Proposition 16
Tag 0997, permanent. Cite it as [Bourbaki, Tag 0997].
Integration, chapter VI, § 1. Integration of vector-valued functions
— Let be two Hausdorff locally convex spaces, with assumed to be quasi-complete. Let be a mapping of into such that is a continuous mapping of into . Then the bilinear mapping of into is hypocontinuous relative to the equicontinuous subsets of and the compact subsets of (which implies that the linear mapping of into is continuous).
The continuity of as a mapping of into follows from Lemma 3 and Remark 2 following Prop. 14 of No. 6. Thus, it remains to prove that for every closed, balanced, convex neighborhood of 0 in and for every equicontinuous subset of , there exists a neighborhood of 0 in such that the relations imply that . One can suppose that , where is a neighborhood of 0 in , consequently one can suppose that is the set of functions such that on a compact subset of . It suffices to show that
|⟨U(λ) · x, x′⟩| ≤ 1 for x ∈ W, x′ ∈ V° and λ ∈ S°. Now, since U(K) is equicontinuous, there exists a neighborhood W of 0 in G such that the relations t ∈ K, x ∈ W imply U(t) · x ∈ V; the relations x ∈ W, x′ ∈ V° therefore imply that the function t ↦ ⟨U(t) · x, x′⟩ belongs to S, hence that |⟨U(t) · x, x′⟩| = |∫⟨U(t) · x, x′⟩ dλ(t)| ≤ 1 by the definition of S°.
Let us now assume that U is a continuous mapping of T into F and, moreover, that U(T) is equicontinuous. Then, the same reasoning as above shows (since H is quasi-complete) that for every bounded positive measure μ on T, ∫ U dμ ∈ F. One can therefore define, as above, a linear mapping λ ↦ ∫ U dλ = U(λ) of M^1(T) into F that extends the analogous mapping of C'(T) into F. Moreover, for every closed, balanced, convex neighborhood V of 0 in H, there exists by hypothesis a neighborhood W of 0 in G such that for every x ∈ W and every t ∈ T, one has U(t) · x ∈ V, consequently (since V is weakly closed) ∫ (U(t) · x)dλ(t) ∈ ||λ|| · V (No. 2, Prop. 5). In other words:
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.