# Spectral Measures: Scale Operators

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

Suppose the bound: $$\eta\in\mathcal{B}(\mathbb{C}):\quad\|\Lambda_{a}\eta\|_\infty<\infty$$

Then one obtains: $$\overline{\eta(N)}\in\mathcal{B}(\mathcal{H}^s\mathcal{H}^{s+a}):\quad\|\overline{\eta(N)}\|\leq\|\Lambda_a\eta\|_\infty$$

Moreover it holds: $$\overline{\vartheta(N)\eta(N)}=\overline{\vartheta(N)}\cdot\overline{\eta(N)}=\overline{\eta(N)}\cdot\overline{\vartheta(N)}$$

Especially one has: $$\overline{\Lambda_a(N)}\in\mathcal{B}(\mathcal{H}^s,\mathcal{H}^{s-a}):\quad\overline{\Lambda_a(N)}^*=\overline{\Lambda_{-a}(N)}=\overline{\Lambda_a(N)}^{-1}$$ $$\overline{N-z}\in\mathcal{B}(\mathcal{H}^s,\mathcal{H}^{s-1}):\quad(\overline{N-z})^{-1}=\overline{R(z)}\quad(z\in\sigma(N))$$

Moreover it holds: $$\overline{\Lambda_{-a}(N)}\cdot\overline{\eta(N)}^*=\overline{\eta^*(N)}\cdot\overline{\Lambda_a(N)}$$

How can I prove this?

Domain

Denote common domain: $$B^\eta_R:=B_R\cap\{|\eta|\leq R\}:\quad\mathcal{D}_0^\eta:=\bigcup_{R>0}\mathcal{R}E(B^\eta_R)$$

It is included in both since: $$\|\eta(N)\varphi\|^2=\int|\eta|^2\mathrm{d}\nu_\varphi\leq R_\varphi^2\|\varphi\|^2<\infty$$ $$\|\Lambda_s(N)\varphi\|^2=\int\Lambda_s^2\mathrm{d}\nu_\varphi\leq(1+ R_\varphi^2)^{|s|}\|\varphi\|^2<\infty$$

It is dense in scale space: $$\|\varphi-E(B^\eta_R) \varphi\|_s^2=\int(1-1^\eta_R)\Lambda_s^2\mathrm{d}\nu_\varphi\stackrel{R\to\infty}{\to}0$$

They remain invariant: $$E(B^{\eta}_R)\eta(N)\varphi=\eta(N)E(B^{\eta}_R)\varphi=\eta(N)\varphi$$

Concluding preparation.

Operator

Define maximal operator: $$\eta(N):\mathcal{D}\eta(N)\cap\mathcal{D}\Lambda_s(N)\to\mathcal{D}\Lambda_{s+a}(N)$$

It is well defined and bounded: $$\|\eta(N)\varphi\|_{s+a}^2=\int\Lambda_{s+a}^2|\eta|^2\mathrm{d}\nu_\varphi\leq\|\Lambda_a\eta\|_\infty^2\int\Lambda_s^2|\eta|^2\mathrm{d}\nu_\varphi=\|\Lambda_a\eta\|_\infty^2\|\varphi\|_s^2$$

By density it extends to all: $$\overline{\mathcal{D}\eta(N)\cap\mathcal{D}\Lambda_s(N)}^s=\overline{\mathcal{D}^\eta_0}^s=\mathcal{H}^s$$

Concluding operator.

Product

As they remain invariant: $$\omega:=|\eta|+|\vartheta|:\quad\eta(N)\mathcal{D}_0^\omega\subseteq\mathcal{D}^\omega_0\subseteq\mathcal{D}\vartheta(N)$$

By measurable calculus: $$\varphi\in\mathcal{D}_0^\omega:\quad\eta(N)\vartheta(N)\varphi=\vartheta(N)\eta(N)\varphi$$

By density it extends to all: $$\overline{\mathcal{D}\vartheta(N)\eta(N)\cap\mathcal{D}\Lambda_s(N)}^s=\overline{\mathcal{D}_0^\omega}^s=\mathcal{H}^s$$

And it splits into product: $$\overline{\vartheta(N)\eta(N)}\subseteq\overline{\vartheta(N)}\cdot\overline{\eta(N)}\subseteq\overline{\vartheta(N)\eta(N)}$$

Concluding product.

Unitaries

They are isometric since: $$\|\Lambda_a(N)\varphi\|_{s-a}=\|\Lambda_{s-a}(N)\Lambda_a(N)\varphi\|=\|\Lambda_s(N)\varphi\|=\|\varphi\|_s$$

As well as surjective since: $$\overline{\Lambda_a(N)}\cdot\overline{\Lambda_{-a}(N)}=\overline{\Lambda_a(N)\Lambda_{-a}(N)}=\overline{1}=1_{s-a}$$

Concluding unitarity.

Resolvent

It is bounded for both:* $$\delta_-(|z|)\stackrel{z\notin\sigma(N)}{\leq}\frac{|\lambda-z|^2}{1+|\lambda|^2}\leq\delta_+(|z|)$$

So one obtains inverses: $$\overline{R(z)}\cdot\overline{(N-z)}=\overline{R(z)(H-z)}=\overline{1_{\mathcal{D}N}}=1_s$$ $$\overline{(N-z)}\cdot\overline{R(z)}=\overline{(N-z)R(z)}=\overline{1_\mathcal{H}}=1_{s-1}$$

Concluding resolvent.

Regard dense elements: $$\varphi,\psi\in\mathcal{D}_0^\eta\subseteq\mathcal{D}\eta(N),\mathcal{D}\eta(N)^*,\mathcal{D}\Lambda_a(N),\mathcal{D}\Lambda_s(N)$$
By measurable calculus: $$\langle\Lambda_a(N)\eta(N)\varphi,\psi\rangle_s=\langle\Lambda_s(N)\Lambda_a(N)\eta(N)\varphi,\Lambda_s(N)\psi\rangle\\ =\langle\Lambda_s(N)\varphi,\Lambda_s(N)\Lambda_a(N)\eta^*(N)\psi\rangle=\langle\varphi,\Lambda_a(N)\eta^*(N)\psi\rangle_s$$
So the adjoint writes as: $$\overline{\Lambda_a(N)}\cdot\overline{\eta^*(N)}=\bigg(\overline{\Lambda_a(N)}\cdot\overline{\eta(N)}\bigg)^*=\overline{\eta(N)}^*\cdot\overline{\Lambda_a(N)}^*=\overline{\eta(N)}^*\cdot\overline{\Lambda_{-a}(N)}$$