# The Gelfand transformation on $\ell^1(\mathbb Z)$ is not isometric. Do you have an example?

I am looking for an element $(a_n)_{n\in\mathbb Z}\in\ell^1(\mathbb Z)$ with the property $$\lVert(a_n)\rVert_{\ell^1(\mathbb Z)} > \lVert\sum_{n\in\mathbb Z}a_n z^n\rVert_\infty,$$ where the norm on the right hand side denotes the sup-norm on $\mathcal C(\mathbb T)$ ($\mathbb T$ is the 1-Torus).

Motivation: I want to prove that the Gelfand transformation on $\ell^1(\mathbb Z)$ is not isometric.

-
There must be something I am missing, since otherwise it seems that $(1,1,0,0,\ldots)$ is an example: norm on the left is 1, norm on the right is 2. –  damiano Aug 21 '10 at 10:31
@damiano: The norm on the left is given by $\lVert(a_n)\rVert_{\ell^1(\mathbb Z)}=\sum_{n\in\mathbb Z}\lvert a_n\rvert$. –  Rasmus Aug 21 '10 at 10:51
Ah, of course! But then I think that it is not possible: the norm of the $n$-th term on the right is bounded above by the $n$-th term of the norm on the left, so that in fact the inequality is the other way around. It is now easy to find examples where the other inequality is strict. –  damiano Aug 21 '10 at 11:03
@damiano: You are right. The inequality has to be the other way around. That is, I am searching for an element in $\ell^1(\mathbb Z)$ that is mapped to a function in $\mathcal C(\mathbb T)$ with a smaller norm. –  Rasmus Aug 21 '10 at 12:23

Take for instance $(1, -1, -1)$. The $\ell^1$-norm is three. However, there is no real $x$ with $|1 - e^{ix} - e^{2ix}|$ equalling 3. Indeed, this would imply (by the equality case of the triangle inequality) that $-1 = e^{ix} = e^{2ix}$, which is impossible.

-
Thank you for your answer. I've been playing around with 2 summands which apparently doesn't suffice. If one defines an involution on $\ell^1(\mathbb Z)$ by $a^∗(n)=\overline{a(-n)}$, then for your particular element above one gets $\lVert a^∗ ∗a\rVert = 5$ while $\lVert a\rVert^2=9$. This is a counterexample to the C*-equation in $\ell^1(\mathbb Z)$. –  Rasmus Aug 21 '10 at 13:06

You corrected the misprint good!

Suppose $f=(a_n)\in\ell^1$, the Gelfand transform $\hat{f}$ of $f$ is given by $$\hat{f}(t)=\sum a_ne^{int}$$ hence $$\sup_t |\hat{f}(t)|=\sup_t\left|\sum a_ne^{int}\right|\leq \sum |a_n|=\|f\|_{\ell^1}$$

You can also take a closer look at the Dirichlet kernel or the Féjer kernel in order to find examples where you have strict inequality.

Also, given any commutative Banach algebra we have the spectral radius formula $$\sup|\hat{f}|=\inf_{n\geq1} \|f^n\|^{1/n}$$ and by the Banach algebra inequality we have $$\|f^n\|^{1/n}\leq\|f\|$$

-
Thank you, I have already corrected my misprint. –  Rasmus Aug 21 '10 at 12:54