TITLE: Does the Gelfand transformation on $\ell^1(\mathbb Z)$ possess a continuous inverse on its image?
QUESTION [9 upvotes]: I am interested in the Gelfand transformation
$$
\Phi\colon\ell^1(\mathbb Z)\to\mathcal C(\mathbb T),\quad a\mapsto\sum_{n\in\mathbb Z}a_n z^n.
$$
This is an injective homomorphism of Banach algebras. It is neither isometric nor surjective. However, its image---the Wiener algebra $W$ consisting of all continuous functions on $\mathbb T$ whose Fourier series is absolutely convergent---is a subalgebra of $\mathcal C(\mathbb T)$ which is dense in the subspace topology.
Question: Can we prove of disprove that $\Phi$ has a continuous inverse on its image $W$?
In other words: Is $\Phi\colon\ell^1(\mathbb Z)\to W$ an isomorphism of topological algebras?
(Here $W$ carries the topology induced by the sup-norm from $\mathcal C(\mathbb T)$.

REPLY [7 votes]: No, because if there were a continuous inverse, then $\Phi$ would be bounded below, and from this it would follow that $W$ is complete.  But it isn't, because there are continuous functions that have Fourier series that are uniformly but not absolutely convergent, and hence Cauchy sequences (the partial sums of the Fourier series) in $W$ with no limit in $W$.  There are even examples of my claim from the last sentence in the disk algebra, i.e. functions in $C(\mathbb{T})$ whose negatively indexed Fourier coefficients vanish, as was first shown by Hardy and as I discovered when trying to answer a MathOverflow question here.