Spectral radius equality

Question

Let $A: X \rightarrow X$ be a continuous linear operator, and $X$ a Banach space. I want to prove that \begin{equation} r(A^n) = r(A)^n, \end{equation} where $r(A)$ denotes the spectral radius of $A$, that is $r(A) = \max_{\lambda \in \sigma(A)} |\lambda|$. I know that if $p$ is a polynomial over $\mathbb{C}$, then $\sigma(p(A)) = p(\sigma(A))$. I started by writing \begin{equation} r(A^n) = \max_{\lambda \in \sigma(A^n)} |\lambda| = \max_{\lambda \in \sigma(A)^n} |\lambda|, \end{equation} but I'm stuck here.

I was also wondering under what condition we may write $f(\max_{x \in K} |x|) = \max_{x \in K} |f(x)|$, where $K$ is, say, a compact set? Thanks.

Answer 1

Using the fact that $\sigma(p(A)) = p(\sigma(A))$, we have $$ r(A^n) = \max_{\lambda \in \sigma(A^n)} |\lambda| = \max_{\lambda \in [\sigma(A)]^n}|\lambda| = \max_{\mu \in |\sigma(A)|} \mu^n \overset{!}{=} \left[\max_{\mu \in |\sigma(A)|}\mu\right]^n $$ To justify $\overset{!}{=}$, it suffices to note that

• The max is attained since $|\sigma(A)|$ is compact, which is true since $|\cdot|$ is continuous over $\Bbb C$.
• $x \mapsto x^n$ is increasing over $[0,\infty)$. So, for any $S \subset [0,\infty)$: $\max_{x \in S}x^n = [\max_{x \in S}x]^n$.