Let $(M,g)$ be a closed Riemannian manifold, $D$ an essentially self-adjoint differential operator on $L^2(M)$. Then the operator group $\{e^{itD}\}_{t\in\mathbb{R}}$ formed via functional calculus solves the following wave equation:
$$\frac{du}{dt}=iDu.$$
In other words, for any $f\in C^\infty(M)$ we have $$\frac{d}{dt}e^{itD}f=iD(e^{itD}f).$$ From this it follows that for any finite-rank Schwartz kernel $\sum_i f_i\otimes g_i,$ where $f_i,g_i\in C^\infty(M)$, the analogous equation also holds:
$$\sum_i\frac{d}{dt}e^{itD}f_i\otimes g_i,=\sum_i iDe^{itD}f_i\otimes g_i.$$
This equation make sense both as a pointwise equality of functions on $M\times M$ and as a statement in the bounded operators $\mathcal{B}(L^2(M))$.
Now let $k(x,y)$ be a smooth Schwartz kernel on $M\times M$, and let $T_k$ be the corresponding bounded operator. Let $e^{itD}k$ and $De^{itD}k$ denote the (smooth) Schwartz kernels of the respective operators $e^{itD}T_k$ and $De^{itD}T_k=e^{itD}DT_k$.
Question: Does the equation $$\frac{d}{dt}(e^{itD}k)(x,y)=iDe^{itD}k(x,y)$$ hold pointwise for all $x,y\in M$?
Thoughts: I have a feeling this is true, just because both sides of the equation seem to make sense, and the operator $T_k$ can be approximated by finite-rank operators. It would be nice if someone could point me to a proof of this or give an explanation.
Added after: When $M=\mathbb{R}$ and $D=-i\frac{d}{dx}$, $e^{itD}$ is a shift by $t$, so there this looks to be true.
