Advertisement

On a reverse Mulholland’s inequality in the whole plane

Open Access
Research
  • 291 Downloads

Abstract

By introducing multi-parameters, applying the weight coefficients and Hermite–Hadamard’s inequality, we give a reverse of the extended Mulholland inequality in the whole plane with the best possible constant factor. The equivalent forms and a few particular cases are also considered.

Keywords

Mulholland’s inequality Parameter Weight coefficient Equivalent form Reverse 

MSC

26D15 47A07 

1 Introduction

Assuming that \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(a_{n},b_{n}\geq0\), \(0<\sum_{n=1}^{\infty}a_{n}^{p}<\infty\), and \(0<\sum_{n=1}^{\infty }b_{n}^{q}<\infty\), we have the following Hardy–Hilbert inequality (cf. [1]):
$$ \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{m+n}< \frac{\pi}{\sin(\pi/p)} \Biggl( \sum_{n=1}^{\infty}a_{n}^{p} \Biggr) ^{1/p} \Biggl( \sum_{n=1}^{\infty}b_{n}^{q} \Biggr) ^{1/q}, $$
(1)
where the constant factor \(\frac{\pi}{\sin(\pi/p)}\) is the best possible. We still have the following Mulholland inequality with the same best possible constant factor \(\frac{\pi}{\sin(\pi/p)}\) (cf. Theorem 343 of [1], replacing \(\frac{a_{m}}{m}\), \(\frac{b_{n}}{n}\) by \(a_{m}\), \(b_{n}\)):
$$ \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln mn}< \frac{\pi }{\sin(\pi/p)} \Biggl( \sum_{n=2}^{\infty}\frac{a_{m}^{p}}{m^{1-p}} \Biggr) ^{1/p} \Biggl( \sum_{n=2}^{\infty} \frac{b_{n}^{q}}{n^{1-q}} \Biggr) ^{1/q}. $$
(2)
Inequalities (1)–(2) are important in the analysis and its applications (cf. [1, 2]).
In 2007, Yang [3] first gave a Hilbert-type integral inequality in the whole plane as follows:
$$\begin{aligned} & \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\frac{f(x)g(y)}{(1+e^{x+y})^{\lambda}}\,dx\,dy \\ &\quad < B \biggl(\frac{\lambda}{2},\frac{\lambda}{2} \biggr) \biggl( \int_{-\infty}^{\infty }e^{-\lambda x}f^{2}(x)\,dx \int_{-\infty}^{\infty}e^{-\lambda y}f^{2}(y)\,dy \biggr)^{\frac{1}{2}}, \end{aligned}$$
(3)
where the constant factor \(B(\frac{\lambda}{2},\frac{\lambda}{2})\) (\(\lambda>0\)) is the best possible. Some new results on inequalities (1)–(3) were provided by [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24]. In 2016, Yang and Chen [25] gave a more accurate extension of (1) in the whole plane as follows:
$$\begin{aligned} &\sum_{ \vert n \vert =1}^{\infty}\sum _{ \vert m \vert =1}^{\infty}\frac{a_{m}b_{n}}{ ( \vert m-\xi \vert + \vert n-\eta \vert ) ^{\lambda}} \\ &\quad < 2B ( \lambda _{1},\lambda_{2} ) \Biggl[ \sum _{ \vert m \vert =1}^{\infty} \vert m-\xi \vert ^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{ \vert n \vert =1}^{\infty} \vert n-\eta \vert ^{q(1-\lambda_{2})-1}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned}$$
(4)
where the constant factor \(2B ( \lambda_{1},\lambda_{2} ) \) (\(0<\lambda_{1},\lambda_{2}\leq1\), \(\lambda_{1}+\lambda_{2}=\lambda \), \(\xi , \eta\in{[} 0,\frac{1}{2}]\)) is the best possible. Another result was provided by Xin et al. [26].

In this paper, by introducing multi-parameters, applying the weight coefficients and Hermite–Hadamard’s inequality, we give a reverse of the extension of Mulholland’s inequality (2) in the whole plane with the best possible constant factor similar to the reverse of (4). The equivalent forms and a few particular cases are also considered.

2 Some lemmas

In the following, we agree that \(0< p<1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(\lambda_{1},\lambda_{2}>0\), \(\lambda_{1}+\lambda_{2}=\lambda\leq 1\), \(\alpha,\beta\in{[} \arccos\frac{1}{3},\frac{\pi}{2}]\), \(\xi ,\eta \in{[} 0,\frac{1}{2}]\), and
$$ H_{\gamma}(\lambda_{1}):=\frac{2\pi\csc^{2}\gamma}{\lambda\sin (\frac{\pi\lambda_{1}}{\lambda})}\quad (\gamma= \alpha,\beta). $$
(5)

Remark 1

In view of the conditions that \(\alpha,\beta\in{[} \arccos\frac{1}{3},\frac{\pi}{2}]\), \(\xi,\eta\in{[} 0,\frac{1}{2}]\), it follows that \((\frac{3}{2}\pm\eta)(1\mp\cos\beta)\geq1\) and \((\frac{3}{2}\pm\xi)(1\mp\cos\alpha)\geq1\).

We set the following functions:
$$ A_{\alpha}(\xi,x)= \vert x-\xi \vert +(x-\xi)\cos\alpha, $$
\(B_{\beta}(\eta,y)= \vert y-\eta \vert +(y-\eta)\cos\beta\), and
$$ H(x,y):=\frac{1}{\ln^{\lambda}A_{\alpha}(\xi,x)+\ln^{\lambda }B_{\beta }(\eta,y)}\quad \biggl( \vert x \vert , \vert y \vert > \frac{3}{2} \biggr). $$
(6)

Definition 1

Define two weight coefficients as follows:
$$\begin{aligned} &\omega(\lambda_{2},m):=\sum_{ \vert n \vert =2}^{\infty} \frac{H(m,n)}{B_{\beta}(\eta,n)}\frac{\ln^{\lambda_{1}}A_{\alpha}(\xi ,m)}{\ln^{1-\lambda_{2}}B_{\beta}(\eta,n)}, \quad \vert m \vert \in \mathbf{N} \backslash \{1\}, \end{aligned}$$
(7)
$$\begin{aligned} &\varpi(\lambda_{1},n):=\sum_{ \vert m \vert =2}^{\infty} \frac{H(m,n)}{A_{\alpha}(\xi,m)}\frac{\ln^{\lambda_{2}}B_{\beta}(\eta ,n)}{\ln^{1-\lambda_{1}}A_{\alpha}(\xi,m)}, \quad \vert n \vert \in \mathbf{N} \backslash \{1\}, \end{aligned}$$
(8)
where \(\sum_{ \vert j \vert =2}^{\infty}\cdots =\sum_{j=-2}^{-\infty}\cdots+\sum_{j=2}^{\infty}\cdots\) (\(j=m,n\)).

Lemma 1

(cf. [26])

Suppose that\(g(t)\) (>0) is strictly decreasing in\((1,\infty)\), satisfying\(\int_{1}^{\infty}g(t)\,dt\in \mathbf{R}_{+}\). We have
$$ \int_{2}^{\infty}g(t)\,dt< \sum _{n=2}^{\infty}g(n)< \int_{1}^{\infty}g(t)\,dt. $$
(9)
If\((-1)^{i}g(t)>0\) (\(i=0,1,2\); \(t\in(\frac{3}{2},\infty)\)), \(\int_{\frac {3}{2}}^{\infty}g(t)\,dt\in\mathbf{R}_{+}\), then we have the following Hermite–Hadamard inequality (cf. [27]):
$$ \sum_{n=2}^{\infty}g(n)< \int_{\frac{3}{2}}^{\infty}g(t)\,dt. $$
(10)

Lemma 2

For\(0<\lambda\leq1\), \(0<\lambda_{2}<1\), the following inequalities are valid:
$$ H_{\beta}(\lambda_{1}) \bigl(1-\theta(\lambda _{2},m) \bigr)< \omega(\lambda_{2},m)< H_{\beta}( \lambda_{1}),\quad \vert m \vert \in\mathbf{N}\backslash \{1\}, $$
(11)
where
$$\begin{aligned} \theta(\lambda_{2},m) :=&\frac{\lambda}{\pi}\sin \biggl( \frac{\pi\lambda_{1}}{\lambda} \biggr) \int_{0}^{\frac{\ln[(2+\eta)(1+\cos\beta)]}{\ln A_{\alpha }(\xi,m)}}\frac{u^{\lambda_{2}-1}}{1+u^{\lambda}}\,du \\ =&O \biggl( \frac{1}{\ln^{\lambda_{2}}A_{\alpha}(\xi,m)} \biggr) \in(0,1). \end{aligned}$$
(12)

Proof

For \(\vert m \vert \in\mathbf{N}\backslash\{ 1\}\), we set the following functions:
$$\begin{aligned} &H^{(1)}(m,y) :=\frac{1}{\ln^{\lambda}A_{\alpha}(\xi,m)+\ln ^{\lambda }[(y-\eta)(\cos\beta-1)]},\quad y< -\frac{3}{2}, \\ &H^{(2)}(m,y) :=\frac{1}{\ln^{\lambda}A_{\alpha}(\xi,m)+\ln ^{\lambda }[(y-\eta)(\cos\beta+1)]},\quad y>\frac{3}{2}, \end{aligned}$$
wherefrom
$$ H^{(1)}(m,-y)=\frac{1}{\ln^{\lambda}A_{\alpha}(\xi,m)+\ln ^{\lambda }[(y+\eta)(1-\cos\beta)]},\quad y>\frac{3}{2}. $$
We find
$$\begin{aligned} \omega(\lambda_{2},m) =&\sum_{n=-2}^{-\infty} \frac{H^{(1)}(m,n)\ln ^{\lambda_{1}}A_{\alpha}(\xi,m)}{(n-\eta)(\cos\beta-1)\ln ^{1-\lambda _{2}} [ (n-\eta)(\cos\beta-1) ] } \\ &{}+\sum_{n=2}^{\infty}\frac{H^{(2)}(m,n)\ln^{\lambda_{1}}A_{\alpha }(\xi ,m)}{(n-\eta)(1+\cos\beta)\ln^{1-\lambda_{2}} [ (n-\eta)(1+\cos \beta) ] } \\ =&\frac{\ln^{\lambda_{1}}A_{\alpha}(\xi,m)}{1-\cos\beta}\sum_{n=2}^{\infty} \frac{H^{(1)}(m,-n)}{(n+\eta)\ln^{1-\lambda _{2}}[(n+\eta)(1-\cos\beta)]} \\ &{}+\frac{\ln^{\lambda_{1}}A_{\alpha}(\xi,m)}{1+\cos\beta}\sum_{n=2}^{\infty} \frac{H^{(2)}(m,n)}{(n-\eta)\ln^{1-\lambda _{2}}[(n-\eta)(1+\cos\beta)]}. \end{aligned}$$
(13)
In virtue of \(0<\lambda\leq1\), \(0<\lambda_{2}<1\), we find that for \(y>\frac{3}{2}\),
$$\begin{aligned} &\frac{d}{dy}\frac{H^{(i)}(m,(-1)^{i}y)}{(y-(-1)^{i}\eta)\ln ^{1-\lambda _{2}}[(y-(-1)^{i}\eta)(1+(-1)^{i}\cos\beta)]} < 0, \\ &\frac{d^{2}}{dy^{2}}\frac{H^{(i)}(m,(-1)^{i}y)}{(y-(-1)^{i}\eta)\ln ^{1-\lambda_{2}}[(y-(-1)^{i}\eta)(1+(-1)^{i}\cos\beta)]} >0\quad (i=1.2), \end{aligned}$$
it follows that \(\frac{H^{(i)}(m,(-1)^{i}y)}{(y-(-1)^{i}\eta)\ln ^{1-\lambda_{2}}[(y-(-1)^{i}\eta)(1+(-1)^{i}\cos\beta)]}\) (\(i=1.2\)) are strictly decreasing and strictly convex in \((\frac{3}{2},\infty)\). By (10) and (13) we find
$$\begin{aligned} \omega(\lambda_{2},m) < &\frac{\ln^{\lambda_{1}}A_{\alpha}(\xi ,m)}{1-\cos\beta} \int_{3/2}^{\infty}\frac{H^{(1)}(m,-y)\,dy}{(y+\eta)\ln ^{1-\lambda_{2}}[(y+\eta)(1-\cos\beta)]} \\ &{}+\frac{\ln^{\lambda_{1}}A_{\alpha}(\xi,m)}{1+\cos\beta}\int_{3/2}^{\infty}\frac{H^{(2)}(m,y)\,dy}{(y-\eta)\ln^{1-\lambda _{2}}[(y-\eta)(1+\cos\beta)]}. \end{aligned}$$
Setting \(u=\frac{\ln[(y+\eta)(1-\cos\beta)]}{\ln A_{\alpha}(\xi ,m)}\) (\(u=\frac{\ln[(y-\eta)(1+\cos\beta)]}{\ln A_{\alpha}(\xi,m)}\)) in the above first (second) integral, in view of Remark 1, by simplifications, we obtain
$$\begin{aligned} \omega(\lambda_{2},m) < & \biggl( \frac{1}{1-\cos\beta}+ \frac{1}{1+\cos \beta} \biggr) \int_{0}^{\infty}\frac{u^{\lambda_{2}-1}}{1+u^{\lambda}}\,du \\ =&\frac{2\csc^{2}\beta}{\lambda} \int_{0}^{\infty}\frac{v^{(\lambda _{2}/\lambda)-1}\,dv}{1+v}= \frac{2\pi\csc^{2}\beta}{\lambda\sin(\frac{\pi\lambda_{1}}{\lambda})}=H_{\beta}(\lambda_{1}). \end{aligned}$$
By (9) and (13), in the same way, we still have
$$\begin{aligned} \omega(\lambda_{2},m) >&\frac{\ln^{\lambda_{1}}A_{\alpha}(\xi ,m)}{1-\cos\beta} \int_{2}^{\infty}\frac{H^{(1)}(m,-y)\,dy}{(y+\eta)\ln ^{1-\lambda_{2}}[(y+\eta)(1-\cos\beta)]} \\ &{}+\frac{\ln^{\lambda_{1}}A_{\alpha}(\xi,m)}{1+\cos\beta}\int_{2}^{\infty}\frac{H^{(2)}(m,y)\,dy}{(y-\eta)\ln^{1-\lambda _{2}}[(y-\eta)(1+\cos\beta)]} \\ \geq& \biggl( \frac{1}{1-\cos\beta}+\frac{1}{1+\cos\beta} \biggr) \int_{\frac{\ln[(2+\eta)(1+\cos\beta)]}{\ln A_{\alpha}(\xi ,m)}}^{\infty}\frac{u^{\lambda_{2}-1}}{1+u^{\lambda}}\,du \\ =&H_{\beta}(\lambda_{1})-2\csc^{2}\beta \int_{0}^{\frac{\ln[(2+\eta )(1+\cos\beta)]}{\ln A_{\alpha}(\xi,m)}}\frac{u^{\lambda_{2}-1}}{1+u^{\lambda}}\,du \\ =&H_{\beta}(\lambda_{1}) \bigl(1-\theta(\lambda _{2},m) \bigr)>0, \end{aligned}$$
where \(\theta(\lambda_{2},m)\) is indicated by (12). It follows that \(\theta(\lambda_{2},m)<1\) and
$$\begin{aligned} 0 < &\theta(\lambda_{2},m) \\ < &\frac{\lambda}{\pi}\sin \biggl( \frac{\pi\lambda _{1}}{\lambda} \biggr) \int_{0}^{\frac{\ln[(2+\eta)(1+\cos\beta)]}{\ln A_{\alpha}(\xi,m)}}u^{\lambda_{2}-1}\,du \\ =&\frac{\lambda}{\pi\lambda_{2}}\sin \biggl(\frac{\pi\lambda _{1}}{\lambda} \biggr) \biggl( \frac{\ln[(2+\eta)(1+\cos\beta)]}{\ln A_{\alpha}(\xi,m)} \biggr) ^{\lambda_{2}}. \end{aligned}$$
Hence, (11) and (12) are valid. □

In the same way, for \(0<\lambda\leq1\), \(0<\lambda_{1}<1\), we still have the following.

Lemma 3

The following inequalities are valid:
$$ H_{\alpha}(\lambda_{1}) \bigl(1-\widetilde{\theta}(\lambda _{1},n) \bigr)< \varpi(\lambda_{1},n)< H_{\alpha}( \lambda_{1}), \quad \vert n \vert \in\mathbf{N}\backslash\{1\}, $$
(14)
where
$$\begin{aligned} \widetilde{\theta}(\lambda_{1},n) :=&\frac{\lambda}{\pi}\sin \biggl( \frac{\pi \lambda_{1}}{\lambda} \biggr) \int_{0}^{\frac{\ln[(2+\xi)(1+\cos\alpha)]}{\ln B_{\beta}(\eta,n)}}\frac{u^{\lambda_{1}-1}}{1+u^{\lambda}}\,du \\ =&O \biggl( \frac{1}{\ln^{\lambda_{1}}B_{\beta}(\eta,n)} \biggr) \in(0,1). \end{aligned}$$
(15)

Lemma 4

If\((\zeta,\gamma)=(\xi,\alpha)\) (or\((\eta ,\beta)\)), \(\rho>0\), then we have
$$\begin{aligned} h_{\rho}(\zeta,\gamma) :=&\sum_{ \vert n \vert =2}^{\infty} \frac{\ln^{-1-\rho} [ \vert n-\zeta \vert +(n-\zeta)\cos\gamma ] }{ [ \vert n-\zeta \vert +(n-\zeta)\cos\gamma ] } \\ =&\frac{1}{\rho} \bigl(2\csc^{2}\gamma+o(1) \bigr)\quad \bigl( \rho\rightarrow0^{+} \bigr) . \end{aligned}$$
(16)

Proof

Still by (10) we find
$$\begin{aligned} h_{\rho}(\zeta,\gamma) =&\sum_{n=-2}^{-\infty} \frac{\ln^{-1-\rho }[(n-\zeta)(\cos\gamma-1)]}{(n-\zeta)(\cos\gamma-1)} \\ &{}+\sum_{n=2}^{\infty}\frac{\ln^{-1-\rho}[(n-\zeta)(\cos\gamma +1)]}{(n-\zeta)(\cos\gamma+1)} \\ =&\sum_{n=2}^{\infty} \biggl\{ \frac{\ln^{-1-\rho}[(n+\zeta)(1-\cos \gamma)]}{(n+\zeta)(1-\cos\gamma)}+\frac{\ln^{-1-\rho}[(n-\zeta )(\cos \gamma+1)]}{(n-\zeta)(\cos\gamma+1)} \biggr\} \\ \leq& \int_{\frac{3}{2}}^{\infty} \biggl\{ \frac{\ln^{-1-\rho }[(y+\zeta )(1-\cos\gamma)]}{(y+\zeta)(1-\cos\gamma)}+ \frac{\ln^{-1-\rho }[(y-\zeta)(\cos\gamma+1)]}{(y-\zeta)(\cos\gamma+1)} \biggr\} \,dy \\ =&\frac{1}{\rho} \biggl\{ \frac{\ln^{-\rho}[(\frac{3}{2}+\zeta )(1-\cos \gamma)]}{1-\cos\gamma}+\frac{\ln^{-\rho}[(\frac{3}{2}-\zeta )(1+\cos \gamma)]}{1+\cos\gamma} \biggr\} \\ =&\frac{1}{\rho} \biggl\{ \frac{1}{1-\cos\gamma}+\frac {1}{1+\cos\gamma}+o_{1}(1) \biggr\} \quad \bigl(\rho\rightarrow0^{+} \bigr). \end{aligned}$$
By (9), we still can find that
$$\begin{aligned} &h_{\rho}(\zeta,\gamma) \\ &\quad =\sum_{n=2}^{\infty} \biggl\{ \frac{\ln^{-1-\rho}[(n+\zeta)(1-\cos \gamma)]}{(n+\zeta)(1-\cos\gamma)}+\frac{\ln^{-1-\rho}[(n-\zeta )(\cos \gamma+1)]}{(n-\zeta)(\cos\gamma+1)} \biggr\} \\ &\quad \geq \int_{2}^{\infty} \biggl\{ \frac{\ln^{-1-\rho}[(y+\zeta)(1-\cos \gamma)]}{(y+\zeta)(1-\cos\gamma)}+ \frac{\ln^{-1-\rho}[(y-\zeta)(\cos \gamma+1)]}{(y-\zeta)(\cos\gamma+1)} \biggr\} \,dy \\ &\quad = \frac{1}{\rho} \biggl\{ \frac{\ln^{-\rho}[(2+\zeta)(1-\cos \gamma)]}{1-\cos\gamma}+\frac{\ln^{-\rho}[(2-\zeta)(1+\cos\gamma )]}{1+\cos \gamma} \biggr\} \\ &\quad = \frac{1}{\rho} \biggl\{ \frac{1}{1-\cos\gamma}+\frac {1}{1+\cos\gamma}+o_{2}(1) \biggr\} \quad \bigl(\rho\rightarrow0^{+} \bigr). \end{aligned}$$
Hence, we prove that (16) is valid. □

3 Main results and a few particular cases

We also define
$$ H(\lambda_{1}):=H_{\beta}^{1/p}(\lambda _{1})H_{\alpha}^{1/q}(\lambda_{1})= \frac{2\pi\csc^{2/p}\beta\csc^{2/q}\alpha}{\lambda\sin(\frac{\pi\lambda_{1}}{\lambda})}. $$
(17)

Theorem 1

Suppose that\(a_{m},b_{n}\geq0\) (\(\vert m \vert , \vert n \vert \in\mathbf{N}\backslash\{1\}\)) and
$$\begin{aligned} &0 < \sum_{ \vert m \vert =2}^{\infty}\frac{\ln^{p(1-\lambda _{1})-1}A_{\alpha}(\xi,m)}{(A_{\alpha}(\xi,m))^{1-p}}a_{m}^{p}< \infty, \\ &0 < \sum_{ \vert n \vert =2}^{\infty}\frac{\ln^{q(1-\lambda _{2})-1}B_{\beta}(\eta,n)}{(B_{\beta}(\eta,n))^{1-q}}b_{n}^{q}< \infty. \end{aligned}$$
We have the following reverse equivalent inequalities:
$$\begin{aligned} &\begin{aligned}[b] I:={}&\sum_{ \vert n \vert =2}^{\infty} \sum_{ \vert m \vert =2}^{\infty}\frac{1}{\ln^{\lambda}A_{\alpha}(\xi,m)+\ln^{\lambda }B_{\beta}(\eta,n)}a_{m}b_{n} \\ >{}&H(\lambda_{1}) \Biggl[ \sum_{ \vert m \vert =2}^{\infty } \bigl(1-\theta(\lambda_{2},m) \bigr)\frac{\ln^{p(1-\lambda _{1})-1}A_{\alpha}(\xi ,m)}{(A_{\alpha}(\xi,m))^{1-p}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ &{}\times \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln ^{q(1-\lambda_{2})-1}B_{\beta}(\eta,n)}{(B_{\beta}(\eta,n))^{1-q}}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned} \end{aligned}$$
(18)
$$\begin{aligned} &\begin{aligned}[b] J :={}& \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln^{p\lambda _{2}-1}B_{\beta}(\eta,n)}{B_{\beta}(\eta,n)} \Biggl( \sum_{ \vert m \vert =2}^{\infty} \frac{a_{m}}{\ln^{\lambda}A_{\alpha}(\xi ,m)+\ln^{\lambda}B_{\beta}(\eta,n)} \Biggr) ^{p} \Biggr] ^{\frac {1}{p}} \\ >{}&H(\lambda_{1}) \Biggl[ \sum_{ \vert m \vert =2}^{\infty } \bigl(1-\theta(\lambda_{2},m) \bigr)\frac{\ln^{p(1-\lambda _{1})-1}A_{\alpha}(\xi ,m)}{(A_{\alpha}(\xi,m))^{1-p}}a_{m}^{p} \Biggr] ^{1/p}, \end{aligned} \end{aligned}$$
(19)
$$\begin{aligned} &\begin{aligned}[b] L :={}& \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \frac{\ln^{q\lambda _{1}-1}A_{\alpha}(\xi,m)}{(1-\theta(\lambda_{2},m))^{q-1}A_{\alpha }(\xi ,m)} \\ &{}\times \Biggl( \sum_{ \vert n \vert =2}^{\infty} \frac{1}{\ln ^{\lambda}A_{\alpha}(\xi,m)+\ln^{\lambda}B_{\beta}(\eta,n)}b_{n} \Biggr) ^{q} \Biggr] ^{\frac{1}{q}} \\ >{}&H(\lambda_{1}) \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln^{q(1-\lambda_{2})-1}B_{\beta}(\eta,n)}{(B_{\beta}(\eta ,n))^{1-q}}b_{n}^{q} \Biggr] ^{1/q}. \end{aligned} \end{aligned}$$
(20)
In particular, (i) for\(\alpha=\beta=\frac{\pi}{2}\), \(\xi,\eta\in {}[ 0,\frac{1}{2}]\), setting
$$\begin{aligned} \theta_{1}(\lambda_{2},m) :=&\frac{\lambda}{\pi}\sin \biggl(\frac{\pi\lambda _{1}}{\lambda} \biggr) \int_{0}^{\frac{\ln(2+\eta)}{\ln \vert m-\xi \vert }}\frac{u^{\lambda_{2}-1}}{1+u^{\lambda}}\,du \\ =&O \biggl( \frac{1}{\ln^{\lambda_{2}} \vert m-\xi \vert } \biggr) \in(0,1), \end{aligned}$$
(21)
we have the following reverse equivalent inequalities:
$$\begin{aligned} &\begin{aligned}[b] &\sum_{ \vert n \vert =2}^{\infty}\sum _{ \vert m \vert =2}^{\infty}\frac{1}{\ln^{\lambda} \vert m-\xi \vert +\ln ^{\lambda} \vert n-\eta \vert }a_{m}b_{n} \\ &\quad >\frac{2\pi}{\lambda\sin(\frac{\pi\lambda_{1}}{\lambda})} \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \bigl(1-\theta_{1}(\lambda_{2},m) \bigr)\frac{\ln^{p(1-\lambda_{1})-1} \vert m-\xi \vert }{ \vert m-\xi \vert ^{1-p}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ &\qquad {}\times \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln ^{q(1-\lambda_{2})-1} \vert n-\eta \vert }{ \vert n-\eta \vert ^{1-q}}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned} \end{aligned}$$
(22)
$$\begin{aligned} &\begin{aligned}[b] & \Biggl\{ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln^{p\lambda _{2}-1} \vert n-\eta \vert }{ \vert n-\eta \vert } \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \frac{a_{m}}{\ln^{\lambda } \vert m-\xi \vert +\ln^{\lambda} \vert n-\eta \vert } \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}} \\ &\quad >\frac{2\pi}{\lambda\sin(\frac{\pi\lambda_{1}}{\lambda})} \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \bigl(1-\theta_{1}(\lambda_{2},m) \bigr)\frac{\ln^{p(1-\lambda_{1})-1} \vert m-\xi \vert }{ \vert m-\xi \vert ^{1-p}}a_{m}^{p} \Biggr] ^{\frac{1}{p}}, \end{aligned} \end{aligned}$$
(23)
$$\begin{aligned} &\begin{aligned}[b] & \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \frac{\ln^{q\lambda _{1}-1} \vert m-\xi \vert }{(1-\theta_{1}(\lambda_{2},m))^{q-1} \vert m-\xi \vert } \Biggl( \sum_{ \vert n \vert =2}^{\infty} \frac{1}{\ln ^{\lambda} \vert m-\xi \vert +\ln^{\lambda} \vert n-\eta \vert }b_{n} \Biggr) ^{q} \Biggr] ^{\frac{1}{q}} \\ &\quad >\frac{2\pi}{\lambda\sin(\frac{\pi\lambda_{1}}{\lambda})} \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln^{q(1-\lambda _{2})-1} \vert n-\eta \vert }{ \vert n-\eta \vert ^{1-q}}b_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned} \end{aligned}$$
(24)
(ii) For\(\xi,\eta=0\), \(\alpha,\beta\in{[} \arccos\frac{1}{3},\frac{\pi}{2}]\), setting
$$\begin{aligned} \theta_{2}(\lambda_{2},m) :=&\frac{\lambda}{\pi}\sin \biggl(\frac{\pi\lambda _{1}}{\lambda} \biggr) \int_{0}^{\frac{\ln2(1+\cos\beta)}{\ln( \vert m \vert +m\cos \alpha)}}\frac{u^{\lambda_{2}-1}}{1+u^{\lambda}}\,du \\ =&O \biggl( \frac{1}{\ln^{\lambda_{2}}( \vert m \vert +m\cos \alpha)} \biggr) \in(0,1), \end{aligned}$$
(25)
we have the following equivalent inequalities:
$$\begin{aligned} &\begin{aligned}[b] &\sum_{ \vert n \vert =2}^{\infty}\sum _{ \vert m \vert =2}^{\infty}\frac{a_{m}b_{n}}{\ln^{\lambda}( \vert m \vert +m\cos\alpha)+\ln^{\lambda}( \vert n \vert +n\cos\beta)} \\ &\quad >H(\lambda_{1}) \Biggl[ \sum_{ \vert m \vert =2}^{\infty } \bigl(1-\theta_{2}(\lambda_{2},m) \bigr) \frac{\ln^{p(1-\lambda_{1})-1}( \vert m \vert +m\cos\alpha)}{( \vert m \vert +m\cos\alpha)^{1-p}} a_{m}^{p} \Biggr] ^{1/p} \\ &\qquad {}\times \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln ^{q(1-\lambda_{2})-1}( \vert n \vert +n\cos\beta)}{( \vert n \vert +n\cos\beta)^{1-q}}b_{n}^{q} \Biggr] ^{1/q}, \end{aligned} \end{aligned}$$
(26)
$$\begin{aligned} &\begin{aligned}[b] & \Biggl\{ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln^{p\lambda _{2}-1}( \vert n \vert +n\cos\beta)}{ \vert n \vert +n\cos\beta} \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \frac{a_{m}}{\ln^{\lambda}( \vert m \vert +m\cos\alpha)+\ln^{\lambda }( \vert n \vert +n\cos\beta)} \Biggr] ^{p} \Biggr\} ^{\frac {1}{p}} \\ &\quad >H(\lambda_{1}) \Biggl[ \sum_{ \vert m \vert =2}^{\infty } \bigl(1-\theta_{2}(\lambda_{2},m) \bigr) \frac{\ln^{p(1-\lambda_{1})-1}( \vert m \vert +m\cos\alpha)}{( \vert m \vert +m\cos\alpha)^{1-p}} a_{m}^{p} \Biggr] ^{\frac{1}{p}}, \end{aligned} \end{aligned}$$
(27)
$$\begin{aligned} &\begin{aligned}[b] & \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \frac{\ln^{q\lambda _{1}-1}( \vert m \vert +m\cos\alpha)}{(1-\theta_{2}(\lambda _{2},m))^{q-1}( \vert m \vert +m\cos \alpha)} \\ &\qquad {}\times \Biggl( \sum_{ \vert n \vert =2}^{\infty} \frac{1}{\ln ^{\lambda}( \vert m \vert +m\cos\alpha)+\ln^{\lambda}( \vert n \vert +n\cos\beta)}b_{n} \Biggr) ^{q} \Biggr] ^{\frac{1}{q}} \\ &\quad > H(\lambda_{1}) \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln^{q(1-\lambda _{2})-1}( \vert n \vert +n\cos\beta)}{( \vert n \vert +n\cos \beta)^{1-q}}b_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned} \end{aligned}$$
(28)

Proof

By the reverse Hölder inequality with weight (cf. [28]) and (8), we find
$$\begin{aligned} & \Biggl( \sum_{ \vert m \vert =2}^{\infty}H(m,n)a_{m} \Biggr) ^{p} \\ &\quad = \Biggl\{ \sum_{ \vert m \vert =2}^{\infty}H(m,n) \biggl[ \frac{(A_{\alpha}(\xi,m))^{1/q}\ln^{(1-\lambda_{1})/q}A_{\alpha}(\xi ,m)}{\ln ^{(1-\lambda_{2})/p}B_{\beta}(\eta,n)}a_{m} \biggr] \\ &\qquad {}\times \biggl[ \frac{\ln^{(1-\lambda_{2})/p}B_{\beta}(\eta,n)}{(A_{\alpha}(\xi,m))^{1/q}\ln^{(1-\lambda_{1})/q}A_{\alpha}(\xi,m)} \biggr] \Biggr\} ^{p} \\ &\quad \geq \sum_{ \vert m \vert =2}^{\infty}H(m,n) \frac{(A_{\alpha }(\xi,m))^{p/q}\ln^{(1-\lambda_{1})p/q}A_{\alpha}(\xi,m)}{\ln ^{1-\lambda_{2}}B_{\beta}(\eta,n)}a_{m}^{p} \\ &\qquad {}\times \Biggl[ \sum_{ \vert m \vert =2}^{\infty}H(m,n) \frac{\ln ^{(1-\lambda_{2})q/p}B_{\beta}(\eta,n)}{A_{\alpha}(\xi,m)\ln ^{1-\lambda_{1}}A_{\alpha}(\xi,m)} \Biggr] ^{p-1} \\ &\quad =\frac{(\varpi(\lambda_{1},n))^{p-1}B_{\beta}(\eta,n)}{\ln ^{p\lambda _{2}-1}B_{\beta}(\eta,n)}\sum_{ \vert m \vert =2}^{\infty} \frac{H(m,n)(A_{\alpha}(\xi,m))^{\frac{p}{q}}\ln^{(1-\lambda_{1})\frac {p}{q}}A_{\alpha}(\xi,m)}{B_{\beta}(\eta,n)\ln^{1-\lambda_{2}}B_{\beta }(\eta,n)}a_{m}^{p}. \end{aligned}$$
By (14), in view of \(p-1<0\), it follows that
$$\begin{aligned} J >&H_{\alpha}^{1/q}(\lambda_{1}) \Biggl[ \sum _{ \vert n \vert =2}^{\infty}\sum_{ \vert m \vert =2}^{\infty} \frac{H(m,n)(A_{\alpha}(\xi,m))^{p/q}\ln^{(1-\lambda_{1})p/q}A_{\alpha }(\xi ,m)}{B_{\beta}(\eta,n)\ln^{1-\lambda_{2}}B_{\beta}(\eta ,n)}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =&H_{\alpha}^{1/q}(\lambda_{1}) \Biggl[ \sum _{ \vert m \vert =2}^{\infty}\sum_{ \vert n \vert =2}^{\infty} \frac{H(m,n)(A_{\alpha}(\xi,m))^{p/q}\ln^{(1-\lambda_{1})p/q}A_{\alpha }(\xi ,m)}{B_{\beta}(\eta,n)\ln^{1-\lambda_{2}}B_{\beta}(\eta ,n)}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =&H_{\alpha}^{1/q}(\lambda_{1}) \Biggl[ \sum _{ \vert m \vert =2}^{\infty}\omega(\lambda_{2},m) \frac{\ln^{p(1-\lambda _{1})-1}A_{\alpha}(\xi,m)}{(A_{\alpha}(\xi,m))^{1-p}}a_{m}^{p} \Biggr] ^{\frac{1}{p}}. \end{aligned}$$
(29)
By (11) and (17), we have (19).
Using the reverse Hölder inequality again, we have
$$\begin{aligned} I =&\sum_{ \vert n \vert =2}^{\infty} \Biggl[ \frac{\ln^{\lambda _{2}-(1/p)}B_{\beta}(\eta,n)}{(B_{\beta}(\eta,n))^{1/p}}\sum_{ \vert m \vert =2}^{\infty}H(m,n)a_{m} \Biggr] \biggl[ \frac{\ln ^{(1/p)-\lambda_{2}}B_{\beta}(\eta,n)}{(B_{\beta}(\eta ,n))^{-1/p}}b_{n} \biggr] \\ \geq&J \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln ^{q(1-\lambda_{2})-1}B_{\beta}(\eta,n)}{(B_{\beta}(\eta,n))^{1-q}}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned}$$
(30)
and then by (19) we have (18).
On the other hand, assuming that (18) is valid, we set
$$ b_{n}:=\frac{\ln^{p\lambda_{2}-1}B_{\beta}(\eta,n)}{B_{\beta }(\eta,n)} \Biggl( \sum _{ \vert m \vert =2}^{\infty}H(m,n)a_{m} \Biggr) ^{p-1},\quad \vert n \vert \in\mathbf{N}\backslash\{1\}, $$
and find
$$ J= \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln^{q(1-\lambda _{2})-1}B_{\beta}(\eta,n)}{(B_{\beta}(\eta,n))^{1-q}}b_{n}^{q} \Biggr] ^{\frac{1}{p}}. $$
By (29) it follows that \(J>0\). If \(J=\infty\), then (19) is trivially valid; if \(J<\infty\), then we have
$$\begin{aligned} &\begin{aligned} &\sum_{ \vert n \vert =2}^{\infty} \frac{\ln^{q(1-\lambda _{2})-1}B_{\beta}(\eta,n)}{(B_{\beta}(\eta,n))^{1-q}}b_{n}^{q} \\ &\quad =J^{p}=I \\ &\quad >H(\lambda_{1}) \Biggl[ \sum_{ \vert m \vert =2}^{\infty } \bigl(1-\theta(\lambda_{2},m) \bigr)\frac{\ln^{p(1-\lambda _{1})-1}A_{\alpha}(\xi ,m)}{(A_{\alpha}(\xi,m))^{1-p}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ &\qquad {}\times \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln ^{q(1-\lambda_{2})-1}B_{\beta}(\eta,n)}{(B_{\beta}(\eta,n))^{1-q}}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned} \\ &\begin{aligned} J &= \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln ^{q(1-\lambda_{2})-1}B_{\beta}(\eta,n)}{(B_{\beta}(\eta,n))^{1-q}}b_{n}^{q} \Biggr] ^{\frac{1}{p}} \\ &> H(\lambda_{1}) \Biggl[ \sum_{ \vert m \vert =2}^{\infty } \bigl(1-\theta(\lambda_{2},m) \bigr)\frac{\ln^{p(1-\lambda _{1})-1}A_{\alpha}(\xi ,m)}{(A_{\alpha}(\xi,m))^{1-p}}a_{m}^{p} \Biggr] ^{\frac{1}{p}}. \end{aligned} \end{aligned}$$
Hence (19) is valid, which is equivalent to (18).
We have proved that (18) is valid. Then we set
$$ a_{m}:= \frac{\ln^{q\lambda_{1}-1}A_{\alpha}(\xi,m)}{(1-\theta (\lambda _{2},m))^{q-1}A_{\alpha}(\xi,m)} \Biggl( \sum _{ \vert n \vert =2}^{\infty}H(m,n)b_{n} \Biggr) ^{q-1},\quad \vert m \vert \in\mathbf{N}\backslash\{ 1\}, $$
and find
$$ L= \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \bigl(1-\theta( \lambda_{2},m) \bigr)\frac{\ln^{p(1-\lambda_{1})-1}A_{\alpha}(\xi ,m)}{(A_{\alpha }(\xi,m))^{1-p}}a_{m}^{p} \Biggr] ^{\frac{1}{q}}. $$
If \(L=0\), then (20) is impossible, namely \(L>0\). If \(L=\infty \), then (20) is trivially valid; if \(L<\infty\), then we have
$$\begin{aligned} &\begin{aligned} &\sum_{ \vert m \vert =2}^{\infty} \bigl(1- \theta(\lambda _{2},m) \bigr)\frac{\ln^{p(1-\lambda_{1})-1}A_{\alpha}(\xi,m)}{(A_{\alpha}(\xi ,m))^{1-p}}a_{m}^{p} \\ &\quad =L^{q}=I \\ &\quad >H(\lambda_{1}) \Biggl[ \sum_{ \vert m \vert =2}^{\infty } \bigl(1-\theta(\lambda_{2},m) \bigr)\frac{\ln^{p(1-\lambda _{1})-1}A_{\alpha}(\xi ,m)}{(A_{\alpha}(\xi,m))^{1-p}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ &\qquad {}\times \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln ^{q(1-\lambda_{2})-1}B_{\beta}(\eta,n)}{(B_{\beta}(\eta,n))^{1-q}}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned} \\ &\begin{aligned} L &= \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \bigl(1-\theta (\lambda_{2},m) \bigr)\frac{\ln^{p(1-\lambda_{1})-1}A_{\alpha}(\xi ,m)}{(A_{\alpha }(\xi,m))^{1-p}}a_{m}^{p} \Biggr] ^{\frac{1}{q}} \\ &>H(\lambda_{1}) \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln^{q(1-\lambda_{2})-1}B_{\beta}(\eta,n)}{(B_{\beta}(\eta ,n))^{1-q}}b_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned} \end{aligned}$$
Hence, (20) is valid.
On the other hand, assuming that (20) is valid, using the reverse Hölder inequality, we have
$$\begin{aligned} I =&\sum_{ \vert m \vert =2}^{\infty} \biggl[ \frac{\ln ^{(1/q)-\lambda_{1}}A_{\alpha}(\xi,m)}{(1-\theta(\lambda _{2},m))^{-1/p}(A_{\alpha}(\xi,m))^{-1/q}}a_{m} \biggr] \\ &{}\times \Biggl[ \frac{\ln^{\lambda_{1}-(1/q)}A_{\alpha}(\xi,m)}{(1-\theta(\lambda_{2},m))^{1/p}(A_{\alpha}(\xi,m))^{1/q}}\sum _{ \vert n \vert =2}^{\infty}H(m,n)b_{n} \Biggr] \\ \geq& \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \bigl(1- \theta(\lambda_{2},m) \bigr)\frac{\ln^{p(1-\lambda _{1})-1}A_{\alpha}(\xi,m)}{(A_{\alpha }(\xi,m))^{1-p}}a_{m}^{p} \Biggr] ^{\frac{1}{p}}L, \end{aligned}$$
(31)
and then by (20) we have (18), which is equivalent to (20).

Therefore, inequalities (18), (19), and (20) are equivalent.

The theorem is proved. □

Theorem 2

With regards to the assumptions of Theorem 1, the constant factor\(H(\lambda_{1})\)in (18), (19), and (20) is the best possible.

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
$$\begin{aligned} &\widetilde{a}_{m}:=\frac{\ln^{\lambda_{1}-(\varepsilon /p)-1}A_{\alpha }(\xi,m)}{A_{\alpha}(\xi,m)}=\frac{\ln^{\widetilde{\lambda}_{1}-1}A_{\alpha}(\xi,m)}{A_{\alpha}(\xi,m)}\quad \bigl( \vert m \vert \in\mathbf{N}\backslash\{1\} \bigr), \\ &\widetilde{b}_{n}:=\frac{\ln^{\lambda_{2}-(\varepsilon /q)-1}B_{\beta }(\eta,n)}{B_{\beta}(\eta,n)}=\frac{\ln^{\widetilde{\lambda}_{2}-\varepsilon-1}B_{\beta}(\eta,n)}{B_{\beta}(\eta,n)}\quad \bigl( \vert n \vert \in\mathbf{N}\backslash\{ 1\} \bigr). \end{aligned}$$
By (16) and (14) we find
$$\begin{aligned} &\begin{aligned} \widetilde{I}_{1} := {}& \Biggl[ \sum _{ \vert m \vert =2}^{\infty } \bigl(1-\theta(\lambda_{2},m) \bigr)\frac{\ln^{p(1-\lambda _{1})-1}A_{\alpha}(\xi ,m)}{(A_{\alpha}(\xi,m))^{1-p}} \widetilde{a}_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ &{}\times \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln ^{q(1-\lambda_{2})-1}B_{\beta}(\eta,n)}{(B_{\beta}(\eta,n))^{1-q}}\widetilde{b}_{n}^{q} \Biggr] ^{\frac{1}{q}} \\ ={}& \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \frac{\ln ^{-1-\varepsilon}A_{\alpha}(\xi,m)}{A_{\alpha}(\xi,m)}-\sum_{ \vert m \vert =2}^{\infty} \frac{O(\ln^{-1-(\lambda_{2}+\varepsilon )}A_{\alpha}(\xi,m))}{A_{\alpha}(\xi,m)} \Biggr] ^{\frac{1}{p}} \\ &{}\times \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln ^{-1-\varepsilon}B_{\beta}(\eta,n)}{B_{\beta}(\eta,n)} \Biggr] ^{\frac{1}{q}} \\ ={}&\frac{1}{\varepsilon} \bigl(2\csc^{2}\alpha+o(1)-\varepsilon O(1) \bigr)^{1/p} \bigl(2\csc^{2}\beta+\widetilde{o}(1) \bigr)^{1/q}\quad \bigl(\varepsilon\rightarrow0^{+} \bigr), \end{aligned} \\ &\begin{aligned} \widetilde{I} &:=\sum_{ \vert m \vert =2}^{\infty } \sum_{ \vert n \vert =2}^{\infty}H(m,n)\widetilde{a}_{m}\widetilde{b}_{n} \\ &=\sum_{ \vert n \vert =2}^{\infty}\sum _{ \vert m \vert =2}^{\infty}H(m,n)\frac{\ln^{\widetilde{\lambda}_{1}-1}A_{\alpha }(\xi ,m)\ln^{\widetilde{\lambda}_{2}-\varepsilon-1}B_{\beta}(\eta,n)}{A_{\alpha}(\xi,m)B_{\beta}(\eta,n)} \\ &=\sum_{ \vert n \vert =2}^{\infty}\varpi(\widetilde{ \lambda}_{1},n)\frac{\ln^{-1-\varepsilon}B_{\beta}(\eta,n)}{B_{\beta }(\eta,n)}< H_{\alpha}( \widetilde{\lambda}_{1})\sum_{ \vert n \vert =2}^{\infty} \frac{\ln^{-1-\varepsilon}B_{\beta}(\eta,n)}{B_{\beta }(\eta,n)} \\ &=\frac{1}{\varepsilon}H_{\alpha} \biggl(\lambda_{1}- \frac{\varepsilon}{p} \biggr) \bigl(2\csc^{2}\beta+ \widetilde{o}(1) \bigr). \end{aligned} \end{aligned}$$
If there exists a positive number \(K\geq H(\lambda_{1})\) such that (18) is still valid when replacing \(H(\lambda_{1})\) by K, then, in particular, we have
$$ \varepsilon\widetilde{I}=\varepsilon\sum_{ \vert m \vert =2}^{\infty} \sum_{ \vert n \vert =2}^{\infty}H(m,n)\widetilde{a}_{m} \widetilde{b}_{n}>\varepsilon K\widetilde{I}_{1}. $$
In view of the above results, it follows that
$$\begin{aligned} &H_{\alpha} \biggl(\lambda_{1}-\frac{\varepsilon}{p} \biggr) \bigl(2\csc^{2}\beta+\widetilde{o}(1) \bigr) >K\cdot \bigl(2 \csc^{2}\alpha+o(1)-\varepsilon O(1) \bigr)^{1/p} \bigl(2 \csc^{2}\beta+\widetilde{o}(1) \bigr)^{1/q}, \end{aligned}$$
and then
$$ \frac{4\pi\csc^{2}\alpha}{\lambda\sin(\frac{\pi\lambda _{1}}{\lambda})}\csc^{2}\beta\geq2K\csc^{2/p}\alpha \csc^{2/q}\beta \quad \bigl(\varepsilon\rightarrow0^{+} \bigr), $$
namely
$$ H(\lambda_{1})=\frac{2\pi}{\lambda\sin(\frac{\pi\lambda _{1}}{\lambda})}\csc^{2/p}\beta\csc ^{2/q}\alpha\geq K. $$
Hence, \(K=H(\lambda_{1})\) is the best possible constant factor in (18).

The constant factor \(H(\lambda_{1})\) in (19) ((20)) is still the best possible. Otherwise we would reach a contradiction by (30) ((31)) that the constant factor in (18) is not the best possible. □

Remark 2

(i) For \(\xi=\eta=0\) in (22), setting
$$\begin{aligned} \widetilde{\theta}_{1}(\lambda_{2},m) :=& \frac{\lambda}{\pi}\sin \biggl(\frac{\pi\lambda_{1}}{\lambda} \biggr) \int_{0}^{\frac{\ln2}{\ln \vert m \vert }}\frac{u^{\lambda _{2}-1}}{1+u^{\lambda}}\,du \\ =&O \biggl( \frac{1}{\ln^{\lambda_{2}} \vert m \vert } \biggr) \in(0,1), \end{aligned}$$
we have the following new inequality with the best possible constant factor \(\frac{2\pi}{\lambda\sin(\frac{\pi\lambda_{1}}{\lambda})}\):
$$\begin{aligned} &\sum_{ \vert n \vert =2}^{\infty}\sum _{ \vert m \vert =2}^{\infty}\frac{a_{m}b_{n}}{\ln^{\lambda} \vert m \vert +\ln ^{\lambda} \vert n \vert } \\ &\quad >\frac{2\pi}{\lambda\sin(\frac{\pi\lambda_{1}}{\lambda})} \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \bigl(1-\widetilde{\theta}_{1}(\lambda_{2},m) \bigr)\frac{\ln^{p(1-\lambda_{1})-1} \vert m \vert }{ \vert m \vert ^{1-p}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ &\qquad {}\times \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln ^{q(1-\lambda_{2})-1} \vert n \vert }{ \vert n \vert ^{1-q}}b_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$
(32)
It follows that (20) ((22)) is an extension of (29).
(ii) If \(a_{-m}=a_{m}\) and \(b_{-n}=b_{n}\) (\(m,n\in\mathbf{N}\backslash\{1\}\)), setting
$$\begin{aligned} & \theta_{2}(\lambda_{2},m) :=\frac{\lambda}{\pi}\sin \biggl(\frac{\pi\lambda _{1}}{\lambda} \biggr) \int_{0}^{\frac{\ln(2+\eta)}{\ln(m-\xi)}}\frac{u^{\lambda_{2}-1}}{1+u^{\lambda}}\,du =O \biggl( \frac{1}{\ln^{\lambda_{2}}(m-\xi)} \biggr) \in(0,1), \\ & \theta_{3}(\lambda_{2},m) :=\frac{\lambda}{\pi}\sin \biggl(\frac{\pi\lambda _{1}}{\lambda} \biggr) \int_{0}^{\frac{\ln(2+\eta)}{\ln(m+\xi)}}\frac{u^{\lambda_{2}-1}}{1+u^{\lambda}}\,du =O \biggl( \frac{1}{\ln^{\lambda_{2}}(m+\xi)} \biggr) \in(0,1), \end{aligned}$$
then (22) reduces to
$$\begin{aligned} &\sum_{n=2}^{\infty}\sum _{m=2}^{\infty} \biggl[ \frac{1}{\ln^{\lambda }(m-\xi)+\ln^{\lambda}(n-\eta)}+ \frac{1}{\ln^{\lambda}(m-\xi)+\ln ^{\lambda}(n+\eta)} \\ &\qquad {} +\frac{1}{\ln^{\lambda}(m+\xi)+\ln^{\lambda}(n-\eta)}+\frac{1 }{\ln^{\lambda}(m+\xi)+\ln^{\lambda}(n+\eta)} \biggr] a_{m}b_{n} \\ &\quad >\frac{2\pi}{\lambda\sin(\pi\lambda_{1}/\lambda)} \Biggl\{ \sum_{m=2}^{\infty} \biggl[ \bigl(1-\theta_{2}(\lambda_{2},m) \bigr) \frac{\ln ^{p(1-\lambda_{1})-1}(m-\xi)}{(m-\xi)^{1-p}} \\ &\qquad {} + \bigl(1-\theta_{3}(\lambda_{2},m) \bigr) \frac{\ln^{p(1-\lambda _{1})-1}(m+\xi)}{(m+\xi)^{1-p}} \biggr] a_{m}^{p} \Biggr\} ^{\frac{1}{p}} \\ &\qquad {}\times \Biggl\{ \sum_{n=2}^{\infty} \biggl[ \frac{\ln^{q(1-\lambda _{2})-1}(n-\eta)}{(n-\eta)^{1-q}}+\frac{\ln^{q(1-\lambda _{2})-1}(n+\eta)}{(n+\eta)^{1-q}} \biggr] b_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned}$$
(33)
In particular, for \(\xi=\eta=0\), \(\lambda=1\), \(\lambda_{1}=\lambda _{2}=\frac{1}{2}\) in (33), setting
$$ \theta(m):=\frac{1}{\pi} \int_{0}^{\frac{\ln2}{\ln m}}\frac{u^{-1/2}}{1+u}\,du=O \biggl( \frac{1}{\ln^{1/2}m} \biggr) \in(0,1), $$
we have the following reverse Mulholland inequality (2) with the best possible constant π:
$$\begin{aligned} \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln mn} >&\pi \Biggl[ \sum_{m=2}^{\infty} \bigl(1-\theta(m) \bigr) \frac{\ln^{\frac{p}{2}-1}m}{m^{1-p}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ &{} \times \Biggl( \sum_{n=2}^{\infty} \frac{\ln^{\frac{q}{2}-1}n}{n^{1-q}}b_{n}^{q} \Biggr) ^{\frac{1}{q}}. \end{aligned}$$
(34)

4 Conclusions

In this paper, by introducing multi-parameters, applying the weight coefficients, and using Hermite–Hadamard’s inequality, we give a reverse of the extension of Mulholland’s inequality in the whole plane with the best possible constant factor in Theorems 12. The equivalent forms and a few particular cases are considered. The technique of real analysis is very important as it is the key to proving the reverse equivalent inequalities with the best possible constant factor. The lemmas and theorems provide an extensive account of this type of inequalities.

Notes

Acknowledgements

This work is supported by the National Natural Science Foundation (Nos. 61370186, 61640222), and the Science and Technology Planning Project Item of Guangzhou City (No. 201707010229). We are grateful for this help.

Authors’ contributions

BY carried out the mathematical studies, participated in the sequence alignment, and drafted the manuscript. AW participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

References

  1. 1.
    Hardy, G.H., Littlewood, J.E., Pólya, G.: Inequalities. Cambridge University Press, Cambridge (1934) MATHGoogle Scholar
  2. 2.
    Mitrinović, D.S., Pečarić, J.E., Fink, A.M.: Inequalities Involving Functions and Their Integrals and Derivatives. Kluwer Academic, Boston (1991) CrossRefMATHGoogle Scholar
  3. 3.
    Yang, B.: A new Hilbert’s type integral inequality. Soochow J. Math. 33(4), 849–859 (2007) MathSciNetMATHGoogle Scholar
  4. 4.
    Gao, M., Yang, B.: On the extended Hilbert’s inequality. Proc. Am. Math. Soc. 126(3), 751–759 (1998) MathSciNetCrossRefMATHGoogle Scholar
  5. 5.
    Hong, Y.: All-sided generalization about Hardy–Hilbert integral inequalities. Acta Math. Sin. 44(4), 619–626 (2001) MathSciNetMATHGoogle Scholar
  6. 6.
    Laith, E.A.: On some extensions of Hardy–Hilbert’s inequality and applications. J. Inequal. Appl. 2008, Article ID 546828 (2008) MathSciNetGoogle Scholar
  7. 7.
    Krnić, M., Pečarić, J.E.: General Hilbert’s and Hardy’s inequalities. Math. Inequal. Appl. 8(1), 29–51 (2005) MathSciNetMATHGoogle Scholar
  8. 8.
    Perić, I., Vuković, P.: Multiple Hilbert’s type inequalities with a homogeneous kernel. Banach J. Math. Anal. 5(2), 33–43 (2011) MathSciNetCrossRefMATHGoogle Scholar
  9. 9.
    Rassias, M.T., Yang, B.: A Hilbert-type integral inequality in the whole plane related to the hyper geometric function and the beta function. J. Math. Anal. Appl. 428(2), 1286–1308 (2015) MathSciNetCrossRefMATHGoogle Scholar
  10. 10.
    Vandanjav, A., Tserendorj, B., Krnić, M.: Multiple Hilbert-type inequalities involving some differential operators. Banach J. Math. Anal. 10(2), 320–337 (2016) MathSciNetCrossRefMATHGoogle Scholar
  11. 11.
    Yang, B.: On a more accurate Hardy–Hilbert’s type inequality and its applications. Acta Math. Sin. 49(2), 363–368 (2006) MathSciNetMATHGoogle Scholar
  12. 12.
    Li, Y., He, B.: On inequalities of Hilbert’s type. Bull. Aust. Math. Soc. 76(1), 1–13 (2007) CrossRefMATHGoogle Scholar
  13. 13.
    Zhong, W.: A Hilbert-type linear operator with the norm and its applications. J. Inequal. Appl. 2009, Article ID 494257 (2009) MathSciNetCrossRefMATHGoogle Scholar
  14. 14.
    Huang, Q.: On a multiple Hilbert’s inequality with parameters. J. Inequal. Appl. 2010, Article ID 309319 (2010) MathSciNetCrossRefMATHGoogle Scholar
  15. 15.
    Krnić, M., Vukovic, P.: On a multidimensional version of the Hilbert type inequality. Anal. Math. 38(4), 291–303 (2012) MathSciNetCrossRefMATHGoogle Scholar
  16. 16.
    Huang, Q.: A new extension of Hardy–Hilbert-type inequality. J. Inequal. Appl. 2015, 397 (2015) MathSciNetCrossRefMATHGoogle Scholar
  17. 17.
    Huang, Q., Yang, B.: A more accurate half-discrete Hilbert inequality with a nonhomogeneous kernel. J. Funct. Spaces Appl. 2013, Article ID 628250 (2013) MathSciNetMATHGoogle Scholar
  18. 18.
    Huang, Q., Wang, A., Yang, B.: A more accurate half-discrete Hilbert-type inequality with a general nonhomogeneous kernel and operator expressions. Math. Inequal. Appl. 17(1), 367–388 (2014) MathSciNetMATHGoogle Scholar
  19. 19.
    Huang, Q., Wu, S., Yang, B.: Parameterized Hilbert-type integral inequalities in the whole plane. Sci. World J. 2014, Article ID 169061 (2014) Google Scholar
  20. 20.
    Rassias, M.T., Yang, B.: On a multidimensional Hilbert-type integral inequality associated to the gamma function. Appl. Math. Comput. 249, 408–418 (2014) MathSciNetMATHGoogle Scholar
  21. 21.
    Yang, B., Chen, Q.: A more accurate multidimensional Hardy–Hilbert type inequality with a general homogeneous kernel. J. Math. Inequal. 12(1), 113–128 (2018) Google Scholar
  22. 22.
    Batbold, T., Sawano, Y.: Sharp bounds for m-linear Hilbert-type operators on the weighted Morrey spaces. Math. Inequal. Appl. 20, 263–283 (2017) MathSciNetMATHGoogle Scholar
  23. 23.
    Garayev, M.T., Gurudal, M., Okudan, A.: Hardy–Hilbert’s inequality and power inequalities for Berezin numbers of operators. Math. Inequal. Appl. 19, 883–891 (2016) MathSciNetMATHGoogle Scholar
  24. 24.
    He, B., Wang, Q.: A multiple Hilbert-type discrete inequality with a new kernel and best possible constant factor. J. Math. Anal. Appl. 431(2), 889–902 (2015) MathSciNetCrossRefMATHGoogle Scholar
  25. 25.
    Yang, B., Chen, Q.: A new extension of Hardy–Hilbert’s inequality in the whole plane. J. Funct. Spaces 2016, Article ID 9197476 (2016) MathSciNetMATHGoogle Scholar
  26. 26.
    Xin, D., Yang, B., Chen, Q.: A discrete Hilbert-type inequality in the whole plane. J. Inequal. Appl. 2016, 133 (2016) MathSciNetCrossRefMATHGoogle Scholar
  27. 27.
    Chen, Q., Yang, B.: On a more accurate half-discrete Hardy–Hilbert-type inequality related to the kernel of arc tangent function. SpringerPlus 5, 1317 (2016) CrossRefGoogle Scholar
  28. 28.
    Kuang, J.: Applied Inequalities. Shangdong Science Technic Press, Jinan (2010) (in Chinesse) Google Scholar

Copyright information

© The Author(s) 2018

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors and Affiliations

  1. 1.Department of MathematicsGuangdong University of EducationGuangzhouP.R. China

Personalised recommendations