lemmaroutineinferred_dep

Kakeya overlap bound for tubes

Bounds overlap of δ\delta-separated tubes in Ln/(n−1)L^{n/(n-1)} with loss L(1/δ)2AL(1/\delta)^{2A}
passes from caps to tubes; dualizes to Kakeya maximal
Lemma·lem:kakeya-overlap·§9.2 From caps to tubes

Statement (paper)

Assume and set q=p/2=n/(n−1)q=p/2=n/(n-1). Let 0<δ≤1/20<\delta\le1/2, let the ωj\omega_j be δ\delta-separated on Sn−1S^{n-1}, and choose xj∈Rnx_j\in\RR^n arbitrarily. Then for Tj=T2δ(ωj,xj)T_j=T_{2\delta}(\omega_j,x_j) and aj≥0a_j\ge0, ∥∑jaj1Tj∥Lq(Rn)≲L(1/δ)2A(δd∑jajq)1/q. \Big\|\sum_j a_j\mathbf1_{T_j}\Big\|_{L^q(\RR^n)} \lesssim L(1/\delta)^{2A} \Big(\delta^d\sum_j a_j^q\Big)^{1/q}.

In plain English

Assuming the local extension estimate, for q=p/2=n/(n−1)q=p/2=n/(n-1), δ\delta-separated directions ωj\omega_j, arbitrary positions xjx_j, tubes Tj=T2δ(ωj,xj)T_j=T_{2\delta}(\omega_j,x_j) and weights aj≥0a_j\ge0, we have ∥∑jaj1Tj∥Lq≲L(1/δ)2A(δd∑jajq)1/q\|\sum_j a_j\mathbf1_{T_j}\|_{L^q}\lesssim L(1/\delta)^{2A}(\delta^d\sum_j a_j^q)^{1/q}.

Why it's here

This turns the local spherical extension bound into a Kakeya-type overlap estimate for tubes in δ\delta-separated directions, with only a L(1/δ)2AL(1/\delta)^{2A} loss. Dualizing it gives the Kakeya maximal estimate ∥Kδf∥Ln(Sn−1)\|K_\delta f\|_{L^n(S^{n-1})} used in the quantitative Kakeya corollary, so it is the step that passes from caps to tubes.
Classified routine:
This is the standard Bourgain/Stein argument that extension (restriction) estimates imply Kakeya overlap bounds, run here by testing the local extension estimate on randomized modulated cap functions (Khintchine) whose extensions concentrate on tubes. The only change from the textbook version is tracking the L(1/\delta)^{2A} loss.

Sources

PaperOpen
Section
§9.2 From caps to tubes
Page
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Label
lem:kakeya-overlap
Line
sections/kakeya.tex:133
Leannot aligned

Phase 2 aligns this node to a Lean declaration.

Friction

  • info
    inferred_dep
    critique: The lemma assumes . Its proof (kakeya.tex:166-167) explicitly says to "apply on the enclosing ball", so the dependency should be visible in the graph.