I am trying ton understand the proof of The Krein-Milman theorem. Specifically I am reading this webpage. It goes like this:
From math online.wikidot.com:
Theorem: Let $X$ be a locally convex topological vector space and let $K$ be a nonempty, convex, compact subset of $X$. Then $K$ is equal to the closed convex hull of the extreme points of $K.$
Proof: Let $E$ be the set of extreme points of $K$ and let $C$ be the closed convex hull of $E$. Note that since $E\subseteq K$ and we have that $C\subset K$. If $K=C$ we are done. Otherwise, suppose $K\neq C$ and $x_0\in K\setminus C$. Now since $K$ is nonempty, convex, compact subset of $X$ we have by The Krein-Milman lemma that $E\neq\emptyset$ and so $C\neq \emptyset$. Since $x_0\in K\setminus C$ and since $C$ is nonempty closed and convex there exists a continuous linear functional $f$ of $X$ such that:
\begin{equation*}
f(x_0)>\sup_{x\in C}f(x).
\end{equation*}
And the proof goes on…..
The part that I don't understand is the existence of linear functional $f$ of $X$ with $f(x_0)>\sup_{x\in C}f(x)$. What guarantees the existence such linear functional? Maybe Hahn-Banach Theorem helps to see why? And why $f(x_0)>\sup_{x\in C}f(x)$? not $f(x_0)\geq\sup_{x\in C}f(x)$?
Best Answer
$\{x_0\}$ is a compact convex set and $C$ is a closed convex set. These two are disjoint. By Hahn Banach Theorem there exist a continuous linear functional $f$ and a real number $r$ such that $f(x_0) >r>f(x)$ for all $x \in C$. [Ref. Rudin's FA]. This implies that $\sup_{x \in C} f(x) \leq r <f(x_0)$.