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Let $A$ be a unital Banach algebra and $a\in A$. Let $U$ be an open subset of $\mathbb C$ containing $\sigma (a)$. Prove that there is $\delta>0$ such that for every $b\in A$, if $\|b-a\|<\delta$ then $\sigma(b)\subseteq U.$

What I've tried, Since $U$ is open we can find an open ball $B$ with radius $\|a\|+\epsilon$ such that $\sigma(a)\subseteq B\subseteq U$. Let $\lambda \ge \|a\|+\epsilon$, this implies $\lambda-a$ is invertible with inverse $c$, let $\delta=\frac{1}{\|c\|}$ then if $\|b-a\|=\|\lambda-b-(\lambda-a)\|<\delta$ then $\lambda-b$ is inverible which implies $r(b)\le \|a\|+\epsilon$, Thus, $\sigma(b)\subseteq B\subseteq U$ as required.

I have a feeling that I'm missing something or did something worng, can you help me?

i.a.m
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    No, such a $B$ need not exist. For example, imagine $\sigma(a)={0,1}$ and $U$ is a union of two small disks centered at $0$ and $1$. Or imagine that $\sigma(a)$ is the unit circle and $U$ is a thin open annulus containing the circle. – Jonas Meyer Feb 11 '13 at 04:01

1 Answers1

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Such a $B$ need not exist (draw pictures), and I don't see how you have fixed a $\delta>0$ (your $\delta$ depends on which $\lambda$ you chose), but otherwise there are similarities to your work and the approach I'll share here.

Let $\rho(a)=\mathbb C\setminus \sigma(a)$ denote the resolvent set of $a$. Define $f:\rho(a)\to (0,\infty)$ by $f(z)=\|(a-z)^{-1}\|$. Then $f$ is continuous and $\lim\limits_{|z|\to\infty}f(z)=0$. Because each set $\overline{B(0,R)}\setminus U$ is compact, it follows that $f$ takes on a maximum value $C>0$ on $\mathbb C\setminus U\subset \rho(a)$. We will see that $\delta=\dfrac{1}{C}$ suffices.

Suppose that $\|a-b\|<\dfrac{1}{C}$ and $\lambda\in\mathbb C\setminus U$. Then $\|(a-\lambda)^{-1}\|\leq C$, so $$\|(a-\lambda)-(b-\lambda)\|=\|a-b\|<\dfrac{1}{C}\leq\dfrac{1}{\|(a-\lambda)^{-1}\|}.$$

This implies that $b-\lambda$ is invertible. Thus, $\lambda \in \rho(b)$. Since $\lambda\in\mathbb C\setminus U$ was arbitrary, this shows that $\sigma(b)\subseteq U$.

Jonas Meyer
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  • Thank you for this answer, but I have a question, you concluded that $f$ takes a maximum on $\mathbb C\U$, is this because the limit of $f$ tends to zero as $|z|\rightarrow \infty$ so after some point $f$ is going to be small on large disks and so the maximum of $f$ will be on some closed disk in some sense? – i.a.m Feb 11 '13 at 04:44
  • @i.a.m.: Yes, except instead of "small on large disks" you mean "small outside large disks". Take any $z_0\in\mathbb C\setminus U$. Because $\lim_{|z|\to\infty}f(z)=0$, there exists $R>0$ such that $|z|>R$ implies that $f(z)<f(z_0)$. Hence for all $z\in\mathbb C\setminus U$, $f(z)\leq \max{f(w):w\in\overline{B(0,R)}\setminus U}$. Hence the max on the compact subset $\overline{B(0,R)}\setminus U$ is also the max on $\mathbb C\setminus U$. (It wasn't mentioned explicitly, but $R$ can be taken larger than the spectral radius of $a$.) – Jonas Meyer Feb 11 '13 at 04:46
  • Thank you for helping me, I really appreciate it. – i.a.m Feb 11 '13 at 04:51