# The pure-point subspace of a Hilbert space is closed

1. Aug 24, 2013

### 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 $A$?

Last edited: Aug 24, 2013
2. Sep 5, 2013

### 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.

3. Sep 5, 2013

### AxiomOfChoice

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

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

4. Sep 8, 2013

### 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.