For each $\tau\in\Theta(r)$ put
$$
F_\tau=\sum_{\substack{\theta\in\Theta(R)\\\theta\subset\tau}}f_\theta w_{r^{1/3}}.
$$
Multiplication by $w_{r^{1/3}}$ enlarges Fourier support by at most $r^{-1/3}$ because $\widehat w_{r^{1/3}}$ is supported in $B_{r^{-1/3}}(0)$. The supports of the $F_\tau$ therefore have uniformly bounded overlap: separated $r^{-1/3}$-scale moment-curve intervals remain disjoint except for boundedly many neighbors.

The <fourier-support almost orthogonality> supplied by the Plancherel theorem now gives
$$
\int_{\mathbb R^3}\left|\sum_{\theta\in\Theta(R)}f_\theta w_{r^{1/3}}\right|^2
\lesssim
\sum_{\tau\in\Theta(r)}\int_{\mathbb R^3}
\left|\sum_{\substack{\theta\in\Theta(R)\\\theta\subset\tau}}f_\theta w_{r^{1/3}}\right|^2.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 163 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-163/4/b/solution