THE POLYNOMIAL METHOD

THE POLYNOMIAL METHOD

LECTURE 1

THE KAKEYA CONJECTURE IN FINITE FIELDS

MARCO VITTURI

Abstract. This series of notes is intended to provide an introduction to thepolynomial method through examples of its successful application - mainly inHarmonic Analysis, but other elds will also be considered (e.g. combinatorics,transcendence theory, etc). In this rst set of notes we introduce the Kakeyaproblem and then the analogous Kakeya problem in the nite elds. Thisintroduction will necessarily be brief and won't do justice to this vast andbeautiful eld. For a more comprehensive introduction we refer the readerto e.g. [KT02]. We then give the explicit construction of a Besicovitch set.Finally we present Dvir's solution to the Kakeya conjecture in the nite eldsby use of the polynomial method.

Contents

1. The Kakeya conjecture 12. The construction of Besicovitch 93. Dvir's proof 11Appendix A. Proof of lemma 2 14Appendix B. Multiplicity 15References 16

Notation. throughout these notes, |E| will denote the cardinality of the set Ewhen this is discrete and nite, and its Lebesgue measure otherwise. Context willusually suce to determine which is the case.The expression X À Y will mean that there is a constant C ¡ 0 s.t. X ¤ CY .We might specify the dependence of this constant on additional parameters e.g. αby writing X Àα Y when this is the case. By X Y we will mean X À Y andY À X.

1. The Kakeya conjecture

In 1917 Soichi Kakeya asked what was the minimal area of a region in which aunit segment could be turned around continuously by 360. This question becameknown as the Kakeya needle problem, and an answer came from a constructionof Besicovitch which is described in section 2 (originally appearing in [Bes19]).Namely, for every ε ¡ 0 there exists a set that contains a unit segment in everydirection but has area smaller than ε, and therefore there is no minimal area to sucha region - a result that was somewhat surprising for the time1. A generalization ofthe denitions leads to2

1It was believed a deltoid curve was a sharp example.2We drop the requirement that the unit segment can be turned continuously because it won't

be relevant to us.

1

2 MARCO VITTURI

Denition 1. A Kakeya set is a set E Rn that contains a unit segment in everydirection.

Thus the construction of Besicovitch shows that for n 2 these sets can have ar-bitrarily small Lebesgue measure - and indeed there are Kakeya sets with Lebesguemeasure exactly zero3. Moreover, by taking a solid of revolution with cross sectiona Besicovitch set it can be seen that we can build such sets with Lebesgue measurezero in dimensions n ¡ 2 too.A question arises then: what's the dimension of these sets? this question can ofcourse be asked with respect to various notions of dimension - e.g. Hausdor,upper/lower Minkowski, packing, etc. In 1971 Davies proved in [Dav71] that, al-though the Lebesgue measure can be zero, when n 2 a Kakeya set must alwayshave Hausdor dimension 2. This result led to the conjecture that the (Hausdoror Minkowski) dimension would still be n for any Kakeya set:

Conjecture 1 (The Kakeya Conjecture (for Hausdor dimension)). Let K be aKakeya set in Rn. Then dimHauspKq n.

Conjecture 2 (The Kakeya Conjecture (for Minkowski dimension)). Let K be aKakeya set in Rn. Then dimMinkpKq n.

There is an analogous conjecture with upper Minkowski dimension, of course.The conjectures are still wide open for n ¡ 2, although a huge number of partialresults exist in the literature. We don't attempt to survey them in here. See [KT02]for a summary up to the year 2000 (thus somewhat outdated, as it doesn't containthe result of Dvir presented in these notes). It is still open even in the weakercase4 of upper or lower Minkowski dimension. Recall that the upper Minkowskidimension is dened as

dimMinkpEq : lim supδÑ0

nlog |Eδ|

log δ,

where Eδ is the δ-neighbourhood of E, and analogously the lower Minkowski di-mension is dened as

dimMinkpEq : lim infδÑ0

nlog |Eδ|

log δ;

when they coincide, they are commonly referred to as the Minkowski dimension ofE.Here we assume that 0 δ ! 1. IfK is a Kakeya set in Rn then its δ-neighbourhoodcontains a δ-tube in every direction, where by δ-tube we mean exactly the δ-neighbourhood of a unit segment. Since the tubes have cross section of radiusδ it doesn't make sense to distinguish directions that dier by an angle less than δ(they become indistinguishable); therefore we might restrict ourselves to sets thatare unions of N distinct δ-tubes that have δ-separated directions, in the sense thatwhenever T, T 1 are δ-tubes with distinct directions ω, ω1 P Sn1 then =ω, ω1 Á δ.Notice the δ-separation condition forces N to be Opδ1nq, and so we can assume- and we will do so from now on - that the collection is critical in the sense thatN δ1n.With this setup we have that conjecture 2 is equivalent to

3They can be built by specializing the construction in section 2 to nested collections of paral-lelograms, for example, and then taking intersections.

4Weaker because dimHauspEq ¤ dimMinkpEq ¤ dimMinkpEq when they are all dened.

THE KAKEYA PROBLEM 3

Conjecture 2 (The Kakeya Conjecture (for Minkowski dimension)). Let tTjuNj1

be a collection of δ-tubes in Rn with δ-separated directions. Then for every ε ¡ 0

N¤j1

Tj Áε δε. (1)

Remark 1. At the heuristical level, conjecture 1 is saying that the δ-separatedδ-tubes are essentially disjoint, up to a logarithmic error. Indeed, if they wereexactly disjoint, then we'd have

|N¤j1

Tj | Nδn1 1.

We can now sketch the proof of Davies for the weaker conjecture 1. With a littlemore eort it can be turned in a proof for the Hausdor dimension. Observe that¸

j

χTj

2L2

¸j,k

|Tj X Tk|;

let the direction of tube Tj be given by ωj P S1, then

¸j,k

|Tj X Tk|

Opδ1q¸k1

¸j,k :

|=ωj ,ωk|kδ

|Tj X Tk|.

Now, it's a simple geometric fact that if the angle between Tj and Tk is θ then

|Tj X Tk| Àδ2

|θ|,

so that

Opδ1q¸k1

¸j,k :

|=ωj ,ωk|kδ

|Tj X Tk| À

Opδ1q¸k1

δ2

kδN

Opδ1q¸k1

1

k log δ1.

Thus by Cauchy-Schwarz

1 ¸j

χTj

2L1 ¤

¸j

χTj

2L2

χj Tj

2L2 À log δ1

¤j

Tj,

which gives the desired inequality.

There is also a stronger variant of the Kakeya conjecture(s) that is more directlyrelevant to harmonic analysts, namely the Kakeya maximal function conjecture.This can be stated in the language of δ-tubes too:

Conjecture 3 (The Kakeya maximal function conjecture). Let tTjuNj1 be a col-

lection of δ-tubes in Rn with δ-separated directions (as above). Then for everyε ¡ 0 N

j1

χTj

Lnpn1qpRnq

Àε δε. (2)

The conjecture can also be put in the more suggestive form

N

j1

χTj

Lnpn1qpRnq

Àε δε N

j1

|Tj |qpn1qn,

4 MARCO VITTURI

which is another assertion of essential disjointness of the tubes.A few observations are in order. Firstly, for 1 d ¤ n (not necessarily integer)notice that (2) implies the inequality

N

j1

χTj

Ldpd1qpRnq

Àε δ1n

dε. (3)

Indeed, we have the (sharp) trivial inequality

N

j1

χTj

L8pRnq

À δ1n,

thus if δ P r0, 1s is such that d1d p1 θqn1

n θ 18 , then by logarithmic Hölder

inequality

N

j1

χTj

Ldpd1qpRnq

Àε pδεq1θpδ1nqθ δOpεqδ1n

d .

Notice that nn1

dd1 for d n, so n

n1 is the endpoint of a family of estimatesreally.Secondly, it is stronger than conjecture 1 because it implies it; notice in particularthat the expression on the left in the conjectured inequality is not just measuringthe Lebesgue measure of the union but is taking into account the multiplicity ofoverlapping too, so this is to be expected. The implication goes as follows (im-plicitely already used above): by Hölder

N

j1

χTj

L1pRnq

¤ N

j1

χTj

Lnpn1qpRnq

χj T

j

LnpRnq

Àε δε N¤j1

Tj1n;

but then N

j1

χTj

L1pRnq

N

j1

|Tj | 1,

and thus (1) is implied. As it turns out, conjecture 3 also implies the Kakeyaconjecture for Hausdor dimension, but we won't comment further on this here.Thirdly, the name of the conjecture refers to an actual maximal function that ishidden in the statement through duality. The Kakeya maximal function is givenby

fδ pωq : supT δtube s.t. T ||ω

1

|T |

»T

|fpyq| dy,

where ω P Sn1 and T ||ω means that the tube T has direction parallel to thatof ω. We shall show that conjecture 3 is equivalent to the boundedness (up to alogarithm) of this maximal function in an appropriate range; but in order to do sowe introduce a small lemma rst.

Lemma 2. The estimate 3 is equivalent to

N

j1

ajχTj

Ldpd1qpRnq

Àε δε, (4)

uniformly in tajuNj1 P `

dpd1q s.t.°j a

dpd1qj À δ1n.

Proof. See appendix A for the ñ implication. For the full proof see [Mat15].


We see that by duality the estimate (4) is equivalent to

N

j1

aj

»Tj

|g| dy Àε δεgLnpRnq @g P Ln;

since this must hold for all collections of δ-separated δ-tubes then it's equivalent to

suptTjuj

N

j1

aj

»Tj

|g| dy Àε δεgLnpRnq,

where the supremum is taken over such collections. If Ω is a maximal set of δ-separated direction in Sn1 then the last expression on the left hand side is com-parable to ¸

ωPΩ

aω supT δtube s.t. T ||ω

»T

|g| dy,

where we have written aω for aj , with ω direction of Tj ; notice that

supT δtube s.t. T ||ω

»T

|g| dy δn1gδ pωq,

so taking the supremum in the sequences aj the above is equivalent to ¸ωPΩ

gδ pωqn1n

gδ LnpSn1q,

and therefore conjecture 3 is equivalent to

Conjecture 3 (the Kakeya maximal function conjecture). For all ε ¡ 0

fδ LnpSn1q Àε δεfLnpRnq.

Remark 2. By testing it against the characteristic function of a Besicovitch set asconstructed in section 2 we see that the inequality would certainly be false withoutthe logarithmic factor δε.

Finally, conjecture 3 can be restated equivalently in terms of just characteristicfunctions of sets that are supported inside a family of tubes. We include it forcompleteness.

Conjecture 3 (the Kakeya maximal function conjecture). Let tTjuNj1 be a collec-

tion of δ-tubes in Rn with δ-separated directions. Let 0 λ 1; if tEjuNj1 is a

collection of sets s.t. |Ej | ¡ λ|Tj | for all j, then for all ε ¡ 0 and all 1 ¤ d ¤ n

N¤j1

Ej Áε λdδndε.

The advantage of these formulations involving δ-tubes is that the combinatorialnature of the problem is immediately evident: we want to study the overlapping oftubes in space and prove that under our conditions they don't overlap much in onesense or another.

The Kakeya maximal function conjecture was solved for n 2 by Córdoba in[Cor77] by means of geometric arguments a little more sosticated than those in[Dav71], and in a dierent way by Bourgain in [Bou91]. A fundamental featureof the two dimensional problem is that the endpoint exponent npn 1q is just 2,which provides a nice geometrical interpretation as used above, namely that¸

j

χTj

2L2

¸j,k

|Tj X Tk|.

6 MARCO VITTURI

In subsequent years two conceptually simple but clever arguments appeared thatprovided lowerbounds for the dimension of a Kakeya set in any dimension n. Theseare

i) the Bush argument of Bourgain [Bou91], that provided the lowerbound

j Tj Á

δpn1q2; the idea behind the argument is that if

j Tj is small then there

must be a point covered by a high number of δ-tubes; but these tubes haveδ-separated directions and thus at distance 110 from the point they must bedisjoint from each other - thus having large total mass;

ii) the Hairbrush argument of Wol [Wol95], that improved the lowerbound toj Tj

Á δpn2q2; the idea is that instead of looking for just a point of highmultiplicity we look for a δ-tube of high multiplicity, in the sense that a largefraction of its points have high multiplicity. Then if the tubes that intersectsuch a tube are arranged in bushes we might argue somewhat as in the Bushargument (only with way more bushes now); if the tubes in distinct bushes arenot essentially disjoint the argument doesn't work as well, but one can rule outthis case separately.

Notice that by denition of Minkowski dimension the latter result implies that aKakeya set has (lower) Minkowski dimension at least n n2

2 n22 . This gives

52 for a 3-dimensional Kakeya set. Improving upon the Hairbrush argument provedextremely dicult - e.g. Katz, aba and Tao improved the lowerbound for thedimension to 5

2 1010 in [KT00] with a huge amount of eort (compared tothe simplicity of the Hairbrush argument). Since it seemed that the techniquesavailable at the time were not able to push this much further, Wol came up with abrilliant idea: to ask the analogous questions in the simpler setting of nite elds.Indeed, it is possible to dene Kakeya sets in Fnq (where q is a power of a prime) assets that contain an Fq-line in every direction. Lines in Fnq are of the form

`x,v tx tv s.t. t P Fqu

for x P Fnq , 0 v P Fnq Fq (i.e. v is identied with v1 i v tv1 for some t P Fq ).Thus the analogous of the Kakeya conjecture in this setting takes the form

Conjecture 4 (the Kakeya conjecture in nite elds). Let K be a Kakeya set inFnq . Then for every ε ¡ 0 K Áε |Fq|nε.

Since there are no non-trivial neighbourhoods in Fnq (it has the discrete topology),this is the only possible form of a Kakeya conjecture in the nite elds - there areno tubes-equivalents because there are no tubes. An equivalent way to look at thisis to say that in the nite elds there are no scales - as opposed to the Rn case,where there is a continuum of scales.The insight behind this proposal of Wol is that many of the arguments givenfor the real Kakeya case adapt to the nite elds case nicely and become actuallysomewhat cleaner, while mantaining the same role for the main ideas used. Thehope is then that an improvement or solution in the simpler nite elds case wouldshed light on the real case and perhaps translate to a corresponding solution too.To support this point we present briey the aforementioned Bush and Hairbrusharguments in Fnq . We write F in place of Fq with the understanding that the niteeld is xed.

i) the Bush argument: let µ be a xed multiplicity parameter to be chosen later;then either there is a point p P K such that there are µ lines in K passingthrough p or for every point in K there are fewer than µ lines passing throughit. In the former case, since the lines have distinct directions, they must become


disjoint when the point p is removed; this implies

|K| Á µ|F|.

In the latter case, by double counting

|F||F|n1 ¸pPK

¸`PK

1pp P `q ¤ µ|K|,

which gives the lower bound

|K| ¥|F|n

µ.

By optimizing in µ we get (for µ |F|pn1q2)

|K| Á |F|pn1q2.

ii) the Hairbrush argument: rst notice that by repeating almost verbatim5 theargument of Davies given above for the Kakeya conjecture (with Minkowskidimension) in dimension n 2 we get that if K is a Kakeya set in F2 then

|K| Á |F|2.

By being slightly more careful we might see that if Km is a set of F2 thatcontains a line in m distinct dimensions then actually

|Km| Á m|F|. (5)

Let then µ be a xed multiplicity parameter to be chosen later. We say thata line ` has high multiplicity if for at least |F|2 points p P ` there are at leastµ lines (distinct from `) in K passing through p. Then either there exists a

line of high multiplicity or it doesn't. In the former case, let ˜ be such a lineof high multiplicity, and consider the family Π of 2-dimensional planes passingthrough ˜. If a line ` intersects with ˜ then there is a unique π P Π s.t. `, ˜ π.Let Lπ denote the set t` π such that ` intersects ˜u. Then K X π is a setin (a isomorphic copy of) F2 that contains at least |Lπ| lines, and thus by the2-dimensional Kakeya estimate (5) above

|K X π| Á |Lπ||F|;

therefore

|K| ¥¸πPΠ

|pK X πqz˜| Á |F|¸πPΠ

|Lπ| Á µ|F|2,

where the last inequality is due to the fact that since ˜ has high multiplicitythen it intersects at least µ|F|2 lines (they must all be distinct).In the case there are no lines of high multiplicity, let

K 1 : tp P K such that p belongs to at most µ lines in Ku;

notice that, by assumption, for any line ` K we have |K 1 X `| ¡ |F|2.Therefore, by double counting as before

|K| ¥ |K 1| ¥

°pPK1

°`PK 1pp P `q

µ

1

µ

¸`PK

|K 1 X `| Á1

µ|F|n1|F|.

Thus we have the estimates

|K| Á µ|F|2 or |K| Á |F|nµ;

5The bound on the intersection of tubes becomes simply |`X `1| ¤ 1.

8 MARCO VITTURI

optimizing in µ we see that we get

|K| Á |F|pn2q2.

Being very promising, the nite elds case was intensively studied, most notably in[MT04], but still without substantial advancements. It then came as a big surprisethat Dvir, in 2008, was able to prove the Kakeya conjecture in nite elds by asimple application of the polynomial method - without logarithmic factors, too. Insection 3 we will present his beautifully simple proof as an illustration of the powerof the polynomial method.

Remark 3. We haven't yet mentioned one of the major results of the eld, namelythe solution of the Multilinear Kakeya conjecture. We introduce it briey (for now)because it provided a fertile ground for the study of polynomial method techniquesand is thus very much relevant to the topic of these notes.We can imagine writing, in the estimate in the Kakeya maximal function conjecture,the left hand side as¸

j

χTj

npn1q

Lnpn1q

»Rn

n¹k1

¸j

χTj

1pn1q

dx;

the multilinear Kakeya conjecture may be thought of as a somewhat restrictedversion of the Kakeya maximal function conjecture, where the factors in the productabove are allowed to be distinct but each sub-collection of tubes is assumed to beall oriented in roughly the same direction (within Op1q of that of a coordinate axis)and these directions must be distinct for distinct families. More precisely

Conjecture 5 (the Multilinear Kakeya conjecture). Let tTj,auNj

a1 for j 1, . . . , nbe n families of δ-tubes with δ-separated directions, such that the directions of the

tubes in tTj,auNj

a1 form an angle of at most p10nq1 with the coordinate vector ej.Then for every ε ¡ 0»

Rn

n¹j1

Nj¸a1

χTj,a

1pn1q

dx Àε δε

n¹j1

pδN1pn1qj q.

Notice repetitions within a family are allowed. As before, this estimate can beseen as the endpoint of a family of estimates for a range of exponents; namely fornn1 ¤ p ¤ 8 the corresponding estimate would be

n¹j1

Nj¸a1

χTj,a

Lpn

Àε δε

n¹j1

pδnpNjq.

This conjecture was solved by Bennett, Carbery and Tao in [BCT06], except forthe endpoint. Amongst the techniques employed in the proof are the heat owmethod and induction on scales6. The endpoint case was later solved too, byGuth, in [Gut10] - and the solution was directly inspired by Dvir's proof for the(linear) Kakeya conjecture in nite elds! Now, adapting polynomial methods to themultilinear Kakeya case is absolutely not straightforward and Guth ended up usinglarge amounts of advanced algebraic geometry and algebraic topology tools. Sometime later though, Carbery and Valdimarsson were able in [CV13] to successfullyreprove the endpoint result with a proof that reduced all of the algebraic topologyneeded to just the Borsuk-Ulam theorem - which is needed to prove the polynomialHam Sandwich theorem. Informally, the polynomial Ham Sandwich theorem states

6This turns out to be the crucial point, since recently Guth has given a short proof of theMultilinear Kakeya inequality (except for the endpoint) using nothing but an induction on scalesand the Loomis-Whitney inequality. See [Gut15]


that given N bodies in Rn we can nd a polynomial in RrX1, . . . , Xns of degree atmost OpN1nq s.t. the hypersurface given by its zero set ZpP q exactly bisects allof the N bodies.The next set of notes will contain a more extensive discussion of the polynomial HamSandwich theorem and its applications, as it's one of the gems of the polynomialmethod.

Remark 4. We have not mentioned the relationship between the Kakeya conjec-ture and one of the other big open problems in Harmonic Analysis - the Restrictionconjecture. The two are deeply related in extremely interesting ways, but a discus-sion of this topic would take us too much o track. Suces to say that in generalthe Restriction conjecture implies the Kakeya conjecture, but partial implicationsin the other direction are also possible, as the Kakeya conjecture can provide usefulestimates on oscillatory integrals. We direct the interested reader to the surveys in[Tao01] and [Wol99].

2. The construction of Besicovitch

We describe in this section a geometric construction that for every ε ¡ 0 allowsone to produce a set E containing a unit segment in every direction (in a xed arcof S1) with |E| ε in a nite number of operations. The resulting set is calleda Besicovitch set, but the construction presented here is due to Perron7. It's asimplication of Besicovitch's original one.

Fix a parameter 1 ¡ λ ¡ 12. Take then a triangle ABC of height 1, and letAB be its basis and M the middle point of AB. Consider the two sub-trianglesAMC and MBC and slide one onto the other along the direction of the basis untiltheir respective bases partially overlap by a certain amount specied as follows: theratio of the total length of the overlapping bases to the length AB is equal to λ.Call the resulting gure T 1

λpABCq. Notice that T1λpABCq still contains (copies of)

all the directions of unit segments in ABC. See g. 2 The second iteration goes as

A M

C

B A M

C

M' B

C'

E

F G

Figure 1. The rst application, with ABC on the left andT 1λpABCq on the right.

follows. Subdivide ABC into AMC and MBC as before and apply T 1λ separately

to these two sub-triangles; then slide the resulting T 1λpAMCq and T 1

λpMBCq ontoeach other along the direction of AB in such a way that the total length of theoverlapping bases of the gures is λ times the sum of their separate lengths. CallT 2λpABCq the resulting gure. Notice the basis has now length λ2 times that ofAB.For the n-th iteration, do as follows. Subdivide AB into AMC and MBC as

7the gure produced is also referred to as a Perron tree in the literature.

10 MARCO VITTURI

before and apply Tn1λ separately to these two sub-triangles; then slide the resulting

Tn1λ pAMCq and Tn1

λ pMBCq onto each other along the direction of AB in sucha way that the total length of the overlapping bases of the gures is λ times thesum of their separate lengths. Call Tnλ pABCq the resulting gure. And so on.Now, since it's obvious that the set of directions is invariant under Tnλ , it remains

Figure 2. The resulting gure after the second and third iteration(T 2λpABCq and T

3λpABCq respectively; unlabeled).

to prove that the area of Tnλ pABCq goes to 0 as n goes to innity. With referenceto gure 2, consider T 1

λpABCq. Some simple considerations of euclidean geometryshow that ABE is similar to the original triangle ABC and the sidelength ratio isλ; moreover, the triangles C 1EF and CGE are each union of a triangle similar toAMC and one similar to MBC, with ratio p1 λq (draw a line through E parallelto AB). Therefore we have for the area

|T 1λpABCq| pλ2 2p1 λq2q|ABC|.

Analogous simple geometrical considerations allow one to conclude for the generalcase that

|Tnλ pABCq| ¤ pλ2n 2p1 λ2q 2p1 λ2qλ2 . . . 2p1 λ2qλ2n2q|ABC|

¤ pλn 2p1 λqq|ABC|,

which can be made arbitrarily small by choosing λ suciently close to 1 and nsuciently large. See [Ste93], ch. X, for additional detail.

Remark 5. The phenomenal construction of Besicovitch has found many appli-cations in Harmonic Analysis beyond the Kakeya needle problem. Indeed, theconstruction was originally devised by Besicovitch (in 1917) in order to answer aquestion pertaining Riemann-integrability: namely, is it true that if f is Riemann-integrable in the plane then we can nd two orthogonal axes with respect to whichx ÞÑ fpx, yq is Riemann-integrable and y ÞÑ

³fpx, yq dx is Riemann-integrable?

The answer is negative, since one can use the Besicovitch set to build a counterex-ample (see e.g. [Fal86] for details).Another extraordinary application of Besicovitch sets was found by C. Feerman[Fef71], who used one to disprove the ball-multiplier conjecture, proving that the

multiplier dened by xTf χBp0,1qf is bounded only8 on L2. This in particular im-plies (though it's not entirely straightforward it does) the following negative resultfor spherical summation of double (and higher) Fourier series: if f P LppT2q and

f ¸

m,nPNfpm,nqe2πipmxnyq

8It was believed that T would be bounded on Lp for all 43 p 4.


is its Fourier series, then

limRÑ8

¸m,nPN

m2n2 R

fpm,nqe2πipmxnyq

need not converge to f in Lp-norm when p 2. This is in sharp contrast with theone dimensional case, in which one has Lp convergence for all 1 p 8. Evenworse perhaps, by a similar argument, it was shown in [MN73] that for any p 2one can build a function f P LppT2q s.t.

lim supRÑ8

¸m,nPN

m2n2 R

fpm,nqe2πipmxnyq 8 a.e.,

a result again in sharp contrast with the theory for the one dimensional case, whereby Carleson-Hunt theorem we have a.e. convergence for all 1 p 8.Another (much simpler) application of the Besicovitch set is in proving that themaximal function

Mfpxq : supRQx

1

|R|

»R

|fpyq| dy,

where the supremum ranges over the collection R of all rectangles of arbitrary ori-entation and arbitrary sidelength, is unbounded on any Lp except p 8 (see again[Ste93], ch. X for details). This in particular implies the failure of dierentiabilityof Lp functions along the collection R, i.e.

limdiampRqÑ0,

RPR

1

|R|

»R

fpx yq dy

need not tend to fpxq for a.e. x.

3. Dvir's proof

Now we nally come to the anticipated result of Dvir. In order to make thepresentation clearer, we introduce some lemmas beforehand.

Let F be a eld. For P a polynomial in FrX1, . . . , Xns we denote by ZFpP q thezero set of P , i.e.

ZFpP q tx P Fn s.t. P pxq 0u.

For a general eld F we have

Lemma 3 (Fundamental fact). If E Fn has cardinality |E| dnn

then there

exists a (non-zero) polynomial P P FrX1, . . . , Xns of degree degpP q ¤ d such thatE ZFpP q.

Notice thatdnn

is the dimension of the vector space of polynomials in FrX1, . . . , Xns

that have degree at most d, denoted FrX1, . . . , Xns¤d. This simple fact can be seen

as follows: the number of monomials Xj11 Xjn

n with degree k is the number of

solutions to j1 . . . jn k, which isnk1n

. Summing in k ¤ d one obtains the

claim by known properties of binomial coecients.

Proof. This is just the consequence of a simple linear algebra fact. Consider thespace of polynomials in FrX1, . . . , Xns of degree at most d; this has dimension

dnn

.

Therefore the conditions P px1, . . . , xnq 0 for all px1, . . . , xnq P E dene a system

of |E| linear equations indnn

variables (the coecients of the polynomial), and

since |E| dnn

the system has a non-trivial solution.

12 MARCO VITTURI

Remark 6. Observe the trivial but useful estimates

dn

n!¤

d n

n

¤pd nqn

n!.

Another fact needed in the proof is the following estimate on the number ofzeroes of P P FrX1, . . . , Xns, where F is now a nite eld. It generalizes the onedimensional estimate |ZFpP q| ¤ degpP q and agrees with the intuition that ZpP qshould behave like a hypersurface.

Lemma 4. Let F Fq be a nite eld and P P FrX1, . . . , Xns be a polynomial ofdegree d that doesn't vanish identically. Then

|ZFpP q| ¤ d|F|n1.

Proof. The proof is by induction, the case n 1 being well known. Thus, supposethe lemma is true for dimension n 1. For xed t P F, if PtpX1, . . . , Xn1q :P pX1, . . . , Xn1, tq vanishes then P pX1, . . . , Xnq pXn tqQpX1, . . . , Xnq, wheredegpQq ¤ d 1. If t1 t moreover Pt1 vanishes if and only if Qt1 does. So we canobtain a factorization9

P pX1, . . . , Xnq RpX1, . . . , Xnq¹tPE

pXn tq,

where |E| ¤ d and Rt does not vanish identically for all t P E (notice degpRq ¤d |E|). Therefore by inductive hypothesis

|ZFpP q| ¤ |¤tPE

Fn1 ttu| |¤tRE

ZFpRtq ttu|

¤ |F|n1|E| ¸tRE

pd |E|q|F|n2 ¤ d|F|n1.

With these lemmas we can then prove the Kakeya conjecture in nite elds.

Theorem 5 (Dvir, [Dvi09]). Let K be a Kakeya set in Fnq . Then

|K| Án |Fq|n.

The constant arising from Dvir's argument is 1n!. It is worth noticing thatthis constant can be improved to the much larger 2n by repeating the proof belowwhile postulating higher multiplicities for the points in the Kakeya set (i.e. not justcontained in ZpP q but with order much greater than 1). This was done by Dvir,Kopparty, Saraf and Sudan in [DKSS13]; a proof is included in appendix B.

Proof. We write F for Fq for ease of notation. Suppose by contradiction that

|K| |F|2n

n

. Therefore by the fundamental Lemma 3 there exists a polynomial

P P FrX1, . . . , Xns of degree d : degpP q ¤ |F| 2 such that K ZFpP q. Let

cj1,...,jn be the coecient of the monomial Xj11 Xjn

n in P .Now we add a ctious coordinate to our space and consider the homogeneizedpolynomial (of homogeneous degree d)

PhpX0, X1, . . . , Xnq :¸

j1,...,jn

cj1,...,jnXdj1...jn0 Xj1

1 Xjnn .

The homogeneity implies that for λ P F it is Phpλx0, λxq λdPhpx0, xq. SincePhp1, xq P pxq, if x P ZFpP q it follows that pλ, λxq P ZFpPhq for all λ P F. If` ta tv s.t. t P Fu is a line contained in the Kakeya set K then the pointspλ, λa λtvq for t P F, λ P F are all contained in ZFpPhq. Geometrically this is a

9Note some t's may be repeated.


plane with a line removed, but we claim the line is contained in the zero set of Phtoo. Indeed, restrict the polynomial Ph to this plane, thus obtaining a polynomialQpλ, tq of degree at most d s.t. the zero set has cardinality at least |F|p|F|1q. But|F|p|F| 1q ¡ d|F| since d ¤ |F| 2, and thus by the contrapositive of Lemma 4 thepolynomial Q must be the zero polynomial, i.e. ZFpPhq contains the entire planepλ, λa tvq for t, λ P F.We have thus proven that for all directions v the zero set ZFpPhq contains p0, tvqfor all t P F, i.e. the hyperplane t0uFn is all contained in ZFpPhq. In other words,Php0, xq 0 for all x P Fn; but Php0, X1, . . . , Xnq is the sum of the monomials oftop degree in P . This is a (non-trivial) polynomial of degree d |F| that vanisheseverywhere, and this is impossible by Lemma 4.It then follows that

|K| ¥

|F| 2 n

n

¥p|F| 2qn

n!Án |F|n,

as claimed.

Let's briey review the strategy of the proof above, as it contains the essence ofthe polynomial method: we assumed the cardinality of K was small, and thereforewe could nd a polynomial P of small degree that vanished on K; then, using thestructure of K, we proved that the polynomial must vanish somewhere else too;nally, we argued that the polynomial can't vanish on such a large set if it hassmall degree.

The above proof is remarkably simple, ultimately relying on extremely simplefacts of linear algebra such as expressed in Lemma 3 and Lemma 4. Thus theKakeya conjecture in the nite elds turns out to be remarkably simpler than thecorresponding question in the real case F R. The dierence here is in the factthat the space Fnq has a trivial topology (the discrete one), as mentioned before,and in particular it lacks scales completely. Also mentioned before was the factthat scales are not an obstacle but can in fact be put to good use, as has been donefor Multilinear Kakeya ([BCT06],[Gut15]). Thus we are seeing that the nite eldscase and the real case diverge when it comes to their topology - which therefore issuspected to play a big role in understanding the latter.In the next series of notes we will illustrate this point further by considering thefundamental result of incidence theory known as Szemerédi-Trotter theorem10 forboth the nite elds and the real plane. In the former case the best estimate possibleis far weaker than that one can get in R2, and the reason again is topological - therecan be a bigger number of incidences in F2 because there are fewer (none actually)topological obstructions. In particular, we will give a proof of the Szemerédi-Trottertheorem in the real plane by using polynomial partitioning, which is a consequenceof the polynomial Ham Sandwich theorem.

Remark 7. It is worth mentioning that the analogue of the Kakeya maximal func-tion conjecture in nite elds has been solved too, by Ellenberg, Oberlin and Tao(see [EOT10]). As is to be expected, the proof makes heavy use of the polynomialmethod. We state the conjecture briey.The Kakeya maximal function in a nite eld F Fq is given by

fpvq : sup`||v

¸xP`

|fpxq|,

10For a given set of points and a set of lines, the theorem bounds the maximum number ofpoint-line incidences - pp, `q s.t. p P ` - in terms of the number of points and the number of lines.

14 MARCO VITTURI

where v ranges in the directions of Fn, i.e. FnF PFn1 (the projective space).Then the Kakeya maximal function conjecture (or more appropriately theorem)says that

f`npPFnq Àn |F|pn1qnf`npFnq.

Appendix A. Proof of lemma 2

First of all observe that there is a slightly more precise version of the Kakeyamaximal function conjecture which we have left unstated. If N is the number ofδ-separated δ-tubes, then the claim is that for 1 ¤ d ¤ n

N

j1

χTj

Ldpd1q Àε δ

εpNδn1qpd1qd

(notice that Nδn1 would be the volume of the tubes if they were disjoint). Thetwo are equivalent, although this is not immediate. See [Mat15] for details. Assumethe inequality above then.On the one hand

N

j1

δn1χTj

Ldpd1qpRnq

Àε δnndε À 1,

and on the other hand aj ¤ N pd1qd δpd1qpn1qd. Therefore, it suces to

consider those j's such that δn1 ¤ aj ¤ δpn1qpd1qd. Thus, dene Apkq :tj s.t. aj 2ku, so that

¸j

ajχTj

cn,d log δ1¸kcn log δ

2k¸

jPApkq

χTj ;

therefore

N

j1

ajχTj

Ldpd1qpRnq

À


2k ¸jPApkq

χTj

Ldpd1qpRnq

Àε


2kδεp|Apkq|δn1qpd1qd.

Notice that |Apkq|2kdpd1q À°j a

dpd1qj ¤ δ1n, and therefore |Apkq|pd1qd À

δp1nqpd1qd2k. Therefore the right hand side in the last inequality is bounded by


2kδεδp1nqpd1qd2kδpn1qpd1qd


δε n,d,ε δOpεq.


Appendix B. Multiplicity

In one variable the multiplicity of a zero z of P pXq P FrXs is dened as themaximal m s.t. pXzqm divides P pXq. We denote this fact by ordzpP q m. Thisis equivalent to say that the Hasse derivatives11 P D0P,DP,D2P, . . . ,DPm1

all vanish in z, and this denition has the advantage of being well-posed even whenthere are multiple variables; therefore we dene the multiplicity of P in p, denotedordppP q, as the largest m s.t. Di1,...,inP ppq 0 for all i1 . . . in m. Noticethat since by Taylor expansion

P pX1, . . . , Xnq ¸

i1,...,in

Di1,...,inP ppqpX1 p1qi1 pXn pnq

in ,

we can deduce that the multiplicity has the multiplicative property

ordppPQq ordppP q ordppQq.

We can modify the proofs of the Lemmas 3 and 4 very slightly and obtain thefollowing versions with multiplicity:

Lemma 3 (with multiplicity). If E Fn is a set with associated tcpupPE s.t.°pPE

cp1n

n

¤dnn

, then there exists a (non-zero) polynomial P P FrX1, . . . , Xns

of degree degpP q ¤ d s.t. for all p P E it is ordppP q ¥ cp.

Lemma 4 (with multiplicity). Let F Fq be a nite eld and P P FrX1, . . . , Xnsbe a polynomial of degree d that doesn't vanish identically. Then¸

pPFn

ordppP q ¤ d|F|n1.

Proof. Left to the reader.

Finally, with these two tools, we can improve the constant in the estimate forthe cardinality of a Kakeya set as stated above. Indeed, let K be a Kakeya set inFnq and suppose that |K| |F|n2n; we x 1 ¤ l ¤ m d to be chosen later, andassume that

|K|pm 1 nqn

n! dn

n!,

so that in particular it holds that

|K|

m 1 n

n

d n

n

;

so by Lemma 3 with multiplicity we can nd a polynomial P P FrX1, . . . , Xns ofdegree at most d s.t. ordppP q ¥ m for all p P K. For the following, we choose d so

that the above inequality is tight, i.e. we choose d pm nq|K|1n.We denote by i a multi-index pi1, . . . , inq P Nn and by |i| : i1 . . . in. Now, let` be an Fq-line contained in the Kakeya set K, then for every p P ` and for everymulti-index i s.t. |i| ¤ l we have that ordppD

iP q ¥ m |i|. Thus, by Lemma 3with multiplicity applied to the restriction of P to `, either

|F|pm |i|q ¸pP`

ordppDiP q ¤ d |i| (6)

or` ZFpD

iP q.

11Recall that the Hasse derivatives of P P FrX1, . . . , Xns are dened on monomials by

Di1,...,inXj1

1 Xjnn

#0 if jk ik for some k,j1i1

jnin

Xj1i1 Xjnin

n otherwise,

and then extended to arbitrary polynomials by linearity; thus they coincide with the ordinaryderivatives when the eld is R.

16 MARCO VITTURI

By choosing m and l so that (6) is violated then we conclude that K is containedin ZpDiP q for all i s.t. |i| ¤ l. Therefore we can argue as in the proof of Dvir andconclude that for P0, the top degree part of P , it holds thatD

iP0 vanishes identically(in other words, P0 vanishes identically to order l). Now, choose m

X1110 l\and

choose l s.t. (6) is indeed violated. Then by Lemma 4 with multiplicity applied toP0,

l|F|n ¤¸pPFn

ordppP0q ¤ d|F|n1;

but we claim this is a contradiction. Indeed, by our choice of d this would imply

l|F|n p11

10l nq

|F|2|F|n1,

which is false far suciently large l. Therefore we have just proven

Theorem (Improved Dvir bound). Let K be a Kakeya set in Fnq , then

|K| ¥1

2n|F|n.

Remark 8. In the proof above any choice of m tαlu with 1 α 2 would work.

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Marco Vitturi, Room 4606, James Clerk Maxwell Building, University of Edin-

burgh, Peter Guthrie Tait Road, Edinburgh, EH9 3FD.

E-mail address: [email protected]

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