Use $D=-i\partial$, so $P(D)e^{i\lambda\cdot x}=P(\lambda)e^{i\lambda\cdot x}$. A concrete distributional division construction is
$$
T=\operatorname*{FP}_{s=0}\left(\overline{P(\lambda)}|P(\lambda)|^{2(s-1)}\right).
$$
The locally integrable family, initially defined for sufficiently large $\operatorname{Re}s$, has a meromorphic continuation; $\operatorname{FP}$ denotes its finite part at zero. Multiplication before continuation gives
$$
P(\lambda)\overline{P(\lambda)}|P(\lambda)|^{2(s-1)}=|P(\lambda)|^{2s}.
$$
The right side is holomorphic at $s=0$ with value $1$, so comparison of constant Laurent coefficients yields $PT=1$. Therefore
$$
\boxed{E=\mathcal F^{-1}T}
$$
satisfies $P(D)E=\delta_0$. This finite-part formula explicitly realizes the <Malgrange–Ehrenpreis theorem>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 327 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-327/2/a/solution