Lemma 1

The space of functions of $L^1(G)$ whose Fourier transform has compact support is dense in $L^1(G)$.

Since $\mathscr{K}(\widehat{G})$ is dense in $L^2(\widehat{G})$ and the Fourier transform of $L^2(G)$ is an isometry onto $L^2(\widehat{G})$ (th. 1 of II, p. 215), the subspace V of $L^2(G)$ formed by the $f\in L^2(G)$ such that $\mathscr{F}_G(f)\in \mathscr{K}(\widehat{G})$ is dense in $L^2(G)$.

Let $g\in L^1(G)$. There exist $g_1, g_2\in L^2(G)$ such that $g=g_1g_2$ (one may for example take $g_1=|g|^{1/2}$, and $g_2(x) = 0$ if $g(x) = 0,g_2(x) =$ $g(x)/g_1(x)$ otherwise). It follows therefore from what precedes that $g$ is the limit of a sequence of functions of the form $h_1h_2$, where $h_1$ and $h_2$ belong to V. Now $\mathscr{F}_G(h_1h_2) =\mathscr{F}_G(h_1)*\mathscr{F}_G(h_2)$ (II, p. 223, prop. 14), and $\mathscr{F}_G(h_1)*$ $\mathscr{F}_G(h_2)$ belongs to $\mathscr{K}(G)$. The lemma follows.