Proposition 8

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

— Let E be a locally convex space, $\mu$ a promeasure on E, and $\Phi$ the Fourier transform of $\mu$ . One has the inequalities

|\Phi(x')| \leq \Phi(0)

|\Phi(x') - \Phi(y')|^2 \leq 2\Phi(0)\left(\Phi(0) - \Re \Phi(x' - y')\right)

for $x', y'$ in $E'$ .

Formula (5) of No. 3 permits reducing to the case that E is finite-dimensional and $\mu$ is a measure. Then

|\Phi(x')| = \left| \int_E e^{i\langle x, x' \rangle} d\mu(x) \right| \leq \int_E |e^{i\langle x, x' \rangle}| d\mu(x) = \int_E d\mu(x) = \Phi(0),

whence (53). Moreover, if $a$ and $b$ are real numbers, then

|e^{ia} - e^{ib}|^2 = |e^{ib}|^2|e^{i(a-b)} - 1|^2 = (e^{i(a-b)} - 1)(e^{-i(a-b)} - 1) = 2 - 2\cos(a - b);

by the Cauchy–Schwarz inequality, we then have

|\Phi(x') - \Phi(y')|^2 = \left| \int_E (e^{i\langle x, x' \rangle} - e^{i\langle x, y' \rangle}) d\mu(x) \right|^2

\leq \int_E |e^{i\langle x, x' \rangle} - e^{i\langle x, y' \rangle}|^2 d\mu(x) \int_E 1^2 d\mu(x)

= \int_E (2 - 2\cos\langle x, x' - y' \rangle)) d\mu(x) \cdot \Phi(0)

= 2\Phi(0)\left(\Phi(0) - \Re \Phi(x' - y')\right),

whence (54).

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

Cited by

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