§ 7. Barycenters

1. Definition of barycenters

Let E be a Hausdorff locally convex space over $\mathbf{R}$ , $E'$ its dual, and ${E'}^*$ the algebraic dual of $E'$ , E being canonically identified with a linear subspace of ${E'}^*$ . Let K be a compact subset of E; the canonical injection of K into E being continuous with compact support, for every measure $\mu$ on K the integral $\int x\, d\mu(x)$ is therefore defined and is an element of ${E'}^*$ (Ch. III, §3, No. 1). Moreover, on K, the topology induced by the weak topology $\sigma({E'}^*, E')$ is identical with the original topology. Finally, if C is the closed convex envelope of K in ${E'}^*$ equipped with $\sigma({E'}^*, E')$ , then $C \cap E$ is the closed convex envelope of K in E for the original topology (or for the weakened topology $\sigma(E, E')$ ).

Definition 1

Let K be a compact subset of a Hausdorff locally convex space E. For every positive measure $\mu$ on K of total mass equal to 1, the vector $b_\mu = \int x\, d\mu(x)$ (belonging to ${E'}^*$ ) is called the barycenter of $\mu$ .

Example

— Let $\mu$ be a discrete measure on K, positive and of total mass 1; it is thus of the form $\mu = \sum_{i=1}^n \lambda_i \varepsilon_{x_i}$ , where $x_i \in K$ , and the $\lambda_i$ are real numbers such that $\lambda_i \geq 0$ for all $i$ and $\sum_{i=1}^n \lambda_i = 1$ . Since $\int x\, d\varepsilon_y(x) = y$ (Ch. III, §3, No. 1, Example 3), $b_\mu = \int x\, d\mu(x) = \sum_i \lambda_i x_i$ . In particular, for every $x \in K$ , x is the barycenter of the measure $\varepsilon_x$ .

Proposition 1

Let E be a Hausdorff locally convex space, K a compact subset of E, and C the closed convex envelope of K in E. The set C then consists of the points of E that are barycenters of at least one positive measure of mass 1 on K.

This is nothing more than Prop. 5 of Ch. III, §3, No. 2 applied to the canonical injection of K into E.

Corollary

If the closed convex envelope C of K in E is compact, then the barycenter of every positive measure of total mass 1 on K belongs to E.

For, C is then also the closed convex envelope of K in ${E'}^*$ equipped with the weak topology $\sigma(E', E')$ , and it suffices to apply, to the canonical injection of K into E, Prop. 4 of Ch. III, §3, No. 2.

Remark

The Cor. of Prop. 1 is applicable in particular when K is convex or when E is quasi-complete.

Proposition 2

— Let K be a compact convex subset of a Hausdorff locally convex space E, $\mu$ a positive measure of total mass 1 on K, $b_\mu$ its barycenter. For every convex numerical function $f \geq 0$ that is lower semi-continuous on K,

f(b_\mu) \leq \int^* f d\mu.

It is known (TVS, II, §5, No. 4, Prop. 5) that

f

is the upper envelope of a family of restrictions to K of continuous affine linear functions

h_\alpha : x \mapsto c_\alpha + \langle x, z'_\alpha \rangle

. Then

\int h_\alpha(x) d\mu(x) \leq \int^* f(x) d\mu(x)

for every

\alpha

; now,

\int h_\alpha(x) d\mu(x) = c_\alpha + \int \langle x, z'_\alpha \rangle d\mu(x)

since

\mu

has total mass 1; but

\int \langle x, z'_\alpha \rangle d\mu(x) = \langle b_\mu, z'_\alpha \rangle

by the definition of barycenter. Therefore

\sup_\alpha \int h_\alpha(x) d\mu(x) = \sup_\alpha h_\alpha(b_\mu) = f(b_\mu)

, whence the conclusion.

When $\mu$ is a discrete positive measure on K of total mass 1, Prop. 2 yields anew the inequality that defines the convex functions on K.

Corollary

— For every convex numerical function function g on K that is bounded and lower semi-continuous, $g(b_\mu) \leq \int g d\mu$ .

It suffices to note that $\inf_{x \in K} g(x) = a$ is finite and apply Prop. 2 to $g - a$ .

2. Extremal points and barycenters

Proposition 3

— Let K be a compact convex subset of a Hausdorff locally convex space E, x a point of K. Every measure $\mu$ on K, positive, of total mass 1, and admitting x as barycenter, is in the vague closure of the set of discrete positive measures of total mass 1 that admit x as barycenter.

Let U be a neighborhood of $\mu$ for the vague topology; we can suppose that U consists of the measures $\nu$ on K such that

|\mu(f_i) - \nu(f_i)| \leq \delta

for a finite number of functions

f_i \in \mathscr{C}(K; \mathbf{C})

(

1 \leq i \leq p

) and a number

\delta > 0

. For every point

a \in K

, there exists a closed convex neighborhood

V_a

of 0 in E such that

|f_i(y) - f_i(a)| \leq \delta/2

for $1 \leq i \leq p$ and for every $y \in W_a = K \cap (a + V_a)$ . Since $K$ is compact, there exists a finite number of points $a_j$ ( $1 \leq j \leq r$ ) of $K$ such that the $W_{a_j}$ form a covering of $K$ ( $1 \leq j \leq r$ ). Consider a continuous partition of unity $(g_j)_{1 \leq j \leq r}$ on $K$ , subordinate to the covering $(W_{a_j})$ , and set $\alpha_j = \mu(g_j)$ for all $j$ ; if $\alpha_j \neq 0$ set $\mu_j = \alpha_j^{-1} g_j \cdot \mu$ , and if $\alpha_j = 0$ set $\mu_j = \varepsilon_{a_j}$ . Each of the measures $\mu_j$ is positive, of total mass 1, and its support is contained in the compact convex set $W_{a_j}$ ; moreover, by definition,

\mu = \sum_{j=1}^r \alpha_j \mu_j

since $g_j \cdot \mu = 0$ if $\mu(g_j) = 0$ ; the $\alpha_j$ are $\geq 0$ and

\sum_{j=1}^r \alpha_j = \sum_{j=1}^r \mu(g_j) = \mu \left( \sum_{j=1}^r g_j \right) = \mu(1) = 1 .

Let $x_j$ be the barycenter of $\mu_j$ , which belongs to $W_{a_j}$ (No. 1, Prop. 1), and consider the discrete measure $\nu = \sum_{j=1}^r \alpha_j \varepsilon_{x_j}$ ; it is positive and of total mass 1, and its barycenter is $\sum_{j=1}^r \alpha_j x_j$ , which is also the barycenter of $\mu$ by virtue of (3), thus is equal to $x$ . Moreover, by (2), $|f_i(y) - f_i(a_j)| \leq \delta/2$ for all $y \in W_{a_j}$ and for all $i$ , whence, since $\operatorname{Supp}(\mu_j) \subset W_{a_j}$ , $|\mu_j(f_i) - f_i(a_j)| \leq \delta/2$ for $1 \leq i \leq p$ . On the other hand, since $x_j \in W_{a_j}$ , one also has

|\varepsilon_{x_j}(f_i) - f_i(a_j)| \leq \delta/2

for $1 \leq i \leq p$ , whence $|\mu_j(f_i) - \varepsilon_{x_j}(f_i)| \leq \delta$ for all $i$ and $j$ . Since the $\alpha_j$ are $\geq 0$ and have sum 1, it follows from (3) and the definition of $\nu$ that $\nu$ satisfies the inequality (1).

Q.E.D.

Corollary

— *Let $K'$ be a compact subset of $K$ such that $K$ is the closed convex envelope of $K'$ . In order that $x \in K'$ be an extremal point of $K$ , it is necessary and sufficient that $\varepsilon_x$ be the only positive measure on $K'$ , of total mass 1, having $x$ as barycenter.

Suppose $x$ is an extremal point of $K$ ; to prove that $\varepsilon_x$ is the only positive measure on $K'$ , of total mass 1, having $x$ as barycenter, it suffices, by Prop. 3, to see that the set of discrete measures $\nu$ on $K'$ that are positive, of total mass 1, and have $x$ as barycenter, reduces to $\varepsilon_x$ . But such a measure $\nu$ is of the form $\sum_{i=1}^r \lambda_i \varepsilon_{x_i}$ with $\lambda_i > 0$ for $1 \leq i \leq r$ and $\sum_{i=1}^{r} \lambda_i = 1$ , and the hypothesis that $x$ is the barycenter of $\nu$ may be written $x = \sum_{i=1}^{r} \lambda_i x_i$ . Since $x$ is extremal, this implies that $x_i = x$ for all $i$ , whence $\nu = \varepsilon_x$ .

Conversely, let us assume that $\varepsilon_x$ is the only positive measure on $K'$ , of total mass 1, having $x$ as barycenter, and let us show that $x$ is extremal. In the contrary case, there would exist two distinct points $x', x''$ of $K$ and a real number $\lambda$ such that $0 < \lambda < 1$ and $x = \lambda x' + (1 - \lambda) x''$ . By Prop. 1, $x'$ (resp. $x''$ ) is the barycenter of a positive measure $\mu'$ (resp. $\mu''$ ) on $K'$ of total mass 1. Then $x$ is the barycenter of $\lambda \mu' + (1 - \lambda) \mu''$ . Therefore $\lambda \mu' + (1 - \lambda) \mu'' = \varepsilon_x$ . Therefore $\mu'$ and $\mu''$ are proportional to $\varepsilon_x$ , whence $x' = x'' = x$ , which is absurd.

Theorem 1 (Choquet)

— *Let E be a Hausdorff locally convex space over $\mathbf{R}$ , K a metrizable compact convex subset of E, and M the set of extremal points of K. The set M is the intersection of a countable family of open sets in K, and every point of K is the barycenter of a measure $\mu$ on K such that $\mu(K - M) = 0$ .

To prove the first assertion, denote by I the interval $[0, 1]$ of $\mathbf{R}$ ; since K is compact and metrizable, so is $K \times K \times I$ ; the subset U of $K \times K \times I$ formed by the triples $(x, y, \lambda)$ such that $x \neq y$ and $0 < \lambda < 1$ is open in $K \times K \times I$ , therefore there exists a sequence $(F_n)$ of closed sets in $K \times K \times I$ whose union is U (GT, IX, §2, No. 5, Prop. 7). The mapping $q : K \times K \times I \to K$ defined by $q(x, y, \lambda) = \lambda x + (1 - \lambda) y$ is continuous and, by the definition of extremal point, $K - M = q(U) = \bigcup_n q(F_n)$ ; but $F_n$ is compact since it is closed in $K \times K \times I$ , therefore $q(F_n)$ is compact and therefore closed in K; the set $U_n = K - q(F_n)$ is therefore open in K, and we have $M = \bigcap_n U_n$ .

In the continuation of the proof, we shall denote by $u$ a continuous and convex numerical function on K, by $G \subset E \times \mathbf{R}$ the graph of $u$ , and by S the closed convex envelope of G in $E \times \mathbf{R}$ .

Lemma 1

— *Let $\overline{u}$ be the lower envelope of the continuous affine linear functions on E that are $\geq u$ on K. Then S is the set of points $(a, b) \in E \times \mathbf{R}$ such that $a \in K$ and $u(a) \leq b \leq \overline{u}(a)$ .

By the Hahn-Banach theorem, for $(a, b)$ to belong to S, it is necessary and sufficient that $h(a, b) \geq 0$ for every continuous affine linear function $h$ on $E \times \mathbf{R}$ such that $h(x, u(x)) \geq 0$ for $x \in K$ . Setting $h(x, t) = f(x) - \lambda t$ , where $f$ is a continuous affine linear function on E, we see that the relation $(a, b) \in S$ is equivalent to the following property: the relation

f(x) \geq \lambda u(x) \quad \text{for all } x \in K

implies

(5)

f(a) \geq \lambda b .

First set $\lambda = 0$ ; the fact that (4) implies (5) for every continuous affine linear function $f$ on $E$ is equivalent to the relation $a \in K$ by the Hahn-Banach theorem. Next, set $\lambda = -1$ and replace $f$ by $-f$ ; then the relation $f(x) \leq u(x)$ in $K$ must imply $f(a) \leq b$ . But since $u$ is convex and continuous on $K$ , $u(a)$ is equal to the supremum of the $f(a)$ for the continuous affine linear functions $f$ on $E$ such that $f(x) \leq u(x)$ on $K$ (TVS, II, §5, No. 4, Prop. 5); we thus obtain the relation $b \geq u(a)$ . Finally, set $\lambda = 1$ ; to say that (4) implies (5) then means, by definition, that $b \leq \overline{u}(a)$ . This proves the lemma, since the relation (4) (resp. (5)) is equivalent to the one obtained by multiplying both sides by a scalar $> 0$ .

Lemma 2

If $u$ is strictly convex on $K$ , then $u(x) < \overline{u}(x)$ for every non-extremal point $x$ of $K$ .

For, there then exist two distinct points $y, z$ of $K$ such that $x = (y + z)/2$ , whence $u(x) < (u(y) + u(z))/2$ since $u$ is strictly convex. If $f$ is an affine linear function on $E$ , then $f(x) = (f(y) + f(z))/2$ ; applying this relation to the continuous affine linear functions $f$ that are $\geq u$ on $K$ , we obtain

\overline{u}(x) \geq (\overline{u}(y) + \overline{u}(z))/2 \geq (u(y) + u(z))/2 > u(x) .

These lemmas established, we therefore (Lemma 1) have $(a, \overline{u}(a)) \in S$ for all $a \in K$ . Since $G$ is compact, being the image of $K$ under the continuous mapping $x \mapsto (x, u(x))$ , there exists, by Prop. 1 of No. 1, a positive measure $\nu$ on $G$ , of total mass 1 , having $(a, \overline{u}(a))$ as barycenter. Since the restriction of the projection $pr_1$ to $G$ is a homeomorphism of $G$ onto $K$ , the measure $\nu$ may be transported by means of this homeomorphism, which yields a measure $\mu$ on $K$ (positive and of total mass 1 ) such that

(6)

a = \int x \, d\mu(x) \quad \text{and} \quad \overline{u}(a) = \int u(x) \, d\mu(x) .

The first of the relations (6) means that $a$ is the barycenter of $\mu$ . The function $\overline{u}$ is upper semi-continuous and bounded on $K$ , hence is $\mu$ -integrable (§4, No. 4, Cor. 1 of Prop. 5); moreover, since the function $-\overline{u}$ is by definition convex, we have, by the Cor. of Prop. 2 of No. 1,

(7)

\overline{u}(a) \geq \int \overline{u}(x) \, d\mu(x) ,

whence, on comparing with the second of the formulas (6),

\int u(\mathbf{x})\, d\mu(\mathbf{x}) \geq \int \overline{u}(\mathbf{x})\, d\mu(\mathbf{x}) .

But since $u(\mathbf{x}) \leq \overline{u}(\mathbf{x})$ on K, the relation (8) implies that $u(\mathbf{x}) = \overline{u}(\mathbf{x})$ almost everywhere for $\mu$ . Taking into account Lemma 2, one sees that Theorem 1 will be proved once the following lemma has been established:

Lemma 3

Let E be a Hausdorff locally convex space over $\mathbf{R}$ , and K a metrizable compact convex subset of E. Then, there exists a strictly convex numerical function on K.

For, the Banach space $\mathscr{C}(K; \mathbf{R})$ is separable (GT, X, §3, No. 3, Th. 1), therefore so is the subspace $\mathscr{A}$ of $\mathscr{C}(K; \mathbf{R})$ formed by the restrictions to K of the continuous affine linear functions on E. Thus let $(h_n)$ be a sequence dense in $\mathscr{A}$ , and let $\alpha_n > \sup_{\mathbf{x} \in K} |h_n(\mathbf{x})|$ . Then each of the functions $h_n^2 / n^2 \alpha_n^2$ is convex in K (TVS, II, §2, No. 8, Examples), and the series with general term $h_n^2 / n^2 \alpha_n^2$ is normally convergent, therefore its sum $u$ is continuous and convex in K. It remains to see that $u$ is strictly convex, and for this it suffices to prove that for any two distinct points $\mathbf{x}, \mathbf{x}'$ of K, there is an integer $n$ such that the restriction of $h_n^2$ to the segment with endpoints $\mathbf{x}, \mathbf{x}'$ is strictly convex; but for this it suffices that $h_n(\mathbf{x}) \neq h_n(\mathbf{x}')$ (loc. cit.). Now, there exists a function $h \in \mathscr{A}$ such that $h(\mathbf{x}) \neq h(\mathbf{x}')$ (TVS, II, §4, No. 1, Cor. 1 of Prop. 2) and since the sequence $(h_n)$ is dense in $\mathscr{A}$ , there exists an $n$ such that $h_n(\mathbf{x}) \neq h_n(\mathbf{x}')$ .

Q.E.D.

Corollary

Let E be a Hausdorff locally convex space over $\mathbf{R}$ , C a proper convex cone in E with vertex 0, that is complete and metrizable for the uniform structure induced by the weakened uniform structure of E. Let M be the union of the extremal generators of C. For every $\mathbf{x} \in C$ there exist a compact convex subset K of C and a measure $\lambda \geq 0$ on K of total mass 1, such that $K - (M \cap K)$ is $\lambda$ -negligible and the barycenter of $\lambda$ is equal to $\mathbf{x}$ .

For, $\mathbf{x}$ belongs to a cap K of C (TVS, II, §7, No. 2, Prop. 5), and $M \cap K$ contains the set of extremal points of K (loc. cit., Cor. 1 of Prop. 4). It then suffices to apply Th. 1.

3. Applications: I. Vector spaces of continuous real functions

Let X be a nonempty compact space, $\mathscr{H}$ a linear subspace of the Banach space $\mathscr{C}(X; \mathbf{R})$ that contains the constants and separates the points of X (GT, X, §4, No. 1, Def. 1). We equip $\mathscr{H}$ with the normed space topology induced by that of $C(X; \mathbf{R})$ , and denote by $\mathscr{H}'$ the dual of this normed space. For every $x \in X$ , the mapping $f \mapsto f(x)$ is a continuous linear form on $\mathscr{H}$ (the restriction to $\mathscr{H}$ of the Dirac measure $\varepsilon_x$ ), thus is an element of $\mathscr{H}'$ that will be denoted $i_{\mathscr{H}}(x)$ , so that

\langle f, i_{\mathscr{H}}(x) \rangle = f(x)

for every function $f \in \mathscr{H}$ and every $x \in X$ .

The mapping $i_{\mathscr{H}}$ of X into $\mathscr{H}'$ is injective and continuous when $\mathscr{H}'$ is equipped with the weak topology $\sigma(\mathscr{H}', \mathscr{H})$ ; the second assertion follows at once from the definitions and (9); as for the first, note that if $x, x'$ are two distinct points of X, by hypothesis there exists a function $h \in \mathscr{H}$ such that $h(x) \neq h(x')$ , therefore, by (9), $\langle h, i_{\mathscr{H}}(x) \rangle \neq \langle h, i_{\mathscr{H}}(x') \rangle$ and a fortiori $i_{\mathscr{H}}(x) \neq i_{\mathscr{H}}(x')$ . The image $i_{\mathscr{H}}(X)$ is therefore a compact subset of $\mathscr{H}'$ (for the weak topology), and $i_{\mathscr{H}}$ is a homeomorphism of X onto $i_{\mathscr{H}}(X)$ .

Proposition 4

(i) The closed convex envelope C of $i_{\mathscr{H}}(X)$ in $\mathscr{H}'$ (for the weak topology $\sigma(\mathscr{H}', \mathscr{H})$ ) is compact.

(ii) For a point $i_{\mathscr{H}}(x)$ to be an extremal point of C, it is necessary and sufficient that the only positive measure $\lambda$ on X such that

h(x) = \int h \, d\lambda

for every function $h \in \mathscr{H}$ (which implies in particular that $\lambda$ has total mass 1, since $1 \in \mathscr{H}$ ) be the Dirac measure $\varepsilon_x$ .

It follows that, for each $h \in \mathscr{H}$ , the function $z' \mapsto \langle h, z' \rangle$ on C attains its supremum at at least one extremal point of C (TVS, II, §7, No. 1, Prop. 1), and this point belongs to $i_{\mathscr{H}}(X)$ (loc. cit., Cor. of Prop. 2).

(i) By (9), $\| i_{\mathscr{H}}(x) \| \leq 1$ in the normed space $\mathscr{H}'$ , in other words $i_{\mathscr{H}}(X)$ is bounded, and the assertion follows from the fact that $\mathscr{H}'$ , equipped with the weak topology $\sigma(\mathscr{H}', \mathscr{H})$ , is quasi-complete (TVS, III, §4, No. 2, Cor. 5 of Th. 1).

(ii) Every positive measure $\mu$ of mass 1 on $i_{\mathscr{H}}(X)$ arises, by transport of structure by means of the homeomorphism $i_{\mathscr{H}}$ , from a positive measure $\lambda$ of mass 1 on X, the Dirac measure $\varepsilon_{i_{\mathscr{H}}(x)}$ arising from $\varepsilon_x$ . To say that $\mu$ admits $i_{\mathscr{H}}(x)$ as barycenter means, by definition, that

\int_X \langle h, i_{\mathscr{H}}(z) \rangle \, d\lambda(z) = \langle h, i_{\mathscr{H}}(x) \rangle

for every function $h \in \mathscr{H}$ . Taking (9) into account, the assertion (ii) is just the translation of the criterion of No. 2, Cor. of Prop. 3 for $i_{\mathscr{H}}(x)$ to be an extremal point of C.

We shall say that a point $x \in X$ satisfying condition (ii) of Prop. 4 is $\mathscr{H}$ -extremal; we denote by $\mathrm{Ch}_{\mathscr{H}}(X)$ (or simply $\mathrm{Ch}(X)$ ) the set of these points, and by $\check{S}_{\mathscr{H}}(X)$ (or simply $\check{S}(X)$ ) the closure of $\mathrm{Ch}_{\mathscr{H}}(X)$ in $X$ .

Proposition 5

Every function* $h \in \mathscr{H}$ attains its supremum at at least one $\mathscr{H}$ *-extremal point*.

Let $x$ be a point of $X$ , $h$ a function in $\mathscr{H}$ . The relation $h(z) \leq h(x)$ for all $z \in X$ may be written $\langle h, i_{\mathscr{H}}(z) \rangle \leq \langle h, i_{\mathscr{H}}(x) \rangle$ for all $z \in X$ , and therefore means that the weakly closed hyperplane of $\mathscr{H}'$ with equation $\langle h, t' \rangle = \langle h, i_{\mathscr{H}}(x) \rangle$ is a support hyperplane of $i_{\mathscr{H}}(X)$ . It is known (TVS, II, §7, No. 1, Cor. of Prop. 1) that such a hyperplane contains at least one extremal point of the closed convex envelope of $i_{\mathscr{H}}(X)$ , and such a point $i_{\mathscr{H}}(y)$ is the image of an $\mathscr{H}$ -extremal point $y$ by definition; $h(y)$ is therefore equal to the supremum of $h$ in $X$ .

Proposition 6

For every point* $x \in X$ , the following properties are equivalent:

a) $x$ is $\mathscr{H}$ *-extremal*.

b) For every open neighborhood $U$ of $x$ in $X$ and every $\varepsilon > 0$ , there exists a function $h \geq 0$ in $\mathscr{H}$ such that $h(x) \leq \varepsilon$ and $h(y) \geq 1$ for every $y \in X - U$ .

Let $x$ be any point of $X$ , $f$ a function in $\mathscr{C}(X; \mathbf{R})$ ; it is known (TVS, II, §3, No. 1, Prop. 1) that the infimum of the numbers $\lambda(f)$ , for all the positive measures on $X$ such that $\lambda(h) = h(x)$ for every function $h \in \mathscr{H}$ , is equal to the supremum of the numbers $h(x)$ , where $h$ runs over the set of functions $h \in \mathscr{H}$ such that $h \leq f$ . Suppose that $x$ is $\mathscr{H}$ -extremal; it then follows from Prop. 4, (ii) that for every function $f \in \mathscr{C}(X; \mathbf{R})$ ,

f(x) = \sup_{h \in \mathscr{H}, h \leq f} h(x).

To show that a) implies b), we take for $f$ a continuous mapping of $X$ into $[0, 1]$ , with support contained in $U$ , such that $f(x) = 1$ ; then, by (11), there exists a function $h' \in \mathscr{H}$ such that $h' \leq f$ and $h'(x) \geq 1 - \varepsilon$ . Since $1 \in \mathscr{H}$ , the function $h = 1 - h'$ meets the conditions of b).

Conversely, suppose that the condition b) is verified; this condition implies that $1 - h \leq \varphi_U$ ; for every positive measure $\lambda$ on $X$ satisfying the condition (10), we therefore have

\lambda(U) = \lambda(\varphi_U) \geq \lambda(1 - h) = 1 - h(x) \geq 1 - \varepsilon.

Since, by hypothesis, this relation holds for every $\varepsilon > 0$ and every open neighborhood $U$ of $x$ , it follows that

\lambda(\{x\}) = \inf_U \lambda(U) \geq 1 - \varepsilon

for every $\varepsilon > 0$ , therefore $\lambda(\{x\}) = 1$ . Since $\lambda$ is positive and of total mass 1, necessarily $\lambda = \varepsilon_x$ , which proves that $x$ is $\mathscr{H}$ -extremal, by virtue of Prop. 4, (ii).

Proposition 7

Let F be a closed subset of X. The following properties are equivalent:

a) F contains $\check{S}_{\mathscr{H}}(X)$ .

b) For every function $h \in \mathscr{H}$ , the set F intersects the set of points of X where h attains its supremum.

c) For every point $x \in X$ , there exists a positive measure $\mu$ of total mass 1 on X, such that $\operatorname{Supp}(\mu) \subset F$ and $h(x) = \int h \, d\mu$ for every function $h \in \mathscr{H}$ .

Let $G = i_{\mathscr{H}}(F)$ . The condition a) signifies that G contains the set of extremal points of C. The condition b) signifies that G meets the intersection of $i_{\mathscr{H}}(X)$ with each of the closed support hyperplanes of $i_{\mathscr{H}}(X)$ . Finally, the condition c) signifies that every point of $i_{\mathscr{H}}(X)$ is the barycenter of a measure with support contained in G; by No. 1, Prop. 1, this is also equivalent to saying that the closed convex envelope of $i_{\mathscr{H}}(X)$ is equal to the closed convex envelope of G. The equivalence of the conditions a), b) and c) therefore follows from TVS, II, §7, No. 1, Cor. of Prop. 2.

Proposition 8

Suppose X is metrizable. Then the set $\operatorname{Ch}_{\mathscr{H}}(X)$ of $\mathscr{H}$ -extremal points of X is the intersection of a countable family of open sets in X, and for every $x \in X$ , there exists a positive measure $\mu$ of total mass 1 on X such that

\mu(X - \operatorname{Ch}_{\mathscr{H}}(X)) = 0 \quad \text{and} \quad \int h \, d\mu = h(x)

for every $h \in \mathscr{H}$ .

This is the translation of Th. 1 of No. 2, by transport of structure by means of the homeomorphism $x \mapsto i_{\mathscr{H}}(x)$ , as in Prop. 5.

A certain number of results of this No. may be extended when $\mathscr{H}$ is replaced by a set $\mathscr{P}$ of functions defined on X, with values in $\mathbf{R} \cup \{+\infty\}$ , that are lower semi-continuous, $\mathscr{P}$ being assumed to contain the constants and to satisfy $\mathscr{P} + \mathscr{P} \subset \mathscr{P}$ (Exer. 2).

Example

— Take X to be the unit ball $\|x\| \leq 1$ in $\mathbf{R}^3$ , and let $\mathscr{H}$ be a vector space of continuous functions on X, containing the restrictions to X of the affine linear functions on $\mathbf{R}^3$ and satisfying the maximum principle, that is, for every non-constant function $h \in \mathscr{H}$ , the set of points of X where h attains its supremum is contained in the sphere $S_2$ . It then follows easily from Props. 5 and 7 that $\operatorname{Ch}_{\mathscr{H}}(X) = \check{S}_{\mathscr{H}}(X) = S_2$ . An important example of a vector space $\mathscr{H}$ satisfying the preceding conditions is the set of functions continuous on X and harmonic in the open ball $\|x\| < 1$ . For these functions, one proves that the positive measure $\mu$ of mass 1 such that $\operatorname{Supp}(\mu) \subset S_2$ and $h(x) = \int h \, d\mu$ for all $h \in \mathscr{H}$ is given, if $\|x\| < 1$ , by Poisson’s formula

d\mu(z) = \frac{1 - \|z\|^2}{\|z - x\|^3} d\sigma(z),

where $\sigma$ is the measure on $S_2$ invariant under the orthogonal group and such that $\sigma(S_2) = 1$ (Ch. VII, §3, Exer. 8).

4. Applications: II. Vector spaces of continuous complex functions

Let $X$ be a nonempty compact space, $\mathscr{H}$ a linear subspace of the complex Banach space $\mathscr{C}(X; \mathbf{C})$ that contains the constants and separates the points of $X$ . The set of real parts $\Re(f)$ of the functions $f \in \mathscr{H}$ is a linear subspace $\mathscr{H}_r$ of the real vector space $\mathscr{C}(X; \mathbf{R})$ ; for every $f \in \mathscr{H}$ , the set $\mathscr{H}_r$ also contains $\mathscr{I}(f) = \Re(-if)$ ; it follows that $\mathscr{H}_r$ separates the points of $X$ , because the relation $h(x) = h(y)$ for all $h \in \mathscr{H}_r$ implies that $\Re(f(x)) = \Re(f(y))$ and $\mathscr{I}(f(x)) = \mathscr{I}(f(y))$ and so $f(x) = f(y)$ for all $f \in \mathscr{H}$ . The $\mathscr{H}_r$ -extremal points in $X$ are again called $\mathscr{H}$ -extremal, the set of them is denoted $\mathrm{Ch}_{\mathscr{H}}(X)$ , and the closure of the latter set is denoted $\check{S}_{\mathscr{H}}(X)$ . The analogues of Props. 5 and 7 are the following:

Proposition 9

For every function $f \in \mathscr{H}$ , $\mathrm{Ch}_{\mathscr{H}}(X)$ intersects the set of points where $|f|$ attains its supremum.

We may limit ourselves to the case that $f$ is not the constant 0. Let $a$ be a point of $X$ where $|f|$ attains its supremum, and set $g = f/f(a)$ ; then $g(a) = 1$ and $|g(x)| \leq 1$ for all $x \in X$ , whence

\Re(g(a)) = 1 \quad \text{and} \quad \Re(g(x)) \leq 1 \quad \text{for all } x \in X.

By Prop. 5 of No. 3 applied to $\mathscr{H}_r$ , there exists $b \in \mathrm{Ch}_{\mathscr{H}}(X)$ where $\Re((g(x))$ attains its supremum 1, whence $|g(b)| = 1$ since $|g(b)| \leq 1$ ; it follows that $|f(b)| = |(f(a)| \geq |f(x)|$ for all $x \in X$ .

Proposition 10

Let $F$ be a closed subset of $X$ . The following properties are equivalent: a) $F$ contains $\check{S}_{\mathscr{H}}(X)$ . b) For every function $f \in \mathscr{H}$ , $F$ intersects the set of points of $X$ where $|f|$ attains its supremum. c) For every point $x \in X$ , there exists a positive measure $\mu$ of total mass 1 on $X$ such that $\mathrm{Supp}(\mu) \subset F$ and $f(x) = \int f d\mu$ for every function $f \in \mathscr{H}$ . Let us prove the equivalence of the conditions a) and c): let $f = f_1 + i f_2$ with $f_1, f_2$ in $\mathscr{H}_r$ ; the relation $f(x) = \int f d\mu$ is equivalent to the two relations $f_1(x) = \int f_1 d\mu$ and $f_2(x) = \int f_2 d\mu$ ; it thus suffices to apply to $\mathscr{H}_r$ the equivalence of the conditions a) and c) of Prop. 7 of No. 3. The fact that a) implies b) follows from Prop. 9. Let us show that b) implies a); this is a matter of seeing that if b) is verified, then, for every $h \in \mathscr{H}_r$ , F intersects the set of points where $h$ attains its infimum in X. The condition b) implies that F is nonempty; since F is compact, there exists $a \in F$ such that $h(a) \leq h(y)$ for all $y \in F$ . Let $f \in \mathscr{H}$ be such that $h = \mathscr{R}(f)$ ; for every $\varepsilon > 0$ , the function $g = f - h(a) + \varepsilon$ belongs to $\mathscr{H}$ , and

\mathscr{R}(g(y)) = h(y) - h(a) + \varepsilon \geq \varepsilon

for all

y \in F

. Let

c

be the supremum of

|g|

in X, and set

b = c^2 / 2\varepsilon

; for every

y \in F

|g(y) - b|^2 = |g(y)|^2 - 2b \mathscr{R}((g(y))) + b^2 \leq c^2 - 2b\varepsilon + b^2 = b^2,

in other words, the supremum in F of the function

|g - b|

\leq b

. Since

g - b \in \mathscr{H}

, the hypothesis on F implies that

|g - b| \leq b

, whence

b^2 \geq |g - b|^2 = |g|^2 - 2b \mathscr{R}(g) + b^2

and so

\mathscr{R}(g) \geq |g|^2 / 2b \geq 0

; since

\mathscr{R}(g) = h - h(a) + \varepsilon

, and

\varepsilon > 0

is arbitrary, we have

h \geq h(a)

, and

h(a)

is the infimum of

h

in X, which completes the proof.

Remark

If $f$ is a continuous real function, a point where $|f|$ attains its supremum is a point where one of the functions $f, -f$ attains its supremum. For a vector space $\mathscr{H}$ of continuous real functions satisfying the hypotheses of No. 3, the Props. 9 and 10 are thus trivial corollaries of Props. 5 and 7, respectively.

5. Applications: III. Algebras of continuous functions

Lemma 4

Let X be a compact space, $\mathscr{H}$ a closed linear subspace of the Banach space $\mathscr{C}(X; \mathbf{C})$ (resp. $\mathscr{C}(X; \mathbf{R})$ ). Let a be a point of X admitting a countable fundamental system of neighborhoods; assume that, for any numbers c and d such that $0 < c < d < 1$ and any open neighborhood U of a, there exists an $f \in \mathscr{H}$ such that

|f| \leq 1,\quad |f(a)| \geq d,\quad |f(x)| \leq c \text{ for all } x \in X - U.

Then there exists a function

u \in \mathscr{H}

such that

|u(x)| < |u(a)|

for all

x \neq a

Let $(V_n)$ ( $n \geq 1$ ) be a fundamental system of neighborhoods of $a$ , and let $\lambda, \mu, \varepsilon$ be numbers such that

0 < \lambda < 1,\quad 1 < \mu < \mu + \varepsilon \leq 1 + \lambda.

Thus

0 < \lambda / \mu < 1 / \mu < 1

. We are going to define, by induction on

n

(

n \geq 1

), a decreasing sequence

(U_n)

of open neighborhoods of

a

such that

U_n \subset V_n

for all

n

, and a sequence

(h_n)

of functions in

\mathscr{H}

satisfying the relations

\begin{align*} (13_n) &\quad |h_n(x)| \leq \mu \quad \text{for all } x \in X \\ (14_n) &\quad h_n(a) = 1 \\ (15_n) &\quad |h_n(x)| \leq \lambda \quad \text{for all } x \in X - U_n \\ (16_n) &\quad \left| \sum_{j=1}^n \lambda^j h_j(y) \right| < \sum_{j=1}^{n+1} \lambda^j \quad \text{for all } y \in X. \end{align*}

Assume

h_m

and

U_m

defined for

1 \leq m < n

, satisfying the four preceding conditions (with

n

replaced by

m

); on the other hand, set

U_0 = X

. The function

\sum_{j=1}^{n-1} \lambda^j h_j

(equal to 0 if

n = 1

) is continuous and takes the value

\sum_{j=1}^{n-1} \lambda^j

at the point

a

; therefore there exists an open neighborhood

U_n

a

, contained in

U_{n-1} \cap V_n

, such that

\left| \sum_{j=1}^{n-1} \lambda^j h_j(y) \right| < \sum_{j=1}^{n-1} \lambda^j + \varepsilon \lambda^n \quad \text{for all } y \in U_n.

By hypothesis there exists a function

f \in \mathscr{H}

such that

\begin{align*} |f(x)| &\leq 1 \quad \text{for all } x \in X , \quad |f(a)| \geq 1 / \mu , \\ |f(x)| &\leq \lambda / \mu \quad \text{for } x \in X - U_n . \end{align*}

Set

h_n = f / f(a)

; the relations

(13_n)

(14_n)

and

(15_n)

then hold; set

g = \sum_{j=1}^n \lambda^j h_j = \sum_{j=1}^{n-1} \lambda^j h_j + \lambda^n h_n .

By (17) and

(13_n)

, for

y \in U_n

we have

|g(y)| < \sum_{j=1}^{n-1} \lambda^j + \varepsilon \lambda^n + \mu \lambda^n \leq \sum_{j=1}^{n+1} \lambda^j ,

since $\varepsilon + \mu \leq 1 + \lambda$ ; for $x \in X - U_n$ , we have $|h_p(x)| \leq \lambda$ for $1 \leq p \leq n$ , whence also

|g(x)| \leq \sum_{j=2}^{n+1} \lambda^j < \sum_{j=1}^{n+1} \lambda^j,

which completes the proof of

(16_n)

This being so, the series $\sum_{n=1}^{\infty} \lambda^n h_n$ is normally convergent in $X$ since $\lambda < 1$ and $|h_n(x)| \leq \mu$ for all $n$ and all $x \in X$ ; let $u$ be its sum, which belongs to $\mathscr{H}$ since $\mathscr{H}$ is closed. By the relation $(14_n)$ , we have $u(a) = \sum_{n=1}^{\infty} \lambda^n$ ; on the other hand if $x \neq a$ , there exists an integer $n$ such that $x \notin U_{n+1}$ ; therefore $|h_{n+k}(x)| \leq \lambda$ for all $k \geq 1$ by the relation $(15_n)$ ; it follows, using $(16_n)$ , that

|u(x)| \leq \left| \sum_{j=1}^n \lambda^j h_j(x) \right| + \left| \sum_{j=n+1}^{\infty} \lambda^j h_j(x) \right| < \sum_{j=1}^{n+1} \lambda^j + \lambda \sum_{j=n+1}^{\infty} \lambda^j

= \sum_{j=1}^{\infty} \lambda^j = |u(a)|.

Theorem 2 (E. Bishop)

— *Let $X$ be a compact space, $\mathscr{A}$ a closed subalgebra of the complex Banach algebra $\mathscr{C}(X; \mathbf{C})$ . Assume that $\mathscr{A}$ contains the constants and separates the points of $X$ . Let $a$ be a point of $X$ ; the following conditions are equivalent:

a) There exists a function $f \in \mathscr{A}$ such that $|f(x)| < |f(a)|$ for all $x \neq a$ .

b) The point $a$ is $\mathscr{A}$ -extremal and admits a countable fundamental system of neighborhoods.

a) $\Rightarrow$ b): Let $f \in \mathscr{A}$ be such that $|f(a)| > |f(x)|$ for $x \neq a$ ; by Prop. 9 of No. 4, $a$ is an $\mathscr{A}$ -extremal point. On the other hand, if $U_n$ is the set of $x \in X$ such that $|f(x)| > |f(a)| - 1/n$ , then $U_n$ is an open neighborhood of $a$ , and the intersection of the $U_n$ reduces to $a$ ; since $X$ is compact, the $U_n$ form a fundamental system of neighborhoods of $a$ (GT, I, §9, No. 1, Th. 1).

b) $\Rightarrow$ a): It suffices to verify that b) implies the hypotheses of Lemma 4. With the notations of that lemma, set $\varepsilon = \log d / \log c$ ; thus $0 < \varepsilon < 1$ . Since $a$ is an $\mathscr{A}_r$ -extremal point, there exists a function $g \in \mathscr{A}$ such that

\mathscr{R}(g) \geq 0,\quad \mathscr{R}(g(a)) \leq \varepsilon,\quad \mathscr{R}(g(x)) \geq 1 \text{ for } x \in X - U

(No. 3, Prop. 6, b)). Set $f = c^g$ ; since $f$ is the sum of the normally convergent series $\sum_{n=0}^{\infty} (\log c)^n g^n / n!$ , we have $f \in \mathscr{A}$ and

|f| \leq 1,\quad |f(a)| \geq c^{\varepsilon} = d,\quad |f(x)| \leq c \text{ for } x \in X - U.

Q.E.D.

Corollary

— Suppose in addition that $X$ is metrizable. Then the following properties are equivalent:

a) $a$ is an $\mathscr{A}$ -extremal point of $X$ .

b) There exists $u \in \mathscr{A}$ such that $|u(x)| < |u(a)|$ for all $x \neq a$ .

c) Let $\mathfrak{M}$ be the set of subsets $M$ of $X$ such that for every function $f \in \mathscr{A}$ , $|f|$ attains its supremum in $X$ at at least one point of $M$ . Then $a$ belongs to all of the sets $M \in \mathfrak{M}$ .

d) Let $\mathfrak{N}$ be the set of subsets $N$ of $X$ such that, for every function $f \in \mathscr{A}$ , $\mathscr{R}(f)$ attains its supremum in $X$ at at least one point of $N$ . Then $a$ belongs to all of the sets $N \in \mathfrak{N}$ .

In other words,

\mathrm{Ch}_{\mathscr{A}}(X) = \bigcap_{M \in \mathfrak{M}} M = \bigcap_{N \in \mathfrak{N}} N.

Since, in a metrizable space, every point admits a countable fundamental system of neighborhoods, the equivalence of a) and b) follows from Th. 2. Let us show that b) implies c): indeed, $a$ is the unique point where $|u|$ attains its supremum; on the other hand, c) implies a) because, for every $f \in \mathscr{A}$ , $\mathrm{Ch}_{\mathscr{A}}(X)$ intersects the set of points where $|f|$ attains its supremum (No. 4, Prop. 9). The same reasoning, using Prop. 5 of No. 3, shows that d) implies a). Finally, to see that b) implies d), we can restrict ourselves to the case that $X$ does not reduce to the single point $a$ , therefore $u(a) \neq 0$ ; the function $v = u/u(a)$ then belongs to $\mathscr{A}$ , and we have $v(a) = 1$ and $|v(x)| < 1$ for $x \neq a$ , whence $\mathscr{R}(v(a)) = 1$ and $\mathscr{R}(v(x)) < 1$ for $x \neq a$ . Since the function $\mathscr{R}(v)$ attains its supremum only at the point $a$ , we have indeed $a \in N$ for every $N \in \mathfrak{N}$ .

Example

— Let $X_1$ be the set of points $(z_1, z_2) \in \mathbf{C}^2$ such that $|z_1|^2 + |z_2|^2 \leq 1$ (the unit ball in $\mathbf{R}^4$ ) and let $\mathscr{A}_1'$ be the set of restrictions to $X_1$ of the holomorphic functions, with values in $\mathbf{C}$ , defined in a neighborhood of $X_1$ in $\mathbf{C}^2$ (the neighborhood depending on the function considered); let $\mathscr{A}_1$ be the closure of $\mathscr{A}_1'$ in $\mathscr{C}(X_1; \mathbf{C})$ , which is obviously a closed complex subalgebra of $\mathscr{C}(X_1; \mathbf{C})$ and separates the points of $X_1$ . Application of the 'maximum principle' for holomorphic functions shows that $\mathrm{Ch}_{\mathscr{A}_1}(X_1)$ is the sphere $S_3$ .

In the preceding definition, let us replace $X_1$ by the 'polydisk' $X_2$ defined by the relations $|z_1| \leq 1$ and $|z_2| \leq 1$ , which yields subalgebras $\mathscr{A}_2'$ and $\mathscr{A}_2$ (the closure of $\mathscr{A}_2'$ of $\mathscr{C}(X_2; \mathbf{C})$ . Here, the maximum principle shows that $\mathrm{Ch}_{\mathscr{A}_2}(X_2)$ is the 'torus' defined by the relations $|z_1| = 1$ and $|z_2| = 1$ .

From these results, one deduces that there does not exist an analytic isomorphism of an open neighborhood of $X_1$ onto an open neighborhood of $X_2$ that transforms $X_1$ into $X_2$ ; for, if $v$ were the restriction to $X_1$ of such a mapping, one would have $\mathscr{A}_2 = v \mathscr{A}_1 v^{-1}$ and so $v$ would transform $S_3$ into a space homeomorphic to $T^2$ , which is absurd since $S_3$ is simply connected but $T^2$ is not. One will observe, however, that the spaces $X_1$ and $X_2$ are homeomorphic, both being bounded convex sets in $\mathbf{R}^4$ with nonempty interior.

6. Uniqueness of integral representations

Let E be a Hausdorff weak locally convex space (TVS, II, §6, No. 2), C a proper pointed convex cone in E. One knows that C is the set of elements $\geqslant 0$ of E for an order relation compatible with the vector space structure of E. When C is said to be lattice-ordered, it is of course the order induced on C by that of E that is understood.

Lemma 5

Assume that C is weakly complete. Let $\mathscr{A}$ be the set of restrictions to C of the continuous linear forms on E. Let $(f_\lambda)_{\lambda \in \Lambda}$ be a finite family of elements of $\mathscr{A}$ , and $f = \sup(f_\lambda)$ . For every $x \in C$ , define

\overline{f}(x) = \sup \left( f(x_1) + f(x_2) + \cdots + f(x_n) \right),

*the supremum being taken over the set $S_x$ of sequences $(x_1, x_2, \ldots, x_n)$ of elements of C such that $x_1 + x_2 + \cdots + x_n = x$ . Let $\mathrm{Card}\, \Lambda = p$ . Then there exists $(y_1, \ldots, y_p) \in S_x$ such that $\overline{f}(x) = f(y_1) + \cdots + f(y_p)$ .

Denote by $f_1, f_2, \ldots, f_p$ the elements of the family $(f_\lambda)$ . For $k = 1, 2, \ldots, p$ , let $C_k$ be the set of $y \in C$ such that

f_1(y) < f(y),\ f_2(y) < f(y),\ \ldots,\ f_{k-1}(y) < f(y),\ f_k(y) = f(y).

The $C_k$ are disjoint convex cones with union C. Let $x_1, x_2, \ldots, x_n$ in C be such that $x_1 + x_2 + \cdots + x_n = x$ . Let $y_k$ be the sum of the $x_i$ that belong to $C_k$ . Then $y_1 + y_2 + \cdots + y_p = x$ . Since $f$ is affine on $C_k$ , $f(y_1) + \cdots + f(y_p) = f(x_1) + \cdots + f(x_n)$ . Therefore

f(x) = \sup \left( f(y_1) + \cdots + f(y_p) \right),

where $(y_1, y_2, \ldots, y_p)$ runs over the set of sequences of $p$ points of C such that $y_1 + y_2 + \cdots + y_p = x$ . Set $D = C \cap (x - C)$ . Since D is compact (TVS, II, §6, No. 8, Cor. 2 of Prop. 11), so is the set of elements $(y_1, \ldots, y_p)$ of $D^p$ such that $y_1 + \cdots + y_p = x$ , thus the supremum (19) is attained.

Lemma 6

We maintain the hypotheses and notations of Lemma 5, and assume the $f_\lambda$ to be positive. The function $\overline{f}$ is positively homogeneous, concave and upper semi-continuous in $C$ . It is affine if $C$ is lattice-ordered.

It is clear that $\overline{f}$ is positively homogeneous. Let $x, y$ belong to $C$ . If $x_1, \ldots, x_m, y_1, \ldots, y_n$ in $C$ are such that $x_1 + \cdots + x_m = x,\ y_1 + \cdots + y_n = y$ , then $x_1 + \cdots + x_m + y_1 + \cdots + y_n = x + y$ , therefore

f(x_1) + \cdots + f(x_m) + f(y_1) + \cdots + f(y_n) \leq \overline{f}(x + y);

it follows that $\overline{f}(x) + \overline{f}(y) \leq \overline{f}(x + y)$ , therefore $\overline{f}$ is concave. Let $L$ (resp. $L_\lambda$ ) be the set of $(t, x) \in \mathbf{R} \times E$ such that $x \in C$ and $0 \leq t \leq \overline{f}(x)$ (resp. $0 \leq t \leq f_\lambda(x)$ ). Each of the $L_\lambda$ is closed in the weakly complete proper convex cone $\mathbf{R}_+ \times C$ , therefore the sum $\sum_{\lambda \in \Lambda} L_\lambda$ is closed (TVS, II, §6, No. 8, Cor. 2 of Prop. 11). By Lemma 5, this sum is equal to $L$ . Therefore $L$ is closed, which proves that $\overline{f}$ is upper semi-continuous. Finally, assume $C$ is lattice-ordered, and let us prove that $\overline{f}$ is convex. Let $x, y$ belong to $C$ and let $\varepsilon > 0$ . There exist $z_1, z_2, \ldots, z_n$ in $C$ such that $f(z_1) + \cdots + f(z_n) \geq \overline{f}(x + y) - \varepsilon$ and $z_1 + \cdots + z_n = x + y$ . The vector space $C - C$ is lattice-ordered for the order induced by that of $E$ (A, VI, §1, No. 9, Prop. 8). By the decomposition theorem (loc. cit., No. 10, Th. 1), there exist $x_1, \ldots, x_n, y_1, \ldots, y_n$ in $C$ such that

x_1 + y_1 = z_1,\ \ldots,\ x_n + y_n = z_n,\ x_1 + \cdots + x_n = x,\ y_1 + \cdots + y_n = y.

Then, since $f$ is positively homogeneous and convex,

\begin{align*} \overline{f}(x + y) &\leq \varepsilon + f(z_1) + \cdots + f(z_n) \\ &\leq \varepsilon + f(x_1) + f(y_1) + \cdots + f(x_n) + f(y_n) \\ &\leq \varepsilon + \overline{f}(x) + \overline{f}(y). \end{align*}

Since $\varepsilon$ is an arbitrary number $> 0$ , we have proved that $\overline{f}$ is indeed convex.

Theorem 3 (Choquet)

Let $E$ be a Hausdorff weak locally convex space, $C$ a weakly complete proper convex cone with vertex 0 in $E$ , $G$ the union of the extremal generators of $C$ , $K$ a compact convex subset of $C$ , $\lambda$ and $\lambda'$ positive measures of mass 1 on $K$ , admitting the same barycenter, such that $\lambda^*(K - (K \cap G)) = {\lambda'}^*(K - (K \cap G)) = 0$ . Assume that $C$ is lattice-ordered. Then, for every lower semi-continuous, positively homogeneous convex function $f \geq 0$ on $C$ , $\lambda^*(f|K) = {\lambda'}^*(f|K)$ .

Let $\mathscr{A}$ (resp. $\mathscr{A}'$ ) be the set of restrictions to $C$ of the continuous linear forms (resp. affine functions) on $E$ . We know (TVS, II, §5, No. 4, Remark 2) that $f$ is the upper envelope of the set of elements of $\mathscr{A}$ that are $\leq f$ . The set of functions of the form $\sup(f_1, \ldots, f_p)$ , where $f_1, \ldots, f_p$ belong to $\mathscr{A}$ , $f_1 \geq 0, \ldots, f_p \geq 0$ , is an increasing directed set and has $f$ as its upper envelope. Taking into account §1, No. 1, Th. 1, it suffices to verify the equality $\lambda(f|K) = \lambda'(f|K)$ when $f$ is of the preceding form.

Define $\overline{f}$ as in Lemma 5. It is clear that $\overline{f}(y) = f(y)$ if $y \in G$ . Since $\lambda^*(K - (K \cap G)) = 0$ , we have $\lambda(f|K) = \lambda(\overline{f}|K)$ . By Lemma 6, $\overline{f}$ is affine and upper semi-continuous. Therefore $\overline{f}|K$ is the lower envelope of a decreasing directed set of restrictions of elements of $\mathscr{A}'$ to K (TVS, II, §5, No. 4, Prop. 6). Let $x \in K$ be the barycenter of $\lambda$ . If $g \in \mathscr{A}$ then $\lambda(g|K) = g(x)$ . Therefore $\lambda(\overline{f}|K) = \overline{f}(x)$ (§ 4, No. 4, Cor. 2 of Prop. 5). Thus $\lambda(f|K) = \overline{f}(x)$ , and one sees similarly that $\lambda'(f|K) = \overline{f}(x)$ .

Corollary

Let E be a Hausdorff locally convex space, C a proper convex cone with vertex 0 in E, admitting a compact sole M, and let G be the union of the extremal generators of C. Let $x \in M$ . If C is lattice-ordered, then there exists at most one positive measure $\lambda$ of mass 1 on M, such that $\lambda^*(M - (G \cap M)) = 0$ , and admitting x as barycenter.

Replacing the topology of E by the weakened topology (which does not change the topology of M), one can suppose E to be a weak space. Let $\lambda$ and $\lambda'$ be two measures on M having the stated properties, and let h be a continuous linear form on E such that M is the intersection of C and the hyperplane with equation $h(x) = 1$ . Let $\mathscr{S}$ be the subset of $\mathscr{C}(M)$ consisting of the restrictions to M of the positively homogeneous and continuous convex functions $\geq 0$ on C. The cone C is weakly complete (TVS, II, §7, No. 3). By Th. 3, $\lambda(f) = \lambda'(f)$ for every $f \in \mathscr{S}$ .

If $f_1, f_2, f_3, f_4$ belong to $\mathscr{S}$ , then

\begin{align*} \sup(f_1 - f_2, f_3 - f_4) &= \sup(f_1 + f_4, f_3 + f_2) - (f_2 + f_4) \in \mathscr{S} - \mathscr{S} \\ \inf(f_1 - f_2, f_3 - f_4) &= -\sup(f_2 - f_1, f_4 - f_3) \in \mathscr{S} - \mathscr{S}. \end{align*}

Since

h|M \in \mathscr{S}

\mathscr{S} - \mathscr{S}

contains the constant functions. If x and y are two distinct points of M, there exists a continuous linear form on E that separates x and y, and this form is the difference of two continuous linear forms that are positive on C (TVS, II, §6, No. 8, Lemma 1). It follows from the foregoing that for

\alpha, \beta

real, there exists

f \in \mathscr{S} - \mathscr{S}

such that

f(x) = \alpha, f(y) = \beta

. Then

\mathscr{S} - \mathscr{S}

is dense in

\mathscr{C}(M)

for the topology of uniform convergence (GT, X, §4, No. 1, Cor. of Prop. 2). Since

\lambda

and

\lambda'

coincide on

\mathscr{S} - \mathscr{S}

, we have

\lambda = \lambda'

Exercises

See the exercises for § 7.