Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 163 (Fourier Restriction Theory and Applications)

The Khintchine inequality states that for independent Rademacher random variables ${\epsilon }_n$ and every $0<p<\infty$, there are constants $A_p,B_p>0$ such thatThe constants depend only on $p$.

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Taking only one nonzero coefficient gives $C_{p,k}(N)\ge 1$. For the second obstruction, take $b_n=1$. On the boxwith $c=c(k)>0$ sufficiently small, every phase ${\sum }_{j=2}^k{n}^j{x}_j$ lies in a fixed short arc modulo one. The terms therefore exhibit constructive interference, and the exponential sum has modulus at least $c^{\prime }N$ throughout the box.

The box has measureIts contribution to the integral is consequently at least $c{N}^{p-k(k+1)/2+1}$. Since ${\sum }_{n\le N}|b_n{|}^2=N$, division by $N^{p/2}$ givesCombining this with the first obstruction and using $\max (1,X)\ge (1+X)/2$ proves the stated lower bound.

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Let $F(x)={\sum }_{n=1}^N{b}_{n}e(n^{2}x)$. Orthogonality givesThe number $r_2(m)$ of representations of $m$ as two squares is at most a constant times its divisor function. Since $m\le 2N^2$, the supplied divisor bound and Cauchy-Schwarz implyFor $2\le p\le 4$, monotonicity of $L^p$ norms on $[0,1]$ gives $\Vert F{\Vert }_p^p{\lesssim }_{ϵ}{N}^{ϵ}\Vert b{\Vert }_2^p$. For $p\ge 4$, use $\Vert F{\Vert }_{\infty }\le N^{1/2}\Vert b{\Vert }_2$ to obtainAfter enlarging the constant and the harmless epsilon loss, both ranges give

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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

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One form of the trilinear Kakeya inequality in three dimensions is the following. If ${\mathcal{T}}_j$ are finite families of $R^{1/2}\times R^{1/2}\times R$ tubes whose directions lie within $1/30$ of $e_j$, then for every $ϵ>0$,The same statement holds for any three uniformly transverse direction caps, with a constant depending on their transversality.

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For $t_j=j/3$, the three central radial directions are proportional to $(1,t_j,t_j^2)$. Their determinant is the nonzero Vandermonde productChoose $c>0$ small enough that every triple ${\omega }_j\in {\Sigma }_j$ still has determinant bounded away from zero. An invertible linear transformation sends the three central directions to the standard basis; it sends the given tubes to comparable tubes with directions in fixed small neighborhoods of $e_1,e_2,e_3$ and distorts volume and dimensions by fixed factors.

Applying part a after this transformation gives, uniformly in the three families,This is the required moment-curve version; its uniformity is precisely the transversality supplied by the moment curve and the Vandermonde determinant.

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The three unit normals of any selected blocks remain uniformly linearly independent. Hence the intersection of three thickness-one slabs with these normals has bounded volume, and thereforeExpanding the cube of the requested $L^3$ norm and using nonnegativity givesTaking cube roots proves the estimate with a constant independent of $r$, which is stronger than the allowed factor $C_{ϵ}{r}^{ϵ}$.

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Let $\Theta (R)$ partition an $R^{-1}$-neighborhood of the parabola into $R^{-1/2}\times R^{-1}$ caps, and suppose ${\hat{f}}_{\theta }$ is supported in $\theta$. The decoupling inequality for the parabola states that for $2\le p<\infty$ and every $ϵ>0$,

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For each $\tau \in \Theta (r)$ putMultiplication by $w_{r^{1/3}}$ enlarges Fourier support by at most $r^{-1/3}$ because ${\hat{w}}_{r^{1/3}}$ 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

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SetFreeze $x_3$. In the $(x_1,x_2)$ Fourier variables, the functions $F_{\tau }(\cdot ,\cdot ,x_3)$ are supported in caps of tangential length $r^{-1/3}$ and normal width $r^{-2/3}$ along the parabola. Apply part a with decoupling scale $R^{\prime }=r^{2/3}$ and critical exponent $p=6$. Since $(R^{\prime }{)}^{1/2-3/6}=1$, for each fixed $x_3$,Integrate in $x_3$. Minkowski's inequality in $L^3(d{x}_3)$ giveswhich is exactly the claimed inequality after absorbing a change in $ϵ$.

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