Corollary 1

For every closed subset P of $\widehat{G}$ and every $\chi \in \widehat{G}$ not belonging to P, there exists a function $f\in L^1(G)$ such that $\mathscr{F}(f)$ is zero on P and non-zero at $\chi$.

Since, by (12), one has $\mathscr{F}(\chi f) =\varepsilon_{\chi}*\mathscr{F}(f)$ for every $\chi \in \widehat{G}$, it is enough to consider the case where $\chi$ is the identity element of $\widehat{G}$.

Let U be a compact symmetric neighbourhood of $e\in \widehat{G}$ such that $U^2\cap P =\emptyset$. Let $\varphi$ be a positive continuous function on $\widehat{G}$, zero outside U and such that $\varphi (e) = 1$. The function $\varphi_1=\varphi *\varphi$ is then zero on P and $\varphi_1(e)>0$. It is enough therefore to prove that $\varphi_1$ belongs to the image of the Fourier transform on $L^1(G)$. Now $\varphi$ and $\varphi_1$ belong to $L^1(\widehat{G})\cap L^2(\widehat{G})$. Put $f=\overline{\mathscr{F}}(\varphi )\circ \eta$ and $f_1=\overline{\mathscr{F}}(\varphi_1)\circ \eta$. Lemma 6 implies that $f$ and $f_1$ belong to $L^2(G)$ and satisfy $\varphi =\mathscr{F}(f)$ and $\varphi_1=\mathscr{F}(f_1)$. Moreover

and therefore $f_1\in L^1(G)$. Thus $\varphi_1=\mathscr{F}(f_1)$ does indeed belong to the image of $L^1(G)$ by the Fourier transform.