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This thread is only Q&A!

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

Consider a spectral measure: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$$

Regard the domain: $$\int|f(\lambda)|^2\mathrm{d}\|E(\lambda)\varphi\|^2<\infty$$

And the calculus: $$\langle f(E)\varphi,\chi\rangle=\int_\mathbb{C} f(\lambda)\mathrm{d}\langle E(\lambda)\varphi,\chi\rangle$$

Then they are normal: $$|f|^2(E)=f(E)^*f(E)=f(E)f(E)^*$$

How to prove this?

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1 Answer 1

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Remind the domain: $$\mathcal{D}f(E)g(E)=\mathcal{D}(fg)(E)\cap\mathcal{D}g(E)$$

One has the bound: $$\int|f|^2\mathrm{d}\nu_\varphi\leq\int(1+|f|^4)\mathrm{d}\nu_\varphi=\int|f|^4\mathrm{d}\nu_\varphi+\|\varphi\|^2$$

So for the domain: $$\mathcal{D}|f|^2\subseteq\mathcal{D}f(E)=\mathcal{D}\overline{f}(E)$$

Concluding the assertion.

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