Let $\chi$ be a character of $\mathbb Z$ i.e. $\chi : \mathbb Z \longrightarrow S^1$ is a group homomorphism. Define $\psi_{\chi} : \ell^1 (\mathbb Z) \longrightarrow \mathbb C$ by $\psi_{\chi} ((a_n)_{n \geq 1}) = \sum\limits_{n \in \mathbb Z} a_n \chi (n).$ Then it is easy to show that $\psi_{\chi}$ is a multiplicative linear functional on $\ell^1(\mathbb Z),$ when considering $\ell^1(\mathbb Z)$ as a Banach algebra with respect to the convolution product. Let $\widehat {\mathbb Z}$ be the set of all characters of $\mathbb Z.$ Endow $\widehat {\mathbb Z}$ with the compact open topology. Now define a map $\varphi : \widehat {\mathbb Z} \longrightarrow (\ell^1(\mathbb Z))^*$ by $\chi \mapsto \psi_{\chi}.$ Show that $\varphi$ is continuous.
This problem has been left as an exercise in the lecture note I am following. I have tried to show that but I couldn't quite able to do it. Could anyone please help me in this regard?
Thanks a bunch.
Since $\mathbb Z$ is cyclic, group homomorphisms from $\mathbb Z$ to $S^1$ are in bijective correspondence with elements of $S^1$ via $\chi\mapsto \chi(1)$. The inverse map is given by $z\mapsto (n\mapsto z^n)$, and it is not hard to see that it is continuous. Since $S^1$ is compact, this means that it is a homeomorphism. So we may just as well consider $\phi\colon S^1\to \ell^1(\mathbb Z)^\ast,\,z\mapsto (a\mapsto\sum_n a_n z^n)$ instead of the map from the question.
This map is not continuous if we endow $\ell^1(\mathbb Z)^\ast$ with the norm topology: If $z_m=e^{\pi i/m}$ with $m\in\mathbb Z$, then $z_m^m=-1$ and therefore $$ \|\phi(z_m)-\phi(1)\|=\sup_{n\in\mathbb Z}|z_m^n-1|=2. $$ But of course $z_m\to 1$.
The map $\phi$ is continuous if we endow $\ell^1(\mathbb Z)^\ast$ with the weak$^\ast$ topology: If $z_k\to z$ in $S^1$ and $(a_n)\in \ell^1(\mathbb Z)$, then $|a_n z_k^n|\leq |a_n|$ and thus $$ \sum_{n\in\mathbb Z}a_nz_k^n\to\sum_{n\in\mathbb Z}a_n z^n $$ by the dominated convergence theorem.