Minor Detail in the Proof of the Spectral Radius Formula

Question

Theorem: Let $A$ be a Banach Algebra then for $x\in A$ the spectral radius $r(x)$ of $x$ satisfies $$r(x)=\lim_{n\rightarrow\infty}\|x^n\|^\frac{1}{n}.$$

The proofs that I have seen of this (I have mostly been using Murphy: C* algebras and operator theory, and a few sets of online lecture notes which follow his book) show that if $\lambda\in\mathbb{C}$ satisfies $r(x)<|\lambda^{-1}|$ then $\limsup_{n\rightarrow\infty}\|x^n\|^\frac{1}{n}\leq|\lambda^{-1}|$, in fact I am quite happy with showing this inequality. However what I don't understand is why we then can conclude that $\limsup_{n\rightarrow\infty}\|x^n\|^\frac{1}{n}\leq r(x)$, I mean it must be really trivial because everybody just concludes it without a second thought, but I just can't for the life of me see how it follows.

Can anybody help me? I'm sorry that it's probably a very silly question.

Answer 1

Let $S$ be the set of all $\lambda \in \mathbb C$ such that $r(x) < |\lambda^{-1}|$. Then we have $r(x) = \inf_{\lambda \in S} |\lambda^{-1}|$ (think about why this is true).

Since for each $\lambda \in S$ we have $\limsup_{n\rightarrow\infty}\|x^n\|^\frac{1}{n}\leq|\lambda^{-1}|$, combining all these inequalities gives us $\limsup_{n\rightarrow\infty}\|x^n\|^\frac{1}{n}\leq\inf_{\lambda\in S}|\lambda^{-1}| = r(x)$

Answer 2

You know that $$ r_{\sigma}(x) < \rho \implies \limsup_n \|x^n\|^{1/n} \le \rho. $$ Therefore $\limsup_n\|x^n\|^{1/n} \le r_{\sigma}(x)$. If this were not the case, then $r_{\sigma}(x) < \lim_n\|x^n\|^{1/n}$ would hold, and there would exist $\rho$ such that $r_{\sigma} < \rho$ and $\rho < \lim_n\|x^n\|^{1/n}$, which would contradict the given condition.