Proposition 13

Tag 02IU, permanent. Cite it as [Bourbaki, Tag 02IU].

Théories spectrales, chapter II, § 1. Transformation de Fourier

Let $\alpha \in \mathscr{M}^1(G)$ and $\beta \in \mathscr{M}^1(\widehat{G})$ . Then one has (30) $\mathscr{F}_G(\overline{\mathscr{F}}_{\widehat{G}}(\beta )\cdot \alpha ) =\beta *\mathscr{F}_G(\alpha )$

and in particular

\int_G\mathscr{F}_{\widehat{G}}(\beta )(x)d\alpha (x) =\int_{\widehat{G}}\mathscr{F}_G(\alpha )(\chi )d\beta (\chi ) \tag{31}

Formula (30) implies formula (31) by evaluating both sides of the identity at $\chi = 1$ . Let us prove (30). Let $\chi \in \widehat{G}$ . There follows

(\mathscr{F}_G(\overline{\mathscr{F}}_{\widehat{G}}(\beta )\cdot \alpha ))(\chi ) =\int_G\overline{\langle\chi , x\rangle}\overline{\mathscr{F}}_{\widehat{G}}(\beta )(x)d\alpha (x)

=\int_G\overline{\langle\chi , x\rangle}(\int_{\widehat{G}}\langle \xi , x\rangle d\beta (\xi ))d\alpha (x)

The function $(x, \xi )\mapsto \langle \chi , x\rangle \langle \xi , x\rangle$ is continuous and bounded, hence integrable on $G\times \widehat{G}$ with respect to the measure $\alpha \otimes \beta$ . By Lebesgue-Fubini's theorem (INT, V, §8, n $^o4$ , th. 1, a)), one obtains

(\mathscr{F}_G(\overline{\mathscr{F}}_{\widehat{G}}(\beta )\cdot \alpha ))(\chi ) =\int_{\widehat{G}}(\int_G\overline{\langle\chi \xi^{-1}, x\rangle}d\alpha (x))d\beta (\xi )

=\int_{\widehat{G}}\mathscr{F}_G(\alpha )(\chi \xi^{-1})d\beta (\xi ) = (\beta *\mathscr{F}_G(\alpha ))(\chi )

as desired.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

Cited by

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Cites

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