11.3 Compact Convex Sets
Definition 11.3.1 (Extreme Point).label Let $E$ be a vector space over $\real$, $K \subset E$, and $x \in K$, then $x$ is extremal if there exists no $y, z \in K$ such that $x \in (y, z) \subset K$.
Definition 11.3.2 (Extreme Subset).label Let $E$ be a vector space over $\real$, $K \subset E$ be convex, and $A \subset K$, then $A$ is extreme set if for any $x \in A$ and $y, z \in K$ such that $x \in (y, z)$, $y, z \in A$ as well.
Lemma 11.3.3.label Let $E$ be a locally convex space over $\real$, $K \subset E$ be non-empty and compact, and $\phi \in E^{*}$. Let $\alpha = \sup\bracs{\dpn{x, \phi}{E}|x \in K}$, then $A = \bracs{\phi = \alpha}\cap K$ is a non-empty extreme subset of $K$.
Proof. Since $K$ is compact, $\alpha < \infty$ and $A$ is non-empty by Proposition 5.16.2. Let $x \in A$ and $y, z \in K$ such that $x \in (y, z)$. By definition of $\alpha$, $\dpn{y, \phi}{E}= \dpn{z, \phi}{E}= \alpha$. Thus $y, z \in A$ as well.$\square$
Theorem 11.3.4 (Krein-Milman).label Let $E$ be a separated locally convex space over $\real$ and $K \subset E$ be a compact convex set, then $K$ is the closed convex hull of its extreme points.
Proof, [Theorem 1.12.5, BS17]. Assume without loss of generality that $K \ne \emptyset$.
Let $K_{0} \subset K$ be a closed, extreme subset of $K$, and $\mathcal{E}(K_{0}) \subset 2^{K}$ be the collection of all non-empty closed extreme subsets of $K$ contained in $K_{0}$. Since $K_{0} \in \mathcal{E}(K_{0})$, $\mathcal{E}(K_{0}) \ne \emptyset$. Since $K$ is compact, for any chain $\mathcal{C}\subset \mathcal{E}$, $\bigcap_{A \in \mathcal{C}}A$ is also non-empty, closed, and extreme.
By Zorn’s lemma, there exists a minimal element $A$ of $\mathcal{E}(K_{0})$. Let $x, y \in A$, $\phi \in E^{*}$, and $\alpha = \sup_{z \in A}\dpn{z, \phi}{E}$, then $\bracs{\phi = \alpha}\cap K$ is a non-empty, closed, and extreme subset of $K$ by Lemma 11.3.3, so $A \cap \bracs{\phi = \alpha}$ is also extreme. By minimality of $A$, $A \subset \bracs{\phi = \alpha}$. Thus by the Hahn-Banach Theorem, $A$ consists of exactly one point. In which case, for any $y, z \in E$ with $A \subset (y, z) \subset K$, $y = z \in A$. Therefore there exists an extreme point of $K$ in $K_{0}$.
Since $K$ itself is an extreme subset, a minimal element of $\mathcal{E}(K)$ represents an extreme point, so $K$ admits at least one extreme point.
Now, let $C$ be the collection of all extreme points in $K$. Assume for contradiction that $\ol{\conv}(C) \subsetneq K$, then by the Hahn-Banach Theorem, there exists $x \in K$ and $\phi \in E^{*}$ such that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E}< \dpn{x, \phi}{E}$. Let $\alpha = \sup_{z \in K}\dpn{z, \phi}{E}$, then by Lemma 11.3.3, $K \cap \bracs{\phi = \alpha}$ is a non-empty, closed, and extreme subset of $K$. By the preceding discussion, there exists an extreme point of $K$ in $K \cap \bracs{\phi = \alpha}$, which contradicts the fact that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E}< \alpha$.$\square$
Theorem 11.3.5 (Markov-Kakutani Fixed Point Theorem).label Let $E$ be a separated topological vector space over $\real$, $K \subset E$ be a compact convex set, and $\cf \subset C(K; K)$ such that:
- (a)
For any $f, g \in \cf$, $f \circ g = g \circ f$.
- (b)
For each $f \in \cf$, $x, y \in K$, and $t \in [0, 1]$,
\[f(tx + (1 - t)y) = tf(x) + (1 - t)f(y)\]
then there exists $x_{0} \in K$ such that $f(x_{0}) = x_{0}$ for all $f \in \cf$.
Proof, [Theorem 1.12.10, BS17]. For each $f \in \cf$ and $n \in \natp$, define $f^{(n)}= \frac{1}{n}\sum_{k = 0}^{n - 1}f^{k}$, then $f^{(n)}\in C(K; K)$ as well. Via a closure operation, assume without loss of generality that:
- (1)
For any $f \in \cf$ and $n \in \natp$, $f^{(n)}\in \cf$.
- (2)
For any $f, g \in \cf$, $f \circ g \in \cf$.
Let $\bracsn{f_j}_{1}^{N} \subset \cf$ and $\bracsn{n_j}_{1}^{N} \subset \natp$, then
thus any finite intersections of elements in
is non-empty. Since $E$ is separated, by Proposition 5.16.2 and Proposition 5.16.3, $\mathcal{K}$ is a family of closed sets satisfying the finite intersection property. Hence $\bigcap_{n \in \natp}\bigcap_{f \in \cf}f^{(n)}(K) \ne \emptyset$.
Now, let $x \in \bigcap_{n \in \natp}\bigcap_{f \in \cf}f^{(n)}(K)$ and $U \in \cn_{E}(0)$, then there exists $N \in \natp$ such that $NU \supset K$. In which case, there exists $y \in K$ such that
In which case,
As this holds for all $U \in \cn_{E}(0)$ and $E$ is separated, $f(x) = x$.$\square$
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