Corollary 1

Let $f\in L^1(G)$. Then $f\in B(G)$ if and only if $f$ belongs to the image of the Fourier transform $\mathscr{F}_{\widehat{G}}$ on $L^1(\widehat{G})$. In particular, one has $A(G)\subset B(G)$.

Theorem 3 proves that if $f\in B(G)$, then $f=\mathscr{F}_{\widehat{G}}(\overline{\mathscr{F}}_G(f))$, where $\mathscr{F}_G(f)$ belong to $L^1(\widehat{G})$. Conversely, if $f=\mathscr{F}_{\widehat{G}}(g)$, where $g\in L^1(\widehat{G})$, then one has $g\in B(\widehat{G})$ and therefore $f\in B(G)$ by the theorem. The last assertion then results from prop. 11 of II, p. 217.