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Let $A\colon D(A) \to \mathcal H$ be a self-adjoint operator on a separable Hilbert space $\mathcal H$. Prove that $$\sigma_{\text{ess}}(A^2) = \left( \sigma_{\text{ess}}(A) \right)^2,$$where $\sigma_{\text{ess}}(A^2)$ denotes the essential spectrum, that is $$\sigma_{\text{ess}}(A) := \\ \sigma(A) \setminus \{\lambda \in \sigma(A): \lambda \text{ is an eigenvalue of } A \text{ isolated in } \sigma(A) \text{ and has finite mult.}\}$$

I attempted this using Weyl sequences: Let $\{x_n\}_{n\in \mathbb N}$ be a (singular) Weyl sequence, that is $\lVert x_n \rVert = 1, \lVert (A-\lambda) x_n \rVert \to 0$ strongly, $x_n \rightharpoonup 0$ weakly. I want to show that $-\sqrt{\lambda}$ or $\sqrt{\lambda}$ is in the essential spectrum of $A$. To that end, I think I can claim that $\{x_n\}_{n\in \mathbb N}$ is a singular Weyl sequence for $-\sqrt{\lambda}$ or, if not, $\left((A+\sqrt{\lambda})x_n\right)_{n\in \mathbb N}$ is a Weyl sequence for $\sqrt{\lambda}$. However, this ended in messy calculations and I couldn't conclude without knowing $A$ to be bounded. Any hints appreciated.

(Removed wrong proof)

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  • $\begingroup$ If all thats missing is $\sigma_{ess}(A^2)\supseteq (\sigma_{ess}(A))^2$ note that $\sigma(A^2)=\sigma(A)^2$, so all you have to do is check that if $\lambda'$ is an isolated eigenvalue of $A^2$ that it must then be a square of an isolated eigenvalue of $A$ that has a smaller multiplicity. You can prove that by contradiction (if either of $\pm\sqrt{\lambda'}$ are not isolated in $\sigma(A)$ then $\lambda'$ is not isolated in $\sigma(A^2)$, similarly the eigenspace of $\lambda'$ from $A^2$ contains the sum of the eigenspaces of $\pm\sqrt{\lambda'}$ of $A$.). $\endgroup$
    – s.harp
    Dec 16, 2019 at 15:05
  • $\begingroup$ You're right, that's an easy argument. However, my proof for the other inclusion is wrong: The fact that $x_n \rightharpoonup 0$ does not imply $(A+\lambda)x_n \rightharpoonup 0$ if $A$ is not bounded... $\endgroup$
    – lasik43
    Dec 16, 2019 at 17:03
  • $\begingroup$ Can you make the definition of the essential spectrum more precise? In my answer I wrote two possible interpretations of your definition, but there may be more. $\endgroup$
    – s.harp
    Dec 16, 2019 at 18:06
  • $\begingroup$ Your interpretation is correct. I tried to make it more precise. $\endgroup$
    – lasik43
    Dec 16, 2019 at 18:10
  • $\begingroup$ My problem is with the word "isolated eigenvalue": Does this mean "isolated in the set of eigenvalues $\sigma_p(A)$" or "isolated in $\sigma(A)$"? $\endgroup$
    – s.harp
    Dec 16, 2019 at 18:12

1 Answer 1

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Check the following:

  1. If a point $\lambda\in \sigma(A)$ is not an eigenvalue of $A$ then $\lambda$ is not isolated in $\sigma(A)$.

  2. $\lambda$ is "isolated from" $\sigma(A)^2 \iff \pm\sqrt{\lambda}$ are both "isolated from" $\sigma(A)$. This tells you: $\lambda$ is not isolated from $\sigma(A)^2 \iff $ at least one of $\pm\sqrt\lambda$ is not isolated from $\sigma(A)$. Here "isolated from" just means isolated point except that the point is not necessarily in the set.

  3. The eigenspace $E_\lambda(A^2) = E_{+\sqrt\lambda}(A) + E_{-\sqrt\lambda}(A)$, in particular if the lefthand side is an infinite dimensional space one of the spaces on the righthand side is infinite dimensional.

Now make use of $\sigma(A^2)=\sigma(A)^2$.

A point $\lambda$ is in $\sigma_{ess}(A^2)$ if and only if it is either not an eigenvalue of $A^2$ (in which case both of $\pm\sqrt\lambda$ are not an eigenvalue of $A$), it is not isolated from $\sigma(A^2)$ (in which case one of $\pm\sqrt\lambda$ is not isolated from $\sigma(A)$), or its eigenspace is infinite dimensional (in which case one of the eigenspaces of $\pm\sqrt\lambda$ must be infinite dimensional). This gives $\sigma_{ess}(A^2)\subseteq \sigma_{ess}(A)^2$.

A point $\lambda$ is in $\sigma_{ess}(A)^2$ if and only if either one of the roots $\pm\sqrt\lambda$ is not an eigenvalue (in which case that root is not isolated in $\sigma(A)$, hence its square $\lambda$ is not an isolated point of $\sigma(A^2)$), one of the roots is not isolated in $\sigma(A)$ (in which case the square also is not isolated), or the eigenspace to $A$ of one of the roots is infinite dimensional (in which case the eiegenspace to $A^2$ of $\lambda$ is also infinite dimensional). This gives $\sigma_{ess}(A)^2\subseteq \sigma_{ess}(A^2)$.

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  • $\begingroup$ Thanks. Here it seems that the statement is easier to prove it from the very definition rather than using Weyl sequences. $\endgroup$
    – lasik43
    Dec 16, 2019 at 18:13

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