Spectral Measures: Scale Forms

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

Regard a bounded form: $$s\in\mathcal{B}(\mathcal{H}^s,\mathcal{H}^{s'};\mathbb{C})$$

Then it represents as: $$S\in\mathcal{B}(\mathcal{H}^s,\mathcal{H}^{s'}):\quad s(\varphi,\psi)=\langle S\varphi,\psi\rangle_{s'}$$

How can I check this?

Answer 1

Regard the transform: $$s_0(\varphi,\psi):=s(\overline{\Lambda_{-s}(N)}\varphi,\overline{\Lambda_{-s'}(N)}\psi)$$

By unitarity one gets: $$|s_0(\varphi,\psi)|\leq\|s\|\cdot\|\overline{\Lambda_{-s}(N)}^0\varphi\|_s\cdot\|\overline{\Lambda_{-s'}(N)}^0\psi\|_{s'}=\|s\|\cdot\|\varphi\|_0\cdot\|\psi\|_0$$

By Lax-Milgram one has: $$s_0\in\mathcal{B}(\mathcal{H}^0,\mathcal{H}^0;\mathbb{C})\implies S_0\in\mathcal{B}(\mathcal{H}^0,\mathcal{H}^0)$$

Define the operator: $$S:=\overline{\Lambda_{-s'}(N)}S_0\overline{\Lambda_s(N)}\in\mathcal{B}(\mathcal{H}^s,\mathcal{H}^{s'})$$

Regard the backtransform: $$s(\varphi,\psi)=s_0(\overline{\Lambda_s(N)}\varphi,\overline{\Lambda_{s'}(N)}\psi)$$

Then one obtains: $$s(\varphi,\psi)=s_0(\overline{\Lambda_s(N)}\varphi,\overline{\Lambda_{s'}(N)}\psi)=\langle S_0\overline{\Lambda_s(N)}\varphi,\overline{\Lambda_{s'}(N)}\psi\rangle_0\\ =\langle\overline{\Lambda_{-s'}(N)}S_0\overline{\Lambda_s(N)}\varphi,\overline{\Lambda_{-s'}(N)}\cdot\overline{\Lambda_{s'}(N)}\psi\rangle_{s'}=\langle S\varphi,\psi\rangle_{s'}$$

Concluding the assertion.