Let $G$ be a locally compact, metrizable and separable group, $X=L^\infty(G)$ and $A\subseteq X$ and $1\in X$. I know that forall $\varepsilon>0$ and all compact sets $C\subseteq G$ there exists an element $a\in A$ such that $\mid 1-a(g)\mid \leq \varepsilon$ for all $g\in C$.
How can I deduce that $1$ lies in the weak closure of $A$? In which cases does $L^1(G)\equiv L^\infty (G)'$ hold?
It isn't true. Take $G = \mathbb{Z}$, so that $X = \ell^\infty(\mathbb{Z})$, and $A = c_0(\mathbb{Z})$, the set of all $a : G \to \mathbb{R}$ satisfying $\lim_{g \to \pm \infty} a(g) = 0$. Your hypothesis is clearly satisfied, yet $A$ is weakly closed and doesn't contain $1$. (One way to see this is to note that $A$ is norm-closed and convex; see A convex subset of a Banach space is closed if and only if it is weakly closed.)
Generally, $L^1$ is practically never the dual of $L^\infty$, except in trivial cases, since the dual of $L^\infty$ tends to contain Banach limits and such. In the case of metrizable groups, I think you could prove that $L^1(G)\equiv L^\infty (G)'$ iff $G$ is finite.
