# Krein-Milman Theorem

## Theorem

Let $\struct {X, \norm \cdot}$ be a normed vector space over $\R$.

Let $K$ be a non-empty compact convex subset of $X$.

Let $\map E K$ be the set of extreme points in $K$.

Then $K$ is the closed convex hull of $\map E K$.

## Proof

Let $K'$ be the closed convex hull of $\map E K$.

We show that $K' = K$.

From Convex Hull is Smallest Convex Set containing Set, we have:

- $\map E K \subseteq \map {\operatorname {conv} } {\map E K}$

From Set is Subset of its Topological Closure, we therefore have:

- $\map E K \subseteq K'$

We also have, from Convex Hull preserves Subsets:

- $\map {\operatorname {conv} } {\map E K} \subseteq \map {\operatorname {conv} } K$

From Convex Hull is Smallest Convex Set containing Set: Corollary, we have:

- $\map {\operatorname {conv} } {\map E K} \subseteq K$

Then, from Set Closure Preserves Set Inclusion, we have:

- $K' \subseteq K^-$

where $K^-$ is the closure of $K$.

From Compact Subspace of Hausdorff Space is Closed, we have that:

- $K$ is closed.

So we have:

- $K' \subseteq K$

Aiming for a contradiction, suppose suppose that $K' \ne K$.

Then there exists $a \in K \setminus K'$.

Note that $K'$ is closed from Topological Closure is Closed.

Since $K$ is compact and $K' \subseteq K$, we have:

- $K'$ is compact

from Closed Subspace of Compact Space is Compact.

From Closed Convex Hull in Normed Vector Space is Convex, we have that:

- $K'$ is convex.

Finally:

- $\set a$ is closed

from Singleton in Normed Vector Space is Closed, and:

- $\set a$ is convex

from Singleton is Convex Set.

So, from Hahn-Banach Separation Theorem in the case of a compact convex set and a closed convex set, there exists $f \in X^\ast$ such that:

- $\map f x < \map f a$

for each $x \in K'$.

Now let:

- $\ds K^f = \set {x \in K : \map f x = \max_{y \mathop \in K} \map f y}$

From Preimage of Maximum of Bounded Linear Functional on Extreme Set in Convex Compact Set is Extreme Set:

- $K^f$ is an extreme set in $K$.

Since:

- $\map f x < \map f a$ for each $x \in K'$

and $a \in K$, we have:

- $\ds \map f x < \max_{y \mathop \in K} \map f y$ for each $x \in K'$

from the definition of maximum.

So, we have:

- $K^f \cap K' = \O$

However, from Extreme Set in Compact Convex Set contains Extreme Point, we also have:

- $K^f$ contains an extreme point of $K$, say $x \in \map E K$

while:

- $\map E K \subseteq K$

So we have:

- $x \in K^f \cap K'$

contradicting that $K^f \cap K$ is empty.

So, there exists no such $a \in K \setminus K'$, and we have:

- $K = K'$

$\blacksquare$

## Source of Name

This entry was named for Mark Grigorievich Krein and David Pinhusovich Milman.

## Sources

- 2020: James C. Robinson:
*Introduction to Functional Analysis*... (previous) ... (next) $21.6$: The Krein-Milman Theorem