Bourbaki

Proposition 2

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

Integration, chapter III, § 3. Integrals of continuous vector-valued functions

— Let uu be a continuous linear mapping of EE into a locally convex space FF; for every function f∈K~(X;E)f \in \widetilde{\mathscr{K}}(X;E), the function u∘fu \circ f belongs to K~(X;F)\widetilde{\mathscr{K}}(X;F) and

∫u(f(x)) dμ(x)=ttu(∫f(x) dμ(x)). \int u(f(x))\, d\mu(x) = ^{tt} u \left( \int f(x)\, d\mu(x) \right).

For every continuous linear form z′∈F′z' \in F', we have z′∘u∘f=y′∘fz' \circ u \circ f = y' \circ f on setting y′=z′∘u=tu(z′)∈E′y' = z' \circ u = {}^t u(z') \in E', whence the first assertion; moreover, for all z′∈F′z' \in F',

⟨∫(u∘f) dμ,z′⟩=∫⟨u∘f,z′⟩ dμ=∫⟨f,tu(z′)⟩ dμ=⟨∫f dμ,tu(z′)⟩=⟨ttu(∫f dμ),z′⟩, \left\langle \int (u \circ f)\, d\mu, z' \right\rangle = \int \langle u \circ f, z' \rangle\, d\mu = \int \langle f, {}^t u(z') \rangle\, d\mu \\ = \left\langle \int f\, d\mu, {}^t u(z') \right\rangle = \left\langle {}^{tt} u \left( \int f\, d\mu \right), z' \right\rangle,

whence the formula (2).

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

Cited by

Cites

Nothing.

Transcribed from the printed volume. Not translated.