Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 9 / 1 / b / Solution

It is enough to prove the stronger planar Kakeya maximal function boundWe first work with a $\delta$-separated net of unoriented directions $e_1,\dots ,e_M$, with $M\asymp {\delta }^{-1}$. Write ${\alpha }_{ij}$ for angular distance modulo $\pi$. Choose arbitrary length-one, width-$\delta$ rectangles $T_j$ in these directions. The permitted rectangle intersection fact isThe first alternative includes parallel rectangles. Their locations are arbitrary; only separation of their directions matters.

Define $Af(j)={\delta }^{-1}{\int }_{T_j}f$, and put $\Vert a{\Vert }_{{\ell }_{\delta }^2}^2=\delta {\sum }_j|a_j{|}^2$. With these weights the adjoint operator is $A^{\ast }a={\sum }_j{a}_j{1}_{T_j}$. Since a separated angular net has only a bounded number of directions at each distance scale $k\delta$ from $e_i$, its overlap matrix satisfiesThe diagonal term is of size $\delta$. Using $|a_i{a}_j|\le (|a_i{|}^2+|a_j{|}^2)/2$ and symmetry gives the Schur test estimateBy duality of Lp spaces, $\Vert Af{\Vert }_{{\ell }_{\delta }^2}\lesssim \sqrt{\log (2/\delta )}\Vert f{\Vert }_2$. This is uniform over every choice of the translated rectangles. For each direction choose a rectangle approaching the supremum for $|f|$, then take the limit. That gives the same estimate for the discretized Kakeya maximal function.

To recover all directions, partition the direction circle into arcs of length comparable to $\delta$, each with a net direction. A tube in an arc is contained in a rectangle in its net direction with width $C\delta$ and length at most two. A bounded subdivision in the length direction reduces this to the same averaging operators; the wider tubes obey the same overlap estimate with fixed-factor changes. ThereforeFinally $\sqrt{\log (2/\delta )}\le C_{\epsilon }{\delta }^{-\epsilon }$ for every $\epsilon >0$. The planar Kakeya maximal estimate therefore has the required arbitrary small power loss.