3
$\begingroup$

Let $T:[0,1]\to[0,1]$ be a continuous map, which is neither surjective nor injective. Put $$ C([0,1])\ni f\mapsto \Phi(f):=f\circ T\in C([0,1]). $$ Notice that, under the above conditions, $0\in\sigma(\Phi)$.

I'm searching for an example of $\Phi$ as above such that $$ 0<\operatorname{dist}\left(\sigma_{ph}(\Phi),(\sigma(\Phi)\setminus\sigma_{ph}(\Phi))\right)<1, $$ where $\sigma_{ph}$ stands for "peripheral spectrum".

I'm also searching for examples $\Phi$ satisfying, additionally, $$ \sigma_{ph}(\Phi)\supsetneq\{1\}. $$

$\endgroup$
5
  • $\begingroup$ Welcome on MathOverflow! Before I give an answer, may I ask where the questions stems from? $\endgroup$ Dec 20, 2019 at 23:40
  • $\begingroup$ just to find an example of a multiplicative *-map such that the spectrum is neither included in the unit circle, nor consisting of the whole unit disc in the complex plane $\endgroup$
    – fidaleo
    Dec 20, 2019 at 23:43
  • $\begingroup$ Thank you for your response! Alright, but how/in which context did this question occur to you? (I'm asking because it seems a bit like a homework problem in an operator theory course to me.) $\endgroup$ Dec 20, 2019 at 23:49
  • $\begingroup$ If it is an homework, please provide me an answer/reference $\endgroup$
    – fidaleo
    Dec 20, 2019 at 23:53
  • 1
    $\begingroup$ Well, my point was actually whether you were assigned this as a homework problem (which would make the question inappropriate for this site) - but I take your last comment as a "no" :-). My apologies for being maybe a bit over-skeptical. Anyway, see my answer below. $\endgroup$ Dec 20, 2019 at 23:59

1 Answer 1

1
$\begingroup$

EDIT: I adjusted the answer to the new version of the question.

Such an example does not exist. More precisely, for every compact Hausdorff space $K$ and every continuous mapping $T: K \to K$ the associated Koopman operator $\Phi_T: C(K) \to C(K)$ (given by $\Phi_Tf = f \circ T$ for each $f \in C(K)$) has either the closed unit disk $\overline{D}$ as its spectrum, or the spectrum is contained in the union of the unit circle $\mathbb{T}$ and $\{0\}$.

In fact, the following holds:

(i) If $T^{n+1}(K) \not= T^n(K)$ for all $n \in \mathbb{N}_0$, then every complex number $\lambda$ of modulus $|\lambda| < 1$ is an approximate eigenvalue of $\Phi_T$; in particlar, $\sigma(\Phi_T) = \overline{D}$.

(ii) If there exists a number $n \in \mathbb{N}_0$ such that $T^{n+1}(K) = T^n(K)$, then precisely one of the following two assertions holds:

  • The open unit disk is contained in the point spectrum of the dual operator $(\Phi_T)'$ on $C(K)'$; in particular, $\sigma(\Phi_T) = \overline{D}$.

  • We have $\sigma(\Phi_T) \subseteq \mathbb{T} \cup \{0\}$. In this case, we have $0 \in \sigma(\Phi_T)$ if and only if $T(K) \not= K$.

This was proved by E. Scheffold in Theorem 2.7 of his paper "Das Spektrum von Verbandsoperatoren in Banachverbänden" (1971). Unfortunately, I do not know any reference where the result is proved (or merely stated) in English.

$\endgroup$
6
  • $\begingroup$ Many thanks, I was knowing such an example, but I was searching for something more involved $\endgroup$
    – fidaleo
    Dec 21, 2019 at 0:05
  • 2
    $\begingroup$ @fidaleo: Alright, then I suggest that you edit your question in order to describe more precisely what you are looking for / how involved the example should be. $\endgroup$ Dec 21, 2019 at 0:08
  • 1
    $\begingroup$ Now it seems that the question is ok $\endgroup$
    – fidaleo
    Dec 21, 2019 at 0:12
  • 1
    $\begingroup$ Another good example might be something satisfying the question I raised, plus the condition that the peripheral spectrum includes properly 1. $\endgroup$
    – fidaleo
    Dec 21, 2019 at 0:23
  • 1
    $\begingroup$ @fidaleo: Thanks for the edit! This makes the question much more interesting, of course. I'll think about it... $\endgroup$ Dec 21, 2019 at 0:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.