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This thread is Q&A.

Problem

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

Consider a normal operator: $$N:\mathcal{D}(N)\subseteq\mathcal{H}\to\mathcal{H}:\quad N^*N=NN^*$$

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

Regard pushforward: $$E_\eta:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad E_\eta(A):=E(\eta^{-1}A)$$

Then one obtains: $$\eta(N)=\int\lambda\,\mathrm{d}E_\eta(\lambda)$$

And for compositions: $$\vartheta(\eta(N))=(\vartheta\circ\eta)(N)$$

How can I prove this?

Application

For invertibles one gets: $$\mathcal{N}\eta(N)=(0):\quad\eta(N)^{-1}=\frac{1}{\eta}(N)=\frac{1}{\eta(N)}$$

That justifies notation!

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Consider the pushforward: $$E_\eta:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad E_\eta(A):=E(\eta^{-1}A)$$

Their domain agree as:* $$\int|\vartheta\circ\eta|^2\mathrm{d}\nu_\varphi(\lambda)=\int|\vartheta|^2\mathrm{d}\nu^\eta_\varphi$$

Denote for shorthand: $$\mathcal{D}:=\mathcal{D}(\vartheta\circ\eta)(N)=\mathcal{D}(\vartheta\circ\eta)(E)=\mathcal{D}\vartheta(E_\eta)$$

In particular one has:** $$\varphi\in\mathcal{D}:\quad\int|\vartheta\circ\eta|\,\mathrm{d}|\mu_{\varphi\chi}|<\infty\quad\int|\vartheta|\,\mathrm{d}|\mu^\eta_{\varphi\chi}|<\infty$$

For simple functions: $$\int s\circ\eta\,\mathrm{d}\mu_{\varphi\chi}=\sum_kb_k\mu(\eta^{-1}B_k)=\int s\,\mathrm{d}\mu^\eta_{\varphi\chi}$$

By Lebesgue one gets:*** $$\varphi\in\mathcal{D}:\quad\int\vartheta\circ\eta\,\mathrm{d}\mu_{\varphi\chi}=\int\vartheta\,\mathrm{d}\mu^\eta_{\varphi\chi}(\lambda)$$

But note that it was: $$\mathrm{id}(E_\eta)=\eta(E)=\eta(N)$$

Concluding the assertion.

*See the thread: Pushforward (BM)

**Note the thread: Pushforward (CM)

***See the thread: Lebesgue (CM)

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