Proposition 1

Let $B$ be the set of extremal points of $A$, a non-empty compact convex set in a Hausdorff locally convex space $E$, and let $f$ be a convex function defined in $A$ and upper semi-continuous. Then $f$ attains its upper bound in $A$ at one point (at least) of $B$.

Use $\mathscr{F}$ to denote the family of subsets X of A that are non-empty, closed, and such that every open segment that is contained in A and meets X necessarily lies in X. It has the following properties; (i) A belongs to $\mathscr{F}$. (ii) A point $a \in A$ is such that $\{a\} \in \mathscr{F}$, if, and only if, $a$ is an extremal point of A. (iii) Every non-empty intersection X of a family $(X_\alpha)$ of sets of $\mathscr{F}$ also belongs to $\mathscr{F}$.

The properties (i), (ii) and (iii) follow immediately from the definitions.

(iv) Let $X \in \mathscr{F}$, and let $h$ be a function that is convex and upper semi-continuous in A; then the set Y of the points of X where the restriction $h|X$ attains its upper bound in X is such that Y belongs to $\mathscr{F}$.

For, $h|X$ being upper semi-continuous in X attains its upper bound $\alpha$ over X in at least one point of X (GT, IV, § 6.2, th. 3); thus Y is non-empty, it is also closed (GT, IV, § 6.2, prop. 1). On the other hand let $x, y$ be two distinct points of A and let $z = \lambda x + (1 - \lambda) y$ be a point of Y such that $0 < \lambda < 1$; as $Y \subset X$ and $X \in \mathscr{F}$, we have $x \in X$ and $y \in X$; on the other hand, as $h$ is convex, we have

but as $h(x) \leq \alpha, h(y) \leq \alpha$ and $h(z) = \alpha$, of necessity $h(x) = h(y) = \alpha$, that is to say $x \in Y$ and $y \in Y$. Therefore $Y \in \mathscr{F}$.

With these properties established, let M be the set of $x \in A$ where $f$ attains its upper bound in A; by (iv), $M \in \mathscr{F}$. On the other hand, by (iii) and the fact that the sets of $\mathscr{F}$ are closed subsets ot the compact set A, it follows that $\mathscr{F}$ is inductive for the order relation $\supset$. By th. 2 of S, III, § 2.4, M contains a subset N which is a minimal element of $\mathscr{F}$. We shall show that N consists of a single point and this will complete the proof of the proposition. Since E is a Hausdorff locally convex space, it is sufficient to show that every continuous linear form $u$ on E is constant in N (II, p. 38, cor. 1). Now it follows from (iv) that the set $N'$ of the $x \in N$ where $u|N$ attains its upper bound in N is such that $N'$ belongs to $\mathscr{F}$; since N is minimal in $\mathscr{F}$ we necessarily have $N' = N$.