A similar question was already asked in [question](https://mathoverflow.net/questions/14246/spectra-of-sums-and-products-in-banach-algebras-was-spectrum-in-banach-algebr) titled "Spectra of sums and products in (Banach) algebras [was: Spectrum in Banach Algebra]".


Answer there led me to the following question.


If for elements $a,b$ in Banach algebra $A$ hold that $\operatorname{spec}(\lambda a+\mu b)\subseteq \lambda \operatorname{spec}(a)+\mu \operatorname{spec}(b)$ for every $\lambda, \mu\in \mathbb{C}$, can we say something about the subalgebra $B$ of $A$ generated by $a$ and $b$. Might we conclude something about $B/\operatorname{rad}(B)$?


Where to look for (possible) counterexample of $B/\operatorname{rad}(B)$ being commutative?


Thank you for finding a counterexample. Now I would like to add some additional assumptions. Suppose that for a selfadjoint element $a$ and a unitary element $u$ of $C^∗$-algebra holds that $\operatorname{spec}(a+\lambda a^2)=\operatorname{spec}(a+\lambda ua^2u^∗)$ for every $λ∈\mathbb{C}$. Moreover, if $\lambda \in \mathbb{R}$ elements $a+\lambda a^2$ and $a+\lambda ua^2u^∗$ are unitarily equivalent.



spectra of sums in (Banach) algebras
Here is a counterexample to $B/rad(B)$ being commutative. Let $S\colon l^{2}(\Bbb N)\to l^{2}(\Bbb N)$ be the unilateral shift. Then the spectrum of $S$ is $\overline{\Bbb D}=\{z\in \Bbb C:|z|\leq 1\}.$ Now $S^{\*}$ has the same spectrum. Consider the algebra they generate inside $B(l^{2}(\Bbb N)),$ it is a C$^{\*}$-algebra, hence its radical is zero. For any $\lambda,\mu \in \Bbb C$ we have $\|\lambda S+\mu S^{\*}\|\leq |\lambda|+|\mu|.$ Thus the spectrum of $\lambda S+\mu S^{\*}$ is contained in $(|\lambda|+|\mu|)\overline{\Bbb D}.$ It is straigthfoward to show that $(|\lambda|+|\mu|)\overline{\Bbb D}=\lambda\overline{\Bbb D}+\mu \overline{\Bbb D},$ so this does it.


In the case of a $C^{\*}$-algebra $A$ one can at say that the closed convex hull of the spectrum. of $a\in A$ is the set of $\phi(a),$ where $\phi$ is a state. Thus you always get that the closed convex hull of $a+b$ is contained in the sum of the closed convex hull of $a$ and $b.$ I don't know if this is true in arbitrary Banach algebras, but maybe there is a similar statement involving polynomial convex hulls.


EDIT: Okay this last statement about states may not be true, as I think you need $a$ to be normal for this statement about the closed convex hull of the spectrum of $a$ to be true.


