Let $G$ be a group generated by $a,b$ (for the sake of simplicity). Consider the element 
$$S=a+b+a^{-1}+b^{-1}\in{\mathbb C}[G],$$
which may also be interpreted as an operator in $l^2(G)$ (by left regular representation). Obviously $\|S\|\le 4$, and by the old result of Kesten
$\|S\|=4$ iff $G$ is amenable.


Now, suppose that $G$ is not amenable, that is, $\|S\|<4$. Is there an eigenvector in $l^2(G)$ corresponding to the eigenvalue $\|S\|$? (I do not know the answer even for a free group.)



A question about spectral properties of a non-amenable group
Linnell proved in his paper "Zero divisors and $\ell^2(G)$ C. R. Acad. Sci. Paris Sér. I Math. 315 (1992), no. 1, 49-53." that all elements of the group ring of a right orderable groups, containing Abelian and free groups, are nonzero divisors, i.e. if $0\neq\alpha\in\mathbb C[G]$, then for all $0\neq\beta\in\ell^2(G)$ we have $\alpha\beta\neq0\neq\beta\alpha$. So, in particular, the answer to your question in the case of right orderable groups is negative.


