Bourbaki

Exercise 8, § 7

Integration, IV, p. 147

Bourbaki marks this one as harder than the rest.

The general notations and hypotheses being those of Exer. 6, set S=SˇH(X)S = \check{S}_{\mathscr{H}}(X). A function f∈C(S;R)f \in \mathscr{C}(S; \mathbf{R}) is said to be H-resolvent\mathscr{H}\text{-resolvent} if there exists a function g∈Hcg \in \mathscr{H}^c such that f=g∣Sf = g|S (this function is then unique).

a) Show that the following conditions on a function f∈C(S;R)f \in \mathscr{C}(S; \mathbf{R}) are equivalent: (i) ff is H-resolvent\mathscr{H}\text{-resolvent}. (ii) For every x∈Xx \in X, H‾xS(f)=H‾xS(f)\underline{H}_x^S(f) = \overline{H}_x^S(f). (iii) For every x∈Xx \in X and every pair of measures μ1,μ2\mu_1, \mu_2 in MxS\mathscr{M}_x^S, one has ∫fdμ1=∫fdμ2\int f d\mu_1 = \int f d\mu_2. (Make use of Exer. 6 b) and c).) b) Show that the following properties are equivalent: (i) Every function f∈C(S;R)f \in \mathscr{C}(S; \mathbf{R}) is H-resolvent\mathscr{H}\text{-resolvent}. (ii) For every x∈Xx \in X and every function f∈C(S;R)f \in \mathscr{C}(S; \mathbf{R}), H‾xS(f)=H‾xS(f)\underline{H}_x^S(f) = \overline{H}_x^S(f). (iii) The set MxS\mathscr{M}_x^S reduces to a single element. (iv) The set Hc\mathscr{H}^c is lattice-ordered. (v) For every function u∈EHu \in \mathscr{E}_{\mathscr{H}} (Exer. 6 g)), there exists a greatest lower bound huh_u of uu in Hc\mathscr{H}^c. (To show that (iv) implies (i), apply Exer. 7 d) to Hc\mathscr{H}^c, proving that for every point s∈Ss \in S, the linear form h↦h(s)h \mapsto h(s) is latticial in Hc\mathscr{H}^c; then use Stone’s theorem and Exer. 6 c). To see that (i) implies (v), note that Hc⊂EH\mathscr{H}^c \subset \mathscr{E}_{\mathscr{H}} and consider the unique function hu∈Hch_u \in \mathscr{H}^c such that hu∣S=u∣Sh_u|S = u|S. To see that (v) implies (iv), note that if h∈Hch \in \mathscr{H}^c then −h−=inf⁡(h,0)∈EH-h^- = \inf(h, 0) \in \mathscr{E}_{\mathscr{H}}, and apply (v) to u=−h−u = -h^-.)

c) If the conditions of b) are satisfied and if γx\gamma_x is the unique element of MxS\mathscr{M}_x^S, show that the mapping x↦γxx \mapsto \gamma_x of X into M+(X)\mathscr{M}_+(X) (equipped with the vague topology) is continuous; from this, deduce that one then has ChH(X)=SˇH(X)\mathrm{Ch}_{\mathscr{H}}(X) = \check{S}_{\mathscr{H}}(X) (note that γx=εx\gamma_x = \varepsilon_x in ChH(X)\mathrm{Ch}_{\mathscr{H}}(X)).

d) In order that every function f∈C(S;R)f \in \mathscr{C}(S; \mathbf{R}) be the restriction to S of a function h∈Hh \in \mathscr{H}, it is necessary and sufficient that H\mathscr{H} be closed in C(X;R)\mathscr{C}(X; \mathbf{R}) and be lattice-ordered.

Solution

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Transcribed from the printed volume. Not translated.