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Let $a,b$ be two elements of a Banach algebra $A$. Suppose that $\sigma(a+\lambda b)$ contains only one element for all $\lambda \in \mathbb{C}$. Prove that both $\sigma(a)$ and $\sigma(b)$ contain only one point in their respective spectra, and that $\sigma(a+\lambda b) = \sigma(a) + \lambda \sigma(b)$, for all $\lambda \in \mathbb{C}$.

Consider $f: \mathbb{C} \to A$ given by $f(\lambda)=a+\lambda b$, then clearly $f$ is analytic.

By assumption, $\sigma(a+\lambda b)=\sigma(f(\lambda)) = \{\alpha(\lambda)\}$ for all $\lambda \in \mathbb{C}$, where $\alpha$ is a mapping from $\mathbb{C}$ into $\mathbb{C}$.

By Corollary 3.4.18 in Aupetit's A Primer on Spectral Theorey, $\alpha$ is holomorphic on $\mathbb{C}$, and therefore, for every $\lambda \in \mathbb{C}$, \begin{align*} \alpha(\lambda) &= \alpha_0 + \alpha_1\lambda + \alpha_2\lambda^2 + \dots \end{align*}

Taking $\lambda = 0$ shows that $\sigma(a)=\sigma(f(0))= \{\alpha(0)\}=\{\alpha_0\}$, which shows that $\sigma(a)$ contains only one element.

How can I use this to continue with the problem, or am I on the wrong track completely?

user860374
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1 Answers1

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You have a very good approach. I'll contribute a partial solution.

One ingredient is that the spectral radius is bounded by the norm. In this case, that means that $$ |\alpha(\lambda)|\leq\|a+\lambda b\|\leq\|a\|+|\lambda|\,\|b\| $$ Then, using Cauchy's Estimate as in this answer, we get that $$ \alpha(\lambda)=\alpha_0+\alpha_1\,\lambda. $$ So all that remains is to show that $\sigma(b)=\{\alpha_1\}$.

From $$b+\frac1n\,a=\frac1n\,(a+nb)$$ we get that $$ \sigma\Big(b+\frac1n\,a\Big)=\Big\{\frac{\alpha(n)}n\Big\}=\Big\{\frac{\alpha_0}n+\alpha_1\Big\},\qquad\qquad n\in\mathbb N. $$ Since $$ b-\alpha_1\,1=\lim_n\Big[b+\frac1n\,a-\Big(\frac{\alpha_0}n+\alpha_1\Big)\Big] $$ is a limit of non-invertible operators, we get that $b-\alpha_1$ is not invertible and so $\alpha_1\in\sigma(b)$. The missing step is to now show that there are no other points in $\sigma(b)$.

Martin Argerami
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  • Thank you very much for the helpful information. Just confirming one thing, the fact that $\alpha(\lambda)=\alpha_0 + \alpha_1\lambda$, stems from the fact that, for sufficiently large $|\lambda|$ we have that $|\alpha(\lambda)| \le M|\lambda|$, where $M$ is some constant, and therefore $\alpha$ is a polynomial of degree at most one? – user860374 Feb 20 '23 at 15:21
  • Yes indeed. $ \ $ – Martin Argerami Feb 20 '23 at 16:54