Let G be a locally compact group and $\mathcal{H}$ a Hilbert space. Then there is a bijection between strongly continuous unitary representations of G and non-degenerate *-representations of $L^1(G)$ on $\mathcal{H}$ (see Group algebra of locally compact group).
Now if we use apply this bijection to the left regular representation L of G, we get $\pi_L : L^1(G) \to B(L^2(G))$, where $\pi_L (f) = f \ast $, which sends $h \in L^1(G) \cap L^2(G)$ to $f\ast h$ and is then extended by continuity to the whole of $L^2(G)$.
Now suppose G is abelian. In Principles of Harmonic Analysis chapter 3.3, they define the $C^{*}$-algebra of G to be the norm closure of the image of $\pi_L$, which is a $C^{*}$-subalgebra of $B(L^2(G))$. Let's denote it by $\mathcal{C}$ and let $C_0(\hat{G})$ be the $C^{*}$-algebra of continuous functions on $\hat{G}$ that vanish at infinity. They later show that $$\mathcal{C} \cong C_0(\hat{G})$$
Because of this, $C_0(\hat{G})$ can be put inside of $B(L^2(G))$. The fourier transform $\mathcal{F}$ is a Banach *- algebra homomorphism from $L^1(G)$ to $C_0(\hat{G})$, and by the Stone Weirstrass theorem, its image is dense in $C_0(\hat{G})$. At the same time, by the isomorphism mentioned above, the Fourier transform can be seen as a *-representation of $L^1(G)$ on $L^2(G)$.
Under this perspective, I have two questions:
- Is the Fourier transform going to be non-degenerate?
- If it is non-degenerate, what is the unique unitary representation of G associated to it by the bijection mentioned in the first paragraph?