Lemma 1

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

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

For every $\chi \in \widehat{G}$ , the mapping $\mu\mapsto \chi (\mu)$ is a hermitian character of the involutive Banach algebra $\mathscr{M}^1(G)$ .

By INT, VIII, §3, No $^o3$ , Prop. 11, the mapping $\mu\mapsto \chi (\mu)$ is a character of the involutive Banach algebra $\mathscr{M}^1(G)$ . Further, we have:

\chi (\mu^*) =\int_G\chi (x^{-1})d\mu(x) =\int_G\overline{\chi(x)}d\mu(x) =\overline{\chi(\mu)}

and therefore this character is hermitian.

Thus we have defined a mapping of $\widehat{G}$ into $\mathsf{X}(\mathscr{M}^1(G))$ ; it will be called canonical.

Let $\chi \in \widehat{G}$ . The restriction of $\mu\mapsto \chi (\mu)$ to $L^1(G)$ is non-zero (cf. INT, VIII, §2, No $^o7$ , Prop. 10). By restriction to the involutive Banach subalgebra $L^1(G)$ , we therefore obtain a mapping of $\widehat{G}$ into $\mathsf{X}(L^1(G))$ , called canonical. It associates to $\chi \in \widehat{G}$ the hermitian character

f\mapsto \chi (f) =\chi (f\cdot dx) =\int_Gf(x)\chi (x)dx(f\in L^1(G)) \tag{2}

of $L^1(G)$ .

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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