Can the Fourier transform witness complete normality of the dual group? Let $\Gamma$ be a countable Abelian group and let $G$ be its dual (compact) group. Consider the Fourier transform $F$ as a map from $\ell_1(\Gamma)$ into $C(G)$.
Take a non-empty open subset $U$ of $G$. Is there $f$ in $\ell_1(\Gamma)$ such that $F(f)$ is positive exactly on $U$ and zero otherwise?
 A: Edit: The answer is no in general, since $\{f>0\}$ must be an $F_\sigma$. If $S$ is an $F_\sigma$ the answer is yes.
Note In fact there's no significant use of the fact that $G$  is compact below; if $G$ is any LCA group and $\Gamma=\hat G$ then (of course writing $L^1(\Gamma)$ in place of  $\ell_1$) the answer is yes if and only if $S$ is a countable union of compact sets. 
Hint to maybe get started: In this context the basic  way one gets $f\in\ell_1$ such that $\hat f$ does something is to note that Plancherel shows that $$L^2*L^2=\widehat{L^1}.$$
More explicitly:


Lemma Suppose $V$ is a neighborhood of the origin in $G$. There exists $f\in\ell^1(\Gamma)$ such that $\hat f\ge0$, $\hat f(0)>0$, and $\hat f$ vanishes on $G\setminus V$.


Proof: Let $W$ be an open set with $0\in W$ and $W+W\subset V$. Let $$\phi_V=\chi_W*\chi_W.$$ Then $\phi_V\ge0$, $\phi_V(0)>0$, and $\phi_V$ vanishes on $G\setminus V$.
And $\hat\phi_V=(\hat\chi_W)^2$, so $$||\hat\phi_V||_1\le||\chi_W||_2^2=m(W)<\infty.$$Qed.
It's easy to use the lemma to prove this:


Theorem. If $U\subset G$ is open and $K\subset U$ is compact there exists $f\in\ell_1(\Gamma)$ with $\hat f\ge0$, $\hat f>0$ on $K$, and $\hat f=0$ on $G\setminus U$.


Hint: Say the function obtained in the lemma is $f_V$. Say $S(V)$ is the set where $f_V>0$. You can concoct things so $x_j+V_j\subset U$, $K\subset\bigcup_j(x_j+S(V_j))$.
(And then $f=\sum c_jf_j$ gives $\{f>0\}=U$ if  $U$ is an $F_\sigma$...)
