30.5 $\ell^{1}(\integer)$

Proof. For each $f, g \in \ell^{1}(\integer)$, by Fubini’s Theorem,

$\square$

Proof. Let $f = \one_{\bracs{n = 1}}- \one_{\bracs{2 \le n \le 3}}$, then

so $\normn{f^2}_{\ell^1(\integer)}= 7 < \normn{f}_{\ell^1(\integer)}^{2}$. By Proposition 29.8.3, the Gelfand transform is not isometric.$\square$

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.$\square$

Proof. By Theorem 30.5.3 and (4) of Proposition 29.8.2.$\square$