Is $L^1(G)$ weak $*$ dense in $M(G)$

Question

I am attempting to find justifications for why the Fourier Stieltjes transform is a natural generalization of the Fourier transform. I am trying to prove some `unique extension' property, i.e. that the Stieltjes transform is the unique weak $*$ continuous extension of the Fourier transform to $M(G)$. This would be easy to prove if $L^1(G)$ was weak $*$ dense in $M(G)$, but I am unable to prove this. Is it true? Or if not, is there another way to justify why the Stieltjes transform is a natural extension of the Gelfand representation?

Answer 1

Assume that our group is abelian and locally compact, and the proof follows fairly simple from the inversion theorem of the Fourier transform. Indeed, we now assume that the Fourier-Stieltjes transform is injective: if $\widehat{\mu} = 0$, then $\mu = 0$.

Let $f_U$ be a bounded approximate identity at the origin. A simple calculation shows $\widehat{f_U} \to 1$ pointwise. Then, without loss of generality, since $f_U$ is bounded, we may apply the Banach-Alaoglu theorem to conclude that $f_U$ weak $*$ converges to some measure $\mu$. Then $\mu$ is verified to be the Dirac delta function, because $\widehat{f_U * \nu} = \widehat{f_U} * \widehat{\nu}$ converges pointwise to $\widehat{\nu}$, but also by continuity of the Fourier transform to $\widehat{g} * \widehat{\nu}$. After verifying that the convolution operator is weak $*$ continuous-enough, we can conclude that any measure can be weak $*$ approximated by functions in $L^1(G)$.

Answer 2

The weak * topology on $M(G)$ is also called the vague topology. And $L^1(G)$ is dense in $M(G)$. Quoting Folland: "For a measure $\mu$, consider $\phi_t*\mu$, where $\{\phi_t\}_{t>0}$ is an approximate identity."

There is a nice proposition which says that if $\| \mu_n \| \leq C < \infty$ and $\widehat{\mu}_k \rightarrow \widehat{\mu}$ pointwise, then $\mu_k \rightarrow \mu$ vaguely.