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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

• Received: 29 July 2015
• 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. ):
$$\sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{a_{m}b_{n}}{m+n}< \frac{\pi }{\sin (\pi /p)} \Biggl( \sum_{n=1}^{\infty }a_{n}^{p} \Biggr) ^{1/p} \Biggl( \sum_{n=1}^{\infty }b_{n}^{q} \Biggr) ^{1/q},$$
(1)
where the constant factor $$\frac{\pi }{\sin (\pi /p)}$$ is the best possible. We also have the following Hardy-Hilbert-type inequality (cf. ):
$$\sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{\ln (m/n)a_{m}b_{n}}{m-n}< \biggl[ \frac{\pi }{\sin (\pi /p)} \biggr] ^{2} \Biggl( \sum _{n=1}^{\infty }a_{n}^{p} \Biggr) ^{1/p} \Biggl( \sum_{n=1}^{\infty }b_{n}^{q} \Biggr) ^{1/q},$$
(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. ): 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) (). 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. , 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. )
$$\int_{0}^{\infty }\frac{\ln u}{u-1}u^{a-1}\,du= \biggl[ \frac{\pi }{\sin (a\pi )} \biggr] ^{2}\quad (0< a< 1),$$
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. ), 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. ) 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. ), 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).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

## Authors’ Affiliations

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

## References

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