The pure-point subspace of a Hilbert space is closed

[AxiomOfChoice]

(All that follows assumes we are talking about a self-adjoint operator $A$ on a Hilbert space $\mathscr H$.) The first volume of Reed-Simon defines

$$\mathscr H_{\rm pp} = \left\{ \psi \in \mathscr H: \mu_\psi \text{ is pure point} \right\}.$$

The book seems to take for granted that $\mathscr H_{\rm pp}$ is a closed subspace of $\mathscr H$, but this is not at all obvious to me. Can someone please explain why this is the case? Thanks in advance!

I suppose I'd also like to know the following: Is there anything in $\mathscr H_{\rm pp}$ that is not an eigenvector for $A$?

 

[brmath]

I'm having a little trouble with understanding your notation, which I presume is standard, but I am not familiar with it. What are the $$\mu_\Phi ?$$ Are they functions such that $$A\Phi = \mu\Phi?$$

I know this isn't an answer, but I'm just learning Latex and need the previews in order to write a comprehensible comment.

 

[AxiomOfChoice]

I'm having a little trouble with understanding your notation, which I presume is standard, but I am not familiar with it. What are the $$\mu_\Phi ?$$ Are they functions such that $$A\Phi = \mu\Phi?$$

I know this isn't an answer, but I'm just learning Latex and need the previews in order to write a comprehensible comment.

The $\mu_\psi$ are the spectral measures, defined for Borel sets $B$ (using the spectral theorem) as follows:

$$\mu_\psi(B) = (\psi, \mathcal X_B(A) \psi)$$

 

[brmath]

This is not my field, but I've been enjoying reading about it. However, I won't catch up to you any time soon. Still, might I throw out some general suggestions? Re the closure, you have the condition that H is a Hilbert space, that A is self-adjoint, and that you are looking only at pure points. Taken together , and possibly given just one of these, they have to imply $$H_{pp}$$ is closed. Could you try constructing something on say [text]L_1[tex] which does not get you a closed set? Or similarly try it with an A that is not self-adjoint. Or maybe you can find a very simple specific case in which anyone could see it is closed and that might throw some light on it. Re the eigenvector question, it seems to me that the pure points are the simplest, and pretty much defined to generate eigenvectors, whereas the other kinds of points do not seem as if they would, at least not ordinary eigenvectors. Yet surely there has to be at least one eigenvector?

Thanks for introducing me to this. I never did spectral analysis on a non-finite space.

 

[blue_raver22]

I can provide an explanation for why \mathscr H_{\rm pp} is a closed subspace of \mathscr H.

Firstly, let us define what a closed subspace means. A subspace \mathscr S of a Hilbert space \mathscr H is said to be closed if for any sequence \left\{ \psi_n \right\} \subseteq \mathscr S that converges to a limit \psi \in \mathscr H, we have \psi \in \mathscr S as well.

Now, in order to show that \mathscr H_{\rm pp} is closed, we need to show that for any sequence \left\{ \psi_n \right\} \subseteq \mathscr H_{\rm pp} that converges to a limit \psi \in \mathscr H, we have \psi \in \mathscr H_{\rm pp} as well.

To do this, we will use the definition of \mathscr H_{\rm pp}. Let \left\{ \psi_n \right\} \subseteq \mathscr H_{\rm pp} be a sequence that converges to a limit \psi \in \mathscr H. This means that for each n, \mu_{\psi_n} is a pure point measure. Now, since the limit of a sequence of measures is the measure of the limit, we have that \mu_{\psi} is also a pure point measure. This implies that \psi \in \mathscr H_{\rm pp}, and thus, \mathscr H_{\rm pp} is closed.

To answer your second question, it is not necessary that every element in \mathscr H_{\rm pp} is an eigenvector of A. However, every element in \mathscr H_{\rm pp} is associated with a pure point measure, which means that it is a generalized eigenvector of A. This may not be a regular eigenvector, but it is still a part of the spectral decomposition of A.