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On a new discrete Mulholland-type inequality in the whole plane

Journal of Inequalities and Applications20182018:184

https://doi.org/10.1186/s13660-018-1777-9

  • Received: 4 June 2018
  • Accepted: 17 July 2018
  • Published:

Abstract

A new discrete Mulholland-type inequality in the whole plane with a best possible constant factor is presented by introducing multi-parameters, applying weight coefficients, and using Hermite–Hadamard’s inequality. Moreover, the equivalent forms, some particular cases, and the operator expressions are considered.

Keywords

  • Mulholland-type inequality
  • Parameter
  • Weight coefficient
  • Equivalent form
  • Operator expression

MSC

  • 26D15
  • 47A07

1 Introduction

Assume that \(p > 1\), \(\frac{1}{p} + \frac{1}{q} = 1\), \(a_{m},b_{n} \ge0\), \(0 < \sum_{m = 1}^{\infty} a_{m}^{p} < \infty\), and \(0 < \sum_{n = 1}^{\infty} b_{n}^{q} < \infty\), Hardy–Hilbert’s inequality is provided as follows (cf. [1]):
$$ \sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{a_{m}b_{n}}{m + n} < \frac{\pi}{\sin(\pi/p)} \Biggl( \sum_{m = 1}^{\infty} a_{m}^{p} \Biggr)^{\frac{1}{p}} \Biggl( \sum_{n = 1}^{\infty} b_{n}^{q} \Biggr)^{\frac{1}{q}}, $$
(1)
where \(\frac{\pi}{\sin(\pi/p)}\) is the best possible constant factor. By Theorem 343 in [1] (replacing \(\frac{a_{m}}{m} \) and \(\frac {b_{n}}{n} \) by \(a _{m}\) and \(b _{n}\), respectively), it yields the following Mulholland’s inequality:
$$ \sum_{n = 2}^{\infty} \sum _{m = 2}^{\infty} \frac{a_{m}b_{n}}{\ln mn} < \frac{\pi}{\sin(\pi/p)} \Biggl( \sum_{m = 2}^{\infty} \frac{a_{m}^{p}}{m} \Biggr)^{\frac{1}{p}} \Biggl( \sum_{n = 2}^{\infty} \frac{b_{n}^{q}}{n} \Biggr)^{\frac{1}{q}}. $$
(2)
Equations (1) and (2) are important inequalities in analysis and its applications (cf. [1, 2]).
In 2007, Yang [3] firstly provided the following Hilbert-type integral inequality in the whole plane:
$$ \int_{ - \infty}^{\infty} \int_{ - \infty}^{\infty} \frac{f(x)g(y)}{(1 + e^{x + y})^{\lambda}} \,dx\,dy < B\biggl( \frac{\lambda}{2},\frac{\lambda}{ 2}\biggr) \biggl( \int_{ - \infty}^{\infty} e^{ - \lambda x}f^{2}(x)\,dx \int_{ - \infty}^{\infty} e^{ - \lambda y}g^{2}(y)\,dy \biggr)^{\frac{1}{2}}, $$
(3)
where \(B(\frac{\lambda}{2},\frac{\lambda}{2})\) (\(\lambda> 0\)) is the best possible constant factor. Various extensions of (1)–(3) have been presented since then (cf. [415]).
Recently, Yang and Chen [16] presented an extension of (1) in the whole plane as follows:
$$ \begin{aligned}[b] &\sum_{ \vert n \vert = 1}^{\infty} \sum_{ \vert m \vert = 1}^{\infty} \frac{a_{m}b_{n}}{( \vert m - \xi \vert + \vert n - \eta \vert )^{\lambda}} \\&\quad< 2B(\lambda_{1},\lambda_{2}) \Biggl[ \sum _{ \vert m \vert = 1}^{\infty} \vert m - \xi \vert ^{p(1 - \lambda_{1}) - 1}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum _{ \vert n \vert = 1}^{\infty} \vert n - \eta \vert ^{q(1 - \lambda_{2}) - 1} b_{n}^{q} \Biggr]^{\frac{1}{q}}, \end{aligned} $$
(4)
where \(2B(\lambda_{1},\lambda_{2})\) (\(0 < \lambda_{1},\lambda_{2} \le 1\), \(\lambda_{1} + \lambda_{2} = \lambda\), \(\xi,\eta\in[0,\frac{1}{2}]\)) is the best possible constant factor. In addition, Yang et al. [17, 18] also carried out a few similar works.

In this paper, we present a new discrete Mulholland-type inequality in the whole plane with a best possible constant factor that is similar to that in (4) via introducing multi-parameters, applying weight coefficients, and using Hermite–Hadamard’s inequality. Moreover, the equivalent forms, some particular cases, and the operator expressions are considered.

2 An example and two lemmas

In what follows, we assume that \(0 < \lambda_{1},\lambda_{2} < 1\), \(\lambda_{1} + \lambda_{2} = \lambda\le1\), \(\xi,\eta\in[0,\frac{1}{2}]\), \(\alpha,\beta\in[\arccos\frac{1}{3},\frac{\pi}{2}] \), and
$$ k_{\gamma} (\lambda_{1}): = \frac{2\pi^{2}\csc^{2}\gamma}{ \lambda^{2}\sin^{2}(\frac{\pi\lambda_{1}}{\lambda} )}\quad(\gamma= \alpha,\beta). $$
(5)

Remark 1

In view of the assumptions that \(\xi,\eta\in [0,\frac{1}{2}]\), \(\alpha,\beta\in[\arccos\frac{1}{3},\frac{\pi}{2}] \), it follows that
$$\biggl(\frac{3}{2} \pm\eta\biggr) (1 \mp\cos\beta) \ge1\quad\mbox{and} \quad\biggl( \frac{3}{2} \pm \xi\biggr) (1 \mp\cos\alpha) \ge1. $$

Example 1

For \(u > 0 \), we set \(g(u): = \frac{\ln u}{u - 1}\) (\(u > 0\)), \(g(1): = \lim_{u \to1}g(u) = 1 \). Then we have \(g(u) > 0\), \(g'(u) < 0\), \(g''(u) > 0\) (\(u > 0\)). In fact, we find
$$g(u) = \frac{\ln[1 + (u - 1)]}{u - 1} = \sum_{k = 0}^{\infty} ( - 1)^{k}\frac{(u - 1)^{k}}{k + 1} = \sum_{k = 0}^{\infty} \frac{( - 1)^{k}k!}{k + 1} \frac{(u - 1)^{k}}{k!}\quad( - 1 < u - 1 \le1), $$
and then \(g^{(k)}(1) = \frac{( - 1)^{k}k!}{k + 1}\) (\(k = 0,1,2, \ldots\)). Hence, \(g^{(0)}(1) = g(1)\), \(g'(1) = - \frac{1}{2}\), \(g''(1) = \frac{2}{3} \). It is evident that \(g(u) > 0 \). We obtain \(g'(u) = \frac{h(u)}{u(u - 1)^{2}}\), \(h(u): = u - 1 - u\ln u \). Since
$$h'(u) = - \ln u > 0\quad(0 < u < 1);\qquad h'(u) < 0\quad(u > 1), $$
it follows that \(h_{\max} = h(1) = 0 \) and \(h(u) < 0\) (\(u \ne1\)). Then we have \(g'(u) < 0\) (\(u \ne1\)). In view of \(g'(1) = - \frac{1}{2} < 0 \), it follows that \(g'(u) < 0\) (\(u > 0\)). We find
$$g''(u) = \frac{J(u)}{u^{2}(u - 1)^{3}},\qquad J(u): = - (u - 1)^{2} - 2u(u - 1) + 2u^{2}\ln u, $$
\(J'(u) = - 4(u - 1) + 4u\ln u \), and
$$J''(u) = 4\ln u < 0 \quad(0 < u < 1);\qquad J''(u) > 0\quad(u > 1). $$
It follows that \(J'_{\min} = J'(1) = 0 \), \(J'(u) > 0\) (\(u \ne1\)) and \(J(u)\) is strictly increasing. In view of \(J(1) = 0 \), we have
$$J(u) < 0\quad(0 < u < 1);\qquad J(u) > 0\quad(u > 1), $$
and \(g''(u) > 0\) (\(u \ne1\)). Since \(g''(1) = \frac{2}{3} > 0 \), we find \(g''(u) > 0\) (\(u > 0\)).
For \(0 < \lambda\le1\), \(0 < \lambda_{2} < 1 \), setting \(G(u): = g(u^{\lambda} )u^{\lambda_{2} - 1}\) (\(u > 0\)), we still have \(G(u) > 0 \), \(G'(u) = \lambda g'(u^{\lambda} )u^{\lambda+ \lambda_{2} - 2} + (\lambda_{2} - 1)g(u^{\lambda} )u^{\lambda_{2} - 2} < 0 \), and
$$\begin{aligned} G''(u) = {}& \lambda^{2}g''\bigl(u^{\lambda} \bigr)u^{2\lambda+ \lambda_{2} - 3} + \lambda(\lambda+ \lambda_{2} - 2)g'\bigl(u^{\lambda} \bigr)u^{\lambda+ \lambda_{2} - 3} \\ &+ \lambda(\lambda_{2} - 1)g'\bigl(u^{\lambda} \bigr)u^{\lambda+ \lambda_{2} - 3} + (\lambda_{2} - 1) (\lambda_{2} - 2)g\bigl(u^{\lambda} \bigr)u^{\lambda_{2} - 3} > 0. \end{aligned} $$
We set \(F(x,y): = \frac{\ln(x/y)}{x^{\lambda} - y^{\lambda}} (\frac{y}{x})^{\lambda_{2} - 1}\) (\(x,y > 0\)). Since \(F(x,y) = \frac{1}{x^{\lambda}} G(\frac{y}{x})\), we have
$$F(x,y) > 0,\qquad\frac{\partial}{\partial y}F(x,y) < 0,\qquad\frac{\partial^{2}}{\partial y^{2}}F(x,y) > 0. $$
Hence, for \(x,y > 1 \), we still have
$$\frac{1}{y}F(\ln x,\ln y) > 0,\qquad\frac{\partial}{\partial y}\biggl( \frac {1}{y}F(\ln x,\ln y)\biggr) < 0,\qquad\frac{\partial^{2}}{\partial y^{2}}\biggl( \frac{1}{y}F(\ln x,\ln y)\biggr) > 0. $$

Lemma 1

(cf. [19])

If \(f(u) > 0\), \(f'(u) < 0\), \(f''(u) > 0\) (\(u > \frac{3}{2}\)) and \(\int_{\frac{3}{2}}^{\infty} f(u)\,du < \infty \), then we have the following Hermite–Hadamard’s inequality:
$$\int_{k}^{k + 1} f(u)\,du < f(k) < \int_{k - \frac{1}{2}}^{k + \frac{1}{2}} f(u)\,du \quad\bigl(k \in\mathbf{N} \setminus\{ 1\} \bigr), $$
and then
$$ \int_{2}^{\infty} f(u)\,du < \sum _{k = 2}^{\infty} f(k) < \int_{\frac{3}{2}}^{\infty} f(u)\,du. $$
(6)
For \(|x|,|y| \ge\frac{3}{2} \), let the functions
$$A_{\xi,\alpha} (x): = \vert x - \xi \vert + (x - \xi)\cos\alpha, $$
\(A_{\eta,\beta} (y) = |y - \eta| + (y - \eta)\cos\beta \), and
$$ k(x,y): = \frac{\ln(\ln A_{\xi,\alpha} (x)/\ln A_{\eta,\beta} (y))}{\ln^{\lambda} A_{\xi,\alpha} (x) - \ln^{\lambda} A_{\eta,\beta} (y)}. $$
(7)
We define two weight coefficients as follows:
$$\begin{aligned}& \omega(\lambda_{2},m): = \sum_{ \vert n \vert = 2}^{\infty} \frac{k(m,n)}{A_{\eta,\beta} (n)} \cdot \frac{\ln^{\lambda_{1}}A_{\xi,\alpha} (m)}{\ln^{1 - \lambda_{2}}A_{\eta,\beta} (n)}, \quad \vert m \vert \in\mathbf{N} \setminus\{ 1\}, \end{aligned}$$
(8)
$$\begin{aligned}& \varpi(\lambda_{1},n): = \sum_{ \vert m \vert = 2}^{\infty} \frac{k(m,n)}{A_{\xi,\alpha} (m)} \cdot \frac{\ln^{\lambda_{2}}A_{\eta,\beta} (n)}{\ln^{1 - \lambda_{1}}A_{\xi,\alpha} (m)}, \quad \vert n \vert \in\mathbf{N} \setminus\{ 1\}, \end{aligned}$$
(9)
where \(\sum_{|j| = 2}^{\infty} \cdots= \sum_{j = - 2}^{ - \infty} \cdots+ \sum_{j = 2}^{\infty} \cdots\) (\(j = m,n\)).

Lemma 2

The inequalities
$$ k_{\beta} (\lambda_{1}) \bigl(1 - \theta( \lambda_{2},m)\bigr) < \omega (\lambda_{2},m) < k_{\beta} (\lambda_{1}), \quad \vert m \vert \in\mathbf{N} \setminus \{ 1\} $$
(10)
are valid, where
$$\begin{aligned}[b] \theta(\lambda_{2},m)&: = \biggl[\frac{\lambda}{\pi} \sin\biggl( \frac{\pi \lambda_{1}}{\lambda} \biggr)\biggr]^{2} \int_{0}^{\frac{\ln[(2 + \eta)(1 + \cos \beta )]}{\ln A_{\xi,\alpha} (m)}} \frac{\ln u}{u^{\lambda} - 1} u^{\lambda _{2} - 1} \,du \\&= O\biggl(\frac{1}{\ln^{\lambda_{2}/2}A_{\xi,\alpha} (m)}\biggr) \in (0,1).\end{aligned} $$
(11)

Proof

For \(|m| \in\mathbf{N}\setminus\{ 1\} \), let
$$\begin{gathered} k^{(1)}(m,y): = \frac{\ln\ln A_{\xi,\alpha} (m) - \ln \ln [(y - \eta)(\cos\beta- 1)]}{\ln^{\lambda} A_{\xi,\alpha} (m) - \ln^{\lambda} [(y - \eta)(\cos\beta- 1)]},\quad y < - \frac{3}{2}, \\ k^{(2)}(m,y): = \frac{\ln\ln A_{\xi,\alpha} (m) - \ln\ln[(y - \eta )(\cos\beta+ 1)]}{\ln^{\lambda} A_{\xi,\alpha} (m) - \ln^{\lambda} [(y - \eta)(\cos\beta+ 1)]},\quad y > \frac{3}{2}. \end{gathered} $$
Then we have
$$k^{(1)}(m, - y) = \frac{\ln\ln A_{\xi,\alpha} (m) - \ln\ln[(y + \eta)(1 - \cos\beta)]}{\ln^{\lambda} A_{\xi,\alpha} (m) - \ln^{\lambda} [(y + \eta)(1 - \cos\beta)]},\quad y > \frac{3}{2}, $$
yields
$$ \begin{aligned}[b] \omega(\lambda_{2},m) = {}&\sum _{n = - 2}^{ - \infty} \frac{k^{(1)}(m,n)\ln^{\lambda_{1}}A_{\xi,\alpha} (m)}{(n - \eta)(\cos \beta- 1)\ln^{1 - \lambda_{2}}[(n - \eta)(\cos\beta- 1)]} \\ &+ \sum_{n = 2}^{\infty} \frac{k^{(2)}(m,n)\ln^{\lambda_{1}}A_{\xi ,\alpha} (m)}{(n - \eta)(1 + \cos\beta)\ln^{1 - \lambda_{2}}[(n - \eta)(1 + \cos \beta)]} \\ ={}& \frac{\ln^{\lambda_{1}}A_{\xi,\alpha} (m)}{1 - \cos\beta} \sum_{n = 2}^{\infty} \frac{k^{(1)}(m, - n)}{(n + \eta)\ln^{1 - \lambda_{2}}[(n + \eta)(1 - \cos\beta)]} \\ &+ \frac{\ln^{\lambda_{1}}A_{\xi,\alpha} (m)}{1 + \cos\beta} \sum_{n = 2}^{\infty} \frac{k^{(2)}(m,n)}{(n - \eta)\ln^{1 - \lambda_{2}}[(n - \eta)(1 + \cos\beta)]}. \end{aligned} $$
(12)
In virtue of \(0 < \lambda\le1\), \(0 < \lambda_{2} < 1 \), and Example 1, we find that for \(y > \frac{3}{2} \),
$$\begin{gathered} \frac{k^{(i)}(m,( - 1)^{i}y)}{(y - ( - 1)^{i}\eta)\ln^{1 - \lambda_{2}}[(y - ( - 1)^{i}\eta)(1 + ( - 1)^{i}\cos\beta)]} > 0, \\ \frac{d}{dy}\frac{k^{(i)}(m,( - 1)^{i}y)}{(y - ( - 1)^{i}\eta)\ln^{1 - \lambda_{2}}[(y - ( - 1)^{i}\eta)(1 + ( - 1)^{i}\cos\beta)]} < 0, \\ \frac{d^{2}}{dy^{2}}\frac{k^{(i)}(m,( - 1)^{i}y)}{(y - ( - 1)^{i}\eta )\ln^{1 - \lambda_{2}}[(y - ( - 1)^{i}\eta)(1 + ( - 1)^{i}\cos\beta )]} > 0\quad(i = 1,2), \end{gathered} $$
it follows that
$$\frac{k^{(i)}(m,( - 1)^{i}y)}{(y - ( - 1)^{i}\eta)\ln^{1 - \lambda_{2}}[(y - ( - 1)^{i}\eta)(1 + ( - 1)^{i}\cos\beta)]}\quad(i = 1,2) $$
are strictly decreasing and convex in \(( \frac{3}{2},\infty )\). Then, by (5), (12) yields
$$\begin{aligned} \omega(\lambda_{2},m) < {}& \frac{\ln^{\lambda_{1}}A_{\xi,\alpha} (m)}{1 - \cos\beta} \int_{\frac{3}{2}}^{\infty} \frac{k^{(1)}(m, - y)}{(y + \eta)\ln^{1 - \lambda_{2}}[(y + \eta)(1 - \cos\beta)]} \,dy \\ &+ \frac{\ln^{\lambda_{1}}A_{\xi,\alpha} (m)}{1 + \cos\beta} \int_{\frac{3}{2}}^{\infty} \frac{k^{(2)}(m,y)}{(y - \eta)\ln^{1 - \lambda_{2}}[(y - \eta)(1 + \cos\beta)]} \,dy. \end{aligned} $$
Setting \(u = \frac{\ln[(y + \eta)(1 - \cos\beta)]}{\ln A_{\xi,\alpha} (m)}\) (\(u = \frac{\ln[(y - \eta)(1 + \cos\beta)]}{\ln A_{\xi,\alpha} (m)}\)) in the above first (second) integral, in view of Remark 1, we obtain
$$\begin{aligned} \omega(\lambda_{2},m) &< \biggl( \frac{1}{1 - \cos\beta} + \frac{1}{1 + \cos\beta} \biggr) \int_{0}^{\infty} \frac{\ln u}{u^{\lambda} - 1} u^{\lambda_{2} - 1} \,du \\ &= \frac{2\csc^{2}\beta}{\lambda^{2}} \int_{0}^{\infty} \frac{\ln v}{v - 1} v^{(\lambda_{2}/\lambda) - 1}\,dv = \frac{2\pi^{2}\csc^{2}\beta}{ \lambda^{2}\sin^{2}(\frac{\pi\lambda_{1}}{\lambda} )} = k_{\beta} (\lambda_{1}) \end{aligned} $$
by simplifications. Similarly, by (5), (12) also yields
$$\begin{aligned} \omega(\lambda_{2},m) >{}& \frac{\ln^{\lambda_{1}}A_{\xi,\alpha} (m)}{1 - \cos\beta} \int_{2}^{\infty} \frac{k^{(1)}(m, - y)}{(y + \eta)\ln^{1 - \lambda_{2}}[(y + \eta)(1 - \cos\beta)]} \,dy \\ &+ \frac{\ln^{\lambda_{1}}A_{\xi,\alpha} (m)}{1 + \cos\beta} \int_{2}^{\infty} \frac{k^{(2)}(m,y)}{(y - \eta)\ln^{1 - \lambda _{2}}[(y - \eta)(1 + \cos\beta)]} \,dy \\ \ge{}& \biggl(\frac{1}{1 - \cos\beta} + \frac{1}{1 + \cos\beta} \biggr) \int_{\frac{\ln[(2 + \eta)(1 + \cos\beta)]}{\ln A_{\xi,\alpha} (m)}}^{\infty} \frac{\ln u}{u^{\lambda} - 1} u^{\lambda_{2} - 1} \,du \\ ={}& k_{\beta} (\lambda_{1}) - 2\csc^{2}\beta \int_{0}^{\frac{\ln[(2 + \eta )(1 + \cos\beta)]}{\ln A_{\xi,\alpha} (m)}} \frac{\ln u}{u^{\lambda} - 1}u^{\lambda_{2} - 1} \,du \\ ={}& k_{\beta} (\lambda_{1}) \bigl(1 - \theta( \lambda_{2},m)\bigr) > 0, \end{aligned} $$
where \(\theta(\lambda_{2},m)\) (<1) is indicated by (11). Since
$$\frac{\ln u}{u^{\lambda} - 1}u^{\lambda_{2}/2} \to0\quad\bigl(u \to0^{ +} \bigr);\qquad \frac{\ln u}{u^{\lambda} - 1}u^{\lambda_{2}/2} \to\frac{1}{\lambda} \quad(u \to1), $$
there exists a positive constant C such that \(\frac{\ln u}{u^{\lambda} - 1}u^{\lambda_{2}/2} \le C\) (\(0 < u \le1\)), and then for \(A_{\xi,\alpha} (m) \ge(2 + \eta)(1 + \cos\beta)\), we have
$$ \begin{aligned}[b] 0 &< \theta(\lambda_{2},m) \le C\biggl[ \frac{\lambda}{\pi} \sin\biggl(\frac{\pi \lambda_{1}}{\lambda} \biggr)\biggr]^{2} \int_{0}^{\frac{\ln[(2 + \eta)(1 + \cos \beta )]}{\ln A_{\xi,\alpha} (m)}} u^{\frac{\lambda_{2}}{2} - 1} \,du \\ &= \frac{2C}{\lambda_{2}}\biggl[\frac{\lambda}{\pi} \sin\biggl(\frac{\pi \lambda_{1}}{\lambda} \biggr)\biggr]^{2}\biggl\{ \frac{\ln[(2 + \eta)(1 + \cos\beta )]}{\ln A_{\xi,\alpha} (m)}\biggr\} ^{\frac{\lambda_{2}}{2}}. \end{aligned} $$
(13)
Hence, (10) and (11) are valid. □

Similarly, we have the following.

Lemma 3

For \(0 < \lambda\le1\), \(0 < \lambda_{1} < 1 \), the inequalities
$$ k_{\alpha} (\lambda_{1}) \bigl(1 - \tilde{\theta} ( \lambda_{1},n)\bigr) < \varpi (\lambda_{1},n) < k_{\alpha} (\lambda_{1}), \quad \vert n \vert \in\mathbf{N} \setminus \{ 1\} $$
(14)
are valid, where
$$\begin{aligned}[b] \tilde{\theta} (\lambda_{1},n)&: = \biggl[\frac{\lambda}{\pi} \sin \biggl(\frac{\pi \lambda_{1}}{\lambda} \biggr)\biggr]^{2} \int_{0}^{\frac{\ln[(2 + \xi)(1 + \cos \alpha )]}{\ln A_{\eta,\beta} (n)}} \frac{\ln u}{u^{\lambda} - 1} u^{\lambda _{1} - 1} \,du\\& = O\biggl(\frac{1}{\ln^{\lambda_{1}/2}A_{\eta,\beta} (n)}\biggr) \in (0,1).\end{aligned} $$
(15)

Lemma 4

If \((\varsigma,\gamma) = (\xi,\alpha )\) (or \((\eta,\beta )\)), \(\rho> 0 \), then we have
$$ H_{\rho} (\varsigma,\gamma): = \sum_{|k| = 2}^{\infty} \frac{\ln^{ - 1 - \rho} A_{\varsigma,\gamma} (k)}{A_{\varsigma,\gamma} (k)} = \frac {1}{\rho} \bigl(2\csc^{2}\gamma+ o(1) \bigr) \quad\bigl(\rho\to0^{ +} \bigr). $$
(16)

Proof

According to (5), we obtain
$$\begin{aligned} H_{\rho} (\varsigma,\gamma) &= \sum _{k = - 2}^{ - \infty} \frac{\ln^{ - 1 - \rho} [(k - \varsigma)(\cos\gamma- 1)]}{(k - \varsigma )(\cos\gamma- 1)} + \sum _{k = 2}^{\infty} \frac{\ln^{ - 1 - \rho} [(k - \varsigma)(\cos\gamma+ 1)]}{(k - \varsigma)(\cos\gamma+ 1)} \\ &= \sum_{k = 2}^{\infty} \biggl\{ \frac{\ln^{ - 1 - \rho} [(k + \varsigma)(1 - \cos\gamma)]}{(k - \varsigma)(1 - \cos\gamma)} + \frac{\ln^{ - 1 - \rho} [(k - \varsigma)(\cos\gamma+ 1)]}{(k - \varsigma)(\cos\gamma+ 1)}\biggr\} \\ &< \int_{\frac{3}{2}}^{\infty} \biggl\{ \frac{\ln^{ - 1 - \rho} [(y + \varsigma )(1 - \cos\gamma)]}{(y - \varsigma)(1 - \cos\gamma)} + \frac{\ln^{ - 1 - \rho} [(y - \varsigma)(\cos\gamma+ 1)]}{(y - \varsigma)(\cos \gamma+ 1)}\biggr\} \,dy \\ & = \frac{1}{\rho} \biggl\{ \frac{\ln^{ - \rho} [(\frac{3}{2} + \varsigma)(1 - \cos\gamma)]}{1 - \cos\gamma} + \frac{\ln^{ - \rho} [(\frac{3}{2} - \varsigma)(1 + \cos\gamma)]}{1 + \cos\gamma} \biggr\} \\ &= \frac{1}{\rho} \biggl(\frac{1}{1 - \cos\gamma} + \frac{1}{1 + \cos\gamma } + o_{1}(1)\biggr) = \frac{1}{\rho} \bigl(2\csc^{2} \gamma+ o_{1}(1)\bigr)\quad \bigl(\rho\to0^{ +} \bigr), \end{aligned} $$
and
$$\begin{aligned} H_{\rho} (\varsigma,\gamma) &= \sum _{k = 2}^{\infty} \biggl\{ \frac{\ln^{ - 1 - \rho} [(k + \varsigma)(1 - \cos\gamma)]}{(k - \varsigma)(1 - \cos\gamma)} + \frac{\ln^{ - 1 - \rho} [(k - \varsigma )(\cos\gamma+ 1)]}{(k - \varsigma)(\cos\gamma+ 1)}\biggr\} \\ &> \int_{2}^{\infty} \biggl\{ \frac{\ln^{ - 1 - \rho} [(y + \varsigma)(1 - \cos \gamma)]}{(y - \varsigma)(1 - \cos\gamma)} + \frac{\ln^{ - 1 - \rho} [(y - \varsigma)(\cos\gamma+ 1)]}{(y - \varsigma)(\cos\gamma+ 1)}\biggr\} \,dy \\ &= \frac{1}{\rho} \biggl\{ \frac{\ln^{ - \rho} [(2 + \varsigma )(1 - \cos\gamma)]}{1 - \cos\gamma} + \frac{\ln^{ - \rho} [(2 - \varsigma)(1 + \cos\gamma)]}{1 + \cos\gamma} \biggr\} \\ &= \frac{1}{\rho} \biggl(\frac{1}{1 - \cos\gamma} + \frac{1}{1 + \cos\gamma } + o_{2}(1)\biggr) = \frac{1}{\rho} \bigl(2\csc^{2}\gamma+ o_{2}(1)\bigr) \quad\bigl(\rho\to0^{ +} \bigr). \end{aligned} $$
Therefore, (16) is valid. □

3 Main results

Theorem 1

Suppose that \(p > 1\), \(\frac{1}{p} + \frac{1}{q} = 1 \), we set
$$ k(\lambda_{1}): = k_{\beta}^{1/p}( \lambda_{1})k_{\alpha}^{1/q}(\lambda_{1}) = \frac{2\pi^{2}\csc^{2/p}\beta\csc^{2/q}\alpha}{[\lambda\sin(\frac {\pi \lambda_{1}}{\lambda} )]^{2}}. $$
(17)
If \(a_{m},b_{n} \ge0\) (\(|m|,|n| \in\mathbf{N}\setminus\{ 1\} \)) satisfy
$$0 < \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}} a_{m}^{p} < \infty ,\qquad0 < \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} < \infty, $$
then we obtain the following equivalent inequalities:
$$\begin{aligned}& \begin{aligned}[b] I&: = \sum_{|n| = 2}^{\infty} \sum_{|m| = 2}^{\infty} \frac{\ln(\ln A_{\xi,\alpha} (m)/\ln A_{\eta,\beta} (n))}{\ln^{\lambda} A_{\xi,\alpha} (m) - \ln^{\lambda} A_{\eta,\beta} (n)} a_{m}b_{n} \\ &< \frac{2\pi^{2}\csc^{2/p}\beta\csc^{2/q}\alpha}{[\lambda\sin(\frac {\pi \lambda_{1}}{\lambda} )]^{2}} \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}\\ &\quad{}\times \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}}, \end{aligned} \end{aligned}$$
(18)
$$\begin{aligned}& \begin{aligned}[b] J&: = \Biggl\{ \sum_{|n| = 2}^{\infty} \frac{\ln^{p\lambda_{2} - 1}A_{\eta,\beta} (n)}{A_{\eta,\beta} (n)} \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln(\ln A_{\xi,\alpha} (m)/\ln A_{\eta,\beta} (n))}{\ln^{\lambda} A_{\xi,\alpha} (m) - \ln^{\lambda} A_{\eta,\beta} (n)}a_{m} \Biggr]^{p} \Biggr\} ^{\frac{1}{p}} \\ &< \frac{2\pi^{2}\csc^{2/p}\beta\csc^{2/q}\alpha}{[\lambda\sin(\frac {\pi \lambda_{1}}{\lambda} )]^{2}} \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}. \end{aligned} \end{aligned}$$
(19)
Particularly, (i) for \(\alpha= \beta= \frac{\pi}{2}\), \(\xi,\eta\in [0,\frac{1}{2}] \), we have the following equivalent inequalities:
$$\begin{aligned}& \sum_{ \vert n \vert = 2}^{\infty} \sum _{ \vert m \vert = 2}^{\infty} \frac{\ln( \vert m - \xi \vert / \vert n - \eta \vert )a_{m}b_{n}}{\ln^{\lambda} \vert m - \xi \vert - \ln^{\lambda} \vert n - \eta \vert } \\& \quad< \frac{2\pi^{2}}{[\lambda\sin(\frac{\pi\lambda_{1}}{\lambda} )]^{2}} \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1} \vert m - \xi \vert }{ \vert m - \xi \vert ^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1} \vert n - \eta \vert }{ \vert n - \eta \vert ^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}}, \end{aligned}$$
(20)
$$\begin{aligned}& \Biggl[ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{p\lambda_{2} - 1} \vert n - \eta \vert }{ \vert n - \eta \vert } \Biggl( \sum_{ \vert m \vert = 2}^{\infty} \frac{\ln( \vert m - \xi \vert / \vert n - \eta \vert )a_{m}}{\ln^{\lambda} \vert m - \xi \vert - \ln^{\lambda} \vert n - \eta \vert } \Biggr)^{p} \Biggr]^{\frac{1}{p}} \\& \quad< \frac{2\pi^{2}}{[\lambda\sin(\frac{\pi \lambda_{1}}{\lambda} )]^{2}} \Biggl[ \sum _{ \vert m \vert = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1} \vert m - \xi \vert }{ \vert m - \xi \vert ^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}. \end{aligned}$$
(21)
(ii) For \(\xi= \eta= 0\), \(\alpha,\beta\in[\arccos\frac{1}{3},\frac{\pi}{ 2}] \), we have the following equivalent inequalities:
$$\begin{aligned}& \begin{aligned}[b] &\sum_{ \vert n \vert = 2}^{\infty} \sum_{ \vert m \vert = 2}^{\infty} \frac{\ln[\ln( \vert m \vert + m\cos \alpha)/\ln( \vert n \vert + n\cos\beta)]}{\ln^{\lambda} ( \vert m \vert + m\cos\alpha) - \ln^{\lambda} ( \vert n \vert + n\cos\beta)} a_{m}b_{n} \\ &\quad< \frac{2\pi^{2}\csc^{2/p}\beta\csc^{2/q}\alpha}{[\lambda\sin (\frac{\pi \lambda_{1}}{\lambda} )]^{2}}\\&\qquad{}\times \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}( \vert m \vert + m\cos\alpha)}{( \vert m \vert + m\cos\alpha)^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}( \vert n \vert + n\cos\beta)}{( \vert n \vert + n\cos \beta )^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}}, \end{aligned} \end{aligned}$$
(22)
$$\begin{aligned}& \begin{aligned}[b]& \Biggl\{ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{p\lambda _{2} - 1}( \vert n \vert + n\cos \beta)}{ \vert n \vert + n\cos\beta} \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \frac{\ln [\ln ( \vert m \vert + m\cos\alpha)/\ln( \vert n \vert + n\cos\beta)]}{\ln^{\lambda} ( \vert m \vert + m\cos \alpha) - \ln^{\lambda} ( \vert n \vert + n\cos\beta)}a_{m} \Biggr]^{p} \Biggr\} ^{\frac{1}{p}} \\ &\quad< \frac{2\pi^{2}\csc^{2/p}\beta\csc^{2/q}\alpha}{[\lambda\sin (\frac{\pi \lambda_{1}}{\lambda} )]^{2}} \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}( \vert m \vert + m\cos\alpha)}{( \vert m \vert + m\cos\alpha)^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}. \end{aligned} \end{aligned}$$
(23)

Proof

According to Hölder’s inequality with weight (cf. [20]) and (9), we find
$$\begin{gathered} \Biggl( \sum_{|m| = 2}^{\infty} k(m,n)a_{m} \Biggr)^{p} \\ \quad= \Biggl\{ \sum_{|m| = 2}^{\infty} k(m,n) \biggl[ \frac{(A_{\xi,\alpha} (m))^{\frac{1}{q}}\ln^{\frac{1 - \lambda_{1}}{q}}A_{\xi,\alpha} (m)}{\ln^{\frac{1 - \lambda_{2}}{p}}A_{\eta,\beta} (n)}a_{m} \biggr] \biggl[ \frac{\ln^{\frac{1 - \lambda_{2}}{p}}A_{\eta,\beta} (n)}{(A_{\xi,\alpha} (m))^{\frac{1}{q}}\ln^{\frac{1 - \lambda_{1}}{q}}A_{\xi,\alpha} (m)} \biggr] \Biggr\} ^{p} \\ \quad\le\sum_{|m| = 2}^{\infty} k(m,n) \frac{(A_{\xi,\alpha} (m))^{\frac{p}{q}}\ln^{\frac{(1 - \lambda_{1})p}{q}}A_{\xi,\alpha} (m)}{\ln^{1 - \lambda_{2}}A_{\eta,\beta} (n)}\\ \quad\quad{}\times a_{m}^{p} \Biggl[ \sum _{|m| = 2}^{\infty} k(m,n)\frac{\ln^{\frac{(1 - \lambda_{2})q}{p}}A_{\eta,\beta} (n)}{A_{\xi,\alpha} (m)\ln^{1 - \lambda_{1}}A_{\xi,\alpha} (m)} \Biggr]^{p - 1} \\ \quad= \frac{(\varpi(\lambda_{1},n))^{p - 1}A_{\eta,\beta} (n)}{\ln^{p\lambda_{2} - 1}A_{\eta,\beta} (n)}\sum_{|m| = 2}^{\infty} k(m,n)\frac{(A_{\xi,\alpha} (m))^{\frac{p}{q}}\ln^{\frac{(1 - \lambda_{1})p}{q}}A_{\xi,\alpha} (m)}{A_{\eta,\beta} (n)\ln^{1 - \lambda_{2}}A_{\eta,\beta} (n)}a_{m}^{p}. \end{gathered} $$
Then, by (14), it yields
$$ \begin{aligned}[b] J &< k_{\alpha}^{1/q}( \lambda_{1}) \Biggl[ \sum_{|n| = 2}^{\infty} \sum_{|m| = 2}^{\infty} k(m,n)\frac{(A_{\xi,\alpha} (m))^{\frac{p}{q}}\ln^{\frac{(1 - \lambda_{1})p}{q}}A_{\xi,\alpha} (m)}{A_{\eta,\beta} (n)\ln^{1 - \lambda_{2}}A_{\eta,\beta} (n)}a_{m}^{p} \Biggr]^{\frac{1}{p}} \\ &= k_{\alpha}^{1/q}(\lambda_{1}) \Biggl[ \sum _{|m| = 2}^{\infty} \sum _{|n| = 2}^{\infty} k(m,n)\frac{(A_{\xi,\alpha} (m))^{\frac{p}{q}}\ln^{\frac{(1 - \lambda_{1})p}{q}}A_{\xi,\alpha} (m)}{A_{\eta,\beta} (n)\ln^{1 - \lambda_{2}}A_{\eta,\beta} (n)}a_{m}^{p} \Biggr]^{\frac{1}{p}} \\ &= k_{\alpha}^{1/q}(\lambda_{1}) \Biggl[ \sum _{|m| = 2}^{\infty} \omega (\lambda_{2},m) \frac{n^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}. \end{aligned} $$
(24)
Combining (10) and (17), we obtain (19).
Using Hölder’s inequality again, we obtain
$$ \begin{aligned}[b] I &= \sum_{|n| = 2}^{\infty} \Biggl[ \frac{(A_{\eta,\beta} (n))^{\frac{ - 1}{p}}}{\ln^{\frac{1}{p} - \lambda_{2}}A_{\eta,\beta} (n)}\sum_{|m| = 2}^{\infty} k(m,n)a_{m} \Biggr] \biggl[ \frac{\ln^{\frac{1}{p} - \lambda_{2}}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{\frac{ - 1}{p}}}b_{n} \biggr] \\ &\le J \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} \Biggr]^{\frac{1}{q}}. \end{aligned} $$
(25)
Then, according to (19), we obtain (18).
On the other hand, assuming that (18) is valid, we let
$$b_{n}: = \frac{\ln^{p\lambda_{2} - 1}A_{\eta,\beta} (n)}{A_{\eta,\beta} (n)} \Biggl( \sum _{ \vert m \vert = 2}^{\infty} k(m,n)a_{m} \Biggr)^{p - 1},\quad \vert n \vert \in \mathbf{N}\setminus\{ 1\}. $$
According to (24), it follows that \(J < \infty \). If \(J = 0 \), then (20) is trivially valid; if \(J > 0 \), then we have
$$\begin{gathered} \begin{aligned} 0 &< \sum _{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} \\&= J^{p} = I \\ &< k(\lambda_{1}) \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}}, \end{aligned} \\ \begin{aligned}J &= \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} \Biggr]^{\frac{1}{p}} \\&< k( \lambda_{1}) \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}.\end{aligned} \end{gathered} $$
Thus (19) is valid, which is equivalent to (18). □

Theorem 2

With regards to the assumptions in Theorem 1, \(k(\lambda _{1})\) is the best possible constant factor in (18) and (19).

Proof

For \(0 < \varepsilon< \min\{ q(1 - \lambda_{1}),q\lambda_{2}\} \), we let \(\tilde{\lambda}_{1} = \lambda_{1} + \frac{\varepsilon}{q}\) (\(\in(0,1)\)), \(\tilde{\lambda}_{2} = \lambda_{2} - \frac{\varepsilon}{q}\) (\(\in(0,1)\)), and
$$\begin{gathered}\tilde{a}_{m}: = \frac{\ln^{\lambda_{1} - \frac{\varepsilon}{p} - 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} = \frac{\ln^{\tilde{\lambda}_{1} - \varepsilon- 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)}\quad\bigl( \vert m \vert \in\mathbf{N}\setminus\{ 1\} \bigr), \\ \tilde{b}_{n}: = \frac{\ln^{\lambda_{2} - \frac{\varepsilon}{q} - 1}A_{\eta,\beta} (n)}{A_{\eta,\beta} (n)} = \frac{\ln^{\tilde{\lambda}_{2} - 1}A_{\eta,\beta} (n)}{A_{\eta,\beta} (n)}\quad\bigl( \vert n \vert \in \mathbf{N}\setminus\{ 1\} \bigr). \end{gathered} $$
Then (16) and (14) yield
$$\begin{gathered} \begin{aligned} \tilde{I}_{1}&: = \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}\tilde{a}_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{1 - q}} \tilde{b}_{n}^{q} \Biggr]^{\frac{1}{q}} \\ &= \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{ - 1 - \varepsilon} A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} \Biggr]^{\frac{1}{p}} \Biggl[ \sum _{|n| = 2}^{\infty} \frac{\ln^{ - 1 - \varepsilon} A_{\eta,\beta} (n)}{A_{\eta,\beta} (n)} \Biggr]^{\frac{1}{q}} \\ &= \frac{1}{\varepsilon} \bigl(2\csc^{2}\alpha+ o(1) \bigr)^{\frac{1}{p}}\bigl(2\csc^{2}\beta+ \tilde{o}(1) \bigr)^{\frac{1}{q}}\quad\bigl(\varepsilon\to0^{ +} \bigr), \end{aligned} \\ \begin{aligned} \tilde{I}&: = \sum_{|n| = 2}^{\infty} \sum_{|m| = 2}^{\infty} k(m,n) \tilde{a}_{m} \tilde{b}_{n} = \sum_{|m| = 2}^{\infty} \sum_{|n| = 2}^{\infty} k(m,n) \frac{\ln^{\tilde{\lambda}_{1} - \varepsilon - 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} \frac{\ln^{\tilde{\lambda}_{2} - 1}A_{\eta,\beta} (n)}{A_{\eta,\beta} (n)} \\ &= \sum_{|m| = 2}^{\infty} \omega(\tilde{ \lambda}_{2},m)\frac{\ln^{ - \varepsilon- 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} > k_{\beta} (\tilde{ \lambda}_{1})\sum_{|m| = 2}^{\infty} \bigl(1 - \theta (\tilde{\lambda}_{2},m)\bigr)\frac{\ln^{ - \varepsilon- 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} \\ &= k_{\beta} (\tilde{\lambda}_{1}) \Biggl[ \sum _{|m| = 2}^{\infty} \frac {\ln^{ - \varepsilon- 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} - \sum _{|m| = 2}^{\infty} \frac{O(\ln^{ - (\frac{\varepsilon}{p} + \frac{\lambda_{2}}{2}) - 1}A_{\xi,\alpha} (m))}{A_{\xi,\alpha} (m)} \Biggr] \\ &= \frac{1}{\varepsilon} k_{\beta} (\tilde{\lambda}_{1}) \quad\bigl(2\csc ^{2}\alpha+ o(1) - \varepsilon O(1)\bigr). \end{aligned} \end{gathered} $$
If there exists a positive number \(K \le k(\lambda_{1})\) such that (18) is still valid when replacing \(k(\lambda_{1})\) by K, then we obtain
$$\varepsilon\tilde{I} = \varepsilon\sum_{|n| = 2}^{\infty} \sum_{|m| = 2}^{\infty} k(m,n) \tilde{a}_{m} \tilde{b}_{n} < \varepsilon K\tilde{I}_{1}. $$
Hence, in view of the above results, it follows that
$$k_{\beta} \biggl(\lambda_{1} + \frac{\varepsilon}{q}\biggr) \bigl(2\csc^{2}\alpha+ o(1) - \varepsilon O(1)\bigr) < K\bigl(2 \csc^{2}\alpha+ o(1)\bigr)^{\frac{1}{p}}\bigl(2\csc ^{2} \beta + \tilde{o}(1)\bigr)^{\frac{1}{q}}, $$
and then
$$\frac{4\pi^{2}}{[\lambda\sin(\frac{\pi\lambda_{1}}{\lambda} )]^{2}}\csc^{2}\beta\csc^{2}\alpha\le2K \csc^{\frac{2}{p}}\alpha \csc^{\frac{2}{q}}\beta\quad\bigl(\varepsilon \to0^{ +} \bigr), $$
namely
$$k(\lambda_{1}) = \frac{2\pi^{2}}{[\lambda\sin(\frac{\pi \lambda_{1}}{\lambda} )]^{2}}\csc^{\frac{2}{p}}\beta \csc^{\frac{2}{q}}\alpha\le K. $$
Hence, \(K = k(\lambda_{1})\) is the best possible constant factor in (18).

\(k(\lambda_{1})\) in (19) is still the best possible constant factor. Otherwise we would reach a contradiction by (25) that \(k(\lambda_{1})\) in (18) is not the best possible constant factor. □

4 Operator expressions and a remark

Let \(\varphi(m): = \frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}\) (\(|m| \in\mathbf{N}\setminus\{ 1\} \)), and \(\psi(n): = \frac{\ln^{q(1 - \lambda_{2}) - 1}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{1 - q}} \), wherefrom
$$\psi^{1 - p}(n): = \frac{\ln^{p\lambda_{2} - 1}A_{\eta,\beta} (n)}{A_{\eta,\beta} (n)}\quad\bigl( \vert n \vert \in \mathbf{N}\setminus\{ 1\} \bigr). $$
We define the real weighted normed function spaces as follows:
$$\begin{gathered} l_{p,\varphi}: = \Biggl\{ a = \{ a_{m}\}_{ \vert m \vert = 2}^{\infty}; \Vert a \Vert _{p,\varphi} = \Biggl( \sum_{ \vert m \vert = 2}^{\infty} \varphi (m) \vert a_{m} \vert ^{p} \Biggr)^{\frac{1}{p}} < \infty \Biggr\} , \\ l_{q,\psi}: = \Biggl\{ b = \{ b_{n}\}_{ \vert n \vert = 2}^{\infty}; \Vert b \Vert _{q,\psi} = \Biggl( \sum_{ \vert n \vert = 2}^{\infty} \psi(n) \vert b_{n} \vert ^{q} \Biggr)^{\frac {1}{q}} < \infty \Biggr\} , \\ l_{p,\psi^{1 - p}}: = \Biggl\{ c = \{ c_{n}\}_{ \vert n \vert = 2}^{\infty}; \Vert c \Vert _{p,\psi^{1 - p}} = \Biggl( \sum_{ \vert n \vert = 2}^{\infty} \psi^{1 - p}(n) \vert c_{n} \vert ^{p} \Biggr)^{\frac{1}{p}} < \infty \Biggr\} . \end{gathered} $$
For \(a = \{ a_{m}\}_{|m| = 2}^{\infty} \in l_{p,\varphi} \), we let \(c_{n} = \sum_{|m| = 2}^{\infty} k(m,n)a_{m} \) and \(c = \{ c_{n}\}_{|n| = 2}^{\infty} \), it follows by (19) that \(\|c\|_{p,\psi^{1 - p}} < k(\lambda_{1})\|a\|_{p,\varphi} \), namely \(c \in l_{p,\psi^{1 - p}} \).
Further, we define a Mulholland-type operator \(T:l_{p,\varphi} \to l_{p,\psi^{1 - p}} \) as follows: For \(a_{m} \ge0\), \(a = \{ a_{m}\}_{|m| = 2}^{\infty} \in l_{p,\varphi} \), there exists a unique representation \(Ta = c \in l_{p,\psi^{1 - p}} \). We also define the following formal inner product of Ta and \(b = \{ b_{n}\}_{|n| = 2}^{\infty} \in l_{q,\psi}\) (\(b_{n} \ge 0\)):
$$ (Ta,b): = \sum_{|n| = 2}^{\infty} \sum _{|m| = 2}^{\infty} k(m,n) a_{m}b_{n}. $$
(26)
Hence, we can respectively rewrite (18) and (19) as the following operator expressions:
$$\begin{aligned}& (Ta,b) < k(\lambda_{1}) \Vert a \Vert _{p,\varphi} \Vert b \Vert _{q,\psi}, \end{aligned}$$
(27)
$$\begin{aligned}& \Vert Ta \Vert _{p,\psi^{1 - p}} < k(\lambda_{1}) \Vert a \Vert _{p,\varphi}. \end{aligned}$$
(28)
It follows that the operator T is bounded with
$$ \Vert T \Vert : = \sup_{a( \ne\theta) \in l_{p,\varphi}} \frac{ \Vert Ta \Vert _{p,\psi ^{1 - p}}}{ \Vert a \Vert _{p,\varphi}} \le k( \lambda_{1}). $$
(29)
Since \(k(\lambda_{1})\) in (19) is the best possible constant factor, we obtain
$$ \Vert T \Vert = k(\lambda_{1}) = \frac{2\pi^{2}\csc^{2/p}\beta\csc^{2/q}\alpha}{ [\lambda\sin(\frac{\pi\lambda_{1}}{\lambda} )]^{2}}. $$
(30)

Remark 2

(i) For \(\xi= \eta= 0 \) in (20), we have the following new inequality:
$$\begin{aligned}[b] &\sum_{ \vert n \vert = 2}^{\infty} \sum _{ \vert m \vert = 2}^{\infty} \frac{\ln(\ln \vert m \vert /\ln \vert n \vert )a_{m}b_{n}}{\ln^{\lambda} \vert m \vert - \ln^{\lambda} \vert n \vert } \\&\quad< \frac{2\pi^{2}}{[\lambda\sin(\frac{\pi\lambda_{1}}{\lambda} )]^{2}} \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1} \vert m \vert }{ \vert m \vert ^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1} \vert n \vert }{ \vert n \vert ^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}}.\end{aligned} $$
(31)
It follows that (20) is an extension of (31). In particular, for \(\lambda= 1\), \(\lambda_{1} = \frac{1}{q}\), \(\lambda_{2} = \frac{1}{p} \), we have the following simple Mulholland-type inequality in the whole plane with the best possible constant factor \(\frac{2\pi^{2}}{\sin^{2}(\frac{\pi}{p})} \):
$$ \sum_{ \vert n \vert = 2}^{\infty} \sum _{ \vert m \vert = 2}^{\infty} \frac{\ln(\ln \vert m \vert /\ln \vert n \vert )}{\ln( \vert m \vert / \vert n \vert )} a_{m}b_{n} < \frac{2\pi^{2}}{\sin^{2}(\frac{\pi}{p})} \Biggl( \sum_{ \vert m \vert = 2}^{\infty} \frac{a_{m}^{p}}{ \vert m \vert ^{1 - p}} \Biggr)^{\frac{1}{p}} \Biggl( \sum _{ \vert n \vert = 2}^{\infty} \frac{b_{n}^{q}}{ \vert n \vert ^{1 - q}} \Biggr)^{\frac{1}{q}}. $$
(32)
(ii) If \(a_{ - m} = a_{m}\), \(b_{ - n} = b_{n}\) (\(m,n \in\mathbf{N}\setminus \{ 1\} \)), then (20) reduces to
$$ \begin{aligned}[b] &\sum_{n = 2}^{\infty} \sum_{m = 2}^{\infty} \biggl\{ \frac{\ln[\ln(m - \xi)/\ln(n - \eta)]}{\ln^{\lambda} (m - \xi) - \ln^{\lambda} (n - \eta)} + \frac{\ln[\ln(m - \xi)/\ln(n + \eta )]}{\ln^{\lambda} (m - \xi) - \ln^{\lambda} (n + \eta)} \\ &\qquad{} + \frac{\ln[\ln(m + \xi)/\ln(n - \eta)]}{\ln ^{\lambda} (m + \xi ) - \ln^{\lambda} (n - \eta)} + \frac{\ln[\ln(m + \xi)/\ln(n + \eta )]}{\ln^{\lambda} (m + \xi) - \ln^{\lambda} (n + \eta)} \biggr\} a_{m}b_{n} \\ &\quad< \frac{2\pi^{2}}{[\lambda\sin(\frac{\pi\lambda_{1}}{\lambda} )]^{2}} \Biggl\{ \sum_{m = 2}^{\infty} \biggl[ \frac{\ln^{p(1 - \lambda _{1}) - 1}(m - \xi)}{(m - \xi)^{1 - p}} + \frac{\ln^{p(1 - \lambda_{1}) - 1}(m + \xi)}{(m + \xi)^{1 - p}} \biggr]a_{m}^{p} \Biggr\} ^{\frac{1}{p}} \\ &\qquad{}\times \Biggl\{ \sum_{n = 2}^{\infty} \biggl[ \frac{\ln^{q(1 - \lambda_{2}) - 1}(n - \eta)}{(n - \eta)^{1 - q}} + \frac{\ln^{q(1 - \lambda_{2}) - 1}(n + \eta)}{(n + \eta)^{1 - q}} \biggr]b_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned} $$
(33)
In particular, for \(\lambda= 1\), \(\lambda_{1} = \frac{1}{q}\), \(\lambda_{2} = \frac{1}{p}\), \(\xi= \eta\in[0,\frac{1}{2}] \), we obtain
$$ \begin{aligned}[b]& \sum_{n = 2}^{\infty} \sum_{m = 2}^{\infty} \biggl\{ \frac{\ln[\ln(m - \xi)/\ln(n - \xi)]}{\ln[(m - \xi)/(n - \xi)]} + \frac{\ln[\ln(m - \xi)/\ln(n + \xi)]}{\ln[(m - \xi)/(n + \xi)]} \\ &\qquad{} + \frac{\ln[\ln(m + \xi)/\ln(n - \xi)]}{\ln[(m + \xi)/(n - \xi)]} + \frac{\ln[\ln(m + \xi)/\ln(n + \xi)]}{\ln[(m + \xi)/(n + \xi)]} \biggr\} a_{m}b_{n} \\ &\quad< \frac{2\pi^{2}}{\sin^{2}(\frac{\pi}{p})} \Biggl\{ \sum_{m = 2}^{\infty} \biggl[ \frac{1}{(m - \xi)^{1 - p}} + \frac{1}{(m + \xi)^{1 - p}} \biggr]a_{m}^{p} \Biggr\} ^{\frac{1}{p}}\\&\quad{}\times \Biggl\{ \sum_{n = 2}^{\infty} \biggl[ \frac{1}{(n - \xi)^{1 - q}} + \frac{1}{(n + \xi)^{1 - q}} \biggr]b_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned} $$
(34)
For \(\xi= 0 \), (34) reduces to the following simple Mulholland-type inequality with the best possible constant factor \(\frac{\pi^{2}}{\sin^{2}(\frac{\pi}{p})} \):
$$ \sum_{n = 2}^{\infty} \sum _{m = 2}^{\infty} \frac{\ln(\ln m/\ln n)}{\ln (m/n)} a_{m}b_{n} < \frac{\pi^{2}}{\sin^{2}(\frac{\pi}{p})} \Biggl( \sum_{m = 2}^{\infty} \frac{a_{m}^{p}}{m^{1 - p}} \Biggr)^{\frac{1}{p}} \Biggl( \sum _{n = 2}^{\infty} \frac{b_{n}^{q}}{n^{1 - q}} \Biggr)^{\frac{1}{q}}. $$
(35)

5 Conclusions

In this paper, we present a new discrete Mulholland-type inequality in the whole plane with a best possible constant factor that is similar to that in (4) via introducing multi-parameters, applying weight coefficients, and using Hermite–Hadamard’s inequality in Theorem 1 and Theorem 2. Moreover, the equivalent forms, some particular cases, and the operator expressions are considered. The lemmas and theorems provide an extensive account of this type of inequalities.

Declarations

Funding

This work is supported by the National Natural Science Foundation (No. 61772140) and Science and Technology Planning Project Item of Guangzhou City (No. 201707010229).

Authors’ contributions

BY carried out the mathematical studies, participated in the sequence alignment, and drafted the manuscript. QC 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.

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’ Affiliations

(1)
Department of Mathematics, Guangdong University of Education, Guangzhou, P.R. China
(2)
Department of Computer Science, Guangdong University of Education, Guangzhou, P.R. China

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