# Prove that $\sigma_{\text {ess}} (A)$ is a closed subset of $\mathbb R.$

Let $$A$$ be a self-adjoint operator on a Hilbert space $$\mathcal H.$$ Let $$E_A$$ be the unique spectral measure associated to $$A$$ obtained from spectral theory for self-adjoint operators defined on the Borel-$$\sigma$$-algebra of subsets of $$\left [-\|A\|, \|A\| \right ]$$ i.e.

$$A = \int_{\left [-\|A\|, \|A\| \right ]} t\ dE_A(t).$$

Let $$\sigma (A)$$ denote the spectrum of $$A.$$ Then what I know is that for any self-adjoint operator $$A$$ on a Hilbert space $$\mathcal H$$ we have $$\sigma (A) = \text {supp} (E_A),$$ where $$\text {supp} (E_A)$$ denotes the support of $$E_A.$$ This shows that if $$\lambda \in \sigma (A)$$ then $$E_A (\lambda - \varepsilon, \lambda + \varepsilon) \neq 0,$$ for every $$\varepsilon \gt 0.$$ This leads us to the following subdivision of the spectrum $$\sigma (A)$$ of $$A.$$

An element $$\lambda \in \sigma (A)$$ is said to be an essential spectrum of $$A$$ if the range of the projection $$E_A (\lambda - \varepsilon, \lambda + \varepsilon)$$ is infinite dimensional for every $$\varepsilon \gt 0.$$ Otherwise we say that $$\lambda$$ is a discrete spectrum of $$A.$$ The collection of all essential spectrum of $$A$$ is denoted by $$\sigma_{\text {ess}} (A)$$ and the collection of all discrete spectrum of $$A$$ is denoted by $$\sigma_{\text {disc}} (A).$$

Now two results have been left as (easy) exercises which are the following $$:$$

$$(1)$$ $$\sigma_{\text {ess}} (A)$$ is a closed subset of $$\mathbb R$$ for any self-adjoint operator $$A$$ on a Hilbert space $$\mathcal H.$$

$$(2)$$ If $$\lambda \in \sigma_p(A)$$ has infinite multiplicity then $$\lambda \in \sigma_{\text {ess}} (A),$$ where $$\sigma_p (A)$$ denotes the point spectrum (or the collection of eigenvalues) of $$A.$$

But I find it difficult to prove the first one. I have tried by taking a sequence $$\{\lambda_n\}_{n \geq 1}$$ in $$\sigma_{\text {ess}} (A)$$ converging to $$\lambda.$$ Then the range of the projection $$E_A (\lambda_n - \varepsilon, \lambda_n + \varepsilon)$$ is infinite dimensional for every $$\varepsilon \gt 0$$ and for all $$n \geq 1.$$ But how does it guarantee that the range of the projection $$E_A (\lambda - \varepsilon, \lambda + \varepsilon)$$ is also infinite dimensional for all $$\varepsilon \gt 0\$$? I have asked about it to our instructor. He told me that it is an one line argument. But I don't know why can't I able to see the proof. Also I don't have any idea about the second one. May be I am so stupid. Would anybody give me some suggestion here? I am totally confused at thus stage about how to proceed further.

Any help regarding this will be warmly appreciated. Thanks!

EDIT $$:$$ Finally I am able to prove the first one. Let us take $$\varepsilon \gt 0$$ arbitrarily. Let $$\{\lambda_n\}_{n \geq 1}$$ be a sequence in $$\sigma_{\text {ess}} (A)$$ converging to $$\lambda \in \mathbb R.$$ So there exists $$N \geq 1$$ such that $$\lambda_n \in (\lambda - \varepsilon, \lambda + \varepsilon),$$ for all $$n \geq N.$$ In particular $$\lambda_N \in (\lambda - \varepsilon, \lambda + \varepsilon).$$ Choose $$\delta \gt 0$$ small enough so that $$(\lambda_N - \delta, \lambda_N + \delta) \subseteq (\lambda - \varepsilon, \lambda + \varepsilon).$$ This implies that $$E_A ((\lambda_N - \delta, \lambda_N + \delta)) \leq E_A ((\lambda - \varepsilon, \lambda + \varepsilon)).$$ But this in turn implies that $$\text {Range} \left (E_A ((\lambda_N - \delta, \lambda_N + \delta)) \right ) \subseteq \text {Range} \left ( E_A ((\lambda - \varepsilon, \lambda + \varepsilon)) \right ).$$ Now since $$\lambda_N \in \sigma_{\text {ess}} (A)$$ it follows that $$\text {Range} \left (E_A ((\lambda_N - \delta, \lambda_N + \delta)) \right )$$ is infinite dimensional and hence we have $$\text {Range} \left (E_A ((\lambda - \varepsilon, \lambda + \varepsilon)) \right )$$ is infinite dimensional. This completes the proof.

Now how do I prove the second one? Do anybody give any idea about it? Thanks!

• Comments are not for extended discussion; this conversation has been moved to chat.
– Pedro
Jun 20, 2021 at 10:35

Q2. $$\lambda \in \sigma(A)$$ is an eigenvalue if and only if $$E_A(\{\lambda\})\neq 0$$. In this situation, $$E_A(\{\lambda\})$$ is the projection onto the eigenspace with respect to $$A$$ at the point $$\lambda$$. If the eigenvalue $$\lambda$$ has infinite multiplicty, then $$E_A((\lambda - \varepsilon, \lambda + \varepsilon))$$ must have infinite rank for every $$\varepsilon>0$$.

Your comments above are how I would proceed, but I don't follow how you arrived at the important inclusion for this problem: $$\ker(A-\lambda)\subseteq \text{Ran}E_{A}(\lambda)$$. For example, how does $$v$$ not in the range imply $$E_{A}(\lambda)v=0$$? It may have a component in the range and one in the orthogonal complement. Then I also don't follow how you conclude that the measure should concentrate around $$\lambda$$. The other inclusion looks good, however.

Here's how I would prove the inclusion $$\ker(A-\lambda)\subseteq \text{Ran}E_{A}(\lambda)$$. For $$v\in \ker(A-\lambda)$$, then $$d\mu_v=\| v\|^2\delta_\lambda$$. Thus, $$0=\langle v,E_{A}(\mathbb{R}\setminus\{ \lambda\})v\rangle$$ and by orthogonality of the projection, $$v\in \ker(E_{A}(\mathbb{R}\setminus \{\lambda\})=\text{Ran}(I-E_{A}(\mathbb{R}\setminus \{\lambda\}))=\text{Ran}(E_A(\lambda))$$.

So in particular, if $$\dim(\ker(A-\lambda))=+\infty$$, then $$\dim(\text{Ran}(A-\lambda))=+\infty$$ and $$\lambda\in \sigma_{ess}(A)$$.

edit: for future readers, OP had attempted a proof that $$\ker(A-\lambda)=E_{A}(\lambda)$$ in the comments. While this is true, the proof of one of the inclusions was incorrect (and is the important one for solving this problem). I have provided a proof of this above.

• Comments are not for extended discussion; this conversation has been moved to chat.
– Pedro
Jun 20, 2021 at 10:36