# Is it true that core of $C^0$ one-parameter unitary group generator on Hilbert space is invariant under action of this unitary group?

Let $$\mathcal{H}$$ be complex Hilbert space and $$A:\mathcal{H}\supseteq \mathcal{D}(A)\rightarrow\mathcal{H}$$ be self-adjoint operator. As Stone theorem says, operator $$A$$ generates one parameter $$C^0$$ unitary group $$(e^{itA})_{t\in\mathbb{R}}$$ acting on $$\mathcal{H}$$. Is it true that if $$D\subset \mathcal{D}(A)$$ is a core of self-adjoint operator $$A$$ then the following $$e^{itA}D\subseteq D,\quad t\in\mathbb{R}$$ holds?
If yes then reference to a book or paper would be welcome.

No, this is not true. If $$A$$ is bounded, then a core for $$A$$ is nothing but a dense subspace of $$\mathcal H$$. There is no reason why every dense subspace should be invariant under $$e^{itA}$$. To give a concrete example, let $$\mathcal H=L^2([0,1])$$, $$Af(x)=xf(x)$$ and $$D$$ be the set of all polynomials on $$[0,1]$$. Then $$D$$ is dense in $$\mathcal H$$, yet $$e^{itA}1=e^{itx}$$, which is not a polynomial.