Spectral Measures: Scale Forms 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?
 A: 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.
