Applying $D_x^\alpha$ changes only the constants in the defining symbol estimates, while every $D_\theta$ lowers the power of $\langle\theta\rangle$ by one. Hence
$$
\boxed{D_x^\alpha D_\theta^\beta a
\in\operatorname{Sym}(X,\mathbb R^k;N-|\beta|)}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 327 / 3 / a / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-327/3/a/i/solution