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

Regard a projection: $$P\in\mathcal{B}(\mathcal{H}):\quad P^2=P=P^*$$

Then for adjoint: $$PN\subseteq PN\iff PN^*\subseteq N^*P$$

And one has also: $$PN\subseteq NP\implies PE(A)=E(A)P\implies P\eta(N)\subseteq\eta(N)P$$

How can I prove this?

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Meanwhile I got it...

Adjoint

Denote for readability: $$\mathcal{D}:=\mathcal{D}(N)=\mathcal{D}(N^*)=:\mathcal{D}^*$$

By reducibility one has:* $$N(\mathcal{S}\cap\mathcal{D})\subseteq\mathcal{S}\quad N(\mathcal{S}^\perp\cap\mathcal{D})\subseteq\mathcal{S}^\perp\quad(P\mathcal{D}\subseteq\mathcal{D})$$

So for the domain: $$P\mathcal{D}^*=P\mathcal{D}\subseteq\mathcal{D}=\mathcal{D}^*$$

And for invariance: $$\varphi\in\mathcal{S}\cap\mathcal{D}^*:\quad\langle N^*\varphi,\chi\rangle=\langle\varphi,N\chi\rangle=0\quad(\chi\in\mathcal{S}^\perp\cap\mathcal{D})$$ $$\psi\in\mathcal{S}^\perp\cap\mathcal{D}^*:\quad\langle N^*\psi,\chi\rangle=\langle\psi,N\chi\rangle=0\quad(\chi\in\mathcal{S}\cap\mathcal{D})$$

But they were dense:** $$\overline{\mathcal{S}\cap\mathcal{D}}=\mathcal{S}\quad\overline{\mathcal{S}^\perp\cap\mathcal{D}}=\mathcal{S}^\perp$$

Concluding adjoint.*

Equality

Consider the projections: $$N_\Re:=\frac{1}{2}\{N+N^*\}\quad N_\Im:=\frac{1}{2i}\{N-N^*\}$$

Their resolvents reduce: $$R_\alpha(z)P=R_\alpha(z)P(z-N_\alpha)R_\alpha(z) \subseteq R_\alpha(z)(z-N_\alpha)PR_\alpha(z)=PR_\alpha(z)$$

By Stone's formula: $$PE_\alpha(-\infty,\lambda_\alpha]\varphi=P\lim_{\delta\to0^+}\lim_{\varepsilon\to0^+}\frac{1}{2\pi i}\int_{-\infty}^{\lambda_\alpha+\delta}\Delta R_\alpha(s\pm i\varepsilon)\varphi\mathrm{d}s\\ =\lim_{\delta\to0^+}\lim_{\varepsilon\to0^+}\frac{1}{2\pi i}\int_{-\infty}^{\lambda_\alpha+\delta}\Delta R_\alpha(s\pm i\varepsilon)P\varphi\mathrm{d}s=E_\alpha(-\infty,\lambda_\alpha]P\varphi$$

By construction one has: $$E_\Re(A)=E(A\times\mathbb{R})\quad E_\Im(B)=E(\mathbb{R}\times B)$$

Therefore one obtains: $$PE(-\infty,\lambda]=PE\bigg((-\infty,\lambda_\Re]\times\mathbb{R}\bigg)E\bigg(\mathbb{R}\times(-\infty,\lambda_\Im]\bigg)\\ =E\bigg((-\infty,\lambda_\Re]\times\mathbb{R}\bigg)E\bigg(\mathbb{R}\times(-\infty,\lambda_\Im]\bigg)P=E(-\infty,\lambda]P$$

By Dynkin one derives: $$S\in\mathcal{B}(\mathbb{C}):\quad PE(S)=E(S)P$$

Concluding equality.

Inclusion

For the domain issue: $$\int_\mathbb{R}|\eta(\lambda)|^2\mathrm{d}\|E(\lambda)P\varphi\|^2=\int_\mathbb{R}|\eta(\lambda)|^2\mathrm{d}\|PE(\lambda)\varphi\|^2 \leq\int_\mathbb{R}|\eta(\lambda)|^2\mathrm{d}\|E(\lambda)\varphi\|^2<\infty$$

And they act the same: $$\langle P\eta(H)\varphi,\chi\rangle=\langle\eta(H)\varphi,P\chi\rangle=\int_\mathbb{R}\eta(\lambda)\mathrm{d}\langle E(\lambda)\varphi,P\chi\rangle\\ =\int_\mathbb{R}\eta(\lambda)\mathrm{d}\langle PE(\lambda)\varphi,\chi\rangle=\int_\mathbb{R}\eta(\lambda)\mathrm{d}\langle E(\lambda)P\varphi,\chi\rangle=\langle\eta(H)P\varphi,\chi\rangle$$

Concluding inclusion.

Strictness

Consider the case: $$N:\mathcal{D}(N)\to\mathcal{H}:\quad\mathcal{D}(N)\subsetneq\mathcal{H}$$

Then one obtains: $$P=0:\quad\mathcal{D}(PN)=\mathcal{D}(N)\subsetneq\mathcal{H}=\mathcal{D}(0)=\mathcal{D}(NP)$$

Concluding strictness.

*See the thread: Characterization

**See the thread: Denseness

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