Spectral radius in Banach Algebra

Question

Let $A$ be a unital Banach algebra and $a\in A$ and $\lambda \in \rho(a)$. I want to prove that $$r(R(a,\lambda))=\frac{1}{d(\lambda,\sigma(a))}.$$ where $R(a,\lambda)=(\lambda 1-a)^{-1}$ and $r(.)$ is the spectral radius.

I'm giving this hint: prove the result if $A$ is commutative then do it for the general case.

I'm trying to do this but the only thing that I can prove is $\|R(a,\lambda)\|\ge\frac{1}{d(\lambda,\sigma(a))}$, but since $r(.)\le \|.\|$ things don't add up. and when dealing with the general case a'm not sure how to go from the commutative case to the general case I have a feeling that this has to do with $r(.)$ being upper semi continuous. Please help. Thank you.

Answer 1

There is no need to assume $A$ commutative. Consider the holomorphic function $$ f(z)=\frac{1}{\lambda -z} $$ on $\mathbb{C}\setminus\{\lambda\}$, which is an open neighborhood of the spectrum $\sigma(a)$ of $a$. The holomorphic functional calculus makes sense of $$ f(a)=(\lambda 1-a)^{-1}. $$ and yields, by spectral mapping: $$ \sigma(f(a))=f(\sigma(a)). $$ Therefore $$ r(f(a))=\max_{\beta\in \sigma(f(a))}|\beta|=\max_{\alpha \in \sigma(a)}|f(\alpha)|=\max_{\alpha\in\sigma(a)}\frac{1}{|\lambda-\alpha| }=\frac{1}{\min_{\alpha\in \sigma(a)}|\lambda-\alpha|}=\frac{1}{d(\lambda,\sigma(a))} $$ where $d(\lambda,\sigma(a))=\inf_{\alpha\in \sigma(a)} |\lambda -\alpha|=\min_{\alpha\in \sigma(a)} |\lambda -\alpha|$ by compactness of the spectrum.