## 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

 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)

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

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

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 $\frac{1}{p}+\frac{1}{q}=1\,(p>1)$, 则

$\begin{matrix} \sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty}\frac{a_{m}b_{n}}{m+n}\leq \frac{\pi}{\sin(\pi/p)}\|\tilde{a}\|_{p}\|\tilde{b}\|_{q},\end{matrix}$

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

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

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

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

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

$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

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

\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*}

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

## 3 超齐次核 Hilbert 型离散不等式的构建条件

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

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

(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

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

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

(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

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

\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*}

$c>0$ 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

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

\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*}

(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

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

\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*}

\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*}

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

$c>0$ 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

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

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

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

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

$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*}

\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*}

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

\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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