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An exercise of Banach algebra, section of spectrum has wanted the proof of this statement:

Let $A$ be a Banach algebra and $x\in A$. Show that for every open set $U$ in $\mathbb{C}$ that contains $\sigma(x)$, there exist a positive number $\delta$ such that $\sigma(y)\subset U$ whenever $y\in A$ satisfies $||y-x||<\delta$.

How can I prove this statement? Thanks for your guidance.

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  • $\begingroup$ I remember that this proof is not too trivial, the version I know uses the Cauchy integral formula and the fact, that the residue function is holomorphic. You should look up "spectral radius" in some textbook. $\endgroup$ – Daniel Oct 5 '14 at 12:35
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I'm assuming a unit $1$. The resolvent $(x-\lambda 1)^{-1}$ is uniformly bounded near $\infty$ because $$ (x-\lambda 1)^{-1} = -\sum_{n=0}^{\infty}\frac{1}{\lambda^{n}}x^{n},\;\;\; |\lambda| > r_{\sigma}(x). $$ If $U$ is any open set containing $\sigma(x)$, then $M=\sup_{\lambda\in\mathbb{C}\setminus U}\|(x-\lambda 1)^{-1}\| < \infty$ because the resolvent $R(\lambda)=(x-\lambda 1)^{-1}$ is continuous on the closed set $\mathbb{C}\setminus U$ and vanishes at $\infty$.

Consider the following for $\lambda\in\mathbb{C}\setminus U$: $$ \begin{align} (y-\lambda 1) & = \{(y-x)+(x-\lambda 1)\} \\ & =\{(y-x)(x-\lambda 1)^{-1}+1\}(x-\lambda 1). \end{align} $$ The above is invertible if $\|(y-x)(x-\lambda 1)^{-1}\| < 1$, which holds if $\delta < 1/M$ and $\|y-x\| < \delta$. Therefore $$ \|y-x\| < \delta \implies \mathbb{C}\setminus U \subseteq \rho(y) \implies \sigma(y) \subset U. $$

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Lemma: Let $A$ be a unital Banach algebra and $\{a_n\} \subset A$ such that $a_n \to a\in A$. Suppose $\lambda_n \in \sigma(a_n)$ are such that $\lambda_n \to \lambda$ in $\mathbb{C}$, then $\lambda \in \sigma(a)$

Proof: Suppose $\lambda \notin \sigma(a)$, then $(a-\lambda 1) \in GL(A)$, which is open. So $\exists \epsilon > 0$ such that $$ \|y - (a-\lambda 1)\| < \epsilon \Rightarrow y\in GL(A) $$ Now choose $N\in \mathbb{N}$ large enough so that $$ \|a_N - a\| < \epsilon/2 \text{ and } |\lambda_N - \lambda| < \epsilon/2 $$ Then we conclude that $y:= (a_N - \lambda_N 1) \in GL(A)$, which is a contradiction.


Now suppose $A$ is a unital Banach algebra, $x \in A$ and $U$ is an open set containing $\sigma(x)$. Suppose no $\delta > 0$ exists such that $$ \|y-x\| < \delta \Rightarrow \sigma(y) \subset U $$ Then there is a sequence $(y_n)\subset A$ such that $\|y_n - x\| < 1/n$ and $$ \sigma(y_n) \cap (\mathbb{C}\setminus U) \neq \emptyset \quad\forall n\in \mathbb{N} $$ So choose $\lambda_n \in \sigma(y_n)\cap (\mathbb{C}\setminus U)$. Now note that since $y_n \to x$, $\{\|y_n\|\}$ is bounded, and therefore, $\{\lambda_n\}$ is also bounded.

Passing to a subsequence if necessary, we may assume that $\lambda_n \to \lambda$ and $\lambda \in (\mathbb{C}\setminus U)$ since $U$ is open. This contradicts the lemma.

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