Spectral Measures: Scale Spaces 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?
 A: 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.
