## 1 Introduction

If $$p>1,$$ $$\frac{1}{p}+\frac{1}{q}=1,$$ $$a_{m}, b_{n}\ge 0,$$ $$a=\{a_{m}\}_{m=1}^{\infty }\in l^{p},$$ $$b=\{b_{n}\}_{n=1}^{\infty }\in l^{q},$$

\begin{aligned} \Vert a\Vert _{p}=\left( \sum _{m=1}^{\infty }a_{m}^{p}\right) ^{\frac{1}{p}}>0, \quad \Vert b\Vert _{q}>0, \end{aligned}

then we have the following Hardy–Hilbert inequality with the best possible constant factor $$\frac{\pi }{\sin (\pi /p)}$$:

\begin{aligned} \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{m+n}<\frac{\pi }{ \sin (\pi /p)}\Vert a\Vert _{p}\Vert b\Vert _{q}. \end{aligned}
(1)

We also have the following Mulholland inequality

\begin{aligned} \sum _{m=2}^{\infty }\sum _{n=2}^{\infty }\frac{a_{m}b_{n}}{\ln mn}<\frac{\pi }{\sin (\pi /p)}\left( \sum _{m=2}^{\infty }\frac{a_{m}^{p}}{m^{1-p}}\right) ^{\frac{1}{p}}\left( \sum _{n=2}^{\infty }\frac{b_{n}^{q}}{n^{1-q}}\right) ^{ \frac{1}{q}}, \end{aligned}
(2)

with the same best possible constant factor $$\frac{\pi }{\sin (\pi /p)}$$ (cf. [5]). The inequalities (1) and (2) are important in Mathematical Analysis and its various applications (cf. [5, 15, 35,36,37,38]).

If $$\mu _{i},\upsilon _{j}>0\,(i,j\in {\mathbf {N}}=\{1,2,\ldots \}),$$

\begin{aligned} U_{m}:=\sum _{i=1}^{m}\mu _{i},\quad V_{n}:=\sum _{j=1}^{n}\upsilon _{j}\qquad (m,n\in {\mathbf {N}}), \end{aligned}
(3)

then we have the following Hardy–Hilbert-type inequality (cf. Theorem 321 of [5], replacing $$\mu _{m}^{1/q}a_{m}$$ and $$\upsilon _{n}^{1/p}b_{n}$$ with $$a_{m}$$ and $$b_{n}$$) :

\begin{aligned} \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{U_{m}+V_{n}}<\frac{ \pi }{\sin \left( \frac{\pi }{p}\right) }\left( \sum _{m=1}^{\infty }\frac{a_{m}^{p}}{\mu _{m}^{p-1}}\right) ^{\frac{1}{p}}\left( \sum _{n=1}^{\infty }\frac{b_{n}^{q}}{ \upsilon _{n}^{q-1}}\right) ^{\frac{1}{q}}. \end{aligned}
(4)

For $$\mu _{i}=\upsilon _{j}=1\,(i,j\in {\mathbf {N}}),$$ inequality (4) reduces to (1).

In 1998, by introducing an independent parameter $$\lambda \in (0,1]$$, Yang [34] proved an extension of the integral analogous of (1) with the kernel $$\frac{1}{(x+y)^{\lambda }}$$ for $$p=q=2$$. Recently, Yang [36] presented the following extensions of (1) and (2): If $$\lambda _{1},\lambda _{2}\in {\mathbf {R}},\lambda _{1}+\lambda _{2}=\lambda ,k_{\lambda }(x,y)$$ is a finite non-negative homogeneous function of degree $$-\lambda ,$$ with

\begin{aligned} k(\lambda _{1})=\int _{0}^{\infty }k_{\lambda }(t,1)t^{\lambda _{1}-1}\,{\mathrm{d}}t\in {\mathbf {R}}_{+}, \end{aligned}

and $$k_{\lambda }(x,y)x^{\lambda _{1}-1}\,(k_{\lambda }(x,y)y^{\lambda _{2}-1})$$ is decreasing with respect to $$x>0\,(y>0),$$

\begin{aligned} \phi (x)=x^{p(1-\lambda _{1})-1},\quad \psi (x)=x^{q(1-\lambda _{2})-1}, \end{aligned}

then for $$a_{m},b_{n}\ge 0,$$

\begin{aligned} a=\{a_{m}\}_{m=1}^{\infty }\in l_{p,\phi }=\left\{ a;\Vert a\Vert _{p,\phi }:=\left( \sum _{m=1}^{\infty }\phi (m)|a_{m}|^{p}\right) ^{\frac{1}{p}}<\infty \right\} , \end{aligned}

$$b=\{b_{n}\}_{n=1}^{\infty }\in l_{q,\psi },$$ $$\Vert a\Vert _{p,\phi },\Vert b\Vert _{q,\psi }>0,$$ we have

\begin{aligned} \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }k_{\lambda }(m,n)a_{m}b_{n}<k(\lambda _{1})\Vert a\Vert _{p,\phi }\Vert b\Vert _{q,\psi }, \end{aligned}
(5)

where the constant factor $$k(\lambda _{1})$$ is still the best possible.

Clearly, for $$\lambda =1,$$ $$\lambda _{1}=\frac{1}{q},\lambda _{2}=\frac{1}{p} ,$$ $$k_{1}(x,y)=\frac{1}{x+y},$$ the inequality (5) reduces to (1). Some other new results including multidimensional Hilbert-type inequalities, Hardy–Mulholland-type inequalities and Hardy–Hilbert-type inequalities are established in [1,2,3,4, 6,7,8,9,10, 12, 14, 16,17,31, 33, 39,40,48].

In this paper, by the use of weight coefficients and techniques of real analysis, we prove a new Hardy–Mulholland-type inequality with the following mixed kernel

\begin{aligned} \frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}\quad (0\le -2\alpha <\lambda \le 2;m,n\in {\mathbf {N}}\backslash \{1\}). \end{aligned}

and a best possible constant factor expressed in terms of the hypergeometric function. This inequality constitutes an extension of (2). Equivalent forms, operator expressions with the norm and reverses are considered as well.

## 2 An example and some lemmas

In the sequel, we consider that

\begin{aligned} p\ne 0,1,\quad \frac{1}{p}+\frac{1}{q} =1,\quad \lambda _{1}+\lambda _{2}=\lambda ,\quad \mu _{i},\upsilon _{j}>0\ (i,j\in {\mathbf {N}}), \end{aligned}

with $$\mu _{1}=\upsilon _{1}=1,\ U_{m}$$ and $$V_{n}$$ are defined in (3)$$,a_{m},b_{n}\ge 0,$$

\begin{aligned} \Vert a\Vert _{p,\Phi _{\lambda }}:=\left( \sum _{m=2}^{\infty }\Phi _{\lambda }(m)a_{m}^{p}\right) ^{\frac{1}{p}}\quad \text{and}\quad \Vert b\Vert _{q,\Psi _{\lambda }}:=\left( \sum _{n=2}^{\infty }\Psi _{\lambda }(n)b_{n}^{q}\right) ^{\frac{1}{q}}, \end{aligned}

where

\begin{aligned} \Phi _{\lambda }(m):=\frac{(\ln U_{m})^{p(1-\lambda _{1})-1}}{U_{m}^{1-p}\mu _{m}^{p-1}},\quad \Psi _{\lambda }(n):=\frac{(\ln V_{n})^{q(1-\lambda _{2})-1}}{ V_{n}^{1-q}\upsilon _{n}^{q-1}}\qquad (m,n\in {\mathbf {N}}\backslash \{1\}). \end{aligned}

We introduce the following hypergeometric function (cf. [32]):

\begin{aligned} F(\alpha ,\beta ,\gamma ,z):=\frac{\Gamma (\gamma )}{\Gamma (\beta )\Gamma (\gamma -\beta )}\int _{0}^{1}t^{\beta -1}(1-t)^{\gamma -\beta -1}(1-zt)^{-\alpha }\,{\mathrm{d}}t, \end{aligned}
(6)

where $$\mathrm{{Re}}(\gamma )>\mathrm{{Re}}(\beta )>0,|\arg (1-z)|<\pi ;(1-zt)^{-\alpha }=1,$$ for $$z=0.$$

### Example 2.1

For $$-\alpha<\lambda _{1},\lambda _{2}\le 1\,(-2\alpha <\lambda =\lambda _{1}+\lambda _{2}\le 2),$$ we set

\begin{aligned} k_{\lambda }(x,y):=\frac{(\min \{x,y\})^{\alpha }}{(x+y)^{\lambda +\alpha }}\quad ((x,y)\in {\mathbf {R}}_{+}^{2}). \end{aligned}
1. (1)

Since $$\lambda +\alpha >-\alpha ,$$ there exists a constant $$L_{\alpha }=\max \{2^{\alpha },1\}>0,$$ such that

\begin{aligned} \frac{1}{(t+1)^{\lambda +\alpha }}\le (t+1)^{\alpha }\le L_{\alpha }\quad (t\in (0,1)). \end{aligned}

For $$\lambda _{1},\lambda _{2}>-\alpha ,$$ we get that

\begin{aligned} 0< & {} k_{\alpha }(\lambda _{1}):=\int _{0}^{\infty }k_{\lambda }(1,t)t^{\lambda _{2}-1}\,{\mathrm{d}}t=\int _{0}^{\infty }k_{\lambda }(t,1)t^{\lambda _{1}-1}\,{\mathrm{d}}t \\= & {} \int _{0}^{\infty }\frac{(\min \{t,1\})^{\alpha }}{(t+1)^{\lambda +\alpha } }t^{\lambda _{1}-1}\,{\mathrm{d}}t=\int _{0}^{1}\frac{t^{\lambda _{1}+\alpha -1}+t^{\lambda _{2}+\alpha -1}}{(t+1)^{\lambda +\alpha }}\,{\mathrm{d}}t \\\le & {} L_{\alpha }\int _{0}^{1}(t^{\lambda _{1}+\alpha -1}+t^{\lambda _{2}+\alpha -1})\,{\mathrm{d}}t=L_{\alpha }\left( \frac{1}{\lambda _{1}+\alpha }+\frac{1}{ \lambda _{2}+\alpha }\right) <\infty , \end{aligned}
(7)

and thus by (6) and (7), it follows that

\begin{aligned} k_{\alpha }(\lambda _{1})=\sum _{i=1}^{2}\frac{1}{\lambda _{i}+\alpha } F(\lambda +\alpha ,\lambda _{i}+\alpha ,\lambda _{i}+\alpha +1,-1)\in {\mathbf {R}}_{+}. \end{aligned}
(8)
1. (i)

For $$\alpha =0,$$ we obtain $$k_{\lambda }(x,y)=\frac{1}{(x+y)^{\lambda }}\, (0<\lambda \le 2)$$ and

\begin{aligned} k_{0}(\lambda _{1})=\int _{0}^{\infty }\frac{t^{\lambda _{1}-1}\,{\mathrm{d}}t}{ (t+1)^{\lambda }}=B(\lambda _{1},\lambda _{2})=\sum _{i=1}^{2}\frac{1}{ \lambda _{i}}F(\lambda ,\lambda _{i},\lambda _{i}+1,-1); \end{aligned}
(9)
2. (ii)

for $$-\alpha<\lambda +\alpha \le 1\,(<2+\alpha ),$$ we derive that

\begin{aligned} k_{\alpha }(\lambda _{1})= & {} \int _{0}^{1}\frac{t^{\lambda _{1}+\alpha -1}+t^{\lambda _{2}+\alpha -1}}{(t+1)^{\lambda +\alpha }}\,{\mathrm{d}}t \\= & {} \int _{0}^{1}\sum _{k=0}^{\infty }\left( _{_{{}}\,\,\,\,\,\,\,\,k}^{-\lambda -\alpha }\right) t^{k}(t^{\lambda _{1}+\alpha -1}+t^{\lambda _{2}+\alpha -1})\,{\mathrm{d}}t \\= & {} \int _{0}^{1}\sum _{k=0}^{\infty }(-1)^{k}\left( _{_{{}}\,\,\,\,\,\,\,\,k}^{\lambda +\alpha +k-1}\right) t^{k}(t^{\lambda _{1}+\alpha -1}+t^{\lambda _{2}+\alpha -1})\,{\mathrm{d}}t\\= & {} \int _{0}^{1}\sum _{k=0}^{\infty }\left( _{_{{}}\,\,\,\,\,\,\,\,2k}^{\lambda +\alpha +2k-1}\right) \left( 1-\frac{\lambda +\alpha +2k}{2k+1}t\right) t^{2k}(t^{\lambda _{1}+\alpha -1}+t^{\lambda _{2}+\alpha -1})\,{\mathrm{d}}t. \end{aligned}

Since

\begin{aligned} 1-\frac{\lambda +\alpha +2k}{2k+1}t\ge \frac{1-(\lambda +\alpha )+2k(1-t)}{2k+1}\ge 0, \end{aligned}

in view of the Lebesgue term by term integration theorem (cf. [13]), we obtain that

\begin{aligned} k_{\alpha }(\lambda _{1})= & {} \sum _{k=0}^{\infty }\int _{0}^{1}\left( _{_{{}}\,\,\,\,\,\,\,\,2k}^{\lambda +\alpha +2k-1}\right) \left( 1-\frac{\lambda +\alpha +2k }{2k+1}t\right) t^{2k}(t^{\lambda _{1}-1}+t^{\lambda _{2}-1})\,{\mathrm{d}}t \\= & {} \sum _{k=0}^{\infty }\left( _{_{{}}\,\,\,\,\,\,\,\,k}^{-\lambda -\alpha }\right) \int _{0}^{1}t^{k}(t^{\lambda _{1}+\alpha -1}+t^{\lambda _{2}+\alpha -1})\,{\mathrm{d}}t \\= & {} \sum _{k=0}^{\infty }\left( _{_{{}}\,\,\,\,\,\,\,\,k}^{-\lambda -\alpha }\right) \left( \frac{1}{k+\lambda _{1}+\alpha }+\frac{1}{k+\lambda _{2}+\alpha }\right) . \end{aligned}
(10)
2. (2)

Suppose that $$\alpha \le 0\,(\alpha >-1).$$ We have

\begin{aligned} \lambda +\alpha >-\alpha \ge 0,\quad 0<\lambda _{i}+\alpha \le \lambda _{i}\,(i=1,2). \end{aligned}

For $$\lambda _{2}\le 1\,(\lambda _{2}+\alpha \le \lambda _{2}\le 1)$$ and fixed $$x>0,$$ we deduce that

\begin{aligned} k_{\lambda }(x,y)\frac{1}{y^{1-\lambda _{2}}}=\left\{ \begin{array}{ll} \frac{1}{(x+y)^{\lambda +\alpha }y^{1-(\lambda _{2}+\alpha )}},&{}\quad 0<y<x \\ \frac{x^{\alpha }}{(x+y)^{\lambda +\alpha }y^{1-\lambda _{2}}},&{}\quad y\ge x \end{array} \right. \end{aligned}

is strictly decreasing with respect to $$y>0$$. Similarly, for $$\lambda _{1}\le 1$$ and fixed $$y>0,$$

\begin{aligned} k_{\lambda }(x,y)\frac{1}{x^{1-\lambda _{1}}} \end{aligned}

is strictly decreasing with respect to $$x>0$$.

### Lemma 2.2

If $$0\le -\alpha <\lambda _{1},\lambda _{2}\le 1,k_{\alpha }(\lambda _{1})$$ is defined as in (7), and we define the following weight coefficients:

\begin{aligned} \omega (\lambda _{2},m):= & {} \sum _{n=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}\frac{\upsilon _{n}\ln ^{\lambda _{1}}U_{m}}{V_{n}\ln ^{1-\lambda _{2}}V_{n}},\quad m\in {\mathbf {N}}\backslash \{1\}, \end{aligned}
(11)
\begin{aligned} \varpi (\lambda _{1},n):= & {} \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}\frac{\mu _{m}\ln ^{\lambda _{2}}V_{n}}{U_{m}\ln ^{1-\lambda _{1}}U_{m}},\quad n\in {\mathbf {N}}\backslash \{1\}, \end{aligned}
(12)

then the following inequalities hold true:

\begin{aligned} \omega (\lambda _{2},m)< & {} k_{\alpha }\,(\lambda _{1})(-\alpha <\lambda _{2}\le 1,\lambda _{1}>-\alpha ;m\in \mathbf {N\backslash }\{1\}), \end{aligned}
(13)
\begin{aligned} \varpi (\lambda _{1},n)< & {} k_{\alpha }\,(\lambda _{1})(-\alpha <\lambda _{1}\le 1,\lambda _{2}>-\alpha ;n\in \mathbf {N\backslash }\{1\}). \end{aligned}
(14)

### Proof

In view of (3), we set

\begin{aligned} \mu (t):=\mu _{m},\ t\in (m-1,m]\,\,(m\in {\mathbf {N}});\quad \upsilon (t):=\upsilon _{n},\ t\in (n-1,n]\,\,(n\in {\mathbf {N}}), \end{aligned}

and

\begin{aligned} U(x):=\int _{0}^{x}\mu (t)\,{\mathrm{d}}t\,(x\ge 0),\quad V(y):=\int _{0}^{y}\upsilon (t)\,{\mathrm{d}}t\,(y\ge 0). \end{aligned}
(15)

Then it follows that

\begin{aligned} U(m)=U_{m},\quad V(n)=V_{n}\qquad (m,n\in {\mathbf {N}}). \end{aligned}

For $$x\in (m-1,m),$$

\begin{aligned} U^{\prime }(x)=\mu (x)=\mu _{m}\quad (m\in {\mathbf {N}}); \end{aligned}

for $$y\in (n-1,n),$$

\begin{aligned} V^{\prime }(y)=\upsilon (y)=\upsilon _{n}\quad (n\in \mathbf { N}). \end{aligned}

Since V(y) is strictly increasing in $$(n-1,n](n\in {\mathbf {N}})$$, and $$-\alpha <\lambda _{2}\le 1,$$ $$\lambda _{1}>-\alpha$$, in view of Example 2.1(2), we derive that

\begin{aligned} \omega (\lambda _{2},m)= & {} \sum _{n=2}^{\infty }\int _{n-1}^{n}\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }} \frac{\ln ^{\lambda _{1}}U_{m}}{V_{n}\ln ^{1-\lambda _{2}}V_{n}}V^{\prime }(y)\,{\mathrm{d}}y \\< & {} \sum _{n=2}^{\infty }\int _{n-1}^{n}\frac{(\min \{\ln U_{m},\ln V(y)\})^{\alpha }}{(\ln U_{m}V(y))^{\lambda +\alpha }}\frac{\ln ^{\lambda _{1}}U_{m}}{V(y)\ln ^{1-\lambda _{2}}V(y)}V^{\prime }(y)\,{\mathrm{d}}y \\= & {} \int _{1}^{\infty }\frac{(\min \{\ln U_{m},\ln V(y)\})^{\alpha }}{(\ln U_{m}V(y))^{\lambda +\alpha }}\frac{\ln ^{\lambda _{1}}U_{m}}{V(y)\ln ^{1-\lambda _{2}}V(y)}V^{\prime }(y)\,{\mathrm{d}}y. \end{aligned}

Setting $$t=\frac{\ln V(y)}{\ln U_{m}},$$ we obtain that

\begin{aligned} \frac{V^{\prime }(y)}{V(y) }\,{\mathrm{d}}y=\ln U_{m}\,{\mathrm{d}}t, \quad \ln V(1)=\ln \upsilon _{1}=0\ (\upsilon _{1}=1) \end{aligned}

and

\begin{aligned} \omega (\lambda _{2},m)< & {} \int _{0}^{\frac{\ln V(\infty )}{\ln U_{m}}}\frac{ (\min \{1,t\})^{\alpha }}{(1+t)^{\lambda +\alpha }}t^{\lambda _{2}-1}\,{\mathrm{d}}t \\\le & {} \int _{0}^{\infty }\frac{(\min \{1,t\})^{\alpha }}{(1+t)^{\lambda +\alpha }}t^{\lambda _{2}-1}\,{\mathrm{d}}t=k_{\alpha }(\lambda _{1}). \end{aligned}

Hence, we obtain (13). Similarly, for $$-\alpha <\lambda _{1}\le 1,\lambda _{2}>-\alpha ,$$ we have (14). $$\square$$

### Lemma 2.3

If $$0\le -\alpha <\lambda _{1},\lambda _{2}\le 1,k_{\alpha }(\lambda _{1})$$ is defined in (7), $$U_{\infty }=V_{\infty }=\infty ,$$ there exist $$m_{0},n_{0}\in {\mathbf {N}},$$ such that $$\{\mu _{m}\}_{m=m_{0}}^{\infty }$$ and $$\{\upsilon _{n}\}_{n=n_{0}}^{\infty }$$ are decreasing, then:

1. (i)

for $$m,n\in {\mathbf {N}}\backslash \{1\},$$ we have

\begin{aligned} k_{\alpha }(\lambda _{1})(1-\theta (\lambda _{2},m))< & {} \omega (\lambda _{2},m)\quad (-\alpha <\lambda _{2}\le 1,\lambda _{1}>-\alpha ), \end{aligned}
(16)
\begin{aligned} k_{\alpha }(\lambda _{1})(1-\vartheta (\lambda _{1},n))< & {} \varpi (\lambda _{1},n)\quad (-\alpha <\lambda _{1}\le 1,\lambda _{2}>-\alpha ), \end{aligned}
(17)

where

\begin{aligned} \theta (\lambda _{2},m):= & {} \frac{1}{k_{\alpha }(\lambda _{1})}\int _{0}^{ \frac{\ln V_{n_{0}+1}}{\ln U_{m}}}\frac{(\min \{t,1\})^{\alpha }}{ (t+1)^{\lambda +\alpha }}t^{\lambda _{2}-1}\,{\mathrm{d}}t \\= & {} O\left( \frac{1}{\ln ^{\alpha +\lambda _{2}}U_{m}}\right) \in (0,1), \end{aligned}
(18)
\begin{aligned} \vartheta (\lambda _{1},n):= & {} \frac{1}{k_{\alpha }(\lambda _{1})}\int _{0}^{ \frac{\ln U_{m_{0}+1}}{\ln V_{n}}}\frac{(\min \{t,1\})^{\alpha }}{ (t+1)^{\lambda +\alpha }}t^{\lambda _{1}-1}\,{\mathrm{d}}t \\= & {} O\left( \frac{1}{\ln ^{\alpha +\lambda _{1}}V_{n}}\right) \in (0,1); \end{aligned}
(19)
2. (ii)

for any $$a>0,$$ we have

\begin{aligned} \sum _{m=2}^{\infty }\frac{\mu _{m}}{U_{m}\ln ^{1+a}U_{m}}= & {} \frac{1}{a} \left[ \frac{1}{\ln ^{a}U_{m_{0}+1}}+aO_{1}(1)\right] , \end{aligned}
(20)
\begin{aligned} \sum _{n=2}^{\infty }\frac{\upsilon _{n}}{V_{n}\ln ^{1+a}V_{n}}= & {} \frac{1}{a} \left[ \frac{1}{\ln ^{a}V_{n_{0}+1}}+aO_{2}(1)\right] . \end{aligned}
(21)

### Proof

Since $$\upsilon _{n}\ge \upsilon _{n+1}\,\,(n\ge n_{0}),-\alpha <\lambda _{2}\le 1,\lambda _{1}>-\alpha$$ and $$V(\infty )=\infty ,$$ by Example 2.1(2), we have

\begin{aligned} \omega (\lambda _{2},m)\ge & {} \sum _{n=n_{0}+1}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}\frac{ \upsilon _{n+1}\ln ^{\lambda _{1}}U_{m}}{V_{n}\ln ^{1-\lambda _{2}}V_{n}} \\> & {} \sum _{n=n_{0}+1}^{\infty }\int _{n}^{n+1}\frac{(\min \{\ln U_{m},\ln V(y)\})^{\alpha }}{(\ln U_{m}V(y))^{\lambda +\alpha }}\frac{V^{\prime }(y)\ln ^{\lambda _{1}}U_{m}}{V(y)\ln ^{1-\lambda _{2}}V(y)}\,{\mathrm{d}}y \\= & {} \int _{n_{0}+1}^{\infty }\frac{(\min \{\ln U_{m},\ln V(y)\})^{\alpha }}{ (\ln U_{m}V(y))^{\lambda +\alpha }}\frac{V^{\prime }(y)\ln ^{\lambda _{1}}U_{m}}{V(y)\ln ^{1-\lambda _{2}}V(y)}\,{\mathrm{d}}y\,\,\left( t=\frac{\ln V(y)}{\ln U_{m}}\right) \\= & {} \int _{\frac{\ln V_{n_{0}+1}}{\ln U_{m}}}^{\infty }\frac{(\min \{1,t\})^{\alpha }}{(1+t)^{\lambda +\alpha }} t^{\lambda _{2}-1}\,{\mathrm{d}}t=k_{\alpha }(\lambda _{1})(1-\theta (\lambda _{2},m))>0. \end{aligned}

For $$U_{m}\ge V_{n_{0}+1}\,(m\ge 2),$$ by Example 2.1(1), we obtain that

\begin{aligned} 0< & {} \theta (\lambda _{2},m)=\frac{1}{k_{\alpha }(\lambda _{1})}\int _{0}^{ \frac{\ln V_{n_{0}+1}}{\ln U_{m}}}\frac{(\min \{1,t\})^{\alpha }}{ (1+t)^{\lambda +\alpha }}t^{\lambda _{2}-1}\,{\mathrm{d}}t\\= & {} \frac{1}{k_{\alpha }(\lambda _{1})}\int _{0}^{\frac{\ln V_{n_{0}+1}}{\ln U_{m}}}\frac{t^{\alpha }}{(1+t)^{\lambda +\alpha }}t^{\lambda _{2}-1}\,{\mathrm{d}}t \\\le & {} \frac{L_{\alpha }}{k_{\alpha }(\lambda _{1})}\int _{0}^{\frac{\ln V_{n_{0}+1}}{\ln U_{m}}}t^{\alpha +\lambda _{2}-1}\,{\mathrm{d}}t=\frac{L_{\alpha }}{ (\alpha +\lambda _{2})k_{\alpha }(\lambda _{1})}\left( \frac{\ln V_{n_{0}+1} }{\ln U_{m}}\right) ^{\alpha +\lambda _{2}}, \end{aligned}

namely, (16) and (18) follow. Similarly, for $$-\alpha <\lambda _{1}\le 1,\lambda _{2}>-\alpha ,$$ we obtain (17) and (19).

For $$a>0,$$ we deduce that

\begin{aligned}&\sum _{m=2}^{\infty }\frac{\mu _{m}}{U_{m}\ln ^{1+a}U_{m}} =\sum _{m=2}^{m_{0}+1}\frac{\mu _{m}}{U_{m}\ln ^{1+a}U_{m}} +\sum _{m=m_{0}+2}^{\infty }\frac{\mu _{m}}{U_{m}\ln ^{1+a}U_{m}}\\&\quad <\sum _{m=2}^{m_{0}+1}\frac{\mu _{m}}{U_{m}\ln ^{1+a}U_{m}} +\sum _{m=m_{0}+2}^{\infty }\int _{m-1}^{m}\frac{U^{\prime }(x)}{U(x)\ln ^{1+a}U(x)}\,{\mathrm{d}}x \\&\quad =\sum _{m=2}^{m_{0}+1}\frac{\mu _{m}}{U_{m}\ln ^{1+a}U_{m}} +\int _{m_{0}+1}^{\infty }\frac{U^{\prime }(x)}{U(x)\ln ^{1+a}U(x)}\,{\mathrm{d}}x \\&\quad =\frac{1}{a}\left( \frac{1}{\ln ^{a}U_{m_{0}+1}}+a\sum _{m=2}^{m_{0}+1} \frac{\mu _{m}}{U_{m}\ln ^{1+a}U_{m}}\right) ,\\&\sum _{m=2}^{\infty }\frac{\mu _{m}}{U_{m}\ln ^{1+a}U_{m}}\ge \sum _{m=m_{0}+1}^{\infty }\frac{\mu _{m+1}}{U_{m}\ln ^{1+a}U_{m}} >\sum _{m=m_{0}+1}^{\infty }\int _{m}^{m+1}\frac{U^{\prime }(x)\,{\mathrm{d}}x}{U(x)\ln ^{1+a}U(x)} \\&\quad =\int _{m_{0}+1}^{\infty }\frac{U^{\prime }(x)\,{\mathrm{d}}x}{U(x)\ln ^{1+a}U(x)}=\frac{ 1}{a\ln ^{a}U_{m_{0}+1}}. \end{aligned}

Hence we obtain (20). Similarly, we derive (21). $$\square$$

### Lemma 2.4

If $$0\le -\alpha <\lambda _{1},\lambda _{2}\le 1,k_{\alpha }(\lambda _{1})$$ is defined as in (7), then for

\begin{aligned} 0<\delta <\min \{\alpha +\lambda _{1},\alpha +\lambda _{2}\}, \end{aligned}

we have

\begin{aligned} k_{\alpha }(\lambda _{1}\pm \delta )=k_{\alpha }(\lambda _{1})+o(1)(\delta \rightarrow 0^{+}). \end{aligned}
(22)

### Proof

For $$0<\delta <\min \{\alpha +\lambda _{1},\alpha +\lambda _{2}\},$$ we get that

\begin{aligned}&|k_{\alpha }(\lambda _{1}+\delta )-k_{\alpha }(\lambda _{1})|\le \int _{0}^{\infty }\frac{(\min \{t,1\})^{\alpha }t^{\lambda _{1}-1}|t^{\delta }-1|}{(t+1)^{\lambda +\alpha }}\,{\mathrm{d}}t \\&\quad =\int _{0}^{1}\frac{t^{\alpha +\lambda _{1}-1}(1-t^{\delta })}{ (t+1)^{\lambda +\alpha }}\,{\mathrm{d}}t+\int _{0}^{1}\frac{t^{\alpha +\lambda _{2}-1}(t^{-\delta }-1)}{(t+1)^{\lambda +\alpha }}\,{\mathrm{d}}t \\&\quad \le L_{\alpha }\left[ \int _{0}^{1}t^{\alpha +\lambda _{1}-1}(1-t^{\delta })\,{\mathrm{d}}t+\int _{0}^{1}t^{\alpha +\lambda _{2}-1}(t^{-\delta }-1)\,{\mathrm{d}}t\right] \\&\quad =L_{\alpha }\left[ \frac{1}{\alpha +\lambda _{1}}-\frac{1}{\alpha +\lambda _{1}+\delta }+\frac{1}{\alpha +\lambda _{2}-\delta }-\frac{1}{\alpha +\lambda _{2}}\right] \\&\quad \rightarrow 0(\delta \rightarrow 0^{+}). \end{aligned}

Similarly, we obtain

\begin{aligned}&|k_{\alpha }(\lambda _{1}-\delta )-k_{\alpha }(\lambda _{1})| \\&\quad \le L_{\alpha }\left[ \frac{1}{\alpha +\lambda _{1}-\delta }-\frac{1}{ \alpha +\lambda _{1}}+\frac{1}{\alpha +\lambda _{2}}-\frac{1}{\alpha +\lambda _{2}+\delta }\right] \\&\quad \rightarrow 0\,(\delta \rightarrow 0^{+}). \end{aligned}

Hence, we derive (22). $$\square$$

## 3 Main results and operator expressions

We also set the following functions:

\begin{aligned} {\widetilde{\Phi }}_{\lambda }(m):= & {} \omega (\lambda _{2},m)\frac{(\ln U_{m})^{p(1-\lambda _{1})-1}}{U_{m}^{1-p}\mu _{m}^{p-1}}, \\ {\widetilde{\Psi }}_{\lambda }(n):= & {} \varpi (\lambda _{1},n)\frac{(\ln V_{n})^{q(1-\lambda _{2})-1}}{V_{n}^{1-q}\upsilon _{n}^{q-1}}\quad (m,n\in {\mathbf {N}}\backslash \{1\}). \end{aligned}
(23)

### Theorem 3.1

If $$0\le -\alpha <\lambda _{1},\lambda _{2}\le 1,$$ then:

1. (i)

for $$p>1,$$ we have the following equivalent inequalities:

\begin{aligned} I:= & {} \sum _{n=2}^{\infty }\sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}b_{n}\le \Vert a\Vert _{p,{\widetilde{\Phi }}_{\lambda }}\Vert b\Vert _{q,{\widetilde{\Psi }}_{\lambda }}, \end{aligned}
(24)
\begin{aligned} J:= & {} \left\{ \sum _{n=2}^{\infty }\frac{\upsilon _{n}\ln ^{p\lambda _{2}-1}V_{n}}{(\varpi (\lambda _{1},n))^{p-1}V_{n}}\left[ \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}\right] ^{p}\right\} ^{\frac{1}{p}} \\\le & {} \Vert a\Vert _{p,{\widetilde{\Phi }}_{\lambda }}; \end{aligned}
(25)
2. (ii)

for $$0<p<1$$ (or $$p<0),$$ we have the equivalent reverses of (24) and (25).

### Proof

(i) By Hölder’s inequality with weight (cf. [11]) and (12), we have

\begin{aligned}&\left[ \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{ (\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}\right] ^{p} \\&\quad =\left[ \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha } }{(\ln U_{m}V_{n})^{\lambda +\alpha }}\right. \\&\qquad \times \left. \left( \frac{U_{m}^{1/q}(\ln U_{m})^{(1-\lambda _{1})/q}}{ (\ln V_{n})^{(1-\lambda _{2})/p}\mu _{m}^{1/q}}a_{m}\right) \left( \frac{ (\ln V_{n})^{(1-\lambda _{2})/p}\mu _{m}^{1/q}}{U_{m}^{1/q}(\ln U_{m})^{(1-\lambda _{1})/q}}\right) \right] ^{p} \\&\quad \le \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{ (\ln U_{m}V_{n})^{\lambda +\alpha }}\frac{U_{m}^{p-1}(\ln U_{m})^{(1-\lambda _{1})p/q}}{(\ln V_{n})^{1-\lambda _{2}}\mu _{m}^{p/q}}a_{m}^{p} \\&\qquad \times \left[ \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}\frac{(\ln V_{n})^{(1-\lambda _{2})(q-1)}\mu _{m}}{U_{m}(\ln U_{m})^{1-\lambda _{1}}}\right] ^{p-1} \\&\quad =\frac{(\varpi (\lambda _{1},n))^{p-1}V_{n}}{(\ln V_{n})^{p\lambda _{2}-1}\upsilon _{n}} \\&\qquad \times \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{ (\ln U_{m}V_{n})^{\lambda +\alpha }}\frac{\upsilon _{n}U_{m}^{p-1}(\ln U_{m})^{(1-\lambda _{1})(p-1)}a_{m}^{p}}{V_{n}(\ln V_{n})^{1-\lambda _{2}}\mu _{m}^{p-1}}. \end{aligned}
(26)

Hence by (11), we deduce that

\begin{aligned} J\le & {} \left[ \sum _{n=2}^{\infty }\sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}\frac{ \upsilon _{n}U_{m}^{p-1}(\ln U_{m})^{(1-\lambda _{1})(p-1)}}{V_{n}(\ln V_{n})^{1-\lambda _{2}}\mu _{m}^{p-1}}a_{m}^{p}\right] ^{\frac{1}{p}} \\= & {} \left[ \sum _{m=2}^{\infty }\sum _{n=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}\frac{ \upsilon _{n}(\ln U_{m})^{\lambda _{1}}}{V_{n}(\ln V_{n})^{1-\lambda _{2}}} \frac{(\ln U_{m})^{p(1-\lambda _{1})-1}}{U_{m}^{1-p}\mu _{m}^{p-1}}a_{m}^{p} \right] ^{\frac{1}{p}} \\= & {} \left[ \sum _{m=2}^{\infty }\omega (\lambda _{2},m)\frac{(\ln U_{m})^{p(1-\lambda _{1})-1}}{U_{m}^{1-p}\mu _{m}^{p-1}}a_{m}^{p}\right] ^{ \frac{1}{p}}, \end{aligned}
(27)

and then (25) follows.

By Hölder’s inequality (cf. [11]), we have

\begin{aligned} I= & {} \sum _{n=2}^{\infty }\left[ \frac{(\ln V_{n})^{\lambda _{2}-\frac{1}{p} }\upsilon _{n}^{1/p}}{(\varpi (\lambda _{1},n))^{\frac{1}{q}}V_{n}^{\frac{1}{ p}}}\sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}\right] \\&\times \left[ (\varpi (\lambda _{1},n))^{\frac{1}{q}}\frac{(\ln V_{n})^{ \frac{1}{p}-\lambda _{2}}}{V_{n}^{\frac{-1}{p}}\upsilon _{n}^{\frac{1}{p}}} b_{n}\right] \le J\Vert b\Vert _{q,\widetilde{\Psi }_{\lambda }}. \end{aligned}
(28)

Then by (25), we derive (24).

On the other hand, assuming that (24) holds true, we set

\begin{aligned} b_{n}:=\frac{(\ln V_{n})^{p\lambda _{2}-1}\upsilon _{n}}{(\varpi (\lambda _{1},n))^{p-1}V_{n}}\left[ \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}\right] ^{p-1},\quad n\in {\mathbf {N}}\backslash \{1\}. \end{aligned}
(29)

Then we obtain that $$J^{p}=\Vert b\Vert _{q,{\widetilde{\Psi }}_{\lambda }}^{q}.$$ If $$J=0,$$ then (25) is trivially valid; if $$J=\infty ,$$ then by (27), (25) takes the form of equality ($$=\infty$$). Suppose that $$0<J<\infty .$$ By (24), it follows that

\begin{aligned} \Vert b\Vert _{q,{\widetilde{\Psi }}_{\lambda }}^{q}= & {} J^{p}=I\le \Vert a\Vert _{p, {\widetilde{\Phi }}_{\lambda }}\Vert b\Vert _{q,{\widetilde{\Psi }}_{\lambda }}, \end{aligned}
(30)
\begin{aligned} \Vert b\Vert _{q,{\widetilde{\Psi }}_{\lambda }}^{q-1}= & {} J\le \Vert a\Vert _{p,{\widetilde{\Phi }}_{\lambda }}, \end{aligned}
(31)

and then (25) follows, which is equivalent to (24).

(ii) For $$0<p<1$$ (or $$p<0),$$ by the reverse Hölder inequality with weight (cf. [11]) and (12), we obtain the reverse of (26) (or (26)). Then by (11), we derive the reverse of (27), and thus the reverse of (25) follows. By Hölder’s inequality (cf. [11]), we obtain the reverse of (28) and then by the reverse of (25), we deduce the reverse of (24).

On the other hand, assuming that the reverse of (24) holds true, we set $$b_{n}$$ as in (29). Then we obtain that $$J^{p}=\Vert b\Vert _{q,{\widetilde{\Psi }} _{\lambda }}^{q}.$$ If $$J=\infty ,$$ then the reverse of (25) is trivially valid; if $$J=0,$$ then by the reverse of (27), (25) takes the form of equality ($$=0$$). Suppose that $$0<J<\infty .$$ By the reverse of (24), it follows that the reverses of (30) and (31) are valid and then the reverse of (25) follows, which is equivalent to the reverse of (24). $$\square$$

### Theorem 3.2

If $$0\le -\alpha <\lambda _{1},\lambda _{2}\le 1,k_{\alpha }(\lambda _{1})$$ is defined as in (8), there exist $$m_{0},n_{0}\in {\mathbf {N}},$$ such that $$\{\mu _{m}\}_{m=m_{0}}^{\infty }$$ and $$\{\upsilon _{n}\}_{n=n_{0}}^{\infty }$$ are decreasing, $$U_{\infty }=V_{\infty }=\infty$$, then for $$p>1$$, $$\Vert a\Vert _{p,\Phi _{\lambda }}\in {\mathbf {R}}_{+}$$ and $$\Vert b\Vert _{q,\Psi _{\lambda }}\in {\mathbf {R}}_{+},$$ we have the following equivalent inequalities:

\begin{aligned}&\sum _{n=2}^{\infty }\sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}b_{n}<k_{\alpha }(\lambda _{1})\Vert a\Vert _{p,\Phi _{\lambda }}\Vert b\Vert _{q,\Psi _{\lambda }}, \end{aligned}
(32)
\begin{aligned}&J_{1} :=\left\{ \sum _{n=2}^{\infty }\frac{\upsilon _{n}\ln ^{p\lambda _{2}-1}V_{n}}{V_{n}}\left[ \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}\right] ^{p}\right\} ^{\frac{1}{p}} \\&\quad <k_{\alpha }(\lambda _{1})\Vert a\Vert _{p,\Phi _{\lambda }}, \end{aligned}
(33)

where the constant factor $$k_{\alpha }(\lambda _{1})$$ is the best possible.

### Proof

Applying (13) and (14) in (24) and (25), since

\begin{aligned} (\omega (\lambda _{2},m))^{\frac{1}{p}}<(k_{\alpha }(\lambda _{1}))^{\frac{1 }{p}}\quad (p>1),\qquad (\varpi (\lambda _{1},n))^{\frac{1}{q}}<(k_{\alpha }(\lambda _{1}))^{\frac{1}{q}}\quad (q>1), \end{aligned}

and

\begin{aligned} \frac{1}{(k_{\alpha }(\lambda _{1}))^{p-1}}<\frac{1}{(\varpi (\lambda _{1},n))^{p-1}}\quad (p>1), \end{aligned}

by simplification, we obtain the equivalent inequalities (32) and (33).

For $$\varepsilon \in (0,q(\alpha +\lambda _{2})),$$ we set $${\widetilde{\lambda }}_{1}=\lambda _{1}+\frac{\varepsilon }{q}\,(>-\alpha ),{\widetilde{\lambda }}_{2}=\lambda _{2}-\frac{\varepsilon }{q}\,(\in (-\alpha ,1)),$$ and $${\widetilde{a}}=\{{\widetilde{a}}_{m}\}_{m=2}^{\infty },$$ $${\widetilde{b}}=\{ {\widetilde{b}}_{n}\}_{n=2}^{\infty },$$

\begin{aligned} {\widetilde{a}}_{m}:= & {} \frac{\mu _{m}}{U_{m}}\ln ^{\widetilde{\lambda } _{1}-\varepsilon -1}U_{m}=\frac{\mu _{m}}{U_{m}}\ln ^{\lambda _{1}-\frac{\varepsilon }{p}-1}U_{m}, \\ {\widetilde{b}}_{n}= & {} \frac{\upsilon _{n}}{V_{n}}\ln ^{{\widetilde{\lambda }} _{2}-1}V_{n}=\frac{\upsilon _{n}}{V_{n}}\ln ^{\lambda _{2}-\frac{\varepsilon }{q}-1}V_{n}. \end{aligned}
(34)

Then by (20), (21) and (19), we have

\begin{aligned}&\Vert {\widetilde{a}}\Vert _{p,\Phi _{\lambda }}\Vert {\widetilde{b}}\Vert _{q,\Psi _{\lambda }}=\left( \sum _{m=2}^{\infty }\frac{\mu _{m}}{U_{m}\ln ^{1+\varepsilon }U_{m} }\right) ^{\frac{1}{p}}\left( \sum _{n=2}^{\infty }\frac{\upsilon _{n}}{ V_{n}\ln ^{1+\varepsilon }V_{n}}\right) ^{\frac{1}{q}} \\&\quad =\frac{1}{\varepsilon }\left[ \frac{1}{\ln ^{\varepsilon }U_{m_{0}+1}} +\varepsilon O_{1}(1)\right] ^{\frac{1}{p}}\left[ \frac{1}{\ln ^{\varepsilon }V_{n_{0}+1}}+\varepsilon O_{2}(1)\right] ^{\frac{1}{q}},\\&{\widetilde{I}} :=\sum _{n=2}^{\infty }\sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}\widetilde{ a}_{m}{\widetilde{b}}_{n} \\&\quad =\sum _{m=2}^{\infty }\left[ \sum _{n=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}\frac{ \upsilon _{n}\ln ^{{\widetilde{\lambda }}_{1}}U_{m}}{V_{n}\ln ^{1-{\widetilde{\lambda }}_{2}}V_{n}}\right] \frac{\mu _{m}}{U_{m}\ln ^{\varepsilon +1}U_{m}}\\&\quad =\sum _{m=2}^{\infty }\omega ({\widetilde{\lambda }}_{2},m)\frac{\mu _{m}}{ U_{m}\ln ^{\varepsilon +1}U_{m}} \\&\quad \ge k_{\alpha }({\widetilde{\lambda }}_{1})\sum _{m=2}^{\infty }\left( 1-O\left( \frac{1 }{\ln ^{{\widetilde{\lambda }}_{2}+\alpha }U_{m}}\right) \right) \frac{\mu _{m}}{U_{m}\ln ^{\varepsilon +1}U_{m}} \\&\quad = k_{\alpha }({\widetilde{\lambda }}_{1})\left[ \sum _{m=2}^{\infty }\frac{\mu _{m}}{U_{m}\ln ^{\varepsilon +1}U_{m}}-\sum _{m=2}^{\infty }O\left( \frac{\mu _{m}}{U_{m}(\ln U_{m})^{\left( \frac{\varepsilon }{p}+\alpha +\lambda _{2}\right) +1}} \right) \right] \\&\quad =\frac{1}{\varepsilon }k_{\alpha }\left( \lambda _{1}+\frac{\varepsilon }{q}\right) \left[ \frac{1}{\ln ^{\varepsilon }U_{m_{0}+1}}+\varepsilon (O_{1}(1)-O(1)) \right] . \end{aligned}

If there exists a positive constant $$K\le k_{\alpha }(\lambda _{1}),$$ such that (32) is valid when replacing $$k_{\alpha }(\lambda _{1})$$ by K,  then in particular, we have $$\varepsilon {\widetilde{I}}<\varepsilon K\Vert {\widetilde{a}}\Vert _{p,\Phi _{\lambda }}\Vert {\widetilde{b}}\Vert _{q,\Psi _{\lambda }},$$ namely

\begin{aligned}&k_{\alpha }\left( \lambda _{1}+\frac{\varepsilon }{q}\right) \left[ \frac{1}{\ln ^{\varepsilon }U_{m_{0}+1}}+\varepsilon (O_{1}(1)-O(1))\right] \\&\quad <K\left[ \frac{1}{\ln ^{\varepsilon }U_{m_{0}+1}}+\varepsilon O_{1}(1) \right] ^{\frac{1}{p}}\left[ \frac{1}{\ln ^{\varepsilon }V_{n_{0}+1}} +\varepsilon O_{2}(1)\right] ^{\frac{1}{q}}. \end{aligned}
(35)

In view of (22), it follows that $$k_{\alpha }(\lambda _{1})\le K\,(\varepsilon \rightarrow 0^{+}).$$ Hence, $$K=k_{\alpha }(\lambda _{1})$$ is the best possible constant factor of (32).

Similarly to (28), we can still obtain that

\begin{aligned} I\le J_{1}\Vert b\Vert _{q,\Psi _{\lambda }}. \end{aligned}
(36)

Hence, we can prove that the constant factor $$k_{\alpha }(\lambda _{1})$$ in (33) is the best possible. Otherwise, we would reach a contradiction by (36) that the constant factor in (32) is not the best possible.

For $$p>1,$$

\begin{aligned} \Psi _{\lambda }^{1-p}(n)=\frac{\upsilon _{n}}{V_{n}}(\ln V_{n})^{p\lambda _{2}-1}\quad (n\in {\mathbf {N}}\backslash \{1\}), \end{aligned}

we define the following normed spaces:

\begin{aligned} l_{p,\Phi _{\lambda }}:= & {} \{a=\{a_{m}\}_{m=2}^{\infty };\Vert a\Vert _{p,\Phi _{\lambda }}<\infty \}, \\ l_{q,\Psi _{\lambda }}:= & {} \{b=\{b_{n}\}_{n=2}^{\infty };\Vert b\Vert _{q,\Psi _{\lambda }}<\infty \}, \\ l_{p,\Psi _{\lambda }^{1-p}}:= & {} \{c=\{c_{n}\}_{n=2}^{\infty };\Vert c\Vert _{p,\Psi _{\lambda }^{1-p}}<\infty \}. \end{aligned}

Assuming that $$a=\{a_{m}\}_{m=2}^{\infty }\in l_{p,\Phi _{\lambda }},$$ setting

\begin{aligned} c=\{c_{n}\}_{n=2}^{\infty },\quad c_{n}:=\sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m},\ n\in {\mathbf {N}}\backslash \{1\}, \end{aligned}

we can rewrite (33) as follows:

\begin{aligned}\Vert c\Vert _{p,\Psi _{\lambda }^{1-p}}<k_{\alpha }(\lambda _{1})\Vert a\Vert _{p,\Phi _{\lambda }}<\infty ,\end{aligned}

namely, $$c\in l_{p,\Psi _{\lambda }^{1-p}}.$$

### Definition 3.3

Define a Hardy–Mulholland-type operator $$T:l_{p,\Phi _{\lambda }}\rightarrow l_{p,\Psi _{\lambda }^{1-p}}$$ as follows: For any $$a=\{a_{m}\}_{m=2}^{\infty }\in l_{p,\Phi _{\lambda }},$$ there exists a unique representation $$Ta=c\in l_{p,\Psi _{\lambda }^{1-p}}$$. Define the formal inner product of Ta and $$b=\{b_{n}\}_{n=2}^{\infty }\in l_{q,\Psi _{\lambda }}$$ as follows:

\begin{aligned} (Ta,b):=\sum _{n=2}^{\infty }\left[ \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m} \right] b_{n}. \end{aligned}
(37)

Then we can rewrite (32) and (33) as bellow:

\begin{aligned} (Ta,b)< & {} k_{\alpha }(\lambda _{1})\Vert a\Vert _{p,\Phi _{\lambda }}\Vert b\Vert _{q,\Psi _{\lambda }}, \end{aligned}
(38)
\begin{aligned} \Vert Ta\Vert _{p,\Psi _{\lambda }^{1-p}}< & {} k_{\alpha }(\lambda _{1})\Vert a\Vert _{p,\Phi _{\lambda }}. \end{aligned}
(39)

Define the norm of operator T as follows:

\begin{aligned} \Vert T\Vert :=\sup _{a(\ne \theta )\in l_{p,\Phi _{\lambda }}}\frac{\Vert Ta\Vert _{p,\Psi _{\lambda }^{1-p}}}{\Vert a\Vert _{p,\Phi _{\lambda }}}. \end{aligned}

Then by (39), we derive that $$\Vert T\Vert \le k_{\alpha }(\lambda _{1}).$$ Since the constant factor in (39) is the best possible, we have

\begin{aligned} \Vert T\Vert =k_{\alpha }(\lambda _{1})=\sum _{i=1}^{2}\frac{1}{\lambda _{i}+\alpha } F(\lambda +\alpha ,\lambda _{i}+\alpha ,\lambda _{i}+\alpha +1,-1). \end{aligned}
(40)

## 4 Some equivalent reverses

In the following, we also set

\begin{aligned} {\widetilde{\Omega }}_{\lambda }(m):= & {} (1-\theta (\lambda _{2},m))\frac{(\ln U_{m})^{p(1-\lambda _{1})-1}}{U_{m}^{1-p}\mu _{m}^{p-1}}, \\ {\widetilde{\Upsilon }}_{\lambda }(n):= & {} (1-\vartheta (\lambda _{1},n))\frac{ (\ln V_{n})^{q(1-\lambda _{2})-1}}{V_{n}^{1-q}\upsilon _{n}^{q-1}}\quad (m,n\in {\mathbf {N}}\backslash \{1\}). \end{aligned}
(41)

For $$0<p<1$$ or $$p<0,$$ we still use $$\Vert a\Vert _{p,\Phi _{\lambda }}$$, $$\Vert b\Vert _{q,\Psi _{\lambda }},\Vert a\Vert _{p,{\widetilde{\Omega }}_{\lambda }}$$ and $$\Vert b\Vert _{q,{\widetilde{\Upsilon }}_{\lambda }}$$ as the formal symbols.

### Theorem 4.1

With regard to the assumptions of Theorem 3.2, if $$0<p<1,$$ $$\Vert a\Vert _{p,\Phi _{\lambda }}\in {\mathbf {R}}_{+}$$ and $$\Vert b\Vert _{q,\Psi _{\lambda }}\in {\mathbf {R}}_{+},$$ then we have the following equivalent inequalities with the best possible constant factor $$k_{\alpha }(\lambda _{1}):$$

\begin{aligned} \sum _{n=2}^{\infty }\sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}b_{n}>k_{\alpha }(\lambda _{1})\Vert a\Vert _{p,\widetilde{\Omega }_{\lambda }}\Vert b\Vert _{q,\Psi _{\lambda }}, \end{aligned}
(42)
\begin{aligned} \left\{ \sum _{n=2}^{\infty }\frac{\upsilon _{n}\ln ^{p\lambda _{2}-1}V_{n}}{ V_{n}}\left[ \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}\right] ^{p}\right\} ^{\frac{1}{p}}>k_{\alpha }(\lambda _{1})\Vert a\Vert _{p,{\widetilde{\Omega }}_{\lambda }}. \end{aligned}
(43)

### Proof

Using (16) and (14) in the reverses of (24) and (25), since

\begin{aligned} (\omega (\lambda _{2},m))^{\frac{1}{p}}> & {} (k_{\alpha }(\lambda _{1}))^{ \frac{1}{p}}(1-\theta (\lambda _{2},m))^{\frac{1}{p}}\quad (0<p<1), \\ (\varpi (\lambda _{1},n))^{\frac{1}{q}}> & {} (k_{\alpha }(\lambda _{1}))^{ \frac{1}{q}}\quad (q<0), \end{aligned}

and

\begin{aligned} \frac{1}{(k_{\alpha }(\lambda _{1}))^{p-1}}>\frac{1}{(\varpi (\lambda _{1},n))^{p-1}}\quad (0<p<1), \end{aligned}

by simplification, we obtain the equivalent inequalities (42) and (43).

For $$\varepsilon \in (0,p(\alpha +\lambda _{1})),$$ we set

\begin{aligned}&\widetilde{\lambda }_{1}=\lambda _{1}-\frac{\varepsilon }{p}\,(\in (-\alpha ,1)),\quad {\widetilde{\lambda }}_{2}=\lambda _{2}+\frac{\varepsilon }{p}\,(>-\alpha ),\quad \text{and} \\&{\widetilde{a}}=\{{\widetilde{a}}_{m}\}_{m=2}^{\infty },\quad {\widetilde{b}}=\{ {\widetilde{b}}_{n}\}_{n=2}^{\infty },\\&{\widetilde{a}}_{m} :=\frac{\mu _{m}}{U_{m}}\ln ^{\widetilde{\lambda } _{1}-1}U_{m}=\frac{\mu _{m}}{U_{m}}\ln ^{\lambda _{1}-\frac{\varepsilon }{p}-1}U_{m}, \\&{\widetilde{b}}_{n} =\frac{\upsilon _{n}}{V_{n}}\ln ^{{\widetilde{\lambda }} _{2}-\varepsilon -1}V_{n}=\frac{\upsilon _{n}}{V_{n}}\ln ^{\lambda _{2}- \frac{\varepsilon }{q}-1}V_{n}. \end{aligned}

Then by (20), (21) and (14), we obtain that

\begin{aligned}&\Vert a\Vert _{p,{\widetilde{\Omega }}_{\lambda }}\Vert b\Vert _{q,\Psi _{\lambda }} \\&\quad =\left[ \sum _{m=2}^{\infty }(1-\theta (\lambda _{2},m))\frac{\mu _{m}}{ U_{m}\ln ^{1+\varepsilon }U_{m}}\right] ^{\frac{1}{p}}\left( \sum _{n=2}^{\infty }\frac{\upsilon _{n}}{V_{n}\ln ^{1+\varepsilon }V_{n}}\right) ^{\frac{1}{q}} \\&\quad =\left( \sum _{m=2}^{\infty }\frac{\mu _{m}}{U_{m}\ln ^{1+\varepsilon }U_{m} }-\sum _{m=2}^{\infty }O(\frac{\mu _{m}}{U_{m}\ln ^{1+\alpha +\lambda _{2}+\varepsilon }U_{m}})\right) ^{\frac{1}{p}} \\&\qquad \times \left( \sum _{n=2}^{\infty }\frac{\upsilon _{n}}{V_{n}\ln ^{1+\varepsilon }V_{n}}\right) ^{\frac{1}{q}} \\&\quad =\frac{1}{\varepsilon }\left[ \frac{1}{\ln ^{\varepsilon }U_{m_{0}+1}} +\varepsilon (O_{1}(1)-O(1))\right] ^{\frac{1}{p}}\left[ \frac{1}{\ln ^{\varepsilon }V_{n_{0}+1}}+\varepsilon O_{2}(1)\right] ^{\frac{1}{q}},\\&{\widetilde{I}} :=\sum _{n=2}^{\infty }\sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}\widetilde{ a}_{m}{\widetilde{b}}_{n} \\&\quad =\sum _{n=2}^{\infty }\left[ \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}\frac{\mu _{m}\ln ^{{\widetilde{\lambda }}_{2}}V_{n}}{U_{m}\ln ^{1-{\widetilde{\lambda }} _{1}}U_{m}}\right] \frac{\upsilon _{n}}{V_{n}\ln ^{\varepsilon +1}V_{n}}\\&\quad =\sum _{n=2}^{\infty }\varpi (\widetilde{\lambda }_{1},n)\frac{\upsilon _{n} }{V_{n}\ln ^{\varepsilon +1}V_{n}}\le k_{\alpha }({\widetilde{\lambda }} _{1})\sum _{n=2}^{\infty }\frac{\upsilon _{n}}{V_{n}\ln ^{\varepsilon +1}V_{n}} \\&\quad =\frac{1}{\varepsilon }k_{\alpha }\left( \lambda _{1}-\frac{\varepsilon }{p}\right) \left[ \frac{1}{\ln ^{\varepsilon }V_{n_{0}+1}}+\varepsilon O_{2}(1)\right] . \end{aligned}

If there exists a positive constant $$K\ge k_{\alpha }(\lambda _{1}),$$ such that (42) is valid when replacing $$k_{\alpha }(\lambda _{1})$$ by K,  then in particular, we have $$\varepsilon {\widetilde{I}}>\varepsilon K\Vert {\widetilde{a}}\Vert _{p,\widetilde{\Omega }_{\lambda }}\Vert {\widetilde{b}}\Vert _{q,\Psi _{\lambda }},$$ namely

\begin{aligned}&k_{\alpha }\left( \lambda _{1}-\frac{\varepsilon }{p}\right) \left[ \frac{1}{\ln ^{\varepsilon }V_{n_{0}+1}}+\varepsilon O_{2}(1)\right] \\&\quad >K\left[ \frac{1}{\ln ^{\varepsilon }U_{m_{0}+1}}+\varepsilon (O_{1}(1)-O(1))\right] ^{\frac{1}{p}}\left[ \frac{1}{\ln ^{\varepsilon }V_{n_{0}+1}}+\varepsilon O_{2}(1)\right] ^{\frac{1}{q}}. \end{aligned}

In view of (22), it follows that $$k_{\alpha }(\lambda _{1})\ge K\,(\varepsilon \rightarrow 0^{+}).$$ Hence, $$K=k_{\alpha }(\lambda _{1})$$ is the best possible constant factor of (42). The constant factor $$k_{\alpha }(\lambda _{1})$$ in (43) is still the best possible. Otherwise, we would reach a contradiction by the reverse of (36) that the constant factor in (42) is not the best possible.

### Theorem 4.2

With regard to the assumptions of Theorem 3.2, if $$p<0,\ \Vert a\Vert _{p,\Phi _{\lambda }}\in {\mathbf {R}}_{+}$$ and $$\Vert b\Vert _{q,\Psi _{\lambda }}\in {\mathbf {R}}_{+},$$ then we have the following equivalent inequalities with the best possible constant factor $$B(\lambda _{1},\lambda _{2})$$:

\begin{aligned}&\sum _{n=2}^{\infty }\sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}b_{n}>k_{\alpha }(\lambda _{1})\Vert a\Vert _{p,\Phi _{\lambda }}\Vert b\Vert _{q,{\widetilde{\Upsilon }} _{\lambda }}, \end{aligned}
(44)
\begin{aligned}&J_{2} :=\left\{ \sum _{n=2}^{\infty }\frac{\upsilon _{n}\ln ^{p\lambda _{2}-1}V_{n}}{(1-\vartheta (\lambda _{1},n))^{p-1}V_{n}}\left[ \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}\right] ^{p}\right\} ^{\frac{1}{p}} \\&\quad >k_{\alpha }(\lambda _{1})\Vert a\Vert _{p,\Phi _{\lambda }}. \end{aligned}
(45)

### Proof

Using (13) and (17) in the reverses of (24) and (25), since

\begin{aligned} (\omega (\lambda _{2},m))^{\frac{1}{p}}> & {} (k_{\alpha }(\lambda _{1}))^{ \frac{1}{p}}\quad (p<0), \\ (\varpi (\lambda _{1},n))^{\frac{1}{q}}> & {} (k_{\alpha }(\lambda _{1}))^{ \frac{1}{q}}(1-\vartheta (\lambda _{1},n))^{\frac{1}{q}}\quad (0<q<1), \end{aligned}

and

\begin{aligned} \left[ \frac{1}{(k_{\alpha }(\lambda _{1}))^{p-1}(1-\vartheta (\lambda _{1},n))^{p-1}}\right] ^{\frac{1}{p}}>\left[ \frac{1}{(\varpi (\lambda _{1},n))^{p-1}}\right] ^{ \frac{1}{p}}\quad (p<0), \end{aligned}

by simplification, we obtain equivalent inequalities (44) and (45).

For $$\varepsilon \in (0,q(\alpha +\lambda _{2})),$$ we set

\begin{aligned} {\widetilde{\lambda }}_{1}= & {} \lambda _{1}+\frac{\varepsilon }{q}\,(>-\alpha ),\quad {\widetilde{\lambda }}_{2}=\lambda _{2}-\frac{\varepsilon }{q}\,(\in (-\alpha ,1)),\quad \text{and}\quad {\widetilde{a}}=\{{\widetilde{a}}_{m}\}_{m=2}^{\infty },\quad {\widetilde{b}}=\{ {\widetilde{b}}_{n}\}_{n=2}^{\infty },\\ {\widetilde{a}}_{m}= & {} \frac{\mu _{m}}{U_{m}}\ln ^{\widetilde{\lambda } _{1}-\varepsilon -1}U_{m}=\frac{\mu _{m}}{U_{m}}\ln ^{\lambda _{1}-\frac{ \varepsilon }{p}-1}U_{m}, \\ {\widetilde{b}}_{n}= & {} \frac{\upsilon _{n}}{V_{n}}\ln ^{{\widetilde{\lambda }} _{2}-1}V_{n}=\frac{\upsilon _{n}}{V_{n}}\ln ^{\lambda _{2}-\frac{\varepsilon }{q}-1}V_{n}. \end{aligned}

Then by (20), (21) and (12), we have

\begin{aligned}&\Vert {\widetilde{a}}\Vert _{p,\Phi _{\lambda }}\Vert {\widetilde{b}}\Vert _{q,{\widetilde{q}}_{\lambda }} \\&\quad =\left( \sum _{m=2}^{\infty }\frac{\mu _{m}}{U_{m}\ln ^{\varepsilon +1}U_{m} }\right) ^{\frac{1}{p}}\left[ \sum _{n=2}^{\infty }(1-\vartheta (\lambda _{1},n))\frac{\upsilon _{n}}{V_{n}\ln ^{\varepsilon +1}V_{n}}\right] ^{\frac{1}{q}} \\&\quad =\left( \sum _{m=2}^{\infty }\frac{\mu _{m}}{U_{m}\ln ^{\varepsilon +1}U_{m} }\right) ^{\frac{1}{p}} \\&\qquad \times \left[ \sum _{n=2}^{\infty }\frac{\upsilon _{n}}{V_{n}\ln ^{\varepsilon +1}V_{n}}-\sum _{n=2}^{\infty }O\left( \frac{\upsilon _{n}}{V_{n}\ln ^{1+(\alpha +\lambda _{1}+\varepsilon )}V_{n}}\right) \right] ^{\frac{1}{q}} \\&\quad =\frac{1}{\varepsilon }\left[ \frac{1}{\ln ^{\varepsilon }U_{m_{0}+1}} +\varepsilon O_{1}(1)\right] ^{\frac{1}{p}}\left[ \frac{1}{\ln ^{\varepsilon }V_{n_{0}+1}}+\varepsilon (O_{2}(1)-O(1))\right] ^{\frac{1}{q}},\\&{\widetilde{I}} =\sum _{m=2}^{\infty }\sum _{n=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}{\widetilde{a}}_{m}{\widetilde{b}}_{n} \\&\quad =\sum _{m=2}^{\infty }\left[ \sum _{n=2}^{\infty }\frac{\ln ^{{\widetilde{\lambda }}_{1}}U_{m}}{\ln ^{\lambda }(U_{m}V_{n})}\frac{\upsilon _{n}}{V_{n}} \ln ^{\widetilde{\lambda }_{2}-1}V_{n}\right] \frac{\mu _{m}}{U_{m}\ln ^{\varepsilon +1}U_{m}}\\&\quad =\sum _{m=2}^{\infty }\omega ({\widetilde{\lambda }}_{2},m)\frac{\mu _{m}}{ U_{m}\ln ^{\varepsilon +1}U_{m}}\le k_{\alpha }({\widetilde{\lambda }} _{1})\sum _{n=2}^{\infty }\frac{\mu _{m}}{U_{m}\ln ^{\varepsilon +1}U_{m}} \\&\quad =\frac{1}{\varepsilon }k_{\alpha }\left( \lambda _{1}+\frac{\varepsilon }{q}\right) \left[ \frac{1}{\ln ^{\varepsilon }U_{m_{0}+1}}+\varepsilon O_{1}(1)\right] . \end{aligned}

If there exists a positive constant $$K\ge k_{\alpha }(\lambda _{1}),$$ such that (44) is satisfied when replacing $$k_{\alpha }(\lambda _{1})$$ by K,  then in particular, we have $$\varepsilon {\widetilde{I}}>\varepsilon K\Vert {\widetilde{a}}\Vert _{p,\Phi _{\lambda }}\Vert {\widetilde{b}}\Vert _{q,{\widetilde{\Upsilon }}_{\lambda }},$$ namely

\begin{aligned}&k_{\alpha }\left( \lambda _{1}+\frac{\varepsilon }{q}\right) \left[ \frac{1}{\ln ^{\varepsilon }U_{m_{0}+1}}+\varepsilon O_{1}(1)\right] >K\left[ \frac{1}{ \ln ^{\varepsilon }U_{m_{0}+1}}+\varepsilon O_{1}(1)\right] ^{\frac{1}{p}} \\&\quad \times \left[ \frac{1}{\ln ^{\varepsilon }V_{n_{0}+1}}+\varepsilon (O_{2}(1)-O(1))\right] ^{\frac{1}{q}}. \end{aligned}

In view of (22), it follows that $$k_{\alpha }(\lambda _{1})\ge K(\varepsilon \rightarrow 0^{+}).$$ Hence, $$K=k_{\alpha }(\lambda _{1})$$ is the best possible constant factor of (44). Similarly to the reverse of (28), we still obtain that

\begin{aligned} I\ge J_{2}\Vert b\Vert _{q,{\widetilde{\Upsilon }}_{\lambda }}. \end{aligned}
(46)

Hence, the constant factor $$k_{\alpha }(\lambda _{1})$$ in (45) is still the best possible. Otherwise, we would reach a contradiction by (46) that the constant factor in (44) is not the best possible. $$\square$$

### Remark 4.3

For $$\alpha =0$$ in (32), by (9), we have

\begin{aligned} \sum _{n=2}^{\infty }\sum _{m=2}^{\infty }\frac{a_{m}b_{n}}{(\ln U_{m}V_{n})^{\lambda }}<B(\lambda _{1},\lambda _{2})\Vert a\Vert _{p,\Phi _{\lambda }}\Vert b\Vert _{q,\Psi _{\lambda }}. \end{aligned}
(47)

For $$\lambda =1,\lambda _{1}=\frac{1}{q},\lambda _{2}=\frac{1}{p},\mu _{i}=\upsilon _{i}=1(i\in {\mathbf {N}})$$ in (47), we deduce (2). Hence, (47) is an extension of (2); so is (32).