Bourbaki

Theorem 1

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

Integration, chapter IV, § 6. Convexity inequalities

— Let XX be a locally compact space, μ\mu a positive measure on XX, FF a real Banach space, DD a closed convex set in FF, and ff a function on XX such that f(X)⊂Df(X) \subset D. For every non-negligible integrable numerical function g⩾0g \geqslant 0 such that fgfg is integrable, the point ∫fg dμ∫g dμ\frac{\int fg\, d\mu}{\int g\, d\mu} belongs to DD.

For, let F′F' be the dual of FF and let ⟨z,a′⟩⩽α\langle z, a' \rangle \leqslant \alpha (a′∈F′,α∈Ra' \in F', \alpha \in \mathbf{R}) be a relation defining a closed half-space containing DD. Since fgfg is integrable, so is the numerical function ⟨fg,a′⟩=⟨f,a′⟩g\langle fg, a' \rangle = \langle f, a' \rangle g, and

∫⟨fg,a′⟩ dμ=⟨∫fg dμ,a′⟩ \int \langle fg, a' \rangle\, d\mu = \left\langle \int fg\, d\mu, a' \right\rangle

(§ 4, No. 2, Cor. 1 of Th. 1); but, by hypothesis, ⟨f(x),a′⟩⩽α\langle f(x), a' \rangle \leqslant \alpha for all x∈Xx \in X, therefore ⟨f(x)g(x),a′⟩⩽αg(x)\langle f(x)g(x), a' \rangle \leqslant \alpha g(x); on integrating, we have

⟨∫fg dμ,a′⟩⩽α∫g dμ. \left\langle \int fg\, d\mu, a' \right\rangle \leqslant \alpha \int g\, d\mu .

This proves that the point ∫fg dμ∫g dμ\frac{\int fg\, d\mu}{\int g\, d\mu} belongs to every closed half-space containing DD; but, by the Hahn–Banach theorem, DD is the intersection of the closed half-spaces containing it (TVS, II, §5, No. 3, Cor. 1 of Prop. 4), whence the theorem.

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

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Transcribed from the printed volume. Not translated.