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

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

(1.2)

where the constant factor $\frac{\pi }{sin(\pi /p)}$ 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 $H_{p,q}(r,s)$ be the set of functions $k(x,y)$ 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 $k(x,y)$ is continuous in $(0,\mathrm{\infty })\times (0,\mathrm{\infty })$ and satisfies:

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

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

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

$${\overline{k}}_l(\epsilon ,x):={\int }_0^{\mathrm{\infty }}k(x,t){\left(\frac{x}{t}\right)}^{\frac{1+\epsilon }{l}}dt(l=r,s),$$

where ${\overline{k}}_l(\epsilon ,x)$ is independent of x, ${\overline{k}}_l(0,x):={\int }_0^{\mathrm{\infty }}k(x,t){(\frac{x}{t})}^{\frac{1}{l}}dt=k_p$ ($l=r,s$), $k_p$ is a positive constant independent of x, and ${\overline{k}}_l(\epsilon ,x)=k_p(\epsilon )=k_p+o(1)$ ($\epsilon \to 0^{+}$).

$$\begin{array}{c}(3)\sum _{m=1}^{\mathrm{\infty }}\frac{1}{m^{1+\epsilon }}{\int }_0^{1}k(m,t){\left(\frac{m}{t}\right)}^{\frac{1+\epsilon (s/q)}{s}}dt=O(1)(\epsilon \to 0^{+}),\hfill \\ \phantom{(3)}\sum _{m=1}^{\mathrm{\infty }}\frac{1}{m^{1+\epsilon }}{\int }_0^{1}k(m,t){\left(\frac{m}{t}\right)}^{\frac{1+\epsilon (r/p)}{r}}dt=O(1)(\epsilon \to 0^{+}).\hfill \end{array}$$

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 $k(x,y)\in H_{p,q}(r,s)$, $a_n,b_n\ge 0$, 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

$$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}k(m,n)a_m{b}_n<k_r{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{a}_n^p\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{s}-1}{b}_n^q\right\}}^{\frac{1}{q}},$$

(1.3)

$$\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{[\sum _{m=1}^{\mathrm{\infty }}k(m,n)a_m]}^p<{(k_r)}^p\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{a}_n^p.$$

(1.4)

Here the constant factors $k_r$ and ${(k_r)}^p$ 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 $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $k(x,y)\in H(p,q)$, 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

$$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}k(m,n)a_n{b}_n<k_p\left\{\sum _{n=1}^{\mathrm{\infty }}{a}_n^p\right\}{\left\{\sum _{n=1}^{\mathrm{\infty }}{b}_n^q\right\}}^{\frac{1}{q}},$$

(1.5)

$$\sum _{n=1}^{\mathrm{\infty }}{[\sum _{n=1}^{\mathrm{\infty }}k(m,n)a_m]}^p<{(k_p)}^p\sum _{n=1}^{\mathrm{\infty }}{a}_n^p,$$

(1.6)

where the constant factors $k_p$ and ${(k_p)}^p$ 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 $H_{p,q}(r,s)$ be the set of functions $k(x,y)$ satisfying the following conditions:

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

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

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

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

$${\overline{k}}_l(\epsilon ,x):={\int }_0^{\mathrm{\infty }}k(x,t){\left(\frac{x}{t}\right)}^{\frac{1+\epsilon }{l}}dt(l=r,s),$$

where ${\overline{k}}_l(\epsilon ,x)$ is independent of x, and ${\overline{k}}_l(0,x):={\int }_0^{\mathrm{\infty }}k(x,t){(\frac{x}{t})}^{\frac{1}{l}}dt=k_p$ ($l=r,s$), $k_p$ is a positive constant independent of x, and ${\overline{k}}_l(\epsilon ,x)=k_p(\epsilon )=k_p+o(1)$ ($\epsilon \to 0^{+}$).

• (3) There exists a positive constant ${\lambda }^{\prime }$ such that

  $$\begin{array}{c}{\theta }_{\lambda }(s,m)=\frac{1}{k_r}{\int }_0^{1}k(m,t){\left(\frac{m}{t}\right)}^{\frac{1}{s}}dt=O(1/m^{{\lambda }^{\prime }})\in (0,1)(m\to \mathrm{\infty }),\hfill \\ {\theta }_{\lambda }(r,n)=\frac{1}{k_r}{\int }_0^{1}k(t,n){\left(\frac{n}{t}\right)}^{\frac{1}{r}}dt=O(1/n^{{\lambda }^{\prime }})\in (0,1)(n\to \mathrm{\infty }).\hfill \end{array}$$

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

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

(2.1)

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

(2.2)

then we have

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

(2.3)

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

$$\begin{array}{rl}\omega (r,p,m)& =m^{\frac{p}{r}-1}\sum _{n=1}^{\mathrm{\infty }}k(m,n){\left(\frac{m}{n}\right)}^{\frac{1}{s}}\\ \le m^{\frac{p}{r}-1}{\int }_0^{\mathrm{\infty }}k(m,t){\left(\frac{m}{t}\right)}^{\frac{1}{s}}dt\\ =m^{\frac{p}{r}-1}{k}_r.\end{array}$$

However, we find

$$\begin{array}{rl}\omega (r,p,m)& =m^{\frac{p}{r}-1}\sum _{n=1}^{\mathrm{\infty }}k(m,n){\left(\frac{m}{n}\right)}^{\frac{1}{s}}\ge m^{\frac{p}{r}-1}{\int }_1^{\mathrm{\infty }}k(m,t){\left(\frac{m}{t}\right)}^{\frac{1}{s}}dt\\ =m^{\frac{p}{r}-1}{\int }_0^{\mathrm{\infty }}k(m,t){\left(\frac{m}{t}\right)}^{\frac{1}{s}}dt-m^{\frac{p}{r}-1}{\int }_0^{1}k(m,t){\left(\frac{m}{t}\right)}^{\frac{1}{s}}dt\\ =m^{\frac{p}{r}-1}{k}_r-m^{\frac{p}{r}-1}{\int }_0^{1}k(m,t){\left(\frac{m}{t}\right)}^{\frac{1}{s}}dt\\ =m^{\frac{p}{r}-1}{k}_r(1-\frac{1}{k_r}{\int }_0^{1}k(m,t){\left(\frac{m}{t}\right)}^{\frac{1}{s}}dt)\\ =m^{\frac{p}{r}-1}{k}_r(1-{\theta }_{\lambda }(s,m)).\end{array}$$

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

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

$$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}k(m,n)m^{-\frac{1}{r}-\frac{\epsilon }{p}}{n}^{-\frac{1}{s}-\frac{\epsilon }{q}}<(kr+o(1))\sum _1^{\mathrm{\infty }}{m}^{-1-\epsilon }(\epsilon \to 0^{+}).$$

(2.4)

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

$$\begin{array}{c}\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}k(m,n)m^{-\frac{1}{r}-\frac{\epsilon }{p}}{n}^{-\frac{1}{s}-\frac{\epsilon }{q}}\hfill \\ =\sum _{m=1}^{\mathrm{\infty }}{m}^{-1-\epsilon }\epsilon \sum _{n=1}^{\mathrm{\infty }}k(m,n){\left(\frac{m}{n}\right)}^{\frac{1+\epsilon (s/q)}{s}}\hfill \\ \le \sum _{m=1}^{\mathrm{\infty }}{m}^{-1-\epsilon }{\int }_0^{\mathrm{\infty }}k(m,t){\left(\frac{m}{t}\right)}^{\frac{1+\epsilon (s/q)}{s}}dt\hfill \\ =\sum _{m=1}^{\mathrm{\infty }}{m}^{-1-\epsilon }kr\left(\frac{\epsilon s}{q}\right)=(kr+o(1))\sum _1^{\mathrm{\infty }}{m}^{-1-\epsilon }.\hfill \end{array}$$

The lemma is proved. □

3 Main results

Theorem 3.1 If $0<p<1$, $\frac{1}{p}+\frac{1}{q}=1$, $r>1$, $\frac{1}{r}+\frac{1}{s}=1$, and $k(x,y)\in H_{p,q}(r,s)$, $a_n,b_n\ge 0$, 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

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

(3.1)

$$\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{[\sum _{m=1}^{\mathrm{\infty }}k(m,n)a_m]}^p>{(k_r)}^p\sum _{n=1}^{\mathrm{\infty }}[1-{\theta }_{\lambda }(s,n)]n^{\frac{p}{r}-1}{a}_n^p,$$

(3.2)

where the constant factor $k_r$ and ${(k_r)}^p$ are the best possible. Inequality (3.1) is equivalent to (3.2).

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

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

(3.3)

Then, by (2.3), in view of $0<p<1$ 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

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

(3.4)

By virtue of (2.4), we have

$$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}k(m,n){\overline{a}}_m{\overline{b}}_n\hfill \\ =\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}k(m,n)m^{-\frac{1}{r}-\frac{\epsilon }{p}}{n}^{-\frac{1}{s}-\frac{\epsilon }{q}}<(kr+o(1))\sum _1^{\mathrm{\infty }}{m}^{-1-\epsilon }(\epsilon \to 0^{+}).\hfill \end{array}$$

(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

$$(kr+o(1))\sum _1^{\mathrm{\infty }}{n}^{-1-\epsilon }>K\sum _{n=1}^{\mathrm{\infty }}{n}^{-1-\epsilon }{[1-{\left(\sum _{n=1}^{\mathrm{\infty }}{n}^{-1-\epsilon }\right)}^{-1}\sum _{n=1}^{\mathrm{\infty }}O(1/n^{{\lambda }^{\prime }})n^{-1-\epsilon }]}^{\frac{1}{p}},$$

that is,

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

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

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

by (3.1), we have

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

(3.6)

Hence, we obtain

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

(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

$$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}k(m,n)a_m{b}_n\hfill \\ =\sum _{n=1}^{\mathrm{\infty }}[n^{\frac{1}{q}-\frac{1}{s}}\sum _{m=1}^{\mathrm{\infty }}k(m,n)a_m]\left[n^{\frac{1}{s}-\frac{1}{q}}{b}_n\right]\hfill \\ \ge {\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{[\sum _{m=1}^{\mathrm{\infty }}k(m,n)a_m]}^p\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{s}-1}{b}_n^q\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

(3.8)

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

If the constant factor ${(k_r)}^p$ 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) Setting

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

for $0\le \epsilon <min\{p(\frac{\lambda +1}{2}-\frac{1}{r}),q(\frac{\lambda +1}{2}-\frac{1}{s})\}$, and for fixed $x>0$, we find (see [4])

$${\overline{k}}_s(\epsilon ,x)\to B(\frac{s(\lambda +1)-2}{2s},\frac{r(\lambda +1)-2}{2r})=k_r(\epsilon \to 0^{+}),$$

and ${\overline{k}}_s(\epsilon ,x)\to k_r$ ($\epsilon \to 0^{+}$);

$$\begin{array}{rcl}0& <& {\int }_0^{1}k(m,t){\left(\frac{m}{t}\right)}^{\frac{1}{s}}dt={\int }_0^1\frac{{(mt)}^{\frac{\lambda -1}{2}}}{{(m+t)}^{\lambda }}{\left(\frac{m}{t}\right)}^{\frac{1}{s}}dt\\ \le & {\int }_0^1\frac{{(mt)}^{\frac{\lambda -1}{2}}}{m^{\lambda }}{\left(\frac{m}{t}\right)}^{\frac{1}{s}}dt=\frac{1}{(\frac{\lambda -1}{2}+\frac{1}{r})}\frac{1}{m^{\frac{1+\lambda }{2}-\frac{1}{s}}}.\end{array}$$

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

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

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

Corollary 4.1 If $0<p<1$, $1/p+1/q=1$, $1/r+1/s=1$, $1-2min\{\frac{1}{r},\frac{1}{s}\}<\lambda \le 1+2min\{\frac{1}{r},\frac{1}{s}\}$, 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

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

(4.1)

$$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{{(mn)}^{\frac{\lambda -1}{2}}}{{(m+n)}^{\lambda }}{a}_m\right]}^p\hfill \\ >{\left[B(\frac{s(\lambda +1)-2}{2s},\frac{r(\lambda +1)-2}{2r})\right]}^p\sum _{n=1}^{\mathrm{\infty }}[1-{\theta }_{\lambda }(s,n)]n^{\frac{p}{r}-1}{a}_n^p,\hfill \end{array}$$

(4.2)

where the constant factors

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

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

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

$$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{{(mn)}^{\frac{\lambda -1}{2}}}{{(m+n)}^{\lambda }}{a}_m{b}_n\hfill \\ >B(\frac{p(\lambda +1)-2}{2p},\frac{q(\lambda +1)-2}{2q}){\left\{\sum _{n=1}^{\mathrm{\infty }}[1-{\theta }_{\lambda }(p,n)]n^{p-2}{a}_n^p\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{q-2}{b}_n^q\right\}}^{\frac{1}{q}},\hfill \end{array}$$

(4.3)

$$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}{n}^{p-2}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{{(mn)}^{\frac{\lambda -1}{2}}}{{(m+n)}^{\lambda }}{a}_m\right]}^p\hfill \\ >{\left[B(\frac{p(\lambda +1)-2}{2p},\frac{q(\lambda +1)-2}{2q})\right]}^p\sum _{n=1}^{\mathrm{\infty }}[1-{\theta }_{\lambda }(p,n)]n^{p-2}{a}_n^p.\hfill \end{array}$$

(4.4)

• (b) For $r=s=2$ and $0<\lambda \le 2$, we have

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

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

  (4.6)

• (2) Let

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

For $0\le \epsilon <min\{p(\frac{\lambda +1}{2}-\frac{1}{r}),q(\frac{\lambda +1}{2}-\frac{1}{s})\}$ and $x>0$, we find (see [4])

$${\overline{k}}_s(\epsilon ,x)\to \frac{1}{\lambda }B(\frac{s(\lambda +1)-2}{2s\lambda },\frac{r(\lambda +1)-2}{2r\lambda })=k_r(\epsilon \to 0^{+}),$$

and ${\overline{k}}_s(\epsilon ,x)\to k_r$ ($\epsilon \to 0^{+}$);

$$\begin{array}{rcl}{\int }_0^{1}k(m,t){\left(\frac{m}{t}\right)}^{\frac{1}{s}}dt& =& {\int }_0^1\frac{{(mt)}^{\frac{\lambda -1}{2}}}{m^{\lambda }+t^{\lambda }}{\left(\frac{m}{t}\right)}^{\frac{1}{s}}dt\\ \le & {\int }_0^1\frac{{(mt)}^{\frac{\lambda -1}{2}}}{m^{\lambda }}{\left(\frac{m}{t}\right)}^{\frac{1}{s}}dt=\frac{1}{(\frac{\lambda -1}{2}+\frac{1}{r})}{m}^{\frac{1}{s}-\frac{1+\lambda }{2}}.\end{array}$$

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

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

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

Corollary 4.2 If $0<p<1$, $\frac{1}{p}+\frac{1}{q}=1$, $r>1$, $\frac{1}{r}+\frac{1}{s}=1$, $1-2min\{\frac{1}{r},\frac{1}{s}\}<\lambda \le 1+2min\{\frac{1}{r},\frac{1}{s}\}$, 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

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

(4.7)

$$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{{(mn)}^{\frac{\lambda -1}{2}}}{m^{\lambda }+n^{\lambda }}{a}_m\right]}^p\hfill \\ >{\left[\frac{1}{\lambda }B(\frac{s(\lambda +1)-2}{2s\lambda },\frac{r(\lambda +1)-2}{2r\lambda })\right]}^p\sum _{n=1}^{\mathrm{\infty }}[1-{\theta }_{\lambda }(s,n)]n^{\frac{p}{r}-1}{a}_n^p,\hfill \end{array}$$

(4.8)

where the constant factors

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

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

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

$$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{{(mn)}^{\frac{\lambda -1}{2}}}{m^{\lambda }+n^{\lambda }}{a}_m{b}_n\hfill \\ >\frac{1}{\lambda }B(\frac{p(\lambda +1)-2}{2p\lambda },\frac{q(\lambda +1)-2}{2q\lambda })\hfill \\ \times {\left\{\sum _{n=1}^{\mathrm{\infty }}[1-{\theta }_{\lambda }(p,n)]n^{p-2}{a}_n^p\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{q-2}{b}_n^q\right\}}^{\frac{1}{q}},\hfill \end{array}$$

(4.9)

$$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}{n}^{p-2}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{{(mn)}^{\frac{\lambda -1}{2}}}{{(m+n)}^{\lambda }}{a}_m\right]}^p\hfill \\ >{\left[\frac{1}{\lambda }B(\frac{p(\lambda +1)-2}{2p\lambda },\frac{q(\lambda +1)-2}{2q\lambda })\right]}^p\sum _{n=1}^{\mathrm{\infty }}[1-{\theta }_{\lambda }(p,n)]n^{p-2}{a}_n^p.\hfill \end{array}$$

(4.10)

• (b) For $r=s=2$ and $0<\lambda \le 2$, we have

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

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

  (4.12)

• (3) Let

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

for $0\le \epsilon <min\{p(\frac{\lambda +1}{2}-\frac{1}{r}),q(\frac{\lambda +1}{2}-\frac{1}{s})\}$ and $x>0$, then we find (see [4])

$${\overline{k}}_s(\epsilon ,x)\to \frac{4rs\lambda }{[r(\lambda +1)-2][s(\lambda +1)-2]}=k_r(\epsilon \to 0^{+}),$$

and ${\overline{k}}_r(\epsilon ,x)\to k_r$ ($\epsilon \to 0^{+}$)

$$\begin{array}{rcl}0& <& {\int }_0^{1}k(m,t){\left(\frac{m}{t}\right)}^{\frac{1}{s}}dt={\int }_0^1\frac{{(mt)}^{\frac{\lambda -1}{2}}}{{(max\{m,t\})}^{\lambda }}{\left(\frac{m}{t}\right)}^{\frac{1}{s}}dt\\ =& {\int }_0^1\frac{{(mt)}^{\frac{\lambda -1}{2}}}{m^{\lambda }}{\left(\frac{m}{t}\right)}^{\frac{1}{s}}dt=\frac{1}{(\frac{\lambda -1}{2}+\frac{1}{r})}\frac{1}{m^{\frac{1+\lambda }{2}-\frac{1}{s}}}.\end{array}$$

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

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

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

Corollary 4.3 If $0<p<1$, $\frac{1}{p}+\frac{1}{q}=1$, $r>1$, $\frac{1}{r}+\frac{1}{s}=1$, $1-2min\{\frac{1}{r},\frac{1}{s}\}<\lambda \le 1+2min\{\frac{1}{r},\frac{1}{s}\}$, 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

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

(4.13)

$$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{r}-1}{\left[\frac{{(mn)}^{\frac{\lambda -1}{2}}}{{(max\{m,n\})}^{\lambda }}{a}_m\right]}^p\hfill \\ >{\left(\frac{4rs\lambda }{[r(\lambda +1)-2][s(\lambda +1)-2]}\right)}^p\sum _{n=1}^{\mathrm{\infty }}[1-{\theta }_{\lambda }(s,n)]n^{\frac{p}{r}-1}{a}_n^p.\hfill \end{array}$$

(4.14)

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

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

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

(4.15)

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

(4.16)

• (b) For $r=s=2$ and $0<\lambda \le 2$, we have

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

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

  (4.18)