We consider $X$ a normed space, $K$ a compact and convex subset of $X$ and $E \subseteq K$. We need to show that the following are equivalent:
- $K= \overline{conv(E)}$
- for every continuous linear functional $f:X\rightarrow \mathbb{R} $ we have $ \sup_{x \in K}f(x)= \sup_{x \in E}f(x)$
Also, by $conv(E)$, we denote the convex hull of $E$, meaning the set of all convex combinations of points in $E$.
I know that $K = conv(Ext(K))$, by the Krein-Milman theorem, where $Ext(K)$ is the set of the extreme points of $K$.
I don't know where to go from there. Any help would be greatly appreciated.
Best Answer
The Krein Milman theorem does not say only that. It also says that, if $K=\overline{\text{co}}(E)$ for some subset $E$, then $E$ contains the extreme points of $K$. See for example Murphy's approach in appendix A.
So assume the first bullet holds. Fix a functional $f\in X^*$. By our earlier remark it (is necessary and) suffices to show that $\sup_{x\in K}f(x)=\sup_{x\in\text{ext}(K)}f(x)$. Note that since $K$ is compact, this supremum is actually a maximum. Set $$M=\{y\in K: f(y)=\sup_{x\in K}f(y)\}$$ and therefore $M$ is non-empty. Note that $M$ is a face of $K$, so it contains an extreme point, by the Krein-Milman theorem. But extreme points of a face of $K$ are extreme points in $K$ as well, so $M\cap\text{ext}(K)\neq\emptyset$. Therefore, if $c\in M\cap\text{ext}(K)$ we have that $$\sup_{x\in K}f(x)=f(c)\leq\sup_{x\in\text{ext}(K)}f(x)$$ and the other inequality is obvious.
For the converse, we use the Hahn-Banach theorem: if $\overline{\text{co}}(E)\neq K$, then find a point $x_0\in K\setminus\overline{\text{co}}(E)$. Since $\overline{\text{co}}(E)$ is a closed subset of $K$, it is compact. It is obviously convex, so by the Hahn-Banach theorem there exists a continuous functional $f$ and a real number $t$ so that $f(x)<t<f(x_0)$ for all $x\in\overline{\text{co}}(E)$. But then $\sup_{x\in E}f(x)\leq t<f(x_0)\leq\sup_{x\in K}f(x)$, a contradiction to our hypothesis.