Proposition 6

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

Integration, VIII, p. 20, § 3. Convolution of measures on groups

Let $\mu$ and $\nu$ be two convolvable measures on G. Let $\chi$ be a continuous representation of G in $\mathbf{C}^*$ . Then $\chi \cdot \mu$ and $\chi \cdot \nu$ are convolvable and $(\chi \cdot \mu) * (\chi \cdot \nu) = \chi \cdot (\mu * \nu)$ .

Let $f \in \mathscr{K}(G)$ . Then $f\chi \in \mathscr{K}(G)$ , therefore the function

(x, y) \mapsto f(xy)\chi(xy) = f(xy)\chi(x)\chi(y)

on $G \times G$ is integrable for $\mu \otimes \nu$ ; therefore the function $(x, y) \mapsto f(xy)$ is integrable for $(\chi \cdot \mu) \otimes (\chi \cdot \nu)$ ; therefore $\chi \cdot \mu$ and $\chi \cdot \nu$ are convolvable. Moreover,

\langle \chi \cdot \mu * \chi \cdot \nu, f \rangle = \int f(xy)\chi(x)\chi(y)\, d\mu(x)\, d\nu(y) = \int (f\chi)(xy)\, d\mu(x)\, d\nu(y) = \langle \mu * \nu, \chi f \rangle ,

whence $(\chi \cdot \mu) * (\chi \cdot \nu) = \chi \cdot (\mu * \nu)$ .

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

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