Let $A$ be a Banach algebra(with unit $e$), $x,y \in A$ and $xy=yx$. Prove that $\sigma(xy) \subseteq \sigma (x) \sigma (y)$, also $\sigma (x+y)\subseteq \sigma(x)+\sigma(y)$. Where $\sigma$ means the spectrum.
Here $\sigma(x) \sigma(y)=\{ \lambda_1 \lambda_2 \in \mathcal{C}: \lambda_1 \in \sigma(x), \lambda_2 \in \sigma(y)\} $, and similarly for $\sigma(x)+\sigma(y)$.
For my attempt, let $B$ be the closed subalgebra generate by $e$, $x$ and $y$. Then $B$ is commutative. Denote $\Delta_{B}$ by the maximal ideal space of $B$. Then we have $\sigma_B(x)=\{ \phi(x): \phi \in \Delta_{B} \}$, and similarly for $y$,$x+y$, $xy$. Use $\phi$ is multiple linear functional,it’s clear $\sigma_{B}(x+y)\subseteq \sigma_{B}(x)+\sigma_{B}(y)$ and $\sigma_{B}(xy)\subseteq \sigma_{B}(x)\sigma_{B}(y)$. But $\sigma_{B}(x)$ maybe larger than $\sigma(x)$. And I don’t know how to continue. I know that the spectrum radical is keep invariant for subalgebra, so $r(x+y)\leq r(x)+r(y)$ and $r(xy)\leq r(x)r(y)$, this is a standard result on Rudin’s Functional Analysis.
This is an exercise from another book. But I heavily doubt the claim of this exercise is not true...
Any help or hint? Thanks!
Instead of subalgebra generated by $e, x, y$, consider some maximal commutative subalgebra $C$ that contains $x, y$ (it exists because union of ascending chain of such subalgebras is also commutative subalgebra).
We can show that $a \in C \leftrightarrow \forall b \in C: ab = ba$ - this implies $C$ is closed and if $a \in C$ is inversible then $a^{-1} \in C$.
So if $a \in C$, then $a - \lambda e$ is inversible in $C$ iff it's inversible in $A$, and your proof works.