# Show that $\lim\limits_{n\rightarrow\infty}\|T^n\|^{1/n}$ exists.

Let $T$ be a bounded linear operator on a complex Banach space. Assume that $\|T^{m+n}\|\le\|T^m\|\|T^n\|$. (Hint: $a_n=\|T^n\|$, $b_n=\ln a_n,$ $\alpha=\inf(b_n/n)$) and show that $b_n/n\rightarrow \alpha.)$ Show that $\lim\limits_{n\rightarrow\infty}\|T^n\|^{1/n}$ exits.

So if I am understanding this correctly I am to show that $$\lim\limits_{n\rightarrow\infty}\|T^n\|^{1/n}\le \lim\limits_{n\rightarrow\infty}\frac{1}{n}\ln \|T^n\| = \alpha.$$

However I don't see how I can show that either the inequality or the equality hold. Any clarification would be greatly appreciated!

• Have you ever heard of Fekete's lemma? – Sangchul Lee Apr 18 '17 at 4:56
• Just curious, but does using submultiplicativity (such as $\lim_{n\to\infty} ||T^n||^{1/n}\leq \lim_{n\to\infty} (||T||^n)^{1/n} = \lim_{n\to\infty}||T|| = ||T||$) work? – Mark Apr 18 '17 at 5:04
• @Mark This shows the sequence is bounded but doesn't guarantee the limit exists. – Jacky Chong Apr 18 '17 at 5:09

Fix $m \in \mathbb{N}$. The standard trick is to observe \begin{align} \log \|T^n\| = \log\|T^{mk+r}\| \end{align} where $0\leq r <m$. Then we see that \begin{align} \frac{1}{n}\log \|T^{n}\| \leq \frac{1}{n}\log\|T^{mk}\| + \frac{1}{n}\log \|T^r\| \leq \frac{k}{n}\log \|T^m\|+\frac{1}{n}\log \|T^r\| \end{align} In particular, \begin{align} \limsup_{n\rightarrow \infty} \frac{\log \|T^n\|}{n} \leq \frac{\log\|T^m\|}{m} \end{align} since for a fixed $m$ we have $\frac{k}{n} \rightarrow \frac{1}{m}$. Hence taking the infimum on the right hand side yields \begin{align} \limsup_{n\rightarrow \infty} \frac{\log \|T^n\|}{n} \leq \liminf_{m\rightarrow \infty} \frac{\log \|T^m\|}{m} \end{align} Hence the limit exists.