A new extension of a Hardy-Hilbert-type inequality

Abstract

By introducing independent parameters, and applying weight coefficients and the technique of real analysis, we give a new extension of a Hardy-Hilbert-type inequality with a best possible constant factor. Furthermore, the equivalent forms, the operator expressions, and the reverses are considered.

1 Introduction

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

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

(1)

where the constant factor $\frac{\pi }{\sin (\pi /p)}$ is the best possible. We also have the following Hardy-Hilbert-type inequality (cf. [2]):

$$ \sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{\ln (m/n)a_{m}b_{n}}{m-n}< \biggl[ \frac{\pi }{\sin (\pi /p)} \biggr] ^{2} \Biggl( \sum _{n=1}^{\infty }a_{n}^{p} \Biggr) ^{1/p} \Biggl( \sum_{n=1}^{\infty }b_{n}^{q} \Biggr) ^{1/q}, $$

(2)

where the constant factor $[ \frac{\pi }{\sin (\pi /p)} ] ^{2}$ is still the best possible. In 2008, by introducing some parameters, Yang gave an extension of inequality (2) (cf. [3]): If $0<\lambda _{1},\lambda _{2}\leq 1$, $\lambda _{1}+\lambda _{2}=\lambda$, $a_{n},b_{n}\geq 0$, $0<\sum_{n=1}^{\infty }n^{p(1-\lambda _{1})-1}a_{n}^{p}<\infty$, and $0<\sum_{n=1}^{\infty }n^{q(1-\lambda _{2})-1}b_{n}^{q}<\infty$, then the following inequality holds:

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

(3)

where the constant factor $[ \frac{\pi }{\lambda \sin (\pi \lambda _{1}/\lambda )} ] ^{2}$ is the best possible. There are lots of improvements, generalizations, and applications of inequality (2) ([3–11]). For more details, Yang gives a summary of introducing independent parameters ([12, 13]).

In this article, by introducing independent parameters, and applying weight coefficients and the technique of real analysis, we give a new extension of (2) with a best possible constant factor. Furthermore, the equivalent forms, the operator expressions, and the reverses are considered.

2 Some lemmas

We agree on the following assumptions in this paper: $p\neq 0$, $\frac{1}{p}+\frac{1}{q}=1$, $\lambda >0$, $0<\lambda _{i}\leq 1$ (${i=1,2}$), $\lambda _{1}+\lambda _{2}=\lambda$, $k_{\lambda }(\lambda _{2})=k_{\lambda }(\lambda _{1})= [ \frac{\pi }{\lambda \sin (\pi \lambda _{1}/\lambda )} ] ^{2}$, $\{\mu _{m}\}_{m=1}^{\infty }$ and $\{\nu _{n}\}_{n=1}^{\infty }$ are positive sequences, $U_{m}=\sum_{i=1}^{m}\mu _{i}$, $V_{n}=\sum_{i=1}^{n}\nu _{i}$, and $a_{n},b_{n}\geq 0$ ($m,n\in \mathbf{N}=\{1,2,\ldots \}$),

$$ 0< \sum_{m=1}^{\infty }\frac{U_{m}^{p(1-\lambda _{1})-1}}{\mu _{m}^{p-1}}a_{m}^{p}< \infty ,\qquad 0< \sum_{n=1}^{\infty }\frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}b_{n}^{q}< \infty . $$

Lemma 1

Define the weight coefficients as follows:

$$\begin{aligned}& \omega (\lambda _{2},m):=\sum_{n=1}^{\infty } \frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}\frac{U_{m}^{\lambda _{1}}}{V_{n}^{1-\lambda _{2}}}\nu _{n},\quad m\in \mathbf{N}, \end{aligned}$$

(4)

$$\begin{aligned}& \varpi (\lambda _{1},n):=\sum_{m=1}^{\infty } \frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}\frac{V_{n}^{\lambda _{2}}}{U_{m}^{1-\lambda _{1}}}\mu _{m},\quad n\in \mathbf{N}. \end{aligned}$$

(5)

We have the following inequalities:

$$\begin{aligned}& \omega (\lambda _{2},m)< k_{\lambda }(\lambda _{1})\quad (m \in \mathbf{N};0< \lambda _{2}\leq 1,\lambda _{1}>0), \end{aligned}$$

(6)

$$\begin{aligned}& \varpi (\lambda _{1},n)< k_{\lambda }(\lambda _{1}) \quad (n \in \mathbf{N};0< \lambda _{1}\leq 1,\lambda _{2}>0). \end{aligned}$$

(7)

Proof

Putting $\mu (t):=\mu _{m}$, $t\in (m-1,m]$ ($m=1,2,\ldots$), $\nu (t):=\nu _{n}$, $t\in (n-1,n]$ ($n=1,2,\ldots$),

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

Then we have $U(m)=U_{m}$, $V(n)=V_{n}$ ($m,n\in \mathbf{N}$). $U^{\prime} (x)=\mu (x)=\mu _{m}$ when $x\in (m-1,m]$; $V^{\prime}(y)=\nu (y)=\nu _{n}$ when $y\in (n-1,n]$. Since the function $V(y)$ ($y>0$) is strictly increasing and $f(x)=\frac{\ln (m/x)}{m^{\lambda }-x^{\lambda }}$ ($x>0$) is strictly decreasing (cf. [4], Example 2.2.1), in view of $1-\lambda _{2}\geq 0$, we have

$$\begin{aligned} \omega (\lambda _{2},m) =&\sum_{n=1}^{\infty } \int_{n-1}^{n}\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}\frac{U_{m}^{\lambda _{1}}}{V_{n}^{1-\lambda _{2}}}V^{\prime}(t) \,dt \\ < &\sum_{n=1}^{\infty } \int_{n-1}^{n}\frac{\ln (U_{m}/V(t))}{U_{m}^{\lambda }-V^{\lambda }(t)}\frac{U_{m}^{\lambda _{1}}}{V^{1-\lambda _{2}}(t)}V^{\prime}(t) \,dt. \end{aligned}$$

Putting $u=\frac{V^{\lambda }(t)}{U_{m}^{\lambda }}$ in the above integral, and in view of the fact that (cf. [2])

$$ \int_{0}^{\infty }\frac{\ln u}{u-1}u^{a-1}\,du= \biggl[ \frac{\pi }{\sin (a\pi )} \biggr] ^{2}\quad (0< a< 1), $$

it follows that

$$\begin{aligned} \omega (\lambda _{2},m) < &\frac{1}{\lambda ^{2}}\sum _{n=1}^{\infty } \int_{\frac{V^{\lambda }(n-1)}{U_{m}^{\lambda }}}^{\frac{V^{\lambda }(n)}{U_{m}^{\lambda }}}\frac{\ln u}{u-1}u^{\frac{\lambda _{2}}{\lambda }-1}\,du \\ =&\frac{1}{\lambda ^{2}} \int_{0}^{\frac{V^{\lambda }(\infty )}{U_{m}^{\lambda }}}\frac{\ln u}{u-1}u^{\frac{\lambda _{2}}{\lambda }-1}\,du \leq \frac{1}{\lambda ^{2}} \int_{0}^{\infty }\frac{\ln u}{u-1}u^{\frac{\lambda _{2}}{\lambda }-1}\,du \\ =& \biggl[ \frac{\pi }{\lambda \sin (\pi \lambda _{2}/\lambda )} \biggr] ^{2}= \biggl[ \frac{\pi }{\lambda \sin (\pi \lambda _{1}/\lambda )} \biggr] ^{2}=k_{\lambda }(\lambda _{1}). \end{aligned}$$

Hence we prove that (6) is valid. In the same way, we can prove that (7) is valid too. □

Lemma 2

Suppose that $\{\mu _{m}\}_{m=1}^{\infty }$ and $\{\nu _{n}\}_{n=1}^{\infty }$ are decreasing sequences, and $U(\infty )=V(\infty )=\infty$, then we have the following inequalities:

$$\begin{aligned}& k_{\lambda }(\lambda _{1}) \bigl(1-\theta _{1}(\lambda _{2},m)\bigr)< \omega (\lambda _{2},m) \quad (m\in \mathbf{N};0< \lambda _{2}\leq 1,\lambda _{1}>0), \end{aligned}$$

(8)

$$\begin{aligned}& k_{\lambda }(\lambda _{1}) \bigl(1-\theta _{2}(\lambda _{1},n)\bigr)< \varpi (\lambda _{1},n)\quad (n\in \mathbf{N};0< \lambda _{1}\leq 1,\lambda _{2}>0), \end{aligned}$$

(9)

where $\theta _{1}(\lambda _{2},m)=O(\frac{1}{U_{m}^{\lambda _{2}/2}})\in (0,1)$ and $\theta _{2}(\lambda _{1},n)=O(\frac{1}{V_{n}^{\lambda _{1}/2}})\in (0,1)$. Moreover, we get

$$\begin{aligned}& \sum_{m=1}^{\infty }\frac{\mu _{m}}{U_{m}^{1+\varepsilon }} = \frac{1}{\varepsilon }\bigl(1+o_{1}(1)\bigr) \quad \bigl(\varepsilon \rightarrow 0^{+}\bigr), \end{aligned}$$

(10)

$$\begin{aligned}& \sum_{n=1}^{\infty }\frac{\nu _{n}}{V_{n}^{1+\varepsilon }} = \frac{1}{\varepsilon }\bigl(1+o_{2}(1)\bigr) \quad \bigl(\varepsilon \rightarrow 0^{+}\bigr). \end{aligned}$$

(11)

Proof

By the decreasing property of $\{\nu _{n}\}_{n=1}^{\infty }$, and in view of $1-\lambda _{2}\geq 0$, $V(\infty )=\infty$, we find

$$\begin{aligned} \omega (\lambda _{2},m) \geq &\sum_{n=1}^{\infty } \frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}\frac{U_{m}^{\lambda _{1}}}{V_{n}^{1-\lambda _{2}}}\nu _{n+1} \\ =&\sum_{n=1}^{\infty } \int_{n}^{n+1}\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}\frac{U_{m}^{\lambda _{1}}}{V_{n}^{1-\lambda _{2}}}V^{\prime}(t) \,dt \\ >&\sum_{n=1}^{\infty } \int_{n}^{n+1}\frac{\ln (U_{m}/V(t))}{U_{m}^{\lambda }-V^{\lambda }(t)}\frac{U_{m}^{\lambda _{1}}}{V^{1-\lambda _{2}}(t)}V^{\prime}(t) \,dt \\ =&\frac{1}{\lambda ^{2}}\sum_{n=1}^{\infty } \int_{\frac{V^{\lambda }(n)}{U_{m}^{\lambda }}}^{\frac{V^{\lambda }(n+1)}{U_{m}^{\lambda }}}\frac{\ln u}{u-1}u^{\frac{\lambda _{2}}{\lambda }-1} \,du=\frac{1}{\lambda ^{2}} \int_{\frac{V^{\lambda }(1)}{U_{m}^{\lambda }}}^{\infty }\frac{\ln u}{u-1}u^{\frac{\lambda _{2}}{\lambda }-1}\,du \\ =&k_{\lambda }(\lambda _{1})-\frac{1}{\lambda ^{2}} \int_{0}^{\frac{\nu _{1}^{\lambda }}{U_{m}^{\lambda }}}\frac{\ln u}{u-1}u^{\frac{\lambda _{2}}{\lambda }-1} \,du=k_{\lambda }(\lambda _{1}) \bigl(1-\theta _{1}( \lambda _{2},m)\bigr), \end{aligned}$$

where

$$ \theta _{1}(\lambda _{2},m):=\frac{1}{\lambda ^{2}k_{\lambda }(\lambda _{1})}\int_{0}^{\frac{\nu _{1}^{\lambda }}{U_{m}^{\lambda }}}\frac{\ln u}{u-1}u^{\frac{\lambda _{2}}{\lambda }-1}\,du \in (0,1). $$

In virtue of

$$\begin{aligned}& \lim_{x\rightarrow \infty } \frac{\int_{0}^{\nu _{1}^{\lambda }/x^{\lambda }}\frac{\ln u}{u-1}u^{\frac{\lambda _{2}}{\lambda }-1}\,du}{x^{-\lambda _{2}/2}} \\& \quad = \lim_{x\rightarrow \infty }\frac{2\lambda ^{2}\nu _{1}^{\lambda _{2}}}{\lambda _{2}}\biggl(\frac{\nu _{1}^{\lambda }}{x^{\lambda }}-1 \biggr)^{-1}\biggl(\frac{1}{x^{\lambda _{2}/2}}\ln \frac{\nu _{1}}{x}\biggr)=0, \end{aligned}$$

it is obvious that $\theta _{1}(\lambda _{2},m)=O(\frac{1}{U_{m}^{\lambda _{2}/2}})$. Hence (8) is valid. In the same way, we can prove that (9) is valid too. Moreover, we have

$$\begin{aligned}& \begin{aligned} \sum_{m=1}^{\infty }\frac{\mu _{m}}{U_{m}^{1+\varepsilon }} &= \frac{1}{\mu _{1}^{\varepsilon }}+\sum_{m=2}^{\infty } \int_{m-1}^{m}\frac{U^{\prime}(t)}{U_{m}^{1+\varepsilon }}\,dt \\ &\leq \frac{1}{\mu _{1}^{\varepsilon }}+\sum_{m=2}^{\infty } \int_{m-1}^{m}\frac{U^{\prime}(t)}{U^{1+\varepsilon }(t)}\,dt \\ &=\frac{1}{\mu _{1}^{\varepsilon }}+\sum_{m=2}^{\infty } \int_{U(m-1)}^{U(m)}\frac{1}{u^{1+\varepsilon }}\,du= \frac{1}{\mu _{1}^{\varepsilon }}+ \int_{\mu _{1}}^{\infty }\frac{1}{u^{1+\varepsilon }}\,du \\ &=\frac{1}{\varepsilon }\biggl[1+\biggl(\frac{1}{\mu _{1}^{\varepsilon }}+\frac{\varepsilon }{\mu _{1}^{\varepsilon }}-1 \biggr)\biggr], \end{aligned}\\& \begin{aligned} \sum_{m=1}^{\infty }\frac{\mu _{m}}{U_{m}^{1+\varepsilon }} &\geq \sum_{m=1}^{\infty } \int_{m}^{m+1}\frac{\mu _{m+1}}{U_{m}^{1+\varepsilon }}\,dt \\ &=\sum_{m=1}^{\infty } \int_{m}^{m+1}\frac{U^{\prime}(t)}{U_{m}^{1+\varepsilon }}\,dt>\sum _{m=1}^{\infty } \int_{m}^{m+1}\frac{U^{\prime}(t)}{U^{1+\varepsilon }(t)}\,dt \\ &=\sum_{m=1}^{\infty } \int_{U(m)}^{U(m+1)}\frac{1}{u^{1+\varepsilon }}\,du= \int_{\mu _{1}}^{\infty }\frac{1}{u^{1+\varepsilon }}\,du \\ &=\frac{1}{\varepsilon }\biggl[1+\biggl(\frac{1}{\mu _{1}^{\varepsilon }}-1\biggr)\biggr]. \end{aligned} \end{aligned}$$

Then we have (10). In the same way, we have (11). □

Remark 1

Taking $\varepsilon =a>0$, we write by (10) and (11) that

$$ \sum_{m=1}^{\infty }\frac{\mu _{m}}{U_{m}^{1+a}}=O_{1}(1),\qquad \sum_{n=1}^{\infty }\frac{\nu _{n}}{V_{n}^{1+a}}=O_{2}(1). $$

3 Equivalent forms and operator expressions

Theorem 1

Suppose that $p>1$, then we have the following equivalent inequalities:

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

(12)

$$\begin{aligned}& \begin{aligned}[b] J &:= \Biggl\{ \sum_{n=1}^{\infty } \frac{\nu _{n}}{V_{n}^{1-p\lambda _{2}}} \Biggl( \sum_{m=1}^{\infty } \frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}a_{m} \Biggr) ^{p} \Biggr\} ^{\frac{1}{p}} \\ &< \biggl[ \frac{\pi }{\lambda \sin (\pi \lambda _{1}/\lambda )} \biggr] ^{2} \Biggl( \sum _{m=1}^{\infty }\frac{U_{m}^{p(1-\lambda _{1})-1}}{\mu _{m}^{p-1}}a_{m}^{p} \Biggr) ^{1/p}. \end{aligned} \end{aligned}$$

(13)

Proof

By Hölder’s inequality with weight (cf. [14]), we find

$$\begin{aligned} & \Biggl( \sum_{m=1}^{\infty } \frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}a_{m} \Biggr) ^{p} \\ &\quad = \Biggl\{ \sum _{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}\biggl[ \frac{U_{m}^{(1-\lambda _{1})/q}\nu _{n}^{1/p}}{V_{n}^{(1-\lambda _{2})/p}\mu _{m}^{1/q}}a_{m}\biggr] \biggl[\frac{V_{n}^{(1-\lambda _{2})/p}\mu _{m}^{1/q}}{U_{m}^{(1-\lambda _{1})/q}\nu _{n}^{1/p}} \biggr] \Biggr\} ^{p} \\ & \quad \leq \sum_{m=1}^{\infty } \frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}\frac{U_{m}^{(1-\lambda _{1})p/q}\nu _{n}^{{}}}{V_{n}^{1-\lambda _{2}}\mu _{m}^{p/q}}a_{m}^{p} \Biggl[ \sum _{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }} \frac{V_{n}^{(1-\lambda _{2})(q-1)}\mu _{m}^{{}}}{U_{m}^{1-\lambda _{1}}\nu _{n}^{q-1}} \Biggr] ^{p-1} \\ & \quad =\bigl(\varpi (\lambda _{1},n)\bigr)^{p-1} \frac{V_{n}^{1-p\lambda _{2}}}{\nu _{n}^{{}}}\sum_{m=1}^{\infty } \frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}\frac{U_{m}^{(1-\lambda _{1})p/q}\nu _{n}^{{}}}{V_{n}^{1-\lambda _{2}}\mu _{m}^{p/q}}a_{m}^{p}. \end{aligned}$$

(14)

By (7), it follows that

$$\begin{aligned} J < &\bigl(k_{\lambda }(\lambda _{1})\bigr)^{\frac{1}{q}} \Biggl[ \sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}\frac{U_{m}^{(1-\lambda _{1})p/q}\nu _{n}^{{}}}{V_{n}^{1-\lambda _{2}}\mu _{m}^{p/q}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k_{\lambda }(\lambda _{1})\bigr)^{\frac{1}{q}} \Biggl[ \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}\frac{U_{m}^{(1-\lambda _{1})(p-1)}\nu _{n}^{{}}}{V_{n}^{1-\lambda _{2}}\mu _{m}^{p-1}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k_{\lambda }(\lambda _{1})\bigr)^{\frac{1}{q}} \Biggl[ \sum_{m=1}^{\infty }\omega (\lambda _{2},m)\frac{U_{m}^{p(1-\lambda _{1})-1}}{\mu _{m}^{p-1}}a_{m}^{p} \Biggr] ^{\frac{1}{p}}. \end{aligned}$$

(15)

Combining (8) and (15), we have (13).

Using Hölder’s inequality again, we have

$$\begin{aligned} I =&\sum_{n=1}^{\infty } \Biggl[ \frac{\nu _{n}^{1/p}}{V_{n}^{\frac{1}{p}-\lambda _{2}}}\sum_{m=1}^{\infty } \frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}a_{m} \Biggr] \biggl[ \frac{V_{n}^{\frac{1}{p}-\lambda _{2}}}{\nu _{n}^{1/p}}b_{n} \biggr] \\ \leq &J \Biggl[ \sum_{n=1}^{\infty } \frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned}$$

(16)

and then we have (12) by using (13). On the other hand, assuming that (12) is valid, setting

$$ b_{n}=\frac{\nu _{n}}{V_{n}^{1-p\lambda _{2}}} \Biggl[ \sum_{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}a_{m} \Biggr] ^{p-1}, \quad n\in \mathbf{N}, $$

then we find $J= [ \sum_{n=1}^{\infty }\frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}b_{n}^{q} ] ^{1/p}$. By (15), it follows that $J<\infty$. If $J=0$, then (13) is trivially valid. If $0< J<\infty$, then we have

$$\begin{aligned}& \begin{aligned} 0 &< \sum_{n=1}^{\infty }\frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}b_{n}^{q}=J^{p}=I \\ &< k_{\lambda }(\lambda _{1}) \Biggl[ \sum _{m=1}^{\infty }\frac{U_{m}^{p(1-\lambda _{1})-1}}{\mu _{m}^{p-1}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty } \frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}b_{n}^{q} \Biggr] ^{\frac{1}{q}}< \infty , \end{aligned}\\& J= \Biggl[ \sum_{n=1}^{\infty } \frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}b_{n}^{q} \Biggr] ^{1/p}< k_{\lambda }( \lambda _{1}) \Biggl[ \sum_{m=1}^{\infty } \frac{U_{m}^{p(1-\lambda _{1})-1}}{\mu _{m}^{p-1}}a_{m}^{p} \Biggr] ^{\frac{1}{p}}. \end{aligned}$$

Hence (13) is valid, which is equivalent to (12). □

Theorem 2

Suppose that $p>1$, $\{\mu _{m}\}_{m=1}^{\infty }$ and $\{\nu _{n}\}_{n=1}^{\infty }$ are decreasing positive sequences, and $U(\infty )=V(\infty )=\infty$, then the constant factor $k_{\lambda }(\lambda _{1})= [ \frac{\pi }{\lambda \sin (\lambda _{1}\pi /\lambda )} ] ^{2}$ is the best possible in (12) and (13).

Proof

For $0<\varepsilon <p\lambda _{1}$, we set $\widetilde{\lambda }_{1}=\lambda _{1}-\frac{\varepsilon }{p}$ ($\in (0,1)$), $\widetilde{\lambda }_{2}=\lambda _{2}+\frac{\varepsilon }{p}$ (>0), $\widetilde{a}_{m}=U_{m}^{\widetilde{\lambda }_{1}-1}\mu _{m}$, $\widetilde{b}_{n}=V_{n}^{\widetilde{\lambda }_{2}-\varepsilon -1}\nu _{n}$. By (10), (11), and (9), in view of Remark 1, we find

$$\begin{aligned}& \sum_{m=1}^{\infty }\frac{U_{m}^{p(1-\lambda _{1})-1}}{\mu _{m}^{p-1}}\widetilde{a}_{m}^{p} = \sum_{m=1}^{\infty } \frac{\mu _{m}}{U_{m}^{1+\varepsilon }}=\frac{1}{\varepsilon }\bigl(1+o_{1}(1)\bigr), \\& \sum_{n=1}^{\infty }\frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}\widetilde{b}_{n}^{q} = \sum_{n=1}^{\infty } \frac{\nu _{n}}{V_{n}^{1+\varepsilon }}=\frac{1}{\varepsilon }\bigl(1+o_{2}(1)\bigr) , \\& \begin{aligned} \widetilde{I} &:=\sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }} \widetilde{a}_{m}\widetilde{b}_{n} \\ &=\sum_{n=1}^{\infty } \Biggl[ \sum _{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}\frac{V_{n}^{\widetilde{\lambda }_{2}}\mu _{m}}{U_{m}^{1-\widetilde{\lambda }_{1}}} \Biggr] \frac{\nu _{n}}{V_{n}^{\varepsilon +1}} \\ &=\sum_{n=1}^{\infty }\varpi (\widetilde{\lambda }_{1},n)\frac{\nu _{n}}{V_{n}^{\varepsilon +1}}\geq k_{\lambda }(\widetilde{\lambda }_{1})\sum_{n=1}^{\infty } \bigl(1-\theta _{2}(\widetilde{\lambda }_{1},n)\bigr) \frac{\nu _{n}}{V_{n}^{\varepsilon +1}} \\ &=k_{\lambda }(\widetilde{\lambda }_{1}) \Biggl[ \sum _{n=1}^{\infty }\frac{\nu _{n}}{V_{n}^{\varepsilon +1}}-\sum _{n=1}^{\infty }O\biggl(\frac{\nu _{n}}{V_{n}^{\frac{1}{2}(\frac{\varepsilon }{q}+\varepsilon +\lambda _{1})+1}}\biggr) \Biggr] \\ &=\frac{1}{\varepsilon } \biggl[ \frac{\pi }{\lambda \sin (\pi \widetilde{\lambda }_{1}/\lambda )} \biggr] ^{2} \bigl[1+o_{2}(1)-\varepsilon O(1)\bigr]. \end{aligned} \end{aligned}$$

If there exists a positive number $K\leq k_{\lambda }(\lambda _{1})$, such that (12) is still valid when replacing $k_{\lambda }(\lambda _{1})$ by K, then, in particular, we have

$$\begin{aligned} \varepsilon \widetilde{I} =&\varepsilon \sum_{n=1}^{\infty } \sum_{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }} \widetilde{a}_{m}\widetilde{b}_{n} \\ < &\varepsilon K \Biggl[ \sum_{m=1}^{\infty } \frac{U_{m}^{p(1-\lambda _{1})-1}}{\mu _{m}^{p-1}}\widetilde{a}_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty } \frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}\widetilde{b}_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$

We obtain from the above results

$$ \biggl[ \frac{\pi }{\lambda \sin (\pi \widetilde{\lambda }_{1}/\lambda )} \biggr] ^{2} \bigl[1+o_{2}(1)-\varepsilon O(1)\bigr]< K \bigl( 1+o_{1}(1) \bigr) ^{\frac{1}{p}} \bigl( 1+o_{2}(1) \bigr) ^{\frac{1}{q}}, $$

and then it follows that $k_{\lambda }(\lambda _{1})\leq K$ (for $\varepsilon \rightarrow 0^{+}$). Hence $K=k_{\lambda }(\lambda _{1})$ is the best value of (12).

We conform that the constant factor $k_{\lambda }(\lambda _{1})$ in (13) is the best possible. Otherwise we can get a contradiction by (16): that the constant factor in (12) is not the best value. □

For $p>1$, setting

$$ \varphi (m):=\frac{U_{m}^{p(1-\lambda _{1})-1}}{\mu _{m}^{p-1}},\qquad \psi (n):=\frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}} \quad (n,m\in \mathbf{N}), $$

then it follows that $[ \psi (n) ] ^{1-p}=\frac{\nu _{n}}{V_{n}^{1-p\lambda _{2}}}$, and we define the real weighted normed function spaces as follows:

$$\begin{aligned}& l_{p,\varphi }:= \Biggl\{ a=\{a_{m}\}_{m=1}^{\infty }; \Vert a\Vert _{p,\varphi }= \Biggl\{ \sum_{m=1}^{\infty } \frac{U_{m}^{p(1-\lambda _{1})-1}}{\mu _{m}^{p-1}}\vert a_{m}\vert ^{p} \Biggr\} ^{\frac{1}{p}}< \infty \Biggr\} , \\& l_{q,\psi }:= \Biggl\{ b=\{b_{n}\}_{n=1}^{\infty }; \Vert b\Vert _{q,\psi }= \Biggl\{ \sum_{n=1}^{\infty } \frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}\vert b_{n}\vert ^{q} \Biggr\} ^{\frac{1}{q}}< \infty \Biggr\} , \\& l_{p,\psi ^{1-p}}:= \Biggl\{ c=\{c_{n}\}_{n=1}^{\infty }; \Vert c\Vert _{p,\psi ^{1-p}}= \Biggl\{ \sum_{n=1}^{\infty } \frac{\nu _{n}}{V_{n}^{1-p\lambda _{2}}}\vert c_{n}\vert ^{p} \Biggr\} ^{\frac{1}{p}}< \infty \Biggr\} . \end{aligned}$$

For $a=\{a_{m}\}_{m=1}^{\infty }\in l_{p,\varphi }$, putting $h_{n}:=\sum_{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}a_{m}$, $h=\{h_{n}\}_{n=1}^{\infty }$, then it follows by (13) that $\Vert h\Vert _{p,\psi ^{1-p}}< k_{\lambda }(\lambda _{1})\Vert a\Vert _{p,\varphi }$, and $h\in l_{p,\psi ^{1-p}}$.

Definition 1

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

$$ (Ta,b):=\sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}a_{m}b_{n}. $$

(17)

Hence (12) and (13) may be rewritten in terms of the following operator expressions:

$$\begin{aligned}& (Ta,b) < k_{\lambda }(\lambda _{1})\Vert a\Vert _{p,\varphi }\Vert b\Vert _{q,\psi }, \end{aligned}$$

(18)

$$\begin{aligned}& \Vert Ta\Vert _{p,\psi ^{1-p}} < k_{\lambda }(\lambda _{1}) \Vert a\Vert _{p,\varphi }. \end{aligned}$$

(19)

It follows that the operator T is bounded with

$$ \Vert T\Vert :=\sup_{a(\neq \theta )\in l_{p,\varphi }} \frac{\Vert Ta\Vert _{p,\psi ^{1-p}}}{\Vert a\Vert _{p,\varphi }}\leq k_{\lambda }(\lambda _{1}). $$

Since the constant factor $k_{\lambda }(\lambda _{1})$ in (19) is the best possible, we have

$$ \Vert T\Vert =k_{\lambda }(\lambda _{1})= \biggl[ \frac{\pi }{\lambda \sin (\lambda _{1}\pi /\lambda )} \biggr] ^{2}. $$

(20)

4 Some reverses

We set $\widetilde{\varphi }(m):=(1-\theta _{1}(\lambda _{2},m))\frac{U_{m}^{p(1-\lambda _{1})-1}}{\mu _{m}^{p-1}}$, $\widetilde{\psi }(n):=(1-\theta _{2}(\lambda _{1},m))\frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}$ ($n,m\in \mathbf{N}$). For $0< p<1$ or $p<0$, we still use the formal symbol of the norm in this part for convenience.

Theorem 3

Suppose that $0< p<1$, $\{\mu _{m}\}_{m=1}^{\infty }$ and $\{\nu _{n}\}_{n=1}^{\infty }$ are decreasing positive sequences, and $U(\infty )=V(\infty )=\infty$, then we have the following equivalent inequalities:

$$\begin{aligned}& I =\sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}a_{m}b_{n}> \biggl[ \frac{\pi }{\lambda \sin (\pi \lambda _{1}/\lambda )} \biggr] ^{2}\Vert a\Vert _{p,\widetilde{\varphi }}\Vert b\Vert _{q,\psi }, \end{aligned}$$

(21)

$$\begin{aligned}& \begin{aligned}[b] J &= \Biggl\{ \sum_{n=1}^{\infty } \frac{\nu _{n}}{V_{n}^{1-p\lambda _{2}}} \Biggl( \sum_{m=1}^{\infty } \frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}a_{m} \Biggr) ^{p} \Biggr\} ^{\frac{1}{p}} \\ &> \biggl[ \frac{\pi }{\lambda \sin (\pi \lambda _{1}/\lambda )} \biggr] ^{2}\Vert a\Vert _{p,\widetilde{\varphi }} \end{aligned} \end{aligned}$$

(22)

where the constant factor $[ \frac{\pi }{\lambda \sin (\pi \lambda _{1}/\lambda )} ] ^{2}$ is the best possible.

Proof

By the reverse Hölder inequality with weight (cf. [14]) and (7), we obtain the reverse forms of (14) and (15). It follows that (22) is valid by (8). Using the reverse Hölder inequality (cf. [14]), we find

$$ I=\sum_{n=1}^{\infty } \Biggl[ \frac{\nu _{n}^{1/p}}{V_{n}^{\frac{1}{p}-\lambda _{2}}} \sum_{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}a_{m} \Biggr] \biggl[ \frac{V_{n}^{\frac{1}{p}-\lambda _{2}}}{\nu _{n}^{1/p}}b_{n} \biggr] \geq J\Vert b\Vert _{q,\psi }. $$

(23)

Hence (21) is valid by using (22). Setting

$$ b_{n}=\frac{\nu _{n}}{V_{n}^{1-p\lambda _{2}}} \Biggl[ \sum_{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}a_{m} \Biggr] ^{p-1},\quad n\in \mathbf{N}, $$

then we have $J= [ \sum_{n=1}^{\infty }\frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}b_{n}^{q} ] ^{1/p}$. Assume that (21) is valid. By the reverse of (15), it follows that $J>0$. If $J=\infty$, then (22) is trivially valid. If $0< J<\infty$, then we find

$$\begin{aligned}& \sum_{n=1}^{\infty }\frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}b_{n}^{q}=J^{p}=I>k_{\lambda }( \lambda _{1})\Vert a\Vert _{p,\widetilde{\varphi }} \Biggl[ \sum _{n=1}^{\infty }\frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \\& J= \Biggl[ \sum_{n=1}^{\infty } \frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}b_{n}^{q} \Biggr] ^{1/p}>k_{\lambda }( \lambda _{1})\Vert a\Vert _{p,\widetilde{\varphi }}. \end{aligned}$$

Hence (22) is valid, which is equivalent to (21).

For $0<\varepsilon <p\lambda _{1}$, we set $\widetilde{\lambda }_{1}=\lambda _{1}-\frac{\varepsilon }{p}$ ($\in (0,1)$), $\widetilde{\lambda }_{2}=\lambda _{2}+\frac{\varepsilon }{p}$ (>0), $\widetilde{a}_{m}=U_{m}^{\widetilde{\lambda }_{1}-1}\mu _{m}$, $\widetilde{b}_{n}=V_{n}^{\widetilde{\lambda }_{2}-\varepsilon -1}\nu _{n}$. By (10), (11), and (7), in view of Remark 1, we find

$$\begin{aligned}& \sum_{m=1}^{\infty }\bigl(1-\theta _{1}(\lambda _{2},m)\bigr)\frac{U_{m}^{p(1-\lambda _{1})-1}}{\mu _{m}^{p-1}} \widetilde{a}_{m}^{p}\\& \quad =\sum _{m=1}^{\infty }\biggl(1-O\biggl(\frac{1}{U_{m}^{\lambda _{2}/2}}\biggr) \biggr)\frac{\mu _{m}}{U_{m}^{1+\varepsilon }} \\& \quad =\sum_{m=1}^{\infty } \frac{\mu _{m}}{U_{m}^{1+\varepsilon }}-\sum_{m=1}^{\infty }O \biggl(\frac{\mu _{m}}{U_{m}^{1+\varepsilon +(\lambda _{2}/2)}}\biggr)=\frac{1}{\varepsilon }\bigl(1+o_{1}(1)- \varepsilon O(1)\bigr), \\& \sum_{n=1}^{\infty }\frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}} \widetilde{b}_{n}^{q}=\sum_{n=1}^{\infty } \frac{\nu _{n}}{V_{n}^{1+\varepsilon }}=\frac{1}{\varepsilon }\bigl(1+o_{2}(1)\bigr) , \\& \begin{aligned} \widetilde{I} &=\sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }} \widetilde{a}_{m}\widetilde{b}_{n}=\sum _{n=1}^{\infty } \Biggl[ \sum_{m=1}^{\infty } \frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}\frac{V_{n}^{\widetilde{\lambda }_{2}}\mu _{m}}{U_{m}^{1-\widetilde{\lambda }_{1}}} \Biggr] \frac{\nu _{n}}{V_{n}^{\varepsilon +1}} \\ &=\sum_{n=1}^{\infty }\varpi (\widetilde{\lambda }_{1},n)\frac{\nu _{n}}{V_{n}^{\varepsilon +1}}< k_{\lambda }(\widetilde{\lambda }_{1})\sum_{n=1}^{\infty } \frac{\nu _{n}}{V_{n}^{\varepsilon +1}} \\ &=\frac{1}{\varepsilon } \biggl[ \frac{\pi }{\lambda \sin (\pi \widetilde{\lambda }_{1}/\lambda )} \biggr] ^{2} \bigl(1+o_{2}(1)\bigr). \end{aligned} \end{aligned}$$

If there exists a positive number $K\geq k_{\lambda }(\lambda _{1})$, such that (21) is still valid when replacing $k_{\lambda }(\lambda _{1})$ by K, then in particular, we have

$$\begin{aligned} \varepsilon \widetilde{I} =&\varepsilon \sum_{n=1}^{\infty } \sum_{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }} \widetilde{a}_{m}\widetilde{b}_{n} \\ >&\varepsilon K \Biggl[ \sum_{m=1}^{\infty } \bigl(1-\theta _{1}(\lambda _{2},m)\bigr)\frac{U_{m}^{p(1-\lambda _{1})-1}}{\mu _{m}^{p-1}}\widetilde{a}_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty } \frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}\widetilde{b}_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$

We obtain from the above results that

$$ \biggl[ \frac{\pi }{\lambda \sin (\pi \widetilde{\lambda }_{1}/\lambda )} \biggr] ^{2} \bigl(1+o_{2}(1)\bigr)>K \bigl( 1+o_{1}(1)-\varepsilon O(1) \bigr) ^{\frac{1}{p}} \bigl( 1+o_{2}(1) \bigr) ^{\frac{1}{q}}, $$

and then $k_{\lambda }(\lambda _{1})\geq K$ (for $\varepsilon \rightarrow 0^{+}$). Hence $K=k_{\lambda }(\lambda _{1})$ is the best value of (21).

We conform that the constant factor $k_{\lambda }(\lambda _{1})$ in (22) is the best possible. Otherwise we can get a contradiction by (23): that the constant factor in (21) is not the best value. □

Theorem 4

Suppose that $p<0$, $\{\mu _{m}\}_{m=1}^{\infty }$ and $\{\nu _{n}\}_{n=1}^{\infty }$ are decreasing positive sequences, and $U(\infty )=V(\infty )=\infty$, then we have the following equivalent inequalities:

$$\begin{aligned}& I =\sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}a_{m}b_{n}> \biggl[ \frac{\pi }{\lambda \sin (\pi \lambda _{1}/\lambda )} \biggr] ^{2}\Vert a\Vert _{p,\varphi }\Vert b\Vert _{q,\widetilde{\psi }}, \end{aligned}$$

(24)

$$\begin{aligned}& \begin{aligned}[b] J_{1} &= \Biggl\{ \sum_{n=1}^{\infty } \frac{(1-\theta _{2}(\lambda _{1},n))^{1-p}\nu _{n}}{V_{n}^{1-p\lambda _{2}}} \Biggl( \sum_{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}a_{m} \Biggr) ^{p} \Biggr\} ^{\frac{1}{p}} \\ &> \biggl[ \frac{\pi }{\lambda \sin (\pi \lambda _{1}/\lambda )} \biggr] ^{2}\Vert a\Vert _{p,\varphi }, \end{aligned} \end{aligned}$$

(25)

where the constant factor $[ \frac{\pi }{\lambda \sin (\pi \lambda _{1}/\lambda )} ] ^{2}$ is the best possible.

Proof

Using the same way of obtaining (14) and (15), by the reverse Hölder inequality with weight and (9), we have

$$ J_{1}>\bigl(k_{\lambda }(\lambda _{1}) \bigr)^{\frac{1}{q}} \Biggl[ \sum_{m=1}^{\infty } \omega (\lambda _{2},m)\frac{U_{m}^{p(1-\lambda _{1})-1}}{\mu _{m}^{p-1}}a_{m}^{p} \Biggr] ^{\frac{1}{p}}, $$

(26)

then we obtain (25) by (6). Using the reverse Hölder inequality, we have

$$\begin{aligned} I =&\sum_{n=1}^{\infty } \Biggl[ \frac{(1-\theta _{2}(\lambda _{1},n))^{-\frac{1}{q}}\nu _{n}^{1/p}}{V_{n}^{\frac{1}{p}-\lambda _{2}}}\sum_{m=1}^{\infty } \frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}a_{m} \Biggr] \\ &{}\times \biggl[ \bigl(1-\theta _{2}(\lambda _{1},n) \bigr)^{\frac{1}{q}}\frac{V_{n}^{\frac{1}{p}-\lambda _{2}}}{\nu _{n}^{1/p}}b_{n} \biggr] \\ \geq &J_{1}\Vert b\Vert _{q,\widetilde{\psi }}. \end{aligned}$$

(27)

Hence (24) is valid by (25). Assuming that (24) is valid, setting

$$ b_{n}=\frac{(1-\theta _{2}(\lambda _{1},n))^{1-p}\nu _{n}}{V_{n}^{1-p\lambda _{2}}} \Biggl[ \sum_{m=1}^{\infty } \frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}a_{m} \Biggr] ^{p-1},\quad n\in \mathbf{N}, $$

we find

$$ J_{1}= \Biggl[ \sum_{n=1}^{\infty } \bigl(1-\theta _{2}(\lambda _{1},n)\bigr) \frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}b_{n}^{q} \Biggr] ^{1/p}. $$

It follows that $J_{1}>0$ by (26). If $J_{1}=\infty$, then (25) is trivially valid. If $0< J_{1}<\infty$, then we find

$$\begin{aligned}& \sum_{n=1}^{\infty }\bigl(1-\theta _{2}(\lambda _{1},n)\bigr)\frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}b_{n}^{q}=J_{1}^{p}=I \\& \hphantom{\sum_{n=1}^{\infty }\bigl(1-\theta _{2}(\lambda _{1},n)\bigr)\frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}b_{n}^{q}}>k_{\lambda }(\lambda _{1})\Vert a\Vert _{p,\varphi } \Biggl[ \sum_{n=1}^{\infty } \bigl(1-\theta _{2}(\lambda _{1},n)\bigr) \frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \\& J_{1}= \Biggl[ \sum_{n=1}^{\infty } \bigl(1-\theta _{2}(\lambda _{1},n)\bigr) \frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}b_{n}^{q} \Biggr] ^{1/p}>k_{\lambda }( \lambda _{1})\Vert a\Vert _{p,\varphi }. \end{aligned}$$

Hence (25) is valid, which is equivalent to (24).

For $0<\varepsilon <q\lambda _{2}$, we set $\widetilde{\lambda }_{1}=\lambda _{1}+\frac{\varepsilon }{q}$ (>0), $\widetilde{\lambda }_{2}=\lambda _{2}-\frac{\varepsilon }{q}$ ($\in (0,1)$), $\widetilde{a}_{m}=U_{m}^{\widetilde{\lambda }_{1}-\varepsilon -1}\mu _{m}$, $\widetilde{b}_{n}=V_{n}^{\widetilde{\lambda }_{2}-1}\nu _{n}$. By (10), (11), and (6), in view of Remark 1, we have

$$\begin{aligned}& \sum_{m=1}^{\infty }\frac{U_{m}^{p(1-\lambda _{1})-1}}{\mu _{m}^{p-1}}\widetilde{a}_{m}^{p}=\sum_{m=1}^{\infty } \frac{\mu _{m}}{U_{m}^{1+\varepsilon }}=\frac{1}{\varepsilon }\bigl(1+o_{1}(1)\bigr), \\& \sum_{n=1}^{\infty }\bigl(1-\theta _{2}(\lambda _{1},n)\bigr)\frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}} \widetilde{b}_{n}^{q}\\& \quad =\sum _{n=1}^{\infty }\biggl(1-O\biggl(\frac{1}{V_{n}^{\lambda _{1}/2}}\biggr) \biggr)\frac{\nu _{n}}{V_{n}^{1+\varepsilon }} \\& \quad =\frac{1}{\varepsilon }\bigl(1+o_{2}(1)-\varepsilon O(1)\bigr) , \\& \begin{aligned} \widetilde{I} &= \sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }} \widetilde{a}_{m}\widetilde{b}_{n} \\ &= \sum_{m=1}^{\infty } \Biggl[ \sum _{n=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}\frac{U_{m}^{\widetilde{\lambda }_{1}}\nu _{n}}{V_{n}^{1-\widetilde{\lambda }_{2}}} \Biggr] \frac{\mu _{m}}{U_{m}^{1+\varepsilon }} \\ &= \sum_{m=1}^{\infty }\varpi (\widetilde{\lambda }_{2},m)\frac{\mu _{m}}{U_{m}^{1+\varepsilon }}< k_{\lambda }(\widetilde{\lambda }_{1})\sum_{n=1}^{\infty } \frac{\mu _{m}}{U_{m}^{1+\varepsilon }} \\ &= \frac{1}{\varepsilon } \biggl[ \frac{\pi }{\lambda \sin (\pi \widetilde{\lambda }_{1}/\lambda )} \biggr] ^{2} \bigl(1+o_{1}(1)\bigr). \end{aligned} \end{aligned}$$

If there exists a positive number $K\geq k_{\lambda }(\lambda _{1})$, such that (24) is still valid as we replace $k_{\lambda }(\lambda _{1})$ by K, then, in particular, we have

$$\begin{aligned} \varepsilon \widetilde{I} =&\varepsilon \sum_{n=1}^{\infty } \sum_{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }} \widetilde{a}_{m}\widetilde{b}_{n} \\ >&\varepsilon K \Biggl[ \sum_{m=1}^{\infty } \frac{U_{m}^{p(1-\lambda _{1})-1}}{\mu _{m}^{p-1}}\widetilde{a}_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty } \bigl(1-\theta _{2}(\lambda _{1},n)\bigr)\frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}} \widetilde{b}_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$

From the above results, we have

$$ \biggl[ \frac{\pi }{\lambda \sin (\pi \widetilde{\lambda }_{1}/\lambda )} \biggr] ^{2} \bigl(1+o_{1}(1)\bigr)>K \bigl( 1+o_{1}(1) \bigr) ^{\frac{1}{p}} \bigl( 1+o_{2}(1)-\varepsilon O(1) \bigr) ^{\frac{1}{q}}. $$

It follows that $k_{\lambda }(\lambda _{1})\geq K$ (for $\varepsilon \rightarrow 0^{+}$). Hence $K=k_{\lambda }(\lambda _{1})$ is the best value of (24). We conform that the constant factor $k_{\lambda }(\lambda _{1})$ in (25) is the best possible. Otherwise we can get a contradiction by (27): that the constant factor in (24) is not the best value. □

Remark 2

For $\mu _{i}=\nu _{i}=1$ ($i=1,2,\ldots$), (12) reduces to (3); for $\lambda =1$, $\lambda _{1}=\frac{1}{q}$, $\lambda _{2}=\frac{1}{p}$, it follows by (12) that

$$ \sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}-V_{n}}a_{m}b_{n}< \biggl[ \frac{\pi }{\sin (\pi /p)} \biggr] ^{2} \Biggl[ \sum _{m=1}^{\infty }\frac{1}{\mu _{m}^{p-1}}a_{m}^{p} \Biggr] ^{1/p} \Biggl[ \sum_{n=1}^{\infty } \frac{1}{\nu _{n}^{q-1}}b_{n}^{q} \Biggr] ^{1/q}; $$

(28)

for $\lambda =1$, $\lambda _{1}=\frac{1}{p}$, $\lambda _{2}=\frac{1}{q}$, (12) reduces to the dual form of (28) as follows:

$$ \sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{{}}-V_{n}^{{}}}a_{m}b_{n}< \biggl[ \frac{\pi }{\sin (\pi /p)} \biggr] ^{2} \Biggl[ \sum _{m=1}^{\infty }\frac{U_{m}^{p-2}}{\mu _{m}^{p-1}}a_{m}^{p} \Biggr] ^{1/p} \Biggl[ \sum_{n=1}^{\infty } \frac{V_{n}^{q-2}}{\nu _{n}^{q-1}}b_{n}^{q} \Biggr] ^{1/q}. $$

(29)