Continuity of evaluation map for characters of a LCA group First I will remind this definition. If $G$ is a LCA (i.e. locally compact, abelian and Hausdorff) topological group, then its dual group $\Gamma$ is the set of all continuous group homomorphisms (characters) $\gamma:\Gamma\to S$ (where $S$ is the unit sphere, with its operation given by complex multiplication), with its own operation given by $(\gamma_1+\gamma_2)(s)=\gamma_1(s)\gamma_2(s)$ and its topology is the induced topology from the weak-star topology of $L^{\infty}(G)$. In other words, if $(\gamma_i)$ is a net in $\Gamma$, we have $\gamma_i\to \gamma$ iff $\hat{f}(\gamma_i)=\int_G f(t)\gamma_i(-t)dt\to \int_G f(t)\gamma(-t)dt=\hat{f}(\gamma)$ for all $f\in L^1(G)$.
I am having trouble with the following result:

Let $G$ be a LCA group and let $\Gamma$ be its dual group. Then the map
  \begin{align*}G\times \Gamma &\to S \\ 
(t,\gamma)&\mapsto \gamma(t)
\end{align*}
  is continuous.

In Rudin's Fourier analysis on groups, the author simply says that since the map  $(x,\gamma)\mapsto\hat{f}_x(\gamma)=\int_{G}f(t-x)\gamma(-t)dt$ is continuous (so in particular, $\gamma \mapsto \hat{f}(\gamma)$ is continuous), and the relation 
$$\hat{f}(\gamma)\gamma(x) =\hat{f}_x(\gamma)$$
holds, then($*$) $(x,\gamma)\mapsto \gamma(x)$ is continuous. But I fail to understand the implication ($*$). In general, if $(a_n)$ and $(b_n)$ are two sequences of complex numbers such that $a_nb_n\to ab$ and $a_n\to a$, then I can deduce that either $b_n\to b$ or $a=0$. However I cannot rule out the possibility that $a=\hat{f}(\gamma)=0$, so how do I proceed? Perhaps it depends on the fact that $\hat{f}(\gamma)$ depends only on $\gamma$ and not on $x$?
I would be satisfied if I could only get an intuitive idea of why this is true.
 A: The solution lies on the fact that the relation 
$$\hat{f}(\gamma)\gamma(x)=\hat{f}_x(\gamma) $$
holds for all $f\in L^1(G)$. Now, suppose $(x_i)$, $(\gamma_i)$ are nets such that $x_i\to x$, $\gamma_i\to \gamma$. We know that $\hat{f}(\gamma_i)\gamma_i(x_i)\to \hat{f}(\gamma)\gamma(x)$ for all $f$. But $\hat{f_0}(\gamma)\neq 0$ for some $f_0\in L^1(G)$ - for instance, let $f_0(x)=\gamma(x)\chi_{K}(x)$ where $K$ is a compact subset of $G$. (On a more theoretical side, if $\hat{f}(\gamma)=0$ for all $f$, then the character $\gamma$ could only be identified with a multiplicative functional that is identically $0$ - but $0$ is excluded from the spectrum of $L^1(G)$ - in other words, its kernel is not a maximal ideal of $L^1(G)$).
We now have:
$$|\hat{f}_0(\gamma_i)\gamma_i(x_i)-\hat{f}_0(\gamma)\gamma(x)|\geq |\hat{f}_0(\gamma)\gamma_i(x_i)-\hat{f}_0(\gamma)\gamma(x)|-|\hat{f}_0(\gamma_i)\gamma_i(x_i)-\hat{f}_0(\gamma)\gamma_i(x_i)|= \\ 
=|\hat{f}_0(\gamma)||\gamma_i(x_i)-\gamma(x)|-|\gamma_i(x_i)||\hat{f}_0(\gamma_i)-\hat{f}_0(\gamma)|
 $$
as $i\to \infty$,  (notice that $|\gamma_i(x_i)|=1$ for all $i$)
$$0\geq |\hat{f}_0(\gamma)|\limsup_{i\to \infty}|\gamma_i(x_i)-\gamma(x)|-0  $$
and since $\hat{f}_0(\gamma)\neq 0$, we are done.
