Problem

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

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

By the previous threads: $$Z=N\sqrt{(1+N^*N)^{-1}}\quad N=Z\left(\sqrt{1-Z^*Z}\right)^{-1}$$

Especially one had: $$Z=\int\lambda\mathrm{d}F:\quad F(\overline{\mathbb{D}})=1\quad F(\mathbb{S})=0$$

Define the function: $$\eta\in\mathcal{B}(\mathbb{D}):\quad\eta(\lambda):=\frac{\lambda}{\sqrt{1-|\lambda|^2}}$$

Construct as measure: $$E(A):=F_\eta(A):=F(\eta^{-1}A)$$

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

How can I prove this?

Reference

This builds up on: Tranform, Retransform

up vote 1 down vote accepted

Meanwhile I got it...

Define the functions: $$f(\lambda):=\lambda\quad \iota(\lambda):=\frac{1}{\lambda}\quad\chi(\lambda):=\sqrt{\lambda}\quad g(\lambda):=1-|\lambda|^2$$

As it was bounded: $$Z\in\mathcal{B}(\mathcal{H}):\quad 1-Z^*Z=g(Z)$$

For compositions:* $$\iota(\chi(g(Z))=(\iota\circ\chi\circ g)(Z)=:g'(Z)$$

By measurable calculus: $$N=Z\sqrt{1-Z^*Z}^{-1}=f(Z)g'(Z)\subseteq(fg')(Z)=\eta(Z)$$

For normal extensions: $$N\subseteq\eta(Z)\implies N=\eta(Z)$$

So one arrives at:* $$N=\eta(Z)=\int\eta(\lambda)\mathrm{d}F(\lambda)=\int\lambda\mathrm{d}F_\eta(\lambda)=\int\lambda\mathrm{d}E(\lambda)$$

Concluding existence.

*See the thread: Pushforward

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.