# On reverse Hilbert-type inequalities

## Abstract

By introducing two pairs of conjugate exponents and estimating the weight coefficients, we establish reverse versions of Hilbert-type inequalities, as described by Jin (J. Math. Anal. Appl. 340:932-942, 2008), and we prove that the constant factors are the best possible. As applications, some particular results are considered.

## 1 Introduction

If both ${a}_{n}$ and ${b}_{n}\ge 0$, 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 (see [1])

$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{{a}_{m}{b}_{n}}{m+n}<\pi {\left\{\sum _{n=1}^{\mathrm{\infty }}{a}_{n}^{2}\right\}}^{\frac{1}{2}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{b}_{n}^{2}\right\}}^{\frac{1}{2}},$
(1.1)

where the constant factor π has the best possible value. Inequality (1.1) is the well-known Hilbert inequality, introduced in 1925; inequality (1.1) has been generalized by Hardy as follows.

If $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, and both ${a}_{n}$ and ${b}_{n}\ge 0$, such that $0<{\sum }_{n=1}^{\mathrm{\infty }}{a}_{n}^{p}<\mathrm{\infty }$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{b}_{n}^{q}<\mathrm{\infty }$, then we have

$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{{a}_{m}{b}_{n}}{m+n}<\frac{\pi }{sin\left(\pi /p\right)}{\left\{\sum _{n=1}^{\mathrm{\infty }}{a}_{n}^{p}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{b}_{n}^{q}\right\}}^{\frac{1}{q}},$
(1.2)

where the constant factor $\frac{\pi }{sin\left(\pi /p\right)}$ is the best possible. Inequality (1.2) is the well-known Hardy-Hilbert inequality, which is important in analysis and applications (see [2]). In recent years, many results with generalizations of this type of inequality have been obtained (see [3]).

Under the same conditions as in (1.2), there are some Hilbert-type inequalities that are similar to (1.2), which also have been studied and generalized by some mathematicians.

Recently, by studying a Hilbert-type operator, Jin [4] obtained a new bilinear operator inequality with the norm, and he provided some new Hilbert-type inequalities with the best constant factor. First, we repeat the results of [4].

Definition 1.1 Let ${H}_{p,q}\left(r,s\right)$ be the set of functions $k\left(x,y\right)$ satisfying the following conditions.

Let $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $r>1$, $\frac{1}{r}+\frac{1}{s}=1$, suppose that $k\left(x,y\right)$ is continuous in $\left(0,\mathrm{\infty }\right)×\left(0,\mathrm{\infty }\right)$ and satisfies:

1. (1)

$k\left(x,y\right)=k\left(y,x\right)>0$, where $x,y\in \left(0,\mathrm{\infty }\right)$.

2. (2)

For $\epsilon \ge 0$ and $x>0$, the function $k\left(x,t\right){\left(\frac{x}{t}\right)}^{\frac{1+\epsilon }{l}}$ ($l=r,s$) is decreasing in $t\in \left(0,\mathrm{\infty }\right)$.

For $\epsilon \ge 0$ small enough, $x>0$, and ${\overline{k}}_{l}\left(\epsilon ,x\right)$ can be written as

${\overline{k}}_{l}\left(\epsilon ,x\right):={\int }_{0}^{\mathrm{\infty }}k\left(x,t\right){\left(\frac{x}{t}\right)}^{\frac{1+\epsilon }{l}}\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{1em}{0ex}}\left(l=r,s\right),$

where ${\overline{k}}_{l}\left(\epsilon ,x\right)$ is independent of x, ${\overline{k}}_{l}\left(0,x\right):={\int }_{0}^{\mathrm{\infty }}k\left(x,t\right){\left(\frac{x}{t}\right)}^{\frac{1}{l}}\phantom{\rule{0.2em}{0ex}}dt={k}_{p}$ ($l=r,s$), ${k}_{p}$ is a positive constant independent of x, and ${\overline{k}}_{l}\left(\epsilon ,x\right)={k}_{p}\left(\epsilon \right)={k}_{p}+o\left(1\right)$ ($\epsilon \to {0}^{+}$).

$\begin{array}{c}\left(3\right)\phantom{\rule{1em}{0ex}}\sum _{m=1}^{\mathrm{\infty }}\frac{1}{{m}^{1+\epsilon }}{\int }_{0}^{1}k\left(m,t\right){\left(\frac{m}{t}\right)}^{\frac{1+\epsilon \left(s/q\right)}{s}}\phantom{\rule{0.2em}{0ex}}dt=O\left(1\right)\phantom{\rule{1em}{0ex}}\left(\epsilon \to {0}^{+}\right),\hfill \\ \phantom{\left(3\right)\phantom{\rule{1em}{0ex}}}\sum _{m=1}^{\mathrm{\infty }}\frac{1}{{m}^{1+\epsilon }}{\int }_{0}^{1}k\left(m,t\right){\left(\frac{m}{t}\right)}^{\frac{1+\epsilon \left(r/p\right)}{r}}\phantom{\rule{0.2em}{0ex}}dt=O\left(1\right)\phantom{\rule{1em}{0ex}}\left(\epsilon \to {0}^{+}\right).\hfill \end{array}$

We have Jin’s result as follows.

Theorem 1.1 If $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $r>1$, $\frac{1}{r}+\frac{1}{s}=1$, and $k\left(x,y\right)\in {H}_{p,q}\left(r,s\right)$, ${a}_{n},{b}_{n}\ge 0$, such that $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{a}_{n}^{p}<\mathrm{\infty }$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{s}-1}{b}_{n}^{q}<\mathrm{\infty }$, then we have

$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}k\left(m,n\right){a}_{m}{b}_{n}<{k}_{r}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{a}_{n}^{p}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{s}-1}{b}_{n}^{q}\right\}}^{\frac{1}{q}},$
(1.3)
$\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{\left[\sum _{m=1}^{\mathrm{\infty }}k\left(m,n\right){a}_{m}\right]}^{p}<{\left({k}_{r}\right)}^{p}\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{a}_{n}^{p}.$
(1.4)

Here the constant factors ${k}_{r}$ and ${\left({k}_{r}\right)}^{p}$ are the best possible. Inequality (1.3) is equivalent to (1.4).

If $p=r$ and $q=s$ in Theorem 1.1, then Theorem 1.1 reduces to Yang’s result [5] as follows.

Theorem 1.2 If $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $k\left(x,y\right)\in H\left(p,q\right)$, and both ${a}_{n}$ and ${b}_{n}\ge 0$, such that $0<{\sum }_{n=1}^{\mathrm{\infty }}{a}_{n}^{p}<\mathrm{\infty }$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{a}_{n}^{p}<\mathrm{\infty }$, then we have

$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}k\left(m,n\right){a}_{n}{b}_{n}<{k}_{p}\left\{\sum _{n=1}^{\mathrm{\infty }}{a}_{n}^{p}\right\}{\left\{\sum _{n=1}^{\mathrm{\infty }}{b}_{n}^{q}\right\}}^{\frac{1}{q}},$
(1.5)
$\sum _{n=1}^{\mathrm{\infty }}{\left[\sum _{n=1}^{\mathrm{\infty }}k\left(m,n\right){a}_{m}\right]}^{p}<{\left({k}_{p}\right)}^{p}\sum _{n=1}^{\mathrm{\infty }}{a}_{n}^{p},$
(1.6)

where the constant factors ${k}_{p}$ and ${\left({k}_{p}\right)}^{p}$ are the best possible. Inequality (1.3) is equivalent to (1.4).

In this paper, by introducing some parameters, we establish a reverse version of the inequality (1.3). As applications, some particular results are considered.

## 2 Some lemmas

Definition 2.1 Let ${H}_{p,q}\left(r,s\right)$ be the set of functions $k\left(x,y\right)$ satisfying the following conditions:

Let $0, $\frac{1}{p}+\frac{1}{q}=1$, $r>1$, $\frac{1}{r}+\frac{1}{s}=1$, suppose that $k\left(x,y\right)$ is continuous in $\left(0,\mathrm{\infty }\right)×\left(0,\mathrm{\infty }\right)$ and satisfies:

1. (1)

$k\left(x,y\right)=k\left(y,x\right)>0$, $x,y\in \left(0,\mathrm{\infty }\right)$.

2. (2)

For $\epsilon \ge 0$ and $x>0$, the function $k\left(x,t\right){\left(\frac{x}{t}\right)}^{\frac{1+\epsilon }{l}}$ ($l=r,s$) is decreasing in $t\in \left(0,\mathrm{\infty }\right)$.

For $\epsilon \ge 0$ small enough, for $x>0$, ${\overline{k}}_{l}\left(\epsilon ,x\right)$ can be described as

${\overline{k}}_{l}\left(\epsilon ,x\right):={\int }_{0}^{\mathrm{\infty }}k\left(x,t\right){\left(\frac{x}{t}\right)}^{\frac{1+\epsilon }{l}}\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{1em}{0ex}}\left(l=r,s\right),$

where ${\overline{k}}_{l}\left(\epsilon ,x\right)$ is independent of x, and ${\overline{k}}_{l}\left(0,x\right):={\int }_{0}^{\mathrm{\infty }}k\left(x,t\right){\left(\frac{x}{t}\right)}^{\frac{1}{l}}\phantom{\rule{0.2em}{0ex}}dt={k}_{p}$ ($l=r,s$), ${k}_{p}$ is a positive constant independent of x, and ${\overline{k}}_{l}\left(\epsilon ,x\right)={k}_{p}\left(\epsilon \right)={k}_{p}+o\left(1\right)$ ($\epsilon \to {0}^{+}$).

1. (3)

There exists a positive constant ${\lambda }^{\prime }$ such that

$\begin{array}{c}{\theta }_{\lambda }\left(s,m\right)=\frac{1}{{k}_{r}}{\int }_{0}^{1}k\left(m,t\right){\left(\frac{m}{t}\right)}^{\frac{1}{s}}\phantom{\rule{0.2em}{0ex}}dt=O\left(1/{m}^{{\lambda }^{\prime }}\right)\in \left(0,1\right)\phantom{\rule{1em}{0ex}}\left(m\to \mathrm{\infty }\right),\hfill \\ {\theta }_{\lambda }\left(r,n\right)=\frac{1}{{k}_{r}}{\int }_{0}^{1}k\left(t,n\right){\left(\frac{n}{t}\right)}^{\frac{1}{r}}\phantom{\rule{0.2em}{0ex}}dt=O\left(1/{n}^{{\lambda }^{\prime }}\right)\in \left(0,1\right)\phantom{\rule{1em}{0ex}}\left(n\to \mathrm{\infty }\right).\hfill \end{array}$

Lemma 2.2 If $0, $\frac{1}{p}+\frac{1}{q}=1$, $r>1$, $\frac{1}{r}+\frac{1}{s}=1$, $k\left(x,y\right)\in {H}_{p,q}\left(r,s\right)$, and the weight coefficients $w\left(r,p,m\right)$ and $w\left(s,q,n\right)$ are defined as

$\omega \left(r,p,m\right)=\sum _{n=1}^{\mathrm{\infty }}k\left(m,n\right)\frac{{m}^{\frac{p-1}{r}}}{{n}^{\frac{1}{s}}},$
(2.1)
$\omega \left(s,q,n\right)=\sum _{m=1}^{\mathrm{\infty }}k\left(m,n\right)\frac{{n}^{\frac{q-1}{s}}}{{m}^{\frac{1}{r}}},$
(2.2)

then we have

$\begin{array}{r}{m}^{\frac{p}{r}-1}{k}_{r}\left(1-{\theta }_{\lambda }\left(s,m\right)\right)<\omega \left(r,p,m\right)<{m}^{\frac{p}{r}-1}{k}_{r},\\ {n}^{\frac{q}{s}-1}{k}_{r}\left(1-{\theta }_{\lambda }\left(r,n\right)\right)<\omega \left(s,q,n\right)<{n}^{\frac{q}{s}-1}{k}_{r}.\end{array}$
(2.3)

Proof By the assumption of the lemma, because $k\left(x,t\right){\left(\frac{x}{t}\right)}^{\frac{1}{s}}$ ($t\in \left(0,\mathrm{\infty }\right)$) is decreasing, then we find

$\begin{array}{rl}\omega \left(r,p,m\right)& ={m}^{\frac{p}{r}-1}\sum _{n=1}^{\mathrm{\infty }}k\left(m,n\right){\left(\frac{m}{n}\right)}^{\frac{1}{s}}\\ \le {m}^{\frac{p}{r}-1}{\int }_{0}^{\mathrm{\infty }}k\left(m,t\right){\left(\frac{m}{t}\right)}^{\frac{1}{s}}\phantom{\rule{0.2em}{0ex}}dt\\ ={m}^{\frac{p}{r}-1}{k}_{r}.\end{array}$

However, we find

$\begin{array}{rl}\omega \left(r,p,m\right)& ={m}^{\frac{p}{r}-1}\sum _{n=1}^{\mathrm{\infty }}k\left(m,n\right){\left(\frac{m}{n}\right)}^{\frac{1}{s}}\ge {m}^{\frac{p}{r}-1}{\int }_{1}^{\mathrm{\infty }}k\left(m,t\right){\left(\frac{m}{t}\right)}^{\frac{1}{s}}\phantom{\rule{0.2em}{0ex}}dt\\ ={m}^{\frac{p}{r}-1}{\int }_{0}^{\mathrm{\infty }}k\left(m,t\right){\left(\frac{m}{t}\right)}^{\frac{1}{s}}\phantom{\rule{0.2em}{0ex}}dt-{m}^{\frac{p}{r}-1}{\int }_{0}^{1}k\left(m,t\right){\left(\frac{m}{t}\right)}^{\frac{1}{s}}\phantom{\rule{0.2em}{0ex}}dt\\ ={m}^{\frac{p}{r}-1}{k}_{r}-{m}^{\frac{p}{r}-1}{\int }_{0}^{1}k\left(m,t\right){\left(\frac{m}{t}\right)}^{\frac{1}{s}}\phantom{\rule{0.2em}{0ex}}dt\\ ={m}^{\frac{p}{r}-1}{k}_{r}\left(1-\frac{1}{{k}_{r}}{\int }_{0}^{1}k\left(m,t\right){\left(\frac{m}{t}\right)}^{\frac{1}{s}}\phantom{\rule{0.2em}{0ex}}dt\right)\\ ={m}^{\frac{p}{r}-1}{k}_{r}\left(1-{\theta }_{\lambda }\left(s,m\right)\right).\end{array}$

It is easy to show that the above inequalities take the form of a strict inequality. Hence, we have ${m}^{\frac{p}{r}-1}{k}_{r}\left(1-{\theta }_{\lambda }\left(s,m\right)\right)<\omega \left(r,p,m\right)<{m}^{\frac{p}{r}-1}{k}_{r}$. Similarly, we can obtain ${n}^{\frac{q}{s}-1}{k}_{r}\left(1-{\theta }_{\lambda }\left(r,n\right)\right)<\omega \left(s,q,n\right)<{n}^{\frac{q}{s}-1}{k}_{r}$. The lemma is proved. □

Lemma 2.3 If $0, $\frac{1}{p}+\frac{1}{q}=1$, $r>1$, $\frac{1}{r}+\frac{1}{s}=1$, and $k\left(x,y\right)\in {H}_{p,q}\left(r,s\right)$, for $\epsilon >0$ small enough, we have

$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}k\left(m,n\right){m}^{-\frac{1}{r}-\frac{\epsilon }{p}}{n}^{-\frac{1}{s}-\frac{\epsilon }{q}}<\left(kr+o\left(1\right)\right)\sum _{1}^{\mathrm{\infty }}{m}^{-1-\epsilon }\phantom{\rule{1em}{0ex}}\left(\epsilon \to {0}^{+}\right).$
(2.4)

Proof For $\epsilon >0$, by Definition 2.1, we have

$\begin{array}{c}\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}k\left(m,n\right){m}^{-\frac{1}{r}-\frac{\epsilon }{p}}{n}^{-\frac{1}{s}-\frac{\epsilon }{q}}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{m=1}^{\mathrm{\infty }}{m}^{-1-\epsilon }\epsilon \sum _{n=1}^{\mathrm{\infty }}k\left(m,n\right){\left(\frac{m}{n}\right)}^{\frac{1+\epsilon \left(s/q\right)}{s}}\hfill \\ \phantom{\rule{1em}{0ex}}\le \sum _{m=1}^{\mathrm{\infty }}{m}^{-1-\epsilon }{\int }_{0}^{\mathrm{\infty }}k\left(m,t\right){\left(\frac{m}{t}\right)}^{\frac{1+\epsilon \left(s/q\right)}{s}}\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{m=1}^{\mathrm{\infty }}{m}^{-1-\epsilon }kr\left(\frac{\epsilon s}{q}\right)=\left(kr+o\left(1\right)\right)\sum _{1}^{\mathrm{\infty }}{m}^{-1-\epsilon }.\hfill \end{array}$

The lemma is proved. □

## 3 Main results

Theorem 3.1 If $0, $\frac{1}{p}+\frac{1}{q}=1$, $r>1$, $\frac{1}{r}+\frac{1}{s}=1$, and $k\left(x,y\right)\in {H}_{p,q}\left(r,s\right)$, ${a}_{n},{b}_{n}\ge 0$, such that $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{a}_{n}^{p}<\mathrm{\infty }$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{s}-1}{b}_{n}^{q}<\mathrm{\infty }$, then we have

$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}k\left(m,n\right){a}_{m}{b}_{n}>{k}_{r}{\left\{\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(s,n\right)\right]{n}^{\frac{p}{r}-1}{a}_{n}^{p}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{s}-1}{b}_{n}^{q}\right\}}^{\frac{1}{q}},$
(3.1)
$\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{\left[\sum _{m=1}^{\mathrm{\infty }}k\left(m,n\right){a}_{m}\right]}^{p}>{\left({k}_{r}\right)}^{p}\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(s,n\right)\right]{n}^{\frac{p}{r}-1}{a}_{n}^{p},$
(3.2)

where the constant factor ${k}_{r}$ and ${\left({k}_{r}\right)}^{p}$ are the best possible. Inequality (3.1) is equivalent to (3.2).

Proof By Hölder’s inequality, we have (see [6])

$\begin{array}{rcl}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}k\left(m,n\right){a}_{m}{b}_{n}& =& \sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\left\{{\left[k\left(m,n\right)\right]}^{\frac{1}{q}}\frac{{m}^{\frac{1}{qr}}}{{n}^{\frac{1}{ps}}}{a}_{m}\right\}\left\{{\left[k\left(m,n\right)\right]}^{\frac{1}{q}}\frac{{n}^{\frac{1}{ps}}}{m\frac{1}{qr}}{b}_{n}\right\}\\ \ge & {\left\{\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}k\left(m,n\right)\frac{{m}^{\frac{p-1}{r}}}{{n}^{\frac{1}{s}}}{a}_{m}^{p}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}k\left(m,n\right)\frac{{n}^{\frac{q-1}{s}}}{m\frac{1}{r}}{b}_{n}^{q}\right\}}^{\frac{1}{q}}\\ =& {\left\{\sum _{m=1}^{\mathrm{\infty }}\omega \left(r,p,m\right){a}_{m}^{p}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}\omega \left(s,q,n\right){b}_{n}^{q}\right\}}^{\frac{1}{q}}.\end{array}$
(3.3)

Then, by (2.3), in view of $0 and $q<0$, we have (3.1).

For $\epsilon >0$, setting ${\overline{a}}_{n}={n}^{-\frac{1}{r}-\frac{\epsilon }{q}}$ and ${\overline{b}}_{n}={n}^{-\frac{1}{s}-\frac{\epsilon }{q}}$, we find

$\begin{array}{c}{\left\{\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(s,n\right)\right]{n}^{\frac{p}{r}-1}{\overline{a}}_{n}^{p}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{s}-1}{\overline{b}}_{n}^{q}\right\}}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}={\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{-1-\epsilon }-\sum _{n=1}^{\mathrm{\infty }}O\left(1/{n}^{{\lambda }^{\prime }}\right){n}^{-1-\epsilon }\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{-1-\epsilon }\right\}}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{n=1}^{\mathrm{\infty }}{n}^{-1-\epsilon }{\left[1-{\left(\sum _{n=1}^{\mathrm{\infty }}{n}^{-1-\epsilon }\right)}^{-1}\sum _{n=1}^{\mathrm{\infty }}O\left(1/{n}^{{\lambda }^{\prime }}\right){n}^{-1-\epsilon }\right]}^{\frac{1}{p}}.\hfill \end{array}$
(3.4)

By virtue of (2.4), we have

$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}k\left(m,n\right){\overline{a}}_{m}{\overline{b}}_{n}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}k\left(m,n\right){m}^{-\frac{1}{r}-\frac{\epsilon }{p}}{n}^{-\frac{1}{s}-\frac{\epsilon }{q}}<\left(kr+o\left(1\right)\right)\sum _{1}^{\mathrm{\infty }}{m}^{-1-\epsilon }\phantom{\rule{1em}{0ex}}\left(\epsilon \to {0}^{+}\right).\hfill \end{array}$
(3.5)

If the constant factor ${k}_{r}$ in (3.1) is not the best possible factor, then there exists a positive number K (with $K>{k}_{r}$), such that (3.1) is still valid if the constant factor ${k}_{r}$ is replaced by K. In particular, by (3.4) and (3.5), we have

$\left(kr+o\left(1\right)\right)\sum _{1}^{\mathrm{\infty }}{n}^{-1-\epsilon }>K\sum _{n=1}^{\mathrm{\infty }}{n}^{-1-\epsilon }{\left[1-{\left(\sum _{n=1}^{\mathrm{\infty }}{n}^{-1-\epsilon }\right)}^{-1}\sum _{n=1}^{\mathrm{\infty }}O\left(1/{n}^{{\lambda }^{\prime }}\right){n}^{-1-\epsilon }\right]}^{\frac{1}{p}},$

that is,

$\left(kr+o\left(1\right)\right)>K{\left[1-{\left(\sum _{n=1}^{\mathrm{\infty }}{n}^{-1-\epsilon }\right)}^{-1}\sum _{n=1}^{\mathrm{\infty }}O\left(1/{n}^{{\lambda }^{\prime }}\right){n}^{-1-\epsilon }\right]}^{\frac{1}{p}}.$

For $\epsilon \to {0}^{+}$, it follows that $K\le {k}_{r}$, which contradicts the fact that $K>{k}_{r}$. Hence, the constant factor ${k}_{r}$ in (3.1) is the best possible.

Setting ${b}_{n}$ as

${b}_{n}:={n}^{\frac{p}{r}-1}{\left[\sum _{m=1}^{\mathrm{\infty }}k\left(m,n\right){a}_{m}\right]}^{p-1},$

by (3.1), we have

$\begin{array}{rcl}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{s}-1}{b}_{n}^{q}\right\}}^{p}& =& {\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{\left[\sum _{m=1}^{\mathrm{\infty }}k\left(m,n\right){a}_{m}\right]}^{p}\right\}}^{p}\\ =& {\left\{\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}k\left(m,n\right){a}_{m}{b}_{n}\right\}}^{p}\\ \ge & {\left({k}_{r}\right)}^{p}\left\{\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(s,n\right)\right]{n}^{\frac{p}{r}-1}{a}_{n}^{p}\right\}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{s}-1}{b}_{n}^{q}\right\}}^{p-1}.\end{array}$
(3.6)

Hence, we obtain

$\mathrm{\infty }>\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{\left[\sum _{m=1}^{\mathrm{\infty }}k\left(m,n\right){a}_{m}\right]}^{p}\ge {\left({k}_{r}\right)}^{p}\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(s,n\right)\right]{n}^{\frac{p}{r}-1}{a}_{n}^{p}>0.$
(3.7)

By (3.1), both (3.6) and (3.7) take the form of a strict inequality, and we have (3.2).

However, if (3.2) is valid, by Hölder’s inequality, we find

$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}k\left(m,n\right){a}_{m}{b}_{n}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{n=1}^{\mathrm{\infty }}\left[{n}^{\frac{1}{q}-\frac{1}{s}}\sum _{m=1}^{\mathrm{\infty }}k\left(m,n\right){a}_{m}\right]\left[{n}^{\frac{1}{s}-\frac{1}{q}}{b}_{n}\right]\hfill \\ \phantom{\rule{1em}{0ex}}\ge {\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{\left[\sum _{m=1}^{\mathrm{\infty }}k\left(m,n\right){a}_{m}\right]}^{p}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{s}-1}{b}_{n}^{q}\right\}}^{\frac{1}{q}}.\hfill \end{array}$
(3.8)

Then, by (3.2), we have (3.1). Hence (3.2) and (3.1) are equivalent.

If the constant factor ${\left({k}_{r}\right)}^{p}$ in (3.2) is not the best possible, by using (3.8), we find the contradiction that the constant factor ${k}_{r}$ in (3.1) is not the best possible. The theorem is completed. □

## 4 Some particular results

1. (1)

Setting

$k\left(x,y\right)=\frac{{\left(xy\right)}^{\frac{\lambda -1}{2}}}{{\left(x+y\right)}^{\lambda }}\left(1-2min\left\{\frac{1}{r},\frac{1}{s}\right\}<\lambda \le 1+2min\left\{\frac{1}{r},\frac{1}{s}\right\}\right),$

for $0\le \epsilon , and for fixed $x>0$, we find (see [4])

${\overline{k}}_{s}\left(\epsilon ,x\right)\to B\left(\frac{s\left(\lambda +1\right)-2}{2s},\frac{r\left(\lambda +1\right)-2}{2r}\right)={k}_{r}\phantom{\rule{1em}{0ex}}\left(\epsilon \to {0}^{+}\right),$

and ${\overline{k}}_{s}\left(\epsilon ,x\right)\to {k}_{r}$ ($\epsilon \to {0}^{+}$);

$\begin{array}{rcl}0& <& {\int }_{0}^{1}k\left(m,t\right){\left(\frac{m}{t}\right)}^{\frac{1}{s}}\phantom{\rule{0.2em}{0ex}}dt={\int }_{0}^{1}\frac{{\left(mt\right)}^{\frac{\lambda -1}{2}}}{{\left(m+t\right)}^{\lambda }}{\left(\frac{m}{t}\right)}^{\frac{1}{s}}\phantom{\rule{0.2em}{0ex}}dt\\ \le & {\int }_{0}^{1}\frac{{\left(mt\right)}^{\frac{\lambda -1}{2}}}{{m}^{\lambda }}{\left(\frac{m}{t}\right)}^{\frac{1}{s}}\phantom{\rule{0.2em}{0ex}}dt=\frac{1}{\left(\frac{\lambda -1}{2}+\frac{1}{r}\right)}\frac{1}{{m}^{\frac{1+\lambda }{2}-\frac{1}{s}}}.\end{array}$

Hence, ${\theta }_{\lambda }\left(s,m\right)=O\left({m}^{\frac{1}{s}-\frac{1+\lambda }{2}}\right)$. Similarly, we obtain ${\theta }_{\lambda }\left(r,n\right)=O\left({n}^{\frac{1}{r}-\frac{1+\lambda }{2}}\right)$. For $\epsilon \ge 0$, $1-2min\left\{\frac{1}{r},\frac{1}{s}\right\}<\lambda \le 1+2min\left\{\frac{1}{r},\frac{1}{s}\right\}$, and fixed $x>0$, the function

$k\left(x,t\right){\left(\frac{x}{t}\right)}^{\frac{1+\epsilon }{l}}=\frac{{\left(xt\right)}^{\frac{\lambda -1}{2}}}{{\left(x+t\right)}^{\lambda }}{\left(\frac{x}{t}\right)}^{\frac{1+\epsilon }{l}}=\frac{{x}^{\frac{1+\epsilon }{l}+\frac{\lambda -1}{2}}}{{\left(x+t\right)}^{\lambda }}{t}^{\frac{\lambda -1}{2}-\frac{1+\epsilon }{l}}\phantom{\rule{1em}{0ex}}\left(l=r,s\right)$

is decreasing in $\left(0,\mathrm{\infty }\right)$. Hence, $k\left(x,y\right)\in {H}_{p}\left(r,s\right)$. By Theorem 3.1, we have the following.

Corollary 4.1 If $0, $1/p+1/q=1$, $1/r+1/s=1$, $1-2min\left\{\frac{1}{r},\frac{1}{s}\right\}<\lambda \le 1+2min\left\{\frac{1}{r},\frac{1}{s}\right\}$, and both ${a}_{n},{b}_{n}\ge 0$ such that $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{a}_{n}^{p}<\mathrm{\infty }$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{s}-1}{b}_{n}^{q}<\mathrm{\infty }$, then we have

$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(mn\right)}^{\frac{\lambda -1}{2}}}{{\left(m+n\right)}^{\lambda }}{a}_{m}{b}_{n}\hfill \\ \phantom{\rule{1em}{0ex}}>B\left(\frac{s\left(\lambda +1\right)-2}{2s},\frac{r\left(\lambda +1\right)-2}{2r}\right){\left\{\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(s,n\right)\right]{n}^{\frac{p}{r}-1}{a}_{n}^{p}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{b}_{n}^{q}\right\}}^{\frac{1}{q}},\hfill \end{array}$
(4.1)
$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(mn\right)}^{\frac{\lambda -1}{2}}}{{\left(m+n\right)}^{\lambda }}{a}_{m}\right]}^{p}\hfill \\ \phantom{\rule{1em}{0ex}}>{\left[B\left(\frac{s\left(\lambda +1\right)-2}{2s},\frac{r\left(\lambda +1\right)-2}{2r}\right)\right]}^{p}\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(s,n\right)\right]{n}^{\frac{p}{r}-1}{a}_{n}^{p},\hfill \end{array}$
(4.2)

where the constant factors

$B\left(\frac{s\left(\lambda +1\right)-2}{2s},\frac{r\left(\lambda +1\right)-2}{2r}\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{\left[B\left(\frac{s\left(\lambda +1\right)-2}{2s},\frac{r\left(\lambda +1\right)-2}{2r}\right)\right]}^{p}$

are the best possible. Inequality (4.1) is equivalent to (4.2).

In particular, (a) for $r=q$, $s=p$, and $1-2min\left\{\frac{1}{p},\frac{1}{q}\right\}<\lambda \le 1+2min\left\{\frac{1}{p},\frac{1}{q}\right\}$, we have

$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(mn\right)}^{\frac{\lambda -1}{2}}}{{\left(m+n\right)}^{\lambda }}{a}_{m}{b}_{n}\hfill \\ \phantom{\rule{1em}{0ex}}>B\left(\frac{p\left(\lambda +1\right)-2}{2p},\frac{q\left(\lambda +1\right)-2}{2q}\right){\left\{\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(p,n\right)\right]{n}^{p-2}{a}_{n}^{p}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{q-2}{b}_{n}^{q}\right\}}^{\frac{1}{q}},\hfill \end{array}$
(4.3)
$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}{n}^{p-2}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(mn\right)}^{\frac{\lambda -1}{2}}}{{\left(m+n\right)}^{\lambda }}{a}_{m}\right]}^{p}\hfill \\ \phantom{\rule{1em}{0ex}}>{\left[B\left(\frac{p\left(\lambda +1\right)-2}{2p},\frac{q\left(\lambda +1\right)-2}{2q}\right)\right]}^{p}\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(p,n\right)\right]{n}^{p-2}{a}_{n}^{p}.\hfill \end{array}$
(4.4)
1. (b)

For $r=s=2$ and $0<\lambda \le 2$, we have

$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(mn\right)}^{\frac{\lambda -1}{2}}}{{\left(m+n\right)}^{\lambda }}{a}_{m}{b}_{n}>B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right){\left\{\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(2,n\right)\right]{n}^{\frac{p}{2}-1}{a}_{n}^{p}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{2}-1}{b}_{n}^{q}\right\}}^{\frac{1}{q}},$
(4.5)
$\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{2}-1}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(mn\right)}^{\frac{\lambda -1}{2}}}{{\left(m+n\right)}^{\lambda }}{a}_{m}\right]}^{p}>{\left[B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right)\right]}^{p}\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(2,n\right)\right]{n}^{\frac{p}{2}-1}{a}_{n}^{p}.$
(4.6)
2. (2)

Let

$k\left(x,y\right)=\frac{{\left(xy\right)}^{\frac{\lambda -1}{2}}}{{x}^{\lambda }+{y}^{\lambda }}\left(1-2min\left\{\frac{1}{r},\frac{1}{s}\right\}<\lambda \le 1+2min\left\{\frac{1}{r},\frac{1}{s}\right\}\right).$

For $0\le \epsilon and $x>0$, we find (see [4])

${\overline{k}}_{s}\left(\epsilon ,x\right)\to \frac{1}{\lambda }B\left(\frac{s\left(\lambda +1\right)-2}{2s\lambda },\frac{r\left(\lambda +1\right)-2}{2r\lambda }\right)={k}_{r}\phantom{\rule{1em}{0ex}}\left(\epsilon \to {0}^{+}\right),$

and ${\overline{k}}_{s}\left(\epsilon ,x\right)\to {k}_{r}$ ($\epsilon \to {0}^{+}$);

$\begin{array}{rcl}{\int }_{0}^{1}k\left(m,t\right){\left(\frac{m}{t}\right)}^{\frac{1}{s}}\phantom{\rule{0.2em}{0ex}}dt& =& {\int }_{0}^{1}\frac{{\left(mt\right)}^{\frac{\lambda -1}{2}}}{{m}^{\lambda }+{t}^{\lambda }}{\left(\frac{m}{t}\right)}^{\frac{1}{s}}\phantom{\rule{0.2em}{0ex}}dt\\ \le & {\int }_{0}^{1}\frac{{\left(mt\right)}^{\frac{\lambda -1}{2}}}{{m}^{\lambda }}{\left(\frac{m}{t}\right)}^{\frac{1}{s}}\phantom{\rule{0.2em}{0ex}}dt=\frac{1}{\left(\frac{\lambda -1}{2}+\frac{1}{r}\right)}{m}^{\frac{1}{s}-\frac{1+\lambda }{2}}.\end{array}$

Hence, ${\theta }_{\lambda }\left(s,m\right)=O\left({m}^{\frac{1}{s}-\frac{1+\lambda }{2}}\right)$. Similarly, we can obtain ${\theta }_{\lambda }\left(r,n\right)=O\left({n}^{\frac{1}{r}-\frac{1+\lambda }{2}}\right)$. For $\epsilon \ge 0$, $1-2min\left\{\frac{1}{r},\frac{1}{s}\right\}<\lambda \le 1+2min\left\{\frac{1}{r},\frac{1}{s}\right\}$, and $x>0$, the function

$k\left(x,t\right){\left(\frac{x}{t}\right)}^{\frac{1+\epsilon }{l}}=\frac{{\left(xt\right)}^{\frac{\lambda -1}{2}}}{{x}^{\lambda }+{t}^{\lambda }}{\left(\frac{x}{t}\right)}^{\frac{1+\epsilon }{l}}=\frac{{x}^{\frac{1+\epsilon }{l}+\frac{\lambda -1}{2}}}{{x}^{\lambda }+{t}^{\lambda }}{t}^{\frac{\lambda -1}{2}-\frac{1+\epsilon }{l}}\phantom{\rule{1em}{0ex}}\left(l=r,s\right)$

is decreasing in $\left(0,\mathrm{\infty }\right)$. Hence $k\left(x,y\right)\in {H}_{p}\left(r,s\right)$. By Theorem 3.1, we have the following corollary.

Corollary 4.2 If $0, $\frac{1}{p}+\frac{1}{q}=1$, $r>1$, $\frac{1}{r}+\frac{1}{s}=1$, $1-2min\left\{\frac{1}{r},\frac{1}{s}\right\}<\lambda \le 1+2min\left\{\frac{1}{r},\frac{1}{s}\right\}$, and both ${a}_{n},{b}_{n}\ge 0$ such that $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{a}_{n}^{p}<\mathrm{\infty }$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{s}-1}{b}_{n}^{q}<\mathrm{\infty }$, then we have

$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(mn\right)}^{\frac{\lambda -1}{2}}}{{m}^{\lambda }+{n}^{\lambda }}{a}_{m}{b}_{n}\hfill \\ \phantom{\rule{1em}{0ex}}>\frac{1}{\lambda }B\left(\frac{s\left(\lambda +1\right)-2}{2s\lambda },\frac{r\left(\lambda +1\right)-2}{2r\lambda }\right){\left\{\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(s,n\right)\right]{n}^{\frac{p}{r}-1}{a}_{n}^{p}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{b}_{n}^{q}\right\}}^{\frac{1}{q}},\hfill \end{array}$
(4.7)
$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(mn\right)}^{\frac{\lambda -1}{2}}}{{m}^{\lambda }+{n}^{\lambda }}{a}_{m}\right]}^{p}\hfill \\ \phantom{\rule{1em}{0ex}}>{\left[\frac{1}{\lambda }B\left(\frac{s\left(\lambda +1\right)-2}{2s\lambda },\frac{r\left(\lambda +1\right)-2}{2r\lambda }\right)\right]}^{p}\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(s,n\right)\right]{n}^{\frac{p}{r}-1}{a}_{n}^{p},\hfill \end{array}$
(4.8)

where the constant factors

$\frac{1}{\lambda }B\left(\frac{s\left(\lambda +1\right)-2}{2s\lambda },\frac{r\left(\lambda +1\right)-2}{2r\lambda }\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{\left[\frac{1}{\lambda }B\left(\frac{s\left(\lambda +1\right)-2}{2s\lambda },\frac{r\left(\lambda +1\right)-2}{2r\lambda }\right)\right]}^{p}$

are the best possible. Inequality (4.7) is equivalent to (4.8).

In particular, (a) for $r=q$, $s=p$, and $1-2min\left\{\frac{1}{p},\frac{1}{q}\right\}<\lambda \le 1+2min\left\{\frac{1}{p},\frac{1}{q}\right\}$, we have

$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(mn\right)}^{\frac{\lambda -1}{2}}}{{m}^{\lambda }+{n}^{\lambda }}{a}_{m}{b}_{n}\hfill \\ \phantom{\rule{1em}{0ex}}>\frac{1}{\lambda }B\left(\frac{p\left(\lambda +1\right)-2}{2p\lambda },\frac{q\left(\lambda +1\right)-2}{2q\lambda }\right)\hfill \\ \phantom{\rule{2em}{0ex}}×{\left\{\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(p,n\right)\right]{n}^{p-2}{a}_{n}^{p}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{q-2}{b}_{n}^{q}\right\}}^{\frac{1}{q}},\hfill \end{array}$
(4.9)
$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}{n}^{p-2}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(mn\right)}^{\frac{\lambda -1}{2}}}{{\left(m+n\right)}^{\lambda }}{a}_{m}\right]}^{p}\hfill \\ \phantom{\rule{1em}{0ex}}>{\left[\frac{1}{\lambda }B\left(\frac{p\left(\lambda +1\right)-2}{2p\lambda },\frac{q\left(\lambda +1\right)-2}{2q\lambda }\right)\right]}^{p}\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(p,n\right)\right]{n}^{p-2}{a}_{n}^{p}.\hfill \end{array}$
(4.10)
1. (b)

For $r=s=2$ and $0<\lambda \le 2$, we have

$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(mn\right)}^{\frac{\lambda -1}{2}}}{{m}^{\lambda }+{n}^{\lambda }}{a}_{m}{b}_{n}>\frac{\pi }{\lambda }{\left\{\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(2,n\right)\right]{n}^{\frac{p}{2}-1}{a}_{n}^{p}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{2}-1}{b}_{n}^{q}\right\}}^{\frac{1}{q}},$
(4.11)
$\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{2}-1}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(mn\right)}^{\frac{\lambda -1}{2}}}{{m}^{\lambda }+{n}^{\lambda }}{a}_{m}\right]}^{p}>{\left[\frac{\pi }{\lambda }\right]}^{p}\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(2,n\right)\right]{n}^{\frac{p}{2}-1}{a}_{n}^{p}.$
(4.12)
2. (3)

Let

$k\left(x,y\right)=\frac{{\left(xy\right)}^{\frac{\lambda -1}{2}}}{{\left(max\left\{x,y\right\}\right)}^{\lambda }}\left(1-2min\left\{\frac{1}{r},\frac{1}{s}\right\}<\lambda \le 1+2min\left\{\frac{1}{r},\frac{1}{s}\right\}\right),$

for $0\le \epsilon and $x>0$, then we find (see [4])

${\overline{k}}_{s}\left(\epsilon ,x\right)\to \frac{4rs\lambda }{\left[r\left(\lambda +1\right)-2\right]\left[s\left(\lambda +1\right)-2\right]}={k}_{r}\phantom{\rule{1em}{0ex}}\left(\epsilon \to {0}^{+}\right),$

and ${\overline{k}}_{r}\left(\epsilon ,x\right)\to {k}_{r}$ ($\epsilon \to {0}^{+}$)

$\begin{array}{rcl}0& <& {\int }_{0}^{1}k\left(m,t\right){\left(\frac{m}{t}\right)}^{\frac{1}{s}}\phantom{\rule{0.2em}{0ex}}dt={\int }_{0}^{1}\frac{{\left(mt\right)}^{\frac{\lambda -1}{2}}}{{\left(max\left\{m,t\right\}\right)}^{\lambda }}{\left(\frac{m}{t}\right)}^{\frac{1}{s}}\phantom{\rule{0.2em}{0ex}}dt\\ =& {\int }_{0}^{1}\frac{{\left(mt\right)}^{\frac{\lambda -1}{2}}}{{m}^{\lambda }}{\left(\frac{m}{t}\right)}^{\frac{1}{s}}\phantom{\rule{0.2em}{0ex}}dt=\frac{1}{\left(\frac{\lambda -1}{2}+\frac{1}{r}\right)}\frac{1}{{m}^{\frac{1+\lambda }{2}-\frac{1}{s}}}.\end{array}$

Hence, ${\theta }_{\lambda }\left(s,m\right)=O\left(\frac{1}{{m}^{\frac{1+\lambda }{2}-\frac{1}{s}}}\right)$. Similarly, we can obtain ${\theta }_{\lambda }\left(r,n\right)=O\left(\frac{1}{{n}^{\frac{1+\lambda }{2}-\frac{1}{r}}}\right)$. For $\epsilon \ge 0$, $1-2min\left\{\frac{1}{r},\frac{1}{s}\right\}<\lambda \le 1+2min\left\{\frac{1}{r},\frac{1}{s}\right\}$, and $x>0$, the function

$k\left(x,t\right){\left(\frac{x}{t}\right)}^{\frac{1+\epsilon }{l}}=\frac{{\left(mt\right)}^{\frac{\lambda -1}{2}}}{{\left(max\left\{m,t\right\}\right)}^{\lambda }}{\left(\frac{x}{t}\right)}^{\frac{1+\epsilon }{l}}=\frac{{x}^{\frac{1+\epsilon }{l}+\frac{\lambda -1}{2}}}{{\left(max\left\{m,t\right\}\right)}^{\lambda }}{t}^{\frac{\lambda -1}{2}-\frac{1+\epsilon }{l}}\phantom{\rule{1em}{0ex}}\left(l=r,s\right)$

is decreasing in $\left(0,\mathrm{\infty }\right)$. Hence, $k\left(x,y\right)\in {H}_{p,q}\left(r,s\right)$. By Theorem 3.1, we have the following corollary.

Corollary 4.3 If $0, $\frac{1}{p}+\frac{1}{q}=1$, $r>1$, $\frac{1}{r}+\frac{1}{s}=1$, $1-2min\left\{\frac{1}{r},\frac{1}{s}\right\}<\lambda \le 1+2min\left\{\frac{1}{r},\frac{1}{s}\right\}$, and both ${a}_{n},{b}_{n}\ge 0$, such that $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{a}_{n}^{p}<\mathrm{\infty }$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{s}-1}{b}_{n}^{q}<\mathrm{\infty }$, then we have

$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(mn\right)}^{\frac{\lambda -1}{2}}}{{\left(max\left\{m,n\right\}\right)}^{\lambda }}{a}_{m}{b}_{n}\hfill \\ \phantom{\rule{1em}{0ex}}>\frac{4rs\lambda }{\left[r\left(\lambda +1\right)-2\right]\left[s\left(\lambda +1\right)-2\right]}{\left\{\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(s,n\right)\right]{n}^{\frac{p}{r}-1}{a}_{n}^{p}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{s}-1}{b}_{n}^{q}\right\}}^{\frac{1}{q}},\hfill \end{array}$
(4.13)
$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{\left[\frac{{\left(mn\right)}^{\frac{\lambda -1}{2}}}{{\left(max\left\{m,n\right\}\right)}^{\lambda }}{a}_{m}\right]}^{p}\hfill \\ \phantom{\rule{1em}{0ex}}>{\left(\frac{4rs\lambda }{\left[r\left(\lambda +1\right)-2\right]\left[s\left(\lambda +1\right)-2\right]}\right)}^{p}\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(s,n\right)\right]{n}^{\frac{p}{r}-1}{a}_{n}^{p}.\hfill \end{array}$
(4.14)

Here the constant factors $\frac{4rs\lambda }{\left[r\left(\lambda +1\right)-2\right]\left[s\left(\lambda +1\right)-2\right]}$ and ${\left(\frac{4rs\lambda }{\left[r\left(\lambda +1\right)-2\right]\left[s\left(\lambda +1\right)-2\right]}\right)}^{p}$ are the best possible. Inequality (4.13) is equivalent to (4.14).

In particular, (a) for $r=q$, $s=p$, and $1-2min\left\{\frac{1}{p},\frac{1}{q}\right\}<\lambda \le 1+2min\left\{\frac{1}{p},\frac{1}{q}\right\}$, we have

$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(mn\right)}^{\frac{\lambda -1}{2}}}{{\left(max\left\{m,n\right\}\right)}^{\lambda }}{a}_{m}{b}_{n}\hfill \\ \phantom{\rule{1em}{0ex}}>\frac{4pq\lambda }{\left[p\left(\lambda +1\right)-2\right]\left[q\left(\lambda +1\right)-2\right]}{\left\{\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(p,n\right)\right]{n}^{p-2}{a}_{n}^{p}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{q-2}{b}_{n}^{q}\right\}}^{\frac{1}{q}},\hfill \end{array}$
(4.15)
$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}{n}^{p-2}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(mn\right)}^{\frac{\lambda -1}{2}}}{{\left(max\left\{m,n\right\}\right)}^{\lambda }}{a}_{m}\right]}^{p}\hfill \\ \phantom{\rule{1em}{0ex}}>{\left(\frac{4pq\lambda }{\left[p\left(\lambda +1\right)-2\right]\left[q\left(\lambda +1\right)-2\right]}\right)}^{p}\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(p,n\right)\right]{n}^{p-2}{a}_{n}^{p}.\hfill \end{array}$
(4.16)
1. (b)

For $r=s=2$ and $0<\lambda \le 2$, we have

$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(mn\right)}^{\frac{\lambda -1}{2}}}{{\left(max\left\{m,n\right\}\right)}^{\lambda }}{a}_{m}{b}_{n}>\frac{4}{\lambda }{\left\{\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(2,n\right)\right]{n}^{\frac{p}{2}-1}{a}_{n}^{p}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{2}-1}{b}_{n}^{q}\right\}}^{\frac{1}{q}},$
(4.17)
$\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{2}-1}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(mn\right)}^{\frac{\lambda -1}{2}}}{{\left(max\left\{m,n\right\}\right)}^{\lambda }}{a}_{m}\right]}^{p}>{\left(\frac{4}{\lambda }\right)}^{p}\sum _{n=1}^{\mathrm{\infty }}\left[1-{\theta }_{\lambda }\left(2,n\right)\right]{n}^{\frac{p}{2}-1}{a}_{n}^{p}.$
(4.18)

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## Acknowledgements

The work is supported by Scientific Research Program Funded by Shaanxi Provincial Education Department (No. 2013JK1139), China Postdoctoral Science Foundation (No. 2013M542370), NNSFC (No. 11326161), key projects of Science and Technology Research of the Henan Education Department (No. 14A110011). The authors deeply appreciate the support.

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All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

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Xu, B., Wang, XH., Wei, W. et al. On reverse Hilbert-type inequalities. J Inequal Appl 2014, 198 (2014). https://doi.org/10.1186/1029-242X-2014-198