%
\right\vert , \\
\text{ for }r\neq -\frac{1}{p} \\
\\
t^{r}x^{r}\left\vert \ln x-\ln t\right\vert ,\text{ for }r=-\frac{1}{p}%
\end{array}%
\right. \\
\text{if }f^{\prime }\ell -rf\in L_{p}\left[ a,b\right]%
\end{array}
&  \\
&  \\
\left\Vert f^{\prime }\ell -rf\right\Vert _{1}\frac{t^{r}x^{r}}{\min \left\{
x^{r+1},t^{r+1}\right\} } &
\end{array}%
\right.  \notag
\end{align}%
or, equivalently%
\begin{align}
& \left\vert \frac{f\left( x\right) }{x^{r}}-\frac{f\left( t\right) }{t^{r}}%
\right\vert  \label{e.opp.3.a} \\
& \leq \left\{
\begin{array}{ll}
\frac{1}{\left\vert r\right\vert }\left\Vert f^{\prime }\ell -rf\right\Vert
_{\infty }\left\vert \frac{1}{x^{r}}-\frac{1}{t^{r}}\right\vert ,\text{ if }%
f^{\prime }\ell -rf\in L_{\infty }\left[ a,b\right] &  \\
&  \\
\begin{array}{l}
\left\Vert f^{\prime }\ell -rf\right\Vert _{p}\left\{
\begin{array}{l}
\frac{1}{\left\vert 1-q\left( r+1\right) \right\vert }\left\vert \frac{1}{%
x^{1-q\left( r+1\right) }}-\frac{1}{t^{1-q\left( r+1\right) }}\right\vert ,%
\text{ for }r\neq -\frac{1}{p} \\
\\
\left\vert \ln x-\ln t\right\vert ,\text{ for }r=-\frac{1}{p}%
\end{array}%
\right. \\
\text{if }f^{\prime }\ell -rf\in L_{p}\left[ a,b\right]%
\end{array}
&  \\
&  \\
\left\Vert f^{\prime }\ell -rf\right\Vert _{1}\frac{1}{\min \left\{
x^{r+1},t^{r+1}\right\} } &
\end{array}%
\right.  \notag
\end{align}%
where $p>1,\frac{1}{p}+\frac{1}{q}=1.$
\end{lemma}

\begin{proof}
If $f$ is absolutely continuous, then $f/\left( \cdot \right) ^{r}$ is
absolutely continuous on the interval $\left[ a,b\right] $ and
\begin{equation*}
\int_{t}^{x}\left( \frac{f\left( s\right) }{s^{r}}\right) ^{\prime }ds=\frac{%
f\left( x\right) }{x^{r}}-\frac{f\left( t\right) }{t^{r}}
\end{equation*}%
for any $t,x\in \left[ a,b\right] $ with $x\neq t.$

Since%
\begin{equation*}
\int_{t}^{x}\left( \frac{f\left( s\right) }{s^{r}}\right) ^{\prime
}ds=\int_{t}^{x}\frac{f^{\prime }\left( s\right) s^{r}-rs^{r-1}f\left(
s\right) }{s^{2r}}ds=\int_{t}^{x}\frac{f^{\prime }\left( s\right) s-rf\left(
s\right) }{s^{r+1}}ds
\end{equation*}%
then we get the following identity%
\begin{equation}
t^{r}f\left( x\right) -x^{r}f\left( t\right) =x^{r}t^{r}\int_{t}^{x}\frac{%
f^{\prime }\left( s\right) s-rf\left( s\right) }{s^{r+1}}ds  \label{e.opp.4}
\end{equation}%
for any $t,x\in \left[ a,b\right] .$

Taking the modulus in (\ref{e.opp.4}) we have%
\begin{eqnarray}
\left\vert t^{r}f\left( x\right) -x^{r}f\left( t\right) \right\vert
&=&x^{r}t^{r}\left\vert \int_{t}^{x}\frac{f^{\prime }\left( s\right)
s-rf\left( s\right) }{s^{r+1}}ds\right\vert  \label{e.opp.5} \\
&\leq &x^{r}t^{r}\left\vert \int_{t}^{x}\frac{\left\vert f^{\prime }\left(
s\right) s-rf\left( s\right) \right\vert }{s^{r+1}}ds\right\vert :=I  \notag
\end{eqnarray}%
and utilizing H\"{o}lder's integral inequality we deduce
\begin{eqnarray}
I &\leq &x^{r}t^{r}\left\{
\begin{array}{ll}
\sup_{s\in \left[ t,x\right] \left( \left[ x,t\right] \right) }\left\vert
f^{\prime }\left( s\right) s-rf\left( s\right) \right\vert \left\vert
\int_{t}^{x}\frac{1}{s^{r+1}}ds\right\vert &  \\
&  \\
\left\vert \int_{t}^{x}\left\vert f^{\prime }\left( s\right) s-rf\left(
s\right) \right\vert ^{p}ds\right\vert ^{1/p}\left\vert \int_{t}^{x}\frac{1}{%
s^{q\left( r+1\right) }}ds\right\vert ^{1/q} &  \\
&  \\
\left\vert \int_{t}^{x}\left\vert f^{\prime }\left( s\right) s-rf\left(
s\right) \right\vert ds\right\vert \sup_{s\in \left[ t,x\right] \left( \left[
x,t\right] \right) }\left\{ \frac{1}{s^{r+1}}\right\} &
\end{array}%
\right.  \label{e.opp.6} \\
&\leq &x^{r}t^{r}\left\{
\begin{array}{ll}
\frac{1}{\left\vert r\right\vert }\left\Vert f^{\prime }\ell -rf\right\Vert
_{\infty }\left\vert \frac{1}{x^{r}}-\frac{1}{t^{r}}\right\vert &  \\
&  \\
\left\Vert f^{\prime }\ell -rf\right\Vert _{p}\left\{
\begin{array}{l}
\frac{1}{\left\vert 1-q\left( r+1\right) \right\vert }\left\vert \frac{1}{%
x^{1-q\left( r+1\right) }}-\frac{1}{t^{1-q\left( r+1\right) }}\right\vert ,
\\
r\neq -\frac{1}{p} \\
\\
\left\vert \ln x-\ln t\right\vert ,r=-\frac{1}{p}%
\end{array}%
\right. &  \\
&  \\
\left\Vert f^{\prime }\ell -rf\right\Vert _{1}\frac{1}{\min \left\{
x^{r+1},t^{r+1}\right\} }. &
\end{array}%
\right.  \notag
\end{eqnarray}%
where $p>1,\frac{1}{p}+\frac{1}{q}=1,$ and the inequality (\ref{e.opp.3}) is
proved.
\end{proof}

\subsection{Some Ostrowski Type Results}

The following new result also holds.

\begin{theorem}[Dragomir, 2013 \protect\cite{oppSSD1}]
\label{t.opp.3.2}Let $f:\left[ a,b\right] \rightarrow \mathbb{C}$ be an
absolutely continuous function on the interval $\left[ a,b\right] $ with $%
b>a>$ $0.$ If $r\in \mathbb{R}$, $r\neq 0,$ and $f^{\prime }\ell -rf\in
L_{\infty }\left[ a,b\right] ,$ then for any $x\in \left[ a,b\right] $ we
have
\begin{align}
& \left\vert \frac{b^{r+1}-a^{r+1}}{r+1}f\left( x\right)
-x^{r}\int_{a}^{b}f\left( t\right) dt\right\vert  \label{e.opp.4.1} \\
& \leq \frac{1}{\left\vert r\right\vert }\left\Vert f^{\prime }\ell
-rf\right\Vert _{\infty }  \notag \\
& \times \left\{
\begin{array}{l}
\frac{2rx^{r+1}-x^{r}\left( a+b\right) \left( r+1\right) +b^{r+1}+a^{r+1}}{%
r+1},\text{ if }r>0 \\
\\
\frac{x^{r}\left( a+b\right) \left( r+1\right) -2rx^{r+1}-b^{r+1}-a^{r+1}}{%
r+1},\text{ if }r\in \left( -\infty ,0\right) \diagdown \left\{ -1\right\} .%
\end{array}%
\right.  \notag
\end{align}%
Also, for $r=-1,$ we have
\begin{equation}
\left\vert f\left( x\right) \ln \frac{b}{a}-\frac{1}{x}\int_{a}^{b}f\left(
t\right) dt\right\vert \leq 2\left\Vert f^{\prime }\ell +f\right\Vert
_{\infty }\left( \ln \frac{x}{\sqrt{ab}}+\frac{\frac{a+b}{2}-x}{x}\right)
\label{e.opp.4.1.a}
\end{equation}%
for any $x\in \left[ a,b\right] ,$ provided $f^{\prime }\ell +f\in L_{\infty
}\left[ a,b\right] $

The constant $2$ in (\ref{e.opp.4.1.a}) is best possible.
\end{theorem}

\begin{proof}
Utilising the first inequality in (\ref{e.opp.3}) for $r\neq -1$ we have%
\begin{eqnarray}
\left\vert \frac{b^{r+1}-a^{r+1}}{r+1}f\left( x\right)
-x^{r}\int_{a}^{b}f\left( t\right) dt\right\vert &\leq
&\int_{a}^{b}\left\vert t^{r}f\left( x\right) -x^{r}f\left( t\right)
\right\vert dt  \label{e.opp.4.2} \\
&\leq &\frac{1}{\left\vert r\right\vert }\left\Vert f^{\prime }\ell
-rf\right\Vert _{\infty }\int_{a}^{b}\left\vert t^{r}-x^{r}\right\vert dt.
\notag
\end{eqnarray}%
Observe that
\begin{equation*}
\int_{a}^{b}\left\vert t^{r}-x^{r}\right\vert dt=\left\{
\begin{array}{l}
\int_{a}^{x}\left( x^{r}-t^{r}\right) dt+\int_{x}^{b}\left(
t^{r}-x^{r}\right) dt,\text{ if }r>0 \\
\\
\int_{a}^{x}\left( t^{r}-x^{r}\right) dt+\int_{x}^{b}\left(
x^{r}-t^{r}\right) dt,\text{ if }r\in \left( -\infty ,0\right) \diagdown
\left\{ -1\right\} .%
\end{array}%
\right.
\end{equation*}%
Then for $r>0$ we have%
\begin{equation*}
\int_{a}^{x}\left( x^{r}-t^{r}\right) dt+\int_{x}^{b}\left(
t^{r}-x^{r}\right) dt=\frac{2rx^{r+1}-x^{r}\left( a+b\right) \left(
r+1\right) +b^{r+1}+a^{r+1}}{r+1}
\end{equation*}%
and for $r\in \left( -\infty ,0\right) \diagdown \left\{ -1\right\} $ we
have
\begin{equation*}
\int_{a}^{x}\left( t^{r}-x^{r}\right) dt+\int_{x}^{b}\left(
x^{r}-t^{r}\right) dt=-\frac{2rx^{r+1}-x^{r}\left( a+b\right) \left(
r+1\right) +b^{r+1}+a^{r+1}}{r+1}.
\end{equation*}%
Making use of (\ref{e.opp.4.2}) we get (\ref{e.opp.4.1}).

Utilizing the inequality (\ref{e.opp.3}) for $r=-1$ we have%
\begin{equation*}
\left\vert t^{-1}f\left( x\right) -x^{-1}f\left( t\right) \right\vert \leq
\left\Vert f^{\prime }\ell +f\right\Vert _{\infty }\left\vert
t^{-1}-x^{-1}\right\vert
\end{equation*}%
if $f^{\prime }\ell +f\in L_{\infty }\left[ a,b\right] $.

Integrating this inequality, we have%
\begin{align}
\left\vert f\left( x\right) \ln \frac{b}{a}-x^{-1}\int_{a}^{b}f\left(
t\right) dt\right\vert & \leq \int_{a}^{b}\left\vert t^{-1}f\left( x\right)
-x^{-1}f\left( t\right) \right\vert dt  \label{e.opp.4.3} \\
& \leq \left\Vert f^{\prime }\ell +f\right\Vert _{\infty
}\int_{a}^{b}\left\vert t^{-1}-x^{-1}\right\vert dt.  \notag
\end{align}%
Since%
\begin{equation*}
\int_{a}^{b}\left\vert \frac{1}{x}-\frac{1}{t}\right\vert dt=2\left( \ln
\frac{x}{\sqrt{ab}}+\frac{\frac{a+b}{2}-x}{x}\right) ,
\end{equation*}%
then by (\ref{e.opp.4.3}) we get the desired inequality (\ref{e.opp.4.1.a}).

Now, assume that (\ref{e.opp.4.1.a}) holds with a constant $C>0,$ i.e.%
\begin{equation}
\left\vert f\left( x\right) \ln \frac{b}{a}-x^{-1}\int_{a}^{b}f\left(
t\right) dt\right\vert \leq C\left\Vert f^{\prime }\ell +f\right\Vert
_{\infty }\left( \ln \frac{x}{\sqrt{ab}}+\frac{\frac{a+b}{2}-x}{x}\right)
\label{e.opp.4.4}
\end{equation}%
for any $x\in \left[ a,b\right] .$

If we take in (\ref{e.opp.4.4}) $f\left( t\right) =1,t\in \left[ a,b\right]
, $ then we get%
\begin{equation}
\left\vert \ln \frac{b}{a}-\frac{b-a}{x}\right\vert \leq C\left( \ln \frac{x%
}{\sqrt{ab}}+\frac{\frac{a+b}{2}-x}{x}\right)  \label{e.opp.4.5}
\end{equation}%
for any for any $x\in \left[ a,b\right] .$

Making $x=a$ in (\ref{e.opp.4.4}) produces the inequality%
\begin{equation*}
\left\vert \ln \frac{b}{a}-\frac{b-a}{a}\right\vert \leq C\left( \frac{b-a}{%
2a}-\frac{1}{2}\ln \frac{b}{a}\right)
\end{equation*}%
which implies that $C\geq 2.$

This proves the sharpness of the constant $2$ in (\ref{e.opp.4.1.a}).
\end{proof}

\begin{remark}
\label{r.opp.3.5}Consider the $r$-Logarithmic mean%
\begin{equation*}
L_{r}=L_{r}\left( a,b\right) :=\left[ \frac{b^{r+1}-a^{r+1}}{\left(
r+1\right) \left( b-a\right) }\right] ^{1/r}
\end{equation*}%
defined for $r\in \mathbb{R\diagdown }\left\{ 0,-1\right\} $ and the
Logarithmic mean, defined as%
\begin{equation*}
L=L\left( a,b\right) :=\frac{b-a}{\ln b-\ln a}.
\end{equation*}%
If $A=A\left( a,b\right) :=\frac{a+b}{2},$ then from (\ref{e.opp.4.1}) we
get for $x=A$ the inequality
\begin{align}
& \left\vert L_{r}^{r}\left( b-a\right) f\left( A\right)
-A^{r}\int_{a}^{b}f\left( t\right) dt\right\vert  \label{e.opp.4.5.a} \\
& \leq \frac{2}{\left\vert r\right\vert }\left\Vert f^{\prime }\ell
-rf\right\Vert _{\infty }\left\{
\begin{array}{l}
\frac{A\left( b^{r+1},a^{r+1}\right) -A^{r+1}}{r+1},\text{ if }r>0 \\
\\
\frac{A^{r+1}-A\left( b^{r+1},a^{r+1}\right) }{r+1},\text{ if }r\in \left(
-\infty ,0\right) \diagdown \left\{ -1\right\} ,%
\end{array}%
\right.  \notag
\end{align}%
while from (\ref{e.opp.4.1.a}) we get
\begin{equation}
\left\vert L^{-1}\left( b-a\right) f\left( A\right)
-A^{-1}\int_{a}^{b}f\left( t\right) dt\right\vert \leq 2\left\Vert f^{\prime
}\ell +f\right\Vert _{\infty }\ln \frac{A}{G}  \label{e.opp.4.5.b}
\end{equation}
\end{remark}

The following related result holds.

\begin{theorem}[Dragomir, 2013 \protect\cite{oppSSD1}]
\label{t.opp.3.1}Let $f:\left[ a,b\right] \rightarrow \mathbb{C}$ be an
absolutely continuous function on the interval $\left[ a,b\right] $ with $%
b>a>$ $0.$ If $r\in \mathbb{R}$, $r\neq 0,$ then for any $x\in \left[ a,b%
\right] $ we have
\begin{align}
& \left\vert \frac{f\left( x\right) }{x^{r}}\left( b-a\right) -\int_{a}^{b}%
\frac{f\left( t\right) }{t^{r}}dt\right\vert  \label{e.opp.3.1} \\
& \leq \frac{1}{\left\vert r\right\vert }\left\Vert f^{\prime }\ell
-rf\right\Vert _{\infty }  \notag \\
& \times \left\{
\begin{array}{l}
\frac{2x^{1-r}-a^{1-r}-b^{1-r}}{1-r}+\frac{1}{x^{r}}\left( b+a-2x\right) ,%
\text{ }r\in \left( 0,\infty \right) \diagdown \left\{ 1\right\} \\
\\
\frac{a^{1-r}+b^{1-r}-2x^{1-r}}{1-r}+\frac{1}{x^{r}}\left( 2x-a-b\right) ,%
\text{ if }r<0.%
\end{array}%
\right. ,  \notag
\end{align}%
Also, for $r=1,$ we have
\begin{equation}
\left\vert \frac{f\left( x\right) }{x}\left( b-a\right) -\int_{a}^{b}\frac{%
f\left( t\right) }{t}dt\right\vert \leq 2\left\Vert f^{\prime }\ell
-f\right\Vert _{\infty }\left( \ln \frac{x}{\sqrt{ab}}+\frac{\frac{a+b}{2}-x%
}{x}\right)  \label{e.opp.3.1.a}
\end{equation}%
for any $x\in \left[ a,b\right] ,$ provided $f^{\prime }\ell -f\in L_{\infty
}\left[ a,b\right] .$

The constant $2$ is best possible in (\ref{e.opp.3.1.a}).
\end{theorem}

\begin{proof}
From the first inequality in (\ref{e.opp.3.a}) we have%
\begin{equation}
\left\vert \frac{f\left( x\right) }{x^{r}}-\frac{f\left( t\right) }{t^{r}}%
\right\vert \leq \frac{1}{\left\vert r\right\vert }\left\Vert f^{\prime
}\ell -rf\right\Vert _{\infty }\left\vert \frac{1}{x^{r}}-\frac{1}{t^{r}}%
\right\vert ,\text{ }  \label{e.opp.3.2}
\end{equation}%
for any $t,x\in \left[ a,b\right] ,$ provided $f^{\prime }\ell -rf\in
L_{\infty }\left[ a,b\right] .$

Integrating over $t\in \left[ a,b\right] $ we get%
\begin{align}
\left\vert \frac{f\left( x\right) }{x^{r}}\left( b-a\right) -\int_{a}^{b}%
\frac{f\left( t\right) }{t^{r}}dt\right\vert & \leq \int_{a}^{b}\left\vert
\frac{f\left( x\right) }{x^{r}}-\frac{f\left( t\right) }{t^{r}}\right\vert dt
\label{r.opp.3.3} \\
& \leq \frac{1}{\left\vert r\right\vert }\left\Vert f^{\prime }\ell
-rf\right\Vert _{\infty }\int_{a}^{b}\left\vert \frac{1}{x^{r}}-\frac{1}{%
t^{r}}\right\vert dt,\text{ }  \notag
\end{align}%
for $r\in \mathbb{R}$, $r\neq 0.$

For $r\in \left( 0,\infty \right) \diagdown \left\{ 1\right\} $ we have%
\begin{equation*}
\int_{a}^{b}\left\vert \frac{1}{x^{r}}-\frac{1}{t^{r}}\right\vert dt=\frac{%
2x^{1-r}-a^{1-r}-b^{1-r}}{1-r}+\frac{1}{x^{r}}\left( b+a-2x\right)
\end{equation*}%
for any $x\in \left[ a,b\right] .$

For $r<0,$ we also have%
\begin{equation*}
\int_{a}^{b}\left\vert \frac{1}{x^{r}}-\frac{1}{t^{r}}\right\vert dt=\frac{%
a^{1-r}+b^{1-r}-2x^{1-r}}{1-r}+\frac{1}{x^{r}}\left( 2x-a-b\right)
\end{equation*}%
for any $x\in \left[ a,b\right] .$

For $r=1$ we have
\begin{equation*}
\int_{a}^{b}\left\vert \frac{1}{x}-\frac{1}{t}\right\vert dt=2\left( \ln
\frac{x}{\sqrt{ab}}+\frac{\frac{a+b}{2}-x}{x}\right)
\end{equation*}%
for any $x\in \left[ a,b\right] ,$ and the inequality (\ref{e.opp.3.1.a}) is
obtained.

The sharpness of the constant $2$ follows as in the proof of Theorem \ref%
{t.opp.3.2} and the details are omitted.
\end{proof}

\begin{remark}
\label{r.opp.3.6}If we take $x=A$ in Theorem \ref{t.opp.3.1}, then we we
have
\begin{align}
& \left\vert \frac{f\left( A\right) }{A^{r}}\left( b-a\right) -\int_{a}^{b}%
\frac{f\left( t\right) }{t^{r}}dt\right\vert  \label{e.opp.3.4} \\
& \leq \frac{2}{\left\vert r\right\vert }\left\Vert f^{\prime }\ell
-rf\right\Vert _{\infty }\left\{
\begin{array}{l}
\frac{A^{1-r}-A\left( a^{1-r},b^{1-r}\right) }{1-r},\text{ }r\in \left(
0,\infty \right) \diagdown \left\{ 1\right\} \\
\\
\frac{A\left( a^{1-r},b^{1-r}\right) -A^{1-r}}{1-r},\text{ if }r<0.%
\end{array}%
\right. ,  \notag
\end{align}%
Also, for $r=1,$ we have
\begin{equation}
\left\vert \frac{f\left( A\right) }{A}\left( b-a\right) -\int_{a}^{b}\frac{%
f\left( t\right) }{t}dt\right\vert \leq 2\left\Vert f^{\prime }\ell
-f\right\Vert _{\infty }\ln \frac{A}{G}.  \label{e.opp.3.5}
\end{equation}
\end{remark}

\begin{remark}
\label{r.opp.3.4}The interested reader may obtain other similar results in
terms of the $p$-norms $\left\Vert f^{\prime }\ell -rf\right\Vert _{p}$ with
$p\geq 1.$ However, since some calculations are too complicated, the details
are not presented here.
\end{remark}

\section{Ostrowski Via an Exponential Pompeiu's Inequality}

\subsection{An Exponential Pompeiu's Inequality}

We can provide some similar results for complex-valued functions with the
exponential instead of $\ell .$

\begin{lemma}[Dragomir, 2013 \protect\cite{opeSSD1}]
\label{l.ope.2}Let $f:\left[ a,b\right] \rightarrow \mathbb{C}$ be an
absolutely continuous function on the interval $\left[ a,b\right] $ and $%
\alpha \in \mathbb{C}$ with ${\rm Re}\left( \alpha \right) \neq 0.$ Then
for any $t,x\in \left[ a,b\right] $ we have%
\begin{multline}
\left\vert \frac{f\left( x\right) }{\exp \left( \alpha x\right) }-\frac{%
f\left( t\right) }{\exp \left( \alpha t\right) }\right\vert  \label{e.ope.3}
\\
\\
\leq \left\{
\begin{array}{ll}
\begin{array}{l}
\left\vert {\rm Re}\left( \alpha \right) \right\vert \left\Vert f^{\prime
}-\alpha f\right\Vert _{\infty } \\
\times \left\vert \frac{1}{\exp \left( t {\rm Re}\left( \alpha \right)
\right) }-\frac{1}{\exp \left( x{\rm Re}\left( \alpha \right) \right) }%
\right\vert%
\end{array}
&
\begin{array}{l}
\text{if }{\small f}^{\prime }{\small -\alpha f} \\
{\small \in L}_{\infty }\left[ a,b\right] ,%
\end{array}
\\
&  \\
\begin{array}{l}
q^{1/q}\left\vert {\rm Re}\left( \alpha \right) \right\vert
^{1/q}\left\Vert f^{\prime }-\alpha f\right\Vert _{p} \\
\times \left\vert \frac{1}{\exp \left( tq{\rm Re}\left( \alpha \right)
\right) }-\frac{1}{\exp \left( xq{\rm Re}\left( \alpha \right) \right) }%
\right\vert ^{1/q}%
\end{array}
&
\begin{array}{l}
\text{if }{\small f}^{\prime }{\small -\alpha f} \\
{\small \in L}_{p}\left[ a,b\right] \\
\begin{array}{c}
{\small p>1,} \\
\frac{1}{p}{\small +}\frac{1}{q}{\small =1,}%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f^{\prime }-\alpha f\right\Vert _{1}\frac{1}{\min \left\{ \exp
\left( t{\rm Re}\left( \alpha \right) \right) ,\exp \left( x{\rm Re}\left(
\alpha \right) \right) \right\} }, &
\end{array}%
\right.
\end{multline}%
or, equivalently%
\begin{multline}
\left\vert \exp \left( \alpha t\right) f\left( x\right) -f\left( t\right)
\exp \left( \alpha x\right) \right\vert  \label{e.ope.3.a} \\
\\
\leq \left\{
\begin{array}{ll}
\begin{array}{l}
\left\vert {\rm Re}\left( \alpha \right) \right\vert \left\Vert f^{\prime
}-\alpha f\right\Vert _{\infty } \\
\times \left\vert \exp \left( x{\rm Re}\left( \alpha \right) \right) -\exp
\left( t{\rm Re}\left( \alpha \right) \right) \right\vert%
\end{array}
&
\begin{array}{l}
\text{if }{\small f}^{\prime }{\small -\alpha f} \\
{\small \in L}_{\infty }\left[ a,b\right] ,%
\end{array}
\\
&  \\
\begin{array}{l}
q^{1/q}\left\vert {\rm Re}\left( \alpha \right) \right\vert
^{1/q}\left\Vert f^{\prime }-\alpha f\right\Vert _{p} \\
\times \left\vert \exp \left( xq{\rm Re}\left( \alpha \right) \right) -\exp
\left( tq{\rm Re}\left( \alpha \right) \right) \right\vert ^{1/q}%
\end{array}
&
\begin{array}{l}
\text{if }{\small f}^{\prime }{\small -\alpha f} \\
{\small \in L}_{p}\left[ a,b\right] \\
\begin{array}{c}
{\small p>1,} \\
\frac{1}{p}{\small +}\frac{1}{q}{\small =1,}%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f^{\prime }-\alpha f\right\Vert _{1}\max \left\{ \exp \left( t%
{\rm Re}\left( \alpha \right) \right) ,\exp \left( x{\rm Re}\left( \alpha
\right) \right) \right\} . &
\end{array}%
\right.
\end{multline}
\end{lemma}

\begin{proof}
If $f$ is absolutely continuous, then $f/\exp \left( \alpha \cdot \right) $
is absolutely continuous on the interval $\left[ a,b\right] $ and
\begin{equation*}
\int_{t}^{x}\left( \frac{f\left( s\right) }{\exp \left( \alpha s\right) }%
\right) ^{\prime }ds=\frac{f\left( x\right) }{\exp \left( \alpha x\right) }-%
\frac{f\left( t\right) }{\exp \left( \alpha t\right) }
\end{equation*}%
for any $t,x\in \left[ a,b\right] $ with $x\neq t.$

Since%
\begin{align*}
\int_{t}^{x}\left( \frac{f\left( s\right) }{\exp \left( \alpha s\right) }%
\right) ^{\prime }ds& =\int_{t}^{x}\frac{f^{\prime }\left( s\right) \exp
\left( \alpha s\right) -\alpha f\left( s\right) \exp \left( \alpha s\right)
}{\exp \left( 2\alpha s\right) }ds \\
& =\int_{t}^{x}\frac{f^{\prime }\left( s\right) -\alpha f\left( s\right) }{%
\exp \left( \alpha s\right) }ds,
\end{align*}%
then we get the following identity%
\begin{equation}
\frac{f\left( x\right) }{\exp \left( \alpha x\right) }-\frac{f\left(
t\right) }{\exp \left( \alpha t\right) }=\int_{t}^{x}\frac{f^{\prime }\left(
s\right) -\alpha f\left( s\right) }{\exp \left( \alpha s\right) }ds
\label{e.ope.4}
\end{equation}%
for any $t,x\in \left[ a,b\right] $ with $x\neq t.$

Taking the modulus in (\ref{e.ope.4}) we have%
\begin{align}
\left\vert \frac{f\left( x\right) }{\exp \left( \alpha x\right) }-\frac{%
f\left( t\right) }{\exp \left( \alpha t\right) }\right\vert & =\left\vert
\int_{t}^{x}\frac{f^{\prime }\left( s\right) -\alpha f\left( s\right) }{\exp
\left( \alpha s\right) }ds\right\vert  \label{e.ope.5} \\
& \leq \left\vert \int_{t}^{x}\frac{\left\vert f^{\prime }\left( s\right)
-\alpha f\left( s\right) \right\vert }{\left\vert \exp \left( \alpha
s\right) \right\vert }ds\right\vert :=I  \notag
\end{align}%
and utilizing H\"{o}lder's integral inequality we deduce%
\begin{align}
I& \leq \left\{
\begin{array}{ll}
\sup_{s\in \left[ t,x\right] \left( \left[ x,t\right] \right) }\left\vert
f^{\prime }\left( s\right) -\alpha f\left( s\right) \right\vert \left\vert
\int_{t}^{x}\frac{1}{\left\vert \exp \left( \alpha s\right) \right\vert }%
ds\right\vert , &  \\
&  \\
\left\vert \int_{t}^{x}\left\vert f^{\prime }\left( s\right) -\alpha f\left(
s\right) \right\vert ^{p}ds\right\vert ^{1/p}\left\vert \int_{t}^{x}\frac{1}{%
\left\vert \exp \left( \alpha s\right) \right\vert ^{q}}ds\right\vert ^{1/q},
&  \\
&  \\
\left\vert \int_{t}^{x}\left\vert f^{\prime }\left( s\right) -\alpha f\left(
s\right) \right\vert ds\right\vert \sup_{s\in \left[ t,x\right] \left( \left[
x,t\right] \right) }\left\{ \frac{1}{\left\vert \exp \left( \alpha s\right)
\right\vert }\right\} , &
\end{array}%
\right.  \label{e.ope.5.1} \\
& \leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }-\alpha f\right\Vert _{\infty }\left\vert \int_{t}^{x}%
\frac{1}{\left\vert \exp \left( \alpha s\right) \right\vert }ds\right\vert ,
&  \\
&  \\
\left\Vert f^{\prime }-\alpha f\right\Vert _{p}\left\vert \int_{t}^{x}\frac{1%
}{\left\vert \exp \left( \alpha s\right) \right\vert ^{q}}ds\right\vert
^{1/q}, &  \\
&  \\
\left\Vert f^{\prime }-\alpha f\right\Vert _{1}\sup_{s\in \left[ t,x\right]
\left( \left[ x,t\right] \right) }\left\{ \frac{1}{\left\vert \exp \left(
\alpha s\right) \right\vert }\right\} . &
\end{array}%
\right.  \notag
\end{align}%
Now, since $\alpha ={\rm Re}\left( \alpha \right) +i{\rm Im}\left( \alpha
\right) $ and $s\in \left[ a,b\right] $, then%
\begin{equation*}
\left\vert \exp \left( \alpha s\right) \right\vert =\exp \left( s{\rm Re}%
\left( \alpha \right) \right) .
\end{equation*}%
We have%
\begin{equation*}
\int_{t}^{x}\frac{1}{\left\vert \exp \left( \alpha s\right) \right\vert }ds=%
{\rm Re}\left( \alpha \right) \left[ \frac{1}{\exp \left( t{\rm Re}\left(
\alpha \right) \right) }-\frac{1}{\exp \left( x{\rm Re}\left( \alpha
\right) \right) }\right]
\end{equation*}%
and by (\ref{e.ope.5}) and (\ref{e.ope.5.1}) we get%
\begin{equation*}
\left\vert \frac{f\left( x\right) }{\exp \left( \alpha x\right) }-\frac{%
f\left( t\right) }{\exp \left( \alpha t\right) }\right\vert \leq \left\Vert
f^{\prime }-\alpha f\right\Vert _{\infty }\left\vert {\rm Re}\left( \alpha
\right) \right\vert \left\vert \frac{1}{\exp \left( t{\rm Re}\left( \alpha
\right) \right) }-\frac{1}{\exp \left( x{\rm Re}\left( \alpha \right)
\right) }\right\vert
\end{equation*}%
and the first part of (\ref{e.ope.3}) is proved.

We have%
\begin{equation*}
\int_{t}^{x}\frac{1}{\left\vert \exp \left( \alpha s\right) \right\vert ^{q}}%
ds=q{\rm Re}\left( \alpha \right) \left[ \frac{1}{\exp \left( tq{\rm Re}%
\left( \alpha \right) \right) }-\frac{1}{\exp \left( xq{\rm Re}\left(
\alpha \right) \right) }\right]
\end{equation*}%
and by (\ref{e.ope.5}) and (\ref{e.ope.5.1}) we get the second part of (\ref%
{e.ope.3}).

We have
\begin{equation*}
\sup_{s\in \left[ t,x\right] \left( \left[ x,t\right] \right) }\left\{ \frac{%
1}{\left\vert \exp \left( \alpha s\right) \right\vert }\right\} =\frac{1}{%
\min \left\{ \exp \left( t{\rm Re}\left( \alpha \right) \right) ,\exp
\left( x{\rm Re}\left( \alpha \right) \right) \right\} }
\end{equation*}%
and by (\ref{e.ope.5}) and (\ref{e.ope.5.1}) we get the last part of (\ref%
{e.ope.3}).

The inequality (\ref{e.ope.3.a}) follows by (\ref{e.ope.3}) on multiplying
with $\left\vert \exp \left( \alpha x\right) \exp \left( \alpha t\right)
\right\vert $ and performing the required calculation.
\end{proof}

The following particular case is of interest.

\begin{corollary}
\label{c.ope.2}Let $f:\left[ a,b\right] \rightarrow \mathbb{C}$ be an
absolutely continuous function on the interval $\left[ a,b\right] .$ Then
for any $t,x\in \left[ a,b\right] $ we have%
\begin{multline}
\left\vert \frac{f\left( x\right) }{\exp \left( x\right) }-\frac{f\left(
t\right) }{\exp \left( t\right) }\right\vert  \label{e.ope.6} \\
\\
\leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }-f\right\Vert _{\infty }\left\vert \frac{1}{\exp
\left( t\right) }-\frac{1}{\exp \left( x\right) }\right\vert &
\begin{array}{l}
\text{if }f^{\prime }-f \\
\in L_{\infty }\left[ a,b\right] ,%
\end{array}
\\
&  \\
q^{1/q}\left\Vert f^{\prime }-f\right\Vert _{p}\left\vert \frac{1}{\exp
\left( tq\right) }-\frac{1}{\exp \left( xq\right) }\right\vert ^{1/q} &
\begin{array}{l}
\text{if }f^{\prime }-f \\
\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1,%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f^{\prime }-f\right\Vert _{1}\frac{1}{\min \left\{ \exp \left(
t\right) ,\exp \left( x\right) \right\} }, &
\end{array}%
\right.
\end{multline}%
or, equivalently%
\begin{multline}
\left\vert \exp \left( t\right) f\left( x\right) -f\left( t\right) \exp
\left( x\right) \right\vert  \label{e.ope.7} \\
\\
\leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }-f\right\Vert _{\infty }\left\vert \exp \left(
x\right) -\exp \left( t\right) \right\vert &
\begin{array}{l}
\text{if }f^{\prime }-f \\
\in L_{\infty }\left[ a,b\right] ,%
\end{array}
\\
&  \\
q^{1/q}\left\Vert f^{\prime }-f\right\Vert _{p}\left\vert \exp \left(
xq\right) -\exp \left( tq\right) \right\vert ^{1/q} &
\begin{array}{l}
\text{if }f^{\prime }-f \\
\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1,%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f^{\prime }-f\right\Vert _{1}\max \left\{ \exp \left( t\right)
,\exp \left( x\right) \right\} . &
\end{array}%
\right.
\end{multline}
\end{corollary}

\begin{remark}
\label{r.ope.1}If ${\rm Re}\left( \alpha \right) =0$ then the inequality (%
\ref{e.ope.5.1}) becomes
\begin{align*}
I& \leq \left\{
\begin{array}{ll}
\sup_{s\in \left[ t,x\right] \left( \left[ x,t\right] \right) }\left\vert
f^{\prime }\left( s\right) -i{\rm Im}\left( \alpha \right) f\left( s\right)
\right\vert \left\vert \int_{t}^{x}\frac{1}{\left\vert \exp \left( i{\rm Im}%
\left( \alpha \right) s\right) \right\vert }ds\right\vert , &  \\
&  \\
\left\vert \int_{t}^{x}\left\vert f^{\prime }\left( s\right) -i{\rm Im}%
\left( \alpha \right) f\left( s\right) \right\vert ^{p}ds\right\vert
^{1/p}\left\vert \int_{t}^{x}\frac{1}{\left\vert \exp \left( i{\rm Im}%
\left( \alpha \right) s\right) \right\vert ^{q}}ds\right\vert ^{1/q}, &  \\
&  \\
\left\vert \int_{t}^{x}\left\vert f^{\prime }\left( s\right) -i{\rm Im}%
\left( \alpha \right) f\left( s\right) \right\vert ds\right\vert \sup_{s\in %
\left[ t,x\right] \left( \left[ x,t\right] \right) }\left\{ \frac{1}{%
\left\vert \exp \left( i{\rm Im}\left( \alpha \right) s\right) \right\vert }%
\right\} , &
\end{array}%
\right. \\
& \leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }-i{\rm Im}\left( \alpha \right) f\right\Vert _{\infty
}\left\vert \int_{t}^{x}ds\right\vert , &  \\
&  \\
\left\Vert f^{\prime }-i{\rm Im}\left( \alpha \right) f\right\Vert
_{p}\left\vert \int_{t}^{x}ds\right\vert ^{1/q}, &  \\
&  \\
\left\Vert f^{\prime }-i{\rm Im}\left( \alpha \right) f\right\Vert _{1}, &
\end{array}%
\right. =\left\{
\begin{array}{ll}
\left\Vert f^{\prime }-i{\rm Im}\left( \alpha \right) f\right\Vert _{\infty
}\left\vert x-t\right\vert , &  \\
&  \\
\left\Vert f^{\prime }-i{\rm Im}\left( \alpha \right) f\right\Vert
_{p}\left\vert x-t\right\vert ^{1/q}, &  \\
&  \\
\left\Vert f^{\prime }-i{\rm Im}\left( \alpha \right) f\right\Vert _{1}. &
\end{array}%
\right.
\end{align*}%
Therefore we have%
\begin{equation}
\left\vert \frac{f\left( x\right) }{\exp \left( i{\rm Im}\left( \alpha
\right) x\right) }-\frac{f\left( t\right) }{\exp \left( i{\rm Im}\left(
\alpha \right) t\right) }\right\vert \leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }-i{\rm Im}\left( \alpha \right) f\right\Vert _{\infty
}\left\vert x-t\right\vert , &  \\
&  \\
\left\Vert f^{\prime }-i{\rm Im}\left( \alpha \right) f\right\Vert
_{p}\left\vert x-t\right\vert ^{1/q}, &  \\
&  \\
\left\Vert f^{\prime }-i{\rm Im}\left( \alpha \right) f\right\Vert _{1}, &
\end{array}%
\right.  \label{e.ope.8}
\end{equation}%
or, equivalently%
\begin{eqnarray}
&&\left\vert \exp \left( i{\rm Im}\left( \alpha \right) t\right) f\left(
x\right) -f\left( t\right) \exp \left( i{\rm Im}\left( \alpha \right)
x\right) \right\vert  \label{e.ope.9} \\
&\leq &\left\{
\begin{array}{ll}
\left\Vert f^{\prime }-i{\rm Im}\left( \alpha \right) f\right\Vert _{\infty
}\left\vert x-t\right\vert , &  \\
&  \\
\left\Vert f^{\prime }-i{\rm Im}\left( \alpha \right) f\right\Vert
_{p}\left\vert x-t\right\vert ^{1/q}, &  \\
&  \\
\left\Vert f^{\prime }-i{\rm Im}\left( \alpha \right) f\right\Vert _{1} &
\end{array}%
\right.  \notag
\end{eqnarray}%
for any $t,x\in \left[ a,b\right] .$

In particular, we have%
\begin{equation}
\left\vert \frac{f\left( x\right) }{\exp \left( ix\right) }-\frac{f\left(
t\right) }{\exp \left( it\right) }\right\vert \leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }-if\right\Vert _{\infty }\left\vert x-t\right\vert , &
\\
&  \\
\left\Vert f^{\prime }-if\right\Vert _{p}\left\vert x-t\right\vert ^{1/q}, &
\\
&  \\
\left\Vert f^{\prime }-if\right\Vert _{1}, &
\end{array}%
\right.  \label{e.ope.10}
\end{equation}%
or, equivalently%
\begin{equation}
\left\vert \exp \left( it\right) f\left( x\right) -f\left( t\right) \exp
\left( ix\right) \right\vert \leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }-if\right\Vert _{\infty }\left\vert x-t\right\vert , &
\\
&  \\
\left\Vert f^{\prime }-if\right\Vert _{p}\left\vert x-t\right\vert ^{1/q}, &
\\
&  \\
\left\Vert f^{\prime }-if\right\Vert _{1}, &
\end{array}%
\right.  \label{e.ope.11}
\end{equation}%
for any $t,x\in \left[ a,b\right] .$
\end{remark}

\subsection{Inequalities of Ostrowski Type}

The following result holds:

\begin{theorem}[Dragomir, 2013 \protect\cite{opeSSD1}]
\label{t.ope.1}Let $f:\left[ a,b\right] \rightarrow \mathbb{C}$ be an
absolutely continuous function on the interval $\left[ a,b\right] $ and $%
\alpha \in \mathbb{C}$ with ${\rm Re}\left( \alpha \right) >0.$ Then for
any $x\in \left[ a,b\right] $ we have%
\begin{multline}
\left\vert f\left( x\right) \frac{\exp \left( \alpha b\right) -\exp \left(
\alpha a\right) }{\alpha }-\exp \left( \alpha x\right) \int_{a}^{b}f\left(
t\right) dt\right\vert  \label{e.ope.12} \\
\\
\leq \left\{
\begin{array}{ll}
\left\vert {\rm Re}\left( \alpha \right) \right\vert \left\Vert f^{\prime
}-\alpha f\right\Vert _{\infty }B_{1}(a,b,x,\alpha ) &
\begin{array}{l}
\text{if }{\small f}^{\prime }{\small -\alpha f} \\
{\small \in L}_{\infty }\left[ a,b\right] ,%
\end{array}
\\
&  \\
\begin{array}{l}
q^{1/q}\left\vert {\rm Re}\left( \alpha \right) \right\vert ^{1/q}\left(
b-a\right) ^{1/p} \\
\times \left\Vert f^{\prime }-\alpha f\right\Vert _{p}\left\vert
B_{q}(a,b,x,\alpha )\right\vert ^{1/q}%
\end{array}
&
\begin{array}{l}
\text{if }{\small f}^{\prime }{\small -\alpha f} \\
{\small \in L}_{p}\left[ a,b\right] \\
\begin{array}{c}
{\small p>1,} \\
\frac{1}{p}{\small +}\frac{1}{q}{\small =1,}%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f^{\prime }-\alpha f\right\Vert _{1}B_{\infty }(a,b,x,\alpha ) &
\end{array}%
\right.
\end{multline}%
where%
\begin{multline*}
B_{q}(a,b,x,\alpha ):=2\left[ \exp \left( xq{\rm Re}\left( \alpha \right)
\right) \left( x-\frac{a+b}{2}\right) \right. \\
+\left. \frac{1}{q{\rm Re}\left( \alpha \right) }\left( \frac{\exp \left( bq%
{\rm Re}\left( \alpha \right) \right) +\exp \left( aq{\rm Re}\left( \alpha
\right) \right) }{2}-\exp \left( xq{\rm Re}\left( \alpha \right) \right)
\right) \right]
\end{multline*}%
for $q\geq 1$ and
\begin{equation*}
B_{\infty }(a,b,x,\alpha ):=\exp \left( x{\rm Re}\left( \alpha \right)
\right) \left( x-a\right) +\frac{\exp \left( b{\rm Re}\left( \alpha \right)
\right) -\exp \left( x{\rm Re}\left( \alpha \right) \right) }{{\rm Re}%
\left( \alpha \right) }.
\end{equation*}
\end{theorem}

\begin{proof}
Utilising the first inequality in (\ref{e.ope.3.a}) we have%
\begin{align}
& \left\vert f\left( x\right) \int_{a}^{b}\exp \left( \alpha t\right)
dt-\exp \left( \alpha x\right) \int_{a}^{b}f\left( t\right) dt\right\vert
\label{e.ope.13} \\
& \leq \int_{a}^{b}\left\vert \exp \left( \alpha t\right) f\left( x\right)
-f\left( t\right) \exp \left( \alpha x\right) \right\vert dt  \notag \\
& \leq \left\vert {\rm Re}\left( \alpha \right) \right\vert \left\Vert
f^{\prime }-\alpha f\right\Vert _{\infty }\int_{a}^{b}\left\vert \exp \left(
x{\rm Re}\left( \alpha \right) \right) -\exp \left( t{\rm Re}\left( \alpha
\right) \right) \right\vert dt  \notag
\end{align}%
for any $x\in \left[ a,b\right] .$

Observe that, since ${\rm Re}\left( \alpha \right) >0,$ then
\begin{align*}
& \int_{a}^{b}\left\vert \exp \left( x{\rm Re}\left( \alpha \right) \right)
-\exp \left( t{\rm Re}\left( \alpha \right) \right) \right\vert dt \\
& =2\left[ \exp \left( x{\rm Re}\left( \alpha \right) \right) \left( x-%
\frac{a+b}{2}\right) \right. \\
& +\left. \frac{1}{{\rm Re}\left( \alpha \right) }\left( \frac{\exp \left( b%
{\rm Re}\left( \alpha \right) \right) +\exp \left( a{\rm Re}\left( \alpha
\right) \right) }{2}-\exp \left( x{\rm Re}\left( \alpha \right) \right)
\right) \right]
\end{align*}%
for any $x\in \left[ a,b\right] .$

Also%
\begin{align*}
& f\left( x\right) \int_{a}^{b}\exp \left( \alpha t\right) dt-\exp \left(
\alpha x\right) \int_{a}^{b}f\left( t\right) dt \\
& =f\left( x\right) \frac{\exp \left( \alpha b\right) -\exp \left( \alpha
a\right) }{\alpha }-\exp \left( \alpha x\right) \int_{a}^{b}f\left( t\right)
dt
\end{align*}%
for any $x\in \left[ a,b\right] $ and by (\ref{e.ope.13}) we get the first
inequality in (\ref{e.ope.12}).

Using the second inequality in (\ref{e.ope.3.a}) we have%
\begin{align}
& \left\vert f\left( x\right) \int_{a}^{b}\exp \left( \alpha t\right)
dt-\exp \left( \alpha x\right) \int_{a}^{b}f\left( t\right) dt\right\vert
\label{e.ope.14} \\
& \leq \int_{a}^{b}\left\vert \exp \left( \alpha t\right) f\left( x\right)
-f\left( t\right) \exp \left( \alpha x\right) \right\vert dt  \notag \\
& \leq q^{1/q}\left\vert {\rm Re}\left( \alpha \right) \right\vert
^{1/q}\left\Vert f^{\prime }-\alpha f\right\Vert _{p}\int_{a}^{b}\left\vert
\exp \left( xq{\rm Re}\left( \alpha \right) \right) -\exp \left( tq{\rm Re}%
\left( \alpha \right) \right) \right\vert ^{1/q}dt  \notag
\end{align}%
for any $x\in \left[ a,b\right] .$

By H\"{o}lder's integral inequality we also have%
\begin{align*}
& \int_{a}^{b}\left\vert \exp \left( xq{\rm Re}\left( \alpha \right)
\right) -\exp \left( tq{\rm Re}\left( \alpha \right) \right) \right\vert
^{1/q}dt \\
& \leq \left( b-a\right) ^{1/p}\left[ \int_{a}^{b}\left\vert \exp \left( xq%
{\rm Re}\left( \alpha \right) \right) -\exp \left( tq{\rm Re}\left( \alpha
\right) \right) \right\vert dt\right] ^{1/q},
\end{align*}%
for any $x\in \left[ a,b\right] .$

Observe that, as above, we have%
\begin{align*}
& \int_{a}^{b}\left\vert \exp \left( xq{\rm Re}\left( \alpha \right)
\right) -\exp \left( tq{\rm Re}\left( \alpha \right) \right) \right\vert dt
\\
& =2\left[ \exp \left( xq{\rm Re}\left( \alpha \right) \right) \left( x-%
\frac{a+b}{2}\right) \right. \\
& +\left. \frac{1}{q{\rm Re}\left( \alpha \right) }\left( \frac{\exp \left(
bq{\rm Re}\left( \alpha \right) \right) +\exp \left( aq{\rm Re}\left(
\alpha \right) \right) }{2}-\exp \left( xq{\rm Re}\left( \alpha \right)
\right) \right) \right] \\
& =B_{q}(a,b,x,\alpha )
\end{align*}%
for any $x\in \left[ a,b\right] $ and by (\ref{e.ope.14}) we get the second
part of (\ref{e.ope.12}).

Using the third inequality in (\ref{e.ope.3.a}) we have%
\begin{align}
& \left\vert f\left( x\right) \int_{a}^{b}\exp \left( \alpha t\right)
dt-\exp \left( \alpha x\right) \int_{a}^{b}f\left( t\right) dt\right\vert
\label{e.ope.15} \\
& \leq \int_{a}^{b}\left\vert \exp \left( \alpha t\right) f\left( x\right)
-f\left( t\right) \exp \left( \alpha x\right) \right\vert dt  \notag \\
& \leq \left\Vert f^{\prime }-\alpha f\right\Vert _{1}\int_{a}^{b}\max
\left\{ \exp \left( t{\rm Re}\left( \alpha \right) \right) ,\exp \left( x%
{\rm Re}\left( \alpha \right) \right) \right\} dt  \notag
\end{align}%
for any $x\in \left[ a,b\right] .$

Observe that,%
\begin{align*}
& \int_{a}^{b}\max \left\{ \exp \left( t{\rm Re}\left( \alpha \right)
\right) ,\exp \left( x{\rm Re}\left( \alpha \right) \right) \right\} dt \\
& =\int_{a}^{x}\max \left\{ \exp \left( t{\rm Re}\left( \alpha \right)
\right) ,\exp \left( x{\rm Re}\left( \alpha \right) \right) \right\} dt \\
& +\int_{x}^{b}\max \left\{ \exp \left( t{\rm Re}\left( \alpha \right)
\right) ,\exp \left( x{\rm Re}\left( \alpha \right) \right) \right\} dt \\
& =\int_{a}^{x}\exp \left( x{\rm Re}\left( \alpha \right) \right)
dt+\int_{x}^{b}\exp \left( t{\rm Re}\left( \alpha \right) \right) dt= \\
& =\exp \left( x{\rm Re}\left( \alpha \right) \right) \left( x-a\right) +%
\frac{\exp \left( b{\rm Re}\left( \alpha \right) \right) -\exp \left( x%
{\rm Re}\left( \alpha \right) \right) }{{\rm Re}\left( \alpha \right) }
\end{align*}%
and by (\ref{e.ope.15}) we get the third part of (\ref{e.ope.12}).
\end{proof}

\begin{remark}
\label{r.ope.2}If ${\rm Re}\left( \alpha \right) <0,$ then a similar result
may be stated. However the details are left to the interested reader.
\end{remark}

\begin{corollary}
\label{c.ope.3}Let $f:\left[ a,b\right] \rightarrow \mathbb{C}$ be an
absolutely continuous function on the interval $\left[ a,b\right] $. Then
for any $x\in \left[ a,b\right] $ we have%
\begin{multline}
\left\vert f\left( x\right) \left[ \exp \left( b\right) -\exp \left(
a\right) \right] -\exp \left( x\right) \int_{a}^{b}f\left( t\right)
dt\right\vert  \label{e.ope.15.a} \\
\\
\leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }-f\right\Vert _{\infty }B_{1}(a,b,x) & \text{if }%
{\small f}^{\prime }{\small -f\in L}_{\infty }\left[ a,b\right] , \\
&  \\
\begin{array}{l}
q^{1/q}\left( b-a\right) ^{1/p}\left\Vert f^{\prime }-f\right\Vert _{p} \\
\times \left\vert B_{q}(a,b,x)\right\vert ^{1/q}%
\end{array}
&
\begin{array}{l}
\text{if }{\small f}^{\prime }{\small -f\in L}_{p}\left[ a,b\right] \\
\begin{array}{c}
{\small p>1,} \\
\frac{1}{p}{\small +}\frac{1}{q}{\small =1,}%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f^{\prime }-f\right\Vert _{1}B_{\infty }(a,b,x) &
\end{array}%
\right.
\end{multline}%
where%
\begin{align*}
& B_{q}(a,b,x) \\
& :=2\left[ \left( x-\frac{a+b}{2}\right) \exp \left( xq\right) +\frac{1}{q}%
\left( \frac{\exp \left( bq\right) +\exp \left( aq\right) }{2}-\exp \left(
xq\right) \right) \right]
\end{align*}%
for $q\geq 1$ and
\begin{equation*}
B_{\infty }(a,b,x):=\left( x-a\right) \exp \left( x\right) +\exp \left(
b\right) -\exp \left( x\right) .
\end{equation*}
\end{corollary}

\begin{remark}
\label{r.ope.3}The midpoint case is as follows:%
\begin{multline}
\left\vert f\left( \frac{a+b}{2}\right) \left[ \exp \left( b\right) -\exp
\left( a\right) \right] -\exp \left( \frac{a+b}{2}\right)
\int_{a}^{b}f\left( t\right) dt\right\vert  \label{e.ope.15.b} \\
\\
\leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }-f\right\Vert _{\infty }B_{1}(a,b) & \text{if }{\small %
f}^{\prime }{\small -f\in L}_{\infty }\left[ a,b\right] , \\
&  \\
\begin{array}{l}
q^{1/q}\left( b-a\right) ^{1/p}\left\Vert f^{\prime }-f\right\Vert _{p} \\
\times \left\vert B_{q}(a,b)\right\vert ^{1/q}%
\end{array}
&
\begin{array}{l}
\text{if }{\small f}^{\prime }{\small -f\in L}_{p}\left[ a,b\right] \\
\begin{array}{c}
{\small p>1,} \\
\frac{1}{p}{\small +}\frac{1}{q}{\small =1,}%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f^{\prime }-f\right\Vert _{1}B_{\infty }(a,b) &
\end{array}%
\right.
\end{multline}%
where%
\begin{equation*}
B_{q}(a,b,x):=\frac{2}{q}\left( \frac{\exp \left( bq\right) +\exp \left(
aq\right) }{2}-\exp \left( \frac{a+b}{2}q\right) \right)
\end{equation*}%
for $q\geq 1$ and
\begin{equation*}
B_{\infty }(a,b):=\frac{b-a}{2}\exp \left( \frac{a+b}{2}\right) +\exp \left(
b\right) -\exp \left( \frac{a+b}{2}\right) .
\end{equation*}
\end{remark}

The case ${\rm Re}\left( \alpha \right) =0$ is different and may be stated
as follows.

\begin{theorem}[Dragomir, 2013 \protect\cite{opeSSD1}]
\label{t.ope.2}Let $f:\left[ a,b\right] \rightarrow \mathbb{C}$ be an
absolutely continuous function on the interval $\left[ a,b\right] $ and $%
\alpha \in \mathbb{C}$ with ${\rm Re}\left( \alpha \right) =0$ and ${\rm Im}
\left( \alpha \right) \neq 0.$ Then for any $x\in \left[ a,b\right] $ we
have%
\begin{multline}
\left\vert f\left( x\right) \frac{\exp \left( i{\rm Im}\left( \alpha
\right) b\right) -\exp \left( i{\rm Im}\left( \alpha \right) a\right) }{i%
{\rm Im}\left( \alpha \right) }-\exp \left( i{\rm Im}\left( \alpha \right)
x\right) \int_{a}^{b}f\left( t\right) dt\right\vert  \label{e.ope.16} \\
\\
\leq \left\{
\begin{array}{ll}
\begin{array}{l}
\left\Vert f^{\prime }-i{\rm Im}\left( \alpha \right) f\right\Vert _{\infty
} \\
\left[ \frac{1}{4}+\left( \frac{x-\frac{a+b}{2}}{b-a}\right) ^{2}\right]
\left( b-a\right) ^{2}%
\end{array}
&
\begin{array}{l}
\text{if }{\small f}^{\prime }{\small -i{\rm Im}\left( \alpha \right) f} \\
{\small \in L}_{\infty }\left[ a,b\right] ,%
\end{array}
\\
&  \\
\begin{array}{l}
\frac{q}{q+1}\left\Vert f^{\prime }-i{\rm Im}\left( \alpha \right)
f\right\Vert _{p} \\
\times \left[ \left( \frac{b-x}{b-a}\right) ^{\frac{q+1}{q}}+\left( \frac{x-a%
}{b-a}\right) ^{\frac{q+1}{q}}\right] \left( b-a\right) ^{\frac{q+1}{q}}%
\end{array}
&
\begin{array}{l}
\text{if }{\small f}^{\prime }{\small -i{\rm Im}\left( \alpha \right) f} \\
{\small \in L}_{p}\left[ a,b\right] \\
\begin{array}{c}
{\small p>1,} \\
\frac{1}{p}{\small +}\frac{1}{q}{\small =1,}%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f^{\prime }-i{\rm Im}\left( \alpha \right) f\right\Vert
_{1}\left( b-a\right) . &
\end{array}%
\right.
\end{multline}
\end{theorem}

\begin{proof}
Utilizing the inequality (\ref{e.ope.9}) we have%
\begin{align}
& \left\vert f\left( x\right) \frac{\exp \left( i{\rm Im}\left( \alpha
\right) b\right) -\exp \left( i{\rm Im}\left( \alpha \right) a\right) }{i%
{\rm Im}\left( \alpha \right) }-\exp \left( i{\rm Im}\left( \alpha \right)
x\right) \int_{a}^{b}f\left( t\right) dt\right\vert  \label{e.ope.17} \\
&  \notag \\
& \leq \int_{a}^{b}\left\vert \exp \left( i{\rm Im}\left( \alpha \right)
t\right) f\left( x\right) -f\left( t\right) \exp \left( i{\rm Im}\left(
\alpha \right) x\right) \right\vert dt  \notag \\
&  \notag \\
& \leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }-i{\rm Im}\left( \alpha \right) f\right\Vert _{\infty
}\int_{a}^{b}\left\vert x-t\right\vert dt, &  \\
&  \\
\left\Vert f^{\prime }-i{\rm Im}\left( \alpha \right) f\right\Vert
_{p}\int_{a}^{b}\left\vert x-t\right\vert ^{1/q}dt, &  \\
&  \\
\left\Vert f^{\prime }-i{\rm Im}\left( \alpha \right) f\right\Vert
_{1}\int_{a}^{b}dt. &
\end{array}%
\right.  \notag
\end{align}%
Since%
\begin{equation*}
\int_{a}^{b}\left\vert x-t\right\vert dt=\left[ \frac{1}{4}+\left( \frac{x-%
\frac{a+b}{2}}{b-a}\right) ^{2}\right] \left( b-a\right) ^{2}
\end{equation*}%
and%
\begin{equation*}
\int_{a}^{b}\left\vert x-t\right\vert ^{1/q}dt=\frac{q}{q+1}\left[ \left(
\frac{b-x}{b-a}\right) ^{\frac{q+1}{q}}+\left( \frac{x-a}{b-a}\right) ^{%
\frac{q+1}{q}}\right] \left( b-a\right) ^{\frac{q+1}{q}},
\end{equation*}%
then we get from (\ref{e.ope.17}) the desired result (\ref{e.ope.16}).
\end{proof}

\begin{corollary}
\label{c.ope.4}Let $f:\left[ a,b\right] \rightarrow \mathbb{C}$ be an
absolutely continuous function on the interval $\left[ a,b\right] .$ Then
for any $x\in \left[ a,b\right] $ we have%
\begin{multline}
\left\vert f\left( x\right) \frac{\exp \left( ib\right) -\exp \left(
ia\right) }{i}-\exp \left( ix\right) \int_{a}^{b}f\left( t\right)
dt\right\vert  \label{e.ope.18} \\
\\
\leq \left\{
\begin{array}{ll}
\begin{array}{l}
\left\Vert f^{\prime }-if\right\Vert _{\infty } \\
\times \left[ \frac{1}{4}+\left( \frac{x-\frac{a+b}{2}}{b-a}\right) ^{2}%
\right] \left( b-a\right) ^{2}%
\end{array}
&
\begin{array}{l}
\text{if }{\small f}^{\prime }{\small -if} \\
{\small \in L}_{\infty }\left[ a,b\right] ,%
\end{array}
\\
&  \\
\begin{array}{l}
\frac{q}{q+1}\left\Vert f^{\prime }-if\right\Vert _{p} \\
\times \left[ \left( \frac{b-x}{b-a}\right) ^{\frac{q+1}{q}}+\left( \frac{x-a%
}{b-a}\right) ^{\frac{q+1}{q}}\right] \left( b-a\right) ^{\frac{q+1}{q}}%
\end{array}
&
\begin{array}{l}
\text{if }{\small f}^{\prime }{\small -if} \\
{\small \in L}_{p}\left[ a,b\right] \\
\begin{array}{c}
{\small p>1,} \\
\frac{1}{p}{\small +}\frac{1}{q}{\small =1,}%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f^{\prime }-if\right\Vert _{1}\left( b-a\right) . &
\end{array}%
\right.
\end{multline}
\end{corollary}

\begin{remark}
\label{r.ope.4}The midpoint case is as follows%
\begin{align}
& \left\vert f\left( \frac{a+b}{2}\right) \frac{\exp \left( ib\right) -\exp
\left( ia\right) }{i}-\exp \left( i\frac{a+b}{2}\right) \int_{a}^{b}f\left(
t\right) dt\right\vert  \label{e.ope.19} \\
&  \notag \\
& \leq \left\{
\begin{array}{l}
\frac{1}{4}\left\Vert f^{\prime }-if\right\Vert _{\infty }\left( b-a\right)
^{2},\text{ if }{\small f}^{\prime }{\small -if\in L}_{\infty }\left[ a,b%
\right] , \\
\\
\frac{q}{\left( q+1\right) 2^{1/q}}\left\Vert f^{\prime }-if\right\Vert
_{p}\left( b-a\right) ^{\frac{q+1}{q}},\text{ if }{\small f}^{\prime }%
{\small -if\in L}_{p}\left[ a,b\right] .%
\end{array}%
\right.  \notag
\end{align}
\end{remark}

Similar inequalities may be stated if one uses (\ref{e.ope.3}) and
integrates over $t$ on $\left[ a,b\right] .$ The details are left to the
interested reader.

\section{Ostrowski Via a Two Functions Pompeiu's Inequality}

\subsection{A General Pompeiu's Inequality}

We start with the following generalization of Pompeiu's inequality:

\begin{theorem}[Dragomir, 2013 \protect\cite{opgSSD1}]
\label{t.opg.2}Let $f,g:\left[ a,b\right] \rightarrow \mathbb{C}$ be
absolutely continuous functions on the interval $\left[ a,b\right] $ with $%
g\left( t\right) \neq 0$ for all $t\in \left[ a,b\right] .$ Then for any $%
t,x\in \left[ a,b\right] $ we have%
\begin{multline}
\left\vert \frac{f\left( x\right) }{g\left( x\right) }-\frac{f\left(
t\right) }{g\left( t\right) }\right\vert  \label{e.opg.3} \\
\\
\leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{\infty }\left\vert
\int_{t}^{x}\frac{1}{\left\vert g\left( s\right) \right\vert ^{2}}%
ds\right\vert & \text{if }f^{\prime }g-fg^{\prime }\in L_{\infty }\left[ a,b%
\right] , \\
&  \\
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{p}\left\vert \int_{t}^{x}%
\frac{1}{\left\vert g\left( s\right) \right\vert ^{2q}}ds\right\vert ^{1/q}
&
\begin{array}{l}
\text{if }f^{\prime }g-fg^{\prime }\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1,%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{1}\sup_{s\in \left[ t,x%
\right] \left( \left[ x,t\right] \right) }\left\{ \frac{1}{\left\vert
g\left( s\right) \right\vert ^{2}}\right\} &
\end{array}%
\right.
\end{multline}%
or, equivalently%
\begin{multline}
\left\vert g\left( t\right) f\left( x\right) -f\left( t\right) g\left(
x\right) \right\vert  \label{e.opg.3.a} \\
\\
\leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{\infty }\left\vert g\left(
t\right) g\left( x\right) \right\vert \left\vert \int_{t}^{x}\frac{1}{%
\left\vert g\left( s\right) \right\vert ^{2}}ds\right\vert & \text{if }%
f^{\prime }g-fg^{\prime }\in L_{\infty }\left[ a,b\right] , \\
&  \\
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{p}\left\vert g\left(
t\right) g\left( x\right) \right\vert \left\vert \int_{t}^{x}\frac{1}{%
\left\vert g\left( s\right) \right\vert ^{2q}}ds\right\vert ^{1/q} &
\begin{array}{l}
\text{if }f^{\prime }g-fg^{\prime }\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1,%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{1}\left\vert g\left(
t\right) g\left( x\right) \right\vert \sup_{s\in \left[ t,x\right] \left( %
\left[ x,t\right] \right) }\left\{ \frac{1}{\left\vert g\left( s\right)
\right\vert ^{2}}\right\} . &
\end{array}%
\right.
\end{multline}
\end{theorem}

\begin{proof}
If $f$ and $g$ are absolutely continuous and $g\left( t\right) \neq 0$ for
all $t\in \left[ a,b\right] $, then $f/g$ is absolutely continuous on the
interval $\left[ a,b\right] $ and
\begin{equation*}
\int_{t}^{x}\left( \frac{f\left( s\right) }{g\left( s\right) }\right)
^{\prime }ds=\frac{f\left( x\right) }{g\left( x\right) }-\frac{f\left(
t\right) }{g\left( t\right) }
\end{equation*}%
for any $t,x\in \left[ a,b\right] $ with $x\neq t.$

Since%
\begin{equation*}
\int_{t}^{x}\left( \frac{f\left( s\right) }{g\left( s\right) }\right)
^{\prime }ds=\int_{t}^{x}\frac{f^{\prime }\left( s\right) g\left( s\right)
-f\left( s\right) g^{\prime }\left( s\right) }{g^{2}\left( s\right) }ds,
\end{equation*}%
then we get the following identity%
\begin{equation}
\frac{f\left( x\right) }{g\left( x\right) }-\frac{f\left( t\right) }{g\left(
t\right) }=\int_{t}^{x}\frac{f^{\prime }\left( s\right) g\left( s\right)
-f\left( s\right) g^{\prime }\left( s\right) }{g^{2}\left( s\right) }ds
\label{e.opg.4}
\end{equation}%
for any $t,x\in \left[ a,b\right] .$

Taking the modulus in (\ref{e.opg.4}) we have%
\begin{align}
\left\vert \frac{f\left( x\right) }{g\left( x\right) }-\frac{f\left(
t\right) }{g\left( t\right) }\right\vert & =\left\vert \int_{t}^{x}\frac{%
f^{\prime }\left( s\right) g\left( s\right) -f\left( s\right) g^{\prime
}\left( s\right) }{g^{2}\left( s\right) }ds\right\vert  \label{e.opg.5} \\
& \leq \left\vert \int_{t}^{x}\frac{\left\vert f^{\prime }\left( s\right)
g\left( s\right) -f\left( s\right) g^{\prime }\left( s\right) \right\vert }{%
\left\vert g\left( s\right) \right\vert ^{2}}ds\right\vert :=I  \notag
\end{align}%
and utilizing H\"{o}lder's integral inequality we deduce%
\begin{align*}
I& \leq \left\{
\begin{array}{ll}
\sup_{s\in \left[ t,x\right] \left( \left[ x,t\right] \right) }\left\vert
f^{\prime }\left( s\right) g\left( s\right) -f\left( s\right) g^{\prime
}\left( s\right) \right\vert \left\vert \int_{t}^{x}\frac{1}{\left\vert
g\left( s\right) \right\vert ^{2}}ds\right\vert , &  \\
&  \\
\left\vert \int_{t}^{x}\left\vert f^{\prime }\left( s\right) g\left(
s\right) -f\left( s\right) g^{\prime }\left( s\right) \right\vert
^{p}ds\right\vert ^{1/p}\left\vert \int_{t}^{x}\frac{1}{\left\vert g\left(
s\right) \right\vert ^{2q}}ds\right\vert ^{1/q} &
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1,%
\end{array}
\\
&  \\
\left\vert \int_{t}^{x}\left\vert f^{\prime }\left( s\right) g\left(
s\right) -f\left( s\right) g^{\prime }\left( s\right) \right\vert
ds\right\vert \sup_{s\in \left[ t,x\right] \left( \left[ x,t\right] \right)
}\left\{ \frac{1}{\left\vert g\left( s\right) \right\vert ^{2}}\right\} , &
\end{array}%
\right. \\
& \leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{\infty }\left\vert
\int_{t}^{x}\frac{1}{\left\vert g\left( s\right) \right\vert ^{2}}%
ds\right\vert , &  \\
&  \\
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{p}\left\vert \int_{t}^{x}%
\frac{1}{\left\vert g\left( s\right) \right\vert ^{2q}}ds\right\vert ^{1/q}
&
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1,%
\end{array}
\\
&  \\
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{1}\sup_{s\in \left[ t,x%
\right] \left( \left[ x,t\right] \right) }\left\{ \frac{1}{\left\vert
g\left( s\right) \right\vert ^{2}}\right\} &
\end{array}%
\right.
\end{align*}%
and the inequality (\ref{e.opg.3}) is proved.
\end{proof}

The following particular case extends Pompeiu's inequality to other $p$%
-norms than $p=\infty $ obtained in (\ref{c.opg.2}).

\begin{corollary}
\label{c.opg.2}Let $f:\left[ a,b\right] \rightarrow \mathbb{C}$ be an
absolutely continuous function on the interval $\left[ a,b\right] $ with $%
b>a>$ $0.$ Then for any $t,x\in \left[ a,b\right] $ we have%
\begin{multline}
\left\vert \frac{f\left( x\right) }{x}-\frac{f\left( t\right) }{t}\right\vert
\label{e.opg.6} \\
\\
\leq \left\{
\begin{array}{ll}
\left\Vert f-\ell f^{\prime }\right\Vert _{\infty }\left\vert \frac{1}{t}-%
\frac{1}{x}\right\vert & \text{if }f-\ell f^{\prime }\in L_{\infty }\left[
a,b\right] , \\
&  \\
\frac{1}{2q-1}\left\Vert f-\ell f^{\prime }\right\Vert _{p}\left\vert \frac{1%
}{t^{2q-1}}-\frac{1}{x^{2q-1}}\right\vert ^{1/q} &
\begin{array}{c}
\text{if }f-\ell f^{\prime }\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1,%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f-\ell f^{\prime }\right\Vert _{1}\frac{1}{\min \left\{
t^{2},x^{2}\right\} } &
\end{array}%
\right.
\end{multline}%
or, equivalently%
\begin{multline}
\left\vert tf\left( x\right) -xf\left( t\right) \right\vert  \label{e.opg.7}
\\
\\
\leq \left\{
\begin{array}{ll}
\left\Vert f-\ell f^{\prime }\right\Vert _{\infty }\left\vert x-t\right\vert
& \text{if }f-\ell f^{\prime }\in L_{\infty }\left[ a,b\right] , \\
&  \\
\frac{1}{2q-1}\left\Vert f-\ell f^{\prime }\right\Vert _{p}\left\vert \frac{%
x^{q}}{t^{q-1}}-\frac{t^{q}}{x^{q-1}}\right\vert ^{1/q} &
\begin{array}{l}
\text{if }f-\ell f^{\prime }\in L_{p}\left[ a,b\right] \\
p>1,\frac{1}{p}+\frac{1}{q}=1,%
\end{array}
\\
&  \\
\left\Vert f-\ell f^{\prime }\right\Vert _{1}\frac{\max \left\{ t,x\right\}
}{\min \left\{ t,x\right\} }, &
\end{array}%
\right.
\end{multline}%
where $\ell \left( t\right) =t,t\in \left[ a,b\right] .$
\end{corollary}

The proof follows by (\ref{e.opg.3}) for $g\left( t\right) =\ell \left(
t\right) =t,$ $t\in \left[ a,b\right] .$

The general case for power functions is as follows.

\begin{corollary}
\label{c.opg.3}Let $f:\left[ a,b\right] \rightarrow \mathbb{C}$ be an
absolutely continuous function on the interval $\left[ a,b\right] $ with $%
b>a>$ $0.$ If $r\in \mathbb{R}$, $r\neq 0,$ then for any $t,x\in \left[ a,b%
\right] $ we have%
\begin{multline}
\left\vert \frac{f\left( x\right) }{x^{r}}-\frac{f\left( t\right) }{t^{r}}%
\right\vert  \label{e.opg.8} \\
\\
\leq \left\{
\begin{array}{ll}
\frac{1}{\left\vert r\right\vert }\left\Vert f^{\prime }\ell -rf\right\Vert
_{\infty }\left\vert \frac{1}{x^{r}}-\frac{1}{t^{r}}\right\vert ,\text{ if }%
f^{\prime }\ell -rf\in L_{\infty }\left[ a,b\right] , &  \\
&  \\
\begin{array}{l}
\left\Vert f^{\prime }\ell -rf\right\Vert _{p} \\
\times \left\{
\begin{array}{l}
\frac{1}{\left\vert 1-q\left( r+1\right) \right\vert }\left\vert \frac{1}{%
x^{1-q\left( r+1\right) }}-\frac{1}{t^{1-q\left( r+1\right) }}\right\vert ,%
\text{ for }r\neq -\frac{1}{p} \\
\\
\left\vert \ln x-\ln t\right\vert ,\text{ for }r=-\frac{1}{p}%
\end{array}%
\right. \\
\text{if }f^{\prime }\ell -rf\in L_{p}\left[ a,b\right] ,%
\end{array}
&  \\
&  \\
\left\Vert f^{\prime }\ell -rf\right\Vert _{1}\frac{1}{\min \left\{
x^{r+1},t^{r+1}\right\} }, &
\end{array}%
\right.
\end{multline}%
or, equivalently%
\begin{multline}
\left\vert t^{r}f\left( x\right) -x^{r}f\left( t\right) \right\vert
\label{e.opg.9} \\
\\
\leq \left\{
\begin{array}{ll}
\frac{1}{\left\vert r\right\vert }\left\Vert f^{\prime }\ell -rf\right\Vert
_{\infty }\left\vert t^{r}-x^{r}\right\vert ,\text{if }f^{\prime }\ell
-rf\in L_{\infty }\left[ a,b\right] , &  \\
&  \\
\begin{array}{l}
\left\Vert f^{\prime }\ell -rf\right\Vert _{p} \\
\times \left\{
\begin{array}{l}
\frac{1}{\left\vert 1-q\left( r+1\right) \right\vert }\left\vert \frac{t^{r}%
}{x^{1-q\left( r+1\right) -r}}-\frac{x^{r}}{t^{1-q\left( r+1\right) -r}}%
\right\vert ,\text{ for }r\neq -\frac{1}{p} \\
\\
t^{r}x^{r}\left\vert \ln x-\ln t\right\vert ,\text{ for }r=-\frac{1}{p}%
\end{array}%
\right. \\
\text{if }f^{\prime }\ell -rf\in L_{p}\left[ a,b\right] ,%
\end{array}
&  \\
&  \\
\left\Vert f^{\prime }\ell -rf\right\Vert _{1}\frac{t^{r}x^{r}}{\min \left\{
x^{r+1},t^{r+1}\right\} }, &
\end{array}%
\right.
\end{multline}%
where $p>1,\frac{1}{p}+\frac{1}{q}=1.$
\end{corollary}

The proof follows by (\ref{e.opg.3}) for $g\left( t\right) =t^{r},$ $t\in %
\left[ a,b\right] .$ The details for calculations are omitted.

We have the following result for exponential.

\begin{corollary}
\label{c.opg.2.1}Let $f:\left[ a,b\right] \rightarrow \mathbb{C}$ be an
absolutely continuous function on the interval $\left[ a,b\right] $ and $%
\alpha \in \mathbb{R}$, $\alpha \neq 0.$ Then for any $t,x\in \left[ a,b%
\right] $ we have%
\begin{multline}
\left\vert \frac{f\left( x\right) }{\exp \left( i\alpha x\right) }-\frac{%
f\left( t\right) }{\exp \left( i\alpha t\right) }\right\vert
\label{e.opg.10} \\
\\
\leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }-i\alpha f\right\Vert _{\infty }\left\vert
x-t\right\vert & \text{if }f^{\prime }-i\alpha f\in L_{\infty }\left[ a,b%
\right] , \\
&  \\
\left\Vert f^{\prime }-i\alpha f\right\Vert _{p}\left\vert x-t\right\vert
^{1/q} &
\begin{array}{l}
\text{if }f^{\prime }-i\alpha f\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1,%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f^{\prime }-i\alpha f\right\Vert _{1} &
\end{array}%
\right.
\end{multline}%
or, equivalently%
\begin{multline}
\left\vert \exp \left( i\alpha t\right) f\left( x\right) -f\left( t\right)
\exp \left( i\alpha x\right) \right\vert  \label{e.opg.11} \\
\\
\leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }-i\alpha f\right\Vert _{\infty }\left\vert
x-t\right\vert & \text{if }f^{\prime }-i\alpha f\in L_{\infty }\left[ a,b%
\right] , \\
&  \\
\left\Vert f^{\prime }-i\alpha f\right\Vert _{p}\left\vert x-t\right\vert
^{1/q} &
\begin{array}{l}
\text{if }f^{\prime }-i\alpha f\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1,%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f^{\prime }-i\alpha f\right\Vert _{1}. &
\end{array}%
\right.
\end{multline}
\end{corollary}

\subsection{An Inequality Generalizing Ostrowski's}

The following result holds:

\begin{theorem}[Dragomir, 2013 \protect\cite{opgSSD1}]
\label{t.opg.1}Let $f,g:\left[ a,b\right] \rightarrow \mathbb{C}$ be
absolutely continuous functions on the interval $\left[ a,b\right] .$ If $%
0<m\leq \left\vert g\left( t\right) \right\vert \leq M<\infty $ for any $%
t\in \left[ a,b\right] ,$ then%
\begin{multline}
\left\vert f\left( x\right) \int_{a}^{b}g\left( t\right) dt-g\left( x\right)
\int_{a}^{b}f\left( t\right) dt\right\vert  \label{e.opg.3.0} \\
\\
\leq \left( \frac{M}{m}\right) ^{2}\left\{
\begin{array}{ll}
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{\infty }\left( b-a\right)
^{2}\left[ \frac{1}{4}+\left( \frac{t-\frac{a+b}{2}}{b-a}\right) ^{2}\right]
& \text{if }f^{\prime }g-fg^{\prime }\in L_{\infty }\left[ a,b\right] , \\
&  \\
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{p}\left[ \frac{\left(
b-x\right) ^{1+1/q}+\left( x-a\right) ^{1+1/q}}{1+1/q}\right] &
\begin{array}{l}
\text{if }f^{\prime }g-fg^{\prime }\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1,%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{1}\left( b-a\right) &
\end{array}%
\right.
\end{multline}%
for any $x\in \left[ a,b\right] .$
\end{theorem}

\begin{proof}
Utilizing (\ref{e.opg.3.a}) we have
\begin{multline}
\left\vert f\left( x\right) \int_{a}^{b}g\left( t\right) dt-g\left( x\right)
\int_{a}^{b}f\left( t\right) dt\right\vert  \label{e.opg.3.1} \\
\\
\leq \int_{a}^{b}\left\vert g\left( t\right) f\left( x\right) -f\left(
t\right) g\left( x\right) \right\vert dt \\
\\
\leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{\infty }\left\vert g\left(
x\right) \right\vert \int_{a}^{b}\left( \left\vert g\left( t\right)
\right\vert \left\vert \int_{t}^{x}\frac{1}{\left\vert g\left( s\right)
\right\vert ^{2}}ds\right\vert \right) dt,\text{ } &  \\
&  \\
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{p}\left\vert g\left(
x\right) \right\vert \int_{a}^{b}\left( \left\vert g\left( t\right)
\right\vert \left\vert \int_{t}^{x}\frac{1}{\left\vert g\left( s\right)
\right\vert ^{2q}}ds\right\vert ^{1/q}\right) dt,\text{ } &  \\
&  \\
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{1}\left\vert g\left(
x\right) \right\vert \int_{a}^{b}\left( \left\vert g\left( t\right)
\right\vert \sup_{s\in \left[ t,x\right] \left( \left[ x,t\right] \right)
}\left\{ \frac{1}{\left\vert g\left( s\right) \right\vert ^{2}}\right\}
\right) dt &
\end{array}%
\right.
\end{multline}%
for any $x\in \left[ a,b\right] ,$ which is of interest in itself.

Since $0<m\leq \left\vert g\left( t\right) \right\vert \leq M<\infty $ for
any $t\in \left[ a,b\right] ,$ then%
\begin{align*}
\left\vert g\left( x\right) \right\vert \int_{a}^{b}\left( \left\vert
g\left( t\right) \right\vert \left\vert \int_{t}^{x}\frac{1}{\left\vert
g\left( s\right) \right\vert ^{2}}ds\right\vert \right) dt& \leq \left(
\frac{M}{m}\right) ^{2}\int_{a}^{b}\left\vert x-t\right\vert dt \\
& =\left( \frac{M}{m}\right) ^{2}\left[ \frac{1}{4}+\left( \frac{t-\frac{a+b%
}{2}}{b-a}\right) ^{2}\right] ,
\end{align*}

\begin{align*}
& \left\vert g\left( x\right) \right\vert \int_{a}^{b}\left( \left\vert
g\left( t\right) \right\vert \left\vert \int_{t}^{x}\frac{1}{\left\vert
g\left( s\right) \right\vert ^{2q}}ds\right\vert ^{1/q}\right) dt \\
& \leq \left( \frac{M}{m}\right) ^{2}\int_{a}^{b}\left\vert x-t\right\vert
^{1/q}dt=\left( \frac{M}{m}\right) ^{2}\frac{\left( b-x\right)
^{1+1/q}+\left( x-a\right) ^{1+1/q}}{1+1/q}
\end{align*}%
and%
\begin{equation*}
\left\vert g\left( x\right) \right\vert \int_{a}^{b}\left( \left\vert
g\left( t\right) \right\vert \sup_{s\in \left[ t,x\right] \left( \left[ x,t%
\right] \right) }\left\{ \frac{1}{\left\vert g\left( s\right) \right\vert
^{2}}\right\} \right) dt\leq \left( \frac{M}{m}\right)
^{2}\int_{a}^{b}dt=\left( \frac{M}{m}\right) ^{2}\left( b-a\right)
\end{equation*}%
for any $x\in \left[ a,b\right] $ and by (\ref{e.opg.3.1}) we get the
desired result (\ref{e.opg.3.0}).
\end{proof}

\begin{remark}
\label{r.opg.1}If we take $g\left( t\right) =1,t\in \left[ a,b\right] $ in
the first inequality (\ref{e.opg.3.0}) we recapture Ostrowski's inequality.
\end{remark}

\begin{corollary}
\label{c.opg.4}With the assumptions in Theorem \ref{t.opg.1} we have the
midpoint inequalities%
\begin{multline}
\left\vert f\left( \frac{a+b}{2}\right) \int_{a}^{b}g\left( t\right)
dt-g\left( \frac{a+b}{2}\right) \int_{a}^{b}f\left( t\right) dt\right\vert
\label{e.opg.3.1.a} \\
\\
\leq \left( \frac{M}{m}\right) ^{2}\left\{
\begin{array}{ll}
\frac{1}{4}\left( b-a\right) ^{2}\left\Vert f^{\prime }g-fg^{\prime
}\right\Vert _{\infty } & \text{if }f^{\prime }g-fg^{\prime }\in L_{\infty }%
\left[ a,b\right] , \\
&  \\
&  \\
&  \\
\frac{1}{2^{1/q}\left( 1+1/q\right) }\left( b-a\right) ^{1+1/q}\left\Vert
f^{\prime }g-fg^{\prime }\right\Vert _{p} &
\begin{array}{l}
\text{if }f^{\prime }g-fg^{\prime }\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1.%
\end{array}%
\end{array}%
\end{array}%
\right.
\end{multline}
\end{corollary}

The following result also holds:

\begin{theorem}[Dragomir, 2013 \protect\cite{opgSSD1}]
\label{t.opg.3}Let $f,g:\left[ a,b\right] \rightarrow \mathbb{C}$ be
absolutely continuous functions on the interval $\left[ a,b\right] ,g\left(
x\right) \neq 0$ for $x\in \left[ a,b\right] $ and $g^{-2}\in L_{\infty }%
\left[ a,b\right] .$ Then
\begin{multline}
\left\vert \frac{f\left( x\right) }{g\left( x\right) }\int_{a}^{b}g\left(
t\right) dt-\int_{a}^{b}f\left( t\right) dt\right\vert  \label{e.opg.3.2} \\
\\
\leq \left\Vert g^{-2}\right\Vert _{\infty }\times \left\{
\begin{array}{ll}
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{\infty
}\int_{a}^{b}\left\vert g\left( t\right) \right\vert \left\vert
x-t\right\vert dt, & \text{if }f^{\prime }g-fg^{\prime }\in L_{\infty }\left[
a,b\right] , \\
&  \\
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{p}\int_{a}^{b}\left\vert
g\left( t\right) \right\vert \left\vert x-t\right\vert ^{1/q}dt &
\begin{array}{l}
\text{if }f^{\prime }g-fg^{\prime }\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1,%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{1}\int_{a}^{b}\left\vert
g\left( t\right) \right\vert dt &
\end{array}%
\right.
\end{multline}%
for any $x\in \left[ a,b\right] .$
\end{theorem}

\begin{proof}
Utilizing (\ref{e.opg.3.a}) we have
\begin{multline}
\left\vert \frac{f\left( x\right) }{g\left( x\right) }\int_{a}^{b}g\left(
t\right) dt-\int_{a}^{b}f\left( t\right) dt\right\vert  \label{e.opg.3.3} \\
\\
\leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{\infty }\int_{a}^{b}\left(
\left\vert g\left( t\right) \right\vert \left\vert \int_{t}^{x}\frac{1}{%
\left\vert g\left( s\right) \right\vert ^{2}}ds\right\vert \right) dt,\text{
} &  \\
&  \\
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{p}\int_{a}^{b}\left(
\left\vert g\left( t\right) \right\vert \left\vert \int_{t}^{x}\frac{1}{%
\left\vert g\left( s\right) \right\vert ^{2q}}ds\right\vert ^{1/q}\right) dt,%
\text{ } &  \\
&  \\
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{1}\int_{a}^{b}\left(
\left\vert g\left( t\right) \right\vert \sup_{s\in \left[ t,x\right] \left( %
\left[ x,t\right] \right) }\left\{ \frac{1}{\left\vert g\left( s\right)
\right\vert ^{2}}\right\} \right) dt &
\end{array}%
\right.
\end{multline}%
for any $x\in \left[ a,b\right] $.

Since%
\begin{equation*}
\left\vert \int_{t}^{x}\frac{1}{\left\vert g\left( s\right) \right\vert ^{2}}%
ds\right\vert \leq \left\Vert g^{-2}\right\Vert _{\infty }\left\vert
x-t\right\vert ,\left\vert \int_{t}^{x}\frac{1}{\left\vert g\left( s\right)
\right\vert ^{2q}}ds\right\vert ^{1/q}\leq \left\Vert g^{-2}\right\Vert
_{\infty }\left\vert x-t\right\vert
\end{equation*}%
and%
\begin{equation*}
\sup_{s\in \left[ t,x\right] \left( \left[ x,t\right] \right) }\left\{ \frac{%
1}{\left\vert g\left( s\right) \right\vert ^{2}}\right\} \leq \left\Vert
g^{-2}\right\Vert _{\infty }
\end{equation*}%
for any $x,t\in \left[ a,b\right] ,$ then on making use of (\ref{e.opg.3.3})
we get the desired result (\ref{e.opg.3.2}).
\end{proof}

We have the midpoint inequalities:

\begin{corollary}
\label{c.opg.5}With the assumptions of Theorem \ref{t.opg.3} we have
\begin{multline}
\left\vert \frac{f\left( \frac{a+b}{2}\right) }{g\left( \frac{a+b}{2}\right)
}\int_{a}^{b}g\left( t\right) dt-\int_{a}^{b}f\left( t\right) dt\right\vert
\label{e.opg.3.2.a} \\
\\
\leq \left\Vert g^{-2}\right\Vert _{\infty }\times \left\{
\begin{array}{ll}
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{\infty
}\int_{a}^{b}\left\vert g\left( t\right) \right\vert \left\vert \frac{a+b}{2}%
-t\right\vert dt, & \text{if }f^{\prime }g-fg^{\prime }\in L_{\infty }\left[
a,b\right] , \\
&  \\
&  \\
&  \\
\left\Vert f^{\prime }g-fg^{\prime }\right\Vert _{p}\int_{a}^{b}\left\vert
g\left( t\right) \right\vert \left\vert \frac{a+b}{2}-t\right\vert ^{1/q}dt
&
\begin{array}{l}
\text{if }f^{\prime }g-fg^{\prime }\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1.%
\end{array}%
\end{array}%
\end{array}%
\right.
\end{multline}
\end{corollary}

We have the following exponential version of Ostrowski's inequality as well:

\begin{theorem}[Dragomir, 2013 \protect\cite{opgSSD1}]
\label{t.opg.4}Let $f:\left[ a,b\right] \rightarrow \mathbb{C}$ be an
absolutely continuous function on the interval $\left[ a,b\right] $ and $%
\alpha \in \mathbb{R}$, $\alpha \neq 0.$ Then for any $x\in \left[ a,b\right]
$ we have%
\begin{multline}
\left\vert \frac{\exp \left( i\alpha \left( b-x\right) \right) -\exp \left(
-i\alpha \left( x-a\right) \right) }{i\alpha }f\left( x\right)
-\int_{a}^{b}f\left( t\right) dt\right\vert  \label{e.opg.3.4} \\
\\
\leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }-i\alpha f\right\Vert _{\infty }\left( b-a\right) ^{2}
\left[ \frac{1}{4}+\left( \frac{t-\frac{a+b}{2}}{b-a}\right) ^{2}\right] ,%
\text{ if }f^{\prime }-i\alpha f\in L_{\infty }\left[ a,b\right] , &  \\
&  \\
\left\Vert f^{\prime }-i\alpha f\right\Vert _{p}\frac{\left( b-x\right)
^{1+1/q}+\left( x-a\right) ^{1+1/q}}{1+1/q},\text{ }%
\begin{array}{l}
\text{if }f^{\prime }-i\alpha f \\
\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1,%
\end{array}%
\end{array}
&  \\
&  \\
\left\Vert f^{\prime }-i\alpha f\right\Vert _{1}. &
\end{array}%
\right.
\end{multline}
\end{theorem}

\begin{proof}
If we write the inequality (\ref{e.opg.3.1}) for $g\left( t\right) =\exp
\left( i\alpha t\right) ,$ $t\in \left[ a,b\right] ,$ then we get
\begin{multline*}
\left\vert f\left( x\right) \int_{a}^{b}\exp \left( i\alpha t\right) dt-\exp
\left( i\alpha x\right) \int_{a}^{b}f\left( t\right) dt\right\vert \\
\\
\leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }-i\alpha f\right\Vert _{\infty }\int_{a}^{b}\left\vert
x-t\right\vert dt,\text{ if }f^{\prime }-i\alpha f\in L_{\infty }\left[ a,b%
\right] &  \\
&  \\
\left\Vert f^{\prime }-i\alpha f\right\Vert _{p}\left\vert g\left( x\right)
\right\vert \int_{a}^{b}\left\vert x-t\right\vert ^{1/q}dt,\text{ }%
\begin{array}{l}
\text{if }f^{\prime }-i\alpha f \\
\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1%
\end{array}%
\end{array}
&  \\
&  \\
\left\Vert f^{\prime }-i\alpha f\right\Vert _{1}, &
\end{array}%
\right.
\end{multline*}%
which, after simple calculation, is equivalent with (\ref{e.opg.3.4}).

The details are omitted.
\end{proof}

\begin{corollary}
\label{c.opg.4.1}With the assumptions of Theorem \ref{t.opg.4} we have the
midpoint inequalities%
\begin{multline}
\left\vert \frac{\exp \left( i\alpha \left( \frac{b-a}{2}\right) \right)
-\exp \left( -i\alpha \left( \frac{b-a}{2}\right) \right) }{i\alpha }f\left(
\frac{a+b}{2}\right) -\int_{a}^{b}f\left( t\right) dt\right\vert
\label{e.opg.3.5} \\
\\
\leq \left\{
\begin{array}{ll}
\frac{1}{4}\left\Vert f^{\prime }-i\alpha f\right\Vert _{\infty }\left(
b-a\right) ^{2},\text{ if }f^{\prime }-i\alpha f\in L_{\infty }\left[ a,b%
\right] , &  \\
&  \\
&  \\
&  \\
\frac{1}{2^{1/q}\left( 1+1/q\right) }\left( b-a\right) ^{1+1/q}\left\Vert
f^{\prime }-i\alpha f\right\Vert _{p},\text{ }%
\begin{array}{l}
\text{if }f^{\prime }-i\alpha f\in L_{p}\left[ a,b\right] \\
p>1,\frac{1}{p}+\frac{1}{q}=1,%
\end{array}
&
\end{array}%
\right.
\end{multline}%
or, equivalently%
\begin{multline}
\left\vert \frac{2\sin \left( \alpha \left( \frac{b-a}{2}\right) \right) }{%
\alpha }f\left( \frac{a+b}{2}\right) -\int_{a}^{b}f\left( t\right)
dt\right\vert  \label{e.opg.3.6} \\
\\
\leq \left\{
\begin{array}{ll}
\frac{1}{4}\left\Vert f^{\prime }-i\alpha f\right\Vert _{\infty }\left(
b-a\right) ^{2},\text{ if }f^{\prime }-i\alpha f\in L_{\infty }\left[ a,b%
\right] , &  \\
&  \\
&  \\
&  \\
\frac{1}{2^{1/q}\left( 1+1/q\right) }\left( b-a\right) ^{1+1/q}\left\Vert
f^{\prime }-i\alpha f\right\Vert _{p},\text{ }%
\begin{array}{l}
\text{if }f^{\prime }-i\alpha f\in L_{p}\left[ a,b\right] \\
p>1,\frac{1}{p}+\frac{1}{q}=1.%
\end{array}
&
\end{array}%
\right.
\end{multline}
\end{corollary}

\subsection{An Application for CBS-Inequality}

The following inequality is well known in the literature as the
Cauchy-Bunyakovsky-Schwarz inequality, or the CBS-inequality, for short:%
\begin{equation}
\left\vert \int_{a}^{b}f\left( t\right) g\left( t\right) dt\right\vert
^{2}\leq \int_{a}^{b}\left\vert f\left( t\right) \right\vert
^{2}dt\int_{a}^{b}\left\vert g\left( t\right) \right\vert ^{2}dt,
\label{e.opg.4.1}
\end{equation}%
provided that $f,g\in L_{2}\left[ a,b\right] .$

We have the following result concerning some reverses of the CBS-inequality:

\begin{theorem}[Dragomir, 2013 \protect\cite{opgSSD1}]
\label{t.opg.4.1}Let $f,g:\left[ a,b\right] \rightarrow \mathbb{C}$ be
absolutely continuous functions on the interval $\left[ a,b\right] $ with $%
g\left( t\right) \neq 0$ for all $t\in \left[ a,b\right] .$ Then%
\begin{multline}
0\leq \int_{a}^{b}\left\vert g\left( t\right) \right\vert
^{2}dt\int_{a}^{b}\left\vert f\left( t\right) \right\vert ^{2}dt-\left\vert
\int_{a}^{b}f\left( t\right) g\left( t\right) dt\right\vert ^{2}
\label{e.opg.4.2} \\
0000000000000000\\
\leq \frac{1}{2}\times \left\{
\begin{array}{ll}
\left\Vert f^{\prime }\overline{g}-f\overline{g}^{\prime }\right\Vert
_{\infty }^{2}\left( \int_{a}^{b}\left\vert g\left( t\right) \right\vert
^{2}dt\right) ^{2}\left( \int_{a}^{b}\frac{1}{\left\vert g\left( t\right)
\right\vert ^{2}}dt\right) ^{2},\text{ if }%
\begin{array}{c}
f^{\prime }\overline{g}-f\overline{g}^{\prime }\in L_{\infty }\left[ a,b%
\right] , \\
\frac{1}{\left\vert g\right\vert ^{2}}\in L\left[ a,b\right]%
\end{array}
&  \\
&  \\
\left\Vert f^{\prime }\overline{g}-f\overline{g}^{\prime }\right\Vert
_{p}^{2}\left( \int_{a}^{b}\left\vert g\left( t\right) \right\vert
^{2}dt\right) ^{2}\left( \int_{a}^{b}\frac{1}{\left\vert g\left( t\right)
\right\vert ^{2q}}dt\right) ^{2/q},\text{ if }%
\begin{array}{l}
f^{\prime }\overline{g}-f\overline{g}^{\prime }\in L_{p}\left[ a,b\right] ,
\\
\frac{1}{\left\vert g\right\vert ^{2q}}\in L\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1,%
\end{array}%
\end{array}
&  \\
&  \\
\left\Vert f^{\prime }\overline{g}-f\overline{g}^{\prime }\right\Vert
_{1}^{2}\left( \int_{a}^{b}\left\vert g\left( t\right) \right\vert
^{2}dt\right) ^{2}ess\sup_{t\in \left[ a,b\right] }\left\{ \frac{1}{%
\left\vert g\left( t\right) \right\vert ^{4}}\right\} ,\text{ if }\frac{1}{%
\left\vert g\right\vert }\in L_{\infty }\left[ a,b\right] . &
\end{array}%
\right.\\
\end{multline}
\end{theorem}

\begin{proof}
Utilising the inequality (\ref{e.opg.3.a}) we have
\begin{multline}
\left\vert \overline{g\left( t\right) }f\left( x\right) -f\left( t\right)
\overline{g\left( x\right) }\right\vert  \label{e.opg.4.3} \\
\\
\leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }\overline{g}-f\overline{g}^{\prime }\right\Vert
_{\infty }\left\vert g\left( t\right) g\left( x\right) \right\vert
\left\vert \int_{t}^{x}\frac{1}{\left\vert g\left( s\right) \right\vert ^{2}}%
ds\right\vert &
\begin{array}{l}
\text{if }f^{\prime }\overline{g}-f\overline{g}^{\prime } \\
\in L_{\infty }\left[ a,b\right] ,%
\end{array}
\\
&  \\
\left\Vert f^{\prime }\overline{g}-f\overline{g}^{\prime }\right\Vert
_{p}\left\vert g\left( t\right) g\left( x\right) \right\vert \left\vert
\int_{t}^{x}\frac{1}{\left\vert g\left( s\right) \right\vert ^{2q}}%
ds\right\vert ^{1/q} &
\begin{array}{l}
\text{if }f^{\prime }\overline{g}-f\overline{g}^{\prime } \\
\in L_{p}\left[ a,b\right] \\
\begin{array}{c}
p>1, \\
\frac{1}{p}+\frac{1}{q}=1,%
\end{array}%
\end{array}
\\
&  \\
\left\Vert f^{\prime }\overline{g}-f\overline{g}^{\prime }\right\Vert
_{1}\left\vert g\left( t\right) g\left( x\right) \right\vert \sup_{s\in %
\left[ t,x\right] \left( \left[ x,t\right] \right) }\left\{ \frac{1}{%
\left\vert g\left( s\right) \right\vert ^{2}}\right\} . &
\end{array}%
\right.
\end{multline}%
for any $t,x\in \left[ a,b\right] .$

Taking the square in (\ref{e.opg.4.3}) and integrating over $\left(
t,x\right) \in \left[ a,b\right] ^{2}$ we have%
\begin{multline}
\int_{a}^{b}\int_{a}^{b}\left\vert \overline{g\left( t\right) }f\left(
x\right) -f\left( t\right) \overline{g\left( x\right) }\right\vert ^{2}dtdx
\label{e.opg.4.4} \\
\\
\leq \left\{
\begin{array}{ll}
\left\Vert f^{\prime }\overline{g}-f\overline{g}^{\prime }\right\Vert
_{\infty }^{2}\int_{a}^{b}\int_{a}^{b}\left\vert g\left( t\right) g\left(
x\right) \right\vert ^{2}\left\vert \int_{t}^{x}\frac{1}{\left\vert g\left(
s\right) \right\vert ^{2}}ds\right\vert ^{2}dtdx,\text{ } &  \\
&  \\
\left\Vert f^{\prime }\overline{g}-f\overline{g}^{\prime }\right\Vert
_{p}^{2}\int_{a}^{b}\int_{a}^{b}\left\vert g\left( t\right) g\left( x\right)
\right\vert ^{2}\left\vert \int_{t}^{x}\frac{1}{\left\vert g\left( s\right)
\right\vert ^{2q}}ds\right\vert ^{2/q}dtdx,\text{ } &  \\
&  \\
\left\Vert f^{\prime }\overline{g}-f\overline{g}^{\prime }\right\Vert
_{1}^{2}\int_{a}^{b}\int_{a}^{b}\left\vert g\left( t\right) g\left( x\right)
\right\vert ^{2}\sup_{s\in \left[ t,x\right] \left( \left[ x,t\right]
\right) }\left\{ \frac{1}{\left\vert g\left( s\right) \right\vert ^{4}}%
\right\} dtdx. &
\end{array}%
\right.
\end{multline}%
Observe that%
\begin{align*}
& \int_{a}^{b}\int_{a}^{b}\left\vert \overline{g\left( t\right) }f\left(
x\right) -f\left( t\right) \overline{g\left( x\right) }\right\vert ^{2}dtdx
\\
& =\int_{a}^{b}\int_{a}^{b}\left( \left\vert g\left( t\right) \right\vert
^{2}\left\vert f\left( x\right) \right\vert ^{2}-2{\rm Re}\left[ \overline{%
g\left( t\right) }f\left( x\right) \overline{f\left( t\right) \overline{%
g\left( x\right) }}\right] +\left\vert g\left( x\right) \right\vert
^{2}\left\vert f\left( t\right) \right\vert ^{2}\right) dtdx \\
& =2\left[ \int_{a}^{b}\left\vert g\left( t\right) \right\vert
^{2}dt\int_{a}^{b}\left\vert f\left( t\right) \right\vert ^{2}dt-\left\vert
\int_{a}^{b}f\left( t\right) g\left( t\right) dt\right\vert ^{2}\right] ,
\end{align*}%
\begin{equation*}
\int_{a}^{b}\int_{a}^{b}\left[ \left\vert g\left( t\right) g\left( x\right)
\right\vert ^{2}\left\vert \int_{t}^{x}\frac{1}{\left\vert g\left( s\right)
\right\vert ^{2}}ds\right\vert ^{2}\right] dtdx\leq \left(
\int_{a}^{b}\left\vert g\left( t\right) \right\vert ^{2}dt\right) ^{2}\left(
\int_{a}^{b}\frac{1}{\left\vert g\left( t\right) \right\vert ^{2}}dt\right)
^{2},
\end{equation*}%
\begin{align*}
& \int_{a}^{b}\int_{a}^{b}\left[ \left\vert g\left( t\right) g\left(
x\right) \right\vert ^{2}\left\vert \int_{t}^{x}\frac{1}{\left\vert g\left(
s\right) \right\vert ^{2q}}ds\right\vert ^{2/q}\right] dtdx \\
& \leq \left( \int_{a}^{b}\left\vert g\left( t\right) \right\vert
^{2}dt\right) ^{2}\left( \int_{a}^{b}\frac{1}{\left\vert g\left( t\right)
\right\vert ^{2q}}dt\right) ^{2/q}
\end{align*}%
and%
\begin{align*}
& \int_{a}^{b}\int_{a}^{b}\left[ \left\vert g\left( t\right) g\left(
x\right) \right\vert ^{2}\sup_{s\in \left[ t,x\right] \left( \left[ x,t%
\right] \right) }\left\{ \frac{1}{\left\vert g\left( s\right) \right\vert
^{4}}\right\} \right] dtdx \\
& \leq \left( \int_{a}^{b}\left\vert g\left( t\right) \right\vert
^{2}dt\right) ^{2}ess\sup_{t\in \left[ a,b\right] }\left\{ \frac{1}{%
\left\vert g\left( t\right) \right\vert ^{4}}\right\} ,
\end{align*}%
then by (\ref{e.opg.4.4}) we get the desired result (\ref{e.opg.4.2}).
\end{proof}
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\end{document}
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