Iterate the correction: after constructing $b_{j-1}\in S^{-(j-1)}$, set
$$
a_j=a\,b_{j-1}\in S^{-M-(j-1)}.
$$
The same calculation improves the remainder by one order:
$$
Q(x,\lambda+D_x)\sum_{j=1}^Na_j=1-b_N,\qquad b_N\in S^{-N}.
$$
Let $A_j$ and $B_N$ be the corresponding inverse <oscillatory integral>[oscillatory integrals]. Then
$$
\boxed{Q(x,D)(A_1+\cdots+A_N)=\delta_0-B_N}.
$$
When $N>n$, the frequency integral defining $B_N$ converges absolutely and depends continuously on $x$, so $B_N\in C(X)$. This completes the finite-order <parametrix> construction.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 327 / 3 / c / vi / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-327/3/c/vi/solution