Mourre Adjoint: Approximation I will provide an answer later...
Given a Hilbert space $\mathcal{H}$.
Consider a Hamiltonian:
$$H:\mathcal{D}(H)\to\mathcal{H}:\quad H=H^*$$
And an operator:
$$A:\mathcal{H}\to\mathcal{H}:\quad\|A\|<\infty$$
Regard the Mourre adjoint:
$$\mathrm{ad}(A)\in\mathcal{B}(\mathcal{H}):\quad\langle\mathrm{ad}(A)\varphi,\psi\rangle:=\langle iA\varphi,H\psi\rangle-\langle iAH\varphi,\psi\rangle$$
And the adjoint variation:
$$\delta H_\varepsilon:=\frac{1}{i\varepsilon}\{e^{i\varepsilon H}-1\}:\quad\mathrm{ad}_\varepsilon(A):=i[\delta H_\varepsilon,A]$$
Denote for shorthand:
$$\|\chi_\lambda\|_{\lambda\in\Lambda}:=\sup_{\lambda\in\Lambda}\|\chi_\lambda\|$$

Then one has:
  $$\mathrm{ad}^N(A)\in\mathcal{B}(\mathcal{H})\iff\|\mathrm{ad}_\varepsilon^N(A)\|_{\varepsilon\neq0}<\infty$$
Especially it holds:
  $$\mathrm{ad}^N(A)\varphi=\lim_{\varepsilon\to0}\mathrm{ad}_\varepsilon^N(A)\varphi\quad(\varphi\in\mathcal{H})$$

How can I prove this?
 A: Adjoint Formula
It holds the relations:
$$\mathrm{ad}_\varepsilon(A)=\delta\tau_{+\varepsilon}[A]e^{i\varepsilon H}=e^{i\varepsilon H}\delta\tau_{-\varepsilon}[A]$$
They are derivations:
$$\mathrm{ad}_\varepsilon(AB)=\mathrm{ad}_\varepsilon(A)B+A\mathrm{ad}_\varepsilon(B)$$
And they vanish on:
$$\mathrm{ad}_\varepsilon(e^{itH})=i[\delta H_\varepsilon,e^{itH}]=0$$
By iteration one gets:
$$\mathrm{ad}_\varepsilon^N(A)=\delta\tau_{+\varepsilon}^N[A]e^{Ni\varepsilon H}=e^{Ni\varepsilon H}\delta\tau_{-\varepsilon}^N[A]$$
Also they commute:
$$\tau_\varepsilon,\mathrm{id}\in\mathcal{B}(\mathcal{B}(\mathcal{H})):\quad\tau_\varepsilon\circ\mathrm{id}=\mathrm{id}\circ\tau_\varepsilon$$
And they preserve:
$$\tau:\mathbb{R}\to\mathcal{B}(\mathcal{B}(\mathcal{H})):\quad\tau^{\varepsilon+\varepsilon'}=\tau^\varepsilon\circ\tau^{\varepsilon'}$$
By Newton's formula:
$$\delta\tau_\varepsilon^N=\frac{1}{\varepsilon^N}\sum_{n=0}^N\binom{N}{n}(-1)^{N-n}\tau^{n\varepsilon}$$
So one derives at:
$$\mathrm{ad}_\varepsilon^N(A)=\frac{1}{\varepsilon^N}\sum_{n=0}^N\binom{N}{n}(-1)^{N-n}\tau^{n\varepsilon}[A]e^{iN\varepsilon H}$$
Concluding formula.
Taylor Expansion
Regard an expansion:
$$F_\varepsilon\in\mathcal{C}^N(\mathbb{R},E):\quad F_\varepsilon=P^\varepsilon_K+R^\varepsilon_K$$
For Taylor polynomial:*
$$\frac{1}{\varepsilon^N}\sum_{n=0}^N\binom{N}{n}(-1)^{N-n}P^\varepsilon_K(n\varepsilon)\stackrel{K=N-1}{=}0$$
Suppose one has:
$$\|F_\varepsilon^{(N)}(n\varepsilon s)\|^{\varepsilon\neq0}_{s\in[0,1]}<\infty:\quad F_\varepsilon^{(N)}(n\varepsilon s)\stackrel{\varepsilon\to0}{\to} F_0^{(N)}(0)$$
For Taylor remainder:*
$$\lim_{\varepsilon\to0}\frac{1}{\varepsilon^N}\sum_{n=0}^N\binom{N}{n}(-1)^{N-n}R_K(n\varepsilon)\stackrel{K=N-1}{=}F_0^{(N)}(0)$$
Concluding expansion.
Adjoint Variation
By the previous thread:
$$\left.\frac{\mathrm{d}^N}{\mathrm{d}t^N}\right|_{t=n\varepsilon s}\tau^t[A]e^{iN\varepsilon H}\varphi=\tau^{n\varepsilon s}[\mathrm{ad}^N(A)]e^{iN\varepsilon H}\varphi$$
They admit a dominant:
$$\|\tau^{n\varepsilon s}[\mathrm{ad}^N(A)]e^{iN\varepsilon H}\varphi\|\leq\|\mathrm{ad}^N(A)\|\cdot\|\varphi\|$$
And converge pointwise:
$$\tau^{n\varepsilon s}[\mathrm{ad}^N(A)]e^{iN\varepsilon H}\varphi\stackrel{\varepsilon\to0}{\to}\mathrm{ad}^N(A)\varphi$$
So the above gives:
$$\mathrm{ad}^N(A)\varphi=\lim_{\varepsilon\to0}\mathrm{ad}_\varepsilon^N(A)\varphi=:\mathrm{ad}_0^N(A)\varphi$$
The dominant bounds:
$$\|\mathrm{ad}_\varepsilon(A)\|_{\varepsilon\neq0}\leq\|\mathrm{ad}_0^N(A)\|=\|\mathrm{ad}^N(A)\|<\infty$$
Concluding adjoint variation.
Mourre Adjoint
Regard the core:
$$\mathcal{D}^M:=\bigcap_{m=0}^M\mathcal{D}(H^m):\quad\overline{(H^m)_{\mathcal{D}^M}}=H$$
And regular functions:
$$\eta(\varphi,\psi):=\langle\tau[A]\varphi,\psi\rangle\in\mathcal{C}^M(\mathbb{R},\mathbb{C})$$
By induction one gets:
$$\eta^{(M)}_0(\varphi,\psi)=i^M\sum_{m=0}^M\binom{M}{m}(-1)^{M-m}\langle AH^m\varphi,H^{M-m}\psi\rangle$$
Note that it holds:
$$\eta^{(m)}_{n\varepsilon s}(\varphi,\psi)=\eta^{(m)}_0(e^{-in\varepsilon sH}\varphi,e^{-in\varepsilon sH}\psi)$$
They admit a dominant:
$$|\langle\tau^{n\varepsilon s}[A]e^{iN\varepsilon H}H^m\varphi,H^{M-m}\psi\rangle|\leq\|A\|\cdot\|H^m\varphi\|\cdot\|H^{M-m}\psi\|$$
And converge pointwise:
$$\langle\tau^{n\varepsilon s}[A]e^{iN\varepsilon H}H^m\varphi,H^{M-m}\psi\rangle\stackrel{\varepsilon\to0}{\to}\langle AH^m\varphi,H^{M-m}\psi\rangle$$
So the above gives:
$$\eta_0^{(N)}(\varphi,\psi)=\lim_{\varepsilon\to0}\langle\mathrm{ad}_\varepsilon^N(A)\varphi,\psi\rangle=:\langle\mathrm{ad}_0^N\varphi,\psi\rangle$$
That gives the bound:
$$|\eta^{(N)}_\theta(\varphi,\psi)|=\lim_{\varepsilon\to0}|\langle\mathrm{ad}_\varepsilon^N(A)e^{-i\theta H}\varphi,e^{-i\theta H}\psi\rangle|\leq\|\mathrm{ad}_\varepsilon^N(A)\|_{\varepsilon\neq0}\|\varphi\|\cdot\|\psi\|$$
Set another expansion:
$$\eta_\theta(\varphi,\psi)=\sum_{l=0}^{L=N-1}\frac{1}{l!}\eta^{(l)}_0(\varphi,\psi)\theta^l+\frac{N}{N!}\theta^{N}\int_0^1(1-s)^{(N-1)}\eta^{(N)}_{\theta s}(\varphi,\psi)\mathrm{d}s$$
Note the trivial bound:
$$|\eta_\theta(\varphi,\psi)|=|\langle\tau^\theta[A]\varphi,\psi\rangle|\leq\|A\|\cdot\|\varphi\|\cdot\|\psi\|$$
That implies bounds:**
$$|\eta^{(l)}_0(\varphi,\psi)|\leq\|\eta^{(l)}_0\|\cdot\|\varphi\|\cdot\|\psi\|$$
Especially one has:
$$|\langle iA\varphi,H\psi\rangle-\langle iAH\varphi,\psi\rangle|=|\eta^{(1)}_0(\varphi,\psi)|\leq\|\eta^{(1)}_0\|\cdot\|\varphi\|\cdot\|\psi\|$$
Concluding Mourre adjoint.
*See the thread: Binomial
**Here is a flaw: Summands not positive!
