Proposition 2

Let I be a closed ideal of $L^1(G)$, and let $f\in L^1(G)$. If $\overline{\mathscr{F}}_G(f)$ vanishes on a neighbourhood of V(I), then $f$ belongs to I.

Let $\varepsilon >0$. There exists a function $g\in L^1(G)$ such that $\|f-f*g\|_1<$ $\varepsilon$ (prop. 8 of II, p. 211 (iv)). Let $h\in L^1(G)$ be such that the support of $\overline{\mathscr{F}}_G(h)$ is compact and $\|f\|_1\|g-h\|_1< \varepsilon$ (lemma 1). We have

By the assumption on $f$, the function $\overline{\mathscr{F}}_G(f*h) =\overline{\mathscr{F}}_G(f)\overline{\mathscr{F}}_G(h)$ has compact support disjoint from V(I), which implies that $f*h\in I$ (remark 2 above). Since $\varepsilon$ is arbitrarily small, we have $f\in$ I = I.