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\parbox{2.75cm}{\includegraphics*[keepaspectratio=true,scale=1.6]{BJMA.jpg}}
\parbox{4.75in}{\hspace{0.1mm}\\[1.5cm]\noindent
Banach J. Math. Anal. 2 (2008), no. 1, 16--20\\
$\frac{\rule{4in}{0.05in}}{{}}$\\
{\footnotesize
\textcolor[rgb]{0.65,0.00,0.95}{\textsc{\textbf{\large{B}}anach
\textbf{\large{J}}ournal of \textbf{\large{M}}athematical
\textbf{\large{A}}nalysis}}\\
ISSN: 1735-8787 (electronic)\\
\textcolor[rgb]{0.00,0.00,0.84}{\textbf{http://www.math-analysis.org }}\\
$\frac{\rule{4in}{0.05in}}{{}}$}\\[.5in]}


\title[Hardy--Hilbert's
integral inequalities]{A study On some new types of Hardy--Hilbert's
integral inequalities}

\author[W.T. Sulaiman]{Waad T. Sulaiman$^1$}


\address{$^{1}$ Department of Mathematics, College of Computer Sciences and Mathematics, University
of Mosul, Iraq.}

\email{\textcolor[rgb]{0.00,0.00,0.84}{waadsulaiman@hotmail.com}}

\dedicatory{{\rm  Submitted by F. Kittaneh}}

\subjclass[2000]{Primary 54C05; Secondary 46A03, 46A55.}

\keywords{Convex, balanced, absorbing, basis, empty interior,
non-continuous linear functional.}

\date{Received: 8 February 2007; Accepted: 20 February 2007.}

\begin{abstract}
Some new kinds of Hardy--Hilbert's integral inequality in which the
weight function is homogeneous function are given. Other results are
also obtained.
\end{abstract} \maketitle

\section{Introduction}

\noindent Let $f,g\geq0$ satisfy
\begin{eqnarray*}
0 < \int_0^\infty f ^{2}(t)dt<\infty \text{ and } 0<\int_0^\infty
g ^{2}(t)dt<\infty,
\end{eqnarray*}
then
\begin{eqnarray}\label{H.eq}
\int_0^\infty \int_0^\infty
\frac{f(x)g(y)}{x+y}dxdy<\pi\left(\int_0^\infty f^2(t)dt
\int_0^\infty g^2(t)dt \right)^{1/2}\, ,
\end{eqnarray}
where the constant factor $\pi$ is the best possible (cf. Hardy et
al. \cite{H-L-P}). Inequality (\ref{H.eq}) is well known as
Hilbert's integral inequality. This inequality had been extended by
Hardy \cite{HAR} as follows: If $p>1$, $\frac{1}{p}+\frac{1}{q}=1$,
$f,g\geq0$ satisfy
\begin{eqnarray*}
0<\int_0^\infty f^p(t)dt<\infty \text{ and } 0<\int_0^\infty
g^q(t)dt<\infty,
\end{eqnarray*}
then
\begin{eqnarray}\label{H-H.ii}
\int_0^\infty \int_0^\infty
\frac{f(x)g(y)}{x+y}dxdy<\frac{\pi}{sin({\pi/p})}\left(\int_0^\infty
f^p(t)dt\right)^{1/p}\left(\int_0^\infty g^q(t)dt\right)^{1/q},
\end{eqnarray}
where the constant factor $\frac{\pi}{sin(\pi/p)}$ is the best
possible. Inequality (\ref{H-H.ii}) is called Hardy--Hilbert's
integral inequality and is important in analysis and applications
(cf. Mitrinovic et al. \cite{M-P-F}).

B. Yang gave the following extension of (\ref{H-H.ii}) as follows :

Theorem \cite{YAN}. If $ \lambda > 2-\min\{p,q\}$ and $f,g\geq0$
satisfy
\begin{eqnarray*}
0<\int_0^\infty t^{1-\lambda}f^p(t)dt<\infty \text{ and }
0<\int_0^\infty t^{1-\lambda}g^q(t)dt<\infty,
\end{eqnarray*}
then
\begin{eqnarray*}
\int_0^\infty \int_0^\infty
\frac{f(x)g(y)}{(x+y)^\lambda}dxdy<k_\lambda(p)\left(\int_0^\infty
t^{1-\lambda}f^p(t)dt\right)^{1/p}\left(\int_0^\infty
t^{1-\lambda}g^q(t)dt\right)^{1/q},
\end{eqnarray*}
where the constant factor
$k_\lambda(p)=B(\frac{p+\lambda-2}{p},\frac{q+\lambda-2}{q})$ is the
best possible $B$ is the beta function. The function $f(x,y)$ is
said to be homogeneous of degree $\lambda$, if

\[f(tx,ty)=t^\lambda f(x,y) \qquad (t>0)\,.\]

The object of this paper is that to give some new inequalities
similar to that of Hardy--Hilbert's integral inequality.

\section{Main result}

\begin{lemma}\label{lem}
Let $K(t,1), K(1,t)$ be positive increasing functions, $0<
\mu+1\leq\alpha$. Set for $s\geq1$,
\begin{eqnarray*}
f(s)=s^{-\alpha}\int_0^s\frac{t^\mu}{K(1,t)}dt,\qquad
g(s)=s^{-\alpha}\int_0^s\frac{t^\mu}{K(t,1)}dt.
\end{eqnarray*}
Then
\[f(s)\leq f(1), g(s)\leq g(1).\]
\end{lemma}
\begin{proof}
We have
\begin{eqnarray*}
   f'(s)&=& s^{-\alpha}\frac{s^\mu}{K(1,s)}-\alpha
   s^{-\alpha-1}\int_0^s{\frac{t^\mu}{K(1,t)}dt}\\
   &\leq& \frac{s^{\mu-\alpha}}{K(1,s)}- \frac{\alpha s^{-\alpha-1}}{K(1,s)}
   \int_0^s{t^\mu dt}\\
   &=& \frac{s^{\mu-\alpha}}{K(1,s)} \left(
   1-\frac{\alpha}{\mu+1}\right)\leq 0.
\end{eqnarray*}
This shows that $f$ is nonincreasing and hence $f(s)\leq f(1)$. The
other part has a similar proof.
\end{proof}
The following is our main result

\begin{theorem}\label{main}
Let $f,g\geq0$, $K(u,v)$ be positive, increasing, homogeneous
function of degree $\lambda$, $0<\lambda \leq\min\{(1-b){q/p},
(1-a){p/q}\}$, $a,b>0$, $p>1$, $\frac{1}{p}+\frac{1}{q}=1.$ Set
\[F(u)=\int_0^uf(t)dt, \qquad G(v)=\int_0^vg(t)dt.\]
Then
\begin{eqnarray*}
\int_0^T\int_0^T\frac{F(u)G(v)}{K(u,v)}dudv&\leq&
T^\alpha\sqrt[p]{pK_1}\sqrt[q]{qK_2}\left(\int_0^T(T-t)F^{p-1}(t)f(t)dt\right)^{1/p}\nonumber \\
&&\times \left(\int_0^T(T-t)G^{q-1}(t)g(t)dt\right)^{1/q}\,,
\end{eqnarray*}
where
\[K_1=\int_0^1\frac{t^{a-1}}{K(1,t)}dt, K_2=\int_0^1\frac{t^{b-1}}{K(t,1)}dt.\]
\end{theorem}
\begin{proof}
\begin{eqnarray*}
\int_0^T\int_0^T\frac{F(u)G(v)}{K(u,v)}dudv&=
\int_0^T\int_0^T\frac{F(u)v^{\frac{a-1}{p}}}{u^{\frac{b-1}{q}}K^{1/p}(u,v)}\times
\frac{G(v)u^{\frac{b-1}{q}}}{v^{\frac{a-1}{p}}K^{1/q}(u,v)}dudv \\
&\leq\left(\int_0^T\int_0^T\frac{F^p(u)v^{a-1}}{u^{(b-1)p/q}K(u,v)}dudv\right)^{1/p}\\
&\times\left(\int_0^T\int_0^T\frac{G^q(v)u^{b-1}}{v^{(a-1)q/p}K(u,v)}dudv\right)^{1/q} \\
&=M^{1/p}N^{1/q}.
\end{eqnarray*}
We first consider
\[M=\int_0^Tu^{(1-b)p/q}F^p(u)du\int_0^T\frac{v^{a-1}}{K(u,v)}dv.\]
Observe that on putting $v=uy, dv=udy, 0\leq y\leq  t/u$, we have,
in view of Lemma \ref{lem}, by writing $\alpha=a+(1-b)p/q-\lambda$,
\begin{eqnarray*}
\int_0^T\frac{v^{a-1}}{K(u,v)}dv &=\int_0^{T/u}\frac{(uy)^{a-1}u}{K(u,uy)}dy
=u^{a-\lambda}\int_0^{T/u}\frac{y^{a-1}}{K(1,y)}dy\\
&= u^{a-\lambda}\left(\frac{T}{u}\right)^\alpha\left(\frac{T}{u}\right)^{-\alpha}\int_0^{T/u}\frac{y^{a-1}}{K(1,y)}dy\\
&\leq T^\alpha
u^{a-\lambda-\alpha}\int_0^1\frac{y^{a-1}}{K(1,y)}dy=T^\alpha
K_1u^{a-\lambda-\alpha}.
\end{eqnarray*}
By above we obtain
\begin{eqnarray*}
M &\leq T^\alpha K_1 \int_0^T
u^{a+(1-b)p/q-\lambda-\alpha}F^p(u)du\\
&= T^\alpha K_1 \int_0^T F^p(u)du.\\
\end{eqnarray*}
As
\[F^p(u)=\int_0^u\left(F^p(t)\right)'dt=p\int_0^u F^{p-1}(t)f(t)dt,\]
we have
\begin{eqnarray*}
M&\leq pT^\alpha K_1\int_0^T\int_0^u F^{p-1}(t)f(t)dtdu\\
&= pT^\alpha K_1 \int_0^T(T-t)F^{p-1}(t)f(t)dt.\\
\end{eqnarray*}
Similarly, the other part follows by using Lemma \ref{lem},
replacing $\alpha$ by $\beta$, where $\beta=b+(1-a)q/p-\lambda$ to
obtain
\[N\leq qT^\alpha K_2\int_0^T(T-t)G^{q-1}(t)g(t)dt.\]
This completes the proof of the theorem.
\end{proof}

\section{Applications}
\begin{corollary}
By an application of Theorem {\rm \ref{main}}, for the special case
$a=b =\lambda/2$, we have
\begin{eqnarray*}
\begin{split}
\int_0^T\int_0^T\frac{F(u)G(v)}{K(u,v)}dudv\leq
T^\alpha\sqrt[p]{pK_1}\sqrt[q]{qK_3}\left(\int_0^T(T-t)F^{p-1}(t)f(t)dt
\right)^{1/p}\times\\
\left(\int_0^T(T-t)G^{q-1}(t)g(t)dt\right)^{1/q}\,,
\end{split}
\end{eqnarray*}
where
\[K_3=\int_1^\infty\frac{t^{\frac{\lambda}{2}-1}}{K(1,t)}dt,\]
Furthermore, when $K(u,v)=(u+v)^\lambda$, we have
\begin{eqnarray*}
\begin{split}
\int_0^T\int_0^T\frac{F(u)G(v)}{(u+v)^\lambda}dudv\leq T^\alpha
B\left(\frac{\lambda}{2},\frac{\lambda}{2}\right)\sqrt[p]{p}
\sqrt[q]{q}\left(\int_0^T(T-t)F^{p-1}(t)f(t)dt\right)^{1/p}
\times\\
\left(\int_0^T(T-t)G^{q-1}(t)g(t)dt\right)^{1/q}\,.
\end{split}
\end{eqnarray*}
\begin{proof}
For $a=b=\lambda/2$, we have $K_2=K_3$ as
\[ \int_0^1\frac{t^{\frac{\lambda}{2}-1}}{K(t,1)}=\int_0^1\frac{t^{\frac{\lambda}{2}-1}}{K(t,tt^{-1})}dt=
\int_0^1\frac{t^{-\frac{\lambda}{2}-1}}{K(1,t^{-1})}dt=\int_1^\infty\frac{t^{\frac{\lambda}{2}-1}}{K(1,t)}dt.\]
The other part follows from the fact that for
$K(1,t)=(1+t)^\lambda,$
\[K_1=K_2=K_3=B\left(\frac{\lambda}{2},\frac{\lambda}{2}\right)\,.\]
\end{proof}
\end{corollary}
\begin{corollary}
By an application of Theorem {\rm \ref{main}} with
$K(u,v)=u^\lambda+v^\lambda$, we have
\begin{eqnarray*}
\begin{split}
\int_0^T\int_0^T\frac{F(u)G(v)}{u^\lambda+v^\lambda}dudv\leq
T^\alpha \sqrt[p]{pK_a}\sqrt[q]{qK_b}
\left(\int_0^T(T-t)F^{p-1}(t)f(t)dt\right)^{1/p}\times\\
\left(\int_0^T(T-t)G^{q-1}(t)g(t)dt\right)^{1/q},
\end{split}
\end{eqnarray*}
where
\[K_r=\int_0^1\frac{t^{r-1}}{1+t^\lambda}dt\qquad (r\in\{a,b\}).\]
\end{corollary}

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\bibliographystyle{amsplain}
\begin{thebibliography}{10}

\bibitem{HAR} G. H. Hardy, \textit{Note on a theorem of Hilbert concerning series of positive
terms}, Proc. Math. Soc. \textbf{23(2)} (1925), Records of Proc.
XLV-XLVI.

\bibitem{H-L-P} G. H. Hardy, J. E. Littlewood and G. Polya, \textit{Inequalities}, Cambridge
University Press, Cambridge, 1952.

\bibitem{M-P-F}D. S. Mitrinovic, J. E. Pecaric and A. M. Fink, \textit{Inequalities involving
functions and their integrals and derivatives}, Kluwer Academic
Publishers, Bosten, 1991.

\bibitem{YAN}B. Yang, \textit{On Hardy--Hilbert's integral inequality}, J. Math. Anal. Appl.
\textbf{261} (2001), 295--306.

\end{thebibliography}

\end{document}

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