Let $G$ be a compact metric abelian group. $T$ be the circle group. Let $\mathcal{A}$ be the set of all finite linear combinations of continuous homomorphisms from $G \to T$. I want to show that the its uniform closure contains all complex continuous functions.
It is easy to see that $\mathcal{A}$ is a self adjoint algebra. Also the identity homomorphism is non zero for all elements of $G$. The only thing that remains to show is that $\mathcal{A}$ separates points. Let $a,b \in G$, $a\neq b$. I want to find an element $\phi$ in $\mathcal{A}$ for which $\phi(a)\neq \phi(b)$. I firmly believe that we can find a continuous homomorphism $\phi$. Then it is enough to show $\phi(ab^{-1}) \neq 1$. So it boils down to show that
$x\neq e$ implies there exist a continuous homomorphism $\phi$ such that $\phi(x)\neq 1$
I am stuck here. I dont know if it is true or not. Counterexamples are welcomed.