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 $T_f$?
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2 Answers
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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.
answered May 14, 2018 at 8:14
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$\begingroup$ I do not understand the last line of your answer. Does Fourier transform image $C_c(G)$ on $C_c(\hat G)$? $\endgroup$ Commented May 14, 2018 at 10:40
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$\begingroup$ @MeisamSoleimaniMalekan No, it doesn't, but that is not what Robert is claiming. $\widehat{f}$ is a bounded continuous function on $\widehat{G}$, since $\Vert f \Vert_1 < \infty$. $\endgroup$ Commented May 14, 2018 at 16:14
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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).
answered May 14, 2018 at 17:23
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1$\begingroup$ ... and this is essentially the only example on $\mathbb Z$. That is, if $f$ has finite support in $\mathbb Z$, $\widehat{f}$ extends to an entire function, so if $\{x \in \mathbb R:\; \widehat{f}(x) = t\}$ has positive measure it must be all of $\mathbb R$, and $T_f$ is a constant multiple of the identity. $\endgroup$ Commented May 14, 2018 at 20:14
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$\begingroup$ @RobertIsrael: Good point! I can't resist, though, to point out the following two things: (i) The dual group of $\mathbb{Z}$ is the complex unit circle $\mathbb{T}$, so I guess you'd rather consider the set $\{z \in \mathbb{T}: \hat f(z) = t\}$; (ii) The Fourier transform $\hat f$ is a rational function which might have a pole at $0$. (Or did I missunderstand you and you would like to consider the function $\hat f(x) := \sum_{k=-\infty}^\infty f(k) e^{ikx}$ rather than $\hat f(z) := \sum_{k=-\infty}^\infty f(k)z^k$)? $\endgroup$ Commented May 14, 2018 at 20:49
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$\begingroup$ I was considering $\widehat{f}(x) = \sum_{k=-\infty}^\infty f(k) e^{ikx}$ which extends from $[0,2\pi)$ to an entire function on $\mathbb C$. Of course the two points of view are equivalent. $\endgroup$ Commented May 15, 2018 at 8:28