Bourbaki

Theorem 2 (E. Bishop)

Tag 08Z7, permanent. Cite it as [Bourbaki, Tag 08Z7].

Integration, chapter IV, § 7. Barycenters

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

a) There exists a function f∈Af \in \mathscr{A} such that ∣f(x)∣<∣f(a)∣|f(x)| < |f(a)| for all x≠ax \neq a.

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

a) ⇒\Rightarrow b): Let f∈Af \in \mathscr{A} be such that ∣f(a)∣>∣f(x)∣|f(a)| > |f(x)| for x≠ax \neq a; by Prop. 9 of No. 4, aa is an A\mathscr{A}-extremal point. On the other hand, if UnU_n is the set of x∈Xx \in X such that ∣f(x)∣>∣f(a)∣−1/n|f(x)| > |f(a)| - 1/n, then UnU_n is an open neighborhood of aa, and the intersection of the UnU_n reduces to aa; since XX is compact, the UnU_n form a fundamental system of neighborhoods of aa (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 ε=log⁡d/log⁡c\varepsilon = \log d / \log c; thus 0<ε<10 < \varepsilon < 1. Since aa is an Ar\mathscr{A}_r-extremal point, there exists a function g∈Ag \in \mathscr{A} such that

R(g)≥0,R(g(a))≤ε,R(g(x))≥1 for x∈X−U \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=cgf = c^g; since ff is the sum of the normally convergent series ∑n=0∞(log⁡c)ngn/n!\sum_{n=0}^{\infty} (\log c)^n g^n / n!, we have f∈Af \in \mathscr{A} and

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

Q.E.D.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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