Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 9 (The Kakeya Universe and Incidence Problems in Rn)

For a unit direction $e$, let $T_{\delta }(a,e)$ be a length-one Kakeya tube with transverse radius $\delta$, centered at $a$, and define the Kakeya maximal function byUsing normalized surface measure on the direction sphere, the Kakeya maximal conjecture is the following family of estimates, in the formulation relevant to this paper:The constant is independent of $\delta$ and $f$. Replacing round Kakeya tubes by comparable rectangular tubes changes only dimensional constants.

A bounded Kakeya set contains a unit line segment in every direction. The Kakeya Minkowski dimension conjecture says that every such set has full Minkowski dimension $n$. The maximal estimate in fact gives full lower as well as upper Minkowski dimension.

To prove that implication, let $E_{\delta }=\{x:\mathrm{dist}(x,E)<\delta \}$. Each unit segment in $E$ has a thinner tube contained in $E_{\delta }$, so ${\mathcal{K}}_{c\delta }{1}_{E_{\delta }}(e)\ge 1$ for every $e$, with a fixed dimensional $c>0$. Apply the maximal estimate to this indicator function:If $N_{\delta }(E)$ is the smallest number of radius-$\delta$ balls covering $E$, that cover, enlarged by a fixed factor, covers $E_{\delta }$. Thus $|E_{\delta }|{\lesssim }_n{\delta }^n{N}_{\delta }(E)$ andTaking the lower limit of $\log N_{\delta }(E)/\log (1/\delta )$ and then letting $\epsilon \downarrow 0$ gives lower Minkowski dimension at least $n$. Bounded subsets of ${\mathbb{R}}^n$ have upper Minkowski dimension at most $n$. Consequently both dimensions equal $n$, which proves the requested implication.

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.

Use the Fourier transform convention $\hat{\nu }(\xi )=\int e^{-ix\cdot \xi }d\nu (x)$ throughout this question. The decay assumption implies square integrability, since polar coordinates giveNear zero the integrand is bounded by $r$, and at infinity it is bounded by $r^{-1-2ϵ}$. The positive $ϵ$ is what makes the latter integrable.

By the Plancherel theorem, there is a function $g\in L^2({\mathbb{R}}^2)$ whose Fourier transform is $\hat{\mu }$. It is the density of $\mu$ with respect to Lebesgue measure. To justify this step rather than assume a density, for every Schwartz function $\phi$, Fourier inversion givesThe finite measure and the locally integrable function thus define the same tempered distribution, so they agree as measures: $d\mu =gdx$. In particular $g\ge 0$ almost everywhere and $\int g=\mu ({\mathbb{R}}^2)<\infty$. This is the L2 density from a square-integrable Fourier transform principle.

For $f\in L^{\infty }(\mu )$, the density of $fd\mu$ is $fg$. The inequality $|f|\le \Vert f{\Vert }_{L^{\infty }(\mu )}$ holds wherever $g>0$, apart from a Lebesgue measure zero set, soA second application of the Plancherel theorem yields the explicit boundThe implicit constant in the requested estimate may depend on the measure and its Fourier-decay bound, but is independent of $f$.

Parametrize the unit circle by $\omega (\theta )=(\cos \theta ,\sin \theta )$, with $-\pi \le \theta <\pi$ and $d\sigma =d\theta /(2\pi )$. Extract the constant phase at $\omega (0)$:On the support, $|\theta |\le 2\delta /C$. The frequency rectangle and the elementary bounds $|1-\cos \theta |\le {\theta }^2/2$, $|\sin \theta |\le |\theta |$ implyChoose the fixed constant $C$ large enough that this is less than $\pi /3$. Every phase then has real part at least $1/2$. Nonnegativity of $\psi$ prevents cancellation after this phase rotation, while its central plateau gives $\int \psi d\sigma \ge \delta /(\pi C)$. Hence the Fourier transform satisfiesThis is the circle cap Fourier lower bound. No upper bound on the values of $\psi$ is needed here; the support, plateau and nonnegativity suffice. The long radial scale ${\delta }^{-2}$ comes from the quadratic term $\cos \theta -1$, whereas the transverse scale ${\delta }^{-1}$ comes from the linear term $\sin \theta$.

Write $R_j$ for the rectangles, $x_j$ for their centers and $e_j$ for their long-axis directions. For the finite exponent $p$ in the displayed estimate, take smooth rotated cap functions ${\psi }_j$, with $0\le {\psi }_j\le 1$, equal to one on angular distance at most $\delta /C$ from $e_j$ and supported within $2\delta /C$. Choose $C$ sufficiently large once and for all. The direction separation makes these cap supports disjoint, and $\int |{\psi }_j{|}^{p}d\sigma \lesssim \delta$.

DefineThe Fourier modulation and translation identity gives the second equality. Rotating the circle cap Fourier lower bound then gives $|h_j(x)|\gtrsim \delta$ on $R_j$. The half-side lengths of $R_j$ are no larger than the two frequency bounds used in part (b).

Let ${\epsilon }_j$ be independent Rademacher random variables. Because the input cap supports are disjoint, for every choice of signswhere $M=\#\mathcal{R}$. Apply the assumed Fourier extension estimate to the sum. Average over signs and use the Khintchine inequality pointwise, followed by the Tonelli theorem:There is no requirement that the spatial rectangles be disjoint; disjointness is used only for the input caps on the unit circle. Their spatial overlaps are precisely what the square function measures. The cap lower bounds now implySince each rectangle has area ${\delta }^{-3}$, the restriction-to-rectangle overlap principle givesThe constants are independent of $\delta$, the centers and the collection. The finite-$p$ interpretation is the one for which the printed power integral is defined. A single cap also shows that the assumed diagonal Fourier extension estimate can hold only for $p\ge 4$: its output contributes at least ${\delta }^{p-3}$ to the $p$th-power norm, whereas its input contributes at most a constant times $\delta$.

Set $m=\lceil p/100\rceil$ and $d=m-1$, so that $d<p/100<p$. The dimension of a bounded-total-degree polynomial space in $n$ variables over ${\mathbb{F}}_p$ is $\left(\genfrac{}{}{0ex}{}{n+d}{n}\right)$. If $\#N$ were smaller than this dimension, evaluation at the points of $N$ would impose fewer homogeneous linear conditions than unknown coefficients. The rank-nullity theorem would give a nonzero multivariate polynomial $P$ of total degree at most $d$, vanishing on $N$.

For every $x$, select one of the promised rich affine lines in a vector space through $x$, and write it as $\ell =\{a+tv:t\in {\mathbb{F}}_p\}$ with $v\ne 0$. The polynomial restriction to a line $Q(t)=P(a+tv)$ has degree at most $d$, and at least $m=d+1$ distinct roots of a polynomial from $N\cap \ell$. By the root bound for a polynomial, $Q$ is identically zero, so $P(x)=0$. Since this works for every $x$, $P$ vanishes at all $p^n$ points of ${\mathbb{F}}_p^n$.

The Schwartz-Zippel lemma says that a nonzero multivariate polynomial of total degree $d$ has at most $d{p}^{n-1}$ zeros on this grid. Here $d<p$, so that count is strictly less than $p^n$, a contradiction. The distinction between a formal polynomial and its function on a finite field is crucial: our degree bound is what rules out a nonzero polynomial vanishing everywhere.

We have proved the stronger quantitative rich line covering bound over a finite fieldThus $\#N{\gtrsim }_n{p}^n$, as required. The implied constant may depend on the fixed dimension $n$. The very large numerical lower bound on $p$ is more than this proof needs.

For a finite collection $\mathcal{L}$ of distinct affine lines in a vector space in ${\mathbb{R}}^n$, $n\ge 2$, a joint is a point incident to $n$ lines whose direction vectors are linearly independent. The joints theorem assertsIn the customary three-dimensional formulation this is $\#J\lesssim L^{3/2}$, with three noncoplanar incident lines at every joint. We prove the general form, which includes that formulation.

Let $L=\#\mathcal{L}$ and $M=\#J$. The conclusion is immediate when $M=0$. Otherwise set $D=\lceil n{M}^{1/n}\rceil$. Suppose for contradiction that $M>LD$. Repeatedly delete any line incident to at most $D$ of the currently retained joints, deleting those joints at the same time. Each deleted line loses at most $D$ current joints, so even deleting all $L$ lines could lose at most $LD<M$ joints. Therefore the process must stop with a nonempty set $J^{\prime }$ and a line collection ${\mathcal{L}}^{\prime }$ such that each retained line contains more than $D$ retained joints. Every retained joint still has its original $n$ independent incident lines: if any line through it had been deleted, the joint would have been deleted too.

There is a nonzero multivariate polynomial of total degree at most $D$ vanishing on $J^{\prime }$, becauseChoose such a polynomial $P$ of smallest possible total degree $d\le D$. This is an application of the polynomial method in combinatorics. Every line of ${\mathcal{L}}^{\prime }$ contains more than $D\ge d$ roots of a polynomial of its polynomial restriction to a line, so $P$ vanishes identically on every such line.

At a retained joint $x$, differentiating along each of its independent line directions $v_1,\dots ,v_n$ gives $v_i\cdot \nabla P(x)=0$. Their linear independence therefore forces $\nabla P(x)=0$. Each partial derivative of $P$ vanishes on all of $J^{\prime }$ and has smaller total degree. Minimality of $d$ forces every partial derivative to be the zero polynomial. Over the real numbers, a polynomial with all partial derivatives zero is constant; a nonzero constant cannot vanish on the nonempty $J^{\prime }$. This is the required contradiction.

It follows that $M\le LD$. Since $D\le (n+1)M^{1/n}$ for $M\ge 1$,This proves the joints theorem by the pruning and minimal-degree polynomial argument.

The exponent is sharp. Take all axis-parallel lines passing through the grid $\{1,\dots ,m{\}}^n$. There are $n{m}^{n-1}$ distinct lines and $m^n$ joints; the coordinate directions span ${\mathbb{R}}^n$ at every grid point. Thus no smaller power of the number of lines can bound all joint configurations.