拟齐次核的Hilbert型级数不等式的最佳搭配参数条件及应用

拟齐次核的Hilbert型级数不等式的最佳搭配参数条件及应用

洪勇^,¹, 陈强^,²

¹ 广州华商学院应用数学系广州 511300

² 广东第二师范学院计算机学院广州 510303

The Best Matching Parameters Conditions of Hilbert-Type Series Inequality with Quasi-Homogeneous Kernel and Applications

Hong Yong^,¹, Chen Qiang^,²

¹ Department of Applied Mathematics, Guangzhou Huashang College, Guangzhou 511300

² School of Computer, Guangdong University of Education, Guangzhou 510303

收稿日期: 2020-10-28

基金资助:

国家自然科学基金. 61772140

Received: 2020-10-28

Fund supported:

the NSFC. 61772140

作者简介 About authors

洪勇,E-mail:hongyonggdcc@yeah.net , E-mail：hongyonggdcc@yeah.net

陈强,E-mail:cq_c120m3@163.com , E-mail：cq_c120m3@163.com

Abstract

Choosing $a, b$ as the matching parameters, we can sue the weight function method to obtain Hilbert-type series inequality

$\sum\limits_{m=1}^{\infty}\sum\limits_{n=1}^{\infty}K(m, n)a_{m}b_{n}\leq M(a, b)\|\tilde{a}\|_{p, \alpha(a, b)}\|\tilde{b}\|_{q, \beta(a, b)}.$

in the paper, the problem of how to choose

$a, b$

in order to make

$M(a, b)$

the best constant factor in inequality with quasi-homogeneous kernels are discussed, necessary and sufficient conditions are obtained for

$a, b$

to the best matching parameters, the formula for the best constant factor is obtained. Finally, their applications to solving operator morn are discussed.

Keywords： Hilbert-type series inequality ; Quasi-homogeneous kernel ; The best constant factor ; The best matching parameter ; Operator norm

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本文引用格式

洪勇, 陈强. 拟齐次核的Hilbert型级数不等式的最佳搭配参数条件及应用. 数学物理学报[J], 2022, 42(1): 26-34 doi:

Hong Yong, Chen Qiang. The Best Matching Parameters Conditions of Hilbert-Type Series Inequality with Quasi-Homogeneous Kernel and Applications. Acta Mathematica Scientia[J], 2022, 42(1): 26-34 doi:

1 引言与预备知识

设 $\frac{1}{p}+\frac{1}{q}=1\ (p>1)$ , $K(m, n)\geq0$ , 选择搭配参数 $a, b$ , 利用权函数方法可得下面形式的Hilbert型级数不等式

$\begin{equation} \sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty}K(m, n)a_{m}b_{n}\leq M(a, b)\|\tilde{a}\|_{p, \alpha(a, b)}\|\tilde{b}\|_{q, \beta(a, b)}, \end{equation}$

(1.1)

其中 $\tilde{a}=\{a_{m}\}\in l_{p}^{\alpha(a, b)}$ , $\tilde{b}=\{b_{n}\}\in l_{q}^{\beta(a, b)}$ , 而

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

一般地, 任意选取的搭配参数 $a, b$ 并不能使式(1.1)的常数因子 $M(a, b)$ 最佳, 针对不同类型的核 $K(m, n)$ , 只有精心选取适当的 $a, b$ , 才有可能得到具有最佳常数因子的Hilbert型不等式(1.1). 如何选取最佳的搭配参数来获得具有最佳常数因子的Hilbert型不等式一直是Hilbert型不等式研究的重要课题.

令 $T$ 为级数算子

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

根据Hilbert型不等式的基本理论, 式(1.1)等价于算子不等式

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

因此选取最佳搭配参数, 实质上就是讨论算子 $T: l_{p}^{\alpha(a, b)}\rightarrow l_{p}^{\beta(a, b)(1-p)}$ 的算子范数.

国内外的学者凭借自身的经验和实分析技巧, 针对各种各样的核, 已获得了许多具有最佳常数因子的Hilbert型不等式^[1-11]. 显然, 寻求最佳搭配参数 $a, b$ 所满足的条件, 探究其基本特征是一件很有意义的工作.

设 $\lambda_{1}\lambda_{2}>0$ , $G(u, v)$ 是 $\lambda$ 阶齐次函数, 则称 $K(m, n)=G(m^{\lambda_{1}}, n^{\lambda_{2}})$ 为拟齐次函数. 显然对于 $t>0$ , 有

$K(tm, n)=t^{\lambda_{1}\lambda}K(m, t^{-\frac{\lambda_{1}}{\lambda_{2}}}n), \ \ \ K(m, tn)=t^{\lambda_{2}\lambda}K(t^{-\frac{\lambda_{2}}{\lambda_{1}}}m, n).$

2012年, 文献[12]针对 $\lambda=1$ 情况下的拟齐次核获得了最佳搭配参数 $a, b$ 的充分条件是 $\lambda_{1}bp+\lambda_{2}aq=\lambda_{1}\lambda_{2}+\lambda_{1}+\lambda_{2}$ , 但没有讨论此条件是否必要, 而且还限制了 $a>0$ , $b>0$ . 2016年, 文献[13]针对齐次核, 在更宽泛的条件下证明了 $a, b$ 为最佳搭配参数的充分必要条件是 $aq+bp=\lambda+2$ , 结果比较完美. 本文将在文献[12, 13]的基础上, 讨论一般情况下针对拟齐次核的Hilbert级数不等式的最佳搭配参数, 得到了 $a, b$ 为最佳搭配参数的充分必要条件.

引理1.1 设 $\frac{1}{p}+\frac{1}{q}=1\, (p>1)$ , $a, b\in \mathbb{R}$ , $\lambda_{1}\lambda_{2}>0$ , $G(u, v)$ 是 $\lambda$ 阶齐次非负可测函数, $K(m, n)=G(m^{\lambda_{1}}, n^{\lambda_{2}})$ , $\frac{1}{\lambda_{1}}aq+\frac{1}{\lambda_{2}}bp=\lambda+\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}$ , $K(t, 1)t^{-aq}$ 及 $K(1, t)t^{-bp}$ 都在 $(0, +\infty)$ 上递减, 记

$W_{1}(b, p)=\int_{0}^{+\infty}G(1, t^{\lambda_{2}})t^{-bp}\, {\rm d}t, \ \ W_{2}(a, q)=\int_{0}^{+\infty}G(t^{\lambda_{1}}, 1)t^{-aq}\, {\rm d}t,$

则 $\lambda_{2}W_{1}(b, p)=\lambda_{1}W_{2}(a, q)$ , 且

$\omega_{1}(b, p, m)=\sum\limits_{n=1}^{\infty}G(m^{\lambda_{1}}, n^{\lambda_{2}})n^{-bp}\leq m^{\lambda_{1}(\lambda-\frac{1}{\lambda_{2}}bp+\frac{1}{\lambda_{2}})}W_{1}(b, p),$

$\omega_{2}(a, q, n)=\sum\limits_{m=1}^{\infty}G(m^{\lambda_{1}}, n^{\lambda_{2}})m^{-aq}\leq n^{\lambda_{2}(\lambda-\frac{1}{\lambda_{1}}aq+\frac{1}{\lambda_{1}})}W_{2}(a, q).$

证作积分变换 $t^{-\frac{\lambda_{2}}{\lambda_{1}}}=u$ 并根据 $\frac{1}{\lambda_{1}}aq+\frac{1}{\lambda_{2}}bp=\lambda+\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}$ , 有

$\begin{eqnarray*} W_{1}(b, p)&=&\int_{0}^{+\infty}K(1, t)t^{-bp}\, {\rm d}t=\int_{0}^{+\infty} K(t^{-\frac{\lambda_{2}}{\lambda_{1}}}, 1)t^{\lambda_{2}\lambda-bp}\, {\rm d}t\\ & =&\frac{\lambda_{1}}{\lambda_{2}}\int_{0}^{+\infty}K(u, 1) u^{-\lambda_{1}\lambda+\frac{\lambda_{1}}{\lambda_{2}}bp-\frac{\lambda_{1}}{\lambda_{2}}-1}\, {\rm d}u =\frac{\lambda_{1}}{\lambda_{2}}\int_{0}^{+\infty}K(u, 1)u^{-aq}\, {\rm d}u\\ & =&\frac{\lambda_{1}}{\lambda_{2}}W_{2}(a, q), \end{eqnarray*}$

故 $\lambda_{2}W_{1}(b, p)=\lambda_{1}W_{2}(a, q)$ .

根据 $K(1, t)t^{-bp}$ 在 $(0, +\infty)$ 上递减, 有

$\begin{eqnarray*} \omega_{1}(b, p, m)&=&\sum\limits_{n=1}^{\infty}K(m, n)n^{-bp}=m^{\lambda_{1}\lambda}\sum\limits_{n=1}^{\infty}K(1, m^{-\frac{\lambda_{1}}{\lambda_{2}}}n)n^{-bp}\\ & \leq& m^{\lambda_{1}\lambda-\frac{\lambda_{1}}{\lambda_{2}}bp}\int_{0}^{+\infty} K(1, m^{-\frac{\lambda_{1}}{\lambda_{2}}}u)(m^{-\frac{\lambda_{1}}{\lambda_{2}}}u)^{-bp}{\rm d}u\\ & =&m^{\lambda_{1}\lambda-\frac{\lambda_{1}}{\lambda_{2}}bp+\frac{\lambda_{1}}{\lambda_{2}}}\int_{0}^{+\infty} K(1, t)t^{-bp}{\rm d}t=m^{\lambda_{1}(\lambda-\frac{1}{\lambda_{2}}bp+\frac{1}{\lambda_{2}})}W_{1}(b, p). \end{eqnarray*}$

同理, 根据 $K(t, 1)t^{-aq}$ 在 $(0, +\infty)$ 上递减, 可证 $\omega_{2}(a, q, n)\leq n^{\lambda_{2}(\lambda-\frac{1}{\lambda_{1}}aq+\frac{1}{\lambda_{1}})}W_{2}(a, q)$ .

2 最佳搭配参数的充要条件

定理2.1 设 $\frac{1}{p}+\frac{1}{q}=1\, (p>1)$ , $a, b\in \mathbb{R}$ , $\lambda_{1}\lambda_{2}>0$ , $G(u, v)$ 是 $\lambda$ 阶齐次非负可测函数, $\frac{1}{\lambda_{1}}aq+\frac{1}{\lambda_{2}}bp-(\lambda+\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}})=c$ , $G(1, t^{\lambda_{2}})t^{-bp}, G(t^{\lambda_{1}}, 1)t^{-aq}, G(1, t^{\lambda_{2}})t^{-bp+\frac{\lambda_{2}c}{q}}$ 及 $G(t^{\lambda_{1}}, 1)t^{-aq+\frac{\lambda_{1}c}{p}}$ 都在 $(0, +\infty)$ 上递减, 且

$W_{1}(b, p)=\int_{0}^{+\infty}G(1, t^{\lambda_{2}})t^{-bp}\, {\rm d}t, \ \ W_{2}(a, q)=\int_{0}^{+\infty}G(t^{\lambda_{1}}, 1)t^{-aq}\, {\rm d}t$

收敛, 那么

${\rm (i)}$ 记 $\alpha=\lambda_{1}[\lambda+\frac{1}{\lambda_{2}}+p(\frac{a}{\lambda_{1}}-\frac{b}{\lambda_{2}})]$ , $\beta=\lambda_{2}[\lambda+\frac{1}{\lambda_{1}}+q(\frac{b}{\lambda_{2}}-\frac{a}{\lambda_{1}})]$ , 有

$\begin{equation} \sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty}G(m^{\lambda_{1}}, n^{\lambda_{2}})a_{m}b_{n} \leq W_{1}^{\frac{1}{p}}(b, p)W_{2}^{\frac{1}{q}}(a, q)\|\tilde{a}\|_{p, \alpha}\|\tilde{b}\|_{q, \beta}, \end{equation}$

(2.1)

其中 $\tilde{a}=\{a_{m}\}\in l_{p}^{\alpha}$ , $\tilde{b}=\{b_{n}\}\in l_{}^{\beta}$ .

${\rm (ii)}$ 当且仅当 $c=0$ , 即 $\frac{1}{\lambda_{1}}aq+\frac{1}{\lambda_{2}}bp=\lambda+\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}$ 时, (2.1)式中常数因子 $W_{1}^{\frac{1}{p}}(b, p)W_{2}^{\frac{1}{q}}(a, q)$ 是最佳的, 且当 $c=0$ 时, (2.1)式化为

$\begin{equation} \sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty}G(m^{\lambda_{1}}, n^{\lambda_{2}})a_{m}b_{n} \leq \frac{W_{0}}{|\lambda_{1}|^{\frac{1}{q}}|\lambda_{2}|^{\frac{1}{p}}}\|\tilde{a}\|_{p, apq-1}\|\tilde{b}\|_{q, bpq-1}, \end{equation}$

(2.2)

其中 $W_{0}=|\lambda_{1}|W_{2}(a, q)=|\lambda_{2}|W_{1}(b, p)$ .

证 (i) 选取 $a, b$ 为搭配参数, 根据Hölder不等式及引理1.1, 有

$\begin{eqnarray*} &&\sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty}G(m^{\lambda_{1}}, n^{\lambda_{2}})a_{m}b_{n} \leq \sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty}\left(\frac{m^{a}}{n^{b}}|a_{m}|\right)\left(\frac{n^{b}}{m^{a}}|b_{n}|\right)G(m^{\lambda_{1}}, n^{\lambda_{2}})\\ & \leq&\left(\sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty}\left(\frac{m^{ap}}{n^{bp}}|a_{m}|^{p}\right)G(m^{\lambda_{1}}, n^{\lambda_{2}})\right)^{\frac{1}{p}} \left(\sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty}\left(\frac{n^{bq}}{m^{aq}}|b_{n}|^{q}\right)G(m^{\lambda_{1}}, n^{\lambda_{2}})\right)^{\frac{1}{q}}\\ & =&\left(\sum\limits_{m=1}^{\infty}m^{ap}|a_{m}|^{p}\omega_{1}(b, p, m)\right)^{\frac{1}{p}} \left(\sum\limits_{n=1}^{\infty}n^{bq}|b_{n}|^{q}\omega_{2}(a, q, n)\right)^{\frac{1}{p}}\\ & \leq& W_{1}^{\frac{1}{p}}(b, p)W_{2}^{\frac{1}{q}}(a, q) \left(\sum\limits_{m=1}^{\infty}m^{ap+\lambda_{1}(\lambda-\frac{1}{\lambda_{2}}bp+\frac{1}{\lambda_{2}})}|a_{m}|^{p}\right)^{\frac{1}{p}} \left(\sum\limits_{n=1}^{\infty}n^{bq+\lambda_{2}(\lambda-\frac{1}{\lambda_{1}}aq+\frac{1}{\lambda_{1}})}|b_{n}|^{q}\right)^{\frac{1}{q}}\\ & =&W_{1}^{\frac{1}{p}}(b, p)W_{2}^{\frac{1}{q}}(a, q)\|\tilde{a}\|_{p, \alpha}\|\tilde{b}\|_{q, \beta}, \end{eqnarray*}$

故(2.2)式成立.

(ii) 充分性: 设 $c=0$ , 则 $\alpha=apq-1$ , $\beta=bpq-1$ . 根据引理1.1, 有 $\lambda_{2}W_{1}(b, p)=\lambda_{1}W_{2}(a, q)$ , 于是(2.1)式化为(2.2)式.

若(2.2)式的常数因子不是最佳的, 则存在常数 $M_{0}<\frac{W_{0}}{|\lambda_{1}|^{\frac{1}{q}}|\lambda_{2}|^{\frac{1}{p}}}$ , 使

$\begin{eqnarray*} \sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty}G(m^{\lambda_{1}}, n^{\lambda_{2}})a_{m}b_{n} \leq M_{0}\|\tilde{a}\|_{p, apq-1}\|\tilde{b}\|_{q, bpq-1}. \end{eqnarray*}$

取充分小的 $\varepsilon>0$ 的及足够大的自然数 $N$ , 令 $b_{n}=n^{(-bpq-|\lambda_{2}|\varepsilon)/q}\ (n=1, 2, \cdots)$ ,

$a_{m}= \left\{\begin{array}{ll} 0, & m=1, 2, \cdots, N-1, \\ m^{\frac{-apq-|\lambda_{1}|\varepsilon}{p}}, & m=N, N+1, \cdots \end{array}\right.$

则有

$\begin{eqnarray*} \|\tilde{a}\|_{p, apq-1}\|\tilde{b}\|_{q, bpq-1} &=&\left(\sum\limits_{m=N}^{\infty}m^{-1-|\lambda_{1}|\varepsilon}\right)^{\frac{1}{p}} \left(\sum\limits_{n=1}^{\infty}n^{-1-|\lambda_{2}|\varepsilon}\right)^{\frac{1}{q}}\\ &\leq& \left(1+\int_{1}^{+\infty}t^{-1-|\lambda_{1}|\varepsilon}{\rm d}t\right)^{\frac{1}{p}} \left(1+\int_{1}^{+\infty}t^{-1-|\lambda_{2}|\varepsilon}{\rm d}t\right)^{\frac{1}{q}}\\ &=&\frac{1}{\varepsilon|\lambda_{1}|^{\frac{1}{p}}|\lambda_{2}|^{\frac{1}{q}}} \left(1+|\lambda_{1}|\varepsilon\right)^{\frac{1}{p}}\left(1+|\lambda_{2}|\varepsilon\right)^{\frac{1}{q}}. \end{eqnarray*}$

记 $K(m, n)=G(m^{\lambda_{1}}, n^{\lambda_{2}})$ , 根据 $K(1, t)t^{-bp}$ 在 $(0, +\infty)$ 上的递减性, 有

$\begin{eqnarray*} && \sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty}G(m^{\lambda_{1}}, n^{\lambda_{2}})a_{m}b_{n} =\sum\limits_{m=N}^{\infty}m^{-aq-\frac{|\lambda_{1}|\varepsilon}{p}}\left(\sum\limits_{n=1}^{\infty}K(m, n)n^{-bp-\frac{|\lambda_{2}|\varepsilon}{q}}\right)\\ &=&\sum\limits_{m=N}^{\infty}m^{\lambda_{1}\lambda-aq-\frac{|\lambda_{1}|\varepsilon}{p}} \left(\sum\limits_{n=1}^{\infty}K(1, m^{-\frac{\lambda_{1}}{\lambda_{2}}}n)n^{-bp-\frac{|\lambda_{2}|\varepsilon}{q}}\right)\\ &=&\sum\limits_{m=N}^{\infty}m^{\lambda_{1}\lambda-aq-\frac{|\lambda_{1}|\varepsilon}{p}-\frac{\lambda_{1}}{\lambda_{2}}(bp+\frac{|\lambda_{2}|\varepsilon}{q})} \left(\sum\limits_{n=1}^{\infty}K(1, m^{-\frac{\lambda_{1}}{\lambda_{2}}}n) (m^{-\frac{\lambda_{1}}{\lambda_{2}}}n)^{-bp-\frac{|\lambda_{2}|\varepsilon}{q}}\right)\\ &\geq& \sum\limits_{m=N}^{\infty}m^{\lambda_{1}\lambda-aq-\frac{|\lambda_{1}|\varepsilon}{p}-\frac{\lambda_{1}}{\lambda_{2}}(bp+\frac{|\lambda_{2}|\varepsilon}{q})} \left(\int_{1}^{+\infty}K(1, m^{-\frac{\lambda_{1}}{\lambda_{2}}}u) (m^{-\frac{\lambda_{1}}{\lambda_{2}}}u)^{-bp-\frac{|\lambda_{2}|\varepsilon}{q}}{\rm d}u\right)\\ &=&\sum\limits_{m=N}^{\infty}m^{\lambda_{1}\lambda-aq-\frac{|\lambda_{1}|\varepsilon}{p}- \frac{\lambda_{1}}{\lambda_{2}}(bp+\frac{|\lambda_{2}|\varepsilon}{q})+\frac{\lambda_{1}}{\lambda_{2}}} \left(\int_{m^{-\frac{\lambda_{1}}{\lambda_{2}}}}^{+\infty}K(1, t)t^{-bp-\frac{|\lambda_{2}|\varepsilon}{q}}{\rm d}t\right)\\ &=&\sum\limits_{m=N}^{\infty}m^{-1-|\lambda_{1}|\varepsilon} \left(\int_{m^{-\frac{\lambda_{1}}{\lambda_{2}}}}^{+\infty}G(1, t^{\lambda_{2}})t^{-bp-\frac{|\lambda_{2}|\varepsilon}{q}}{\rm d}t\right)\\ &\geq& \int_{N}^{\infty}t^{-1-|\lambda_{1}|\varepsilon}{\rm d}t \int_{N^{-\frac{\lambda_{1}}{\lambda_{2}}}}^{+\infty}G(1, t^{\lambda_{2}})t^{-bp-\frac{|\lambda_{2}|\varepsilon}{q}}{\rm d}t\\ &=&\frac{1}{|\lambda_{1}|\varepsilon}N^{-|\lambda_{1}|\varepsilon} \int_{N^{-\frac{\lambda_{1}}{\lambda_{2}}}}^{+\infty}G(1, t^{\lambda_{2}})t^{-bp-\frac{|\lambda_{2}|\varepsilon}{q}}{\rm d}t. \end{eqnarray*}$

于是可得

$\begin{eqnarray*} \frac{1}{|\lambda_{1}|}N^{-|\lambda_{1}|\varepsilon} \int_{N^{-\frac{\lambda_{1}}{\lambda_{2}}}}^{+\infty}G(1, t^{\lambda_{2}})t^{-bp-\frac{|\lambda_{2}|\varepsilon}{q}}{\rm d}t \leq \frac{M_{0}}{|\lambda_{1}|^{\frac{1}{p}}|\lambda_{2}|^{\frac{1}{q}}} \left(1+|\lambda_{1}|\varepsilon\right)^{\frac{1}{p}}\left(1+|\lambda_{2}|\varepsilon\right)^{\frac{1}{q}}. \end{eqnarray*}$

先令 $\varepsilon\rightarrow0^{+}$ , 然后再令 $N\rightarrow +\infty$ , 得

$\begin{eqnarray*} \frac{1}{|\lambda_{1}|}\int_{0}^{+\infty}G(1, t^{\lambda_{2}})t^{-bp}{\rm d}t \leq \frac{M_{0}}{|\lambda_{1}|^{\frac{1}{p}}|\lambda_{2}|^{\frac{1}{q}}}. \end{eqnarray*}$

由此得到 $\frac{W_{0}}{|\lambda_{1}|^{\frac{1}{q}}|\lambda_{2}|^{\frac{1}{p}}}\leq M_{0}$ , 这与 $M_{0}<\frac{W_{0}}{|\lambda_{1}|^{\frac{1}{q}}|\lambda_{2}|^{\frac{1}{p}}}$ 相矛盾, 故式(2.2)中的常数因子是最佳的.

必要性: 设(2.1)式的常数因子 $W_{1}^{\frac{1}{p}}(b, p)W_{2}^{\frac{1}{q}}(a, q)$ 是最佳的. 记 $a_{1}=a-\frac{\lambda_{1}c}{pq}$ , $b_{1}=b-\frac{\lambda_{2}c}{pq}$ , 则

$\frac{1}{\lambda_{1}}a_{1}q+\frac{1}{\lambda_{2}}b_{1}p=\frac{1}{\lambda_{1}}q-\frac{c}{p}+\frac{1}{\lambda_{2}}bp-\frac{c}{q}=\lambda+ \frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}},$

从而

$\alpha=\lambda_{1}\left[\lambda+\frac{1}{\lambda_{2}}+p(\frac{a}{\lambda_{1}}-\frac{b}{\lambda_{2}})\right] =\lambda_{1}\left[\lambda+\frac{1}{\lambda_{2}}+p(\frac{a_{1}}{\lambda_{1}}-\frac{b_{1}}{\lambda_{2}})\right]=a_{1}pq-1,$

$\beta=\lambda_{2}\left[\lambda+\frac{1}{\lambda_{1}}+q(\frac{b}{\lambda_{2}}-\frac{a}{\lambda_{1}})\right] =\lambda_{2}\left[\lambda+\frac{1}{\lambda_{1}}+q(\frac{b_{1}}{\lambda_{2}}-\frac{a_{1}}{\lambda_{1}})\right]=b_{1}pq-1,$

且计算可得

$W_{2}(a, q)=\int_{0}^{+\infty}G(t^{\lambda_{1}}, 1)t^{-aq}{\rm d}t =\frac{\lambda_{2}}{\lambda_{1}}\int_{0}^{+\infty}G(1, t^{\lambda_{2}})t^{-bp+\lambda_{2}c}{\rm d}t.$

于是(2.1)式可等价于

$\begin{eqnarray} \sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty}G(m^{\lambda_{1}}, n^{\lambda_{2}})a_{m}b_{n} & \leq& W_{1}^{\frac{1}{p}}(b, p)\left(\frac{\lambda_{2}}{\lambda_{1}}\int_{0}^{+\infty}G(1, t^{\lambda_{2}}) t^{-bp+\lambda_{2}c}{\rm d}t\right)^{\frac{1}{q}}\\ &&\times\|\tilde{a}\|_{p, a_{1}pq-1}\|\tilde{b}\|_{q, b_{1}pq-1}. \end{eqnarray}$

(2.3)

根据假设, (2.3)式的最佳常数因子为

$W_{1}^{\frac{1}{p}}(b, p)\left(\frac{\lambda_{2}}{\lambda_{1}}\int_{0}^{+\infty}G(1, t^{\lambda_{2}})t^{-bp+\lambda_{2}c}{\rm d}t\right)^{\frac{1}{q}}.$

又因为 $\frac{1}{\lambda_{1}}a_{1}q+\frac{1}{\lambda_{2}}b_{1}p=\lambda+\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}$ , 且 $G(1, t^{\lambda_{2}})t^{-b_{1}p}=G(1, t^{\lambda_{2}})t^{-bp+\frac{\lambda_{2}c}{q}}$ 及 $G(t^{\lambda_{1}}, 1)t^{-a_{1}q}=G(t^{\lambda_{1}}, 1)t^{-aq+\frac{\lambda_{1}c}{p}}$ 都在 $(0, +\infty)$ 上递减, 根据前面充分性的证明, (2.3)式的最佳常数因子应为

$\begin{eqnarray*} \frac{\bar{W}_{0}}{|\lambda_{1}|^{\frac{1}{q}}|\lambda_{2}|^{\frac{1}{p}}} =\left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{\frac{1}{q}}\int_{0}^{+\infty}G(1, t^{\lambda_{2}})t^{-b_{1}p}{\rm d}t =\left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{\frac{1}{q}}\int_{0}^{+\infty}G(1, t^{\lambda_{2}})t^{-bp+\frac{\lambda_{2}c}{q}}{\rm d}t. \end{eqnarray*}$

于是有

$\begin{eqnarray} \int_{0}^{+\infty}G(1, t^{\lambda_{2}})t^{-bp+\frac{\lambda_{2}c}{q}}{\rm d}t =W_{1}^{\frac{1}{p}}(b, p)\left(\int_{0}^{+\infty}G(1, t^{\lambda_{2}})t^{-bp+\lambda_{2}c}{\rm d}t\right)^{\frac{1}{q}}. \end{eqnarray}$

(2.4)

针对函数 $1$ 和 $t^{\frac{\lambda_{2}c}{q}}$ , 利用Hölder不等式, 有

$\begin{eqnarray} &&\int_{0}^{+\infty}G(1, t^{\lambda_{2}})t^{-bp+\frac{\lambda_{2}c}{q}}{\rm d}t =\int_{0}^{+\infty}1\cdot t^{\frac{\lambda_{2}c}{q}}G(1, t^{\lambda_{2}})t^{-bp}{\rm d}t\\ &\leq &\left(\int_{0}^{+\infty}1^{p}G(1, t^{\lambda_{2}})t^{-bp}{\rm d}t\right)^{\frac{1}{p}} \left(\int_{0}^{+\infty}t^{\lambda_{2}c}G(1, t^{\lambda_{2}})t^{-bp}{\rm d}t\right)^{\frac{1}{q}}\\ &=&W_{1}^{\frac{1}{p}}(b, p)\left(\int_{0}^{+\infty}G(1, t^{\lambda_{2}})t^{-bp+\lambda_{2}c}{\rm d}t\right)^{\frac{1}{q}}. \end{eqnarray}$

(2.5)

由(2.4)式可知(2.5)式取等号, 再根据Hölder不等式取等号的条件, 可得 $t^{\lambda_{2}c}=$ 常数, 故 $c=0$ . 定理2.1证毕.

注2.1 记 $\Delta=\frac{1}{\lambda_{1}}aq+\frac{1}{\lambda_{2}}bp-(\lambda+\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}})$ , 则 $\Delta=0$ 等价于 $a, b$ 是最佳搭配参数. 称 $\Delta$ 是 $a, b$ 为最佳搭配参数的判别式.

例2.1 设 $\frac{1}{p}+\frac{1}{q}=1\, (p>1)$ , $\frac{1}{r}+\frac{1}{s}=1\, (r>1)$ , $\lambda_{1}>0$ , $\lambda_{2}>0$ , $0<\lambda<\min\{r(\frac{1}{\lambda_{1}}-1)+1, s(\frac{1}{\lambda_{2}}-1)+1\}$ , $\alpha=\lambda_{1}p(\frac{1}{\lambda_{1}}-\frac{\lambda-1}{r})-1$ , $\beta=\lambda_{2}q(\frac{1}{\lambda_{2}}-\frac{\lambda-1}{s})-1$ , 试讨论取怎样的搭配参数, 可以由权函数方法得到具有最佳常数因子的Hilbert型级数不等式

$\begin{eqnarray} \sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty}\frac{\min\{m^{\lambda_{1}}, n^{\lambda_{2}}\}}{(m^{\lambda_{1}}+n^{\lambda_{2}})^{\lambda}}a_{m}b_{n} \leq M_{0}\|\tilde{a}\|_{p, \alpha}\|\tilde{b}\|_{q, \beta}, \end{eqnarray}$

(2.6)

并求出最佳常数因子 $M_{0}$ , 其中 $\tilde{a}=\{a_{m}\}\in l_{p}^{\alpha}$ , $\tilde{b}=\{b_{n}\}\in l_{q}^{\beta}$ .

记 $G(m^{\lambda_{1}}, n^{\lambda_{2}})=\frac{\min\{m^{\lambda_{1}}, n^{\lambda_{2}}\}}{(m^{\lambda_{1}}+n^{\lambda_{2}})^{\lambda}}$ , 则 $G(u, v)$ 是 $1-\lambda$ 阶齐次非负函数. 令 $apq-1=\alpha$ , $bpq-1=\beta$ , 可得 $a=\frac{\lambda_{1}}{q}(\frac{1}{\lambda_{1}}-\frac{\lambda-1}{r})$ , $b=\frac{\lambda_{2}}{p}(\frac{1}{\lambda_{2}}-\frac{\lambda-1}{s})$ . 经计算可得知 $\Delta=\frac{1}{\lambda_{1}}aq+\frac{1}{\lambda_{2}}bp-(1-\lambda+\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}})=0$ , 故 $a, b$ 是式(2.6)的最佳搭配参数.

又根据已知条件, 易知 $G(1, t^{\lambda_{2}})t^{-bp}$ 及 $G(t^{\lambda_{1}}, 1)t^{-aq}$ 都在 $(0, +\infty)$ 上递减, 且

$\begin{eqnarray*} W_{0}&=&|\lambda_{2}|W_{1}(b, p)=\lambda_{2}\int_{0}^{+\infty}G(1, t^{\lambda_{2}})t^{-bp}{\rm d}t\\ &=&\int_{0}^{+\infty}G(1, u)u^{-\frac{1}{\lambda_{1}}bp+\frac{1}{\lambda_{2}}-1}{\rm d}u =\int_{0}^{+\infty}\frac{\min\{1, u\}}{(1+u)^{\lambda}}u^{\frac{\lambda-1}{s}-1}{\rm d}u\\ &=&\int_{0}^{1}\frac{1}{(1+t)^{\lambda}}\left(t^{\frac{1}{s}(\lambda-1)}+t^{\frac{1}{r}(\lambda-1)}\right){\rm d}t<+\infty. \end{eqnarray*}$

根据定理1.1, 取搭配参数 $a=\frac{\lambda_{1}}{q}(\frac{1}{\lambda_{1}}-\frac{\lambda-1}{r})$ , $b=\frac{\lambda_{2}}{p}(\frac{1}{\lambda_{2}}-\frac{\lambda-1}{s})$ . 可得最佳Hilbert型不等式(2.6), 且它的最佳常数因子为

$M_{0}=\frac{1}{\lambda_{1}^{\frac{1}{q}}\lambda_{2}^{\frac{1}{p}}} \int_{0}^{1}\frac{1}{(1+t)^{\lambda}}\left(t^{\frac{1}{s}(\lambda-1)}+t^{\frac{1}{r}(\lambda-1)}\right){\rm d}t.$

3 应用

根据Hilbert型级数不等式与相应级数算子不等式的关系, 由定理2.1可得到下列定理.

定理3.1 设 $\frac{1}{p}+\frac{1}{q}=1\, (p>1)$ , $a, b\in \mathbb{R}$ , $\lambda_{1}\lambda_{2}>0$ , $G(u, v)$ 是 $\lambda$ 阶齐次非负可测函数, $\frac{1}{\lambda_{1}}aq+\frac{1}{\lambda_{2}}bp=\lambda+\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}$ , $G(t^{\lambda_{1}}, 1)t^{-aq}, G(1, t^{\lambda_{2}})t^{-bp}$ 在 $(0, +\infty)$ 上递减, $\alpha=apq-1$ , $\beta=bpq-1$ , 且

$W_{1}(b, p)=\int_{0}^{+\infty}G(1, t^{\lambda_{2}})t^{-bp}{\rm d}t$

收敛, 则算子

$T(\tilde{a})_{n}=\sum\limits_{m=1}^{\infty}G(m^{\lambda_{1}}, n^{\lambda_{2}})a_{m}, \ \ \tilde{a}=\{a_{m}\}\in l_{p}^{\alpha}$

是 $l_{p}^{\alpha}$ 到 $l_{p}^{\beta(1-p)}$ 的有界算子, 且 $T$ 的算子范数为 $\|T\|=(\frac{\lambda_{2}}{\lambda_{1}})^{\frac{1}{q}}W_{1}(b, p)$ .

例3.1 设 $\frac{1}{p}+\frac{1}{q}=1\, (p>1)$ , $\frac{1}{r}+\frac{1}{s}=1\, (r>1)$ , $\lambda>0$ , $\lambda_{1}>0$ , $\lambda_{2}>0$ , $\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}\geq \max\{\frac{s}{r}\lambda, \frac{r}{s}\lambda\}$ , $-\frac{\lambda}{s}<\frac{1}{\lambda_{1}r}-\frac{1}{\lambda_{2}s}<\frac{\lambda}{r}$ , $\alpha=\frac{p}{r}(1+\frac{\lambda_{1}}{\lambda_{2}})-\frac{p}{s}\lambda_{1}\lambda-1$ , $\beta=\frac{q}{s}(1+\frac{\lambda_{2}}{\lambda_{1}})-\frac{q}{r}\lambda_{2}\lambda-1$ , 则级数算子

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

是 $l_{p}^{\alpha}$ 到 $l_{p}^{\beta(1-p)}$ 的有界算子, 且 $T$ 的算子范数为

$\|T\|=\frac{1}{\lambda_{1}^{\frac{1}{q}}\lambda_{2}^{\frac{1}{p}}} B\left(\frac{\lambda}{r}-\frac{1}{\lambda_{1}r}+\frac{1}{\lambda_{2}s}, \frac{\lambda}{s}+\frac{1}{\lambda_{1}r}-\frac{1}{\lambda_{2}s}\right).$

证记 $G(u, v)=\frac{1}{(u+v)^{\lambda}}$ , 则 $G(u, v)$ 是 $-\lambda$ 阶齐次非负函数. 令

$apq-1=\alpha=\frac{p}{r}(1+\frac{\lambda_{1}}{\lambda_{2}})-\frac{p}{s}\lambda_{1}\lambda-1, \ \ bpq-1=\beta=\frac{q}{s}(1+\frac{\lambda_{2}}{\lambda_{1}})-\frac{q}{r}\lambda_{2}\lambda-1,$

则可得 $a=\frac{1}{rq}(1+\frac{\lambda_{1}}{\lambda_{2}})-\frac{1}{sp}\lambda_{1}\lambda$ , $b=\frac{1}{sp}(1+\frac{\lambda_{2}}{\lambda_{1}})-\frac{1}{rp}\lambda_{2}\lambda$ , 且有

$\frac{1}{\lambda_{1}}aq+\frac{1}{\lambda_{2}}bp=-\lambda+\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}},$

故 $a, b$ 是最佳搭配参数.

由 $\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}\geq \max\{\frac{s}{r}\lambda, \frac{r}{s}\lambda\}$ , 可得 $\frac{1}{r}\lambda_{2}\lambda-\frac{1}{s}(1+\frac{\lambda_{2}}{\lambda_{1}})\leq0$ , $\frac{1}{s}\lambda_{1}\lambda-\frac{1}{r}(1+\frac{\lambda_{1}}{\lambda_{2}})\leq0$ , 故

$G(1, t^{\lambda_{2}})t^{-bp}=\frac{1}{(1+t^{\lambda_{2}})^{\lambda}} t^{\frac{1}{r}\lambda_{2}\lambda-\frac{1}{s}(1+\frac{\lambda_{2}}{\lambda_{1}})},$

$G(t^{\lambda_{1}}, 1)t^{-aq}=\frac{1}{(t^{\lambda_{1}}+1)^{\lambda}} t^{\frac{1}{s}\lambda_{1}\lambda-\frac{1}{r}(1+\frac{\lambda_{1}}{\lambda_{2}})}$

都在 $(0, +\infty)$ 上递减.

由 $-\frac{\lambda}{s}<\frac{1}{\lambda_{1}r}-\frac{1}{\lambda_{2}s}<\frac{\lambda}{r}$ , 可得 $\frac{\lambda}{r}-\frac{1}{\lambda_{1}r}+\frac{1}{\lambda_{2}s}>0$ , $\frac{\lambda}{s}+\frac{1}{\lambda_{1}r}-\frac{1}{\lambda_{2}s}>0$ . 于是有

$\begin{eqnarray*} W_{1}(b, p)&=&\int_{0}^{+\infty}G(1, t^{\lambda_{2}})t^{-bp}{\rm d}t =\int_{0}^{+\infty}\frac{1}{(1+t^{\lambda_{2}})^{\lambda}}t^{-bp}{\rm d}t\\ &=&\frac{1}{\lambda_{2}}\int_{0}^{+\infty}\frac{1}{(1+u)^{\lambda}}u^{-\frac{1}{\lambda_{2}}bp+\frac{1}{\lambda_{2}}-1}{\rm d}u =\frac{1}{\lambda_{2}}\int_{0}^{+\infty}\frac{1}{(1+u)^{\lambda}} u^{\frac{\lambda}{r}-\frac{1}{\lambda_{1}r}+\frac{1}{\lambda_{2}s}-1}{\rm d}u\\ &=&\frac{1}{\lambda_{2}}B\left(\frac{\lambda}{r}-\frac{1}{\lambda_{1}r}+\frac{1}{\lambda_{2}s}, \lambda-\frac{\lambda}{r}+\frac{1}{\lambda_{1}r}-\frac{1}{\lambda_{2}s}\right)\\ &=&\frac{1}{\lambda_{2}}B\left(\frac{\lambda}{r}-\frac{1}{\lambda_{1}r}+\frac{1}{\lambda_{2}s}, \frac{\lambda}{s}+\frac{1}{\lambda_{1}r}-\frac{1}{\lambda_{2}s}\right)<+\infty. \end{eqnarray*}$

根据定理3.1, $T$ 是 $l_{p}^{\alpha}$ 到 $l_{p}^{\beta(1-p)}$ 的有界算子, 且 $T$ 的算子范数为

$\|T\|=(\frac{\lambda_{2}}{\lambda_{1}})^{\frac{1}{q}}W_{1}(b, p)=\frac{1}{\lambda_{1}^{\frac{1}{q}}\lambda_{2}^{\frac{1}{p}}} B\left(\frac{\lambda}{r}-\frac{1}{\lambda_{1}r}+\frac{1}{\lambda_{2}s}, \frac{\lambda}{s}+\frac{1}{\lambda_{1}r}-\frac{1}{\lambda_{2}s}\right).$

例题得证.

例3.2 设 $\frac{1}{p}+\frac{1}{q}=1\, (p>1)$ , $\frac{1}{r}+\frac{1}{s}=1\, (r>1)$ , $\lambda_{1}>0$ , $\lambda_{2}>0$ , $0<\lambda<\frac{1}{\lambda_{1}s}+\frac{1}{\lambda_{2}r}$ , $0<\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}<s\lambda$ , $\alpha=p(\frac{1}{r}+\frac{1}{s}\frac{\lambda_{1}}{\lambda_{2}})-1$ , $\beta=q(\frac{1}{r}+\frac{1}{s}\frac{\lambda_{2}}{\lambda_{1}})-1$ , 则级数算子

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

是 $l_{p}^{\alpha}$ 到 $l_{p}^{\beta(1-p)}$ 的有界算子, 且 $T$ 的算子范数为

$\|T\|=\frac{1}{\lambda_{1}^{\frac{1}{q}}\lambda_{2}^{\frac{1}{p}}} B\left(\frac{1}{s}(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}), \lambda-\frac{1}{s}(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}})\right).$

证记 $G(u, v)=\frac{1}{(1+\frac{u}{v})^{\lambda}}$ , 则 $G(u, v)$ 是 $0$ 阶齐次非负函数. 令

$apq-1=\alpha=p(\frac{1}{r}+\frac{1}{s}\frac{\lambda_{1}}{\lambda_{2}})-1, \ \ bpq-1=\beta=q(\frac{1}{r}+\frac{1}{s}\frac{\lambda_{2}}{\lambda_{1}})-1,$

则可得 $a=\frac{1}{q}(\frac{1}{r}+\frac{1}{s}\frac{\lambda_{1}}{\lambda_{2}})$ , $b=\frac{1}{p}(\frac{1}{r}+\frac{1}{s}\frac{\lambda_{2}}{\lambda_{1}})$ , 且 $\frac{1}{\lambda_{1}}aq+\frac{1}{\lambda_{2}}bp=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}$ , 故 $a, b$ 是最佳搭配参数.

由 $0<\lambda<\frac{1}{\lambda_{1}s}+\frac{1}{\lambda_{2}r}$ , 可得 $\lambda_{2}\lambda-(\frac{1}{r}+\frac{1}{s}\frac{\lambda_{1}}{\lambda_{2}})\leq0$ , $-(\frac{1}{r}+\frac{1}{s}\frac{\lambda_{2}}{\lambda_{1}})\leq0$ , 故

$G(1, t^{\lambda_{2}})t^{-bp}=\frac{1}{(1+t^{\lambda_{2}})^{\lambda}} t^{\lambda_{2}\lambda-(\frac{1}{r}+\frac{1}{s}\frac{\lambda_{1}}{\lambda_{2}})},$

$G(t^{\lambda_{1}}, 1)t^{-aq}=\frac{1}{(1+t^{\lambda_{1}})^{\lambda}}t^{-(\frac{1}{r}+\frac{1}{s}\frac{\lambda_{2}}{\lambda_{1}})}$

都在 $(0, +\infty)$ 上递减.

由 $0<\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}<s\lambda$ , 有 $\frac{1}{s}(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}})>0$ , $\lambda-\frac{1}{s}(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}})>0$ , 故

$\begin{eqnarray*} W_{1}(b, p)&=&\int_{0}^{+\infty}G(1, t^{\lambda_{2}})t^{-bp}{\rm d}t =\int_{0}^{+\infty}\frac{1}{(1+t^{-\lambda_{2}})^{\lambda}}t^{-bp}{\rm d}t\\ &=&\frac{1}{\lambda_{2}}\int_{0}^{+\infty}\frac{1}{(1+u)^{\lambda}}u^{\frac{1}{\lambda_{2}}bp-\frac{1}{\lambda_{2}}-1}{\rm d}u =\frac{1}{\lambda_{2}}\int_{0}^{+\infty}\frac{1}{(1+u)^{\lambda}} u^{\frac{1}{s}(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}})-1}{\rm d}u\\ &=&\frac{1}{\lambda_{2}}B\left(\frac{1}{s}(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}), \lambda-\frac{1}{s}(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}})\right)<+\infty. \end{eqnarray*}$

根据定理3.1, $T$ 是 $l_{p}^{\alpha}$ 到 $l_{p}^{\beta(1-p)}$ 的有界算子, 且 $T$ 的算子范数为

$\|T\|=(\frac{\lambda_{2}}{\lambda_{1}})^{\frac{1}{q}}W_{1}(b, p)=\frac{1}{\lambda_{1}^{\frac{1}{q}}\lambda_{2}^{\frac{1}{p}}} B\left(\frac{1}{s}(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}), \lambda-\frac{1}{s}(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}})\right).$

例题得证.

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... 国内外的学者凭借自身的经验和实分析技巧, 针对各种各样的核, 已获得了许多具有最佳常数因子的Hilbert型不等式^[1-11]. 显然, 寻求最佳搭配参数

$a, b$

所满足的条件, 探究其基本特征是一件很有意义的工作. ...

关于零阶齐次核的Hardy-Hilbert型不等式

2013

关于零阶齐次核的Hardy-Hilbert型不等式

2013

含变量可转移函数核的Hilbert型级数不等式

2014

含变量可转移函数核的Hilbert型级数不等式

2014

On new extensions of Hilbert's integral inequality

1999

一个混合核Hilbert型积分不等式及其算子范数表达式

2020

一个混合核Hilbert型积分不等式及其算子范数表达式

2020

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2000

一个新的Hilbert型不等式及其推广

2005

一个新的Hilbert型不等式及其推广

2005

关于一个推广的Hardy-Hilbert积分不等式

2002

关于一个推广的Hardy-Hilbert积分不等式

2002

Hardy-Riesz拓广的Hilbert不等式的一个改进

1994

Hardy-Riesz拓广的Hilbert不等式的一个改进

1994

On a Hilbert-type inequality in the plane related to the extended Riemann zeta function

2019

On multidimensional version of the Hilbert-type inequality

2012

... 国内外的学者凭借自身的经验和实分析技巧, 针对各种各样的核, 已获得了许多具有最佳常数因子的Hilbert型不等式^[1-11]. 显然, 寻求最佳搭配参数

$a, b$

所满足的条件, 探究其基本特征是一件很有意义的工作. ...

准齐次核的Hardy-Hilbert型级数不等式

2012

... 2012年, 文献[12]针对

$\lambda=1$

情况下的拟齐次核获得了最佳搭配参数

$a, b$

的充分条件是

$\lambda_{1}bp+\lambda_{2}aq=\lambda_{1}\lambda_{2}+\lambda_{1}+\lambda_{2}$

, 但没有讨论此条件是否必要, 而且还限制了

$a>0$

$b>0$

. 2016年, 文献[13]针对齐次核, 在更宽泛的条件下证明了

$a, b$

为最佳搭配参数的充分必要条件是

$aq+bp=\lambda+2$

, 结果比较完美. 本文将在文献[12, 13]的基础上, 讨论一般情况下针对拟齐次核的Hilbert级数不等式的最佳搭配参数, 得到了

$a, b$

为最佳搭配参数的充分必要条件. ...

... , 结果比较完美. 本文将在文献[12, 13]的基础上, 讨论一般情况下针对拟齐次核的Hilbert级数不等式的最佳搭配参数, 得到了

$a, b$

为最佳搭配参数的充分必要条件. ...

准齐次核的Hardy-Hilbert型级数不等式

2012

... 2012年, 文献[12]针对

$\lambda=1$

情况下的拟齐次核获得了最佳搭配参数

$a, b$

的充分条件是

$\lambda_{1}bp+\lambda_{2}aq=\lambda_{1}\lambda_{2}+\lambda_{1}+\lambda_{2}$

, 但没有讨论此条件是否必要, 而且还限制了

$a>0$

$b>0$

. 2016年, 文献[13]针对齐次核, 在更宽泛的条件下证明了

$a, b$

为最佳搭配参数的充分必要条件是

$aq+bp=\lambda+2$

, 结果比较完美. 本文将在文献[12, 13]的基础上, 讨论一般情况下针对拟齐次核的Hilbert级数不等式的最佳搭配参数, 得到了

$a, b$

为最佳搭配参数的充分必要条件. ...

... , 结果比较完美. 本文将在文献[12, 13]的基础上, 讨论一般情况下针对拟齐次核的Hilbert级数不等式的最佳搭配参数, 得到了

$a, b$

为最佳搭配参数的充分必要条件. ...

齐次核的Hilbert型级数不等式取最佳常数因子的充要条件

2016

... 2012年, 文献[12]针对

$\lambda=1$

情况下的拟齐次核获得了最佳搭配参数

$a, b$

的充分条件是

$\lambda_{1}bp+\lambda_{2}aq=\lambda_{1}\lambda_{2}+\lambda_{1}+\lambda_{2}$

, 但没有讨论此条件是否必要, 而且还限制了

$a>0$

$b>0$

. 2016年, 文献[13]针对齐次核, 在更宽泛的条件下证明了

$a, b$

为最佳搭配参数的充分必要条件是

$aq+bp=\lambda+2$

, 结果比较完美. 本文将在文献[12, 13]的基础上, 讨论一般情况下针对拟齐次核的Hilbert级数不等式的最佳搭配参数, 得到了

$a, b$

为最佳搭配参数的充分必要条件. ...

... , 13]的基础上, 讨论一般情况下针对拟齐次核的Hilbert级数不等式的最佳搭配参数, 得到了

$a, b$

为最佳搭配参数的充分必要条件. ...

齐次核的Hilbert型级数不等式取最佳常数因子的充要条件

2016

... 2012年, 文献[12]针对

$\lambda=1$

情况下的拟齐次核获得了最佳搭配参数

$a, b$

的充分条件是

$\lambda_{1}bp+\lambda_{2}aq=\lambda_{1}\lambda_{2}+\lambda_{1}+\lambda_{2}$

, 但没有讨论此条件是否必要, 而且还限制了

$a>0$

$b>0$

. 2016年, 文献[13]针对齐次核, 在更宽泛的条件下证明了

$a, b$

为最佳搭配参数的充分必要条件是

$aq+bp=\lambda+2$

, 结果比较完美. 本文将在文献[12, 13]的基础上, 讨论一般情况下针对拟齐次核的Hilbert级数不等式的最佳搭配参数, 得到了

$a, b$

为最佳搭配参数的充分必要条件. ...

... , 13]的基础上, 讨论一般情况下针对拟齐次核的Hilbert级数不等式的最佳搭配参数, 得到了

$a, b$

为最佳搭配参数的充分必要条件. ...

〈

〉