Let $G$ be a locally compact Hausdorff abelian group and $E$ be a non-empty open subset of $\hat{G}$. Show that there exists $f \in L^1(G)$ such that $\hat{f} \ne 0$ and $\hat{f}=0$ on $\hat{G} \setminus E$
First of all, since $\hat{G}$ is locally compact Hausdorff, one has open subset $U\subset E $ such that $U \subset \bar{U} \subset E$ and $\bar{U}$ is Compact. Thus can choose $g \in C_c(\hat{G})$ with $Supp(g) \subset E$ and then look at it's Fourier transform and do an identification via Pontryagin duality, but the problem here is that Fourier transform of $C_c$ functions on general locally compact groups can be very far from being $L^1$ functions!
By Stone-Weierstrass, I showed that $\mathscr A(\hat{G})=\{\hat{f}: f \in L^1(G)\}$ is dense in $C_0(\hat{G})$. But how to proceed from here?
Inverse Fourier transform, Plancherel identity and convolution theorem ?
With $h\in C^0(G)$ the inverse Fourier transform of $\hat{h}\in C^0_c(\hat{G})$ then $h^2= F^{-1}(\hat{h}\ast \hat{h})$ and $$\|h^2\|_{L^1(G)}=\|h\|_{L^2(G)}^2=\|\hat{h}\|_{L^2(\hat{G})}^2$$ So $h^2\in L^1(G)$.
Its Fourier transform $\hat{h}\ast \hat{h}$ can be chosen to be supported on an arbitrary small neighborhood of $0$.
Then we conclude with the shift theorem.