# Banach空间上凸集的k凸性K-Convexity of Convex Sets in Banach Spaces

• 全文下载: PDF(624KB)    PP.580-583   DOI: 10.12677/PM.2018.85077
• 下载量: 526  浏览量: 2,157

This paper breaks through the shackles of the unit ball, and generalizes the convexity theory of Banach space to the convex set which is not empty internally. It is more extensive than the k-convexity research of Banach Spaces.

1. 引言

1977年，Sullian引入了k一致凸空间的概念，开始了对Banach空间的k凸性的研究 [1] 。Banach空间的各种k凸性有一个共同的特点，即以Banach空间的单位球作为研究对象 [2] 。本文把Banach空间的凸性理论推广到内部不空的凸集上。首先，引入下面定义 [3]

2. 主要结果

1) 当 ${\mu }_{A}\left({\sum }_{i=1}^{k+1}{x}_{i}\right)={\sum }_{i=1}^{k+1}{\mu }_{A}\left({x}_{i}\right),\text{\hspace{0.17em}}{x}_{1},{x}_{2},\cdots ,{x}_{k+1}\in \partial A$ 时，存在不全为0的 $k+1$ 个实数 ${r}_{1},{r}_{2},\cdots ,{r}_{k+1}$ ，使得 ${\sum }_{i=1}^{k+1}{r}_{i}{x}_{i}=0$ 成立。其中 ${\mu }_{A}$ 是A上的Minkowski泛函。

2) 任意 $x\in \partial \left(A\right)$ ，则x是A的k端点。

3) A是k严格凸集。

${x}_{1},{x}_{2},\cdots ,{x}_{k+1}\in A$$0<{t}_{i}<1$${\sum }_{i=1}^{k+1}{t}_{i}=1$ ，使得 ${z}_{0}={\sum }_{i=1}^{k+1}{t}_{i}{x}_{i}$

${\mu }_{A}$ 为A上的Minkowski泛函，且 $0\in \mathrm{int}\left(A\right)$ ，故 ${\mu }_{A}$ 为X上的连续泛函，且 $\mathrm{int}\left(A\right)=\left\{x:{\mu }_{A}\left(x\right)<1\right\}$ ，且有 $\stackrel{¯}{\mathrm{int}\left(A\right)}=\stackrel{¯}{A}$ 成立。所以

$1={\mu }_{A}\left({z}_{0}\right)={\mu }_{A}\left(\underset{i=1}{\overset{k+1}{\sum }}{t}_{i}{x}_{i}\right)\le \underset{i=1}{\overset{k+1}{\sum }}{t}_{i}{\mu }_{A}\left({x}_{i}\right)\le \underset{i=1}{\overset{k+1}{\sum }}{t}_{i}=1$

$\begin{array}{c}{\mu }_{A}\left({z}_{0}\right)={\mu }_{A}\left({\left({t}_{1}+\cdots +{t}_{k+1}\right)}^{k}{z}_{0}\right)={\mu }_{A}\left({\left({t}_{1}+\cdots +{t}_{k+1}\right)}^{k}\left(\underset{i=1}{\overset{k+1}{\sum }}{t}_{i}{x}_{i}\right)\right)\\ ={\mu }_{A}\left({t}_{1}^{k+1}{x}_{1}+\cdots +{t}_{k+1}^{k+1}{x}_{k+1}+\cdots +k!{t}_{1}{t}_{2}\cdots {t}_{k+1}\left({x}_{1}+\cdots +{x}_{k+1}\right)\right)\\ <{t}_{1}^{k+1}+\cdots +{t}_{k+1}^{k+1}+\cdots +k!{t}_{1}{t}_{2}\cdots {t}_{k+1}\left(k+1\right)\\ ={\left({t}_{1}+\cdots +{t}_{k+1}\right)}^{k}\left({t}_{1}+\cdots +{t}_{k+1}\right)=1\end{array}$

${\mu }_{A}\left({z}_{0}\right)=1$ ，即 $1<1$ ，矛盾。故 ${\mu }_{A}\left({\sum }_{i=1}^{k+1}{x}_{i}\right)={\sum }_{i=1}^{k+1}{\mu }_{A}\left({x}_{i}\right)$ 成立。但不存在一组不全为0的实数 ${r}_{1},{r}_{2},\cdots ,{r}_{k+1}$ ，使得 ${\sum }_{i=1}^{k+1}{r}_{i}{x}_{i}=0$ ，同1)矛盾，故2)成立。

2)⟹3)任意 $x\in \partial \left(A\right)$ ，及实泛函f，满足 $f\left(x\right)\ge \mathrm{sup}f\left(\mathrm{int}A\right)$ ，任取 ${x}_{1},{x}_{2},\cdots ,{x}_{k+1}\in \left\{y:f\left(y\right)=f\left(x\right)\right\}$ ，令 $z=\frac{1}{k+1}\left({\sum }_{i=1}^{k+1}{x}_{i}\right)$ ，易证 $z\in \left\{y:f\left(y\right)=f\left(x\right)\right\}$ ，则 $z\in \partial A$ (事实上，假设 $z\in \mathrm{int}A$ ，则存在 $r\in R$ ，且 $r>1$

s.t $rz\in \mathrm{int}\left(A\right)$ ，由 $f\left(x\right)\ge \mathrm{sup}f\left(\mathrm{int}A\right)$$\stackrel{¯}{\mathrm{int}A}=\stackrel{¯}{A}$$f\left(x\right)\ge \mathrm{sup}f\left(A\right)$$r\left(z\right)\in A$ ，故 $f\left(x\right)=f\left(z\right)\ge f\left(rz\right)$ ，故 $1\ge r$ ，矛盾)通过2)知z是A的k端点。故不存在一组不全为0的实数 ${r}_{1},{r}_{2},\cdots ,{r}_{k+1}$ ，使得 ${\sum }_{i=1}^{k+1}{r}_{i}{x}_{i}=0$ ，故 $\left\{y=f\left(y\right)=f\left(x\right)\right\}$ 张成的实线性子空间维数不超过k。故3)成立。

3)⟹1)假设1)不成立，则存在 ${x}_{1},{x}_{2},\cdots ,{x}_{k+1}\in \partial A$ ，使得 ${\mu }_{A}\left({\sum }_{i=1}^{k+1}{x}_{i}\right)={\sum }_{i=1}^{k+1}{\mu }_{A}\left({x}_{i}\right)$ ，但不存在一组不全为0的实数 ${r}_{1},{r}_{2},\cdots ,{r}_{k+1}$ ，使得 ${\sum }_{i=1}^{k+1}{r}_{i}{x}_{i}=0$ ，令 ${x}_{0}=\frac{1}{k+1}{\sum }_{i=1}^{k+1}{x}_{i}⇒{\mu }_{A}\left({x}_{0}\right)=\frac{1}{k+1}{\sum }_{i=1}^{k+1}{\mu }_{A}\left({x}_{i}\right)=1$ ，故 ${x}_{0}\in \partial A$ 。由分离定理存在实泛函f，使得 $f\left({x}_{0}\right)\ge \mathrm{sup}f\left(\mathrm{int}A\right)$ ，故 $f\left({x}_{i}\right)=f\left({x}_{0}\right),i=1,2,\cdots ,k+1$ ，因此 ${\left\{{x}_{i}\right\}}_{i=1}^{k+1}\subset \left\{y:f\left(y\right)=f\left({x}_{0}\right)\right\}$ 。由3)知 $\left\{y=f\left(y\right)=f\left({x}_{0}\right)\right\}$ 张成的实线性子空间维数不超过k，矛盾。故1)成立。

$A\left({x}_{1},{x}_{2},\cdots ,{x}_{k+1}\right)=\mathrm{sup}\left\{‖\begin{array}{cccc}1& 1& \cdots & 1\\ {f}_{1}\left({x}_{1}\right)& {f}_{1}\left({x}_{2}\right)& \cdots & {f}_{1}\left({x}_{k+1}\right)\\ ⋮& ⋮& \ddots & ⋮\\ {f}_{k+1}\left({x}_{1}\right)& {f}_{k+1}\left({x}_{1}\right)& \cdots & {f}_{k+1}\left({x}_{k+1}\right)\end{array}‖:{f}_{i}\in S\left({X}^{*}\right)\right\}<\epsilon$

$\left\{\left({x}_{1}^{\left(n\right)},{x}_{2}^{\left(n\right)},\cdots ,{x}_{k+2}^{\left(n\right)}\right)|n=1,2,\cdots \right\}$ 是属于A边界的 $\left(k+2\right)$ 个元列，且存在实数列 ${\left\{{\delta }_{n}\right\}}_{n=1}^{\infty }$${\delta }_{n}\to 0\left(n\to \infty \right)$${B}_{{\delta }_{n}}\left(\frac{1}{k+1}\left({x}_{1}+{x}_{2}+\cdots +{x}_{k+1}\right)\right)\not\subset A$ 成立，则存在实数列 ${\left\{{{\delta }^{\prime }}_{n}\right\}}_{n=1}^{\infty }$${{\delta }^{\prime }}_{n}\to 0\left(n\to \infty \right)$ ，使得 ${B}_{{{\delta }^{\prime }}_{n}}\left(\frac{1}{k+1}\left({x}_{1}+{x}_{2}+\cdots +{x}_{k+1}\right)\right)\not\subset A$ 成立。(事实上，假设上述结论不成立，则存在 $\delta >0$ ，使得 ${B}_{\delta }\left(\frac{1}{k+1}\left({x}_{1}+{x}_{2}+\cdots +{x}_{k+1}\right)\right)\subset A$ 成立。即 $\frac{1}{k+1}\left({x}_{1}+{x}_{2}+\cdots +{x}_{k+1}\right)\in \mathrm{int}\left(A\right)$ ，不失一般性，不妨令 $0\in \mathrm{int}\left(A\right)$ ，令 ${\mu }_{A}$ 为A上的Minkowski泛函，则 ${\mu }_{A}$ 连续且 $\mathrm{int}\left(A\right)=\left\{x:{\mu }_{A}<1\right\}$ ，则

$\begin{array}{l}{\mu }_{A}\left(\frac{1}{k+2}\left({x}_{1}+{x}_{2}+\cdots +{x}_{k+2}\right)\right)={\mu }_{A}\left(\frac{k+1}{k+2}\left(\frac{{x}_{1}^{\left({n}_{k}\right)}+\cdots +{x}_{k+1}^{\left({n}_{k}\right)}}{k+1}\right)+\frac{1}{k+2}{x}_{k+2}^{\left({n}_{k}\right)}\right)\\ =\frac{k+1}{k+2}{\mu }_{A}\left(\frac{1}{k+1}\left({x}_{1}^{\left({n}_{k}\right)}+\cdots +{x}_{k+2}^{\left({n}_{k}\right)}\right)+\frac{1}{k+2}{x}_{k+2}^{\left({n}_{k}\right)}\right)=m<1\end{array}$

${y}_{{n}_{k}}=\frac{1}{k+2}\left({x}_{1}^{\left({n}_{k}\right)}+\cdots +{x}_{k+2}^{\left({n}_{k}\right)}\right)$ ，由 ${\mu }_{A}$ 的连续性知，存在0点的领域 ${B}_{{\delta }_{1}}\left(0\right)$ ，使得 ${\mu }_{A}\left({B}_{{\delta }_{1}}\left(0\right)\right)<\frac{1}{2}\left(1-m\right)$ ，取 ${y}_{{n}_{k}}$ 的领域 ${y}_{{n}_{k}}+{B}_{{\delta }_{1}}\left(0\right)$ ，当 $y\in {y}_{{n}_{k}}+{B}_{{\delta }_{1}}\left(0\right)$ 时， ${\mu }_{A}\left(y\right)\le {\mu }_{A}\left(y-{y}_{{n}_{k}}\right)+{\mu }_{A}\left({y}_{{n}_{k}}\right)\le \frac{1}{2}\left(1-m\right)+m<1$ ，故 ${y}_{{n}_{k}}+{B}_{{\delta }_{1}}\left(0\right)\subset \mathrm{int}\left(A\right)\subset A$ 。这与存在实数列 ${\left\{{\delta }_{n}\right\}}_{n=1}^{\infty }$${\delta }_{n}\to 0\left(n\to \infty \right)$${B}_{{\delta }_{n}}\left(\frac{1}{k+1}\left({x}_{1}+{x}_{2}+\cdots +{x}_{k+1}\right)\right)\not\subset A$ 矛盾。)由A的k一致凸性知 $A\left({x}_{1}^{\left(n\right)},\cdots ,{x}_{k+1}^{\left(n\right)}\right)\to 0\left(n\to \infty \right)$ ，同理当 $\left(1\le i\le k+2\right)$ 时， $A\left({x}_{1}^{\left(n\right)},\cdots ,{x}_{i-1}^{\left(n\right)},{x}_{i+1}^{\left(n\right)},\cdots ,{x}_{k+1}^{\left(n\right)}\right)\to 0\left(n\to \infty \right)$

$\begin{array}{l}\mathrm{sup}\left\{‖\begin{array}{cccc}1& 1& \cdots & 1\\ {f}_{1}\left({x}_{1}\right)& {f}_{1}\left({x}_{2}\right)& \cdots & {f}_{1}\left({x}_{k+1}\right)\\ ⋮& ⋮& \ddots & ⋮\\ {f}_{k}\left({x}_{k}\right)& {f}_{k}\left({x}_{k}\right)& \cdots & {f}_{k}\left({x}_{k+1}\right)\end{array}‖:{f}_{i}\in S{\left(X,{‖\text{ }\cdot \text{ }‖}_{1}\right)}^{*}\right\}\\ =‖{f}_{1}‖\cdots ‖{f}_{k}‖\mathrm{sup}\left\{‖\begin{array}{cccc}1& 1& \cdots & 1\\ \frac{{f}_{1}}{{f}_{1}}\left({x}_{1}\right)& \frac{{f}_{1}}{{f}_{1}}\left({x}_{2}\right)& \cdots & \frac{{f}_{1}}{{f}_{1}}\left({x}_{k+1}\right)\\ ⋮& ⋮& \ddots & ⋮\\ \frac{{f}_{k}}{{f}_{k}}\left({x}_{k}\right)& \frac{{f}_{k}}{{f}_{k}}\left({x}_{k}\right)& \cdots & \frac{{f}_{k}}{{f}_{k}}\left({x}_{k+1}\right)\end{array}‖:{f}_{i}\in S{\left(X,{‖\text{ }\cdot \text{ }‖}_{1}\right)}^{*}\right\}\\ \le ‖{f}_{1}‖\cdots ‖{f}_{k}‖A\left({x}_{1},\cdots ,{x}_{k+1}\right)\le {\left(\frac{1}{{m}^{\prime }}\right)}^{k}{‖{f}_{1}‖}_{1}\cdots {‖{f}_{k}‖}_{1}\frac{\epsilon }{M}=\frac{1}{M}{\left(\frac{1}{{m}^{\prime }}\right)}^{k}\epsilon \end{array}$

$\left(X,{‖\text{ }\cdot \text{ }‖}_{1}\right)$ 是k一致凸。

 [1] 方习年. Banach空间的凸性和光滑性及其应用[J]. 安徽工程大学学报, 2001, 16(3): 1-8. [2] 乌日娜. Banach空间的某些凸性、光滑性与可凹性的研究[D]: [硕士学位论文]. 内蒙古: 内蒙古师范大学, 2008. [3] 商绍强. Banach空间上凸集的凸性[J]. 应用泛函分析学报, 2009, 11(2): 170-177.