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

Choose $\delta =c{N}^{-2}$ with a sufficiently small absolute $c$. If an integer $b$ satisfies $\mathrm{dist}(b\sqrt{2},\mathbb{Z})<\delta$, then for every $|n|\le N$ the phaselies within $c$ of an integer. All summands defining $f(0,b)$ therefore have positive real part bounded below, and $|f(0,b)|\ge a_0{N}^2$.

The supplied equidistribution estimate produces $\gtrsim \delta T\asymp T/N^2$ such integers in $[0,T]$, and distinct integer times are separated by at least one. For $T\lesssim N^2$, include instead one time $t=c^{\prime }{N}^{-2}\in (0,T)$, at which both quadratic phases are uniformly small. Discarding endpoints and, if necessary, every other selected integer leaves a set $\mathcal{T}\subset (0,T)$ with

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The spacetime Fourier support of $g$ lies where $|\xi |\lesssim N$ and $|\tau |\lesssim N^2$. Choose a Schwartz function $\psi$ equal to one on this support and write $g=g\ast K_N$, whereThis anisotropic reproducing kernel is the quantitative local constancy principle. Its Schwartz decay gives, for every $M$,Split the convolution at $(x_0,t_0)$ into the stated box $B$ and its complement. On $B$ the kernel is at most $C_0{N}^4$. Outside $B$, choosing $M$ in terms of $ϵ$ makes its $L^1$ tail at most $C_{ϵ}{N}^{-1000}$. Therefore

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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

Solved by gpt-5.6-sol high.