TITLE: Invertibility of group Laplacian in $\ell^1$
QUESTION [7 upvotes]: Let $G$ be a discrete group and let $S$ be a generating set for $G$; assume that $S$ is symmetric (i.e., $g\in S$ iff $g^{-1}\in S$).  Let $L=L_S=\frac{1}{|S|}(\sum_{g\in S} g-1)$ be an element of the group algebra.  For $F\in \ell^p(G)$ and $g,h\in G$ let $\lambda_p(g) F$ be defined by $\lambda_p(g) F (h) = F(hg^{-1})$; this is the regular representation of $G$ on $\ell^p(G)$.  Finally let $L_p = \pi_p(L)$.
It is a classical result of Kesten that $G$ is non-amenable iff $L_2$ has a bounded inverse (i.e., $0$ is not in the spectrum of $L_2$).
What is known in the case of $p=1$? Clearly if $G$ is amenable, then there are functions $f_n\in \ell^1(G)$ with the property that $\Vert L_1 f_n \Vert_1 / \Vert f_n \Vert_1 \to 0$ showing that $0$ is in the spectrum of $L_1$ and so $L_1$ is not invertible.  Also, if $G$ is not amenable, then such $f_n$ cannot exist under the additional assumption that $f_n\geq 0$. 
Given $G$ non-amenable, can one find a generating set $S$ with the property that $L_1$ is invertible?

REPLY [2 votes]: The Laplacian is never invertible on $\ell^1(G)$.
There is already a correct answer by ARG. 
I am writing this answer as there is a demand for more information. 
I will thus try to provide a more conceptual point of view. But I should remark that my answer is essentially the same as ARG's answer.

Let $\mu$ be a probability measure on $G$ and assume the Laplacian $\Delta=1-\mu$ is an automorphism of $\ell^1(G)$.
By taking the double dual, $\Delta$ extends to an automorphism of $\ell^\infty(G)^*$.
Recall that a mean on $G$ is a norm 1 positive element of $\ell^\infty(G)^*$ and let $M$ be the space of all means on $G$. It is easy to see that $M$ is convex and compact for the w*-topology. Note that $G$ acts on $M$ preserving its structure. In particular, $M$ is $\mu$-invariant. By Markov–Kakutani fixed-point theorem there is a $\mu$-invariant mean $m\in M$. Thus $\ker \Delta\neq 0$ as 
$$ \Delta m=m-\mu m =0. $$
This is a contradiction.

Note that, by the same argument, $\Delta$ is never an automorphism of $\ell^\infty(G)$.
For completeness let me add that for $1<p<\infty$ and $\mu$ with a generating support, $\Delta$ is an automorphism of $\ell^p(G)$ iff $G$ is not amenable.
For this recall that the Mazur map $f \mapsto \text{sgn}(f)|f|^{2/p}$ is a $G$-equivariant uniformly continuous homeomorphism from the unit sphere of $\ell^2(G)$ to the unit sphere of $\ell^p(G)$. It follows that there exist almost invariant vectors in $\ell^p(G)$ iff there exist almost invariant vectors in $\ell^2(G)$, and this happens iff $G$ is amenable (by Kesten). Thus, if $G$ is amenable $\Delta$ is not an automorphism (and this works also for $p=1$). If $G$ is not amenable, by the uniform convexity of the unit ball of $\ell^p(G)$ (which fails for $p=1$), we get that $\|\mu\|<1$, thus the sum $\sum_{n=0}^\infty \mu^n$ exists in the space of bounded operators, and since $(1-\mu)\sum_{n=0}^\infty \mu^n=1$, $\Delta$ is indeed an automorphism.