Mourre Adjoint: Regularity This thread is only Q&A!
Given a Hilbert space $\mathcal{H}$.
Consider a Hamiltonian:
$$H:\mathcal{D}(H)\to\mathcal{H}:\quad H=H^*$$
Regard the domain:
$$A\in\mathcal{B}(\mathcal{H}):\quad\mathcal{D}(s_A):=\mathcal{D}(H)\times\mathcal{D}(H)$$
Construct the form:
$$s_A(\varphi,\psi):=\langle iA\varphi,H\psi\rangle-\langle iAH\varphi,\psi\rangle$$
Suppose it is bounded:
$$|s_A(\varphi,\varphi)|\leq\|s\|\cdot\|\varphi\|^2\implies|\overline{s}_A(\varphi,\varphi)|\leq\|\overline{s}_A\|\cdot\|\varphi\|^2$$
By Lax-Milgram one has:
$$\mathrm{ad}(A)\in\mathcal{B}(\mathcal{H}):\quad \overline{s}_A(\varphi,\psi)=\langle\mathrm{ad}(A)\varphi,\psi\rangle$$
Introduce the evolution:
$$A\in\mathcal{B}(\mathcal{H}):\quad\tau^t[A]:=e^{itH}Ae^{-itH}$$

Then equivalence holds:
  $$\mathrm{ad}^n(A)\in\mathcal{B}(\mathcal{H})\iff\tau[A]\varphi,\tau[A^*]\varphi\in\mathcal{C}^n(\mathbb{R},\mathcal{H})^\quad(\varphi\in\mathcal{H})$$
Especially one has then:
  $$\frac{\mathrm{d}^n}{\mathrm{d}t^n}\tau^t[A]\varphi=\tau^t[\mathrm{ad}^n(A)]\varphi\quad(\varphi\in\mathcal{H})$$

How can I prove this?
 A: Evolution
Regard dense elements:
$$\overline{\mathcal{D}(H)}=\mathcal{H}:\quad\varphi\in\mathcal{D}(H)$$
By the previous thread:
$$\mathrm{ad}(A)\in\mathcal{B}(\mathcal{H})\implies\frac{\mathrm{d}}{\mathrm{d}t}\tau^t[A]\varphi=\tau^t[\mathrm{ad}(A)]\varphi$$
That gives the identity:
$$\frac{1}{h}\int_0^h\tau^{t+s}[\mathrm{ad}(A)]\varphi\mathrm{d}s=\frac{1}{h}\int_0^h\frac{\mathrm{d}}{\mathrm{d}s}\tau^{t+s}[A]\varphi\mathrm{d}s=\frac{1}{h}\{\tau^{t+h}[A]-\tau^t[A]\}\varphi$$
Now regard elements
$$\varphi\in\mathcal{H}:\quad\varphi=\lim_n\varphi_n\quad(\varphi_n\in\mathcal{D}(H))$$
One has the dominant:
$$\|\tau^{t+s}[\mathrm{ad}(A)]\varphi_n\|\leq\|\mathrm{ad}(A)\|(1+\|\varphi\|)$$
So by dominated convergence:
$$\frac{1}{h}\{\tau^{t+h}[A]-\tau^t[A]\}\varphi=\lim_n\frac{1}{h}\int_0^h\tau^{t+s}[\mathrm{ad}(A)]\varphi_n\mathrm{d}s\\
=\frac{1}{h}\int_0^h\lim_n\tau^{t+s}[\mathrm{ad}(A)]\varphi_n\mathrm{d}s=\frac{1}{h}\int_0^h\tau^{t+s}[\mathrm{ad}(A)]\varphi\mathrm{d}s\stackrel{h\to0}{\to}\tau^{t}[\mathrm{ad}(A)]\varphi$$
Concluding one direction.
Commutator
Denote the derivative:
$$\varphi\in\mathcal{H}:\quad D(A)\varphi:=\left(\frac{\mathrm{d}}{\mathrm{d}t}\tau^t[A]\varphi\right)_{t=0}$$
Regard elements:
$$\varphi,\psi\in\mathcal{D}(H)=\mathcal{D}(H^*)$$
Then invariance follows:
$$\langle\varphi,D(A)\psi\rangle=\langle H\varphi,iA\psi\rangle-\langle\varphi,iAH\psi\rangle\implies iA\psi\in\mathcal{D}(H^*)=\mathcal{D}(H)$$
So one obtains:
$$\varphi\in\mathcal{D}(H):\quad D(A)\varphi=\{iHA-iAH\}\varphi$$
For arbitrary elements:
$$\langle D(A)\varphi,\psi\rangle=\lim_{h\to0}\langle\frac{1}{h}\{\tau^h[A]-\tau^0[A]\}\varphi,\psi\rangle\\
=\lim_{h\to0}\langle\varphi,\frac{1}{h}\{\tau^h[A^*]-\tau^0[A^*]\}\psi\rangle=\langle\varphi,D(A^*)\psi\rangle$$
So one obtains:
$$D(A^*)=D(A)^*\implies D(A)=\overline{D(A)}\implies D(A)\in\mathcal{B}(\mathcal{H})$$
Concluding other direction
