The Krein-Milman Theorem

If $(X, \Omega , \mu )$ is a $\sigma$-finite measure space and $1 < p < \infty$, then the set of extreme points of $\ball L^p(\mu )$ is $\left\{ f \in L^p(\mu ) : \left\| f\right\|_{p} = 1 \right\}$.

If $(X, \Omega , \mu )$ is a $\sigma$-finite measure space, the set of extreme points of $\ball L^1(\mu )$ is $\left\{ \alpha \chi_E : E \text{ is an atom of } \mu , \ \alpha \in \mathbb {F}, \text{ and } \left|\alpha \right| = \mu (E)^{-1} \right\}$.

If $(X, \Omega , \mu )$ is a $\sigma$-finite measure space, the set of extreme points of $\ball L^\infty (\mu )$ is $\left\{ f \in L^\infty (\mu ) : \left|f(x)\right| = 1 \text{ a.e. } [\mu ] \right\}$.

If $X$ is completely regular, the set of extreme points of $\ball C_b(X)$ is $\left\{ f \in C_b(X) : \left|f(x)\right| = 1 \text{ for all } x \right\}$. So $\ball C_{\mathbb {R}}[0,1]$ has only two extreme points.

Let $X$ be a totally disconnected compact space. (That is, $X$ is compact and if $x \in X$ and $U$ is a neighborhood of $x$, then there is a subset $V$ of $X$ that is both open and closed and such that $x \in V \subseteq U$. The Cantor set is an example of such a space.) Show that $\ball C(X)$ is the norm closure of the convex hull of its extreme points.

Show that $\ball l^1$ is the norm closure of the convex hull of its extreme points.

Show that if $X$ is locally compact but not compact, then $\ball C_0(X)$ has no extreme points.

If $\mathcal{X}$ is a LCS and $K_1, \dots , K_n$ are compact convex subsets of $\mathcal{X}$, then $\overline{\co }(K_1 \cup \dots \cup K_n) = \co (K_1 \cup \dots \cup K_n)$ and this convex hull is compact.

Let $K$ be convex and let $T : K \to \mathcal{Y}$ be an affine map. If $y$ is an extreme point of $T(K)$ and $x$ is an extreme point of $T^{-1}(y)$, then $x$ is an extreme point of $K$.

If $\mathcal{H}$ is a Hilbert space and either $T$ or $T^*$ is an isometry, show that $T$ is an extreme point of the closed unit ball of $\mathcal{B}(\mathcal{H})$.