How to prove that the convolution operator associated to a discrete measure on a LCA group has natural spectrum?

Question

Let $\mu$ be a Borel measure with finite variation on a locally compact abelian group $G$, let $\Gamma$ denote the dual group of $G$, and let $\hat \mu: \Gamma \to \mathbb{C}$ be the Fourier-Stieltjes transform of $\mu$. The measure $\mu$ induces a convolution operator $T_\mu$ on the space $L_1(G)$ defined by $T_\mu f = \mu \star f$, where $G$ is endowed with its Haar measure $m$.

The set $\hat\mu(\Gamma)$ is contained in the spectrum of $T$, and $T_\mu$ is said to have \emph{natural spectrum} when $\sigma(T_\mu)$ coincides with the closure of $\hat\mu(\Gamma)$. It is well-known that $T_\mu$ has natural spectrum when $\mu$ is discrete (its support is a countable union of atoms for $m$). See the Comment in page 619 of [M. Zafran. On the spectra of multipliers. Pacific J. Math. 47 (1973), 609-626].

I would like to have a direct proof of the last result, or a reference where I can find this proof.

Answer 1

Discrete measures form a unital subalgebra of the algebra of measures on $G$, isomorphic to $L^{1}(G_d)$, where $G_d$ is the group $G$ endowed with the discrete topology. The spectrum in a subalgebra can only be bigger, so we just need to show that the spectrum of $\mu$ considered as an element of $L^{1}(G_d)$ is contained in $\overline{\hat{\mu}(\Gamma)}$. But the spectrum of the algebra $L^{1}(G_d)$ is precisely the Bohr compactification $B\Gamma$ of $\Gamma$ in which $\Gamma$ is dense. So, if $\varphi \in B\Gamma$ then $\varphi(\mu)$ is contained in the closure of $\hat{\mu}(\Gamma)$. Since the spectrum in $L^{1}(G_d)$ is precisely the set of values $\varphi(\mu)$ for $\mu \in B\Gamma$, we get the result.