The Krein-Milman Lemma

# The Krein-Milman Lemma

Recall from the Extreme Subsets and Extreme Points of a Set in a LCTVS page that if $X$ is a locally convex topological vector space and if $K$ is a nonempty convex subset of $X$ then a subset $E$ of $K$ is said to be an extreme subset of $K$ if $E$ is nonempty, closed, convex, and whenever $x \in E$ is such that $x = \lambda u + (1 - \lambda)v$ for some $\lambda \in [0,1]$ and $u, v \in K$ we have that $u, v \in E$.

We said that a point $x \in K$ is an extreme point of $K$ if $\{ x \}$ is an extreme subset of $K$.

Lemma 1: Let $X$ be a locally convex topological vector space and let $K$ be a nonempty, convex, compact subset of $X$. If $f : X \to \mathbb{R}$ is linear and continuous then the set of points in $K$ for which $f$ attains its maximum value is an extreme subset of $K$. |

**Proof:**Since $K$ is nonempty and compact, and $f : X \to \mathbb{R}$ is continuous, we have by the extreme value theorem that $f$ attains a maximum value on $K$. Let:

\begin{align} \quad m = \max_{k \in K} f(k) \end{align}

- Let:

\begin{align} \quad M = \{ k \in K : f(k) = m \} \end{align}

- Then $M$ is nonempty from above. Furthermore, since the singleton set $\{ m \}$ is closed in $\mathbb{R}$ and $f$ is continuous, $f^{-1}(\{ m \}) = M$ is closed. We now show that $M$ is convex.

- Let $k_1, k_2 \in M$ and let $\lambda \in [0, 1]$. Let $k = \lambda k_1 + (1 - \lambda)k_2$. Then by the linearity of $f$:

\begin{align} \quad f(k) = f(\lambda k_1) + f((1 - \lambda)k_2) = \lambda f(k_1) + (1 - \lambda) f(k_2) = \lambda m + (1 - \lambda) m = m \end{align}

- So $k \in M$, i.e., $M$ is convex.

- Lastly, let $x \in M$ and suppose that $x = \lambda u + (1 - \lambda) v$ for some $u, v \in K$. Since $u, v \in K$ we must have that $f(u) \leq m$ and $f(v) \leq m$. So:

\begin{align} \quad m = f(x) = f(\lambda u + (1 - \lambda) v) = \lambda f(u) + (1 - \lambda) f(v) \leq \lambda m + (1 - \lambda) m \end{align}

- Therefore:

\begin{align} \quad \lambda f(u) + (1 - \lambda) f(v) = m \end{align}

- Which implies that $f(u) = m$ and $f(v) = m$, that is, $u, v \in M$. So $M$ is an extreme subset of $K$. $\blacksquare$

Lemma 2 (The Krein-Milman Lemma): Let $X$ be a locally convex topological vector space and let $K$ be a nonempty convex, compact subset of $X$. Then $K$ has an extreme point. |

The Krein-Milman lemma tells us that every nonempty convex and compact subset of a locally convex topological vector space has an extreme point.