Skip to main content

A new Hardy–Mulholland-type inequality with a mixed kernel

Abstract

By the use of weight coefficients and techniques of real analysis, we establish a new Hardy–Mulholland-type inequality with a mixed kernel and a best possible constant factor in terms of the hypergeometric function. Equivalent forms, an operator expression with the norm and reverses are also considered.

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.

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). \(\square \)

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)

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).

Availability of data and materials

Not applicable.

References

  1. 1.

    Arpad, B., Choonghong, O.: Best constant for certain multilinear integral operator. J. Inequal. Appl. 28582 (2006). https://doi.org/10.1155/JIA/2006/28582

  2. 2.

    Azar, L.: On some extensions of Hardy–Hilbert’s inequality and applications. J. Inequal. Appl. 546829 (2009). https://doi.org/10.1155/2008/546829

  3. 3.

    Chen, Q., He, B., Hong, Y., Zhen, L.: Equivalent parameter conditions for the validity of half-discrete Hilbert-type multiple integral inequality with generalized homogeneous kernel. J. Funct. Spaces. 2020 (2020). Article ID 7414861

  4. 4.

    Gao, P.: On weight Hardy inequalities for non-increasing sequence. J. Math. Inequal. 12(2), 551–557 (2018)

    MathSciNet  Article  Google Scholar 

  5. 5.

    Hardy, G.H., Littlewood, J.E., Pólya, G.: Inequalities. Cambridge University Press, Cambridge (1934)

    MATH  Google Scholar 

  6. 6.

    Hong, Y.: On Hardy–Hilbert integral inequalities with some parameters. J. Inequal. Pure Appl. Math. 6(4), 1–10 (2005). Art. 92

  7. 7.

    Huang, Q.L., Yang, B.C.: A more accurate Hardy–Hilbert-type inequality. J. Guangdong Univ. Educ. 35(5), 27–35 (2015)

    Google Scholar 

  8. 8.

    Krnić, M., Pečarić, J.E.: Hilbert’s inequalities and their reverses. Publ. Math. Debr. 67(3–4), 315–331 (2005)

    MathSciNet  MATH  Google Scholar 

  9. 9.

    Krnić, M., Pečarić, J.E., Vuković, P.: On some higher-dimensional Hilbert’s and Hardy–Hilbert’s type integral inequalities with parameters. Math. Inequal. Appl. 11, 701–716 (2008)

    MathSciNet  MATH  Google Scholar 

  10. 10.

    Krnić, M., Vuković, P.: On a multidimensional version of the Hilbert-type inequality. Anal. Math. 38, 291–303 (2012)

    MathSciNet  Article  Google Scholar 

  11. 11.

    Kuang, J.C.: Applied Inequalities. Shangdong Science Technic Press, Jinan (2004)

    Google Scholar 

  12. 12.

    Kuang, J.C., Debnath, L.: On Hilbert’s type inequalities on the weighted Orlicz spaces. Pac. J. Appl. Math. 1(1), 95–103 (2007)

    MathSciNet  MATH  Google Scholar 

  13. 13.

    Kuang, J.C.: Real Analysis and Functional Analysis. Higher Education Press, Beijing (2014)

    Google Scholar 

  14. 14.

    Liu, Q.: A Hilbert-type integral inequality under configuring free power and its applications. J. Inequal. Appl. 2019, 91 (2019)

    MathSciNet  Article  Google Scholar 

  15. 15.

    Mitrinović, D.S., Pečarič, J.E., Fink, A.M.: Inequalities Involving Functions and Their Integrals and Derivatives. Kluwer Academic Publishers, Boston (1991)

    Book  Google Scholar 

  16. 16.

    Rassias, M.Th., Yang, B.C.: On half-discrete Hilbert’s inequality. Appl. Math. Comput. 220, 75–93 (2013)

    MathSciNet  MATH  Google Scholar 

  17. 17.

    Rassias, M.Th., Yang, B.C.: A multidimensional half-discrete Hilbert-type inequality and the Riemann zeta function. Appl. Math. Comput. 225, 263–277 (2013)

    MathSciNet  MATH  Google Scholar 

  18. 18.

    Rassias, M.Th., Yang, B.C.: On a multidimensional Hilbert-type integral inequality associated to the gamma function. Appl. Math. Comput. 249, 408–418 (2014)

    MathSciNet  MATH  Google Scholar 

  19. 19.

    Rassias, M.Th., Yang, B.C.: On a multidimensional half-discrete Hilbert-type inequality related to the hyperbolic cotangent function. Appl. Math. Comput. 242, 800–813 (2014)

    MathSciNet  MATH  Google Scholar 

  20. 20.

    Rassias, M.Th., Yang, B.C.: A half-discrete Hilbert-type inequality in the whole plane related to the Riemann zeta function. Appl. Anal. 97(9), 1505–1525 (2018). https://doi.org/10.1080/00036811.2017.1313411

    MathSciNet  Article  MATH  Google Scholar 

  21. 21.

    Rassias, M.Th., Yang, B.C.: Equivalent conditions of a Hardy-type integral inequality related to the extended Riemann zeta function. Adv. Oper. Theory 2(3), 237–256 (2017)

    MathSciNet  MATH  Google Scholar 

  22. 22.

    Rassias, M.Th., Yang, B.C.: On a Hilbert-type integral inequality in the whole plane related to the extended Riemann zeta function. Complex Anal. Oper. Theory 13(4), 1765–1782 (2018). https://doi.org/10.1007/s11785-018-0830-5

    MathSciNet  Article  MATH  Google Scholar 

  23. 23.

    Rassias, M.Th., Yang, B.C.: On a Hilbert-type integral inequality related to the extended Hurwitz zeta function in the whole plane. Acta Appl. Math. 160, 67–80 (2019). https://doi.org/10.1007/s10440-018-0195-9

    MathSciNet  Article  MATH  Google Scholar 

  24. 24.

    Rassias, M.Th., Yang, B.C.: Equivalent properties of a Hilbert-type integral inequality with the best constant factor related the Hurwitz zeta function. Ann. Funct. Anal. 9(2), 282–295 (2018)

    MathSciNet  Article  Google Scholar 

  25. 25.

    Rassias, M.Th., Yang, B.C., Raigorodskii, A.: Two kinds of the reverse Hardy-type integral inequalities with the equivalent forms related to the extended Riemann zeta function. Appl. Anal. Discret. Math. 12, 273–296 (2018)

    MathSciNet  Article  Google Scholar 

  26. 26.

    Rassias, M.Th., Yang, B.C.: On an equivalent property of a reverse Hilbert-type integral inequality related to the extended Hurwitz-zeta function. J. Math. Inequal. 13(2), 315–334 (2019)

    MathSciNet  Article  Google Scholar 

  27. 27.

    Rassias, M.Th., Yang, B.C.: A reverse Mulholland-type inequality in the whole plane with multi-parameters. Appl. Anal. Discret. Math. 13, 290–308 (2019)

    MathSciNet  Article  Google Scholar 

  28. 28.

    Rassias, M.Th., Yang, B.C., Raigorodskii, A.: On Hardy-type integral inequality in the whole plane related to the extended Hurwitz-zeta function. J. Inequal. Appl. 2020, 94 (2020)

    Article  Google Scholar 

  29. 29.

    Rassias, M.Th., Yang, B.C., Raigorodskii, A.: On the reverse Hardy-type integral inequalities in the whole plane with the extended Riemann-zeta function. J. Math. Inequal. 14(2), 525–546 (2020)

    MathSciNet  Article  Google Scholar 

  30. 30.

    Shi, Y.P., Yang, B.C.: On a multidimensional Hilbert-type inequality with parameters. J. Inequal. Appl. 2015, 371 (2015)

    MathSciNet  Article  Google Scholar 

  31. 31.

    Shi, Y.P., Yang, B.C.: A new Hardy–Hilbert-type inequality with multiparameters and a best possible constant factor. J. Inequal. Appl. 2015, 380 (2015)

    MathSciNet  Article  Google Scholar 

  32. 32.

    Wang, Z.X., Guo, D.R.: Introduction to Special Functions. Science Press, Beijing (1979)

    Google Scholar 

  33. 33.

    Wang, A.Z., Huang, Q.L., Yang, B.C.: A strengthened Mulholland-type inequality with parameters. J. Inequal. Appl. 2015, 329 (2015)

    MathSciNet  Article  Google Scholar 

  34. 34.

    Yang, B.C.: On Hilbert’s integral inequality. J. Math. Anal. Appl. 220, 778–785 (1998)

    MathSciNet  Article  Google Scholar 

  35. 35.

    Yang, B.C.: Hilbert-Type Integral Inequalities. Bentham Science Publishers Ltd., Sharjah (2009)

    Google Scholar 

  36. 36.

    Yang, B.C.: The Norm of Operator and Hilbert-Type Inequalities. Science Press, Beijing (2009)

    Google Scholar 

  37. 37.

    Yang, B.C.: Discrete Hilbert-Type Inequalities. Bentham Science Publishers Ltd., Sharjah (2011)

    Google Scholar 

  38. 38.

    Yang, B.C.: Two Types of Multiple Half-Discrete Hilbert-Type Inequalities. Lambert Academic Publishing, Saarbrücken (2012)

    Book  Google Scholar 

  39. 39.

    Yang, B.C.: An extension of a Hardy–Hilbert-type inequality. J. Guangdong Univ. Educ. 35(3), 1–7 (2015)

    Google Scholar 

  40. 40.

    Yang, B.C., Chen, Q.: On a Hardy–Hilbert-type inequality with parameters. J. Inequal. Appl. 2015, 339 (2015)

    MathSciNet  Article  Google Scholar 

  41. 41.

    Yang, B.C., Krnić, M.: On the norm of a multi-dimensional Hilbert-type operator. Sarajevo J. Math. 7(20), 223–243 (2011)

    MathSciNet  MATH  Google Scholar 

  42. 42.

    Yang, B.C., Rassias, Th.M.: On the way of weight coefficient and research for Hilbert-type inequalities. Math. Inequal. Appl. 6(4), 625–658 (2003)

    MathSciNet  MATH  Google Scholar 

  43. 43.

    Yang, B.C., Rassias, Th.M.: On a Hilbert-type integral inequality in the subinterval and its operator expression. Banach J. Math. Anal. 4(2), 100–110 (2010)

    MathSciNet  Article  Google Scholar 

  44. 44.

    Yang, B.C., Brnetić, I., Krnić, M., Pečarić, J.E.: Generalization of Hilbert and Hardy–Hilbert integral inequalities. Math. Inequal. Appl. 8(2), 259–272 (2005)

    MathSciNet  MATH  Google Scholar 

  45. 45.

    You, M.H., Guan, Y.: On a Hilbert-type integral inequality with non-homogeneous kernel of mixed hyperbolic functions. J. Math. Inequal. 13(4), 1197–1208 (2019)

    MathSciNet  Article  Google Scholar 

  46. 46.

    Zhao, C.Z., Cheung, W.S.: On Hilbert’s inequalities with alternating signs. J. Math. Inequal. 12(1), 191–200 (2018)

    MathSciNet  Article  Google Scholar 

  47. 47.

    Zhong, W.Y.: The Hilbert-type integral inequality with a homogeneous kernel of Lambda-degree. J. Inequal. Appl. 917392 (2008). https://doi.org/10.1155/2008/917392

  48. 48.

    Zhong, W.Y., Yang, B.C.: On multiple Hardy–Hilbert’s integral inequality with kernel. J. Inequal. Appl. 2007 (2007). https://doi.org/10.1155/2007/27. Art.ID 27962

Download references

Acknowledgements

We are thankful to the mathematicians who have read the manuscript of the paper, for their constructive comments that helped improve its presentation.

Funding

Open Access funding provided by University of Zurich. B. Yang: This work is supported by the National Natural Science Foundation (no. 61772140), and Science and Technology Planning Project Item of Guangzhou City (no. 201707010229). We are grateful for their support. A. M. Raigorodskii: The research was partially supported by the grant NSh-2540.2020.1 of the Russian President supporting leading scientific schools of Russia

Author information

Affiliations

Authors

Contributions

The authors have contributed equally for the preparation of the present paper.

Corresponding author

Correspondence to Michael Th. Rassias.

Ethics declarations

Competing interests

The authors of the present paper do not have any competing interests.

Additional information

Communicated by Mario Krnic.

Rights and permissions

Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.

Reprints and Permissions

About this article

Verify currency and authenticity via CrossMark

Cite this article

Rassias, M.T., Yang, B. & Raigorodskii, A. A new Hardy–Mulholland-type inequality with a mixed kernel. Adv. Oper. Theory 6, 27 (2021). https://doi.org/10.1007/s43036-020-00123-0

Download citation

  • Received:

  • Accepted:

  • Published:

  • DOI: https://doi.org/10.1007/s43036-020-00123-0

Keywords

  • Hardy–Mulholland-type inequality
  • Weight coefficients
  • Equivalent form
  • Hypergeometric function
  • Operator
  • Reverse

Mathematics Subject Classification

  • 26D15
  • 47A07