# Essential spectrum of squared operator

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)

• 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 $A$.). Dec 16, 2019 at 15:05
• 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... Dec 16, 2019 at 17:03
• 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. Dec 16, 2019 at 18:06
• Your interpretation is correct. I tried to make it more precise. Dec 16, 2019 at 18:10
• My problem is with the word "isolated eigenvalue": Does this mean "isolated in the set of eigenvalues $\sigma_p(A)$" or "isolated in $\sigma(A)$"? Dec 16, 2019 at 18:12

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 $$\sigma_{ess}(A)^2\subseteq \sigma_{ess}(A^2)$$.

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