Bourbaki

Exercise 4, § 7

Integration, IV, p. 144

Bourbaki marks this one as harder than the rest.

Let E be a Hausdorff locally convex space, X a compact convex set in E, P\mathscr{P} the set of finite numerical functions continuous and convex in X; μ≺ν\mu \prec \nu denotes the pre-order relation defined on M+(X)\mathscr{M}_+(X) by P\mathscr{P} (Exer. 2).

a) Show that μ≺ν\mu \prec \nu is an order relation on M+(X)\mathscr{M}_+(X) (use Exer. 29 of TVS, II, §5) and that, for every x∈Xx \in X, the relation εx≺μ\varepsilon_x \prec \mu is equivalent to saying that μ\mu has mass 1 and xx is the barycenter of μ\mu. For every maximal measure μ\mu (for the preceding order relation) of mass 1, there therefore exists one and only one x∈Xx \in X such that εx≺μ\varepsilon_x \prec \mu. Conversely, every z∈Xz \in X is the barycenter of at least one maximal measure. For x∈Xx \in X to be an extremal point of X, it is necessary and sufficient that εx\varepsilon_x be a maximal measure.

b) Let μ\mu be a maximal measure on X, ff a function in C+(X)\mathscr{C}_+(X). Show that for every ε>0\varepsilon > 0, there exists a continuous convex function gg on X such that 0≤g≤f0 \leq g \leq f and μ(g)≥μ(f)−ε\mu(g) \geq \mu(f) - \varepsilon (use Exers. 3 b) and 2 d)). From this, deduce that the support of μ\mu is contained in the closure of the set of extremal points of X.

c) Let μ\mu be a maximal measure on X, (fn)n≥1(f_n)_{n \geq 1} a decreasing sequence of functions in C+(X)\mathscr{C}_+(X). Show that if, for every extremal point x∈Xx \in X, the sequence (fn(x))(f_n(x)) tends to 0, then lim⁡n→∞μ(fn)=0\lim_{n \to \infty} \mu(f_n) = 0 (use b) to construct a decreasing sequence (gn)(g_n) of continuous convex functions such that 0≤gn≤fn0 \leq g_n \leq f_n and μ(gn)≥μ(fn)−ε\mu(g_n) \geq \mu(f_n) - \varepsilon for all nn, and show that the sequence (gn(y))(g_n(y)) tends to 0 for every y∈Xy \in X.

d) Deduce from c) that if A⊂XA \subset X contains no extremal point and is the union of a sequence (Kn)(K_n) of compact subsets of X, each of which is a countable intersection of open sets in X, then μ(A)=0\mu(A) = 0 for every maximal measure μ\mu.

e) Let A\mathscr{A} be the vector space of continuous functions on X that are the restrictions to X of continuous affine linear functions on E. The relation λ≺μ\lambda \prec \mu implies that λ(h)=μ(h)\lambda(h) = \mu(h) for every function h∈Ah \in \mathscr{A}; for every function f∈C(X;R)f \in \mathscr{C}(X; \mathbf{R}), the lower envelope S(f)S(f) of the functions h∈Ah \in \mathscr{A} such that f≤hf \leq h is also the lower envelope of the functions g∈−Pg \in -\mathscr{P} such that f≤gf \leq g. For every x∈Xx \in X, the relation εx≺μ\varepsilon_x \prec \mu is equivalent to the relation h(x)=⟨h,μ⟩h(x) = \langle h, \mu \rangle for all h∈Ah \in \mathscr{A}; the extremal points of X are identical to the A\mathscr{A}-extremal points. If f∈Pf \in \mathscr{P} then, for every x∈Xx \in X,

(*)

S(f)(x)=sup⁡εx≺μ∫f(y) dμ(y) S(f)(x) = \sup_{\varepsilon_x \prec \mu} \int f(y) \, d\mu(y)

(use Exer. 2 b)). This formula is no longer necessarily valid if it is only assumed that ff is convex and upper semi-continuous on X (with the notations of Exer. 1 b), consider the function ff equal to 1 on the image of I in K, and to 0 elsewhere.

f) Show that the formula (*) is again valid when in the second member one limits oneself to the discrete measures μ\mu having xx as barycenter.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.