Bourbaki

Lemma 3

Tag 09AS, permanent. Cite it as [Bourbaki, Tag 09AS].

Integration, VI, p. 29, § 2. Vectorial measures . "

Let F be a Hausdorff locally convex space, ff an element of LFs′∞\mathscr{L}_{F'_s}^\infty. For every z∈Fz \in F, one has ⟨z,f⟩∈L∞\langle z, f \rangle \in \mathscr{L}^\infty, and the linear mapping z↦π(⟨z,f⟩)z \mapsto \pi(\langle z, f \rangle) of F into L∞L^\infty is continuous; if, moreover, F is a normed space, then N∞(⟨z,f⟩)≤∣z∣⋅sup⁡t∈T∣f(t)∣N_\infty(\langle z, f \rangle) \leq |z| \cdot \sup_{t \in T} |f(t)|.

It is clear by definition that ⟨z,f⟩\langle z, f \rangle is μ\mu-measurable and bounded in measure; replacing if necessary ff by a function belonging to the same class of LFs′∞L_{F'_s}^\infty, we can suppose in addition that f(T)⊂V∘f(T) \subset V^\circ, where V is a balanced convex neighborhood of 0 in F (which does not modify ⟨z,f⟩\langle z, f \rangle except on a locally negligible set, depending on zz). Then the relation z∈Vz \in V implies that ∣⟨z,f(t)⟩∣≤1|\langle z, f(t) \rangle| \leq 1 for all t∈Tt \in T, which proves the continuity of z↦π(⟨z,f⟩)z \mapsto \pi(\langle z, f \rangle). The final assertion is obvious.

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

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