spectrum of multiplicative morphisms

Question

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, $$ \sigma_{ph}(\Phi)\supsetneq\{1\}. $$

Answer 1

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.