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. [1]). 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. [2]). Inequalities (1) and (2) are important in analysis and its applications (cf. [24]).

In 2011, Yang gave an extension of (2) as follows (cf. [5]): 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 [623].

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. [24])

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. [25]) 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. [25]), 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.