Let $D:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ be a densely defined, first-order elliptic differential operator that is essentially self-adjoint. Then for each $t\geq 0$, one can define the functional calculus operator $e^{itD}$, which is bounded $L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$.
I have seen it said that $e^{itD}$ is the solution operator for the differential equation $\dfrac{\partial u}{\partial t}=iDu$. This seems to imply that for any compactly supported smooth $u\in C_c^\infty(\mathbb{R})$, the function $e^{itD}u$ is also smooth. But I cannot see why this is true.
Question: In the above situation, is $e^{itD}u$ always smooth and why?