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?