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# On a more accurate multidimensional Mulholland-type inequality

Journal of Inequalities and Applications20142014:322

https://doi.org/10.1186/1029-242X-2014-322

• Received: 17 March 2014
• Accepted: 29 July 2014
• Published:

## Abstract

In this paper, by using the way of weight coefficients and technique of real analysis, a more accurate multidimensional discrete Mulholland-type inequality with the best possible constant factor is given, which is an extension of the Mulholland inequality. The equivalent form, the operator expression with the norm as well as a few particular cases are also considered.

MSC:26D15, 47A07.

## Keywords

• Mulholland-type inequality
• weight coefficient
• equivalent form
• operator
• norm

## 1 Introduction

Suppose that $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $f\left(x\right),g\left(y\right)\ge 0$, $f\in {L}^{p}\left({\mathbf{R}}_{+}\right)$, $g\in {L}^{q}\left({\mathbf{R}}_{+}\right)$, ${\parallel f\parallel }_{p}={\left\{{\int }_{0}^{\mathrm{\infty }}{f}^{p}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\right\}}^{\frac{1}{p}}>0$, ${\parallel g\parallel }_{q}>0$. We have the following Hardy-Hilbert integral inequality (cf. ):
${\int }_{0}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}\frac{f\left(x\right)g\left(y\right)}{x+y}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy<\frac{\pi }{sin\left(\pi /p\right)}{\parallel f\parallel }_{p}{\parallel g\parallel }_{q},$
(1)
where the constant factor $\frac{\pi }{sin\left(\pi /p\right)}$ is the best possible. Assuming that ${a}_{m},{b}_{n}\ge 0$, $a={\left\{{a}_{m}\right\}}_{m=1}^{\mathrm{\infty }}\in {l}^{p}$, $b={\left\{{b}_{n}\right\}}_{n=1}^{\mathrm{\infty }}\in {l}^{q}$, ${\parallel a\parallel }_{p}={\left\{{\sum }_{m=1}^{\mathrm{\infty }}{a}_{m}^{p}\right\}}^{\frac{1}{p}}>0$, ${\parallel b\parallel }_{q}>0$, we have the following Hardy-Hilbert inequality with the same best constant $\frac{\pi }{sin\left(\pi /p\right)}$ (cf. ):
$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{{a}_{m}{b}_{n}}{m+n}<\frac{\pi }{sin\left(\pi /p\right)}{\parallel a\parallel }_{p}{\parallel b\parallel }_{q}.$
(2)
Inequalities (1) and (2) are important in analysis and its applications (cf. ). Also we have the following Mulholland inequality (cf. ):
$\sum _{m=2}^{\mathrm{\infty }}\sum _{n=2}^{\mathrm{\infty }}\frac{{a}_{m}{b}_{n}}{lnmn}<\frac{\pi }{sin\left(\pi /p\right)}{\left\{\sum _{m=2}^{\mathrm{\infty }}\frac{{a}_{m}^{p}}{{m}^{1-p}}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=2}^{\mathrm{\infty }}\frac{{b}_{n}^{q}}{{n}^{1-q}}\right\}}^{\frac{1}{q}}.$
(3)
In 1998, by introducing an independent parameter $\lambda \in \left(0,1\right]$, Yang  gave an extension of (1) for $p=q=2$. Yang  gave some extensions of (1) and (2) as follows: If ${\lambda }_{1},{\lambda }_{2},\lambda \in \mathbf{R}$, ${\lambda }_{1}+{\lambda }_{2}=\lambda$, ${k}_{\lambda }\left(x,y\right)$ is a non-negative homogeneous function of degree −λ, with
$k\left({\lambda }_{1}\right)={\int }_{0}^{\mathrm{\infty }}{k}_{\lambda }\left(t,1\right){t}^{{\lambda }_{1}-1}\phantom{\rule{0.2em}{0ex}}dt\in {\mathbf{R}}_{+},$
$\varphi \left(x\right)={x}^{p\left(1-{\lambda }_{1}\right)-1}$, $\psi \left(x\right)={x}^{q\left(1-{\lambda }_{2}\right)-1}$, $f\left(x\right),g\left(y\right)\ge 0$,
$f\in {L}_{p,\varphi }\left({\mathbf{R}}_{+}\right)=\left\{f;{\parallel f\parallel }_{p,\varphi }:={\left\{{\int }_{0}^{\mathrm{\infty }}\varphi \left(x\right)|f\left(x\right){|}^{p}\phantom{\rule{0.2em}{0ex}}dx\right\}}^{\frac{1}{p}}<\mathrm{\infty }\right\},$
$g\in {L}_{q,\psi }\left({\mathbf{R}}_{+}\right)$, ${\parallel f\parallel }_{p,\varphi },{\parallel g\parallel }_{q,\psi }>0$, then
${\int }_{0}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}{k}_{\lambda }\left(x,y\right)f\left(x\right)g\left(y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy
(4)
where the constant factor $k\left({\lambda }_{1}\right)$ is the best possible. Moreover, if ${k}_{\lambda }\left(x,y\right)$ is finite and ${k}_{\lambda }\left(x,y\right){x}^{{\lambda }_{1}-1}\left({k}_{\lambda }\left(x,y\right){y}^{{\lambda }_{2}-1}\right)$ is decreasing with respect to $x>0$ ($y>0$), then for ${a}_{m},{b}_{n}\ge 0$,
$a\in {l}_{p,\varphi }=\left\{a;{\parallel a\parallel }_{p,\varphi }:={\left\{\sum _{n=1}^{\mathrm{\infty }}\varphi \left(n\right){|{a}_{n}|}^{p}\right\}}^{\frac{1}{p}}<\mathrm{\infty }\right\},$
$b={\left\{{b}_{n}\right\}}_{n=1}^{\mathrm{\infty }}\in {l}_{q,\psi }$, ${\parallel a\parallel }_{p,\varphi },{\parallel b\parallel }_{q,\psi }>0$, it follows that
$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{k}_{\lambda }\left(m,n\right){a}_{m}{b}_{n}
(5)

where the constant factor $k\left({\lambda }_{1}\right)$ is still the best possible.

Clearly, for $\lambda =1$, ${k}_{1}\left(x,y\right)=\frac{1}{x+y}$, ${\lambda }_{1}=\frac{1}{q}$, ${\lambda }_{2}=\frac{1}{p}$, inequality (3) reduces to (1), while (5) reduces to (2). Some other results including the multidimensional Hilbert-type integral inequalities are provided by .

About half-discrete Hilbert-type inequalities with the non-homogeneous kernels, Hardy et al. provided a few results in Theorem 351 of . But they did not prove that the constant factors are the best possible. However, Yang  gave a result with the kernel $\frac{1}{{\left(1+nx\right)}^{\lambda }}$ ($0<\lambda \le 2$) by introducing a variable and proved that the constant factor is the best possible. In 2011 Yang  gave a half-discrete Hardy-Hilbert inequality with the best possible constant factor. Zhong et al.  investigated several half-discrete Hilbert-type inequalities with particular kernels. Using the way of weight functions and the techniques of discrete and integral Hilbert-type inequalities with some additional conditions on the kernel, a half-discrete Hilbert-type inequality with a general homogeneous kernel of degree $-\lambda \in \mathbf{R}$ and a best constant factor $k\left({\lambda }_{1}\right)$ is obtained as follows:
${\int }_{0}^{\mathrm{\infty }}f\left(x\right)\sum _{n=1}^{\mathrm{\infty }}{k}_{\lambda }\left(x,n\right){a}_{n}\phantom{\rule{0.2em}{0ex}}dx
(6)

(see Yang and Chen ). At the same time, a half-discrete Hilbert-type inequality with a general non-homogeneous kernel and the best constant factor is given by Yang .

In this paper, by using the way of weight coefficients and technique of real analysis, a more accurate multidimensional discrete Mulholland-type inequality with the best possible constant factor is given, which is an extension of (3). The equivalent form, the operator expression with the norm as well as a few particular cases are also considered.

## 2 Some lemmas

Lemma 1 If ${\left(-1\right)}^{i}{h}^{\left(i\right)}\left(t\right)>0$ ($t>0$; $i=1,2$), then for $b>0$, $0<\alpha \le 1$,
${\left(-1\right)}^{i}\frac{{d}^{i}}{d{x}^{i}}h\left({\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}\right)>0\phantom{\rule{1em}{0ex}}\left(x>1;i=1,2\right).$
(7)
Proof We find
$\begin{array}{c}\frac{d}{dx}h\left({\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}\right)=\frac{1}{x}{h}^{\prime }\left({\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}\right){\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }-1}{ln}^{\alpha -1}x<0,\hfill \\ \frac{{d}^{2}}{d{x}^{2}}h\left({\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}\right)=\frac{d}{dx}\left[\frac{1}{x}{h}^{\prime }\left({\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}\right){\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }-1}{ln}^{\alpha -1}x\right]\hfill \\ \phantom{\frac{{d}^{2}}{d{x}^{2}}h\left({\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}\right)}=-\frac{1}{{x}^{2}}{h}^{\prime }\left({\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}\right){\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }-1}{ln}^{\alpha -1}x\hfill \\ \phantom{\frac{{d}^{2}}{d{x}^{2}}h\left({\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}\right)=}+\frac{1}{{x}^{2}}{h}^{″}\left({\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}\right){\left(b+{ln}^{\alpha }x\right)}^{\frac{2}{\alpha }-2}{ln}^{2\alpha -2}x\hfill \\ \phantom{\frac{{d}^{2}}{d{x}^{2}}h\left({\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}\right)=}+\alpha \left(\frac{1}{\alpha }-1\right)\frac{1}{{x}^{2}}{h}^{\prime }\left({\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}\right){\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }-2}{ln}^{2\alpha -2}x\hfill \\ \phantom{\frac{{d}^{2}}{d{x}^{2}}h\left({\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}\right)=}+\left(\alpha -1\right)\frac{1}{{x}^{2}}{h}^{\prime }\left({\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}\right){\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }-1}{ln}^{\alpha -2}x\hfill \\ \phantom{\frac{{d}^{2}}{d{x}^{2}}h\left({\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}\right)}=\left[-{h}^{\prime }\left({\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}\right)\left(b+{ln}^{\alpha }x\right)lnx\hfill \\ \phantom{\frac{{d}^{2}}{d{x}^{2}}h\left({\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}\right)=}+{h}^{″}\left({\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}\right){\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}{ln}^{\alpha }x\hfill \\ \phantom{\frac{{d}^{2}}{d{x}^{2}}h\left({\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}\right)=}+b\left(\alpha -1\right){h}^{\prime }\left({\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }}\right)\right]\frac{1}{{x}^{2}}{\left(b+{ln}^{\alpha }x\right)}^{\frac{1}{\alpha }-2}{ln}^{\alpha -2}x>0.\hfill \end{array}$

Then we have (7). □

If ${i}_{0},{j}_{0}\in \mathbf{N}$ (N is the set of positive integers), $\alpha ,\beta >0$, we set
${\parallel x\parallel }_{\alpha }:={\left(\sum _{k=1}^{{i}_{0}}{|{x}_{k}|}^{\alpha }\right)}^{\frac{1}{\alpha }}\phantom{\rule{1em}{0ex}}\left(x=\left({x}_{1},\dots ,{x}_{{i}_{0}}\right)\in {\mathbf{R}}^{{i}_{0}}\right),$
(8)
${\parallel y\parallel }_{\beta }:={\left(\sum _{k=1}^{{j}_{0}}{|{y}_{k}|}^{\beta }\right)}^{\frac{1}{\beta }}\phantom{\rule{1em}{0ex}}\left(y=\left({y}_{1},\dots ,{y}_{{j}_{0}}\right)\in {\mathbf{R}}^{{j}_{0}}\right).$
(9)
Lemma 2 If $s\in \mathbf{N}$, $\gamma ,M>0$, $\mathrm{\Psi }\left(u\right)$ is a non-negative measurable function in $\left(0,1\right]$, and
${D}_{M}:=\left\{x\in {\mathbf{R}}_{+}^{s};\sum _{i=1}^{s}{x}_{i}^{\gamma }\le {M}^{\gamma }\right\}=\left\{x;\sum _{i=1}^{s}{\left(\frac{{x}_{i}}{M}\right)}^{\gamma }\le 1\right\},$
then we have (cf. )
$\begin{array}{c}\int \cdots {\int }_{{D}_{M}}\mathrm{\Psi }\left(\sum _{i=1}^{s}{\left(\frac{{x}_{i}}{M}\right)}^{\gamma }\right)\phantom{\rule{0.2em}{0ex}}d{x}_{1}\cdots \phantom{\rule{0.2em}{0ex}}d{x}_{s}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{{M}^{s}{\mathrm{\Gamma }}^{s}\left(\frac{1}{\gamma }\right)}{{\gamma }^{s}\mathrm{\Gamma }\left(\frac{s}{\gamma }\right)}{\int }_{0}^{1}\mathrm{\Psi }\left(u\right){u}^{\frac{s}{\gamma }-1}\phantom{\rule{0.2em}{0ex}}du.\hfill \end{array}$
(10)
Lemma 3 If $s\in \mathbf{N}$, $\gamma >0$, $\epsilon >0$, $d=\left({d}_{1},\dots ,{d}_{s}\right)\in {\left[\frac{1}{2},1\right]}^{s}$, then
$\begin{array}{rcl}{A}_{s}\left(\epsilon \right)& :=& \sum _{m}{\parallel ln\left(m+d\right)\parallel }_{\gamma }^{-s-\epsilon }\frac{1}{{\prod }_{i=1}^{s}\left({m}_{i}+{d}_{i}\right)}\\ =& \frac{{\mathrm{\Gamma }}^{s}\left(\frac{1}{\gamma }\right)}{\epsilon {s}^{\epsilon /\gamma }{\gamma }^{s-1}\mathrm{\Gamma }\left(\frac{s}{\gamma }\right)}+O\left(1\right)\phantom{\rule{1em}{0ex}}\left(\epsilon \to {0}^{+}\right).\end{array}$
(11)
Proof For $M>{s}^{1/\gamma }$, we set
$\mathrm{\Psi }\left(u\right)=\left\{\begin{array}{ll}0,& 0
Then by the decreasing property and (10), it follows that
$\begin{array}{rcl}{A}_{s}\left(\epsilon \right)& \ge & {\int }_{\left\{x\in {\mathbf{R}}_{+}^{s};{x}_{i}\ge e-{d}_{i}\right\}}{\parallel ln\left(x+d\right)\parallel }_{\gamma }^{-s-\epsilon }\frac{dx}{{\prod }_{i=1}^{s}\left({x}_{i}+{d}_{i}\right)}\\ \stackrel{{u}_{i}=ln\left({x}_{i}+{d}_{i}\right)}{=}& {\int }_{\left\{u\in {\mathbf{R}}_{+}^{s};{u}_{i}\ge 1\right\}}{\parallel u\parallel }_{\gamma }^{-s-\epsilon }\phantom{\rule{0.2em}{0ex}}du\\ =& \underset{M\to \mathrm{\infty }}{lim}\int \cdots {\int }_{{D}_{M}}\mathrm{\Psi }\left(\sum _{i=1}^{s}{\left(\frac{{x}_{i}}{M}\right)}^{\gamma }\right)\phantom{\rule{0.2em}{0ex}}d{x}_{1}\cdots \phantom{\rule{0.2em}{0ex}}d{x}_{s}\\ =& \underset{M\to \mathrm{\infty }}{lim}\frac{{M}^{s}{\mathrm{\Gamma }}^{s}\left(\frac{1}{\gamma }\right)}{{\gamma }^{s}\mathrm{\Gamma }\left(\frac{s}{\gamma }\right)}{\int }_{s/{M}^{\gamma }}^{1}{\left(M{u}^{1/\gamma }\right)}^{-s-\epsilon }{u}^{\frac{s}{\gamma }-1}\phantom{\rule{0.2em}{0ex}}du=\frac{{\mathrm{\Gamma }}^{s}\left(\frac{1}{\gamma }\right)}{\epsilon {s}^{\epsilon /\gamma }{\gamma }^{s-1}\mathrm{\Gamma }\left(\frac{s}{\gamma }\right)}.\end{array}$
In the following, by mathematical induction we prove that, for any $s\in \mathbf{N}$,
${A}_{s}\left(\epsilon \right)\le {O}_{s}\left(1\right)+\frac{{\mathrm{\Gamma }}^{s}\left(\frac{1}{\gamma }\right)}{\epsilon {s}^{\epsilon /\gamma }{\gamma }^{s-1}\mathrm{\Gamma }\left(\frac{s}{\gamma }\right)}\phantom{\rule{1em}{0ex}}\left(\epsilon \to {0}^{+}\right).$
(12)
For $s=1$, by the Hermite-Hadamard inequality (cf. ), it follows that
$\begin{array}{rcl}{A}_{1}\left(\epsilon \right)& =& \sum _{{m}_{1}=1}^{2}\frac{{ln}^{-1-\epsilon }\left({m}_{1}+{d}_{1}\right)}{{m}_{1}+{d}_{1}}+\sum _{{m}_{1}=3}^{\mathrm{\infty }}\frac{{ln}^{-1-\epsilon }\left({m}_{1}+{d}_{1}\right)}{{m}_{1}+{d}_{1}}\\ \le & {O}_{1}\left(1\right)+{\int }_{\frac{5}{2}}^{\mathrm{\infty }}\frac{{ln}^{-1-\epsilon }\left(x+{d}_{1}\right)\phantom{\rule{0.2em}{0ex}}dx}{x+{d}_{1}}\le {O}_{1}\left(1\right)+{\int }_{e-{d}_{1}}^{\mathrm{\infty }}\frac{{ln}^{-1-\epsilon }\left(x+{d}_{1}\right)\phantom{\rule{0.2em}{0ex}}dx}{x+{d}_{1}}\\ \stackrel{u=ln\left(x+{d}_{1}\right)}{=}& {O}_{1}\left(1\right)+{\int }_{1}^{\mathrm{\infty }}{u}^{-1-\epsilon }\phantom{\rule{0.2em}{0ex}}du={O}_{1}\left(1\right)+\frac{1}{\epsilon },\end{array}$
and then (12) is valid. Assuming that (12) is valid for $s-1\in \mathbf{N}$, then for s, we set
$\begin{array}{rcl}{A}_{s}\left(\epsilon \right)& =& \sum _{\left\{m\in {\mathbf{N}}^{s};\mathrm{\exists }{i}_{0},{m}_{{i}_{0}}=1,2\right\}}{\parallel ln\left(m+d\right)\parallel }_{\gamma }^{-s-\epsilon }\frac{1}{{\prod }_{i=1}^{s}\left({m}_{i}+{d}_{i}\right)}\\ +\sum _{\left\{m\in {\mathbf{N}}^{s};{m}_{i}\ge 3\right\}}{\parallel ln\left(m+d\right)\parallel }_{\gamma }^{-s-\epsilon }\frac{1}{{\prod }_{i=1}^{s}\left({m}_{i}+{d}_{i}\right)}.\end{array}$
There exist constants $a,b\in {\mathbf{R}}_{+}$, such that
$\begin{array}{c}\sum _{\left\{m\in {\mathbf{N}}^{s};\mathrm{\exists }{i}_{0},{m}_{{i}_{0}}=1,2\right\}}{\parallel ln\left(m+d\right)\parallel }_{\gamma }^{-s-\epsilon }\frac{1}{{\prod }_{i=1}^{s}\left({m}_{i}+{d}_{i}\right)}\hfill \\ \phantom{\rule{1em}{0ex}}\le a+b\sum _{\left\{m\in {\mathbf{N}}^{s-1};{m}_{i}\ge 1\right\}}{\parallel ln\left(m+d\right)\parallel }_{\gamma }^{-\left(s-1\right)-\left(1+\epsilon \right)}\frac{1}{{\prod }_{i=1}^{s-1}\left({m}_{i}+{d}_{i}\right)}.\hfill \end{array}$
By the assumption of mathematical induction for $s-1$, we find
$\begin{array}{c}\sum _{\left\{m\in {\mathbf{N}}^{s-1};{m}_{i}\ge 1\right\}}{\parallel ln\left(m+d\right)\parallel }_{\gamma }^{-\left(s-1\right)-\left(1+\epsilon \right)}\frac{1}{{\prod }_{i=1}^{s-1}\left({m}_{i}+{d}_{i}\right)}\hfill \\ \phantom{\rule{1em}{0ex}}\le {O}_{s-1}\left(1\right)+\frac{{\mathrm{\Gamma }}^{s-1}\left(\frac{1}{\gamma }\right)}{\left(1+\epsilon \right){\left(s-1\right)}^{\left(1+\epsilon \right)/\gamma }{\gamma }^{s-2}\mathrm{\Gamma }\left(\frac{s-1}{\gamma }\right)},\hfill \end{array}$
and then
$\sum _{\left\{m\in {\mathbf{N}}^{s};\mathrm{\exists }{i}_{0},{m}_{{i}_{0}}=1,2\right\}}{\parallel ln\left(m+d\right)\parallel }_{\gamma }^{-s-\epsilon }\frac{1}{{\prod }_{i=1}^{s}\left({m}_{i}+{d}_{i}\right)}\le {O}_{s}\left(1\right).$
By Lemma 1 and the Hermite-Hadamard inequality (cf. ), we obtain
$\begin{array}{c}\sum _{\left\{m\in {\mathbf{N}}^{s};{m}_{i}\ge 3\right\}}{\parallel ln\left(m+d\right)\parallel }_{\gamma }^{-s-\epsilon }\frac{1}{{\prod }_{i=1}^{s}\left({m}_{i}+{d}_{i}\right)}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\int }_{\left\{x\in {\mathbf{R}}_{+}^{s};{x}_{i}\ge \frac{5}{2}\right\}}{\parallel ln\left(x+d\right)\parallel }_{\gamma }^{-s-\epsilon }\frac{1}{{\prod }_{i=1}^{s}\left({x}_{i}+{d}_{i}\right)}\phantom{\rule{0.2em}{0ex}}dx\hfill \\ \phantom{\rule{1em}{0ex}}\le {\int }_{\left\{x\in {\mathbf{R}}_{+}^{s};{x}_{i}\ge e-{d}_{i}\right\}}{\parallel ln\left(x+d\right)\parallel }_{\gamma }^{-s-\epsilon }\frac{1}{{\prod }_{i=1}^{s}\left({x}_{i}+{d}_{i}\right)}\phantom{\rule{0.2em}{0ex}}dx\hfill \\ \phantom{\rule{1em}{0ex}}\stackrel{{u}_{i}=ln\left({x}_{i}+{d}_{i}\right)}{=}{\int }_{\left\{u\in {\mathbf{R}}_{+}^{s};{u}_{i}\ge 1\right\}}{\parallel u\parallel }_{\gamma }^{-s-\epsilon }\phantom{\rule{0.2em}{0ex}}du=\frac{{\mathrm{\Gamma }}^{s}\left(\frac{1}{\gamma }\right)}{\epsilon {s}^{\epsilon /\gamma }{\gamma }^{s-1}\mathrm{\Gamma }\left(\frac{s}{\gamma }\right)}.\hfill \end{array}$

Hence we prove that (12) is valid for $s\in \mathbf{N}$. Therefore, we have (11). □

Lemma 4 If C is the set of complex numbers and ${\mathbf{C}}_{\mathrm{\infty }}=\mathbf{C}\cup \left\{\mathrm{\infty }\right\}$, ${z}_{k}\in \mathbf{C}\mathrm{\setminus }\left\{z|Rez\ge 0,Imz=0\right\}$ ($k=1,2,\dots ,n$) are different points, the function $f\left(z\right)$ is analytic in ${\mathbf{C}}_{\mathrm{\infty }}$ except for ${z}_{i}$ ($i=1,2,\dots ,n$), and $z=\mathrm{\infty }$ is a zero point of $f\left(z\right)$ whose order is not less than 1, then for $\alpha \in \mathbf{R}$, we have
${\int }_{0}^{\mathrm{\infty }}f\left(x\right){x}^{\alpha -1}\phantom{\rule{0.2em}{0ex}}dx=\frac{2\pi i}{1-{e}^{2\pi \alpha i}}\sum _{k=1}^{n}Res\left[f\left(z\right){z}^{\alpha -1},{z}_{k}\right],$
(13)
where $0. In particular, if ${z}_{k}$ ($k=1,\dots ,n$) are all poles of order 1, setting ${\phi }_{k}\left(z\right)=\left(z-{z}_{k}\right)f\left(z\right)$ (${\phi }_{k}\left({z}_{k}\right)\ne 0$), then
${\int }_{0}^{\mathrm{\infty }}f\left(x\right){x}^{\alpha -1}\phantom{\rule{0.2em}{0ex}}dx=\frac{\pi }{sin\pi \alpha }\sum _{k=1}^{n}{\left(-{z}_{k}\right)}^{\alpha -1}{\phi }_{k}\left({z}_{k}\right).$
(14)
Proof By [, p.118], we have (13). We find
$\begin{array}{rcl}1-{e}^{2\pi \alpha i}& =& 1-cos2\pi \alpha -isin2\pi \alpha \\ =& -2isin\pi \alpha \left(cos\pi \alpha +isin\pi \alpha \right)=-2i{e}^{i\pi \alpha }sin\pi \alpha .\end{array}$
In particular, since $f\left(z\right){z}^{\alpha -1}=\frac{1}{z-{z}_{k}}\left({\phi }_{k}\left(z\right){z}^{\alpha -1}\right)$, it is obvious that
$Res\left[f\left(z\right){z}^{\alpha -1},-{a}_{k}\right]=z_{k}{}^{\alpha -1}{\phi }_{k}\left({z}_{k}\right)=-{e}^{i\pi \alpha }{\left(-{z}_{k}\right)}^{\alpha -1}{\phi }_{k}\left({z}_{k}\right).$

Then by (13), we obtain (14). □

Example 1 For $s\in \mathbf{N}$, we set
${k}_{\lambda }\left(x,y\right)=\prod _{k=1}^{s}\frac{1}{\left({x}^{\lambda /s}+{c}_{k}{y}^{\lambda /s}\right)}\phantom{\rule{1em}{0ex}}\left(0<{c}_{1}<\cdots <{c}_{s},0<\lambda \le s\right).$
For $0<{\lambda }_{1}\le {i}_{0}$, $0<{\lambda }_{2}\le {j}_{0}$, ${\lambda }_{1}+{\lambda }_{2}=\lambda$, by (14), we find
$\begin{array}{rcl}{k}_{s}\left({\lambda }_{1}\right)& :=& {\int }_{0}^{\mathrm{\infty }}\prod _{k=1}^{s}\frac{1}{{t}^{\lambda /s}+{c}_{k}}{t}^{{\lambda }_{1}-1}\phantom{\rule{0.2em}{0ex}}dt\\ \stackrel{u={t}^{\lambda /s}}{=}& \frac{s}{\lambda }{\int }_{0}^{\mathrm{\infty }}\prod _{k=1}^{s}\frac{1}{u+{c}_{k}}{u}^{\frac{s{\lambda }_{1}}{\lambda }-1}\phantom{\rule{0.2em}{0ex}}du\\ =& \frac{\pi s}{\lambda sin\left(\frac{\pi s{\lambda }_{1}}{\lambda }\right)}\sum _{k=1}^{s}{c}_{k}^{\frac{s{\lambda }_{1}}{\lambda }-1}\prod _{j=1\phantom{\rule{0.2em}{0ex}}\left(j\ne k\right)}^{s}\frac{1}{{c}_{j}-{c}_{k}}\in {\mathbf{R}}_{+}.\end{array}$
(15)
In particular, for $s=1$, we obtain
${k}_{1}\left({\lambda }_{1}\right)=\frac{1}{\lambda }{\int }_{0}^{\mathrm{\infty }}\frac{{u}^{\left({\lambda }_{1}/\lambda \right)-1}}{u+{c}_{1}}\phantom{\rule{0.2em}{0ex}}du=\frac{\pi }{\lambda sin\left(\frac{\pi {\lambda }_{1}}{\lambda }\right)}{c}_{1}^{\frac{{\lambda }_{1}}{\lambda }-1}.$
Definition 1 For $s\in \mathbf{N}$, $0<\alpha ,\beta \le 1$, $0<{c}_{1}<\cdots <{c}_{s}$, $0<\lambda \le s$, $0<{\lambda }_{1}\le {i}_{0}$, $0<{\lambda }_{2}\le {j}_{0}$, ${\lambda }_{1}+{\lambda }_{2}=\lambda$, $\tau =\left({\tau }_{1},\dots ,{\tau }_{{i}_{0}}\right)\in {\left[\frac{1}{2},1\right]}^{{i}_{0}}$, $\sigma =\left({\sigma }_{1},\dots ,{\sigma }_{{j}_{0}}\right)\in {\left[\frac{1}{2},1\right]}^{{j}_{0}}$, $ln\left(m+\tau \right)=\left(ln\left({m}_{1}+{\tau }_{1}\right),\dots ,ln\left({m}_{{i}_{0}}+{\tau }_{{i}_{0}}\right)\right)\in {\mathbf{R}}_{+}^{{i}_{0}}$, $ln\left(n+\sigma \right)=\left(ln\left({n}_{1}+{\sigma }_{1}\right),\dots ,ln\left({n}_{{j}_{0}}+{\sigma }_{{j}_{0}}\right)\right)\in {\mathbf{R}}_{+}^{{j}_{0}}$, we define two weight coefficients ${w}_{\lambda }\left({\lambda }_{2},n\right)$ and ${W}_{\lambda }\left({\lambda }_{1},m\right)$ as follows:
$\begin{array}{r}{w}_{\lambda }\left({\lambda }_{2},n\right):=\sum _{m}\frac{{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{{\lambda }_{2}}{\parallel ln\left(m+\tau \right)\parallel }_{\alpha }^{{\lambda }_{1}-{i}_{0}}}{{\prod }_{k=1}^{s}\left[{\parallel ln\left(m+\tau \right)\parallel }_{\alpha }^{\lambda /s}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}\right]{\prod }_{i=1}^{{i}_{0}}\left({m}_{i}+{\tau }_{i}\right)},\\ {W}_{\lambda }\left({\lambda }_{1},m\right):=\sum _{n}\frac{{\parallel ln\left(m+\tau \right)\parallel }_{\alpha }^{{\lambda }_{1}}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{{\lambda }_{2}-{j}_{0}}}{{\prod }_{k=1}^{s}\left[{\parallel ln\left(m+\tau \right)\parallel }_{\alpha }^{\lambda /s}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}\right]{\prod }_{j=1}^{{j}_{0}}\left({n}_{j}+{\sigma }_{j}\right)},\end{array}$
(16)

where ${\sum }_{m}={\sum }_{{m}_{{i}_{0}}=1}^{\mathrm{\infty }}\cdots {\sum }_{{m}_{1}=1}^{\mathrm{\infty }}$ and ${\sum }_{n}={\sum }_{{n}_{{j}_{0}}=1}^{\mathrm{\infty }}\cdots {\sum }_{{n}_{1}=1}^{\mathrm{\infty }}$.

Lemma 5 Let the assumptions as in Definition  1 be fulfilled. Then:
1. (i)
we have
${w}_{\lambda }\left({\lambda }_{2},n\right)<{K}_{2}\phantom{\rule{1em}{0ex}}\left(n\in {\mathbf{N}}^{{j}_{0}}\right),$
(17)

${W}_{\lambda }\left({\lambda }_{1},m\right)<{K}_{1}\phantom{\rule{1em}{0ex}}\left(m\in {\mathbf{N}}^{{i}_{0}}\right),$
(18)
where
${K}_{1}:=\frac{{\mathrm{\Gamma }}^{{j}_{0}}\left(\frac{1}{\beta }\right)}{{\beta }^{{j}_{0}-1}\mathrm{\Gamma }\left(\frac{{j}_{0}}{\beta }\right)}{k}_{s}\left({\lambda }_{1}\right),\phantom{\rule{2em}{0ex}}{K}_{2}:=\frac{{\mathrm{\Gamma }}^{{i}_{0}}\left(\frac{1}{\alpha }\right)}{{\alpha }^{{i}_{0}-1}\mathrm{\Gamma }\left(\frac{{i}_{0}}{\alpha }\right)}{k}_{s}\left({\lambda }_{1}\right),$
(19)
and ${k}_{s}\left({\lambda }_{1}\right)$ is indicated by (15);
1. (ii)
for $p>1$, $0<\epsilon , setting ${\stackrel{˜}{\lambda }}_{1}={\lambda }_{1}-\frac{\epsilon }{p}$, ${\stackrel{˜}{\lambda }}_{2}={\lambda }_{2}+\frac{\epsilon }{p}$, we have
$0<{\stackrel{˜}{K}}_{2}\left(1-{\stackrel{˜}{\theta }}_{\lambda }\left(n\right)\right)<{w}_{\lambda }\left({\stackrel{˜}{\lambda }}_{2},n\right),$
(20)

where
$\begin{array}{c}{\stackrel{˜}{\theta }}_{\lambda }\left(n\right):=\frac{1}{{k}_{s}\left({\stackrel{˜}{\lambda }}_{1}\right)}{\int }_{0}^{{i}_{0}^{\lambda /\left(\alpha s\right)}/{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}}\frac{{v}^{\frac{s{\lambda }_{1}}{\lambda }-1}}{{\prod }_{k=1}^{s}\left(v+{c}_{k}\right)}\phantom{\rule{0.2em}{0ex}}dv\hfill \\ \phantom{{\stackrel{˜}{\theta }}_{\lambda }\left(n\right)}=O\left(\frac{1}{{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{{\stackrel{˜}{\lambda }}_{1}}}\right),\hfill \end{array}$
(21)
${\stackrel{˜}{K}}_{2}=\frac{{\mathrm{\Gamma }}^{{i}_{0}}\left(\frac{1}{\alpha }\right){k}_{s}\left({\stackrel{˜}{\lambda }}_{1}\right)}{{\alpha }^{{i}_{0}-1}\mathrm{\Gamma }\left(\frac{{i}_{0}}{\alpha }\right)}\in {\mathbf{R}}_{+}.$
(22)
Proof By Lemma 1, the Hermite-Hadamard inequality (cf. ), (10), and (15), it follows that
$\begin{array}{c}{w}_{\lambda }\left({\lambda }_{2},n\right)\hfill \\ \phantom{\rule{1em}{0ex}}<{\int }_{{\left(\frac{1}{2},\mathrm{\infty }\right)}^{{i}_{0}}}\frac{{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{{\lambda }_{2}}{\parallel ln\left(x+\tau \right)\parallel }_{\alpha }^{{\lambda }_{1}-{i}_{0}}\phantom{\rule{0.2em}{0ex}}dx}{{\prod }_{k=1}^{s}\left[{\parallel ln\left(x+\tau \right)\parallel }_{\alpha }^{\lambda /s}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}\right]{\prod }_{i=1}^{{i}_{0}}\left({x}_{i}+{\tau }_{i}\right)}\hfill \\ \phantom{\rule{1em}{0ex}}\stackrel{{u}_{i}=ln\left({x}_{i}+{\tau }_{i}\right)}{=}{\int }_{\left\{u\in {\mathbf{R}}_{+}^{{i}_{0}};{u}_{i}>ln\left(\frac{1}{2}+{\tau }_{i}\right)\right\}}\frac{{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{{\lambda }_{2}}{\parallel u\parallel }_{\alpha }^{{\lambda }_{1}-{i}_{0}}}{{\prod }_{k=1}^{s}\left[{\parallel u\parallel }_{\alpha }^{\lambda /s}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}\right]}\phantom{\rule{0.2em}{0ex}}du\hfill \\ \phantom{\rule{1em}{0ex}}\le {\int }_{{\mathbf{R}}_{+}^{{i}_{0}}}\frac{{\parallel n-\sigma \parallel }_{\beta }^{{\lambda }_{2}}{\parallel u\parallel }_{\alpha }^{{\lambda }_{1}-{i}_{0}}}{{\prod }_{k=1}^{s}\left[{\parallel u\parallel }_{\alpha }^{\lambda /s}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}\right]}\phantom{\rule{0.2em}{0ex}}du\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{M\to \mathrm{\infty }}{lim}{\int }_{{\mathbf{D}}_{M}}\frac{{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{{\lambda }_{2}}{M}^{{\lambda }_{1}-{i}_{0}}{\left[{\sum }_{i=1}^{{j}_{0}}{\left(\frac{{u}_{i}}{M}\right)}^{\alpha }\right]}^{\left({\lambda }_{1}-{i}_{0}\right)/\alpha }}{{\prod }_{k=1}^{s}\left\{{M}^{\frac{\lambda }{s}}{\left[{\sum }_{i=1}^{{i}_{0}}{\left(\frac{{u}_{i}}{M}\right)}^{\alpha }\right]}^{\frac{\lambda }{\alpha s}}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\frac{\lambda }{s}}\right\}}\phantom{\rule{0.2em}{0ex}}du\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{M\to \mathrm{\infty }}{lim}\frac{{M}^{{i}_{0}}{\mathrm{\Gamma }}^{{i}_{0}}\left(\frac{1}{\alpha }\right)}{{\alpha }^{{i}_{0}}\mathrm{\Gamma }\left(\frac{{i}_{0}}{\alpha }\right)}{\int }_{0}^{1}\frac{{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{{\lambda }_{2}}{M}^{{\lambda }_{1}-{i}_{0}}{t}^{\left({\lambda }_{1}-{i}_{0}\right)/\alpha }{t}^{\frac{{i}_{0}}{\alpha }-1}}{{\prod }_{k=1}^{s}\left({M}^{\frac{\lambda }{s}}{t}^{\frac{\lambda }{\alpha s}}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\frac{\lambda }{s}}\right)}\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{M\to \mathrm{\infty }}{lim}\frac{{M}^{{\lambda }_{1}}{\mathrm{\Gamma }}^{{i}_{0}}\left(\frac{1}{\alpha }\right)}{{\alpha }^{{i}_{0}}\mathrm{\Gamma }\left(\frac{{i}_{0}}{\alpha }\right)}{\int }_{0}^{1}\frac{{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{{\lambda }_{2}}{t}^{\frac{{\lambda }_{1}}{\alpha }-1}}{{\prod }_{k=1}^{s}\left({M}^{\frac{\lambda }{s}}{t}^{\frac{\lambda }{\alpha s}}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\frac{\lambda }{s}}\right)}\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}\stackrel{t={\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\alpha }{M}^{-\alpha }{v}^{\alpha s/\lambda }}{=}\frac{s{\mathrm{\Gamma }}^{{i}_{0}}\left(\frac{1}{\alpha }\right)}{\lambda {\alpha }^{{i}_{0}-1}\mathrm{\Gamma }\left(\frac{{i}_{0}}{\alpha }\right)}{\int }_{0}^{\mathrm{\infty }}\frac{{v}^{\frac{s{\lambda }_{1}}{\lambda }-1}}{{\prod }_{k=1}^{s}\left(v+{c}_{k}\right)}\phantom{\rule{0.2em}{0ex}}dv\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{{\mathrm{\Gamma }}^{{i}_{0}}\left(\frac{1}{\alpha }\right)}{{\alpha }^{{i}_{0}-1}\mathrm{\Gamma }\left(\frac{{i}_{0}}{\alpha }\right)}{k}_{s}\left({\lambda }_{1}\right)={K}_{2}.\hfill \end{array}$

Hence, we have (17). In the same way, we have (18).

By the decreasing property and (10), similarly to the proof of (11), we find
$\begin{array}{c}{w}_{\lambda }\left({\stackrel{˜}{\lambda }}_{2},n\right)\hfill \\ \phantom{\rule{1em}{0ex}}>{\int }_{\left\{x\in {\mathbf{R}}_{+}^{{i}_{0}};{x}_{i}\ge e-{\tau }_{i}\right\}}\frac{{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{{\stackrel{˜}{\lambda }}_{2}}{\parallel ln\left(x+\tau \right)\parallel }_{\alpha }^{{\stackrel{˜}{\lambda }}_{1}-{i}_{0}}\phantom{\rule{0.2em}{0ex}}dx}{{\prod }_{k=1}^{s}\left[{\parallel ln\left(x+\tau \right)\parallel }_{\alpha }^{\lambda /s}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}\right]{\prod }_{i=1}^{{i}_{0}}\left({x}_{i}+{\tau }_{i}\right)}\hfill \\ \phantom{\rule{1em}{0ex}}\stackrel{{u}_{i}=ln\left({x}_{i}+{\tau }_{i}\right)}{=}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{{\stackrel{˜}{\lambda }}_{2}}{\int }_{\left\{u\in {\mathbf{R}}_{+}^{{i}_{0}};{u}_{i}\ge 1\right\}}\frac{{\parallel u\parallel }_{\alpha }^{{\stackrel{˜}{\lambda }}_{1}-{i}_{0}}\phantom{\rule{0.2em}{0ex}}du}{{\prod }_{k=1}^{s}\left[{\parallel u\parallel }_{\alpha }^{\lambda /s}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}\right]}\hfill \\ \phantom{\rule{1em}{0ex}}={\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{{\stackrel{˜}{\lambda }}_{2}}\underset{M\to \mathrm{\infty }}{lim}\int \cdots {\int }_{{D}_{M}}\frac{{\left\{{\sum }_{i=1}^{{i}_{0}}{\left(\frac{{u}_{i}}{M}\right)}^{\alpha }\right\}}^{\frac{{\stackrel{˜}{\lambda }}_{1}-{i}_{0}}{\alpha }}{M}^{{\stackrel{˜}{\lambda }}_{1}-{i}_{0}}\phantom{\rule{0.2em}{0ex}}d{u}_{1}\cdots \phantom{\rule{0.2em}{0ex}}d{u}_{{i}_{0}}}{{\prod }_{k=1}^{s}\left[{\left\{{\sum }_{i=1}^{{i}_{0}}{\left(\frac{{u}_{i}}{M}\right)}^{\alpha }\right\}}^{\frac{\lambda }{\alpha s}}{M}^{\frac{\lambda }{s}}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}\right]}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{{M}^{{i}_{0}}{\mathrm{\Gamma }}^{{i}_{0}}\left(\frac{1}{\alpha }\right)}{{\alpha }^{{i}_{0}}\mathrm{\Gamma }\left(\frac{{i}_{0}}{\alpha }\right)}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{{\stackrel{˜}{\lambda }}_{2}}\underset{M\to \mathrm{\infty }}{lim}{\int }_{\frac{{i}_{0}}{{M}^{\alpha }}}^{1}\frac{{t}^{\frac{{\stackrel{˜}{\lambda }}_{1}-{i}_{0}}{\alpha }}{M}^{{\stackrel{˜}{\lambda }}_{1}-{i}_{0}}{t}^{\frac{{i}_{0}}{\alpha }-1}}{{\prod }_{k=1}^{s}\left[{t}^{\frac{\lambda }{\alpha s}}{M}^{\frac{\lambda }{s}}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}\right]}\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{s{\mathrm{\Gamma }}^{{i}_{0}}\left(\frac{1}{\alpha }\right)}{\lambda {\alpha }^{{i}_{0}-1}\mathrm{\Gamma }\left(\frac{{i}_{0}}{\alpha }\right)}{\int }_{\frac{{i}_{0}^{\lambda /\left(\alpha s\right)}}{{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}}}^{\mathrm{\infty }}\frac{{v}^{\frac{s{\stackrel{˜}{\lambda }}_{1}}{\lambda }-1}\phantom{\rule{0.2em}{0ex}}dv}{{\prod }_{k=1}^{s}\left(v+{c}_{k}\right)}={\stackrel{˜}{K}}_{2}\left(1-{\stackrel{˜}{\theta }}_{\lambda }\left(n\right)\right)>0,\hfill \\ 0<{\stackrel{˜}{\theta }}_{\lambda }\left(n\right)=\frac{s}{\lambda {k}_{s}\left({\stackrel{˜}{\lambda }}_{1}\right)}{\int }_{0}^{{i}_{0}^{\lambda /\left(\alpha s\right)}/{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}}\frac{{v}^{\frac{s{\lambda }_{1}}{\lambda }-1}}{{\prod }_{k=1}^{s}\left(v+{c}_{k}\right)}\phantom{\rule{0.2em}{0ex}}dv\hfill \\ \phantom{0}\le \frac{s}{\lambda {k}_{s}\left({\stackrel{˜}{\lambda }}_{1}\right){\prod }_{k=1}^{s}{c}_{k}}{\int }_{0}^{{i}_{0}^{\lambda /\left(\alpha s\right)}/{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}}{v}^{\frac{s{\stackrel{˜}{\lambda }}_{1}}{\lambda }-1}\phantom{\rule{0.2em}{0ex}}dv\hfill \\ \phantom{0}=\frac{1}{{\stackrel{˜}{\lambda }}_{1}{k}_{s}\left({\stackrel{˜}{\lambda }}_{1}\right){\prod }_{k=1}^{s}{c}_{k}}\frac{{i}_{0}^{{\stackrel{˜}{\lambda }}_{1}/\alpha }}{{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{{\stackrel{˜}{\lambda }}_{1}}}.\hfill \end{array}$

Hence, we have (20) and (21). □

## 3 Main results and operator expressions

Setting $\mathrm{\Phi }\left(m\right):={\prod }_{i=1}^{{i}_{0}}{\left({m}_{i}+{\tau }_{i}\right)}^{p-1}{\parallel ln\left(m+\tau \right)\parallel }_{\alpha }^{p\left({i}_{0}-{\lambda }_{1}\right)-{i}_{0}}$ ($m\in {\mathbf{N}}^{{i}_{0}}$) and $\mathrm{\Psi }\left(n\right):={\prod }_{j=1}^{{j}_{0}}{\left({n}_{j}+{\sigma }_{j}\right)}^{q-1}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{q\left({j}_{0}-{\lambda }_{2}\right)-{j}_{0}}$ ($n\in {\mathbf{N}}^{{j}_{0}}$), wherefrom
${\left[\mathrm{\Psi }\left(n\right)\right]}^{1-p}=\prod _{j=1}^{{j}_{0}}{\left({n}_{j}+{\sigma }_{j}\right)}^{-1}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{p{\lambda }_{2}-{j}_{0}},$

we have the following.

Theorem 1 If $s\in \mathbf{N}$, $0<\alpha ,\beta \le 1$, $0<{c}_{1}<\cdots <{c}_{s}$, $0<\lambda \le s$, $0<{\lambda }_{1}\le {i}_{0}$, $0<{\lambda }_{2}\le {j}_{0}$, ${\lambda }_{1}+{\lambda }_{2}=\lambda$, $\tau \in {\left[\frac{1}{2},1\right]}^{{i}_{0}}$, $\sigma \in {\left[\frac{1}{2},1\right]}^{{j}_{0}}$, then for $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, ${a}_{m},{b}_{n}\ge 0$, $0<{\parallel a\parallel }_{p,\mathrm{\Phi }},{\parallel b\parallel }_{q,\mathrm{\Psi }}<\mathrm{\infty }$, we have
$\begin{array}{rcl}I& :=& \sum _{n}\sum _{m}\frac{{a}_{m}{b}_{n}}{{\prod }_{k=1}^{s}\left[{\parallel ln\left(m+\tau \right)\parallel }_{\alpha }^{\lambda /s}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}\right]}\\ <& {K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}{\parallel a\parallel }_{p,\mathrm{\Phi }}{\parallel b\parallel }_{q,\mathrm{\Psi }},\end{array}$
(23)
where the constant factor
${K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}={\left[\frac{{\mathrm{\Gamma }}^{{j}_{0}}\left(\frac{1}{\beta }\right)}{{\beta }^{{j}_{0}-1}\mathrm{\Gamma }\left(\frac{{j}_{0}}{\beta }\right)}\right]}^{\frac{1}{p}}{\left[\frac{{\mathrm{\Gamma }}^{{i}_{0}}\left(\frac{1}{\alpha }\right)}{{\beta }^{{i}_{0}-1}\mathrm{\Gamma }\left(\frac{{i}_{0}}{\alpha }\right)}\right]}^{\frac{1}{q}}{k}_{s}\left({\lambda }_{1}\right)$
(24)

is the best possible (${k}_{s}\left({\lambda }_{1}\right)$ is indicated by (15)).

Proof By the Hölder inequality (cf. ), we have
$\begin{array}{rcl}I& =& \sum _{n}\sum _{m}\frac{1}{{\prod }_{k=1}^{s}\left[{\parallel ln\left(m+\tau \right)\parallel }_{\alpha }^{\lambda /s}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}\right]}\\ ×\left[\frac{{\parallel ln\left(m+\tau \right)\parallel }_{\alpha }^{\left({i}_{0}-{\lambda }_{1}\right)/q}}{{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\left({j}_{0}-{\lambda }_{2}\right)/p}}\frac{{\prod }_{i=1}^{{i}_{0}}{\left({m}_{i}+{\tau }_{i}\right)}^{1/q}}{{\prod }_{j=1}^{{j}_{0}}{\left({n}_{j}+{\sigma }_{j}\right)}^{1/p}}{a}_{m}\right]\\ ×\left[\frac{{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\left({j}_{0}-{\lambda }_{2}\right)/p}}{{\parallel ln\left(m+\tau \right)\parallel }_{\alpha }^{\left({i}_{0}-{\lambda }_{1}\right)/q}}\frac{{\prod }_{j=1}^{{j}_{0}}{\left({n}_{j}+{\sigma }_{j}\right)}^{1/p}}{{\prod }_{i=1}^{{i}_{0}}{\left({m}_{i}+{\tau }_{i}\right)}^{1/q}}{b}_{n}\right]\\ \le & {\left\{\sum _{m}{W}_{\lambda }\left({\lambda }_{1},m\right)\prod _{i=1}^{{i}_{0}}{\left({m}_{i}+{\tau }_{i}\right)}^{p-1}{\parallel ln\left(m+\tau \right)\parallel }_{\alpha }^{p\left({i}_{0}-{\lambda }_{1}\right)-{i}_{0}}{a}_{m}^{p}\right\}}^{\frac{1}{p}}\\ ×{\left\{\sum _{n}{w}_{\lambda }\left({\lambda }_{2},n\right)\prod _{j=1}^{{j}_{0}}{\left({n}_{j}+{\sigma }_{j}\right)}^{q-1}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{q\left({j}_{0}-{\lambda }_{2}\right)-{j}_{0}}{b}_{n}^{q}\right\}}^{\frac{1}{q}}.\end{array}$

Then by (17) and (18), we have (23).

For $0<\epsilon , ${\stackrel{˜}{\lambda }}_{1}={\lambda }_{1}-\frac{\epsilon }{p}$, ${\stackrel{˜}{\lambda }}_{2}={\lambda }_{2}+\frac{\epsilon }{p}$, we set
$\begin{array}{c}{\stackrel{˜}{a}}_{m}={\parallel ln\left(m+\tau \right)\parallel }_{\alpha }^{-{i}_{0}+{\lambda }_{1}-\frac{\epsilon }{p}}\frac{1}{{\prod }_{i=1}^{{i}_{0}}\left({m}_{i}+{\tau }_{i}\right)},\hfill \\ {\stackrel{˜}{b}}_{n}={\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{-{j}_{0}+{\lambda }_{2}-\frac{\epsilon }{q}}\frac{1}{{\prod }_{j=1}^{{j}_{0}}\left({n}_{j}+{\sigma }_{j}\right)}\phantom{\rule{1em}{0ex}}\left(m\in {\mathbf{N}}^{{i}_{0}},n\in {\mathbf{N}}^{{j}_{0}}\right).\hfill \end{array}$
Then by (11) and (20)-(22), we obtain
$\begin{array}{c}{\parallel \stackrel{˜}{a}\parallel }_{p,\mathrm{\Phi }}{\parallel \stackrel{˜}{b}\parallel }_{q,\mathrm{\Psi }}={\left\{\sum _{m}\prod _{i=1}^{{i}_{0}}{\left({m}_{i}+{\tau }_{i}\right)}^{p-1}{\parallel ln\left(m+\tau \right)\parallel }_{\alpha }^{p\left({i}_{0}-{\lambda }_{1}\right)-{i}_{0}}{\stackrel{˜}{a}}_{m}^{p}\right\}}^{\frac{1}{p}}\hfill \\ \phantom{{\parallel \stackrel{˜}{a}\parallel }_{p,\mathrm{\Phi }}{\parallel \stackrel{˜}{b}\parallel }_{q,\mathrm{\Psi }}=}×{\left\{\sum _{n}\prod _{j=1}^{{j}_{0}}{\left({n}_{j}+{\sigma }_{j}\right)}^{q-1}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{q\left({j}_{0}-{\lambda }_{2}\right)-{j}_{0}}{\stackrel{˜}{b}}_{n}^{q}\right\}}^{\frac{1}{q}}\hfill \\ \phantom{{\parallel \stackrel{˜}{a}\parallel }_{p,\mathrm{\Phi }}{\parallel \stackrel{˜}{b}\parallel }_{q,\mathrm{\Psi }}}={\left\{\sum _{m}{\parallel ln\left(m+\tau \right)\parallel }_{\alpha }^{-{i}_{0}-\epsilon }\frac{1}{{\prod }_{i=1}^{{i}_{0}}\left({m}_{i}+{\tau }_{i}\right)}\right\}}^{\frac{1}{p}}\hfill \\ \phantom{{\parallel \stackrel{˜}{a}\parallel }_{p,\mathrm{\Phi }}{\parallel \stackrel{˜}{b}\parallel }_{q,\mathrm{\Psi }}=}×{\left\{\sum _{n}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{-{j}_{0}-\epsilon }\frac{1}{{\prod }_{j=1}^{{j}_{0}}\left({n}_{j}+{\sigma }_{j}\right)}\right\}}^{\frac{1}{q}}\hfill \\ \phantom{{\parallel \stackrel{˜}{a}\parallel }_{p,\mathrm{\Phi }}{\parallel \stackrel{˜}{b}\parallel }_{q,\mathrm{\Psi }}}=\frac{1}{\epsilon }{\left[\frac{{\mathrm{\Gamma }}^{{i}_{0}}\left(\frac{1}{\alpha }\right)}{{i}_{0}^{\epsilon /\alpha }{\alpha }^{{i}_{0}-1}\mathrm{\Gamma }\left(\frac{{i}_{0}}{\alpha }\right)}+\epsilon O\left(1\right)\right]}^{\frac{1}{p}}{\left[\frac{{\mathrm{\Gamma }}^{{j}_{0}}\left(\frac{1}{\beta }\right)}{{j}_{0}^{\epsilon /\beta }{\beta }^{{j}_{0}-1}\mathrm{\Gamma }\left(\frac{{j}_{0}}{\beta }\right)}+\epsilon \stackrel{˜}{O}\left(1\right)\right]}^{\frac{1}{q}},\hfill \end{array}$
(25)
$\begin{array}{c}\stackrel{˜}{I}:=\sum _{n}\left[\sum _{m}\frac{{\stackrel{˜}{a}}_{m}}{{\prod }_{k=1}^{s}\left({\parallel ln\left(m+\tau \right)\parallel }_{\alpha }^{\lambda /s}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}\right)}\right]{\stackrel{˜}{b}}_{n}\hfill \\ \phantom{\stackrel{˜}{I}}=\sum _{n}{w}_{\lambda }\left({\stackrel{˜}{\lambda }}_{2},n\right){\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{-{j}_{0}-\epsilon }\frac{1}{{\prod }_{j=1}^{{j}_{0}}\left({n}_{j}+{\sigma }_{j}\right)}\hfill \\ \phantom{\stackrel{˜}{I}}>{\stackrel{˜}{K}}_{2}\sum _{n}\left(1-O\left(\frac{1}{{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{{\stackrel{˜}{\lambda }}_{1}}}\right)\right){\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{-{j}_{0}-\epsilon }\frac{1}{{\prod }_{j=1}^{{j}_{0}}\left({n}_{j}+{\sigma }_{j}\right)}\hfill \\ \phantom{\stackrel{˜}{I}}={\stackrel{˜}{K}}_{2}\left[\frac{{\mathrm{\Gamma }}^{{j}_{0}}\left(\frac{1}{\beta }\right)}{\epsilon {j}_{0}^{\epsilon /\beta }{\beta }^{{j}_{0}-1}\mathrm{\Gamma }\left(\frac{{j}_{0}}{\beta }\right)}+\stackrel{˜}{O}\left(1\right)-O\left(1\right)\right].\hfill \end{array}$
(26)
If there exists a constant $K\le {K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}$, such that (23) is valid when replacing ${K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}$ by K, then we have
$\begin{array}{c}\left({K}_{2}+o\left(1\right)\right)\left[\frac{{\mathrm{\Gamma }}^{{j}_{0}}\left(\frac{1}{\beta }\right)}{{j}_{0}^{\epsilon /\beta }{\beta }^{{j}_{0}-1}\mathrm{\Gamma }\left(\frac{{j}_{0}}{\beta }\right)}+\epsilon \stackrel{˜}{O}\left(1\right)-\epsilon O\left(1\right)\right]\hfill \\ \phantom{\rule{1em}{0ex}}<\epsilon \stackrel{˜}{I}<\epsilon K{\parallel \stackrel{˜}{a}\parallel }_{p,\phi }{\parallel \stackrel{˜}{b}\parallel }_{q,\psi }=K{\left[\frac{{\mathrm{\Gamma }}^{{i}_{0}}\left(\frac{1}{\alpha }\right)}{{i}_{0}^{\epsilon /\alpha }{\alpha }^{{i}_{0}-1}\mathrm{\Gamma }\left(\frac{{i}_{0}}{\alpha }\right)}+\epsilon O\left(1\right)\right]}^{\frac{1}{p}}{\left[\frac{{\mathrm{\Gamma }}^{{j}_{0}}\left(\frac{1}{\beta }\right)}{{j}_{0}^{\epsilon /\beta }{\beta }^{{j}_{0}-1}\mathrm{\Gamma }\left(\frac{{j}_{0}}{\beta }\right)}+\epsilon \stackrel{˜}{O}\left(1\right)\right]}^{\frac{1}{q}}.\hfill \end{array}$
For $\epsilon \to {0}^{+}$, we find
$\frac{{\mathrm{\Gamma }}^{{j}_{0}}\left(\frac{1}{\beta }\right){\mathrm{\Gamma }}^{{i}_{0}}\left(\frac{1}{\alpha }\right){k}_{s}\left({\lambda }_{1}\right)}{{\beta }^{{j}_{0}-1}\mathrm{\Gamma }\left(\frac{{j}_{0}}{\beta }\right){\alpha }^{{i}_{0}-1}\mathrm{\Gamma }\left(\frac{{i}_{0}}{\alpha }\right)}\le K{\left[\frac{{\mathrm{\Gamma }}^{{i}_{0}}\left(\frac{1}{\alpha }\right)}{{\alpha }^{{i}_{0}-1}\mathrm{\Gamma }\left(\frac{{i}_{0}}{\alpha }\right)}\right]}^{\frac{1}{p}}{\left[\frac{{\mathrm{\Gamma }}^{{j}_{0}}\left(\frac{1}{\beta }\right)}{{\beta }^{{j}_{0}-1}\mathrm{\Gamma }\left(\frac{{j}_{0}}{\beta }\right)}\right]}^{\frac{1}{q}},$

and then ${K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}\le K$. Hence, $K={K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}$ is the best possible constant factor of (23). □

Theorem 2 With the assumptions of Theorem  1, for $0<{\parallel a\parallel }_{p,\mathrm{\Phi }}<\mathrm{\infty }$, we have the following inequality with the best constant factor ${K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}$:
$\begin{array}{rcl}J& :=& {\left\{\sum _{n}{\left[\mathrm{\Psi }\left(n\right)\right]}^{1-p}{\left(\sum _{m}\frac{{a}_{m}}{{\prod }_{k=1}^{s}\left[{\parallel ln\left(m+\tau \right)\parallel }_{\alpha }^{\lambda /s}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}\right]}\right)}^{p}\right\}}^{\frac{1}{p}}\\ <& {K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}{\parallel a\parallel }_{p,\mathrm{\Phi }},\end{array}$
(27)

which is equivalent to (23).

Proof We set ${b}_{n}$ as follows:
${b}_{n}:={\left[\mathrm{\Psi }\left(n\right)\right]}^{1-p}{\left(\sum _{m}\frac{{a}_{m}}{{\prod }_{k=1}^{s}\left({\parallel ln\left(m+\tau \right)\parallel }_{\alpha }^{\lambda /s}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}\right)}\right)}^{p-1}.$
Then it follows that ${J}^{p}={\parallel b\parallel }_{q,\mathrm{\Psi }}^{q}$. If $J=0$, then (27) is trivially valid, since $0<{\parallel a\parallel }_{p,\mathrm{\Phi }}<\mathrm{\infty }$; if $J=\mathrm{\infty }$, then it is a contradiction since the right hand side of (27) is finite. Suppose that $0. Then by (23), we find
${\parallel b\parallel }_{q,\mathrm{\Psi }}^{q}={J}^{p}=I<{K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}{\parallel a\parallel }_{p,\mathrm{\Phi }}{\parallel b\parallel }_{q,\mathrm{\Psi }},$

namely, ${\parallel b\parallel }_{q,\mathrm{\Psi }}^{q-1}=J<{K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}{\parallel a\parallel }_{p,\mathrm{\Phi }}$, and then (27) follows.

On the other hand, assuming that (27) is valid, by the Hölder inequality, we have
$\begin{array}{rcl}I& =& \sum _{n}{\left(\mathrm{\Psi }\left(n\right)\right)}^{\frac{-1}{q}}\left[\sum _{m}\frac{{a}_{m}}{{\prod }_{k=1}^{s}\left({\parallel ln\left(m+\tau \right)\parallel }_{\alpha }^{\lambda /s}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}\right)}\right]\\ ×\left[{\left(\mathrm{\Psi }\left(n\right)\right)}^{\frac{1}{q}}{b}_{n}\right]\le J{\parallel b\parallel }_{q,\mathrm{\Psi }}.\end{array}$
(28)

Then by (27), we have (23). Hence (27) and (23) are equivalent.

By the equivalency, the constant factor ${K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}$ in (27) is the best possible. Otherwise, we would reach a contradiction by (28) that the constant factor ${K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}$ in (23) is not the best possible. □

For $p>1$, we define two real weight normal discrete spaces ${l}_{p,\phi }$ and ${l}_{q,\psi }$ as follows:
$\begin{array}{c}{l}_{p,\phi }:=\left\{a=\left\{{a}_{m}\right\};{\parallel a\parallel }_{p,\mathrm{\Phi }}={\left\{\sum _{m}\mathrm{\Phi }\left(m\right){a}_{m}^{p}\right\}}^{\frac{1}{p}}<\mathrm{\infty }\right\},\hfill \\ {l}_{q,\psi }:=\left\{b=\left\{{b}_{n}\right\};{\parallel b\parallel }_{q,\mathrm{\Psi }}={\left\{\sum _{n}\mathrm{\Psi }\left(n\right){b}_{n}^{q}\right\}}^{\frac{1}{q}}<\mathrm{\infty }\right\}.\hfill \end{array}$

With the assumptions of Theorem 2, in view of $J<{K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}{\parallel a\parallel }_{p,\mathrm{\Phi }}$, we have the following definition.

Definition 2 Define a multidimensional Hilbert-type operator $T:{l}_{p,\mathrm{\Phi }}\to {l}_{p,{\mathrm{\Psi }}^{1-p}}$ as follows: For $a\in {l}_{p,\mathrm{\Phi }}$, there exists an unique representation $Ta\in {l}_{p,{\mathrm{\Psi }}^{1-p}}$, satisfying for $n\in {\mathbf{N}}^{{j}_{0}}$,
$\left(Ta\right)\left(n\right):=\sum _{m}\frac{{a}_{m}}{{\prod }_{k=1}^{s}\left[{\parallel ln\left(m+\tau \right)\parallel }_{\alpha }^{\lambda /s}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}\right]}.$
(29)
For $b\in {l}_{q,\mathrm{\Psi }}$, we define the following formal inner product of Ta and b as follows:
$\left(Ta,b\right):=\sum _{n}\sum _{m}\frac{{a}_{m}{b}_{n}}{{\prod }_{k=1}^{s}\left[{\parallel ln\left(m+\tau \right)\parallel }_{\alpha }^{\lambda /s}+{c}_{k}{\parallel ln\left(n+\sigma \right)\parallel }_{\beta }^{\lambda /s}\right]}.$
(30)
Then by Theorem 1 and Theorem 2, for $0<{\parallel a\parallel }_{p,\phi },{\parallel b\parallel }_{q,\psi }<\mathrm{\infty }$, we have the following equivalent inequalities:
$\left(Ta,b\right)<{K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}{\parallel a\parallel }_{p,\mathrm{\Phi }}{\parallel b\parallel }_{q,\mathrm{\Psi }},$
(31)
${\parallel Ta\parallel }_{p,{\mathrm{\Psi }}^{1-p}}<{K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}{\parallel a\parallel }_{p,\mathrm{\Phi }}.$
(32)
It follows that T is bounded since
$\parallel T\parallel :=\underset{a\phantom{\rule{0.2em}{0ex}}\left(\ne \theta \right)\in {l}_{p,\mathrm{\Phi }}}{sup}\frac{{\parallel Ta\parallel }_{p,{\mathrm{\Psi }}^{1-p}}}{{\parallel a\parallel }_{p,\mathrm{\Phi }}}\le {K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}.$
(33)

Since the constant factor ${K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}$ in (32) is the best possible, we have:

Corollary 1 With the assumptions of Theorem  2, T is defined by Definition  2, it follows that
$\parallel T\parallel ={K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}={\left[\frac{{\mathrm{\Gamma }}^{{j}_{0}}\left(\frac{1}{\beta }\right)}{{\beta }^{{j}_{0}-1}\mathrm{\Gamma }\left(\frac{{j}_{0}}{\beta }\right)}\right]}^{\frac{1}{p}}{\left[\frac{{\mathrm{\Gamma }}^{{i}_{0}}\left(\frac{1}{\alpha }\right)}{{\alpha }^{{i}_{0}-1}\mathrm{\Gamma }\left(\frac{{i}_{0}}{\alpha }\right)}\right]}^{\frac{1}{q}}{k}_{s}\left({\lambda }_{1}\right).$
(34)
Remark 1 (i) Setting ${\mathrm{\Phi }}_{1}\left(m\right):={\prod }_{i=1}^{{i}_{0}}{\left({m}_{i}+1\right)}^{p-1}{\parallel ln\left(m+1\right)\parallel }_{\alpha }^{p\left({i}_{0}-{\lambda }_{1}\right)-{i}_{0}}$ ($m\in {\mathbf{N}}^{{i}_{0}}$) and ${\mathrm{\Psi }}_{1}\left(n\right):={\prod }_{j=1}^{{j}_{0}}{\left({n}_{j}+1\right)}^{q-1}{\parallel ln\left(n+1\right)\parallel }_{\beta }^{q\left({j}_{0}-{\lambda }_{2}\right)-{j}_{0}}$ ($n\in {\mathbf{N}}^{{j}_{0}}$), then putting $\tau =\sigma =1$ in (23) and (27), we have the following equivalent inequalities with the best constant factor ${K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}$:
$\sum _{n}\sum _{m}\frac{{a}_{m}{b}_{n}}{{\prod }_{k=1}^{s}\left[{\parallel ln\left(m+1\right)\parallel }_{\alpha }^{\lambda /s}+{c}_{k}{\parallel ln\left(n+1\right)\parallel }_{\beta }^{\lambda /s}\right]}<{K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}{\parallel a\parallel }_{p,{\mathrm{\Phi }}_{1}}{\parallel b\parallel }_{q,{\mathrm{\Psi }}_{1}},$
(35)
$\begin{array}{c}{\left\{\sum _{n}{\left[{\mathrm{\Psi }}_{1}\left(n\right)\right]}^{1-p}{\left(\sum _{m}\frac{{a}_{m}}{{\prod }_{k=1}^{s}\left[{\parallel ln\left(m+1\right)\parallel }_{\alpha }^{\lambda /s}+{c}_{k}{\parallel ln\left(n+1\right)\parallel }_{\beta }^{\lambda /s}\right]}\right)}^{p}\right\}}^{\frac{1}{p}}\hfill \\ \phantom{\rule{1em}{0ex}}<{K}_{1}^{\frac{1}{p}}{K}_{2}^{\frac{1}{q}}{\parallel a\parallel }_{p,{\mathrm{\Phi }}_{1}}.\hfill \end{array}$
(36)
Hence, (23) and (27) are more accurate inequalities than (35) and (36).
1. (ii)
Putting ${i}_{0}={j}_{0}=1$, $\lambda =s$, ${\varphi }_{1}\left(m\right):={\left(m+1\right)}^{p-1}{ln}^{p\left(1-{\lambda }_{1}\right)-1}\left(m+1\right)$ ($m\in \mathbf{N}$) and ${\psi }_{1}\left(n\right):={\left(n+1\right)}^{q-1}{ln}^{q\left(1-{\lambda }_{2}\right)-1}\left(n+1\right)$ ($n\in \mathbf{N}$), in (32), we have the following new inequality:
$\begin{array}{c}\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{{a}_{m}{b}_{n}}{{\prod }_{k=1}^{s}ln\left(m+1\right){\left(n+1\right)}^{{c}_{k}}}\hfill \\ \phantom{\rule{1em}{0ex}}<\frac{\pi }{sin\left(\pi {\lambda }_{1}\right)}\sum _{k=1}^{s}\prod _{j=1\phantom{\rule{0.2em}{0ex}}\left(j\ne k\right)}^{s}\frac{{c}_{k}^{{\lambda }_{1}-1}}{{c}_{j}-{c}_{k}}{\parallel a\parallel }_{p,{\varphi }_{1}}{\parallel b\parallel }_{q,{\psi }_{1}}.\hfill \end{array}$
(37)

In particular, for $s={c}_{k}=1$, ${\lambda }_{1}=\frac{1}{q}$, ${\lambda }_{2}=\frac{1}{p}$ in (37), we can deduce (4). Hence, (23) is an extension of (4).

## Declarations

### Acknowledgements

This work is supported by the National Natural Science Foundation of China (No. 61370186), and 2013 Knowledge Construction Special Foundation Item of Guangdong Institution of Higher Learning College and University (No. 2013KJCX0140).

## Authors’ Affiliations

(1)
Department of Computer Science, Guangdong University of Education, Guangzhou, Guangdong, 510303, P.R. China
(2)
Department of Mathematics, Guangdong University of Education, Guangzhou, Guangdong, 510303, P.R. China

## References 