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# A half-discrete Hilbert-type inequality with the non-monotone kernel and the best constant factor

Journal of Inequalities and Applications20122012:184

https://doi.org/10.1186/1029-242X-2012-184

• Received: 19 December 2011
• Accepted: 9 August 2012
• Published:

## Abstract

By introducing two pairs of conjugate exponents and using the improved Euler-Maclaurin summation formula, we estimate the weight functions and obtain a half-discrete Hilbert-type inequality with the non-monotone kernel and the best constant factor. We also consider its equivalent forms.

MSC:26D15.

## Keywords

• Hilbert-type inequality
• conjugate exponent
• Hölder’s inequality
• best constant factor
• equivalent form

## 1 Introduction

If ${a}_{n},{b}_{n}\ge 0$, such that $0<{\sum }_{n=1}^{\mathrm{\infty }}{a}_{n}^{2}<\mathrm{\infty }$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{b}_{n}^{2}<\mathrm{\infty }$, then we have the famous Hilbert’s inequality as follows (cf. ):
$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{{a}_{m}{b}_{n}}{m+n}<\pi {\left(\sum _{n=1}^{\mathrm{\infty }}{a}_{n}^{2}\sum _{n=1}^{\mathrm{\infty }}{b}_{n}^{2}\right)}^{\frac{1}{2}},$
(1)

where the constant factor π is the best possible.

Under the same condition of (1), Xin et al.  gave the following inequality:
$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{|ln\left(m/n\right)|}{m+n}{a}_{m}{b}_{n}<{c}_{0}{\left(\sum _{n=1}^{\mathrm{\infty }}{a}_{n}^{2}\sum _{n=1}^{\mathrm{\infty }}{b}_{n}^{2}\right)}^{\frac{1}{2}},$
(2)

where the constant factor ${c}_{0}=8{\sum }_{n=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n-1}}{{\left(2n-1\right)}^{2}}={7.3277}^{+}$ is the best possible. And Yang  gave the integral analogues of (2).

In 1934, Hardy et al.  established a few results on the half-discrete Hilbert-type inequalities with the non-homogeneous kernel (see Theorem 351). But they did not prove that the constant factors are the best possible. However, Yang  gave a result by introducing an interval variable and proved that the constant factor is the best possible. Recently, Yang et al.  gave some half-discrete Hilbert-type inequalities and their reverses with the monotone kernels and best constant factors.

Recently, Yang  gave the following half-discrete Hilbert-type inequality with the non-monotone kernel and the best constant factor 8:
$\sum _{n=1}^{\mathrm{\infty }}{\int }_{1}^{\mathrm{\infty }}\frac{|ln\left(x/n\right)|{a}_{n}}{max\left\{x,n\right\}}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx<8{\left(\sum _{n=1}^{\mathrm{\infty }}{a}_{n}^{2}{\int }_{1}^{\mathrm{\infty }}{f}^{2}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\right)}^{\frac{1}{2}}.$
(3)

Obviously, for a half-discrete Hilbert-type inequality with the monotone kernel, it is easy to build the relating inequality by estimating the series form and the integral form of weight functions. However, for a half-discrete Hilbert-type inequality with the non-monotone kernel, it is much more difficult to prove.

In this paper, by using the way of weight functions, we give a new half-discrete Hilbert-type inequality with the non-monotone kernel as follows:
$\sum _{n=1}^{\mathrm{\infty }}{a}_{n}{\int }_{1}^{\mathrm{\infty }}\frac{|ln\left(\frac{x}{n}\right)|f\left(x\right)}{x+n}\phantom{\rule{0.2em}{0ex}}dx<8\sum _{k=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k}}{{\left(2k+1\right)}^{2}}{\left(\sum _{n=1}^{\mathrm{\infty }}{a}_{n}^{2}{\int }_{1}^{\mathrm{\infty }}{f}^{2}\left(x\right)\right)}^{\frac{1}{2}},$
(4)

where the constant factor $8{\sum }_{k=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k}}{{\left(2k+1\right)}^{2}}$ is the best possible. The main objective of this paper is to build the best extension of (4) with parameters and equivalent forms.

## 2 Some lemmas

Lemma 2.1 If${x}_{1}\in \mathbf{R}$, ${n}_{1}\in \mathbf{Z}$ (Z is the set of non-negative integers), $\left[{x}_{1}\right]={n}_{1}$, $\rho \left(y\right)=y-\left[y\right]-\frac{1}{2}$ ($y\in \mathbf{R}$) is the Bernoulli function of first order, then we have (cf. )
${\int }_{{n}_{1}}^{{x}_{1}}\rho \left(y\right)\phantom{\rule{0.2em}{0ex}}dy=-\frac{{\epsilon }_{1}}{8}\phantom{\rule{1em}{0ex}}\left({\epsilon }_{1}\in \left[0,1\right]\right).$
(5)
Lemma 2.2 If$r>1$, $\frac{1}{r}+\frac{1}{s}=1$, $f\left(x,y\right):=\frac{|ln\left(\frac{x}{y}\right)|}{x+y}{\left(\frac{x}{y}\right)}^{\frac{1}{r}}$ ($x,y\in \left(0,\mathrm{\infty }\right)$), N is the set of positive integers, define the weight functions as follows:
Then we have
$\omega \left(n\right)<{c}_{r}:=\sum _{k=0}^{\mathrm{\infty }}{\left(-1\right)}^{k}\left[\frac{1}{{\left(k+\frac{1}{r}\right)}^{2}}+\frac{1}{{\left(k+\frac{1}{s}\right)}^{2}}\right],\phantom{\rule{1em}{0ex}}\varpi \left(x\right)<{c}_{r}.$
(8)
Proof Setting $u=\frac{x}{n}$ in (6), we have
$\omega \left(n\right)={\int }_{\frac{1}{n}}^{\mathrm{\infty }}\frac{|lnu|}{u+1}{u}^{-\frac{1}{s}}\phantom{\rule{0.2em}{0ex}}du<{\int }_{0}^{\mathrm{\infty }}\frac{|lnu|}{u+1}{u}^{-\frac{1}{s}}\phantom{\rule{0.2em}{0ex}}du={c}_{r}.$
Setting $u=\frac{y}{x}$, then it follows
${\int }_{0}^{1}f\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dy={\int }_{0}^{\frac{1}{x}}\frac{\left(-lnu\right)}{u+1}{u}^{-\frac{1}{r}}\phantom{\rule{0.2em}{0ex}}du>s{\int }_{0}^{\frac{1}{x}}\frac{\left(-lnu\right)}{\frac{1}{x}+1}\phantom{\rule{0.2em}{0ex}}d{u}^{-\frac{1}{r}+1}=\frac{{x}^{\frac{1}{r}}}{x+1}\left(slnx+{s}^{2}\right).$
(9)
For $1\le y, $f\left(x,y\right)=-\frac{ln\left(\frac{y}{x}\right)}{x+y}{\left(\frac{x}{y}\right)}^{\frac{1}{r}}$, we have
$\begin{array}{rcl}{f}_{y}^{\prime }\left(x,y\right)& =& {x}^{\frac{1}{r}}\left[-\frac{1}{\left(y+x\right){y}^{1+\frac{1}{r}}}+\frac{ln\left(\frac{y}{x}\right)}{{\left(y+x\right)}^{2}{y}^{\frac{1}{r}}}+\frac{ln\left(\frac{y}{x}\right)}{r\left(y+x\right){y}^{1+\frac{1}{r}}}\right]\\ =& -{x}^{\frac{1}{r}}\left[\frac{1}{\left(y+x\right){y}^{1+\frac{1}{r}}}+\frac{2xln\left(\frac{y}{x}\right)}{\left(y-x\right){\left(y+x\right)}^{2}{y}^{\frac{1}{r}}}+\frac{2xln\left(\frac{y}{x}\right)}{r\left(y-x\right)\left(y+x\right){y}^{1+\frac{1}{r}}}\right]\\ +{x}^{\frac{1}{r}}\left[\frac{ln\left(\frac{y}{x}\right)}{\left(y-x\right)\left(y+x\right){y}^{\frac{1}{r}}}+\frac{ln\left(\frac{y}{x}\right)}{r\left(y-x\right){y}^{1+\frac{1}{r}}}\right].\end{array}$
For $y\ge x$, $f\left(x,y\right)=\frac{ln\left(\frac{y}{x}\right)}{x+y}{\left(\frac{x}{y}\right)}^{\frac{1}{r}}$, we find
$\begin{array}{rcl}{f}_{y}^{\prime }\left(x,y\right)& =& {x}^{\frac{1}{r}}\left[\frac{1}{\left(y+x\right){y}^{1+\frac{1}{r}}}-\frac{ln\left(\frac{y}{x}\right)}{{\left(y+x\right)}^{2}{y}^{\frac{1}{r}}}-\frac{ln\left(\frac{y}{x}\right)}{r\left(y+x\right){y}^{1+\frac{1}{r}}}\right]\\ =& {x}^{\frac{1}{r}}\left[\frac{1}{\left(y+x\right){y}^{1+\frac{1}{r}}}+\frac{2xln\left(\frac{y}{x}\right)}{\left(y-x\right){\left(y+x\right)}^{2}{y}^{\frac{1}{r}}}+\frac{2xln\left(\frac{y}{x}\right)}{r\left(y-x\right)\left(y+x\right){y}^{1+\frac{1}{r}}}\right]\\ -{x}^{\frac{1}{r}}\left[\frac{ln\left(\frac{y}{x}\right)}{\left(y-x\right)\left(y+x\right){y}^{\frac{1}{r}}}+\frac{ln\left(\frac{y}{x}\right)}{r\left(y-x\right){y}^{1+\frac{1}{r}}}\right].\end{array}$
Then we have $-{f}_{y}^{\prime }\left(x,y\right)=g\left(y\right)-h\left(y\right)$. Setting $a=\frac{1}{2{x}^{2}}$, $b=-\frac{1}{2{x}^{2}}$, then $a-b=\frac{1}{{x}^{2}}$. Define two functions as follows:
$\stackrel{˜}{g}\left(y\right)=\left\{\begin{array}{cc}{g}_{1}\left(y\right)-a,\hfill & y
Since ${g}_{1}\left(x-0\right)-a={g}_{2}\left(x\right)$, ${h}_{1}\left(x-0\right)-b={h}_{2}\left(x\right)$, then both $\stackrel{˜}{g}\left(y\right)$ and $\stackrel{˜}{h}\left(y\right)$ ($y\in \left[1,\mathrm{\infty }\right)$) are decreasing and continuous. Besides $y=x$, we have ${\left(-1\right)}^{i}{\stackrel{˜}{g}}^{\left(i\right)}\left(y\right)\ge 0$, ${\left(-1\right)}^{i}{\stackrel{˜}{h}}^{\left(i\right)}\left(y\right)\ge 0$ ($i=0,1$), and $\stackrel{˜}{g}\left(\mathrm{\infty }\right)=\stackrel{˜}{h}\left(\mathrm{\infty }\right)=0$. By the improved Euler-Maclaurin summation formula (cf. , Theorem 2.2.2) and (5), for ${\epsilon }_{1}\in \left[0,1\right]$, ${\epsilon }_{i}\in \left(0,1\right)$ ($i=2,3$), it follows
Since ${g}_{1}\left(1\right)-a\ge {g}_{1}\left(x-0\right)-a={g}_{2}\left(x\right)>0$, ${h}_{1}\left(1\right)-b\ge {h}_{1}\left(x-0\right)-b={h}_{2}\left(x\right)>0$, then we have
$\begin{array}{rcl}-{\int }_{1}^{\mathrm{\infty }}\rho \left(y\right){f}_{y}^{\prime }\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dy& >& -\frac{1}{8}\left({g}_{1}\left(1\right)-a\right)-\frac{1}{8}\left(a-b\right)\\ =& -\frac{{x}^{\frac{1}{r}}}{8\left(x+1\right)}-\frac{{x}^{1+\frac{1}{r}}lnx}{4{\left(x+1\right)}^{2}\left(x-1\right)}-\frac{{x}^{1+\frac{1}{r}}lnx}{4r\left(x+1\right)\left(x-1\right)}-\frac{1}{16{x}^{2}}.\end{array}$
(10)
By the improved Euler-Maclaurin summation formula , we have
$\begin{array}{rcl}\varpi \left(x\right)& =& \sum _{k=1}^{\mathrm{\infty }}f\left(x,k\right)={\int }_{1}^{\mathrm{\infty }}f\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dy+\frac{1}{2}f\left(x,1\right)+{\int }_{1}^{\mathrm{\infty }}\rho \left(y\right){f}_{y}^{\prime }\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dy\\ =& {\int }_{0}^{\mathrm{\infty }}f\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dy-\left({\int }_{0}^{1}f\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dy-\frac{1}{2}f\left(x,1\right)-{\int }_{1}^{\mathrm{\infty }}\rho \left(y\right){f}_{y}^{\prime }\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dy\right)={c}_{r}-\theta \left(x\right),\end{array}$
where
$\theta \left(x\right):={\int }_{0}^{1}f\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dy-\frac{1}{2}f\left(x,1\right)-{\int }_{1}^{\mathrm{\infty }}\rho \left(y\right){f}_{y}^{\prime }\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dy.$
Since $-\frac{1}{2}f\left(x,1\right)=-\frac{{x}^{\frac{1}{r}}lnx}{2\left(x+1\right)}$, in view of (9), (10), (i) for $1\le x<2$, $-\frac{lnx}{x-1}\ge -1$, we have
$\begin{array}{rcl}\theta \left(x\right)& >& \frac{{x}^{\frac{1}{r}}}{x+1}\left(slnx+{s}^{2}\right)-\frac{{x}^{\frac{1}{r}}lnx}{2\left(x+1\right)}-\frac{{x}^{\frac{1}{r}}}{8\left(x+1\right)}\\ -\frac{{x}^{1+\frac{1}{r}}lnx}{4{\left(x+1\right)}^{2}\left(x-1\right)}-\frac{{x}^{1+\frac{1}{r}}lnx}{4r\left(x+1\right)\left(x-1\right)}-\frac{1}{16{x}^{2}}\\ \ge & \frac{{x}^{\frac{1}{r}}lnx}{\left(x+1\right)}\left(s-\frac{1}{2}\right)+\frac{{x}^{\frac{1}{r}}}{x+1}\left[{s}^{2}-\frac{1}{8}-\frac{x}{4\left(x+1\right)}-\frac{x}{4r}-\frac{x+1}{16{x}^{2+\frac{1}{r}}}\right]\\ >& \frac{{x}^{\frac{1}{r}}}{\left(x+1\right)}\left(1-\frac{1}{8}-\frac{1}{4}-\frac{1}{2}-\frac{1}{8}\right)=0;\end{array}$
(ii) for $x\ge 2$, $-\frac{1}{x-1}\ge -\frac{2}{x}$, we have
$\begin{array}{rcl}\theta \left(x\right)& \ge & \frac{{x}^{\frac{1}{r}}lnx}{\left(x+1\right)}\left[s-\frac{1}{2}-\frac{1}{2\left(x+1\right)}-\frac{1}{2r}\right]+\frac{{x}^{\frac{1}{r}}}{x+1}\left({s}^{2}-\frac{1}{8}-\frac{x+1}{16{x}^{2+\frac{1}{r}}}\right)\\ >& \frac{{x}^{\frac{1}{r}}lnx}{\left(x+1\right)}\left(s-\frac{1}{2}-\frac{1}{6}-\frac{1}{2r}\right)=\frac{\left(6s+2r-3\right){x}^{\frac{1}{r}}lnx}{6r\left(x+1\right)}>0.\end{array}$

Hence, for $x\ge 1$, we have $\theta \left(x\right)>0$, it follows $\varpi \left(x\right)<{c}_{r}$. The lemma is proved. □

Lemma 2.3 As the assumption of Lemma  2.2, if$0<\epsilon <\frac{p}{r}$, $R={\left(\frac{1}{r}-\frac{\epsilon }{p}\right)}^{-1}$, $S={\left(\frac{1}{s}+\frac{\epsilon }{p}\right)}^{-1}$, then we have
$\overline{I}:=\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{1}{S}-\epsilon -1}{\int }_{1}^{\mathrm{\infty }}\frac{|ln\left(\frac{x}{n}\right)|}{x+n}{x}^{\frac{1}{R}-1}\phantom{\rule{0.2em}{0ex}}dx\ge \frac{{c}_{R}}{\epsilon }-O\left(1\right).$
(11)
Proof It is obvious that $R>1$, $\frac{1}{R}+\frac{1}{S}=1$. Setting $u=\frac{x}{n}$, we have
$\begin{array}{rcl}\overline{I}& =& \sum _{n=1}^{\mathrm{\infty }}{n}^{-\epsilon -1}{\int }_{\frac{1}{n}}^{\mathrm{\infty }}\frac{|lnu|}{u+1}{u}^{\frac{1}{R}-1}\phantom{\rule{0.2em}{0ex}}du\\ =& \sum _{n=1}^{\mathrm{\infty }}{n}^{-\epsilon -1}\left[{\int }_{0}^{\mathrm{\infty }}\frac{|lnu|}{u+1}{u}^{\frac{1}{R}-1}\phantom{\rule{0.2em}{0ex}}du+{\int }_{0}^{\frac{1}{n}}\frac{lnu}{u+1}{u}^{\frac{1}{R}-1}\phantom{\rule{0.2em}{0ex}}du\right]\\ >& {c}_{R}{\int }_{1}^{\mathrm{\infty }}{x}^{-\epsilon -1}\phantom{\rule{0.2em}{0ex}}dx+R\sum _{n=1}^{\mathrm{\infty }}{n}^{-\epsilon -1}{\int }_{0}^{\frac{1}{n}}lnu\phantom{\rule{0.2em}{0ex}}d{u}^{\frac{1}{R}}\\ =& \frac{{c}_{R}}{\epsilon }-\left[R\sum _{n=1}^{\mathrm{\infty }}\frac{lnn}{{n}^{1+\epsilon +\frac{1}{R}}}+{R}^{2}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{1+\epsilon +\frac{1}{R}}}\right]\\ =& \frac{{c}_{R}}{\epsilon }-O\left(1\right)\phantom{\rule{1em}{0ex}}\left(\epsilon \to {0}^{+}\right).\end{array}$

The lemma is proved. □

Lemma 2.4 If$p,r>1$, $\frac{1}{r}+\frac{1}{s}=\frac{1}{p}+\frac{1}{q}=1$, ${a}_{n}\ge 0$, $f\left(x\right)$is a non-negative measurable function, then we have

Then we have (13). The lemma is proved. □

## 3 Main results and applications

Theorem 3.1 If$p,r>1$, $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}+\frac{1}{s}=1$, ${a}_{n},f\left(x\right)\ge 0$such that$0<{\int }_{1}^{\mathrm{\infty }}{x}^{\frac{p}{s}-1}{f}^{p}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{a}_{n}^{q}<\mathrm{\infty }$, then we have the following equivalent inequalities:

where the constant factors${c}_{r}={\sum }_{k=0}^{\mathrm{\infty }}{\left(-1\right)}^{k}\left[\frac{1}{{\left(k+\frac{1}{r}\right)}^{2}}+\frac{1}{{\left(k+\frac{1}{s}\right)}^{2}}\right]$, ${c}_{r}^{p}$, ${c}_{r}^{q}$are the best possible.

Proof By the Lebesgue term-by-term integration theorem , there are two kinds of representation in (14). By the conditions of Theorem 3.1, (12) takes the form of a strict inequality, and we have (15). By Hölder’s inequality , we have
$I=\sum _{n=1}^{\mathrm{\infty }}\left[{n}^{\frac{1}{s}-\frac{1}{p}}{\int }_{1}^{\mathrm{\infty }}\frac{|ln\left(\frac{x}{n}\right)|f\left(x\right)}{x+n}\phantom{\rule{0.2em}{0ex}}dx\right]\left[{n}^{-\frac{1}{s}+\frac{1}{p}}{a}_{n}\right]\le {J}_{1}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{a}_{n}^{q}\right\}}^{\frac{1}{q}}.$
(17)
By (15), we have (14). On the other hand, suppose that (14) is valid. Setting ${a}_{n}:={n}^{\frac{p}{s}-1}{\left[{\int }_{1}^{\mathrm{\infty }}\frac{|ln\left(\frac{x}{n}\right)|f\left(x\right)}{x+n}\phantom{\rule{0.2em}{0ex}}dx\right]}^{p-1}$, $n\in \mathbf{N}$, then it follows ${J}_{1}={\sum }_{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{a}_{n}^{q}$. By (12), we have $J<\mathrm{\infty }$. If $J=0$, then (15) is obvious value; if $0, then by (14), we obtain
$\begin{array}{r}\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{a}_{n}^{q}={J}_{1}=I<{c}_{r}{\left\{{\int }_{1}^{\mathrm{\infty }}{x}^{\frac{p}{s}-1}{f}^{p}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{a}_{n}^{q}\right\}}^{\frac{1}{q}},\\ {J}_{1}^{\frac{1}{p}}={\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{a}_{n}^{q}\right\}}^{\frac{1}{p}}<{c}_{r}{\left\{{\int }_{1}^{\mathrm{\infty }}{x}^{\frac{p}{s}-1}{f}^{p}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\right\}}^{\frac{1}{p}}.\end{array}$
(18)

Hence we have (15), which is equivalent to (14).

By Hölder’s inequality , we have
$I={\int }_{1}^{\mathrm{\infty }}\left[{x}^{\frac{1}{r}-\frac{1}{q}}f\left(x\right)\right]\left[{x}^{-\frac{1}{r}+\frac{1}{q}}\sum _{n=1}^{\mathrm{\infty }}\frac{|ln\left(\frac{x}{n}\right)|{a}_{n}}{x+n}\right]\phantom{\rule{0.2em}{0ex}}dx\le {\left\{{\int }_{1}^{\mathrm{\infty }}{x}^{\frac{p}{s}-1}{f}^{p}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\right\}}^{\frac{1}{p}}{J}_{2}^{\frac{1}{q}}.$
(19)
By (16), we have (14). On the other hand, suppose that (14) is valid. Setting $f\left(x\right):={x}^{\frac{q}{r}-1}{\left[{\sum }_{n=1}^{\mathrm{\infty }}\frac{|ln\left(\frac{x}{n}\right)|{a}_{n}}{x+n}\right]}^{q-1}$, $x\in \left[1,\mathrm{\infty }\right)$, then it follows ${J}_{2}={\int }_{1}^{\mathrm{\infty }}{x}^{\frac{p}{s}-1}{f}^{p}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx$. By (13), we have ${J}_{2}<\mathrm{\infty }$. If ${J}_{2}=0$, then (16) is obvious value; if $0<{J}_{2}<\mathrm{\infty }$, then by (14), we obtain
$\begin{array}{r}{\int }_{1}^{\mathrm{\infty }}{x}^{\frac{p}{s}-1}{f}^{p}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx={J}_{2}=I<{c}_{r}{\left\{{\int }_{1}^{\mathrm{\infty }}{x}^{\frac{p}{s}-1}{f}^{p}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{a}_{n}^{q}\right\}}^{\frac{1}{q}},\\ {J}_{2}^{\frac{1}{q}}={\left\{{\int }_{1}^{\mathrm{\infty }}{x}^{\frac{p}{s}-1}{f}^{p}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\right\}}^{\frac{1}{q}}<{c}_{r}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{a}_{n}^{q}\right\}}^{\frac{1}{q}}.\end{array}$
(20)

Hence we have (16), which is equivalent to (14). Therefore (14), (15) and (16) are equivalent.

If the constant factor ${c}_{r}$ in (14) is not best possible, then there exists a positive number K, with $0, such that (14) is still valid if we replace ${c}_{r}$ by K. For $0<\epsilon <{\epsilon }_{0}$, setting $\overline{f}\left(x\right)={n}^{\frac{1}{r}-\frac{\epsilon }{p}-1}$, ${\overline{a}}_{n}={n}^{\frac{1}{s}-\frac{\epsilon }{q}-1}$ ($n\in N$), we have
$\begin{array}{rcl}\overline{I}& =& \sum _{n=1}^{\mathrm{\infty }}{\overline{a}}_{n}{\int }_{1}^{\mathrm{\infty }}\frac{|ln\left(\frac{x}{n}\right)|\overline{f}\left(x\right)}{x+n}\phantom{\rule{0.2em}{0ex}}dx
(21)

By (11) and (21), we have ${c}_{R}-\epsilon O\left(1\right) and for $\epsilon \to {0}^{+}$, by Fatou lemma , we have ${c}_{r}\le {lim}_{\epsilon \to {0}^{+}}\left({c}_{R}-\epsilon O\left(1\right)\right)\le K$. This is a contradiction. Hence we can conclude that the constant ${c}_{r}$ in (14) is the best possible. If the constant factors in (15) and (16) are not the best possible, then we can imply a contradiction that the constant factor in (14) is not the best possible by (17) and (19). The theorem is proved. □

Remark For $p=q=r=s=2$, (14) reduces to (4). Inequality (4) is a new basic half-discrete Hilbert-type inequality with the non-monotone kernel.

## Declarations

### Acknowledgements

This work was supported by the Emphases Natural Science Foundation of Guangdong Institutions of Higher Learning, College and University (No. 05Z026) and the Natural Science Foundation of Guangdong (7004344).

## Authors’ Affiliations

(1)
Department of Mathematics, Guangdong Education University, Guangzhou, Guangdong, 510303, People’s Republic of China

## References 