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\footnote[0]{{\zihao{-5}supported by the key projects of Jiaying University(2014KJZ02) and College students' innovative training program of Guangdong(201510582021)}}\\
\footnote[0]{{\zihao{-5}corresponding author Weiliang Wu, E-mail:weilianggood@126.com}
}
\begin{center} {\Large
A  Hilbert-type Integral Inequality with its best Extension
 }
\end{center}
\begin{center} Weiliang Wu and Donglan Lian \\{\scriptsize (School of Mathematics Jiaying University,Guangdong Meizhou 514031, China)}
\end{center}
\par\vspace{0pt}\setlength{\baselineskip}{2\baselineskip}{{\bf Abstract:}By using the way of weight function and the technique of real analysis, a new  Hilbert-type integral inequality with a  kernel as $min\{x^{\lambda_1},y^{\lambda_2}\}$ and its equivalent form are established. As application, the constant factor on the plane are the best value and its best extension form with some parameters and the reverse forms are also considered.
\par\vspace{0pt}
 {\bf Keywords:}weight function；Hilbert-type integral inequality；best extension；reverse}
\par
{\bf Mathematics Subject Classification:26D15}\\
\par{\normalsize\noindent{\bf\zihao{4} 1. Introduction}\setcounter{section}{1}
%\setcounter{equation}{0}\setlength{\jot}{0pt}

If $f(x),g(x)\geq0,0<\int_{0}^{\infty}f^p(x)dx<\infty$ and $0<\int_{0}^{\infty}g^q(x)dx<\infty$,then we have$[1]$
\beq\int_0^\infty\int_0^\infty\frac{f(x)g(y)}{x+y}
dxdy<\pi(\int_0^\infty f^2(x)dx
\int_0^\infty g^q(y)dy )^\frac{1}{2} ,\eeq
where the constant factor $\pi$is the best possible. we call (1) Hilbert's integral inequality.(1) is important in analysis and its application$[2-3]$. In recent years, by using the way of weight function,a number of extensions (1) were given by Yang et al.$[4-6]$. In 2010, Yang $[7]$ gave a new inequality with a non-homogeneous kernel as follows:
 $$\int_0^\infty\int_0^\infty min\{x,y\}^\lambda f(x)g(y)dxdy<\frac{rs}{\lambda}(\int_0^\infty x^{p(1+\frac{\lambda}{r})-1}f^p(x)dx)^{\frac{1}{p}}(\int_0^\infty y^{q(1+\frac{\lambda}{s})-1}g^q(y)dy)^{\frac{1}{q}}\eqno(2)$$
 where the constant factor$\frac{rs}{\lambda}$ is the best possible.\par
In this paper, by introducing some parameters and using the way of weight function and the technic of real analysis and complex analysis,we give a new Hilbert-type integral inequality with a kernel as
$$k(x,y):=min\{x^{\lambda_1},y^{\lambda_2}\}$$
which is an extensions of (2). As application, the equivalent form and the reverse forms are obtained.



\par{\normalsize\noindent{\bf\zihao{4} 2. Some Lemmas}\\
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{\par\bf Lemma 1}\quad If $\lambda_{1},\lambda_{2}>0,v_1,v_2>0,v_1+v_2=1,\eta(v_1,v_2):=\frac{1}{v_1v_2}
$,define the following weight function：$$\omega_{\lambda_{1}，\lambda_{2}}(v_1,v_2,y):=\int_0^\infty k(x,y)\frac{y^{-\lambda_{2}v_2}}{x^{1+\lambda_{1}v_1}}dx
\quad (y>0),$$
$$\omega_{\lambda_{1}，\lambda_{2}}(v_1,v_2,x):=\int_0^\infty k(x,y)\frac{x^{-\lambda_{1}v_1}}{y^{1+\lambda_{2}v_2}}dy
\quad (x>0),$$
then we have $$\omega_{\lambda_{1}，\lambda_{2}}(v_1,v_2,y)=\frac{1}{\lambda_{1}}\eta(v_1,v_2),$$$$\omega_{\lambda_{1}，\lambda_{2}}(v_1,v_2,y)=\frac{1}{\lambda_{2}}\eta(v_1,v_2).$$

{\bf Proof}\quad Setting $u=x^{\lambda_{1}}/y^{\lambda_{2}},$ we obtain
\begin{eqnarray*}
\omega_{\lambda_{1},\lambda_{2}}(v_1,v_2,y) & = & \frac{1}{\lambda_{1}}\int_0^\infty min\{1,u\}u^{-v_1-1}du\\
& = & \frac{1}{\lambda_{1}}[\int_0^1u^{v_2-1}du+\int_1^\infty u^{-v_1-1}du]\\
& = &\frac{1}{\lambda_{1}}(\frac{1}{v_2}+\frac{1}{v_1})=\frac{1}{\lambda_{1}}\eta(v_1,v_2).\end{eqnarray*}
Similarly, we can calculate that $$\omega_{\lambda_{1}，\lambda_{2}}(v_1,v_2,y)=\frac{1}{\lambda_{2}}\eta(v_1,v_2).$$

{\par\bf Lemma 2}\quad As the assumption of Lemma1, if $p>1,\frac{1}{p}+\frac{1}{q}=1,f(x)\geq0,$we have
{$$J:=\int_0^\infty y^{-p\lambda_{2}v_2-1}\left[\int_0^\infty k(x,y)f(x)dx \right]^pdy
\leq\left[\frac{\eta(v_1,v_2)}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}\right]^p\int_0^\infty x^{p(1+\lambda_{1}v_1)-1}f^p(x)dx.\eqno(4)$$}

\par {\bf Proof}\quad By H\"{o}leder's inequality with weight$^{[8]}$,we obtain
\begin{eqnarray*}
& &\int_0^\infty f(x)k(x,y)dx=\int_0^\infty k(x,y)[\frac{x^{(1+\lambda_{1}v_1)/q}}{y^{(1+\lambda_{2}v_2)/p}}f(x)][\frac{y^{(1+\lambda_{2}v_2)/p}}{x^{(1+\lambda_{1}v_1)/q}}]dx  \\
& \leq &\{\int_0^\infty k(x,y)\frac{x^{(1+\lambda_{1}v_1)(p-1)}}{y^{(1+\lambda_{2}v_2)}}f^p(x)dx\}^{1/p}
\{\int_0^\infty k(x,y)\frac{y^{(1+\lambda_{2}v_2)(q-1)}}{x^{(1+\lambda_{1}v_1)}}dx\}^{1/q}\\
& = &y^{\frac{1}{p}+\lambda_{2}v_2}[\omega_{\lambda_{1}，\lambda_{2}}(v_1,v_2,y)]^{1/q}\{\int_0^\infty k(x,y)\frac{x^{(1+\lambda_{1}v_1)(p-1)}}{y^{(1+\lambda_{2}v_2)}}f^p(x)dx\}^{1/p}.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5) \end{eqnarray*}
By the result of Lemma 1 and Fubini's theorem$^{[9]}$, we obtain
\begin{eqnarray*}
J& \leq &[\frac{\eta(v_1,v_2)}{\lambda_{1}}]^{p-1}\int_0^\infty\int_0^\infty k(x,y)\frac{x^{(1+\lambda_{1}v_1)(p-1)}}{y^{(1+\lambda_{2}v_2)}}f^p(x)dxdy\\
& = &[\frac{\eta(v_1,v_2)}{\lambda_{1}}]^{p-1}\int_0^\infty\omega_{\lambda_{1},\lambda_{2}}(\alpha,v_1,v_2,x)x^{p(1+\lambda_{1}v_1)-1}f^p(x)dx.\end{eqnarray*}
Hence by the above results, we have (4). The lemma is proved.\\

\par{\normalsize\noindent{\bf\zihao{4} 3. Main results}\\
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{\par\bf Theorem 1} As the assumption of Lemma1, if $p>1,\frac{1}{p}+\frac{1}{q}=1,f(x),g(y)\geq 0,
0<\int_0^\infty x^{p(1+\lambda_{1}v_1)-1}f^p(x)dx<\infty,0<\int_0^\infty y^{q(1+\lambda_{2}v_2)-1}g^q(y)dy<\infty,$ then we have two equivalent inequalities as
\begin{eqnarray*}
 I& :=&\int_0^\infty\int_0^\infty k(x,y)f(x)g(y)dxdy\\
& <&\frac{\eta(v_1,v_2,)}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}\left\{\int_0^\infty x^{p(1+\lambda_{1}v_1)-1}f^p(x)dx\right\}^{\frac{1}{p}}\left\{\int_0^\infty y^{q(1+\lambda_{2}v_2)-1}g^q(y)dy\right\}^{\frac{1}{q}}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6) \end{eqnarray*}
$$
J=\int_0^\infty y^{-p\lambda_{2}v_2-1}\left[\int_0^\infty k(x,y)f(x)dx \right]^pdy<[\frac{\eta(v_1,v_2)}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}]^{p}\int_0^\infty x^{p(1+\lambda_{1}v_1)-1}f^p(x)dx.\eqno(7)$$
\par {\bf Proof} We conform that the middle of (5) keeps the form of strict inequality. Otherwise,there exist constants A and B, such that they are not all zero and $[9]$
$$A\frac{x^{(1+\lambda_{1}v_1)(p-1)}}{y^{(1+\lambda_{2}v_2)}}f^p(x)=B\frac{y^{(1+\lambda_{2}v_2)(q-1)}}{x^{(1+\lambda_{1}v_1)}}g^q(y) \quad a.e. in (0,\infty)\times(0,\infty).$$
It follows $Ax^{p(1+\lambda_{1}v_1)}f^p(x)=By^{q(1+\lambda_{2}v_2)}g^q(y)\quad a.e. in (0,\infty)\times(0,\infty).$ Assuming that $A\neq0$, there exists $y>0$, such that $x^{p(1+\lambda_{1}v_1)-1}f^p(x)=\frac{By^{q(1+\lambda_{2}v_2)}g^q(y)}{Ax} a.e. in x\in(0,\infty).$This contradicts the fact that $0<\int_0^\infty x^{p(1+\lambda_{1}v_1)-1}f^p(x)dx<\infty$.Hence(5) takes the strict sign-inequality,so does (4). In addition, we have (7). By H\"{o}lder's inequality$^{[8]}$ ,we find
\begin{eqnarray*}
I&=&\int_0^\infty[y^{-\lambda_{2}v_2-1/p}\int_0^\infty k(x,y)f(x)dx][y^{\lambda_{2}v_2+1/p}g(y)]dy\\
&\leq &J^{1/p}[\int_0^\infty y^{q(1+\lambda_{2}v_2)-1}g^q(y)dy]^{1/q}.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (8)\end{eqnarray*}
Then by (7), we have (6).On the other hand,assuming that (7) is valid ,setting $$g(y):=y^{-p\lambda_{2}v_2-1}[\int_0^\infty k(x,y)f(x)]dx]^{p-1},
$$then we have $J=\int_0^\infty y^{q(1+\lambda_{2}v_2)-1}g^{q}(y)dy,$Through (4),it follows $J<\infty$.If$J=0$,then  （7）is naturally valid.If $0<J<\infty$, then by(6),we find
\begin{eqnarray*}
0& <&\int_0^\infty y^{q(1+\lambda_{2}v_2)-1}g^{q}(y)dy=J=I<\\
& &\frac{\eta(v_1,v_2)}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}\left\{\int_0^\infty x^{p(1+\lambda_{1}v_1)-1}f^p(x)dx\right\}^{\frac{1}{p}}\left\{\int_0^\infty y^{q(1+\lambda_{2}v_2)-1}g^q(y)dy\right\}^{\frac{1}{q}}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (9) \end{eqnarray*}
$$ J^{1/p}=\left\{\int_0^\infty y^{q(1+\lambda_{2}v_2)-1}g^q(y)dy\right\}^{\frac{1}{p}}<\frac{\eta(v_1,v_2}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}\left\{\int_0^\infty x^{p(1+\lambda_{1}v_1)-1}f^p(x)dx\right\}^{\frac{1}{p}}.\eqno(10)$$
and then we have (7),which is equivalent to (6). The Theorem is proved.


{\par\bf Theorem 2}\quad Under the conditions of Theorem 1，the constants $\frac{\eta(v_1,v_2)}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}$and $\left[\frac{\eta(v_1,v_2)}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}\right]^p$in (6)and (7) are the best possible.\par
{\bf Proof}\quad For $0<\varepsilon<pv_1$,if\par
$$\tilde{f}(x):=\left\{\begin{array}{cc}
                                                              0, & \quad x\in(0,1) \\
                                                              x^{-\lambda_{1}v_1-\lambda_{1}\varepsilon/p-1}, &\quad x\in[1,\infty)
                                                            \end{array},\right.
\tilde{g}(y):=\left\{\begin{array}{cc}
                                                              0, & \quad y\in(0,1) \\
                                                              y^{-\lambda_{2}v_2-\lambda_{2}\varepsilon/q-1}, &\quad y\in[1,\infty)
                                                            \end{array}.\right.                                                            $$
we can calculate $$\tilde{J}:=\left\{\int_0^\infty x^{p(1+\lambda_{1}v_1)-1}\tilde{f}^{p}(x)dx\right\}^{\frac{1}{p}}\left\{\int_0^\infty y^{q(1+\lambda_{2}v_2)-1}\tilde{g}^{q}(y)dy\right\}^{\frac{1}{q}}
=\frac{1}{\lambda_{1}^{1/p}\lambda_{2}^{1/q}\varepsilon}.$$ by Fubini's theorem$^{[9]}$,we obtain
\begin{eqnarray*}
I:& =&\int_0^\infty \int_0^\infty k(x,y)\tilde{f}(x)\tilde{g}(x)dxdy\\
&= &\int_1^\infty y^{-\lambda_{2}v_2-\lambda_{2}\varepsilon/q-1}\left[\int_1^\infty k(x,y) x^{-\lambda_{1}v_1-\lambda_{1}\varepsilon/p-1}dx\right]dy
\\
&\underset {=}{u=(x^{\lambda_{1}}/y^{\lambda_{2}})} &\frac{1}{\lambda_{1}}\int_1^\infty  y^{-\lambda_{2}\varepsilon-1}\left[\int_{y^{-\lambda_{2}}}^{\infty}min\{1,u\}u^{-v_1-\varepsilon/p-1}du\right]dy \\
&= &\frac{1}{\lambda_{1}}\left\{\int_1^\infty y^{-\lambda_{2}\varepsilon-1}\left[\int_{y^{-\lambda_{2}}}^1u^{v_2-\varepsilon/p-1}du+\int_1^\infty u^{-v_1-\varepsilon/p-1}du\right]dy\right\}
\\
&= &\frac{1}{\lambda_{1}}\left\{\int_0^1(\int_{u^{-1/\lambda_{2}}}^\infty y^{-\lambda_2\varepsilon-1}dy)u^{v_2-\varepsilon/p-1}du+
\frac{1}{\lambda_{2}\varepsilon}\int_1^\infty u^{-v_1-\varepsilon/p-1}du\right\}\\
&= &\frac{1}{\lambda_{1}\lambda_{2}\varepsilon}\left[\int_0^1u^{v_2+\varepsilon/q-1}du+\int_1^\infty u^{-v_1-\varepsilon/p-1}du\right]. \end{eqnarray*}
If there exists a positive number $k\leq \frac{\eta(v_1,v_2,)}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}},$such that (6)is still valid when we replace $\frac{\eta(v_1,v_2)}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}$ by $k$，then in particular $\tilde{f},\tilde{g},$ by the above results ,we find
$$ \frac{1}{\lambda_{1}\lambda_{2}}\left[\int_0^1u^{v_2+\varepsilon/q-1}du+\int_1^\infty u^{-v_1-\varepsilon/p-1}du\right]=\varepsilon \tilde{I}<
\varepsilon k\tilde{J}=k\frac{1}{\lambda_{1}^{1/p}\lambda_{2}^{1/q}}.\eqno(11)$$
By $Fatou$ lemma$^{[9]}$and（11），we obtain\par$$
\frac{\eta(v_1,v_2)}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}=\frac{1}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}\left[\int_0^1\underset{\varepsilon\rightarrow0^+}{\underline{lim}}u^{v_2+\varepsilon/q-1}du+\int_1^\infty \underset{\varepsilon\rightarrow0^+}{\underline{lim}}u^{-v_1-\varepsilon/p-1}du\right]\leq$$$$\frac{1}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}\underset{\varepsilon\rightarrow0^+}{\underline{lim}}
\left[\int_0^1u^{v_2+\varepsilon/q-1}du+\int_1^\infty u^{-v_1-\varepsilon/p-1}du\right]\leq k.$$\\
Hence $k=\frac{\eta(v_1,v_2)}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}$is the best value of (6).We conform that $\left[\frac{\eta(v_1,v_2)}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}\right]^p$ in (7)is the best possible, otherwise we can get a contradiction by （8）that the constant in (6) is not the best possible.The theorem is proved.

{\par\bf Theorem 3}\quad As the assumption of Theorem 1，if $0<p<1,I,J$are defined by (6)and (4),then we have equivalent inequalities as
$$I>\frac{\eta(v_1,v_2,)}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}\left\{\int_0^\infty x^{p(1+\lambda_{1}v_1)-1}f^p(x)dx\right\}^{\frac{1}{p}}\left\{\int_0^\infty y^{q(1+\lambda_{2}v_2)-1}g^q(y)dy\right\}^{\frac{1}{q}}\eqno(12)$$
$$J>\left[\frac{\eta(v_1,v_2)}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}\right]^p\int_0^\infty x^{p(1+\lambda_{1}v_1)-1}f^p(x)dx.\eqno(13)$$
where the constants in (12)and (13) are the best possible.\par
{\bf Proof}\quad By the reverse H\"{o}lder's inequality$^{[9]}$, we have the reverse of (5),(4),(8). By the same way of Theorem 1,we show that the reverse of(5) keeps the strict sign-inequality and then we have (13).By the reverse of (8) ,we have (12).Assuming that (12)is valid,setting $g(y)$ as Theorem 1.In view of (4), we have $J>0$.If $J=\infty$, then (13) is valid；if$0<J<\infty$,then by（12），we easily find the reverses of (9) and (10),hence the reverse of （7）is valid ，which is equivalent to (12).\
\par
If there exists a positive number $k\leq \frac{\eta(v_1,v_2,)}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}},$such that (12)is still valid when we replace $\frac{\eta(v_1,v_2)}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}$ by $k$，then by the reverse of (11),we find
$$\frac{1}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}
\left[\int_0^1u^{v_2+\varepsilon/q-1}du+\int_1^\infty u^{-v_1-\varepsilon/p-1}du\right]> k\eqno(14)$$
If $0<\varepsilon_0<\left|q\right|v_2$, when $0<\varepsilon\leq\varepsilon_0$，$u\in(0,1],$$u^{\varepsilon/q}\leq u^{\varepsilon_0/q}$ is valid ,and we obtain \\$$\int_0^1
u^{v_2+\varepsilon_0/q-1}du\leq \frac{1}{\varepsilon_0/q+v_2}$$by Lebesgue's control convergence theorem $^{[10]}$, we obtain $$\int_0^1 u^{v_2+\varepsilon/q-1}du=\int_0^1 u^{v_2-1}du+o(1)\quad (\varepsilon\rightarrow0^+).$$We find $\frac{\eta(v_1,v_2,\alpha)}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}\geq k (\varepsilon\rightarrow0^+)of (14)$,
hence $\frac{\eta(v_1,v_2,\alpha)}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}= k$ is the best value of (12). We conform that the constant factor of (13 ) is the best possible,otherwise we can get a contradiction that the constant in reverse of (8) is not the best possible.The theorem is proved.\\\\
{\bf Remark} For $\lambda_1=\lambda_2=\lambda,v_1=\frac{1}{r},v_2=\frac{1}{s}$ in (6), it deduces to (2). Hence inequality (6) is the best extension of (2).\\\\
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