Limit of nth power of operator norm

Question

I am given a compact operator $A$ which lives in a Banach algebra and whose spectral radius obeys $\rho(A)=\lim_{n\rightarrow\infty}||A^n||^{\frac{1}{n}}<1$. Now I want to prove that this implies that $||A^n||\rightarrow 0$ as $n\rightarrow \infty$ (if true).

I tried the following. We know that for $m$ big enough, $||A^k||^{\frac{1}{k}}<1$ for all $k\geq m$, so (choosing such a $k$) also $||A^k||<1$. Therefore $\lim_{n\rightarrow\infty} ||A^k||^n=0$. But since $||A^{nk}||\leq||A^n||^k$, we have $\lim_{n\rightarrow\infty} ||A^{kn}||=0$. But then I have only proven that a subsequence of $||A^n||$ converges to 0!

Is my approach completely futile? Or is limit I want to prove incorrect, and if so, what's a counterexample?

Answer 1

Since $\rho(A) = \lim\limits_{n\to\infty} \lVert A^n\rVert^{1/n} < 1$, you know that for all large enough $n$ you have

$$\lVert A^n\rVert^{1/n} < q := \frac{1+\rho(A)}{2} < 1,$$

whence $\lVert A^n\rVert < q^n \to 0$.

Answer 2

You had a part of the idea: the is $N_0$ and $\delta\gt 0$ such that $\lVert A^n\rVert^{1/n}\leqslant 1-\delta$ for $n\geqslant N_0$.