I dont understand a particular step in the proof of Stone's Theorem [ B.C. Hall, "Quantum Theory for Mathematicians",p.210-213]. Let me state the Theorem and explain where I got stuck.
Stone's Theorem:
Suppose $U(\cdot)$ is a strongly continuous one-parameter unitary group on a Hilbert space $H$. Then the infinitesimal generator of $U(\cdot)$ is densely defined and self adjoint and $U(t)= e^{itA}$ for all $t\in\mathbb{R}$.
Proof:
After proving that $A$ is densely defined one shows that $A$ is symmetric and essentially self adjoint. Then one defines $V(t):=e^{it\bar{A}}$, where $\bar{A}$ is the closure of $A$. Now it suffices to show that for an arbitrary $\psi\in Dom(A)$ the function $w(t):= U(t)\psi-V(t)\psi=0$ for all $t\in \mathbb{R}$.
Now comes the part where I have problems. In the book it is claimed that
$$
\frac{d}{dt}w(t)=iAU(t)\psi-iAV(t)\psi
$$
Since $A$ is the infinitesimal generator of $U$ it is clear that $\frac{d}{dt}(U(t)\psi)=iAU(t)\psi$. The infinitesimal generator of $V(t)$ however is $\bar{A}$, so that $\frac{d}{dt}(V(t)\psi)=i\bar{A}V(t)\psi$. The equation above would follow if we knew that $V(t)\psi\in Dom(A)$.
It is also true that $\frac{d}{dt}(V(t)\psi)=iV(t)\bar{A}\psi=iV(t)A\psi$, so it would also suffice to show that $V(t)$ and $A$ commute.
I do not see why either of these conditions should be true. Maybe there is something else that I am missing here.