Theorem 3

The restriction of the Fourier transform to B(G) induces an isomorphism of vector spaces of B(G) onto $B(\widehat{G})$, whose inverse is induced by the restriction of the co-Fourier transform to $B(\widehat{G})$.

Let $f\in B(G)$. Put $g=\mathscr{F}_G(f)$. Then $g\in L^1(\widehat{G})\cap \mathscr{C}_0(\widehat{G})\subset$ $L^1(\widehat{G})\cap L^2(\widehat{G})$. Put $f_1=\overline{\mathscr{F}}_{\widehat{G}}(g)\in L^2(G)$. For every continuous function with compact support $h\in \mathscr{K}(\widehat{G})$, one has $h\in L^1(\widehat{G})\cap L^2(\widehat{G})$ and

by using the Plancherel theorem and formula (31). Since $\mathscr{K}(\widehat{G})$ is dense in $L^2(\widehat{G})$, its image under the Fourier transform is dense in $L^2(G)$. Consequently, one has $f_1=f$ in $L^1(G)$; this proves that $g\in B(\widehat{G})$.

The formula $f_1=f$ means that the restriction to B(G) of the composition $\overline{\mathscr{F}}_{\widehat{G}}\circ \mathscr{F}_G$ is the identity mapping of B(G). By exchanging the roles of G and $\widehat{G}$, one sees that $\mathscr{F}_G\circ \mathscr{F}_{\widehat{G}}$ is the identity mapping of $B(\widehat{G})$, which completes the proof of the theorem.