Proposition 14

Let $f$ and $g$ be in $L^2(G)$. Then $\mathscr{F}_G(f g) =$ $\mathscr{F}_G(f)*\mathscr{F}_G(g)$.

The equality is true if $f$ and $g$ belong to B(G) (Corollary 2), and in particular if $f$ and $g$ belong to A(G) since $A(G)\subset B(G)$ (Cor. 1). Since A(G) is dense in $L^2(G)$ (Cor. 1 of II, p. 212), it is enough to prove that the two sides of the equality are continuous functions of $(f, g)\in L^2(G)\times L^2(G)$ with values in $\mathscr{C}_0(\widehat{G})$. Now the map $(f, g)\mapsto \mathscr{F}_G(f g)$ is obtained by composing the continuous map $(f, g)\mapsto f g$ of $L^2(G)\times L^2(G)$ into $L^1(G)$ and the Fourier transform $\mathscr{F}_G$ of $L^1(G)$ into $\mathscr{C}_0(\widehat{G})$, which is also continuous. Analogously, the map $(f, g)\mapsto \mathscr{F}_G(f)*\mathscr{F}_G(g)$ is obtained by composing the continuous maps $(f, g)\mapsto (\mathscr{F}_G(f),\mathscr{F}_G(g))$ of $L^2(G)\times L^2(G)$ into $L^2(\widehat{G})\times L^2(\widehat{G})$ and $(h_1, h_2)\mapsto h_1*h_2$ of $L^2(\widehat{G})\times L^2(\widehat{G})$ into $\mathscr{C}_0(\widehat{G})$ (INT, VIII, §4, n$^o5$, prop. 15).