Spectrum in a Banach Algebra

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 $$\vert z-t \vert for $$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?

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

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 $$k\notin\sigma(y)$$.

• Thank you very much. This is a lot easier to follow than what I tried to do. :-)
– Miep
Commented Feb 26, 2020 at 15:30

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.