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# A new extension of a Hardy-Hilbert-type inequality

Journal of Inequalities and Applications20152015:397

https://doi.org/10.1186/s13660-015-0918-7

• Accepted: 29 November 2015
• Published:

## 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.

## Keywords

• Hardy-Hilbert-type inequality
• parameter
• weight coefficient
• equivalent form
• reverse

• 26D15
• 47A07

## 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) ([311]). 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)

## Declarations

### Acknowledgements

This work is supported by the 2013 Knowledge Construction Special Foundation Item of Guangdong Institution of Higher Learning College and University (No. 2013KJCX0140).

## Authors’ Affiliations

(1)
Department of Mathematics, Guangdong University of Education, Guangzhou, Guangdong, 510303, P.R. China

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