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Suppose $A$ and $B$ are linear compact operators on a Hilbert space with $\sigma(A)$ and $\sigma(B)$ as their spectrum.

Is it possible to obtain some continuity result of $\sigma(A+\epsilon B)$ as $\epsilon\downarrow 0$ towards $\sigma(A)$? Is the limiting behavior of the form "$\sigma(A)+\epsilon \sigma(B)$''?

Thanks in advance!

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2 Answers 2

up vote 1 down vote accepted

The first question has an answer and is "yes", see Dunford-Schwartz, Linear operators, part II, XI-9, Lemma 5; it reads:

"Let $A_n, A$ be compact operators and $A_n\rightarrow A$ in the uniform operator topology. Let $\lambda_m(A)$ be an enumeration of the non-zero eigenvalues of $A$, each repeated according to its multiplicity. Then, there exist enumerations $\lambda_m(A_n)$ of the non-zero eigenvalues of $A_n$, with repetitions according to multiplicity, such that $$\lim_{n\rightarrow\infty} \lambda_m(A_n)=\lambda_m(A), \quad m\geq 1,$$ the limit being uniform in $m$."

So, under compact perturbations $\epsilon B$ of a compact operator $A$, the perturbed spectrum moves continuously. I don't know about your second question, the order of convergence might be a little bit tricky to obtain, some Rellich result might be useful. Consider checking Kato's book "Perturbation Theory for Linear Operators".

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A partial answer. I think it is true if $\sigma(A)=\{0\}$. For $\lambda\neq 0$, $A-\lambda I$ is invertible. However, the set of isomorphisms is open in $\mathcal{B}(H)$, so there exist $\varepsilon>0$ such that $A-\lambda I+\varepsilon B$ is also invertible. This means that $\lambda\notin\sigma(A+\varepsilon B)$, for $\varepsilon$ small enough. This should imply that $\sigma(A+\varepsilon B)\to 0$.

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