Taking only one nonzero coefficient gives $C_{p,k}(N)\geq1$. For the second obstruction, take $b_n=1$. On the box
$$
0\leq x_j\leq cN^{-j}\qquad(2\leq j\leq k),
$$
with $c=c(k)>0$ sufficiently small, every phase $\sum_{j=2}^kn^jx_j$ lies in a fixed short arc modulo one. The terms therefore exhibit <constructive interference>, and the exponential sum has modulus at least $c'N$ throughout the box.

The box has measure
$$
\asymp N^{-\sum_{j=2}^kj}=N^{-k(k+1)/2+1}.
$$
Its contribution to the integral is consequently at least $cN^{p-k(k+1)/2+1}$. Since $\sum_{n\leq N}|b_n|^2=N$, division by $N^{p/2}$ gives
$$
C_{p,k}(N)\gtrsim N^{p/2-k(k+1)/2+1}.
$$
Combining this with the first obstruction and using $\max(1,X)\geq(1+X)/2$ proves the stated lower bound.

Solved by gpt-5.6-sol high.

Solution

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

Constructive interference occurs when many oscillatory phases nearly agree, making their exponential terms add with comparable arguments and producing a large value of the sum.


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