Question on the Spectral Theorem. If $A:D(A)\subseteq H\to H$ is a densely defined self-adjoint operator, we can make sense of the unitary group $e^{itA}$ with the Borel functional calculus. 


*

*I'm really struggling to understand why $e^{itA}$ propogates elements of $D(A)$ back into $D(A)$. That is, if $x\in D(A)$ why must $e^{itA}x\in D(A)$?


I can reason from abstract semigroup theory that $D(A)=\{x: \lim_{t\to 0}t^{-1}(e^{itA}x-x)\text{ exists}\}$, which then must include $e^{isA}$ for every $s>0$. I was hoping for something a bit more direct (directly from the spectral theorem).
 A: If $A=\int_{\sigma}\lambda dE(\lambda)$, where $E$ is the spectral resolution of the identity for the densely-defined selfadjoint linear operator $A : \mathcal{D}(A)\subseteq H\rightarrow H$, then
$$
       x \in \mathcal{D}(A) \iff \int \lambda^{2}d\|E(\lambda)x\|^{2} < \infty.
$$
This is a useful characterization of the domain of $A$ which is often overlooked.

Example 1: $f\in L^{2}[0,2\pi]$ is periodic and absolutely continuous on $[0,2\pi]$, with $f' \in L^{2}[0,2\pi]$ iff
  $$
         \sum_{n=\infty}^{\infty}n^{2}\left|\int_{0}^{2\pi}f(t)e^{-int}\,dt\right|^{2} < \infty.
$$
  Example 2: $f \in L^{2}(\mathbb{R})$ is absolutely continuous on $\mathbb{R}$ with $f' \in L^{2}(\mathbb{R})$ iff
  $$
      \int_{-\infty}^{\infty}\lambda^{2}|\hat{f}(\lambda)|^{2}\,d\lambda < \infty,
$$
  where $\hat{f}$ is the Fourier transform of $f$.

In your case: If $f$ is a bounded Borel function, then $f(A)x \in \mathcal{D}(A)$ whenever $x \in \mathcal{D}(A)$ because
$$
       \int \lambda^{2}d\|E(\lambda)f(A)x\|^{2}=\int \lambda^{2}|f(\lambda)|^{2}d\|E(\lambda)x\|^{2} < \infty.
$$
Furthermore, in this case,
$$
                  Af(A)x = \int \lambda f(\lambda)dE(\lambda)x.
$$
Your case is even nicer because $f_{t}(\lambda)=e^{it\lambda}$ is unimodular on the spectrum.
