Spectral Measures: Spectral Spaces (II) 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)
 A: 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!
