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\noindent\parbox{2.95cm}{\includegraphics*[keepaspectratio=true,scale=0.125]{AFA.jpg}}
\noindent\parbox{4.85in}{\hspace{0.1mm}\\[1.5cm]\noindent
\qquad Ann. Funct. Anal. 3 (2012), no. 1, 142--150\\
{\footnotesize \qquad \textsc{\textbf{$\mathscr{A}$}nnals of
\textbf{$\mathscr{F}$}unctional
\textbf{$\mathscr{A}$}nalysis}\\
\qquad ISSN: 2008-8752 (electronic)\\
\qquad URL:
\textcolor[rgb]{0.00,0.00,0.99}{www.emis.de/journals/AFA/}
}\\[.5in]}

\title[Half-Discrete Mulholland-Type Inequality with Parameters]{ A New Half-Discrete Mulholland-Type Inequality with Parameters}

\author[B. Yang]{Bicheng Yang}

\address{Department of Mathematics, Guangdong University of
\and Education, Guangzhou, Guangdong 510303, P. R. China.}
\email{\textcolor[rgb]{0.00,0.00,0.84}{ bcyang@gdei.edu.cn}}

\dedicatory{{\rm Communicated by J. Soria}}

\subjclass[2010]{Primary 26D15; Secondary  47A07.}

\keywords{Mulholland-type inequality, weight function, equivalent
form.}

\date{Received: 6 December 2011; Accepted: 3 March 2012.}

\begin{abstract}
 By means of weight functions and Hadamard's inequality, a new half-discrete
Mulholland-type inequality with a best constant factor is given. A best
extension with parameters, some equivalent forms, the operator expressions
as well as some particular cases are also considered.
\end{abstract} \maketitle

\section{Introduction}

\noindent Assuming that $f,g\in L^{2}(R_{+}),||f||$ $=\{\int_{0}^{\infty
}f^{2}(x)dx\}^{\frac{1}{2}}$ $>0,||g||>0,$ we have the following Hilbert's
integral inequality (cf. \cite{HLP}):
\begin{equation}
\int_{0}^{\infty }\int_{0}^{\infty }\frac{f(x)g(y)}{x+y}dxdy<\pi ||f||||g||,
\label{1}
\end{equation}%
where the constant factor $\pi $ is best possible. If $a=\{a_{n}\}_{n=1}^{%
\infty },b=\{b_{n}\}_{n=1}^{\infty }\in l^{2},||a||=\{\sum_{n=1}^{\infty
}a_{n}^{2}\}^{\frac{1}{2}}>0,||b||>0,$ then we have the following analogous
discrete Hilbert's inequality
\begin{equation}
\sum_{m=1}^{\infty }\sum_{n=1}^{\infty }\frac{a_{m}b_{n}}{m+n}<\pi
||a||||b||,  \label{2}
\end{equation}%
with the same best constant factor $\pi $. Inequalities (\ref{1}) and (\ref%
{2}) are important in analysis and its applications (cf. \cite{MPF,
Y1, Y2}). On the other-hand, we have the following Mulholland's
inequality with the same best constant factor (cf. \cite{HLP, Y7}):%
\begin{equation}
\sum_{m=2}^{\infty }\sum_{n=2}^{\infty }\frac{a_{m}b_{n}}{\ln mn}<\pi
\left\{ \sum_{m=2}^{\infty }ma_{m}^{2}\sum_{n=2}^{\infty }nb_{n}^{2}\right\}
^{\frac{1}{2}}.  \label{3}
\end{equation}

In 1998, by introducing an independent parameter $\lambda \in (0,1]$, Yang
\cite{Y3} gave an extension of (\ref{1}). Refinement
the results from \cite{Y3}, Yang \cite{Y4} gave some best extensions of (\ref%
{1}) and (\ref{2}): If $p>1,\frac{1}{p}+\frac{1}{q}=1,\lambda _{1}+\lambda
_{2}=\lambda ,k_{\lambda }(x,y)$ is a non-negative homogeneous function of
degree $-\lambda $ satisfying $k(\lambda _{1})=\int_{0}^{\infty }k_{\lambda
}(t,1)t^{\lambda _{1}-1}dt$ \\$\in R_{+},$ $\phi (x)=x^{p(1-\lambda _{1})-1}$, $%
\psi (x)=x^{q(1-\lambda _{2})-1},$ $$f(\geq 0)\in L_{p,\phi }(R_{+})
=\{f|||f||_{p,\phi }:=\{\int_{0}^{\infty }\phi (x)|f(x)|^{p}dx\}^{\frac{1}{p}%
}<\infty \},$$ $g(\geq 0)\in L_{q,\psi }(R_{+}),$ and $||f||_{p,\phi },$ $%
||g||_{q,\psi }>0,$ then
\begin{equation}
\int_{0}^{\infty }\int_{0}^{\infty }k_{\lambda }(x,y)f(x)g(y)dxdy<k(\lambda
_{1})||f||_{p,\phi }||g||_{q,\psi },  \label{4}
\end{equation}%
where the constant factor $k(\lambda _{1})$ is best possible. Moreover if $%
k_{\lambda }(x,y)$ is finite and $k_{\lambda }(x,y)x^{\lambda
_{1}-1}(k_{\lambda }(x,y)y^{\lambda _{2}-1})$ is decreasing for $x>0(y>0),$
then for $a_{m,}b_{n}\geq 0,$ $a=\{a_{m}\}_{m=1}^{\infty }\in l_{p,\phi
}=\{a|||a||_{p,\phi }:=\{\sum_{n=1}^{\infty }\phi (n)|a_{n}|^{p}\}^{\frac{1}{%
p}}<\infty \},$ and $b=\{b_{n}\}_{n=1}^{\infty }\in l_{q,\psi },$ $%
||a||_{p,\phi },||b||_{q,\psi }>0,$ we have
\begin{equation}
\sum_{m=1}^{\infty }\sum_{n=1}^{\infty }k_{\lambda
}(m,n)a_{m}b_{n}<k(\lambda _{1})||a||_{p,\phi }||b||_{q,\psi },  \label{5}
\end{equation}%
where the constant $k(\lambda _{1})$ is still best value. Clearly, for $%
p=q=2,\lambda =1,k_{1}(x,y)=\frac{1}{x+y},\lambda _{1}=\lambda _{2}=\frac{1}{%
2},$ (\ref{4}) reduces to (\ref{1}), while (\ref{5}) reduces to (\ref{2}).

Some other results about Hilbert-type inequalities can be found in
\cite{LH}-\cite{YMP}. On half-discrete Hilbert-type inequalities
with the general non-homogeneous kernels, Hardy et al. provided a
few results in Theorem 351 of \cite{HLP}. But they did not prove
that the constant factors are best
possible. In 2005, Yang \cite{Y5} gave a result with the kernel $\frac{1}{%
(1+nx)^{\lambda }}$ by introducing a variable and proved that the
constant factor is best possible. Very recently, Yang \cite{Y6, Y7}
gave
the following half-discrete Hilbert's inequality with best constant factor:%
\begin{equation}
\int_{0}^{\infty }f(x)\sum_{n=1}^{\infty }\frac{a_{n}}{(x+n)^{\lambda }}%
dx<\pi ||f||||a||.  \label{6}
\end{equation}

In this paper, by means of weight functions and Hadamard's inequality, a new
half-discrete Mulholland-type inequality similar to (\ref{3}) and (\ref{6})
with a best constant factor is given as follows:%
\begin{eqnarray}
&&\int_{\frac{3}{2}}^{\infty }f(x)\sum_{n=2}^{\infty }\frac{a_{n}}{1+\ln (x-%
\frac{1}{2})\ln (n-\frac{1}{2})}dx  \notag \\
&<&\pi \left\{ \int_{\frac{3}{2}}^{\infty }(x-\frac{1}{2})f^{2}(x)dx%
\sum_{n=2}^{\infty }(n-\frac{1}{2})a_{n}^{2}\right\} ^{\frac{1}{2}}.
\label{7}
\end{eqnarray}%
Moreover, a best extension of (\ref{7}) with multi-parameters, some
equivalent forms, the operator expressions as well as some particular
inequalities are considered.
\bigskip

\section{Some lemmas}

\begin{lemma}
\textbf{} If $0<\lambda \leq 2,\alpha \in \mathbf{R},\beta \leq
\frac{1}{2},$ and $\omega (n)$ and $\varpi (x)$ are weight functions given by%
\begin{eqnarray}
\omega (n) &:&=\int_{1+\alpha }^{\infty }\frac{\ln ^{\frac{\lambda }{2}%
}(n-\beta )\ln ^{\frac{\lambda }{2}-1}(x-\alpha )dx}{(x-\alpha )[1+\ln
(x-\alpha )\ln (n-\beta )]^{\lambda }},n\in \mathbf{N\backslash \{}1\mathbf{%
\},}  \label{8} \\
\varpi (x) &:&=\sum_{n=2}^{\infty }\frac{\ln ^{\frac{\lambda }{2}}(x-\alpha
)\ln ^{\frac{\lambda }{2}-1}(n-\beta )}{(n-\beta )[1+\ln (x-\alpha )\ln
(n-\beta )]^{\lambda }},x>1+\alpha ,  \label{9}
\end{eqnarray}%
then we have
\begin{equation}
\varpi (x)<\omega (n)=B(\frac{\lambda }{2},\frac{\lambda }{2}).  \label{10}
\end{equation}\end{lemma}

\begin{proof}
\emph{ }Substituting $t=\ln (x-\alpha )\ln (n-\beta )$ in (\ref{8}),
and by simple calculation, we have%
\begin{equation*}
\omega (n)=\int_{0}^{\infty }\frac{1}{(1+t)^{\lambda }}t^{\frac{\lambda }{2}%
-1}dt=B(\frac{\lambda }{2},\frac{\lambda }{2}).
\end{equation*}%
For fixed $x>1+\alpha $, in view of the conditions, it is easy to see that%
\begin{equation*}
h(x,y):=\frac{\ln ^{\frac{\lambda }{2}-1}(y-\beta )}{(y-\beta )[1+\ln
(x-\alpha )\ln (y-\beta )]^{\lambda }}
\end{equation*}%
is decreasing and strictly convex with $h_{y}^{\prime }(x,y)<0$ and $%
h_{y^{2}}^{\prime \prime }(x,y)>0,$ for $y\in (\frac{3}{2},\infty ).$ Hence
by (\ref{9}) and Hadamard's inequality (cf. \cite{K1}), we find%
\begin{equation*}
\varpi (x)<\ln ^{\frac{\lambda }{2}}(x-\alpha )\int_{\frac{3}{2}}^{\infty }%
\frac{\ln ^{\frac{\lambda }{2}-1}(y-\beta )dy}{(y-\beta )[1+\ln (x-\alpha
)\ln (y-\beta )]^{\lambda }}
\end{equation*}%
\begin{equation*}
\overset{t=\ln (x-\alpha )\ln (y-\beta )}{=}\int_{\ln (x-\alpha )\ln (\frac{3%
}{2}-\beta )}^{\infty }\frac{t^{\frac{\lambda }{2}-1}}{(1+t)^{\lambda }}%
dt\leq B(\frac{\lambda }{2},\frac{\lambda }{2}),  \notag
\end{equation*}%
and (\ref{10}) follows.
\end{proof}


\begin{lemma}
\textbf{} Let the assumptions of Lemma 1 be fulfilled and
additionally, let $p>1,\frac{1}{p}+\frac{1}{q}=1,a_{n}\geq 0,n\in \mathbf{%
N\backslash \{}1\mathbf{\}},f(x)$ is a non-negative measurable function in $%
(1+\alpha ,\infty ).$ Then we have the following inequalities:%
\begin{equation*}
J:=\left\{ \sum_{n=2}^{\infty }\frac{\ln ^{\frac{p\lambda }{2}-1}(n-\beta )}{%
n-\beta }\left[ \int_{1+\alpha }^{\infty }\frac{f(x)dx}{[1+\ln (x-\alpha
)\ln (n-\beta )]^{\lambda }}\right] ^{p}\right\} ^{\frac{1}{p}}
\end{equation*}%
\begin{equation}
\leq \left[ B(\frac{\lambda }{2},\frac{\lambda }{2})\right] ^{\frac{1}{q}%
}\left\{ \int_{1+\alpha }^{\infty }\varpi (x)(x-\alpha )^{p-1}\ln ^{p(1-%
\frac{\lambda }{2})-1}(x-\alpha )f^{p}(x)dx\right\} ^{\frac{1}{p}},
\label{11}
\end{equation}%
\begin{eqnarray}
L_{1} &:&=\left\{ \int_{1+\alpha }^{\infty }\frac{(x-\alpha )^{\frac{%
q\lambda }{2}-1}}{[\varpi (x)]^{q-1}}\left[ \sum_{n=2}^{\infty }\frac{a_{n}}{%
[1+\ln (x-\alpha )\ln (n-\beta )]^{\lambda }}\right] ^{q}dx\right\} ^{\frac{1%
}{q}}  \notag \\
&\leq &\left\{ B(\frac{\lambda }{2},\frac{\lambda }{2})\sum_{n=2}^{\infty
}(n-\beta )^{q-1}\ln ^{q(1-\frac{\lambda }{2})-1}(n-\beta )a_{n}^{q}\right\}
^{\frac{1}{q}}.  \label{12}
\end{eqnarray}
\end{lemma}

\begin{proof}
\emph{} Setting $k(x,n):=\frac{1}{[1+\ln (x-\alpha )\ln (n-\beta
)]^{\lambda }},$ by H\"{o}lder's inequality (cf. \cite{K1}) and (\ref{10}),
it follows
\begin{eqnarray*}
&&\left[ \int_{1+\alpha }^{\infty }\frac{f(x)dx}{[1+\ln (x-\alpha )\ln
(n-\beta )]^{\lambda }}\right] ^{p} \\
&=&\left\{ \int_{1+\alpha }^{\infty }k(x,n)\left[ \frac{\ln ^{(1-\frac{%
\lambda }{2})/q}(x-\alpha )}{\ln ^{(1-\frac{\lambda }{2})/p}(n-\beta )}\frac{%
(x-\alpha )^{\frac{1}{q}}f(x)}{(n-\beta )^{\frac{1}{p}}}\right] \right.
\end{eqnarray*}%
\begin{eqnarray*}
&&\times \left. \left[ \frac{\ln ^{(1-\frac{\lambda }{2})/p}(n-\beta )}{\ln
^{(1-\frac{\lambda }{2})/q}(x-\alpha )}\frac{(n-\beta )^{\frac{1}{p}}}{%
(x-\alpha )^{\frac{1}{q}}}\right] dx\right\} ^{p} \\
&\leq &\int_{1+\alpha }^{\infty }k(x,n)\frac{(x-\alpha )^{p-1}\ln ^{(1-\frac{%
\lambda }{2})(p-1)}(x-\alpha )}{(n-\beta )\ln ^{1-\frac{\lambda }{2}%
}(n-\beta )}f^{p}(x)dx
\end{eqnarray*}%
\begin{equation*}
\times \left\{ \int_{1+\alpha }^{\infty }k(x,n)\frac{(n-\beta )^{q-1}\ln
^{(1-\frac{\lambda }{2})(q-1)}(n-\beta )}{(x-\alpha )\ln ^{1-\frac{\lambda }{%
2}}(x-\alpha )}dx\right\} ^{p-1}
\end{equation*}%
\begin{eqnarray*}
&=&\left\{ \omega (n)\frac{\ln ^{q(1-\frac{\lambda }{2})-1}(n-\beta )}{%
(n-\beta )^{1-q}}\right\} ^{p-1} \\
&&\times \int_{1+\alpha }^{\infty }k(x,n)\frac{(x-\alpha )^{p-1}\ln ^{(1-%
\frac{\lambda }{2})(p-1)}(x-\alpha )}{(n-\beta )\ln ^{1-\frac{\lambda }{2}%
}(n-\beta )}f^{p}(x)dx \\
&=&\frac{\left[ B(\frac{\lambda }{2},\frac{\lambda }{2})\right]
^{p-1}(n-\beta )}{\ln ^{\frac{p\lambda }{2}-1}(n-\beta )}\int_{1+\alpha
}^{\infty }k(x,n)\frac{(x-\alpha )^{p-1}\ln ^{(1-\frac{\lambda }{2}%
)(p-1)}(x-\alpha )f^{p}(x)}{(n-\beta )\ln ^{1-\frac{\lambda }{2}}(n-\beta )}%
dx.
\end{eqnarray*}

Then by the Lebesgue term by term integration theorem (cf. \cite{K2}), we have%
\begin{eqnarray*}
J &\leq &\left[ B(\frac{\lambda }{2},\frac{\lambda }{2})\right] ^{\frac{1}{q}%
}\left\{ \sum_{n=2}^{\infty }\int_{1+\alpha }^{\infty }k(x,n)\frac{(x-\alpha
)^{p-1}\ln ^{(1-\frac{\lambda }{2})(p-1)}(x-\alpha )}{(n-\beta )\ln ^{1-%
\frac{\lambda }{2}}(n-\beta )}f^{p}(x)dx\right\} ^{\frac{1}{p}} \\
&=&\left[ B(\frac{\lambda }{2},\frac{\lambda }{2})\right] ^{\frac{1}{q}%
}\left\{ \int_{1+\alpha }^{\infty }\sum_{n=2}^{\infty }k(x,n)\frac{(x-\alpha
)^{p-1}\ln ^{(1-\frac{\lambda }{2})(p-1)}(x-\alpha )}{(n-\beta )\ln ^{1-%
\frac{\lambda }{2}}(n-\beta )}f^{p}(x)dx\right\} ^{\frac{1}{p}} \\
&=&\left[ B(\frac{\lambda }{2},\frac{\lambda }{2})\right] ^{\frac{1}{q}%
}\left\{ \int_{1+\alpha }^{\infty }\varpi (x)(x-\alpha )^{p-1}\ln ^{p(1-%
\frac{\lambda }{2})-1}(x-\alpha )f^{p}(x)dx\right\} ^{\frac{1}{p}},
\end{eqnarray*}
hence, (\ref{11}) follows. By H\"{o}lder's inequality again, we have
\begin{equation*}
\left[ \sum_{n=2}^{\infty }k(x,n)a_{n}\right] ^{q}=\left\{
\sum_{n=2}^{\infty }k(x,n)\left[ \frac{(x-\alpha )^{\frac{1}{q}}\ln ^{(1-%
\frac{\lambda }{2})/q}(x-\alpha )}{(n-\beta )^{\frac{1}{p}}\ln ^{(1-\frac{%
\lambda }{2})/p}(n-\beta )}\right] \right.
\end{equation*}%
\begin{eqnarray*}
&&\times \left. \left[ \frac{\ln ^{(1-\frac{\lambda }{2})/p}(n-\beta )}{\ln
^{(1-\frac{\lambda }{2})/q}(x-\alpha )}\frac{(n-\beta )^{\frac{1}{p}}a_{n}}{%
(x-\alpha )^{\frac{1}{q}}}\right] \right\} ^{q} \\
&\leq &\left\{ \sum_{n=2}^{\infty }k(x,n)\frac{(x-\alpha )^{p-1}\ln ^{(1-%
\frac{\lambda }{2})(p-1)}(x-\alpha )}{(n-\beta )\ln ^{1-\frac{\lambda }{2}%
}(n-\beta )}\right\} ^{q-1}
\end{eqnarray*}%
\begin{eqnarray*}
&&\times \sum_{n=2}^{\infty }k(x,n)\frac{(n-\beta )^{q-1}\ln ^{(1-\frac{%
\lambda }{2})(q-1)}(n-\beta )}{(x-\alpha )\ln ^{1-\frac{\lambda }{2}%
}(x-\alpha )}a_{n}^{q} \\
&=&\frac{[\varpi (x)]^{q-1}(x-\alpha )}{\ln ^{\frac{q\lambda }{2}%
-1}(x-\alpha )}\sum_{n=2}^{\infty }k(x,n)\frac{\ln ^{\frac{\lambda }{2}%
-1}(x-\alpha )}{(x-\alpha )(n-\beta )^{1-q}}\ln ^{(1-\frac{\lambda }{2}%
)(q-1)}(n-\beta )a_{n}^{q}.
\end{eqnarray*}%
By Lebesgue term by term integration theorem, we have%
\begin{equation*}
L_{1}\leq \left\{ \int_{1+\alpha }^{\infty }\sum_{n=2}^{\infty }k(x,n)\frac{%
\ln ^{\frac{\lambda }{2}-1}(x-\alpha )}{(x-\alpha )}(n-\beta )^{q-1}\ln ^{(1-%
\frac{\lambda }{2})(q-1)}(n-\beta )a_{n}^{q}dx\right\} ^{\frac{1}{q}}
\end{equation*}%
\begin{equation*}
=\left\{ \sum_{n=2}^{\infty }\left[ \int_{1+\alpha }^{\infty }k(x,n)\frac{%
\ln ^{\frac{\lambda }{2}}(n-\beta )\ln ^{\frac{\lambda }{2}-1}(x-\alpha )dx}{%
x-\alpha }\right] \frac{\ln ^{q(1-\frac{\lambda }{2})-1}(n-\beta )}{(n-\beta
)^{1-q}}a_{n}^{q}\right\} ^{\frac{1}{q}}
\end{equation*}%
\begin{equation*}
=\left\{ \sum_{n=2}^{\infty }\omega (n)(n-\beta )^{q-1}\ln ^{q(1-\frac{%
\lambda }{2})-1}(n-\beta )a_{n}^{q}\right\} ^{\frac{1}{q}},
\end{equation*}%
and in view of (\ref{10}), inequality (\ref{12}) follows.
\end{proof}

\section{Main results}

We introduce the functions%
\begin{eqnarray*}
\Phi (x) &:&=(x-\alpha )^{p-1}\ln ^{p(1-\frac{\lambda }{2})-1}(x-\alpha
)^{{}}(x\in (1+\alpha ,\infty )), \\
\Psi (n) &:&=(n-\beta )^{q-1}\ln ^{q(1-\frac{\lambda }{2})-1}(n-\beta
)^{{}}(n\in \mathbf{N\backslash \{}1\mathbf{\}}),
\end{eqnarray*}%
wherefrom $[\Phi (x)]^{1-q}$ $=\frac{1}{x-\alpha }\ln ^{\frac{q\lambda }{2}%
-1}(x-\alpha ),$ and $[\Psi (n)]^{1-p}=\frac{1}{n-\beta }\ln ^{\frac{%
p\lambda }{2}-1}(n-\beta )$.

\begin{theorem}
\textbf{} If $0<\lambda \leq 2,\alpha \in \mathbf{R},\beta \leq
\frac{1}{2},$ $p>1,\frac{1}{p}+\frac{1}{q}=1,f(x),a_{n}\geq 0,$ $f\in
L_{p,\Phi }(1+\alpha ,\infty ),a=\{a_{n}\}_{n=2}^{\infty }\in l_{q,\Psi },$ $%
||f||_{p,\Phi }>0$ and $||a||_{q,\Psi }>0,$ then we have the following
equivalent inequalities:%
\begin{eqnarray}
I &:&=\sum_{n=2}^{\infty }\int_{1+\alpha }^{\infty }\frac{a_{n}f(x)dx}{%
[1+\ln (x-\alpha )\ln (n-\beta )]^{\lambda }}  \notag \\
&=&\int_{1+\alpha }^{\infty }\sum_{n=2}^{\infty }\frac{a_{n}f(x)dx}{[1+\ln
(x-\alpha )\ln (n-\beta )]^{\lambda }}<B(\frac{\lambda }{2},\frac{\lambda }{2%
})||f||_{p,\Phi }||a||_{q,\Psi },  \label{13}
\end{eqnarray}%
\begin{eqnarray}
J &=&\left\{ \sum_{n=2}^{\infty }[\Psi (n)]^{1-p}\left[ \int_{1+\alpha
}^{\infty }\frac{f(x)dx}{[1+\ln (x-\alpha )\ln (n-\beta )]^{\lambda }}\right]
^{p}\right\} ^{\frac{1}{p}}  \notag \\
&<&B(\frac{\lambda }{2},\frac{\lambda }{2})||f||_{p,\Phi },  \label{14}
\end{eqnarray}%
\begin{eqnarray}
L &:&=\left\{ \int_{1+\alpha }^{\infty }[\Phi (x)]^{1-q}\left[
\sum_{n=2}^{\infty }\frac{a_{n}}{[1+\ln (x-\alpha )\ln (n-\beta )]^{\lambda }%
}\right] ^{q}dx\right\} ^{\frac{1}{q}}  \notag \\
&<&B(\frac{\lambda }{2},\frac{\lambda }{2})||a||_{q,\Psi },  \label{15}
\end{eqnarray}%
where the constant $B(\frac{\lambda }{2},\frac{\lambda }{2})$ is the best
possible in the above inequalities.
\end{theorem}

\begin{proof}
\emph{} The two expressions for $I$ in (\ref{13}) follow from
Lebesgue's term by term integration theorem. By (\ref{11}) and (\ref{10}),
we have (\ref{14}). By H\"{o}lder's inequality, we have%
\begin{equation}
I=\sum_{n=2}^{\infty }\left[ \Psi ^{\frac{-1}{q}}(n)\int_{1+\alpha }^{\infty
}\frac{f(x)dx}{[1+\ln (x-\alpha )\ln (n-\beta )]^{\lambda }}\right] [\Psi ^{%
\frac{1}{q}}(n)a_{n}]\leq J||a||_{q,\Psi }.  \label{16}
\end{equation}%
Then by (\ref{14}), we have (\ref{13}). On the other-hand, assume that (\ref%
{13}) is valid. Setting
\begin{equation*}
a_{n}:=[\Psi (n)]^{1-p}\left[ \int_{1+\alpha }^{\infty }\frac{f(x)dx}{[1+\ln
(x-\alpha )\ln (n-\beta )]^{\lambda }}\right] ^{p-1},n\in \mathbf{%
N\backslash \{}1\mathbf{\}},
\end{equation*}%
where $J^{p-1}=||a||_{q,\Psi }.$ By (\ref{11}), we find $J<\infty .$ If $%
J=0, $ then (\ref{14}) is trivially valid; if $J>0,$ then by (\ref{13}), we
have
\begin{equation*}
||a||_{q,\Psi }^{q}=J^{q(p-1)}=J^{p}=I<B(\frac{\lambda }{2},\frac{\lambda }{2%
})||f||_{p,\Phi }||a||_{q,\Psi },\text{ }
\end{equation*}%
therefore $||a||_{q,\Psi }^{q-1}=J<B(\frac{\lambda }{2},\frac{\lambda }{2}%
)||f||_{p,\Phi },$ that is, (\ref{14}) is equivalent to (\ref{13}). On the
other-hand, by (\ref{10}) we have $[\varpi (x)\,]^{1-q}>[B(\frac{\lambda }{2},%
\frac{\lambda }{2})]^{1-q}.$ Then in view of (\ref{12}), we have (\ref{15}).
By H\"{o}lder's inequality, we find%
\begin{equation}
I=\int_{1+\alpha }^{\infty }[\Phi ^{\frac{1}{p}}(x)f(x)]\left[ \Phi ^{\frac{%
-1}{p}}(x)\sum_{n=2}^{\infty }\frac{a_{n}}{[1+\ln (x-\alpha )\ln (n-\beta
)]^{\lambda }}\right] dx\leq ||f||_{p,\Phi }L.  \label{17}
\end{equation}%
Then by (\ref{15}), we have (\ref{13}). On the other-hand, assume that (\ref%
{13}) is valid. Setting%
\begin{equation*}
f(x):=[\Phi (x)]^{1-q}\left[ \sum_{n=2}^{\infty }\frac{a_{n}}{[1+\ln
(x-\alpha )\ln (n-\beta )]^{\lambda }}\right] ^{q-1},x\in (1+\alpha ,\infty
),
\end{equation*}%
then $L^{q-1}=||f||_{p,\Phi }.$ By (\ref{12}), we find $L<\infty .$ If $L=0,$
then (\ref{15}) is trivially valid; if $L>0,$ then by (\ref{13}), we have%
\begin{equation*}
||f||_{p,\Phi }^{p}=L^{p(q-1)}=I<B(\frac{\lambda }{2},\frac{\lambda }{2}%
)||f||_{p,\Phi }||a||_{q,\Psi },\text{ }
\end{equation*}%
therefore $||f||_{p,\Phi }^{p-1}=L<B(\frac{\lambda }{2},\frac{\lambda }{2}%
)||a||_{q,\Psi },$ that is, (\ref{15}) is equivalent to (\ref{13}). Hence, (%
\ref{13}), (\ref{14}) and (\ref{15}) are equivalent.

For $0<\varepsilon <\frac{p\lambda }{2},$ setting $\widetilde{f}(x)=\frac{1}{%
x-\alpha }\ln ^{\frac{\lambda }{2}+\frac{\varepsilon }{p}-1}(x-\alpha ),x\in
(1+\alpha ,e+\alpha );\widetilde{f}(x)=0,x\in \lbrack e+\alpha ,\infty ),$
and $\widetilde{a}_{n}=\frac{1}{n-\beta }\ln ^{\frac{\lambda }{2}-\frac{%
\varepsilon }{q}-1}(n-\beta ),n\in \mathbf{N\backslash \{}1\mathbf{\}},$ if
there exists a positive number $k(\leq B(\frac{\lambda }{2},\frac{\lambda }{2%
})),$ such that (\ref{13}) is valid as we replace $B(\frac{\lambda }{2},%
\frac{\lambda }{2})$ with $k,$ then in particular, it follows that%
\begin{eqnarray}
\widetilde{I} &:=&\sum_{n=2}^{\infty }\int_{1+\alpha }^{\infty }\frac{%
\widetilde{a}_{n}\widetilde{f}(x)dx}{[1+\ln (x-\alpha )\ln (n-\beta
)]^{\lambda }}<k||\widetilde{f}||_{p,\Phi }||\widetilde{a}||_{q,\Psi }\nonumber \\
&=&k\left\{ \int_{1+\alpha }^{e+\alpha }\frac{dx}{(x-\alpha )\ln
^{-\varepsilon +1}(x-\alpha )}\right\} ^{\frac{1}{p}}\nonumber \\
&&\times \left\{ \frac{1}{(2-\beta )\ln ^{\varepsilon +1}(2-\beta )}%
+\sum_{n=3}^{\infty }\frac{1}{(n-\beta )\ln ^{\varepsilon +1}(n-\beta )}%
\right\} ^{\frac{1}{q}}\nonumber\\
&<&k(\frac{1}{\varepsilon })^{\frac{1}{p}}\left\{ \frac{1}{(2-\beta
)\ln ^{\varepsilon +1}(2-\beta )}+\int_{2}^{\infty
}\frac{dy}{(y-\beta )\ln
^{\varepsilon +1}(y-\beta )}\right\} ^{\frac{1}{q}}  \notag \\
&=&\frac{k}{\varepsilon }\left\{ \frac{\varepsilon }{(2-\beta )\ln
^{\varepsilon +1}(2-\beta )}+\frac{1}{\ln ^{\varepsilon }(2-\beta )}\right\}
^{\frac{1}{q}},  \label{18}
\end{eqnarray}%
\begin{equation*}
\widetilde{I}=\sum_{n=2}^{\infty }\frac{\ln ^{\frac{\lambda }{2}-\frac{%
\varepsilon }{q}-1}(n-\beta )}{n-\beta }\int_{1+\alpha }^{e+\alpha }\frac{%
\ln ^{\frac{\lambda }{2}+\frac{\varepsilon }{p}-1}(x-\alpha )dx}{(x-\alpha
)[1+\ln (x-\alpha )\ln (n-\beta )]^{\lambda }}
\end{equation*}%
\begin{eqnarray}
&&\overset{t=\ln (x-\alpha )\ln (n-\beta )}{=}\sum_{n=2}^{\infty }\frac{1}{%
(n-\beta )\ln ^{\varepsilon +1}(n-\beta )}\int_{0}^{\ln (n-\beta )}\frac{t^{%
\frac{\lambda }{2}+\frac{\varepsilon }{p}-1}}{(1+t)^{\lambda }}dt  \notag \\
&=&B\left( \frac{\lambda }{2}+\frac{\varepsilon }{p},\frac{\lambda }{2}-%
\frac{\varepsilon }{p}\right) \sum_{n=2}^{\infty }\frac{1}{(n-\beta )\ln
^{\varepsilon +1}(n-\beta )}-A(\varepsilon )  \notag \\
&>&B\left( \frac{\lambda }{2}+\frac{\varepsilon }{p},\frac{\lambda }{2}-%
\frac{\varepsilon }{p}\right) \int_{2}^{\infty }\frac{dy}{(y-\beta )\ln
^{\varepsilon +1}(y-\beta )}-A(\varepsilon )  \notag \\
&=&\frac{1}{\varepsilon \ln ^{\varepsilon }(2-\beta )}B\left( \frac{\lambda
}{2}+\frac{\varepsilon }{p},\frac{\lambda }{2}-\frac{\varepsilon }{p}\right)
-A(\varepsilon ),  \notag \\
A(\varepsilon ) &:&=\sum_{n=2}^{\infty }\frac{1}{(n-\beta )\ln ^{\varepsilon
+1}(n-\beta )}\int_{\ln (n-\beta )}^{\infty }\frac{1}{(t+1)^{\lambda }}t^{%
\frac{\lambda }{2}+\frac{\varepsilon }{p}-1}dt.  \label{19}
\end{eqnarray}%
We find%
\begin{eqnarray*}
0 &<&A(\varepsilon )\leq \sum_{n=2}^{\infty }\frac{1}{(n-\beta )\ln
^{\varepsilon +1}(n-\beta )}\int_{\ln (n-\beta )}^{\infty }\frac{1}{%
t^{\lambda }}t^{\frac{\lambda }{2}+\frac{\varepsilon }{p}-1}dt \\
&=&\frac{1}{\frac{\lambda }{2}-\frac{\varepsilon }{p}}\sum_{n=2}^{\infty }%
\frac{1}{(n-\beta )\ln ^{\frac{\lambda }{2}+\frac{\varepsilon }{q}%
+1}(n-\beta )}<\infty ,
\end{eqnarray*}%
and so $A(\varepsilon )=O(1)(\varepsilon \rightarrow 0^{+}).$ Hence by (\ref%
{18}) and (\ref{19}), it follows that%
\begin{eqnarray*}
&&\frac{1}{\ln ^{\varepsilon }(2-\beta )}B\left( \frac{\lambda }{2}+\frac{%
\varepsilon }{p},\frac{\lambda }{2}-\frac{\varepsilon }{p}\right)
-\varepsilon O(1) \\
&<&k\left\{ \frac{\varepsilon }{(2-\beta )\ln ^{\varepsilon +1}(2-\beta )}+%
\frac{1}{\ln ^{\varepsilon }(2-\beta )}\right\} ^{\frac{1}{q}},
\end{eqnarray*}%
and\ $B(\frac{\lambda }{2},\frac{\lambda }{2})\leq k(\varepsilon \rightarrow
0^{+}).$ Hence $k=B(\frac{\lambda }{2},\frac{\lambda }{2})$ is the best
value of (\ref{13}).

By the equivalence of the inequalities, in view of (\ref{16}) and (\ref{17}), the constant factor $B(\frac{\lambda
}{2},\frac{\lambda }{2})$ in (\ref{14}) and (\ref{15}) is the best possible.
 \end{proof}

\begin{remark}
\textbf{ } (i) Define the first type half-discrete Hilbert-type
operator $T_{1}:L_{p,\Phi }(1+\alpha ,\infty )\rightarrow l_{p,\Psi ^{1-p}}$
as follows: For $f\in L_{p,\Phi }(1+\alpha ,\infty ),$ we define $T_{1}f$ $%
\in l_{p,\Psi ^{1-p}}$ by%
\begin{equation*}
T_{1}f(n)=\int_{1+\alpha }^{\infty }\frac{1}{[1+\ln (x-\alpha )\ln (n-\beta
)]^{\lambda }}f(x)dx,n\in \mathbf{N\backslash \{}1\mathbf{\}}.
\end{equation*}%
Then by (\ref{14}), $||T_{1}f||_{p.\Psi ^{1-p}}\leq B(\frac{\lambda }{2},%
\frac{\lambda }{2})||f||_{p,\Phi }$ and so $T_{1}$ is a bounded operator
with $||T_{1}||\leq B(\frac{\lambda }{2},\frac{\lambda }{2}).$ Since by
Theorem 3.1, the constant factor in (\ref{14}) is best possible, we have $%
||T_{1}||=B(\frac{\lambda }{2},\frac{\lambda }{2}).$

(ii) Define the second type half-discrete Hilbert-type operator $%
T_{2}:l_{q,\Psi }\rightarrow L_{q,\Phi ^{1-q}}(1+\alpha ,\infty )$ as
follows: For $a\in l_{q,\Psi },$ we define $T_{2}a$ $\in L_{q,\Phi
^{1-q}}(1+\alpha ,\infty )$ by%
\begin{equation*}
T_{2}a(x)=\sum_{n=2}^{\infty }\frac{1}{[1+\ln (x-\alpha )\ln (n-\beta
)]^{\lambda }}a_{n},x\in (1+\alpha ,\infty ).
\end{equation*}%
Then by (\ref{15}), $||T_{2}a||_{q.\Phi ^{1-q}}\leq B(\frac{\lambda }{2},%
\frac{\lambda }{2})||a||_{q,\Psi }$ and so $T_{2}$ is a bounded operator
with $||T_{2}||\leq B(\frac{\lambda }{2},\frac{\lambda }{2}).$ Since by
Theorem 3.1, the constant factor in (\ref{15}) is best possible, we have $%
||T_{2}||=B(\frac{\lambda }{2},\frac{\lambda }{2}).$
\end{remark}

\begin{remark}
\textbf{ }For $p=q=2,\lambda =1$ in (\ref{13}), (\ref{14}) and (\ref%
{15}), (i) if $\alpha =\beta =\frac{1}{2},$ then we have (\ref{7}) and the
following equivalent inequalities:%
\begin{equation*}
\sum_{n=2}^{\infty }\frac{1}{n-\frac{1}{2}}\left[ \int_{\frac{3}{2}}^{\infty
}\frac{f(x)dx}{1+\ln (x-\frac{1}{2})\ln (n-\frac{1}{2})}\right] ^{2}<\pi
^{2}\int_{\frac{3}{2}}^{\infty }(x-\frac{1}{2})f^{2}(x)dx,  \label{20}
\end{equation*}%
\begin{equation*}
\int_{\frac{3}{2}}^{\infty }\frac{1}{x-\frac{1}{2}}\left[ \sum_{n=2}^{\infty
}\frac{a_{n}}{1+\ln (x-\frac{1}{2})\ln (n-\frac{1}{2})}\right] ^{2}dx<\pi
^{2}\sum_{n=2}^{\infty }(n-\frac{1}{2})a_{n}^{2};  \label{21}
\end{equation*}%
(ii) if $\alpha =\beta =0,$ then we have the following equivalent
inequalities%
\begin{equation*}
\int_{1}^{\infty }f(x)\sum_{n=2}^{\infty }\frac{a_{n}}{1+\ln x\ln n}dx<\pi
\left\{ \int_{1}^{\infty }xf^{2}(x)dx\sum_{n=2}^{\infty }na_{n}^{2}\right\}
^{\frac{1}{2}},  \label{22}
\end{equation*}%
\begin{equation*}
\sum_{n=2}^{\infty }\frac{1}{n}\left[ \int_{1}^{\infty }\frac{f(x)}{1+\ln
x\ln n}dx\right] ^{2}<\pi ^{2}\int_{1}^{\infty }xf^{2}(x)dx,  \label{23}
\end{equation*}%
\begin{equation*}
\int_{1}^{\infty }\frac{1}{x}\left[ \sum_{n=2}^{\infty }\frac{a_{n}}{1+\ln
x\ln n}\right] ^{2}dx<\pi ^{2}\sum_{n=2}^{\infty }na_{n}^{2}.  \label{24}
\end{equation*}
\end{remark}

\textbf{Acknowledgement.}  This work is supported by Chinese
Guangdong Natural Science Foundation (No. 7004344).

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