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

  • 1
    $\begingroup$ 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. $\endgroup$
    – Conifold
    Commented Apr 21, 2020 at 6:34
  • $\begingroup$ @Conifold Thank you so much! $\endgroup$
    – user709182
    Commented Apr 22, 2020 at 2:04

1 Answer 1


$\{x_0\}$ 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 $\sup_{x \in C} f(x) \leq r <f(x_0)$.


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