Proving $\lim_{n\to\infty} \|T^n\|^{\frac{1}{n}}$ exists and equals $\inf_n \|T^n\|^{\frac{1}{n}}$ for linear operator $T$ on Banach space

Question

This is an exercise I have encountered in my functional analysis class.

Let $T$ be a bounded, linear operator on the Banach space $X$. We are asked to prove $\lim_{n\to\infty} \|T^n\|^{\frac{1}{n}}$ exists and equals $\inf_n \|T^n\|^{\frac{1}{n}}$ using the following steps

We define $a_n = \log \|T^n\|$ and we are asked to show $a_{m+n} \leq a_m + a_n$. This is simple using the fact that the operator norm is submultiplicative and we can take $log$ to get addition. I managed to do this.

For a fixed positive integer $m$ we define $n=mq+r$ for $q,r$ positive integers and $0 \leq r \leq m-1$. We are asked to use the previous step to establish $$ \limsup_n \frac{a_n}{n} \leq \frac{a_m}{m}. $$ This I could not do and I have no idea how to do this.

Finally, we are asked to show $\lim_{n \to \infty} \frac{a_n}{n} = \inf_n \frac{a_n}{n}$ and from this we are asked to conclude that $\lim_{n\to\infty} \|T^n\|^{\frac{1}{n}}$ exists and equals $\inf_n \|T^n\|^{\frac{1}{n}}$. This I also could not do and I have no idea how to do this.

Steps 2 and 3 have me stumped and I have no idea how to actually do them. I would appreciate all help with this. I thank all helpers.

****** Progress: I thought perhaps to normalize the operator $T$ so that $\|T\|=1$ and the powers have norm at most 1 so that the sequence $a_n \leq 0$ after taking $\log$. Can this make things simpler, and is this even allowed?

Answer 1

For your step 2, write $n=mq+r$, $1 \leq r \leq m$, then by induction $a_n \leq qa_m+a_r$, thus $\frac{a_n}{n} \leq \frac{a_r}{n} + \frac{qm}{n}\frac{a_m}{m}$.

The first term is dominated, as $n$ goes to infinity, by some $C/n$ for a large $C > 0$, and the second term is smaller than $\frac{n \pm r}{n}\frac{a_m}{m}$ (depending on whether $a_m \geq 0$), which goes as $n$ goes to infinity to $\frac{a_m}{m}$. That solves your step 2.

Can you do the rest from there?

Answer 2

By induction show that $a_{mq} \leq qa_m$. This gives $\frac {a_n} n \leq \frac {qa_m+c} {mq+r}$ where $c=\max \{a_0,a_1,..,a_{m-1}\}$. Thus $\frac {a_n} n \leq \frac {a_m} m \frac {mq} {mq+r}+\frac c n$. This gives $\lim \sup \frac {a_n} n \leq (\frac {a_m} m )(1)+0$. Thus $\lim \sup \frac {a_n} n \leq \frac {a_m} m$. This is true for each $m$.

Now we get $\lim \sup \frac {a_n} n \leq \inf_m \frac {a_m} m \leq \lim \inf \frac {a_m} m \leq \lim \sup \frac {a_m} m$. So equality holds throughout this chain of inequalities. This completes the proof.