Theorem 30.5.3.label Let $\ell^{1}(\integer)$ be the convolution algebra on $\integer$, then the mapping
is a homeomorphism. Under the identification $\partial B_{\complex}(0, 1) = \Omega(\ell^{1}(\integer))$, the Gelfand transform has the explicit form
Proof, [Theorem 6.3, Zhu93]. Let $z \in \mathbf{S}^{1}$ and $f, g \in \ell^{1}(\integer)$, then by Fubini’s Theorem,
so $F(z)$ is a multiplicative functional for all $z \in \mathbf{S}^{1}$.
For each $n \in \integer$, let $\delta_{n} = \one_{\bracs{k = n}}$, then for any $m, n \in \integer$, $\delta_{m} * \delta_{n} = \delta_{m + n}$. Let $\phi \in \Omega(\ell^{1}(\integer))$ and $z = \phi(\delta_{1})$, then $\phi(\delta_{n}) = z^{n}$ for all $n \in \integer$. Since the span of $\bracsn{\delta_n|n \in \integer}$ is dense in $\ell^{1}(\integer)$, $\phi = F(z)$ by the Dominated Convergence Theorem.
Finally, by the Dominated Convergence Theorem, $F: \textbf{S}^{1} \to \Omega(\ell^{1}(\integer))$ is continuous. Since $\textbf{S}^{1}$ is compact, $\Omega(\ell^{1}(\integer))$ is Hausdorff, and $F$ is a bijection, $F$ is a homeomorphism by Proposition 5.16.4.$\square$
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