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?
Best Answer
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 $k\notin\sigma(y)$.