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I want to prove that if $\mathcal{A}$ is a unital Banach algebra and $r>0$, $x,y$ commuting elements in $\mathcal{A}$ such that $\Vert x - y \Vert < r$ then $\sigma_\mathcal{A}(y) \subseteq B_r (\sigma_\mathcal{A}(x)):=\bigcup_{t\in \sigma_\mathcal{A}(x)}B_r(t).$

I have found a counterexample when the elements are non-commutative, so I am aware that the commutativity is important. I think the way to do it is to prove that $\Vert x-y \Vert < r $ implies that $\vert z-t \vert <r$ for $z \in \sigma_\mathcal{A}(y) $ and $ t \in \sigma_\mathcal{A}(x)$. However I can't make it work, so how would one proceed with this kind of problem? Is there an obvious and easy approach that I have missed?

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    $\begingroup$ What do you mean by $B_r (\sigma_\mathcal{A}(x))$? $\endgroup$
    – cqfd
    Commented Feb 26, 2020 at 12:15
  • $\begingroup$ Oh sorry. The union over t in the spectrum of x of open balls in C with radius r. $\endgroup$
    – Miep
    Commented Feb 26, 2020 at 12:17
  • $\begingroup$ $$\bigcup_{t\in \sigma_\mathcal{A}(x)}B_r(t)?$$ $\endgroup$
    – cqfd
    Commented Feb 26, 2020 at 12:20
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    $\begingroup$ yes exactly that $\endgroup$
    – Miep
    Commented Feb 26, 2020 at 12:26

2 Answers 2

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In a unital Banach algebra, for every commuting elements $ab=ba$, the spectra of $a+b$ and $ab$ satisfy $$ \sigma(a+b)\subseteq \sigma(a)+\sigma(b)\qquad \sigma(ab)\subseteq \sigma(a)\sigma(b).$$

A proof can be found here.


Since $x$ and $y$ are commuting elements, so are $x$ and $(y-x)$. Thus $$\sigma(y)\subseteq\sigma(y-x)+\sigma(x)\tag1\label1.$$

Suppose $k\notin\bigcup_{t\in \sigma(x)}B_r(t)$. Then $|k-t|\geq r$ for all $t\in\sigma(x)$. For the sake of argument, assume that $k\in\sigma(y)$. From $\eqref1$, we see that $k=s+t_0$ for some $s\in\sigma(y-x)$ and $t_0\in \sigma(x)$. So $|k-t_0|=|s|$. Note that $|s|\leq \|y-x\|\lt r.$ Thus $|k-t_0|\lt r$, a contradiction. Hence $k\notin\sigma(y)$.

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    $\begingroup$ Thank you very much. This is a lot easier to follow than what I tried to do. :-) $\endgroup$
    – Miep
    Commented Feb 26, 2020 at 15:30
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Suppose $\lambda\not\in B_r(\sigma(x))$, we must show $\lambda\not\in \sigma(y)$.

We start informally. If $\lambda\not\in \sigma(y)\cup\sigma(x)$ then \begin{eqnarray} (y-\lambda)^{-1} &=& (y-x+x-\lambda)^{-1} = ((x-\lambda)(1-(x-\lambda)^{-1}(x-y))^{-1}\\ &=&(x-\lambda)^{-1}(1-(x-\lambda)^{-1}(x-y))^{-1} \end{eqnarray} Now, for the right hand side is well defined because the assumption together with the spectral radius theorem shows that the geometric series $$\sum_{k=0}^\infty(x-\lambda)^{-n}(x-y)^n$$ converges to $(1-(x-\lambda)^{-1}(x-y))^{-1}$, which proves that $(y-\lambda)^{-1}$ exists.

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