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

(1.1) $ \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} $

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) $.

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

(2.1) $ \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} $

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

(2.2) $ \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} $

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

(2.3) $ \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} $

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

(2.4) $ \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} $

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

(2.5) $ \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} $

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

(2.6) $ \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} $

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. $

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.