Non-density of invertible elements in $\ell_1(\mathbb{N}_0)$ Consider the Banach algebra $\ell_1(\mathbb{N}_0)$ (with convolution / Cauchy product of series). I am looking for an elementary proof of the fact that the group of invertible elements in this algebra is not dense.
If one wishes to use some hammers, here is a way to do so. Call this algebra $A$; it is not too hard to show that $A$ is semi-simple. Thus the Gelfand transform $g\colon A\to C(\Phi_A)$ is injective and has dense range. However $\Phi_A = \overline{\mathbb D}$ and it is well known that invertible elements in $C(\overline{\mathbb D})$ are not dense.

Q1. Is there a more elementary way to see this (preferably that would work for the real scalars too)?
Q2. Is there a characterisation of semigroups $S$ for which invertible elements in $\ell_1(S)$ are (not) dense?

 A: Q1 can be answered with the following result.

An element in the boundary of the invertible group of a unital Banach algebra is a topological divisor of zero.

I must confess I remembered the statement but not the proof, but it can be shown using elementary arguments related to the fact that the invertible group is open. It should be in e.g. Allan's book on Banach algebras (since it is in the lecture notes which evolved into that book) and probably also in Bonsall and Duncan. If I have time in the next day or two I can include the argument here.
Regarding Q2: if $S$ is a (discrete) group with polynomial growth, then I think the invertible elements should be dense in $\ell^1(S)$, using the above criterion and Corollary 1.9 in this paper of Tessera the results of https://doi.org/10.1016/j.jfa.2010.07.014 (sorry for being sketchy right now, please remind me if I don't update this answer within the next couple of days)
I have a feeling that Q2 is open even when S is a discrete solvable group of exponential growth, but I have not kept up with progress in this area for a while. The case of commutative monoids might be tractable
A: For Q1:
Every $f=\sum a_nz^n\in A$ induces a continuous function $\hat{f}$ on the closed disk of radius $1$ and $f\mapsto \hat{f}$ is a continuous operator (for $\ell^1$-norm $\|f\|=\sum|a_n|$ on the left and sup norm $g=\sum_{|z|\le 1}|g(z)|$ on the right), which in addition is multiplicative.
The for $u(z)=z$, it is standard ($*$) from differential/algebraic topology that every $g$ close enough to $\hat{u}$ (the identity of the closed disk) in the supremum norm has a zero. It follows that $u$ (which is indeed $\delta_1=(0,1,0,\dots)$ is in the interior of the set of non-invertible elements of $A$. The same argument works for $z^n$, $n\ge 1$.
($*$) indeed if $g$ is close enough to the identity then it is homotopic to the identity, and if $g$ does not vanish, then we can deduce a homotopy from the identity of the circle to a constant map.
