Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 163 / 1

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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