Let $F(x)=\sum_{n=1}^Nb_ne(n^2x)$. Orthogonality gives
$$
\|F\|_4^4
=\sum_m\left|\sum_{n_1^2+n_2^2=m}b_{n_1}b_{n_2}\right|^2.
$$
The number $r_2(m)$ of representations of $m$ as two squares is at most a constant times its divisor function. Since $m\leq2N^2$, the supplied divisor bound and Cauchy-Schwarz imply
$$
\|F\|_4^4
\lesssim_\epsilon N^\epsilon
\sum_{n_1,n_2}|b_{n_1}|^2|b_{n_2}|^2
=N^\epsilon\|b\|_2^4.
$$
For $2\leq p\leq4$, monotonicity of $L^p$ norms on $[0,1]$ gives $\|F\|_p^p\lesssim_\epsilon N^\epsilon\|b\|_2^p$. For $p\geq4$, use $\|F\|_\infty\leq N^{1/2}\|b\|_2$ to obtain
$$
\|F\|_p^p
\leq\|F\|_\infty^{p-4}\|F\|_4^4
\lesssim_\epsilon N^{p/2-2+\epsilon}\|b\|_2^p.
$$
After enlarging the constant and the harmless epsilon loss, both ranges give
$$
C_{p,2}(N)\lesssim_\epsilon N^\epsilon(1+N^{p/2-2}).
$$

Solved by gpt-5.6-sol high.


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-163/1/c/solution