Limit of Powers of Operators in Banach Space Equals Infimum

Question

I am trying to show that for $X$ a Banach space and $A \in B(X)$ we have that $\lim_{n \to \infty} \|A^n\|^{1/n}$ exists and is equal to $\text{inf}_n \|A^n\|^{1/n}.$ My intuition is that the Banach-Steinhaus theorem is lurking in the background, but I have been unable to leverage it properly. Should I be thinking of the problem differently?

Answer 1

• The series $$ R(\lambda)=\sum_{n=0}^{\infty}\lambda^{n}A^{n} $$ diverges in operator norm if $$ |\lambda|\limsup_{n}\|A^{n}\|^{1/n} > 1. $$ This is because the general term cannot converge to $0$ in such a case.

• The series $R(\lambda)$ converges in operator norm if $$ |\lambda|\inf_{n}\|A^{n}\|^{1/n} < 1, $$ which can be proved using a finite rearrangement of the series: $$ R(\lambda)=(I+\lambda A+\cdots \lambda^{N-1}A^{N-1})(I+\lambda^{N}A^{N}+\lambda^{2N}A^{2N}+\cdots), $$ where $N$ is any positive integer for which $|\lambda|\|A^{N}\|^{1/N} < 1$.

Therefore, $$ |\lambda|\inf_{n}\|A^{n}\|^{1/n} < 1 \implies |\lambda|\limsup_{n}\|A^{n}\|^{1/n} \le 1. $$ Setting $\lambda=1/(\epsilon+\inf_{n}\|A^{n}\|^{1/n})$ for some $\epsilon > 0$ can be used in the above to give $$ \frac{1}{\epsilon+\inf_{n}\|A^{n}\|^{1/n}}\limsup_{n}\|A^{n}\|^{1/n} < 1,\;\;\; \epsilon > 0,\\ \implies \limsup_{n}\|A^{n}\|^{1/n} \le \inf_{n}\|A^{n}\|^{1/n} \\ \implies \limsup_{n}\|A^{n}\|^{1/n} = \liminf_{n}\|A^{n}\|^{1/n}=\inf_{n}\|A^{n}\|^{1/n}. $$