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# A discrete Hilbert-type inequality in the whole plane

Journal of Inequalities and Applications20162016:133

https://doi.org/10.1186/s13660-016-1075-3

• Received: 6 March 2016
• Accepted: 19 April 2016
• Published:

## Abstract

By the use of weight coefficients and technique of real analysis, a discrete Hilbert-type inequality in the whole plane with multi-parameters and a best possible constant factor is given. The equivalent forms, the operator expressions, and a few particular inequalities are considered.

## Keywords

• Hilbert-type inequality
• weight coefficient
• equivalent form
• operator expression

• 26D15
• 47A05

## 1 Introduction

Suppose that $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$a_{m},b_{n}\geq 0$$, $$a=\{a_{m}\}_{m=1}^{\infty}\in l^{p}$$, $$b=\{b_{n}\}_{n=1}^{\infty}\in l^{q}$$, $$\Vert a\Vert _{p}=(\sum_{m=1}^{\infty}a_{m}^{p})^{\frac{1}{p}}>0$$, $$\Vert b\Vert _{q}>0$$. We have the following well-known Hardy-Hilbert inequality:
$$\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{m+n}< \frac {\pi}{\sin(\frac{\pi}{p})}\Vert a \Vert _{p}\Vert b\Vert _{q},$$
(1)
where the constant factor $$\frac{\pi}{\sin(\pi/p)}$$ is the best possible (cf. ). Also we have the following Hilbert-type inequality:
$$\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{\max\{m,n\}}< pq\Vert a\Vert _{p}\Vert b\Vert _{q},$$
(2)
with the best possible constant factor pq (cf. ). Inequalities (1) and (2) are important in analysis and its applications (cf. ).
In 2011, Yang gave an extension of (2) as follows (cf. ): If $$0<\lambda_{1},\lambda_{2}\leq1$$, $$\lambda_{1}+\lambda_{2}=\lambda$$, $$a_{m},b_{n}\geq0$$, $$0<\Vert a\Vert _{p,\varphi}=\{\sum_{m=1}^{\infty }m^{p(1-\lambda_{1})-1}a_{m}^{p}\}^{\frac{1}{p}}<\infty$$, $$0<\Vert b\Vert _{q,\psi}=\{\sum_{n=1}^{\infty}n^{q(1-\lambda_{2})-1} b_{n}^{q}\}^{\frac{1}{q}}<\infty$$, then
$$\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{(\max \{m,n\})^{\lambda}}< \frac{\lambda}{\lambda_{1}\lambda_{2}}\Vert a\Vert _{p,\varphi}\Vert b\Vert _{q,\psi},$$
(3)
where the constant factor $$\frac{\lambda}{\lambda_{1}\lambda_{2}}$$ is the best possible.

For $$\lambda=1$$, $$\lambda_{1}=\frac{1}{q}$$, $$\lambda_{2}=\frac{1}{p}$$, inequality (3) reduces to (2). Some other results relate to (1)-(3) are provided by .

In this paper, by the use of weight coefficients and the technique of real analysis, an extension of (3) in the whole plane is given as follows: For $$0<\lambda_{1},\lambda_{2}\leq1$$, $$\lambda_{1}+\lambda _{2}=\lambda$$, $$a_{m},b_{n}\geq0$$, $$0<\sum_{|m|=1}^{\infty }|m|^{p(1-\lambda _{1})-1}a_{m}^{p}<\infty$$, $$0<\sum_{|n|=1}^{\infty}|n|^{q(1-\lambda _{2})-1}b_{n}^{q}<\infty$$, we have
\begin{aligned} &\sum_{|n|=1}^{\infty}\sum _{|m|=1}^{\infty}\frac{1}{(\max \{|m|,|n|\})^{\lambda}}a_{m}b_{n} \\ &\quad< \frac{2\lambda}{\lambda_{1}\lambda_{2}} \Biggl[ \sum_{|m|=1}^{\infty }|m|^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{|n|=1}^{\infty}|n|^{q(1-\lambda_{2})-1}b_{n}^{q} \Biggr] ^{\frac {1}{q}}. \end{aligned}
(4)
Moreover, a generation of (4) with multi-parameters and a best possible constant factor is proved. The equivalent forms, the operator expressions and a few particular inequalities are also considered.

## 2 Some lemmas

In the following, we agree that $$\mathbf{N}=\{1,2,\ldots\}$$, $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$\alpha,\beta\in(0,\pi)$$, $$\lambda _{1},\lambda _{2}>-\eta$$, $$\lambda_{1}+\lambda_{2}=\lambda$$, and for $$|x|,|y|>0$$,
$$k(x,y):=\frac{(\min\{|x|+x\cos\alpha,|y|+y\cos\beta\})^{\eta }}{(\max \{|x|+x\cos\alpha,|y|+y\cos\beta\})^{\lambda+\eta}}.$$
(5)

### Lemma 1

(cf. )

Suppose that $$g(t)$$ (>0) is decreasing in $$\mathbf{R}_{+}$$ and strictly decreasing in $$[n_{0},\infty)$$ ($$n_{0}\in \mathbf{N}$$), satisfying $$\int_{0}^{\infty}g(t)\,dt\in\mathbf{R}_{+}$$. We have
$$\int_{1}^{\infty}g(t)\,dt< \sum _{n=1}^{\infty}g(n)< \int_{0}^{\infty}g(t)\,dt.$$
(6)

### Definition 1

Define the following weight coefficients:
\begin{aligned}& \omega(\lambda_{2},m) : =\sum _{|n|=1}^{\infty}k(m,n)\frac{(|m|+m\cos \alpha)^{\lambda_{1}}}{(|n|+n\cos\beta)^{1-\lambda_{2}}},\quad |m|\in \mathbf{N}, \end{aligned}
(7)
\begin{aligned}& \varpi(\lambda_{1},n) : =\sum _{|m|=1}^{\infty}k(m,n)\frac{(|n|+n\cos \beta)^{\lambda_{2}}}{(|m|+m\cos\alpha)^{1-\lambda_{1}}},\quad |n|\in \mathbf{N}, \end{aligned}
(8)
where $$\sum_{|j|=1}^{\infty}\cdots=\sum_{j=-1}^{-\infty}\cdots +\sum_{j=1}^{\infty}\cdots$$ ($$j=m,n$$).

### Lemma 2

If $$\lambda_{2}\leq1-\eta$$, then for $$k_{\beta }(\lambda_{1}):=\frac{2(\lambda+2\eta)\csc^{2}\beta}{(\lambda _{1}+\eta )(\lambda_{2}+\eta)}$$, we have
$$k_{\beta}(\lambda_{1}) \bigl(1-\theta( \lambda_{2},m)\bigr)< \omega(\lambda _{2},m)< k_{\beta}( \lambda_{1}),\quad |m|\in\mathbf{N},$$
(9)
where
\begin{aligned} \theta(\lambda_{2},m) :=&\frac{(\lambda_{1}+\eta)(\lambda _{2}+\eta)}{\lambda+2\eta} \int_{0}^{\frac{1+\cos\beta}{|m|+m\cos\alpha}}\frac{ (\min\{1,u\})^{\eta}u^{\lambda_{2}-1}}{(\max\{1,u\})^{\lambda+\eta}}\,du \\ =&O \biggl( \frac{1}{(|m|+m\cos\alpha)^{\eta+\lambda_{2}}} \biggr) \in (0,1),\quad |m|\in\mathbf{N}. \end{aligned}
(10)

### Proof

For $$|x|>0$$, we set
\begin{aligned}& k^{(1)}(x,y) := \frac{[\min\{|x|+x\cos\alpha,y(\cos\beta-1)\} ]^{\eta}}{[\max\{|x|+x\cos\alpha,y(\cos\beta-1)\}]^{\lambda+\eta}},\quad y< 0, \\& k^{(2)}(x,y) := \frac{[\min\{|x|+x\cos\alpha,y(1+\cos\beta)\} ]^{\eta}}{[\max\{|x|+x\cos\alpha,y(1+\cos\beta)\}]^{\lambda+\eta}},\quad y>0, \end{aligned}
from which we have
$$k^{(1)}(x,-y)=\frac{[\min\{|x|+x\cos\alpha,y(1-\cos\beta)\}]^{\eta }}{[\max\{|x|+x\cos\alpha,y(1-\cos\beta)\}]^{\lambda+\eta}},\quad y>0.$$
We obtain
\begin{aligned} \omega(\lambda_{2},m) =&\sum _{n=-1}^{-\infty}k^{(1)}(m,n)\frac{(|m|+m\cos\alpha)^{\lambda_{1}}}{[n(\cos\beta-1)]^{1-\lambda_{2}}} \\ &{}+\sum_{n=1}^{\infty}k^{(2)}(m,n) \frac{(|m|+m\cos\alpha)^{\lambda _{1}}}{[n(1+\cos\beta)]^{1-\lambda_{2}}} \\ =&\frac{(|m|+m\cos\alpha)^{\lambda_{1}}}{(1-\cos\beta)^{1-\lambda _{2}}}\sum_{n=1}^{\infty} \frac{k^{(1)}(m,-n)}{n^{1-\lambda_{2}}} \\ &{}+\frac{(|m|+m\cos\alpha)^{\lambda_{1}}}{(1+\cos\beta)^{1-\lambda _{2}}}\sum_{n=1}^{\infty} \frac{k^{(2)}(m,n)}{n^{1-\lambda_{2}}}. \end{aligned}
(11)
For fixed $$|m|\in\mathbf{N}$$, $$\lambda_{2}\leq1-\eta$$, we find that
\begin{aligned} \frac{k^{(1)}(m,-y)}{y^{1-\lambda_{2}}} =&\frac{[\min\{|m|+m\cos \alpha ,y(1-\cos\beta)\}]^{\eta}}{y^{1-\lambda_{2}}[\max\{|m|+m\cos\alpha ,y(1-\cos\beta)\}]^{\lambda+\eta}} \\ =& \left\{ \textstyle\begin{array}{@{}l@{\quad}l@{}} \frac{(1-\cos\beta)^{\eta}}{(|m|+m\cos\alpha)^{\lambda +\eta}}\frac{1}{y^{1-(\lambda_{2}+\eta)}},&0< y< \frac{|m|+m\cos\alpha}{1-\cos \beta} , \\ \frac{(|m|+m\cos\alpha)^{\eta}}{(1-\cos\beta)^{\lambda +\eta}}\frac{1}{y^{1+(\lambda_{1}+\eta)}},&y\geq\frac{|m|+m\cos\alpha }{1-\cos \beta}\end{array}\displaystyle \right . \end{aligned}
is decreasing for $$y>0$$ and strictly decreasing for $$y\geq\frac {|m|+m\cos \alpha}{1-\cos\beta}$$. Under the same assumptions, it is evident that
\begin{aligned} \frac{k^{(2)}(m,y)}{y^{1-\lambda_{2}}} =&\frac{[\min\{|m|+m\cos \alpha ,y(1+\cos\beta)\}]^{\eta}}{y^{1-\lambda_{2}}[\max\{|m|+m\cos\alpha ,y(1+\cos\beta)\}]^{\lambda+\eta}} \\ =& \left\{ \textstyle\begin{array}{@{}l@{\quad}l@{}} \frac{(1+\cos\beta)^{\eta}}{(|m|+m\cos\alpha)^{\lambda +\eta}}\frac{1}{y^{1-(\lambda_{2}+\eta)}},&0< y< \frac{|m|+m\cos\alpha}{1+\cos \beta} , \\ \frac{(|m|+m\cos\alpha)^{\eta}}{(1+\cos\beta)^{\lambda +\eta}}\frac{1}{y^{1+(\lambda_{1}+\eta)}},&y\geq\frac{|m|+m\cos\alpha }{1+\cos \beta}\end{array}\displaystyle \right . \end{aligned}
is decreasing for $$y>0$$ and strictly decreasing for $$y\geq\frac {|m|+m\cos \alpha}{1+\cos\beta}$$.
By (11) and (6), we have
\begin{aligned} \omega(\lambda_{2},m) < &\frac{(|m|+m\cos\alpha)^{\lambda _{1}}}{(1-\cos \beta)^{1-\lambda_{2}}} \int_{0}^{\infty}\frac {k^{(1)}(m,-y)}{y^{1-\lambda _{2}}}\,dy \\ &{}+\frac{(|m|+m\cos\alpha)^{\lambda_{1}}}{(1+\cos\beta)^{1-\lambda _{2}}} \int_{0}^{\infty}\frac{k^{(2)}(m,y)}{y^{1-\lambda_{2}}}\,dy. \end{aligned}
Setting $$u=\frac{y(1-\cos\beta)}{|m|+m\cos\alpha}(\frac{y(1+\cos \beta)}{|m|+m\cos\alpha})$$ in the above first (second) integral, by simplifications, we find
\begin{aligned} \omega(\lambda_{2},m) < & \biggl(\frac{1}{1-\cos\beta}+ \frac{1}{1+\cos \beta} \biggr) \int_{0}^{\infty}\frac{(\min\{1,u\})^{\eta}u^{\lambda _{2}-1}}{(\max\{1,u\})^{\lambda+\eta}}\,du \\ =&2\csc^{2}\beta \biggl( \int_{0}^{1}u^{\eta+\lambda _{2}-1}\,du+ \int_{1}^{\infty}\frac{u^{\lambda_{2}-1}}{u^{\lambda+\eta }}\,du \biggr) \\ =&\frac{2(\lambda+2\eta)\csc^{2}\beta}{(\lambda_{1}+\eta)(\lambda _{2}+\eta)}=k_{\beta}(\lambda_{1}). \end{aligned}
Still by (11) and (6), we have
\begin{aligned} \omega(\lambda_{2},m) >&\frac{(|m|+m\cos\alpha)^{\lambda _{1}}}{(1-\cos \beta)^{1-\lambda_{2}}} \int_{1}^{\infty}\frac {k^{(1)}(m,-y)}{y^{1-\lambda _{2}}}\,dy \\ &{}+\frac{(|m|+m\cos\alpha)^{\lambda_{1}}}{(1+\cos\beta)^{1-\lambda _{2}}} \int_{1}^{\infty}\frac{k^{(2)}(m,y)}{y^{1-\lambda_{2}}}\,dy \\ \geq&\frac{1}{1-\cos\beta} \int_{\frac{1+\cos\beta}{|m|+m\cos \alpha}}^{\infty}\frac{(\min\{1,u\})^{\eta}u^{\lambda_{2}-1}}{(\max \{1,u\})^{\lambda+\eta}}\,du \\ &{}+\frac{1}{1+\cos\beta} \int_{\frac{1+\cos\beta}{|m|+m\cos\alpha}}^{\infty}\frac{(\min\{1,u\})^{\eta}u^{\lambda_{2}-1}}{(\max \{1,u\})^{\lambda+\eta}}\,du \\ =&k_{\beta}(\lambda_{1}) \bigl(1-\theta( \lambda_{2},m)\bigr)>0. \end{aligned}
We obtain for $$|m|+m\cos\alpha\geq1+\cos\beta$$
\begin{aligned} 0 < &\theta(\lambda_{2},m)=\frac{(\lambda_{1}+\eta)(\lambda _{2}+\eta)}{\lambda+2\eta} \int_{0}^{\frac{1+\cos\beta}{|m|+m\cos\alpha}}\frac{ (\min\{1,u\})^{\eta}u^{\lambda_{2}-1}}{(\max\{1,u\})^{\lambda+\eta}}\,du \\ =&\frac{(\lambda_{1}+\eta)(\lambda_{2}+\eta)}{\lambda+2\eta} \int _{0}^{\frac{1+\cos\beta}{|m|+m\cos\alpha}}u^{\eta+\lambda_{2}-1}\,du \\ =&\frac{\lambda_{1}+\eta}{\lambda+2\eta} \biggl( \frac{1+\cos\beta }{|m|+m\cos\alpha} \biggr) ^{\eta+\lambda_{2}}. \end{aligned}
Then we have (9) and (10). □

In the same way, we have the following.

### Lemma 3

If $$\lambda_{1}\leq1-\eta$$, then for $$k_{\alpha }(\lambda_{1})=\frac{2(\lambda+2\eta)\csc^{2}\alpha}{(\lambda _{1}+\eta )(\lambda_{2}+\eta)}$$, we have
$$k_{\alpha}(\lambda_{1}) \bigl(1-\vartheta( \lambda_{1},n)\bigr)< \varpi(\lambda _{1},n)< k_{\alpha}( \lambda_{1}),\quad |n|\in\mathbf{N},$$
(12)
where
\begin{aligned} \vartheta(\lambda_{1},n) :=&\frac{(\lambda_{1}+\eta)(\lambda _{2}+\eta)}{\lambda+2\eta} \int_{0}^{\frac{1+\cos\alpha}{|n|+n\cos\beta }}\frac{(\min\{1,u\})^{\eta}u^{\lambda_{1}-1}}{(\max\{1,u\})^{\lambda+\eta}}\,du \\ =&O \biggl( \frac{1}{(|n|+n\cos\beta)^{\eta+\lambda_{1}}} \biggr) \in (0,1),\quad |n|\in\mathbf{N}. \end{aligned}
(13)

### Lemma 4

If $$\theta\in(0,\pi)$$, then for $$\rho>0$$, $$H_{\rho }(\theta):=\sum_{|n|=1}^{\infty}\frac{1}{(|n|+n\cos\theta)^{1+\rho}}$$, we have
$$H_{\rho}(\theta)= \biggl[ \frac{1}{(1+\cos\theta)^{1+\rho}}+ \frac{1}{ (1-\cos\theta)^{1+\rho}} \biggr] \frac{1+\rho O(1)}{\rho} \quad\bigl(\rho \rightarrow0^{+}\bigr).$$
(14)

### Proof

We have
\begin{aligned} H_{\rho}(\theta) =&\sum_{n=-1}^{-\infty} \frac{1}{[n(\cos\theta -1)]^{1+\rho}}+\sum_{n=1}^{\infty} \frac{1}{[n(\cos\theta +1)]^{1+\rho}} \\ =& \biggl[ \frac{1}{(1-\cos\theta)^{1+\rho}}+\frac{1}{(1+\cos\theta )^{1+\rho}} \biggr] \sum _{n=1}^{\infty}\frac{1}{n^{1+\rho}}. \end{aligned}
By (6), we find
\begin{aligned}& \begin{aligned} H_{\rho}(\theta) &= \biggl[ \frac{1}{(1-\cos\theta)^{1+\rho}}+ \frac {1}{(1+\cos\theta)^{1+\rho}} \biggr] \Biggl( 1+\sum_{n=2}^{\infty} \frac {1}{n^{1+\rho}} \Biggr) \\ &< \biggl[ \frac{1}{(1-\cos\theta)^{1+\rho}}+\frac{1}{(1+\cos\theta )^{1+\rho}} \biggr] \biggl( 1+ \int_{1}^{\infty}\frac{dy}{y^{1+\rho }} \biggr) \\ &=\frac{1}{\rho} \biggl[ \frac{1}{(1-\cos\theta)^{1+\rho}}+\frac{1}{(1+\cos\theta)^{1+\rho}} \biggr] (1+ \rho), \end{aligned} \\& \begin{aligned} H_{\rho}(\theta) &> \biggl[ \frac{1}{(1-\cos\theta)^{1+\rho}}+\frac {1}{(1+\cos\theta)^{1+\rho}} \biggr] \int_{1}^{\infty}\frac {dy}{y^{1+\rho}} \\ &= \frac{1}{\rho} \biggl[ \frac{1}{(1-\cos\theta)^{1+\rho}}+\frac{1}{(1+\cos\theta)^{1+\rho}} \biggr] . \end{aligned} \end{aligned}
Hence we have (14). □

## 3 Main results

### Theorem 1

If $$\lambda_{1},\lambda_{2}\leq1-\eta$$, $$a_{m},b_{n}\geq0$$ ($$|m|,|n|\in\mathbf{N}$$),
\begin{aligned} &0 < \sum_{|m|=1}^{\infty}\bigl(|m|+m\cos \alpha\bigr)^{p(1-\lambda _{1})-1}a_{m}^{p}< \infty, \\ &0 < \sum_{|n|=1}^{\infty}\bigl(|n|+n\cos \beta\bigr)^{q(1-\lambda _{2})-1}b_{n}^{q}< \infty, \\ &k_{\alpha,\beta}(\lambda_{1}):=k_{\beta}^{\frac{1}{p}}( \lambda _{1})k_{\alpha}^{\frac{1}{q}}(\lambda_{1})= \frac{2(\lambda+2\eta )\csc^{\frac{2}{p}}\beta\csc^{\frac{2}{q}}\alpha}{(\lambda_{1}+\eta )(\lambda _{2}+\eta)}, \end{aligned}
(15)
then we have the following equivalent inequalities:
\begin{aligned} &\begin{aligned}[b] I :={}&\sum_{|n|=1}^{\infty} \sum_{|m|=1}^{\infty}k(m,n)a_{m}b_{n} \\ < {}&k_{\alpha,\beta}(\lambda_{1}) \Biggl[ \sum _{|m|=1}^{\infty }\bigl(|m|+m\cos \alpha\bigr)^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ &{}\times \Biggl[ \sum_{|n|=1}^{\infty}\bigl(|n|+n\cos \beta\bigr)^{q(1-\lambda _{2})-1}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned} \end{aligned}
(16)
\begin{aligned} &\begin{aligned}[b] J :={}& \Biggl[ \sum _{|n|=1}^{\infty}\bigl(|n|+n\cos\beta\bigr)^{p\lambda _{2}-1} \Biggl( \sum _{|m|=1}^{\infty}k(m,n)a_{m} \Biggr) ^{p} \Biggr] ^{\frac{1}{p}} \\ < {}&k_{\alpha,\beta}(\lambda_{1}) \Biggl[ \sum _{|m|=1}^{\infty }\bigl(|m|+m\cos \alpha\bigr)^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}}. \end{aligned} \end{aligned}
(17)
In particular, for $$\alpha=\beta=\frac{\pi}{2}$$, we have the following equivalent inequalities:
\begin{aligned} &\begin{aligned}[b] &\sum_{|n|=1}^{\infty} \sum_{|m|=1}^{\infty}\frac{(\min\{|m|,|n|\} )^{\eta}}{(\max\{|m|,|n|\})^{\lambda+\eta}}a_{m}b_{n} \\ &\quad< \frac{2(\lambda+2\eta)}{(\lambda_{1}+\eta)(\lambda_{2}+\eta )} \Biggl[ \sum_{|m|=1}^{\infty}|m|^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac {1}{p}} \Biggl[ \sum_{|n|=1}^{\infty}|n|^{q(1-\lambda_{2})-1}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned} \end{aligned}
(18)
\begin{aligned} &\begin{aligned}[b] & \Biggl[ \sum_{|n|=1}^{\infty}|n|^{p\lambda_{2}-1} \Biggl( \sum_{|m|=1}^{\infty}\frac{(\min\{|m|,|n|\})^{\eta}}{(\max \{|m|,|n|\})^{\lambda+\eta}}a_{m} \Biggr) ^{p} \Biggr] ^{\frac{1}{p}} \\ &\quad< \frac{2(\lambda+2\eta)}{(\lambda_{1}+\eta)(\lambda_{2}+\eta )} \Biggl[ \sum_{|m|=1}^{\infty}|m|^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac {1}{p}}. \end{aligned} \end{aligned}
(19)

### Proof

By Hölder’s inequality (cf. ) and (8), we have
\begin{aligned} \Biggl( \sum_{|m|=1}^{\infty}k(m,n)a_{m} \Biggr) ^{p} =& \Biggl[ \sum_{|m|=1}^{\infty}k(m,n) \frac{(|m|+m\cos\alpha )^{(1-\lambda _{1})/q}a_{m}}{(|n|+n\cos\beta)^{(1-\lambda_{2})/p}}\frac{(|n|+n\cos \beta)^{(1-\lambda_{2})/p}}{(|m|+m\cos\alpha)^{(1-\lambda_{1})/q}} \Biggr] ^{p} \\ \leq& \sum_{|m|=1}^{\infty}k(m,n) \frac{(|m|+m\cos\alpha)^{(1-\lambda _{1})p/q}}{(|n|+n\cos\beta)^{1-\lambda_{2}}}a_{m}^{p} \\ &{} \times \Biggl[ \sum_{|m|=1}^{\infty}k(m,n) \frac{(|n|+n\cos\beta )^{(1-\lambda_{2})q/p}}{(|m|+m\cos\alpha)^{1-\lambda_{1}}} \Biggr] ^{p-1} \\ =&\frac{(\varpi(\lambda_{1},n))^{p-1}}{(|n|+n\cos\beta)^{p\lambda _{2}-1}}\sum_{|m|=1}^{\infty}k(m,n) \frac{(|m|+m\cos\alpha)^{(1-\lambda _{1})p/q}}{(|n|+n\cos\beta)^{1-\lambda_{2}}}a_{m}^{p}. \end{aligned}
By (12), we have
\begin{aligned} J < &k_{\alpha}^{\frac{1}{q}}(\lambda_{1}) \Biggl[ \sum_{|n|=1}^{\infty }\sum _{|m|=1}^{\infty}k(m,n)\frac{(|m|+m\cos\alpha)^{(1-\lambda _{1})(p-1)}}{(|n|+n\cos\beta)^{1-\lambda_{2}}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =&k_{\alpha}^{\frac{1}{q}}(\lambda_{1}) \Biggl[ \sum _{|m|=1}^{\infty }\sum_{|n|=1}^{\infty}k(m,n) \frac{(|m|+m\cos\alpha)^{(1-\lambda _{1})(p-1)}}{(|n|+n\cos\beta)^{1-\lambda_{2}}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =&k_{\alpha}^{\frac{1}{q}}(\lambda_{1}) \Biggl[ \sum _{|m|=1}^{\infty }\omega (\lambda_{2},m) \bigl(|m|+m\cos \alpha\bigr)^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}}. \end{aligned}
(20)
By (9), we have (17).
By Hölder’s inequality (cf. ), we have
\begin{aligned} I =&\sum_{|n|=1}^{\infty} \Biggl[ \bigl(|n|+n\cos\beta\bigr)^{\lambda_{2}-\frac {1}{p}}\sum_{|m|=1}^{\infty}k(m,n)a_{m} \Biggr] \bigl(|n|+n\cos\beta\bigr)^{\frac {1}{p}-\lambda_{2}}b_{n} \\ \leq&J \Biggl[ \sum_{|n|=1}^{\infty}\bigl(|n|+n\cos \beta\bigr)^{q(1-\lambda _{2})-1}b_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}
(21)
Then by (17), we have (16).
On the other hand, assuming that (16) is valid, we set
$$b_{n}:=\bigl(|n|+n\cos\beta\bigr)^{p\lambda_{2}-1} \Biggl( \sum _{|m|=1}^{\infty }k(m,n)a_{m} \Biggr) ^{p-1},\quad |n|\in\mathbf{N}.$$
Then it follows that
$$J= \Biggl[ \sum_{|n|=1}^{\infty}\bigl(|n|+n\cos \beta\bigr)^{q(1-\lambda _{2})-1}b_{n}^{q} \Biggr] ^{\frac{1}{p}}.$$
By (20), we find $$J<\infty$$. If $$J=0$$, then (17) is evidently valid; if $$J>0$$, then by (16), we have
\begin{aligned}& \begin{aligned} 0 < {}&\sum_{|n|=1}^{\infty}\bigl(|n|+n \cos\beta\bigr)^{q(1-\lambda _{2})-1}b_{n}^{q}=J^{p}=I \\ < {}&k_{\alpha,\beta}(\lambda_{1}) \Biggl[ \sum _{|m|=1}^{\infty }\bigl(|m|+m\cos \alpha\bigr)^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ &{}\times \Biggl[ \sum_{|n|=1}^{\infty}\bigl(|n|+n\cos \beta\bigr)^{q(1-\lambda _{2})-1}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned} \\& \begin{aligned} J ={}& \Biggl[ \sum_{|n|=1}^{\infty}\bigl(|n|+n \cos\beta\bigr)^{q(1-\lambda _{2})-1}b_{n}^{q} \Biggr] ^{\frac{1}{p}} \\ < {}&k_{\alpha,\beta}(\lambda_{1}) \Biggl[ \sum _{|m|=1}^{\infty }\bigl(|m|+m\cos \alpha\bigr)^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}}, \end{aligned} \end{aligned}
namely, (17) follows, which is equivalent to (16). □

### Theorem 2

As regards the assumptions of Theorem  1, the constant factor $$k_{\alpha,\beta}(\lambda_{1})$$ in (16) and (17) is the best possible.

### Proof

For any $$\varepsilon\in(0,q(\lambda_{2}+\eta))$$, we set $$\widetilde{\lambda}_{1}=\lambda_{1}+\frac{\varepsilon}{q}$$ ($$>-\eta$$), $$\widetilde{\lambda}_{2}=\lambda_{2}-\frac{\varepsilon}{q}$$ ($$\in(-\eta ,1-\eta)$$), and
\begin{aligned}& \widetilde{a}_{m} : =\bigl(|m|+m\cos\alpha\bigr)^{(\lambda_{1}-\frac {\varepsilon}{p})-1}=\bigl(|m|+m\cos \alpha\bigr)^{\widetilde{\lambda}_{1}-\varepsilon -1}\quad\bigl(|m|\in \mathbf{N}\bigr), \\& \widetilde{b}_{n} : =\bigl(|n|+n\cos\beta\bigr)^{(\lambda_{2}-\frac {\varepsilon}{q})-1}=\bigl(|n|+n\cos \beta\bigr)^{\widetilde{\lambda}_{2}-1}\quad\bigl(|n|\in\mathbf{N}\bigr). \end{aligned}
Then by (14) and (9), we find
\begin{aligned}& \begin{aligned} \widetilde{I}_{1} :={}& \Biggl[ \sum _{|m|=1}^{\infty}\bigl(|m|+m\cos\alpha \bigr)^{p(1-\lambda_{1})-1} \widetilde{a}_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ &{}\times \Biggl[ \sum_{|n|=1}^{\infty}\bigl(|n|+n\cos \beta\bigr)^{q(1-\lambda _{2})-1}\widetilde{b}_{n}^{q} \Biggr] ^{\frac{1}{q}} \\ ={}& \Biggl[ \sum_{|m|=1}^{\infty} \frac{1}{(|m|+m\cos\alpha )^{1+\varepsilon}} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum _{|n|=1}^{\infty}\frac{1}{(|n|+n\cos \beta)^{1+\varepsilon}} \Biggr] ^{\frac{1}{q}} \\ ={}&\frac{1}{\varepsilon} \biggl[ \frac{1}{(1+\cos\alpha )^{1+\varepsilon}}+\frac{1}{(1-\cos\alpha)^{1+\varepsilon}} \biggr] ^{\frac{1}{p}}\bigl(1+\varepsilon O_{1}(1)\bigr)^{\frac{1}{p}} \\ &{}\times \biggl[ \frac{1}{(1+\cos\beta)^{1+\varepsilon}}+\frac {1}{(1-\cos \beta)^{1+\varepsilon}} \biggr] ^{\frac{1}{q}} \bigl(1+\varepsilon O_{2}(1)\bigr)^{\frac{1}{q}}, \end{aligned} \\& \begin{aligned} \widetilde{I} :={}&\sum_{|n|=1}^{\infty} \sum_{|m|=1}^{\infty}k(m,n)\widetilde{a}_{m}\widetilde{b}_{n} \\ ={}&\sum_{|m|=1}^{\infty}\sum _{|m|=1}^{\infty}k(m,n)\frac{(|m|+m\cos \alpha )^{\widetilde{\lambda}_{1}-\varepsilon-1}}{(|n|+n\cos\beta)^{1-\widetilde{\lambda}_{2}}} \\ ={}&\sum_{|m|=1}^{\infty}\frac{\omega(\widetilde{\lambda}_{2},m)}{(|m|+m\cos\alpha)^{\varepsilon+1}}\geq k_{\beta}(\widetilde{\lambda }_{1})\sum _{|m|=1}^{\infty}\frac{1-\theta(\widetilde{\lambda}_{2},m)}{ (|m|+m\cos\alpha)^{\varepsilon+1}} \\ ={}&k_{\beta}(\widetilde{\lambda}_{1}) \Biggl[ \sum _{|m|=1}^{\infty }\frac{1}{(|m|+m\cos\alpha)^{\varepsilon+1}}-\sum _{|m|=1}^{\infty}\frac{1}{O((|m|+m\cos\alpha)^{(\frac{\varepsilon}{p}+\lambda_{2}+\eta)+1})} \Biggr] \\ ={}&\frac{k_{\beta}(\widetilde{\lambda}_{1})}{\varepsilon}\biggl\{ \biggl[\frac{1}{ (1+\cos\alpha)^{1+\varepsilon}}+\frac{1}{(1-\cos\alpha )^{1+\varepsilon}} \biggr]\bigl(1+\varepsilon O_{1}(1)\bigr)-\varepsilon O(1)\biggr\} . \end{aligned} \end{aligned}
If there exists a constant $$k\leq k_{\alpha,\beta}(\lambda_{1})$$, such that (16) is valid when replacing $$k_{\alpha,\beta}(\lambda_{1})$$ by k, then in particular, we have $$\varepsilon\widetilde {I}<\varepsilon k\widetilde{I}_{1}$$, namely,
\begin{aligned}& k_{\beta}(\widetilde{\lambda}_{1})\biggl\{ \biggl[ \frac{1}{(1+\cos\alpha )^{1+\varepsilon}}+\frac{1}{(1-\cos\alpha)^{1+\varepsilon}}\biggr]\bigl(1+\varepsilon O_{1}(1)\bigr)-\varepsilon O(1)\biggr\} \\& \quad < k \biggl[ \frac{1}{(1+\cos\alpha)^{1+\varepsilon}}+\frac{1}{(1-\cos \alpha)^{1+\varepsilon}} \biggr] ^{\frac{1}{p}} \bigl(1+\varepsilon O_{1}(1)\bigr)^{\frac{1}{p}} \\& \qquad{} \times \biggl[ \frac{1}{(1+\cos\beta)^{1+\varepsilon}}+\frac {1}{(1-\cos \beta)^{1+\varepsilon}} \biggr] ^{\frac{1}{q}}\bigl(1+\varepsilon O_{2}(1)\bigr)^{\frac{1}{q}}. \end{aligned}
It follows that
$$\frac{4(\lambda+2\eta)}{(\lambda_{1}+\eta)(\lambda_{2}+\eta)}\csc ^{2}\beta\csc^{2}\alpha\leq2k \csc^{\frac{2}{p}}\alpha\csc^{\frac {2}{q}}\beta \quad\bigl( \varepsilon\rightarrow0^{+}\bigr),$$
namely,
$$k_{\alpha,\beta}(\lambda_{1})=\frac{2(\lambda+2\eta)\csc^{\frac {2}{p}}\beta\csc^{\frac{2}{q}}\alpha}{(\lambda_{1}+\eta)(\lambda _{2}+\eta)}\leq k.$$
Hence, $$k=k_{\alpha,\beta}(\lambda_{1})$$ is the best possible constant factor of (16).

The constant factor $$k_{\alpha,\beta}(\lambda_{1})$$ in (17) is still the best possible. Otherwise, we would reach a contradiction by (21) that the constant factor in (16) is not the best possible. □

## 4 Operator expressions

We set functions $$\Phi(m)$$ and $$\Psi(n)$$ as follows:
\begin{aligned}& \Phi(m) : =\bigl(|m|+m\cos\alpha\bigr)^{p(1-\lambda_{1})-1} \quad\bigl(|m|\in\mathbf {N}\bigr), \\& \Psi(n) : =\bigl(|n|+n\cos\beta\bigr)^{q(1-\lambda_{2})-1} \quad\bigl(|n|\in\mathbf{N}\bigr), \end{aligned}
from which we have
$$\Psi^{1-p}(n)=\bigl(|n|+n\cos\beta\bigr)^{p\lambda_{2}-1} \quad\bigl(|n|\in\mathbf{N}\bigr).$$
We also set the following weight normed spaces:
\begin{aligned}& l_{p,\Phi} : =\Biggl\{ a=\{a_{m}\}_{|m|=1}^{\infty}; \Vert a\Vert _{p,\Phi}= \Biggl\{ \sum_{|m|=1}^{\infty} \Phi(m)|a_{m}|^{p}\Biggr\} ^{\frac{1}{p}}< \infty \Biggr\} , \\& l_{q,\Psi} : =\Biggl\{ b=\{b_{n}\}_{|n|=1}^{\infty}; \Vert b\Vert _{q,\Psi}= \Biggl\{ \sum_{|n|=1}^{\infty} \Psi(n)|b_{n}|^{q}\Biggr\} ^{\frac{1}{q}}< \infty \Biggr\} , \\& l_{p,\Psi^{1-p}} : =\Biggl\{ c=\{c_{n}\}_{|n|=1}^{\infty}; \Vert c\Vert _{p,\Psi ^{1-p}}= \Biggl\{ \sum_{|n|=1}^{\infty} \Psi^{1-p}(n)|c_{n}|^{p}\Biggr\} ^{\frac {1}{p}}< \infty \Biggr\} . \end{aligned}
Then for $$a=\{a_{m}\}_{|m|=1}^{\infty}\in l_{p,\Phi }$$, $$c=\{c_{n}\}_{|n|=1}^{\infty}$$, $$c_{n}=\sum_{|m|=1}^{\infty }k(m,n)a_{m}$$, in view of (17), we have $$\Vert c\Vert _{p,\Psi^{1-p}}< k_{\alpha,\beta }(\lambda_{1})\Vert a\Vert _{p,\Phi}<\infty$$, namely, $$c\in l_{p,\Psi^{1-p}}$$.

### Definition 2

Define a Hilbert-type operator $$T:l_{p,\Phi }\rightarrow l_{p,\Psi^{1-p}}$$ as follows: For any $$a=\{a_{m}\} _{|m|=1}^{\infty}\in l_{p,\Phi}$$, there exists a unique representation $$c=Ta\in l_{p,\Psi^{1-p}}$$. We also define the formal inner product of Ta and $$b=\{b_{n}\}_{|n|=1}^{\infty}\in l_{q,\Psi}$$ ($$b_{n}\geq0$$) as follows:
$$(Ta,b):=\sum_{|n|=1}^{\infty}\sum _{|m|=1}^{\infty}k(m,n)a_{m}b_{n}.$$
(22)
Then for $$a_{m}\geq0$$ ($$|m|\in\mathbf{N}$$), we may rewrite (16) and (17) as follows:
\begin{aligned}& (Ta,b) < k_{\alpha,\beta}(\lambda_{1})\Vert a\Vert _{p,\Phi}\Vert b\Vert _{q,\Psi}, \end{aligned}
(23)
\begin{aligned}& \Vert Ta\Vert _{p,\Psi^{1-p}} < k_{\alpha,\beta}( \lambda_{1})\Vert a\Vert _{p,\Phi}. \end{aligned}
(24)
We define the norm of operator T as follows:
$$\Vert T\Vert :=\sup_{a\ (\neq\theta)\in l_{p,\Phi}}\frac{\Vert Ta\Vert _{p,\Psi ^{1-p}}}{\Vert a\Vert _{p,\Phi}}.$$
(25)
Then $$\Vert Ta\Vert _{p,\Psi^{1-p}}\leq\Vert T\Vert \cdot\Vert a\Vert _{p,\Phi}$$. Since by Theorem 2, the constant factor $$k_{\alpha,\beta}(\lambda_{1})$$ in (24) is the best possible, we have
$$\Vert T\Vert =k_{\alpha,\beta}(\lambda_{1})= \frac{2(\lambda+2\eta)\csc ^{\frac{2}{p}}\beta\csc^{\frac{2}{q}}\alpha}{(\lambda_{1}+\eta)(\lambda _{2}+\eta)}.$$
(26)

### Remark 1

(i) For $$\eta=0$$, (16) reduces to the following inequality:
\begin{aligned} & \sum_{|n|=1}^{\infty}\sum _{|m|=1}^{\infty}\frac{1}{(\max\{|m|+m\cos \alpha,|n|+n\cos\beta\})^{\lambda}}a_{m}b_{n} \\ & \quad< \frac{2\lambda}{\lambda_{1}\lambda_{2}}\csc^{\frac{2}{p}}\beta \csc^{\frac{2}{q}}\alpha \Biggl[ \sum_{|m|=1}^{\infty}\bigl(|m|+m\cos\alpha \bigr)^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ & \qquad{} \times \Biggl[ \sum_{|n|=1}^{\infty}\bigl(|n|+n \cos\beta\bigr)^{q(1-\lambda _{2})-1}b_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}
(27)
In particular, for $$\alpha=\beta=\frac{\pi}{2}$$, (27) reduces to (4). If $$a_{-m}=a_{m}$$, $$b_{-n}=b_{n}$$ ($$m,n\in\mathbf{N}$$), then (4) reduces to (3). Hence, (16) is an extension of (4) with multi-parameters.
(ii) For $$\eta=-\lambda$$, $$-1\leq\lambda_{1}$$, $$\lambda_{2}<0$$ in (16), we have
\begin{aligned} & \sum_{|n|=1}^{\infty}\sum _{|m|=1}^{\infty}\frac{1}{(\min\{|m|+m\cos \alpha,|n|+n\cos\beta\})^{\lambda}}a_{m}b_{n} \\ &\quad < \frac{2(-\lambda)}{\lambda_{1}\lambda_{2}}\csc^{\frac{2}{p}}\beta \csc^{\frac{2}{q}}\alpha \Biggl[ \sum_{|m|=1}^{\infty}\bigl(|m|+m\cos\alpha \bigr)^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ & \qquad{} \times \Biggl[ \sum_{|n|=1}^{\infty}\bigl(|n|+n \cos\beta\bigr)^{q(1-\lambda _{2})-1}b_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}
(28)
In particular, for $$\alpha=\beta=\frac{\pi}{2}$$, we have
\begin{aligned} &\sum_{|n|=1}^{\infty}\sum _{|m|=1}^{\infty}\frac{1}{(\min \{|m|,|n|\})^{\lambda}}a_{m}b_{n} \\ &\quad< \frac{2(-\lambda)}{\lambda_{1}\lambda_{2}} \Biggl[ \sum_{|m|=1}^{\infty }|m|^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{|n|=1}^{\infty}|n|^{q(1-\lambda_{2})-1}b_{n}^{q} \Biggr] ^{\frac {1}{q}}. \end{aligned}
(29)
(iii) For $$\lambda=0$$ in (16), we have $$\lambda_{2}=-\lambda _{1}$$, $$|\lambda_{1}|<\eta$$ ($$\eta>0$$) and
\begin{aligned}[b] &\sum_{|n|=1}^{\infty}\sum _{|m|=1}^{\infty} \biggl( \frac{\min\{ |m|+m\cos \alpha,|n|+n\cos\beta\}}{\max\{|m|+m\cos\alpha,|n|+n\cos\beta\}} \biggr) ^{\eta}a_{m}b_{n} \\ &\quad< \frac{4\eta}{\eta^{2}-\lambda_{1}^{2}}\csc^{\frac{2}{p}}\beta \csc^{\frac{2}{q}}\alpha \Biggl[ \sum_{|m|=1}^{\infty}\bigl(|m|+m\cos\alpha \bigr)^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ &\qquad{}\times \Biggl[ \sum_{|n|=1}^{\infty}\bigl(|n|+n \cos\beta\bigr)^{q(1+\lambda _{1})-1}b_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}
(30)
In particular, for $$\alpha=\beta=\frac{\pi}{2}$$, we have
\begin{aligned} &\sum_{|n|=1}^{\infty}\sum _{|m|=1}^{\infty} \biggl( \frac{\min\{ |m|,|n|\}}{\max\{|m|,|n|\}} \biggr) ^{\eta}a_{m}b_{n} \\ &\quad< \frac{4\eta}{\eta^{2}-\lambda_{1}^{2}} \Biggl[ \sum_{|m|=1}^{\infty }|m|^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{|n|=1}^{\infty}|n|^{q(1-\lambda_{2})-1}b_{n}^{q} \Biggr] ^{\frac {1}{q}}. \end{aligned}
(31)

The above particular inequalities are all with the best possible constant factors.

## Declarations

### Acknowledgements

This work is supported by the Science and Technology Planning Project of Guangdong Province (No. 2013A011403002), and Appropriative Researching Fund for Professors and Doctors, Guangdong University of Education (No. 2015ARF25). 