Spectral Measures: Scale Spaces

Question

This thread is only Q&A.

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

Construct scale functions: $$\Lambda_s:=\sqrt{1+|\mathrm{id}|^2}^s\in\mathcal{B}(\mathbb{C})$$

As well as scale norms: $$\varphi\in\mathcal{D}\Lambda_s(N):\quad\|\varphi\|_s:=\|\Lambda_s(N)\varphi\|$$

And the scale spaces: $$\mathcal{H}_s:=\overline{\mathcal{D}\Lambda_s(N)}^s:=\widehat{\mathcal{D}\Lambda_s(N)}^s$$

Then one obtains: $$\langle\varphi,\psi\rangle_s:=\frac{1}{4}\sum_{\alpha=0\ldots3}i^\alpha\|\varphi+i^\alpha\psi\|_s$$

They are ordered: $$\mathcal{D}\Lambda_s(N)\supseteq\mathcal{D}\Lambda_{s'}(N)\quad(s\leq s')$$

As well as all dense: $$\overline{\mathcal{D}\Lambda_{s'}(N)}^s=\overline{\mathcal{D}\Lambda_s(N)}^s=\mathcal{H}^s\quad(s\leq s')$$

For original space: $$\mathcal{H}^r=\mathcal{D}\Lambda_r(N)\subseteq\mathcal{H}\quad(r\geq0)$$ $$\mathcal{H}=\mathcal{D}\Lambda_{-r}(N)\subseteq\mathcal{H}^{-r}\quad(r\geq0)$$

For original operator: $$\mathcal{H}^r=\mathcal{D}\Lambda_r(N)=\mathcal{D}N^r\quad(r\geq0)$$

How can I prove this?

Answer 1

Hilbert Spaces

They are norms since: $$\Lambda_s(\lambda)\neq0\quad(\lambda\in\mathbb{C})\implies\mathcal{N}\Lambda_s(N)=(0)$$

Parallelogram identity: $$\nu_{\varphi+\psi}(A)+\nu_{\varphi-\psi}(A)=\|E(A)(\varphi+\psi)\|^2+\|E(A)(\varphi-\psi)\|^2\\ =2\|E(A)\varphi\|^2+2\|E(A)\psi\|^2=\nu_\varphi(A)+\nu_\psi(A)$$

Concluding Hilbert space.

Ordered Domains

The integrand is estimated: $$\varphi\in\mathcal{D}\Lambda_{s'}(N)\implies\int\Lambda_{s}^2\mathrm{d}\nu_\varphi\stackrel{s\leq s'}{\leq}\int\Lambda_{s'}^2\mathrm{d}\nu_\varphi<\infty\implies\varphi\in\mathcal{D}\Lambda_s(N)$$

Concluding ordered domains.

Dense Domains

Denote common domain: $$\mathcal{D}_0:=\bigcup_{R>0}\mathcal{R}E(B_R)\subseteq\mathcal{D}\Lambda_s(N)$$

By dominated convergence: $$\varphi\in\mathcal{D}\Lambda_s(E):\quad\|\varphi-E(B_R)\varphi\|_s^2=\int\Lambda_s^2(1-\chi_R)^2\mathrm{d}\nu_\varphi\to0$$

So one obtains: $$\mathcal{D}\Lambda_s(E)\subseteq\overline{\mathcal{D}_0}^s\implies\mathcal{H}^s=\overline{\mathcal{D}\Lambda_s(E)}^s=\overline{\mathcal{D}_0}^s$$

Concluding dense domains.

Original Space

For negative ones: $$\|\Lambda_{-r}\|_\infty<\infty\implies\mathcal{D}\Lambda_{-r}(N)=\mathcal{H}\quad(r\geq0)$$

For positive ones: $$|\Lambda_r|\geq1\implies\|\Lambda_r(N)\varphi\|\geq\|\varphi\|\quad(r\geq0)$$

So it is equivalent to: $$\|\Lambda_r(N)\varphi\|^2\leq\|\Lambda_r(N)\varphi\|^2+\|\varphi\|^2\leq2\|\Lambda_r(N)\varphi\|^2\quad(r\geq0)$$

But they are closed: $$\Lambda_r(N)=\overline{\Lambda_r(N)}\implies\mathcal{D}\Lambda_r(N)=\overline{\mathcal{D}\Lambda_r(N)}^r\quad(r\geq0)$$

Concluding original space.

Original Operator

For operator domain: $$\int|\lambda|^{2r}\mathrm{d}\nu_\varphi(\lambda)\leq\int(1+|\lambda|^2)^r\mathrm{d}\nu_\varphi(\lambda)$$ $$\int(1+|\lambda|^2)^r\mathrm{d}\nu_\varphi(\lambda)\leq2^r\int_{|\lambda|<1}1\mathrm{d}\nu_\varphi(\lambda)+2^r\int_{|\lambda|\geq1}|\lambda|^{2r}\mathrm{d}\nu_\varphi(\lambda)$$

Concluding original operator.