Invertible elements in $l^1(\mathbb Z)$

Question

The vector space $l^1(\mathbb Z)$ with $\|x\| = \sum_{n \in \mathbb Z} |x_n|$ and $x * y(t) = \sum_{k \in \mathbb Z} x(k)y(t-k)$ forms a unital complex Banach algebra, with the unit being $\mathbf 1(0) = 1$ and $\mathbf 1(z) = 0$ for all $z \neq 1$.

I need to find the invertible elements of this Banach algebra.

The first thing we do is note that if $\|h\| < 1$ then $1-h$ is invertible. Furthermore, $f$ is invertible if and only if $\alpha f$ is invertible for some scalar $\alpha$ non-zero. Therefore, combining these, if there is some non-zero $\alpha \in \mathbb C$ and $h \in l^1(\mathbb Z)$ with $\|h\| < 1$ such that $f = \frac{1}{\alpha} (\mathbf 1 - h)$ then $f$ is invertible.

This simplifies to $\mathbf 1 - \alpha f$ being of norm $<1$ for some $\alpha$ non-zero. By definition of the norm , $$\sum_{k \in \mathbb Z} |(\mathbf 1(k) - \alpha f(k))| < 1 \iff |1-\alpha f(0)| + \sum_{k \neq 0\in \mathbb Z} |\alpha||f(k)| < 1 $$

It is not clear to me how I should proceed further on from this point : this gives some condition on $f$ in terms of $\alpha,h$ and I want to claim that this is sufficient, but no progress has been possible in the other direction because $f * g = \mathbf 1$, from the assumption of $f$ being invertible is not workable because of too many equations in the unknowns $f(k)$.

I believe that this is down to which elements in the Banach algebra don't have zero in their spectrum, so if there is any result in that direction (i.e. results about the spectrum) I would like to know about that as well.

Answer 1

Banach algebras have a generalized Fourier transform, called the Gelfand transform, which maps elements $x$ to continuous functions $\hat{x}$ (acting on the "character space"). A standard result is that, for commutative Banach algebras such as $\ell^1(\mathbb{Z})$, the image of this function is equal to the spectrum of the element.

For $\ell^1(\mathbb{Z})$, the continuous function associated with the sequence $\boldsymbol{x}=(a_n)$ is $\sum_{n\in\mathbb{Z}}a_ne^{in\theta}$. Now $\boldsymbol{x}$ is invertible iff its spectrum does not contain $0$, so by the theorem, this is the same as the function $\hat{x}$ not having $0$ in its image, that is, $\sum_{n\in\mathbb{Z}}a_ne^{in\theta}\ne0$ for all $\theta$.