Krein–Milman closed-convex-hull form

If $K=\mathrm{\varnothing }$, then $\mathrm{ext}K=\mathrm{\varnothing }$ and both sides are empty by the stated convention. Hence suppose $K\mathrm{\ne }\mathrm{\varnothing }$ and put $E=\mathrm{ext}K$ and $C=\stackrel{}{\mathrm{co}}(E)$. By [F1], $E$ and therefore $C$ are nonempty.

The set $K$ is closed by [F5] and convex by hypothesis, and it contains $E$; therefore it contains $\mathrm{co}(E)$ and its closure $C$. The set $C$ is closed by definition and convex by [F6].

Suppose for contradiction that $x\in K\setminus C$. Apply [F2] to the compact convex singleton $\{x\}$ and the nonempty closed convex set $C$, using HB supplied from AC by [F4]. After naming $u=\mathrm{Re}f$, the resulting inequalities give $u(x)<{\inf }_{c\in C}u(c)$.

By [F3], the minimizer set $M=\{y\in K:u(y)={\min }_{K}u\}$ is a nonempty compact face of $K$. By [F1], $M$ has an extreme point $e$. Then $\{e\}$ is a face of $M$, so face transitivity in [F3] makes $\{e\}$ a face of $K$; hence $e\in E\subseteq C$.

Since $e$ minimizes $u$ on $K$ and $x\in K$, one has $u(e)\le u(x)$; step 3.1 gives $u(x)<{\inf }_{C}u$, whereas $e\in C$ gives $u(e)\ge {\inf }_{C}u$, a contradiction. Thus no $x\in K\setminus C$ exists, so $K\subseteq C$.

Step 2.1 gives $C\subseteq K$ and step 5.1 gives the reverse inclusion; together with the empty case in step 1.1 this proves the asserted equality in every case.