In this post I would like to make some technical notes on a standard reduction used in the (Euclidean, maximal) Kakeya problem, known as the *two ends reduction*. This reduction (which takes advantage of the approximate scale-invariance of the Kakeya problem) was introduced by Wolff, and has since been used many times, both for the Kakeya problem and in other similar problems (e.g. by Jim Wright and myself to study curved Radon-like transforms). I was asked about it recently, so I thought I would describe the trick here. As an application I give a proof of the case of the Kakeya maximal conjecture.

As discussed in the previous post, the *Kakeya maximal function conjecture* in can be formulated as follows:

Conjecture 1 (Kakeya maximal function conjecture)If , , and is a collection of tubes oriented in a -separated set of directions, then

A standard duality argument shows that (1) is equivalent to the estimate

for arbitrary non-negative measurable functions ; breaking up into level sets via dyadic decomposition, this estimate is in turn equivalent to the estimate

for arbitrary measurable sets . This estimate is then equivalent to the following:

Conjecture 2 (Kakeya maximal function conjecture, second version)If , , is a collection of tubes oriented in a -separated set of directions, and is a measurable set such that for all , thenfor all .

Indeed, to deduce (2) from Conjecture 2 one can perform another dyadic decomposition, this time based on the dyadic range of the densities . Conversely, (2) implies Conjecture 2 in the case , and the remaining case can then be deduced by the random rotations trick (discussed in this earlier post).

We can reformulate the conjecture again slightly:

Conjecture 3 (Kakeya maximal function conjecture, third version)Let , , and is a collection of tubes oriented in a -separated set of directions with . For each , let be a set with . Thenfor all .

We remark that (the Minkowski dimension version of) the Kakeya set conjecture essentially corresponds to the case of Conjecture 3, while the Hausdorff dimension can be shown to be implied by the case where (actually any lower bound here which is dyadically summable in would suffice). Thus, while the Kakeya set conjecture is concerned with how small one can make unions of tubes , the Kakeya maximal function conjecture is concerned with how small one can make unions of *portions* of tubes , where the density of the tubes are fixed.

A key technical problem in the Euclidean setting (which is not present in the finite field case), is that the portions of may be concentrated in only a small portion of the tube, e.g. they could fill up a subtube, rather than being dispersed uniformly throughout the tube. Because of this, the set could be crammed into a far tighter space than one would ideally like. Fortunately, the *two ends reduction* allows one to eliminate this possibility, letting one only consider portions which are not concentrated on just one end of the tube or another, but occupy both ends of the tube in some sense. A more precise version of this is as follows.

Definition 4 (Two ends condition)Let be a subset of , and let . We say that obeys thetwo ends conditionwith exponent if one has the boundfor all balls in (note that the bound is only nontrivial when ).

Informally, the two ends condition asserts that cannot concentrate in a small ball; it implies for instance that the diameter of is .

We now have

Proposition 5 (Two ends reduction)To prove Conjecture 3 for a fixed value of and , it suffices to prove it under the assumption that the sets all obey the two ends condition with exponent , for any fixed value of .

The key tool used to prove this proposition is

Lemma 6 (Every set has a large rescaled two-ends piece)Let be a set of positive measure and diameter , and let . Then there exists a ball of radius such thatand

for all other balls .

*Proof:* Consider the problem of maximising the quantity among al balls of radius at most the diameter of . On the one hand, this quantity can be at least , simply by taking equal to the smallest ball containing . On the other hand, using the trivial bound we see that the quantity is bounded. Thus the supremum of the is finite. If we pick a ball which comes within a factor of (say) of realising this supremum then the claim easily follows. (Actually one can even attain the supremum exactly by a compactness argument, though this is not necessary for our applications.)

One can view the quantity in the above lemma as describing the “width” of the set ; this is the viewpoint taken for instance in my paper with Jim Wright.

Now we prove Proposition 5.

*Proof:* Suppose Conjecture 3 has already been proven (assuming the two ends condition with exponent ) for some value of , and some small value of . Now suppose we have the setup of Conjecture 3 without the two-ends condition.

The first observation is that the claim is easy when . Indeed, in this case we can just bound from below the volume of a single tube. So we may assume that is much greater than .

Let be arbitrary. We apply Lemma 6 to each , to find a ball such that

for all . From (3) and the fact that , as well as the trivial bound , we obtain the lower bound . Thus there are only about possible dyadic ranges . Using the pigeonhole principle (refining the number of tubes by a factor of ), we may assume that there is a single such that *all* of the lie in the same dyadic range .

The intersection of with is then contained in a tube , and occupies a fraction

of . If we then rescale each of the and by , we can locate subsets of -tubes of density . These tubes have cardinality (the loss here is due to the use of the pigeonhole principle earlier) and occupy a -separated set of directions, but after refining these tubes a bit we may assume that they instead occupy a -separated set of directions, at the expense of cutting the cardinality down to or so. Furthermore, by construction the obey the two-ends condition at exponent . Applying the hypothesis that Conjecture 3 holds for such sets, we conclude that

which on undoing the rescaling by gives

Since was arbitrary, the claim follows.

To give an idea of how this two-ends reduction is used, we give a quick application of it:

Proposition 7The Kakeya maximal function conjecture is true for .

*Proof:* We use the “bush” argument of Bourgain. By the above reductions, it suffices to establish the bound

whenever , and are subsets of tubes in -separated directions with density and obeying the two-ends condition with exponent .

Let be the maximum multiplicity of the , i.e. . On the one hand, we clearly have

This bound is good when is small. What if is large? Then there exists a point which is contained in of the , and hence also contained in (at least) of the tubes . These tubes form a “bush” centred at , but the portions of that tube near the centre of the bush have high overlap. However, the two-ends condition can be used to finesse this issue. Indeed, that condition ensures that for each involved in this bush, we have

for some , and thus

The -separated nature of the tubes implies that the maximum overlap of the portion of the tubes in the bush away from the origin is , and so

Thus we have two different lower bounds for , namely and . Taking the geometric mean of these bounds to eliminate the unknown multiplicity , we obtain

which certainly implies the desired bound since .

Remark 1Note that the two-ends condition actually proved abetterbound than what was actually needed for the Kakeya conjecture, in that the power of was more favourable than necessary. However, this gain disappears under the rescaling argument used in the proof of Proposition 5. Nevertheless, this does illustrate one of the advantages of employing the two-ends reduction; the bounds one gets upon doing so tend to be better (especially for small values of ) than what one would have had without it, and so getting the right bound tends to be a bit easier in such cases. Note though that for the Kakeya set problem, where is essentially , the two-ends reduction is basically redundant.

Remark 2One technical drawback to using the two-ends reduction is that if at some later stage one needs to refine the sets to smaller sets, then one may lose the two-ends property. However, one could invoke the arguments used in Proposition 5 to recover this property again by refining further. One may then lose some other property by this further refinement, but one convenient trick that allows one to take advantage of multiple refinements simultaneously is to iteratively refine the various sets involved and use the pigeonhole principle to find some place along this iteration where all relevant statistics of the system (e.g. the “width” of the ) stabilise (here one needs some sort of monotonicity property to obtain this stabilisation). This type of trick was introduced by Wolff and has been used in several subsequent papers, for instance in this paper of myself and Izabella Laba.

## 12 comments

Comments feed for this article

15 May, 2009 at 3:12 pm

AnonThanks for the great post. two quick questions:

1. Proposition 7 states the conjecture is true when d > (n+1)/2. If I am reading this correctly one hopes for n> d > n- \epsilon. What is the best known today?

2. Is there a nice way to see the bound (1) fails when d=n? (i.e. the conjecture is best possible)

17 May, 2009 at 11:32 am

Terence TaoOops, there was a typo: Proposition 7 applies for , not . As for the “world records”, the survey of Nets and I at http://front.math.ucdavis.edu/math.CA/0010069 covers most of the known Euclidean results, though it predates the heat flow methods, the algebraic geometry methods, and the algebraic topology methods mentioned in my earlier post.

The existence of Besicovitch sets with arbitrarily small measure can be used to disprove the d=n endpoint of (1) (with no epsilon). Because of the presence of the epsilon, the conjecture is closed in d: if it holds for all , it will also hold for (by interpolating with some trivial estimate that loses some power of ).

19 May, 2009 at 1:58 am

small boyi have no idea what this post was about :)

20 May, 2009 at 11:04 am

shuanglin shaoDear Terry,

Two questions and one corrections.

1. Can you say a little more on the derivation of the estimate (2) from Conjecture 2?

2. If formulating this conjecture in terms of

for the maximal function

.

Then the translation and rotation symmetries are quickly seen. How to reflect these informaton in the formulatons of Conjecture 1, ect? Does it suffice by saying that: fixing a set E; then for any collection of tubes, …?

(3) In the last 8 lines in the argument of Proposition 5, and

may need to be exchanged.

20 May, 2009 at 12:15 pm

Terence TaoDear Shuanglin,

To deduce (2) from conjecture 2, first observe that the claim is easy if is very big or very small (e.g. larger than or less than for some large C depending only on n) because one can use trivial estimates such as and in those cases.

For similar reasons, one can also throw away those tubes for which is very small (less than for some large C). Once one does all this, there are now only a logarithmic number of ranges that the could fall into, where is a dyadic number. For each such range, one applies Conjecture 2, which will establish (2) for that particular subcollection of tubes after some algebra (at one point one has to use the fact that ); then one sums in the dyadic ranges, absorbing the logarithmic loss into the factor on the right-hand side.

To see the translation and rotation symmetry of the Kakeya problem in this setting, one has to translate and rotate both E and the tubes .

Thanks for the correction!

21 May, 2009 at 7:26 pm

arie israelHi Terry,

Thanks for the wonderful post. A few possible corrections:

I’m a bit confused with the change from to going from (2) to Conjecture 2. Also, might be enlarged to become .

Also, shouldn't the random rotation trick be used to show Conjecture 2 Conjecture 3? I don’t see why it is needed to prove (2) => Conjecture 2, can’t that be proved directly?

Thanks again.

22 May, 2009 at 6:55 pm

jzahlThe Wolff paper whose abstract you link to in the first paragraph can be found here:

23 May, 2009 at 5:18 am

arie israelSorry, what I was trying to say was that it should be in Conjecture 3.

23 May, 2009 at 10:23 am

Terence TaoDear Arie,

Thanks for the corrections! The random rotations trick is used where it is because of the placement of the exponents. If one replaced the factor in Conjecture 2 by the weaker factor , then (2) and Conjecture 2 would be equivalent without the random rotations trick, but then one would need that trick to deduce Conjecture 3 from Conjecture 2.

23 May, 2009 at 5:41 pm

arie israelDear Terry,

Thanks for the reply, I didn’t notice the difficulty with the exponent of d at first glance. Speaking of exponents, I think there should be a in the statement of Conjecture 3, as well as in the first line of the proof of Proposition 7.

Thanks again for providing such clear concise expositions regarding Kakeya, they’re truly wonderful to read.

20 May, 2017 at 1:23 am

daoguo zhouDear Prof Tao,

Can you give more details on the procedure from estimate (2) to

It seems to me that just breaking up into level sets via dyadic decomposition will not work. In fact, we have

where the estimate (2) has been used. Our target is to bound the above quantity by

Now, we need Hölder’s inequality from to , which is not aviable.

20 May, 2017 at 4:41 pm

Terence TaoNormalise $\|F\|_{L^d({\bf R}^n)} = 1$. The above arguments will take care of the range when (say), since the number of such $k$ is only logarithmic in and one can move from to with a loss that can be absorbed in the factor. The remaining extreme ranges or can then be handled by trivial estimates (the former by bounding by ; the latter by bounding by ).