Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathcal C(\mathbb T)$.

share|improve this question
What's the topology on $W$? If the sum of the absolute values of Fourier coefficients is used, then yes---tautologically. If the sup-norm is used, then no, because the sup-norm is strictly weaker than the sum of the Fourier-coefficient norm; otherwise, it would follow that every continuous $2\pi$-periodic function had an absolutely convergent Fourier series, which is false. –  Akhil Mathew Aug 22 '10 at 16:01
As Akhil say. $W$ is often identified with $\ell^1(\mathbb{Z})$ - it is just different words for the same objects. –  AD. Aug 22 '10 at 19:40
@Akhil Mathew: I would like to consider the topology on $W$ considered as a subalgebra of $\mathcal{C}(\mathbb T)$, that is, the topology given by the supremum-norm. –  Rasmus Aug 22 '10 at 23:12
@Akhil Mathew: In response to the last sentence of your comment: How to you conclude that $W=\mathcal C(\mathbb T)$ from $\Phi\colon\ell^1(\mathbb Z)\to W$ being a homeomorphism? –  Rasmus Aug 22 '10 at 23:16
Well, if $W \to \ell^1(\mathbb{Z})$ were continuous (so bounded), there would be a constant $C$ such that $\sum |\hat{f}(n)| \leq C ||f||_\infty$ (sum of Fourier coefficients) for all sufficiently smooth functions $f$ (which belong to $W$). Hence this boundedness statement is true for all functions, by density of $W$ in $C(\mathbb{T})$. (This implies what I said about every continuous function having an absolutely convergent Fourier series, which is false.) –  Akhil Mathew Aug 22 '10 at 23:59

1 Answer 1

up vote 7 down vote accepted

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.

share|improve this answer
Actually $W$ is very much complete. The topology on $W$ is not given by the sup-norm as is said in a previous comment - see my answer. –  AD. Aug 23 '10 at 5:08
Thank you for your answer and, in particular, for the keyword "bounded below" that I should have recalled an applied myself. –  Rasmus Aug 23 '10 at 9:04
@Rasmus, You're welcome. I could have made the rest simpler along the lines Akhil mentioned. As you said, $W$ is dense in $C(\mathbb{T})$ (as follows e.g. from Stone-Weierstrass or from Cesàro summation of Fourier series) so the fact that it is not all of $C(\mathbb{T})$ shows that it is not complete. However, it was too tempting to mention the examples of Hardy et al. @AD: Rasmus had made clear before I answered that he is considering the sup norm on $W$. This is standard when considering the Gelfand tranform, and regardless was cleared up after Akhil's question. –  Jonas Meyer Aug 25 '10 at 2:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.