第二类与Gamma函数有关的Hardy型不等式的等价条件

钟建华; 曾志红; 陈燕清; 陈强

doi:10.6054/j.jscnun.2020050

第二类与Gamma函数有关的Hardy型不等式的等价条件

1.
广东第二师范学院数学系，广州 510303
2.
广东第二师范学院学报编辑部，广州 510303
3.
广州市聚德中学，广州 510305
4.
广东第二师范学院计算机科学系，广州 510303

基金项目:

国家自然科学基金项目 61772140

广东省科技计划项目 2017A010101021

详细信息

通讯作者:
曾志红，编审，Email:zhzeng@gdei.edu.cn

中图分类号: O178
计量
- 文章访问数: 1144
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出版历程
- 收稿日期: 2019-09-24
- 网络出版日期: 2021-03-21
- 刊出日期: 2020-06-24

The Equivalent Conditions for a Second Kind of Hardy-Type Inequality Related to Gamma Function

1.
Department of Mathematics, Guangdong University of Education, Guangzhou 510303, China
2.
Editorial Department of the University Journal, Guangdong University of Education, Guangzhou 510303, China
3.
Guangzhou Jude Middle School, Guangzhou 510305, China
4.
Department of Computer Science, Guangdong University of Education, Guangzhou 510303, China

摘要

摘要: 引入恰当的参数，运用实分析方法以及权函数的技巧, 获得2个与权函数估算及证明最佳常数因子有关的引理, 并据此建立了几个关于带参数、最佳常数因子与Gamma函数有关的第二类非齐次核的Hardy型积分不等式的等价条件.同时, 得到了几个关于第二类齐次核的Hardy型积分不等式的等价条件.
- Hardy型积分不等式 /
- 权函数 /
- 等价式 /
- Gamma函数
Abstract: Two lemmas that are related to the estimate of weight function and the proving of best constant are obtained with the introduction of suitable parameters and the use of the way and techniques of real analysis and weight functions. On this basis, a few equivalent conditions for a second kind of Hardy-type integral inequality with a non-homogeneous kernel, related to a parameter and a best constant factor expressed in terms of Gamma function, are built. Meanwhile, some equivalent conditions for a second kind of Hardy-type integral inequality with the homogeneous kernel are deduced.
- Hardy-type integral inequality /
- weight function /
- equivalent form /
- Gamma function

HTML全文

设(p, q)为一对共轭指数, 即 $\frac{1}{p}+\frac{1}{q}=1$ , p>1, 对于非负的-1齐次函数k₁(x, y)及常数k_p= $\int_{0}^{\infty} k_{1}$ k₁(u, 1)u^-1/pdu $\in \mathbb{R}_{+}$ , 使得当f(x), g(y)≥0, 0 < $\int_{0}^{\infty} f^{p}$ (x)dx < ∞, 0 < $\int_{0}^{\infty} g^{q}$ (x)dx < ∞时, 获得Hardy-Hilbert积分不等式^[1]:

$\begin{array}{l} \int_0^\infty {\int_0^\infty {{k_1}} } (x,y)f(x)g(y){\rm{d}}x{\rm{d}}y < \\ \ \ \ \ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {k_p}{\left( {\int_0^\infty {{f^p}} (x){\rm{d}}x} \right)^{1/p}}{\left( {\int_0^\infty {{g^q}} (x){\rm{d}}x} \right)^{1/q}}, \end{array}$

(1)

其中，常数因子k_p为最佳值.

当k₁(x, y)=1/(x+y)时, k_p=π/sin(π/p), 式(1)变为经典Hardy-Hilbert积分不等式^[2]:

$\begin{array}{l} \int_0^\infty {\int_0^\infty {\frac{{f(x)g(y)}}{{x + y}}} } {\rm{d}}x{\rm{d}}y < \\ \ \ \ \ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \frac{\pi }{{{\rm{sin}}(\pi /p)}}{\left( {\int_0^\infty {{f^p}} (x){\rm{d}}x} \right)^{1/p}}{\left( {\int_0^\infty {{g^q}} (x){\rm{d}}x} \right)^{1/q}}. \end{array}$

(2)

近10余年来，根据分析理论及Hilbert型不等式理论，式(2)有很多重要推广和应用^[3-8].

设h(u)>0, 令φ(σ)= $\int_{0}^{\infty} h$ (u)u^σ-1du $\in \mathbb{R}_{+}$ , 由式(1)对偶地得到一个非齐次核Hilbert型积分不等式^[1]:

$\begin{array}{l} \int_0^\infty {\int_0^\infty h } (xy)f(x)g(y){\rm{d}}x{\rm{d}}y < \\ \ \ \ \ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \varphi \left( {\frac{1}{p}} \right){\left[ {\int_0^\infty {{x^{p - 2}}} {f^p}(x){\rm{d}}x} \right]^{1/p}}{\left[ {\int_0^\infty {{g^q}} (y){\rm{d}}y} \right]^{1/q}}, \end{array}$

(3)

其中, 常数因子φ(1/p)为最佳值.

2009年, 文献[9-10]给出了式(1)、(2)的一个重要的推广:

设λ₁+λ₂=λ $\in \mathbb{R}$ =(-∞, ∞), k_λ(x, y)是-λ齐次核的非负函数, 且

${k_\lambda }(ux,uy) = {u^{ - \lambda }}{k_\lambda }(x,y){\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} (u,x,y > 0)$

及k(λ₁)= $\int_{0}^{\infty} k_{\lambda}$ (u, 1)u^λ₁-1du $\in \mathbb{R}_{+}$ =(0, ∞), 则

$\begin{array}{*{20}{l}} {\int_0^\infty {\int_0^\infty {{k_\lambda }} } (x,y)f(x)g(y){\rm{d}}x{\rm{d}}y < }\\ {k({\lambda _1}){{\left[ {\int_0^\infty {{x^{p(1 - {\lambda _1}) - 1}}} {f^p}(x){\rm{d}}x} \right]}^{1/p}}{{\left[ {\int_0^\infty {{y^{q(1 - {\lambda _2}) - 1}}} {g^q}(y){\rm{d}}y} \right]}^{1/q}},} \end{array}$

(4)

其中, 常数k(λ₁)是最佳值.

当λ=1, λ₁=1/q, λ₂=1/p时, 式(4)的一个特殊情形为式(1)；当k_λ(x, y)=1/(x+y) (λ=1)时, 式(4)的另一个特殊情形为式(2).此外, 文献[11]给出式(3)的一个推广:

$\begin{array}{*{20}{l}} {\int_0 {\int_0 h } (xy)f(x)g(y){\rm{d}}x{\rm{d}}y < }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \varphi (\sigma ){{\left[ {\int_0^\infty {{x^{p(1 - \sigma ) - 1}}} {f^p}(x){\rm{d}}x} \right]}^{1/p}}{{\left[ {\int_0^\infty {{y^{q(1 - \sigma ) - 1}}} {g^q}(y){\rm{d}}y} \right]}^{1/q}},} \end{array}$

(5)

其中, 常数因子φ(σ)为最佳值.当σ=1/p时, 式(5)变成式(3).

2013年, 文献[11]通过增加一个特殊条件研究了式(4)及式(5)的等价形式:

设h(xy)>0, xy < 1, 令φ(σ)= $\int_{1}^{\infty} h$ (u)u^σ-1du=φ₂(σ) $\in \mathbb{R}_{+}$ , 则式(5)对偶地变成以下具有非齐次核的第二类Hardy型积分不等式:

$\begin{array}{l} \int_0^\infty g (y)\left( {\int_{1/y}^\infty h (xy)f(x){\rm{d}}x} \right){\rm{d}}y < \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\varphi _2}(\sigma ){\left[ {\int_0^\infty {{x^{p(1 - \sigma ) - 1}}} {f^p}(x){\rm{d}}x} \right]^{1/p}}{\left[ {\int_0^\infty {{y^{q(1 - \sigma ) - 1}}} {g^q}(y){\rm{d}}y} \right]^{1/q}}. \end{array}$

(6)

本文运用Hilbert理论以及权函数方法, 引进参数λ、μ、β、γ和σ, 且γ>-1, σ, μ>-β, σ+μ=λ, 得到2个引理, 据此建立几个具有非齐次核的Hardy型积分不等式的等价条件, 这个核为

$h(u) = {k_\lambda }(u,1) = \frac{{{{({\rm{min}}\{ u,1\} )}^\beta }|{\rm{ln}}{\kern 1pt} {\kern 1pt} {\kern 1pt} u{|^\gamma }}}{{{{({\rm{max}}\{ u,1\} )}^{\lambda + \beta }}}}.$

(7)

结合Γ函数^[12]的性质, 得到了一些与Γ函数有关的且为最佳常数因子的第二类Hardy型积分不等式的几个等价式.

1. 重要引理

引理1 若参数λ, μ, β, γ, σ满足γ>-1, σ, μ>-β, σ+μ=λ, h(u)如式(7)所定义, 则

${k_2}(\sigma ): = \int_1^\infty {{u^{\sigma - 1}}} h(u){\rm{d}}u = \frac{1}{{{{(\mu + \beta )}^{\gamma + 1}}}}\Gamma (\gamma + 1),$

(8)

其中, Γ(z)= $\int_{0}^{\infty}$ e^-tt^z-1 dt也称为第二类欧拉积分^[12], Re(z)>0.

证明由引理条件及函数h(u)的性质, 在积分过程中作变换u=e^v, t=(μ+β)v, 有

$\begin{array}{l} {k_2}(\sigma ) = \int_1^\infty {{u^{\sigma - 1}}} h(u){\rm{d}}u = \int_1^\infty {{u^{ - \mu - \beta - 1}}} {({\rm{ln}}{\kern 1pt} {\kern 1pt} {\kern 1pt} u)^\gamma }{\rm{d}}u = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \int_0^\infty {{v^\gamma }} {{\rm{e}}^{ - \mu v - \beta v - v}}{{\rm{e}}^v}{\rm{d}}v = \frac{1}{{{{(\mu + \beta )}^{\gamma + 1}}}}\int_0^\infty {{t^{\gamma + 1 - 1}}} {{\rm{e}}^{ - t}}{\rm{d}}t = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \frac{1}{{{{(\mu + \beta )}^{\gamma + 1}}}}\Gamma (\gamma + 1). \end{array}$

证毕.

当σ=λ时, 有μ=0, k₂(σ)=Γ(γ+1)/β^γ+1.

下文总假定一对共轭指数为(p, q), $\frac{1}{p}+\frac{1}{q}=1$ , p>1, σ₁, $\mu_{1} \in \mathbb{R}$ .

引理2 设h(u)如式(7)所定义, 若λ, μ, β, γ, σ满足引理1条件, 如果存在常数M₂>0, 且任意函数f(x)和g(y)在(0, ∞)上非负可测, 则不等式

$\begin{array}{*{20}{l}} {\int_0^\infty g (y)\left[ {\int_{1/y}^\infty h (xy){\rm{d}}x} \right]{\rm{d}}y \le }\\ {{M_2}{{\left[ {\int_0^\infty {{x^{p(1 - \sigma ) - 1}}} {f^p}(x){\rm{d}}x} \right]}^{1/p}}{{\left[ {\int_0^\infty {{y^{q(1 - {\sigma _1}) - 1}}} {g^q}(y){\rm{d}}y} \right]}^{1/q}}} \end{array}$

(9)

成立, 从而有σ=σ₁, 且M₂≥k₂(σ), 这里, k₂(σ)为式(8)所定义.

证明 (应用反证法)首先，假设σ₁ < σ, 对于n≥1/(σ-σ₁) $(n \in \mathbb{N})$ , 考察2个非负可测函数

${\tilde f_n}(x): = \left\{ {\begin{array}{*{20}{l}} {0\quad (x{\kern 1pt} \in {\kern 1pt} (0,1)),}\\ {{x^{\sigma - (1/(pn)) - 1}}\quad (x{\kern 1pt} \in {\kern 1pt} [1,\infty )),} \end{array}} \right.$

${\tilde g_n}(y): = \left\{ {\begin{array}{*{20}{l}} {{y^{{\sigma _1} + (1/(qn)) - 1}}\quad (y{\kern 1pt} \in {\kern 1pt} (0,1]),}\\ {0\quad (y{\kern 1pt} \in {\kern 1pt} (1,\infty )),} \end{array}} \right.$

则

$\begin{array}{*{20}{l}} {{{\tilde J}_2}: = {{\left[ {\int_0^\infty {{x^{p(1 - \sigma ) - 1}}} \tilde f_n^p(x){\rm{d}}x} \right]}^{1/p}}{{\left[ {\int_0^\infty {{y^{q(1 - {\sigma _1}) - 1}}} \tilde g_n^q(y){\rm{d}}y} \right]}^{1/q}} = }\\ \ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {{\left[ {\int_1^\infty {{x^{( - 1/n) - 1}}} {\rm{d}}x} \right]}^{1/p}}{{\left[ {\int_0^1 {{y^{(1/n) - 1}}} {\rm{d}}y} \right]}^{1/q}} = n < \infty .} \end{array}$

设u=xy, 得

$\begin{array}{l} {{\tilde I}_2}: = \int_0^\infty {{{\tilde g}_n}} (y)\left[ {\int_{1/y}^\infty h (xy){{\tilde f}_n}(x){\rm{d}}x} \right]{\rm{d}}y = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \int_0^1 {{y^{{\sigma _1} + (1/(qn)) - 1}}} \left[ {\int_{1/y}^\infty {\frac{{{{({\rm{min}}\{ xy,1\} )}^\beta }|{\rm{ln}}{\kern 1pt} {\kern 1pt} xy{|^\gamma }}}{{{{({\rm{max}}\{ xy,1\} )}^{\lambda + \beta }}}}} {x^{\sigma - (1/(pn)) - 1}}{\rm{d}}x} \right]{\rm{d}}y = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \int_0^1 {{y^{({\sigma _1} - \sigma ) + (1/n) - 1}}} {\rm{d}}y\int_1^\infty {\frac{{{{({\rm{min}}\{ u,1\} )}^\beta }{{({\rm{ln}}{\kern 1pt} {\kern 1pt} u)}^\gamma }}}{{{{({\rm{max}}\{ u,1\} )}^{\lambda + \beta }}}}} {u^{\sigma - (1/(pn)) - 1}}{\rm{d}}u = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \int_0^1 {{y^{({\sigma _1} - \sigma ) + (1/n) - 1}}} {\rm{d}}y\int_1^\infty {{u^{ - \mu - \beta - (1/pn)) - 1}}} {({\rm{ln}}{\kern 1pt} {\kern 1pt} u)^\gamma }{\rm{d}}u. \end{array}$

由式(9), 对于符合条件的 $\tilde{f}_{n}(x)$ 和 $\tilde{g}_{n}(y)$ , 有

$\begin{array}{*{20}{l}} {{{\tilde I}_2} = \int_0^\infty {{{\tilde g}_n}} (y)\left[ {\int_{1/y}^\infty h (xy){{\tilde f}_n}(x){\rm{d}}x} \right]{\rm{d}}y = }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \int_0^1 {{y^{({\sigma _1} - \sigma ) + (1/n) - 1}}} {\rm{d}}y\int_1^\infty {{u^{ - \mu - \beta - (1/(pn)) - 1}}} {{({\rm{ln}}{\kern 1pt} {\kern 1pt} u)}^\gamma }{\rm{d}}u \le }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {M_2}{{\tilde J}_2} = {M_2}n < \infty .} \end{array}$

(10)

由 $\sigma_{1}-\sigma+\frac{1}{n} \leqslant 0$ , 故得 $\int_{0}^{1} y^{\left(\sigma_{1}-\sigma\right)+(1 / n)-1} \mathrm{d} y=\infty$ .另外, 由于式(10)中的 $\int_{1}^{\infty} u^{-\mu-\beta-(1 /(p n))-1}(\ln u)^{\gamma} \mathrm{d} u$ 为正数, 可得 $\tilde{I}_{2}=\infty$ , 从而有 $M_{2} \tilde{J}_{2}=M_{2} n \geqslant \infty(n \in \mathbb{N})$ , 与式(10)的 $M_{2} \tilde{J}_{2}=M_{2} n < \infty$ 矛盾.

其次, 假设σ₁>σ, 则对于n≥1/(σ₁-σ) $(n \in \mathbb{N})$ , 设2个非负可测函数为:

${f_n}(x): = \left\{ {\begin{array}{*{20}{l}} {{x^{\sigma + (1/(pn)) - 1}}\quad (x{\kern 1pt} \in (0,1]),}\\ {0\quad (x{\kern 1pt} \in (1,\infty )),} \end{array}} \right.$

${g_n}(y): = \left\{ {\begin{array}{*{20}{l}} {0\quad (y{\kern 1pt} \in (0,1)),}\\ {{y^{{\sigma _1} - (1/(qn)) - 1}}\quad (y{\kern 1pt} \in [1,\infty )),} \end{array}} \right.$

则

$\begin{array}{l} {J_2}: = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\left[ {\int_0^\infty {{x^{p(1 - \sigma ) - 1}}} f_n^p(x){\rm{d}}x} \right]^{1/p}}{\left[ {\int_0^\infty {{y^{q(1 - {\sigma _1}) - 1}}} g_n^q(y){\rm{d}}y} \right]^{1/q}} = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\left[ {\int_0^1 {{x^{(1/n) - 1}}} {\rm{d}}x} \right]^{1/p}}{\left[ {\int_1^\infty {{y^{( - 1/n) - 1}}} {\rm{d}}y} \right]^{1/q}} = n < \infty . \end{array}$

设u=xy, μ₁=λ-σ₁, 根据Fubini定理^[13]及式(9), 有

$\begin{array}{l} {I_2}: = \int_0^\infty {{g_n}} (y)\left[ {\int_{1/y}^\infty h (xy){f_n}(x){\rm{d}}x} \right]{\rm{d}}y = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \int_0^\infty {{f_n}} (x)\left[ {\int_{1/x}^\infty h (xy){g_n}(y){\rm{d}}y} \right]{\rm{d}}x = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \int_0^1 {{x^{\sigma + (1/(pn)) - 1}}} \left[ {\int_{1/x}^\infty {\frac{{{{({\rm{min}}\{ xy,1\} )}^\beta }|{\rm{ln}}{\kern 1pt} xy{|^\gamma }}}{{{{({\rm{max}}\{ xy,1\} )}^{\lambda + \beta }}}}} {y^{{\sigma _1} - (1/(qn)) - 1}}{\rm{d}}y} \right]{\rm{d}}x = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \int_0^1 {{x^{(\sigma - {\sigma _1}) + (1/n) - 1}}} {\rm{d}}x\int_1^\infty {\frac{{{{({\rm{min}}\{ u,1\} )}^\beta }{{({\rm{ln}}{\kern 1pt} u)}^\gamma }}}{{{{({\rm{max}}\{ u,1\} )}^{\lambda + \beta }}}}} {u^{{\sigma _1} - (1/(qn)) - 1}}{\rm{d}}u = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \int_0^1 {{x^{(\sigma - {\sigma _1}) + (1/n) - 1}}} {\rm{d}}x\int_1^\infty {{u^{ - {\mu _1} - \beta - (1/(qn)) - 1}}} {({\rm{ln}}{\kern 1pt} u)^\gamma }{\rm{d}}u \le \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {M_2}{J_2} = {M_2}n < \infty . \end{array}$

(11)

因为 $\sigma-\sigma_{1}+\frac{1}{n} \leqslant 0$ , 所以 $\int_{0}^{1} x^{\left(\sigma-\sigma_{1}\right)+(1 / n)-1} \mathrm{d} x=\infty$ .在式(11)中, $\int_{1}^{\infty} u^{-\mu_{1}-\beta-(1 /(q n))-1}(\ln u)^{\gamma} \mathrm{d} u>0$ , 故I₂=∞, 从而有M₂J₂=M₂n≥∞ $(n \in \mathbb{N})$ , 与式(11)的M₂J₂=M₂n < ∞矛盾.

综上所述, 可得σ=σ₁，则μ=μ₁, 因此由式(11)推得

${M_2} \ge \int_1^\infty {{u^{ - \mu - \beta - (1/(qn)) - 1}}} {({\rm{ln}}{\kern 1pt} {\kern 1pt} u)^\gamma }{\rm{d}}u.$

(12)

由于{u^{-μ-β-(1/(qn))-1} (ln u)^γ}_n=1^∞在[1, ∞)上非负单调递增, 由Levi定理^[13]及式(8), 有

$\begin{array}{l} {M_2} \ge \mathop {{\rm{lim}}}\limits_{n \to \infty } \int_1^\infty {{u^{ - \mu - \beta - (1/(qn)) - 1}}} {({\rm{ln}}{\kern 1pt} {\kern 1pt} u)^\gamma }{\rm{d}}u = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \int_1^\infty {\mathop {{\rm{lim}}}\limits_{n \to \infty } } {\kern 1pt} {\kern 1pt} {u^{ - \mu - \beta - (\nu (qn)) - 1}}{({\rm{ln}}{\kern 1pt} {\kern 1pt} u)^\gamma }{\rm{d}}u = {k_2}(\sigma ) = \frac{1}{{{{(\mu + \beta )}^{\gamma + 1}}}}\Gamma (\gamma + 1). \end{array}$

证毕.

2. 主要结果

定理1 若参数λ, μ, β, γ, σ满足引理1条件, h(u)如式(7)所定义, 则以下命题等价：

(ⅰ)存在一个正常数M₂, 使当(0, ∞)上任意非负可测函数f(x)满足 $0 < \int_{0}^{\infty} x^{p(1-\sigma)-1} f^{p}(x) \mathrm{d} x<\infty$ 时, 以下具有非齐次核的第二类Hardy型积分不等式成立:

$\begin{array}{*{20}{l}} {J: = {{\left\{ {\int_0^\infty {{y^{p{\sigma _1} - 1}}} {{\left[ {\int_{1/y}^\infty h (xy)f(x){\rm{d}}x} \right]}^p}{\rm{d}}y} \right\}}^{1/p}} < }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {M_2}{{\left\{ {\int_0^\infty {{x^{p(1 - \sigma ) - 1}}} {f^p}(x){\rm{d}}x} \right\}}^{1/p}};} \end{array}$

(13)

(ⅱ)存在一个正常数M₂, 使当(0, ∞)上任意非负可测函数f(x)和g(y)满足 $0 < \int_{0}^{\infty} x^{p(1-\sigma)-1} f^{p}(x) \mathrm{d} x < \infty$ 和 $0 < \int_{0}^{\infty} y^{q\left(1-\sigma_{1}\right)-1} g^{q}(y) \mathrm{d} y < \infty$ 时, 以下不等式成立：

$\begin{array}{*{20}{l}} {I: = \int_0^\infty g (y)\left[ {\int_{1/y}^\infty h (xy)f(x){\rm{d}}x} \right]{\rm{d}}y < }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {M_2}{{\left[ {\int_0^\infty {{x^{p(1 - \sigma ) - 1}}} {f^p}(x){\rm{d}}x} \right]}^{1/p}}{{\left[ {\int_0^\infty {{y^{q(1 - {\sigma _1}) - 1}}} {g^q}(y){\rm{d}}y} \right]}^{1/q}};} \end{array}$

(14)

(ⅲ) σ=σ₁，M₂≥k₂(σ), 则式(13)及式(14)的常数因子M₂=k₂(σ)= $\frac{1}{(\mu+\beta)^{\gamma+1}}$ Γ(γ+1)为最佳值.

证明 (ⅰ)→(ⅱ):若式(13)成立, 由Hölder不等式^[14], 有

$\begin{array}{*{20}{l}} {I = \int_0^\infty {{y^{{\sigma _1} - (1/p)}}} \left[ {\int_{1/y}^\infty h (xy)f(x){\rm{d}}x} \right] \cdot {y^{(1/p) - {\sigma _1}}}g(y){\rm{d}}y \le }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} J{{\left[ {\int_0^\infty {{y^{q(1 - {\sigma _1}) - 1}}} {g^q}(y){\rm{d}}y} \right]}^{1/q}},} \end{array}$

(15)

则由式(15)、(13)可得到式(14).

(ⅱ)→(ⅲ):由于式(14)成立, 则(ⅱ)满足引理2的条件, 则由引理2及式(9)可得到σ=σ₁.

(ⅲ)→(ⅰ):若σ=σ₁成立, 设u=xy, y>0, 结合式(8), 对以下的权函数进行运算:

$\begin{array}{*{20}{l}} {{\omega _2}(\sigma ,y): = {y^\sigma }\int_{1/y}^\infty {{x^{\sigma - 1}}} h(xy){\rm{d}}x = }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \int_1^\infty {{u^{\sigma - 1}}} h(u){\rm{d}}u = {k_2}(\sigma ) = \frac{1}{{{{(\mu + \beta )}^{\gamma + 1}}}}\Gamma (\gamma + 1).} \end{array}$

(16)

由加权的Hölder不等式^[14]及式(16), 当y>0时, 有

$\begin{array}{l} {\left[ {\int_{1/y}^\infty h (xy)f(x){\rm{d}}x} \right]^p} = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\left\{ {\int_{1/y}^\infty h (xy)\left[ {\frac{{{y^{(\sigma - 1)/p}}}}{{{x^{(\sigma - 1)/q}}}}f(x)} \right]\left[ {\frac{{{x^{(\sigma - 1)/q}}}}{{{y^{(\sigma - 1)/p}}}}} \right]{\rm{d}}x} \right\}^p} \le \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \int_{1/y}^\infty h (xy)\frac{{{y^{\sigma - 1}}}}{{{x^{(\sigma - 1)p/q}}}}{f^p}(x){\rm{d}}x \cdot {\left[ {\int_{1/y}^\infty h (xy)\frac{{{x^{\sigma - 1}}}}{{{y^{(\sigma - 1)//p}}}}{\rm{d}}x} \right]^{p - 1}} = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\left[ {{\omega _2}(\sigma ,y){y^{q(1 - \sigma ) - 1}}} \right]^{p - 1}} \cdot \int_{1/y}^\infty h (xy)\frac{{{y^{\sigma - 1}}}}{{{x^{(\sigma - 1)p/q}}}}{f^p}(x){\rm{d}}x = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {[{k_2}(\sigma )]^{p - 1}}{y^{ - p\sigma + 1}} \cdot \int_{1/y}^\infty h (xy)\frac{{{y^{\sigma - 1}}}}{{{x^{({\sigma ^{ - 1}})p/q}}}}{f^p}(x){\rm{d}}x. \end{array}$

(17)

假设存在y>0, 使式(17)取等号, 则由文献[15], 必存在不全为零的常数A和B, 使得

$\frac{{A{y^{\sigma - 1}}}}{{{x^{(\sigma - 1)p/q}}}}{f^p}(x) = \frac{{B{x^{\sigma - 1}}}}{{{y^{(\sigma - 1)q/p}}}}{\rm{ a}}{\rm{.e}}{\rm{. 于}}(0,\infty ).$

不妨设A≠0(否则, A=B=0), 则有x^p(1-σ)-1f^p(x)=y^q(1-σ) $\frac{B}{A x}$ a.e.于(0, ∞).这与 $0 < \int_{0}^{\infty} x^{p(1-\sigma)-1} f^{p}(x) \mathrm{d} x < \infty$ 矛盾.因此, 式(17)取严格不等号, 则由σ=σ₁、式(17)和Fubini定理^[13], 得

$\begin{array}{l} J = {\left\{ {\int_0^\infty {{y^{p\sigma - 1}}} {{\left[ {\int_{1/y}^\infty h (xy)f(x){\rm{d}}x} \right]}^p}{\rm{d}}y} \right\}^{1/p}} < \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\left[ {{k_2}(\sigma )} \right]^{1/q}}{\left\{ {\mathop \smallint \nolimits_0^\infty \left[ {\int_{1/y}^\infty h (xy)\frac{{{y^{\sigma - 1}}}}{{{x^{(\sigma - 1)p/q}}}}{f^p}(x){\rm{d}}x} \right]{\rm{d}}y} \right\}^{1/p}} = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {[{k_2}(\sigma )]^{1/q}}{\left\{ {\mathop \smallint \nolimits_0^\infty \left[ {\int_{1/x}^\infty h (xy)\frac{{{y^{\sigma - 1}}}}{{{x^{(\sigma - 1)(p - 1)}}}}{\rm{d}}y} \right]{f^p}(x){\rm{d}}x} \right\}^{1/p}} = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {[{k_2}(\sigma )]^{1/q}}{\left[ {\mathop \smallint \nolimits_0^\infty {\omega _2}(\sigma ,x){x^{p(1 - \sigma ) - 1}}{f^p}(x){\rm{d}}x} \right]^{1/p}} = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {k_2}(\sigma ){\left[ {\int_0^\infty {{x^{p(1 - \sigma ) - 1}}} {f^p}(x){\rm{d}}x} \right]^{1/p}}, \end{array}$

即式(13)成立, 所以命题(ⅰ)~(ⅲ)均等价.

当σ=σ₁时，假设(ⅰ)、(ⅱ)的常数因子M₂≤k₂(σ), 则式(14)成立, 从而式(9)成立.这时由引理2得到M₂≥k₂(σ), 由假设推出M₂=k_λ(σ), 即式(14)的常数因子M₂=k₂(σ)= $\frac{1}{(\mu+\beta)^{\gamma+1}}$ Γ(γ+1)为最佳值.式(13)的常数因子M₂=k₂(σ)也为最佳值, 否则由式(15)及σ=σ₁, 得到式(14)的常数因子M₂=k₂(σ)不是最佳的, 这是矛盾的.证毕.

在定理1中, 作变换y=1/Y, 设G(Y)=Y^λ-2×g(1/Y), μ=λ-σ, μ₁=λ-σ₁, 然后把符号对[G(Y), Y]重新换成[g(y), y], 则有以下关于齐次核:

$h(x,y) = \frac{{{{({\rm{min}}\{ x,y\} )}^\beta }|{\rm{ln}}(x/y){|^\gamma }}}{{{{({\rm{max}}\{ x,y\} )}^{\lambda + \beta }}}}$

的第二类Hardy型积分不等式：

推论1 若参数λ, μ, β, γ, σ满足引理1条件, 则以下命题等价：

(ⅰ)存在一个正常数M₂, 使当(0, ∞)上任意非负可测函数f(x)满足 $0 < \int_{0}^{\infty} x^{p(1-\sigma)-1} f^{p}(x) \mathrm{d} x < \infty$ 时, 以下具有齐次核的第二类Hardy型积分不等式成立：

$\begin{array}{l} {\left\{ {\int_0^\infty {{y^{p{\mu _1} - 1}}} {{\left[ {\int_y^\infty {\frac{{{{({\rm{min}}\{ x,y\} )}^\beta }|{\rm{ln}}(x/y){|^\gamma }}}{{{{({\rm{max}}\{ x,y\} )}^{\lambda + \beta }}}}} f(x){\rm{d}}x} \right]}^p}{\rm{d}}y} \right\}^{1/p}} < \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {M_2}{\left\{ {\int_0^\infty {{x^{p(1 - \sigma ) - 1}}} {f^p}(x){\rm{d}}x} \right\}^{1/p}}; \end{array}$

(18)

(ⅱ)存在一个正常数M₂, 使当(0, ∞)上任意非负可测函数f(x)和g(y)满足 $0 < \int_{0}^{\infty} x^{p(1-\sigma)-1} f^{p}(x) \mathrm{d} x < \infty$ 和 $0 < \int_{0}^{\infty} y^{q\left(1-\mu_{1}\right)-1} g^{q}(y) \mathrm{d} y < \infty$ 时, 以下不等式成立：

$\begin{array}{l} \int_0^\infty g (y)\left[ {\int_y^\infty {\frac{{{{({\rm{min}}\{ x,y\} )}^\beta }|{\rm{ln}}(x/y){|^\gamma }}}{{{{({\rm{max}}\{ x,y\} )}^{\lambda + \beta }}}}} {\rm{d}}x} \right]{\rm{d}}y < \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {M_2}{\left[ {\int_0^\infty {{x^{p(1 - \sigma ) - 1}}} {f^p}(x){\rm{d}}x} \right]^{1/p}}{\left[ {\int_0^\infty {{y^{q(1 - {\mu _1}) - 1}}} {g^q}(y){\rm{d}}y} \right]^{1/q}}; \end{array}$

(19)

(ⅲ) μ=μ₁.

如果命题(ⅲ)成立, 则M₂≥k₂(σ), 且式(18)及式(19)的常数因子M₂=k₂(σ)= $\frac{1}{(\mu+\beta)^{\gamma+1}}$ Γ(γ+1)为最佳值.

致谢: 衷心感谢杨必成教授的细心指导.

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