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 $N' = N$.

For, if $f(x) = \alpha$ is an equation of H and $f(x) \leq \alpha$ in A, it is sufficient to apply prop. 1 to $f$.

For, let C be the closed convex envelope of the set of extremal points of A; clearly $C \subset A$. To see that $A \subset C$, it is sufficient to prove that, if $u$ is an affine linear function, continuous in E and if $u(x) \geq 0$ in C then also $u(x) \geq 0$ in A (II, p. 39, cor. 4); but this follows from prop. 1 applied to $-u$.

For every open half-space D of E containing x, the set $A \cap \overline{D}$ is a compact neighbourhood of x in A, and the intersection of all these neighbourhoods is precisely the point x (any two distinct points can be strictly separated by a closed hyperplane (II, p. 38, prop. 4). By prop. 1 of GT, I, § 9.2, it is sufficient to prove that the sets $A \cap \overline{D}$ form a filter base. Now if we write $L_D = A \cap (E - D)$, the set $L_D$ is convex, compact and contained in the convex set $A - \{x\}$; if $D_1, D_2$ are two open half-spaces of E containing x, the convex envelope B of $L_{D_1} \cup L_{D_2}$ is therefore contained in $A - \{x\}$; but B is a compact set (II, p. 14, prop. 15), therefore there exists a closed hyperplane H that separates x strictly from B (II, p. 38, prop. 4) and if the open half-space determined by H and containing x is D, then we have $L_{D_1} \cup L_{D_2} \subset L_D$, therefore $A \cap \overline{D} \subset (A \cap \overline{D_1}) \cap (A \cap \overline{D_2})$.

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.

The condition is clearly necessary. Conversely, suppose that it is satisfied; suppose that there is a line D' not containing D, passing through x₀ and such that D' ∩ C contains an open segment to which x₀ belongs. Let y ≠ 0 be a direction vector of D'; the hypotheses imply that the point (1 + λ)x₀ + μy belongs to C for |λ| and |μ| sufficiently small. But then, in the plane P determined by D and D' and carrying the canonical topology, x₀ is an interior point of P ∩ C, and it follows that the line P ∩ H contains an open segment contained in H ∩ C and to which x₀ belongs. This contradicts the hypothesis.

a) Let A be a cap of C. Let p be the restriction to C of the gauge of A (II, p. 20). The set of the x ∈ C such that p(x) ≤ 1 is the set A. The function p is lower semi-continuous and has the following properties : (i) For any x, y in C, we have p(x + y) = p(x) + p(y). (ii) For any x ∈ C and λ ∈ R_+^*, we have p(λx) = λp(x). (iii) If x ∈ C, the relation p(x) = 0 is equivalent to x = 0.

b) Conversely, let p be a function defined in C with values in [0, +∞], satisfying the conditions (i), (ii) of a). Let A be the set of the x ∈ C such that p(x) ≤ 1. Then A and C − A are convex. A is a cap, if and only if A is compact and non-empty.

The statement b) is obvious. The properties stated in a) are consequences of the remarks preceding prop. 4 and of the prop. 22 of II, p. 20 and prop. 23 of II, p. 20 with the exception of

It is sufficient to prove this last when x ≠ 0 and y ≠ 0; we have therefore p(x) > 0, p(y) > 0. Let μ, λ be two numbers > 0 such that λ < p(x), μ < p(y), and denote the complement of A in C by B. We have x ∈ λB, y ∈ μB, therefore x + y ∈ λB + μB; by the convexity of B, we have λB + μB ⊂ (λ + μ)B, whence p(x + y) > λ + μ, which implies the inequality stated above.

It is clear that 0 is an extremal point of A. Let x be a point on L an extremal generator of C and such that p(x) = 1. Let y, z be two points of A such that x = $\frac{1}{2}(y + z)$. As L is extremal, we have y = λx and z = μx, where λ and μ are numbers ≥ 0 such that $\frac{1}{2}(\lambda + \mu) = 1$, $\lambda = \lambda p(x) = p(y) \leq 1$ and $\mu = \mu p(x) = p(z) \leq 1$, from which $\lambda = \mu = 1$ and hence y = z = x; so, x is an extremal point of A. Conversely, let x ≠ 0 be an extremal point of A. Obviously p(x) = 1. Let y, y' be two points of C such that x = y + y', and we shall show that y, y' are proportional to x. Without loss of generality we can suppose that the numbers $\lambda = p(y)$ and $\lambda' = p(y')$ are finite and > 0. Then $\lambda^{-1}y \in A, {\lambda'}^{-1}y' \in A, \lambda + \lambda' = 1$ by prop. 4, (i) and the equality $x = \lambda(\lambda^{-1}y) + \lambda'({\lambda'}^{-1}y')$ implies, by hypothesis that

This follows immediately from cor. 1 and the Krein-Milman theorem (II, p. 55, th. 1).

\* Example. — Let X be a locally compact space that is σ-compact. Let C be a closed convex cone of vertex 0 in $\mathscr{M}_+(X)$ with the vague topology. We shall show that C is the union of its caps. Let $(X_n)$ be an increasing sequence of open, relatively compact sets of X whose union is X. Let μ be an element $\neq 0$ of C. There exist $\alpha_n > 0$ such that $\sum_n \alpha_n \mu(X_n) = 1$.

For every measure $v \in C$, put $p(v) = \sum_n \alpha_n v(X_n) \in [0, +\infty]$. The function p on C satisfies conditions (i) and (ii) of prop. 4. It is lower semi-continuous for the vague topology (INT, IV, 2nd ed., § 1, No. 1, prop. 4). The set A of the $\gamma \in C$ such that $p(\gamma) \leq 1$ is therefore closed and non-empty. On the other hand, every compact set of X is contained in one of the $X_n$, thus A being vaguely bounded is also vaguely compact (INT, III, 2nd ed., § 1, No. 9, prop. 15). The set A is therefore a cap of C containing μ. \*

The second statement follows from the first and from cor. 2 above. Using prop. 11 of II, p. 52 reduces the proposition to the case when $E = \mathbf{R}^l$ and $C \subset \mathbf{R}_+^l$. For all $\alpha \in I$, denote the projection $pr_\alpha$ in E by $f_\alpha$; then $f_\alpha$ is a continuous linear form. On the other hand let $(V_n)_{n \in \mathbf{N}}$ be an enumerable fundamental system of neighbourhoods of 0 in C. By the definition of the topology of E, for each $n \in \mathbf{N}$, there exists a finite subset $J_n$ of I and a number $\varepsilon_n > 0$ such that $V_n$ contains the set $W_n$ of the $x \in C$ such that $f_\alpha(x) \leq \varepsilon_n$ for all $\alpha \in J_n$; put $J = \bigcup_{n \in \mathbf{N}} J_n$. Let $y \neq 0$ be a point of C, and p be the function $\sum_{\alpha \in J} \lambda_\alpha (f_\alpha|C)$ where the $\lambda_\alpha > 0$ are chosen so that $p(y) = 1$; this is possible, since if $f_\alpha(y) = 0$ for all $\alpha \in J$, then $y \in V_n$ for all $n$, which implies that $y = 0$, and this is contrary to hypothesis. Now we remark that for all $\alpha \in I$, the function $f_\alpha|C$ is continuous at the point 0, therefore there is an $n \in \mathbf{N}$, such that $f_\alpha$ is bounded in a $W_n$, therefore bounded above in C by a linear combination of a finite number of functions $f_\beta|C$, where $\beta \in J$. It follows that if A in the set of $x \in C$ such that $p(x) \leq 1$, then $f_\alpha$ is bounded in A for all $\alpha \in I$. As p is lower semi-continuous in C, it follows that A is closed and non-empty in C and therefore is compact. Since it is clear that p verifies the conditions (i) and (ii) of prop. 4 of II, p. 58, we see that A is a cap in C and contains y.

By prop. 4 of II, p. 38, there exists a closed hyperplane $H$ which separates 0 strictly from $K$. Now, the convex envelope $A$ of the union of $\{0\}$ and of $K$ is compact (II, p. 14, prop. 15) and is the union of the $\lambda K$ with $0 \leq \lambda \leq 1$. As 0 and $K$ are strictly on opposite sides of $H$, for every $x \in K$ there exists $\lambda$ such that $0 < \lambda < 1$ and $\lambda x \in H$. As $C$ is the union of the $\lambda A$ for $\lambda \geq 1$, we see that $H$ meets every half-line originating at 0 contained in $C$ and that $H \cap A = H \cap C$ is compact. Further, $C$ is also the union of the $\lambda(H \cap C)$ for $\lambda \geq 0$; let $C_n$ be the union of the $\lambda(H \cap C)$ for $0 \leq \lambda \leq n$. Clearly $C_n$ is the convex envelope of the union of $\{0\}$ and of $n(H \cap C)$, therefore it is compact. Also, for all $x \in E$, there is a closed neighbourhood $V$ of $x$ in $E$ and an integer $n$ such that $V \cap C \subset C_n$; in fact, if $H$ is defined by the equation $f(z) = \alpha$, where $\alpha > 0$, it is sufficient to take for $V$ the closed half-space determined by $nH$ and containing 0, where $n$ is so large that $n\alpha > f(x)$. This shows that $C$ is locally compact (taking $x \in C$), and that it is closed in $E$. We can also consider $K$ as a subset of the completion $\hat{E}$, therefore $C$ is also closed in $\hat{E}$ and therefore complete.

Given a cone $C$ and a closed hyperplane $H$ in a Hausdorff topological vector space $E$, such that $H$ does not contain the vertex $s$ of $C$ and $C$ is the smallest cone with vertex $s$ containing $H \cap C$, then we call the intersection $H \cap C$ a « sole » of the cone $C$. Prop. 6 shows that in a Hausdorff locally convex space $E$, the smallest cone of vertex 0, containing a compact convex set $K$ to which 0 does not belong, is a cone of compact sole, and that every convex cone having a compact sole $S$, is locally compact and complete.