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数学物理学报, 2024, 44(4): 837-846

构建超齐次核有界离散算子的参数条件及算子范数估计

洪勇,, 赵茜,*

广州华商学院数据科学学院 广州 511300

Parameter Conditions for Constructing Bounded Discrete Operators with Super-Homogeneous Kernel and Estimation of the Operator Norm

Hong Yong,, Zhao Qian,*

College of Data Science, Guangzhou Huashang College, Guangzhou 511300

通讯作者: *赵茜, E-mail:eunicezhao_777@163.com

收稿日期: 2023-07-17   修回日期: 2024-01-10  

基金资助: 广东省基础与应用基础研究基金(2022A1515012429)
广州华商学院科研团队项目(2021HSKT03)

Received: 2023-07-17   Revised: 2024-01-10  

Fund supported: Fundamental and Applied Basic Research Program of Guangdong Province(2022A1515012429)
Guangzhou Huashang College Research Team Program(2021HSKT03)

作者简介 About authors

洪勇,E-mail:hongyonggdcc@yeah.net

摘要

引入超齐次函数的概念, 首先构建超齐次核的 Hilbert 型离散不等式, 然后利用 Hilbert 型不等式与同核算子的关系, 讨论超齐次核离散算子,得到加权赋范序列空间中超齐次核有界离散算子的构建条件, 解决了算子是否有界的判定方法和算子范数的估计问题.

关键词: 超齐次核; 有界算子; 加权序列空间; 有界算子的判定; Hilbert型离散不等式; 算子范数

Abstract

Introducing the concept of super-homogeneous function, firstly constructing the Hilbert-type discrete inequality with super-homogeneous kernel, and then discussing discrete operator with super-homogeneous kernel by using the relationship between Hilbert-type inequality and operator with the same kernel, obtained the construction conditions for bounded discrete operators with super-homogeneous kernel in weighted normed sequence spaces, and a method for determining whether an operator is bounded or not and the estimation of operator norm are solved.

Keywords: Super-homogeneous function; Bounded discrete operator; Weighted sequence spaces; Determination of bounded operators; Hilbert-type discrete inequality; Operator norm

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本文引用格式

洪勇, 赵茜. 构建超齐次核有界离散算子的参数条件及算子范数估计[J]. 数学物理学报, 2024, 44(4): 837-846

Hong Yong, Zhao Qian. Parameter Conditions for Constructing Bounded Discrete Operators with Super-Homogeneous Kernel and Estimation of the Operator Norm[J]. Acta Mathematica Scientia, 2024, 44(4): 837-846

1 引言与超齐次函数概念

1925 年, Hardy 在文献[1]中得到了推广的具有最佳常数因子的 Hilbert 离散不等式: 若 1p+1q=1(p>1), 则

n=1m=1ambnm+nπsin(π/p)
(1.1)

其中 \tilde{a}=\{a_{m}\}\in l_{p} , \tilde{b}=\{b_{n}\}\in l_{q} . p=q=2 时, (1) 式为经典的 Hilbert 不等式[2]. 由式 (1) 可导出具有相同核 \frac{1}{m+n} 的离散算子

T(\tilde{a})_{n}=\sum\limits_{m=1}^{\infty}\frac{a_{m}}{m+n},\ \ \tilde{a}=\{a_{m}\}

l_{p} 中的有界算子, 且 T 的算子范数 \|T\|=\frac{\pi}{\sin(\pi/p)} .

讨论算子的有界性及算子范数是调和分析等分析类学科的一个基本问题, 为了进一步讨论离散算子, 人们先后研究了齐次核、拟齐次核和若干非齐次核的情形, 并将 l_{p} 作了带权推广: 若 p>1 , \alpha\in \mathbb{R} , 称

l_{p}^{\alpha}=\bigg\{\tilde{a}=\{a_{m}\}:\|\tilde{a}\|_{p,\alpha}=\bigg(\sum\limits_{m=1}^{\infty}m^{\alpha}|a_{m}|^{p}\bigg)^{\frac{1}{p}}<+\infty\bigg\}

为加权赋范序列空间, 目前关于离散算子及与之相关的 Hilbert 型离散不等式的研究已取得了丰硕的成果[3-17].

本文将在已获得的众多成果的基础上, 引入更宽更广的超齐次函数概念, 在加权赋范序列空间中, 探讨具有超齐次核的有界离散算子的构建条件及算子范数的估计, 得到此类算子有界的判别方法及算子范数的估计式, 并在一定的条件下得到算子范数的精确计算公式.

设实参数 \sigma_{1},\sigma_{2},\tau_{1},\tau_{2} 及二元实函数 K(u,v) 满足: 对任意 t>0 , 有

K(tu,v)=t^{\sigma_{1}}K(u,t^{\tau_{1}}v),\ \ K(u,tv)=t^{\sigma_{2}}K(t^{\tau_{2}}u,v),

则称 K(u,v) 是具有参数 \{\sigma_{1},\sigma_{2},\tau_{1},\tau_{2}\} 的超齐次函数.

显然超齐次函数具有性质

K(t,1)=t^{\sigma_{1}}K(1,t^{\tau_{1}}), \ \ K(1,t)=t^{\sigma_{2}}K(t^{\tau_{2}},1).

K(u,v) \lambda 阶齐次函数, 则 K(u,v) 是具有参数 \{\lambda,\lambda,-1,-1\} 的超齐次函数, K_{1}(u,v)=K(u^{\lambda_{1}},v^{\lambda_{2}}) 是具有参数 \{\lambda\lambda_{1},\lambda\lambda_{2}, -\frac{\lambda_{1}}{\lambda_{2}},-\frac{\lambda_{2}}{\lambda_{1}}\} 的超齐次函数; 若 G(u) 是一元实函数, 则 K_{2}(u,v)=G(u^{\lambda_{1}}v^{\lambda_{2}}) 是具有参数 \{0,0,\frac{\lambda_{1}}{\lambda_{2}},\frac{\lambda_{2}}{\lambda_{1}}\} 的超齐次函数, K_{3}(u,v)=G(u^{\lambda_{1}}/v^{\lambda_{2}}) 是具有参数 \{0,0,-\frac{\lambda_{1}}{\lambda_{2}},-\frac{\lambda_{2}}{\lambda_{1}}\} 的超齐次函数. 可见超齐次函数是齐次函数的推广, 它包含了许多常用基本算子的核函数.

根据超齐次函数的定义, 有

K(tu,v)=t^{\sigma_{1}}K(u,t^{\tau_{1}}v)=t^{\sigma_{1}+\tau_{1}\sigma_{2}}K(t^{\tau_{1}\tau_{2}}u,v),

由此可知, 一般而言, 超齐次函数的参数应满足条件: \tau_{1}\tau_{2}=1 , \sigma_{1}+\tau_{1}\sigma_{2}=0 .

2 预备引理

本文中记

W_{1}(s)=\int_{0}^{+\infty}K(1,t)t^{s}\mathrm{d}t,\ \ W_{2}(s)=\int_{0}^{+\infty}K(t,1)t^{s}\mathrm{d}t.
\bar{A}(K,\tilde{a},\tilde{b})=\sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty}K(m,n)a_{m}b_{n}.

引理 2.1 当且仅当 \tau_{1}\tau_{2}=1 , \sigma_{1}+\tau_{1}\sigma_{2}=0 时, \tau_{1}\frac{\beta+1}{q}-\frac{\alpha+1}{p}=\tau_{1}-\sigma_{1}-1 \tau_{2}\frac{\alpha+1}{p}-\frac{\beta+1}{q}=\tau_{2}-\sigma_{2}-1 等价.

\frac{\beta+1}{q}=x_{1} , \frac{\alpha+1}{p}=x_{2} , 则问题化为线性方程组

\begin{cases} \tau_{1}x_{1}-x_{2}=\tau_{1}-\sigma_{1}-1, \\ -x_{1}+\tau_{2}x_{2}=\tau_{2}-\sigma_{2}-1. \end{cases}

有无穷多解, 即 \frac{\tau_{1}}{-1}=\frac{-1}{\tau_{2}}=\frac{\tau_{1}-\sigma_{1}-1}{\tau_{2}-\sigma_{2}-1} , 而此式等价于 \tau_{1}\tau_{2}=1 , \sigma_{1}+\tau_{1}\sigma_{2}=0 , 故引理 2.1 成立.

引理 2.2 \tau_{1}\tau_{2}=1 , \sigma_{1}+\tau_{1}\sigma_{2}=0 , \tau_{1}\frac{\beta+1}{q}-\frac{\alpha+1}{p}=\tau_{1}-\sigma_{1}-1 , K(u,v) 是具有参数 \{\sigma_{1},\sigma_{2},\tau_{1} , \tau_{2}\} 的超齐次函数, 则 W_{1}(-\frac{\beta+1}{q})=\frac{1}{|\tau_{2}|}W_{2}(-\frac{\alpha+1}{p}) .

根据已知条件和引理 2.1, 有 \tau_{2}\frac{\alpha+1}{p}-\frac{\beta+1}{q}=\tau_{2}-\sigma_{2}-1 , 于是

\begin{align*} W_{1}(-\frac{\beta+1}{q})& =\int_{0}^{+\infty}K(1,t)t^{-\frac{\beta+1}{q}}\mathrm{d}t =\int_{0}^{+\infty}K(t^{\tau_{2}},1)t^{\sigma_{2}-\frac{\beta+1}{q}}\mathrm{d}t\\ & =\frac{1}{|\tau_{2}|}\int_{0}^{+\infty}K(u,1)u^{\frac{1}{\tau_{2}}(\sigma_{2}-\frac{\beta+1}{q})+\frac{1}{\tau_{2}}-1}\mathrm{d}u\\ & =\frac{1}{|\tau_{2}|}\int_{0}^{+\infty}K(u,1)u^{-\frac{\alpha+1}{p}}\mathrm{d}u=\frac{1}{|\tau_{2}|}W_{2}(-\frac{\alpha+1}{p}).\end{align*}

证毕.

引理 2.3 K(u,v) 是具有参数 \{\sigma_{1},\sigma_{2},\tau_{1} , \tau_{2}\} 的超齐次函数, c_{1},c_{2}\in \mathbb{R} , K(1,t)t^{-c_{1}} K(t,1)t^{-c_{2}} (0,+\infty) 上递减, 则

\bar{\omega}_{1}(m,c_{1})=\sum\limits_{n=1}^{\infty}K(m,n)n^{-c_{1}} \leq m^{\sigma_{1}+\tau_{1}(c_{1}-1)}W_{1}(-c_{1}),
\bar{\omega}_{2}(n,c_{2})=\sum\limits_{m=1}^{\infty}K(m,n)m^{-c_{2}} \leq n^{\sigma_{2}+\tau_{2}(c_{2}-1)}W_{2}(-c_{2}).

因为 K(1,t)t^{-c_{1}} (0,+\infty) 上递减, 故有

\begin{align*} \bar{\omega}_{1}(m,c_{1})& =m^{\sigma_{1}}\sum\limits_{n=1}^{\infty}K(1,m^{\tau_{1}}n)n^{-c_{1}} =m^{\sigma_{1}+\tau_{1}c_{1}}\sum\limits_{n=1}^{\infty}K(1,m^{\tau_{1}}n)(m^{\tau_{1}}n)^{-c_{1}}\\ & \leq m^{\sigma_{1}+\tau_{1}c_{1}}\int_{0}^{+\infty}K(1,m^{\tau_{1}}u)(m^{\tau_{1}}u)^{-c_{1}}\mathrm{d}u\\ & =m^{\sigma_{1}+\tau_{1}(c_{1}-1)}\int_{0}^{+\infty}K(1,t)t^{-c_{1}}\mathrm{d}t =m^{\sigma_{1}+\tau_{1}(c_{1}-1)}W_{1}(-c_{1}).\end{align*}

类似地, 由 K(t,1)t^{-c_{2}} (0,+\infty) 上递减, 可得 \bar{\omega}_{2}(n,c_{2}) \leq n^{\sigma_{2}+\tau_{2}(c_{2}-1)}W_{2}(-c_{2}) .

3 超齐次核 Hilbert 型离散不等式的构建条件

定理 3.1 \frac{1}{p}+\frac{1}{q}=1\,(p>1) , \alpha,\beta\in\mathbb{R} , K(u,v)\geq0 是具有参数 \{\sigma_{1},\sigma_{2},\tau_{1},\tau_{2}\} 的超齐次可测函数, K(1,t) K(t,1) 几乎处处大于0, \tau_{1}\tau_{2}=1 , \sigma_{1}+\tau_{1}\sigma_{2}=0 , \tau_{1}\frac{\beta+1}{q}-\frac{\alpha+1}{p}-(\tau_{1}-\sigma_{1}-1)=c , K(t,1)t^{-\frac{\alpha+1}{p}},K(1,t)t^{-\frac{\beta+1}{q}},K(t,1)t^{-\frac{\alpha+1}{p}-c} K(1,t)t^{-\frac{\beta+1}{q}+\tau_{2}c} 都在 (0,+\infty) 上递减, 0<W_{2}(-\frac{\alpha+1}{p})<+\infty , 0<W_{2}(-\frac{\alpha+1}{p}-c)<+\infty , \tilde{a}=\{a_{m}\}\in l_{p}^{\alpha} , \tilde{b}=\{b_{n}\}\in l_{q}^{\beta} . 那么

(i) 若 \tau_{1}>0 , 则不论 c 为何值, 都存在常数 M_{1}>0 , 使得

\bar{A}(K,\tilde{a},\tilde{b})\leq M_{1}\|\tilde{a}\|_{p,\alpha}\|\tilde{b}\|_{q,\beta}.
(3.1)

(ii) 若 \tau_{1}<0 , 则当且仅当 c\leq 0 , 即 \tau_{1}\frac{\beta+1}{q}-\frac{\alpha+1}{p}\leq \tau_{1}-\sigma_{1}-1 时, 存在常数 M_{2}>0 , 使得

\bar{A}(K,\tilde{a},\tilde{b})\leq M_{2}\|\tilde{a}\|_{p,\alpha}\|\tilde{b}\|_{q,\beta},
(3.2)

且当 \tau_{1}\frac{\beta+1}{q}-\frac{\alpha+1}{p}=\tau_{1}-\sigma_{1}-1 时, (3.2) 式的最佳常数因子为

M_{0}=\inf\{M_{2}\}=\bigg(\frac{1}{|\tau_{1}|}\bigg)^{\frac{1}{q}}W_{1}(-\frac{\beta+1}{q})=\bigg(\frac{1}{|\tau_{2}|}\bigg)^{\frac{1}{p}}W_{2}(-\frac{\alpha+1}{p}).

(i) 若 \tau_{1}>0 , 则 \tau_{2}>0 . c\leq 0 时, 由 \tau_{1}\frac{\beta+1}{q}-\frac{\alpha+1}{p}-(\tau_{1}-\sigma_{1}-1)=c , 可得

\tau_{1}\frac{\beta+1}{q}-\frac{(\alpha+pc)+1}{p}=\tau_{1}-\sigma_{1}-1.

根据 Hölder 不等式、引理 2.3 和引理 2.2, 并注意 pc\leq 0 , 有

\begin{align*}\bar{A}(K,\tilde{a},\tilde{b})&=\sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty} \left(\frac{m^{\frac{\alpha+pc+1}{pq}}}{n^{\frac{\beta+1}{pq}}}a_{m}\right) \left(\frac{n^{\frac{\beta+1}{pq}}}{m^{\frac{\alpha+pc+1}{pq}}}b_{n}\right)K(m,n)\\ & \leq \left(\sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty} \frac{m^{\frac{\alpha+pc+1}{q}}}{n^{\frac{\beta+1}{q}}}|a_{m}|^{p}K(m,n)\right)^{\frac{1}{p}} \left(\sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty}\frac{n^{\frac{\beta+1}{p}}}{m^{\frac{\alpha+pc+1}{p}}}|b_{n}|^{q}K(m,n)\right)^{\frac{1}{q}} \\ & =\left(\sum\limits_{m=1}^{\infty}m^{\frac{\alpha+pc+1}{q}}|a_{m}|^{p}\bar{\omega}_{1}(m,\frac{\beta+1}{q})\right)^{\frac{1}{p}} \left(\sum\limits_{n=1}^{\infty}n^{\frac{\beta+1}{p}}|b_{n}|^{q}\bar{\omega}_{2}(n,\frac{\alpha+pc+1}{p})\right)^{\frac{1}{q}}\\ & \leq W_{1}^{\frac{1}{p}}(-\frac{\beta+1}{q})W_{2}^{\frac{1}{q}}(-\frac{\alpha+pc+1}{p}) \left(\sum\limits_{m=1}^{\infty}m^{\frac{\alpha+pc+1}{q}+\sigma_{1}+\tau_{1}(\frac{\beta+1}{q}-1)}|a_{m}|^{p}\right)^{\frac{1}{p}}\\ & \ \ \ \ \times\left(\sum\limits_{n=1}^{\infty}n^{\frac{\beta+1}{p}+\sigma_{2}+\tau_{2}(\frac{\alpha+pc+1}{p}-1)}|b_{n}|^{q}\right)^{\frac{1}{q}} \\ & =\bigg(\frac{1}{|\tau_{2}|}\bigg)^{\frac{1}{p}}W_{2}(-\frac{\alpha+1}{p}-c) \left(\sum\limits_{m=1}^{\infty}m^{\alpha+pc}|a_{m}|^{p}\right)^{\frac{1}{p}} \left(\sum\limits_{n=1}^{\infty}n^{\beta}|b_{n}|^{q}\right)^{\frac{1}{q}}\\ & \leq \bigg(\frac{1}{|\tau_{2}|}\bigg)^{\frac{1}{p}}W_{2}(-\frac{\alpha+1}{p}-c) \left(\sum\limits_{m=1}^{\infty}m^{\alpha}|a_{m}|^{p}\right)^{\frac{1}{p}} \left(\sum\limits_{n=1}^{\infty}n^{\beta}|b_{n}|^{q}\right)^{\frac{1}{q}}\\ & =\bigg(\frac{1}{|\tau_{2}|}\bigg)^{\frac{1}{p}}W_{2}(-\frac{\alpha+1}{p}-c)\|\tilde{a}\|_{p,\alpha}\|\tilde{b}\|_{q,\beta},\end{align*}

此时, 取常数 M_{1}\geq (\frac{1}{|\tau_{2}|})^{\frac{1}{p}}W_{2}(-\frac{\alpha+1}{q}-c) , 可得 (3.1) 式.

c>0 时, 由 \tau_{1}\frac{\beta+1}{q}-\frac{\alpha+1}{p}-(\tau_{1}-\sigma_{1}-1)=c , \tau_{1}\tau_{2}=1 , 可得

\tau_{1}\frac{\beta-\tau_{2}qc+1}{q}-\frac{\alpha+1}{p}=\tau_{1}-\sigma_{1}-1.

同样地, 根据 Hölder 不等式、引理 2.3 和引理 2.2, 并注意 \tau_{2}qc>0 , 有

\begin{align*}\bar{A}(K,\tilde{a},\tilde{b})&=\sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty} \left(\frac{m^{\frac{\alpha+1}{pq}}}{n^{\frac{\beta-\tau_{2}qc+1}{pq}}}a_{m}\right) \left(\frac{n^{\frac{\beta-\tau_{2}qc+1}{pq}}}{m^{\frac{\alpha+1}{pq}}}b_{n}\right)K(m,n)\\ & \leq \left(\sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty} \frac{m^{\frac{\alpha+1}{q}}}{n^{\frac{\beta-\tau_{2}qc+1}{q}}}|a_{m}|^{p}K(m,n)\right)^{\frac{1}{p}} \left(\sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty}\frac{n^{\frac{\beta-\tau_{2}qc+1}{p}}}{m^{\frac{\alpha+1}{p}}}|b_{n}|^{q}K(m,n)\right)^{\frac{1}{q}}\\ & =\left(\sum\limits_{m=1}^{\infty}m^{\frac{\alpha+1}{q}}|a_{m}|^{p}\bar{\omega}_{1}(m,\frac{\beta-\tau_{2}qc+1}{q})\right)^{\frac{1}{p}} \left(\sum\limits_{n=1}^{\infty}n^{\frac{\beta-\tau_{2}qc+1}{p}}|b_{n}|^{q}\bar{\omega}_{2}(n,\frac{\alpha+1}{p})\right)^{\frac{1}{q}}\\ & \leq W_{1}^{\frac{1}{p}}(-\frac{\beta-\tau_{2}qc+1}{q})W_{2}^{\frac{1}{q}}(-\frac{\alpha+1}{p}) \left(\sum\limits_{m=1}^{\infty}m^{\frac{\alpha+1}{q}+\sigma_{1}+\tau_{1}(\frac{\beta-\tau_{2}qc+1}{q}-1)}|a_{m}|^{p}\right)^{\frac{1}{p}}\\ & \ \ \ \ \times\left(\sum\limits_{n=1}^{\infty}n^{\frac{\beta-\tau_{2}qc+1}{p}+\sigma_{2}+\tau_{2}(\frac{\alpha+1}{p}-1)}|b_{n}|^{q}\right)^{\frac{1}{q}}\\ & =\bigg(\frac{1}{|\tau_{2}|}\bigg)^{\frac{1}{p}}W_{2}(-\frac{\alpha+1}{p}) \left(\sum\limits_{m=1}^{\infty}m^{\alpha}|a_{m}|^{p}\right)^{\frac{1}{p}} \left(\sum\limits_{n=1}^{\infty}n^{\beta-\tau_{2}qc}|b_{n}|^{q}\right)^{\frac{1}{q}}\\ & \leq \bigg(\frac{1}{|\tau_{2}|}\bigg)^{\frac{1}{p}}W_{2}(-\frac{\alpha+1}{p})\|\tilde{a}\|_{p,\alpha}\|\tilde{b}\|_{q,\beta},\end{align*}

此时, 取常数 M_{1}\geq (\frac{1}{|\tau_{2}|})^{\frac{1}{p}}W_{2}(-\frac{\alpha+1}{q}) , 可得 (3.1) 式.

(ii) 充分性: 若 \tau_{1}<0 , 则 \tau_{2}<0 . c\leq0 , 利用与 (i) 类似的证明方法, 可得

\bar{A}(K,\tilde{a},\tilde{b})\leq \bigg(\frac{1}{|\tau_{2}|}\bigg)^{\frac{1}{p}}W_{2}(-\frac{\alpha+1}{p}-c)\|\tilde{a}\|_{p,\alpha}\|\tilde{b}\|_{q,\beta},

取常数 M_{2}\geq (\frac{1}{|\tau_{2}|})^{\frac{1}{p}}W_{2}(-\frac{\alpha+1}{q}-c) , 可得 (3.2) 式.

必要性: 设 (3.2) 式成立. 若 c>0 , 取

\begin{align*}& a_{m}= \begin{cases} 0, & m=1, \\ m^{-(\alpha+1+c)/p}, & m=2,3,\cdots, \end{cases}\\& b_{n}= \begin{cases} 0, & n=1,\\ n^{-(\beta+1-\frac{c}{\tau_{1}})/q}, & n=2,3,\cdots. \end{cases}\end{align*}

因为 c>0 , \frac{c}{\tau_{1}}<0 , 故

\begin{align*} \|\tilde{a}\|_{p,\alpha}\|\tilde{b}\|_{q,\beta}& =\left(\sum\limits_{n=2}^{\infty}n^{-1-c}\right)^{\frac{1}{p}} \left(\sum\limits_{n=2}^{\infty}n^{-1+\frac{c}{\tau_{1}}}\right)^{\frac{1}{q}}\\ & \leq \left(\int_{1}^{+\infty}t^{-1-c}\mathrm{d}t\right)^{\frac{1}{p}} \left(\int_{1}^{+\infty}t^{-1+\frac{c}{\tau_{1}}}\mathrm{d}t\right)^{\frac{1}{q}} =\frac{1}{c}|\tau_{1}|^{\frac{1}{q}}.\end{align*}

因为 K(1,t)t^{-\frac{\beta+1}{p}} (0,+\infty) 上递减, \tau_{1}<0 , 故有

\begin{align*} \bar{A}(K,\tilde{a},\tilde{b})& =\sum\limits_{m=2}^{\infty}m^{-\frac{\alpha+1}{p}-\frac{c}{p}} \left(\sum\limits_{n=2}^{\infty}K(m,n)n^{-\frac{\beta+1}{q}+\frac{c}{\tau_{1}q}}\right)\\ & =\sum\limits_{m=2}^{\infty}m^{\sigma_{1}-\frac{\alpha+1}{p}-\frac{c}{p}} \left(\sum\limits_{n=2}^{\infty}K(1,m^{\tau_{1}}n)n^{-\frac{\beta+1}{q}+\frac{c}{\tau_{1}q}}\right)\\ & =\sum\limits_{m=2}^{\infty}m^{\sigma_{1}-\frac{\alpha+1}{p}-\frac{c}{p}+\tau_{1}(\frac{\beta+1}{q}-\frac{c}{\tau_{1}q})} \left(\sum\limits_{n=2}^{\infty}K(1,m^{\tau_{1}}n)(m^{\tau_{1}}n)^{-\frac{\beta+1}{q}+\frac{c}{\tau_{1}q}}\right)\\ & \geq \sum\limits_{m=2}^{\infty}m^{\sigma_{1}-\frac{\alpha+1}{p}-\frac{c}{p}+\tau_{1}\frac{\beta+1}{q}-\frac{c}{q}} \left(\int_{2}^{+\infty}K(1,m^{\tau_{1}}u)(m^{\tau_{1}}u)^{-\frac{\beta+1}{q}+\frac{c}{\tau_{1}q}}\mathrm{d}u\right)\\ & =\sum\limits_{m=2}^{\infty}m^{\sigma_{1}+\tau_{1}\frac{\beta+1}{q}-\frac{\alpha+1}{p}-c-\tau_{1}} \left(\int_{2m^{\tau_{1}}}^{+\infty}K(1,t)t^{-\frac{\beta+1}{q}+\frac{c}{\tau_{1}q}}\mathrm{d}t\right)\\ & \geq \sum\limits_{m=2}^{\infty}m^{-1} \int_{2^{\tau_{1}+1}}^{+\infty}K(1,t)t^{-\frac{\beta+1}{q}+\frac{c}{\tau_{1}q}}\mathrm{d}t,\end{align*}

于是得到

\begin{align*} \sum\limits_{m=2}^{\infty}m^{-1}\int_{2^{\tau_{1}+1}}^{+\infty}K(1,t)t^{-\frac{\beta+1}{q}+\frac{c}{\tau_{1}q}}\mathrm{d}t \leq \frac{M_{2}}{c}|\tau_{1}|^{\frac{1}{q}}<+\infty,\end{align*}

\sum\limits_{m=2}^{\infty}m^{-1}=+\infty , K(1,t) 几乎处处大于 0 , 这就得到了矛盾, 故 c>0 不成立, 从而 c\leq0 .

\tau_{1}\frac{\beta+1}{q}-\frac{\alpha+1}{p}=\tau_{1}-\sigma_{1}-1 时, c=0 . 设 (3.2) 式的最佳常数因子为 M_{0}>0 , 则 \bar{A}(K,\tilde{a},\tilde{b})\leq M_{0}\|\tilde{a}\|_{p,\alpha}\|\tilde{b}\|_{q,\beta} , 且由前面充分性的证明, 可知

M_{0}\leq \bigg(\frac{1}{|\tau_{2}|}\bigg)^{\frac{1}{p}}W_{2}(-\frac{\alpha+1}{p}) =\bigg(\frac{1}{|\tau_{1}|}\bigg)^{\frac{1}{q}}W_{1}(-\frac{\beta+1}{q}).

取充分小的 \varepsilon>0 及足够大的自然数 N , 令

\begin{align*}& a_{m}= \begin{cases} 0, & m=1,2,\cdots,N-1, \\ m^{-(\alpha+1-\tau_{1}\varepsilon)/p}, & n=N,N+1,\cdots, \end{cases}\\& b_{n}= \begin{cases} 0, & n=1,\\ n^{-(\beta+1+\varepsilon)/q}, & n=2,3,\cdots, \end{cases}\end{align*}

\begin{align*} \|\tilde{a}\|_{p,\alpha}\|\tilde{b}\|_{q,\beta} & =\left(\sum\limits_{m=N}^{\infty}m^{-1+\tau_{1}\varepsilon}\right)^{\frac{1}{p}}\left(\sum\limits_{n=2}^{\infty}n^{-1-\varepsilon}\right)^{\frac{1}{q}}\\ & \leq \left(\int_{1}^{+\infty}t^{-1+\tau_{1}\varepsilon}\mathrm{d}t\right)^{\frac{1}{p}} \left(\int_{1}^{+\infty}t^{-1-\varepsilon}\mathrm{\mathrm{d}}t\right)^{\frac{1}{q}} =\frac{1}{\varepsilon}\bigg(\frac{1}{|\tau_{1}|}\bigg)^{\frac{1}{p}},\end{align*}
\begin{align*} \bar{A}(K,\tilde{a},\tilde{b})& =\sum\limits_{m=N}^{\infty}m^{-\frac{\alpha+1}{p}+\frac{\tau_{1}\varepsilon}{p}} \left(\sum\limits_{n=2}^{\infty}K(m,n)n^{-\frac{\beta+1}{q}-\frac{\varepsilon}{q}}\right)\\ & =\sum\limits_{m=N}^{\infty}m^{\sigma_{1}-\frac{\alpha+1}{p}+\frac{\tau_{1}\varepsilon}{p}} \left(\sum\limits_{n=2}^{\infty}K(1,m^{\tau_{1}}n)n^{-\frac{\beta+1}{q}-\frac{\varepsilon}{q}}\right)\\ & =\sum\limits_{m=N}^{\infty}m^{\sigma_{1}-\frac{\alpha+1}{p}+\frac{\tau_{1}\varepsilon}{p}+\tau_{1}(\frac{\beta+1}{q}+\frac{\varepsilon}{q})} \left(\sum\limits_{n=2}^{\infty}K(1,m^{\tau_{1}}n)(m^{\tau_{1}}n)^{-\frac{\beta+1}{q}-\frac{\varepsilon}{q}}\right)\\ & \geq \sum\limits_{m=N}^{\infty}m^{\sigma_{1}+\tau_{1}\frac{\beta+1}{q}-\frac{\alpha+1}{p}+\tau_{1}\varepsilon} \left(\int_{2}^{\infty}K(1,m^{\tau_{1}}u)(m^{\tau_{1}}u)^{-\frac{\beta+1}{q}-\frac{\varepsilon}{q}}\mathrm{d}u\right)\\ & =\sum\limits_{m=N}^{\infty}m^{\sigma_{1}+\tau_{1}\frac{\beta+1}{q}-\frac{\alpha+1}{p}+\tau_{1}\varepsilon-\tau_{1}} \left(\int_{2m^{\tau_{1}}}^{+\infty}K(1,t)t^{-\frac{\beta+1}{q}-\frac{\varepsilon}{q}}\mathrm{d}t\right)\\ & \geq \sum\limits_{m=N}^{\infty}m^{-1+\tau_{1}\varepsilon} \int_{2N^{\tau_{1}}}^{+\infty}K(1,t)t^{-\frac{\beta+1}{q}-\frac{\varepsilon}{q}}\mathrm{d}t\\ & \geq \int_{N}^{+\infty}u^{-1+\tau_{1}\varepsilon}\mathrm{d}u \int_{2N^{\tau_{1}}}^{+\infty}K(1,t)t^{-\frac{\beta+1}{q}-\frac{\varepsilon}{q}}\mathrm{d}t\\ & =\frac{1}{|\tau_{1}|\varepsilon}N^{\tau_{1}\varepsilon} \int_{2N^{\tau_{1}}}^{+\infty}K(1,t)t^{-\frac{\beta+1}{q}-\frac{\varepsilon}{q}}\mathrm{d}t,\end{align*}

于是得到

\frac{1}{|\tau_{1}|\varepsilon}N^{\tau_{1}\varepsilon} \int_{2N^{\tau_{1}}}^{+\infty}K(1,t)t^{-\frac{\beta+1}{q}-\frac{\varepsilon}{q}}\mathrm{d}t \leq \frac{M_{0}}{\varepsilon}\bigg(\frac{1}{|\tau_{1}|}\bigg)^{\frac{1}{p}},

从而

\begin{matrix} \bigg(\frac{1}{|\tau_{1}|}\bigg)^{\frac{1}{q}}\int_{2N^{\tau_{1}}}^{+\infty}K(1,t)t^{-\frac{\beta+1}{q}-\frac{\varepsilon}{q}}\mathrm{d}t \leq M_{0}N^{-\tau_{1}\varepsilon}.\end{matrix}

不失科学性, 视 \varepsilon 为一个趋于 0 的正项数列 \{c_{k}\} , 根据著名的 Fatou 引理, 有

\begin{align*} \int_{2N^{\tau_{1}}}^{+\infty}K(1,t)t^{-\frac{\beta+1}{q}}\mathrm{d}t & =\int_{2N^{\tau_{1}}}^{+\infty}\liminf_{k\rightarrow\infty}K(1,t)t^{-\frac{\beta+1}{q}-\frac{c_{k}}{q}}\mathrm{d}t\\ & \leq \liminf_{k\rightarrow\infty}\int_{2N^{\tau_{1}}}^{+\infty}K(1,t)t^{-\frac{\beta+1}{q}-\frac{c_{k}}{q}}\mathrm{d}t,\end{align*}

据此, 在式 (3.3) 中令 \varepsilon\rightarrow0^{+} , 有

\begin{align*} \bigg(\frac{1}{|\tau_{1}|}\bigg)^{\frac{1}{q}}\int_{2N^{\tau_{1}}}^{+\infty}K(1,t)t^{-\frac{\beta+1}{q}}\mathrm{d}t\leq M_{0},\end{align*}

再令 N\rightarrow+\infty , 并注意 \tau_{1}<0 , 得到

\bigg(\frac{1}{|\tau_{1}|}\bigg)^{\frac{1}{q}}W_{1}(-\frac{\beta+1}{q}) =\bigg(\frac{1}{|\tau_{1}|}\bigg)^{\frac{1}{q}}\int_{0}^{+\infty}K(1,t)t^{-\frac{\beta+1}{q}}\mathrm{d}t\leq M_{0},

于是有 M_{0}=(\frac{1}{|\tau_{1}|})^{\frac{1}{q}}W_{1}(-\frac{\beta+1}{q}) . 所以式 (3.2) 的最佳常数因子为

M_{0}=\bigg(\frac{1}{|\tau_{1}|}\bigg)^{\frac{1}{q}}W_{1}(-\frac{\beta+1}{q})=\bigg(\frac{1}{|\tau_{2}|}\bigg)^{\frac{1}{p}}W_{2}(-\frac{\alpha+1}{p}).

证毕.

4 超齐次核有界离散算子的构建条件及算子范数估计

定理 4.1 p>1 , \alpha,\gamma\in\mathbb{R} , \tau_{1}\tau_{2}=1 , \sigma_{1}+\tau_{1}\sigma_{2}=0 , K(u,v)\geq0 是具有参数 \{\sigma_{1},\sigma_{2},\tau_{1},\tau_{2}\} 的超齐次可测函数, K(1,t) K(t,1) 几乎处处大于0, \sigma_{1}+1-\tau_{1}\frac{\gamma+1}{p}-\frac{\alpha+1}{p}=c , 0<W_{2}(-\frac{\alpha+1}{p})<+\infty , 0<W_{2}(-\frac{\alpha+1}{p}-c)<+\infty , 且 K(t,1)t^{-\frac{\alpha+1}{p}} , K(1,t)t^{\frac{\gamma+1}{p}-1} , K(t,1)t^{-\frac{\alpha+1}{p}-c} K(1,t)t^{\frac{\gamma+1}{p}-1+\tau_{2}c} 都在 (0,+\infty) 上递减, 离散算子 T

T(\tilde{a})_{n}=\sum\limits_{m=1}^{\infty}K(m,n)a_{m},\ \ \tilde{a}=\{a_{m}\}\in l_{p}^{\alpha}.

(i) 若 \tau_{1}>0 , 则不论 c 为何值, T 都是 l_{p}^{\alpha} l_{p}^{\gamma} 的有界算子, 即存在常数 M>0 , 使得

\|T(\tilde{a})\|_{p,\gamma}\leq M\|\tilde{a}\|_{p,\alpha},

c>0 时, \|T\|\leq (\frac{1}{|\tau_{2}|})^{\frac{1}{p}}W_{2}(-\frac{\alpha+1}{p}) ; c\leq0 时, \|T\|\leq (\frac{1}{|\tau_{2}|})^{\frac{1}{p}}W_{2}(-\frac{\alpha+1}{p}-c) .

(ii) 若 \tau_{1}<0 , 则当且仅当 c\leq0 , 即 \sigma_{1}+1\leq \tau_{1}\frac{\gamma+1}{p}+\frac{\alpha+1}{p} 时, T l_{p}^{\alpha} l_{p}^{\gamma} 的有界算子. 当 \sigma_{1}+1=\tau_{1}\frac{\gamma+1}{p}+\frac{\alpha+1}{p} 时, T 的算子范数为

\|T\|=\bigg(\frac{1}{|\tau_{2}|}\bigg)^{\frac{1}{p}}W_{2}(-\frac{\alpha+1}{p}) =|\tau_{1}|^{\frac{1}{p}}\int_{0}^{+\infty}K(t,1)t^{-\frac{\alpha+1}{p}}\mathrm{d}t.

q=\frac{p}{p-1} , \beta=\frac{\gamma}{1-p} , 则 \frac{1}{p}+\frac{1}{q}=1 , \gamma=\beta(1-p) , -\frac{\beta+1}{q}=\frac{\gamma+1}{p}-1 , 且

\tau_{1}\frac{\beta+1}{q}-\frac{\alpha+1}{p}-(\tau_{1}-\sigma_{1}-1)=c\ \ \Leftrightarrow\ \ \sigma_{1}+1-\tau_{1}\frac{\gamma+1}{p}-\frac{\alpha+1}{p}=c.

又根据 Hilbert 型不等式的基本理论[18], 下列 Hilbert 型离散不等式及算子不等式

\bar{A}(K,\tilde{a},\tilde{b})\leq M\|\tilde{a}\|_{p,\alpha}\|\tilde{b}\|_{q,\beta}, \ \ \|T(\tilde{a})\|_{p,\beta(1-p)}\leq M\|\tilde{a}\|_{p,\alpha}

等价, 于是由定理 3.1, 知定理 4.1 成立.

在定理 4.1 中, 取 \alpha=\gamma=0 , 则可得下面推论.

命题 4.1 p>1 , \tau_{1}\tau_{2}=1 , \sigma_{1}+\tau_{1}\sigma_{2}=0 , K(u,v) 是具有参数 \{\sigma_{1},\sigma_{2},\tau_{1},\tau_{2}\} 的超齐次非负可测函数, \sigma_{1}+1-\frac{\tau_{1}+1}{p}=c , 0<W_{2}(-\frac{1}{p})<+\infty , 0<W_{2}(-\frac{1}{p}-c)<+\infty , 且 K(t,1)t^{-\frac{1}{p}} , K(1,t)t^{\frac{1}{p}-1} , K(t,1)t^{-\frac{1}{p}-c} K(1,t)t^{\frac{1}{p}-1+\tau_{2}c} 都在 (0,+\infty) 上递减, 离散算子 T

T(\tilde{a})_{n}=\sum\limits_{m=1}^{\infty}K(m,n)a_{m},\ \ \tilde{a}=\{a_{m}\}\in l_{p}.

(i) 若 \tau_{1}>0 , 则不论 c 为何值, T 都是 l_{p} 中的有界算子, 即存在常数 M>0 , 使得

\|T(\tilde{a})\|_{p}\leq M\|\tilde{a}\|_{p},

c>0 时, \|T\|\leq (\frac{1}{|\tau_{2}|})^{\frac{1}{p}}W_{2}(-\frac{1}{p}) ; c\leq0 时, \|T\|\leq (\frac{1}{|\tau_{2}|})^{\frac{1}{p}}W_{2}(-\frac{1}{p}-c) .

(ii) 若 \tau_{1}<0 , 则当且仅当 c\leq0 , 即 \sigma_{1}+1\leq \frac{\tau_{1}+1}{p} 时, T l_{p} 中的有界算子. 当 \sigma_{1}+1=\frac{\tau_{1}+1}{p} 时, T 的算子范数为

\|T\|=\Big(\frac{1}{|\tau_{2}|}\Big)^{\frac{1}{p}}W_{2}(-\frac{1}{p})=|\tau_{1}|^{\frac{1}{p}}\int_{0}^{+\infty}K(t,1)t^{-\frac{1}{p}}\mathrm{d}t.

命题 4.2 p>1 , 0<\lambda_{1}\leq 1 , 0<\lambda_{2}\leq 1 , c>0 , 离散算子 T

T(\tilde{a})_{n}=\sum\limits_{m=1}^{\infty}\frac{a_{m}}{(m^{\lambda_{1}}+cn^{\lambda_{2}})^{2}+n^{2\lambda_{2}}},\ \ \tilde{a}=\{a_{m}\},

T l_{p}^{p(1-\lambda_{1})-1} l_{p}^{\lambda_{2}p-1} 的有界算子, 且 T 的算子范数为\|T\|=\frac{1}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}\Big(\frac{\pi}{2}-\arctan(c)\Big).

K(u,v)=\frac{1}{(u^{\lambda_{1}}+cv^{\lambda_{2}})^{2}+v^{2\lambda_{2}}}\ (u>0,v>0), K(u,v) 是具有参数 \{-2\lambda_{1},-2\lambda_{2},-\frac{\lambda_{1}}{\lambda_{2}},-\frac{\lambda_{2}}{\lambda_{1}}\} 的超齐次非负函数. 因为 \sigma_{1}=-2\lambda_{1} , \sigma_{2}=-2\lambda_{2} , \tau_{1}=-\frac{\lambda_{1}}{\lambda_{2}} , \tau_{2}=-\frac{\lambda_{2}}{\lambda_{1}} , 故 \tau_{1}\tau_{2}=1 , \sigma_{1}+\tau_{1}\sigma_{2}=0 . \alpha=p(1-\lambda_{1})-1 , \gamma=\lambda_{2}p-1 , 则

\sigma_{1}+1-\tau_{1}\frac{\gamma+1}{p}-\frac{\alpha+1}{p} =-2\lambda_{1}+1+\frac{\lambda_{1}}{\lambda_{2}}\frac{\lambda_{2}p}{p}-\frac{p(1-\lambda_{1})}{p}=0.

0<\lambda_{1}\leq 1 , 0<\lambda_{2}\leq 1 , c>0 , 可知

\begin{align*}& K(t,1)t^{-\frac{\alpha+1}{p}}=\frac{1}{(t^{\lambda_{1}}+c)^{2}+1}t^{\lambda_{1}-1},\\& K(1,t)t^{\frac{\gamma+1}{p}-1}=\frac{1}{(1+ct^{\lambda_{2}})^{2}+t^{2\lambda_{2}}}t^{\lambda_{2}-1},\end{align*}

都在 (0,+\infty) 上递减. 又因为 \tau_{1}=-\frac{\lambda_{1}}{\lambda_{2}}<0 , 且

\begin{align*} W_{2}(-\frac{\alpha+1}{p})&=\int_{0}^{+\infty}K(t,1)t^{-\frac{\alpha+1}{p}}\mathrm{d}t =\int_{0}^{+\infty}\frac{1}{(t^{\lambda_{1}}+c)^{2}+1}t^{\lambda_{1}-1}\mathrm{d}t\\ & =\frac{1}{\lambda_{1}}\int_{0}^{+\infty}\frac{1}{(u+c)^{2}+1}u^{\frac{1}{\lambda_{1}}(\lambda_{1}-1)+\frac{1}{\lambda_{1}}-1}\mathrm{d}u\\ & =\frac{1}{\lambda_{1}}\int_{0}^{+\infty}\frac{1}{(u+c)^{2}+1}\mathrm{d}u =\frac{1}{\lambda_{1}}\bigg(\frac{\pi}{2}-\arctan(c)\bigg)<+\infty, \end{align*}

从而根据定理 4.1(ii), 可知 T l_{p}^{p(1-\lambda_{1})-1} l_{p}^{\lambda_{2}p-1} 的有界算子, 且 T 的算子范数为

\|T\|=\bigg(\frac{1}{|\tau_{2}|}\bigg)^{\frac{1}{p}}W_{2}(-\frac{\alpha+1}{p}) =\frac{1}{\lambda_{1}^{1/q}\lambda_{2}^{1/p}}\bigg(\frac{\pi}{2}-\arctan(c)\bigg).

证毕.

在命题 4.2 中, 取 \lambda_{1}=1-\frac{1}{p} , \lambda_{2}=\frac{1}{p} , 则 0<\lambda_{1}<1 , 0<\lambda_{2}<1 , 且 \alpha=p(1-\lambda_{1})-1=0 , \gamma=\lambda_{2}p-1=0 , 于是可得下面推论.

命题 4.3 \frac{1}{p}+\frac{1}{q}=1\,(p>1) , c>0 , 离散算子 T

\begin{align*} T(\tilde{a})_{n}=\sum\limits_{m=1}^{\infty}\frac{a_{m}}{(m^{1/q}+cn^{1/p})^{2}+n^{2/p}},\ \ \tilde{a}=\{a_{m}\} \end{align*}

T l_{p} 中的有界算子, 且 T 的算子范数为

\|T\|=q^{\frac{1}{q}}p^{\frac{1}{p}}\bigg(\frac{\pi}{2}-\arctan(c)\bigg).

在命题 4.3 中, 再取 p=q=2 , 则可得下列结果.

命题 4.4 c>0 , 离散算子 T

\begin{align*} T(\tilde{a})_{n}=\sum\limits_{m=1}^{\infty}\frac{a_{m}}{(\sqrt{m}+c\sqrt{n})^{2}+n},\ \ \tilde{a}=\{a_{m}\} \end{align*}

T l_{2} 中的有界算子, 且 T 的算子范数为

\|T\|=2\bigg(\frac{\pi}{2}-\arctan(c)\bigg)=\pi-2\arctan(c).

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