本原不可幂广义符号矩阵的若干结构指数的界

黄宇飞

doi:10.6054/j.jscnun.2020014

本原不可幂广义符号矩阵的若干结构指数的界

黄宇飞^,

广州民航职业技术学院人文社科学院, 广州 510403

基金项目:

国家自然科学基金项目 11501139

广航院国家自然科学基金校级重点培育项目 18X0429

详细信息

通讯作者:
黄宇飞, 副教授, Email:fayger@qq.com

中图分类号: O151.21
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出版历程
- 收稿日期: 2019-01-24
- 网络出版日期: 2021-03-21
- 刊出日期: 2020-02-24

Bounds on the Structural Indices of Primitive Non-powerful Generalized Sign Pattern Matrices

HUANG Yufei^,

The School of Humanities and Social Sciences, Guangzhou Civil Aviation College, Guangzhou 510403, China

摘要

摘要: 鉴于“环”在结构指数问题研究中的特殊功效, 定义了2类特殊的广义带号有向图:含交圈结构/含违规交圈结构的本原不可幂广义带号有向图.利用有向图的模拟、模糊可达集的分析以及Frobenius数的若干性质, 研究了k点τ-基指数、k点τ-同位基指数、第k重下τ-基指数、第k重上τ-基指数及ω-不可分基指数等结构指数分别在含交圈结构/含违规交圈结构的本原不可幂广义带号有向图类限制下的上界估值问题.
- 本原 /
- 不可幂 /
- (广义)符号矩阵 /
- (广义)带号有向图 /
- 结构指数
Abstract: In view of the special effects of "loop" in the study of structural index problems, two classes of special generalized signed digraphs are defined: primitive non-powerful generalized signed digraphs with intersecting cycles structure and that with distinguished intersecting cycles structure, respectively. With restriction on primitive non-powerful generalized signed digraphs with intersecting cycles structure and those with distinguished intersecting cycles structure, upper bounds on the structural indices, e.g. kth local τ-base, kth same τ-base, kth lower τ-base, kth upper τ-base and ω-indecomposable base, are discussed by imitating the digraphs, analyzing the ambiguous reachable set and using the properties of Frobenius numbers, respectively.
- primitive /
- non-powerful /
- (generalized) sign pattern matrix /
- (generalized) signed digraph /
- structural indices

HTML全文

设 $\frac{1}{p}+\frac{1}{q}=1 \quad(p>1)$ ，α, β ∈ $\mathbb{R}$ ，K(x, y)非负可测，若 $L_{p}^{\alpha}(0, +\infty)=\left\{f(x) \geqslant 0:\|f\|_{p, \alpha}=\left(\int_{0}^{+\infty} x^{\alpha} f^{p}(x) \mathrm{d} x\right)^{1 /p} < +\infty\right\}$ , 则称不等式

$\int_{0}^{+\infty} \int_{0}^{+\infty} K(x, y) f(x) g(y) \mathrm{d} x \mathrm{d} y \leqslant M\|f\|_{p, \alpha}\|g\|_{q, \beta}$

为Hilbert型积分不等式. 由于此类不等式与积分算子T：

$T(f)(y)=\int_{0}^{+\infty} K(x, y) f(x) \mathrm{d} x$

有密切的联系，故而Hilbert型积分不等式对于研究算子T的有界性与算子范数有重要意义.

1991年，XU和GAO^[1]首次提出了研究Hilbert型不等式的权系数方法. 该方法的核心是：引入2个搭配参数a、b，利用Hölder不等式，可得到如下形式的不等式：

$\begin{aligned} &\int_{0}^{+\infty} \int_{0}^{+\infty} K(x, y) f(x) g(y) \mathrm{d} x \mathrm{d} y \leqslant \\ &\quad W_{1}^{1 / p}(b, p) W_{2}^{1 / q}(a, q)\left(\int_{0}^{+\infty} x^{\alpha(a, b)} f^{p}(x) \mathrm{d} x\right)^{1 / p} \times \\ &\quad\left(\int_{0}^{+\infty} y^{\beta(a, b)} g^{q}(y) \mathrm{d} x\right)^{1 / q}. \end{aligned}$

(1)

一般地，随意选取的搭配参数a、b并不能使式(1)的常数因子W₁^1/p(b, p)W₂^1/q(a, q)最佳. 已有的相关研究^[2-13]基本上都是凭借丰富的经验和娴熟的分析技巧选取适当的搭配参数a、b，从而获得最佳的Hilbert型不等式.

若选取的搭配参数a、b能够使式(1)的常数因子最佳，则称其为适配参数或适配数. 文献[14]曾讨论了齐次核的Hilbert型级数不等式的适配参数问题，本文将对拟齐次核的Hilbert型积分不等式讨论搭配参数a、b成为适配数的充分必要条件，并讨论其应用.

1. 预备知识

设G(u, v)是λ阶齐次函数，λ₁λ₂>0，则称K(x, y)=G(x^λ₁, y^λ₂)为拟齐次函数. 显然K(x, y)为拟齐次函数等价于：对t>0，有

$\begin{gathered} K(t x, y)=t^{\lambda_{1} \lambda} K\left(x, t^{-\lambda_{1} / \lambda_{2}} y\right), \\ K(x, t y)=t^{\lambda_{2} \lambda} K\left(t^{-\lambda_{2} / \lambda_{1}} x, y\right). \end{gathered}$

下面给出本文证明过程中所需的引理.

引理1 设1/p+1/q=1 (p>1)，a, b, λ∈ $\mathbb{R}$ ，λ₁λ₂>0，G(u, v)是λ阶齐次非负函数，K(x, y)=G(x^λ₁, y^λ₂)，aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ，记

$\begin{gathered} W_{1}(b, p)=\int_{0}^{+\infty} K(1, t) t^{-b p} \mathrm{~d} t, \\ W_{2}(a, q)=\int_{0}^{+\infty} K(t, 1) t^{-a q} \mathrm{~d} t, \end{gathered}$

则W₁(b, p)/λ₁=W₂(a, q)/λ₂，且

$\begin{aligned} &\omega_{1}(b, p, x)=\int_{0}^{+\infty} K(x, y) y^{-b p} \mathrm{~d} y=x^{\lambda_{1}\left(\lambda-b p / \lambda_{2}+1 / \lambda_{2}\right)} W_{1}(b, p), \\ &\omega_{2}(a, q, y)=\int_{0}^{+\infty} K(x, y) x^{-a q} \mathrm{~d} x=y^{\lambda_{2}\left(\lambda-a q / \lambda_{1}+1 / \lambda_{1}\right)} W_{2}(a, q) . \end{aligned}$

证明由aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ，可得- λ₁λ + λ₁bp/λ₂-λ₁/λ₂－1=－aq. 则有

$\begin{aligned} &W_{1}(b, p)=\int_{0}^{+\infty} t^{\lambda_{2} \lambda} K\left(t^{-\lambda_{2} / \lambda_{1}}, 1\right) t^{-b{p}} \mathrm{~d} t= \\ &\ \ \ \ \ \ \frac{\lambda_{1}}{\lambda_{2}} \int_{0}^{+\infty} K(u, 1) u^{-\lambda_{1} \lambda+\lambda_{1} b p / \lambda_{2}-\lambda_{1} / \lambda_{2}-1} \mathrm{~d} u= \\ &\ \ \ \ \ \ \frac{\lambda_{1}}{\lambda_{2}} \int_{0}^{+\infty} K(u, 1) u^{-a q} \mathrm{~d} u=\frac{\lambda_{1}}{\lambda_{2}} W_{2}(a, q), \end{aligned}$

故W₁(b, p)/λ₁=W₂(a, q)/λ₂.

作变换y=x^λ₁/λ₂t，有

$\begin{aligned} &\omega_{1}(b, p, x)=\int_{0}^{+\infty} x^{\lambda_{1} \lambda} K\left(1, x^{-\lambda_{1} / \lambda_{2}} y\right) y^{-b p} \mathrm{~d} y= \\ &\ \ \ \ \ \ x^{\lambda_{1}\left(\lambda-b p / \lambda_{2}+1 / \lambda_{2}\right)} \int_{0}^{+\infty} K(1, t) t^{-b p} \mathrm{~d} t=\\ &\ \ \ \ \ \ x^{\lambda_{1}\left(\lambda-b p / \lambda_{2}+1 / \lambda_{2}\right)} W_{1}(b, p). \end{aligned}$

同理可证ω₂(a, q, y)=y^{λ₂(λ－aq/λ₁+1/λ₁)}W₂(a, q). 证毕.

2. 适配参数的充分必要条件

定理1 设1/p+1/q=1 (p>1)，a, b, λ∈ $\mathbb{R}$ ，λ₁λ₂>0，G(u, v)是λ阶齐次非负可测函数，K(x, y)=G(x^λ₁, y^λ₂)，W₁(b, p)与W₂(a, q)如引理1所定义. 那么

(1) 若 $\alpha=\lambda_{1}\left[\lambda+\frac{1}{\lambda_{2}}+p\left(\frac{a}{\lambda_{1}}-\frac{b}{\lambda_{2}}\right)\right], \beta=\lambda_{2}\left[\lambda+\frac{1}{\lambda_{1}}+\right.$ $\left.p\left(\frac{b}{\lambda_{2}}-\frac{a}{\lambda_{1}}\right)\right]$ , 则有

$\begin{aligned} &\int_{0}^{+\infty} \int_{0}^{+\infty} K(x, y) f(x) g(y) \mathrm{d} x \mathrm{d} y \leqslant \\ &\quad W_{1}^{1 / p}(b, p) W_{2}^{1 / q}(a, q)\|f\|_{p, \alpha}\|g\|_{q, \beta}, \end{aligned}$

(2)

其中, $f(x) \in L_{p}^{\alpha}(0, +\infty), g(y) \in L_{q}^{\beta}(0, +\infty)$ .

(2) 式(2)中的常数因子W₁^1/p(b, p)W₂^1/q(a, q)是最佳的，当且仅当aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ，W₁(b, p)和W₂(a, q)都收敛. 当aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ时，式(2)化为

$\begin{aligned} &\int_{0}^{+\infty} \int_{0}^{+\infty} K(x, y) f(x) g(y) \mathrm{d} x \mathrm{d} y \leqslant \\ &\ \ \ \ \ \ \ \ \frac{W_{0}}{\left|\lambda_{1}\right|^{1 / q}\left|\lambda_{2}\right|^{1 / p}}\|f\|_{p, a p q-1}\|g\|_{q, b p q-1}, \end{aligned}$

(3)

其中, W₀=|λ₁|W₂(a, q)= |λ₂|W₁(b, p).

证明 (i)选择a、b为搭配参数. 根据Hölder不等式和引理1，利用权系数方法，有

$\begin{aligned} &\int_{0}^{+\infty} \int_{0}^{+\infty} K(x, y) f(x) g(y) \mathrm{d} x \mathrm{d} y= \\ &\qquad \int_{0}^{+\infty} \int_{0}^{+\infty}\left(\frac{x^{a}}{y^{b}} f(x)\right)\left(\frac{y^{b}}{x^{a}} g(y)\right) K(x, y) \mathrm{d} x \mathrm{d} y \leqslant \\ &\qquad \left(\int_{0}^{+\infty} \int_{0}^{+\infty} \frac{x^{a p}}{y^{b p}} f^{p}(x) K(x, y) \mathrm{d} x \mathrm{d} y\right)^{1 / p} \times \\ &\qquad \left(\int_{0}^{+\infty} \int_{0}^{+\infty} \frac{y^{b q}}{x^{a q}} g^{q}(y) K(x, y) \mathrm{d} x \mathrm{d} y\right)^{1 / q}=\\ &\qquad \left(\int_{0}^{+\infty} x^{a p} f^{p}(x) \omega_{1}(b, p, x) \mathrm{d} x\right)^{1 / p} \times \\ &\qquad \left(\int_{0}^{+\infty} y^{b q} g^{q}(y) \omega_{2}(a, q, y) \mathrm{d} y\right)^{1 / q}= \\ &\qquad W_{1}^{1 / p}(b, p) W_{2}^{1 / q}(a, q) \times \\ &\qquad \left(\int_{0}^{+\infty} x^{a p+\lambda_{1}\left(\lambda-b p / \lambda_{2}+1 / \lambda_{2}\right)} f^{p}(x) \mathrm{d} x\right)^{1 / p} \times \\ &\qquad \left(\int_{0}^{+\infty} y^{b q+\lambda_{2}\left(\lambda-a q / \lambda_{1}+1 / \lambda_{1}\right)} g^{q}(y) \mathrm{d} x\right)^{1 / q}= \\ &\qquad W_{1}^{1 / p}(b, p) W_{2}^{1 / q}(a, q)\|f\|_{p, \alpha}\|g\|_{q, \beta}, \end{aligned}$

故式(2)成立.

(ii) 充分性：设aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ，W₁(b, p)和W₂(a, q)收敛. 由引理1，有W₁(b, p)/λ₁=W₂(a, q)/λ₂，故

$W_{1}^{1 / p}(b, p) W_{2}^{1 / q}(a, q)=\left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{1 / q} W_{1}(b, p)=\frac{W_{0}}{\left|\lambda_{1}\right|^{1 / q}\left|\lambda_{2}\right|^{1 / p}},$

且α=apq－1，β=bpq－1，于是式(2)可化为式(3).

设式(3)的最佳常数因子为M₀，则M₀≤W₀/(|λ₁^1/q|λ₂|^1/p)，且用M₀取代式(3)中的常数因子后，式(3)仍然成立.

取充分小的ε>0及δ>0，令

$\begin{aligned} f(x) =&\left\{\begin{array}{l} x^{\left(-a p q-\left|\lambda_{1}\right| \varepsilon\right) / p}\quad (x \geqslant 1), \\ 0 \quad (0<x<1); \end{array} \right.\\ g(y) = &\begin{cases}y^{\left(-b p q-\left|\lambda_{2}\right| \varepsilon\right) / q}\quad (y \geqslant \delta) ,\\ 0 \quad (0<y<\delta).\end{cases} \end{aligned}$

则

$\begin{array}{c} \begin{aligned} &\|f\|_{p, a p q-1}\|g\|_{q, b p q-1}= \\ &\quad\left(\int_{1}^{+\infty} x^{-1-\left|\lambda_{1}\right| \varepsilon} \mathrm{d} x\right)^{1 / p}\left(\int_{\delta}^{+\infty} y^{-1-\left|\lambda_{2}\right| \varepsilon} \mathrm{d} y\right)^{1 / q}= \\ &\quad\left(\frac{1}{\left|\lambda_{1} \varepsilon\right|}\right)^{1 / p}\left(\frac{1}{\left|\lambda_{2}\right| \varepsilon} \delta^{-\left|\lambda_{2}\right| \varepsilon}\right)^{1 / q}= \\ &\quad\frac{1}{\varepsilon\left|\lambda_{1}\right|^{1 / p}\left|\lambda_{2}\right|^{1 / q}} \delta^{-\left|\lambda_{2}\right| \varepsilon / q}, \end{aligned}\\ \begin{aligned} &\int_{0}^{+\infty} \int_{0}^{+\infty} K(x, y) f(x) g(y) \mathrm{d} x \mathrm{d} y= \\ &\quad\int_{1}^{+\infty} x^{-a q-\left|\lambda_{1}\right| \varepsilon / p}\left(\int_{\delta}^{+\infty} y^{-b p-\left|\lambda_{2}\right| \varepsilon / q} K(x, y) \mathrm{d} y\right) \mathrm{d} x= \\ &\quad\int_{1}^{+\infty} x^{-a q-\left|\lambda_{1}\right| \varepsilon / p+\lambda \lambda_{1}}\left(\int_{\delta}^{+\infty} y^{-b p-\left|\lambda_{2}\right| \varepsilon / q} K\left(1, x^{-\lambda_{1} / \lambda_{2}} y\right) \mathrm{d} y\right) \mathrm{d} x= \\ &\quad\int_{1}^{+\infty} x^{-1-\left|\lambda_{1}\right| \varepsilon}\left(\int_{x^{-\lambda_{1} / \lambda_{2}} \delta}^{+\infty} t^{-b p-\left|\lambda_{2}\right| \varepsilon / q} K(1, t) \mathrm{d} t\right) \mathrm{d} x \geqslant \\ &\quad\int_{1}^{+\infty} x^{-1-\left|\lambda_{1}\right| \varepsilon}\left(\int_{\delta}^{+\infty} t^{-b p-\left|\lambda_{2}\right| \varepsilon / q} K(1, t) \mathrm{d} t\right) \mathrm{d} x= \\ &\quad\frac{1}{\left|\lambda_{1}\right| \varepsilon} \int_{\delta}^{+\infty} t^{-b p-\left|\lambda_{2}\right| \varepsilon / q} K(1, t) \mathrm{d} t. \end{aligned} \end{array}$

于是

$\frac{1}{\left|\lambda_{1}\right|} \int_{\delta}^{+\infty} t^{-b p-\left|\lambda_{2}\right| \varepsilon / q} K(1, t) \mathrm{d} t \leqslant \frac{M_{0}}{\left|\lambda_{1}\right|^{1 / p}\left|\lambda_{2}\right|^{1 / q}} \delta^{-\left|\lambda_{2}\right| \varepsilon / q}.$

先令ε→0⁺，再令δ→0⁺，得

$W_{1}(b, p)=\int_{0}^{+\infty} t^{-b p} K(1, t) \mathrm{d} t \leqslant \frac{M_{0}}{\left|\lambda_{1}\right|^{1 / p}\left|\lambda_{2}\right|^{1 / q}}.$

再根据引理1，可得到W₀/(|λ₁|^1/q|λ₂|^1/p)≤M₀. 所以式(3)的最佳常数因子M₀=W₀/(|λ₁|^1/q|λ₂|^1/p).

必要性：设式(2)的常数因子W₁^1/p(b, p)W₂^1/q(a, q)是最佳的，则W₁(b, p)和W₂(a, q)是收敛的. 下证aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ.

记 $\frac{1}{\lambda_{1}} a q+\frac{1}{\lambda_{2}} b p-\left(\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}+\lambda\right)=c, a_{1}=a-\frac{\lambda_{1} c}{p q}, b_{1}=$ $b-\frac{\lambda_{2} c}{p q}$ ，则

$\begin{gathered} \alpha=\lambda_{1}\left[\lambda+\frac{1}{\lambda_{2}}+p\left(\frac{a_{1}}{\lambda_{1}}-\frac{b_{1}}{\lambda_{2}}\right)\right]=\alpha_{1}, \\ \beta=\lambda_{2}\left[\lambda+\frac{1}{\lambda_{1}}+p\left(\frac{b_{1}}{\lambda_{2}}-\frac{a_{1}}{\lambda_{1}}\right)\right]=\beta_{1}, \\ W_{2}(a, q)=\int_{0}^{+\infty} K(t, 1) t^{-a q} \mathrm{~d} t=\frac{\lambda_{2}}{\lambda_{1}} \int_{0}^{+\infty} K(1, t) t^{-b p+\lambda_{2} c} \mathrm{~d} t . \end{gathered}$

于是可知式(2)等价于

$\begin{aligned} &\int_{0}^{+\infty} \int_{0}^{+\infty} K(x, y) f(x) g(y) \mathrm{d} x \mathrm{d} y \leqslant \\ &\quad W_{1}^{1 / p}(b, p)\left(\frac{\lambda_{2}}{\lambda_{1}} \int_{0}^{+\infty} K(1, t) t^{-b p+\lambda_{2} c} \mathrm{~d} t\right)^{1 / q}\|f\|_{p, \alpha_{1}}\|g\|_{q, \beta_{1}} . \end{aligned}$

又经计算有a₁q/λ₁+b₁p/λ₂=1/λ₁+1/λ₂+ λ，α₁=a₁pq－1，β₁=b₁pq－1，故式(2)进一步等价于

$\begin{aligned} &\int_{0}^{+\infty} \int_{0}^{+\infty} K(x, y) f(x) g(y) \mathrm{d} x \mathrm{d} y \leqslant \\ &\qquad W_{1}^{1 / p}(b, p)\left(\frac{\lambda_{2}}{\lambda_{1}} \int_{0}^{+\infty} K(1, t) t^{-b p+\lambda_{2} c} \mathrm{~d} t\right)^{1 / q} \times \\ &\qquad\|f\|_{p, a_{1} p q-1}\|g\|_{q, b_{1} p q-1} . \end{aligned}$

(4)

根据假设，式(4)的最佳常数因子是W₁^1/p(b, p)× ${\left({\frac{{{\lambda _2}}}{{{\lambda _1}}}\int_0^{ + \infty } K (1, t){t^{ - bp + {\lambda _2}c}}{\rm{d}}t} \right)^{1/q}}$ . 又由 $\frac{1}{{{\lambda _1}}}{a_1}q + \frac{1}{{{\lambda _2}}}{b_1}p = \frac{1}{{{\lambda _1}}} + \frac{1}{{{\lambda _2}}} + \lambda$ 及充分性的证明，可知式(4)的最佳常数因子为

$\begin{gathered} \frac{1}{\left|\lambda_{1}\right|^{1 / q}\left|\lambda_{2}\right|^{1 / p}}\left(\left|\lambda_{2}\right| \int_{0}^{+\infty} K(1, t) t^{-b_{1} p} \mathrm{~d} t\right)= \\ \left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{1 / q} \int_{0}^{+\infty} K(1, t) t^{-b p+\lambda_{2} c / q} \mathrm{~d} t \end{gathered}$

于是得到

$\begin{aligned} &\int_{0}^{+\infty} K(1, t) t^{-b p+\lambda_{2} c / q} \mathrm{~d} t= \\ &\quad W_{1}^{1 / p}(b, p)\left(\int_{0}^{+\infty} K(1, t) t^{-b p+\lambda_{2} c} \mathrm{~d} t\right)^{1 / q}. \end{aligned}$

(5)

对于1和t^λ₂c/q，应用Hölder不等式，有

$\begin{aligned} &\int_{0}^{+\infty} K(1, t) t^{-b p+\lambda_{2} c / q} \mathrm{~d} t=\int_{0}^{+\infty} t^{\lambda_{2} c / q} K(1, t) t^{-b p} \mathrm{~d} t \leqslant \\ &\qquad\left(\int_{0}^{+\infty} 1^{p} K(1, t) t^{-b p} \mathrm{~d} t\right)^{1 / p}\left(\int_{0}^{+\infty} t^{\lambda_{2} c} K(1, t) t^{-b p} \mathrm{~d} t\right)^{1 / q}= \\ &\qquad W_{1}^{1 / p}(b, p)\left(\int_{0}^{+\infty} K(1, t) t^{-b p+\lambda_{2} c} \mathrm{~d} t\right)^{1 / q}. \end{aligned}$

(6)

根据式(5)，可知式(6)取等号. 又根据Hölder不等式取等号的条件，可得t^λ₂c/q=常数，故c=0，即aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ₁. 证毕.

注1 定理1表明: 当且仅当aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ时，搭配参数a、b是适配参数. 因此，只要选取a、b满足aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ，就可以得到各种各样的具有最佳常数因子的Hilbert型积分不等式.

推论1 设1/p+1/q=1 (p>1)，λ₁λ₂>0，λ >0，1/r+1/s=1 (r>1)，α=p(1－ λλ₁/r)－1，β=q(1－ λλ₂/s)－1，则

$\begin{aligned} &\int_{0}^{+\infty} \int_{0}^{+\infty} \frac{f(x) g(y)}{\left(x^{\lambda_{1}}+y^{\lambda_{2}}\right)^{\lambda}} \mathrm{d} x \mathrm{d} y \leqslant \\ &\qquad\frac{1}{\left|\lambda_{1}\right|^{1 / q}\left|\lambda_{2}\right|^{1 / p}} \mathrm{~B}\left(\frac{\lambda}{r}, \frac{\lambda}{s}\right)\|f\|_{p, \alpha}\|g\|_{q, \beta}, \end{aligned}$

(7)

其中的常数因子是最佳的，f(x)∈L_p^α(0, +∞)，g(y)∈L_q^β(0, +∞).

证明记K(x, y)=G(x^λ₁, y^λ₂)=1/(x^λ₁+y^λ₂)^λ，则G(u, v)是－λ阶齐次非负函数. 选取搭配参数a= $\frac{1}{q}\left(1-\frac{\lambda \lambda_{1}}{r}\right), b=\frac{1}{p}\left(1-\frac{\lambda \lambda_{2}}{s}\right)$ , 可得

$\frac{1}{\lambda_{1}} a q+\frac{1}{\lambda_{2}} b p=\frac{1}{\lambda_{1}}\left(1-\frac{\lambda \lambda_{1}}{r}\right)+\frac{1}{\lambda_{2}}\left(1-\frac{\lambda \lambda_{2}}{s}\right)=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}-\lambda,$

故a、b是适配参数. 又因为apq－1=p(1－ λλ₁/r)－1=α，bpq－1=q(1－ λλ₂/s)－1=β，且

$\begin{aligned} W_{0}=&\left|\lambda_{2}\right| W_{1}(b, p)=\left|\lambda_{2}\right| \int_{0}^{+\infty} \frac{1}{\left(1+t^{\lambda_{2}}\right)^{\lambda}} t^{\lambda \lambda_{2} / s-1} \mathrm{~d} t=\\ & \int_{0}^{+\infty} \frac{1}{(1+u)^{\lambda}} u^{\lambda / s-1} \mathrm{~d} u=\mathrm{B}\left(\frac{\lambda}{s}, \lambda-\frac{\lambda}{s}\right)=\mathrm{B}\left(\frac{\lambda}{r}, \frac{\lambda}{s}\right). \end{aligned}$

根据定理1，式(7)成立，且其常数因子是最佳的. 证毕.

3. 在求积分算子范数中的应用

根据Hilbert型不等式与相应积分算子的关系理论，由定理1可得如下定理.

定理2 设1/p+1/q=1 (p>1)，a, b, λ ∈ $\mathbb{R}$ ，λ₁λ₂>0，α=apq－1，β=bpq－1，G(u, v)是λ阶齐次非负可测函数，K(x, y)=G(x^λ₁, y^λ₂)，且

$\begin{aligned} &W_{1}(b, p)=\int_{0}^{+\infty} K(1, t) t^{-b p} \mathrm{~d} t<+\infty, \\ &W_{2}(a, q)=\int_{0}^{+\infty} K(t, 1) t^{-a q} \mathrm{~d} t<+\infty, \end{aligned}$

则当aq/λ₁+bp/λ₂=1/λ₁+1/λ₂+ λ时，积分算子T：

$T(f)(y)=\int_{0}^{+\infty} K(x, y) f(x) \mathrm{d} x, f(x) \in L_{p}^{\alpha}(0,+\infty)$

是从L_p^α(0, +∞)到L_p^β(1－p)(0, +∞)的有界算子，且T的算子范数为

$\|T\|=\frac{\left|\lambda_{2}\right| W_{1}(b, p)}{\left|\lambda_{1}\right|^{1 / q}\left|\lambda_{2}\right|^{1 / p}}=\left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{1 / q} \int_{0}^{+\infty} K(1, t) t^{-b p} \mathrm{~d} t.$

推论2 设1/p+1/q=1 (p>1)，λ₁λ₂>0，－1 < λ < min{1±4/λ₁, 1±4/λ₂}，α=p[1+ λ₁(λ －1)/2]－1，β=p[1+ λ₂(λ －1)/2]－1，则积分算子T：

$T(f)(y)=\int_{0}^{+\infty} \frac{\left|x^{\lambda_{1}}-y^{\lambda_{2}}\right|^{\lambda}}{\max \left\{x^{\lambda_{1}}, y^{\lambda_{2}}\right\}} f(x) \mathrm{d} x, f(x) \in L_{p}^{\alpha}(0,+\infty)$

是从L_p^α(0, +∞)到L_p^β(1－p)(0, +∞)的有界算子，且T的算子范数为

$\begin{aligned} &\|T\|=\frac{1}{\left|\lambda_{1}\right|^{1 / q}\left|\lambda_{2}\right|^{1 / p}}\left[\mathrm{~B}\left(\lambda+1, \frac{1-\lambda}{2}-\frac{2}{\lambda_{2}}\right)+\right. \\ &\ \ \ \ \ \ \left.\mathrm{B}\left(\lambda+1, \frac{1-\lambda}{2}+\frac{2}{\lambda_{2}}\right)\right] . \end{aligned}$

证明记K(x, y)=G(x^λ₁, y^λ₂)= |x^λ₁－y^λ₂|^λ/max{x^λ₁, y^λ₂}，则G(u, v)是λ －1阶齐次函数. 取a= $\frac{1}{q}\left[1+\frac{\lambda_{1}}{2}(\lambda-1)\right], b=\frac{1}{p}\left[1+\frac{\lambda_{2}}{2}(\lambda-1)\right]$ ，则

$\begin{aligned} &\frac{1}{\lambda_{1}} a q+\frac{1}{\lambda_{2}} b p=\frac{1}{\lambda_{1}}\left[1+\frac{\lambda_{1}}{2}(\lambda-1)\right]+\frac{1}{\lambda_{2}}\left[1+\frac{\lambda_{2}}{2}(\lambda-1)\right]= \\ &\ \ \ \ \ \ \frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}+\lambda-1, \end{aligned}$

故a、b是适配参数. 又 $a p q-1=p\left[1+\frac{\lambda_{1}}{2}(\lambda-1)\right]-1=\alpha$ ， $b p q-1=q\left[1+\frac{\lambda_{2}}{2}(\lambda-1)\right]-1=\beta$ . 则

$\begin{aligned} &\left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{1 / q} \int_{0}^{+\infty} K(1, t) t^{-b p} \mathrm{~d} t= \\ &\qquad\left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{1 / q} \int_{0}^{+\infty} \frac{\left|1-t^{\lambda_{2}}\right|^{\lambda}}{\max \left\{1, t^{\lambda_{2}}\right\}} t^{-\left[1+\lambda_{2}(\lambda-1) / 2\right]} \mathrm{d} t= \\ &\qquad\frac{1}{\left|\lambda_{1}\right|^{1 / q}\left|\lambda_{2}\right|^{1 / p}}\left[\mathrm{~B}\left(\lambda+1, \frac{1-\lambda}{2}-\frac{2}{\lambda_{2}}\right)+\right. \\ &\qquad\left.\mathrm{B}\left(\lambda+1, \frac{1-\lambda}{2}+\frac{2}{\lambda_{2}}\right)\right]<+\infty . \end{aligned}$

根据定理2，知推论2成立. 证毕.

推论3 设1/p+1/q=1 (p>1)，1/r+1/s=1 (r>1)，λ₁λ₂>0，α=p(1－ λ₁/r)－1，β=q(1－ λ₂/s)－1. 则积分算子T：

$T(f)(y)=\int_{0}^{+\infty} \frac{\ln \left(x^{\lambda_{1}} / y^{\lambda_{2}}\right)}{x^{\lambda_{1}}-y^{\lambda_{2}}} f(x) \mathrm{d} x, f(x) \in L_{p}^{\alpha}(0,+\infty)$

是从L_p^α(0, +∞)到L_p^β(1－p)(0, +∞)的有界算子，且T的算子范数为

$\|T\|=\frac{1}{\left|\lambda_{1}\right|^{1 / q}\left|\lambda_{2}\right|^{1 / p}}\left[\zeta\left(2, \frac{1}{r}\right)+\zeta\left(2, \frac{1}{s}\right)\right],$

其中ζ(t, a)是Riemann函数.

证明记

$K(x, y)=G\left(x^{\lambda_{1}}, y^{\lambda_{2}}\right)=\frac{\ln \left(x^{\lambda_{1}} / y^{\lambda_{2}}\right)}{x^{\lambda_{1}}-y^{\lambda_{2}}},$

则G(u, v)是－1阶齐次非负函数.

取搭配参数 $a=\frac{1}{q}\left(1-\frac{\lambda_{1}}{r}\right), b=\frac{1}{p}\left(1-\frac{\lambda_{2}}{s}\right)$ ，则

$\frac{1}{\lambda_{1}} a q+\frac{1}{\lambda_{2}} b p=\frac{1}{\lambda_{1}}\left(1-\frac{\lambda_{1}}{r}\right)+\frac{1}{\lambda_{2}}\left(1-\frac{\lambda_{2}}{s}\right)=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}-1,$

故a、b是适配参数. 又apq－1=p(1－ λ₁/r)－1=α，bpq－1=q(1－ λ₂/s)－1=β，且

$\begin{gathered} \left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{1 / q} \int_{0}^{+\infty} K(1, t) t^{-b p} \mathrm{~d} t=\left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{1 / q} \int_{0}^{+\infty} \frac{\ln \left(t^{-\lambda_{2}}\right)}{1-t^{\lambda_{2}}} t^{\lambda_{2} / s-1} \mathrm{~d} t= \\ \frac{1}{\left|\lambda_{1}\right|^{1 / q}\left|\lambda_{2}\right|^{1 / p}}\left[\zeta\left(2, \frac{1}{r}\right)+\zeta\left(2, \frac{1}{s}\right)\right]<+\infty \end{gathered}$

根据定理2，知推论3成立. 证毕.

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