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This question is only Q&A!

Problem

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: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad N=\int\lambda\mathrm{d}E(\lambda)$$

Suppose one has: $$\varphi_0\in\mathcal{H}:\quad\mathcal{H}=\overline{\langle\{E(A)\varphi_0:A\in\mathcal{B}(\mathbb{C})\}\rangle}$$

Denote for shorthand: $$\nu_0:=\nu_{\varphi_0}(A)=\|E(A)\varphi_0\|^2$$

Then one has: $$M_\mathrm{id}:\mathcal{D}(M_\mathrm{id})\to\mathcal{L}^2(\nu_0):\quad VM_\mathrm{id}=NV$$

How can I prove this?

Reference

This is a prep-up for: Multi Version (III)

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  • $\begingroup$ This thread deals with the more precise reducibility of normal unbounded operators instead of cyclicity. (For further details see: Reducing Spaces) $\endgroup$ – C-Star-W-Star Jun 24 '15 at 4:07
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Unitary Map

Construct the map: $$V:\mathcal{L}^2(\nu_0)\to\mathcal{H}:\quad Vh:=h(N)\varphi_0$$

By Borel calculus: $$h\in\mathcal{L}^2(\nu_0):\quad\|h(N)\varphi_0\|^2=\int|h|^2\mathrm{d}\nu_0$$

Especially one has: $$V\chi_A=E(A)\varphi_0:\quad\mathcal{H}=\overline{\{V\chi_A:A\in\mathcal{B}(\mathbb{C})\}}$$

Concluding unitarity.

Multiplication

Remind that it holds: $$f(N)g(N)\subseteq(fg)(N)$$

Observe also that: $$h\in\mathcal{L}^2(\nu_0)\iff\varphi_0\in\mathcal{D}h(N)$$

Then one checks: $$NVh=Nh(N)\varphi_0=(\mathrm{id}h)(N)\varphi_0=V(\mathrm{id}h)=VM_\mathrm{id}h$$

Concluding equivalence.

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