# Spectral Measures: Multi Version (III)

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

Denote for shorthand: $$\mathcal{S}_\varphi:=\overline{\langle\{E(A)\varphi_0:A\in\mathcal{B}(\mathbb{C})\}\rangle}$$

By the previous thread: $$\mathcal{A}\subseteq\mathcal{H}:\quad\mathcal{H}=\sum_{\alpha\in\mathcal{A}}\mathcal{S}_\alpha=:\sum_\alpha\mathcal{S}_\alpha$$

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

Introduce the measure space:* $$\Omega:=\coprod_\alpha\mathbb{C}_\alpha\cong\mathbb{C}\times\mathcal{A}:\quad\nu(A):=\sum_\alpha\nu_\alpha(\iota_\alpha^{-1}A)$$

Then one has: $$M_\eta:\mathcal{D}(M_\eta)\to\mathcal{L}^2(\nu):\quad N=U^{-1}M_\eta U$$

How can I prove this?

Reference

*See the thread: Borel Measures: Coproduct

• If you're making like a series or something you should at least motivate each post and link the previous parts! All I see in here is formulas + formulas... The titles don't help either. Commented Jun 24, 2015 at 3:46
• @hjhjhj57: Hmm yes good idea. Thanks! Commented Jun 24, 2015 at 3:50
• @hjhjhj57: I gave a motivation as comment. Do you think this is okay so? Commented Jun 24, 2015 at 4:03
• This thread deals with the more precise reducibility of normal unbounded operators instead of cyclicity. (For further details see: Reducing Spaces) Commented Jun 24, 2015 at 4:07
• Then do it as you like it best in your own style :) I just shared how I like it. Commented Jun 24, 2015 at 4:17

Embedding

Consider the embeddings: $$J_\alpha:\mathcal{S}_\alpha\to\mathcal{H}:\quad J_\beta^*J_\alpha=\delta_{\beta\alpha}1_\alpha\quad J_\alpha J_\alpha^*=P_\alpha$$

Construct the unitary map: $$U\varphi:=(J_\alpha^*\varphi)_\alpha\quad V(\varphi_\alpha)_\alpha:=\sum_\alpha J_\alpha\varphi_\alpha$$

Indeed they are inverses:* $$VU\varphi=\sum_\alpha J_\alpha J_\alpha^*\varphi=\sum_\alpha P_\alpha\varphi=\varphi$$ $$(UV(\varphi_\alpha)_\alpha)_\beta=(J_\beta^*\sum_\alpha J_\alpha\varphi_\alpha)_\beta=(\varphi_\beta)_\beta$$

(Continuity can been used!)

Reducibility

Denote for readability: $$N_\alpha:=J_\alpha^*NJ_\alpha\quad E_\alpha(A):=J_\alpha^*E(A)J_\alpha$$

By reducibility: $$E(A)P_\alpha=P_\alpha E(A)\implies N_\alpha=\int\lambda_\alpha\mathrm{d}E_\alpha(\lambda_\alpha)$$

By the previous thread: $$M_\mathrm{id}^\alpha:\mathcal{D}(M_\mathrm{id}^\alpha)\to\mathcal{L}^2(\nu_\alpha):\quad N_\alpha=U_\alpha^{-1}M_\mathrm{id}^\alpha U_\alpha$$

By reducibility: $$E(A)P_\alpha=P_\alpha\implies P_\alpha N\subseteq NP_\alpha$$

That gives also:* $$\mathcal{D}(N)=\sum_\alpha\mathcal{D}(N)\cap\mathcal{S}_\alpha:\quad N\mathcal{S}_\alpha\subseteq\mathcal{S}_\alpha$$

So one obtains:* $$(UNV(\varphi_\alpha)_\alpha)_\beta=(J_\beta^*N\sum_\alpha J_\alpha\varphi_\alpha)_\beta=(J_\beta^*NJ_\beta\varphi_\beta)_\beta=(N_\beta\varphi_\beta)_\beta$$

In terms of operators: $$UNV=\bigoplus_\alpha N_\alpha=\bigoplus_\alpha U_\alpha^{-1}M_\mathrm{id}^\alpha U_\alpha=\left(\bigoplus_\alpha U_\alpha^{-1}\right)\left(\bigoplus_\alpha M_\mathrm{id}^\alpha\right)\left(\bigoplus_\alpha U_\alpha\right)$$

(This is a checkpoint!)

Identification

Consult the unitary map: $$\Phi:\mathcal{L}^2(\Omega,\nu)\to\bigoplus_\alpha\mathcal{L}^2(\mathbb{C}_\alpha,\nu_\alpha):\quad h\mapsto(h\circ\iota_\alpha)_\alpha$$

Moreover it is algebraic: $$\Phi(hh')=\Phi(h)\Phi(h')\quad\Phi(\overline{h})=(\overline{\Phi(h)_\alpha})_\alpha$$

Define the function: $$\eta:=\Phi^{-1}(\mathrm{id_\alpha})_\alpha:\quad M_\eta h=\Phi^{-1}\left(\bigoplus_\alpha M_\mathrm{id}^\alpha\right)\Phi h$$

All together gives: $$N=V\left(\bigoplus_\alpha U_\alpha^{-1}\right)\Phi M_\eta\Phi^{-1}\left(\bigoplus_\alpha U_\alpha\right)U$$

Concluding the assertion.

*See the thread: Reducing Spaces: Decomposition