Spectral Measures: Spectral Spaces (II)

Question

Problem

Given a Hilbert space $\mathcal{H}$.

Consider a spectral measure: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$$

Denote its probability measures by: $$\nu_\varphi(A):=\|E(A)\varphi\|^2$$

Introduce the pure-point space: $$\mathcal{H}_0(E):=\{\varphi:\exists\#\Lambda_0\leq\aleph_0:\nu_\varphi(\Lambda_0)=\nu_\varphi(\Omega)\}$$

Construct its normal operator: $$\varphi\in\mathcal{D}(N):\quad\langle N\varphi,\chi\rangle=\int_\mathbb{C}\lambda\mathrm{d}\langle E(\lambda)\varphi,\chi\rangle\quad(\chi\in\mathcal{H})$$

Regard its eigenspace: $$\mathcal{E}_\lambda=\{\varphi:N\varphi=\lambda\varphi\}:\quad\mathcal{E}(N):=\cup_{\lambda}\mathcal{E}_\lambda$$

Then one has: $$\mathcal{H}_0(E)=\overline{\langle\mathcal{E}(N)\rangle}$$

How to prove this?

Reference

This thread is related to: Spectral Spaces (I)

Answer 1

Prework

On the one side one has: $$\nu_\varphi(\Lambda_0)=\nu_\varphi(\Omega)\iff E(\Lambda_0)\varphi=\varphi$$

On the other side one has: $$N\varphi=\lambda\varphi\iff E(\lambda)\varphi=\varphi$$

(So both are translated!)

Subwork

So eigenspaces are closed: $$\overline{\mathcal{E}_\lambda}=\overline{\mathcal{R}E(\lambda)}=\mathcal{R}E(\lambda)=\mathcal{E}_\lambda$$ and mutually orthogonal: $$\langle\varphi,\psi\rangle=\langle E(\kappa)\varphi,E(\lambda)\psi\rangle=0\quad(\kappa\neq\lambda)$$

Thus, they admit an ONB: $$\mathcal{S}(N):=\bigcup_\lambda\mathcal{S}_\lambda:\quad\overline{\langle\mathcal{E}(T)\rangle}=\overline{\langle\mathcal{S}(N)\rangle}\quad\left(\mathcal{E}_\lambda=\overline{\langle\mathcal{S}_\lambda\rangle}\right)$$ (Note it may be uncountable!)

Main Work

By strong countable additivity: $$\varphi\in\mathcal{H}_0(E)\implies\varphi=E(\Lambda_0)\varphi=\sum_{\lambda\in\Lambda_0}E(\lambda)\varphi=:\sum_{\lambda\in\Lambda_0}\varphi_\lambda\implies\varphi\in\overline{\langle\mathcal{E}(N)\rangle}$$

Conversely one writes: $$\varphi\in\overline{\langle\mathcal{E}(T)\rangle}:\quad\varphi=\sum_{\chi\in\mathcal{S}(T)}a_\chi\chi\quad(N\chi=\lambda_\chi\chi)$$

By summability one has: $$\Lambda_0:=\{\lambda_\chi:a_\chi\neq0\}:\quad\#\Lambda_0\leq\aleph_0$$

Therefore one checks: $$E(\Lambda_0)\varphi=\sum_{\lambda\in\Lambda_0}\sum_{\chi\in\mathcal{S}(T)}E(\lambda)a_\chi\chi=\sum_{\chi\in\mathcal{S}(T)}a_\chi\chi=\varphi$$

That gives the converse!