= Solution
A sufficient geometric form of the <Hahn-Banach separation theorem> is this: if $C$ is a nonempty closed <convex set> in a real <normed vector space> and $x\notin C$, there exists a <continuous linear functional> $\ell$ such that
$$
\sup_{c\in C}\ell(c)<\ell(x).
$$
We use this form without proving it, as permitted. We will prove the <Krein-Milman theorem> for a nonempty compact <convex set> $K$: \b[$K$ is the closed convex hull of its extreme points].
First prove the existence of <extreme points> by the <minimal compact face argument>. A <face of a convex set> $K$ is a convex subset $F$ such that, whenever $ta+(1-t)b\in F$ with $a,b\in K$ and $0<t<1$, both $a$ and $b$ belong to $F$. Consider all nonempty compact faces of $K$, ordered by reverse inclusion. The set $K$ itself is such a face. A chain has a nonempty compact intersection, by the <finite intersection property> and compactness of $K$. This intersection is convex and a face, since membership in every face forces the same membership of endpoints. It is an upper bound in the reverse order. The <Zorn lemma> therefore gives a minimal nonempty compact face $F$.
If $F$ contained distinct $a,b$, geometric separation of one point from the other would give a continuous real linear $\ell$ with $\ell(a)\ne\ell(b)$. Compactness supplies a maximum $m=\max_F\ell$. The set $F'=\{v\in F:\ell(v)=m\}$ is a nonempty proper compact convex subset. It is also a face of $K$: if a strict convex combination belongs to $F'$, first the face property of $F$ puts both endpoints in $F$, and then linearity and maximality of $m$ force both their values to equal $m$. Thus $F'$ contradicts minimality of $F$. Hence $F=\{e\}$, whose face property says precisely that $e$ is an <extreme point> of $K$. The same argument starts with any nonempty compact face, so every such face contains an <extreme point> of $K$.
Let $E$ be the set of all <extreme points> of $K$ and $C=\overline{\operatorname{conv}}E$. It is nonempty, and $C\subseteq K$ because $K$ is closed and convex. Suppose $x\in K\setminus C$. Geometric separation gives a continuous real linear $\ell$ with $\ell(x)>\sup_C\ell$. Let $m=\max_K\ell$ and consider the maximizer face $F=\{v\in K:\ell(v)=m\}$. It is nonempty, compact and a face of $K$, so the previous argument gives an extreme point $e\in F$. But $e\in E\subseteq C$, while
$$
\ell(e)=m\ge\ell(x)>\sup_C\ell,
$$
a contradiction. Therefore \b[every point of $K$ belongs to the closed convex hull of its extreme points], proving
$$
\boxed{K=\overline{\operatorname{conv}}\operatorname{ext}K.}
$$
Neither completeness of the ambient normed space nor finite dimensionality was used; compactness is the key hypothesis.
Back to article page
