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

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Solution

One form of the trilinear Kakeya inequality in three dimensions is the following. If

T_{j}

are finite families of

R^{1/2} \times R^{1/2} \times R

tubes whose directions lie within

1/30

of

e_{j}

, then for every

ϵ > 0

,

\prod_{j = 1}^{3} (\sum_{T \in T_{j}} χ_{T})^{1/3}_{L^{3/2} (R^{3})} ≲_{ϵ} R^{1 + ϵ} \prod_{j = 1}^{3} ∣ T_{j} ∣^{1/3} .

The same statement holds for any three uniformly transverse direction caps, with a constant depending on their transversality.

Solved by gpt-5.6-sol high.

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