LECTURE NOTES 6 FOR 254B

TERENCE TAO

1. Two dimensional Kakeya theorems

In the last set of notes we introduced the Kakeya conjectures, which we reproduce here for reference:

Conjecture 1.1 (Kakeya maximal operator conjecture). Let T_{ω} be any collection of tubes in R^{n} of eccentricity 1/δ and equal sizes, whose orientations ω are δ-separated. Then we have

∥ ∑_{ω∈Ω} χ_{T_{ω}}∥_{n/(n−1)} ≲ δ^{−ε}(∑_{ω∈Ω} |T_{ω}|)^{(n−1)/n}. (1)

Conjecture 1.2 (Kakeya set conjecture). All Besicovitch sets have Hausdorff and Minkowski dimension n. (A Besicovitch set is defined to be a subset of R^{n} which has a unit line segment in every direction; it solves the n-dimensional analogue of the Kakeya needle problem).

The restriction conjecture implies the Kakeya maximal operator conjecture, which in turn implies the Kakeya set conjecture.

We have already proven the restriction conjecture in two dimensions, so we already have a proof of the Kakeya conjecture in two dimensions also. But it is also instructive to see the proof directly. This argument is due to Córdoba. Apart from a technical issue involving small angles, the point is that every two lines intersect in at most one point, and every two thin rectangles intersect in a small square.

Theorem 1.3 (Córdoba). Let T_{ω} be any collection of rectangles in the plane of eccentricity 1/δ and equal sizes, whose orientations ω are δ-separated. Then we have

∥ ∑_{ω∈Ω} χ_{T_{ω}}∥_{2} ≲ (log 1/δ)^{1/2}(∑_{ω∈Ω} |T_{ω}|)^{1/2}. (2)

Proof We square the left-hand side:

∥ ∑_{ω∈Ω} χ_{T_{ω}}∥_{2}^{2} = ∫ (∑_{ω∈Ω} χ_{T_{ω}}(x))^{2} dx

= ∫ ∑_{ω∈Ω} ∑_{ω′∈Ω} χ_{T_{ω}}(x)χ_{T_{ω′}} (x) dx

= ∑_{ω∈Ω} ∑_{ω′∈Ω} |T_{ω} ∩ T_{ω′}|.

Comparing this with the right hand side of (2), we see that we just need to show

∑_{ω′∈Ω} |T_{ω} ∩ T_{ω′}| ≲ (log 1/δ)|T_{ω}|.

Clearly we may remove the term ω′ = ω from this sum:

∑_{ω′∈Ω,ω′≠ω} |T_{ω} ∩ T_{ω′}| ≲ (log 1/δ)|T_{ω}|.

Suppose ω and ω′ differed in angle by an amount comparable to 2^{−k} for some δ ≲ 2^{−k} ≲ 1. Then some elementary geometry gives

|T_{ω} ∩ T_{ω′}| ≲ 2^{k}δ|T_{ω}|.

Thus we reduce to showing that

∑_{k=0}^{log(1/δ)} ∑_{ω′∈Ω,∠(ω′,ω)∼2^{−k}} 2^{k}δ ≲ log(1/δ).

But for each k, there are only O(δ^{−1}2^{−k}) directions ω′ which are within O(2^{−k}) of ω. The claim then follows.

Informally, Córdoba’s estimate states that any collection of eccentric rectangles in the plane with different orientations behave as if they were disjoint.

As a consequence, we see that Besicovitch sets in the plane have to have dimension 2. However, it is possible for them to have measure zero. One way to do this is by modifying the construction given in Lecture Notes 3 (see [3] for details). Another Besicovitch set with zero measure comes from the following construction of Kahane. Let C be the Cantor set of all numbers in [0, 1] whose base 4 expansion contains only 0s and 1s. Let E be the union of all line segments which connect (−x, 0) to (2y, 1) for x, y ∈ C. Since every number in [0, 1] can be written as x + 2y for some x, y ∈ C, we see that this set contains a unit line segment of every slope from 0 to 1; by taking eight rotated copies of this set we thus obtain a Besicovitch set.

It is much less trivial to see that this set E has measure 0. Since the cross-section with the line y = t is basically 2tC −(1−t)C, it is easy to see that this statement is equivalent to the fact that a random rotation of the Garnett set C × C around the origin has measure zero almost always when projected to the x-axis. Unfortunately, I do not know an elementary proof of this; if someone does know, I would greatly appreaciate knowing!

2. Partial Kakeya estimates

In higher dimensions n ≥ 2 the Kakeya conjecture is still open. Even the simplest problem, that of estimating the Minkowski dimension of Besicovitch sets, is not fully solved. However, it is relatively easy to get some partial progress on this problem.

At this point we shall normalize the Kakeya maximal operator conjecture so that the tubes T_{ω} have length 1 and thickness δ. The conjecture then becomes

∥ ∑_{ω∈Ω} χ_{T_{ω}}∥_{n/(n−1)} ≲ δ^{−ε}(δ^{n−1}#Ω)^{(n−1)/n}. (3)

Since the ω are δ-separated, we have #Ω ≲ δ^{1−n}. So it appears that (3) is stronger than the estimate

∥ ∑_{ω∈Ω} χ_{T_{ω}}∥_{n/(n−1)} ≲ δ^{−ε}. (4)

However, we have the amazing observation that the two are equivalent. This is a completely “soft” argument, using no geometry other than the rotational symmetry of the sphere, and is part of a wider theory known as factorization theory, which was developed by Nikishin, Pisier, and others.

Proposition 2.1. If (4) holds, then (3) holds.

Proof Fix δ. For each integer N ≲ δ^{1−n}, let c(N) denote the smallest constant such that

∥ ∑_{ω∈Ω} χ_{T_{ω}}∥_{n/(n−1)} ≲ c(N)

holds for all δ-separated sets Ω of cardinality at most N. From (4) we see that c(N) ≲ δ^{−ε}; our task is to improve this to

c(N) ≲ δ^{−ε}(δ^{n−1}N)^{(n−1)/n}. (5)

Now suppose that N ≪ δ^{1−n}. By definition we can find a δ-separated set Ω of cardinality at most N such that

∥ ∑_{ω∈Ω} χ_{T_{ω}}∥_{n/(n−1)} = c(N).

Actually, it is clear that Ω will have cardinality exactly N, otherwise we could freely add some more tubes to top up the cardinality.

Let U be a random rotation in R^{n} (using the uniform distribution on the orthogonal group), and consider the set U(Ω). This set will be disjoint from Ω almost always. Define T_{U(ω)} to be the rotated tube U(T_{ω}). Then by symmetry we have

∥ ∑_{ω∈U(Ω)} χ_{T_{ω}}∥_{n/(n−1)} = c(N).

From the trivial observation that ∥f + g∥_{p}^{p} ≥ ∥f∥_{p}^{p} + ∥g∥_{p}^{p} for 1 ≤ p < ∞ and non-negative f, g, we thus have

∥ ∑ ; ω∈Ω ⋃ U(Ω) χ_{T_{ω}}∥_{n/(n−1)} ≥ 2^{(n−1)/n}c(N). (6)

The set Ω ∪ U(Ω) is not likely to be δ-separated; however, if N is small and U is random, then we can make this set δ-separated after removing a small number of elements.

More precisely, let a(Ω, U(Ω)) denote the number of points in Ω, U(Ω) which are within δ of each other:

a(Ω, U(Ω)) = #{(ω, ω′) ∈ Ω × O(Ω) : |ω − ω′| ≤ δ}.

We may rewrite this as

a(Ω, U(Ω)) = ∑_{ω∈Ω} ∑_{ω′∈Ω} χ_{B(0,δ)}(ω − O(ω′)).

Now let O range over the orthogonal group O(n) with normalized Haar measure dO(n). We have

∫_{O(n)} a(Ω, U(Ω)) dO(n) = ∑_{ω∈Ω} ∑_{ω′∈Ω} f(ω, ω′)

where

f(ω, ω′) = ∫_{O(n)} χ_{B(0,δ)}(ω − U(ω′)) dO(n).

From symmetry considerations we see that f is independent of the choice of ω′. Thus we have

f(ω, ω′) = \frac{1}{|S^{n−1}|} ∫_{O(n)} ∫_{S^{n−1}} χ_{B(0,δ)}(ω − U(ω′)) dω′dO(n).

The inner integral is roughly δ^{n−1}, so we thus have f ∼ δ^{n−1} uniformly. Inserting this into the previous we have

∫_{O(n)} a(Ω, U(Ω)) dO(n) ∼ δ^{n−1}N^{2}.

Thus we can find a rotation U such that a(Ω, U(Ω)) ≲ δ^{n−1}N^{2}. This implies that we may separate Ω ∪ U(Ω) into a large δ-separated set (of cardinality at most 2N), and O(1) smaller δ-separated sets with cardinality O(δ^{n−1}N^{2}) each. This implies that

∥ ∑ ; ω∈Ω ⋃ U(Ω) χ_{T_{ω}}∥_{n/(n−1)} ≥ c(2N) + Cc(Cδ^{n−1}N^{2})

for some constants C. Combining this with (6) we get the inequality

c(N) ≤ 2^{−(n−1)/n}c(2N) + Cc(Cδ^{n−1}N^{2}).

If we define c_{k} = 2^{k(n−1)/n}c(2^{−k}C^{−1}δ^{n−1}), then this becomes

c_{k} ≤ c_{k−1} + C2^{−k(n−1)/n}c_{2k}.

Our task is to show that c_{k} = O(δ^{−ε}) for all k ≥ 0. Note that c_{k} vanishes if k > C log(1/δ).

We modify c_{k} slightly to eliminate the annoying error term. Define the sequence d_{k} by

d_{k} = c_{k}(1 + C_{0}2^{−k(n−1)/n}).

where C_{0} is a constant to be chosen later. If C_{0} is large enough, then we see that

d_{k} < d_{k−1} + C2^{−k(n−1)/n}((d_{2k} − d_{k}) + (d_{k−1} − d_{k})).

Let k_{max} be the place where d_{k} is maximized over the range k ≥ 0. Applying the above estimate to k_{max} we obtain a contradiction unless k_{max} = 0. Thus d_{k} ≤ d_{0} for all k. From (3) we have d_{0} = O(δ^{−ε}), so we are done.

(N.B. I could have saved myself some trouble if I had weakened (4) slightly to allow a factor of (δ^{n−1}#Ω)^{ε}, which is of course easy to do since I already had a δ^{−ε} factor. But I did want to illustrate some techniques for handling recursive inequalities.)

So anyway, we can reduce ourselves to (4), which basically amounts to assuming that Ω is maximal.

On the other hand, we have the trivial estimate

∥ ∑_{ω∈Ω} χ_{T_{ω}}∥_{∞} ≲ δ^{1−n}.

Interpolating between the two, we obtain (if the Kakeya conjecture is true)

∥ ∑_{ω∈Ω} χ_{T_{ω}}∥_{p} ≲ δ^{\frac{n}{p} −(n−1)+ε} (7)

for all p ≥ n/(n − 1). These statements can be viewed as intermediate progress on the way to the full conjecture. p = ∞ is trivial, p = n/(n − 1) is the full conjecture, so the problem is to push p down as low as possible.

For each n, let d(n) denote the smallest Hausdorff dimension of a Besicovitch set in R^{n}. The Kakeya set conjecture then states that d(n) = n for all n.

Exercise 1: show that (7) implies that d(n) ≥ p/(p − 1). (Hint: modify the arguments in the previous set of notes. As a warm up, do the Minkowski dimension before the Hausdorff).

To get a feel for how results of the form (7), and lower bounds on d(n), let’s make some trivial observations.

Exercise: show that d(n) ≤ d(n+1). (Hint: The projection of an n+1 dimensional Besicovitch set to R^{n} is still essentially a Besicovitch set).

Exercise 2: show that d(n + m) ≤ d(n) + d(m). (Hint: the Cartesian product of Besicovitch sets is another Besicovitch set).

We know that d(1) = 1, d(2) = 2, but unfortunately the above results are too trivial to conclude anything about the higher dimensional case other than 2 ≤ d(n) ≤ n for n ≥ 2.

Somewhat similar statements hold for (7), but we won’t prove them here.

Let’s review some basic arguments which give some progress on (7). Firstly, we can modify Córdoba’s argument to higher dimensions.

Exercise 3: Use Córdoba’s argument to prove (7) for p = 2.

Unfortunately, it only gets down as far as 2, which for n large is increasingly unimpressive. (On the set level, it reproduces the d(n) ≥ 2 bound).

A better argument, due to Drury will get down to (n + 1)/(n − 1); this implies that Besicovitch sets have dimension at least (n + 1)/2. Let’s give a slightly informal argument to see where this number comes from.

Let E be a Besicovitch set. Thus for every direction ω ∈ S^{n−1}/± there exists a line segment, i.e. an affine map l_{ω} : [0, 1] → E.

Now consider the map ϕ : S^{n−1}/ ± ×[0, 1] × [0, 1] → E × E defined by

ϕ(ω, s, t) = (l_{ω}(s), l_{ω}(t)).

Any two non-parallel lines intersect in at most one point, so this map is one-to-one. Hence the dimension of the range is at least as large as the dimension of the domain. Since the domain has dimension n + 1 and the range dimension 2dim(E), this proves the claim.

Exercise 4: Make this argument rigorous for the Minkowski dimension of E.

That was fairly easy. Now let’s turn to the harder question of how to reproduce this argument for (7). The key geometric fact is the following: if a bunch of lines intersect at a point, then away from that point the lines are all disjoint. (Think about how this fact is related to the fact that ϕ is one-to-one). We’ll also need to apply a lot more combinatorics. However, the combinatorics that we’ll need will be very useful later, and it’s worth learning. By the way, the following arguments are all due to Bourgain.

Okay, we want to prove (7) for p = (n + 1)/(n − 1). At this point we’re going to make some handy notation. We’re already using the ≲ notation: A ≲ B if there exists a constant C such that A ≤ CB. We’ll go one step further, and define a ⪅ symbol: A ⪅ B if one has A ≤ C_{ε}δ^{−ε}B for every ε > 0. Thus A ⪅ B is weaker than A ≲ B. This is very convenient as it allows us to treat logarithmic quantities like log(1/δ) as if they were O(1).

Anyway, (7) has now become

∥ ∑_{ω∈Ω} χ_{T_{ω}}∥_{(n+1)/(n−1)} ⪅ δ^{\frac{n−1}{n+1}} ,

or

∫ (∑_{ω∈Ω} χ_{T_{ω}}(x))^{(n+1)/(n−1)} dx ⪅ δ. (8)

Now (n + 1)/(n − 1) is almost certainly not an integer, so we can’t play the types of games we did in the Còrdoba argument. Instead, we’re going to use more robust combinatorial tools that don’t care about integral exponents.

The first step is the pigeonhole principle - a very useful tool whenever one is allowed to ignore logarithmic factors, as one shall soon see.

The quantity ∑_{ω∈Ω} χ_{T_{ω}}(x) can range in value from 0 to O(δ^{1−n}). Thus, if one lets μ range dyadically from 1 to O(δ^{1−n}), and defines the sets E_{μ} by

E_{μ} = {x : μ ≤ ∑_{ω∈Ω} χ_{T_{ω}}(x) < 2μ},

then

∫ (∑_{ω∈Ω} χ_{T_{ω}}(x))^{(n+1)/(n−1)} dx = ∑_{μ} ∫_{E_{μ}}(∑_{ω∈Ω} χ_{T_{ω}}(x))^{(n+1)/(n−1)} dx

∼ ∑_{μ} μ^{(n+1)/(n−1)}|E_{μ}|.

Now, the number of dyadic μ out there is logarithmic in δ, and is thus ⪅ 1. Therefore to show (8), it suffices to show

μ^{(n+1)/(n−1)}|E_{μ}| ⪅ δ (9)

for each μ. Basically, we only need to consider the region of space where the tubes intersect with multiplicity about μ. This is a step forward because we now have a very useful quantity μ that we can manipulate. We’ll see a lot more quantities like this being produced in the sequel.

Before we actually do the rigorous proof of (9), let’s do a completely non-rigorous argument to show the way. Every point x in E_{μ} is in about μ tubes. There are about δ^{1−n} tubes in all, so on the average each tube contains about μδ^{n−1} of E_{μ}. Since each tube has volume about δ^{n−1}, we thus see that the average proportion λ of a tube which is occupied by E_{μ} is about

λ = μ|E_{μ}|. (10)

If λ is small enough, then this equation will be enough to give (9). Now suppose λ is large - i.e. each tube contains a fair amount of E_{μ}. Consider a typical point x ∈ E_{μ}, it is contained in about μ tubes. Each of these tubes is filled with density λ with E_{μ}. The worst case is if E_{μ} occupies a sub-tube of length λ - this is the most concentrated case. Because these tubes all intersect at x, E_{μ} then looks like this “bush” object with many λ × δ-tubes sticking out from x. In theory there are μ such tubes, but if λ is short then many of these tubes will overlap. If one wants to count essentially disjoint tubes one can only guarantee μλ^{n−1} such tubes (this is a quantitative version of the statement that two non-parallel lines only intersect in one point). This gives a total volume of μλ^{n−1}λδ^{n−1} for the bush, which gives the bound

|E_{μ}| ≥ μλ^{n}δ^{n−1}.

Combining this with (10) and some algebra we get (9).

Okay, now let’s do the rigorous version of the above argument.

If |E_{μ}| ≤ δ^{10n} then we are done, so we may assume |E_{μ}| ≥ δ^{10n}.

Fix μ. From the definition of E_{μ} we have

∫_{E_{μ}} ∑_{ω∈Ω} χ_{T_{ω}}(x) dx ∼ μ|E_{μ}|.

Re-arranging the left-hand side, we get

∑_{ω∈Ω} |T_{ω} ∩ E_{μ}| ∼ μ|E_{μ}|.

The set |T_{ω} ∩ E_{μ}| can range anywhere from 0 to |T_{ω}|. Let’s pigeonhole again. For each dyadic λ from δ^{20n} to 1, define Ω_{λ} to be the set

Ω_{λ} = \{ω ∈ Ω : \frac{λ}{2}|T_{ω}| < |T_{ω} ∩ E_{μ}| ≤ λ|T_{ω}|\}.

Then we have

∑_{ω∈Ω} |T_{ω} ∩ E_{μ}| = ∑_{λ} ∑_{ω∈Ω_{λ}} |T_{ω} ∩ E_{μ}| + ∑_{ω:|T_{ω}∩E_{μ}|≤δ^{20n}|T_{ω}|} |T_{ω} ∩ E_{μ}|.

The second term is much smaller than δ^{10n}, and so can be absorbed into the righthand side. Since the summand in the first term is about δ^{n−1}λ, we thus obtain

∑_{λ} λδ^{n−1}#Ω_{λ} ∼ μ|E_{μ}|.

Since the number of λ is ⪅ 1, the pigeonhole principle then forces the existence of a single λ such that

λδ^{n−1}#Ω_{λ} ≈ μ|E_{μ}|. (11)

In some sense, λ is the dominant density of E_{μ} with respect to a generic tube T_{ω}. In particular, since #Ω_{λ} = O(δ^{1−n}) we have

λ ⪆ μ|E_{μ}| (12)

(cf. (10)).

Having obtained these two quantities λ and μ, which we can now easily prove (9). If we reverse the steps leading to the derivation of (11), we obtain

∑_{ω∈Ω_{λ}} |T_{ω} ∩ E_{μ}| ≈ μ|E_{μ}|,

and thus

∫_{E_{μ}} ∑_{ω∈Ω_{λ}} χ_{T_{ω}}(x) dx ≈ μ|E_{μ}|.

In particular, we see that there must exist at least one x ∈ E_{μ} for which

∑_{ω∈Ω_{λ}} χ_{T_{ω}}(x) ⪆ μ.

(Compare this with the definition of E_{μ}). In other words, this x is contained in ⪆ μ tubes T_{ω} belonging to Ω_{λ}.

These tubes are δ-separated in direction. Unfortunately, if λ is small this will cause a lot of tubes to be essentially identical, and we must refine this collection of tubes. Let’s select a maximal δ/λ-direction-separated subset T of these tubes; it’s easy to see that this set has cardinality ⪆ μλ^{n−1}.

For each tube T_{ω} ∈ T, we have by definition of Ω_{λ} that

|{y ∈ T_{ω} : y ∈ E_{μ}}| ≳ δ^{n−1}λ.

In particular, we have

|{y ∈ T_{ω}\B(x, C^{−1}λ) : y ∈ E_{μ}}| ≳ δ^{n−1}λ (13)

for some sufficiently large C.

Now for the quantitative version of the statement that two non-parallel lines intersect in at most one point: as T_{ω} ranges over the δ/λ-direction-separated subset T, the sets T_{ω}\B(x, C^{−1}λ) are essentially disjoint. Since all the sets in (13) are disjoint, and we have μλ^{n−1} of them, we thus see that

|E_{μ}| ⪆ μλ^{n−1}δ^{n−1}λ.

Combining this with (12) and dreary algebra we obtain (9) as desired.

This argument is known as the “bush” argument. Basically, it says that if you try to make the multiplicity of tubes too large at one point, you will be forced to create a low-multiplicity bush of tubes from that point. This prevents the average multiplicity from getting too high (roughly speaking, it cannot exceed the square root of the number of tubes).

3. Wolff’s Kakeya argument

Remember d(n), the minimal dimension of a Besicovitch set in R^{n}? Currently we have two separate bounds for d(n) in the n ≥ 2 case. The first is d(n) ≥ 2, from Córdoba’s argument. The second is d(n) ≥ (n + 1)/2. These bounds are both very simple, but they are not optimal. The first (small) improvement to these bounds was obtained by Bourgain in 1991; then in 1995 Wolff gave a new argument that gave d(n) ≥ (n + 2)/2, and correspondingly showed (7) for p = (n + 2)/n. Until last year, this bound was the best known in any dimension.

Basically, Wolff’s argument combines Córdoba’s argument and the bush argument into one. The bush, which was used to obtain a lower bound for |E_{μ}|, is now replaced by a larger object which is commonly known as a “hairbrush”: instead of taking all the tubes through a point, we take all the tubes through a central tube. This modest change allows us to gain an extra 1/2 of a dimension in our bound for d(n). The new geometric fact used is the following: if a bunch of lines pass through a line (not a point), then away from that line the lines behave as if they were disjoint (thanks to Córdoba’s estimate, and the fact that the lines can be grouped into planes).

Okay, let’s prove (7) for p = (n + 2)/n, since this implies the corresponding bound for d(n). We have to show

∥ ∑_{ω∈Ω} χ_{T_{ω}}∥_{p} ⪅ δ^{\frac{n}{p} −(n−1)}

for this value of p. We could plunge in and do the pigeonhole stuff we did earlier - and this is what Wolff did in his original paper - but it pays to wait a little and do another trick first, namely this bilinear argument which we employed earlier with regard to restriction theorems.

First, let’s use the triangle inequality and restrict all directions ω to be within 1/10 of the vertical. This is pretty harmless, but we’ll avoid some nastiness involving antipodal directions later on if we do this now.

Let’s square the estimate as

∥ ∑_{ω∈Ω} ∑_{ω′∈Ω} χ_{T_{ω}}χ_{T_{ω′}} ∥_{p/2} ⪅ δ^{\frac{2n}{p} −2n+2}.

Note that p/2 is less than 1, so we have to tread with caution here. - the triangle inequality becomes more subtle, for instance. (When n = 2, though, we’re well on our way to reproducing Córdoba’s argument). Let’s raise things to the power p/2:

∥ ∑_{ω∈Ω} ∑_{ω′∈Ω} χ_{T_{ω}}χ_{T_{ω′}} ∥_{p/2}^{p/2} ⪅ δ^{n−p(n−1)}.

In the bilinear setting a new variable becomes available, namely the angular separation |ω − ω′| between the two directions ω and ω′. So we can split the summation as

∑_{ω∈Ω} ∑_{ω′∈Ω} = ∑_{k=0}^{log(1/δ)} ∑_{ω,ω′:|ω−ω′|∼2^{−k}} + ∑_{ω,ω′:ω′=ω} .

The latter term behaves very much like the case k = log(1/δ), and we’ll ignore it. (The diagonal term is usually the easiest case, never the hardest).

As I said before, we don’t have a triangle inequality for exponents p < 1. However, we still do have the pseudo-triangle inequality

∥ ∑_{i=1}^{n} f_{i}∥_{p} ≤ (∑_{i=1}^{n} ∥f_{i}∥_{p}^{p})^{1/p}

which comes from the observation that a + b ≤ (a^{p} + b^{p})^{1/p} when p ≤ 1. So it suffices to show that

∑_{k=0}^{log(1/δ)} ∥ ∑_{ω,ω′:|ω−ω′|∼2^{−k}} χ_{T_{ω}}χ_{T_{ω′}} ∥_{p/2}^{p/2} ⪅ δ^{n−p(n−1)}.

Since the number of k is ⪅ 1, we therefore just need to show

∥ ∑_{ω,ω′:|ω−ω′|∼2^{−k}} χ_{T_{ω}}χ_{T_{ω′}} ∥_{p/2}^{p/2} ⪅ δ^{n−p(n−1)}

for each k.

Fix k. Let’s cover Ω by about 2^{(n−1)k} finitely overlapping caps C of width C2^{−k}. If one does this correctly, we see that for each ω, ω′ such that |ω − ω′| ∼ 2^{−k}, we can find a cap C which contains them both. So by the p < 1 version of the triangle inequality again, we reduce to

∑_{C} ∥ ∑_{ω,ω′∈C∩Ω:|ω−ω′|∼2^{−k}} χ_{T_{ω}}χ_{T_{ω′}} ∥_{p/2}^{p/2} ⪅ δ^{n−p(n−1)}.

Since there are 2^{(n−1)k} caps, we thus have to prove

∥ ∑_{ω,ω′∈C∩Ω:|ω−ω′|∼2^{−k}} χ_{T_{ω}}χ_{T_{ω′}} ∥_{p/2}^{p/2} ⪅ 2^{−(n−1)k}δ^{n−p(n−1)} (14)

for each cap.

This k is annoying. Fortunately, there is a rescaling argument that allows one to make it go away. In fact, once we show (14) for k = 0, it automatically implies the same estimate for all k (with some improvement thrown in, as we shall see.

Let’s see how this works. Assume we’ve proven (14) for k = 0, now let’s tackle the case k > 0. We can rotate C to be the cap of width 2^{−k} around the vertical direction e_{n}. Now consider the transformation L : R^{n} → R^{n} defined by

L(\underline{x}, x_{n}) = (2^{k}\underline{x}, x_{n})

and the corresponding stretching operator T defined by

T f(x) = f(L^{−1}x).

Since det L = 2^{(n−1)k}, we have ∥T f∥_{q} = 2^{(n−1)k/q}∥f∥_{q}, so we have to show

∥T ( ∑_{ω,ω′∈C∩Ω:|ω−ω′|∼2^{−k}} χ_{T_{ω}}χ_{T_{ω′}})∥_{p/2}^{p/2} ⪅ δ^{n−p(n−1)}

for each cap. We can re-arrange this as

∥ ∑_{ω,ω′∈C∩Ω:|ω−ω′|∼2^{−k}} χ_{L(T_{ω})}χ_{L(T_{ω′})}∥_{p/2}^{p/2} ⪅ δ^{n−p(n−1)}.

Now the expression on the left-hand side is of the form treated in (14), but with δ replaced by 2^{k}δ, Ω replaced by what is essentially 2^{k}Ω, and k replaced by 0. So we can bound the left-hand side by (2^{k}δ)^{n−p(n−1)}. Since p > n/(n − 1), we see that we are done (with quite a bit of room to spare).

So we just need to show (14) when k = 0, i.e.

∥ ∑_{ω,ω′:|ω−ω′|∼1} χ_{T_{ω}}χ_{T_{ω′}} ∥_{p/2}^{p/2} ⪅ δ^{n−p(n−1)}.

Note that this “bilinear reduction” didn’t have much to do with the exact choice of p; it’s quite universal.

By the quasi-triangle inequality we may quickly reduce to

∥ ∑_{ω∈Ω_{1}} ∑_{ω′∈Ω_{2}} χ_{T_{ω}}χ_{T_{ω′}} ∥_{p/2}^{p/2} ⪅ δ^{n−p(n−1)}

where Ω_{1}, Ω_{2} are subsets of Ω which are separated by ∼ 1. We can factor this as

∥( ∑_{ω∈Ω_{1}} χ_{T_{ω}})( ∑_{ω′∈Ω_{2}} χ_{T_{ω′}})∥_{p/2}^{p/2} ⪅ δ^{n−p(n−1)}.

Now we revive the μ and λ quantities we used for the bush argument. This time, though, there will be two of everything, because of the bilinear nature of things. (That’s a bit annoying, but that’s more than made up for by the angular separation condition that we’ve gained. Otherwise we would have to deal with another parameter (generally called σ), the mean angle of separation).

Let E_{μ,μ′} denote the set

E_{μ,μ′} = {x : ∑_{ω∈Ω_{1}} χ_{T_{ω}}(x) ∼ μ, ∑_{ω′∈Ω_{2}} χ_{T_{ω′}} (x) ∼ μ′.}

In other words, E_{μ,μ′} consists of those points which are contained in about μ tubes from Ω_{1} and μ′ tubes from Ω_{2}. By repeating the argument from before, it suffices to show that

μ^{p/2}μ′^{p/2}|E_{μ,μ′}| ⪅ δ^{n−p(n−1)}

for all dyadic μ, μ′ between δ^{10n} and 1. Since we have p = (n + 2)/n we may as well insert this in:

(μμ′)^{(n+2)/2n}|E_{μ,μ′}| ⪅ δ^{(2−n)/n}. (15)

As before, we may assume that |E_{μ,μ′}| ⪆ δ^{20n}.

Fix μ, μ′. Now for the lambdas. Define Ω_{1}^{λ} and Ω_{2}^{λ′} by

Ω_{1}^{λ} = {ω ∈ Ω_{1} : |T_{ω} ∩ E_{μ,μ′}| ∼ λ|T_{ω}|}
Ω_{2}^{λ′} = {ω′ ∈ Ω_{2} : |T_{ω′} ∩ E_{μ,μ′}| ∼ λ′|T_{ω′}|}.

From the definition of E_{μ,μ′} we have

∫_{E_{μ,μ′}} ∑_{ω∈Ω_{1}} ∑_{ω′∈Ω_{2}} χ_{T_{ω}}χ_{T_{ω′}} ∼ μμ′|E_{μ,μ′}|.

Thus by repeating the same pigeonhole argument as before, we can find dyadic λ, λ′ between δ^{20n} and 1 such that

∫_{E_{μ,μ′}} ∑_{ω∈Ω_{1}^{λ}} ∑_{ω′∈Ω_{2}^{λ′}} χ_{T_{ω}}χ_{T_{ω′}} ⪆ μμ′|E_{μ,μ′}|. (16)

Fix λ, λ′. We first show a lower bound on λ, λ′ analogous to (12). Since

∑_{ω′∈Ω_{2}^{λ′}} χ_{T_{ω′}} (x) ≲ μ′

on E_{μ,μ′} by construction, we see from (16) that

∫_{E_{μ,μ′}} ∑_{ω∈Ω_{1}^{λ}} χ_{T_{ω}} ⪆ μ|E_{μ,μ′}|.

Arguing as in the previous section, we thus obtain

λ ⪆ μ|E_{μ,μ′}|. (17)

Now we return to (16). This clearly implies

∑_{ω′∈Ω_{2}^{λ′}} ∫_{T_{ω′}} ∑_{ω∈Ω_{1}^{λ}} χ_{T_{ω}} ⪆ μμ′|E_{μ,μ′}|.

The number of directions in Ω_{2}^{λ′} is O(δ^{1−n}). So there must exist an ω′ such that

∫_{T_{ω′}} ∑_{ω∈Ω_{1}^{λ}} χ_{T_{ω}} ⪆ δ^{n−1}μμ′|E_{μ,μ′}|,

Fix this ω′. We rewrite the above as

∑_{ω∈Ω_{1}^{λ}} |T_{ω} ∩ T_{ω′}| ⪆ δ^{n−1}μμ′|E_{μ,μ′}|.

Now we use the hard-won angular separation condition on ω and ω′ to observe that |T_{ω} ∩ T_{ω′}| ≲ δ^{n}. Thus we have

δ^{n}#{ω ∈ Ω_{1}^{λ} : T_{ω} ∩ T_{ω′} ≠ ∅} ⪆ δ^{n−1}μμ′|E_{μ,μ′}|.

To summarize: we have located a tube T_{ω′} which intersects at least ⪆ δ^{−1}μμ′|E_{μ,μ′}| tubes T_{ω}, each of are at angle ∼ 1 to the stem tube T_{ω′}, and are filled with density λ with E_{μ,μ′}.

We’d like to convert this hairbrush object into a quantitative lower bound for E_{μ,μ′}, as in the previous section.

Let T denote the collection of all the T_{ω} which intersect T_{ω′}, so that

#T ⪆ δ^{−1}μμ′|E_{μ,μ′}|. (18)

For each T_{ω} ∈ T, we have

∫_{E_{μ,μ′}} χ_{T_{ω}} ∼ λδ^{n−1}.

As before, it will be convenient to cut out a small portion of the integral, and obtain

∫_{E_{μ,μ′}} χ_{T_{ω}∩Σ} ∼ λδ^{n−1}

where Σ is the exterior cylinder

Σ = {x : d(x, T_{ω′}) > C^{−1}λ}

and C is a sufficiently large constant. (Recall that T_{ω} and T_{ω′} intersect at angles ∼ 1). Summing this up over all T_{ω} ∈ T we have

∫_{E_{μ,μ′}} ∑_{T_{ω}∈T} χ_{T_{ω}∩Σ} ≳ λδ^{n−1}#T.

Now we’re going to apply Córdoba’s estimate, which means that we’re going to have to introduce the L^{2} norm. From Cauchy-Schwarz, the left-hand side is bounded by

|E_{μ,μ′}|^{1/2}∥ ∑_{T_{ω}∈T} χ_{T_{ω}∩Σ}∥_{2},

so we obtain

∥ ∑_{T_{ω}∈T} χ_{T_{ω}∩Σ}∥_{2} ⪆ λδ^{n−1}#T |E_{μ,μ′}|^{−1/2}.

We will shortly show the estimate

∥ ∑_{T_{ω}∈T} χ_{T_{ω}∩Σ}∥_{2} ⪅ (#T λ^{2−n}δ^{n−1})^{1/2}. (19)

Combining this with the previous, we obtain after some algebra

λ^{n}δ^{n−1}#T ⪅ |E_{μ,μ′}|.

Inserting (18), this becomes

λ^{n}δ^{n−2}μμ′ ⪅ 1.

Now inserting (17), we get

μ^{n+1}μ′|E_{μ,μ′}| ⪅ δ^{2−n}.

By a completely symmetric argument we obtain

μμ′^{n+1}|E_{μ,μ′}| ⪅ δ^{2−n}.

Taking the geometric mean of the two estimates we obtain (15), as desired.

It remains to show (19). We square this as

∫ ∑_{T_{ω_{1}} ∈T} ∑_{T_{ω_{2}} ∈T} χ_{T_{ω_{1}} ∩Σ}χ_{T_{ω_{2}} ∩Σ} ⪅ #T λ^{2−n}δ^{n−1}.

By Fubini’s theorem, we can simplify the left-hand side:

∑_{T_{ω_{1}} ∈T} ∑_{T_{ω_{2}} ∈T} |T_{ω_{1}} ∩ T_{ω_{2}} ∩ Σ| ⪅ #T λ^{2−n}δ^{n−1}.

Thus it suffices to show that

∑_{T_{ω_{2}} ∈T:T_{ω_{1}} ∩T_{ω_{2}} ∩Σ≠∅} |T_{ω_{1}} ∩ T_{ω_{2}} ∩ Σ| ⪅ λ^{2−n}δ^{n−1}

for all T_{ω_{1}} ∈ T.

Fix T_{ω_{1}}. We first observe that the case ω_{1} = ω_{2} is trivial to handle, so we restrict the summation to ω_{1} ≠ ω_{2}. We can then subdivide further based on the angle between ω_{1} and ω_{2}:

∑_{k=0}^{log(1/δ)} ∑_{T_{ω_{2}} ∈T:∠ω_{1},ω_{2}∼2^{−k},T_{ω_{1}} ∩T_{ω_{2}} ∩Σ≠∅} |T_{ω_{1}} ∩ T_{ω_{2}} ∩ Σ| ⪅ λ^{2−n}δ^{n−1}.

Since the number of k here is ≈ 1, it suffices to show this estimate for each k individually.

Fix k. Now we use the Córdoba observation, that if ω_{1} and ω_{2} make an angle of about 2^{−k}, then the intersection of T_{ω_{1}} and T_{ω_{2}} is basically contained in a δ × 2^{k}δ tube and so has measure at most 2^{k}δ^{n}. So we reduce to showing that

∑_{T_{ω_{2}} ∈T:∠ω_{1},ω_{2}∼2^{−k},T_{ω_{1}} ∩T_{ω_{2}} ∩Σ≠∅} 2^{k}δ^{n} ⪅ λ^{2−n}δ^{n−1},

or in other words that

#{T_{ω_{2}} ∈ T : ∠ω_{1}, ω_{2} ∼ 2^{−k}, T_{ω_{1}} ∩ T_{ω_{2}} ∩ Σ ≠ ∅} ≲ 2^{−k}δ^{−1}λ^{2−n}. (20)

But this bound follows from the fact that the tubes in T are δ-separated, and the following geometric fact (which is a quantitative version of the fact that every triangle lies in a plane):

Lemma 3.1. If T_{ω_{1}}, T_{ω_{2}} both intersect T_{ω′} with an angle ∼ 1, and intersect each other in Σ at an angle ∼ 2^{−k}, then T_{ω_{2}} lies within O(δ/λ) of the 2-dimensional plane π generated by the axes of T_{ω′} and T_{ω_{2}}, and when projected to π it makes an angle of ≲ 2^{−k} with T_{ω_{1}}.

We leave the proof as an exercise. Anyway, this shows that of the n − 1 degrees of freedom in ω_{2}, the degree along π is restricted to an 2^{−k}-arc, and the degrees along the other n − 2 directions are restricted to a δ/λ arc. Counting the number of δ-separated directions in this restricted region one gets (20).

References

1. J. Bourgain, Besicovitch-type maximal operators and applications to Fourier analysis, Geom. and Funct. Anal. 22 (1991), 147–187.

2. S. Drury, L^{p} estimates for the x-ray transform, Ill. J. Math. 27 (1983), 125–129.

3. E. M. Stein, Harmonic Analysis, Princeton University Press, 1993.

4. T. H. Wolff, An improved bound for Kakeya type maximal functions, Revista Mat. Iberoamericana. 11 (1995). 651–674.

5. T. Wolff, Recent work connected with the Kakeya problem, to appear, anniversary proceedings, Princeton 1996.

Department of Mathematics, UCLA, Los Angeles CA 90095-1555

E-mail address: tao@math.ucla.edu