THE KAKEYA UNIVERSE AND INCIDENCE THEOREMS
IN Rn
MICHAEL BATEMAN
1. Introduction
In this course we will explore a variety of problems connected to the
so-called “Kakeya conjecture”. To start with we’ll discuss a number of
questions, some of them seemingly unrelated, and throughout the course we
will see how these questions are in fact related.
1.1. Geometric questions. The first question was proposed by Kakeya
almost 100 years ago:
Question 1. What is the minimal area needed to continuously rotate a
needle in the plane by 180 degrees?
We will see shortly that this question is closely related to the following:
Question 2. What is the minimal area of a planar set with a unit line
segment in every direction?
In fact there is no minimal area, since any > 0 is enough, a fact we will
prove shortly. By elaborating a bit more on our proof, one could actually
construct such a set with measure 0. In much the same spirit as the last
question, we have
Question 3. What is the minimal dimension of a set in Rn with a unit line
segment in every direction?
These questions gives us occasion to introduce an important definition:
Definition 4. A Besicovitch set in Rn is a set containing a unit line segment
in every direction.
Question 3 asks for the minimal dimension of a Besicovitch set. Dimension is a notion of size that is a bit more nuanced than area/measure. What
options are available for dimension? A line segment in the plane has area
zero, but it also has dimension one. The middle one-half Cantor set has measure zero but dimension 12 . One of our early tasks will be to define a suitable
notion of dimension. There are several notions of fractional dimension; we
will discuss Minkowski dimension since it is quite easy to understand and
good enough for our purposes.
Conjecture 5 (Kakeya conjecture). Any Besicovitch in Rn must have Minkowski
dimension n.
1
2
MICHAEL BATEMAN
1.2. Differentiation theorems. An important fact in real analysis is the
Lebesgue Differentation Theorem:
Theorem 6. Suppose f : Rn → R is an integrable function. Let B denote
the ball of radius centered at the origin. Then
Z
1
lim
f (x − y)dy = f (x)
→0 |B | B
except for x in a set of measure zero.
This fact follows from the Hardy-Littlewood maximal theorem, and can
be found in, e.g., [6]. In other words, we can recover a function by averaging
it on smaller and smaller balls. Notice that this fact is a triviality for
continuous functions; the interesting part is that it holds for all functions
with finite integral. Is there anything special about averaging over a ball?
Why not cubes? Why not rectangles?
Question 7. Suppose f : Rn → R is an integrable function. Is it the case
that
Z
1
lim
f (x − y)dy = f (x)
→0 |R | R
(except for x in a set of measure zero) where R is any rectangle of length
and width less than ?
The answer is negative, although if we had demanded that R have bounded
eccentricity the answer would be positive.
Theorems of this form are called “Differentiation Theorems” because they
are generalizations of the Fundamental Theorem of Calculus, which says
Z x
Z
1 x+
∂
f (t)dt = lim
f (t)dt = f (x).
→0 x
∂x
0
1.3. Fourier Series. Given a periodic function f : [0, 1] → R, we can cook
up its Fourier series
∞
X
cn e2πinx
n=−∞
and the corresponding truncations
SN f (x) =
N
X
cn e2πinx .
n=−N
The coefficients are given by the usual formula
Z 1
cn =
e−2πinx f (x)dx.
0
So we have a function f and a magic formula given by the Fourier series. Is
the magic formula correct? In what sense could that statement be true? The
magic formula is actually a series– does this series converge to f ? pointwise?
3
in a mean-square sense? in some other appropriate sense? The answer
is “Probably”. As long as f is reasonable, the series converges in most
reasonable senses. But:
Question 8. Suppose f is a function of TWO variables. Define the truncation
X
cn e2πi (n1 x1 +n2 x2 ) .
SN f (x) =
|n|≤N
Again
Z
cn =
1
e−2πinx f (x)dx,
0
but notice that n is a two-tuple, and x is an element of R2 . Again we ask,
do these truncations converge to the orginal function f in any reasonable
sense?
The answer to this question is more difficult to obtain, and is much more
negative than in the 1 − D case. Suffice it to say that there are reasonable
functions f and reasonable notions of convergence for which the truncations
do not converge to f .
1.4. Subrings of the reals. In a seemingly different direction, we ask
Question 9. Do there exist Borel subrings of the real numbers having Hausdorff dimension strictly between zero and one?
Certainly the rationals are a subring, and the rationals have Hausdorff
dimension zero. Also the reals are certainly a subring of the reals, so we have
reasonable-looking subrings of dimension zero and one. But what about inbetween? The answer is that there are no Borel subrings of intermediate
dimension, but there are non-Borel subrings of any dimension. This latter
fact requires the Continuum Hypothesis. (See a paper of Edgar and Miller...
Borel subrings of the reals . ) Although we will not explore this question
further in these notes, the solution to this question is related to the solution
to the 2D Kakeya conjecture and the Szemeredi-Trotter theorem, which is
discussed below.
1.5. Point-line incidence theorems. Let P ⊆ R2 be a finite collection of
points. Let L be a finite collection of lines. Define the number of incidences
between P and L:
I(P, L) = |{(p, l) : p ∈ l}|.
Question 10. Can we reasonably control the size of I(P,L)?
Conceivably every point in P could intersect every line in L, giving us
only that
I(P, L) ≤ |P ||L|.
But we can actually show that
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MICHAEL BATEMAN
Theorem 11 (Szemeredi-Trotter).
2
2
1
1
I(P, L) . |L| 3 |P | 3 + |L| 2 |P | + |P | 2 |L|
You should ignore the last two terms in the estimate above; they are only
needed when the sizes of P and L are way out of balance. Of course the
2
2
precise numerology here may look silly– the important point is that |L| 3 |P | 3
is far smaller than the trivial estimate of |L||P | when P and L are large.
Further, this estimate is sharp.
1.6. Finite field Kakeya. As an analog to the Kakeya problems in Rn , we
have the following question:
Question 12. What is the minimum size of a Besicovitch set in Fnp ? Here
Fp is the field with p elements, p is prime, and Fnp is the vector space over
this field.
Dvir answered this question using some basic algebraic geometry, showing
that if E is a Besicovitch set, then |E| & pn . Notice that a d dimensional
subspace Fnp would have size pd , which is substantially smaller than pn as
long as p is large.
1.7. Arithmetic approach to Kakeya.
1.8. Restriction conjecture.
1.9. Local smoothing conjecture.
2. Acknowledgement
Thanks very much to Heiki Niglas for pointing out many typos, miscalculations, and other errors.
3. Besicovitch sets and the needle question
In this section we show how to construct a planar set containing a line
segment in every direction. Begin by partitioning the circle S1 as follows:
for a large integer N , divide the circle into closed intervals of length 2π
N.
(Technically, this is not a partition since the intervals are closed and hence
contain their endpoints. We want it this way.) Call the resulting collection
of intervals IN .
besicovitch
Theorem 13. Fix > 0. There exists N = N (), and a collection of
triangles T such that for each I ∈ IN there is a triangle T ∈ T containing
a unit line segment in each direction in I; further
[ T < .
T ∈T
5
3.1. Rotating the needle. Now we show how the existence of small Besicovitch sets allows us to find small sets in which we can rotate a needle
(unit line segment). We already have a small set with a line segment in each
direction; what we still need to show is that we can move the line segment
continuously through the set. (For example, the segments corresponding to
some angle θ and some very nearby angle θ0 could be far apart.) Here is a
basic lemma to help us.
Lemma 14. Let L and L̃ be two unit line segments with the same slope,
at distance at most 100 apart. Let δ > 0. Then there exists a continuous
translation of L to L̃ using area ≤ δ.
Proof. For simplicity, assume the lines are horizontal. Also without loss
of generality, we assume the line segments are a vertical translation of one
another. (If not, perform a horizontal translation of one of them to make it
so, at the cost of zero area.) Now suppose the vertical distance between the
lines is D. Translate the top line by R to the right, where R is a number
to be specified momentarily. Then rotate the line counterclockwise by angle
θ, and translate it back to the left by the same number R. Now rotate
clockwise by angle θ. Notice that this procedure translated the line L down
by some distance, and that by choosing θ appropriately (depending on R
and D) we can make the distance equal to D. A quick calculation shows
that we should have
D
tan θ = .
R
We see then that the area swept out by the needle is within a constant factor
of D
R . By selecting R large enough, we can make this number less than the
required δ.
Proof we can rotate needle in less than 2 area given Theorem 13. . Now consider two consecutive intervals Ij , Ij+1 in the partition of S1 described above.
To each of these intervals Ij we are guaranteed a triangle Tj such that for
each θ ∈ Ij , there is a unit line segment in direction θ contained in Tj .
Specifically, if θ is the shared endpoint of Ij and Ij+1 , then both Tj and
Tj+1 contain a line segment in the direction θ; call these segments Lj and
L̃j respectively. By the lemma we can translate Lj onto L̃j at minimal cost–
we choose δ in the previous lemma to be N , where is the area of the Besicovitch set provided to us by the Theorem, and N is the number of triangles
required by the theorem to achieve this area.
This is all we need to do. We have shown that for each Ij , there is a
triangle Tj containing all lines pointing in directions in Ij , and we have
shown that we can continuously translate a line segment from triangle Tj to
triangle Tj+1 at cost of area N . Hence
Area required to rotate needle
≤ Area of Besicovitch set
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MICHAEL BATEMAN
+
N
−1
X
Area required to translate a segment from Tj to Tj+1
j=1
≤ +
N
−1
X
j=0
N
≤ 2.
3.2. Constructing Besicovitch sets. In this section we prove Theorem
13. The construction we give is classical and is called a Perron tree. We
willl iterate the following sprouting procedure, which we first describe in a
simple situation. Observe two points about the sprouting:
1. The area of the figure decreases after sprouting.
2. The line segments connecting the top vertex to the base remain undamaged.
3.2.1. Sprouting. Consider a triangle T of height y and width x. For the sake
of discussion assume the base is horizontal. Perform the following operation
to this triangle: connect the top vertex to the base to divide T into two equal
parts TL and TR . Now translate TR left by δx so that the two triangles now
overlap. The figure now looks like a smaller triangle, which we call the body,
together with a bowtie-shaped region.
Claim 15.
|body| ≤ (1 − δ)2 |T |
Proof. The height and width are now equal to (1 − δ)y and (1 − δ)x, respectively.
Claim 16.
|bowtie| ≤ δ 2 |T |
Proof. Note that this set has the shape of a bowtie– i.e., it is the union of
two congruent triangles. Elementary geometry shows that each of these two
triangles is contained in a parallelogram that has width ≤ δx
2 and height
≤ 2δy. This proves the claim.
3.2.2. Constructing the tree. Let N be a large number to be determined
later. Assume it is a power of two. Consider the triangle T0 with height
1 and width 1. Divide the base into N intervals of length N1 , and divide
T0 into corresponding triangles TI , with TI having base I and top vertex
equal to the top vertex of T0 . Pair them up, TI with TI 0 , I and I 0 neighbors.
Then carry out the sprouting procedure for each pair, with x = N2 . Slide
the resulting figures together to form a new triangle with width and height
7
1 − δ, together with
we have
N
2
of the bowties. Call this new triangle T1 . Notice that
Total area of bowties after stage 1 ≤ (Number of bowties)(Area of bowtie)
N δ2
≤
2 N
≤ δ2.
Repeat this procedure, but now subdivide the base into only N2 subintervals,
each having length N2 (1 − δ). Again we have N2 triangles formed by these
intervals and the top vertex of the bigger triangle. We group them into N4
pairs of adjacent triangles and translate again by a factor of δ. Once again,
we form some new bowties, and we have
Area of bowties formed during stage 2 ≤ (Number of bowties)( Area of bowtie)
N 2δ 2 (1 − δ)2
≤
4
N
≤ δ 2 (1 − δ)2 .
The factor of (1 − δ)2 in the above calculation comes from the fact that the
big triangle at this stage has area smaller by a factor of (1 − δ)2 .
At each stage we repeat this process, which results in the main triangle
becoming smaller and in more bowties being formed. So the total area of
the set after j iterations is
Total Area ≤ (Area of main triangle) + (Total area of bowties)
≤ (Area of main triangle) +
j
X
(Area of bowties created at stage i)
i=1
≤ (1 − δ)2j +
j
X
δ 2 (1 − δ)2(i−1) .
i=1
Notice that the sum in the expression above is . δ because
j
X
1
(1 − δ)2(i−1) . .
δ
i=1
Hence
Total Area . (1 − δ)2j + δ.
If j is taken to be &
1
δ
log 1δ then
Total Area . δ.
To ensure we can take j this large, we just need to ensure that the number of
1
1
triangles in the first subdivision is large enough, i.e., we need N ≥ 2C δ log δ .
8
MICHAEL BATEMAN
Remark 17. It is important to note that in all of our shuffling around of
triangles, we never broke any of the unit line segments connecting the base of
T0 to the top of T0 . This ensures that the final set still contains a segment in
every direction. (Or at least in a full 60 degree span of directions; by taking
six rotated copies of this set we get all 360 degrees.)
3.3. A small set with a large reach. In this section, we modify our
Perron tree construction to exhibit collections of rectangles with a different
strange property.
reachthm
Theorem 18. Fix > 0. There exists a collection of rectangles R such that
[ R < R∈R
but
[
5R ≥ 1,
R∈R
where 5R is the rectangle with same center and orientation as R but five
times the length.
Remark 19. The number 5 is not special. Any number strictly bigger than
1 is enough.
In fact we will prove a slightly stronger statement. First we define the
reach of a rectangle as follows. A rectangle R is pointed in a direction v(R),
where v(R) is a unit vector. Then the reach is defined to be the set
Reach(R) = R + 2Length(R)v,
i.e., the reach is a translate of R along its long axis. Again, the 2 in the
definition of reach is not important. We just want something a bit bigger
than 1.
Theorem 20. Fix > 0. There exists a collection of rectangles R such that
[ R < ,
R∈R
such that
X
|R| ≥ 1,
R∈R
and such that the sets {Reach(R)}R∈R are pairwise disjoint.
Remark 21. We proceed very much as in the previous section, but with
different details to reflect different priorities. Recall that in the last section
we were interested in preserving unit line segments in certain directions. We
no longer care about that, but we now need to pay attention to the reaches
of the rectangles (triangles) we are moving.
9
3.3.1. A different kind of sprouting. Let h0 , h1 , . . . be a sequence of heights
to be specified momentarily.
We briefly describe a different method of sprouting a triangle. Consider a
triangle T with height hj and width X. Suppose its vertices are called A,B,
C, with C being the top. Assume the segment AB is horizontal. Define two
new points D and E as follows: extend the segment AC until it reaches a
height of hj+1 above the base, and call the endpont D. Similarly, extend
the segment BC until it reaches a height of hj+1 above the base, and call
the endpont E. Now let M be the midpoint of the base AB. Form two
new triangles AM D and BM E. Once again we have a triangle (the original
triangle T ) together with a bowtie-shaped region. Again by using basic
triangle geometry, we see that the width of the bowtie is
.X
(hj+1 − hj )
.
hj+1
The height of this region is hj+1 − hj , giving us that
Area of bowtie . X
(hj+1 − hj )2
.
hj+1
3.3.2. Constructing the set. We will prove a rescaled version ofTheorem 18;
i.e., we will obtain a set of size ∼ 1 whose reach has lage size. A simple
rescaling proves the result claimed there. Starting then with an original
triangle having width 1 and height equal to h0 = 1, we sprout T0 into two
triangles TL and TR of height h1 and width 21 . Now apply the sprouting
process to both TL and TR to obtain 4 total triangles of height h2 and width
1
j
−j and height h . By our
j
4 . After j steps, we have 2 triangles of width 2
−j
calculation earlier (now using X = 2 ),
Area of single bowtie at stage j . 2−j
(hj+1 − hj )2
,
hj+1
but there are 2j new bowties created at each stage, giving
Total Area of bowties =
.
j
X
i=0
j
X
i=0
Area of bowties from step i
(hj+1 − hj )2
.
hj+1
We now make our selection of the heights: Let h0 = 1, and for j ≥ 1 let
1
hj = hj−1 +
.
j+1
Hence for large j, we have hj ∼ log j, so
2
j X
1
1
Total Area of bowties .
i+2
log i
i=0
10
MICHAEL BATEMAN
. 1.
Since the area of the whole figure is equal to the area of the bowties plus
the area of the original triangle, which has area ≤ 1, we have that the whole
figure has area . 1.
3.3.3. Showing the set has large reach. Consider a final triangle T arising
in the last stage of our construction, stage j. Define the rectangle R(T ) to
h
be any rectangle contained in T having height approximately 2j and having
width 2−j−2 . These requirements guarantee that R(T ) covers a substantial
portion of the lower half of T , and is pointed in the same direction as T .
Now translate R(T ) downward along its long axis a distance of 2Length(R).
Call the resulting rectangle R̃(T ).
Claim 22. The rectangles R̃(T ) are pairwise disjoint.
Proof. Notice that in the sprouting procedure, when the triangle T sprouted
into TL and TR , the top of TL is on the right and its base is on the left– hence
the line joining the base of TL to its top are clockwise to the corresponding
line segments from TR . This guarantees that, at least for this simple one
stage picture, R̃(TL ) ∩ R̃(TR ) = ∅. The claim for the rectangles in the final
stage follows in exactly the same way.
Notice that the height of each triangle in the construction is approximately
log j, where j is the final stage in the construction, and the width of each
triangle is approximately 2−j. Hence each triangle has area approximately
log j
. There are 2j such triangles, so the union of the reaches has area
2j
log j
= log j.
2j
Since the set constructed above (without reaches) has area . 1, and since
the reaches have collective area & log j, we have proven the theorem (after
rescaling).
Number of triangles × area of each triangle & 2j ×
4. Differentiation Non-Theorems
In this section we discuss analogs of the Lebesgue Differentiation theorem.
Specifically, we give a negative answer to the question
Question 23. Suppose f : Rn → R is an integrable function. Is it the case
that
Z
1
lim
f (x − y)dy = f (x)
→0 |R | R
(except for x in a set of measure zero) where R is any rectangle of length
(and width) less than ?
In fact we can find counterexamples that are bounded, and hence in every
Lp space, p ≤ ∞, rather than merely integrable. The key point is the
11
existence of small sets with large reach, as described above. Here is the
statement we proved:
Theorem 24. Fix > 0. There exists a collection of rectangles R such that
[ R < R∈R
but
[
5R ≥ 1,
R∈R
where 5R is the rectangle with same center and orientation as R but five
times the length.
We will need a sequence of such constructions. Fix a small number δ.
For each n = 1, 2, . . . , apply the previous theorem to find a collection Rn of
rectangles such that
[ R < 4−n δ
R∈Rn
but
[
5R
≥ 1.
R∈Rn
Furthermore, by scaling, we can assume that all the rectangles have length
≤ 2−n .
Remark 25. Notice that if we scale down the collection by 2−n it may have
area ∼ 2−n as well. This can be repaired by taking the union of 2n translated
copies of the original collection.
Also, we can arrange that all of the rectangles 5R with R ∈ Rn lie in
the square [0, 10]2 . Let xn be a sequence of points to be specified later. (I
encourage you to pretend xn is the zero vector.) Define
[
Kn = xn +
R
R∈Rn
En = xn +
[
5R.
R∈Rn
Now we are ready to define the counterexample function: Let
f (x) = 1∪∞
(x).
n=1 Kn
Remark 26. In previous versions of these notes, the function was defined
a little differently. That version was OK, but a little more complicated with
no benefit.
12
MICHAEL BATEMAN
Notice that
|{x : f (x) 6= 0}| ≤
∞
X
n=1
|Kn | ≤
∞
X
4−n δ < δ.
n=1
The really important point is
Proposition 27. If x ∈ En , then there exists a rectangle R0 3 x with
Length(R0 ) . 2−n such that
Z
1
1
f (y)dy ≥ .
0
|R | y∈R0
25
S
Proof. By definition, En = R∈R 5R. So there exists R ∈ Rn such that
x ∈ 5R. But then
Z
Z
1
1
f (y)dy ≥
1K (y)dy
|5R| y∈5R
|5R| y∈5R n
Z
1
≥
1R (y)dy
|5R| y∈R
1
≥
.
25
Now take R0 = 5R, and notice that length(5R) ≤ 5 · 2−n .
(In other words, En is in the reach of Kn .) This claim has the following
important corollary:
Corollary 28. Fix x. Suppose there exists an (infinite) sequence n1 < n2 <
n3 , . . . such that x ∈ Enj for each nj . Then
Z
1
1
f (y)dy ≥ .
lim sup
5
x∈R,|R|→0 |R| R
Proof. For each nk , we get a corresponding rectangle Rk by the previous
claim, with length . 2−nk , such that
Z
1
1
f (y)dy ≥ .
|Rk | Rk
5
But now we have a sequence of such rectangles, with lengths tending to zero,
and the average of f on Rk is ≥ 51 . This is exactly what we claimed.
It remains to show that there are lots of points x outside the support of
f that still satisfy the hypotheses of the previous corollary. We prove that
now:
sizeofA
Proposition 29. Let A = {x : x ∈ infinitely many En }. Then |A| ≥
1
10 .
Remark 30. The important point of this proposition is that the set A is
much much larger than the support of f , which has size at most δ, where δ
is a number we have chosen to be small.
13
We actually prove the following more general assertion, which easily applies in our setting. We write S = [0, 1]2 .
Lemma 31. Suppose A1 , A2 , . . . is a sequence of sets satisfying An ⊆ S and
|An | ≥ c for each j. Then there exists a sequence of translates xn satisfying
X
1xn +An (x) = ∞
n
for almost every x ∈ S.
Proof. First we note that N contains infinitely many disjoint subsequences
of infinite length: for example,
[
{pk }k≥1 ,
N⊇
p prime
where the union is disjoint and where each {pk }k≥1 is infinite. Hence for
any sequence xn ,
∞
X X
X
1xpk +Apk (x).
1xn +An (x) ≥
n
p prime k=1
Hence we can prove the lemma by proving the following similar claim:
Claim 32. If B1 , B2 , . . . is a sequence of sets satisfying Bk ⊆ [0, 1]2 and
|Bk | ≥ c for each j. Then there exists a sequence of translates xn satisfying
X
1xk +Bk (x) ≥ 1
n
for almost every x ∈ S.
Suppose we have defined translates t1 , t2 , . . . tj−1 . Write
Fj = S \ (t1 + B1 ) ∪ · · · ∪ (tj−1 + Bj−1 ) .
Our goal is to show that there is a translate tj + Bj such that
(tj + Bj ) ∩ Fj
is somewhat large, i.e., that a substantial part of tj + Bj lies outside the
translates of B1 , B2 , . . . Bj−1 . To see this, just note that
Z
Z
Z
|(t + Bj ) ∩ Fj | =
1t+Bj (x)1Fj (x)dxdt
t∈[−1,1]2
t∈[−1,1]2 x∈R2
Z
Z
=
1x−Bj (t)1Fj (x)dtdx
x∈R2
t∈[−1,1]2
= |Bj ||Fj |.
Here we have used that fact that if x ∈ Fj ⊆ S, then x − Bj ⊆ [−1, 1]2 .
Hence by the pigeonhole principle there exists t ∈ [−1, 1]2 such that
1
|(t + Bj ) ∩ Fj | ≥ |Bj ||Fj |.
4
14
MICHAEL BATEMAN
Denote this particular t by tj .
Now we just need to verify that the set
[
(ti + Bi )
i
is filling out the square S. Write
j
[
pj = (ti + Bi ) .
i=1
Note that |Fj | = 1 − pj−1 and that
j
j−1
[
[
pj − pj−1 = (ti + Bi ) − (ti + Bi ) = |(tj + Bj ) ∩ Fj |.
i=1
i=1
We have shown that
1
c
pj − pj−1 ≥ |Bj |(1 − pj−1 ) ≥ (1 − pj−1 ).
4
4
If sup pj < 1 − , then the right side above is bigger than c
4 , implying that
pj → ∞, which is a contradiction. Hence sup pj = 1, which is what we
wanted to prove.
5. The Kakeya conjecture in two dimensions
In this section we prove the Kakeya conjecture in R2 . To do this we must
first discuss Minkowski dimension so that the conjecture makes sense.
5.1. Minkowski dimension. Let
N (S) = {x : dist(x, S) ≤ }
= -neighborhood of S.
Definition 33 (Minkowski dimension). The upper Minkowski dimension of
S is defined to be
log |N (S)|
lim sup n −
log →0
The lower Minkowski dimension of S is defined to be
log |N (S)|
.
lim inf n −
→0
log If the limit exists, the Minkowski dimension of S is defined to be
log |N (S)|
lim n −
.
→0
log Let’s consider some simple examples to check that this notion of dimension
seems sensible.
15
Example 34. Consider the square S = [0, 1] × [0, 1] × {0} as a subset of
R3 , and let > 0 be small. We have
≤ |N (S)| ≤ 4.
Hence
log |N (S)|
log lim inf n −
≥ lim 3 −
→0
→0
log log = 3−1
= 2.
Hence S has lower Minkowski dimension ≥ 2. Similarly,
log 4
log |N (S)|
≤ lim 3 −
lim sup n −
→0
log log →0
= 3−1
= 2.
So S has upper Minkowski dimension ≤ 2, which implies it has Minkowski
dimension 2.
A similar calculation would show that the unit line segment [0, 1] × {0} ×
{0} has Minkowski dimension 1.
Remark 35. If we viewed S as a subset of Rn , we would still get dim(S) = 2,
but the calculation would look slightly different.
Example 36. Let C be the middle
iterating the following procedure:
C0
C1
C2
Cn
1
2
Cantor set, i.e., the set formed by
=
[0, 1]
3
1
= [0, ] ∪ [ , 1]
4
4
1
3 1
3 13
15
= = [0, ] ∪ [ , ] ∪ [ , ] ∪ [ , 1]
16
16 4
4 16
16
···
[
=
s + [0, 4−n ]
s∈Sn
where
Sn = {.a1 a2 . . . an : aj ∈ {0, 3}}
and the numbers in Sn are interpreted base 4. Finally define
C=
∞
\
Cn .
n=1
We see that if ∈ [4−n , 4−n+1 ], then
|N (C)| ≥ |Cn | = 2n 4−n = 2−n ≥
1√
,
2
16
MICHAEL BATEMAN
because Cn is the union of 2n intervals of width 4−n . Similarly, we have
√
|N (C)| ≤ 2n 4−n+2 = 16 · 2−n ≤ 16 .
This means that
lim inf
→0
and
√ !
log | 12 |
1
1−
≥
log 2
√ 1
log |16 |
≤ .
lim sup 1 −
log 2
→0
Hence dim(C) = 12 .
5.2. 2-D Kakeya. With this definition, we now have
Theorem 37. Let E ⊆ R2 be a Besicovitch set. Then E has Minkowski
dimension 2.
Fix a small number > 0, and for convenience assume = 2−n for some
n. We will show that
1
.
|N (E)| &
log 1
From this we conclude that


C
log
1
log |N (E)|
log 
lim inf 2 −
≥ lim inf 2 −
→0
→0
log log !
log log 1
= lim inf 2 −
→0
log 1
= 2.
Our assumption is that E contains a unit line segment in every direction.
In particular, if we let
1
Ω = Ω = {j : j = 0, 1, . . . , },
then there exists a line segment Lω with slope equal to ω such that Lω ⊆ E
for each ω ∈ Ω. Further, if Rω = N (Lω ), then certainly
Rω ⊆ N (E),
and hence
[
Rω ⊆ N (E).
ω∈Ω
So it suffices to show
[
1
Rω &
,
log 1
ω∈Ω
which is what we’ll do.
17
Remark 38. The proof below gives an important but simple method for
proving statements of the following form: “We have a lot of big sets. They
don’t overlap much. Therefore their union is large.” The “They don’t overlap
much.” part is encoded in Claim 39 below, and is the only part of the proof
where we really use anything about our specific situation. In particular, the
first estimate in the proof immediately below could be translated into the
language of any collection of sets in any measure space.
Proof. For the purposes of compactifying notation a bit, write
[
K=
Rω .
ω∈Ω
Also write 1S to denote the characteristic function of a set S. A key estimate
is
X
XZ
|Rω | =
1Rω
ω∈Ω
ω∈Ω
=
Z X
1Rω
K ω∈Ω
v
!
u
p u Z X
≤
|K|t
(
1Rω )2 ,
K ω∈Ω
by the Cauchy-Schwarz inequality. But we know
X
|Rω | ∼ 1
ω∈Ω
since the sum counts ∼
terms, we see that
1
rectangles each having area ∼ . Rearranging
|K| & R
1
P
.
2
K ( ω∈Ω 1Rω )
It remains to prove
cordoba
Claim 39.
Z
(
X
1Rω )2 . log
K ω∈Ω
1
Proof. To estimate the integral we enlarge the range of integration and then
expand the square:
Z X
Z X
2
(
1Rω ) =
(
1Rω )2
R2 ω∈Ω
K ω∈Ω
Z
=
X X
R2 ω ∈Ω ω ∈Ω
1
2
=
X X
ω1 ∈Ω ω2 ∈Ω
1Rω1 1Rω2
|Rω1 ∩ Rω2 |.
18
MICHAEL BATEMAN
To estimate this double sum, we finally use some geometry. We need the
following lemma:
Lemma 40. Suppose R1 and R2 are rectangles of dimensions 1 by δ with
an angle of θ between them. Then
δ2
.
θ
Proof. Without loss of generality, assume R1 is horizontal and that R2 makes
an angle θ with the horizontal. The intersection is (contained in) a parallelogram P of height δ (since it is contained in R1 ) and width θδ . To see
the claim about the width, draw the relevant picture and label the vertices
of the shorter diagonal of the parallelogram P by b and t, with b on the
bottom of R1 and t on top. Let s be the other vertex on the top of the
parallelogram. We are interested in the distance between s and t. Let r be
the unique point on the side of P connecting s to b at distance δ from t.
Consider the triangle with vertices r, s, t. This is a right triangle and the
angle at s is θ. Hence
|R1 ∩ R2 | .
sin θ =
δ
,
dist(s, t)
so
δ
δ
∼ .
sin θ
θ
width of P = dist(s, t) =
Now that we have estimated the height and width as δ and
we get the estimate claimed in the statement of the lemma.
δ
θ
respectively,
We return to the proof of Claim 39. Now for a fixed ω1 ∈ Ω, let
Aj (ω1 ) = {ω2 : |ω1 − ω2 | ∼ 2j }.
Notice that for each ω2 ∈ Aj (ω1 ) we have
|Rω1 ∩ Rω2 | .
.
2j
Also notice that
#Aj (ω1 ) ∼ 2j .
Hence for each fixed ω1 ,
log
X
|Rω1 ∩ Rω2 | =
ω2
1
X
X
|Rω1 ∩ Rω2 |
j=0 ω2 ∈Aj (ω1 )
log
.
1
X
#Aj (ω1 )
j=0
1
. log .
2j
19
Finally summing over ω1 , we get (since there are ∼
XX
1
|Rω1 ∩ Rω2 | . log ,
ω
ω
1
1
such values of ω1 )
2
which is what we needed to show.
5.3. The maximal Kakeya conjecture. In the previous section we actually proved the so-called maximal Kakeya conjecture in two dimensions.
Conjecture 41. Suppose Ω ⊆ Sn−1 is a collection of -separated directions,
and let Rω ⊆ Rn be a tube pointed in the direction ω. Then for each p > 0
! n
Z X
n−1
X
. −p
|Rω |.
1Rω
ω∈Ω
ω∈Ω
We can easily deduce claim about the Minkowski dimension of Besicovitch
sets from the maximal Kakeya conjecture as in the previous section:
If E is a Besicovitch set, then find a -separated set of directions Ω together with lines Lω for each ω ∈ Ω. Then let Rω = N (Lω ). As before,
[
Rω ⊆ N (E),
ω∈Ω
so it suffices to show for every p > 0 that
[
Rω & p .
ω∈Ω
Write K = ∪ω∈Ω Rω . But we have that |Ω| ∼ −n+1 , and |Rω | ∼ n−1 , so
X
1 ∼
|Rω |
ω∈Ω
=
Z X
1Rω
K ω∈Ω
≤ |K|
1
n
!1− 1
Z X
n
n
(
1Rω ) n−1
,
ω∈Ω
with the last line following from Holder’s inequality. But the maximal conjecture tells us this last quantity is controlled by
n−1
X
n−1
1
1
n
. |K| n −p n .
|K| n −p
|Rω |
Rearranging terms we have that
|K| & (n−1)p ,
20
MICHAEL BATEMAN
with p > 0 arbitrary. But this shows that the dimension of K is arbitrarily
close to n, which proves that the dimension of the Besicovitch set E is
actually equal to n.
6. Arithmetic Approach to Kakeya
In this section we start to obtain estimates on the dimension of Kakeya
sets in higher dimensions. Bourgain has shown (with lots of extra work)
that the arguments below can be used to make progress on the maximal
Kakeya conjecture.
6.1. Version Zero. The argument in this section (which is due to Drury?)
hardly deserves any fancy name, and does not really use any substantial additive combinatorics. Nevertheless we will see that it is essentially a primitive
verson of the approach considered in the next section.
Theorem 42. Suppose E ⊆ Rn is a Besicovitch set. Then E has lower
Minkowski dimension at least n+1
2 .
Proof. Certainly we can restrict attention to an even smaller set of directions
since if we can prove a lower bound on the dimension of the smaller set
then we can also prove a lower bound on the dimension of the bigger set.
So assume all lines make an angle ≤ π4 with the x1 -axis. We make two
simplifications before proceeding with the main proof. First, we assume
that each line intersects the hyperplane {x ∈ Rn : x1 = 0} as well as the
hyperplane {x ∈ Rn : x1 = 1}. Since all of the lines point more or less in
the x1 direction,
this assumption can be attained by merely scaling by a
√
factor of 2 and by assuming that all tubes lie near the strip 0 ≤ x1 ≤ 1.
See the remark immediately below for why we can assume this. Second, we
now parametrize the rectangles by a set Ω ⊆ Rn−1 . More specifically, since
the Besicovitch set contains a segment in every direction, we can assume
that for every z ∈ [0, 1]2 , there exists a segment l such that the slope of l
is z. (A line has slope ω ∈ Rn−1 if it contains the points (0, z0 ) and (1, z1 )
and z1 − z0 = ω.) As usual, we consider an -separted set of slopes Ω and
associated tubes Rω .
Remark 43. The only (as yet) unjustified assumption above is the assumption that all lines lie near the strip where 0 ≤ x1 ≤ 1. But examination of
the proof below shows that we do not use the size of Ω except in the final
calculation. So what we actually show is that
|K|2 2(1−n) & #Ω.
So if the tubes are spread over many different strips, we prove the above
result in each slice, and then notice that

2
X
X
|K|2 ≥ 
|Kstrip | ≥
|Kstrip |2 & 2(n−1) #Ω
strips
strips
21
Let’s write K = ∪Rω . Write K(t) = {x ∈ K : x1 = t}. Our simplifying
assumption tells us that
Z 1
|K| ≥
voln−1 (K(t))dt,
0
Our strategy is to show that K is large by showing that K(t) is large for
a typical t, and that the size of K(t) is related to the size of K itself. By
Chebyshev/Markov/obviousness, we have that
1
|{t : voln−1 (K(t)) ≥ M |K|}| ≤
M
for any M ; let’s use M = 100. In particular,
99
,
|{t : voln−1 (K(t)) ≤ 100|K|}| ≥
100
so there exist t1 , t2 such that t1 + 12 ≤ t2 and such that
voln−1 (K(tj )) ≤ 100|K|
for j = 1, 2. We now discretize the slices corresponding to t1 and t2 . Specifically, let
Aj = {z ∈ cZn−1 : (tj , z) ∈ K},
where c is a dimensional constant. We will see below that c = 2√1 n is good
enough. The point of this definition is that the sets Aj are discrete and
cleaner, but they still are related to the sets K(tj ). Specifically, we have
Claim 44. voln−1 (K(tj )) & #Aj n−1 .
c
such that Bz ⊆
Proof. For each z ∈ Aj , there exists a ball Bz of radius 100
K(tj ). But also the balls {Bz }z∈Aj are pairwise disjoint, and each has
volume ∼ n−1 , so
c
voln−1 (K(tj )) ≥ #Aj voln−1 (B(
)) ∼ #Aj n−1 .
100
Further, we can now relate Ω to these discrete sets as well. Recall the
definition
A1 − A2 := {a1 − a2 : aj ∈ Aj }.
Claim 45. We have
#(A1 − A2 ) & #Ω.
Proof. To prove this we appeal to two subclaims. First, for every ω ∈ Ω,
there exists (a1 , a2 ) ∈ A1 × A2 such that Rω 3 (tj , aj ) for j = 1, 2. To
see this, note that we assumed for each ω ∈ Ω there was a tube Rω with
“slope” ω. (A tube has slope ω ∈ Rn−1 if its central axis contains the points
(0, z0 ) and (1, z1 ) and z1 − z0 = ω.) Since each tube intersects K(tj ) in an
n − 1 dimensional ball of radius , all we need to show is that any such ball
22
MICHAEL BATEMAN
intersects the lattice cZn−1 . But as long as c < √1n then this holds, since
√
the diameter of a “unit” cube in the lattice is n. This gives us the points
aj ∈ Aj such that (tj , aj ) ∈ Rω , and so proves the first subclaim. Note that
this subclaim required an upper bound on the parameter c. The next claim
requires only a constant lower bound on c. The particular value 2√1 n meets
both requirements.
The second subclaim is that for any a1 , a2 , there are . 1 values of ω such
that Rω 3 (tj , aj ) for j = 1, 2. To see this, suppose that Rω 3 (tj , aj ) for
1
j = 1, 2. First note that since Rω has width , there are at most cn−1
points
n−1
of cZ
that could possibly lie in the intersection of Rω with K(tj ). Fix
such a point a1 in A1 . If Rω also passes through (t2 , a2 ), then
a1 + ω(t2 − t1 ) ∈ a2 + Bn−1 ().
In words: if we start at t1 with height essentially a1 and go in direction ω,
then when we reach tj we must be at height essentially a2 . But this implies
ω(t2 − t1 ) ∈ a2 − a1 + Bn−1 (),
which implies
ω∈
a2 − a1
+ Bn−1 (2),
t2 − t1
since t2 − t1 ≥ 12 . So all possible ω lie in a ball of radius . , and hence
there are only . 1 of them since they are -separated. This proves the
second subclaim.
So by the first subclaim we have a map φ : Ω → A2 −A1 , and by the second
subclaim the map φ is (.one)-to-one. This implies #(A2 − A1 ) & #Ω, as
was claimed.
Hence
(|K|1−n )2 & #A1 · #A2
≥ #(A1 − A2 )
& −n+1 .
This implies
|K| & n−1
2
.
Ths is precisely the estimate we need, since now we have
log |N (E)|
log |K|
lim inf n −
≥ lim inf n −
→0
→0
log log !
n−1
log 2
≥ lim inf n −
→0
log =
n+1
.
2
23
Now we use the same strategy together with a bit more additive combinatorics to obtain even better Kakeya estimates.
6.2. Using three slices. Now we use the same strategy together with a
bit more additive combinatorics to obtain even better Kakeya estimates.
We proceed as in the previous section, again finding a set of -separated
directions, and constructing the tubes Rω and the set K. Again we see that
99
|{t : voln−1 (K(t)) ≤ 100|K|}| ≥
.
100
But instead of finding two good slices, we now find three good slices in
arithmetic progression. I.e.,
Claim 46. There exist t1 , t2 , t3 ∈ [0, 1] such that t2 − t1 = t3 − t2 , such that
t3 ≥ t2 + 31 ≥ t1 + 23 , and such that
voln−1 (K(tj )) ≤ 100|K|
for j = 1, 2, 3.
This easily follows from the following basic lemma:
Lemma 47. Suppose T ⊆ [0, 1] and |T | ≥ .99. Then T contains a threeterm arithmetic progression with common difference at least 13 .
Proof. Since |T | ≥ .99, there is t2 ∈ T ∩ [.49, .51]. Choose any such t2 . For
convenience of notation, say t2 = 21 . Let
B = {t ∈ T ∩ [0, .1] : 1 − t 6∈ T }.
Of course |B| ≤ .01 since (1 − B) ⊆ [0, 1] \ T . Hence
|{t ∈ T ∩ [0, .1] : 1 − t ∈ T }| ≥ .09.
In particular, there is a t ≤ .1 such that t, 21 , 1 − t ∈ T . Finally, t, 21 , 1 − t is
a progression so this completes the proof.
Now by a simple rescaling we will assume that t1 = 0, t2 =
Again let
1
2 , t3
= 1.
Aj = {z ∈ cZn−1 : (tj , z) ∈ K}
for j = 1, 2, 3. As before we have the estimates |K| & voln−1 (K(tj )) ∼
#Aj n−1 , and A3 − A1 has size ∼ #Ω ∼ −n+1 since K contains a tube in
each of the directions in Ω. Now define
G = {(a, b) ∈ A1 × A3 : there is a tube connecting (0, a) to (1, b)}.
Write
G+ = {a + b : (a, b) ∈ G},
G− = {a − b : (a, b) ∈ G}.
Since we have tubes in every direction, we know
#G− & |Ω| ∼ −n+1 .
24
MICHAEL BATEMAN
But since A2 is small, we also know that
#G+ . #A2 . |K|1−n .
The key, but difficult, lemma proved by Bourgain is
sumdiff
Lemma 48. Suppose A, A0 are subsets of Zn−1 . Suppose #A ≤ L, #A0 ≤
L, and let G ⊆ A × A0 be such that
#G+ ≤ L.
Then
#G− ≤ L2−β ,
where β is universal.
Remark 49. This lemma holds in any abelian group, a fact that is a consequence of the style of proof, which is mostly very elementary. Please note
that “elementary” does not mean “easy to read” or “easy to discover”, and
is in fact not pejorative in any sense.
If we apply this lemma with A = A1 , A0 = A3 , and L = |K|1−n , then we
get
#G− . L2−β = |K|2−β (1−n)(2−β) .
But since we already know
#G− & |Ω| ∼ −n+1 ,
we have
|K|2−β (1−n)(2−β) & −n+1 .
Unraveling all of this implies
|K| & (n−1)(1−β)
(2−β)
.
Remark 50. Bourgain proves Lemma 48 with β =
mensional bound for Besicovitch sets of
13n + 12
.
25
1
13 ,
which gives a di-
7. Points and Lines
Let P ⊆ R2 and let L be a collection of lines. Define
I(P, L) = |{(p, l) : p ∈ l}|.
We immediately see that
I(P, L) ≤ |P ||L|
since at most each point can be on each line. We will see soon that there is
actually a simple way to improve this inequality to
p
I(P, L) ≤ |P | |L| + |L|.
25
(The second term above is only needed when the number of points and lines
is way out of balance; you may wish to ignore it for now.) This particular
estimate will appear as an intermediate step on the way to proving the
following:
Theorem 51 (Szemeredi-Trotter). Using notation as defined immediately
above,
2
I(P, L) . (|P ||L|) 3 + |P | + |L|.
In addition to having a direct relation to the Kakeya-type problems we’ve
been discussing, this theorem can be used to prove the sum-product theorem,
a connection we explore in the next section. The most noteworthy point
about the exponent 23 in the theorem is that it is the best possible.
Example 52. In this example we write [1, L] to indicate only the integers
in the interval from 1 to L. Fix large positive integers M ≥ N . Also assume
M
N is an integer. Consider a grid of points P = [1, N ] × [1, M ]. Let L be the
M
collection of lines with y-intercept in [1, M
2 ] and slope in [1, N ].
Claim 53. Each line l ∈ L contains
&N
2
elements of P
Proof. A generic l is given by the equation y = ax + b for b ∈ [1, M
2 ] and
N
M N
M
M
a ∈ [1, N ] , so for each x ∈ [1, 2 ] we have y ≤ N 2 + 2 ≤ M . Hence each
point (x, y) satifying y = ax + b with x ∈ [1, N2 ] is on the line la,b , and there
are b N2 c such points.
Notice |L| =
M M
2 N
=
M2
2N
and |P | = M N . Hence
I(P, L) ≥ |L|
N
M2 N
M2
=
=
.
2
2N 2
4
But notice that
2
(|P ||L|) 3 = (M N
M2 2
M2
)3 = 2 .
2N
23
So
M2
+ M N ≤ 3M 2 ≤ 12I(P, L)
N
This proves the Szemeredi-Trotter theorem is sharp.
2
(|P ||L|) 3 + |P | + |L| ≤ M 2 +
7.1. The sum-product phenomenon.
Definition 54. Let A ⊆ R be a finite set. Define
A + A = {a + b : a, b ∈ A}
AA = {ab : a, b ∈ A}.
The ratios
|A + A|
|A|
26
MICHAEL BATEMAN
and
|AA|
|A|
are called the additive and multiplicative doubling constants, respectively.
Definition 55. A generalized arithmetic progression of rank R is a set of
the form
{a +
R
X
dj hj : hj ∈ [0, Lj ]}.
j=1
The number Lj is the length in the j direction, and dj is the separation in
the j direction. A generalized geometric progression of rank R is a set of the
form
{a
R
Y
h
dj j : hj ∈ [0, Lj ]}
j=1
with dj 6= 0 6= a.
Example 56. Verify that a generalized arithmetic (geometric) progression
of rank R has additive (multiplicative) doubling constant less than 2R .
An important theorem from additive combinatorics is
Theorem 57 (Freiman). Suppose |A + A| ≤ K|A|. Then there exists a
generalized arithmetic progression P of rank RK and size ≤ CK |A| such
that A ⊆ P . Here RK ,CK are constants depending only on K; in particular
they do not depend on the set A.
In other words, a set with small additive doubling looks like an arithmetic
progression. By restating this theorem in multiplicative language rather
than additive language, we also have
Theorem 58. Suppose |AA| ≤ K|A|. Then there exists a generalized geometric progression P of rank RK and size ≤ CK |A| such that A ⊆ P . Here
RK ,CK are constants depending only on K; in particular they do not depend
on the set A.
In other words, a set with small multiplicative doubling looks like a geometric progression. The sum-product phenomenon is that a set may not
simultaneously have small additive and multiplicative doubling. Precisely,
Theorem 59. Either
4
|A + A| & |A| 3
or
4
|AA| & |A| 3 .
27
We will prove this theorem by appealing to the Szemeredi-Trotter theorem
in a very clever way. Actually, we will prove this theorem with exponent
of 45 rather than 43 . The 34 result is in the same spirit but technically a bit
more involved. The proof given here is due to Elekes, and can also be found
in [17]. First we present a similar result whose proof is a bit less tricky.
Theorem 60 (Sum-product flavor).
3
|A + AA| & |A| 2 .
Proof. Notice that
A + AA =
[
A + xA,
x∈A
so to prove our estimate it suffices to show there exists an x ∈ A such that
3
|A + xA| & |A| 2
We will do this with an appeal to the Szemeredi-Trotter theorem. Define
la,b = {(x, y) : y = ax + b},
and let
L = {la,b : a, b ∈ A}.
Let P = {(x, y) : x ∈ A, y ∈ A + xA}. Note that each line l ∈ L contains
exactly |A| points in P ; specifically the line la,b contains all the points
(x, ax + b) where x ∈ A.
Hence
2
4
2
|A|3 = |L||A| ≤ I(P, L) . (|L||P |) 3 = |A| 3 |P | 3 ,
which implies
5
|A| 2 . |P |.
3
But this means there is some x such that |A + xA| & |A| 2 , as claimed.
Now we prove the sum-product theorem.
Szemeredi-Trotter implies sum-product. Recall that we have A ⊆ R, a finite
set. Define P to be the set of points formed by the Cartesian product
P = (A + A) × AA = {(x, y) : x ∈ A + A, y ∈ AA}.
For a pair a, b ∈ A, define the corresponding line
la,b = {(x, y) : y = a(x − b)}.
Define L to be the set of lines
L = {la,b : a, b ∈ A}.
Notice that |L| =
|A|2 .
The key observation is that
Claim 61. Each line L contains at least |A| elements of P .
28
MICHAEL BATEMAN
Proof. For each line l in L there is a pair a, b such that l = la,b . Just notice
that each point
(b + c, ac) ∈ P
by definition of P . Also notice that a((b + c) − b) = ac, so
(b + c, ac) ∈ la,b
as well. This proves the claim.
With this claim, we already have that
2
|L||A| ≤ I(P, L) . (|P ||L|) 3 .
Inserting known values for |P | and |L| gives us
2
4
|A|3 . (|A + A||AA|) 3 |A| 3 .
Rearranging yields that either
5
|A + A| & |A| 4
or
5
|AA| & |A| 4 .
8. Proving Szemeredi-Trotter
In this section we prove the Szemeredi-Trotter theorem using the cell
decomposition technique. The idea behind this technique is sort of like an
induction-on-scales or bootstrapping argument. We carve the plane into
subregions; we show that a typical subregion is not intersected by too many
lines; then we use a weaker, already known incidence theorem inside each
subregion.
We begin by proving the above-mentioned weaker incidence theorem.
easyinc
Lemma 62 (Easy incidence theorem). Suppose P ⊆ R2 is a finite set and
suppose L is a finite set of lines. Then
p
I(P, L) . |P | |L| + |L|.
Proof. We use Cauchy-Schwarz to see that
X
I(P, L) =
|P ∩ l|
l∈L
p sX
|L|
|P ∩ l|2 .
≤
l∈L
29
We estimate
2

X
|P ∩ l|2 =
X
X

p∈P
l∈L
l∈L
=
1l (p)
X X
|{l ∈ L : p1 , p2 ∈ l}|
p1 ∈P p2 ∈P
=
X
|{l ∈ L : p1 ∈ l} +
p1 ∈P
X X
|{l ∈ L : p1 , p2 ∈ l}|.
p1 ∈P p2 6=p1
We notice that the first sum is just the number I(P, L) we are trying to
estimate. To control the second (double) sum, notice that for any pair
p1 6= p2 there is at most one line through both points. Hence
X
|P ∩ l|2 . I(P, L) + |P |2 .
l∈L
Combining this with our first Cauchy-Schwarz estimate implies that
p p
I(P, L) . |L| I(P, L) + |P |2 .
(This is still a meaningful estimate even though
√ the quantity I(P, L) appears
on both sides, for the same reason that X . X is a meaningful estimate.)
We now consider two cases: either I(P, L) . |P |2 , in which case we can plug
this into the RHS of the previous display to get
p
p p
I(P, L) . |L| I(P, L) + |P |2 . |L||P |,
which is precisely what we want. The other case is I(P, L) ≥ |P |2 : in this
case, we proceed similarly to get
p p
p p
I(P, L) . |L| I(P, L) + |P |2 . |L| I(P, L),
which implies
I(P, L) . |L|.
In either case we have
I(P, L) .
p
|L||P | + |L|,
which is what we wanted to show.
Next we describe how to prove the Szemeredi-Trotter theorem given a
nice cell decomposition theorem, and then we see how to prove the cell
decomposition theorem.
induction
Lemma 63 (Controlling incidences with a cell decomposition). Suppose P 0 ⊆
R2 is a finite set and suppose L is a finite set of lines. Suppose we have a
partition
R2 =
M
[
i=1
Ci
30
MICHAEL BATEMAN
such that |P 0 ∩ Ci | .
|P 0 |
M
for each i. Further suppose that
X
|{i : l ∩ Ci 6= ∅}| ≤ N.
l∈L
Then
r
N
+ N.
M
The number N in the previous lemma is deliberately vague. In our application of this lemma we will have a particular, meaningful number in place
of N .
I(P 0 , L) . |P 0 |
Proof. Let Li = {l ∈ L : l ∩ Ci 6= ∅} and let Pi = P ∩ Ci . Again we have
that
M
X
0
I(P , L) =
I(Pi , Li );
i=1
we now apply Lemma 62 to each of these summands I(Pi , Li ) to obtain
M
X
I(P 0 , L) .
(|Pi |
p
|Li | + |Li |).
i=1
Fortunately the cell decomposition has given us the bound |Pi | .
also can use Cauchy-Schwarz to see that
v
uM
M p
X
√ uX
|L | ≤ M t
|L |.
i
|P 0 |
M .
We
i
i=1
i=1
Finally, note that
M
X
|Li | =
i=1
M
X
|{l ∈ L : l ∩ Ci 6= ∅}|
i=1
=
X
|{i : l ∩ Ci 6= ∅}|
l∈L
≤ N
where the second equality is by double-counting, and the inequality is by
assumption. So now we have
m
X
M
p
(|Pi | |Li | + |Li |) .
i=1
.
m
X
|P 0 | X p
|L| +
|Li |
M
i=1
i=1
v
um
M
0
X
√
uX
|P |
Mt
|Li | +
|Li |
M
i=1
.
|P 0 | √
M
√
M N +N
i=1
31
=
√
|P 0 | N
√
+ N.
M
Now we present the cell decomposition theorem. There are several different ways of proving a reasonable cell decomposition theorem. The method
we present here is rather modern and uses the algebraic techniques that have
been quite popular and useful recently.
Theorem 64 (Cell Decomposition). Let P ⊆ Rn be a collection of points.
Fix M ≥ 1. Then there is a nontrivial polynomial f : Rn → R of degree
1
. M n and a partition
Rn = C1 ∪ · · · ∪ CM ∪ {f = 0}
|
such that |P ∩ Ci | . |P
M , each Ci is open, and such that the boundary of each
set Ci is the set {x : f (x) = 0}.
32
MICHAEL BATEMAN
The proof of the cell decomposition theorem follows from the so-called
(Discrete) Polynomial Ham Sandwich Theorem.
Theorem 65 (Discrete Polynomial Ham Sandwich). Let S1 , . . . , Sm be finite
1
sets of points in Rn . There exists a polynomial g : Rn → R of degree . m n
such that for each i = 1, . . . , m,
#(Si ∩ {x : g(x) > 0}) ≤
#(Si ∩ {x : g(x) < 0}) ≤
#Si
2
#Si
.
2
This is proved by appealing to a version where the sets Si are sets of
positive measure and using a limiting argument. We will not prove it here.
In an early form the polynomials were hyperplanes and the number of sets
m limited to three. The terminology (according to legend, and wikipedia)
comes from the case when the three sets are two slices of bread and a piece
of ham. The theorem proves that the sandwich can be cut into two parts,
with each part containing half the ham, and half of each piece of bread.
Proof of Cell Decomposition via Discrete Polynomial Ham Sandwich. We start
with our collection of points P , and apply the ham sandwich theorem with
m = 1. This gives us a polynomial g1 of degree . 1 such that,
P + = {x : g(x) > 0}
P − = {x : g(x) < 0}
Z1 = {x : g(x) = 0}
we have
#P
2
#P
≤
2
= P + ∪ P − ∪ (P ∩ Z1 ).
#P + ≤
#P −
P
Now having sets P1 , . . . , P2k satisfying #Pi ≤
. (2k )
1
n
#P
,
2k
find gk : Rn → R of degree
such that
#Pi+ ≤
#Pi− ≤
#P
#Pi
≤ k+1
2
2
#Pi
#P
≤ k+1 ,
2
2
where Pi± are defined in analogy with P ± . This gives us

 

k
2k
[
[
P =  (P ∩ Zk ) ∪  (Pi+ ∪ Pi− )
j=1
i=1
33
where Zk = {x : gk (x) = 0}. Further, the set
Q
0}, and the polynomial ki=1 gi has degree
≤
k
X
deg(gj ) .
j=1
k
X
j
Sk
i=1 Zk
⊆ {x :
Qk
i=1 gi (x)
=
k
2n ∼ 2n .
j=1
Taking k such that 2k = M , and setting
f=
log
YM
gj
j=1
gives us the required polynomial.
8.1. Proof of Szemeredi-Trotter. We combine the results stated above
to deduce the Szemeredi-Trotter theorem. We safely assume that
p
|L| ≤ |P | ≤ |L|2 ,
for otherwise our theorem can already be proved by an application of Lemma
62. First, let M be a large number to be determined later. We apply the cell
decomposition√
theorem to our collection P to obtain a nontrivial polynomial
f of degree . M and a partition
R2 = C1 ∪ · · · ∪ CM ∪ {f = 0}
such that |P ∩ Ci | .
in the set
|P |
M.
Every point p ∈ P is either in one of the cells or
Z = {(x, y) : f (x, y) = 0}.
Let P0 = P ∩ Z and let P 0 = P \ P0 . Certainly I(P, L) = I(P0 , L) + I(P 0 , L)
8.1.1. Points in the cells. We first notice that we can assume |P 0 |2 ≥ |L|,
for otherwise by Lemma 62
√
I(P 0 , L) ≤ |P 0 | L + |L| ≤ 2|L|,
which would be good enough. This hypothesis will come into play at the
end of the argument when we optimize the number M of cells used in our
cell decomposition. To control I(P 0 , L) we appeal to Lemma 63. To do this
we need an upper bound on
X
|{i : l ∩ Ci 6= ∅}|;
l∈L
to get such a bound we will estimate the number of cells any line can intersect:
Claim 66. For any l ∈ L that meets a point in P 0 ,
√
|{i : l ∩ Ci 6= ∅}| . M .
Proof. To see this just note that the cells Ci √
are bounded by the set Z,
which is the zero set of a polynomial of degree M . Further,
34
MICHAEL BATEMAN
Claim 67. Any line l can meet Z in at most deg(f ) .
l ⊆ Z.
√
M points unless
This claim follows from a basic lemma:
Lemma 68. Let f : Rn → R be a polynomial of degree d. Let l be a line in
Rn . Let Z = {x : f (x) = 0}. Then either #Z ∩ l ≤ d or l ⊆ Z.
Proof. We first note a fact from algebra: If p : R → R is a polynomial of
degree d, then p has ≤ d zeros or p ≡ 0.
Now suppose l = {at + b : t ∈ R} for some a, b ∈ Rn . Define ϕ : Rn → R
by ϕ(t) = at + b. Now consider a term
xd11 · · · xdnn
appearing in the definition of f , where d1 + · · · + dn ≤ d. Evaluating this
term at the point at + b = (a1 t + b, . . . , an t + b) yields
(a1 t + b)d1 · · · (an t + bn )dn
which is a polynomial of degree at most d in the variable t. Hence f ◦ϕ : R →
R is a polynomial of degree ≤ d. This implies that either f ◦ ϕ has ≤ d zeros
or l ⊆ Z.
So now we can apply Lemma 63 using the P 0 and M as above, and using
the estimate
X
√
|{i : l ∩ Ci 6= ∅}| . |L| M ;
l∈L
√
i.e., we use N ∼ |L| M . This results in the incidence bound
s
1
1
√
1
|L|M 2
|P |L 2
I(P 0 , L) . |P 0 |
+ |L|M 2 .
+ |L| M =
1
M
M4
8.1.2. Points in the zero set. Now we estimate I(P0 , L). We saw earlier that
a given line l ∈ L is either entirely
√ contained in Z = {(x, y) : f (x, y) = 0},
or √
it can intersect Z at most ∼ M times since f is a polynomial of degree
. M . Hence if
L0 = {l ∈ L : l ⊆ Z}
L̃ = L \ L0 ,
we already see that
√
√
I(P0 , L̃) . |L̃| M . |L| M .
√
So it remains to consider I(P0 , L0 ). We will show there are fewer than ∼ M
lines contained in Z. Write d for the degree of f . We will show there are no
more than d lines in Z. For suppose there are d+1 lines l1 , . . . , ld+1 contained
in Z. Without loss of generality (by rotating the picture), none of them is
horizontal. Since there are only finitely many lines, all intersections between
35
the lines l1 , . . . , ld+1 happen in a bounded box. Hence any horizontal line h
outside of this box intersects all lines l1 , . . . , ld+1 in different points. Since
f has degree d, this implies h is also contained in Z. But h was rather
arbitrary, so there exists an entire half-plane contained in Z, whch forces f
to be identically zero, contradicting our knowledge of f .
Using the weakpincidence bound established earlier (actually, the dual
version I(P, L) . |P ||L| from Example Sheet 2,) we have
p
√
I(P, L) . |P ||L̃| ≤ |L| M ,
p
where we haved used the estimate |P | . |L|, for otherwise we could have
proved the theorem initially with this weak bound.
8.1.3. Optimizing the number of cells. Recall that our previous arguments
have been carried out with a generic value of M , the number of cells. We
now optimize the estimates above by choosing
4
M=
|P | 3
2
|L| 3
,
which yields
2
I(P 0 , L) . (|P ||L|) 3 .
Notice that of course we need M to be at least 1. Fortunately, this is
guaranteed by our assumption that |P |2 ≥ |L|, which (again) is available to
us because of the “weak” bound. This finishes our proof of the SzemerediTrotter theorem.
8.2. The joints problem. A three-dimensional incidence problem that has
been solved using similar methods is the so-called joints problem. The joints
problem arose as a simple model for the higher dimensional Kakeya problem,
although it has no direct implication for the Kakeya problem. Its importance
at the time of its solution was that it was one of the the first examples
of the algebraic method being used in the Kakeya family of problems in
Euclidean space. Of course it was not the first example of the algebraic
method being used in the Kakeya family of problems – that was Dvir’s
proof of the finite field Kakeya conjecture, which we shall see in the next
section. More recently, it was discovered that certain variants of the joints
problem arise in the study of the Erdos distance problem, which Guth-Katz
solved using, in part, their ideas from the joints problem. The Erdos distance
problem asks, given a finite set A in the plane, what is
#{|a − b| : a, b ∈ A}?
Definition 69. Let L be a collection of lines in R3 . A joint is a point at
which there are at least three non-coplanar lines intersecting.
How many joints can a collection of lines have?
36
MICHAEL BATEMAN
Theorem 70 (Guth-Katz). A collection of N lines in R3 can form at most
3
. N 2 joints.
The particular numerology here is, once again,
perhaps
most remarkable
√
√
because it is sharp. To see this, consider a N × N grid of lines passing
through the integer lattice and pointed in the x direction, another similar
grid in the y-direction, and a third in the
There
are totally 3N
√ z-direction.
√
√
lines, but each point in the integer grid [ N ] × [ N ] × [ N ] is a joint, since
each such point is contained in one of the lines from each of the collections.
3
This gives us a total of N 2 joints, proving sharpness in the theorem above.
We also remark that embedding this problem in three dimensions is necessary. For example, if we consider a collection L of lines in R2 , and define
a 2-joint to be a point in R2 at which three distinct lines of L intersect,
then we still cannot get a non-trivial estimate. To see this just consider
a collection of N horizontal lines with equations y = i for i = 1, . . . , N ,
together with N vertical lines with equations x = i for i = 1, . . . , N , and
∼ N oblique lines with equations y = x + i for i = −N, . . . , −1, 0, 1, . . . , N .
Then we have again ∼ N lines but we have N 2 2-joints, namely every point
in the grid [N ] × [N ]. So the problem needs to be three-dimensional.
We now turn our attention to the proof of the joints theorem. We will need
the following algebraic lemma, which is similar in spirit to the polynomial
ham sandwich theorem.
zeropoly
Lemma 71. Suppose P ⊆ Rn is a finite set. Then there exists a polynomial
1
f of degree . |P | n such that f vanishes on each point of P .
Of course we are interested here in the case n = 3.
Proof. A direct proof of this fact is rather simple, and we will see one in the
next section in the proof of the finite field Kakeya conjecture. But for now,
notice that it follows from the cell decomposition theorem by taking the
parameter M = C|P | for some C sufficiently large to swamp the . notation
in the cell decomposition theorem. Specifically, we are given a constant C̃
|
such that the number of points in each cell is ≤ C̃ |P
M . Choose M = 2C̃|P |
1
so that there are ≤ 2 points in each cell– i.e., every point is in the zero set
of the polynomial f . Notice that the resulting polynomial can be taken to
1
1
have degree . M n . |P | n .
3
We will assume that the lines L form CN 2 joints for some large value
of C. We will see that taking C large enough yields a contradiction. The
outline of the proof is as follows:
1. Essentially every line contains many joints.
2. There exists a low degree (nontrivial) polynomial that vanishes on all
the joints. Let f be such a polynomial with minimal degree.
3. The previous points imply every line contains many zeros of f .
4. Hence every line is entirely contained in the zero set of f .
37
5. Therefore the gradient of f is zero on all the joints. Since the gradient
has lower degree than f , we have contradicted minimality of degree of f .
3
Proof. Let J be the set of joints. Assume |J| = C|L| 2 . We will find a large
subset of the joints J 0 and a large subset of the lines L0 such that each joint
is hit by many lines in L0 . Precisely, run the following loop:
1
WHILE
UPDATE
C|L| 2
there is l ∈ L such that |l ∩ J| ≤
2
L := L \ {l}
J := J \ (J ∩ l).
Call the remaining sets L0 and J 0 . By construction, every line in l ∈ L0 will
|J|
. Further, we have
satisfy |l ∩ J 0 | ≥ 2|L|
|J 0 | ≥
|J|
2
|J|
since at each iteration of the loop we remove at most 2|L|
points, and there
are at most |L| iterations of the loop. Also, every joint in J 0 is formed by
lines from L0 .
1
By Lemma 71, there exists a nonzero polynomial of degree . |J 0 | ≤ |J| 3 =
1
1
C 3 |L| 2 that vanishes on every point of J 0 ; let f be such a polynomial of
minimal degree. Write Z = {x : f (x) = 0}. Notice that for each line l ∈ L0 ,
we have
1
C|L| 2
|J|
=
.
|l ∩ Z| ≥ |l ∩ J | ≥
2|L|
2
0
But for C large enough we have
1
1
1
C|L| 2
>> C 3 |L| 2 ,
2
and hence if C is large enough we have
|l ∩ Z| > deg(f )
for each l ∈ L0 . This implies that in fact l ⊆ Z for each l ∈ L0 .
So far we have shown that for each x ∈ J 0 , there are three non-coplanar
lines in L0 through x such that f vanishes on all three lines. To finish the
proof, we only need to establish the following two claims:
Claim 72. For x ∈ J 0 , ∇(f )(x) = 0.
Proof. Suppose vj is a unit vector in the direction of the line lj , where
l1 , l2 , l3 are three non-coplanar lines in L passing through x. Using a Taylor
expansion we can write for each j,
f (x + tvj ) = f (x) + t∇(f )(x) · vj + O(t2 ).
38
MICHAEL BATEMAN
By taking t → 0, and using that f (x+tvj ) = 0 for all t since f vanishes on all
of lj , we have that ∇(f )(x) · vj = 0. Since v1 , v2 , v3 are linearly independent,
this proves ∇(f )(x) = 0.
Claim 73. One of the partial derivatives p1 , p2 , p3 is a nonzero polynomial
that vanishes on every point of J and has degree strictly less than that of f .
This claim contradicts our choice of f as the nontrivial polynomial of least
degree vanishing on J.
Proof. We just showed that ∇(f )(x) = 0 for x ∈ J 0 . This proves that each
of the partials vanishes on J. It is clear that each partial has degree strictly
less than that of f by standard rules of differentiation. Finally, the only way
that pj is identically zero is if f is constant in xj . The only way that f is
constant in each of x1 , x2 , x3 is if f is constant, and hence identically zero,
on all of R3 . But this would contradict our assumption on f . So one of the
pj is nonzero.
8.3. The finite field Kakeya problem. In this section we prove the finite field Kakeya conjecture. We write Fp to denote the finite field with p
elements.
Definition 74. A set K ⊆ Fnp is a Besicovitch set if for every a ∈ Fnp there
exists b ∈ Fnp such that
{at + b : t ∈ Fp } ⊆ K.
In other words, K contains a line in every direction.
Conjecture 75. For every > 0, there is a constant cn, such that if K ⊆ Fnp
is a Besicovitch set (with p a prime), then
#K ≥ cn, pn− .
The finite field Kakeya conjecture is easily deduced from a formally weaker
statement:
dvir
Theorem 76. There exists a constant cn such that if K ⊆ Rnp is a Besicovitch set, then
#K ≥ cn pn−1 .
To see how this proves the conjecture, just notice that if K ⊆ Fnp is a
Besicovitch set, then K × · · · × K ⊆ Fnd
p is also a Besicovitch set. Here d ≥ 1
is an integer that we will take to be large. Hence
(#K)d = #(K × · · · × K) ≥ cdn pdn−1 ,
which implies
1
1
d
#K ≥ cdn
pn− d .
For any fixed > 0, we can take
1
d
< to recover the required estimate.
39
Remark 77. This is an example of the “tensor-power trick”, which is typically of the form:
1. Prove a weak estimate that holds in any universe.
2. Embed your universe in a bigger universe.
3. Use weak estimate in bigger universe to imply strong estimate in your
universe.
For more on this trick, have a look at [16].
Besicovitch sets are closely related to a different class of strange sets:
Definition 78. We say N ⊆ Fnp is a Nikodym set if for every x ∈ Fnp there
exists a line l such that l ⊆ {x}∪N . (So the line is contained in the Nikodym
set, except for (possibly) the point x.)
ffzeropoly
The strategy for proving Theorem 76 is as follows:
1. We prove that if K is a Kakeya set, then there is a corresponding
Nikodym set N of similar size.
2. We will find a nontrivial polynomial of low degree that vanishes on N .
3. Every point x ∈ Fnp is contained in a line having large intersection with
N , since N is a Nikodym set.
4. Hence every point x ∈ Fnp is contained in a line having large intersection
with the zero set of the polynomial.
5. The polynomial must be identically zero along this line, since it has so
many zeros already; this proves that the polynomial is zero at every point.
This contradicts nontriviality of the polynomial.
The precise result we use to find the polynomial is
Lemma 79. Let E ⊆ Fnp . Assume #E < n+d
d . Then there exists a nontrivial polynomial f of degree ≤ d that vanishes on every point of E.
Proof. We are looking for a polynomial
X
f (x) =
cd1 ,...,dn xd11 · · · xdnn
d1 +···+dn ≤d
such that f (p) = 0 for each p ∈ P . This means we want
X
cd1 ,...,dn pd11 · · · pdnn = 0
d1 +···+dn ≤d
for each p ∈ P . In other words, we want a solution to a system of |P|
homogeneous linear equations in a large number of variables–namely n+d
d
variables. This number n+d
appeared because it is the number of terms
d
in an n-variable polynomial of degree d. To see this, note that n+d
is
d
precisely the number of integer solutions of the equation
d0 + d1 + . . . dn = d
under the constraints di ≥ 0. and each such solution corresponds to a
particular monomial term
1d0 xd11 · · · xdnn .
40
MICHAEL BATEMAN
There exists a nontrivial soluton to a system
of |P | linear homogensou equa
n+d
n+d
tions in d variables as long as d > |P |. (Note a system of nonhomogeneous linear equations need not have a solution. But the claim in the
homogeneous case is just the claim that a linear operator from a space of
dimension R to a space of dimension < R has a nontrivial kernel.) This
proves the lemma.
Remark 80. This fact is the analogue of Lemma 71 that we used in the
real setting in the last section to prove the joints conjecture. The proof we
give here works equally well in that setting.
nikodym
Lemma 81. Suppose K is a Besicovitch set. Then the set N = {tx : t ∈
Fp , x ∈ K} is a Nikodym set. Note this implies #N ≤ p#K.
Proof. We fix x ∈ Fnp . Because K is a Besicovitch set, there exists a line
in K pointing in direction x. This means there is y ∈ Fnp such that {y +
sx : s ∈ Fp } ⊆ K. But for each such point y + sx, the set N contains
s−1 (y + sx) = s−1 y + x for each 0 6= s ∈ Fp . But
{s−1 y + x : 0 6= s ∈ Fp }
is a line, which proves N is a Nikodym set.
Lemma 82 (Schwartz-Zippel). Suppose f ∈ Fp [x1 , . . . , xn ] is a nontrivial
polynomial of degree d. Then f has ≤ dpn−1 zeros.
Proof. We proceed by induction on n. If n = 1, then the result follows from
the fundamental theorem of algebra. Without loss of generality, we may
write
f (x) = xk1 g(x2 , . . . , xn ) + h(x),
for some value of k, where h(x) has degree ≤ k − 1 in x1 , and where g is a
nonzero polynomial of degree ≤ d−k. (Such a decomposiiton exists unless f
is constant in x1 ; in that case, rename the variables.) For fixed (x2 , . . . , xn ),
define
px2 ,...,xn (x1 ) = f (x1 , x2 , . . . , xn ).
By the one-dimensional version of this theorem, either px2 ,...,xn is identically
zero or it has no more than deg(px2 ,...,xn ) zeros. With this in mind, write
Z1 = {(x1 , x2 , . . . , xn ) : f (x1 , x2 , . . . , xn ) = 0 and g(x2 , . . . , xn ) = 0}
Z2 = {(x1 , x2 , . . . , xn ) : f (x1 , x2 , . . . , xn ) = 0 but g(x2 , . . . , xn ) 6= 0}.
Then #Z = #Z1 + #Z2 .
First, notice that
|Z1 | ≤ |{(x1 , x2 , . . . , xn ) : g(x2 , . . . , xn ) = 0}|
= p|{(x2 , . . . , xn ) : g(x2 , . . . , xn ) = 0}|
≤ p(d − k)pn−2
≤ (d − k)pn−1 .
41
where the second inequality holds by the induction hypothesis and the fact
that g is nonzero of degree ≤ d − k.
Second, we have that if g(x2 , . . . , xn ) 6= 0, then with (x2 , . . . , xn ) fixed,
the polynomial f becomes a nonzero polynomial in the variable x1 of degree
k. (Notice that we need g(x2 , . . . , xn ) 6= 0 to ensure that f is a nonzero polynomial in x1 .) Further, there are ≤ pn−1 possible (n − 1)-tuples (x2 , . . . , xn ),
so
|Z2 | ≤ pn−1 deg(fx2 ,...,xn ) ≤ pn−1 k.
Hence
|Z| = |Z1 | + |Z2 | ≤ (d − k)pn−1 + kpn−1 = dpn−1 .
8.3.1. Wrapping up the proof. We now combine the foregoing lemmas to
prove Dvir’s theorem. Suppose we have a Besicovitch set K with
pn−1
C
for some value of C. We will reach a contradiction if C is too large. By
n
Lemma 81, there exists a Nikodym set N such that #N ≤ pC . By Lemma
79, there exists a nontrivial polynomial f of degree d vanishing at every
1
point of N provided dn ∼ d+n
> #N . Fix such a degree d ∼ (#N ) n
n
and polynomial f . Write Z = {x : f (x) = 0}. Now pick any x ∈ Fnp . By
definition of Nikodym sets, there exists a line l such that l ⊆ {x} ∪ N .
Since |l ∩ Z| ≥ |l ∩ N | ≥ p − 1, we see that f vanishes on all of l provided
deg(f ) < p − 1. But we see that this happens if the number C in the
definition of K is too large, for then
1
p
deg(f ) ∼ (#N ) n = 1 <<< p − 1.
Cn
Finally, if f vanishes on all of l, then it vanishes in particular at x. Since
x was arbitrary, the polynomial f vanishes everywhere. But this contradicts the Schwartz-Zippel lemma, which says that f may have at most
deg(f )pn−1 < (p − 1)pn−1 < pn zeros. Hence the number C could not
have been too large. This completes the proof.
#K =
8.4. Fourier transform basics. In this section we record some fundamental facts about the Fourier transform. This section is far from a thorough
treatment of the Fourier transform. We begin with the definition, for sufficiently nice functions f : Rn → C, of a function fˆ: Rn → C:
Z
fˆ(ξ) = e−2πix·ξ f (x)dx.
To start with, we demand f ∈ L1 so there is no question about convergence
in the above integral. After some work, one can extend this definition to all
f ∈ L2 (Rn ). Recall
42
MICHAEL BATEMAN
Definition 83. We write Lp = Lp (Rn ) to denote the space of functions
satisfying
Z
|f |p < ∞.
We write
Z
||f ||p =
p
1
p
|f |
to denote the “Lp norm of f ”. (Note: this is actually a norm, provided we
think of functions being defined only up to sets of measure zero.)
The Lp norms are a convenient way to measure the size of a function, or
equivalently, the distance between two functions.
Some basic properties of the Fourier transform include:
Proposition 84. For functions f and g and a, b ∈ C,
\
af
+ bg = afˆ + bĝ
2πix·ξ ˆ
τd
f (ξ)
x f (ξ) = e
f[
∗ g(ξ) = fˆ(ξ)ĝ(ξ)
fcg(ξ) = fˆ ∗ ĝ(ξ)
d
∂f
(ξ) = 2πiξj fˆ(ξ)
∂xj
∂ fˆ
\
2πix
(ξ).
j f (ξ) = −
∂ξj
There is a class of functions called the Schwartz class , denoted S =
S(Rn ), that is particularly nicely behaved under the Fourier transform. S
is the set of functions that are infinitely differentiable and that satisfy, for
each p1 , . . . , pn , q1 , . . . , qn , the bound
sup xp11 · · · xpnn D1q1 · · · Dnqn f (x1 , . . . , xn ) < ∞.
x∈Rn
Example 85. C0∞ (Rn ) ( S(Rn ). Gaussian functions are Schwartz but not
compactly supported.
We will need in the inversion theorem that the Fourier transform of a
Schwartz function is again a Schwartz function.
Lemma 86. If f ∈ S, then fˆ ∈ S.
Proof. Estimate using the fact that the Fourier transform sends derivatives
to multiplication and vice versa together with the definition of Schwartzness.
The Fourier transform is useful for many reasons, not the least of which
is that it provides a useful description of a function. More specifically, there
43
exists a “Fourier inversion formula”. We will explore to what extent this
inversion formula actually holds. To get a feel for why it ought to be true,
we first prove it in some simple cases.
Example 87 (Finite setting). Consider the group G = Z/N Z for some
positive integer N . For f : G → C, define
1 X −2πix·ξ
fˆ(ξ) =
e N f (x).
|G|
x∈G
This is the Fourier transform in the group G.
Claim 88. We always have
f (x) =
X
e
2πix·ξ
N
fˆ(ξ).
ξ∈G
Proof. To see this, just expand the RHS, using the definition of fˆ:
X 2πix·ξ
X 2πix·ξ 1 X −2πiy·ξ
e N fˆ(ξ) =
e N f (y)
e N
|G|
ξ∈G
ξ∈G
=
X
y∈G
y∈G
1 X 2πi(x−y)·ξ
f (y)
e N
.
|G|
ξ∈G
Notice that
1 X 2πi(x−y)·ξ
e N
=
|G|
ξ∈G
(
1 if x = y
0 otherwise.
(This is just the fact that the N th roots of unity sum to zero, which can be
verified by explicitly computing the geometric series.) Hence
X 2πix·ξ
e N fˆ(ξ) = f (x).
ξ∈G
Notice that in this last calculation it was important that we understand
2πi(x−y)·ξ
1 P
N
. The analogue in the continuous case is
the quantity |G|
ξ∈G e
Z
e2πi(x−y)·ξ dξ.
Rn
Loosely speaking, this quantity is a delta function evaluated at the point
x − y. The following argument makes this idea rigorous. We now prove the
same kind of inversion formula for very nice functions:
Theorem 89 (Fourier inversion for nice functions). For f ∈ S, we have
Z
f (x) = e2πix·ξ fˆ(ξ)dξ.
We will prove several other useful facts on our way to a proof of this
theorem.
44
MICHAEL BATEMAN
Theorem 90 (Plancherel’s Theorem). For f, g ∈ S, we have
Z
Z
f ĝ = fˆg.
Proof. Because f and g are well-behaved, we can interchange the order of
integration below:
Z
Z
Z
f (x)ĝ(x)dx =
f (x) e−2πix·z g(z)dzdx
Z
Z
=
g(z) e−2πix·z f (x)dxdz
Z
=
g(z)fˆ(z)dz.
Suppose we could take g ≡ 1 and we knew ĝ = δ0 . Then
Z
Z
Z
f (0) = f δ0 = fˆ = fˆ(ξ)e−2πi0·ξ .
Further, using the fact that the Fourier transform turns translation into
modulation,
Z
Z
Z
d
f (x) = τx f (0) = τx f δ0 = τx f = e2πix·ξ fˆ(ξ),
which is precisely the Fourier inversion formula we want. To make this
rigorous, we will consider a family of functions that form an “approximation
of the identity”– in other words, a sequence of functions that looks more and
more like a delta function. We will use scaled copies of a Gaussian function.
We need one more lemma, which tells us the Fourier transform of a Gaussian function.
2
Lemma 91. Suppose f : Rn → R is given by f (x) = e−π|x| . Then
fˆ(ξ) = f (ξ).
Proof. First, we reduce to the one-dimensional case. Define g : R → R by
2
g(t) = e−πt . First, note that since
2
2
2
e−π|x| = e−πx1 · · · e−πxn ,
we also have
fˆ(ξ1 , . . . , ξn ) =
=
=
Z
2
2
e−2πi(x1 ξ1 +···+xn ξn ) e−πx1 · · · e−πxn dx1 · · · dxn
n Z
Y
j=1
n
Y
j=1
2
e−2πixj ξj e−πxj dxj
ĝ(ξj ).
45
2
So the lemma will be proved once we see that ĝ(η) = e−πη , because then
fˆ(ξ1 , . . . , ξn ) =
n
Y
2
2
e−πξj = e−π|ξ| .
j=1
So now we handle the one-dimensional case. Our proof of this fact is
especially slick: We show that there is an initial value problem (ODE) with
solution equal to g and also equal to ĝ. By uniqueness of such solutions,
ĝ = g.
The ODE is
u0 + 2πxu = 0
u(0) = 1.
First, note that g solves the IVP because g(0) = 1 and
2
g 0 (x) = e−πx (−2πx) = −2πxg(ξ).
On the other hand,
\
(ĝ)0 (ξ) = (−2πixg)(ξ)
and
\
2πiξĝ(ξ) = gb0 (ξ) = −2πxg(ξ).
Inserting these identities into the ODE above shows that in fact ĝ solves it
as well. Finally, notice ĝ satisfies the initial condition since
Z
Z
2
ĝ(0) = e−2πi0ξ g(x) = e−πx = 1.
We can finally prove the inversion formula.
Proof. Write f to denote the function given to us in the hypotheses of the
2
theorem, and let g(x) = e−π|x| . Write gλ (x) = g( λx ).
Then
Z x
f (0) = lim
f
ĝ(x)dx
λ→∞
λ
Z
= lim
λn f (x)ĝ(λx)dx
λ→∞
Z
= lim
f (x)gbλ (x)dx
λ→∞
Z
= lim
fˆ(ξ)gλ (ξ)dξ
λ→∞
Z
=
fˆ(ξ) lim gλ (ξ)dξ
λ→∞
Z
=
fˆ(ξ)dξ.
46
MICHAEL BATEMAN
Finally, we can apply the previous calculation to the function τa f to conclude
Z
Z
[
f (a) = (τa f )(0) = (τa f )(ξ)dξ = fˆ(ξ)e2πiaξ dξ
for any a. This completes our proof of Fourier inversion for nice functions.
As a nice corollary to the “inversion” formula, we see that the Fourier
transform is actually invertible on S:
ˆ
Corollary 92. For f ∈ S, we have fˆ(x) = f (−x).
We would like to extend this theorem to a broader class of functions. We
will first prove that the Fourier transform is a continuous operator on L2 ;
from there, we can conclude an L2 inversion formula.
Theorem 93 (Parseval’s Theorem). For f ∈ S, we have ||f ||2 = ||fˆ||2 .
¯(−ξ). We want to apply Plancherel’s theorem to f
Proof. Define h(ξ) = fˆ
and h, so we need to know h is well behaved. Fortunately, since f ∈ S, we
also have h ∈ S. Then
Z
Z
Z
Z
Z
Z
Z
¯
|f |2 = f f¯ = f ĥ = fˆh = fˆ(ξ)fˆ¯(−ξ) = fˆfˆ = |fˆ|2 ,
where the third equality is by Parseval’s theorem and the last equality follows
from a simple computation that shows
¯
fˆ¯(−ξ) = fˆ(ξ).
Now that we know the Fourier transform is continuous with respect to
the L2 norm, we can extend our definition of the Fourier transform to all
functions in L2 . In other words, if f ∈ L2 , find a sequence {fj } of Schwartz
functions such that ||fj − f ||2 → 0 (which can be done since Schwartz functions are dense in L2 ), and define
fˆ = lim fˆj .
j→∞
By continuity of the Fourier transform, this definition is independent of the
choice of converging sequence {fj }.
8.5. Norm convergence. We have seen that for nice compactly supported
functions we have the pointwise convergence
Z
Z
2πixξ
ˆ
f (x) = f (ξ)e
dξ = lim
fˆ(ξ)e2πixξ dξ.
R→∞ |ξ|≤R
Define
Z
TR f (x) =
|ξ|≤R
fˆ(ξ)e2πixξ dξ.
47
Do we have
lim ||TR f − f ||p = 0?
R→∞
To address this question we use the following lemma:
Lemma 94. We have
lim ||TR f − f ||p = 0
R→∞
for every f ∈ Lp (Rn ) as long as
||TR f ||p ≤ Cp ||f ||p
for every f ∈
Lp (Rn ).
Proof of ⇐. Fix f ∈ Lp . Find a compactly supported smooth g such that
||f − g|| < . Notice that
||f − TR f ||p ≤ ||f − g||p + ||g − TR g||p + ||TR g − TR f ||p
By assumption, TR has operator norm ≤ Cp independent of R, so ||TR g −
TR f ||p ≤ Cp ||f − g|| ≤ Cp . Also, notice that if g is compactly supported,
then TR g → g pointwise. Since g is compactly supported, (and since the R
needed to get close to g(x) is a continuous function of x,) we have that in
fact ||TR g − g||p → 0 as R → ∞ as well. Hence for R large enough, we have
||f − TR f ||p ≤ + + Cp . ,
which is what we wanted.
Corollary 95. For f ∈ L2 , we have
lim ||TR f − f ||2 = 0.
R→∞
Proof. Notice that
Z
TR f (x) =
fˆ(ξ)e2πixξ dξ =
|ξ|≤R
Z
1BR (ξ)fˆ(ξ)e2πixξ dξ
where BR is the ball of radius R centered at the origin. So we could have
equivalently defined TR on the Fourier side by
ˆ
Td
R f (ξ) = 1B (ξ)f (ξ),
R
because then by Fourier inversion we have
Z
Z
2πixξ
d
TR f (x) = TR f (ξ)e
= 1BR (ξ)fˆ(ξ)e2πixξ .
Hence the corollary follows from the estimate
ˆ
||TR f ||2 = ||Td
R f ||2 ≤ ||f ||2 = ||f ||2 ,
ˆ ˆ
where the inequality is just the pointwise fact Td
R f = 1BR f ≤ f .
There is a converse to this lemma, although stating it precisely is a bit
clumsy. We say only that it is basically a consequence of the uniform boundedness principle
48
MICHAEL BATEMAN
Lemma 96. Suppose {TR } is a family of operators satisfying
sup sup ||TR f ||p = ∞.
R ||f ||p =1
Then there exists f ∈
Lp
with ||f ||p = 1 such that
sup ||TR f ||p = ∞.
R
x
Claim 97. Given a function f , define fR (x) = f ( R
). Then
||T1 fR ||p
||TR f ||p
=
.
||f ||p
||fR ||p
Proof. Compute from the definitions.
From this it follows that the operator norm of TR is independent of R.
So we will focus on T1 .
The important result of this section is
Theorem 98 (C. Fefferman). The operator T1 is unbounded on Lp (Rn ) for
every p 6= 2, n ≥ 2.
We remark quickly that it is enough to prove the theorem for p > 2 by
duality. Specifically, we can use Plancherel to check that
Z
Z
T f · h = f˜ · T h̃,
where f˜(x) = f (−x), and similarly for h. The point is that T is almost
self-dual, so that
Z
Tf · h
||T ||Lp →Lp =
sup
||f ||p =1,||h||p0 =1
Z
=
sup
f˜ · T h̃
||f ||p =1,||h||p0 =1
= ||T ||Lp0 →Lp0 .
Hence it is enough to estmate ||T ||Lp →Lp for p > 2.
8.6. Fefferman’s argument. The argument we give here is packaged a
bit differently than in the original paper, but essentially the same. This
argument is more or less in [15]. We begin with an outline of the argument.
Recall a key geometric fact from earlier in the notes:
Lemma 99. For every > 0, there exists a collection of rectangles R such
that
[ X
R < |R|,
R∈R
R∈R
but the rectangles {R̃} are pairwise disjoint, where R̃ is the “reach” of R;
i.e., R̃ is a translation of R by a distance of 2length(R) along its long
49
axis. Without loss of generality, by dilating the plane, the rectangles can be
assumed to have height CL and length L2 for some number L, and a fixed
moderately large consant C.
Remark 100. Our simplification to consider only the operator T1 requires
that we consider collections of longer and longer rectangles. We can use
fixed length rectangles if we instead consider the operators TL2 with L → ∞.
Fix some > 0 and a corresponding collection R. For each R ∈ R, define
a smooth function fR̃ that is identically 1 on the set 31 R̃ and supported in
R̃. (Here 13 R is the rectangle with same center and orientation as R but
sides 31 times as long.) We will essentially show :
1. For each R ∈ R, |T fR̃ (x)| > c for x ∈ R.
P
2. Since the P
rectangles R pile up a lot, if we define f = R fR̃ , we should
have T f (x) = T fR̃ (x) large in ∪R.
3. Since the rectangles R̃ are disjoint, we know |f | ≤ 1.
4. So ||T f ||p >> ||f ||p .
We now precisely define the functions fR̃ . Let
R ψ : R → R be a smooth
1 1
nonnegative function supported in [− 2 , 2 ] with ψ = 1. Now let ϕ : R2 → R
be given by ϕ(x1 , x2 ) = ψ(x1 )ψ(x2 ). We use this function to generate a
smoothed-out characteristic function of each rectangle R̃. The relevance of
the smoothing out will be come apparent in our proof of Lemma 101 below.
For the moment, assume R is axis parallel and centered at the origin to
make the following definition easier. Then define
ϕR̃ (x1 , x2 ) = ϕ(
x1 x2
,
)
L2 CL
for some moderately large constant C. The role of C will also become
apparent in the proof of Lemma 101; it is really only present to control an
annoying error term. This is just a bump function scaled to fit the rectangle
R. Now if R is any rectangle, let ϕR̃ be defined analogously but with the
bump rotated and translated to fit R̃.
Finally define
fR̃ (x1 , x2 ) = ϕR̃ (x1 , x2 )e2πivR ·x
where vR is a unit vector in the direction of R. The modulation e2πivR ·x is
a bit more important. We now record the first point above as a lemma. It
is very important, but its proof is a bit technical so we postpone it until the
rest of the proof is complete.
halfplane
Lemma 101. There is c > 0 such that for each R ∈ R, |T fR̃ (x)| > c for
x ∈ R.
To make the second point rigorous we appeal to Khintchine’s inequality:
50
MICHAEL BATEMAN
Lemma 102. Let a1 , a2 , . . . aN ∈ C. Consider a random selection of signs
σj ∈ {−1, 1}, and let p ∈ (1, ∞). Then

1 
1
p
2
N
N
X
X
p
2
E|

σj aj |
∼
|aj |
.
j=1
j=1
where E denotes expectation over all selection of signs σ1 , . . . , σN , which are
independent.
Remark 103. For our purposes, we only need one direction of this inequality, and that only in the case p ≥ 2, which we prove below. The proof is
more difficult in the case p ≤ 2 or in the case of proving the inequality in
the other direction. The typical procedure is to prove one direction of the
inequality for all p, then use duality to conclude the opposite inequality for
all p.
Proof of & when p ≥ 2. Just notice that


!
N
N
N
X
X
X
2
E|
σj aj | = E 
σj aj 
σk ak
j=1
j=1
=
XX
j
=
X
k=1
aj ak Eσj σk
k
|aj |2 .
j
But by Holder, we have
E|
N
X
j=1

2
σj aj | ≤ E|
N
X
2
p
p
σj aj |
.
j=1
P
Using the notation from this lemma, we define
the function fσ = R σR fR̃ .
P
We are interested in the quantity T fσ =
σR T fR̃ . As a result of this we
get a statement about the expected value of the norm of our function f :
Corollary 104.
sX
p
Eσ ||T fσ ||p & |T fR̃ |2 R
p
The reason the right-hand side of this corollary is nicer than the left is that
we only know T fR̃ is large inP
absolute value , but we know nothing about
its sign. So it is possible that R T fR̃ is small because of cancellation. The
object inside the norm on the right is called a square function .
51
Proof. Applying the lemma pointwise in x to our functions T fR̃ in place of
the sequence aj , we have

E|
N
X

1
2
N
X
2
p

∼
|T fR̃ (x)|
.
σR T fR̃ (x)|
1
p
j=1
j=1
Now integrating over x yields
Z
E|
N
X

Z
p
σR T fR̃ (x)| dx &
N
X

j=1
p
2
|T fR̃ (x)|
2
.
j=1
We now collect the pieces P
and prove Fefferman’s theorem.
Recall that we have fσ = R σR fR̃ . Since the reaches R̃ are disjoint, we
know
X
||fσ ||pp = ||
σj fR̃ ||pp
R
=
X
∼
X
||fR̃ ||pp
R
Now we estimate E||T
chine, we have that
E||T
P
σR fR̃ ||pp from below. Using the corollary to Khint-
X
R
|R|.
σR fR̃ ||pp
p
s
X
2
& |T fR̃ | R
p
!p
2
Z
X
=
2
|T fR̃ |
R
!p
2
Z
X
&
|1R |2
,
R
where the last inequality is by Lemma 101. But using Holder in the form
Z
2
Z
p
p
1− 2
g2
g≤
|support(g)| p ,
we have that this last is greater than
p
RP
2
1
R
R
p
| ∪R∈R R| 2 −1
.
52
MICHAEL BATEMAN
Collecting everything, we have
E||T
X
σR fR̃ ||pp
&
R
=
p
P
( |R|) 2
P
p
( |R|) 2 −1
P
|R|
p
2 −1
X
>>>
|R|
∼
||fσ ||pp .
Of course ||fσ ||pp ∼ 1 independently of σ, so there exists a choice of σ such
that
X
σR fR̃ ||pp >>> ||fσ ||pp .
||T
R
This proves the theorem assuming Lemma 101.
8.7. Proof of Lemma 101. We recall the definition of f :
fR̃ (x1 , x2 ) = ϕR̃ (x1 , x2 )e2πivR ·x
where vR is a unit vector in the direction of R. The location and direction of
R̃ are not important for these calculations, so we will assume R̃ is centered
at the origin and parallel to the axes. (If R is pointed in a different direction,
we use a half-plane oriented in the direction of R.) We will have proved the
lemma if we can show that
|T fR̃ (x)| > c for x ∈ R.
We split the operator T into two pieces, each corresponding to a Fourier
multiplier. Write H = {(ξ1 , ξ2 ) : ξ1 ≤ 1} to denote a left half-plane, and
E = H \ B1 (0). Then
T g = TH g − TE g,
Tcg = 1H ĝ − 1E ĝ.
Loosely speaking, we will show that T fR̃ ∼ TH fR̃ . The operator TH is easier
to understand, especially acting on the function fR̃ . More precisely, we have
firstclaim
Claim 105.
1
|TE fR̃ (x)| . √ ,
C
where C is the constant appearing in the definition of ϕR̃ .
This first claim says the error term is negligible. The next claim says that
the important term is the size we want it to be.
secondclaim
Claim 106.
|TH fR̃ (x)| & 1
for x ∈ R.
53
Certainly combining the previous two claims proves the lemma.
Proof of first claim. By definition,
Z
2πixξ
TE fR̃ (x) = 1E (ξ)fc
dξ.
R̃ (ξ)e
By the basic properties of the Fourier transform, we have
3
fc
b 1 − 1)L2 , CLξ2 ).
R̃ (ξ1 , ξ2 ) = CL ϕ((ξ
The factors of L and C come from the usual scaling properties and the
translation in the ξ1 variable comes from the modulation in the definition of
fR̃ . Because of this formula we have that fc
R̃ is essentially supported (because
1
ϕ is Schwartz) in a rectangle of height CL
and width L12 and centered at the
point (1, 0). The point is that the set E has very little intersection with this
rectangle, and the function fc
R̃ is very small outside this rectangle.
More specifically, write
D = {(ξ1 , ξ2 ) : |ξ1 − 1| ≤
1
1
and |ξ2 | ≤
}.
2
L
CL
We have
3 2
−20
|fc
|CLξ2 |−20 .
R̃ (ξ1 , ξ2 )| . CL |L (ξ1 − 1)|
by the decay properties of the Fourier transform of smooth compactly supported functions (see example sheet 3). In other words, if ξ is way outside
of D, then fc
R̃ (ξ1 , ξ2 ) is very small and decaying as ξ moves away from D.
Hence
Z
1
2πixξ c
1
(ξ)
f
(ξ)e
dξ
≤ 5,
E
√
R̃
C
R2 \ CD
since the area of D is CL3 . Further,
Z
√
2πixξ
c
dξ ≤ |E ∩ CD| · CL3 ||ϕ̂||∞
√ 1E (ξ)fR̃ (ξ)e
CD
.
≤
1
3
2
L3
C
1
√
C
CL3 ||ϕ̂||∞
provided C is taken large enough compared to ||ϕ̂||∞ and the implicit constant. Here we have used the estimate
Z √C
√
CL
1
|E ∩ CD| .
t2 dt . 3 ,
0
C 2 L3
which holds because the circle is parabolic near the point (1, 0). Combining
these last two estimates proves the claim.
54
MICHAEL BATEMAN
Proof of second claim. We first prove a relevant fact about a one-dimensional
operator. Let ψ : R → RR be a smooth nonnegative bump function supported
in [− 21 , 12 ] and having ψ = 1. Define
Z 0
ĝ(ξ)e2πixξ dξ.
P− g(x) =
−∞
A slightly tedious calculation shows that
x x x x 1
2
1
2
TH e2πix1 ψ
ψ
(x1 , x2 ) = eπix1 (P− ψ)
ψ
a
b
a
b
2
for any a, b. Applying this with a = L and b = CL gives us
x x 1
2
|TH (fR̃ )(x1 , x2 )| = |TH e2πix1 ψ
ψ
(x1 , x2 )|
2
L
CL
x x 1
2
= (P− ψ)
ψ
,
L2
CL
so Claim 106 follows from this fact:
Claim 107. There is c > 0 such that
c
|P− ψ(x)| >
for |x| > 1.
|x|
Proof.
Z
P− f (x) =
=
=
=
=
0
fˆ(ξ)e2πixξ e2πξ dξ
lim
→0+ −∞
Z 0 Z ∞
e−2πiyξ f (y)dye2πixξ e2πξ dξ
lim
→0+ −∞
Z ∞
−∞
0
Z
lim
e2πiξ(x−y) e2πξ dξdy
f (y)
→0+ −∞
Z ∞
−∞
0
Z
lim
e2πiξ(x−y−i) dξdy
f (y)
→0+ −∞
Z ∞
−∞
lim
f (y)
→0+ −∞
1
dy.
2πi(x − y − i)
Hence for |x| > 1,
Z
∞
|P− ψ(x)| = | lim
→0+ −∞
Z
= | lim
1
2
→0+ − 1
2
ψ(y)
ψ(y)
Z
= |Re
1
2
Z
&
1
2
lim
1
2
→0+ − 1
2
ψ(y)
1
dy
|x|
1
dy|
2πi(x − y − i)
1
dy|
2πi(x − y − i)
!
1
dy |
ψ(y)
2πi(x − y − i)
55
=
1
.
|x|
We have only used the fact |x| > 1 to obtain the estimate |x − y − i| ∼ |x|,
since in the limit is quite small, and since |y| < 12 .
8.8. Aside on the Hilbert transform and the 1-D case. In our understanding of Fefferman’s theorem we considered a related one-dimensional
operator. Given a nice g : R → C, define
Z 0
ĝ(ξ)e2πixξ dξ
P− g(x) =
Z−∞
∞
ĝ(ξ)e2πixξ dξ
P+ g(x) =
0
P+ and P− are the projection operators onto positive and negative frequencies, respectively. Notice that
g = P+ g + P− g,
so that if we define yet another operator by
Hg = P+ g − P− g,
we have
Hg + g = 2P+ g.
We can equivalently describe these operators on the Fourier side:
Pd
+ g(ξ) = 1(0,∞) (ξ)ĝ(ξ)
Pd
− g(ξ) = 1(−∞,0) (ξ)ĝ(ξ)
c
Hg(ξ)
= sgn(ξ)ĝ(ξ).
This formula shows that the operator H is quite closely related to the projection operators P+ and P− . The significance of this is that the operator
H has a particurly nice description:
Z
g(x − y)
Hg(x) = c lim
dy,
→0+ |y|≥
y
where c is just some constant like −2πi or something. Ignoring the limit, we
notice that H is a bit similar to the Hardy-Littlewood maximal operator:
Z r
1
M g(x) = sup
g(x − y)dy.
r>0 2r −r
The operator M is bounded on all Lp for p > 1 and satisfies a weak (1,1)
estimate. (The weak (1,1) estimate is in the first example sheet, and the
other Lp estimates follow from the trivial L∞ estimate and interpolation.)
Using an understanding of the operator M , one can actually prove that H
56
MICHAEL BATEMAN
is bounded on Lp for every 1 < p < ∞. Notice that the L2 estimate for H
follows trivially from its definition, since it is given by a bounded Fourier
multiplier. Finally, the Fourier partial sum operators
Z R
SR g(x) =
ĝ(ξ)e2πixξ dξ
−R
can easily be seen to satisfy
Z
Z R
2πixξ
ĝ(ξ)e
dξ −
SR g(x) =
−∞
= P− (e
2πiR·
−R
ĝ(ξ)e2πixξ dξ
−∞
−2πiR·
g)(x) − P− (e
g)(x).
Hence convergence in Lp for the one-dimensional Fourier integrals follows
from boundedness of P− , which follows from boundedness of H.
9. Connections to the restriction conjecture
In this section we essentially prove that the Kakeya maximal conjecture
follows from the “Restriction conjecture”. For now we will focus on the
connection to the Kakeya problem and otherwise limit our discussion of
the restriction conjecture somewhat. The argument in this section is quite
similar to the argument in the last section showing unboundedness of the
multiplier operator T .
To start with, we note that it makes sense to define the Fourier transform
of a finite measure µ:
Z
µ̂(ξ) = e−2πixξ dµ(ξ)
In what follows, let σ be surface measure on the sphere Sn−1 . The Fourier
transform of σ plays a distinguished role in Fourier analysis. An example is
the following:
RIK
Theorem 108 ((A dualized version of) the restriction conjecture implies the Kakeya conjecture). Su
that for every smooth function f we have the estimate
||fd
dσ||Lp (Rn ) ≤ Cp ||f ||Lp (σ) .
Then the dimension of Kakeya sets in Rn is at least
n−
In particular, if p =
least
2n
n−1
4n
2(n − 1)p
+
.
p−2
p−2
+ , then the dimension of Kakeya sets in Rn is at
n−
2(n − 1)
.
2
n−1 + Notice that this last quantity tends to n as → 0.
57
An inspection of the proof shows that we are essentially proving the
Kakeya maximal conjecture (assuming the restriction estimate, of course),
although it is not precisely the same formulation we gave above. To clear
away all possible distractions, we write the proof here only for n = 2; the
argument goes through verbatim in higher dimensions. In this situation we
will show that Kakeya sets have Minkowski dimension at least
2
2−
,
2+
which tends to 2 as → 0.
9.1. Warm-up stuff. To see why the word “restriction” is used, let’s formally dualize the estimate in the hypothesis of the theorem above:
Z
||b||Lp (σ)→Lp (Rn ) =
sup
fd
dσ(ξ)g(ξ)dξ
||f ||Lp (σ) =1,||g||Lq (Rn ) =1
Z
=
f (y)dσ(y)ĝ(y)dy
sup
||f ||Lp (σ) =1,||g||Lq (Rn ) =1
Z
=
sup
f (y)ĝ(y)dσ(y)
||f ||Lp (σ) =1,||g||Lq (Rn ) =1
= ||b||Lq (Rn )→Lq (σ)
where p1 + 1q = 1. So the estimate in the hypothesis of Theorem 108 is dual
to the estimate
||ĝ||Lq (σ) . ||g||Lq (Rn ) .
(9.1)
The question is whether, given a function g defined on all of Rn , we can
meaningfully say something about its Fourier transform on the unit sphere.
A simple example of a valid inequality of this form is
||ĝ||L∞ (σ) ≤ ||g||L1 (Rn ) ,
which follows straight from the definition, since
Z
2πix·ξ
dx
|ĝ(ξ)| = g(x)e
Z
≤
|g(x)|dx
= ||g||L1 (Rn ) .
If we could establish inequalities of this form for all smooth functions, whose
Fourier transforms are necessarily defined on the unit sphere, we could use
density of smooth functions together with boundedness of the “restriction
operation” to define the Fourier transform of any Lq function on a sphere.
The question “Can we meaningfully restrict the Fourier transform of a function to the unit sphere?” is addressed by the availability of an estimate like
10.1.
58
MICHAEL BATEMAN
It is worth mentioning that restriction comes in many varieties. We need
not have the same exponent q on both sides of the estimate. We also could
consider the restriction problem for different surfaces, such as parabolas or
hyperplanes.
9.2. Restriction implies Kakeya. In this section we prove Theorem 108.
Consider a collection R̃ of 1 × δ rectangles with δ-separated directions. We
will show that

p
2
Z
X
X


. δ 4−p
|R̃|,
(9.2)
1R̃
R̃∈R̃
R̃∈R̃
and then the claim about dimension of Besicovitch sets will follow from the
usual kind of argument. Now let L : R2 → R2 be the dilation given by
L(p) =
1
p,
δ2
and define
R = {L(R) : R ∈ R̃}.
So R is just the collection of rectangles that results after dilating the plane
by a factor of δ12 . Notice that
Z
!p
2
X
1R
. δ 4−p
X
|R|,
∈RR
R∈R
holds if and only if (10.2) holds– this is because the estimate is invariant
under any dilation of the plane. The reason for the rescaling is that we
want to consider intervals of length ∼ δ on the circle, because such intervals
correspond to δ-separted directions, and a rectangle of dimensions δ12 × 1δ is
the natural dual object. For each R ∈ R, let vR be the unit vector indicating
the orientation of R. Also define the interval
IR = {v ∈ S1 : ∠(v, vR ) ≤
δ
},
100
and a smooth bump function ϕR : S1 → R that is supported inside IR with
0 ≤ ϕR ≤ 1. Define ψR (x) = e2πicR ·x ϕR (x), where cR is the center of the
rectangle R. Notice that the function ϕR does not take into account the
location of the rectangle R, only its orientation. This is why we actually
want to deal with the function ψR . Just as in the Fefferman argument, we
will construct a random function fα : S1 → C, where α denotes a vector of
length #R of random signs ±1, of the form
X
fα (x) =
αR ψR (x).
R∈R
59
Since the intervals IR are pairwise disjoint, the functions ψR have disjoint
support, and hence
X
||fα ||pLp (σ) =
||ψR ||pLp (σ) . #R · |IR | ∼ #R · δ,
R
and the estimate is independent of the choice of signs α. Appealing to our
hypothesis gives us for every α that
p
p
||f[
α dσ||Lp (Rn ) . ||fα ||Lp (σ) .
(9.3)
On the other hand, by Khintchine’s inequality again, we have that
p
s
X
X
2
\
\
|
ψ
dσ|
αR
ψR dσ||pLp (R2 )
.
E
||
α
R
p 2
R∈R
R∈R
(9.4)
L (R )
n
= Eα ||f[
α dσ||Lp (Rn ) .
We
Pnow show that the left side of this last display resembles the quantity
|| R∈R 1R ||pp from the Kakeya maximal conjecture. More precisely, we show
Claim 109.
\
ψ
R dσ(ξ) ≥ cδ
for ξ ∈ R.
Assuming the claim and combining it with 10.3 and 10.4 , we can finish
the proof as follows:
p
s X
2
\
#Rδ & |ψR dσ| R∈R
p 2
L (R )
p
s X
& |δ1R |2 p 2
R∈R
L (R )
= δp
!p
2
Z
X
1R
.
R∈R
Rearranging gives us
Z
X
!p
2
1R
δ −p δδ 3 (δ −3 #R)
.
R∈R
δ −p+4
P
|R|,
which was our goal. Now we insert this into the same argument we used in
deducing the Kakeya conjecture from the Kakeya maximal conjecture:
Z X
X
|R̃| =
1R̃
R̃∈R̃
R̃∈R̃
60
MICHAEL BATEMAN
≤ |
[
1− p2
R̃|
 
 p  p2
2
Z
 X


1R̃


R̃∈R̃
R̃∈R̃
2

. |
[
1− p2
R̃|
p
δ −p+4
R̃∈R̃
X
|R̃| .
R̃∈R̃
Rearranging yields
|
[
R̃| & δ
2
(p−4) p−2
X
|R̃|.
R̃∈R̃
R̃∈R̃
This completes the proof of the theorem modulo the claim.
Proof of claim. As usual, the important part of the claim can be seen by
considering a model case: specifically, we assume R is horizontal and we
assume it is centered at the origin. In this case,
Z
cR (ξ) =
ψ
ψR (x)e−2πix·ξ dx
I0
Z
−2πi(1,0)·ξ
= e
ψR (x)e−2πi(x−(1,0))·ξ dx.
I0
The point of comparing with the reference point (1, 0) is that the vector
(x − (1, 0)) · ξ is essentially constant for ξ in the rectangle R. Precisely, if
δ
x ∈ I0 then x − (1, 0) has y-coordinate less than 100
and x-coordinate less
2
1
1
δ
than 100 . Hence if |ξ1 | ≤ δ and |ξ2 | ≤ δ2 then
δ
δ2
2
|ξ1 | +
|ξ2 | ≤
.
100
100
100
Hence for such ξ and x ∈ I0 we also have
1
Re e−2πi(x−(1,0))·ξ ≥ .
2
Hence for such ξ we have
Z
−2πi(x−(1,0))·ξ
c
|ψR (ξ)| ≥ Re
ψR (x)e
dx
I0
Z
1
≥
ψR (x)
2 I0
1
≥
δ.
2
This proves the claim for the special case of a horizontal rectangle centered
at the origin. For a general horizontal rectangle, we note that
|[x − (1, 0) · ξ| ≤
\
cR (ξ) = e2πic
R x ϕ (ξ) = ϕ
ψ
cR (ξ − cR ),
R
and hence is large in a δ12 × 1δ axis-parallel neighborhood centered at cR . For
a rectangle with a different orientation but centered at the origin, follow the
61
computation above to see that ϕ
cR is large on a δ12 × 1δ rectangle centered
at the origin and pointed in the direction of vR . For an arbitrary rectangle
combine the translation and the rotation.
10. Connection to the Wave equation
We restrict our attention to the case of three dimensions here because
the solution formula is easier to manipulate. We discuss the initial value
problem for the wave equation:
utt (x, t) − ∆u(x, t) = 0
u(x, 0) = f (x)
ut (x, 0) = 0
for some suitable function f : R3 → R. A common question has the form
Question 110. If the initial condition f is small in a suitable sense, must
the solution u also be small in a suitable sense?
By computing the solution u on the Fourier side, and invoking Parseval’s
theorem, one can show that in fact for every t ≥ 0, we have
||u(·, t)||2 ≤ ||f ||2 .
By considering a function supported on a thin spherical shell, one can show
that this estimate does not persist for any higher value of p. In other words,
the estimate
||u(·, t)||p . ||f ||p
does not hold for general f and general t. One feature of counterexamples
to such an equality is that they focus heavily at a very small range of times.
This might lead one to ask whether a reasonable estimate could be obtained
by averaging in time. But even this cannot hold: in other words, an estimate
of the form
Z 3
||u(·, t)||pp . ||f ||pp
2
still does not hold if p > 2. (Of course the time interval [2, 3] is not special.)
This follows from Theorem 111, but for p > 3 (this numerology depends
on our choice n = 3) the spherical example mentioned above is all that is
needed.
waveblowup
Theorem 111. Fix p > 2. For any C > 0, there exists a function f so that
Z 3
||u(·, t)||pp > C||f ||pp ,
2
where u is the solution to the wave equation with initial condition f .
62
MICHAEL BATEMAN
Let us recall the solution to the wave equation in 3 dimensions. See, e.g.,
[5] for a derivation of the solution. We have
Z
u(x, t) =
f (y) + ∇f (y) · (y − x)dσ(y),
St (x)
where σ is the normalized surface measure on the sphere St (x) of radius t
centered at x. We follow the same outline as in Fefferman’s argument and
in our proof that restriction implies Kakeya. Fix some number > 0 that is
very small– exactly how small depends on the parameter C in the statement
of the theorem. We have a collection of tubes R such that each tube is
1 × δ × δ and
[ X
R ≤ |R|,
R∈R
R∈R
and such that the reaches R̃ are disjoint. For each tube R we have a bump
function fR̃ defined on R̃, the reach of R. This will give us a solution uR
to the wave equation with initial condition fR̃ . We will then show that for
1
t ∈ [2.45, 2.55] and x ∈ 10
R, |uR (x, P
t)| > c. Notice that by linearity of
the wave equation,
if
we
define
f
=
α
R∈R αR fR̃ for any signs αR , and if
P
we define uα = R∈R αR uR , then u solves the wave equation with initial
condition fα . As usual, by disjointness of reaches, it is easy to see that
X
||fα ||pp .
|R|.
R∈R
Also as before, we can apply Khintchine to see that for a fixed time t ∈
[2.45, 2.55], we have
X
Eα ||u(·, t)||pp
=
Eα ||
αR uR (·, t)||pp
R
&
sX
|uR (·, t)|2 ||pp
||
&
sX
||
|1 1 R |2 ||pp
R
10
R
Z
=
!p
2
X
|1 1 R |2
10
R
>>>>
&
X
|R|
||fα ||pp .
we have used the notation >>>> to indicate that the first quantity is much
p
2 norm with p > 2 of a
larger than the second,
because
the
first
is
an
L
P
function (namely R 1R ) that piles up a lot, and the second quantity is the
L1 norm of the same function. Specifically, >>>> means “bigger than the
63
constant C even after accounting for all of the implicit constants in the &
signs, which can be arranged by choosing small enough”. We have seen
this calculation twice already and skip the details. All that remains is to
give the precise details about the functions fR̃ and how they evolve with the
wave equation.
Let ϕ(x1 , x2 , x3 ) = ψ(x1 )ψ(x2 )ψ(x3 ), where ψ : R → R is a smooth function with support in the interval [− 21 , 12 ] and equal to 1 on the interval
[− 14 , 14 ]. Now define ϕR̃ to be the function ϕ adapted to the tube R̃.
For those interested, the precise construction of such a bump “adapted
to” a tube T are as follows: First let T be any 1 × δ × δ tube in R3 . Let cT
be the center of T . Define the translation τT : R3 → R3 by τT (x) = x − cT .
Given any tube T 0 centered at the origin, there exists a linear map LT 0 that
sends T 0 to the unit square [− 12 , 12 ] × [− 21 , 12 ] × [− 12 , 12 ]. (First rescale the
tube to unit dimensions; this is linear. Then rotating the cube to be axis
parallel; this is also linear.) Now define
ϕT (x) = ϕ(LτT (T ) (x − cT )).
Now back to the main story: define the function
fR̃ (x) = ϕR̃ (x)e
2πx·vR
Cδ 2
,
where vR is the unit vector in the direction of R, and where C is a moderate
constant whose precise value will be specified later. fR̃ is a bump function
adapted to R̃ that is highly oscillatory. Notice that the exponential has
a denominator of δ 2 ; this means the functions is oscillating over intervals
of length ∼ δ 2 in the direction of the rectangle. This will be key in the
following:
mainwave
1
Claim 112. For t ∈ [2.45, 2.55] and x ∈ 10
R, we have uR (x, t) ≥ c, where
uR is the solution to the wave equation with initial condition fR̃ , and where
1
1
10 R is the rectangle with same center as R but 10 the length and width.
Given the outline above, to prove theorem 111 it is enough to prove Claim
112.
Proof. As usual, to simplifiy the calculations we assume that R̃ is the tube
δ δ
δ δ
R̃ = {(x1 , x2 , x3 ) : x1 ∈ [2, 3], x2 ∈ [− , ], x3 ∈ [− , ]}.
2 2
2 2
For ease of notation, let x be the origin. (The argument for x slightly away
from the origin is the same.) Then by the solution formula given above, we
have
Z
uR (0, t) =
fR̃ (y) + ∇fR̃ (y) · (x − y)dσ(y).
St (x)
We will see that the dominant term is the one involving the gradient. Precisely,
64
ABC
MICHAEL BATEMAN
Claim 113.
Z
fR̃ (y)dσ(y) . δ 2
St (x)
Proof. Just notice that
Z
Z
1R̃ (y)dσ . δ 2
f (y)dσ(y) ≤
St (x) R̃
St (x)
The first inequality holds because fR̃ is supported in R̃, and is less than 1.
The second holds because R̃ intersects the sphere of radius 2.5 centered at
the origin in a set of (surface) measure ∼ δ 2 . (The cross sectional slice of R̃
has area δ 2 .)
DEF
Claim 114.
Z
∇f (y) · (y)dσ(y) ≥ c.
St (x)
Proof. With our choice vR = (1, 0, 0), and the placement of R̃, we have
y2
y3 2πiy·(1,0,0)
)ψ( )e Cδ2
δ
δ
y3 2πiy21
y2
= ψ(y1 − 2.5)ψ( )ψ( )e Cδ
δ
δ
fR̃ (y1 , y2 , y3 ) = ψ(y1 − 2.5)ψ(
so that
∂fR̃
y3 2πiy1
y2
(y1 , y2 , y3 ) = ψ(y1 − 2.5)ψ( )ψ( )e Cδ2
∂y1
δ
δ
y3 2πiy1
y2
= ψ(y1 − 2.5)ψ( )ψ( )e Cδ2
δ
δ
2πi
y2
y3 2πiy21
0
+
ψ
(y
−
2.5)ψ(
)ψ(
)e Cδ
1
Cδ 2
δ
δ
2πi
Cδ 2
if |y1 − 2.5| ≤ 41 , in which case the second summand immediately above is
1
zero because ψ is constant on the interval [ −1
4 , 4 ]. (Notice that the range of
applicability for this last equality is why we could have moved x away from
the origin, and why the range of allowable t is also somewhat large.) Define
T = St (x) ∩ R̃; T is the part of the tube R̃ intersecting the sphere we are
interested in. Notice that if |y1 − 2.5| ≤ 14 and |y2 | ≤ 2δ , |y3 | ≤ 2δ , then
∂fR̃
& 1.
(y
,
y
,
y
)
1
2
3
∂x1
δ2
We only need to verify that all of these numbers are pointed in essentially
the same direction as we integrate over y ∈ T , which we do now.
For all y ∈ T , we have |y2 | ≤ 2δ and |y3 | ≤ 2δ , which implies |y1 − t| . δ 2
(because spheres are locally quadratic). Notice that for any y we have
e
2πiy1
Cδ 2
2πit
= e Cδ2 e
2πi(y1 −t)
Cδ 2
,
65
Also for y ∈ T , we have
(y1 − t)
1
.
2
Cδ
C
2
since |y1 − 2.5| . δ . By taking C large enough (but finite, and fixed) we
have
2πi(y1 −t)
1
|1 − e Cδ2 | < ,
10
and hence
2π(y1 −t)
1
2
Re e Cδ
> .
2
|1 − e
2πi(y1 −t)
Cδ 2
|.
We can now use this to see that
Z
2πit Z
2πi(y1 −t)
∂fR̃
y
y
2πi
Cδ2
2
3
e Cδ2 ψ(y1 − 2.5)ψ( )ψ( ) 2 (y1 , y2 , y3 ) = e
St (x) ∂y1
δ
δ Cδ St (x)
Z
2πi(y1 −t)
2π y
y
2
3 Cδ 2
=
ψ(y
−
2.5)ψ(
)ψ(
)
e
1
Cδ 2 St (x)
δ
δ !
Z
2πi(y1 −t)
2π
y2
y3
2
≥
Re
e Cδ ψ(y1 − 2.5)ψ( )ψ( )
Cδ 2
δ
δ
St (x)
!
Z
y2
y3
2π 1
ψ(y1 − 2.5)ψ( )ψ( )
≥
Cδ 2 2
δ
δ
St (x)
1
σ(T )
δ2
∼ 1,
&
because σ(T ) is the cross-sectional area of the tube R̃, which is ∼ δ 2 . On
the other hand,
∂fR̃ 1
∂x2 . δ
since the derivative of ψ is bounded (it is smooth and compactly supported)
and since the vertical direction has been scaled by a factor of 1δ . Similarly,
∂fR̃ 1
∂x3 . δ
Hence the horizontal component
ponent of the gradient. Hence
Z
∇f (y) · (y) =
R̃
of the gradient is the only noticeable com-
Z
Z
≥ ∂fR̃
∂fR̃
∂fR̃ y1 +
ty2 +
ty3
∂x1
∂x2
∂x3 Z
∂fR̃ 1
y1 − 2
∂x1
δ
& 1 − O(δ)
T
66
MICHAEL BATEMAN
& 1.
This is precisely what we claimed.
11. The bush and hairbrush arguments
In this section we discuss the so-called “bush”argument of Bourgain, followed by the “hairbrush”argument”of Wolff. The bush argument will not
give us an improvement on the dimension of Besicovitch sets, but it is a
natural precursor to the hairbrush argument. Also, if we paid a bit more
attention in the argument below, the bush argument would give us the maximal estimate corresponding to the n+1
2 dimensional estimate from before.
11.1. Bushes. Loosely speaking, the idea of the bush argument is this:
Either
1. There are no points where lots of tubes overlap, in which case the
union of the tubes is large, or
2. There exists a point x at which many tubes overlap. Since two tubes
only intersect in a small region, these tubes must be disjoint away from x,
and hence collectively they take up a lot of space.
We encode this heuristic as follows.
Theorem 115. The dimension of Besicovitch sets in Rn is at least
n+1
2 .
As already mentioned, this estimate is not new, since we obtained precisely this numerology by doing the warm-up for the arithmetic approach to
Kakeya (i.e., without appealing to the sums-differences lemma).
Proof. As usual, it is enough to show that if R is a collection of ∼ δ −n+1
tubes of dimensions 1 × δ × · · · × δ in 100δ separated directions, then
[ n−1
R & δ 2
R∈R
since the δ neighborhood of a Besicovitch set contains such a collection of
tubes. (The factor of 100 added to the sepration is just to help us a little
bit later. Of course it is fair for us to assume this since we are decreasing
the number of tubes under consideration and still showing they occupy a
lot of space.) Then a straightforward calculation results in the dimensional
estimate.
S
Fix a number M , and write K := R∈R R. We will show that
bush
Claim 116.
|K| & min(
1
, M δ n−1 ).
M
67
This was for any M ; choosing M = δ −
n−1
2
|K| & δ
n−1
2
yields
,
which finishes the proof. It remains to prove the claim.
P
Proof. If R∈R 1R (x) ≤ M for every x, then we are already done, since
then
X
1 ∼
|R|
R∈R
=
Z X
1R
R∈R
Z
=
X
1R
K R∈R
≤ M |K|,
1
.
i.e., |K| ≥ M
P
So we assume there exists an x such that R∈R 1R (x) ≥ M . This means
there are M tubes intersecting at x, call them R1 , R2 , . . . , RM . Certainly
K⊃
M
[
Rj .
j=1
For each j, write Aj = Rj \ Bx ( 41 ). In words, Aj is the part of Rj that is at
least 14 away from x. Note Aj is still at least half as big as Rj (since each
Rj has length 1 and since we only threw away a ball of radius 41 ). Since any
two tubes Rj , Rk have slopes different by ≥ 100δ, they cannot overlap over
an interval as long as 14 , so that the regions Aj are pairwise disjoint (since
all the tubes pass through a common point x). Hence
M
[ |K| ≥ Rj j=1 [
M
≥ Aj j=1 =
M
X
|Aj |
j=1
& M |R|
∼ M δ n−1 .
This proves the claim.
68
MICHAEL BATEMAN
11.2. Hairbrushes. The hairbrush argument allowed Wolff to push the
lower bound on the dimension of Besicovitch sets to n+2
2 . Notice that in
n+1
low dimensions this is a substantial improvement over 2 . For example
3+1
= 2
2
5
3+2
=
.
2
2
Wolff actually proved the maximal version:
wolffmax
Theorem 117. Suppose R is a collection of δ separated tubes of dimensions
1 × δ × · · · × δ in Rn . Then
! n
Z
n−1
X
X
− n−2
|R|.
1R
. δ 2(n−1)
R∈R
R∈R
wolffdim
Corollary 118. The Hausdorff and (lower) Minkowski dimensions of a
Besicovitch set in Rn is at least n+2
2 .
We avoid discussion of Hausdorff dimension here and quickly recall
S why
Theorem 117 implies the Minkowski dimension estimate. If K = R∈R R,
then
Z X
X
|R| =
1R
R
K R
≤ |K|
. |K|
1
n
1
n
Z X
δ
1R
n−2
− 2(n−1)
n
n−1
n−1
n
! n−1
n
X
|R|
.
R∈R
Applying this when R has essentially the maximal number
and rearranging, gives us
n−2
n−2 X
|R| & δ 2 ,
|K| & δ 2
1
δ n−1
of tubes,
R∈R
which implies the dimensional estimate is
n−2
n+2
≥n−
=
.
2
2
11.2.1. Heuristic argument. Below we give a variant of Wolff’s argument
that does not yield Theorem 117, but still yields the Minkowski dimension
estimate in Corollary 118 (without appealing to Theorem 117). To begin
with, we give quick argument that contains a simplifying assumption that
we will indicate below.
Suppose R is a set of 1 × δ × · · · × δ tubes. Further suppose that their
1
directions are ∼ δ separated, and that #R ∼ δn−1
, which is basically the
maximal size of a δ separated set of directions in n dimensions.
69
Loosely speaking, either
1. Each tube doesn’t have very many bushes on it, i.e., there isn’t very
much piling up, in which case the the union of the tubes is large, or
2. There exists a tube with lots of bushes on it (a hairbrush!), in which
case the union of tubes is bigger than the collective size of all the bushes on
the tube.
More precisely: as in the bush argument, we fix a threshold M and consider two cases. We’ll show
1
|K| & min( , M δ n−2 ),
M
S
wherePK = R∈R R as usual. Either
1.
1R ≤ M for half of every tube,
P or
2. there exists a tube T such that
1R ≥ M on half of T .
In case 1., we knowPthat for each R ∈ R, there exists AR ⊆ R with
|AR | ≥ 12 |R| such that R0 1R0 (x) ≤ M for x ∈ AR . Hence
X
1 ∼
|R|
R∈R
∼
X
|AR |
R∈R
=
Z X
1A R
R∈R
Z
X
=
S
Z
X
=
S
1A R
AR R∈R
1R
AR R∈R
[
≤ M AR ≤ M |K|.
where the second-to-last estimate holds since we defined AR to be the part
of R that was not covered higher than M .
P So now it remains to consider case 2., in which we have a tube T such that
1R ≥ M on half of T . This means that there are lots of bushes covering
T . At this point we make a serious simplifying assumption: namely, that
all tubes intersecting T do so at a rather large angle, say angle at least
∼ 1. This means that each tube R intersecting T does so over an interval
of length . δ. (If this seems puzzling, look back to our solution of the
2D Kakeya problem in which we computed the area of intersection of two
rectangles at a certain angle. If the tubes were perfectly orthgonal, the
length of intersection would be δ.) The point is that we can divide the tube
T into ∼ 1δ subcubes, most of which form the base of a bush at height M .
The important point is that, since the tubes R intersect the base T at a
70
MICHAEL BATEMAN
noticeable angle, the tubes in one bush are distinct from the tubes in the
neighboring bush. If we assume that the bushes themselves are disjoint (not
just formed from distinct tubes!) then we get
|K| ≥ (# of bushes based on T ) × (volume of one bush)
≥ (# of bushes based on T ) × (# of tubes in one bush) × (volume of one tube)
1
&
M δ n−1
δ
= M δ n−2 .
This is the desired estimate, but of course we have made two serious assumptions here. The first is that all tubes intersect the base transversally,
and the second is that the bushes are disjoint. We will show in the next
section that, although the tubes in the bushes may not be disjoint, they
are essentially so. We will see in the next section that if T is intersected
by lots of tubes at a smaller angle, then our hairbrush will be contained in
a somewhat narrow tube, and we can consequently find lots of essentially
disjoint hairbrushes.
11.3. Direct proof of the Minkowski dimension bound in Corollary
118. We now carry out the argument inthe previous section more carefully.
We start with a collection R of 1×δ×· · ·×δ tubes, and assume that the tubes
have directions separated by, say, 100δ. Also assume that R is more or less
1
.
maximal with respect to this property – in particular, assume #R & δn−1
We will again fix a threshold parameter M and prove the following
Claim 119.
|K| & min(
1 M δ n−2
,
).
M (log 1δ )5
As before, this gives us a dimensional estimate of n+2
2 for Besicovitch sets
n
in R . (The logarithmic term in the estimate above should be ignored, since
it does not affect the calculation of the dimension of Besicovitch sets; such
calculations only see powers of δ. Also, the exponent 5 may be overkill; I
am not really paying close attention.)
We will iteratively remove tubes from R using the following lemma.
dichotomy
Lemma 120. Either
1
P 1. for every T ∈ R, we have a set AT ⊆ T such that |AT | ≥ 2 |T | and
R∈R 1R (x) ≤ M for every x ∈ AT , or
2. there exists a tube T together
with
other tubes T (T ) ⊆ R and an
S
some
M δ n−2
integer k = k(T ) such that
R∈T R ≥ (log( 1 )2 2k , and such that all tubes
δ
in T have angle ∼ 2−k with T . (In other words, there exists a hairbrush of
diameter 2−k .)
We now iteratively apply the lemma. If we are ever in case 1., proceed as
in case 1. of the heuristic section above. (Notice that we did not make any
71
simplifying assumptions there.) If we are in case 2., then we define
E(T ) = {R : ∠(T, R) . 2−k },
and remove the tubes E(T ) from R. We continue applying the lemma until
we are in case 1. or until we have removed half of the original tubes. Since
we stop the iteration once we have removed more than half of the tubes, we
know that whenever we are in case 1. of the lemma there are at least half of
the tubes remaining; this ensures that the first ∼ in the calculation for case
1. is still correct. (For example, if we only had an fraction of the tubes
remaining, then the calculation would only yield an fraction of what we
want.)
We first prove Lemma 120, and then show how to use it to finish the proof
of Corollary 118. Essentially, if we are in case 2. of the lemma, we will find
lots of hairbrushes.
11.3.1. Proof of Lemma 120.
P
Proof. We may assume that we are in the case where R 1R (x) ≥ M for
half of T , for otherwise we are already done. We divide into categories
depending on the angle between T and R. Specifically, for these x,
log
M≤
X
1R (x) =
1
Xδ
X
1R (x),
k=0 R : ∠(R,T )∼2−k
R
and pigeonholing gives us an integer k = k(x) such that
X
M
1R (x) ≥
.
log 1δ
−k(x)
R : ∠(R,T )∼2
Note that this number k(x) depends on x, but since there are only log 1δ
possibilities for k(x), we know there exists BT ⊆ T and an integer k with
|T |
|BT | & log
1 such that
δ
X
1R (x) ≥
R : ∠(R,T )∼2−k
M
log 1δ
for all x ∈ BT . Notice that by construction of T , all tubes in T have angle
∼ 2−k with the base T .
Now let T = {R : ∠(R, T ) ∼ 2−k and R ∩ T 6= ∅}. Divide T into δ21k
intervals of length 2k δ. We know a given R ∈ T intersects at most . 1 such
subintervals. Define
[
HT =
R.
R∈T
All that remains is to estimate the size of HT :
72
hairbrushsize
MICHAEL BATEMAN
Claim 121.
|HT | &
M δ n−2
.
2k (log 1δ )2
Proof. Loosely speaking, we proceed as in the heuristic section above: we
just count the number of bushes, the volume of the bush, and hope that
they don’t overlap. This would give us an estimate of
(# of bushes based on T ) × (volume of one bush)
≥ (# of bushes based on T ) × (# of tubes in one bush) × (volume of one tube)
1
M n−1
&
δ
1 k
log δ 2 δ log 1δ
=
M δ n−2
.
(log 1δ )2 2k
The two logarithmic factors arose from the two pigeonholings we did to find
the value of k that worked for every x. The factor of 2k comes in because
all tubes R ∈ T have angle 2−k with T , and hence the base of each bush has
width 2k δ ( rather than δ, as in the transverse intersection case); this means
there are only 21k as many bushes along T . To make this rigorous, we only
need to justify our claim that the bushes are disjoint. We won’t actually
prove that the bushes are disjoint, we’ll just go through the usual argument
showing that the bushes don’t overlap too much and hence take up almost
as much space as they would if they were in fact disjoint. Specifically, as
in the bush argument, we consider the parts of tubes R that are away from
the base T . Let
AR = {x ∈ R : d(x, T ) ≥
The key point is that
X X
|AR ∩ AR0 | . log
R∈T R0 ∈T
2−k
}.
2
1 X
|AR |.
δ
R∈T
This follows from the fact that for a fixed R, any tube R0 ∈ T that also
intersects R must essentially lie in the plane determined by R and T . In
other words,
#{R0 ∈ T : ∠(R0 , R) ∼ 2j δ} . 2j .
But for such R0 , we have
|AR ∩ AR0 | . 2−j |AR |.
Plugging this into the usual calculation yields that
1
!1
2
[
2 X X
X
|AR | . AR |AR ∩ AR0 |
0
R∈T
R∈T
R∈T R ∈T
73
1
[
2
. AR R∈T
1 X
log
|AR |
δ
!1
2
,
R∈T
which rearranges to
P
[
|AR |
,
AR & R∈T 1
log δ
R∈T
which is a precise version of the claim that the tubes from the different
bushes are essentially disjoint. This completes the proof of the claimed
estimate on the size of a hairbrush.
11.3.2. What to do in case 2. (the hairbrush case). Remember that we may
have applied Lemma 120 many times. In other words, we kept removing
hairbrushes until we had removed at least half of all tubes. Write
#(Ck ) = #{ hairbrushes T (T ) with k(T ) = k }.
By construction, given a T , every R ∈ T (T ) has slope within a 2−k radius
of that of T ; Hence
k n−1
2
.
#E(T ) ≤
δ
(Why? Because a 2−k neighborhood on the (n − 1)-sphere contains at most
that many disjoint δ neighborhoods.) Hence
log 1δ
n−1
−k n−1
X
N
2
1
∼
≤
#Ck
,
δ
2
δ
k=0
which implies there exists a k such that
#Ck &
2k(n−1)
.
log 1δ
(We just pigeonholed over the parameter k to find a value of k accounting
for most of the sum.) The point is that there exists a k such that there are
at least
2k(n−1)
log 1δ
hairbrushes of diameter 2−k . Ideally we could just add up the areas of all the
hairbrushes and be done. Sadly, the hairbrushes themselves might overlap.
Fortunately, we can control how much they overlap by using the same old
trick from the 2D Kakeya problem. Specifically, we have the following:
74
MICHAEL BATEMAN
Claim 122. Given two hairbrushes HT and HT 0 of diameter 2−k whose
central axes have an angle of 2−k+l , we have that
1
|HT ∩ HT 0 | . 2−l (log )2 · |HT |.
δ
Proof. Two such hairbrushes intersect over an interval I of length 2−l since
their angle is a factor of 2l times their width. Since the hairbrushes are
not the entire tubes but instead somewhat sparse subsets, it could be the
case that HT and HT 0 are both concentrated in this interval I. Fortunately,
this is not the case. So to prove the claim it suffices to demonstrate that
the area of HT restricted to I is at most 2−l log 1δ |HT |. But this follows
M
immediately from the construction of HT : it is the union of δ2
k tubes, each
n−1
having volume δ
. More specifically, each tube has volume at most 2−l
when restricted to an interval of length 2−l . Hence the volume of HT in I
is less than
2−l
M
1
n−2
(log )2
1 2 kδ
δ
(log δ ) 2
1
= 2−l |HT |(log )2 .
δ
M n−1
δ
.
δ2k
Notice that we have used here the lower bound on |HT | given to us by
Lemma 120.
We can now use the philosophy: we have lots of stuff, it doesn’t overlap
too much, therefore it has large collective area.
Z
X
X
|HT | = S
1HT
HT T ∈C
k
T ∈Ck
1
1 Z
2
[
2
X

2
≤ HT (
1HT )
.
T ∈Ck
T
Using the claim above we can estimate
Z X
XX
(
1HT )2 =
|HT ∩ HT 0 |
T ∈Ck
T
=
T0
k
XX
T
l=0
X
T0:
|HT ∩ HT 0 |
∠(T,T 0 )∼2−k+l
k
1 XX
. log
δ
T
X
l=0 T 0 : ∠(T,T 0 )∼2−k+l
2−l |HT |.
75
Now just notice that the number of T 0 in the inner sum is at most ∼ 2l(n−1) ,
so that
Z X
k
1 X X l(n−1) −l
2
(
1HT ) . log
2
2 |HT |
δ
T ∈Ck
T
l=0
X
1
. log 2k(n−2)
|HT |.
δ
T
Hence
X
Z
X
|HT | =
S
T ∈Ck
1H T
HT T ∈C
k
!1
2
1 k(n−2) X
log 2
|HT |
,
δ
1
[
2
≤ HT T
T
and after rearranging we see that
[
|K| ≥ HT T
P
T ∈Ck |HT |
&
.
log 1δ 2k(n−2)
Fortunately, we know that #Ck &
2k(n−1)
,
log 1δ
so by Claim 121 we have
|K| & (# of hairbrushes ) · (# volume of a hairbrush)
&
∼
M δ n−2
2k(n−1)
log 1δ (log( 1δ ))2 2k
M δ n−2
(log 1δ )4
1
log
1 k(n−2)
δ2
1
log
1 k(n−2)
δ2
,
which is what we claimed.
12. A simple restriction estimate
In this section we present a simple example of a restriction estimate.
Specifically,
babyrestiction
Theorem 123. For any f ∈ L∞ (S1 ), we have
||fd
dσ||L6 (R2 ) . ||f ||L∞ (σ) ,
where σ is normalized arclength measure on the unit circle.
The exponent 6 is not special here and certainly not optimal. Essentially,
6 is the smallest even integer strictly larger than 4. Notice below that we
use the fact that 6 is an even integer in the calculation |X|6 = X 3 X̄ 3 . The
exponent 4 almost works, but we would need to limit ourselves to integrating
76
MICHAEL BATEMAN
over a large ball on the LHS and allow a small error on the right depending
on the radius of the ball. The ∞ exponent is important only because it
allows us to give a simple proof. To prove this theorem we need a basic
lemma, whose proof follows that of the theorem:
decay
Lemma 124. For all y ∈ R2 ,
c
|dσ(y)|
.
1
1
1 + |y| 2
.
2
Proof of Theorem 123 assuming Lemma 124. Recall that if G(x) = e−π|x| ,
2
x
then also Ĝ(ξ) = e−π|ξ| . Define GR (x) = G( R
), and notice that GR (R) =
−π
e . Since the Gaussian is radially decreasing, this implies that if |x| ≤ R,
then
GR (x) ≥ e−π & 1.
Hence for any R > 0, we have
Z
Z
6
d
|f dσ(y)| dy .
GR (y)|fd
dσ(y)|6 dy.
|y|≤R
(We will take R → ∞.) Expanding the definition of fd
dσ(y), this last term
is controlled by
Z
6
Z
−2πiθy
GR (y) e
f (θ)dσ(θ) dy
Z
Z
Z
=
· · · f (θ1 ) · · · f (θ6 ) GR (y)e−2πiy[θ1 +θ2 +θ3 −θ4 −θ5 −θ6 ] dydθ1 · · · dθ6
Z
Z
=
· · · f (θ1 ) · · · f (θ6 )GˆR (θ1 + θ2 + θ3 − θ4 − θ5 − θ6 )dθ1 · · · dθ6
Z
Z
6
≤ ||f ||∞ · · · GˆR (θ1 + θ2 + θ3 − θ4 − θ5 − θ6 )dθ1 · · · dθ6 ,
and this last estimate is made possible since we know what GˆR is– and
specifically, that it is always positive. Unwinding the last display, we have
that it is controlled by
Z
Z Z
||f ||6∞ · · ·
GR (y)e−2πiy[θ1 +θ2 +θ3 −θ4 −θ5 −θ6 ] dydθ1 · · · dθ6
Z
c 6
6
= ||f ||∞ GR (y) dσ(y) dy.
What we have gained is that we have removed the f from fd
dσ so that we
c
can work directly with dσ.
(If we had not inserted the cutoff
R parameter R, we could have proceeded
as above, and used the heuristic e−2πiy[θ1 +θ2 +θ3 −θ4 −θ5 −θ6 ] dy = δ0 (θ1 + θ2 +
θ3 − θ4 − θ5 − θ6 ), where δ0 is just the δ function. Inserting the cutoff
parameter with the Gaussian lets us do this rigorously, much like we used
Gaussians to help us in our proof of the Fourier inversion theorem.)
77
c
But by the lemma, we have |dσ(y)|
.
1
1
. Combining this with our
1+|y| 2
previous work, we have that for any R > 0,
Z
Z
c 6
6
6
d
|f dσ(y)| dy . ||f ||∞ GR (y) dσ(y)
dy
|y|≤R
.
||f ||6∞
Z
!6
1
1
1 + |y| 2
dy
. ||f ||6∞ .
It remains to prove Lemma 124. It basically follows from a classical
result called Van der Corput’s lemma. We give here a rather elementary
formulation:
Lemma 125. Suppose ϕ : [a, b] → R satisfies |ϕ0 (t)| ≥ L for t ∈ [a, b]. Also
assume that ϕ0 is monotone on [a, b]. Then
Z b
1
iϕ(t) e
dt . .
L
a
One should imagine that L is a rather large number. Also notice that the
estimate is independent of the endpoints [a, b].
Example 126. A simple example of (something like) the previous lemma
is the following fact:
Z b
1
sin(Lt)dt . ,
L
a
which can be verified by noticing that the integral of sin(Lt) is zero over any
interval of length 2π
L and then estimating the remainder.
We postpone the proof of Van der Corput’s lemma and see how we can
use it to prove Lemma 124. Our goal is to estimate
Z
c −2πiy·x
dσ(x) .
dσ(y) = e
Because the circle is radially symmetric, we can assume that y = (N, 0) for
some value of N . We will prove the required decay in the paramter N . First
note that
c
dσ(y)
≤1
for any y, so we can focus on N ≥ 1
simplification, we have
Z
−2πiy·x
e
dσ(x) =
c
and prove |dσ(y)|
.
Z
−2πiN
x
1
e
dσ(x)
√1 .
N
With this
78
MICHAEL BATEMAN
Z
1
= e−2πiN cos t dt
2π
We want to apply Van der Corput’s lemma to the phase function ϕ(t) =
−2πiN cos t, so we compute
ϕ0 (t) = 2πiN sin t.
Unfortunately sin t is zero in a few places, but we can control it elsewhere.
Fix a small number > 0; we will make a convenient choice of momentarily.
More precisely, if
G = [, π − ] ∪ [π + , 2π − ]
B = [0, 2π] \ G,
then |ϕ0 (t)| & N for t ∈ G. Further, although ϕ0 is not monotone on each of
these intervals in G, they can be divided into two pieces each (by splitting at
π
3π
0
2 and 2 respectively) on which ϕ is monotone. On each of these intervals,
we can apply Van der Corput’s lemma, resulting in the estimate
Z
1
−2πiN cos t e
dt
. N.
t∈G
But since B is just the union of three intervals of width . each, we have
Z
−2πiN
cos
t
e
dt ≤ |B| . .
t∈B
Choosing =
√1
N
gives us the estimate claimed in Lemma 124.
12.0.3. Proof of Van der Corput’s lemma. The proof given here is probably
quite standard; it is taken from [2]. Integrate by parts to get
Z b
Z b
1
iϕ(t)
e
dt =
iϕ0 (t)eiϕ(t) 0 dt
iϕ (t)
a
a
Z
iϕ(b)
iϕ(a)
e
e
1
iϕ(t) d
=
−
− e
dt.
iϕ0 (b) iϕ0 (a)
dt iϕ0 (t)
The first two terms are ≤ L1 in absolute value by our hypothesis on ϕ0 , so it
remains to estimate the integral in the last expression. Notice that
1
−ϕ00 (t)
d
=
,
dt iϕ0 (t)
ϕ0 (t)2
which is either always positive or always negative, by our assumption that
ϕ0 is monotone. Hence
Z
Z iϕ(t) d
1
1
eiϕ(t) d
dt
dt ≤
e
0
0
dt iϕ (t)
dt iϕ (t) Z d
1
≤
dt iϕ0 (t) dt
Z
d
1
= dt
0
dt iϕ (t)
79
1
1 = 0
− 0 ϕ (b) ϕ (a)
2
≤
,
L
where the last equality follows from the fundamental theorem of calculus.
This completes the proof of Van der Corput’s lemma.
13. Other sources
Essentially nothing in these notes is due to the author, except for organization and possibly presentation. Much material and inspiration was taken
from a variety of sources, many of which are still not the original sources.
• An excellent survey is the Notices article by Tao [13].
• The construction of Besicovitch sets can be found in Green’s notes
[1]. The construction of small sets with large reach (which is very
similar) appears in [6] and [12].
• The so-called “Borel-Cantelli”- type lemma used in the proof of the
Non-Differentiation theorem comes from [12].
• The arithmetic approach to Kakeya is due to Bourgain. Presentations similar to the one given here can be found in [1] and [15]; these
presentations also include proofs of the sums-differences lemma. A
cleaner proof of the sums-differences lemma can be found in [9].
• The proof of the sum-product theorem given here is due to Elekes,
and can also be found in [17]. The world record exponent 43 can be
found in [11].
• Proofs of the Szemeredi-Trotter theorem can be found in [4] and at
[14]. Further browsing in [4] and on Tao’s blog will yield more about
the cell decomposition theorem and the polynomial ham sandwich
theorem.
• The proof given here of the joints conjecture is slightly simpler than
the original Guth-Katz argument, but is still essentially theirs. The
simplified argument can be found in [4] and [8]. An independent
proof in higher dimensions is in [10].
• The original solution of the finite field Kakeya conjecture is in [3]; a
similar presentation, that we followed here, is in [4].
• Basics about measure theory and Lp spaces can be found in [7], for
example, as well as many other places.
• Basics about the Fourier transform can be found in [2], [12] and also
[7]. [2] is far less encyclopedic than [12], but far easier to read. The
arguments given here are essentially from [2].
• Our approach to Fefferman’s theorem is from [15], but is not that
much different from the original [6]. A reworking of the original can
also be found in [12].
• Material on the restriction phenomenon can be found in [15], and
also [18]. The proof given here that restriction implies Kakeya is
80
MICHAEL BATEMAN
from [18], but it is also in [15] in basically the same form, and I
expect neither of these sources is original.
• Material on the connection between Kakeya and the wave equation
can be found in [18]. The argument given here for the initial value
problem is in analogy with an example in [18] for a nonhomogenous
wave equation.
• The bush argument is due to Bourgain [?] and the hairbrush argument is due to Wolff [?]. The presentation here is inspired by the
lecture notes of Tao [15], although our arguments are a bit cleaner
(especially in the hairbrush case) since we only aim for bounds on
the Minkowski dimension rather than the corresponding maximal
version.
• The baby 2D restriction theorem proved here is essentially found
from the lecture notes of Green, who found it in the thesis of Mockenhapt. The estimates on the decay of the Fourier transform of the
measure on the circle are classical, as is Van der Corput’s lemma.
References
G
D
Dpaper
Dsurvey
evans
F
Folland
KSS
KT
Q
Sol
S
Tsurvey
Tblog
T
[1] Ben Green’s lecture notes, available at https://www.dpmms.cam.ac.uk/˜
bjg23/rkp.html (Cited on 83.)
[2] Duoandikoetxea, Javier. Fourier analysis, AMS Graduate Studies in Mathematics,
Vol. 29. (Cited on 82, 84.)
[3] Dvir, Zeev. On the size of Kakeya sets in finite fields J. Amer. Math. Soc. (Cited on
84.)
[4] Dvir, Zeev.
Incidence theorems and their applications Available online at
http://arxiv.org/abs/1208.5073 (Cited on 83, 84.)
[5] Evans, Lawrence. Partial Differential Equations, AMS Graduate Studies in Mathematics, Vol. ? (Cited on 64.)
[6] Fefferman, Charles. The Multiplier Problem for the Ball. The Annals of Mathematics,
2nd Ser., V01. 94, No. 2. (Sep., 1971), pp. 330-336. (Cited on 2, 42, 83, 84.)
[7] Folland, Gerald. (Cited on 84.)
[8] Kaplan, Haim, Sharir, Micha, and Shustin, Eugenii. On lines and joints, Discrete and
Computational Geometry, December 2010, Volume 44, Issue 4, pp 838-843 (Cited on
84.)
[9] Katz, Nets Hawk, and Terence Tao. Bounds on arithmetic projections and applications
... (Cited on 83.)
[10] Quilodrán, René. The joints problem in Rn , SIAM J. of Discrete Math., Vol. 23,
No. 4, pp. 22112213 (Cited on 84.)
[11] Solymosi, Joszef. ...Sums and products ... (Cited on 83.)
[12] Stein, Elias. Harmonic Analysis: (Cited on 83, 84.)
[13] Tao, Terence. From rotating needles to stability of waves: Emerging connections
between combinatorics, analysis, and PDE, Notices Amer. Math. Soc. 48 (2001),
294303. (Cited on 83.)
[14] Tao, Terence. The Szemeredi-Trotter theorem via the polynomial ham sandwich theorem, Available online at http://terrytao.wordpress.com/2011/02/18/the-szemereditrotter-theorem-via-the-polynomial-ham-sandwich-theorem/ (Cited on 83.)
[15] Terry Tao’s lecture notes, available at http://www.math.ucla.edu/˜tao/254b.1.99s/
(Cited on 50, 83, 84.)
81
Ttensor
TV
W
[16] Tao, Terence. Tricks Wiki article: The tensor power trick, available at (Cited on
39.)
http://terrytao.wordpress.com/2008/08/25/tricks-wiki-article-the-tensor-producttrick/
[17] Tao, Terence, and Van Vu. Additive Combinatorics (Cited on 27, 83.)
[18] Wolff, Thomas. Recent work connected with the Kakeya problem Available online at
http://amathe.web.elte.hu/modern/wolff review.pdf (Cited on 84.)
Michael Bateman, University of Cambridge
E-mail address: [email protected]