# On a more accurate Hardy-Mulholland-type inequality

## Abstract

By using weight coefficients, technique of real analysis, and Hermite-Hadamard’s inequality, we give a more accurate Hardy-Mulholland-type inequality with multiparameters and a best possible constant factor related to the beta function. The equivalent forms, the reverses, the operator expressions, and some particular cases are also considered.

## Introduction

Assuming that $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$a_{m},b_{n}\geq 0$$, $$a=\{a_{m}\}_{m=1}^{\infty}\in l^{p}$$, $$b=\{b_{n}\}_{n=1}^{\infty }\in l^{q}$$, $$\|a\|_{p}=(\sum_{m=1}^{\infty}a_{m}^{p})^{\frac{1}{p}}>0$$, and $$\|b\|_{q}>0$$, we have the following Hardy-Hilbert inequality with the best possible constant factor $$\frac{\pi}{\sin(\pi/p)}$$ (see , Theorem 315):

$$\sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{a_{m}b_{n}}{m+n}< \frac {\pi}{\sin(\pi/p)}\|a \|_{p}\|b\|_{q}.$$
(1)

The more accurate and extended inequality of (1) is given as follows (see , Theorem 323 and ):

$$\sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{a_{m}b_{n}}{m+n-\alpha }< \frac{\pi}{\sin(\pi/p)}\|a \|_{p}\|b\|_{q} \quad (0\leq\alpha\leq1),$$
(2)

where the constant factor $$\frac{\pi}{\sin(\pi/p)}$$ is the best possible. Also, we have the following Mulholland inequality similar to (1) with the same best possible constant factor $$\frac{\pi}{\sin (\pi/p)}$$ (see  or , Theorem 343, replacing $$\frac {a_{m}}{n}$$, $$\frac{b_{n}}{n}$$ by $$a_{m}$$, $$b_{n}$$):

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

Inequalities (1)-(3) are important in analysis and its applications (see [1, 2, 420]).

Suppose that $$\mu_{i},\upsilon_{j}>0$$ ($$i,j\in\mathbf{N}=\{1,2,\ldots \}$$) and

$$U_{m}:=\sum_{i=1}^{m} \mu_{i},\qquad V_{n}:=\sum_{j=1}^{n} \upsilon_{j}\quad (m,n\in \mathbf{N}).$$
(4)

Then we have the following Hardy-Hilbert-type inequality (, Theorem 321):

$$\sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{\mu_{m}^{1/q}\upsilon _{n}^{1/p}a_{m}b_{n}}{U_{m}+V_{n}}< \frac{\pi}{\sin(\pi/p)}\|a\|_{p}\|b\|_{q}.$$
(5)

For $$\mu_{i}=\upsilon_{j}=1$$ ($$i,j\in\mathbf{N}$$), inequality (5) reduces to (1). Replacing $$\mu_{m}^{1/q}a_{m}$$ and $$\upsilon _{n}^{1/p}b_{n}$$ by $$a_{m}$$ and $$b_{n}$$ in (5), respectively, we obtain the equivalent form of (5) as follows:

$$\sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac {a_{m}b_{n}}{U_{m}+V_{n}}< \frac{\pi}{\sin(\frac{\pi}{p})} \Biggl( \sum_{m=1}^{\infty}\frac {a_{m}^{p}}{\mu _{m}^{p-1}} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=1}^{\infty} \frac {b_{n}^{q}}{\upsilon_{n}^{q-1}} \Biggr) ^{\frac{1}{q}}.$$
(6)

In 2015, Yang  gave the following extension of (6). For $$0<\lambda_{1},\lambda_{2}\leq1$$, $$\lambda_{1}+\lambda_{2}=\lambda$$, decreasing sequences $$\{\mu_{m}\}_{m=1}^{\infty}$$ and $$\{\upsilon_{n}\}_{n=1}^{\infty}$$, and $$U_{\infty}=V_{\infty}=\infty$$, we have the following inequality with the best possible constant factor $$B(\lambda _{1},\lambda _{2})$$:

\begin{aligned}& \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{a_{m}b_{n}}{(U_{m}+V_{n})^{\lambda}} \\& \quad < B(\lambda_{1},\lambda_{2}) \Biggl[ \sum _{m=1}^{\infty}\frac{U_{m}^{p(1-\lambda_{1})-1}a_{m}^{p}}{\mu_{m}^{p-1}} \Biggr] ^{\frac {1}{p}} \Biggl[ \sum_{n=1}^{\infty} \frac{V_{n}^{q(1-\lambda_{2}-1)}b_{n}^{q}}{\upsilon_{n}^{q-1}} \Biggr] ^{\frac{1}{q}}, \end{aligned}
(7)

where $$B(u,v)$$ is the beta function (see ):

$$B(u,v):= \int_{0}^{\infty}\frac{t^{u-1}}{(1+t)^{u+v}}\,dt\quad (u,v>0).$$
(8)

In this paper, by using weight coefficients, technique of real analysis, and the Hermite-Hadamard inequality, we give a Hardy-Mulholland-type inequality with a best possible constant factor $$\frac{\pi}{\sin(\pi/p)}$$ as follows.

For $$\mu_{1}=\upsilon_{1}=$$1, decreasing sequences $$\{\mu_{m}\} _{m=1}^{\infty}$$ and $$\{\upsilon _{n}\}_{n=1}^{\infty}$$, and $$U_{\infty}=V_{\infty}=\infty$$, we have

\begin{aligned}& \sum_{m=2}^{\infty}\sum _{n=2}^{\infty}\frac{a_{m}b_{n}}{\ln(U_{m}V_{n})} \\& \quad < \frac{\pi}{\sin(\pi/p)} \Biggl[ \sum_{m=2}^{\infty} \biggl(\frac {U_{m}}{\mu _{m+1}} \biggr)^{p-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=2}^{\infty } \biggl(\frac{V_{n}}{\upsilon_{n+1}} \biggr)^{q-1}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned}
(9)

which is an extension of (3). So, we have obtained a more accurate and extended inequality of (9) with multiparameters and a best possible constant factor $$B(\lambda _{1},\lambda_{2})$$. We also consider the equivalent forms, the reverses, the operator expressions, and some particular cases.

## Some lemmas and an example

In the following, we make appointment that $$p\neq0,1$$, $$\frac{1}{p}+\frac {1}{q}=1$$, $$0<\lambda_{1},\lambda_{2}\leq1$$, $$\lambda_{1}+\lambda _{2}=\lambda$$, $$\mu_{i},\upsilon_{j}>0$$ ($$i,j\in\mathbf{N}$$), with $$\mu _{1}=\upsilon _{1}=1$$, $$U_{m}$$ and $$V_{n}$$ are defined by (4), $$\frac {1}{1+\frac{\mu_{2}}{2}}\leq\alpha\leq1$$, $$\frac{1}{1+\frac{\upsilon _{2}}{2}}\leq \beta\leq1$$, $$a_{m},b_{n}\geq0$$, $$\|a\|_{p,\Phi_{\lambda }}:=(\sum_{m=2}^{\infty}\Phi_{\lambda}(m)a_{m}^{p})^{\frac{1}{p}}$$, and $$\|b\|_{q,\Psi_{\lambda}}:=(\sum_{n=2}^{\infty}\Psi_{\lambda }(n)b_{n}^{q})^{\frac{1}{q}}$$, where

\begin{aligned} &\Phi_{\lambda}(m) := \biggl(\frac{U_{m}}{\mu_{m+1}} \biggr)^{p-1}(\ln\alpha U_{m})^{p(1-\lambda_{1})-1}\quad \bigl(m\in \mathbf{N}\backslash\{1\} \bigr), \\ &\Psi_{\lambda}(n) := \biggl(\frac{V_{n}}{\upsilon_{n+1}} \biggr)^{q-1}(\ln \beta V_{n})^{q(1-\lambda_{2})-1}\quad \bigl(n\in\mathbf{N}\backslash\{1\} \bigr). \end{aligned}
(10)

### Lemma 1

If $$a\in\mathbf{R}$$, $$f(x)$$ is continuous in $$[a-\frac{1}{2},a+\frac{1}{2}]$$, and $$f^{\prime}(x)$$ is strictly increasing in the intervals $$(a-\frac{1}{2},a)$$ and $$(a,a+\frac{1}{2})$$ and satisfying

$$\lim_{x\rightarrow a-}f^{\prime}(x)=f^{\prime}(a-0)\leq f^{\prime }(a+0)=\lim_{x\rightarrow a+}f^{\prime}(x),$$

then we have the following Hermite-Hadamard inequality (cf. ):

$$f(a)< \int_{a-\frac{1}{2}}^{a+\frac{1}{2}}f(x)\,dx.$$
(11)

### Proof

Since $$f^{\prime}(a-0)$$ ($$\leq f^{\prime}(a+0)$$) is finite, we define the linear function $$g(x)$$ as follows:

$$g(x):=f^{\prime}(a-0) (x-a)+f(a),\quad x\in \biggl[ a- \frac{1}{2},a+ \frac{1}{2} \biggr].$$

Since $$f^{\prime}(x)$$ is strictly increasing in $$(a-\frac{1}{2},a)$$, we have that, for $$x\in(a-\frac{1}{2},a)$$,

$$\bigl(f(x)-g(x) \bigr)^{\prime}=f^{\prime}(x)-f^{\prime}(a-0)< 0.$$

Since $$f(a)-g(a)=0$$, it follows that $$f(x)-g(x)>0$$, $$x\in(a-\frac{1}{2},a)$$. In the same way, we obtain $$f(x)-g(x)>0$$, $$x\in(a,a+\frac{1}{2})$$. Hence, we find

$$\int_{a-\frac{1}{2}}^{a+\frac{1}{2}}f(x)\,dx> \int_{a-\frac{1}{2}}^{a+\frac {1}{2}}g(x)\,dx=f(a),$$

that is, (11) follows. □

### Example 1

If $$\{\mu_{m}\}_{m=1}^{\infty}$$ and $$\{\upsilon _{n}\}_{n=1}^{\infty}$$ are decreasing, then we define the functions $$\mu(t):=\mu_{m}$$, $$t\in(m-1,m]$$ ($$m\in\mathbf{N}$$); $$\upsilon(t):=\upsilon_{n}$$, $$t\in (n-1,n]$$ ($$n\in\mathbf{N}$$), and

$$U(x):= \int_{0}^{x}\mu(t)\,dt\quad (x\geq0),\qquad V(y):= \int_{0}^{y}\upsilon (t)\,dt\quad (y\geq 0).$$
(12)

Then it follows that $$U(m)=U_{m}$$, $$V(n)=V_{n}$$, $$U(\infty)=U_{\infty }$$, $$V(\infty)=V_{\infty}$$, and

\begin{aligned}& U^{\prime}(x) = \mu(x)=\mu_{m},\quad x\in(m-1,m), \\& V^{\prime}(y) = \upsilon(y)=\upsilon_{n},\quad y\in(n-1,n)\ (m,n\in \mathbf{N}). \end{aligned}

For fixed $$m,n\in\mathbf{N}\backslash\{1\}$$, we also define the function

$$f(x):=\frac{\ln^{\lambda_{2}-1}\beta V(x)}{V(x)(\ln\alpha U_{m}+\ln \beta V(x))^{\lambda}},\quad x\in \biggl[ n-\frac{1}{2},n+ \frac{1}{2} \biggr].$$

Then $$f(x)$$ is continuous in $$[n-\frac{1}{2},n+\frac{1}{2}]$$. For $$x\in (n-\frac{1}{2},n)$$ ($$n\in\mathbf{N}\backslash\{1\}$$), we find

\begin{aligned} f^{\prime}(x) =&- \biggl[ \frac{\ln^{\lambda_{2}-1}\beta V(x)}{V(x)}+\frac{\lambda\ln^{\lambda_{2}-1}\beta V(x)}{\ln\alpha U_{m}+\ln\beta V(x)}+ \frac{1-\lambda_{2}}{V^{2-\lambda_{2}}(x)} \biggr] \\ &{}\times\frac {\upsilon_{n}}{V(x)(\ln\alpha U_{m}+\ln\beta V(x))^{\lambda}}. \end{aligned}

Since $$1-\lambda_{2}\geq0$$, it follows that $$f^{\prime}(x)$$ (<0) is strictly increasing in $$(n-\frac{1}{2},n)$$ and

\begin{aligned} \lim_{x\rightarrow n-}f^{\prime}(x) =&f^{\prime}(n-0) \\ =&- \biggl[ \frac {\ln^{\lambda_{2}-1}\beta V_{n}}{V_{n}}+\frac{\lambda\ln^{\lambda _{2}-1}\beta V_{n}}{\ln\alpha U_{m}+\ln\beta V_{n}}+\frac{1-\lambda_{2}}{V_{n}^{2-\lambda_{2}}} \biggr] \\ &{}\times\frac{\upsilon_{n}}{V_{n}(\ln\alpha U_{m}+\ln\beta V_{n})^{\lambda}}. \end{aligned}

In the same way, for $$x\in(n,n+\frac{1}{2})$$ ($$n\in\mathbf {N}\backslash\{1\}$$), we find

\begin{aligned} f^{\prime}(x) =&- \biggl[ \frac{\ln^{\lambda_{2}-1}\beta V(x)}{V(x)}+\frac{\lambda\ln^{\lambda_{2}-1}\beta V(x)}{\ln\alpha U_{m}+\ln\beta V(x)}+ \frac{1-\lambda_{2}}{V^{2-\lambda_{2}}(x)} \biggr] \\ &{}\times\frac {\upsilon _{n+1}}{V(x)(\ln\alpha U_{m}+\ln\beta V(x))^{\lambda}}, \end{aligned}

so that $$f^{\prime}(x)$$ (<0) is strict increasing in $$(n,n+\frac {1}{2})$$. In view of $$\upsilon_{n+1}\leq\upsilon_{n}$$, it follows that

$$\lim_{x\rightarrow n+}f^{\prime}(x)=f^{\prime}(n+0)\geq f^{\prime}(n-0).$$

Then by (11), for $$m,n\in\mathbf{N}\backslash\{1\}$$, we have

$$f(n)< \int_{n-\frac{1}{2}}^{n+\frac{1}{2}}f(x)\,dx= \int_{n-\frac{1}{2}}^{n+ \frac{1}{2}}\frac{\ln^{\lambda_{2}-1}\beta V(x)}{V(x)(\ln\alpha U_{m}+\ln \beta V(x))^{\lambda}}\,dx.$$
(13)

### Definition 1

Define the following weight coefficients:

\begin{aligned}& \omega(\lambda_{2},m) : =\sum_{n=2}^{\infty} \frac{1}{\ln^{\lambda }(\alpha\beta U_{m}V_{n})}\frac{\upsilon_{n+1}\ln^{\lambda _{1}}\alpha U_{m}}{V_{n}\ln^{1-\lambda_{2}}\beta V_{n}},\quad m\in\mathbf {N}\backslash\{1\}, \end{aligned}
(14)
\begin{aligned}& \varpi(\lambda_{1},n) : =\sum_{m=2}^{\infty} \frac{1}{\ln^{\lambda }(\alpha\beta U_{m}V_{n})}\frac{\mu_{m+1}\ln^{\lambda_{2}}\beta V_{n}}{U_{m}\ln^{1-\lambda_{1}}\alpha U_{m}},\quad n\in\mathbf{N}\backslash\{ 1\}. \end{aligned}
(15)

### Lemma 2

If $$\{\mu_{m}\}_{m=1}^{\infty}$$ and $$\{\upsilon _{n}\}_{n=1}^{\infty}$$ are decreasing and $$U_{\infty}=V_{\infty }=\infty$$, then for $$m,n\in\mathbf{N}\backslash\{1\}$$, we have the following inequalities:

\begin{aligned}& \omega(\lambda_{2},m) < B(\lambda_{1}, \lambda_{2})\quad (0< \lambda _{2}\leq1,\lambda_{1}>0), \end{aligned}
(16)
\begin{aligned}& \varpi(\lambda_{1},n) < B(\lambda_{1}, \lambda_{2})\quad (0< \lambda _{1}\leq1, \lambda_{2}>0). \end{aligned}
(17)

### Proof

For $$x\in(n-\frac{1}{2},n+\frac{1}{2})\backslash \{n\}$$, $$\upsilon_{n+1}\leq V^{\prime}(x)$$, by (13) we find

\begin{aligned} \omega(\lambda_{2},m) < &\sum_{n=2}^{\infty} \upsilon_{n+1} \int _{n-\frac{1}{2}}^{n+\frac{1}{2}}\frac{\ln^{\lambda_{1}}\alpha U_{m}\ln^{\lambda _{2}-1}\beta V(x)}{V(x)(\ln\alpha U_{m}+\ln\beta V(x))^{\lambda}}\,dx \\ \leq&\sum_{n=2}^{\infty} \int_{n-\frac{1}{2}}^{n+\frac{1}{2}}\frac{\ln ^{\lambda_{1}}\alpha U_{m}\ln^{\lambda_{2}-1}\beta V(x)}{(\ln\alpha U_{m}+\ln\beta V(x))^{\lambda}}\frac{V^{\prime}(x)}{V(x)} \,dx \\ =& \int_{\frac{3}{2}}^{\infty}\frac{\ln^{\lambda_{1}}\alpha U_{m}\ln ^{\lambda_{2}-1}\beta V(x)}{(\ln\alpha U_{m}+\ln\beta V(x))^{\lambda }}\frac{V^{\prime}(x)}{V(x)} \,dx. \end{aligned}

Setting $$t=\frac{\ln\beta V(x)}{\ln\alpha U_{m}}$$, since $$\beta V(\frac{3}{2})=\beta(1+\frac{\upsilon_{2}}{2})\geq1$$ and $$\frac{V^{\prime }(x)}{V(x)}\,dx=(\ln\alpha U_{m})\,dt$$, we find

$$\omega(\lambda_{2},m)< \int_{0}^{\infty}\frac{1}{(1+t)^{\lambda}}t^{\lambda_{2}-1} \,dt=B(\lambda_{1},\lambda_{2}).$$

Hence, we obtain (16). In the same way, we obtain (17). □

### Note

For example, $$\mu_{n},\upsilon_{n}=\frac{1}{n^{\sigma}}$$ ($$0\leq\sigma\leq1$$) satisfy the conditions of Lemma 2.

### Lemma 3

With the assumptions of Lemma  2, (i) for $$m,n\in\mathbf{N}\backslash\{1\}$$, we have

\begin{aligned}& B(\lambda_{1},\lambda_{2}) \bigl(1-\theta( \lambda_{2},m) \bigr) < \omega (\lambda _{2},m)\quad (0< \lambda_{2}\leq1,\lambda_{1}>0), \end{aligned}
(18)
\begin{aligned}& B(\lambda_{1},\lambda_{2}) \bigl(1-\vartheta( \lambda_{1},n) \bigr) < \varpi (\lambda_{1},n)\quad (0< \lambda_{1}\leq1,\lambda_{2}>0), \end{aligned}
(19)

where

\begin{aligned}& \theta(\lambda_{2},m) = \frac{1}{B(\lambda_{1},\lambda_{2})}\frac {\ln ^{\lambda_{2}}\beta(1+\upsilon_{2})}{\lambda_{2}[1+\frac{\ln\beta (1+\theta(m)\upsilon_{2})}{\ln\alpha U_{m}}]^{\lambda}} \frac{1}{\ln ^{\lambda_{2}}\alpha U_{m}} \\& \hphantom{\theta(\lambda_{2},m)} = O \biggl(\frac{1}{\ln^{\lambda_{2}}\alpha U_{m}} \biggr) \\& \hphantom{\theta(\lambda_{2},m)}\in(0,1)\quad \biggl(\theta(m)\in \biggl( \frac{1-\beta}{\beta\upsilon_{2}},1 \biggr) \biggr), \end{aligned}
(20)
\begin{aligned}& \vartheta(\lambda_{1},n) = \frac{1}{B(\lambda_{1},\lambda _{2})}\frac{\ln^{\lambda_{1}}\alpha(1+\mu_{2})}{\lambda_{1}[1+\frac{\ln\alpha (1+\vartheta(n)\mu_{2})}{\ln\beta V_{n}}]^{\lambda}} \frac{1}{\ln ^{\lambda_{1}}\beta V_{n}} \\& \hphantom{\vartheta(\lambda_{1},n)} = O \biggl(\frac{1}{\ln^{\lambda_{1}}\beta V_{n}} \biggr) \\& \hphantom{\vartheta(\lambda_{1},n)}\in(0,1)\quad \biggl(\vartheta (n)\in \biggl(\frac{1-\alpha}{\alpha\mu_{2}},1 \biggr) \biggr); \end{aligned}
(21)

(ii) for any $$c>0$$, we have

\begin{aligned}& \sum_{m=2}^{\infty}\frac{\mu_{m+1}}{U_{m}\ln^{1+c}\alpha U_{m}} = \frac{1}{c} \biggl( \frac{1}{\ln^{c}\alpha(1+\mu_{2})}+cO(1) \biggr) , \end{aligned}
(22)
\begin{aligned}& \sum_{n=2}^{\infty}\frac{\upsilon_{n+1}}{V_{n}\ln^{1+c}\beta V_{n}} = \frac{1}{c} \biggl( \frac{1}{\ln^{c}\beta(1+\upsilon_{2})}+c\widetilde {O}(1) \biggr) . \end{aligned}
(23)

### Proof

In view of $$\beta\leq1$$ and $$\beta\geq\frac{1}{ 1+\upsilon_{2}/2}>\frac{1}{1+\upsilon_{2}}$$, it follows that $$1\leq \frac{1-\beta}{\beta\upsilon_{2}}+1<2$$. Since, by Example 1, $$f(x)$$ is strictly decreasing in $$[n,n+1]$$, for $$m\in\mathbf{N}\backslash\{1\}$$, we find

\begin{aligned} \omega(\lambda_{2},m) >&\sum_{n=2}^{\infty} \int_{n}^{n+1}\upsilon _{n+1}\frac{\ln^{\lambda_{1}}\alpha U_{m}\ln^{\lambda_{2}-1}\beta V(x)}{V(x)(\ln\alpha U_{m}+\ln\beta V(x))^{\lambda}}\,dx \\ =& \int_{2}^{\infty}\frac{\ln^{\lambda_{1}}\alpha U_{m}\ln^{\lambda _{2}-1}\beta V(x)}{(\ln\alpha U_{m}+\ln\beta V(x))^{\lambda}}\frac{V^{\prime}(x)}{V(x)}\,dx \\ =& \int_{\frac{1-\beta}{\beta\upsilon_{2}}+1}^{\infty}\frac{\ln ^{\lambda_{1}}\alpha U_{m}\ln^{\lambda_{2}-1}\beta V(x)}{(\ln\alpha U_{m}+\ln\beta V(x))^{\lambda}}\frac{V^{\prime}(x)}{V(x)} \,dx \\ &{}- \int_{\frac{1-\beta}{\beta\upsilon_{2}}+1}^{2}\frac{\ln^{\lambda _{1}}\alpha U_{m}\ln^{\lambda_{2}-1}\beta V(x)}{(\ln\alpha U_{m}+\ln \beta V(x))^{\lambda}}\frac{V^{\prime}(x)}{V(x)} \,dx. \end{aligned}

Setting $$t=\frac{\ln\beta V(x)}{\ln\alpha U_{m}}$$, we have $$\ln\beta V(\frac{1-\beta}{\beta\upsilon_{2}}+1)=\ln\beta(1+\frac{1-\beta }{\beta\upsilon_{2}}\upsilon_{2})=0$$ and

\begin{aligned} \omega(\lambda_{2},m) >& \int_{0}^{\infty}\frac{1}{(1+t)^{\lambda}}t^{\lambda_{2}-1} \,dt \\ &{}- \int_{\frac{1-\beta}{\beta\upsilon_{2}}+1}^{2}\frac{\ln^{\lambda _{1}}\alpha U_{m}\ln^{\lambda_{2}-1}\beta V(x)}{(\ln\alpha U_{m}+\ln \beta V(x))^{\lambda}}\frac{V^{\prime}(x)}{V(x)} \,dx \\ =&B(\lambda_{1},\lambda_{2}) \bigl(1-\theta( \lambda_{2},m) \bigr), \end{aligned}

where

\begin{aligned} \theta(\lambda_{2},m) :=&\frac{1}{B(\lambda_{1},\lambda_{2})} \int_{ \frac{1-\beta}{\beta\upsilon_{2}}+1}^{2}\frac{\ln^{\lambda _{1}}\alpha U_{m}\ln^{\lambda_{2}-1}\beta V(x)}{(\ln\alpha U_{m}+\ln\beta V(x))^{\lambda}}\frac{V^{\prime}(x)}{V(x)} \,dx \\ \in&(0,1). \end{aligned}

There exists $$\theta(m)\in(\frac{1-\beta}{\beta\upsilon_{2}},1)$$ such that

\begin{aligned} \theta(\lambda_{2},m) =&\frac{1}{B(\lambda_{1},\lambda_{2})}\frac {\ln ^{\lambda_{1}}\alpha U_{m}}{(\ln\alpha U_{m}+\ln\beta V(1+\theta (m)))^{\lambda}} \\ &{}\times \int_{\frac{1-\beta}{\beta\upsilon_{2}}+1}^{2}\ln^{\lambda _{2}-1}\beta V(x) \frac{V^{\prime}(x)}{V(x)}\,dx \\ =&\frac{1}{B(\lambda_{1},\lambda_{2})}\frac{\ln^{\lambda_{1}}\alpha U_{m}\ln^{\lambda_{2}}\beta(1+\upsilon_{2})}{\lambda_{2}(\ln\alpha U_{m}+\ln\beta V(1+\theta(m)))^{\lambda}} \\ =&\frac{1}{B(\lambda_{1},\lambda_{2})}\frac{\ln^{\lambda_{2}}\beta (1+\upsilon_{2})}{\lambda_{2}[1+\frac{\ln\beta(1+\theta(m)\upsilon _{2})}{\ln\alpha U_{m}}]^{\lambda}}\frac{1}{\ln^{\lambda_{2}}\alpha U_{m}}. \end{aligned}

Since we find

$$0< \theta(\lambda_{2},m)\leq\frac{\ln^{\lambda_{2}}\beta(1+\upsilon _{2})}{\lambda_{2}B(\lambda_{1},\lambda_{2})}\frac{1}{\ln^{\lambda _{2}}\alpha U_{m}},$$

namely, $$\theta(\lambda_{2},m)=O(\frac{1}{\ln^{\lambda_{2}}\alpha U_{m}})$$, we obtain (18) and (20). In the same way, we obtain (19) and (21).

For any $$c>0$$, we find

\begin{aligned}& \sum_{m=2}^{\infty}\frac{\mu_{m+1}}{U_{m}\ln^{1+c}\alpha U_{m}}\leq \sum _{m=2}^{\infty}\frac{\mu_{m}}{U_{m}\ln^{1+c}\alpha U_{m}} \\& \hphantom{\sum_{m=2}^{\infty}\frac{\mu_{m+1}}{U_{m}\ln^{1+c}\alpha U_{m}}} = \frac{\mu_{2}}{U_{2}\ln^{1+c}\alpha U_{2}}+\sum_{m=3}^{\infty } \frac{\mu_{m}}{U_{m}\ln^{1+c}\alpha U_{m}} \\& \hphantom{\sum_{m=2}^{\infty}\frac{\mu_{m+1}}{U_{m}\ln^{1+c}\alpha U_{m}}} = \frac{\mu_{2}}{U_{2}\ln^{1+c}\alpha U_{2}}+\sum_{m=3}^{\infty } \int_{m-1}^{m}\frac{U^{\prime}(x)}{U_{m}\ln^{1+c}\alpha U_{m}}\,dx \\& \hphantom{\sum_{m=2}^{\infty}\frac{\mu_{m+1}}{U_{m}\ln^{1+c}\alpha U_{m}}} < \frac{\mu_{2}}{U_{2}\ln^{1+c}\alpha U_{2}}+\sum_{m=3}^{\infty } \int_{m-1}^{m}\frac{U^{\prime}(x)}{U(x)\ln^{1+c}\alpha U(x)}\,dx \\& \hphantom{\sum_{m=2}^{\infty}\frac{\mu_{m+1}}{U_{m}\ln^{1+c}\alpha U_{m}}} = \frac{\mu_{2}}{U_{2}\ln^{1+c}\alpha U_{2}}+ \int_{2}^{\infty}\frac{ U^{\prime}(x)}{U(x)\ln^{1+c}\alpha U(x)}\,dx \\& \hphantom{\sum_{m=2}^{\infty}\frac{\mu_{m+1}}{U_{m}\ln^{1+c}\alpha U_{m}}} = \frac{\mu_{2}}{U_{2}\ln^{1+c}\alpha U_{2}}+\frac{1}{c\ln^{c}\alpha (1+\mu_{2})} \\& \hphantom{\sum_{m=2}^{\infty}\frac{\mu_{m+1}}{U_{m}\ln^{1+c}\alpha U_{m}}} = \frac{1}{c} \biggl[ \frac{1}{\ln^{c}\alpha(1+\mu_{2})}+c \frac{\mu _{2}}{U_{2}\ln^{1+c}\alpha U_{2}} \biggr] , \\& \sum_{m=2}^{\infty}\frac{\mu_{m+1}}{U_{m}\ln^{1+c}\alpha U_{m}} =\sum _{m=2}^{\infty} \int_{m}^{m+1}\frac{U^{\prime}(x)\,dx}{U_{m}\ln ^{1+c}\alpha U_{m}} \\& \hphantom{\sum_{m=2}^{\infty}\frac{\mu_{m+1}}{U_{m}\ln^{1+c}\alpha U_{m}}} > \sum_{m=2}^{\infty} \int_{m}^{m+1}\frac{U^{\prime}(x)}{U(x)\ln ^{1+c}\alpha U(x)}\,dx \\& \hphantom{\sum_{m=2}^{\infty}\frac{\mu_{m+1}}{U_{m}\ln^{1+c}\alpha U_{m}}} = \int_{2}^{\infty}\frac{U^{\prime}(x)\,dx}{U(x)\ln^{1+c}\alpha U(x)}= \frac{1}{c} \frac{1}{\ln^{c}\alpha(1+\mu_{2})}. \end{aligned}

Hence, we obtain (22). In the same way, we obtain (23). □

## Main results and operator expressions

In the following, we also set

\begin{aligned} &\widetilde{\Phi}_{\lambda}(m) : =\omega( \lambda_{2},m) \biggl(\frac {U_{m}}{\mu _{m+1}} \biggr)^{p-1}(\ln \alpha U_{m})^{p(1-\lambda_{1})-1}, \\ &\widetilde{\Psi}_{\lambda}(n) : =\varpi(\lambda_{1},n) \biggl( \frac {V_{n}}{\upsilon_{n+1}} \biggr)^{q-1}(\ln\beta V_{n})^{q(1-\lambda _{2})-1}\quad \bigl(m,n\in \mathbf{N}\backslash\{1\} \bigr). \end{aligned}
(24)

### Theorem 1

(i) For $$p>1$$, we have the following equivalent inequalities:

\begin{aligned}& I : =\sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln ^{\lambda}(\alpha\beta U_{m}V_{n})}\leq\|a\|_{p,\widetilde{\Phi}_{\lambda}} \|b\|_{q,\widetilde{\Psi}_{\lambda}}, \end{aligned}
(25)
\begin{aligned}& J : = \Biggl\{ \sum_{n=2}^{\infty} \frac{\upsilon_{n+1}\ln^{p\lambda _{2}-1}\beta V_{n}}{(\varpi(\lambda_{1},n))^{p-1}V_{n}} \Biggl[ \sum_{m=2}^{\infty} \frac{a_{m}}{\ln^{\lambda}(\alpha\beta U_{m}V_{n})} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}}\leq \|a\|_{p,\widetilde{\Phi }_{\lambda }}; \end{aligned}
(26)

(ii) for $$0< p<1$$ (or $$p<0$$), we have the equivalent reverses of (25) and (26).

### Proof

(i) By Hölder’s inequality with weight (see ) and (15) we have

\begin{aligned}& \Biggl[ \sum_{m=2}^{\infty}\frac{a_{m}}{\ln^{\lambda}(\alpha\beta U_{m}V_{n})} \Biggr] ^{p} \\ & \quad = \Biggl[ \sum_{m=2}^{\infty} \frac{1}{\ln ^{\lambda}(\alpha\beta U_{m}V_{n})} \biggl( \frac{U_{m}^{1/q}(\ln\alpha U_{m})^{(1-\lambda _{1})/q}\upsilon_{n+1}^{1/p}}{(\ln\beta V_{n})^{(1-\lambda_{2})/p}\mu _{m+1}^{1/q}}a_{m} \biggr) \biggl( \frac{(\ln\beta V_{n})^{(1-\lambda _{2})/p}\mu_{m+1}^{1/q}}{U_{m}^{1/q}(\ln\alpha U_{m})^{(1-\lambda _{1})/q}\upsilon_{n+1}^{1/p}} \biggr) \Biggr] ^{p} \\ & \quad \leq \sum_{m=2}^{\infty}\frac{1}{\ln^{\lambda}(\alpha\beta U_{m}V_{n})}\frac{U_{m}^{p-1}(\ln\alpha U_{m})^{(1-\lambda_{1})p/q}\upsilon _{n+1}}{(\ln\beta V_{n})^{1-\lambda_{2}}\mu_{m+1}^{p/q}}a_{m}^{p} \\ & \qquad {}\times \Biggl[ \sum_{m=2}^{\infty} \frac{1}{\ln^{\lambda}(\alpha \beta U_{m}V_{n})}\frac{(\ln\beta V_{n})^{(1-\lambda_{2})(q-1)}\mu_{m}}{U_{m}(\ln\alpha U_{m})^{1-\lambda_{1}}\upsilon_{n+1}^{q-1}} \Biggr] ^{p-1} \\ & \quad = \frac{(\varpi(\lambda_{1},n))^{p-1}V_{n}}{(\ln\beta V_{n})^{p\lambda _{2}-1}\upsilon_{n+1}}\sum_{m=2}^{\infty} \frac{\upsilon _{n+1}U_{m}^{p-1}(\ln\alpha U_{m})^{(1-\lambda_{1})(p-1)}a_{m}^{p}}{V_{n}\ln^{\lambda}(\alpha\beta U_{m}V_{n})(\ln\beta V_{n})^{1-\lambda _{2}}\mu_{m+1}^{p-1}}. \end{aligned}
(27)

Then by (14) we find

\begin{aligned} J \leq& \Biggl[ \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{\upsilon _{n+1}}{\ln^{\lambda}(\alpha\beta U_{m}V_{n})} \frac{U_{m}^{p-1}(\ln\alpha U_{m})^{(1-\lambda_{1})(p-1)}}{V_{n}(\ln\beta V_{n})^{1-\lambda _{2}}\mu _{m+1}^{p-1}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =& \Biggl[ \sum_{m=2}^{\infty}\sum _{n=2}^{\infty}\frac{\upsilon _{n+1}(\ln \alpha U_{m})^{\lambda_{1}}}{\ln^{\lambda}(\alpha\beta U_{m}V_{n})}\frac{U_{m}^{p-1}(\ln\alpha U_{m})^{p(1-\lambda_{1})-1}}{V_{n}(\ln\beta V_{n})^{1-\lambda_{2}}\mu_{m+1}^{p-1}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =& \Biggl[ \sum_{m=2}^{\infty}\omega( \lambda_{2},m) \biggl(\frac{U_{m}}{\mu _{m+1}} \biggr)^{p-1}(\ln \alpha U_{m})^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac {1}{p}}, \end{aligned}
(28)

and then (26) follows.

By Hölder’s inequality we have

\begin{aligned} I =&\sum_{n=2}^{\infty} \Biggl[ \frac{(\ln\beta V_{n})^{\lambda _{2}-\frac{1}{p}}\upsilon_{n+1}^{1/p}}{(\varpi(\lambda_{1},n))^{\frac{1}{q}}V_{n}^{1/p}}\sum_{m=1}^{\infty} \frac{a_{m}}{\ln^{\lambda}(\alpha \beta U_{m}V_{n})} \Biggr] \\ &{}\times \biggl[ \bigl(\varpi(\lambda_{1},n) \bigr)^{\frac{1}{q}} \frac{(\ln\beta V_{n})^{\frac{1}{p}-\lambda_{2}}}{V_{n}^{-1/p}\upsilon _{n+1}^{1/p}}b_{n} \biggr] \leq J\|b \|_{q,\widetilde{\Psi}_{\lambda}}. \end{aligned}
(29)

Then by (26) we have (25).

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

$$b_{n}:=\frac{(\ln\beta V_{n})^{p\lambda_{2}-1}\upsilon_{n+1}}{(\varpi (\lambda_{1},n))^{p-1}V_{n}} \Biggl[ \sum_{m=2}^{\infty} \frac{a_{m}}{\ln ^{\lambda}(\alpha\beta U_{m}V_{n})} \Biggr] ^{p-1},\quad n\in\mathbf {N}\backslash \{1 \}.$$
(30)

Then we find $$J^{p}=\|b\|_{q,\widetilde{\Psi}_{\lambda}}^{q}$$. If $$J=0$$, then (26) is trivially valid; if $$J=\infty$$, then by (28), (26) takes the form of equality. Suppose that $$0< J<\infty$$. By (25) it follows that

\begin{aligned}& \|b\|_{q,\widetilde{\Psi}_{\lambda}}^{q} = J^{p}=I\leq\|a\|_{p,\widetilde{\Phi}_{\lambda}} \|b\|_{q,\widetilde{\Psi}_{\lambda}}, \end{aligned}
(31)
\begin{aligned}& \|b\|_{q,\widetilde{\Psi}_{\lambda}}^{q-1} = J\leq \|a\|_{p,\widetilde{\Phi}_{\lambda}}, \end{aligned}
(32)

and then (26) follows, which is equivalent to (25).

(ii) For $$0< p<1$$ (or $$p<0$$), by the reverse Hölder inequality with weight and (15), we obtain the reverse of (27) (or (27)), then we have the reverse of (28), and then the reverse of (26) follows. By Hölder’s inequality we have the reverse of (29), and then by the reverse of (26) the reverse of (25) follows.

On the other hand, assuming that the reverse of (25) is valid, we set $$b_{n}$$ as in (30). Then we find $$J^{p}=\|b\|_{q,\widetilde{\Psi }_{\lambda}}^{q}$$. If $$J=\infty$$, then the reverse of (26) is trivially valid; if $$J=0$$, then by the reverse of (28), (26) takes the form of equality (=0). Suppose that $$0< J<\infty$$. By the reverse of (25) it follows that the reverses of (31) and (32) are valid, and then the reverse of (26) follows, which is equivalent to the reverse of (25). □

### Theorem 2

If $$p>1$$, $$\{\mu_{m}\}_{m=1}^{\infty}$$ and $$\{\upsilon _{n}\}_{n=1}^{\infty}$$ are decreasing, $$U_{\infty}=V_{\infty}=\infty$$, $$\|a\|_{p,\Phi_{\lambda}}\in\mathbf{R}_{+}$$, and $$\|b\|_{q,\Psi _{\lambda}}\in\mathbf{R}_{+}$$, then we have the following equivalent inequalities:

\begin{aligned}& \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln^{\lambda }(\alpha\beta U_{m}V_{n})}< B(\lambda_{1}, \lambda_{2})\|a\|_{p,\Phi _{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}
(33)
\begin{aligned}& J_{1} : = \Biggl\{ \sum_{n=2}^{\infty} \frac{\upsilon_{n+1}}{V_{n}}\ln ^{p\lambda_{2}-1}\beta V_{n} \Biggl[ \sum _{m=2}^{\infty}\frac{a_{m}}{\ln ^{\lambda}(\alpha\beta U_{m}V_{n})} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}} < B(\lambda_{1},\lambda_{2})\|a \|_{p,\Phi_{\lambda}}, \end{aligned}
(34)

where the constant factor $$B(\lambda_{1},\lambda_{2})$$ is the best possible.

### Proof

Using (16) and (17) in (25) and (26), we obtain equivalent inequalities (33) and (34).

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

\begin{aligned} &\tilde{a}_{m} :=\frac{\mu_{m+1}}{U_{m}}\ln^{\tilde{\lambda} _{1}-1} \alpha U_{m}=\frac{\mu_{m+1}}{U_{m}}\ln^{\lambda_{1}-\frac{\varepsilon}{p}-1}\alpha U_{m}, \\ &\tilde{b}_{n} =\frac{\upsilon_{n+1}}{V_{n}}\ln^{\tilde {\lambda}_{2}-\varepsilon-1}\beta V_{n}=\frac{\upsilon_{n+1}}{V_{n}}\ln ^{\lambda _{2}-\frac{\varepsilon}{q}-1}\beta V_{n}. \end{aligned}
(35)

Then by (22), (23), and (19) we have

\begin{aligned}& \|\tilde{a}\|_{p,\Phi_{\lambda}}\|\tilde{b}\|_{q,\Psi _{\lambda}}= \Biggl( \sum _{m=2}^{\infty}\frac{\mu_{m+1}}{U_{m}\ln^{1+\varepsilon }\alpha U_{m}} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=2}^{\infty} \frac{\upsilon_{n+1}}{V_{n}\ln^{1+\varepsilon}\beta V_{n}} \Biggr) ^{\frac{1}{q}} \\& \hphantom{\|\tilde{a}\|_{p,\Phi_{\lambda}}\|\tilde{b}\|_{q,\Psi _{\lambda}}} = \frac{1}{\varepsilon} \biggl[ \frac{1}{\ln^{\varepsilon}\alpha (1+\mu _{2})}+\varepsilon O(1) \biggr] ^{\frac{1}{p}} \biggl[ \frac{1}{\ln ^{\varepsilon}\beta(1+\upsilon_{2})}+\varepsilon\widetilde {O}(1) \biggr] ^{\frac{1}{q}}, \\& \tilde{I} : =\sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac {\tilde{a}_{m}\tilde{b}_{n}}{\ln^{\lambda}(\alpha\beta U_{m}V_{n})} = \sum_{n=2}^{\infty} \Biggl[ \sum_{m=2}^{\infty}\frac{1}{\ln^{\lambda }(\alpha\beta U_{m}V_{n})} \frac{\mu_{m+1}\ln^{\tilde{\lambda}_{2}}\beta V_{n}}{U_{m}\ln^{1-\tilde{\lambda}_{1}}\alpha U_{m}} \Biggr] \frac{\upsilon_{n+1}}{V_{n}\ln^{\varepsilon+1}\beta V_{n}} \\& \hphantom{\tilde{I}} = \sum_{n=2}^{\infty}\varpi( \tilde{\lambda}_{1},n)\frac{\upsilon _{n+1}}{V_{n}\ln^{\varepsilon+1}\beta V_{n}} \geq B(\tilde{\lambda}_{1},\tilde{ \lambda}_{2})\sum_{n=2}^{\infty} \biggl(1-O \biggl(\frac{1}{\ln^{\tilde{\lambda}_{1}}\beta V_{n}} \biggr) \biggr)\frac {\upsilon_{n+1}}{V_{n}\ln^{\varepsilon+1}\beta V_{n}} \\& \hphantom{\tilde{I}} = B(\tilde{\lambda}_{1},\tilde{\lambda}_{2}) \Biggl[ \sum_{n=2}^{\infty}\frac{\upsilon_{n+1}}{V_{n}\ln^{\varepsilon +1}\beta V_{n}}- \sum_{n=2}^{\infty}O \biggl( \frac{\upsilon_{n+1}}{V_{n}(\ln \beta V_{n})^{(\frac{\varepsilon}{q}+\lambda_{1})+1}} \biggr) \Biggr] \\& \hphantom{\tilde{I}} = \frac{1}{\varepsilon}B(\tilde{\lambda}_{1},\tilde{ \lambda }_{2}) \biggl[ \frac{1}{\ln^{\varepsilon}\beta(1+\upsilon_{2})}+\varepsilon \bigl(\widetilde{O}(1)-O(1) \bigr) \biggr] . \end{aligned}

If there exists a positive constant $$K\leq B(\lambda_{1},\lambda_{2})$$ such that (33) is valid when replacing $$B(\lambda_{1},\lambda_{2})$$ by K, then, in particular, we have $$\varepsilon\tilde {I}<\varepsilon K\|\tilde{a}\|_{p,\Phi_{\lambda}}\|\tilde{b}\|_{q,\Psi _{\lambda}}$$, namely

\begin{aligned}& B \biggl(\lambda_{1}-\frac{\varepsilon}{p},\lambda_{2}+ \frac{\varepsilon }{p} \biggr) \biggl[ \frac{1}{\ln^{\varepsilon}\beta(1+\upsilon_{2})}+\varepsilon \bigl( \widetilde{O}(1)-O(1) \bigr) \biggr] \\& \quad < K \biggl[ \frac{1}{\ln^{\varepsilon}\alpha(1+\mu_{2})}+\varepsilon O(1) \biggr] ^{\frac{1}{p}} \biggl[ \frac{1}{\ln^{\varepsilon}\beta (1+\upsilon _{2})}+\varepsilon\widetilde{O}(1) \biggr] ^{\frac{1}{q}}. \end{aligned}

It follows that $$B(\lambda_{1},\lambda_{2})\leq K$$ ($$\varepsilon \rightarrow 0^{+}$$). Hence, $$K=B(\lambda_{1},\lambda_{2})$$ is the best possible constant factor of (33).

Similarly to (29), we still can find the following inequality:

$$I\leq J_{1}\|b\|_{q,\Psi_{\lambda}}.$$
(36)

Hence, we can prove that the constant factor $$B(\lambda_{1},\lambda_{2})$$ in (34) is the best possible. Otherwise, we would reach a contradiction by (36) that the constant factor in (33) is not the best possible. □

### Remark 1

(i) For $$\alpha=\beta=1$$ in (33) and (34), setting

\begin{aligned}& \varphi_{\lambda}(m) : = \biggl(\frac{U_{m}}{\mu_{m+1}} \biggr)^{p-1}( \ln U_{m})^{p(1-\lambda_{1})-1}, \\& \psi_{\lambda}(n) : = \biggl(\frac{V_{n}}{\upsilon_{n+1}} \biggr)^{q-1}(\ln V_{n})^{q(1-\lambda_{2})-1}\quad \bigl(m,n\in\mathbf{N}\backslash\{1\} \bigr), \end{aligned}

we have the following equivalent Mulholland-type inequalities:

\begin{aligned}& \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln^{\lambda }(U_{m}V_{n})}< B(\lambda_{1}, \lambda_{2})\|a\|_{p,\varphi_{\lambda }}\|b\|_{q,\psi_{\lambda}}, \end{aligned}
(37)
\begin{aligned}& \Biggl\{ \sum_{n=2}^{\infty}\frac{\upsilon_{n+1}}{V_{n}} \ln^{p\lambda _{2}-1}V_{n} \Biggl[ \sum_{m=2}^{\infty} \frac{a_{m}}{\ln^{\lambda }(U_{m}V_{n})} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}}< B( \lambda_{1},\lambda _{2})\|a\|_{p,\varphi_{\lambda}}, \end{aligned}
(38)

which are extensions of (9), and the following inequality:

$$\Biggl\{ \sum_{n=2}^{\infty}\frac{\upsilon_{n+1}}{V_{n}} \Biggl[ \sum_{m=2}^{\infty}\frac{a_{m}}{\ln(U_{m}V_{n})} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}}< \frac{\pi}{\sin(\frac{\pi}{p})} \Biggl[ \sum _{m=2}^{\infty} \biggl(\frac{U_{m}}{\mu_{m+1}} \biggr)^{p-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}}.$$
(39)

(ii) For $$\mu_{i}=\upsilon_{j}=1$$ ($$i,j\in\mathbf{N}$$), $$\lambda =1$$, $$\lambda_{1}=\frac{1}{q}$$, $$\lambda_{2}=\frac{1}{p}$$, (33) reduces to the following more accurate and extended Mulholland’s inequality:

$$\sum_{m=2}^{\infty}\sum _{n=2}^{\infty}\frac{a_{m}b_{n}}{\ln(\alpha \beta mn)}< \frac{\pi}{\sin(\pi/p)} \Biggl( \sum_{m=2}^{\infty}\frac {a_{m}^{p}}{m^{1-p}} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=2}^{\infty} \frac {b_{n}^{q}}{n^{1-q}} \Biggr) ^{\frac{1}{q}},$$
(40)

where $$\frac{2}{3}\leq\alpha,\beta\leq1$$.

For $$p>1$$, $$\Psi_{\lambda}^{1-p}(n)=\frac{\upsilon_{n+1}}{V_{n}}(\ln \beta V_{n})^{p\lambda_{2}-1}$$, we define the following normed spaces:

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

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

$$c=\{c_{n}\}_{n=2}^{\infty},\qquad c_{n}:=\sum _{m=2}^{\infty}\frac{a_{m}}{\ln ^{\lambda}(\alpha\beta U_{m}V_{n})},\quad n\in\mathbf{N} \backslash\{1\},$$

we can rewrite (34) as follows:

$$\|c\|_{p,\Psi_{\lambda}^{1-p}}< B(\lambda_{1},\lambda _{2})\|a \|_{p,\Phi _{\lambda}}< \infty,$$

that is, $$c\in l_{p,\Psi_{\lambda}^{1-p}}$$.

### Definition 2

Define the 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:

$$(Ta,b):=\sum_{n=2}^{\infty} \Biggl[ \sum _{m=2}^{\infty}\frac{a_{m}}{\ln ^{\lambda}(\alpha\beta U_{m}V_{n})} \Biggr] b_{n}.$$
(41)

Then we can rewrite (33) and (34) as follows:

\begin{aligned}& (Ta,b) < B(\lambda_{1},\lambda_{2})\|a\|_{p,\Phi_{\lambda }} \|b\|_{q,\Psi_{\lambda}}, \end{aligned}
(42)
\begin{aligned}& \|Ta\|_{p,\Psi_{\lambda}^{1-p}} < B(\lambda_{1},\lambda _{2})\|a \|_{p,\Phi_{\lambda}}. \end{aligned}
(43)

Define the norm of the operator T as follows:

$$\|T\|:=\sup_{a\, (\neq\theta)\in l_{p,\Phi_{\lambda}}}\frac {\|Ta\|_{p,\Psi _{\lambda}^{1-p}}}{\|a\|_{p,\Phi_{\lambda}}}.$$

Then by (43) we find $$\|T\|\leq B(\lambda_{1},\lambda_{2})$$. Since the constant factor in (43) is the best possible, we have

$$\|T\|=B(\lambda_{1},\lambda_{2}).$$
(44)

## Some reverses

In the following, we also set

\begin{aligned} &\widetilde{\Omega}_{\lambda}(m) : = \bigl(1-\theta( \lambda_{2},m) \bigr) \biggl(\frac {U_{m}}{\mu_{m+1}} \biggr)^{p-1}( \ln\alpha U_{m})^{p(1-\lambda_{1})-1}, \\ &\widetilde{\digamma}_{\lambda}(n) : = \bigl(1-\vartheta(\lambda _{1},n) \bigr) \biggl(\frac{V_{n}}{\upsilon_{n+1}} \biggr)^{q-1}(\ln \beta V_{n})^{q(1-\lambda _{2})-1} \quad \bigl(m,n\in\mathbf{N}\backslash\{1\} \bigr). \end{aligned}
(45)

For $$0< p<1$$ or $$p<0$$, we still use the formal symbols $$\|a\|_{p,\Phi _{\lambda}}$$, $$\|b\|_{q,\Psi_{\lambda}}$$, $$\|a\|_{p,\widetilde {\Omega}_{\lambda}}$$, and $$\|b\|_{q,\widetilde{\digamma}_{\lambda}}$$, and so on.

### Theorem 3

If $$0< p<1$$, $$\{\mu_{m}\}_{m=1}^{\infty}$$ and $$\{\upsilon _{n}\}_{n=1}^{\infty}$$ are decreasing, $$U_{\infty}=V_{\infty}=\infty$$, $$\|a\|_{p,\Phi_{\lambda}}\in\mathbf{R}_{+}$$, and $$\|b\|_{q,\Psi _{\lambda}}\in\mathbf{R}_{+}$$, 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{a_{m}b_{n}}{\ln^{\lambda }(\alpha\beta U_{m}V_{n})}>B(\lambda_{1}, \lambda _{2})\|a\|_{p,\widetilde{\Omega}_{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}
(46)
\begin{aligned}& \Biggl\{ \sum_{n=2}^{\infty}\frac{\upsilon_{n+1}}{V_{n}} \ln ^{p\lambda _{2}-1}\beta V_{n} \Biggl[ \sum _{m=2}^{\infty}\frac{a_{m}}{\ln^{\lambda }(\alpha\beta U_{m}V_{n})} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}} > B(\lambda_{1},\lambda_{2})\|a \|_{p,\widetilde{\Omega}_{\lambda}}. \end{aligned}
(47)

### Proof

Using (18) and (17) in the reverses of (25) and (26), since

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

and

$$\frac{1}{(B(\lambda_{1},\lambda_{2}))^{p-1}}>\frac{1}{(\varpi(\lambda _{1},n))^{p-1}}\quad (0< p< 1),$$

we obtain equivalent inequalities (46) and (47).

For $$\varepsilon\in(0,p\lambda_{1})$$, we set $$\tilde{\lambda }_{1}$$, $$\tilde{\lambda}_{2}$$, $$\tilde{a}_{m}$$, and $$\tilde{b}_{n}$$ as in (35). Then by (22), (23), and (17) we find

\begin{aligned}& \|a\|_{p,\widetilde{\Omega}_{\lambda}}\|b\|_{q,\Psi_{\lambda}} = \Biggl[ \sum _{m=2}^{\infty}\frac{(1-\theta(\lambda_{2},m))\mu _{m+1}}{U_{m}\ln^{1+\varepsilon}\alpha U_{m}} \Biggr] ^{\frac{1}{p}} \Biggl( \sum_{n=2}^{\infty}\frac{\upsilon_{n+1}}{V_{n}\ln^{1+\varepsilon }\beta V_{n}} \Biggr) ^{\frac{1}{q}} \\& \hphantom{\|a\|_{p,\widetilde{\Omega}_{\lambda}}\|b\|_{q,\Psi_{\lambda}}} = \Biggl( \sum_{m=2}^{\infty} \frac{\mu_{m+1}}{U_{m}\ln^{1+\varepsilon }\alpha U_{m}}-\sum_{m=2}^{\infty}O \biggl( \frac{\mu_{m+1}}{U_{m}\ln ^{1+\lambda _{2}+\varepsilon}\alpha U_{m}} \biggr) \Biggr) ^{\frac{1}{p}} \\& \hphantom{\|a\|_{p,\widetilde{\Omega}_{\lambda}}\|b\|_{q,\Psi_{\lambda}}={}} {}\times \Biggl( \sum_{n=2}^{\infty} \frac{\upsilon_{n+1}}{V_{n}\ln ^{1+\varepsilon}\beta V_{n}} \Biggr) ^{\frac{1}{q}} \\& \hphantom{\|a\|_{p,\widetilde{\Omega}_{\lambda}}\|b\|_{q,\Psi_{\lambda}}} = \frac{1}{\varepsilon} \biggl[ \frac{1}{\ln^{\varepsilon}\alpha (1+\mu _{2})}+\varepsilon \bigl(O(1)-O_{1}(1) \bigr) \biggr] ^{\frac{1}{p}} \biggl[ \frac{1}{\ln^{\varepsilon}\beta(1+\upsilon_{2})}+\varepsilon \widetilde{O}(1) \biggr] ^{\frac{1}{q}}, \\& \tilde{I} : =\sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac {\tilde{a}_{m}\tilde{b}_{n}}{\ln^{\lambda}(\alpha\beta U_{m}V_{n})} \\& \hphantom{\tilde{I}} = \sum_{n=2}^{\infty} \Biggl[ \sum_{m=2}^{\infty}\frac{1}{\ln^{\lambda }(\alpha\beta U_{m}V_{n})} \frac{\mu_{m+1}\ln^{\tilde{\lambda}_{2}}\beta V_{n}}{U_{m}\ln^{1-\tilde{\lambda}_{1}}\alpha U_{m}} \Biggr] \frac{\upsilon_{n+1}}{V_{n}\ln^{\varepsilon+1}\beta V_{n}} \\& \hphantom{\tilde{I}} = \sum_{n=2}^{\infty}\varpi( \tilde{\lambda}_{1},n)\frac{\upsilon _{n+1}}{V_{n}\ln^{\varepsilon+1}\beta V_{n}}\leq B(\tilde{\lambda }_{1},\tilde{\lambda}_{2})\sum _{n=2}^{\infty}\frac{\upsilon _{n+1}}{V_{n}\ln^{\varepsilon+1}\beta V_{n}} \\& \hphantom{\tilde{I}} = \frac{1}{\varepsilon}B(\tilde{\lambda}_{1},\tilde{ \lambda }_{2}) \biggl[ \frac{1}{\ln^{\varepsilon}\beta(1+\upsilon_{2})}+\varepsilon \widetilde{O}(1) \biggr] . \end{aligned}

If there exists a positive constant $$K\geq B(\lambda_{1},\lambda_{2})$$ such that (46) is valid when replacing $$B(\lambda_{1},\lambda_{2})$$ by K, then, in particular, we have $$\varepsilon\tilde {I}>\varepsilon K\|\tilde{a}\|_{p,\widetilde{\Omega}_{\lambda}}\|\tilde{b}\|_{q,\Psi_{\lambda}}$$, namely

\begin{aligned}& B \biggl(\lambda_{1}-\frac{\varepsilon}{p},\lambda_{2}+ \frac{\varepsilon }{p} \biggr) \biggl[ \frac{1}{\ln^{\varepsilon}\beta(1+\upsilon_{2})}+\varepsilon \widetilde{O}(1) \biggr] \\& \quad > K \biggl[ \frac{1}{\ln^{\varepsilon}\alpha(1+\mu_{2})}+\varepsilon \bigl(O(1)-O_{1}(1) \bigr) \biggr] ^{\frac{1}{p}} \\& \qquad {} \times \biggl[ \frac{1}{\ln^{\varepsilon}\beta(1+\upsilon_{2})}+\varepsilon \widetilde{O}(1) \biggr] ^{\frac{1}{q}}. \end{aligned}

It follows that $$B(\lambda_{1},\lambda_{2})\geq K$$ ($$\varepsilon \rightarrow 0^{+}$$). Hence, $$K=B(\lambda_{1},\lambda_{2})$$ is the best possible constant factor of (46).

The constant factor $$B(\lambda_{1},\lambda_{2})$$ in (47) is still the best possible. Otherwise, we would reach a contradiction by the reverse of (36) that the constant factor in (46) is not the best possible. □

### Remark 2

For $$\alpha=\beta=1$$, setting

\begin{aligned}& \tilde{\theta}(\lambda_{2},m) = \frac{1}{B(\lambda_{1},\lambda _{2})}\frac{\ln^{\lambda_{2}}(1+\upsilon_{2})}{\lambda_{2}[1+\frac{\ln (1+\theta(m)\upsilon_{2})}{\ln U_{m}}]^{\lambda}}\frac{1}{\ln ^{\lambda _{2}}U_{m}} \\& \hphantom{\tilde{\theta}(\lambda_{2},m)} = O \biggl(\frac{1}{\ln^{\lambda_{2}}U_{m}} \biggr)\in(0,1)\quad \bigl( \theta(m)\in(0,1) \bigr), \\& \tilde{\varphi}_{\lambda}(m):= \bigl(1-\tilde{\theta}(\lambda _{2},m) \bigr) \biggl(\frac{U_{m}}{\mu_{m+1}} \biggr)^{p-1}(\ln U_{m})^{p(1-\lambda_{1})-1}, \end{aligned}

it is evident that (46) and (47) are extensions of the following equivalent inequalities:

\begin{aligned}& \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln^{\lambda }(U_{m}V_{n})}>B(\lambda_{1}, \lambda_{2})\|a\|_{p,\tilde{\varphi} _{\lambda}}\|b\|_{q,\psi_{\lambda}}, \end{aligned}
(48)
\begin{aligned}& \Biggl\{ \sum_{n=2}^{\infty}\frac{\upsilon_{n+1}}{V_{n}} \ln^{p\lambda _{2}-1}V_{n} \Biggl[ \sum_{m=2}^{\infty} \frac{a_{m}}{\ln^{\lambda }(U_{m}V_{n})} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}}>B( \lambda_{1},\lambda _{2})\|a\|_{p,\tilde{\varphi}_{\lambda}}, \end{aligned}
(49)

where the constant factor $$B(\lambda_{1},\lambda_{2})$$ is still the best possible.

### Theorem 4

If $$p<0$$, $$\{\mu_{m}\}_{m=1}^{\infty}$$ and $$\{\upsilon _{n}\}_{n=1}^{\infty}$$ are decreasing, $$U_{\infty}=V_{\infty}=\infty$$, $$\|a\|_{p,\Phi_{\lambda}}\in\mathbf{R}_{+}$$, and $$\|b\|_{q,\Psi _{\lambda}}\in\mathbf{R}_{+}$$, then we have the following equivalent inequalities with the best possible constant factor $$B(\lambda_{1},\lambda_{2})$$:

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{\ln^{\lambda }(\alpha\beta U_{m}V_{n})}>B(\lambda_{1}, \lambda_{2})\|a\|_{p,\Phi _{\lambda}}\|b\|_{q,\widetilde{\digamma}_{\lambda}}, \end{aligned}
(50)
\begin{aligned}& J_{2} : = \Biggl\{ \sum_{n=1}^{\infty} \frac{\upsilon_{n+1}\ln ^{p\lambda _{2}-1}\beta V_{n}}{(1-\vartheta(\lambda_{1},n))^{p-1}V_{n}} \Biggl[ \sum_{m=1}^{\infty} \frac{a_{m}}{\ln^{\lambda}(\alpha\beta U_{m}V_{n})} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}} \\& \hphantom{J_{2}} > B(\lambda_{1},\lambda_{2})\|a \|_{p,\Phi_{\lambda}}. \end{aligned}
(51)

### Proof

Using (16) and (19) in the reverses of (25) and (26), since

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

and

$$\biggl[\frac{1}{(B(\lambda_{1},\lambda_{2}))^{p-1}(1-\vartheta (\lambda _{1},n))^{p-1}} \biggr]^{\frac{1}{p}}> \biggl[\frac{1}{(\varpi(\lambda _{1},n))^{p-1}} \biggr]^{\frac{1}{p}}\quad (p< 0),$$

we obtain equivalent inequalities (50) and (51).

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

\begin{aligned}& \tilde{a}_{m} : =\frac{\mu_{m+1}}{U_{m}}\ln^{\tilde{\lambda} _{1}-\varepsilon-1}\alpha U_{m}=\frac{\mu_{m+1}}{U_{m}}\ln^{\lambda _{1}-\frac{\varepsilon}{p}-1}\alpha U_{m}, \\& \tilde{b}_{n} = \frac{\upsilon_{n+1}}{V_{n}}\ln^{\tilde {\lambda}_{2}-1}\beta V_{n}=\frac{\upsilon_{n+1}}{V_{n}}\ln^{\lambda_{2}-\frac {\varepsilon}{q}-1}\beta V_{n}. \end{aligned}

Then by (22), (23), and (16) we have

\begin{aligned}& \|\tilde{a}\|_{p,\Phi_{\lambda}}\|\tilde{b}\|_{q,\widetilde{ \digamma}_{\lambda}} = \Biggl( \sum _{m=2}^{\infty}\frac{\mu_{m+1}}{U_{m}\ln^{\varepsilon +1}\alpha U_{m}} \Biggr) ^{\frac{1}{p}} \Biggl[ \sum_{n=2}^{\infty} \frac{(1-\vartheta(\lambda_{1},n))\upsilon_{n+1}}{V_{n}\ln^{\varepsilon +1}\beta V_{n}} \Biggr] ^{\frac{1}{q}} \\& \hphantom{\|\tilde{a}\|_{p,\Phi_{\lambda}}\|\tilde{b}\|_{q,\widetilde{ \digamma}_{\lambda}}}= \Biggl( \sum_{m=2}^{\infty} \frac{\mu_{m+1}}{U_{m}\ln^{\varepsilon +1}\alpha U_{m}} \Biggr) ^{\frac{1}{p}} \\& \hphantom{\|\tilde{a}\|_{p,\Phi_{\lambda}}\|\tilde{b}\|_{q,\widetilde{ \digamma}_{\lambda}}={}} {} \times \Biggl[ \sum_{n=2}^{\infty} \frac{\upsilon_{n+1}}{V_{n}\ln ^{\varepsilon+1}\beta V_{n}}-\sum_{n=2}^{\infty}O \biggl( \frac{\upsilon _{n+1}}{V_{n}\ln^{1+(\lambda_{1}+\varepsilon)}\beta V_{n}} \biggr) \Biggr] ^{\frac{1}{q}} \\& \hphantom{\|\tilde{a}\|_{p,\Phi_{\lambda}}\|\tilde{b}\|_{q,\widetilde{ \digamma}_{\lambda}}} = \frac{1}{\varepsilon} \biggl[ \frac{1}{\ln^{\varepsilon}\alpha (1+\mu _{2})}+\varepsilon O(1) \biggr] ^{\frac{1}{p}} \\& \hphantom{\|\tilde{a}\|_{p,\Phi_{\lambda}}\|\tilde{b}\|_{q,\widetilde{ \digamma}_{\lambda}}={}}{} \times \biggl[ \frac{1}{\ln^{\varepsilon}\beta(1+\upsilon_{2})}+\varepsilon \bigl(\widetilde{O}(1)-O_{1}(1) \bigr) \biggr] ^{\frac{1}{q}}, \\& \tilde{I} = \sum_{m=2}^{\infty}\sum _{n=2}^{\infty}\frac {\tilde{a}_{m}\tilde{b}_{n}}{\ln^{\lambda}(\alpha\beta U_{m}V_{n})} \\& \hphantom{\tilde{I}} = \sum_{m=2}^{\infty} \Biggl[ \sum_{n=2}^{\infty}\frac{\ln^{\tilde {\lambda}_{1}}\alpha U_{m}}{\ln^{\lambda}(\alpha\beta U_{m}V_{n})} \frac{\upsilon_{n+1}}{V_{n}}\ln^{\tilde{\lambda}_{2}-1}\beta V_{n} \Biggr] \frac{\mu_{m+1}}{U_{m}\ln^{\varepsilon+1}\alpha U_{m}} \\& \hphantom{\tilde{I}} = \sum_{m=2}^{\infty}\omega( \tilde{\lambda}_{2},m)\frac{\mu _{m+1}}{U_{m}\ln^{\varepsilon+1}\alpha U_{m}} \\& \hphantom{\tilde{I}}\leq B(\tilde{ \lambda}_{1},\tilde{\lambda}_{2})\sum _{n=2}^{\infty}\frac{\mu_{m+1}}{U_{m}\ln ^{\varepsilon+1}\alpha U_{m}} \\& \hphantom{\tilde{I}} = \frac{1}{\varepsilon}B(\tilde{\lambda}_{1},\tilde{ \lambda }_{2}) \biggl[ \frac{1}{\ln^{\varepsilon}\alpha(1+\mu_{2})}+\varepsilon O(1) \biggr] . \end{aligned}

If there exists a positive constant $$K\geq B(\lambda_{1},\lambda_{2})$$ such that (50) is valid when replacing $$B(\lambda_{1},\lambda_{2})$$ by K, then, in particular, we have $$\varepsilon\tilde {I}>\varepsilon K\|\tilde{a}\|_{p,\Phi_{\lambda}}\|\tilde{b}\|_{q,\widetilde{\digamma}_{\lambda}}$$, namely

\begin{aligned}& B \biggl(\lambda_{1}+\frac{\varepsilon}{q},\lambda_{2}- \frac{\varepsilon }{q} \biggr) \biggl[ \frac{1}{\ln^{\varepsilon}\alpha(1+\mu_{2})}+\varepsilon O(1) \biggr] \\& \quad > K \biggl[ \frac{1}{\ln^{\varepsilon}\alpha(1+\mu_{2})}+\varepsilon O(1) \biggr] ^{\frac{1}{p}} \\& \qquad {} \times \biggl[ \frac{1}{\ln^{\varepsilon}\beta(1+\upsilon_{2})}+\varepsilon \bigl( \widetilde{O}(1)-O_{1}(1) \bigr) \biggr] ^{\frac{1}{q}}. \end{aligned}

It follows that $$B(\lambda_{1},\lambda_{2})\geq K$$ ($$\varepsilon \rightarrow 0^{+}$$). Hence, $$K=B(\lambda_{1},\lambda_{2})$$ is the best possible constant factor of (50).

Similarly to the reverse of (29), we still can find that

$$I\geq J_{2}\|b\|_{q,\widetilde{\digamma}_{\lambda}}.$$
(52)

Hence, the constant factor $$B(\lambda_{1},\lambda_{2})$$ in (51) is still the best possible. Otherwise, we would reach a contradiction by (52) that the constant factor in (50) is not the best possible. □

### Remark 3

For $$\alpha=\beta=1$$, setting

\begin{aligned}& \tilde{\vartheta}(\lambda_{1},n) = \frac{1}{B(\lambda _{1},\lambda _{2})} \frac{\ln^{\lambda_{1}}(1+\mu_{2})}{\lambda_{1}[1+\frac{\ln (1+\vartheta(n)\mu_{2})}{\ln V_{n}}]^{\lambda}}\frac{1}{\ln^{\lambda _{1}}V_{n}} \\& \hphantom{\tilde{\vartheta}(\lambda_{1},n)} = O \biggl(\frac{1}{\ln^{\lambda_{1}}V_{n}} \biggr)\in(0,1)\quad \bigl( \vartheta(n)\in(0,1) \bigr), \\& \tilde{\psi}_{\lambda}(n):= \bigl(1-\tilde{\vartheta}(\lambda _{1},n) \bigr) \biggl(\frac{V_{n}}{\upsilon_{n+1}} \biggr)^{q-1}(\ln V_{n})^{q(1-\lambda_{2})-1}, \end{aligned}

it is evident that (50) and (51) are extensions of the following equivalent inequalities:

\begin{aligned}& \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln^{\lambda }(U_{m}V_{n})}>B(\lambda_{1}, \lambda_{2})\|a\|_{p,\varphi_{\lambda }}\|b\|_{q,\tilde{\psi}_{\lambda}}, \end{aligned}
(53)
\begin{aligned}& \Biggl\{ \sum_{n=2}^{\infty}\frac{\upsilon_{n+1}\ln^{p\lambda _{2}-1}V_{n}}{(1-\tilde{\vartheta}(\lambda _{1},n))^{p-1}V_{n}} \Biggl[ \sum_{m=2}^{\infty}\frac{a_{m}}{\ln^{\lambda}(U_{m}V_{n})} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}} > B(\lambda_{1}, \lambda_{2})\|a\|_{p,\varphi_{\lambda}}, \end{aligned}
(54)

where the constant factor $$B(\lambda_{1},\lambda_{2})$$ is still the best possible.

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## Acknowledgements

This work is supported by the National Natural Science Foundation (No. 61370186), and Appropriative Researching Fund for Professors and Doctors, Guangdong University of Education (No. 2015ARF25). We are grateful for their help.

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Correspondence to Bicheng Yang.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

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

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