On a new Hardy-Mulholland-type inequality and its more accurate form

Abstract

Using weight coefficients and applying the well-known Hermite-Hadamard inequality, a new Hardy-Mulholand-type inequality with a best possible constant factor is given. Furthermore, we also consider the more accurate equivalent forms, the operator expressions and some particular inequalities. The lemmas and theorems provide an extensive account of this type of inequalities.

Introduction

Assuming that $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $a_{m},b_{n}\geq 0$, $a=\{a_{m}\}_{m=1}^{\infty}\in l^{p}$, $b=\{b_{n}\}_{n=1}^{\infty}\in l^{q}$, $\|a\|_{p}=(\sum_{m=1}^{\infty}a_{m}^{p})^{\frac{1}{p}}>0$, $\|b\|_{q}>0$, we have the following Hardy-Hilbert inequality with the best possible constant factor $\frac{\pi}{\sin(\pi/p)}$ (cf. [1], 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 inequality of (1) is given as follows (cf. [2] and Theorem 323 of [1]):

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

(2)

which is an extension of (1). We still have the following Mulholland inequality similar to (1) with the same best possible constant factor $\frac{\pi}{\sin(\pi/p)}$ (cf. [3] or Theorem 343 of [1], 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 applications (cf. [2, 4–9]).

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

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

(4)

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

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

(5)

For $\mu_{i}=\upsilon_{j}=1$ ($i,j\in\mathbf{N}$), (5) reduces to (1).

In 2015, Yang [10] gave an extension of (5) as follows: For $0<\lambda_{1},\lambda_{2}\leq1$, $\lambda_{1}+\lambda_{2}=\lambda$, $\{\mu_{m}\}_{m=1}^{\infty}$, and $\{\upsilon_{n}\}_{n=1}^{\infty}$ are decreasing, and $U_{\infty}=V_{\infty}=\infty$, we have the following inequality with the best possible constant factor $B(\lambda _{1},\lambda _{2})$:

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

(6)

where $B(u,v)$ is the beta function defined by (cf. [11])

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

(7)

In a similar way, Huang and Yang [12] gave a more accurate inequality of (6) and Yang and Chen [13] obtained a Hardy-Hilbert-type inequality with another kernel and a best possible constant factor.

In this paper, using the way of weight coefficients and applying Hermite-Hadamard’s inequality, a Hardy-Mulholland-type inequality with a best possible constant factor similar to (6) is proved, which is an extension of (3). Furthermore, the more accurate Hardy-Mulholland-type inequality is built by introducing a few parameters. We also consider the equivalent forms, the operator expressions and some particular inequalities.

Some lemmas and an example

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

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

(8)

Lemma 1

Suppose that $a\in\mathbf{R}$, $f(x)$ in continuous in $[a-\frac{1}{2},a+\frac{1}{2}]$, $f^{\prime}(x)$ is strictly increasing in $(a-\frac {1}{2},a)$ and $(a,a+\frac{1}{2})$, and $f^{\prime}(a-0)\leq f^{\prime }(a+0)$. We have the following Hermite-Hadamard inequality (cf. Lemma 1 of [14]):

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

(9)

Example 1

Assuming that $\{\mu_{m}\}_{m=1}^{\infty}$ and $\{\upsilon _{n}\}_{n=1}^{\infty}$ are decreasing, we set $\mu(t):=\mu_{m}$, $t\in(m-1,m]$ ($m\in\mathbf{N}$); $\upsilon(t):=\upsilon_{n}$, $t\in (n-1,n]$ ($n\in\mathbf{N}$),

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

(10)

Then we have $U(m)=U_{m}$, $V(n)=V_{n}$, $U(\infty)=U_{\infty }$, $V(\infty)=V_{\infty}$ 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 define the function $h(x)$ as follows:

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

Then $h(x)$ in continuous in $[n-\frac{1}{2},n+\frac{1}{2}]$, and, for $x\in(n-\frac{1}{2},n)$ ($n\in\mathbf{N}\backslash\{1\}$),

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

In view of $1-\lambda_{2}\geq0$, $h^{\prime}(x)$ (<0) is strictly increasing in $(n-\frac{1}{2},n)$ and

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

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

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

$h^{\prime}(x)$ (<0) is strictly increasing in $(n,n+\frac{1}{2})$ and

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

Since $\upsilon_{n+1}\leq\upsilon_{n}$, we have $h^{\prime}(n-0)\leq h^{\prime}(n+0)$. Then by (9), for $m,n\in\mathbf{N}\backslash\{1\}$, it follows that

$$\begin{aligned} h(n) < & \int_{n-\frac{1}{2}}^{n+\frac{1}{2}}h(x)\,dx = \int_{n-\frac{1}{2}}^{n+\frac{1}{2}}\frac{\ln^{\lambda _{2}-1}(V(x)-\beta)}{(V(x)-\beta)[\ln(U_{m}-\alpha)+\ln(V(x)-\beta )]^{\lambda}}\,dx. \end{aligned}$$

(11)

Lemma 2

For $m,n\in \mathbf{N}\backslash\{1\}$, we define the following weight coefficients:

$$\begin{aligned}& \omega(\lambda_{2},m) :=\sum_{n=2}^{\infty} \frac{1}{\ln^{\lambda }[(U_{m}-\alpha)(V_{n}-\beta)]}\frac{\upsilon_{n+1}\ln^{\lambda _{1}}(U_{m}-\alpha)}{(V_{n}-\beta)\ln^{1-\lambda_{2}}(V_{n}-\beta)}, \end{aligned}$$

(12)

$$\begin{aligned}& \varpi(\lambda_{1},n) :=\sum_{m=2}^{\infty} \frac{1}{\ln^{\lambda }[(U_{m}-\alpha)(V_{n}-\beta)]}\frac{\mu_{m+1}\ln^{\lambda _{2}}(V_{n}-\beta)}{(U_{m}-\alpha)\ln^{1-\lambda_{1}}(U_{m}-\alpha)}. \end{aligned}$$

(13)

If $\{\mu_{m}\}_{m=1}^{\infty}$ and $\{\upsilon _{n}\}_{n=1}^{\infty}$ are decreasing, and $U(\infty)=V(\infty )=\infty$, then

$$\begin{aligned}& \omega(\lambda_{2},m) < B(\lambda_{1}, \lambda_{2}) \quad\bigl(m\in\mathbf {N}\backslash\{1\}\bigr), \end{aligned}$$

(14)

$$\begin{aligned}& \varpi(\lambda_{1},n) < B(\lambda_{1}, \lambda_{2}) \quad\bigl(n\in\mathbf {N}\backslash\{1\}\bigr). \end{aligned}$$

(15)

Proof

For $x\in(n-\frac{1}{2},n+\frac{1}{2})\backslash \{n\}$, $\upsilon_{n+1}\leq V^{\prime}(x)$, by (11), we obtain

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

Setting $t=\frac{\ln(V(x)-\beta)}{\ln(U_{m}-\alpha)}$, since $V(\frac{3}{2})-\beta=1+\frac{\upsilon_{2}}{2}-\beta\geq1$ and $\frac{V^{\prime }(x)}{V(x)-\beta}\,dx=\ln(U_{m}-\alpha)\,dt$, 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 (14). In the same way, we obtain (15). □

Lemma 3

Suppose that $\{\mu_{m}\}_{m=1}^{\infty}$ and $\{\upsilon _{n}\}_{n=1}^{\infty}$ are decreasing, and $U(\infty)=V(\infty )=\infty$. (i) For $m,n\in\mathbf{N}\backslash\{1\}$, we have

$$\begin{aligned} &B(\lambda_{1},\lambda_{2}) \bigl(1-\theta( \lambda_{2},m)\bigr) < \omega (\lambda _{2},m), \end{aligned}$$

(16)

$$\begin{aligned} &B(\lambda_{1},\lambda_{2}) \bigl(1-\vartheta( \lambda_{1},n)\bigr) < \varpi (\lambda_{1},n), \end{aligned}$$

(17)

where

$$\begin{aligned}& \begin{aligned}[b] \theta(\lambda_{2},m) &=\frac{1}{B(\lambda_{1},\lambda_{2})}\frac {\ln ^{\lambda_{2}-1}(1+\upsilon_{2}-\beta)}{\lambda_{2}[1+\frac{\ln (1+\upsilon_{2}\theta(m)-\beta)}{\ln(U_{m}-\alpha)}]^{\lambda }} \frac{1}{\ln^{\lambda_{2}}(U_{m}-\alpha)} \\ &=O\biggl(\frac{1}{\ln^{\lambda_{2}}(U_{m}-\alpha)}\biggr)\in(0,1) \quad\biggl(\theta (m)\in \biggl(\frac{\beta}{\upsilon_{2}},1\biggr)\biggr), \end{aligned} \end{aligned}$$

(18)

$$\begin{aligned}& \begin{aligned}[b] \vartheta(\lambda_{1},n) &=\frac{1}{B(\lambda_{1},\lambda _{2})}\frac{\ln^{\lambda_{1}}(1+\mu_{2}-\alpha)}{\lambda_{1}[1+\frac{\ln(1+\mu _{2}\vartheta(n)-\alpha)}{\ln(V_{n}-\beta)}]^{\lambda}} \frac{1}{\ln ^{\lambda_{1}}(V_{n}-\beta)} \\ &=O\biggl(\frac{1}{\ln^{\lambda_{1}}(V_{n}-\beta)}\biggr) \in(0,1) \quad\biggl(\vartheta (n)\in \biggl(\frac{\alpha}{\mu_{2}},1\biggr)\biggr); \end{aligned} \end{aligned}$$

(19)

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

$$\begin{aligned}& \sum_{m=2}^{\infty}\frac{\mu_{m+1}}{(U_{m}-\alpha)\ln ^{1+c}(U_{m}-\alpha )} = \frac{1}{c} \biggl( \frac{1}{\ln^{c}(1+\mu_{2}-\alpha )}+cO(1) \biggr) , \end{aligned}$$

(20)

$$\begin{aligned}& \sum_{n=2}^{\infty}\frac{\upsilon_{n+1}}{(V_{n}-\beta)\ln ^{1+c}(V_{n}-\beta)} = \frac{1}{c} \biggl( \frac{1}{\ln^{c}(1+\upsilon _{2}-\beta)}+c\widetilde{O}(1) \biggr) . \end{aligned}$$

(21)

Proof

In view of $0\leq\beta\leq\frac{\upsilon_{2}}{2}<\upsilon_{2}$, it follows that $\frac{\beta}{\upsilon_{2}}+1\geq1$ and $\frac{\beta}{\upsilon_{2}}+1<2$. By Example 1, $h(x)$ is strictly decreasing in $[n,n+1]$, then for $m\in\mathbf{N}\backslash\{1\}$, we obtain

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

Setting $t=\frac{\ln(V(x)-\beta)}{\ln(U_{m}-\alpha)}$, since

$$\ln\biggl(V\biggl(\frac{\beta}{\upsilon_{2}}+1\biggr)-\beta\biggr)=\ln \biggl(1+\upsilon_{2}\frac{\beta }{\upsilon_{2}}-\beta\biggr)=0, $$

we find

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

where

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

In view of the integral mid value theorem, there exists a $\theta (m)\in(\frac{\beta}{\upsilon_{2}},1)$, satisfying

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

Since we find

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

namely, $\theta(\lambda_{2},m)=O(\frac{1}{\ln^{\lambda _{2}}(U_{m}-\alpha )})$, we have (16) and (18). In the same way, we obtain (17) and (19).

For any $c>0$, it follows that

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

Hence we obtain (20). In the same way, we obtain (21). □

Main results

We define the following functions:

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

(22)

Theorem 1

We have the following equivalent inequalities:

$$\begin{aligned}& I :=\sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln ^{\lambda}[(U_{m}-\alpha)(V_{n}-\beta)]}\leq\|a\|_{p,\widetilde {\Phi}_{\lambda}} \|b\|_{q,\widetilde{\Psi}_{\lambda}}, \end{aligned}$$

(23)

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

(24)

Proof

By Hölder’s inequality (cf. [15]) and (13), we find

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

(25)

Then by (12) we obtain

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

(26)

namely, (24) follows. By Hölder’s inequality (cf. [15]), we find

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

(27)

Then by (24), (23) follows.

On the other hand, suppose that (23) is valid. We set

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

(28)

Then we have $J^{p}=\|b\|_{q,\widetilde{\Psi}_{\lambda}}^{q}$. If $J=0$, then (24) is trivially valid; if $J=\infty$, then in view of (26), (24) takes the form of an equality. Suppose that $0< J<\infty$. By (23), we obtain

$$\begin{aligned}& \|b\|_{q,\widetilde{\Psi}_{\lambda}}^{q} =J^{p}=I\leq\|a\|_{p,\widetilde{\Phi}_{\lambda}} \|b\|_{q,\widetilde{\Psi}_{\lambda}}, \end{aligned}$$

(29)

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

(30)

namely, (24) follows, which is equivalent to (23). □

Theorem 2

Assuming that $\{\mu_{m}\}_{m=1}^{\infty}$ and $\{\upsilon _{n}\}_{n=1}^{\infty}$ are decreasing, $U(\infty)=V(\infty)=\infty$, $0<\|a\|_{p,\Phi_{\lambda}},\|b\|_{q,\Psi_{\lambda }}<\infty$, we have the following equivalent inequalities:

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

(31)

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

(32)

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

Proof

Applying (14) and (15) in (23) and (24), we have the equivalent inequalities (31) and (32).

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

$$ \widetilde{a}_{m} :=\frac{\mu_{m+1}}{U_{m}-\alpha}\ln^{\widetilde{\lambda}_{1}-1}(U_{m}- \alpha),\qquad \widetilde{b}_{n} :=\frac{\upsilon_{n+1}}{V_{n}-\beta}\ln^{\widetilde {\lambda}_{2}-\varepsilon-1}(V_{n}- \beta). $$

(33)

Then by (20), (21), and (17), we obtain

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

If there exists a positive constant $K\leq B(\lambda_{1},\lambda_{2})$, such that (31) is valid when replacing $B(\lambda_{1},\lambda_{2})$ by K, then in particular, we have $\varepsilon\widetilde {I}<\varepsilon K\|\widetilde{a}\|_{p,\Phi_{\lambda}}\|\widetilde{b}\|_{q,\Psi _{\lambda }}$, namely

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

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 (31).

Similarly, we can obtain

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

(34)

Hence, we can prove that the constant factor $B(\lambda_{1},\lambda_{2})$ in (32) is the best possible. Otherwise, we would reach a contradiction by (34) that the constant factor in (31) is not the best possible. □

We find $\Psi_{\lambda}^{1-p}(n)=\frac{\upsilon_{n+1}}{V_{n}-\beta }\ln ^{p\lambda_{2}-1}(V_{n}-\beta)$, and we define the following weighted 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 $a=\{a_{m}\}_{m=2}^{\infty}\in l_{p,\Phi_{\lambda}}$, setting

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

we can rewrite (32) as follows:

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

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

Definition 1

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 $Ta=c\in l_{p,\Psi_{\lambda}^{1-p}}$. We set the formal inner product of Ta and $b=\{b_{n}\}_{n=2}^{\infty}\in l_{q,\Psi _{\lambda}}$ as follows:

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

(35)

Then we can rewrite (31) and (32) as follows:

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

(36)

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

(37)

We set the norm of operator T as follows:

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

By (37), we find $\|T\|\leq B(\lambda_{1},\lambda_{2})$. Since the constant factor in (37) is the best possible, it follows that $\|T\|=B(\lambda_{1},\lambda_{2})$.

Remark 1

(i) For $\alpha=\beta=0$ in (31) and (32), setting

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

we have the following equivalent Hardy-Mulholland-type inequalities:

$$\begin{aligned}& \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln^{\lambda }(U_{m}V_{n})}< B(\lambda_{1}, \lambda_{2})\|a\|_{p,\varphi_{\lambda }}\|b\|_{q,\psi_{\lambda}}, \end{aligned}$$

(38)

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

(39)

For $\lambda=1$, $\lambda_{1}=\frac{1}{q}$, $\lambda_{2}=\frac{1}{p}$ in (38) and (39), we have the following equivalent inequality:

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

(40)

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

(41)

Hence, (38) is an extension of (40), and (31) is a more accurate inequality of (38) (for $0<\alpha\leq\frac {\mu_{2}}{2}$, $0<\beta\leq\frac{\upsilon_{2}}{2}$).

(ii) For $\mu_{i}=\upsilon_{j}=1$ ($i,j\in\mathbf{N}$), $\lambda =1$, $\lambda _{1}=\frac{1}{q}$, $\lambda_{2}=\frac{1}{p}$ in (31), we reduce our case to the following inequality: For $\alpha,\beta\leq\frac{1}{2}$,

$$\begin{aligned} &\sum_{m=2}^{\infty}\sum _{n=2}^{\infty}\frac{a_{m}b_{n}}{\ln [(m-\alpha )(n-\beta)]} \\ &\quad< \frac{\pi}{\sin(\pi/p)} \Biggl[ \sum_{m=2}^{\infty} \frac {a_{m}^{p}}{(m-\alpha)^{1-p}} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum _{n=2}^{\infty}\frac {b_{n}^{q}}{(n-\beta)^{1-q}} \Biggr] ^{\frac{1}{q}}, \end{aligned}$$

(42)

Hence, (42) is a more accurate inequality of (3) (for $0<\alpha,\beta\leq\frac{1}{2}$).