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, give
$$
\int_{B_{t_0}}|f(x,t)|\,dxdt\gtrsim N^{-2}.
$$
The box has volume $\asymp N^{-4+3\epsilon}$. Hölder's inequality therefore yields
$$
\int_{B_{t_0}}|f|^p
\gtrsim N^{-2p}N^{(4-3\epsilon)(p-1)}
=N^{2p-4-O_p(\epsilon)}.
$$
The time boxes are disjoint because the selected times are one-separated. Summing over $|\mathcal T|\gtrsim T/N^2$ gives
$$
\int_{[0,1]^2\times[0,T]}|f|^p
\gtrsim T N^{2p-6-O_p(\epsilon)}.
$$
Dividing by $(\sum|b_n|^2)^{p/2}\asymp N^p$ and renaming the epsilon loss proves
$$
D_p(N,T)\gtrsim_\epsilon N^{-\epsilon}TN^{p-6}.
$$

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 163 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-163/2/c/solution