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This question is only Q&A!

Problem

Given a Hilbert space $\mathcal{H}$.

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

Regard the spaces: $$\mathcal{S}_\varphi:=\overline{\langle\{E(A)\varphi:A\in\mathcal{B}(\mathbb{C})\}\rangle}$$

All of them reduce: $$\varphi\in\mathcal{H}:\quad E(A)\mathcal{S}_\varphi\subseteq\mathcal{S}_\varphi$$

Then they decompose: $$\mathcal{A}_\infty\subseteq\mathcal{H}:\quad\mathcal{H}=\sum_{\alpha\in\mathcal{A}_\infty}\mathcal{S}_\alpha$$

How can I prove this?

Reference

This is a start-up for: Multi Version (II)

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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 4:06
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Consider the POSET: $$\lambda\in\Lambda:\quad\mathcal{S}\in\lambda\implies\mathcal{S}=\mathcal{S}_\varphi$$ $$\lambda\in\Lambda:\quad\mathcal{S},\mathcal{S}'\in\lambda\implies\mathcal{S}\perp\mathcal{S}'$$

It is nonempty: $$\mathcal{S}_0=(0):\quad\{\mathcal{S}_0\}\in\Lambda$$

And admits upper bounds: $$\Delta\subseteq\Lambda:\quad\lambda_\Delta:=\bigcup_{\delta\in\Delta}\delta\in\Lambda$$

Indeed for chains: $$\mathcal{S}\in\lambda_\Delta\implies\mathcal{S}\in\delta\implies\mathcal{S}=\mathcal{S}_\varphi$$ $$\mathcal{S},\mathcal{S}'\in\lambda_\Delta\implies\mathcal{S},\mathcal{S}'\in\delta\wedge\delta'\implies\mathcal{S}\perp\mathcal{S}'$$

By Zorn's lemma: $$\lambda_\infty\in\Lambda:\quad\lambda\geq\lambda_\infty\implies\lambda=\lambda_\infty$$

Consider the closed space: $$\mathcal{S}_\infty:=\overline{\langle\bigcup_{\mathcal{S}\in\lambda_\infty}\mathcal{S}\rangle}=\sum_{\alpha\in\mathcal{A}_\infty}\mathcal{S}_\alpha$$

Regard the complement: $$\varphi\in\mathcal{S}_\infty^\perp\implies\varphi\in\mathcal{S}_\alpha^\perp\quad(\alpha\in\mathcal{A}_\infty)$$

On dense elements: $$\langle E(A)\varphi,E(A')\alpha\rangle=\langle\varphi,E(A\cap A')\alpha\rangle=0$$

So one obtains: $$\mathcal{S}_\varphi\perp\mathcal{S}_\alpha\quad(\alpha\in\mathcal{A}_\infty)\implies\lambda_\infty\cup\{\mathcal{S}_\varphi\}\in\Lambda$$

But maximality implies: $$\lambda\cup\{\mathcal{S}_\varphi\}\geq\lambda_\infty\implies\mathcal{S}_\varphi\in\lambda_\infty$$

So one derives at: $$\varphi\in\mathcal{S}_\varphi\subseteq\mathcal{S}_\infty\implies\varphi=0$$

Thus as desired: $$\mathcal{S}_\infty^\perp=(0)\implies\mathcal{S}_\infty=\mathcal{H}$$

Concluding the assertion.

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