Consider the Banach space $B = C([0,1] \to \mathbb R)$ of continuous functions from $[0,1] \to \mathbb R$ with the supremum norm. Let $g$ be a continuous function $g:[0,1] \to [0,1]$.
Then one can check easily that $$\Phi_g : B\to B,\ \ \ \ \Phi_g f := f\circ g,$$ is a bounded linear operator with norm $1$.
Question: How is the spectrum $\sigma(\Phi_g)$ related to $g$?
There are some simple observations:
$1 \in \sigma$ as always: It's the eigenvalue with eigenvectors the constant functions.
If $g$ is not onto, then $0 \in \sigma$ is an eigenvalue (Pick a $f$ with support away from the image of $g$).
There's another observation when $g(0) = 0$, $g$ strictly increasing and $g(1)<1$:
- $[0,1] \subset \sigma$, and all of them are eigenvalues, and with infinite multiplicities when $\lambda \neq 1$. (That result can be found in Does equation $f(g(x))=a f(x)$ have a solution?)
What else can we say to the spectrum? Is there a general theory that deal with this operator?
If $g$ is not one-to-one, then $\Phi_g$ is not onto (since $\Phi_g(f)$ satisfies $\Phi_g(f)(s) = \Phi_g(f)(t)$ whenever $g(s) = g(t)$), so $0 \in \sigma$.
$-1$ is an eigenvalue if $g$ is an involution that is not the identity. Indeed, take any $s \in [0,1]$ such that $g(s) \ne s$, let $h$ be a continuous function on $[0,1]$ with $h(s) \ne h(g(s))$, and take $f = h - h \circ g$. You can generalize this slightly: if $g \circ g$ is the identity on some nonempty open set, then $-1$ is an eigenvalue.