Theorem $2.2.6$ in Principles of Harmonic Analysis by Anton Deitmar and Siegfried Echterhoff proves the well-known formula for the spectral radius in a unital Banach algebra, namely $r(a) = \lim_{n\to\infty} \|a^n\|^{1/n}$. One of the steps involves proving $\limsup_{n\to\infty} \|a^n\|^{1/n} \le r(a)$, about which I have some questions.
To see that $\limsup_{n\to\infty} \|a^n\|^{1/n} \le r(a)$ recall $(\lambda 1 - a)^{-1} = \lambda^{-1}(1- \frac a\lambda)^{-1} = \sum_{n=0}^\infty \frac{a^n}{\lambda^{n+1}}$ for $|\lambda| > \|a\|$, and hence, as the function is holomorphic there, the series converges in the norm-topology for every $|\lambda| > r(a)$ as we derive from Corollary B.6.7 applied to $z = \frac 1 \lambda$.
- Why does the series converge in the norm topology for every $|\lambda| > r(a)$? I've stated Corollary B.6.7 below.
For a fixed $|\lambda| > r(a)$ it follows that the sequence $a^n \frac{1}{\lambda^{n+1}}$ is bounded in $A$, so that there exists a constant $C \ge 0$ such that $\|a^n\| \le C|\lambda|^{n+1}$ for every $n\in \Bbb N$. Taking $n$th roots on both sides and then applying $\limsup$ shows that $\limsup_{n\to\infty} \|a^n\|^{1/n} \le |\lambda|$. Since this holds for every $|\lambda| > r(a)$, we get $\limsup_{n\to\infty} \|a^n\|^{1/n} \le r(a)$.
- Why is the sequence $\left\{a^n \frac{1}{\lambda^{n+1}}\right\}$ bounded in $A$?
Thanks!
Corollary B.6.7:
