Corollary

— In a Hausdorff locally convex space let K be a compact subset of a compact convex set A. Then the following conditions are equivalent.

a) A is the closed convex envelope of K.

b) K meets every set that is the intersection of A with one of its support hyperplanes.

c) K contains the set of extremal points of A.

a) $\Rightarrow$ b). Suppose that there exists a support hyperplane H of A whose equation is $f(x) = \alpha$, such that $(H \cap A) \cap K = \varnothing$ and suppose, for example, that $f(x) \geq \alpha$ in A. As $f(x) - \alpha > 0$ for all $x \in K$ by hypothesis and as K is compact we have

and K is, therefore, contained in the closed half-space $f(x) \geq \beta$; therefore the same is true of the closed convex envelope A of K and this is absurd.

b) $\Rightarrow$ c). Suppose that an extremal point x of A does not belong to K; there is a neighbourhood V of x in E such that $V \cap A \cap K = \varnothing$. But by prop. 2, we can suppose that V is an open half-space defined by a hyperplane H with the equation $f(z) = \alpha$. If for example $f(x) > \alpha$, then for all $y \in K$, we have $f(y) \leq \alpha$, therefore K does not meet the intersection of A and the support hyperplane $f(z) = \gamma > \alpha$ parallel to H (II, p. 37, prop. 2); this is absurd.

c) $\Rightarrow$ a). This is an obvious consequence of the Krein-Milman theorem.