Let $A$ be a self adjoint operator in a Hilbert space $H$ and $D\subseteq D(A)$ a dense subset such that $$ e^{iAt}:D \to D. $$ How can I show that $D$ is a core for $A$?
I need to show that $\overline{A|_D} = A$. So I want to show that $\{(x,Ax)\,|\, x\in D\}$ is a dense subset of $\{(x,Ax)\,|\, x\in D(A)\}$ with the norm topology.
I was thinking about using a sequence $\{x_n\}$, such that $\|x-x_n\|^2 + \|Ax - Ax_n\|^2 \to 0$ as $n\to\infty$ (we could also take the norm without squaring, since they are equivalent), but I can't see how to use that $e^{iAt}$ maps $D$ into $D$.
Any help would be greatly appreciated.
Even though there might be some way to prove it the way you intend to, it's easier to prove the restricted operator, $A|_D$, is essentially self adjoint. This way, its closure is a self adjoint operator equal to $A$ in a dense subset of $D(A)$, so it turns out they are the same operator.
Considering $U(t) = e^{iAt}$, we'll show that $\operatorname{ker}\left((A|_D)^* \pm i\right) = \{0\}$. Take $u\in \operatorname{ker}\left((A|_D)^* \pm i\right)$ and $\varphi\in D$, then: \begin{align*} \frac{d}{dt}\langle U(t)\varphi, u \rangle &= \frac{d}{dt}\langle iAU(t)\varphi, u \rangle\\ &= -i\,\langle (A|_D)U(t)\varphi, u \rangle,\quad\text{since $U(t)\varphi\in D$}\\ &= -i\,\langle U(t)\varphi, (A|_D)^*u \rangle \\ &= -i\,\langle U(t)\varphi, \mp\, i\,u \rangle \\ &= \mp\,\langle U(t)\varphi, u \rangle \end{align*} So $\langle U(t)\varphi, u \rangle = \langle\varphi, u \rangle e^{\mp\,t}$, which is clearly unbounded for $t\in\mathbb{R}$. On the other hand, the Cauchy-Schwarz inequality and the unitarity of $U(t)$ imply: $$ \left|\langle U(t)\varphi, u\rangle\right| \leq \|\varphi\|\|u\|, $$ which is clearly bounded. Hence, $\langle \varphi, u\rangle = 0$. The density of $D$ implies that $u=0$.