# The proof of Krein-Milman Theorem and the reason behind the existence of linear functional

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 $$f(x_0)\geq\sup_{x\in C}f(x)$$?

• These types of results are called hyperplane separation theorems, in this case separating a point and a closed convex set. Geometrically, $f$ is the "normal vector" of a hyperplane that the point and the set are on opposite sides of, so it separates them. Hahn-Banach Theorem is indeed used to construct it. Commented Apr 21, 2020 at 6:34
• @Conifold Thank you so much!
– user709182
Commented Apr 22, 2020 at 2:04

$$\{x_0\}$$ 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 .