# spectral radius.

I am stuck in a problem of Conway's A course in a Functional Analysis. Can anyone give me a hint to solve the problem?

The question is "If $A$ is a Banach Algebra, then show that the function $r:A\to R$ defined by $r(a)=$ spectral radius of $a,$ is a upper semi-continuous function."

Recall that spectral radius of $a,$ $r(a):=\sup\{|c|:c\in\mbox{ spectrum of } a\}$ .

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Which properties of spectral radius do you know? –  Davide Giraudo Sep 23 '12 at 19:36

## 1 Answer

Assume that $r$ is not upper semi-continuous at $a$. Then we can find a sequence $\{a_n\}\subset A$ converging to $a$ such that $\limsup_{n\to +\infty}r(a_n)>r(a)$. Hence there exists a $\delta>0$ and a subsequence $\{b_k\}$ of $\{a_n\}$ such that $r(b_k)\geq r(a)+2\delta$. So for each $k$, let $\lambda_k$ in the spectrum of $b_k$ such that $|\lambda_k|\geq r(a)+\delta$. As $\{b_k\}$ is bounded, the sequence $\{\lambda_k\}$ of complex numbers is bounded. So we extract a converging subsequence (but we will denote it in the same way for simple).

Then $\{b_k-\lambda_k e\}$ is not invertible and converges to $a-\lambda e$. It's not invertible, as the set of invertible elements of $A$ is open. So $\lambda\in \sigma(a)$. But $|\lambda|\geq r(a)+\delta$, a contradiction.

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