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Judgment Conditions and Proof of Set Sequence Compactness in Metric Space C(Rn)
DOI: 10.12677/AAM.2019.84065, PDF, HTML, XML, 下载: 1,013  浏览: 3,225

Abstract: Set sequence compactness is an important concept in functional analysis. Using the sequence compactness, one can turn infinite dimensional problems to finite dimensional problems. In this paper, we give a necessary and sufficient condition for set sequence compactness on metric space C(Rn).

1. 引言

$C\left({R}^{n}\right)$ 空间是 ${R}^{n}$ 上全体连续函数构成的空间。则对任意 $u\left(x\right)\in C\left({R}^{n}\right)$ ，令

$u=\underset{k=1}{\overset{\infty }{\sum }}\frac{1}{{2}^{k}}\frac{\underset{x\le k}{\mathrm{max}}|u\left(x\right)|}{1+\underset{x\le k}{\mathrm{max}}|u\left(x\right)|}$

2. 预备知识

$d\left(u,v\right)=\underset{x\in M}{\mathrm{max}}|u\left(x\right)-v\left(x\right)|,\left(\forall u,v\in C\left(M\right)\right).$

3. 主要结论

1) 对任意 $k$ 属于 $C$ ，当 $‖x‖\le k$ 时，存在 ${C}_{k}>0$ ，使得 $\underset{‖x‖\le k}{\mathrm{sup}}|u\left(x\right)|\le {C}_{k}$ ，对 $\forall u\in U$

2) 对任意的 $k>0$$\left\{u\left(x\right)|u\in U\right\}$$‖x‖\le k$ 上是等度连续的。

$\rho \left({u}_{m}\left(x\right),{u}_{n}\left(x\right)\right)=\underset{i=1}{\overset{\infty }{\sum }}\frac{1}{{2}^{i}}\frac{\underset{‖x‖\le i}{\mathrm{max}}|{u}_{m}\left(x\right)-{u}_{n}\left(x\right)|}{1+\underset{‖x‖\le i}{\mathrm{max}}|{u}_{m}\left(x\right)-{u}_{n}\left(x\right)|}\to 0，\left(m,n\to 0\right)\text{.}$

$\underset{i=1}{\overset{\infty }{\sum }}\frac{1}{{2}^{i}}\frac{\underset{‖x‖\le i}{\mathrm{max}}|{u}_{m}\left(x\right)-{u}_{n}\left(x\right)|}{1+\underset{‖x‖\le i}{\mathrm{max}}|{u}_{m}\left(x\right)-{u}_{n}\left(x\right)|}<\frac{\epsilon }{{2}^{j+1}}，\left(m,n>{N}_{j}\right)\text{.}$

$\frac{1}{{2}^{j}}\frac{\underset{‖x‖\le j}{\mathrm{max}}|{u}_{m}\left(x\right)-{u}_{n}\left(x\right)|}{1+\underset{‖x‖\le j}{\mathrm{max}}|{u}_{m}\left(x\right)-{u}_{n}\left(x\right)|}<\frac{\epsilon }{{2}^{j+1}}⇒\underset{‖x‖\le j}{\mathrm{max}}|{u}_{m}\left(x\right)-{u}_{n}\left(x\right)|<\epsilon .$

${u}^{*}\left(x\right)=\left\{\begin{array}{c}{u}^{*1}\left(x\right),x\in \stackrel{¯}{B}\left(0,1\right);\\ {u}^{*2}\left(x\right),x\in \stackrel{¯}{B}\left(0,2\right);\\ \cdots \\ {u}^{*k}\left(x\right),x\in \stackrel{¯}{B}\left(0,k\right).\end{array}$

$\rho \left({u}_{n}\left(x\right),{u}^{*}\left(x\right)\right)=\underset{i=1}{\overset{\infty }{\sum }}\frac{1}{{2}^{i}}\frac{\underset{‖x‖\le i}{\mathrm{max}}|{u}_{n}\left(x\right)-{u}^{*}\left(x\right)|}{1+\underset{‖x‖\le i}{\mathrm{max}}|{u}_{n}\left(x\right)-{u}^{*}\left(x\right)|}\to 0,\left(n\to \infty \right)$

$\underset{i=1}{\overset{\infty }{\sum }}\frac{1}{{2}^{i}}\frac{\underset{‖x‖\le i}{\mathrm{max}}|{u}_{n}\left(x\right)-{u}^{*}\left(x\right)|}{1+\underset{‖x‖\le i}{\mathrm{max}}|{u}_{n}\left(x\right)-{u}^{*}\left(x\right)|}<\epsilon$

$\begin{array}{l}\underset{i=1}{\overset{\infty }{\sum }}\frac{1}{{2}^{i}}\frac{\underset{‖x‖\le i}{\mathrm{max}}|{u}_{n}\left(x\right)-{u}^{*}\left(x\right)|}{1+\underset{‖x‖\le i}{\mathrm{max}}|{u}_{n}\left(x\right)-{u}^{*}\left(x\right)|}\\ =\underset{i=1}{\overset{{n}_{0}}{\sum }}\frac{1}{{2}^{i}}\frac{\underset{‖x‖\le i}{\mathrm{max}}|{u}_{n}\left(x\right)-{u}^{*}\left(x\right)|}{\underset{‖x‖\le i}{\mathrm{max}}|{u}_{n}\left(x\right)-{u}^{*}\left(x\right)|}+\underset{i={n}_{0}+1}{\overset{\infty }{\sum }}\frac{1}{{2}^{i}}\frac{\underset{‖x‖\le i}{\mathrm{max}}|{u}_{n}\left(x\right)-{u}^{*}\left(x\right)|}{\underset{‖x‖\le i}{\mathrm{max}}|{u}_{n}\left(x\right)-{u}^{*}\left(x\right)|}\end{array}$

$\underset{‖x‖\le n}{\mathrm{max}}|{u}_{n}\left(x\right)-{u}^{*}\left(x\right)|<\frac{\epsilon }{2}$

$\begin{array}{l}\underset{n=1}{\overset{{n}_{0}}{\sum }}\frac{1}{{2}^{n}}\frac{\underset{‖x‖\le n}{\mathrm{max}}|{u}_{n}\left(x\right)-{u}^{*}\left(x\right)|}{1+\underset{‖x‖\le n}{\mathrm{max}}|{u}_{n}\left(x\right)-{u}^{*}\left(x\right)|}\\ <\underset{n=1}{\overset{{n}_{0}}{\sum }}\frac{1}{{2}^{n}}\underset{‖x‖\le n}{\mathrm{max}}|{u}_{n}\left(x\right)-{u}^{*}\left(x\right)|<\underset{n=1}{\overset{{n}_{0}}{\sum }}\frac{1}{{2}^{n}}\cdot \frac{\epsilon }{2}<\frac{\epsilon }{2}\end{array}$

$\underset{i={n}_{0}+1}{\overset{\infty }{\sum }}\frac{1}{{2}^{i}}\frac{\underset{‖x‖\le i}{\mathrm{max}}|{u}_{n}\left(x\right)-{u}^{*}\left(x\right)|}{1+\underset{‖x‖\le i}{\mathrm{max}}|{u}_{n}\left(x\right)-{u}^{*}\left(x\right)|}<\underset{i={n}_{0}+1}{\overset{\infty }{\sum }}\frac{1}{{2}^{i}}<\frac{\epsilon }{2}.$

$\underset{i=1}{\overset{\infty }{\sum }}\frac{1}{{2}^{i}}\frac{\underset{‖x‖\le i}{\mathrm{max}}|{u}_{n}\left(x\right)-{u}^{*}\left(x\right)|}{1+\underset{‖x‖\le i}{\mathrm{max}}|{u}_{n}\left(x\right)-{u}^{*}\left(x\right)|}<\epsilon .$

$\underset{i=1}{\overset{\infty }{\sum }}\frac{1}{{2}^{i}}\frac{\underset{‖x‖\le i}{\mathrm{max}}|u\left(x\right)-{u}_{1}^{k}\left(x\right)|}{1+\underset{‖x‖\le i}{\mathrm{max}}|u\left(x\right)-{u}_{1}^{k}\left(x\right)|}\le \frac{1}{{2}^{k+1}}.$

$i$ 取第 $k$ 项，

$\frac{1}{{2}^{k}}\frac{\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{1}^{k}\left(x\right)|}{1+\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{1}^{k}\left(x\right)|}\le \frac{1}{{2}^{k+1}}.$

$\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{1}^{k}\left(x\right)|\le 1.$

$n=2$ 时， $\rho \left(u,{u}_{2}^{k}\right)\le \frac{1}{{2}^{k+1}}$ ，即

$\underset{i=1}{\overset{\infty }{\sum }}\frac{1}{{2}^{k}}\frac{\underset{‖x‖\le i}{\mathrm{max}}|u\left(x\right)-{u}_{2}^{k}\left(x\right)|}{1+\underset{‖x‖\le i}{\mathrm{max}}|u\left(x\right)-{u}_{2}^{k}\left(x\right)|}\le \frac{1}{{2}^{k+1}}.$

$i$ 取第 $k$ 项，

$\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{2}^{k}\left(x\right)|\le 1.$

$n=k$ 时， $\rho \left(u,{u}_{k}^{k}\right)\le \frac{1}{{2}^{k+1}}$

$\underset{i=1}{\overset{\infty }{\sum }}\frac{1}{{2}^{k}}\frac{\underset{‖x‖\le i}{\mathrm{max}}|u\left(x\right)-{u}_{k}^{k}\left(x\right)|}{1+\underset{‖x‖\le i}{\mathrm{max}}|u\left(x\right)-{u}_{k}^{k}\left(x\right)|}\le \frac{1}{{2}^{k+1}}.$

$i$ 取第 $k$ 项，

$\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{k}^{k}\left(x\right)|\le 1.$

$\underset{‖x‖\le k}{\mathrm{sup}}|{u}_{i}|\le {C}_{k}$

$\rho \left(u\left(x\right),{u}_{i}\left(x\right)\right)=\underset{k=1}{\overset{\infty }{\sum }}\frac{1}{{2}^{k}}\frac{\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{i}\left(x\right)|}{1+\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{i}\left(x\right)|}<\frac{{\epsilon }^{\prime }}{3}$

$\frac{1}{{2}^{k}}\frac{\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{i}\left(x\right)|}{1+\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{i}\left(x\right)|}<\frac{{\epsilon }^{\prime }}{3}.$

$\frac{\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{i}\left(x\right)|}{1+\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{i}\left(x\right)|}<\frac{{\epsilon }^{\prime }}{3}\cdot {2}^{k}.$

$\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{i}\left(x\right)|<\frac{{\epsilon }^{\prime }}{3}\cdot {2}^{k}\cdot \left(1+\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{i}\left(x\right)|\right).$

$\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{i}\left(x\right)|<\frac{\frac{{\epsilon }^{\prime }}{3}\cdot {2}^{k}}{1-\frac{{\epsilon }^{\prime }}{3}\cdot {2}^{k}}<\frac{\epsilon }{3}.$

$\begin{array}{c}|u\left({x}_{1}\right)-u\left({x}_{2}\right)|\le |u\left({x}_{1}\right)-{u}_{i}\left({x}_{1}\right)|\\ +|{u}_{i}\left({x}_{1}\right)-{u}_{i}\left({x}_{2}\right)|+|u\left({x}_{2}\right)-{u}_{i}\left({x}_{2}\right)|\\ <2\cdot \frac{\epsilon }{3}+\frac{\epsilon }{3}=\epsilon .\end{array}$

$\begin{array}{c}\rho \left(u,{u}_{i}\right)=\underset{k=1}{\overset{\infty }{\sum }}\frac{1}{{2}^{k}}\frac{\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{i}\left(x\right)|}{1+\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{i}\left(x\right)|}\\ =\underset{k=1}{\overset{{n}_{0}}{\sum }}\frac{1}{{2}^{i}}\frac{\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{i}\left(x\right)|}{1+\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{i}\left(x\right)|}+\underset{k={n}_{0}+1}{\overset{\infty }{\sum }}\frac{1}{{2}^{i}}\frac{\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{i}\left(x\right)|}{1+\underset{‖x‖\le k}{\mathrm{max}}|u\left(x\right)-{u}_{i}\left(x\right)|}\\ <\frac{\epsilon }{4}\cdot \underset{k=1}{\overset{{n}_{0}}{\sum }}\frac{1}{{2}^{k}}+\frac{\epsilon }{2}\\ <\epsilon .\end{array}$

 [1] 张恭庆, 林源渠. 泛函分析讲义[M]. 北京: 北京大学出版社, 2004: 1-157. [2] 夏道行, 吴卓人, 严绍宗, 等. 实变函数与泛函分析[M]. 北京: 高等教育出版社, 1985: 1-119.