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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?

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  • $\begingroup$ What is $P$...? $\endgroup$ – draks ... Jun 24 '15 at 7:50
  • $\begingroup$ @draks...: Ah sorry a projection: $P^2=P=P^*$ $\endgroup$ – C-Star-W-Star Jun 24 '15 at 7:53
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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!

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