## 1 Introduction

If $$p > 1$$, $$\frac{1}{p} + \frac{1}{q} = 1$$, $$a_{m},b_{n} \ge 0$$, $$0 < \sum_{m = 1}^{\infty} a_{m}^{p} < \infty$$ and $$0 < \sum_{n = 1}^{\infty} b_{n}^{q} < \infty$$, then we have the following Hardy-Hilbert’s inequality (cf. [1]):

$$\sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{a_{m}b_{n}}{m + n} < \frac{\pi}{\sin (\pi /p)} \Biggl( \sum_{m = 1}^{\infty} a_{m}^{p} \Biggr)^{\frac{1}{p}} \Biggl( \sum_{n = 1}^{\infty} b_{n}^{q} \Biggr)^{\frac{1}{q}},$$
(1)

where the constant factor $$\frac{\pi}{\sin (\pi /p)}$$ is the best possible. In 1934, Hardy proved the following more accurate inequality of (1) with the same best possible constant factor (cf. Theorem 343 of [2]):

$$\sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{a_{m}b_{n}}{m + n - 1} < \frac{\pi}{\sin (\pi /p)} \Biggl( \sum_{m = 1}^{\infty} a_{m}^{p} \Biggr)^{\frac{1}{p}} \Biggl( \sum_{n = 1}^{\infty} b_{n}^{q} \Biggr)^{\frac{1}{q}}.$$
(2)

We still have the following Mulholland’s inequality with the same best possible constant factor $$\frac{\pi}{\sin (\pi /p)}$$ (cf. Theorem 343 of [2], replacing $$\frac{a_{m}}{m}$$, $$\frac{b_{n}}{n}$$ by $$a_{m}$$, $$b_{n}$$):

$$\sum_{n = 2}^{\infty} \sum _{m = 2}^{\infty} \frac{a_{m}b_{n}}{\ln mn} < \frac{\pi}{\sin (\pi /p)} \Biggl( \sum_{m = 2}^{\infty} \frac{a_{m}^{p}}{m} \Biggr)^{\frac{1}{p}} \Biggl( \sum_{n = 2}^{\infty} \frac{b_{n}^{q}}{n} \Biggr)^{\frac{1}{q}}.$$
(3)

Inequalities (1)-(3) are important in analysis and its applications (cf. [2, 3]). In 2007, Yang [4] first gave a Hilbert-type integral inequality in the whole plane. Many extensions of this type inequalities and (1)-(3) were provided in [520].

In 2016, Yang and Chen [21] gave a more accurate Hardy-Hilbert’s inequality in the whole plane:

\begin{aligned}[b] &\sum_{|n| = 1}^{\infty} \sum _{|m| = 1}^{\infty} \frac{a_{m}b_{n}}{(|m - \xi | + |n - \eta |)^{\lambda}} \\ &\quad< 2B( \lambda_{1},\lambda_{2}) \Biggl[ \sum _{|m| = 1}^{\infty} |m - \xi |^{p(1 - \lambda_{1}) - 1}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{|n| = 1}^{\infty} |n - \eta |^{q(1 - \lambda_{2}) - 1} b_{n}^{q} \Biggr]^{\frac{1}{q}}, \end{aligned}
(4)

where the constant factor $$2B(\lambda_{1},\lambda_{2})$$ ($$0 < \lambda_{1},\lambda_{2} \le 1$$, $$\lambda_{1} + \lambda_{2} = \lambda$$, $$\xi,\eta \in [0,\frac{1}{2}]$$) is the best possible.

In this paper, by introducing independent parameters and applying the weight coefficients and Hermite-Hadamard’s inequality, we give a new more accurate extension of (3) in the whole plane with a best possible constant factor similar to (4). Furthermore, we consider the equivalent forms, the reverses, a few particular cases, and the operator expressions.

## 2 Some lemmas

In the following, we agree that $$p \ne 0,1$$, $$\frac{1}{p} + \frac{1}{q} = 1$$, $$\lambda_{1},\lambda_{2} > 0$$, $$\lambda_{1} + \lambda_{2} = \lambda$$,

$$\alpha,\beta \in \biggl[\arccos \sqrt{\frac{1}{3}},\pi - \arccos \sqrt{ \frac{1}{3}} \biggr]\quad \bigl( \subseteq (0,\pi ) \bigr),$$

$$\xi,\eta \in ( - \frac{3}{2},\frac{3}{2})$$, satisfying

$$\frac{1}{1 - \cos \gamma} - \frac{3}{2} \le \xi,\eta \le \frac{ - 1}{1 + \cos \gamma} + \frac{3}{2},$$
(5)

and

$$h_{\gamma} (\lambda_{1}): = 2B(\lambda_{1}, \lambda_{2})\csc^{2}\gamma \quad(\gamma = \alpha,\beta ).$$
(6)

### Note 1

For $$\alpha,\beta = \frac{\pi}{2}$$, we find $$\xi,\eta \in [ - \frac{1}{2},\frac{1}{2}]$$. If $$\alpha,\beta \in [\arccos \sqrt{\frac{1}{3}},\pi - \arccos \sqrt{\frac{1}{3}} ]$$, then $${\xi,\eta = 0}$$ satisfy (5).

For $$|x|,|y| \ge \frac{3}{2}$$, we set $$A_{\xi,\alpha} (x): = |x - \xi | + (x - \xi )\cos \alpha$$,

$$B_{\eta,\beta} (y): = |y - \eta | + (y - \eta )\cos \beta,$$

and

$$k(x,y): = \frac{1}{(\ln A_{\xi,\alpha} (x) + \ln B_{\eta,\beta} (y))^{\lambda}} = \frac{1}{\ln^{\lambda} A_{\xi,\alpha} (x)B_{\eta,\beta} (y)}.$$
(7)

### Definition 1

Define two weight coefficients as follows:

\begin{aligned}& \omega (\lambda_{2},m): = \sum_{|n| = 2}^{\infty} \frac{k(m,n)}{B_{\eta,\beta} (n)} \cdot \frac{\ln^{\lambda_{1}}A_{\xi,\alpha} (m)}{\ln^{1 - \lambda_{2}}B_{\eta,\beta} (n)},\quad|m| \in \mathbb{N}\setminus \{ 1 \}, \end{aligned}
(8)
\begin{aligned}& \varpi (\lambda_{1},n): = \sum _{|m| = 2}^{\infty} \frac{k(m,n)}{A_{\xi,\alpha} (m)} \cdot \frac{\ln^{\lambda_{2}}B_{\eta,\beta} (n)}{\ln^{1 - \lambda_{1}}A_{\xi,\alpha} (m)},\quad|n| \in \mathbb{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 1

For $$\lambda_{2} \le 1$$, we have the following inequalities:

$$k_{\beta} (\lambda_{1}) \bigl(1 - \theta ( \lambda_{2},m) \bigr) < \omega (\lambda_{2},m) < k_{\beta} (\lambda_{1}),\quad|m| \in \mathbb{N}\setminus \{ 1 \},$$
(10)

where

$$\theta (\lambda_{2},m): = \frac{1}{B(\lambda_{1},\lambda_{2})} \int_{0}^{\frac{\ln [(2 + \eta )(1 + \cos \beta )]}{\ln A_{\xi,\alpha} (m)}} \frac{u^{\lambda_{2} - 1}}{(1 + u)^{\lambda}} \,du = O \biggl( \frac{1}{\ln^{\lambda_{2}}A_{\xi,\alpha} (m)} \biggr) \in (0,1).$$
(11)

### Proof

For $$|m| \in \mathbb{N}\setminus \{ 1\}$$, we set

$$\begin{gathered} k^{(1)}(m,y): = \frac{1}{\ln^{\lambda} [A_{\xi,\alpha} (m)(y - \eta )(\cos \beta - 1)]},\quad y < - \frac{3}{2}, \\ k^{(2)}(m,y): = \frac{1}{\ln^{\lambda} [A_{\xi,\alpha} (m)(y - \eta )(\cos \beta + 1)]},\quad y > \frac{3}{2}, \end{gathered}$$

wherefrom

$$h^{(1)}(m, - y) = \frac{1}{\ln^{\lambda} [A_{\xi,\alpha} (m)(y + \eta )(1 - \cos \beta )]},\quad y > \frac{3}{2}.$$

We find

\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)

For fixed $$|m| \in \mathbb{N}\setminus \{ 1\}$$, since $$\lambda > 0$$, $$0 < \lambda_{2} \le 1$$,we find that, for $$y > \frac{3}{2}$$,

$$\begin{gathered} \frac{d}{dy}\frac{k^{(i)}(m,( - 1)^{i}y)}{(y - ( - 1)^{i}\eta )\ln^{1 - \lambda_{2}}[(y - ( - 1)^{i}\eta )(1 + ( - 1)^{i}\cos \beta )]} < 0, \\ \frac{d^{2}}{dy^{2}}\frac{k^{(i)}(m,( - 1)^{i}y)}{(y - ( - 1)^{i}\eta )\ln^{1 - \lambda_{2}}[(y - ( - 1)^{i}\eta )(1 + ( - 1)^{i}\cos \beta )]} > 0\quad(i = 1,2),\end{gathered}$$

and 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 strict decreasing and strictly convex in $$(\frac{3}{2},\infty )$$. By Hermite-Hadamard’s inequality (see [22]) and (12) we find

\begin{aligned} \omega (\lambda_{2},m) < {}& \frac{\ln^{\lambda_{1}}A_{\xi,\alpha} (m)}{1 - \cos \beta} \int_{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_{3/2}^{\infty} \frac{k^{(2)}(m,y)}{(y - \eta )\ln^{1 - \lambda_{2}}[(y - \eta )(1 + \cos \beta )]}\, dy.\end{aligned}

In view of (5), it follows that $$(\frac{3}{2} \pm \eta )(1 \mp \cos \beta ) \ge 1$$. 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 first (second) integral, 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}\,du}{(1 + u)^{\lambda}} \\ &= 2B(\lambda_{1},\lambda_{2})\csc^{2}\beta = k_{\beta} (\lambda_{1}).\end{aligned}

By monotonicity and (12) we still have

\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{u^{\lambda_{2} - 1}\,du}{(1 + u)^{\lambda}} \\ ={}& k_{\beta} (\lambda_{1}) - 2\csc^{2}\beta \int_{0}^{\frac{\ln [(2 + |\eta |)(1 + |\cos \beta |)]}{\ln A_{\xi,\alpha} (m)}} \frac{u^{\lambda_{2} - 1}\,du}{(1 + u)^{\lambda}} = k_{\beta} (\lambda_{1}) \bigl(1 - \theta ( \lambda_{2},m)\bigr) > 0,\end{aligned}

where $$\theta (\lambda_{2},m)$$ is indicated by (11). It follows that $$\theta (\lambda_{2},m) < 1$$ and

\begin{aligned} 0 &< \theta (\lambda_{2},m) < \frac{1}{B(\lambda_{1},\lambda_{2})} \int_{0}^{\frac{\ln [(2 + |\eta |)(1 + |\cos \beta |)]}{\ln A_{\xi,\alpha} (m)}} u^{\lambda_{2} - 1} \,du \\ &= \frac{1}{\lambda_{2}B(\lambda_{1},\lambda_{2})} \biggl(\frac{\ln [(2 + |\eta |)(1 + |\cos \beta |)]}{\ln A_{\xi,\alpha} (m)} \biggr)^{\lambda_{2}}. \end{aligned}

Hence, (10) and (11) are valid. □

In the same way, we still have the following:

### Lemma 2

For $$\lambda_{1} \le 1$$, we have the following inequalities:

$$k_{\alpha} (\lambda_{1}) \bigl(1 - \tilde{\theta} ( \lambda_{1},n) \bigr) < \varpi (\lambda_{1},n) < k_{\alpha} (\lambda_{1}),\quad|n| \in \mathbb{N}\setminus \{ 1 \},$$
(13)

where

$$\tilde{\theta} (\lambda_{1},n): = \frac{1}{B(\lambda_{1},\lambda_{2})} \int_{0}^{\frac{\ln [(2 + |\xi |)(1 + |\cos \alpha |)]}{\ln B_{\eta,\beta} (n)}} \frac{u^{\lambda_{1} - 1}}{(1 + u)^{\lambda}} \,du = O \biggl( \frac{1}{\ln^{\lambda_{1}}B_{\eta,\beta} (n)} \biggr) \in (0,1).$$
(14)

### Lemma 3

If $$\rho > 0$$, $$\gamma \in [\arccos \sqrt{\frac{1}{3}},\pi - \arccos \sqrt{\frac{1}{3}} ]$$ ($$\gamma = \alpha,\beta$$), and

$$\frac{1}{1 - \cos \gamma} - \frac{3}{2} \le \varsigma \le \frac{ - 1}{1 + \cos \gamma} + \frac{3}{2}\quad(\varsigma = \xi,\eta ),$$

then for $$(\varsigma,\gamma ) = (\xi,\alpha )$$ (or $$(\eta,\beta )$$), we have

$$H_{\rho} (\varsigma,\gamma ): = \sum_{|n| = 2}^{\infty} \frac{\ln^{ - 1 - \rho} [|n - \varsigma | + (n - \varsigma )\cos \gamma ]}{|n - \varsigma | + (n - \varsigma )\cos \gamma} = \frac{1}{\rho} \bigl(2\csc^{2}\gamma + o(1) \bigr)\quad \bigl(\rho \to 0^{ +} \bigr).$$
(15)

### Proof

\begin{aligned} H_{\rho} (\varsigma,\gamma ) ={}& \sum _{n = - 2}^{ - \infty} \frac{\ln^{ - 1 - \rho} [(n - \varsigma )(\cos \gamma - 1)]}{(n - \varsigma )(\cos \gamma - 1)} + \sum _{n = 2}^{\infty} \frac{\ln^{ - 1 - \rho} [(n - \varsigma )(\cos \gamma + 1)]}{(n - \varsigma )(\cos \gamma + 1)} \\ ={}& \sum_{n = 2}^{\infty} \biggl\{ \frac{\ln^{ - 1 - \rho} [(n + \varsigma )(1 - \cos \gamma )]}{(n - \varsigma )(1 - \cos \gamma )} + \frac{\ln^{ - 1 - \rho} [(n - \varsigma )(\cos \gamma + 1)]}{(n - \varsigma )(\cos \gamma + 1)} \biggr\} \\ \le{}& \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)\quad \bigl(\rho \to 0^{ +} \bigr). \end{aligned}

We still can find that

\begin{aligned} H_{\rho} (\varsigma,\gamma ) &= \sum _{n = 2}^{\infty} \biggl\{ \frac{\ln^{ - 1 - \rho} [(n + \varsigma )(1 - \cos \gamma )]}{(n - \varsigma )(1 - \cos \gamma )} + \frac{\ln^{ - 1 - \rho} [(n - \varsigma )(\cos \gamma + 1)]}{(n - \varsigma )(\cos \gamma + 1)} \biggr\} \\ &\ge \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)\quad \bigl(\rho \to 0^{ +} \bigr). \end{aligned}

Hence, we have (15). □

## 3 Main results and the reverses

We also set

$$K(\lambda_{1}): = h_{\beta}^{1/p}( \lambda_{1})h_{\alpha}^{1/q}(\lambda_{1}) = 2B(\lambda_{1},\lambda_{2})\csc^{2/p}\beta \csc^{2/q}\alpha.$$
(16)

### Theorem 1

Suppose that $$p > 1$$, $$\lambda_{1},\lambda_{2} \le 1$$, $$a_{m},b_{n} \ge 0$$ ($$|m|,|n| \in \mathbb{N}\setminus \{ 1\}$$), and

$$0 < \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}} a_{m}^{p} < \infty,\qquad0 < \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} < \infty.$$

We have the following equivalent inequalities:

\begin{aligned}& \begin{aligned}[b]I&: = \sum_{|n| = 2}^{\infty} \sum _{|m| = 2}^{\infty} k(m,n) a_{m}b_{n} \\&< 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}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}},\end{aligned} \end{aligned}
(17)
\begin{aligned}& \begin{aligned}[b]J&: = \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{p\lambda_{2} - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} \Biggl( \sum_{|m| = 2}^{\infty} k(m,n)a_{m} \Biggr)^{p} \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{aligned}
(18)

In particular, (i) for $$\xi,\eta \in [ - \frac{1}{2},\frac{1}{2}]$$ ($$\alpha = \beta = \frac{\pi}{2}$$), we have the following equivalent inequalities:

\begin{aligned}& \begin{aligned}[b] &\sum_{|n| = 2}^{\infty} \sum _{|m| = 2}^{\infty} \frac{a_{m}b_{n}}{\ln^{\lambda} (|m - \xi ||n - \eta |)} \\&\quad< 2B( \lambda_{1},\lambda_{2}) \Biggl[ \sum _{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}|m - \xi |}{|m - \xi |^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}|n - \eta |}{|n - \eta |^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}},\end{aligned} \end{aligned}
(19)
\begin{aligned}& \begin{aligned}[b]&\Biggl\{ \sum_{|n| = 2}^{\infty} \frac{\ln^{p\lambda_{2} - 1}|n - \eta |}{|n - \eta |} \Biggl[ \sum_{|m| = 2}^{\infty} \frac{a_{m}}{\ln^{\lambda} (|m - \xi ||n - \eta |)} \Biggr]^{p} \Biggr\} ^{\frac{1}{p}} \\&\quad< 2B(\lambda_{1}, \lambda_{2}) \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}|m - \xi |}{|m - \xi |^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}.\end{aligned} \end{aligned}
(20)

(ii) For $$\alpha,\beta \in [\arccos \frac{1}{3},\pi - \arccos \frac{1}{3}]$$ ($$\xi = \eta = 0$$), we have the following equivalent inequalities:

\begin{aligned}& \begin{aligned}[b]&\sum_{|n| = 2}^{\infty} \sum _{|m| = 2}^{\infty} \frac{a_{m}b_{n}}{\ln^{\lambda} [(|m| + m\cos \alpha )(|n| + n\cos \beta )]} \\&\quad< K(\lambda_{1}) \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}(|m| + m\cos \alpha )}{(|m| + m\cos \alpha )^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \\ &\qquad\times\Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}(|n| + n\cos \beta )}{(|n| + n\cos \beta )^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}},\end{aligned} \end{aligned}
(21)
\begin{aligned}& \begin{aligned}[b]&\Biggl\{ \sum_{|n| = 2}^{\infty} \frac{\ln^{p\lambda_{2} - 1}(|n| + n\cos \beta )}{|n| + n\cos \beta} \Biggl[ \sum_{|m| = 2}^{\infty} \frac{a_{m}}{\ln^{\lambda} [(|m| + m\cos \alpha )(|n| + n\cos \beta )]} \Biggr]^{p} \Biggr\} ^{\frac{1}{p}} \\&\quad< K(\lambda_{1}) \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}(|m| + m\cos \alpha )}{(|m| + m\cos \alpha )^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}.\end{aligned} \end{aligned}
(22)

### Proof

By Hölder’s inequality with weight (cf. [22]) and (9) we find

\begin{aligned} \Biggl( \sum_{|m| = 2}^{\infty} k(m,n)a_{m} \Biggr)^{p} ={}& \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}}B_{\eta,\beta} (n)}a_{m} \biggr] \\ &\times\biggl[ \frac{\ln^{\frac{1 - \lambda_{2}}{p}}B_{\eta,\beta} (n)}{(A_{\xi,\alpha} (m))^{\frac{1}{q}}\ln^{\frac{1 - \lambda_{1}}{q}}A_{\xi,\alpha} (m)} \biggr] \Biggr\} ^{p} \\\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}}B_{\eta,\beta} (n)}a_{m}^{p}\\ &\times \Biggl[ \sum _{|m| = 2}^{\infty} k(m,n)\frac{\ln^{\frac{(1 - \lambda_{2})q}{p}}B_{\eta,\beta} (n)}{A_{\xi,\alpha} (m)\ln^{1 - \lambda_{1}}A_{\xi,\alpha} (m)} \Biggr]^{p - 1} \\ ={}& \frac{(\varpi (\lambda_{1},n))^{p - 1}B_{\eta,\beta} (n)}{\ln^{p\lambda_{2} - 1}B_{\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)}{B_{\eta,\beta} (n)\ln^{1 - \lambda_{2}}B_{\eta,\beta} (n)}a_{m}^{p}.\end{aligned}

By (13) it follows that

\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)}{B_{\eta,\beta} (n)\ln^{1 - \lambda_{2}}B_{\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)}{B_{\eta,\beta} (n)\ln^{1 - \lambda_{2}}B_{\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}
(23)

By (10) and (16) we have (18).

Using Hölder’s inequality again, we have

\begin{aligned}[b] I &= \sum_{|n| = 2}^{\infty} \Biggl[ \frac{(B_{\eta,\beta} (n))^{\frac{ - 1}{p}}}{\ln^{\frac{1}{p} - \lambda_{2}}B_{\eta,\beta} (n)}\sum_{|m| = 2}^{\infty} k(m,n)a_{m} \Biggr] \biggl[ \frac{\ln^{\frac{1}{p} - \lambda_{2}}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{\frac{ - 1}{p}}}b_{n} \biggr] \\&\le J \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} \Biggr]^{\frac{1}{q}},\end{aligned}
(24)

and then by (18) we have (17).

On the other hand, assuming that (17) is valid, we set

$$b_{n}: = \frac{\ln^{p\lambda_{2} - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} \Biggl( \sum _{|m| = 2}^{\infty} k(m,n)a_{m} \Biggr)^{p - 1},\quad|n| \in \mathbb{N}\setminus \{ 1\},$$

and find

$$J = \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} \Biggr]^{\frac{1}{p}}.$$

By (23) it follows that $$J < \infty$$. If $$J = 0$$, then (18) 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}B_{\eta,\beta} (n)}{(B_{\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}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}},\end{aligned} \\J = \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\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{gathered}

Hence (18) is valid, which is equivalent to (17). □

### Theorem 2

With regards to the assumptions of Theorem 1, the constant factor $$K(\lambda_{1})$$ in (17) and (18) is the best possible.

### Proof

For $$0 < \varepsilon < q\lambda_{2}$$, we set $$\tilde{\lambda}_{1} = \lambda_{1} + \frac{\varepsilon}{q}$$ (>0), $$\tilde{\lambda}_{2} = \lambda_{2} - \frac{\varepsilon}{q}$$ ($$\in (0,1)$$), and

$$\begin{gathered} \tilde{a}_{m}: = \frac{\ln^{\lambda_{1} - \frac{\varepsilon}{p} - 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} = \frac{\ln^{\tilde{\lambda}_{1} - \varepsilon - 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)}\quad\bigl(|m| \in \mathbb{N}\setminus \{ 1\} \bigr), \\ \tilde{b}_{n}: = \frac{\ln^{\lambda_{2} - \frac{\varepsilon}{q} - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} = \frac{\ln^{\tilde{\lambda}_{2} - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)}\quad\bigl(|n| \in \mathbb{N}\setminus \{ 1\} \bigr).\end{gathered}$$

By (15) and (10) we find

\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}B_{\eta,\beta} (n)}{(B_{\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} B_{\eta,\beta} (n)}{B_{\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 \to 0^{ +} \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}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} \\ &= \sum_{|m| = 2}^{\infty} \omega (\tilde{ \lambda}_{2},m)\frac{\ln^{ - 1 - \varepsilon} A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} \ge k_{\beta} (\tilde{ \lambda}_{1})\sum_{|m| = 2}^{\infty} \bigl(1 - \theta (\tilde{\lambda}_{2},m)\bigr)\frac{\ln^{ - 1 - \varepsilon} A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} \\&= k_{\beta} (\tilde{\lambda}_{1}) \Biggl[ \sum _{|m| = 2}^{\infty} \frac{\ln^{ - 1 - \varepsilon} A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} - \sum _{|m| = 2}^{\infty} \frac{O(\ln^{ - 1 - (\frac{\varepsilon}{p} + \lambda_{2})}A_{\xi,\alpha} (m))}{A_{\xi,\alpha} (m)} \Biggr] \\ &= \frac{1}{\varepsilon} k_{\beta} (\tilde{\lambda}_{1}) \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 (17) is still valid when replacing $$K(\lambda_{1})$$ by k, then in particular we have

$$\varepsilon \tilde{I} = \varepsilon \sum_{|n| = 2}^{\infty} \sum_{|m| = 2}^{\infty} k(m,n) \tilde{a}_{m} \tilde{b}_{n} < \varepsilon k\tilde{I}_{1}.$$

We obtain from the above results 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

$$4B(\lambda_{1},\lambda_{2})\csc^{2}\beta \csc^{2}\alpha \le 2k\csc^{\frac{2}{p}}\alpha \csc^{\frac{2}{q}}\beta \quad\bigl(\varepsilon \to 0^{ +} \bigr),$$

namely, $$K(\lambda_{1}) = 2B(\lambda_{1},\lambda_{2})\csc^{\frac{2}{p}}\beta \csc^{\frac{2}{q}}\alpha \le k$$. Hence, $$k = K(\lambda_{1})$$ is the best value of (17).

The constant factor $$K(\lambda_{1})$$ in (18) is still the best possible. Otherwise, we would reach a contradiction by (24) that the constant factor in (17) is not the best possible. □

### Theorem 3

Suppose that $$0 < p < 1$$, $$\lambda_{1},\lambda_{2} \le 1$$, $$a_{m},b_{n} \ge 0$$ ($$|m|,|n| \in \mathbb{N}\setminus \{ 1\}$$), and

$$0 < \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}} a_{m}^{p} < \infty,\qquad0 < \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} < \infty.$$

We have the following equivalent inequalities:

\begin{aligned}& \begin{aligned}[b]I &= \sum_{|n| = 2}^{\infty} \sum _{|m| = 2}^{\infty} k(m,n) a_{m}b_{n} \\&> K(\lambda_{1}) \Biggl[ \sum_{|m| = 2}^{\infty} \bigl( 1 - \theta (\lambda_{2},m) \bigr)\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}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}},\end{aligned} \end{aligned}
(25)
\begin{aligned}& \begin{aligned}[b]J &= \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{p\lambda_{2} - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} \Biggl( \sum_{|m| = 2}^{\infty} k(m,n)a_{m} \Biggr)^{p} \Biggr]^{\frac{1}{p}} \\&> K(\lambda_{1}) \Biggl[ \sum_{|m| = 2}^{\infty} \bigl( 1 - \theta (\lambda_{2},m) \bigr)\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}
(26)
\begin{aligned}& \begin{aligned}[b]L&: = \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{q\lambda_{1} - 1}A_{\xi,\alpha} (m)}{(1 - \theta (\lambda_{2},m))^{q - 1}A_{\xi,\alpha} (m)} \Biggl( \sum_{|n| = 2}^{\infty} k(m,n)b_{n} \Biggr)^{q} \Biggr]^{\frac{1}{q}} \\&> K(\lambda_{1}) \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}},\end{aligned} \end{aligned}
(27)

where the constant factor $$K(\lambda_{1})$$ in (25), (26), and (27) is the best possible.

### Proof

By the reverse Hölder inequality with weight (cf. [22]), and (9), we find

\begin{aligned} \Biggl( \sum_{|m| = 2}^{\infty} k(m,n)a_{m} \Biggr)^{p} ={}& \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}}B_{\eta,\beta} (n)}a_{m} \biggr] \\ &\times\biggl[ \frac{\ln^{\frac{1 - \lambda_{2}}{p}}B_{\eta,\beta} (n)}{(A_{\xi,\alpha} (m))^{\frac{1}{q}}\ln^{\frac{1 - \lambda_{1}}{q}}A_{\xi,\alpha} (m)} \biggr] \Biggr\} ^{p} \\\ge{}&\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}}B_{\eta,\beta} (n)}a_{m}^{p}\\ &\times \Biggl[ \sum _{|m| = 2}^{\infty} k(m,n)\frac{\ln^{\frac{(1 - \lambda_{2})q}{p}}B_{\eta,\beta} (n)}{A_{\xi,\alpha} (m)\ln^{1 - \lambda_{1}}A_{\xi,\alpha} (m)} \Biggr]^{p - 1} \\ ={}& \frac{(\varpi (\lambda_{1},n))^{p - 1}B_{\eta,\beta} (n)}{\ln^{p\lambda_{2} - 1}B_{\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)}{B_{\eta,\beta} (n)\ln^{1 - \lambda_{2}}B_{\eta,\beta} (n)}a_{m}^{p}.\end{aligned}

Since $$p - 1 < 0$$, by (13) it follows that

\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)}{B_{\eta,\beta} (n)\ln^{1 - \lambda_{2}}B_{\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}
(28)

By (10) and (16) we have (26).

Using the reverse Hölder’s inequality again, we have

\begin{aligned}[b] I &= \sum_{|n| = 2}^{\infty} \Biggl[ \frac{\ln^{\lambda_{2} - \frac{1}{p}}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{\frac{1}{p}}}\sum_{|m| = 2}^{\infty} k(m,n)a_{m} \Biggr] \biggl[ \frac{\ln^{\frac{1}{p} - \lambda_{2}}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{\frac{ - 1}{p}}}b_{n} \biggr] \\&\ge J \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} \Biggr]^{\frac{1}{q}},\end{aligned}
(29)

and then by using (26) we have (25).

On the other hand, assuming that (25) is valid, we set

$$b_{n}: = \frac{\ln^{p\lambda_{2} - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} \Biggl( \sum _{|m| = 2}^{\infty} k(m,n)a_{m} \Biggr)^{p - 1},\quad|n| \in \mathbb{N}\setminus \{ 1\},$$

and find

$$J = \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} \Biggr]^{\frac{1}{p}}.$$

By (28) it follows that $$J > 0$$. If $$J = \infty$$, then (26) is trivially valid; if $$0 < J < \infty$$, then we have

$$\begin{gathered} \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}} b_{n}^{q}\\ \quad = J^{p} = I \\ \quad> K(\lambda_{1}) \Biggl[ \sum_{|m| = 2}^{\infty} \bigl( 1 - \theta (\lambda_{2},m) \bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}},\\J = \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} \Biggr]^{\frac{1}{p}} > K( \lambda_{1}) \Biggl[ \sum_{|m| = 2}^{\infty} \bigl( 1 - \theta (\lambda_{2},m) \bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}.\end{gathered}$$

Hence (26) is valid, which is equivalent to (25).

By the reverse Hölder inequality with weight (cf. [22]), and (9) we find

\begin{aligned} \Biggl( \sum_{|n| = 2}^{\infty} k(m,n)b_{n} \Biggr)^{q} \le {}&\Biggl[ \sum _{|n| = 2}^{\infty} k(m,n)\frac{\ln^{(1 - \lambda_{1})p/q}A_{\xi,\alpha} (m)}{B_{\eta,\beta} (n)\ln^{1 - \lambda_{2}}B_{\eta,\beta} (n)} \Biggr]^{q - 1} \\ &\times \sum_{|n| = 2}^{\infty} k(m,n) \frac{(B_{\eta,\beta} (n))^{q/p}\ln^{(1 - \lambda_{2})q/p}B_{\eta,\beta} (n)}{\ln^{1 - \lambda_{1}}A_{\xi,\alpha} (m)} b_{n}^{q} \\={}& \frac{(\omega (\lambda_{2},m))^{q - 1}A_{\xi,\alpha} (m)}{\ln^{q\lambda_{1} - 1}A_{\xi,\alpha} (m)}\sum_{|n| = 2}^{\infty} k(m,n) \frac{(B_{\eta,\beta} (n))^{\frac{q}{p}}\ln^{\frac{(1 - \lambda_{2})q}{p}}B_{\eta,\beta} (n)}{A_{\xi,\alpha} (m)\ln^{1 - \lambda_{1}}A_{\xi,\alpha} (m)}b_{n}^{q}.\end{aligned}

Since $$q < 0$$, by (10) it follows that

\begin{aligned} L &> k_{\beta}^{1/p}(\lambda_{1}) \Biggl[ \sum _{|n| = 2}^{\infty} \sum _{|m| = 2}^{\infty} k(m,n)\frac{(B_{\eta,\beta} (n))^{\frac{q}{p}}\ln^{\frac{(1 - \lambda_{2})q}{p}}B_{\eta,\beta} (n)}{A_{\xi,\alpha} (m)\ln^{1 - \lambda_{1}}A_{\xi,\alpha} (m)}b_{n}^{q} \Biggr]^{\frac{1}{p}} \\ &= k_{\beta}^{1/p}(\lambda_{1}) \Biggl[ \sum _{|n| = 2}^{\infty} \varpi (\lambda_{1},n) \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{p}}.\end{aligned}

By (13) and (16) we have (27).

In the same way, we find

\begin{aligned}[b] I = {}&\sum_{|m| = 2}^{\infty} \biggl[ \bigl( 1 - \theta (\lambda_{2},m) \bigr)^{\frac{1}{p}}\frac{\ln^{\frac{1}{q} - \lambda_{1}}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{ - 1/q}}a_{m} \biggr]\\&\times \Biggl[ \frac{\ln^{\frac{ - 1}{q} + \lambda_{1}}A_{\xi,\alpha} (m)}{ ( 1 - \theta (\lambda_{2},m) )^{\frac{1}{p}}(A_{\xi,\alpha} (m))^{1/q}}\sum_{|n| = 2}^{\infty} k(m,n) b_{n} \Biggr] \\\ge{}& \Biggl[ \sum_{|m| = 2}^{\infty} \bigl( 1 - \theta (\lambda_{2},m) \bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}L,\end{aligned}
(30)

and then we can prove that (27) and (25) are equivalent.

For $$0 < \varepsilon < \min \{ p\lambda_{1},|q|\lambda_{1}\}$$, we set $$\tilde{\lambda}_{1} = \lambda_{1} - \frac{\varepsilon}{p}$$ ($$\in (0,1)$$), $$\tilde{\lambda}_{2} = \lambda_{2} + \frac{\varepsilon}{p}$$ (>0), and

$$\begin{gathered} \tilde{a}_{m}: = \frac{\ln^{\lambda_{1} - \frac{\varepsilon}{p} - 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} = \frac{\ln^{\tilde{\lambda}_{1} - 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)}\quad\bigl(|m| \in \mathbb{N}\setminus \{ 1\} \bigr), \\ \tilde{b}_{n}: = \frac{\ln^{\lambda_{2} - \frac{\varepsilon}{q} - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} = \frac{\ln^{\tilde{\lambda}_{2} - \varepsilon - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)}\quad\bigl(|n| \in \mathbb{N}\setminus \{ 1\} \bigr).\end{gathered}$$

By (15) and (13) we find

\begin{gathered}\begin{aligned} \tilde{I}_{2}&: = \Biggl[ \sum_{|m| = 2}^{\infty} \bigl(1 - \theta (\lambda_{2},m)\bigr)\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}B_{\eta,\beta} (n)}{(B_{\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)} - \sum_{|m| = 2}^{\infty} \frac{O(\ln^{ - 1 - (\lambda_{2} + \varepsilon )}A_{\xi,\alpha} (m))}{A_{\xi,\alpha} (m)} \Biggr]^{\frac{1}{p}} \Biggl[ \sum _{|n| = 2}^{\infty} \frac{\ln^{ - 1 - \varepsilon} B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} \Biggr]^{\frac{1}{q}} \\ &= \frac{1}{\varepsilon} \bigl(2\csc^{2}\alpha + o(1) - \varepsilon O(1) \bigr)^{\frac{1}{p}}\bigl(2\csc^{2}\beta + \tilde{o}(1) \bigr)^{\frac{1}{q}}\quad\bigl(\varepsilon \to 0^{ +} \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} - 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} \frac{\ln^{\tilde{\lambda}_{2} - \varepsilon - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} \\ &= \sum_{|n| = 2}^{\infty} \varpi (\tilde{ \lambda}_{1},n)\frac{\ln^{ - 1 - \varepsilon} B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} \le k_{\alpha} (\tilde{ \lambda}_{1})\sum_{|n| = 2}^{\infty} \frac{\ln^{ - 1 - \varepsilon} B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} \\ &= \frac{1}{\varepsilon} k_{\alpha} (\tilde{\lambda}_{1}) \bigl(2\csc^{2}\beta + o(1)\bigr).\end{aligned}\end{gathered}

If there exists a positive number $$k \ge K(\lambda_{1})$$, such that (25) is still valid when replacing $$K(\lambda_{1})$$ by k, then in particular we have

$$\varepsilon \tilde{I} = \varepsilon \sum_{|m| = 2}^{\infty} \sum_{|n| = 2}^{\infty} k(m,n) \tilde{a}_{m} \tilde{b}_{n} > \varepsilon k\tilde{I}_{2}.$$

We obtain from the above results that

$$k_{\beta} \biggl(\lambda_{1} + \frac{\varepsilon}{q} \biggr) \bigl(2\csc^{2}\alpha + o(1) \bigr) > k \bigl(2\csc^{2} \alpha + o(1) - \varepsilon O(1) \bigr)^{\frac{1}{p}} \bigl(2\csc^{2} \beta + \tilde{o}(1) \bigr)^{\frac{1}{q}},$$

and then

$$4B(\lambda_{1},\lambda_{2})\csc^{2}\beta \csc^{2}\alpha \ge 2k\csc^{\frac{2}{p}}\alpha \csc^{\frac{2}{q}}\beta\quad \bigl(\varepsilon \to 0^{ +} \bigr),$$

namely, $$K(\lambda_{1}) = 2B(\lambda_{1},\lambda_{2})\csc^{\frac{2}{p}}\beta \csc^{\frac{2}{q}}\alpha \ge k$$. Hence, $$k = K(\lambda_{1})$$ is the best value of (25).

The constant factor $$K(\lambda_{1})$$ in (26) is still the best possible. Otherwise, we would reach a contradiction by (30) that the constant factor in (25) is not the best possible.

In the same way, by (30) we can proved that the constant factor $$K(\lambda_{1})$$ in (27) is still the best possible. □

## 4 Operator expressions

Setting $$\varphi (m): = \frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}$$ ($$|m| \in \mathbb{N}\setminus \{ 1\}$$), and $$\psi (n): = \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}}$$, wherefrom $$\psi^{1 - p}(n) = \frac{\ln^{p\lambda_{2} - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)}$$ ($$|n| \in \mathbb{N}\setminus \{ 1\}$$), we define the real weighted normed function spaces as follows:

$$\begin{gathered} l_{p,\varphi}: = \Biggl\{ a = \{ a_{m}\}_{|m| = 2}^{\infty}; \|a\|_{p,\varphi} = \Biggl( \sum_{|m| = 2}^{\infty} \varphi (m)|a_{m}|^{p} \Biggr)^{\frac{1}{p}} < \infty \Biggr\} , \\ l_{q,\psi}: = \Biggl\{ b = \{ b_{n}\}_{|n| = 2}^{\infty}; \|b\|_{q,\psi} = \Biggl( \sum_{|n| = 2}^{\infty} \psi (n)|b_{n}|^{q} \Biggr)^{\frac{1}{q}} < \infty \Biggr\} , \\ l_{p,\psi^{1 - p}}: = \Biggl\{ c = \{ c_{n}\}_{|n| = 2}^{\infty}; \|c\|_{p,\psi^{1 - p}} = \Biggl( \sum_{|n| = 2}^{\infty} \psi^{1 - p}(n)|c_{n}|^{p} \Biggr)^{\frac{1}{p}} < \infty \Biggr\} .\end{gathered}$$

For $$a = \{ a_{m}\}_{|m| = 2}^{\infty} \in l_{p,\varphi}$$, putting $$c_{n} = \sum_{|m| = 2}^{\infty} k(m,n)a_{m}$$ and $$c = \{ c_{n}\}_{|n| = 2}^{\infty}$$, it follows by (18) that $$\|c\|_{p,\psi^{1 - p}} < K(\lambda_{1})\|a\|_{p,\varphi}$$, namely $$c \in l_{p,\psi^{1 - p}}$$.

### Definition 2

Define the Mulholland-type operator $$T:l_{p,\varphi} \to l_{p,\psi^{1 - p}}$$as follows: For $$a_{m} \ge 0$$, $$a = \{ a_{m}\}_{|m| = 2}^{\infty} \in l_{p,\varphi}$$, there exists a unique representation $$Ta = c \in l_{p,\psi^{1 - p}}$$. We also define the following formal inner product of Ta and $$b = \{ b_{n}\}_{|n| = 2}^{\infty} \in l_{q,\psi}$$ ($$b_{n} \ge 0$$):

$$(Ta,b): = \sum_{|n| = 2}^{\infty} \sum _{|m| = 2}^{\infty} k(m,n) a_{m}b_{n}.$$
(31)

Hence, we may rewrite (17) and (18) in the following operator expressions:

\begin{aligned}& (Ta,b) < K(\lambda_{1})\|a\|_{p,\varphi} \|b \|_{q,\psi}, \end{aligned}
(32)
\begin{aligned}& \|Ta\|_{p,\psi^{1 - p}} < K(\lambda_{1})\|a\|_{p,\varphi}. \end{aligned}
(33)

It follows that the operator T is bounded by

$$\|T\|: = \sup_{a( \ne \theta ) \in l_{p,\varphi}} \frac{\|Ta\|_{p,\psi^{1 - p}}}{\|a\|_{p,\varphi}} \le K( \lambda_{1}).$$
(34)

Since the constant factor $$K(\lambda_{1})$$ in (18) is the best possible, we have

$$\|T\| = K(\lambda_{1}) = 2B(\lambda_{1}, \lambda_{2})\csc^{2/p}\beta \csc^{2/q} \alpha.$$
(35)

### Remark 1

1. (i)

For $$\xi = \eta = 0$$ in (19), we have the following new inequality:

\begin{aligned}[b] &\sum_{|n| = 2}^{\infty} \sum _{|m| = 2}^{\infty} \frac{a_{m}b_{n}}{\ln^{\lambda} |mn|}\\ &\quad < 2B( \lambda_{1},\lambda_{2}) \Biggl[ \sum _{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}|m|}{|m|^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}|n|}{|n|^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}}.\end{aligned}
(36)

It follows that (19) is a more accurate inequalitythan (36); so is (17).

2. (ii)

If $$a_{ - m} = a_{m}$$, $$b_{ - n} = b_{n}$$ ($$m,n \in \mathbb{N}\setminus \{ 1\}$$), $$\xi,\eta \in [0,\frac{1}{2}]$$, then (19) reduces to the following inequality:

\begin{aligned}[b] &\sum_{n = 2}^{\infty} \sum _{m = 2}^{\infty} \biggl[ \frac{1}{\ln^{\lambda} [(m - \xi )(n - \eta )]} + \frac{1}{\ln^{\lambda} [(m - \xi )(n + \eta )]} \\ &\qquad+ \frac{1}{\ln^{\lambda} [(m + \xi )(n - \eta )]} + \frac{1}{\ln^{\lambda} [(m + \xi )(n + \eta )]} \biggr]a_{m}b_{n} \\ &\quad< 2B(\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}
(37)
3. (iii)

If $$\lambda = 1$$, $$\lambda_{1} = \frac{1}{q}$$, $$\lambda_{2} = \frac{1}{p}$$, $$\xi = \eta \in [0,\frac{1}{2}]$$, then (37) reduces to

\begin{aligned}[b] &\sum_{n = 2}^{\infty} \sum _{m = 2}^{\infty} \biggl[ \frac{1}{\ln (m - \xi )(n - \xi )} + \frac{1}{\ln (m - \xi )(n + \xi )}\\ &\quad\quad + \frac{1}{\ln (m + \xi )(n - \xi )} + \frac{1}{\ln (m + \xi )(n + \xi )} \biggr]a_{m}b_{n} \\&\quad< \frac{2\pi}{\sin (\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}} \\ &\qquad\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}
(38)

For $$\xi = 0$$, (38) reduces to (3). Hence, (17) is a more accurate extension of (3).

## 5 Conclusions

In this paper, by introducing independent parameters and applying the weight coefficients and Hermite-Hadamard’s inequality we give a more accurate Mulholland-type inequality in the whole plane with a best possible constant factor in Theorems 1-2. Furthermore, the equivalent forms, the reverses in Theorem 3, a few particular cases, and the operator expressions are considered. The method of real analysis is very important, which is the key to prove the equivalent inequalities with the best possible constant factor. The lemmas and theorems provide an extensive account of this type inequalities.