Holomorphic functional calculus: a fixed point

Question

Suppose $\mathcal{A}$ is a unital Banach algebra, $a \in \mathcal{A}$, and $f$ is analytic on a neighbourhood of the spectrum $\sigma(a)$. Using the holomorphic functional calculus, we have an element $f(a) \in \mathcal{A}$. If $f(a) = a$, then what can be said about $f$?

Answer 1

I realised I had actually solved this some time ago. I think that we can conclude that $\sigma(a) = \sigma(f(a)) = f(\sigma(a)) \subset \mathbb{C}$ using the spectral mapping theorem. I wonder if any more can be said (is it common to have invariant subsets of analytic functions)?

Answer 2

There are many examples of this. What it means is that $g(\lambda)=f(\lambda)-\lambda$ is an annihilating function of $a$, i.e., $g(a)=0$. For a matrix, the minimal polynomial $m(\lambda)$ annihilates $a$, which means $f(\lambda)=m(\lambda)+\lambda$ satisfies $f(a)=a$. For operators on an infinite-dimensional space, there may not exist non-trivial such functions. Even if you can find a function that maps all the spectrum to $0$, that may not be enough because quasinilpotent operators '$a$' exist where $a^{n} \ne 0$ for all $n=1,2,3,\cdots$, even though $\sigma(a)=\{0\}$.