一个涉及多重可变上限函数的半离散Hardy-Mulholland型不等式

吴善和; 黄先勇; 杨必成

doi:10.6054/j.jscnun.2022014

一个涉及多重可变上限函数的半离散Hardy-Mulholland型不等式

1.
龙岩学院数学系，龙岩 364012
2.
广东第二师范学院数学系, 广州 510303

基金项目:

国家自然科学基金项目 12071491

福建省自然科学基金项目 2020J01365

2018年广东省本科高校教学质量与教学改革工程建设项目(金融数学特色专业)

详细信息

通讯作者:
吴善和, Email: shanhewu@163.com

中图分类号: O178
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出版历程
- 收稿日期: 2021-07-30
- 网络出版日期: 2022-03-13
- 刊出日期: 2022-02-24

A Half-discrete Hardy-Mulholland-type Inequality Involving One Multiple Upper Limit Function

1.
Department of Mathematics, Longyan University, Longyan 364012, China
2.
Department of Mathematics, Guangdong University of Education, Guangzhou 510303, China

摘要

摘要: 通过引入适当的核函数构造权函数，运用实分析技巧研究一类半离散Hardy-Mulholland型不等式：首先，建立一个涉及多参数和多重可变上限函数的半离散Hardy-Mulholland型不等式；然后，探讨该不等式的常数因子为最佳值时各参数之间的相关性及等价陈述，建立其等价不等式，刻画一类具有最佳常数因子的Hardy-Mulholland型不等式的结构特征。
- 权函数 /
- 半离散Hardy-Mulholland型不等式 /
- 多重可变上限函数 /
- 参数 /
- 最佳常数因子
Abstract: By constructing the kernel function and weight function and using the real analytical techniques, the half-discrete inequalities of Hardy-Mulholland type are investigated. Firstly, a half-discrete Hardy-Mulholland-type inequality containing several parameters and one multiple upper limit function is established. And then, the equivalent statements on the best possible constant factor associated with several parameters are discussed with the aid of the proposed half-discrete Hardy-Mulholland-type inequality, and several inequalities are derived via their equivalent form. The results obtained can be used to depict the structure character of Hardy-Mulholland-type inequalities for which the constant factors are best possible.
- weight function /
- half-discrete Hardy-Mulholland type inequality /
- multiple upper limit function /
- parameter /
- best possible constant factor

HTML全文

设 $p>1, \frac{1}{p}+\frac{1}{q}=1, a_{m}, b_{n} \geqslant 0,$ $0<\sum\nolimits_{m=1}^{\infty} a_{m}^{p}<\infty, 0<\sum\nolimits_{n=1}^{\infty} b_{n}^{q}<\infty$ , 则有如下具有最佳常数因子 $\frac{{\rm{ \mathsf{ π} }}}{\sin ({\rm{ \mathsf{ π} }} / p)}$ 的Hardy-Hilbert不等式^[1]:

$\sum\limits_{m=1}^{\infty} \sum\limits_{n=1}^{\infty} \frac{a_{m} b_{n}}{m+n}<\frac{{\rm{ \mathsf{ π} }}}{\sin ({\rm{ \mathsf{ π} }} / p)}\left(\sum\limits_{m=1}^{\infty} a_{m}^{p}\right)^{\frac{1}{p}}\left(\sum\limits_{n=1}^{\infty} b_{n}^{q}\right)^{\frac{1}{q}} 。$

(1)

有趣的是，在类似于式(1)的条件下，有如下具有相同最佳常数因子的Hardy-Mulholland不等式^[1]:

$\sum\limits_{m=2}^{\infty} \sum\limits_{n=2}^{\infty} \frac{a_{m} b_{n}}{\ln m n}<\frac{{\rm{ \mathsf{ π} }}}{\sin ({\rm{ \mathsf{ π} }} / p)}\left(\sum\limits_{m=2}^{\infty} \frac{1}{m^{1-p}} a_{m}^{p}\right)^{\frac{1}{p}}\left(\sum\limits_{n=2}^{\infty} \frac{1}{n^{1-q}} b_{n}^{q}\right)^{\frac{1}{q}} 。$

(2)

2006年, 应用Euler-Maclaurin求和公式, KRNIC和PECARIC^[2]建立了式(1)的如下推广式:

$\begin{aligned} &\sum\limits_{m=1}^{\infty} \sum\limits_{n=1}^{\infty} \frac{a_{m} b_{n}}{(m+n)^{\lambda}}< \\ &\ \ B\left(\lambda_{1}, \lambda_{2}\right)\left[\sum\limits_{m=1}^{\infty} m^{p\left(1-\lambda_{1}\right)-1} a_{m}^{p}\right]^{\frac{1}{p}}\left[\sum\limits_{n=1}^{\infty} n^{q\left(1-\lambda_{2}\right)-1} b_{n}^{q}\right]^{\frac{1}{q}}, \end{aligned}$

(3)

其中, λ_i∈(0, 2](i=1, 2), λ₁+λ₂=λ∈(0, 4], 常数因子B(λ₁, λ₂)是最佳值, 其中 $B(u, v)=\int_{0}^{\infty} \frac{t^{u-1} \mathrm{~d} t}{(1+t)^{u+v}}$ (u, v > 0)为Beta函数。当p=q=2, λ₁=λ₂=λ/2, 由式(3)可导出杨必成^[3]关于Hilbert不等式推广的一个结果。2019年, 应用式(2)及Abel求和公式, ADIYASUREN等^[4]给出了核为1/(m+n)^λ且级数的项中涉及部分和的Hardy-Hilbert型不等式。众所周知，式(1)和式(2)及其积分形式在分析学及相关领域有重要的应用^[5-12]。

2020年，黄启亮等^[13]给出一个一般齐次核Hardy-Mulholland型不等式, 推广了Hardy-Mulholland不等式, 即:

$\begin{gathered} \sum\limits_{m=2}^{\infty} \sum\limits_{n=2}^{\infty} k_{\lambda}\left(\ln U_{m}, \ln V_{n}\right) a_{m} b_{n}<k_{\lambda}^{1 / p}\left(\lambda-\lambda_{2}\right) k_{\lambda}^{1 / p}\left(\lambda_{1}\right) \times \\ {\left[\sum\limits_{m=2}^{\infty}\left(\frac{U_{m}}{\mu_{m}}\right)^{p-1}\left(\ln U_{m}\right)^{p\left[1-\left(\left(\lambda-\lambda_{2}\right) / p+\lambda_{1} / q\right)\right]-1} a_{m}^{p}\right]^{\frac{1}{p}} \times} \\ {\left[\sum\limits_{n=2}^{\infty}\left(\frac{V_{n}}{\nu_{n}}\right)^{q-1}\left(\ln V_{n}\right)^{q\left[1-\left(\left(\lambda-\lambda_{1}\right) / q+\lambda_{2} / p\right)\right]-1} b_{n}^{q}\right]^{\frac{1}{q}}}, \end{gathered}$

(4)

其中, $U_{m}=\sum_{i=1}^{m} \mu_{i}, V_{n}=\sum_{j=1}^{m} \nu_{j}, k_{\lambda}(x, y)$ 是-λ齐次函数。

作为对离散型不等式和积分型不等式的拓展研究，1934年，HARDY等^[1]给出了如下半离散Hardy-Hilbert型不等式:

设K(t)(t > 0)为递减函数, p > 1, $\frac{1}{p}+\frac{1}{q}=1$ , 0 < $\varphi(s)=\int_{0}^{\infty} K(t) t^{s-1} \mathrm{~d} t<\infty, a_{n} \geqslant 0$ , 使得 $0<\sum\nolimits_{n=1}^{\infty} a_{n}^{p}<\infty$ , 则有

$\int_{0}^{\infty} x^{p-2}\left(\sum\limits_{n=1}^{\infty} K(n x) a_{n}\right)^{p} \mathrm{~d} x<\varphi^{p}\left(\frac{1}{q}\right) \sum\limits_{n=1}^{\infty} a_{n}^{p} 。$

(5)

有关式(5)的一些推广及应用可见文献[14-18]。

2016年, 洪勇和温雅敏^[19]给出式(1)的推广式中最佳常数因子联系参数的一个等价条件。该结果引起许多学者的兴趣，相关的一些研究成果可见文献[20-26]。2019年, YANG等^[27]建立了逆向的半离散Hardy-Hilbert型不等式。2021年, HUANG等^[28]利用积分上限函数和级数的部分和给出了一些新的半离散Hardy-Hilbert型不等式。

黄启亮等^[13]通过引入多个参数和齐次核函数推广Hardy-Mulholland不等式，建立了形如式(4)的离散型Hardy-Mulholland不等式。本文从研究半离散不等式的视角，通过构造适当的核函数和权函数建立含无穷积分和级数混合形式的Hardy-Mulholland型不等式：首先，通过引入核函数1/(ln ne^x)^λ和权函数来建立一个涉及参数和多重可变上限函数的半离散Hardy-Mulholland型不等式；然后，探讨该不等式的常数因子为最佳值时各参数之间的相关性及等价陈述，建立其等价不等式，刻画一类具有最佳常数因子的Hardy-Mulholland型不等式的结构特征。

1. 若干引理

下文中，约定如下记号和假定条件: p > 1, $\frac{1}{p}+\frac{1}{q}=1$ , λ > 0, λ₁∈(0, λ), λ₂∈(0, 1]∩(0, λ), $\hat{\lambda}_{1}:=\frac{\lambda-\lambda_{2}}{p}+\frac{\lambda_{1}}{q}, \hat{\lambda}_{2}:=\frac{\lambda-\lambda_{1}}{q}+\frac{\lambda_{2}}{p}$ , 函数f(x): =F₀(x)在任意区间(0, b)(b > 0)上非负且Lebesgue可积, 递推地可定义如下非负多重可变上限函数 $F_{i}(x):=\int_{0}^{x} F_{i-1}(t) \mathrm{d} t$ (x≥0), 满足条件F_i(x)=o(e^tx) (t > 0, i=1, 2, …, m, x→∞), 即 $\lim \limits_{x \rightarrow \infty} \frac{F_{i}(x)}{\mathrm{e}^{t x}}=0$ 。对于a_n≥0, 假定

$0<\int_{0}^{\infty} x^{-p\left(m-1+\hat{\lambda}_{1}\right)-1} F_{m}^{p}(x)<\infty,$

$0<\sum\limits_{n=2}^{\infty}(\ln n)^{q\left(1-\hat{\lambda}_{2}\right)-1} n^{q-1} a_{n}^{q}<\infty 。$

引理1 设s∈(0, ∞), s₂∈(0, 1]∩(0, s), k_s(s₂): =B(s₂, s-s₂)。定义如下权函数:

$\tilde{\omega}\left(s_{2}, x\right):=x^{s-s_{2}} \sum\limits_{n=2}^{\infty} \frac{(\ln n)^{s_{2}-1}}{n\left(\ln n \mathrm{e}^{x}\right)^{s}} \quad\left(x \in \mathbb{R}_{+}=(0, \infty)\right),$

(6)

则有如下不等式:

$\tilde{\omega}\left(s_{2}, x\right)<k_{s}\left(s_{2}\right)。$

(7)

证明由Beta函数的定义域和性质, 知k_s(s₂)=B(s₂, s-s₂)收敛于正数。故由式(7)知, 式(6)的级数也收敛于正数。固定x > 0, 定义非负函数g_x(t)如下: $g_{x}(t):=\frac{(\ln t)^{s_{2}-1}}{t\left(\ln \mathrm{e}^{x} t\right)^{s}}(t>1)$ 。因为s∈(0, ∞), s₂∈(0, 1]∩(0, s), 所以, 函数g_x(t)在t∈(1, ∞)严格递减。由级数的递减性质, 有：

$\sum\limits_{n=2}^{\infty} g_{x}(n)<\int_{1}^{\infty} g_{x}(t) \mathrm{d} t。$

(8)

作变换 $v=\frac{\ln t}{x}\left(\frac{1}{t} \mathrm{~d} t=x \mathrm{d} v\right)$ , 得到

$\begin{gathered} \int_{1}^{\infty} g_{x}(t) \mathrm{d} t=\int_{1}^{\infty} \frac{(\ln t)^{s_{2}-1}}{(x+\ln t)^{s} t} \mathrm{d} t=\frac{1}{x^{s}} \int_{0}^{\infty} \frac{(x v)^{s_{2}-1}}{(1+v)^{s}} x \mathrm{d} v= \\ \frac{1}{x^{s-s_{2}}} \int_{0}^{\infty} \frac{v^{s_{2}-1}}{(1+v)^{s}} \mathrm{d} v=\frac{1}{x^{s-s_{2}}} k_{s}\left(s_{2}\right) 。 \end{gathered}$

再由式(8), 有

$\tilde{\omega}\left(s_{2}, x\right)=x^{s-s_{2}} \sum\limits_{n=2}^{\infty} g_{x}(n)<x^{s-s_{2}} \int_{1}^{\infty} g_{x}(t) \mathrm{d} t=k_{s}\left(s_{2}\right) 。$

故式(7)成立。证毕。

引理2 在引理1的条件下, 对于s₁∈(0, s), 若

$0<\int_{0}^{\infty} x^{p\left[1-\left(\left(s-s_{2}\right) / p+s_{1} / q\right)\right]-1} f^{p}(x) \mathrm{d} x<\infty,$

$0<\sum\limits_{n=2}^{\infty}(\ln n)^{q\left[1-\left(s_{2} / p+\left(s-s_{1}\right) / q\right)\right]-1} n^{q-1} a_{n}^{q}<\infty,$

则有如下推广的半离散Hardy-Mulholland不等式:

$\begin{aligned} I= \int_{0}^{\infty}& \sum\limits_{n=2}^{\infty} \frac{a_{n}}{\left(\ln n \mathrm{e}^{x}\right)^{s}} f(x) \mathrm{d} x<\left(k_{s}\left(s_{2}\right)\right)^{1 / p}\left(k_{s}\left(s_{1}\right)\right)^{1 / q} \times \\ &\left\{\int_{0}^{\infty} x^{p\left[1-\left(\left(s-s_{2}\right) / p+s_{1} / q\right)\right]-1} f^{p}(x) \mathrm{d} x\right\}^{\frac{1}{p}} \times \\ &\left\{\sum\limits_{n=2}^{\infty}(\ln n)^{q\left[1-\left(s_{2} / p+\left(s-s_{1}\right) / q\right)\right]-1} n^{q-1} a_{n}^{q}\right\}^{\frac{1}{q}}。 \end{aligned}$

(9)

证明作变换v=x/ln n, 可得到另一权函数表示式:

$\begin{aligned} \omega&\left(s_{1}, n\right):=(\ln n)^{s-s_{1}} \int_{0}^{\infty} \frac{x^{s_{1}-1}}{\left(\ln n \mathrm{e}^{x}\right)^{s}} \mathrm{~d} x=\int_{0}^{\infty} \frac{v^{s_{1}-1}}{(v+1)^{s}} \mathrm{~d} v= \\ &\ \ \ \ k_{s}\left(s_{1}\right)\left(n \in \mathbb{N}_{+} \backslash\{1\}\right) 。 \end{aligned}$

(10)

利用Hölder不等式^[29], 有

$\begin{aligned} I= \int_{0}^{\infty}& \sum\limits_{n=2}^{\infty} \frac{1}{\left(\ln n \mathrm{e}^{x}\right)^{s}}\left[\frac{x^{\left(1-s_{1}\right) / q} n^{-1 / p}}{(\ln n)^{\left(1-s_{2}\right) / p}} f(x)\right]\left[\frac{(\ln n)^{\left(1-s_{2}\right) / p}}{x^{\left(1-s_{1}\right) / q} n^{-1 / p}} a_{n}\right] \mathrm{d} x \leqslant \\ & {\left[\int_{0}^{\infty} \sum\limits_{n=2}^{\infty} \frac{1}{\left(\ln n \mathrm{e}^{x}\right)^{s}} \frac{x^{\left(1-s_{1}\right)(p-1)}}{n(\ln n)^{1-s_{2}}} f^{p}(x) \mathrm{d} x\right]^{\frac{1}{p}} \times } \\ & {\left[\sum\limits_{n=2}^{\infty} \int_{0}^{\infty} \frac{1}{\left(\ln n \mathrm{e}^{x}\right)^{s}} \frac{(\ln n)^{\left(1-s_{2}\right)(q-1)} a_{n}^{q}}{x^{1-s_{1}} n^{1-q}} \mathrm{~d} x\right]^{\frac{1}{q}}=} \\ &\left\{\int_{0}^{\infty} \tilde{\omega}\left(s_{2}, x\right) x^{p\left[1-\left(\left(s-s_{2}\right) / p+s_{1} / q\right)\right]-1} f^{p}(x) \mathrm{d} x\right\}^{\frac{1}{p}} \times \\ &\left\{\sum\limits_{n=2}^{\infty} \omega\left(s_{1}, n\right)(\ln n)^{q\left[1-\left(\left(s-s_{1}\right) / q+s_{2} / p\right)\right]-1} n^{q-1} a_{n}^{q}\right\}^{\frac{1}{q}} 。 \end{aligned}$

(11)

下证式(11)取严格不等号。若式(11)取等号, 则由文献[29]知：有不全为零的常数A和B, 使得

$A \frac{x^{\left(1-s_{1}\right)(p-1)}}{n(\ln n)^{1-s_{2}}} f^{p}(x)=B \frac{(\ln n)^{\left(1-s_{2}\right)(q-1)} a_{n}^{q}}{x^{1-s_{1}} n^{1-q}}$

$\text { a.e. 于 } \mathbb{R}_{+} \times \mathbb{N}_{+} \backslash\{1\} \text { 。 }$

不妨设A≠0，则对于 $n \in \mathbb{N}_{+} \backslash\{1\}$ , 有

$\begin{aligned} &x^{p\left[1-\left(\left(s-s_{2}\right) / p+s_{1} / q\right)\right]-1} f^{p}(x)=\\ &\ \ \ \ \ \ \ \ \frac{B}{A}(\ln n)^{q\left(1-s_{2}\right)} n^{q} a_{n}^{q} \frac{1}{x^{1+\left(s-s_{1}-s_{2}\right)}} \quad \text { a.e. 于 } \mathbb{R}_{+\circ} \end{aligned}$

(12)

对式(12)两边积分, 由引理2中条件 $0<\int_{0}^{\infty} x^{p\left[1-\left(\left(s-s_{2}\right) / p+s_{1} / q\right)\right]-1} f^{p}(x) \mathrm{d} x<\infty$ , 可得式(12)左边积分有限，而右边积分 $\int_{0}^{\infty} \frac{1}{x^{1+\left(s-s_{1}-s_{2}\right)}} \mathrm{d} x=\infty$ ，从而，导出矛盾。故式(11)不取等号。再由式(7)及式(10), 可得到式(9)。证毕。

注1 在式(9)中, 置换f(x)为F_m(x), 并设s=λ+m∈(m, ∞), s₁=λ₁+m∈(m, s), s₂=λ₂∈(0, 1]∩(0, λ), 则有s∈(0, ∞), s₁∈(0, s), s₂∈(0, 1]∩(0, s)。于是, 得到如下含新参数的半离散Hardy-Mulholland不等式:

$\begin{aligned} \int_{0}^{\infty} \sum\limits_{n=2}^{\infty}& \frac{a_{n}}{(x+\ln n)^{\lambda+m}} F_{m}(x) \mathrm{d} x<\left(k_{\lambda+m}\left(\lambda_{2}\right)\right)^{1 / p} \times \\ &\left(k_{\lambda+m}\left(\lambda_{1}+m\right)\right)^{1 / q}\left[\int_{0}^{\infty} x^{-p\left(m-1+\hat{\lambda}_{1}\right)-1} F_{m}^{p}(x) \mathrm{d} x\right]^{1 / p} \times \\ &{\left[\sum\limits_{n=2}^{\infty}(\ln n)^{q\left(1-\hat{\lambda}_{2}\right)-1} n^{q-1} a_{n}^{q}\right]^{1 / q} }。 \end{aligned}$

(13)

引理3 对于t > 0, 有如下表示式:

$\int_{0}^{\infty} \mathrm{e}^{-t x} f(x) \mathrm{d} x=t^{m} \int_{0}^{\infty} \mathrm{e}^{-t x} F_{m}(x) \mathrm{d} x。$

(14)

证明对于m=0, 因F₀(x)=f(x), 式(14)自然成立; 对于 $m \in \mathbb{N}_{+}$ , 由分部积分法及条件F_i(0)=0, F_i(x)=o(e^tx)(t > 0;i=1, 2, …, m; x→∞), 可得

$\begin{aligned} \int_{0}^{\infty} &\mathrm{e}^{-t x} F_{i-1}(x) \mathrm{d} x=\int_{0}^{\infty} \mathrm{e}^{-t x} \mathrm{~d} F_{i}(x)=\left.\mathrm{e}^{-t x} F_{i}(x)\right|_{0} ^{\infty}- \\ &\int_{0}^{\infty} F_{i}(x) \mathrm{de}^{-t x}=\lim \limits_{x \rightarrow \infty} \mathrm{e}^{-t x} F_{i}(x)+t \int_{0}^{\infty} \mathrm{e}^{-t x} F_{i}(x) \mathrm{d} x= \\ &t \int_{0}^{\infty} \mathrm{e}^{-t x} F_{i}(x) \mathrm{d} x。 \end{aligned}$

依次代入i=1, 2, …, m, 即得式(14)。证毕。

2. 主要结果

定理1 设p > 1, $\frac{1}{p}+\frac{1}{q}=1$ , λ > 0, λ₁∈(0, λ), λ₂∈(0, 1]∩(0, λ), $\hat{\lambda}_{1}:=\frac{\lambda-\lambda_{2}}{p}+\frac{\lambda_{1}}{q}, \hat{\lambda}_{2}:=\frac{\lambda-\lambda_{1}}{q}+\frac{\lambda_{2}}{p}$ , 0 < $\int_{0}^{\infty} x^{-p\left(m-1+\hat{\lambda}_{1}\right)-1} F_{m}^{p}(x)<\infty,$ $0<\sum_{n=2}^{\infty}(\ln n)^{q\left(1-\hat{\lambda}_{2}\right)-1} n^{q-1} a_{n}^{q}<\infty$ , 则有如下涉及多重可变上限函数的半离散Hardy-Mulholland不等式:

$\begin{aligned} I:=& \int_{0}^{\infty} \sum\limits_{n=2}^{\infty} \frac{a_{n}}{\left(\ln n \mathrm{e}^{x}\right)^{\lambda}} f(x) \mathrm{d} x<\\ & \frac{\Gamma(\lambda+m)}{\Gamma(\lambda)}\left(k_{\lambda+m}\left(\lambda_{2}\right)\right)^{1 / p}\left(k_{\lambda+m}\left(\lambda_{1}+m\right)\right)^{1 / q} \times \\ & {\left[\int_{0}^{\infty} x^{-p\left(m-1+\hat{\lambda}_{1}\right)-1} F_{m}^{p}(x) \mathrm{d} x\right]^{\frac{1}{p}} \times } \\ & {\left[\sum\limits_{n=2}^{\infty}(\ln n)^{q\left(1-\hat{\lambda}_{2}\right)-1} n^{q-1} a_{n}^{q}\right]^{\frac{1}{q}} }。 \end{aligned}$

(15)

特别地, 若λ₁+λ₂=λ(λ₁∈(0, λ), λ₂∈(0, 1]∩(0, λ)), 则有

$0<\int_{0}^{\infty} x^{-p\left(m-1+\lambda_{1}\right)-1} F_{m}^{p}(x) \mathrm{d} x<\infty ,$

$0<\sum\limits_{n=2}^{\infty}(\ln n)^{q\left(1-\lambda_{2}\right)-1} n^{q-1} a_{n}^{q}<\infty,$

及

$\begin{gathered} I=\int_{0}^{\infty} \sum\limits_{n=2}^{\infty} \frac{a_{n}}{\left(\ln n \mathrm{e}^{x}\right)^{\lambda}} f(x) \mathrm{d} x<\frac{\Gamma(\lambda+m)}{\Gamma(\lambda)} k_{\lambda+m}\left(\lambda_{1}+m\right) \times \\ {\left[\int_{0}^{\infty} x^{-p\left(m-1+\lambda_{1}\right)-1} F_{m}^{p}(x) \mathrm{d} x\right]^{\frac{1}{p}}\left[\sum\limits_{n=2}^{\infty}(\ln n)^{q\left(1-\lambda_{2}\right)-1} n^{q-1} a_{n}^{q}\right]^{\frac{1}{q}}}。 \end{gathered}$

(16)

证明运用换元积分及Gamma函数表示, 得

$\frac{1}{\left(\ln n \mathrm{e}^{x}\right)^{\lambda}}=\frac{1}{(x+\ln n)^{\lambda}}=\frac{1}{\Gamma(\lambda)} \int_{0}^{\infty} t^{\lambda-1} \mathrm{e}^{-(x+\ln n) t} \mathrm{~d} t。$

由Lebesgue逐项积分定理^[30]及式(11), 得到

$\begin{aligned} I=& \frac{1}{\Gamma(\lambda)} \int_{0}^{\infty} \sum\limits_{n=2}^{\infty} a_{n} f(x) \int_{0}^{\infty} t^{\lambda-1} \mathrm{e}^{-(x+\ln n) t} \mathrm{~d} t \mathrm{d} x=\\ & \frac{1}{\Gamma(\lambda)} \int_{0}^{\infty} t^{\lambda-1}\left(\int_{0}^{\infty} \mathrm{e}^{-x t} f(x) \mathrm{d} x\right) \sum\limits_{n=2}^{\infty} \mathrm{e}^{-t \ln n} a_{n} \mathrm{d} t=\\ & \frac{1}{\Gamma(\lambda)} \int_{0}^{\infty} t^{\lambda-1}\left(t^{m} \int_{0}^{\infty} \mathrm{e}^{-x t} F_{m}(x) \mathrm{d} x\right) \sum\limits_{n=2}^{\infty} \mathrm{e}^{-t \ln n} a_{n} \mathrm{d} t=\\ & \frac{1}{\Gamma(\lambda)} \int_{0}^{\infty} \sum\limits_{n=2}^{\infty} a_{n} F_{m}(x) \int_{0}^{\infty} t^{(\lambda+m)-1} \mathrm{e}^{-(x+\ln n) t} \mathrm{~d} t \mathrm{d} x=\\ & \frac{\Gamma(\lambda+m)}{\Gamma(\lambda)} \int_{0}^{\infty} \sum\limits_{n=2}^{\infty} \frac{a_n}{(x+\ln n)^{\lambda+m}} F_{m}(x) \mathrm{d} x 。 \end{aligned}$

再由式(13), 可得到式(15)。证毕。

定理2 若λ₁+λ₂=λ, 则式(15)的常数因子必为最佳值。

证明若λ₁+λ₂=λ, 则式(15)变为式(16)。任给0 < ε < pλ₁, 设

$\tilde{F}_{0}(x)=\tilde{f}(x):=\left\{\begin{array}{l} 0 \quad(0<x<1), \\ x^{\lambda_{1}-\varepsilon / p-1} \quad(x \geqslant 1) ; \end{array}\right.$

$\widetilde{a}_{n}:=\frac{1}{n}(\ln n)^{\lambda_{2}-\varepsilon / p-1} \quad\left(n \in \mathbb{N}_{+} \backslash\{1\}\right) 。$

对于i∈{1, 2, …, m}, 设

$\begin{aligned} \tilde{F}_{i}(x)&:=\int_{0}^{x} \tilde{F}_{i-1}(t) \mathrm{d} t \leqslant \\ &\left\{\begin{array}{l} 0 \quad(0<x<1), \\ \frac{1}{\prod\nolimits_{j=0}^{i-1}\left(\lambda_{1}+j-\varepsilon / p\right)} x^{\lambda_{1}+i-1-\varepsilon / p} \quad(x \geqslant 1), \end{array}\right. \end{aligned}$

满足 $\tilde{F}_{i}(x)=o\left(\mathrm{e}^{t x}\right)(t>0, i=1, 2, \cdots, m ; x \rightarrow \infty)$ 。

若存在 $M\left(0<M \leqslant \frac{\Gamma(\lambda+m)}{\Gamma(\lambda)} k_{\lambda+m}\left(\lambda_{1}+m\right)\right)$ , 使得M取代式(16)的常数因子 $\frac{\Gamma(\lambda+m)}{\Gamma(\lambda)} k_{\lambda+m}\left(\lambda_{1}+m\right)$ 后, 式(16)仍成立, 则有

$\begin{aligned} \tilde{I}:=& \int_{0}^{\infty} \sum\limits_{n=2}^{\infty} \frac{\tilde{a}_{n} \tilde{f}(x)}{(x+\ln n)^{\lambda}} \mathrm{d} x<\\ & M\left[\int_{0}^{\infty} x^{-p\left(m-1+\lambda_{1}\right)-1} \widetilde{F}_{m}^{p}(x) \mathrm{d} x\right]^{\frac{1}{p}} \times \\ & {\left[\sum\limits_{n=2}^{\infty}(\ln n)^{q\left(1-\lambda_{2}\right)-1} n^{q-1} \tilde{a}_{n}^{q}\right]^{\frac{1}{q}} }。 \end{aligned}$

(17)

对于m=0及ε≥0, 以下不妨定义\$\prod\nolimits_{i=0}^{m-1}\left(\lambda_{1}+i-\frac{\varepsilon}{p}\right)=1\$。由式(17)及级数的递减性质, 可得

$\begin{aligned} \tilde{I}<&\frac{M}{\prod\nolimits_{i=0}^{m-1}\left(\lambda_{1}+i-\varepsilon / p\right)}\left[\int_{1}^{\infty} x^{-p\left(m-1+\lambda_{1}\right)-1} x^{p\left(\lambda_{1}+m-1\right)-\varepsilon} \mathrm{d} x\right]^{\frac{1}{p}} \times\\ &\left[\sum\limits_{n=2}^{\infty}(\ln n)^{q\left(1-\lambda_{2}\right)-1} n^{q-1}(\ln n)^{q \lambda_{2}-\varepsilon-q} n^{-q}\right]^{\frac{1}{q}}=\\ &\frac{M}{\prod\nolimits_{i=0}^{m-1}\left(\lambda_{1}+i-\varepsilon / p\right)}\left(\int_{1}^{\infty} x^{-\varepsilon^{-1}} \mathrm{~d} x\right)^{\frac{1}{p}}\left[\frac{1}{2}(\ln 2)^{-\varepsilon{-1}}+\right.\\ &\left.\sum\limits_{n=3}^{\infty}(\ln n)^{-\varepsilon-1} n^{-1}\right]^{\frac{1}{q}} \leqslant \frac{M}{\prod\nolimits_{i=0}^{m-1}\left(\lambda_{1}+i-\varepsilon / p\right)} \times\\ &\left(\int_{1}^{\infty} x^{-\varepsilon-1} \mathrm{~d} x\right)^{\frac{1}{p}}\left[\frac{1}{2}(\ln 2)^{-\varepsilon-1}+\int_{2}^{\infty}(\ln t)^{-\varepsilon-1} t^{-1} \mathrm{~d} t\right]^{\frac{1}{q}}= \\ &\frac{M}{\varepsilon \prod\nolimits_{i=0}^{m-1}\left(\lambda_{1}+i-\varepsilon / p\right)}\left[\frac{\varepsilon}{2}(\ln 2)^{-\varepsilon-1}+(\ln 2)^{-\varepsilon}\right]^{\frac{1}{q}} 。 \end{aligned}$

由式(10), 取s=λ, s₁= $\tilde{\lambda}_{1}$ : =λ₁-ε/p∈(0, λ)(0 < $\tilde{\lambda}_{2}$ : =λ₂+ε/p < λ), 则有

$\begin{aligned} \tilde{I}=& \sum\limits_{n=2}^{\infty}\left[(\ln n)^{\left(\lambda_{2}+\varepsilon / p\right)} \int_{1}^{\infty} \frac{1}{(x+\ln n)^{\lambda}} x^{\left(\lambda_{1}-\varepsilon / p\right)-1} \mathrm{~d} x\right] \times \\ &(\ln n)^{-\varepsilon{-1}} n^{-1}=\sum\limits_{n=2}^{\infty} \omega\left(\tilde{\lambda}_{1}, n\right)(\ln n)^{-\varepsilon{-1}} n^{-1}-\\ & \sum\limits_{n=2}^{\infty}(\ln n)^{\tilde{\lambda}_{2}-\varepsilon{-1}} n^{-1} \int_{0}^{1} \frac{x^{\tilde{\lambda}_{1}-1}}{(x+\ln n)^{\lambda}} \mathrm{d} x>\\ &B\left(\tilde{\lambda}_{1}, \tilde{\lambda}_{2}\right) \sum\limits_{n=2}^{\infty}(\ln n)^{-\varepsilon{-1}} n^{-1}-\\ &\sum\limits_{n=2}^{\infty}(\ln n)^{\left(\tilde{\lambda}_{2}-\varepsilon\right)-1} n^{-1} \int_{0}^{1} \frac{x^{\tilde{\lambda}_{1}-1}}{(\ln n)^{\lambda}} \mathrm{d} x= \\ &B\left(\tilde{\lambda}_{1}, \tilde{\lambda}_{2}\right) \sum\limits_{n=2}^{\infty} \frac{1}{n}(\ln n)^{-\varepsilon-1}- \\ &\frac{1}{\tilde{\lambda}_{1}} \sum\limits_{n=2}^{\infty} \frac{1}{n}(\ln n)^{-\left(\lambda_{1}+\varepsilon / q\right)-1} \geqslant \\ &B\left(\tilde{\lambda}_{1}, \tilde{\lambda}_{2}\right) \int_{2}^{\infty} \frac{1}{y}(\ln y)^{-\varepsilon{-1}} \mathrm{~d} y-O(1)= \\ &\frac{1}{\varepsilon}\left[B\left(\lambda_{1}-\frac{\varepsilon}{p}, \lambda_{2}+\frac{\varepsilon}{p}\right)(\ln 2)^{-\varepsilon}-\varepsilon O(1)\right]。 \end{aligned}$

从而有

$\begin{aligned} B&\left(\lambda_{1}-\frac{\varepsilon}{p}, \lambda_{2}+\frac{\varepsilon}{p}\right)(\ln 2)^{-\varepsilon}-\varepsilon O(1)<\varepsilon \tilde{I}< \\ &\frac{M}{\prod\nolimits_{i=0}^{m-1}\left(\lambda_{1}+i-\varepsilon / p\right)}\left[\frac{\varepsilon}{2}(\ln 2)^{-\varepsilon{-1}}+(\ln 2)^{-\varepsilon}\right]^{\frac{1}{q}} 。 \end{aligned}$

令ε→0⁺, 由Beta函数的连续性, 有

$\begin{gathered} \frac{\Gamma(\lambda+m)}{\Gamma(\lambda)} k_{\lambda+m}\left(\lambda_{1}+m\right)=\frac{\Gamma\left(\lambda_{1}+m\right) \Gamma\left(\lambda_{2}\right)}{\Gamma(\lambda)}= \\ B\left(\lambda_{1}, \lambda_{2}\right) \prod\limits_{i=0}^{m-1}\left(\lambda_{1}+i\right) \leqslant M。 \end{gathered}$

故 $M=\frac{\Gamma(\lambda+m)}{\Gamma(\lambda)} k_{\lambda+m}\left(\lambda_{1}+m\right)$ 为式(16) (即式(15)取λ₁+λ₂=λ时)的最佳值。证毕。

定理3 若式(15)的常数因子 $\frac{\Gamma(\lambda+m)}{\Gamma(\lambda)} \times\left(k_{\lambda+m}\left(\lambda_{2}\right)\right)^{1 / p}\left(k_{\lambda+m}\left(\lambda_{1}+m\right)\right)^{1 / q}$ 为最佳值, 则当λ₂, λ-λ₁≤1时, 必有λ₁+λ₂=λ。

证明对于 $\hat{\lambda}_{1}=\frac{\lambda-\lambda_{2}}{p}+\frac{\lambda_{1}}{q}, \hat{\lambda}_{2}=\frac{\lambda-\lambda_{1}}{q}+\frac{\lambda_{2}}{p}$ , 有

$\hat{\lambda}_{1}+\hat{\lambda}_{2}=\frac{\lambda-\lambda_{2}}{p}+\frac{\lambda_{1}}{q}+\frac{\lambda-\lambda_{1}}{q}+\frac{\lambda_{2}}{p}=\lambda,$

$0<\hat{\lambda}_{1}, \hat{\lambda}_{2}<\frac{\lambda}{p}+\frac{\lambda}{q}=\lambda,$ $\frac{\Gamma(\lambda+m)}{\Gamma(\lambda)} k_{\lambda+m}\left(\hat{\lambda}_{1}+m\right) \in \mathbb{R}_{+}$ , 且因λ₂, λ-λ₁≤1, 则有 $\hat{\lambda}_{2} \leqslant \frac{1}{q}+\frac{1}{p}=1$ 。由式(16), 代入 $\hat{\lambda}_{i}=\lambda_{i}$ (i=1, 2), 有:

$\begin{aligned} &I=\int_{0}^{\infty} \sum\limits_{n=2}^{\infty} \frac{a_{n}}{(x+\ln n)^{\lambda}} f(x) \mathrm{d} x<\frac{\Gamma(\lambda+m)}{\Gamma(\lambda)} k_{\lambda+m}\left(\hat{\lambda}_{1}+m\right) \times \\ &{\left[\int_{0}^{\infty} x^{-p\left(m-1+\hat{\lambda}_{1}\right)-1} F_{m}^{p}(x) \mathrm{d} x\right]^{\frac{1}{p}}\left[\sum\limits_{n=2}^{\infty}(\ln n)^{q\left(1-\hat{\lambda}_{2}\right)-1} n^{q-1} a_{n}^{q}\right]^{\frac{1}{q}} }。 \end{aligned}$

(18)

由Hölder不等式, 可得

$\begin{aligned} k_{\lambda+m}&\left(\hat{\lambda}_{1}+m\right)=k_{\lambda+m}\left(\frac{\lambda-\lambda_{2}}{p}+\frac{\lambda_{1}}{q}+m\right)= \\ &\int_{0}^{\infty} \frac{1}{(1+u)^{\lambda+m}} u^{\left(\lambda-\lambda_{2}+m\right) / p+\left(\lambda_{1}+m\right) / q-1} \mathrm{~d} u= \\ &\int_{0}^{\infty} \frac{1}{(1+u)^{\lambda+m}} u^{\left(\lambda-\lambda_{2}+m-1\right) / p} u^{\left(\lambda_{1}+m-1\right) / q} \mathrm{~d} u \leqslant \\ &{\left[\int_{0}^{\infty} \frac{1}{(1+u)^{\lambda+m}} u^{\lambda-\lambda_{2}+m-1} \mathrm{~d} u\right]^{\frac{1}{p}} \times}\\ &{\left[\int_{0}^{\infty} \frac{1}{(1+u)^{\lambda+m}} u^{\lambda_{1}+m-1} \mathrm{~d} u\right]^{\frac{1}{q}}=} \\ &{\left[\int_{0}^{\infty} \frac{1}{(1+v)^{\lambda+m}} v^{\lambda_{2}-1} \mathrm{~d} v\right]^{\frac{1}{p}} \times} \\ &{\left[\int_{0}^{\infty} \frac{1}{(1+u)^{\lambda+m}} u^{\left(\lambda_{1}+m\right)-1} \mathrm{~d} u\right]^{\frac{1}{q}}=} \\ &\left(k_{\lambda+m}\left(\lambda_{2}\right)\right)^{1 / p}\left(k_{\lambda+m}\left(\lambda_{1}+m\right)\right)^{1 / q} 。 \end{aligned}$

(19)

由于常数因子 $\frac{\Gamma(\lambda+m)}{\Gamma(\lambda)}\left(k_{\lambda+m}\left(\lambda_{2}\right)\right)^{1 / p}\left(k_{\lambda+m}\left(\lambda_{1}+\right)\right)m^{1 / q}$ 为式(15)的最佳值, 故比较式(15)与式(18)的常数因子, 可得:

$\begin{aligned} &\frac{\Gamma(\lambda+m)}{\Gamma(\lambda)}\left(k_{\lambda+m}\left(\lambda_{2}\right)\right)^{1 / p}\left(k_{\lambda+m}\left(\lambda_{1}+m\right)\right)^{1 / q} \leqslant \\ &\quad \frac{\Gamma(\lambda+m)}{\Gamma(\lambda)} k_{\lambda+m}\left(\hat{\lambda}_{1}+m\right), \end{aligned}$

即 $k_{\lambda+m}\left(\hat{\lambda}_{1}+m\right) \geqslant\left(k_{\lambda+m}\left(\lambda_{2}\right)\right)^{1 / p}\left(k_{\lambda+m}\left(\lambda_{1}+m\right)\right)^{1 / q}$ , 因而式(19)取等号。由文献[29]知，式(19)取等号的充分必要条件是存在不全为0的常数A和B, 使得

$A u^{\lambda-\lambda_{2}+m-1}=B u^{\lambda_{1}+m-1} \text { a.e. 于 } \mathbb{R}_{+} \text {。 }$

不妨设A≠0, 则有u^{λ-λ₂-λ₁}=B/A a.e. 于 $\mathbb{R}_{+}$ , 于是λ-λ₂-λ₁=0, 因而有λ₁+λ₂=λ。证毕。

3. 等价形式

定理4 在满足定理1条件下, 有如下与式(15)等价的半离散Hardy-Mulholland不等式:

$\begin{aligned} J:=& {\left[\sum\limits_{n=2}^{\infty} \frac{1}{n}(\ln n)^{p \hat{\lambda}_{2}-1}\left(\int_{0}^{\infty} \frac{f(x)}{\left(\ln n \mathrm{e}^{x}\right)^{\lambda}} \mathrm{d} x\right)^{p}\right]^{\frac{1}{p}}<} \\ & \frac{\Gamma(\lambda+m)}{\Gamma(\lambda)}\left(k_{\lambda+m}\left(\lambda_{2}\right)\right)^{1 / p}\left(k_{\lambda+m}\left(\lambda_{1}+m\right)\right)^{1 / q} \times \\ & {\left[\int_{0}^{\infty} x^{-p\left(m-1+\hat{\lambda}_{1}\right)-1} F_{m}^{p}(x) \mathrm{d} x\right]^{\frac{1}{p}} }。 \end{aligned}$

(20)

特别地, 当λ₁+λ₂=λ∈(0, ∞), λ₁∈(0, λ), λ₂∈(0, 1]∩(0, λ)时, 有如下式(16)的等价式:

$\begin{aligned} &{\left[\sum\limits_{n=2}^{\infty} \frac{1}{n}(\ln n)^{p \lambda_{2}-1}\left(\int_{0}^{\infty} \frac{f(x)}{\left(\ln n \mathrm{e}^{x}\right)^{\lambda}} \mathrm{d} x\right)^{p}\right]^{\frac{1}{p}}<} \\ &\ \ \frac{\Gamma(\lambda+m)}{\Gamma(\lambda)} B\left(\lambda_{1}+m, \lambda_{2}\right)\left[\int_{0}^{\infty} x^{-p\left(m-1+\lambda_{1}\right)-1} F_{m}^{p}(x) \mathrm{d} x\right]^{\frac{1}{p}} 。 \end{aligned}$

(21)

证明假定式(20)成立。由Hölder不等式, 有

$\begin{aligned} I=& \sum\limits_{n=2}^{\infty}\left[(\ln n)^{\hat{\lambda}_{2}-1 / p} n^{-1 / p} \int_{0}^{\infty} \frac{f(x)}{\left(\ln n \mathrm{e}^{x}\right)^{\lambda}} \mathrm{d} x\right] \times \\ & {\left[(\ln n)^{-\hat{\lambda}_{2}+1 / p} n^{1 / p} a_{n}\right] \leqslant } \\ & J\left[\sum\limits_{n=2}^{\infty}(\ln n)^{q\left(1-\hat{\lambda}_{2}\right)-1} n^{q-1} a_{n}^{q}\right]^{\frac{1}{q}}, \end{aligned}$

(22)

则由式(20)可得式(15)。若式(15)成立, 置

$a_{n}:=\frac{1}{n}(\ln n)^{p \hat{\lambda}_2{-1}}\left(\int_{0}^{\infty} \frac{f(x)}{\left(\ln n \mathrm{e}^{x}\right)^{\lambda}} \mathrm{d} x\right)^{p-1} \quad\left(n \in \mathbb{N}_{+} \backslash\{1\}\right)。$

若J=0, 则式(20)显然成立。下设0 < J < ∞。由式(15), 有

$\begin{aligned} 0<& \sum\limits_{n=2}^{\infty}(\ln n)^{q\left(1-\hat{\lambda}_{2}\right)-1} n^{q-1} a_{n}^{q}=J^{p}=I<\\ & \frac{\Gamma(\lambda+m)}{\Gamma(\lambda)}\left(k_{\lambda+m}\left(\lambda_{2}\right)\right)^{1 / p}\left(k_{\lambda+m}\left(\lambda_{1}+m\right)\right)^{1 / q} \times \\ & {\left[\int_{0}^{\infty} x^{-p\left(m-1+\hat{\lambda}_{1}\right)-1} F_{m}^{p}(x) \mathrm{d} x\right]^{\frac{1}{p}} J^{p-1}<\infty }, \end{aligned}$

(23)

$\begin{aligned} J=& {\left[\sum\limits_{n=2}^{\infty}(\ln n)^{q\left(1-\hat{\lambda}_{2}\right)-1} n^{q-1} a_{n}^{q}\right]^{\frac{1}{p}}<} \\ & \frac{\Gamma(\lambda+m)}{\Gamma(\lambda)}\left(k_{\lambda+m}\left(\lambda_{2}\right)\right)^{1 / p}\left(k_{\lambda+m}\left(\lambda_{1}+m\right)\right)^{1 / q} \times \\ & {\left[\int_{0}^{\infty} x^{-p\left(m-1+\hat{\lambda}_{1}\right)-1} F_{m}^{p}(x) \mathrm{d} x\right]^{\frac{1}{p}}, } \end{aligned}$

即式(20)成立。综上, 式(20)等价于式(15)。证毕。

定理5 若λ₁+λ₂=λ, 则式(20)的常数因子 $\frac{\Gamma(\lambda+m)}{\Gamma(\lambda)}\left(k_{\lambda+m}\left(\lambda_{2}\right)\right)^{1 / p}\left(k_{\lambda+m}\left(\lambda_{1}+m\right)\right)^{1 / q}$ 必为最佳值。若式(20)的常数因子 $\frac{\Gamma(\lambda+m)}{\Gamma(\lambda)}\left(k_{\lambda+m}\left(\lambda_{2}\right)\right)^{1 / p} \times\left(k_{\lambda+m}\left(\lambda_{1}+m\right)\right)^{1 / q}$ 为最佳值且λ₂, λ-λ₁≤1, 则有λ₁+λ₂=λ。

证明若λ₁+λ₂=λ, 则由定理2知, 式(16)的常数因子是最佳值, 从而得到式(21)(即式(20)取λ₁+λ₂=λ时)的常数因子是最佳值(注：倘若式(21)的常数因子不是最佳值, 则由式(22)可推导出式(16)的常数因子也不是最佳值, 从而导出矛盾)。

若式(20)的常数因子 $\frac{\Gamma(\lambda+m)}{\Gamma(\lambda)}\left(k_{\lambda+m}\left(\lambda_{2}\right)\right)^{1 / p} \times\left(k_{\lambda+m}\left(\lambda_{1}+m\right)\right)^{1 / q}$ 是最佳值, 则式(15)的常数因子也必是最佳值(注：倘若式(15)的常数因子不是最佳值，则由式(23)可推导出式(20)的常数因子也不是最佳值, 这就导出矛盾)。因此, 进一步运用定理3, 可得λ₁+λ₂=λ。证毕。

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