# spectra of sums in (Banach) algebras

A similar question was already asked in question 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.

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Since you have already accepted an answer to your original question, perhaps you should post the new version as a separate question (with the c-star algebras tag)? – Yemon Choi May 31 '11 at 23:39

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.
Okay, thanks for your correction and clarification. Can one use this to get some sort of statement about the spectrum of $a+b.$ (I'm not familiar with the definition of numerical range as in an arbitrary unital Banach algebra). – Benjamin Hayes May 8 '11 at 7:31