1 Introduction
If \(p>1,\frac{1}{p}+\frac{1}{q}=1\), \(a_{m},b_{n}\geq 0\), \(0<\sum_{m=1}^{\infty }a_{m}^{p}<\infty \), \(0<\sum_{n=1}^{\infty }b_{n}^{q}<\infty \), then we have the following Hardy–Hilbert inequality:
with the best possible constant factor \(\frac{\pi }{\sin (\pi /p)}\) [1]. A more accurate form of (1) with the same best possible constant factor was given in [2, Theorem 323]:
Inequalities (1) and (2) played an important role in analysis and its applications (see [2‐4]).
$$ \sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{a_{m}b_{n}}{m+n}< \frac{ \pi }{\sin (\frac{\pi }{p})}\Biggl( \sum_{m=1}^{\infty }a_{m}^{p} \Biggr)^{ \frac{1}{p}}\Biggl(\sum_{n=1}^{\infty }b_{n}^{q} \Biggr)^{\frac{1}{q}}, $$
(1)
$$ \sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{a_{m}b_{n}}{m+n-1}< \frac{ \pi }{\sin (\frac{\pi }{p})}\Biggl( \sum_{m=1}^{\infty }a_{m}^{p} \Biggr)^{ \frac{1}{p}}\Biggl(\sum_{n=1}^{\infty }b_{n}^{q} \Biggr)^{\frac{1}{q}}. $$
(2)
In 2011, Yang [5] gave the following an extension of (2): If \(0<\lambda_{1}\), \(\lambda_{2}\leq 1\), \(\lambda_{1}+\lambda_{2}=\lambda \), \(a_{m},b_{n}\geq 0\),
then
where the constant factor \(B(\lambda_{1},\lambda_{2})\) is the best possible, and \(B(u,v)\) is the beta function defined as (see [6])
For \(\lambda =1\), \(\lambda_{1}=\frac{1}{q}\), \(\lambda_{2}=\frac{1}{p}\), and \(\alpha =\frac{1}{2}\), (3) reduces to (2). Some other results related to (1)–(3) were provided in [7‐24]. In 2016–17, a few extensions of (1)–(3) with some reverses in the whole plane were obtained in [25‐27].
$$\begin{aligned}& \Vert a \Vert _{p,\varphi } = \Biggl\{ \sum_{m=1}^{\infty }(m- \alpha )^{p(1-\lambda_{1})-1}a _{m}^{p}\Biggr\} ^{\frac{1}{p}}\in (0, \infty ), \\& \Vert b \Vert _{q,\psi } = \Biggl\{ \sum_{n=1}^{\infty }(n- \alpha )^{q(1-\lambda_{2})-1}b _{n}^{q}\Biggr\} ^{\frac{1}{q}}\in (0, \infty ), \end{aligned}$$
$$ \sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{a_{m}b_{n}}{(m+n-2 \alpha )^{\lambda }}< B(\lambda_{1}, \lambda_{2})\Vert a \Vert _{p,\varphi }\Vert b \Vert _{q, \psi } \quad \biggl(0\leq \alpha \leq \frac{1}{2}\biggr), $$
(3)
$$ B(u,v):= \int_{0}^{\infty }\frac{1}{(1+t)^{u+v}}t^{u-1}\,dt\quad (u,v>0). $$
(4)
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In this paper, using weight coefficients, a complex integral formula, and Hermite–Hadamard’s inequality, we give the following extension of the reverse of (1) in the whole plane: If \(0< p<1\) (\(q<0\)), \(\frac{1}{p}+\frac{1}{q}=1\), \(0<\lambda_{1}\), \(\lambda_{2}<1\), \(\lambda_{1}+ \lambda_{2}=\lambda \leq 1\), \(\xi ,\eta \in [ 0,\frac{1}{2}],a _{m},b_{n}\geq 0\),
then setting
we have the following reverse Hilbert-type inequality in the whole plane:
Moreover, we prove an extended inequality of (6) with multiparameters and a best possible constant factor. We also consider equivalent forms and a few particular cases.
$$\begin{aligned}& 0< \sum_{\vert m \vert =1}^{\infty }\vert m-\xi \vert ^{p(1-\lambda_{1})-1}a_{m}^{p}< \infty , \\& 0< \sum _{\vert n \vert =1}^{\infty }\vert n-\eta \vert ^{q(1-\lambda_{2})-1}b_{n}^{q}< \infty , \end{aligned}$$
$$\begin{aligned} \theta_{1}(\lambda_{2},m) &:=\frac{\lambda \sin (\frac{\pi \lambda _{1}}{\lambda })}{\pi } \int_{\frac{\vert m-\xi \vert }{1+\eta }}^{\infty }\frac{u ^{\lambda_{1}-1}}{u^{\lambda }+1}\,du \\ &=O \biggl( \frac{1}{\vert m-\xi \vert ^{\lambda_{2}}} \biggr) \in (0,1),\quad \vert m \vert \in \mathbf{N}, \end{aligned}$$
(5)
$$\begin{aligned}& \sum_{\vert n \vert =1}^{\infty }\sum _{\vert m \vert =1}^{\infty }\frac{a_{m}b_{n}}{ \vert m-\xi \vert ^{\lambda }+ \vert n-\eta \vert ^{\lambda }} \\& \quad > \frac{2\pi }{\lambda \sin (\frac{\pi \lambda_{1}}{\lambda })} \Biggl[ \sum_{\vert m \vert =1}^{\infty } \bigl(1-\theta_{1}(\lambda_{2},m)\bigr)\vert m-\xi \vert ^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\& \quad \quad {}\times \Biggl[ \sum_{\vert n \vert =1}^{\infty }\vert n- \eta \vert ^{q(1-\lambda_{2})-1}b _{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$
(6)
2 Some lemmas and an example
Lemma 1
Let
C
be the set of complex numbers, \(\mathbf{C}_{\infty }=\mathbf{C}\cup \{\infty \}\), and let
\(z_{k}\in \mathbf{C}\backslash \{z\mid \operatorname{Re}z\geq 0, \operatorname{Im}z=0\}\) (\(k=1,2,\ldots ,n\)) be different points. Suppose that a function
\(f(z)\)
is analytic in
\(\mathbf{C}_{ \infty }\)
except for
\(z_{i}\) (\(i=1,2,\ldots ,n\)) and that
\(z=\infty \)
is a zero point of
\(f(z)\)
of order not less than 1. Then, for
\(\alpha \in \mathbf{R}\), we have
where
\(0<\operatorname{Im}\ln z=\arg z<2\pi \). In particular, if
\(z_{k}\) (\(k=1,\ldots ,n\)) are all poles of order 1, then setting
\(\varphi_{k}(z)=(z-z _{k})f(z)\) (\(\varphi_{k}(z_{k})\neq 0\)), we have
$$ \int_{0}^{\infty }f(x)x^{\alpha -1}\,dx= \frac{2\pi i}{1-e^{2\pi \alpha i}}\sum_{k=1}^{n} \operatorname{Re}s\bigl[f(z)z^{\alpha -1},z_{k}\bigr], $$
(7)
$$ \int_{0}^{\infty }f(x)x^{\alpha -1}\,dx= \frac{\pi }{\sin \pi \alpha } \sum_{k=1}^{n}(-z_{k})^{\alpha -1} \varphi_{k}(z_{k}). $$
(8)
Proof
By [28] (p. 118) we have (7). We find
In particular, since \(f(z)z^{\alpha -1}=\frac{1}{z-z_{k}}(\varphi_{k}(z)z ^{\alpha -1})\), it is obvious that
Then by (7) we obtain (8). □
$$\begin{aligned} 1-e^{2\pi \alpha i} &=1-\cos 2\pi \alpha -i\sin 2\pi \alpha \\ &=-2i\sin \pi \alpha (\cos \pi \alpha +i\sin \pi \alpha ) \\ &=-2ie^{i \pi \alpha } \sin \pi \alpha . \end{aligned}$$
$$ \operatorname{Re}s\bigl[f(z)z^{\alpha -1},-z_{k} \bigr]=z_{k}{}^{\alpha -1}\varphi_{k}(z _{k})=-e^{i\pi \alpha }(-z_{k})^{\alpha -1} \varphi_{k}(z_{k}). $$
Example 1
For \(s\in \mathbf{N=\{}1,2,\ldots \}\), \(c_{s}\geq \cdots \geq c_{1}>0, \varepsilon >0\), \(\lambda_{1},\lambda_{2}>0\), \(\lambda_{1}+\lambda_{2}=s \lambda \), we define the function
and constants \(\widetilde{c}_{k}=c_{k}+(k-1)\varepsilon \) (\(k=1,\ldots ,s\)).
$$ k_{s\lambda }(x,y)=\frac{1}{\prod_{k=1}^{s}(x^{\lambda }+c_{k}y^{ \lambda })} $$
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Since \(\widetilde{c}_{s}>\cdots >\widetilde{c}_{1}=c_{1}>0\), by (8) we find
$$\begin{aligned} \widetilde{k}_{s}(\lambda_{1}) &:=\int_{0}^{\infty }\frac{1}{\prod_{k=1}^{s}(t^{\lambda }+\widetilde{c}_{k})}t^{\lambda_{1}-1}\,dt \\ &\overset{u=t^{\lambda /s}}{=}\frac{1}{\lambda } \int_{0}^{\infty }\frac{1}{ \prod_{k=1}^{s}(u+\widetilde{c}_{k})}u^{\frac{\lambda_{1}}{\lambda }-1}\,du \\ &=\frac{\pi }{\lambda \sin (\frac{\pi \lambda_{1}}{\lambda })}\sum_{k=1}^{s} \widetilde{c}_{k}^{\frac{\lambda_{1}}{\lambda }-1} \prod_{j=1(j\neq k)}^{s} \frac{1}{\widetilde{c}_{j}-\widetilde{c}_{k}} \in \mathbf{R}_{+}. \end{aligned}$$
Since
it follows that
$$\begin{aligned} 0 &< \widetilde{k}_{s}(\lambda_{1})=\frac{1}{\lambda } \int_{0}^{ \infty }\prod_{k=1}^{s} \frac{1}{u+\widetilde{c}_{k}}u^{\frac{\lambda _{1}}{\lambda }-1}\,du \\ &\leq \frac{1}{\lambda } \int_{0}^{\infty }\frac{1}{(u+c_{1})^{s}}u ^{\frac{\lambda_{1}}{\lambda }-1}\,du \\ &\overset{u=c_{1}v}{=}\frac{1}{\lambda c_{1}^{\lambda_{2}/\lambda }} \int_{0}^{\infty }\frac{1}{(v+1)^{s}}v^{\frac{\lambda_{1}}{\lambda }-1}\,dv \\ &=\frac{1}{\lambda c_{1}^{\lambda_{2}/\lambda }}B\biggl(\frac{\lambda_{1}}{ \lambda },\frac{\lambda_{2}}{\lambda }\biggr)\in \mathbf{R}_{+}, \end{aligned}$$
$$\begin{aligned} k_{s}(\lambda_{1}) &=\lim_{\varepsilon \rightarrow 0^{+}} \widetilde{k}_{s}(\lambda_{1}) \\ &=\frac{\pi }{\lambda \sin (\frac{\pi \lambda_{1}}{\lambda })}\sum_{k=1}^{s}c_{k}^{\frac{\lambda_{1}}{\lambda }-1} \prod_{j=1(j\neq k)} ^{s}\frac{1}{c_{j}-c_{k}}\in \mathbf{R}_{+}. \end{aligned}$$
(9)
In particular, for \(s=1\), we obtain
for \(c_{s}=\cdots =c_{1}\), we have
$$ k_{1}(\lambda_{1})=\frac{1}{\lambda } \int_{0}^{\infty }\frac{u^{( \lambda_{1}/\lambda )-1}}{u+c_{1}}\,du= \frac{\pi }{\lambda c_{1}^{ \lambda_{2}/\lambda }\sin (\frac{\pi \lambda_{1}}{\lambda })}; $$
(10)
$$ k(\lambda_{1}):= \int_{0}^{\infty }\frac{t^{\lambda_{1}-1}}{(t^{\lambda }+c_{1})^{s}}\,dt= \frac{1}{\lambda c_{1}^{\lambda_{2}/\lambda }}B\biggl(\frac{ \lambda_{1}}{\lambda },\frac{\lambda_{2}}{\lambda }\biggr). $$
(11)
We further assume that \(s\in \mathbf{N}\), \(c_{s}\geq \cdots \geq c_{1}>0\), \(\alpha ,\beta \in (0,\pi )\), \(\xi ,\eta \in [ 0,\frac{1}{2}]\), \(0<\lambda_{1},\lambda_{2},\lambda \leq 1,\lambda_{1}+\lambda_{2}=s \lambda \) (\(s\geq 2\)); \(0<\lambda_{1}\), \(\lambda_{2}<1\), \(0<\lambda_{1}+\lambda_{2}=\lambda \leq 1\) (\(s=1\)). For \(\vert t \vert >\frac{1}{2}\), we set
\(((\zeta ,\theta ,t)=(\xi ,\alpha ,x)\mbox{ or }(\eta ,\beta ,y))\) and
$$ A_{\zeta ,\theta }(t):=\vert t-\zeta \vert +(t-\zeta )\cos \theta $$
$$\begin{aligned} k(x,y) &:=k_{s\lambda }\bigl(A_{\xi ,\alpha }(x),A_{\eta ,\beta }(y)\bigr) \\ &=\frac{1}{\prod_{k=1}^{s}(A_{\xi ,\alpha }^{\lambda }(x)+c_{k}A_{ \eta ,\beta }^{\lambda }(y))}. \end{aligned}$$
Definition 1
Define the following weight coefficients:
where \(\sum_{\vert j \vert =1}^{\infty }\cdots =\sum_{j=-1}^{-\infty }+\cdots +\sum_{j=1}^{\infty }\cdots\) (\(j=m,n\)).
$$\begin{aligned}& \omega (\lambda_{2},m) : =\sum_{\vert n \vert =1}^{\infty }k(m,n) \frac{A_{ \xi ,\alpha }^{\lambda_{1}}(m)}{A_{\eta ,\beta }^{1-\lambda_{2}}(n)},\quad \vert m \vert \in \mathbf{N}, \end{aligned}$$
(12)
$$\begin{aligned}& \varpi (\lambda_{1},n) : =\sum_{\vert m \vert =1}^{\infty }k(m,n) \frac{A_{ \eta ,\beta }^{\lambda_{2}}(n)}{A_{\xi ,\alpha }^{1-\lambda_{1}}(m)},\quad \vert n \vert \in \mathbf{N}, \end{aligned}$$
(13)
Lemma 2
With regards to the above agreement, replacing
\(0<\lambda_{1}\leq 1\) (\(0<\lambda_{1}<1\)) by
\(\lambda_{1}>0\)
and setting
we still have
where
$$ h_{\beta }(\lambda_{1}):=2k_{s}( \lambda_{1})\csc^{2}\beta , $$
$$ h_{\beta }(\lambda_{1}) \bigl(1-\theta (\lambda_{2},m) \bigr)< \omega (\lambda_{2},m)< h _{\beta }(\lambda_{1}),\quad \vert m \vert \in \mathbf{N,} $$
(14)
$$\begin{aligned} \theta (\lambda_{2},m) &:=\frac{1}{k_{s}(\lambda_{1})} \int_{\frac{A_{\xi ,\alpha }(m)}{(1+\eta )(1+\cos \beta )}}^{\infty }\frac{u ^{\lambda_{1}-1}}{\prod_{k=1}^{s}(u^{\lambda }+c_{k})}\,du \\ &=O \biggl( \frac{1}{A_{\xi ,\alpha }^{\lambda_{2}}(m)} \biggr) \in (0,1),\quad \vert m \vert \in \mathbf{N}. \end{aligned}$$
(15)
Proof
For \(\vert x \vert >\frac{1}{2}\), we set
wherefrom, for \(y>\frac{1}{2}\),
$$\begin{aligned}& k^{(1)}(x,y) : =\frac{1}{\prod_{k=1}^{s}\{A_{\xi ,\alpha }^{\lambda }(x)+c_{k}[(y-\eta )(\cos \beta -1)]^{\lambda }\}}, \\& \quad y < -\frac{1}{2}, \\& k^{(2)}(x,y) : =\frac{1}{\prod_{k=1}^{s}\{A_{\xi ,\alpha }^{\lambda }(x)+c_{k}[(y-\eta )(1+\cos \beta )]^{\lambda }\}}, \\& \quad y > \frac{1}{2}, \end{aligned}$$
$$ k^{(1)}(x,-y)=\frac{1}{\prod_{k=1}^{s}\{A_{\xi ,\alpha }^{\lambda }(x)+c _{k}[(y+\eta )(1-\cos \beta )]^{\lambda }\}}. $$
We find
□
$$\begin{aligned}& \omega (\lambda_{2},m)=\sum_{n=-1}^{-\infty }k^{(1)}(m,n) \frac{A_{ \xi ,\alpha }^{\lambda_{1}}(m)}{[(n-\eta )(\cos \beta -1)]^{1-\lambda _{2}}} \\& \hphantom{\omega (\lambda_{2},m)=}{}+\sum_{n=1}^{\infty }k^{(2)}(m,n) \frac{A_{\xi ,\alpha }^{\lambda _{1}}(m)}{[(n-\eta )(1+\cos \beta )]^{1-\lambda_{2}}} \\& \hphantom{\omega (\lambda_{2},m)}{} = \frac{A_{\xi ,\alpha }^{\lambda_{1}}(m)}{(1-\cos \beta )^{1-\lambda _{2}}}\sum_{n=1}^{\infty } \frac{k^{(1)}(m,-n)}{(n+\eta )^{1-\lambda _{2}}} \\& \hphantom{\omega (\lambda_{2},m)=}{} +\frac{A_{\xi ,\alpha }^{\lambda_{1}}(m)}{(1+\cos \beta )^{1-\lambda _{2}}}\sum_{n=1}^{\infty } \frac{k^{(2)}(m,n)}{(n-\eta )^{1-\lambda _{2}}}. \end{aligned}$$
(16)
It is evident that, for fixed \(m\in \mathbf{N}\), \(0<\lambda_{2}\leq 1\), \(0< \lambda \leq 1\), both \(\frac{k^{(1)}(m,-y)}{(y+\eta )^{1-\lambda_{2}}}\) and \(\frac{k^{(2)}(m,y)}{(y- \eta )^{1-\lambda_{2}}}\) are strictly decreasing and strictly convex with respect to \(y\in (\frac{1}{2},\infty )\) and satisfy
and
By Hermite–Hadamard’s inequality (see [29]) we find
Setting \(u=\frac{A_{\xi ,\alpha }(m)}{(y+\eta )(1-\cos \beta )}\) (\(\frac{A_{\xi ,\alpha }(m)}{(y-\eta )(1+\cos \beta )}\)) in the first (second) integral, by simplification we find
$$\begin{aligned}& \frac{k^{(i)}(m,(-1)^{i}y)}{[y-(-1)^{i}\eta ]^{1-\lambda_{2}}}>0, \\& \frac{d}{dy}\frac{k^{(i)}(m,(-1)^{i}y)}{[y-(-1)^{i}\eta ]^{1-\lambda _{2}}}< 0 \end{aligned}$$
$$ \frac{d^{2}}{dy^{2}}\frac{k^{(i)}(m,(-1)^{i}y)}{[y-(-1)^{i}\eta ]^{1- \lambda_{2}}}>0\quad (i=1,2). $$
$$\begin{aligned} \omega (\lambda_{2},m) < &\frac{A_{\xi ,\alpha }^{\lambda_{1}}(m)}{(1- \cos \beta )^{1-\lambda_{2}}} \int_{\frac{1}{2}}^{\infty }\frac{k^{(1)}(m,-y)}{(y+ \eta )^{1-\lambda_{2}}}\,dy \\ &{}+\frac{A_{\xi ,\alpha }^{\lambda_{1}}(m)}{(1+\cos \beta )^{1- \lambda_{2}}} \int_{\frac{1}{2}}^{\infty }\frac{k^{(2)}(m,y)}{(y- \eta )^{1-\lambda_{2}}}\,dy. \end{aligned}$$
$$\begin{aligned} \omega (\lambda_{2},m) &< \biggl(\frac{1}{1-\cos \beta }+ \frac{1}{1+\cos \beta }\biggr) \int_{0}^{\infty }\frac{u^{\lambda_{1}-1}}{\prod_{k=1}^{s}(u ^{\lambda }+c_{k})}\,du \\ &=2k_{s}(\lambda_{1})\csc^{2}\beta =h_{\beta }(\lambda_{1}). \end{aligned}$$
Since both \(\frac{k^{(1)}(m,-y)}{(y+\eta )^{1-\lambda_{2}}}\) and \(\frac{k^{(2)}(m,y)}{(y-\eta )^{1-\lambda_{2}}}\) are strictly decreasing, we still have
where \(\theta (\lambda_{2},m)(<1)\) is indicated by (15). We obtain
Then we have (14) and estimate (15). □
$$\begin{aligned} \omega (\lambda_{2},m) &>\frac{A_{\xi ,\alpha }^{\lambda_{1}}(m)}{(1- \cos \beta )^{1-\lambda_{2}}} \int_{1}^{\infty }\frac{k^{(1)}(m,-y)}{(y+ \eta )^{1-\lambda_{2}}}\,dy \\ &\quad {}+\frac{A_{\xi ,\alpha }^{\lambda_{1}}(m)}{(1+\cos \beta )^{1- \lambda_{2}}} \int_{1}^{\infty }\frac{k^{(2)}(m,y)}{(y-\eta )^{1- \lambda_{2}}}\,dy \\ &=\frac{1}{1-\cos \beta } \int_{0}^{\frac{A_{\xi ,\alpha }(m)}{(1+ \eta )(1-\cos \beta )}}\frac{u^{\lambda_{1}-1}\,du}{\prod_{k=1}^{s}(u ^{\lambda }+c_{k})} \\ &\quad {}+\frac{1}{1+\cos \beta } \int_{0}^{\frac{A_{\xi ,\alpha }(m)}{(1- \eta )(1+\cos \beta )}}\frac{u^{\lambda_{1}-1}\,du}{\prod_{k=1}^{s}(u ^{\lambda }+c_{k})} \\ &\geq 2\csc^{2}\beta \int_{0}^{\frac{A_{\xi ,\alpha }(m)}{(1+\eta )(1+ \cos \beta )}}\frac{u^{\lambda_{1}-1}}{\prod_{k=1}^{s}(u^{\lambda }+c _{k})}\,du \\ &=h_{\beta }(\lambda_{2}) \bigl(1-\theta ( \lambda_{2},m)\bigr)>0, \end{aligned}$$
$$\begin{aligned} 0 &< \theta (\lambda_{2},m)< \frac{1}{k_{s}(\lambda_{1})} \int_{\frac{A_{\xi ,\alpha }(m)}{(1+\eta )(1+\cos \beta )}}^{\infty }\frac{u ^{\lambda_{1}-1}}{u^{s\lambda }}\,du \\ &=\frac{1}{k_{s}(\lambda_{1})} \int_{\frac{A_{\xi ,\alpha }(m)}{(1+\eta )(1+\cos \beta )}}^{\infty }u ^{-\lambda_{2}-1}\,du \\ &=\frac{1}{\lambda_{2}k_{s}(\lambda_{1})} \biggl[ \frac{(1+\eta )(1+ \cos \beta )}{A_{\xi ,\alpha }(m)} \biggr] ^{\lambda_{2}}. \end{aligned}$$
In the same way, we have
Lemma 3
With regards to the above agreement, replacing
\(0<\lambda_{2}\leq 1\) (\(0<\lambda_{2}<1\)) by
\(\lambda_{2}>0\), for
we still have
where
$$ h_{\alpha }(\lambda_{1})=2k_{s}( \lambda_{1})\csc^{2}\alpha , $$
$$ h_{\alpha }(\lambda_{1}) \bigl(1-\vartheta ( \lambda_{1},n)\bigr)< \varpi (\lambda_{1},n)< h _{\alpha }(\lambda_{1}),\quad \vert n \vert \in \mathbf{N,} $$
(17)
$$\begin{aligned} \vartheta (\lambda_{1},n) &:=\frac{1}{k_{s}(\lambda_{1})} \int_{\frac{A_{\eta ,\beta }(n)}{(1+\xi )(1+\cos \alpha)}}^{\infty }\frac{u^{\lambda_{2}-1}}{\prod_{k=1}^{s}(u^{\lambda }+c_{k})}\,du \\ &=O \biggl( \frac{1}{A_{\eta ,\beta }^{\lambda_{1}}(n)} \biggr) \in (0,1),\quad \vert n \vert \in \mathbf{N}. \end{aligned}$$
(18)
Lemma 4
If
\(\zeta \in {}[ 0,\frac{1}{2}],\theta \in (0,\pi )\), \((\zeta , \theta )=(\xi ,\alpha )\) (or
\((\eta ,\beta )\)), then, for
\(\rho >0\),
$$\begin{aligned} H_{\rho }(\zeta ,\theta )&:=\sum_{\vert k \vert =1}^{\infty } \frac{1}{A_{\zeta , \theta }^{1+\rho }(k)}=\frac{1+o(1)}{\rho } \\ &\quad {}\times \biggl[ \frac{1}{(1+\cos \theta )^{1+\rho }}+\frac{1}{(1-\cos \theta )^{1+\rho }} \biggr] \quad \bigl(\rho \rightarrow 0^{+}\bigr). \end{aligned}$$
(19)
Proof
We find
For \(a=\frac{1}{(1-\zeta )^{1+\rho }}>0\), by Hermite–Hadamard’s inequality we have
We still obtain
Hence we have (19). □
$$\begin{aligned} H_{\rho }(\zeta ,\theta ) &=\sum_{k=-1}^{-\infty } \frac{1}{[(k- \zeta )(\cos \theta -1)]^{1+\rho }} \\ &\quad {}+\sum_{k=1}^{\infty }\frac{1}{[(k-\zeta )(\cos \theta +1)]^{1+ \rho }} \\ &=\frac{1}{(1-\cos \theta )^{1+\rho }}\sum_{k=1}^{\infty } \frac{1}{(k+ \zeta )^{1+\rho }} \\ &\quad {}+\frac{1}{(1+\cos \theta )^{1+\rho }}\sum_{k=1}^{\infty } \frac{1}{(k- \zeta )^{1+\rho }}. \end{aligned}$$
$$\begin{aligned} H_{\rho }(\zeta ,\theta ) &\leq \biggl[ \frac{1}{(1-\cos \theta )^{1+ \rho }}+ \frac{1}{(1+\cos \theta )^{1+\rho }} \biggr] \\ &\quad {}\times \Biggl[ a+\sum_{k=2}^{\infty } \frac{1}{(k-\zeta )^{1+\rho }} \Biggr] \\ &< \biggl[ \frac{1}{(1-\cos \theta )^{1+\rho }}+\frac{1}{(1+\cos \theta )^{1+\rho }} \biggr] \\ &\quad {}\times \biggl[ a+ \int_{\frac{3}{2}}^{\infty }\frac{1}{(y-\zeta )^{1+ \rho }}\,dy \biggr] \\ &=\frac{a\rho +(\frac{3}{2}-\zeta )^{-\rho }}{\rho } \biggl[ \frac{1}{(1- \cos \theta )^{1+\rho }}+\frac{1}{(1+\cos \theta )^{1+\rho }} \biggr] . \end{aligned}$$
$$\begin{aligned}& H_{\rho }(\zeta ,\theta )\geq \biggl[ \frac{1}{(1-\cos \theta )^{1+ \rho }}+ \frac{1}{(1+\cos \theta )^{1+\rho }} \biggr] \sum_{k=1}^{\infty }\frac{1}{(k+\zeta )^{1+\rho }} \\& \hphantom{ H_{\rho }(\zeta ,\theta )}{}> \biggl[ \frac{1}{(1-\cos \theta )^{1+\rho }}+\frac{1}{(1+\cos \theta )^{1+\rho }} \biggr] \int_{1}^{\infty }\frac{dy}{(y+\zeta )^{1+ \rho }} \\& \hphantom{ H_{\rho }(\zeta ,\theta )}{}= \frac{(1+\zeta )^{-\rho }}{\rho } \biggl[ \frac{1}{(1-\cos \theta )^{1+ \rho }}+\frac{1}{(1+\cos \theta )^{1+\rho }} \biggr] . \end{aligned}$$
3 Main results and some particular cases
Theorem 5
Suppose that
\(0< p<1\) (\(q<0\)), \(\frac{1}{p}+\frac{1}{q}=1\),
\(a_{m},b_{n}\geq 0\)
\((\vert m\vert ,\vert n\vert \in \mathbf{N})\), and
We have the following reverse equivalent inequalities:
$$ K_{\alpha ,\beta }(\lambda_{1}):=h_{\beta }^{\frac{1}{p}}( \lambda_{1})h _{\alpha }^{\frac{1}{q}}(\lambda_{1})=2k_{s}( \lambda_{1}) \csc^{\frac{2}{p}}\beta \csc^{\frac{2}{q}}\alpha , $$
(20)
$$\begin{aligned}& 0< \sum_{\vert m \vert =1}^{\infty }A_{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m} ^{p}< \infty , \\& 0< \sum_{\vert n \vert =1}^{\infty }A_{\eta ,\beta }^{q(1-\lambda _{2})-1}(n)b_{n}^{q}< \infty . \end{aligned}$$
$$\begin{aligned}& I:=\sum_{\vert n \vert =1}^{\infty }\sum _{\vert m \vert =1}^{\infty }\frac{1}{\prod_{k=1} ^{s}(A_{\xi ,\alpha }^{\lambda }(m)+c_{k}A_{\eta ,\beta }^{\lambda }(n))}a _{m}b_{n} \\& \hphantom{I}{}>K_{\alpha ,\beta }(\lambda_{1}) \Biggl[ \sum _{\vert m \vert =1}^{\infty }\bigl(1- \theta (\lambda_{2},m) \bigr)A_{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m} ^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{\vert n \vert =1}^{\infty }A_{\eta ,\beta }^{q(1-\lambda_{2})-1}(n)b _{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned}$$
(21)
$$\begin{aligned}& J : = \Biggl\{ \sum_{\vert n \vert =1}^{\infty }A_{\eta ,\beta }^{p\lambda_{2}-1}(n) \Biggl[ \sum_{\vert m \vert =1}^{\infty }\frac{a_{m}}{\prod_{k=1}^{s}(A_{\xi , \alpha }^{\lambda }(m)+c_{k}A_{\eta ,\beta }^{\lambda }(n))} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}} \\& \hphantom{J}{}> K_{\alpha ,\beta }(\lambda_{1}) \Biggl[ \sum _{\vert m \vert =1}^{\infty }\bigl(1- \theta (\lambda_{2},m) \bigr)A_{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m} ^{p} \Biggr] ^{\frac{1}{p}}, \end{aligned}$$
(22)
$$\begin{aligned}& L : = \Biggl\{ \sum_{\vert m\vert =1}^{\infty }\frac{A_{ \xi ,\alpha }^{q\lambda_{1}-1}(m)}{(1-\theta (\lambda_{2},m))^{q-1}} \Biggl[ \sum_{\vert n\vert =1}^{\infty } \frac{1}{ \prod_{k=1}^{s}(A_{\xi ,\alpha }^{\lambda }(m)+c_{k}A_{\eta ,\beta } ^{\lambda }(n))}b_{n} \Biggr] ^{q} \Biggr\} ^{\frac{1}{q}} \\& \hphantom{L}{}>K_{\alpha ,\beta }(\lambda_{1}) \Biggl[ \sum _{\vert n\vert =1} ^{\infty }A_{\eta ,\beta }^{q(1-\lambda_{2})-1}(n)b_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$
(23)
In particular, for \(s=c_{1}=1\), \(\alpha =\beta =\frac{\pi }{2}\) (\(0<\lambda_{1}\), \(\lambda_{2}<1\), \(\lambda_{1}+\lambda_{2}=\lambda \leq 1\)), (21) reduces to (6); and (22) and (23) reduce to the equivalent forms of (6) as follows:
$$\begin{aligned}& \Biggl\{ \sum_{\vert n \vert =1}^{\infty }\vert n-\eta \vert ^{p\lambda_{2}-1} \Biggl( \sum_{\vert m \vert =1}^{\infty } \frac{a_{m}}{\vert m-\xi \vert ^{\lambda }+\vert n-\eta \vert ^{ \lambda }} \Biggr) ^{p} \Biggr\} ^{\frac{1}{p}} \\& \quad > \frac{2\pi }{\lambda \sin (\frac{\pi \lambda_{1}}{\lambda })} \Biggl[ \sum_{\vert m \vert =1}^{\infty } \bigl(1-\theta_{1}(\lambda_{2},m)\bigr)\vert m-\xi \vert ^{p(1- \lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}}, \end{aligned}$$
(24)
$$\begin{aligned}& \Biggl[ \sum_{\vert m\vert =1}^{\infty } \frac{\vert m-\xi \vert ^{q \lambda_{1}-1}}{(1-\theta_{1}(\lambda_{2},m))^{q-1}} \Biggl( \sum_{\vert n\vert =1}^{\infty } \frac{b_{n}}{\vert m-\xi \vert ^{ \lambda }+\vert n-\eta \vert ^{\lambda }} \Biggr) ^{q} \Biggr] ^{\frac{1}{q}} \\& \quad > \frac{2\pi }{\lambda \sin (\frac{\pi \lambda_{1}}{\lambda })} \Biggl[ \sum_{\vert n\vert =1}^{\infty } \vert n-\eta \vert ^{q(1-\lambda_{2})-1}b _{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$
(25)
Proof
By the reverse Hölder inequality with weight (see [29]) and (12) we find
By (17), in view of \(p-1<0\), we have
Then by (14) we have (22).
$$\begin{aligned}& \Biggl( \sum_{\vert m \vert =1}^{\infty }k(m,n)a_{m} \Biggr) ^{p} \\& \quad = \Biggl[ \sum_{\vert m \vert =1}^{\infty }k(m,n) \frac{A_{\xi ,\alpha }^{(1- \lambda_{1})/q}(m)a_{m}}{A_{\eta ,\beta }^{(1-\lambda_{2})/p}(n)}\frac{A _{\eta ,\beta }^{(1-\lambda_{2})/p}(n)}{A_{\xi ,\alpha }^{(1-\lambda _{1})/q}(m)} \Biggr] ^{p} \\& \quad \geq \sum_{\vert m \vert =1}^{\infty }h(m,n) \frac{A_{\xi ,\alpha }^{(1-\lambda_{1})p/q}(m)}{A _{\eta ,\beta }^{1-\lambda_{2}}(n)}a_{m}^{p} \Biggl[ \sum _{\vert m \vert =1}^{ \infty }k(m,n)\frac{A_{\eta ,\beta }^{(1-\lambda_{2})q/p}(n)}{A_{ \xi ,\alpha }^{1-\lambda_{1}}(m)} \Biggr] ^{p-1} \\& \quad = \frac{(\varpi (\lambda_{1},n))^{p-1}}{A_{\eta ,\beta }^{p\lambda_{2}-1}(n)} \sum_{\vert m \vert =1}^{\infty }k(m,n) \frac{A_{\xi ,\alpha }^{(1-\lambda_{1})p/q}(m)}{A _{\eta ,\beta }^{1-\lambda_{2}}(n)}a_{m}^{p}. \end{aligned}$$
$$\begin{aligned} J >&h_{\alpha }^{\frac{1}{q}}(\lambda_{1}) \Biggl[ \sum _{\vert n \vert =1}^{ \infty }\sum _{\vert m \vert =1}^{\infty }k(m,n)\frac{A_{\xi ,\alpha }^{(1-\lambda _{1})p/q}(m)}{A_{\eta ,\beta }^{1-\lambda_{2}}(n)}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =&h_{\alpha }^{\frac{1}{q}}(\lambda_{1}) \Biggl[ \sum _{\vert m \vert =1}^{\infty }\sum_{\vert n \vert =1}^{\infty }k(m,n) \frac{A_{\xi ,\alpha }^{(1-\lambda_{1})p/q}(m)}{A _{\eta ,\beta }^{1-\lambda_{2}}(n)}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =&h_{\alpha }^{\frac{1}{q}}(\lambda_{1}) \Biggl[ \sum _{\vert m \vert =1}^{\infty }\omega (\lambda_{2},m)A_{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m} ^{p} \Biggr] ^{\frac{1}{p}}. \end{aligned}$$
(26)
By Hölder’s inequality (see [29]) we have
Then by (22) we have (21).
$$\begin{aligned} I = \sum_{\vert n \vert =1}^{\infty } \Biggl[ A_{\eta ,\beta }^{\lambda_{2}- \frac{1}{p}}(n)\sum_{\vert m \vert =1}^{\infty }k(m,n)a_{m} \Biggr] A_{\eta , \beta }^{\frac{1}{p}-\lambda_{2}}(n)b_{n} \geq J \Biggl[ \sum_{\vert n \vert =1}^{\infty }A_{\eta ,\beta }^{q(1-\lambda_{2})-1}(n)b _{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$
(27)
On the other hand, assuming that (21) is valid, we set
and then
By (26) we find \(J>0\). If \(J=\infty \), then (22) is evidently valid; if \(J<\infty \), then by (21) we have
namely, (22) follows, which is equivalent to (21).
$$ b_{n}:=A_{\eta ,\beta }^{p\lambda_{2}-1}(n) \Biggl( \sum _{\vert m \vert =1}^{ \infty }k(m,n)a_{m} \Biggr) ^{p-1},\quad \vert n \vert \in \mathbf{N,} $$
$$ J= \Biggl[ \sum_{\vert n \vert =1}^{\infty }A_{\eta ,\beta }^{q(1-\lambda_{2})-1}(n)b _{n}^{q} \Biggr] ^{\frac{1}{p}}. $$
$$\begin{aligned}& \sum_{\vert n \vert =1}^{\infty }A_{\eta ,\beta }^{q(1-\lambda_{2})-1}(n)b_{n} ^{q} \\& \quad=J^{p}=I >K_{\alpha ,\beta }(\lambda_{1}) \Biggl[ \sum_{\vert m \vert =1}^{\infty }\bigl(1- \theta (\lambda_{2},m)\bigr)A_{ \xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{\vert n \vert =1}^{\infty }A_{\eta ,\beta }^{q(1-\lambda_{2})-1}(n)b _{n}^{q} \Biggr] ^{\frac{1}{q}}, \\& J = \Biggl[ \sum_{\vert n \vert =1}^{\infty }A_{\eta ,\beta }^{q(1-\lambda_{2})-1}(n)b _{n}^{q} \Biggr] ^{\frac{1}{p}} > K_{\alpha ,\beta }(\lambda_{1}) \Biggl[ \sum _{\vert m \vert =1}^{\infty }\bigl(1- \theta (\lambda_{2},m) \bigr)A_{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m} ^{p} \Biggr] ^{\frac{1}{p}}, \end{aligned}$$
We have proved that (21) is valid. Then we set
and find
If \(L=0\), then (23) is impossible, so that \(L>0\). If \(L=\infty \), then (23) is trivially valid; if \(L<\infty \), then we have
thats is, (23) follows.
$$ a_{m}:= \frac{A_{\xi ,\alpha }^{q\lambda_{1}-1}(m)}{(1-\theta ( \lambda_{2},m))^{q-1}} \Biggl( \sum _{\vert n\vert =1}^{ \infty }k(m,n)b_{n} \Biggr) ^{q-1},\quad \vert m\vert \in \mathbf{N,} $$
$$ L= \Biggl[ \sum_{\vert m\vert =1}^{\infty }\bigl(1-\theta ( \lambda_{2},m)\bigr)A _{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m}^{p} \Biggr] ^{\frac{1}{q}}. $$
$$\begin{aligned}& \sum_{\vert m\vert =1}^{\infty }\bigl(1-\theta ( \lambda_{2},m)\bigr)A _{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m}^{p} \\& \quad =L^{q}=I>K_{\alpha ,\beta }(\lambda_{1}) \Biggl[ \sum _{\vert m\vert =1} ^{\infty }\bigl(1-\theta (\lambda_{2},m) \bigr)A_{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a _{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{\vert n\vert =1}^{\infty }A_{\eta , \beta }^{q(1-\lambda_{2})-1}(n)b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \\& L = \Biggl[ \sum_{\vert m\vert =1}^{\infty }\bigl(1-\theta ( \lambda_{2},m)\bigr)A_{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m}^{p} \Biggr] ^{\frac{1}{q}} > K_{\alpha ,\beta }(\lambda_{1}) \Biggl[ \sum _{\vert n\vert =1}^{\infty }A_{\eta ,\beta }^{q(1- \lambda_{2})-1}(n)b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned}$$
On the other-hand, assuming that (23) is valid, using the reverse Hölder inequality, we have
and then by (23) we have (21), which is equivalent to (23).
$$\begin{aligned} I =&\sum_{\vert m\vert =1}^{\infty } \biggl[ \frac{A_{ \xi ,\alpha }^{(1/q)-\lambda_{1}}(m)}{(1-\theta (\lambda_{2},m))^{-1/p}}a _{m} \biggr] \\ &\times \Biggl[ \frac{A_{\xi ,\alpha }^{\lambda_{1}-(1/q)}(m)}{(1- \theta (\lambda_{2},m))^{1/p}}\sum_{\vert n\vert =1}^{ \infty }k(m,n)b_{n} \Biggr] \\ \geq & \Biggl[ \sum_{\vert m\vert =1}^{\infty }\bigl(1- \theta ( \lambda_{2},m)\bigr)A_{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m}^{p} \Biggr] ^{\frac{1}{p}}L, \end{aligned}$$
(28)
Theorem 6
Proof
For \(0<\varepsilon <p\lambda_{1}\), we set \(\widetilde{\lambda}_{1}=\lambda_{1}-\frac{\varepsilon }{p}\) (\(\in (0,1)\)), \(\widetilde{\lambda }_{2}=\lambda_{2}+\frac{\varepsilon }{p}\) (>0), and
By (19) and (17) we find
$$\begin{aligned}& \widetilde{a}_{m}:=A_{\xi ,\alpha }^{\lambda_{1}-(\varepsilon /p)-1}(m)=A _{\xi ,\alpha }^{\widetilde{\lambda }_{1}-1}(m)\quad \bigl(\vert m\vert \in \mathbf{N}\bigr), \\& \widetilde{b}_{n}:=A_{\eta ,\beta }^{\lambda_{2}-(\varepsilon /q)-1}(n)=A_{\eta ,\beta }^{\widetilde{\lambda }_{2}-\varepsilon -1}(n)\quad \bigl(\vert n\vert \in \mathbf{N}\bigr). \end{aligned}$$
$$\begin{aligned}& \widetilde{I}_{1}:= \Biggl[ \sum_{\vert m\vert =1}^{\infty } \bigl(1- \theta (\lambda_{2},m)\bigr)A_{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m) \widetilde{a}_{m}^{p} \Biggr] ^{\frac{1}{p}}\Biggl[ \sum_{\vert n\vert =1}^{\infty }A_{\eta , \beta }^{q(1-\lambda_{2})-1}(n) \widetilde{b}_{n}^{q} \Biggr] ^{ \frac{1}{q}} \\& \hphantom{\widetilde{I}_{1}} = \Biggl[ \sum_{\vert m\vert =1}^{\infty }A_{\xi ,\alpha } ^{-1-\varepsilon }(m)-\sum_{\vert m\vert =1}^{\infty }O\bigl(A _{\xi ,\alpha }^{-1-(\lambda_{2}+\varepsilon )}(m)\bigr) \Biggr] ^{ \frac{1}{p}} \Biggl( \sum _{\vert n\vert =1}^{\infty }A_{ \eta ,\beta }^{-1-\varepsilon }(n) \Biggr) ^{\frac{1}{q}} \\& \hphantom{\widetilde{I}_{1}} =\frac{1}{\varepsilon }\bigl(2\csc^{2}\alpha +o(1)-\varepsilon O(1) \bigr)^{ \frac{1}{p}}\bigl(2\csc^{2}\beta +\widetilde{o}(1) \bigr)^{\frac{1}{q}}\quad \bigl( \varepsilon \rightarrow 0^{+}\bigr), \\& \widetilde{I}:=\sum_{\vert m\vert =1}^{\infty } \sum _{\vert n\vert =1}^{\infty }k(m,n)\widetilde{a}_{m} \widetilde{b}_{n} \\& \hphantom{\widetilde{I}}=\sum_{\vert n\vert =1}^{\infty } \sum_{\vert m\vert =1}^{\infty }k(m,n)A_{\xi ,\alpha }^{ \widetilde{\lambda }_{1}-1}(m)A_{\eta ,\beta }^{\widetilde{\lambda }_{2}-\varepsilon -1}(n) \\& \hphantom{\widetilde{I}} = \sum_{\vert n\vert =1}^{\infty }\varpi ( \widetilde{ \lambda }_{1},n)A_{\eta ,\beta }^{-1-\varepsilon }(n)< h _{\alpha }( \widetilde{\lambda }_{1})\sum_{\vert n\vert =2} ^{\infty }A_{\eta ,\beta }^{-1-\varepsilon }(n) \\& \hphantom{\widetilde{I}} = \frac{1}{\varepsilon }h_{\alpha }\biggl(\lambda_{1}- \frac{\varepsilon }{p}\biggr) \bigl(2\csc^{2}\beta +\widetilde{o}(1) \bigr). \end{aligned}$$
If there exists a positive number \(K\geq K_{\alpha ,\beta }(\lambda _{1})\) such that (21) is still valid when replacing \(K_{\alpha ,\beta }(\lambda_{1})\) by K, then, in particular, we have
In view of the preceding results, it follows that
and then
namely,
Hence \(K=K_{\alpha ,\beta }(\lambda_{1})\) is the best possible constant factor in (21).
$$ \varepsilon \widetilde{I}=\varepsilon \sum_{\vert m\vert =1} ^{\infty }\sum_{\vert n\vert =1}^{\infty }k(m,n) \widetilde{a}_{m}\widetilde{b}_{n}>\varepsilon K \widetilde{I}_{1}. $$
$$\begin{aligned}& h_{\alpha }\biggl(\lambda_{1}-\frac{\varepsilon }{p}\biggr) \bigl(2\csc^{2}\beta + \widetilde{o}(1)\bigr) >K\cdot \bigl(2\csc^{2}\alpha +o(1)-\varepsilon O(1) \bigr)^{1/p}\bigl(2\csc^{2} \beta +\widetilde{o}(1) \bigr)^{1/q}, \end{aligned}$$
$$ 4k_{s}(\lambda_{1})\csc^{2}\alpha\csc^{2}\beta \geq 2K\csc^{2/p} \alpha \csc^{2/q}\beta \quad \bigl(\varepsilon \rightarrow 0^{+}\bigr), $$
$$ K_{\alpha ,\beta }(\lambda_{1})=2k_{s}(\lambda_{1})\csc^{2/p}\beta \csc^{2/q}\alpha \geq K. $$
Remark 1
(i) For \(\xi =\eta =0\) and \(\alpha =\beta =\frac{\pi }{2}\) in (21), setting
we have the following new reverse inequality with the best possible constant factor \(2k_{s}(\lambda_{1})\):
It follows that (21) is an extension of (29).
$$\begin{aligned} \widetilde{\theta }(\lambda_{2},m) &:=\frac{1}{k_{s}(\lambda_{1})} \int_{\vert m \vert }^{\infty }\frac{u^{\lambda_{1}-1}}{\prod_{k=1}^{s}(u^{ \lambda }+c_{k})}\,du \\ &=O \biggl( \frac{1}{\vert m \vert ^{\lambda_{2}}} \biggr) \in (0,1),\quad \vert m \vert \in \mathbf{N}, \end{aligned}$$
$$\begin{aligned}& \sum_{\vert n \vert =1}^{\infty }\sum _{\vert m \vert =1}^{\infty }\frac{1}{\prod_{k=1} ^{s}(\vert m \vert ^{\lambda }+c_{k}\vert n \vert ^{\lambda })}a_{m}b_{n} \\& \quad >2k_{s}( \lambda _{1}) \Biggl[ \sum_{\vert m \vert =1}^{\infty }\bigl(1- \widetilde{\theta }(\lambda_{2},m)\bigr)\vert m \vert ^{p(1- \lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum _{\vert n \vert =1}^{ \infty }\vert n \vert ^{q(1-\lambda_{2})-1}b_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$
(29)
(ii) If \(a_{-m}=a_{m}\) and \(b_{-n}=b_{n}\) (\(m,n\in \mathbf{N}\)), then for
(29) reduces to the following reverse Hilbert-type inequality:
$$\begin{aligned} \widetilde{\theta }(\lambda_{2},m) =&\frac{1}{k_{s}(\lambda_{1})} \int _{m}^{\infty }\frac{u^{\lambda_{1}-1}}{\prod_{k=1}^{s}(u^{\lambda }+c _{k})}\,du \\ =&O \biggl( \frac{1}{m^{\lambda_{2}}} \biggr) \in (0,1),\quad m\in \mathbf{N}, \end{aligned}$$
$$\begin{aligned}& \sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{1}{\prod_{k=1}^{s}(m ^{\lambda }+c_{k}n^{\lambda })}a_{m}b_{n} \\& \quad > k_{s}(\lambda_{1}) \Biggl[ \sum _{m=1}^{\infty }\bigl(1- \widetilde{\theta }( \lambda_{2},m)\bigr)m^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty }n^{q(1-\lambda_{2})-1}b_{n} ^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$
(30)
(iii) If \(a_{-m}=a_{m}\) and \(b_{-n}=b_{n}\) (\(m,n \in \mathbf{N}\)), then setting
(6) reduces to
$$\begin{aligned} \theta_{2}(\lambda_{2},m) :=&\frac{\lambda \sin (\frac{\pi \lambda _{1}}{\lambda })}{\pi } \int_{\frac{m-\xi }{1+\eta }}^{\infty }\frac{u ^{\lambda_{1}-1}}{u^{\lambda }+1}\,du \\ =&O \biggl( \frac{1}{(m-\xi )^{\lambda_{2}}} \biggr) \in (0,1),\quad m\in \mathbf{N}, \\ \theta_{3}(\lambda_{2},m) :=&\frac{\lambda \sin (\frac{\pi \lambda _{1}}{\lambda })}{\pi } \int_{\frac{m+\xi }{1+\eta }}^{\infty }\frac{u ^{\lambda_{1}-1}}{u^{\lambda }+1}\,du \\ =&O \biggl( \frac{1}{(m+\xi )^{\lambda_{2}}} \biggr) \in (0,1),\quad m\in \mathbf{N}, \end{aligned}$$
$$\begin{aligned}& \sum_{n=1}^{\infty }\sum _{m=1}^{\infty } \biggl[ \frac{1}{(m-\xi )^{ \lambda }+(n-\eta )^{\lambda }}+ \frac{1}{(m-\xi )^{\lambda }+(n+ \eta )^{\lambda }} \\& \quad \quad {}+\frac{1}{(m+\xi )^{\lambda }+(n-\eta )^{\lambda }}+\frac{1}{(m+ \xi )^{\lambda }+(n+\eta )^{\lambda }} \biggr] a_{m}b_{n} \\& \quad > \frac{2\pi }{\lambda \sin (\pi \lambda_{1}/\lambda )} \Biggl\{ \sum_{m=1}^{\infty } \bigl[ \bigl(1-\theta_{2}(\lambda_{2},m)\bigr) (m-\xi )^{p(1- \lambda_{1})-1} \\& \quad \quad {} +\bigl(1-\theta_{3}(\lambda_{2},m) \bigr) (m+\xi )^{p(1-\lambda_{1})-1} \bigr] a _{m}^{p} \Biggr\} ^{\frac{1}{p}} \\& \quad \quad {}\times \Biggl\{ \sum_{n=1}^{\infty } \bigl[ (n- \eta )^{q(1-\lambda_{2})-1}+(n+ \eta )^{q(1-\lambda_{2})-1} \bigr] b_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned}$$
(31)
In particular, for \(\xi =\eta =0\), \(\lambda =1\), \(\lambda_{1}=\lambda_{2}=\frac{1}{2}\) in (31) (or for \(s=\lambda =c_{1}=1\), \(\lambda_{1}=\lambda_{2}=\frac{1}{2}\) in (30)), setting
we have the following reverse Hardy–Hilbert inequality with the best possible constant π:
Hence (21) is an extended reverse Hardy–Hilbert’s inequality in the whole plane.
$$ \theta (m):=\frac{1}{\pi } \int_{m}^{\infty }\frac{u^{-1/2}}{u+1}\,du=O \biggl( \frac{1}{m^{1/2}} \biggr) \in (0,1),\quad m\in \mathbf{N}, $$
$$\begin{aligned}& \sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{a_{m}b_{n}}{m+n} > \pi \Biggl[ \sum_{m=1}^{\infty }\bigl(1-\theta (m)\bigr)m^{\frac{p}{2}-1}a _{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl( \sum_{n=1}^{\infty }n^{ \frac{q}{2}-1}b_{n}^{q} \Biggr) ^{\frac{1}{q}}. \end{aligned}$$
4 Conclusions
In this paper, using the weight coefficients, a complex integral formula, and Hermite–Hadamard’s inequality, we give an extended reverse Hardy–Hilbert’s inequality in the whole plane with multiparameters and a best possible constant factor (Theorems 5 and 6). We consider equivalent forms and a few particular cases. The technique of real analysis is very important, which is the key to prove the reverse equivalent inequalities with the best possible constant factor. The lemmas and theorems provide an extensive account of this type inequalities.
Acknowledgements
This work is supported by the National Natural Science Foundation (No. 61772140) and Science and Technology Planning Project Item of Guangzhou City (No. 201707010229). We are grateful for this help.
Competing interests
The authors declare that they have no competing interests.
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