The three unit normals of any selected blocks remain uniformly linearly independent. Hence the intersection of three thickness-one slabs with these normals has bounded volume, and therefore
$$
|S_1\cap S_2\cap S_3\cap B_r|\lesssim1.
$$
Expanding the cube of the requested $L^3$ norm and using nonnegativity gives
$$
\begin{aligned}
\left\|\prod_{j=1}^3\left(\sum_{S_j}c_{S_j}\chi_{S_j}\right)^{1/3}\right\|_{L^3(B_r)}^3
&=\sum_{S_1,S_2,S_3}c_{S_1}c_{S_2}c_{S_3}|S_1\cap S_2\cap S_3\cap B_r|\\
&\lesssim\prod_{j=1}^3\sum_{S_j\in\mathcal S_j}c_{S_j}.
\end{aligned}
$$
Taking cube roots proves the estimate with a constant independent of $r$, which is stronger than the allowed factor $C_\epsilon r^\epsilon$.

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 163 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-163/3/c/solution