Corollary 2

The vector space B(G) is an algebra both for multiplication and for convolution. The Fourier transform exchanges convolution and multiplication in B(G) and $B(\widehat{G})$.

We have already seen that B(G) is a subalgebra of $L^1(G)$. On the other hand, if $f$ and $g$ belong to B(G), then $f g\in L^1(G)$ since $f\in L^1(G)$ and $g$ belong to the image of the Fourier transform on $L^1(\widehat{G})$ (Corollary 1). Since there exist $f_1$ and $g_1$ in $L^1(\widehat{G})$ such that $f$ = $\mathscr{F}_{\widehat{G}}(f_1)$ and $g=\mathscr{F}_{\widehat{G}}(g_1)$ (loc. cit.), one has $f g=\mathscr{F}_{\widehat{G}}(f_1*g_1)$, and therefore $f g\in B(G)$ again by the preceding corollary.