I've never really studied Pontryagin duals and Haar measures but they just came up in a problem for von Neumann algebras and I'm wondering anyone can help.
The problem: Let $G$ be a discrete abelian group and let $\hat G$ be its Pontryagin dual group, which is a compact abelian group and hence has a finite Haar measure $\mu$. Show that $L(G) \cong L^\infty(\hat G, \mu)$.
My attempt:
I looked up Pontryagin and found that $\hat G = Hom(G, \mathbb{T})$ the group of homomorphisms $\phi: G \rightarrow \mathbb{T}$, where $ \mathbb{T} = \{ \zeta \in \mathbb{C}: |\zeta| = 1 \}$.
I looked up Haar measures and found that these are measures that let us think about integration with groups, but since we're talking about $L^\infty$, the measure $\mu$ will only probably come up relating to the essential supremum $||f||_\infty$ for $f \in L^\infty(\hat G, \mu)$. But I did find out that (left) Haar measures are left-translation invariant, finite on compact subsets, outer and inner regular, etc. I'm confused by the fact that my (Pontryagin dual) group is a group of homomorphisms and I'm not sure how the measure will play out because of that.
I already know that $L(G) = \overline{span(\lambda(G))}^{WOT}$ is generated by $\lambda(g)$ for $g \in G$ where this is the left regular representation and acts like $\lambda(g) \delta_h = \delta_{gh}$. I know some other facts about this guy but I don't know if they're relevant.
My idea was to notice that evaluation maps are nice functions that could act on $\phi$ from $\hat G$ so what if I make a correspondence $$ L(G) \ni \lambda(g) \quad \leftrightarrow \quad ev_g \in L^\infty(\hat G, \mu)$$ where $ev_g(\phi) = \phi(g)$ is evaluation of a group homomorphism acting on $G$ at a particular $g \in G$. Now I'm stuck on showing that maybe the weak closure of the span of such $ev_g$'s would indeed give me all of $L^\infty(\hat G, \mu)$. I have shown previously that if $X$ is a compact Hausdorff space and $\mu \in M(X)$ then the set of continuous functions on $X$ is weak* dense in $L^\infty(X,\mu)$, thought I'm not sure if that helps me here, but WOT closure and weak* closure can be related I believe by thinking of $L^\infty$ in terms of bounded operators on $L^2$ functions instead. Any help would be much appreciated as my advisor is very busy this weekend and we didn't get through this problem in my meeting today.