Operator Algebras – Proof of the Spectral Radius Formula

Question

I was just reading the proof of the Gelfand spectral radius formula and yet failed to understand one step in the proof. Let $A$ be a Banach algebra and $a \in A$. Apparently whenever $|\lambda| > R$, where $R$ denotes the spectral radius of $a$, then the series
$$
\sum_{n=0}^\infty \frac{a^n}{\lambda^n}
$$
should converge and equal $(1 – a/\lambda)^{-1}$. It is clear that this identity holds in case of convergence, but I don't see how one would prove the convergence in this case. I thought I'd multiply the partial sums by $(1 – a/\lambda)$, but one gets the condition
$$
\left\| \frac{a^k}{\lambda^k} \right\| \to 0
$$
so that convergence of the series already assumes the Gelfand spectral radius formula that we want to prove.

Answer 1

I suggest to firstly consider $z\mapsto (I-za)^{-1}$ to avoid your problem of convergence. The operator-valued function $z\mapsto (I-za)^{-1}$ ($z\in\mathbb C$) is holomorphic at whichever point it is defined and admits the power series expansion $\sum_{n=0}^\infty a^nz^n$ on any open disk centered at $0$ where $z\mapsto (I-za)^{-1}$ is defined, and especially for $\{z\mid \|za\|<1\}$. While the operator-version Cauchy-Hadamard theorem entails that $\sum a^nz^n$ diverges if $(\overline{\lim\limits_{n\to\infty}}\|a^n\|^{\frac{1}{n}})^{-1}<|z|$. Therefore for any positive $r<\overline{\lim\limits_{n\to\infty}}\|a^n\|^{\frac{1}{n}}$, there is a $\lambda\in\mathbb C$ such that $r<|\lambda|<\overline{\lim\limits_{n\to\infty}}\|a^n\|^{\frac{1}{n}}$, and $I-\lambda^{-1}a$ has no inverse in $A$ (If $I-\lambda^{-1}a$ is invertible, then by what has been proved it has the expansion $\sum\frac{a^n}{\lambda^n}$, which is a contradiction). That is to say, $\lambda I-a$ has no inverse in $A$ and thus $\lambda$ lies in the spectrum of $a$. Consequently $r<R$ and the arbitrariness of $r$ indicates that $\overline{\lim\limits_{n\to\infty}}\|a^n\|^{\frac{1}{n}}\leqslant R$. This completes half of the spectral radius formula, wheras the remnant half is quite straightforward.