convergence in $C^*(A)$ implies uniform convergence of the fourier transform

Question

Let $A$ be a locally compact abelian group.

Let $f_n$ be a sequence in $L^1(A)$ conveging to $f\in C^*(A)$ with the operator norm

why does the fourier transform $\hat{f_n}$ converge to $\hat{f}$ uniformly in $\hat{A}$?

that is, why do we have $\lVert \hat{f_n} - \hat{f} \rVert_{\hat{A}} = \lVert f_n - f \rVert_{L^1(A)} \rightarrow 0$ ? For all I know the operator norm is less than the $L^1$ norm so I don't see how convergence in the first would imply convergence in the latter.

(the text below is from lemma 3.4.4 of "Principles of Harmonic Analysis", A. Deitmar, S. Echterhoff)

Answer 1

The Fourier transform $$C^*(A)\buildrel \widehat{\kern 5pt }\over \rightarrow C_0(\hat A) $$ coincides with the Gelfand transform and it is an isometric $^*$-isomorphism relative to the $C^*$-norm on $C^*(A)$ and the uniform norm on $C_0(\hat A)$.