Point spectrum of a positive invertible operator Let $G$ be a l.c. group and $f$ belong to $C_c(G)$, the space of continuous functions with compact support. Define an operator$T_f$ on $L^2(G)$ by $T_f(g)=f*g$ (the convolution product). If $T_f$ is positive and invertible, could $\|T_f\|$ belong to the point spectrum of $T_f$?
 A: Suppose $G$ is abelian.  Presumably the measure being used is Haar measure.  $T_f$ is unitarily equivalent (via the Fourier transform) to multiplication by $\widehat{f}$ on $L^2(\widehat{G})$.  This has norm $\|\widehat{f}\|_\infty$, and that is in the point spectrum if $\{x: \widehat{f}(x) = \|\widehat{f}\|_\infty\}$ has positive measure.  That can happen if $G$ is compact.
EDIT: On the other hand, for $T_f$ to be invertible, you want $\epsilon > 0$ such that $|\widehat{f}(x)| > \epsilon$ almost everywhere.  Since $\widehat{f} \in L^2$, that will require $\widehat{G}$ to be compact.  So we're reduced to the case where $G$ is finite.
A: Although the question was already answered by Robert Israel's post, here is a slightly more concrete counterexample:
Let $G$ be the additive group $\mathbb{Z}$ (endowed with the discrete topology). Let $f$ be the function which is $1$ at the point $0$ and $0$ elsewhere. Then $T_f$ is simply the identity operator on $\ell^2(\mathbb{Z})$. In particular, $1$ is an eigenvalue of $T_f$.
You can easily generalise this example to every discrete group $G$ (be it finite or infinite).
