Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 163 / 2 / c / Solution

Take all coefficients equal to one, so $\sum |b_n{|}^2\asymp N^2$. At each $t_0\in \mathcal{T}$, parts a and b, with the negligible tail absorbed, giveThe box has volume $\asymp N^{-4+3ϵ}$. Hölder's inequality therefore yieldsThe time boxes are disjoint because the selected times are one-separated. Summing over $|\mathcal{T}|\gtrsim T/N^2$ givesDividing by $(\sum |b_n{|}^2{)}^{p/2}\asymp N^p$ and renaming the epsilon loss proves