 Solution

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.

Solved by gpt-5.6-sol high.