Bourbaki

Proposition 9

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

Integration, chapter IX, § 6. Promeasures and measures on a locally convex space

— If F is barreled, then the Fourier transform of every bounded measure on F' is a uniformly continuous function on F.

Let μ\mu be a bounded measure on F′F' and Φ\Phi its Fourier transform. Let ε>0\varepsilon > 0. There exists a compact subset K of F′F' such that μ(F′−K)≤ε\mu(F' - K) \leq \varepsilon. Now, K is compact for the weak topology σ(F′,F)\sigma(F', F), hence is equicontinuous because F is barreled (TVS, III, §4, No. 2, Th. 1). Therefore there exists a symmetric neighborhood U of 0 in F whose polar U∘U^\circ contains K. Let x be in εU\varepsilon U; then

Φ(0)−ℜΦ(x)=∫F′(1−cos⁡⟨x,x′⟩)dμ(x′). \Phi(0) - \Re \Phi(x) = \int_{F'} (1 - \cos \langle x, x' \rangle) d\mu(x').

Now, 0≤1−cos⁡⟨x,x′⟩≤20 \leq 1 - \cos \langle x, x' \rangle \leq 2 for every x′∈F′−Kx' \in F' - K, and

1−cos⁡⟨x,x′⟩≤12⟨x,x′⟩2≤ε22 1 - \cos \langle x, x' \rangle \leq \frac{1}{2} \langle x, x' \rangle^2 \leq \frac{\varepsilon^2}{2}

for x′∈K⊂U∘x' \in K \subset U^\circ; it follows that

0≤Φ(0)−ℜΦ(x)≤2μ(F′−K)+ε22μ(K)≤2ε+ε22μ(F′). 0 \leq \Phi(0) - \Re \Phi(x) \leq 2 \mu(F' - K) + \frac{\varepsilon^2}{2} \mu(K) \leq 2\varepsilon + \frac{\varepsilon^2}{2} \mu(F').

The second member of this inequality tends to 0 with ε\varepsilon; thus ℜΦ\Re \Phi is continuous at 0 and the proposition follows from the Cor. of Prop. 8.

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

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