Proposition 11

Let $f\in A(G)$. Then $\mathscr{F}(f)\in L^1(\widehat{G})$ and, for every $x\in G$, one has

In other words, for $f\in A(G)$, one has

where $\eta$ denotes the canonical mapping of G into the bidual group $\widehat{\widehat{G}}$.

By Lemma 4 of II, p. 213 and Proposition 9 of II, p. 214, one has $\mathscr{F}(f)\in L^1(\widehat{G})$ for every function $f\in A(G)$. By the Plancherel formula (24), for $f$ and $g$ in $L^2(G)$, one has

Let $f$ and $g$ be in $L^1(G)\cap L^2(G)$ and let $h=f*\widetilde{g}\in A(G)$. Since the Fourier transform is an involutive morphism, formula (27) is assertion (25) for the function $h$ at the point $x=e$. By linearity, one deduces that formula (25) is valid at the point $x=e$ for every function $h\in A(G)$.

Let $x\in G$ and $h\in A(G)$. Let $h_1=\varepsilon_{x^{-1}}*h$. Then $h_1\in A(G)$ and $h_1(e) =h(x)$. Since moreover $\mathscr{F}(h_1)(\chi ) =\overline{\langle\chi , x\rangle}\mathscr{F}(f)(\chi )$ for every $\chi \in \widehat{G}($cf. formula (11) of II, p. 208), formula (25) for the function $h_1$ at the point $e$ implies formula (25) for $h$ at the point $x$.