Remark

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

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

Let $f\in L^2(G)$ and $g\in L^2(\widehat{G})$ . Applying the Plancherel formula (24) to $f$ and $\overline{\mathscr{F}_{\widehat{G}}(g)}$ , one obtains the formula

\int_Gf(x)\mathscr{F}_{\widehat{G}}(g)(x)dx=\int_{\widehat{G}}\mathscr{F}_G(f)(\chi )g(\chi )d\widehat{x}(\chi ) \tag{29}

since one has $\mathscr{F}_G(\mathscr{F}_{\widehat{G}}(g)) =\mathscr{F}_G(\overline{\mathscr{F}}_{\widehat{G}}(\overline{g})) =\overline{g}$ .

The Fourier transform and cotransform defined on $\mathscr{M}^1(\widehat{G})$ take their values in the space of bounded continuous functions on G. For $\beta \in \mathscr{M}^1(\widehat{G})$ and $x\in G$ , one has

\mathscr{F}_{\widehat{G}}(\beta )(x) =\int_{\widehat{G}}\overline{\langle\chi , x\rangle}d\beta (\chi ),\overline{\mathscr{F}}_{\widehat{G}}(\beta )(x) =\int_{\widehat{G}}\langle \chi , x\rangle d\beta (\chi )

The Fourier transforms of G and $\widehat{G}$ are also transposes of one another. More precisely:

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.

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