The Krein-Milman Theorem

Recall from The Krein-Milman Lemma page that if $X$ is a locally convex topological vector space and $K$ is a nonempty, closed, convex subset of $X$, $f : X \to \mathbb{R}$ is continuous, the set of points in $K$ for which $f$ attains its maximum value on $K$ is an extreme subset of $K$.

Furthermore, the Krein-Milman lemma states that if $X$ is a locally convex topological vector space then every nonempty, convex, compact subset of $X$ has an extreme point.

We will now state and prove the very important Krein-Milman theorem.

Theorem 1 (The Krein-Milman 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 $E \subseteq K$ we have that $C \subseteq K$. If $K = C$ we are done. Otherwise, suppose that $K \neq C$ and let $x_0 \in K \setminus K$.

• 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 a nonempty, closed, and convex, there exists a continuous linear functional $f$ of $X$ such that:

(1)

\begin{align} \quad f(x_0) > \sup_{x \in C} f(x) \end{align}

• Since $K$ is compact and $C \subset K$ is closed, we have that $C$ is compact, and so more specifically:

(2)

\begin{align} \quad f(x_0) > \max_{x \in C} f(x) \end{align}

• And since $E \subset C$ we have that:

(3)

\begin{align} \quad f(x_0) > \max_{x \in C} f(x) \geq \max_{x \in E} f(x) \end{align}

• Let $m = \max_{x \in K} f(x)$. Then $M = f^{-1}(\{ m \})$, the set of all points in $K$ for which the maximum of $f$ is attained is an extreme subset of $K$ (from the lemma mentioned at the top of the page). $M$ is nonempty, convex, and compact. So by the Krein-Milman lemma again, $M$ has an extreme point, call it $z \in M$.

• Since $z$ is an extreme point of $M$ and $M \subseteq K$ we have that $z$ is an extreme point of $K$. So $z \in E$. Hence

(4)

\begin{align} \quad f(z) \geq f(x_0) > \max_{x \in C} f(x) \geq \max_{x \in E} f(x) > f(z) \end{align}

• Therefore $f(z) < f(z)$, a contradiction. So the assumption that $K \neq C$ is false. Hence $K$ is the closed convex hull of its extreme points. $\blacksquare$