Spectral Measures: Embedding This thread is just a note!
Given a Hilbert space $\mathcal{H}$.
Consider a normal operator:
$$N:\mathcal{D}(N)\to\mathcal{H}:\quad N^*N=NN^*$$
And its spectral measure:
$$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad N=\int_\Omega\lambda\mathrm{d}E(\lambda)$$
Regard an embedding:
$$J\in\mathcal{B}(\mathcal{H}_0,\mathcal{H}):\quad J^*J=1_0$$
Suppose it reduces:
$$P:=JJ^*:\quad PN\subseteq NP$$
Then one has:
$$N_0:=J^*NJ:\quad N_0^*N_0=N_0N_0^*$$
And it holds:
$$E_0:=J^*EJ:\quad N_0=\int_\Omega\lambda\mathrm{d}E_0(\lambda)$$
How to prove this from scratch?
 A: Spectral Measure
By reducibility:
$$PN\subseteq NP\implies PE(A)=E(A)P$$
They are projections:
$$E_0(A)^2=J^*E(A)PE(A)J=J^*E(A)J=E_0(A)$$
$$E_0(A)^*=J^*E(A)^*J^{**}=J^*E(A)J=E_0(A)$$
And they are additive:
$$\sum_{k\in\mathbb{N}}E_0(A_k)=J^*\left(\sum_{k\in\mathbb{N}}E(A_k)\right)J=E_0(A)$$
Also they are complete:
$$E_0(\mathbb{C})=J^*E(\mathbb{C})J=J^*1J=1_0$$
Concluding spectral measure.
Normal Operator
For the embedding:
$$(J^*J)^2=J^*J=(J^*J)^*\implies J=JJ^*J$$
By reducibility one checks:
$$\|E(A)J\varphi\|^2=\langle J^*E(A)E(A)J\varphi,\varphi\rangle=\langle J^*E(A)JJ^*E(A)J\varphi,\varphi\rangle=\|E_0(A)\varphi\|^2$$
So one obtains:
$$\int|\lambda|^2\mathrm{d}\|E(A)J\varphi\|^2=\int|\lambda|^2\mathrm{d}\|E_0(A)\varphi\|^2$$
Thus for the domain:
$$\varphi\in\mathcal{D}(N_0)\iff J\varphi\in\mathcal{D}(N)\iff\varphi\in\mathcal{D}\mathrm{id}(E_0)$$
Also they agree formally:
$$\langle N_0\varphi,\chi\rangle=\langle NJ\varphi,J\chi\rangle=\int\lambda\mathrm{d}\langle E(\lambda)J\varphi,J\chi\rangle=\int\lambda\mathrm{d}\langle E_0(\lambda)\varphi,\chi\rangle$$
In particular it is normal!
