Unitary operator that leaves a dense subspace invariant Let $\mathcal{H}$ be a Hilbert space, $\mathcal{D}$ be a dense subspace of $\mathcal{H}$ and $U$ be an unitary operator on $\mathcal{H}$.
Suppose that $U\mathcal{D}\subseteq \mathcal{D}$. Can we say that $U\mathcal{D}= \mathcal{D}$?
If this is not true, do you know a counter example? 

Suppose that base is $\{e_n, n \in\mathbb{Z}\}$ consider the shift operator $S(e_n)=e_{n+1}$ and $D=\{x=(x_i)_{i\in\mathbb{Z}}$ with $x_i\neq 0, i<0$. $S(D)\subset D$, but $x=(x_n)$ with $x_n=1/n, i\neq 0, x_0=0$ is not in $S(D)$.
