Let $A$ be a closed linear operator densely defined in a Banach space $X$. Furthermore, we suppose $\sigma(A)$ as its spectrum. We define $r_{A} := \sup\limits_{\lambda \in \sigma(A)} |\lambda|$ as the spectral radius of $A$. Assume $A$ is bounded linear operator of $X$ and bounded, then it is known that $r_{A} = \max\limits_{\lambda\in\sigma(A)}|\lambda| = \lim\limits_{n\to\infty}||A^{n}||^{\frac{1}{n}}$ and $r_{A} \leq ||A||$. Now, we let $r \in \mathbb{R}$ such that $r_{A} < r < \infty$.
My problem is I want to show that $\lim\limits_{n\to\infty}\frac{||A^{n}||}{r^{n}}=0$ but I don't know where to start. Any help is much appreciated!
Thank you very much!
Let $0<\epsilon <r-r_A$. Then there exists $k$ such that $\|A^{n}\| ^{1/n} <r_A+\epsilon$ for $n \geq k$. Hence $\frac {\|A^{n}\|} {r^{n}} <(\frac {r_A+\epsilon} r)^{n}\to 0$ since $r_A+\epsilon <r$.