The same topologies

Question

Let $L^1 (\mathbb{Z})$ be the space of all functions $f:\mathbb{Z}\rightarrow \mathbb{C}$ such that $\left\{\|f\|=\sum_{k\in \mathbb{Z}}|f(k)|<\infty\right\}$. Clearly, $L^1 (\mathbb{Z})$ is a separable, commutative, unital Banach algebra with usual convolution. It is easy to see that the character space of $L^1 (\mathbb{Z})$ is homeomorphic to ${T}=\{z\in \mathbb{C}:\ |z|=1\}$. Are the Gelfand and norm topologies equal, on the character space of $L^1 (\mathbb{Z})$?

Answer 1

No. The Gelfand topology is the same as the usual topology on $\Bbb T$. The norm topology is discrete:

Say $z,w\in\Bbb T$ and $z\ne w$. The dual of $L^1$ is $L^\infty$, so the norm of $z-w$, as an element of $L^1(\Bbb Z)^*$, is $$||z-w||=\sup_{n\in\Bbb Z}|z^n-w^n|=\sup_{n\in\Bbb Z}|1-(z/w)^n|.$$

It's easy to see that there exists $c>0$ such that if $|\alpha| =1$ and $\alpha\ne0$ then $\sup_n|1-\alpha^n|\ge c$ (I think the worst case is when $\alpha$ is a cube root of $1$). Hence $||z-w||\ge c$.