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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.

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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.

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