On a Conjecture of E. M. Stein on
the Hilbert Transform on Vector
Fields
Michael Lacey and Xiaochun Li
Michael Lacey, School of Mathematics, Georgia Insti-
tute of Technology, Atlanta GA 30332
Xiaochun Li, Department of Mathematics, University
of Illinois, Urbana IL 61801
E-mail address : xcli@math.uiuc.edu
http://arxiv.org/abs/0704.0808v3
1991 Mathematics Subject Classification. Primary 42A50, 42B25
The authors are supported in part by NSF grants. M.L. was
supported in part by the Guggenheim Foundation.
Abstract. Let v be a smooth vector field on the plane, that is a
map from the plane to the unit circle. We study sufficient condi-
tions for the boundedness of the Hilbert transform
Hv,ǫ f(x) := p.v.
f(x− yv(x)) dy
where ǫ is a suitably chosen parameter, determined by the smooth-
ness properties of the vector field. It is a conjecture, due to
E.M. Stein, that if v is Lipschitz, there is a positive ǫ for which
the transform above is bounded on L2. Our principal result gives
a sufficient condition in terms of the boundedness of a maximal
function associated to v, namely that this new maximal function
be bounded on some Lp, for some 1 < p < 2. We show that
the maximal function is bounded from L2 to weak L2 for all Lips-
chitz vector fields. The relationship between our results and other
known sufficient conditions is explored.
Contents
Preface vii
Chapter 1. Overview of Principal Results 1
Chapter 2. Connections to Besicovitch Set and Carleson’s
Theorem 7
Besicovitch Set 7
The Kakeya Maximal Function 7
Carleson’s Theorem 8
The Weak L2 Estimate in Theorem 1.15 is Sharp 10
Chapter 3. The Lipschitz Kakeya Maximal Function 11
The Weak L2 Estimate 11
An Obstacle to an Lp estimate, for 1 < p < 2 22
Bourgain’s Geometric Condition 23
Vector Fields that are a Function of One Variable 27
Chapter 4. The L2 Estimate for Hilbert Transform on Lipschitz
Vector Fields 31
Definitions and Principle Lemmas 31
Truncation and an Alternate Model Sum 36
Proofs of Lemmata 39
Chapter 5. Almost Orthogonality Between Annuli 63
Application of the Fourier Localization Lemma 63
The Fourier Localization Estimate 79
References 87
Preface
This memoir is devoted to a question in planar Harmonic Analysis,
a subject which is a circle of problems all related to the Besicovitch set.
This anomalous set has zero Lebesgue measure, yet contains a line seg-
ment of unit length in each direction of the plane. It is a known, since
the 1970’s, that such sets must necessarily have full Hausdorff dimen-
sion. The existence of these sets, and the full Hausdorff dimension, are
intimately related to other, independently interesting issues [26]. An
important tool to study these questions is the so-called Kakeya Maxi-
mal Function, in which one computes the maximal average of a function
over rectangles of a fixed eccentricity and arbitrary orientation.
Most famously, Charles Fefferman showed [10] that the Besicovitch
set is the obstacle to the boundedness of the disc multiplier in the
plane. But as well, this set is intimately related to finer questions of
Bochner-Riesz summability of Fourier series in higher dimensions and
space-time regularity of solutions of the wave equation.
This memoir concerns one of the finer questions which center around
the Besicovitch set in the plane. (There are not so many of these
questions, but our purpose here is not to catalog them!) It concerns
a certain degenerate Radon transform. Given a vector field v on R2,
one considers a Hilbert transform computed in the one dimensional line
segment determined by v, namely the Hilbert transform of a function
on the plane computed on the line segment {x+ tv(x) | |t| ≤ 1}.
The Besicovitch set itself says that choice of v cannot be just mea-
surable, for you can choose the vector field to always point into the set.
Finer constructions show that one cannot take it to be Hölder continu-
ous of any index strictly less than one. Is the sharp condition of Hölder
continuity of index one enough? This is the question of E. M. Stein,
motivated by an earlier question of A. Zygmund, who asked the same
for the question of differentiation of integrals.
The answer is not known under any condition of just smoothness
of the vector field. Indeed, as is known, and we explain, a positive
answer would necessarily imply Carleson’s famous theorem on the con-
vergence of Fourier series, [6]. This memoir is concerned with reversing
viii PREFACE
this implication: Given the striking recent successes related to Car-
leson’s Theorem, what can one say about Stein’s Conjecture? In this
direction, we introduce a new object into the study, a Lipschitz Kakeya
Maximal Function, which is a variant of the more familiar Kakeya Max-
imal Function, which links the vector field v to the ‘Besicovitch sets’
associated to the vector field. One averages a function over rectan-
gles of arbitrary orientation and—in contrast to the classical setting—
arbitrary eccentricity. But, the rectangle must suitably localize the
directions in which the vector field points. This Maximal Function ad-
mits a favorable estimate on L2, and this is one of the main results of
the Memoir.
On Stein’s Conjecture, we prove a conditional result: If the Lips-
chitz Kakeya Maximal Function associated with v maps is an estimate
a little better than our L2 estimate, then the associated Hilbert trans-
form is indeed bounded. Thus, the main question left open concerns
the behavior of these novel Maximal Functions.
While the main result is conditional, it does contain many of the
prior results on the subject, and greatly narrows the possible avenues
of a resolution of this conjecture.
The principal results and conjectures are stated in the Chapter 1;
following that we collect some of the background material for this sub-
ject, and prove some of the folklore results known about the subject.
The remainder of the Memoir is taken up with the proofs of the The-
orems stated in the Chapter 1.
Acknowledgment. The efforts of a strikingly generous referee has
resulted in corrections of arguments, and improvements in presentation
throughout this manuscript. We are indebted to that person.
Michael T. Lacey and Xiaochun Li
CHAPTER 1
Overview of Principal Results
We are interested in singular integral operators on functions of two
variables, which act by performing a one dimensional transform along
a particular line in the plane. The choice of lines is to be variable.
Thus, for a measurable map, v from R2 to the unit circle in the plane,
that is a vector field, and a Schwartz function f on R2, define
Hv,ǫ f(x) := p.v.
f(x− yv(x)) dy
This is a truncated Hilbert transform performed on the line segment
{x+ tv(x) : |t| < 1}. We stress the limit of the truncation in the defi-
nition above as it is important to different scale invariant formulations
of our questions of interest. This is an example of a Radon transform,
one that is degenerate in the sense that we seek results independent of
geometric assumptions on the vector field. We are primarily interested
in assumptions of smoothness on the vector field.
Also relevant is the corresponding maximal function
(1.1) Mv,ǫ f := sup
0<t≤ǫ
(2t)−1
|f(x− sv(x))| ds
The principal conjectures here concern Lipschitz vector fields.
Zygmund Conjecture 1.2. Suppose that v is Lipschitz. Then, for
all f ∈ L2(R2) we have the pointwise convergence
(1.3) lim
(2t)−1
f(x− sv(x)) ds = f(x) a. e.
More particularly, there is an absolute constant K > 0 so that if ǫ−1 =
K‖v‖Lip, we have the weak type estimate
(1.4) sup
λ|{Mv,ǫ f > λ}|1/2 . ‖f‖2 .
The origins of this question go back to the discovery of the Besicov-
itch set in the 1920s, and in particular, constructions of this set show
that the Conjecture is false under the assumption that v is Hölder con-
tinuous for any index strictly less than 1. These constructions, known
2 1. OVERVIEW OF PRINCIPAL RESULTS
since the 1920’s, were the inspiration for A. Zygmund to ask if integrals
of, say, L2(R2) functions could be differentiated in a Lipschitz choice
of directions. That is, for Lipschitz v, and f ∈ L2, is it the case that
(2ǫ)−1
f(x− yv(x)) dy = f(x) a.e.(x)
These and other matters are reviewed in the next chapter.
Much later, E. M. Stein [25] raised the singular integral variant of
this conjecture.
E.M. Stein Conjecture 1.5. There is an absolute constant K > 0
so that if ǫ−1 = K‖v‖Lip, we have the weak type estimate
(1.6) sup
λ|{|Hv,ǫ f | > λ}|1/2 . ‖f‖2 .
These are very difficult conjectures. Indeed, it is known that if
the Stein Conjecture holds for, say, C2 vector fields, then Carleson’s
Theorem on the pointwise convergence of Fourier series [6] would follow.
This folklore result is recalled in the next Chapter.
We will study these questions using modifications of the phase plane
analysis associated with Carleson’s Theorem [15–20] and a new tool,
which we term a Lipschitz Kakeya Maximal Function.
Associated with the Besicovitch set is the Kakeya Maximal Func-
tion, a maximal function over all rectangles of a given eccentricity.
A key estimate is that the L2 −→ L2,∞ norm of this operator grows
logarithmically in the eccentricity, [27, 28].
Associated with a Lipschitz vector field, we define a class of maxi-
mal functions taken over rectangles of arbitrary eccentricity, but these
rectangles are approximate level sets of the vector field. Perhaps sur-
prisingly, these maximal functions admit an L2 bound that is indepen-
dent of eccentricity. Let us explain.
A rectangle is determined as follows. Fix a choice of unit vectors in
the plane (e, e⊥), with e⊥ being the vector e rotated by π/2. Using these
vectors as coordinate axes, a rectangle is a product of two intervals
R = I × J . We will insist that |I| ≥ |J |, and use the notations
(1.7) L(R) = |I|, W(R) = |J |
for the length and width respectively of R.
The interval of uncertainty of R is the subarc EX(R) of the unit
circle in the plane, centered at e, and of length W(R)/ L(R). See Fig-
ure 1.1.
1. OVERVIEW OF PRINCIPAL RESULTS 3
0 EX(R) R
Figure 1.1. An example eccentricity interval EX(R).
The circle on the left has radius one.
We now fix a Lipschitz map v of the plane into the unit circle. We
only consider rectangles R with
(1.8) L(R) ≤ (100‖v‖Lip)−1 .
For such a rectangle R, set V(R) = R ∩ v−1(EX(R)). It is essential to
impose a restriction of this type on the length of the rectangles, for with
out it, one can modify constructions of the Besicovitch set to provide
examples which would contradict the main results and conjectures of
this work.
For 0 < δ < 1, we consider the maximal functions
(1.9) Mv,δ f(x)
= sup
|V(R)|≥δ|R|
1R(x)
|f(y)| dy.
That is we only form the supremum over rectangles for which the vector
field lies in the interval of uncertainty for a fixed positive proportion δ
of the rectangle, see Figure 1.2.
Weak L2 estimate for the Lipschitz Kakeya Maximal Function
1.10. The maximal function Mδ,v is bounded from L
2(R2) to L2,∞(R2)
with norm at most . δ−1/2. That is, for any λ > 0, and f ∈ L2(R2),
this inequality holds:
(1.11) λ2|{x ∈ R2 : Mδ,v f(x) > λ}| . δ−1‖f‖22 .
The norm estimate in particular is independent of the Lipschitz vector
field v.
A principal Conjecture of this work is:
Conjecture 1.12. For some 1 < p < 2, and some finite N and all
0 < δ < 1 and all Lipschitz vector fields v, the maximal function Mδ,v
is bounded from Lp(R2) to Lp,∞(R2) with norm at most . δ−N .
4 1. OVERVIEW OF PRINCIPAL RESULTS
Figure 1.2. A rectangle, with the vector field pointing
in the long direction of the rectangle at three points.
We cannot verify this Conjecture, only establishing that the norm
of the operator can be controlled by a slowly growing function of ec-
centricity.
In fact, this conjecture is stronger than what is needed below. Let us
modify the definition of the Lipschitz Kakeya Maximal Function, by re-
stricting the rectangles that enter into the definition to have an approx-
imately fixed width. For 0 < δ < 1, and choice of 0 < w < 1
‖v‖Lip,
parameterizing the width of the rectangles we consider, define
(1.13) Mv,δ,w f(x)
= sup
|V(R)|≥δ|R|
w≤W(R)≤2w
1R(x)
|f(y)| dy.
We can restrict attention to this case as the primary interest below
is the Hilbert transform on vector fields applied to functions with fre-
quency support in a fixed annulus. By Fourier uncertainty, the width
of the fixed annulus is the inverse of the parameter w above.
Conjecture 1.14. For some 1 < p < 2, and some finite N and all
0 < δ < 1, all Lipschitz vector fields v and 0 < w < 1
‖v‖Lip the
maximal function Mδ,v,w is bounded from L
p(R2) to Lp,∞(R2) with norm
at most . δ−N .
These conjectures are stated as to be universal over Lipschitz vec-
tor fields. On the other hand, we will state conditional results below
in which we assume that a given vector field satisfies the Conjecture
above, and then derive consequences for the Hilbert transform on vec-
tor fields. We also show that e. g. real-analytic vector fields [3] satisfy
these conjectures.
We turn to the Hilbert transform on vector fields. As it turns out,
it is useful to restrict functions in frequency variables to an annulus.
Such operators are given by
St f(x) =
1/t≤|ξ|≤2/t
f̂(ξ) eiξ·x dξ .
The relevance in part is explained in part by this result of the authors
[15], valid for measurable vector fields.
1. OVERVIEW OF PRINCIPAL RESULTS 5
Theorem 1.15. For any measurable vector field v we have the L2 into
λ|{|Hv,∞ ◦ St f | > λ}|1/2 . ‖f‖2 .
The inequality holds uniformly in t > 0.
It is critical that the Fourier restriction St enters in, for otherwise
the Besicovitch set would provide a counterexample, as we indicate in
the first section of Chapter 2. This is one point at which the differ-
ence between the maximal function and the Hilbert transform is strik-
ing. The maximal function variant of the estimate above holds, and
is relatively easy to prove, yet the Theorem above contains Carleson’s
Theorem on the pointwise convergence of Fourier series as a Corollary.
The weak L2 estimate is sharp for measurable vector fields, and so
we raise the conjecture
Conjecture 1.16. There is a universal constant K for which we have
the inequalities
(1.17) sup
0<t<‖v‖Lip
‖Hv,ǫ ◦ St‖2→2 <∞ ,
where ǫ = ‖v‖Lip/K.
Modern proofs of the pointwise convergence of Fourier series use the
so-called restricted weak type approach, invented by Muscalu, Tao and
Thiele in [21]. This method uses refinements of the weak L2 estimates,
together with appropriate maximal function estimates, to derive Lp
inequalities, for 1 < p < 2. In the case of Theorem 1.15—for which
this approach can not possibly work—the appropriate maximal func-
tion is the maximal function over all possible line segments. This is
the unbounded Kakeya Maximal function for rectangles with zero ec-
centricity. One might suspect that in the Lipschitz case, there is a
bounded maximal function. This is another motivation for our Lips-
chitz Kakeya Maximal Function, and our main Conjecture 1.12. We
illustrate how these issues play out in our current setting, with this
conditional result, one of the main results of this memoir.
Theorem 1.18. Assume that Conjecture 1.14 holds for a choice of
Lipschitz vector field v. Then we have the inequalities
(1.19) ‖Hv,ǫ ◦ St‖2 . 1 , 0 < t < ‖v‖Lip .
Here, ǫ is as in (1.17). Moreover, if the vector field as 1+η derivatives,
we have the estimate
(1.20) ‖Hv,ǫ‖2 . (1 + log‖v‖C1+η)2 .
In this case, ǫ = K/‖v‖C1+η and η > 0.
6 1. OVERVIEW OF PRINCIPAL RESULTS
While this is a conditional result, we shall see that it sheds new
light on prior results, such as one of Bourgain [3] on real analytic vector
fields. See Proposition 3.30, and the discussion of that Proposition.
The authors are not aware of any conceptual obstacles to the fol-
lowing extension of the Theorem above to be true, namely that one can
establish Lp estimates, for all p > 2. As our argument currently stands,
we could only prove this result for p sufficiently close to 2, because of
our currently crude understanding of the underlying orthogonality ar-
guments.
Conjecture 1.21. Assume that Conjecture 1.14 holds for a choice of
vector field v with 1 + η > 1 derivatives, then we have the inequalities
below
(1.22) ‖Hv,ǫ‖p . (1 + log‖v‖C1+η)2 , 2 < p <∞ .
In this case, ǫ = K/‖v‖C1+η .
For a brief remark on what is required to prove this conjecture, see
Remark 4.65.
The results of Christ, Nagel, Stein and Wainger [7] apply to certain
vector fields v. This work is a beautiful culmination of the ‘geometric’
approach to questions concerning the boundedness of Radon trans-
forms. Earlier, a positive result for analytic vector fields followed from
Nagel, Stein and Wainger [22]. E.M. Stein [25] specifically raised the
question of the boundedness of Hv for smooth vector fields v. And
the results of D. Phong and Stein [23, 24] also give results about Hv.
J. Bourgain [3] considered real–analytic vector field. N. H. Katz [13]
has made an interesting contribution to maximal function question.
Also see the partial results of Carbery, Seeger, Wainger and Wright [5].
CHAPTER 2
Connections to Besicovitch Set and Carleson’s
Theorem
Besicovitch Set
The Besicovitch set is a compact set that contains a line segment of
unit length in each direction in the plane. Anomalous constructions of
such sets show that they can have very small measure. Indeed, given
ǫ > 0 one can select rectangles R1, . . . , Rn, with disjoint eccentricities,
|EX(R)| ≃ n−1, and of unit length, so that |B| ≤ ǫ for B :=
n=1Rj.
On the other hand, letting ej ∈ EX(Rj), one has that the rectangles
Rj + ej are essentially disjoint. See Figure 2.1. Call the ‘reach’ of the
Besicovitch set
Reach :=
Rj + ej .
This set has measure about one. On the Reach, one can define a vector
field with points to a line segment contained in the Besicovitch set.
Clearly, one has
|Hv 1B(x)| ≃ 1 , x ∈ Reach .
Further, constructions of this set permit one to take the vector field
to be Lipschitz continuous of any index strictly less than one. And
conversely, if one considers a Besicovitch set associated to a vector field
of sufficiently small Lipschitz norm, of index one, the corresponding
Besicovitch set must have large measure. Thus, Lipschitz estimates
are critical.
The Kakeya Maximal Function
The Kakeya maximal function is typically defined as
(2.1) MK,ǫ f(x) := sup
|EX(R)|≥ǫ
1R(x)
|f(y)| dy , ǫ > 0 .
One is forced to take ǫ > 0 due to the existence of the Besicovitch set.
It is a critical fact that the norm of this operator admits a norm bound
on L2 that is logarithmic in ǫ. See Córdoba and Fefferman [8], and
8 2. CONNECTIONS
Figure 2.1. A Besicovitch Set on the left, and it’s
Reach on the right.
Strömberg [27, 28]. Subsequently, there have been several refinements
of this observation, we cite only Nets H. Katz [12], Alfonseca, Soria
and Vargas [1], and Alfonseca [2]. These papers contain additional
references. For the L2 norm, the following is the sharp result.
Theorem 2.2. We have the estimate below valid for all 0 < ǫ < 1.
‖MK,ǫ‖2→2 . 1 + log 1/ǫ .
The standard example of taking f to be the indicator of a small
disk show that the estimate above is sharp, and that the norm grows
as an inverse power of ǫ for 1 < p < 2.
Carleson’s Theorem
We explain the connection between the Hilbert transform on vector
fields and Carleson’s Theorem on the pointwise convergence of Fourier
series. Since smooth functions have a convergent Fourier expansion,
the main point of Carleson’s Theorem is to provide for the control of
an appropriate maximal function. We recall that maximal function in
this Theorem.
Carleson’s Theorem 2.3. For all measurable functions N : R −→
R, the operator below maps L2 into itself.
CNf(x) := p.v.
eiN(x)y f(x− y)dy
The implied operator norm is independent of the choice of measurable
N(x).
CARLESON’S THEOREM 9
v(x1)
N(x1)
ξ2 = J
σ(ξ1 −N(x1))
Figure 2.2. Deducing Carleson’s Theorem from Stein’s Conjecture.
For fixed function f , an appropriate choice of N will give us
∣∣∣p.v.
eiNy f(x− y)dy
∣∣∣ . |CNf(x)| .
Thus, in the Theorem above we have simply linearized the supremum.
Also, we have stated the Theorem with the un-truncated integral. The
content of the Theorem is unchanged if we make a truncation of the
integral, which we will do below.
Let us now show how to deduce this Theorem from an appropriate
bound on certain bound on Hilbert transforms on vector fields. (This
observation is apparently due to R.Coifman from the 1970’s.)
Proposition 2.4. Assume that we have, say, the bound
‖Hv,1‖2→2 . 1 ,
assuming that ‖v‖C2 ≤ 1. It follows that the Carleson maximal operator
is bounded on L2(R).
Proof. The Proposition and the proof are only given in their most
obvious formulation. Set σ(ξ) =
iξy dy
. For a C2 function N :
R −→ R we deduce that the operator with symbol σ(ξ − N(x)) maps
L2(R) into itself with norm that is independent of the C2 norm of
the function N(x). A standard limiting argument then permits one
to conclude the same bound for all measurable choices of N(x), as is
required for the deduction of Carleson’s inequality.
This argument is indicated in Figure 2.2. Take the vector field to
be v(x1, x2) = (1,−N(x1)/n) where n is chosen much larger than the
10 2. CONNECTIONS
C2 norm of the function N(x1). Then, Hv,1 is bounded on L
2(R2) with
norm bounded by an absolute constant. The symbol of Hv,1 is
σ(ξ1, ξ2) = σ(ξ1 − ξ2N(x1)/n) .
The trace of this symbol along the line ξ2 = J defines a symbol of a
bounded operator on L2(R). Taking J very large, we obtain a very good
approximation to symbol σ(ξ1 − ξ2N(x1)/n), deducing that it maps
L2(R) into itself with a bounded constant. Our proof is complete. �
The Weak L2 Estimate in Theorem 1.15 is Sharp
An example shows that under the assumption that the vector field
is measurable, the sharp conclusion is that Hv ◦ S1 maps L2 into L2,∞.
And a variant of the approach to Carleson’s theorem by Lacey and
Thiele [20] will prove this norm inequality. This method will also show,
under only the measurability assumption, that Hv S1 maps L
p into itself
for p > 2, as is shown by the current authors [15]. The results and
techniques of that paper are critical to this one.
CHAPTER 3
The Lipschitz Kakeya Maximal Function
The Weak L2 Estimate
We prove Theorem 1.10, the weak L2 estimate for the maximal
function defined in (1.9), by suitably adapting classical covering lemma
arguments.
The Covering Lemma Conditions. We adopt the covering lemma
approach of Córdoba and R. Fefferman [8]. To this end, we regard the
choice of vector field v and 0 < δ < 1 as fixed. Let R be any finite
collection of rectangles obeying the conditions (1.8) and |V(R)| ≥ δ|R|.
We show that R has a decomposition into disjoint collections R′ and
R′′ for which these estimates hold.
. δ−1
,(3.1)
R∈R′′
∣∣∣ .
(3.2)
The first of these conditions is the stronger one, as it bounds the L2
norm squared by the L1 norm; the verification of it will occupy most
of the proof.
Let us see how to deduce Theorem 1.10. Take λ > 0 and f ∈ L2
which is non negative and of norm one. Set R to be all the rectangles
R of prescribed maximum length as given in (1.8), density with respect
to the vector field, namely |V(R)| ≥ δ|R|, and
f(y) dy ≥ λ|R| .
We should verify the weak type inequality
(3.3) λ
. δ−1/2 .
12 3. LIPSCHITZ KAKEYA
Apply the decomposition to R. Observe that
. δ−1/2
Here of course we have used (3.1). This implies that
. δ−1/2.
Therefore clearly (3.3) holds for the collection R′.
Concerning the collection R′′, apply (3.2) to see that
R∈R′′
. δ−1/2 .
This completes our proof of (3.3).
The remainder of the proof is devoted to the proof of (3.1) and
(3.2).
The Covering Lemma Estimates.
Construction of R′ and R′′. In the course of the proof, we will need
several recursive procedures. The first of these occurs in the selection
of R′ and R′′.
We will have need of one large constant κ, of the order of say 100,
but whose exact value does not concern us. Using this notation hides
distracting terms.
Let Mκ be a maximal function given as
Mκ f(x) = sup
|f(y)| dy , sup
|f(x+ σω)| dσ
Here, Q is the unit square in plane, and Ω is a set of uniformly dis-
tributed points on the unit circle of cardinality equal to κ. It follows
from the usual weak type bounds that this operator maps L1(R2) into
weak L1(R2).
To initialize the recursive procedure, set
R′ ← ∅ ,
STOCK← R .
THE WEAK L
ESTIMATE 13
R′ κR
Figure 3.1. The rectangle R′ would have been removed
from STOCK upon the selection of R as a member of R′.
The main step is this while loop. While STOCK is not empty, select
R ∈ STOCK subject to the criteria that first it have a maximal length
L(R), and second that it have minimal value of |EX(R)|. Update
R′ ←R′ ∪ {R}.
Remove R from STOCK. As well, remove any rectangle R′ ∈ STOCK
which is also contained in
1κR ≥ κ−1
As the collection R is finite, the while loop will terminate, and at
this point we set R′′ def= R−R′. In the course of the argument below,
we will refer the order in which rectangles were added to R′.
With this construction, it is obvious that (3.3) holds, with a bound
that is a function of κ. Yet, κ is an absolute constant, so this depen-
dence does not concern us. And so the rest of the proof is devoted to
the verification of (3.1).
An important aspect of the qualitative nature of the interval of
eccentricity is encoded into this algorithm. We will choose κ so large
that this is true: Consider two rectangles R and R′ with R ∩ R′ 6= ∅,
L(R) ≥ L(R′), W(R) ≥ W(R′), |EX(R)| ≤ |EX(R′)| and EX(R) ⊂
10EX(R′) then we have
(3.4) R′ ⊂ κR .
See Figure 3.1.
Uniform Estimates. We estimate the left hand side of (3.1). In
so doing we expand the square, and seek certain uniform estimates.
14 3. LIPSCHITZ KAKEYA
Expanding the square on the left hand side of (3.1), we can estimate
l.h.s. of (3.1) ≤
|R|+ 2
(ρ,R)∈P
|ρ ∩R|
where P consists of all pairs (ρ, R) ∈ R′×R′ such that ρ∩R 6= ∅, and
ρ was selected to be a member of R′ before R was. It is then automatic
that L(R) ≤ L(ρ). And since the density of all tiles is positive, it follows
that dist(EX(ρ),EX(R)) ≤ 2‖v‖LipL(ρ) < 150 .
We will split up the collection P into sub-collections {SR : R ∈ R′}
and {Tρ : ρ ∈ R′}.
For a rectangle R ∈ R′, we take SR to consist of all rectangles ρ
such that (a) (ρ, R) ∈ P; and (b) EX(ρ) ⊂ 10EX(R). We assert that
(3.5)
|R ∩ ρ| ≤ |R|, R ∈ R.
This estimate is in fact easily available to us. Since the rectangles
ρ ∈ SR were selected to be in R′ before R was, we cannot have the
inclusion
(3.6) R ⊂
1κρ > κ
Now the rectangle ρ are also longer. Thus, if (3.5) does not hold, we
would compute the maximal function of
in a direction which is close, within an error of 2π/κ, of being orthog-
onal to the long direction of R. In this way, we will contradict (3.6).
The second uniform estimate that we need is as follows. For fixed
ρ, set Tρ to be the set of all rectangles R such that (a) (ρ, R) ∈ P and
(b) EX(ρ) 6⊂ 10EX(R). We assert that
(3.7)
|R ∩ ρ| . δ−1|ρ|, ρ ∈ R′.
This proof of this inequality is more involved, and taken up in the next
subsection.
Remark 3.8. In the proof of (3.7), it is not necessary that ρ ∈ R′.
Writing ρ = Iρ × Jρ, in the coordinate basis e and e⊥, we could take
any rectangle of the form I × Jρ.
THE WEAK L
ESTIMATE 15
. . .
Figure 3.2. Proof of Lemma 3.9
These two estimates conclude the proof of (3.1). For any two dis-
tinct rectangles ρ, R ∈ P, we will have either ρ ∈ SR or R ∈ Tρ. Thus
(3.1) follows by summing (3.5) on R and (3.7) on ρ.
The Proof of (3.7). We do not need this Lemma for the proof of
(3.7), but this is the most convenient place to prove it.
Lemma 3.9. Let S be any finite collection of rectangles with L(R) ≤
2 L(R′), and with |V(R)| ≥ δ|R| for all R,R′ ∈ S. Then it is the case
(3.10)
≤ 2δ−1 .
Proof. Fix a point x at which we give an upper bound on the
sum above. Let C(x) be any circle centered at x. We shall show that
there exists at most one R in S such that V(R) ∩ C(x) 6= ∅. By the
assumption |V(R)| ≥ δ|R| this proves the Lemma.
We prove this last claim by contradiction of the Lipschitz assump-
tion on the vector field v. Assume that there exist at least two rectan-
gles R,R′ ∈ S for which the sets V(R) and V(R′) intersect C(x). Thus
there exist y and y′ in C(x) such that v(y) ∈ EX(R) and v(y′) ∈ EX(R′).
Since v is Lipschitz, we have
|v(y)− v(y′)| ≤ ‖v‖Lip|y − y′|
≤ 4‖v‖Lip L(R)|v(y)− v(y′)| ,
but this is a contradiction to our assumption (1.8). See Figure 3.2. �
We fix ρ, and begin by making a decomposition of the collection Tρ.
Suppose that the coordinate axes for ρ are given by eρ, associated with
the long side of R, and e⊥ρ , with the short side. Write the rectangle
as a product of intervals Iρ × J , where |Iρ| = L(ρ). Denote one of the
endpoints of J as α. See Figure 3.3.
16 3. LIPSCHITZ KAKEYA
Figure 3.3. Notation for the proof of (3.7).
For rectangles R ∈ Tρ, let IR denote the orthogonal projection R
onto the line segment 2Iρ×{α}. Subsequently, we will consider different
subsets of this line segment. The first of these is as follows. For R ∈ Tρ,
let VR be the projection of the set V(R) onto 2Iρ × {α}. The angle θ
between ρ and R is at most |θ| ≤ 2‖v‖Lip L(ρ) ≤ 150 . It follows that
(3.11) 1
L(R) ≤ |IR| ≤ 2 L(R), and δ L(R) . |VR|.
A recursive mechanism is used to decompose Tρ. Initialize
STOCK← Tρ ,
U ← ∅ .
While STOCK 6= ∅ select R ∈ STOCK of maximal length. Update
U ← U ∪ {R},
U(R)← {R′ ∈ STOCK : VR ∩ VR′ 6= ∅}.
STOCK← STOCK− U(R).
(3.12)
When this while loop stops, it is the case that Tρ =
R∈U U(R).
With this construction, the sets {VR : R ∈ U} are disjoint. By
(3.11), we have
(3.13)
L(R) . δ−1 L(ρ) .
The main point, is then to verify the uniform estimate
(3.14)
R′∈U(R)
|R′ ∩ ρ| . L(R) ·W(ρ) , R ∈ U .
Note that both estimates immediately imply (3.7).
THE WEAK L
ESTIMATE 17
Figure 3.4. Proof of Lemma 3.15: The rectangles
R,R′ ∈ U(ρ), and so the angles R and R′ form with
ρ are nearly the same.
Proof of (3.14). There are three important, and more technical,
facts to observe about the collections U(R).
For any rectangle R′ ∈ U(R), denote its coordinate axes as eR′ and
e⊥R′ , associated to the long and short sides of R
′ respectively.
Lemma 3.15. For any rectangle R′ ∈ U(R) we have
|eR′ − eR| ≤ 12 |eρ− eR|
Proof. There are by construction, points x ∈ V(R) and x′ ∈ V(R′)
which get projected to the same point on the line segment Iρ × {α}.
See Figure 3.4. Observe that
|eR′ − eR| ≤ |EX(R′)|+ |EX(R)|+ |v(x′)− v(x)|
≤ |EX(R′)|+ |EX(R)|+ ‖v‖Lip · L(R) · |eρ− eR|
≤ |EX(R′)|+ |EX(R)|+ 1
|eρ− eR|
Now, |EX(R)| ≤ 1
|eρ− eR|, else we would have ρ ∈ SR. Likewise,
|EX(R′)| ≤ 1
|eR′ − eR|. And this proves the desired inequality.
Lemma 3.16. Suppose that there is an interval I ⊂ Iρ such that
(3.17)
R′∈U(R)
L(R′)≥8|I|
|R′ ∩ I × J | ≥ |I × J | .
Then there is no R′′ ∈ U(R) such that L(R′′) < |I| and R′′∩4I×J 6= ∅.
Proof. There is a natural angle θ between the rectangles ρ and R,
which we can assume is positive, and is given by |eρ− eR|. Notice that
we have θ ≥ 10|EX(R)|, else we would have ρ ∈ SR, which contradicts
our construction.
18 3. LIPSCHITZ KAKEYA
4I × J
I × J
Figure 3.5.
Moreover, there is an important consequence of Lemma 3.15: For
any R′ ∈ U(R), there is a natural angle θ′ between R′ and ρ. These
two angles are close. For our purposes below, these two angles can be
regarded as the same.
For any R′ ∈ U(R), we will have
|κR′ ∩ ρ|
|I × J | ≃ κ
W(R′) ·W(ρ)
θ|I|W(ρ)
W(R′)
θ · |I| .
Recall Mκ is larger than the maximal function over κ uniformly
distributed directions. Choose a direction e′ from this set of κ directions
that is closest to e⊥ρ . Take a line segment Λ in direction e
′ of length
κθ|I|, and the center of Λ is in 4I × J . See Figure 3.5. Then we have
|κR′ ∩ Λ|
|Λ| ≥
W(R′)
θ · |I|
Thus by our assumption (3.17),
R′∈U(R)
|κR′ ∩ Λ| & 1 .
That is, any of the lines Λ are contained in the set
1κR′ > κ
Clearly our construction does not permit any rectangle R′′ ∈ U(R)
contained in this set. To conclude the proof of our Lemma, we seek a
contradiction. Suppose that there is an R′′ ∈ U(R) with L(R′′) < |I|
and R′′ intersects 2I × J . The range of line segments Λ we can permit
THE WEAK L
ESTIMATE 19
4I × J
Figure 3.6. The proof of Lemma 3.18
is however quite broad. The only possibility permitted to us is that
the rectangle R′′ is quite wide. We must have
W(R′′) ≥ 1
|Λ| = κ
· θ · |I|.
This however forces us to have |EX(R′′)| ≥ κ
θ. And this implies that
ρ ∈ SR′′ , as in (3.5). This is the desired contradiction.
Our third and final fact about the collection U(R) is a consequence
of Lemma 3.15 and a geometric observation of J.-O. Stromberg [27,
Lemma 2, p. 400].
Lemma 3.18. For any interval I ⊂ IR we have the inequality
(3.19)
R′∈U(R)
L(R′)≤|I|≤
κL(R′)
|R′ ∩ I × J | ≤ 5|I| ·W(ρ) .
Proof. For each point x ∈ 4I × J , consider the square S centered
at x of side length equal to
κ · |I| · |eR− eρ|. See Figure 3.6. It is
Stromberg’s observation that for R′ ∈ U(R) we have
|κR′ ∩ I × J |
|I × J | ≃
|S ∩ κR′|
with the implied constant independent of κ. Indeed, by Lemma 3.15,
we have that
|κR′ ∩ I × J |
|I × J | ≃
κW(R′)
|eR− eρ| · |I|
≃ κW(R
′) · |I| · |eR− eρ|
(|eR− eρ| · |I|)2
≃ |S ∩ κR
|S| ,
20 3. LIPSCHITZ KAKEYA
as claimed.
Now, assume that (3.19) does not hold and seek a contradiction.
Let U ′ ⊂ U(R) denote the collection of rectangles R′ over which the
sum is made in (3.19). The rectangles in U ′ were added in some order
to the collection R′, and in particular there is a rectangle R0 ∈ U ′ that
was the last to be added to U ′. Let U ′′ be the collection U ′−{R0}. We
certainly have ∑
R′∈U ′′
|R′′ ∩ I × J | ≥ 4|I × J |.
Since we cannot have ρ ∈ SR0 , Stromberg’s observation implies that
R′∈U ′′
1κR′ > κ
Here, we rely upon the fact that the maximal function Mκ is larger than
the usual maximal function over squares. But this is a contradiction
to our construction, and so the proof is complete. �
The principal line of reasoning to prove (3.14) can now begin with
it’s initial recursive procedure. Initialize
C(R′)← R′ ∩ ρ .
We are to bound the sum
R′∈U(R)|C(R′)|. Initialize a collection of
subintervals of IR to be
I ← ∅
WHILE there is an interval I ⊂ IR satisfying∑
R′∈V(I)
|C(R′) ∩ I × J | ≥ 40|I| ·W(ρ) ,(3.20)
V(I) = {R′ ∈ U(R) | |C(R′) ∩ I × J | 6= ∅ , L(R′) ≥ 8|I|} ,(3.21)
we take I to be an interval of maximal length with this property, and
update
I ← I ∪ {I} ;
C(R′, I) = C(R′) ∩ I × J , R′ ∈ V(I);
C(R′)← C(R′)− I × J, R′ ∈ V(I) .
[We remark that this last updating is not needed in the most important
special case when all rectangles have the same width. But the case we
are considering, rectangles can have variable widths, so that |C(R)| can
be much larger than any |I| · |J | that would arise from this algorithm.]
Once the WHILE loop stops, we have
R′ ∩ ρ = C(R′) ∪
{C(R′, I) | I ∈ I , R′ ∈ V(I)} .
THE WEAK L
ESTIMATE 21
Here the union is over pairwise disjoint sets.
We first consider the collection of sets {C(R′) | R′ ∈ U(R)} that
remain after the WHILE loop has finished. Since we must not have
R′ ⊂ 1/4κ · ρ, it follows that the minimum value of L(R′) is 1
W(ρ).
Thus, if in (3.20), we consider an interval I of length 1
W(ρ), the
condition L(R′) ≥ 8|I| in the definition of V(I) in (3.21) is vacuous.
Thus, we necessarily have
R′∈V(I)
|C(R′) ∩ I × J | ≤ 40|I| ·W(ρ) .
For if this inequality failed, the WHILE loop would not have stopped.
We can partition IR by intervals of length close to
W(ρ), showing
that we have ∑
R′∈U(R)
|C(R′)| . |IR| ·W(ρ) .
Turning to the central components of the argument, namely the
bound for the terms associated with the intervals in I, consider I ∈ I.
The inequality (3.20) and Lemma 3.18 implies that each I ∈ I must
have length |I| ≤ κ−1/2|Iρ|. But we choose intervals in I to be of
maximal length. Thus,
R′∈V(I)
|C(R′, I)| ≤ 100 · |I| ·W(ρ) .
(3.22)
Indeed, suppose this last inequality fails. Let I ⊂ Ĩ ⊂ Iρ be an interval
twice as long as I. By Lemma 3.18, we conclude that
R′∈V(I)
L(R′)≤8|eI|
|R′ ∩ Ĩ × J | ≤ 10|I| ·W(ρ) .
Notice that we are restricting the sum on the left by the length of |Ĩ|.
Therefore, we have the inequalities
R′∈V(I)
L(R′)>8|eI|
|C(R′, I)| ≥ 90 · |I| ·W(ρ) > 40 · |Ĩ| ·W(ρ) .
That is, Ĩ would have been selected, contradicting our construction.
Lemmas 3.16 and 3.18 place significant restrictions on the collection
of intervals I. If we have I 6= I ′ ∈ I with 3
I ∩ 3
I ′ 6= ∅, then we must
have e.g.
κ|I ′| < |I|, as follows from Lemma 3.18. Moreover, V(I ′)
must contain a rectangle R′ with L(R′) < |I|. But this contradicts
Lemma 3.16.
22 3. LIPSCHITZ KAKEYA
Therefore, we must have
|I| . |IR| . L(R).
With (3.22), this completes the proof of (3.14).
An Obstacle to an Lp estimate, for 1 < p < 2
We address one of the main conjectures of this memoir, namely
Conjecture 1.12. Let us first observe
Proposition 3.23. We have the estimate below valid for all 0 < w <
‖v‖Lip.
λ|{Mv,w f > λ}|2/3 . δ−1/3(1 + logw−1‖v‖Lip)1/3‖f‖3/2
Proof. Let ‖v‖Lip = 1. This just relies upon the fact that with
0 < w < 1
fixed, there are only about log 1/w possible values of L(R).
This leads very easily to the following two estimates. Following the
earlier argument, consider an arbitrary collection of rectangles R with
each R ∈ R satisfying (1.8) and |V(R)| ≥ δ|R|. We can then decompose
R into disjoint collections R′ and R′′ for which these estimates hold.
. δ−1(log 1/w)
,(3.24)
R∈R′′
∣∣∣ .
(3.25)
Compare to (3.1) and (3.2). Following the same line of reasoning that
was used to prove (3.3), we prove our Proposition.
We can devise proofs of smaller bounds on the norm of the maximal
function than that given by this proposition. But no argument that we
can find avoids the some logarithmic term in the width of the rectangle.
Let us illustrate the difficulty in the estimate with an object pointed
out to us by Ciprian Demeter. We term it a pocketknife, and it is
pictured in Figure 3.7.
A pocketknife comes with a handle, namely a rectangle Rhandle that
is longer than any other rectangle in the pocketknife. We call a collec-
tion of rectangles B a set of blades if these two conditions are met. In
the first place,
(3.26) Rhandle ∩
R 6= ∅ .
BOURGAIN’S GEOMETRIC CONDITION 23
handle
blades
. . .
hinges
Figure 3.7. A pocketknife.
In the second place, we have
angle(R,Rhandle) ≃ angle(R′, Rhandle) , R, R′ ∈ B .
Let θ(B) denote the angle between Rhandle and the rectangles in the
blade B. We refer to as a hinge a rectangle of dimensions w/θ(B) by w,
in the same coordinate system of Rhandle that contains the intersection
in (3.26).
Now, let B be a collection of blades for the handle Rhandle. Our proof
of the weak L2 estimate for the Lipschitz Kakeya Maximal function
shows that we can assume
♯B · w2 · θ(B)−1 . |Rhandle| .
This is essentially the estimate (3.5).
But, to follow the covering lemma approach to the L3/2 estimate
for the maximal function, we need to control
(♯B)2 · w2 · θ(B)−1 .
We can only find control of expressions of this type in terms of some
slowly varying function of w−1.
Bourgain’s Geometric Condition
Jean Bourgain [3] gives a geometric condition on the Lipschitz vec-
tor field that is sufficient for the L2 boundedness of the maximal func-
tion associated with v. We describe the condition, and show how it
immediately proves that the corresponding Lipschitz Kakeya maximal
function admits a weak type bound on L1. In particular our Conjec-
ture 1.12 holds for these vector fields.
To motivate Bourgain’s condition, let us recall the earlier condition
considered by Nagel, Stein and Wainger in [22]. This condition imposes
24 3. LIPSCHITZ KAKEYA
a restriction on the maximum and minimum curvatures of the integral
curves of the vector field through the assumption that
supx∈Ω det[∇v(x)v(x), v(x)]
infx∈Ω det[∇v(x)v(x), v(x)]
Here, Ω is a domain in R2, and one can achieve an upper bound on the
norm of the maximal function associated to v, appropriately restricted
to Ω, in terms of this ratio.
Bourgain’s condition permits the vector field to have integral curves
which are flat. Suppose that v is defined on all of R2. Define
(3.27) ω(x; t) := |det[v(x+ tv(x)), v(x)]| , |t| ≤ 1
‖v‖Lip .
Assume a uniform estimate of the following type: For absolute con-
stants 0 < c, C <∞ and 0 < ǫ0 < 12‖v‖Lip,
(3.28) |{|t| ≤ ǫ | ω(x; t) < τ sup
|s|≤ǫ
ω(x, s)}| ≤ Cτ cǫ ,
this condition holding for all x ∈ R2, 0 < τ < 1 and 0 < ǫ < ǫ0.
The interest in this condition stems from the fact [3] that real-
analytic vector fields satisfy it. Also see Remark 3.35. Bourgain proved:
Theorem 3.29. Assume that (3.28) holds. Then, the maximal opera-
tor Mv,ǫ0 defined in (1.1) maps L
2 into itself.
This paper claims that the same methods would prove the bounds
‖Mv,ǫ0‖p . ‖f‖p , 1 < p <∞ .
And suggests that similar methods would apply to the localized Hilbert
transform with respect to these vector fields.
Here, we prove
Proposition 3.30. Assume that (3.28) holds. Then, the Lipschitz
Kakeya Maximal Functions
Mv,δ,w , 0 < δ < 1 , 0 < w < ǫ0
defined in (1.13) satisfy the weak L1 estimate
λ|{Mv,δ,w f > λ}| . δ−1(1 + log 1/δ)‖f‖1 .
The implied constants depend upon the constants in (3.28).
That is, these vector fields easily fall within the scope of our anal-
ysis. As a corollary to Theorem 1.18, we see that Hv maps L
2 into
itself.
BOURGAIN’S GEOMETRIC CONDITION 25
Figure 3.8. Proof of (3.31).
Proof. Let us assume that ‖v‖Lip = 1. Fix δ > 0 and 0 < w < ǫ0.
Let R be the class of rectangles with L(R) < κ and satisfying |V(R)| ≥
δ|R|.
Say that R′ ⊂ R has scales separated by s > 3 iff for R,R′ ∈ R′ the
condition 4 L(R) < L(R′) implies that 2s L(R) < L(R′). One sees that
R can be decomposed into ≃ s sub-collections with scales separated
by s.
The fortunate observation is this: Assuming (3.28), and taking s ≃
log 1/δ, any subset R′ ⊂ R with scales separated by s further enjoys
this property: If R,R′ ∈ R′ with C EX(R) ∩ C EX(R′) = ∅, with C a
fixed constant, then
(3.31) L(R) ≃ L(R′) or R ∩R′ = ∅ .
Let us see why this is true, arguing by contradiction. Thus we
assume that L(R′) ≤ 2−s L(R), R∩R′ 6= ∅ and C EX(R)∩C EX(R′) = ∅.
Since the rectangles have an essentially fixed width, it follows that
2|EX(R′)| ≥ |EX(R)|. Fix a line ℓ in the long direction of 2R with
|{x ∈ ℓ | v(x) ∈ V(R)}| ≥ δ
|ℓ| = δ
L(R) .
Let x0 be in the set above, x
0 ∈ V (R′) and x′0 is the projection of x′′0
onto the line ℓ. See Figure 3.8. Observe that we can estimate
|v(x′′0)− v(x′0)| ≤ 2|v(x0)− v(x′′0)| L(R′)(3.32)
Therefore, for C sufficiently large, we have
|v(x0)− v(x′0)| ≥
∣∣|v(x′0)− v(x′′0)| − |v(x′′0)− v(x0)|
≥ |v(x′′0)− v(x0)|(1− 2 L(R′))
≥ |EX(R′)|
provided C is large enough.
Now, after a moments thought, one sees that
|det[v(x0), v(x′0)]| ≃ angle(v(x0), v(x′0)) .
26 3. LIPSCHITZ KAKEYA
Therefore, for any x ∈ ℓ
s≤L(R)
ω(x; s) & |EX(R′)| .
But the vector field satisfies (3.28), which we will apply with
τ ≃ EX(R)
EX(R′)
≃ L(R
It follows that
L(R) ≤ |{x ∈ ℓ | ω(x; s) ≤ cτ |EX(R′)|}|
≤ |{x ∈ ℓ | ω(x; s) ≤ τ sup
|s|≤ǫ
ω(x, s)}|
. τ c L(R) .
Therefore, we see
(δ/2)1/c .
L(R′)
which is a contradiction to R′ have scales separated by s, and s ≃
1 + log 1/δ.
Let us see how to prove the Proposition now that we have proved
(3.31). Take s ≃ log 1/δ, and a finite sub-collection R′ ⊂ R of rectan-
gles with scales separated by s. We may take a further subset R′′ ⊂ R′
such that ∥∥∥
R′′∈R′′
. δ−1 ,(3.33)
R∈R′−R′′
∣∣∣ .
R∈R′′
|R′′| .(3.34)
These are precisely the covering estimates needed to prove the weak
L1 estimate claimed in the proposition.
But, in choosing R′′ to satisfy (3.33), it is clear that we need only
be concerned about rectangles with a fixed length, and the separation
in scales are (3.31) will control rectangles of distinct lengths.
The procedure that we apply to select R′′ is inductive. Set
R′′ ← ∅ ,
S ← ∅ ,
STOCK←R′ .
WHILE STOCK 6= ∅, select R ∈ STOCK with maximal length, and
update R′′ ← R′′ ∪ {R}, as well as STOCK ← STOCK − {R}. In
addition, for any R′ ∈ STOCK with R′ ⊂ 4CR, where C ≥ 1 is the
VECTOR FIELDS THAT ARE A FUNCTION OF ONE VARIABLE 27
constant that insures that (3.31) holds, remove these rectangles from
STOCK and add them to S.
Once the WHILE loop stops, we will have STOCK = ∅ and we have
our decomposition of R′. By construction, it is clear that (3.34) holds.
We need only check that (3.33) holds. Now, consider R,R′ ∈ R′, with
two rectangles have their scales separated, thus 2s L(R′) < L(R). If it
is the case that R∩R′ 6= ∅ and C EX(R)∩C EX(R′) 6= ∅, then R would
been selected to be in R′ first, whence R′ would have been placed in
Therefore, C EX(R) ∩ C EX(R′) = ∅, but then (3.31) implies that
R ∩ R′ = ∅. Thus, the only contribution to the L∞ norm in (3.33)
can come from rectangles of about the same length. But Lemma 3.9
then implies that such rectangles can overlap only about δ−1 times.
Our proof is complete. (As the interest in (3.28) is in small values of
c, it will be more efficient to use Lemma 3.9 to handle the case of the
rectangles having approximately the same length.) �
Remark 3.35. To conclude that the Hilbert transform on vector fields
is bounded, one could weaken Bourgain’s condition (3.28) to
|{|t| ≤ ǫ | ω(x; t) < τ sup
|s|≤ǫ
ω(x, s)}| ≤ C exp(−(log 1/τ)c)ǫ .
This inequality is to hold universally in x ∈ R2, 0 < τ < 1, and
0 < ǫ < ‖v‖Lip. This is of interest for 0 < c < 1. The proof above
can be modified to show that the maximal functions Mv,δ,w satisfy the
weak L1 inequality, with constant at most . δ−1−1/c.
Vector Fields that are a Function of One Variable
We specialize to the vector fields that are a function of just one real
variable. Assume that the vector field v is of the form
(3.36) v(x1, x2) = (v1(x2), v2(x2)) ,
and for the moment we do not impose the condition that the vector
field take values in the unit circle. The point is simply this: If we are
interested in transforms where the kernel is not localized, the restriction
on the vector field is immaterial. Namely, for any vector field v
Hv,∞ f(x) = p.v.
f(x− yv(x)) dy
= p.v.
f(x− yṽ(x)) dy
, ṽ(x) =
|v(x)| .
We return to a theme implicit in the proof of Proposition 2.4. This
proof only relies upon vector fields that are only a function of one
28 3. LIPSCHITZ KAKEYA
variable. Thus, it is a significant subcase of the Stein Conjecture to
verify it for Lipschitz vector fields of just one variable. Indeed, the
situation is this.
Proposition 3.37. Suppose that a choice of vector field v(x1, x2) =
(1, v1(x1)) is just a function of, say, the first coordinate. Then, Hv,∞
maps L2(R2) into itself.
Proof. The symbol of Hv,∞ is
sgn(ξ1 + ξ2v1(x1)) .
For each fixed ξ2, this is a bounded symbol. And in the special case
of the L2 estimate, this is enough to conclude the boundedness of the
operator. �
It is of interest to extend this Theorem in any Lp, for p 6= 2, for
some reasonable choice of vector fields.
The corresponding questions for the maximal function are also of
interest, and here the subject is much more developed. The paper [5]
studies the maximal function Mv,∞. They proved the boundedness of
this maximal function on Lp, p > 1, assuming that the vector field was
of the form v(x) = (1, v2(x)), that D v2 was positive, and increasing,
and satisfied a third more technical condition. More recently, [14] has
showed that the third condition is not needed. Namely the following is
true.
Theorem 3.38. Assume that v(x) = (1, v2(x)), and moreover that
D v2 ≥ 0 and is monotonically increasing. Then, Mv,∞ is bounded on
Lp, for 1 < p <∞.
These vector fields present far fewer technical difficulties than a gen-
eral Lipschitz vector field, and there are a richer set of proof techniques
that one can bring to bear on them, as indicated in part in the proof
of Proposition 3.37. The papers [5, 14] cleverly exploit the Plancherel
identity (in the independent variable), and other orthogonality consid-
erations to prove their results.
These considerations are not completely consistent with the domi-
nant theme of this monograph, in which the transforms are localized.
Nevertheless, it would be interesting to explore methods, possibly mod-
ifications of this memoir, that could provide an extension of Proposi-
tion 3.37.
In this direction, let us state a possible direction of study. The
definition of the the sets V(R) for vector fields of magnitude 1 is given
as V(R) = R ∩ v−1(EX(R)). For vector fields of arbitrary magnitude,
VECTOR FIELDS THAT ARE A FUNCTION OF ONE VARIABLE 29
we define these sets to be
V(R) = {x ∈ R | v(x)|v(x)| ∈ EX(R)} .
Define a maximal function—an extension of our Lipschitz Kakeya Max-
imal Function—by
(3.39) M̃v,δf(x) = sup
|V(R)|≥δ|R|
|R|−1
f(y) dy .
In this definition, we require the rectangles to have density δ, but do
not restrict their eccentricities, or lengths.
Conjecture 3.40. Assume that the vector field is of the form v(x) =
(1, v2(x2)), and the derivative D v ≥ 0 and is monotone. Then for all
0 < δ < 1, we have the estimate
‖M̃v,δ‖p . δ−1 , 1 < p <∞ .
One can construct examples which show that the L1 to weak L1
norm of the maximal function is not bounded in terms of δ. Indeed,
recalling the ‘pocketknife’ examples of Figure 3.7, we comment that one
can construct examples of vector fields with these properties, which we
describe with the terminology associated with the pocketknife exam-
ples.
• The width of all rectangles are fixed. And all rectangles have
density δ.
• The ‘handle’ of the pocketknife has positive angle θ with the
x1 axis.
• There is ‘hinge’ whose blades have angles which are positive,
and greater than θ. The number of blades can be unbounded,
as the width of the rectangles decreases to zero.
The assumption that the vector field is only a function of x2 then
greatly restricts, but does not completely forbid, the existence of addi-
tional hinges. So the combinatorics of these vector fields, as expressed
in the Lipschitz Kakeya Maximal Function, are not so simple.
CHAPTER 4
The L2 Estimate for Hilbert Transform on
Lipschitz Vector Fields
We prove one of our main conditional results about the Hilbert
transform on Lipschitz vector fields, the inequality (1.19) which is the
estimate at L2, for functions with frequency support in an annulus,
assuming an appropriate estimate for the Lipschitz Kakeya Maximal
Function.
We begin the proof by setting notation appropriate for phase plane
analysis for functions f on the plane supported on an annulus. With
this notation, we can define appropriate discrete analogs of the Hilbert
transform on vector fields. The Lemmas 4.22 and 4.23 are the combi-
natorial analogs of our Theorem 1.15. We then take up the proofs. The
main step in the proof is Lemma 4.50 which combines the (standard)
orthogonality considerations with the conjectures about the Lipschitz
Kakeya Maximal Functions.
Definitions and Principle Lemmas
Throughout this chapter, κ will denote a fixed small positive con-
stant, whose exact value need not concern us. κ of the order of 10−3
would suffice. The following definitions are as in the authors’ previous
paper [15].
Definition 4.1. A grid is a collection of intervals G so that for all
I, J ∈ G, we have I ∩ J ∈ {∅, I, J}. The dyadic intervals are a grid. A
grid G is central iff for all I, J ∈ G, with I ⊂6= J we have 500κ−20I ⊂ J .
The reader can find the details on how to construct such a central
grid structure in [11].
Let ρ be rotation on T by an angle of π/2. Coordinate axes for R2
are a pair of unit orthogonal vectors (e, e⊥) with ρ e = e⊥.
Definition 4.2. We say that ω ⊂ R2 is a rectangle if it is a product
of intervals with respect to a choice of axes (e, e⊥) of R
2. We will
say that ω is an annular rectangle if ω = (−2l−1, 2l−1)× (a, 2a) for an
integer l with 2l < κa, with respect to the axes (e, e⊥). The dimensions
of ω are said to be 2l × a. Notice that the face (−2l−1, 2l−1) × a is
32 4. L
ESTIMATE FOR Hv
es Rs
Figure 4.1. The two rectangles ωs and Rs whose prod-
uct is a tile. The gray rectangles are other possible loca-
tions for the rectangle Rs.
tangent to the circle |ξ| = a at the midpoint to the face, (0, a). We
say that the scale of ω is scl(ω) := 2l and that the annular parameter
of ω is ann(ω) := a. In referring to the coordinate axes of an annular
rectangle, we shall always mean (e, e⊥) as above.
Annular rectangles will decompose our functions in the frequency
variables. But our methods must be sensitive to spatial considerations;
it is this and the uncertainty principle that motivate the next definition.
Definition 4.3. Two rectangles R and R are said to be dual if they
are rectangles with respect to the same basis (e, e⊥), thus R = r1 × r2
and R = r1 × r2 for intervals ri, ri, i = 1, 2. Moreover, 1 ≤ |ri| · |ri| ≤ 4
for i = 1, 2. The product of two dual rectangles we shall refer to as a
phase rectangle. The first coordinate of a phase rectangle we think of
as a frequency component and the second as a spatial component.
We consider collections of phase rectangles AT which satisfy these
conditions. For s, s′ ∈ AT we write s = ωs × Rs, and require that
ωs is an annular rectangle,(4.4)
Rs and ωs are dual,(4.5)
The rectangles Rs are from the product of central grids.(4.6)
{1000κ−100R | ωs × R ∈ AT } covers R2, for all ωs.(4.7)
ann(ωs) = 2
j for some integer j,(4.8)
♯{ωs | scl(s) = scl, ann(s) = ann} ≥ c
,(4.9)
scl(s) ≤ κann(s).(4.10)
DEFINITIONS AND PRINCIPLE LEMMAS 33
0 ρωs1
Figure 4.2. An annular rectangular ωs, and three as-
sociated subintervals of ρωs1, ωs1, and ωs2.
We assume that there are auxiliary sets ωs,ωs1,ωs2 ⊂ T associated to
s—or more specifically ωs—which satisfy these properties.
Ω := {ωs,ωs1,ωs2 | s ∈ AT } is a grid in T,(4.11)
ωs1 ∩ ωs2 = ∅, |ωs| ≥ 32(|ωs1|+ |ωs2|+ dist(ωs1,ωs2))(4.12)
ωs1 lies clockwise from ωs2 on T,(4.13)
|ωs| ≤ K
scl(ωs)
ann(ωs)
,(4.14)
{ ξ|ξ| | ξ ∈ ωs} ⊂ ρωs1.(4.15)
In the top line, the intervals ωs1 and ωs2 are small subintervals of
the unit circle, and we can define their dilate by a factor of 2 in an
obvious way. Recall that ρ is the rotation that takes e into e⊥. Thus,
eωs ∈ ωs1. See the figures Figure 4.1 and Figure 4.2 for an illustration
of these definitions.
Note that |ωs| ≥ |ωs1| ≥ scl(ωs)/ann(ωs). Thus, eωs is in ωs1, and
ωs serves as ‘the angle of uncertainty associated to Rs.’ Let us be
more precise about the geometric information encoded into the angle
of uncertainty. Let Rs = rs × rs⊥ be as above. Choose another set of
coordinate axes (e′, e′⊥) with e
′ ∈ ωs and let R′ be the product of the
intervals rs and rs⊥ in the new coordinate axes. Then K
′ ⊂ Rs ⊂
′ for an absolute constant K0 > 1.
We say that annular tiles are collections AT satisfying the condi-
tions (4.4)—(4.15) above. We extend the definition of e⊥, eω⊥, ann(ω)
and scl(ω) to annular tiles in the obvious way, using the notation es,
es⊥, ann(s) and scl(s).
34 4. L
ESTIMATE FOR Hv
A phase rectangle will have two distinct functions associated to it.
In order to define these functions, set
Ty f(x) := f(x− y), y ∈ R2 (Translation operator)
Modξ f(x) := e
iξ·x f(x), ξ ∈ R2 (Modulation operator)
R1×R2 f(x1, x2) :=
(|R1||R2|)1/p
(Dilation operator).
In the last display, 0 < p ≤ ∞, and R1 × R2 is a rectangle, and the
coordinates (x1, x2) are those of the rectangle. Note that the definition
depends only on the side lengths of the rectangle, and not the location.
And that it preserves Lp norm.
For a function ϕ and tile s ∈ AT set
(4.16) ϕs := Modc(ωs)Tc(Rs)D
We shall consider ϕ to be a Schwartz function for which ϕ̂ ≥ 0 is
supported in a small ball, of radius κ, about the origin in R2, and is
identically 1 on another smaller ball around the origin. (Recall that κ
is a fixed small constant.)
We introduce the tool to decompose the singular integral kernels.
In so doing, we consider a class of functions ψt, t > 0, so that
Each ψt is supported in frequency in [−θ − κ,−θ + κ].(4.17)
|ψt(x)| . CN(1 + |x|)−N , N > 1 .(4.18)
In the top line, θ is a fixed positive constant so that the second half of
(4.19) is true.
Define
φs(x) :=
ϕs(x− yv(x))ψs(y) dy
(v(x))
ϕs(x− yv(x))ψs(sy) dy.
(4.19)
ψs(y) := scl(s)ψscl(s)(scl(s)y).(4.20)
An essential feature of this definition is that the support of the integral
is contained in the set {v(x) ∈ ωs2}, a fact which can be routinely
verified. That is, we can insert the indicator 1
(v(x)) without loss
of generality. The set ωs2 serves to localize the vector field, while ωs1
serves to identify the location of ϕs in the frequency coordinate.
DEFINITIONS AND PRINCIPLE LEMMAS 35
The model operator we consider acts on a Schwartz functions f ,
and it is defined by
(4.21) Cannf :=
s∈AT (ann)
scl(s)≥‖v‖Lip
〈f, ϕs〉φs.
In this display, AT (ann) := {s ∈ AT | ann(s) = ann}, and we have
deliberately formulated the operator in a dilation invariant manner.
Lemma 4.22. Assume that the vector field is Lipschitz, and satisfies
Conjecture 1.14. Then, for all ann ≥ ‖v‖−1Lip, the operator Cann extends
to a bounded map from L2 into itself, with norm bounded by an absolute
constant.
We remind the reader that for 2 < p <∞ the only condition needed
for the boundedness of Cann is the measurability of the vector field, a
principal result of Lacey and Li [15]. It is of course of great importance
to add up the Cann over ann. The method we use for doing this are
purely L2 in nature, and lead to the estimate for C :=
j=1 C2j .
Lemma 4.23. Assume that the vector field is of norm at most one
in Cα for some α > 1, and satisfies Conjecture 1.14. Then C maps
L2 into itself. In addition we have the estimate below, holding for all
values of scl.
(4.24)
ann=−∞
s∈AT (ann)
scl(s)=scl
〈f, ϕs〉φs
. (1 + log(1 + scl−1‖v‖Cα)).
Moreover, these operators are unconditionally convergent in s ∈ AT .
These are the principal steps towards the proof of Theorem 1.15.
In the course of the proof, we shall not invoke the additional notation
needed to account for the unconditional convergence, as it is entirely
notational. They can be added in by the reader.
Observe that (4.24) is only of interest when scl < ‖v‖Cα. This
inequality depends critically on the fact that the kernel sclψ(scly) has
mean zero. Without this assumption, this inequality is certainly false.
The proof of Theorem 1.15 from these two lemmas is an argument
in which one averages over translations, dilations and rotations of grids.
The specifics of the approach are very close to the corresponding argu-
ment in [15]. The details are omitted.
The operators Cann and C are constructed from a a kernel which is a
smooth analog of the truncated kernel p.v. 1
1{|t|≤1}. Nevertheless, our
36 4. L
ESTIMATE FOR Hv
main theorem follows,1 due to the observation that we can choose a
sequence of Schwartz kernels ψ(1+κ)n , for n ∈ Z, which satisfy (4.17)
and (4.18), and so that for
K(t) :=
an(1 + κ)
nψ(1+κ)n((1 + κ)
we have p.v. 1
1{|t|≤1} = K(t)−K(t). Here, for n ≥ 0 we have |an| . 1.
And for n < 0, we have |an| . (1+κ)n. The principal sum is thus over
n ≥ max(0, ‖v‖Cα), and this corresponds to the operator C. For those
n < max(0, ‖v‖Cα), we use the estimate (4.24), and the rapid decay of
the coefficient an.
Truncation and an Alternate Model Sum
There are significant obstacles to proving the boundedness of the
model sum Cann on an Lp space, for 1 < p < 2. In this section, we
rely upon some naive L2 estimates to define a new model sum which is
bounded on Lp, for some 1 < p < 2.
Our next Lemma is indicative of the estimates we need. For choices
of scl < κann, set
AT (ann, scl) := {s ∈ AT (ann) | scl(s) = scl}.
Lemma 4.25. For measurable vector fields v and all choices of ann
and scl. ∥∥∥
s∈AT (ann,scl)
〈f, ϕs〉φs
. ‖f‖2
Proof. The scale and annulus are fixed in this sum, making the
Bessel inequality
s∈AT (ann,scl)
|〈f, ϕs〉|2 . ‖f‖22
evident. For any two tiles s and s′ that contribute to this sum, if ωs 6=
ωs′, then φs and φs′ are disjointly supported. And if ωs = ωs′, then
Rs and R
s are disjoint, but share the same dimensions and orientation
in the plane. The rapid decay of the functions φs then gives us the
1In the typical circumstance, one uses a maximal function to pass back and
forth between truncated and smooth kernels. This route is forbidden to us; there
is no appropriate maximal function to appeal to.
TRUNCATION AND AN ALTERNATE MODEL SUM 37
estimate
s∈AT (ann,scl)
〈f, ϕs〉φs
s∈AT (ann,scl)
|〈f, ϕs〉|2
. ‖f‖2
Consider the variant of the operator (4.21) given by
(4.26) Φf =
s∈AT (ann)
scl(s)≥κ−1‖v‖Lip
〈f, ϕs〉φs.
As ann is fixed, we shall begin to suppress it in our notations for oper-
ators. The difference between Φ and Cann is the absence of the initial
. log(1+ ‖v‖Lip) scales in the former. The L2 bound for these missing
scales is clearly provided by Lemma 4.25, and so it remains for us to
establish
(4.27) ‖Φ‖2 . 1,
the implied constant being independent of ann, and the Lipschitz norm
of v.
It is an important fact, the main result of Lacey and Li [15], that
(4.28) ‖Φ‖p . 1, 2 < p <∞.
This holds without the Lipschitz assumption.
We are now at a point where we can be more directly engaged with
the construction of our alternate model sum. We only consider tiles
with κ−1‖v‖Lip ≤ scl(s) ≤ κann. A parameter is introduced which is
used to make a spatial truncation of the functions ϕs; it is
(4.29) γ2s := 100
−2 scl(s)
‖v‖Lip
Write ϕs = αs + βs where αs = (Tc(Rs)D
ζ)ϕs, and ζ is a smooth
Schwartz function supported on |x| < 1/2, and equal to 1 on |x| < 1/4.
Write for choices of tiles s,
(4.30) ψs(y) = ψs−(y) + ψs+(y)
where ψs−(y) is a Schwartz function on R, with
supp(ψs−) ⊂
γs(scl(s))
−1[−1, 1] ,
38 4. L
ESTIMATE FOR Hv
and equal to ψscl(s)(y) for |y| < 14γs(scl(s))
−1. Then define
(4.31) as±(x) = 1ωs2(v(x))
φs(x− yv(x))ψs±(y) dy.
Thus, φs = as− + as+. Recalling the notation Sann in Theorem 1.15,
define
(4.32) A± f :=
s∈AT (ann)
scl(s)≥κ−1‖v‖Lip
〈Sann f, αs〉as±
We will write Φ = ΦSann = A+ +A− +B, where B is an operator
defined in (4.35) below. The main fact we need concerns A−.
Lemma 4.33. There is a choice of 1 < p0 < 2 so that
‖A−‖p . 1, p0 < p <∞.
The implied constant is independent of the value of ann, and the Lips-
chitz norm of v.
The proof of this Lemma is given in the next section, modulo three
additional Lemmata stated therein. The following Lemma is important
for our approach to the previous Lemma. It is proved below.
Lemma 4.34. For each choice of κ−1‖v‖Lip < scl < κann, we have the
estimate ∑
s∈AT (ann,scl)
|〈Sann f, αs〉|2 . ‖f‖22.
Define
(4.35) B f :=
s∈AT (ann)
scl(s)≥κ−1‖v‖Lip
〈Sann f, βs〉φs
Lemma 4.36. For a Lipschitz vector field v, we have
‖B‖p . 1, 2 ≤ p <∞.
Proof. For choices of integers κ−1‖v‖Lip ≤ scl < κann, consider
the vector valued operator
Tj,k f :=
{〈Sann f, βs〉√
1{v(x)∈ωs2}Tc(Rs)D
(1 + | · |2)N )(x)
| s ∈ AT (ann, scl)
where N is a fixed large integer.
PROOFS OF LEMMATA 39
Recall that βs is supported off of
γsRs. This is bounded linear op-
erator from L∞(R2) to ℓ∞(AT (ann, scl)). It has norm. (scl/‖v‖Lip)−10.
Routine considerations will verify that
Tj,k : L
2(R2) −→ ℓ2(AT (ann, scl))
with a similarly favorable estimate on its norm. By interpolation, we
achieve the same estimate for Tj,k from L
p(R2) into ℓp(AT (ann, scl)),
2 ≤ p <∞.
It is now very easy to conclude the Lemma by summing over scales
in a brute force way, and using the methods of Lemma 4.25. �
We turn to A+, as defined in (4.32).
Lemma 4.37. We have the estimate
‖A+‖p . 1 2 ≤ p <∞.
Proof. We redefine the vector valued operator Tj,k to be
Tj,k f :=
{〈Sann f, αs〉√
1{v(x)∈ωs2}Tc(Rs)D
(1 + |x|2)N
| s ∈ AT (ann, scl)
where N is a fixed large integer. This operator is bounded from
Lp(R2) −→ ℓp(AT (ann, scl)) , 2 ≤ p <∞
Its norm is at most . 1.
But, for s ∈ AT (ann, scl), we have
(4.38) |as+| . (scl/‖v‖Lip)−10|Rs|−1/2(M1Rs)100.
Here M denotes the strong maximal function in the plane in the coordi-
nates determined by Rs. This permits one to again adapt the estimate
of Lemma 4.25 to conclude the Lemma. �
Now we conclude that ‖Φ‖2 . 1. And since Φ = A− +A++B, it
follows from the Lemmata of this section.
Proofs of Lemmata
Proof of Lemma 4.33. We have Φ = A−+A+ +B, so from
(4.28), Lemma 4.36 and Lemma 4.37, we deduce that ‖A−‖p . 1 for
all 2 < p <∞. It remains for us to verify that A− is of restricted weak
type p0 for some choice of 1 < p0 < 2. That is, we should verify that
for all sets F,G ⊂ R2 of finite measure
(4.39) |〈A− 1F , 1G〉| . |F |1/p|G|1−1/p, p0 < p < 2.
40 4. L
ESTIMATE FOR Hv
Since A− maps L
p into itself for 2 < p < ∞, it suffices to consider
the case of |F | < |G|. Since we assume only that the vector field is
Lipschitz, we can use a dilation to assume that 1 < |G| < 2, and so
this set will not explicitly enter into our estimates.
We fix the data F ⊂ R2 of finite measure, ann, and vector field v
with ‖v‖Lip ≤ κann. Take p0 = 2 − κ2. We need a set of definitions
that are inspired by the approach of Lacey and Thiele [20], and are
also used in Lacey and Li [15]. For subsets S ⊂ Av := {s ∈ AT (ann) |
κ−1‖v‖Lip ≤ scl(s) < κann}, set
〈Sann 1F , αs〉as−
Set χ(x) = (1 + |x|)−1000/κ. Define
(4.40) χ
:= χ(p)s = Tc(Rs)D
χ, 0 ≤ p ≤ ∞.
And set χ̃
s = 1γsRsχ
Remark 4.41. It is typical to define a partial order on tiles, following
an observation of C. Fefferman [9]. In this case, there doesn’t seem to
be an appropriate partial order. Begin with this assumption on the
order relation ‘<’ on tiles:
(4.42)
If ωs ×Rs ∩ ωs′ ×Rs′ 6= ∅, then s and s′ are comparable under ‘<’.
It follows from transitivity of a partial order that that one can have
tiles s1, . . . , sJ , with sj+1 < sj for 1 ≤ j < J , J ≃ log(‖v‖Lip · ann), and
yet the rectangles RsJ and Rs1 can be far apart, namely RsJ ∩ (cJ)Rs1 ,
for a positive constant c. See Figure 4.3. (We thank the referee for
directing us towards this conclusion.) Therefore, one cannot make
the order relation transitive, and maintain control of the approximate
localization of spatial variables, as one would wish. The partial order
is essential to the argument of [9], but while it is used in [20], it is not
essential to that argument.
We recall a fact about the eccentricity. There is an absolute con-
stant K ′ so that for any two tiles s, s′
(4.43) ωs ⊃ ωs′ , Rs ∩ Rs′ 6= ∅ implies Rs ⊂ K ′Rs′ .
Figure 3.1 illustrates this in the case where the two rectangles Rs and
Rs′ have different widths, which is not the case here.
We define an order relation on tiles by s . s′ iff ωs ) ωs′ and
Rs ⊂ κ−10Rs′. Thus, (4.42) holds for this order relation, and it is
certainly not transitive.
PROOFS OF LEMMATA 41
Figure 4.3. The rectangles Rs1 , . . . , RsJ of Remark 4.41.
A tree is a collection of tiles T ⊂ Av, for which there is a (non–
unique) tile ωT × RT ∈ AT (ann) with Rs ⊂ 100κ−10RT, and ωs ⊃ ωT
for all s ∈ T. Here, we deliberately use a somewhat larger constant
100κ−10 than we used in the definition of the order relation ‘..’
For j = 1, 2, call T a i–tree if the tiles for all s, s′ ∈ T, if scl(s) 6=
scl(s′), then ωsi∩ωs′i = ∅. 1–trees are especially important. A few tiles
in such a tree are depicted in Figure 4.4.
Remark 4.44. This remark about the partial order ‘.’ and trees is
useful to us below. Suppose that we have two trees T, with top s(T)
and T′ with top s(T′). Suppose in addition that s(T′) . s(T). Then,
it is the case that T ∪ T′ is a tree with top s(T). Indeed, we must
necessarily have ωT ( ωT, since the Rs are from products of a central
grid. Also, 100κ−1RT′ ⊂ 100κ−1RT. And so every tile in T′ could also
be a tile in T.
Our proof is organized around these parameters and functions as-
sociated to tiles and sets of tiles. Of particular note here are the first
definitions of ‘density,’ which have to be formulated to accommodate
the lack of transitivity in the partial order. Note that in the first defi-
nition, the supremum is taken over tiles s′ ∈ AT of the same annular
parameter as s. We choose the collection AT as it is ‘universal,’ cover-
ing all scales in a uniform way, due to different assumptions including
42 4. L
ESTIMATE FOR Hv
(4.7).
dense(S) := sup
s′∈AT
G∩v−1(ωs′ )
s′ dx | ∃ s , s′′ ∈ S :
ωs ⊃ ωs′ ⊃ ωs′′ , Rs ⊂ 100κ−10Rs′ ,
Rs′ ⊂ 100κ−10Rs′
(4.45)
∆(T)2 :=
|〈Sann 1F , αs〉|2
1Rs , T is a 1–tree,(4.46)
size(S) := sup
T is a 1–tree
∆(T) dx.(4.47)
Observe that dense(S) only really applies to ‘tree-like’ sets of tiles, and
that—and this is important—the tile s′ that appear in (4.45) are not
in S, but only assumed to be in AT . Finally, note that
dense(s) ≃
G∩v−1(ωs)
χ̃(1)s dx
with the implied constants only depending upon κ, χ, and other fixed
quantities.
Observe these points about size. First, it is computed relative to
the truncated functions αs, recall (4.29). Second, that for p > 1,
(4.48) ‖∆(T)‖p . |F |1/p ,
because of a standard Lp estimate for a Littlewood-Paley square func-
tion. Third, that size(Av(ann)) . 1. And fourth, that one has an
estimate of John-Nirenberg type.
Lemma 4.49. For a 1-tree T we have the estimate
‖∆(T)‖p . size(T)|RT|1/p , 1 < p <∞ .
Proofs of results of this type are well represented in the literature.
See [4, 11].
Given a set of tiles, say that count(S) < A iff S is a union of trees
T ∈ T for which ∑
|sh(T)| < A.
We will also use the notation count(S) . A, implying the existence of
an absolute constant K for which count(S) ≤ KA.
The principal organizational Lemma is
PROOFS OF LEMMATA 43
Figure 4.4. A few possible tiles in a 1–tree. Rectangles
ωs are on the left in different shades of gray. Possible
locations of Rs are in the same shade of gray.
Lemma 4.50. Any finite collection of tiles S ⊂ Av is a union of four
subsets
Slight, Ssmall, S
large, ℓ = 1, 2.
They satisfy these properties.
size(Ssmall) <
size(S),(4.51)
dense(Slight) <
dense(S),(4.52)
and both Sℓlarge are unions of trees T ∈ T ℓ, for which we have the
estimates
count(S1large) .
size(S)
−2−κ|F |
size(S)−p dense(S)−M |F |
+ size(S)
dense(S)−1
dense(S)−1
(4.53)
count(S2large) .
size(S)
(log 1/ size(S))3|F |
size(S)κ/50 dense(S)−1
(4.54)
What is most important here is the middle estimate in (4.53). Here,
p is as in Conjecture 1.14, and M > 0 is only a function of N in that
Conjecture.
The estimates that involve size(S)−2|F | are those that follow from
orthogonality considerations. The estimates in dense(S)−1 are those
that follow from density considerations which are less complicated.
However, in the second half of (4.54), the small positive power of size
is essential for us. All of these estimates are all variants of those in
[20].
The middle estimate of (4.53) is not of this type, and is the key
ingredient that permits us to obtain an estimate below L2. Note that
44 4. L
ESTIMATE FOR Hv
it gives the best bound for collections with moderate density and size.
For it we shall appeal to our assumed Conjecture 1.14.
Logarithms, such as those that arise in (4.54), arise from our trun-
cation arguments, associated with the parameters γs in (4.29).
For individual trees, we need two estimates.
Lemma 4.55. If T is a 1–tree with −
∆(T) ≥ σ, then we have
(4.56) |F ∩ σ−κRT| & σ1+κ|RT|.
Lemma 4.57. For trees T we have the estimate
(4.58)
|〈Sann1F , αs〉〈as−, 1G〉| . Ψ
dense(T) size(T)
|sh(T)|.
Here Ψ(x) = x|log cx|, and inside the logarithm, c is a small fixed
constant, to insure that c dense(T) · size(T) < 1
, say.
Sum(S) :=
|〈Sann1F , αs〉〈as−, 1G〉|
We want to provide the bound Sum(Av) . |F |1/p for p0 < p < 2. We
have the trivial bound
(4.59) Sum(S) . Ψ
dense(S) size(S)
count(S).
It is incumbent on us to provide a decomposition of Av into sub-
collections for which this last estimate is effective.
By inductive application of our principal organizational Lemma 4.50,
Av is the union of Sℓδ,σ, ℓ = 1, 2 for δ, σ ∈ 2 := {2n | n ∈ Z , n ≤ 0},
satisfying
dense(Sℓδ,σ) . δ,(4.60)
size(Sℓδ,σ) . σ,(4.61)
count(Sℓδ,σ) .
min(σ−2−κ|F |, δ−Mσ−p|F |+ σ1/κδ−1, δ−1) ℓ = 1,
min(σ−2(log 1/σ)3|F |, δ−1σκ/50) ℓ = 2
(4.62)
Using (4.59), we see that
Sum(S1δ,σ) . min(Ψ(δ)σ
−1−κ|F |, δ−M+1σ−p+1|F |+ σ1/κ+1, σ)
Sum(S2δ,σ) . min(Ψ(δ)σ
−1(log 1/σ)4|F |, σ1+κ/50)
(4.63)
One can check that for ℓ = 1, 2,
(4.64)
δ,σ∈2
Sum(Sℓδ,σ) . |F |1/p, p0 < p < 2.
PROOFS OF LEMMATA 45
This completes the proof of Lemma 4.33, aside from the proof of
Lemma 4.50.
Proof of (4.64). We can assume that |G| = 1, and that |F | ≤ 1,
for otherwise the result follows from the known Lp estimates, for p > 2
and measurable vector fields, see Theorem 1.15.
The case of ℓ = 2 in (4.64) is straightforward. Notice that in (4.63),
for ℓ = 2, the two terms in the minimum are roughly comparable,
ignoring logarithmic terms, for
δ|F | ≃ σ2+κ/50 .
Therefore, we set
T1 = {(δ, σ) ∈ 2× 2 | δ|F | ≤ σ2+κ/50 ≤ |F |} ,
T2 = {(δ, σ) ∈ 2× 2 | σ2+κ/50 ≤ δ|F |}
and T3 = 2× 2− T1 − T2.
We can estimate
(δ,σ)∈T1
Sum(S2δ,σ) .
(δ,σ)∈T1
Ψ(δ)σ−1(log 1/σ)4|F |
σ2+κ/50≤|F |
σ1+κ/75
. |F |1/p0 , p0 =
2 + κ/50
1 + κ/75
< 2 .
Notice that we have absorbed harmless logarithmic terms into a slightly
smaller exponent in σ above.
The second term is
(δ,σ)∈T2
Sum(S2δ,σ) .
(δ,σ)∈T2
σ1+κ/50
(δF )1/p1 , p1 =
2 + κ/50
1 + κ/50
< 2 ,
. |F |1/p1 .
46 4. L
ESTIMATE FOR Hv
The third term is
(δ,σ)∈T3
Sum(S2δ,σ) .
(δ,σ)∈T3
Ψ(δ)σ−1(log 1/σ)4|F |
σ2+κ/50≥|F |
σ−1|F |1−κ/75
. |F |1/p0 .
Here, we have again absorbed harmless logarithms into a slightly smaller
power of |F |, and p0 < 2 is as in the first term.
The novelty in this proof is the proof of (4.64) in the case of ℓ = 1.
We comment that if one uses the proof strategy just employed, that
is only relying upon the first and last estimates from the minimum in
(4.63), in the case of ℓ = 1, one will only show that |F |1/2.
In the definitions below, we will have a choice of 0 < τ < 1, where
τ = τ(M, p) ≃ M−1·(2−p) will only depend uponM and p in (4.63). (τ
enters into the definition of T4 and T5 below.) The choice of 0 < κ < τ
will be specified below.
T1 = {(δ, σ) ∈ 2× 2 | |F |
(2+κ)(1+κ) ≤ σ} ,
T2 = {(δ, σ) ∈ 2× 2 | σ < |F |
2−κ , δ ≥ σ1/κ} ,
T3 = {(δ, σ) ∈ 2× 2 | σ < |F |
2−κ , δ > σ1/κ} ,
T4 = {(δ, σ) ∈ 2× 2 | |F |
2−κ ≤ σ < |F |
(2+κ)(1+κ) , δ > στ} ,
T5 = {(δ, σ) ∈ 2× 2 | |F |
2−κ ≤ σ < |F |
(2+κ)(1+κ) , δ ≤ στ} ,
T (T ) =
(δ,σ)∈T
Sum(S1δ,σ) .
Note that for T1 we can use the first term in the minimum in (4.63).
T (T1) .
(δ,σ)∈T1
δσ−1−κ|F |
σ≥|F |
(2+κ)(1+κ)
σ−1−κ|F |
. |F |1−
2+κ .
This last exponent on |F | is strictly larger than 1
as desired. The point
of the definition of T1 is that when it comes time to use the middle term
PROOFS OF LEMMATA 47
of the minimum for ℓ = 1 in (4.63), we can restrict attention to the
δ−M+1σ−p+1|F | .
For the collection T2, use the last term in the minimum in (4.63).
T (T2) .
(δ,σ)∈T2
σ≤|F |
σ log 1/σ
. |F |
2−κ/2 .
Again, for 0 < κ < 1, the exponent on |F | above is strictly greater
than 1/2.
The term T3 can be controlled with the first term in the minimum
in (4.63).
T (T3) .
(δ,σ)∈T3
δσ−1−κ|F |
σ|F | . |F | .
The term T4 is the heart of the matter. It is here that we use the
middle term in the minimum of (4.63), and that the role of τ becomes
clear. We estimate
T (T4) .
(δ,σ)∈T4
δ−Mσ−p+1|F |
δ≥|F |τ
δ−M |F |1−
. |F |1−
−Mτ .
Recall that 1 < p < 2, so that 0 < p − 1 < 1. Therefore, for 0 < κ
sufficiently small, of the order of 2− p, we will have
1− p− 1
2− κ >
+ 2−p
Therefore, choosing τ ≃ (2 − p)/M will leave us with a power on |F |
that is strictly larger than 1
The previous term did not specify κ > 0. Instead it shows that
for 0 < κ < 1 sufficiently small, we can make a choice of τ , that is
48 4. L
ESTIMATE FOR Hv
independent of κ, for which T (T4) admits the required control. The
bound in the last term will specify a choice of κ on us. We estimate
T (T5) .
(δ,σ)∈T5
δσ−1−κ|F |
σ≥|F |
σ−1−κ|F |1+τ
. |F |1+τ+
Choosing κ = τ/6 will result in the estimate
which is as required, so our proof is finished. �
Remark 4.65. The resolution of Conjecture 1.21 would depend upon
refinements of Lemma 4.50, as well as using the restricted weak type
approach of [21].
Proof of Lemma 4.34. We only consider tiles s ∈ AT (ann, scl),
and sets ω ∈ Ω which are associated to one of these tiles. For an
element a = {as} ∈ ℓ2(AT (ann, scl)),
s :ωs=ω
asSannαs
For |ωs| = |ωs′|, note that dist(ωs,ωs′) is measured in units of scl/ann.
By a lemma of Cotlar and Stein, it suffices to provide the estimate
′‖2 . ρ−3, ρ = 1 +
dist(ω,ω′).
Now, the estimate ‖T
‖2 . 1 is obvious. For the case ω 6= ω′, by
Schur’s test, it suffices to see that
(4.66) sup
s′ :ωs′=ω
s :ωs=ω
|〈Sannαs, Sannαs′〉| . ρ−3.
For tiles s′ and s as above, recall that 〈ϕs, ϕs′〉 = 0, note that
|Rs′ ∩ Rs|
ann dist(ω,ω′)
≃ ρ−1,
and in particular, for a fixed s′, let Ss′ be those s for which ρRs∩ρRs′ 6=
∅. Clearly,
card(Ss′) .
|ρRs|
|2ρRs′ ∩ 2ρRs|
ρ ≃ ρ2
PROOFS OF LEMMATA 49
If for r > 1, rRs ∩ rRs′ = ∅, then it is routine to show that
|〈Sannαs, Sannαs′〉| . r−10
And so we may directly sum over those s 6∈ Ss′ ,
s 6∈Ss′
|〈Sannαs, Sannαs′〉| . ρ−3.
For those s ∈ Ss′, we estimate the inner product in frequency vari-
ables. Recalling the definition of αs = (Tc(Rs)D
ζ)ϕs, we have
α̂s = (Mod−c(Rs) D
ζ̂) ∗ ϕ̂s.
Recall that ζ is a smooth compactly supported Schwartz function. We
estimate the inner product
|〈Ŝannαs, Ŝannαs′〉|
without appealing to cancellation. Since we choose the function λ̂ to
be supported in an annulus 1
< |ξ| < 3
so that λ̂ann = λ̂(·/ann) is
supported in the annulus 1
ann < |ξ| < 3
ann. We can restrict our
attention to this same range of ξ. In the region |ξ| > ann/4, suppose,
without loss of generality, that ξ is closer to ωs than ωs′. Then since
ωs and ωs′ are separated by an amount & anndist(ω,ω
|α̂s(ξ)α̂s′(ξ)| . χ(2)ωs (ξ)χ
dist(ω,ω)
. χ(2)ωs (ξ)χ
(ξ)ρ−20.
Here, χ is the non–negative bump function in (4.40). Hence, we have
the estimate ∫
|λ̂ann(ξ)|2|α̂s(ξ)α̂s′(ξ)|dξ . ρ−10.
This is summed over the . ρ2 possible choices of s ∈ Ss′ , giving the
estimate ∑
s∈Ss′
|〈Sannαs, Sannαs′〉| . ρ−8 . ρ−3.
This is the proof of (4.66). And this concludes the proof of Lemma 4.34.
Proof of the Principal Organizational Lemma 4.50. Recall
that we are to decompose S into four distinct subsets satisfying the
favorable estimates of that Lemma. For the remainder of the proof set
dense(S) := δ and size(S) := σ. Take Slight to be all those s ∈ S for
which there is no tile s′ ∈ AT of density at least δ/2 for which s . s′.
It is clear that this set so constructed has density at most δ/2, that
this is a set of tiles, and that S1 := S− Slight is also .
50 4. L
ESTIMATE FOR Hv
The next Lemma and proof comment on the method we use to
obtain the middle estimate in (4.53) which depends upon the Lipschitz
Kakeya Maximal Function Conjecture 1.14. It will be used to obtain
the important inequality (4.82) below.
Lemma 4.67. Suppose we have a collection of trees T ∈ T , with these
conditions.
a: For T ∈ T there is a 1-tree T1 ⊂ T with
(4.68)
∆(T1) dx ≥ κσ ,
b: Each tree has top element s(T) := ωT×RT of density at least
c: The collections of tops {s(T) | T ∈ T } are pairwise incompa-
rable under the order relation ‘.’.
d: For all T ∈ T , γT = γωT×RT ≥ κ−1/2σ−κ/5N . Here, N is the
exponent on δ in Conjecture 1.14.
Then we have
|RT| . δ−Np−1σ−p(1+κ/4)|F |+ σ1/κδ−1.(4.69)
|RT| . δ−1.(4.70)
Concerning the role of γT, recall from the definition, (4.29), that
γs is a quantity that grows as does the ratio scl(s)/‖v‖Lip, hence there
are only . log σ−1 scales of tiles that do not satisfy the assumption d
above.
Proof. Our primary interest is in (4.69), which is a consequence of
our assumption about the Lipschitz Kakeya Maximal Functions, Con-
jecture 1.12.
s(T) := ωT × σ−κ/10NRT .
Let us begin by noting that
κ−1‖v‖Lip ≤ scl(s(T)) ≤ κann(s(T)), T ∈ T ,(4.71)
dense(s(T)) ≥ δσκ/10N , T ∈ T ,(4.72)
|F ∩Rs(T)| ≥ σ1+κ/4N |Rs(T)| .(4.73)
The conclusion (4.71) is straightforward, as is (4.72). The inequality
(4.73) follows from Lemma 4.55.
PROOFS OF LEMMATA 51
Note that the length of σ−κ/10NRT satisfies
σ−κ/10N L(RT) ≤ γT L(RT)
scl(s)
‖v‖Lip
≤ (100‖v‖Lip)−1 .
(4.74)
This is the condition (1.8) that we impose in the definition of the
Lipschitz Kakeya Maximal Functions.
Observe that we can regard ann(s(T)) ≃ σκ/10ann as a constant
independent of T.
The point of these observations is that our assumption about the
Lipschitz Kakeya Maximal Function applies to the maximal function
formed over the set of tiles {s(T) | T ∈ T }. And it will be applied
below.
Let Tk be the collection of trees so that T ∈ Tk if k ≥ 0 is the
smallest integer such that
(4.75) |(2kRT) ∩ v−1(ωT) ∩G| ≥ 220k/κ
2−1δ|RT| .
Then since the density of s(T) for every tree T ∈ T is at least δ, we
have T =
k=0 Tk. We can apply Conjecture 1.12 to these collections,
with the value of δ in that Conjecture being 220k/κ
2−1δ.
For each Tk, we decompose it by the following algorithm. Initialize
T selectedk ← ∅, T stockk ← Tk .
While T stockk 6= ∅, select T ∈ T stockk such that scl(s(T)) is minimal.
Define Tk(T) by
Tk(T) = {T′ ∈ Tk : (2kRT) ∩ (2kRT′) 6= ∅ and ωT ⊂ ωT′} .
Update
T selectedk ← T selectedk ∪ {T} , T stockk ← T stockk \Tk(T) .
Thus we decompose Tk into
T∈T selected
T′∈Tk(T)
{T′} .
And ∑
|RT| =
T∈T selected
T′∈Tk(T)
|RT′| .
Notice that RT′ ’s are disjoint for all T
′ ∈ Tk(T) and they are contained
in 5(2kRT). This is so, since the tops of the trees are assumed to be
incomparable with respect to the order relation ‘.’ on tiles.
52 4. L
ESTIMATE FOR Hv
Thus we have
|RT| .
T∈T selected
22k|RT|
. δ−12−10k/κ
T∈T selected
|(2kRT) ∩ v−1(ωT) ∩G| .
Observe that (2kRT)∩v−1(ωT)’s are disjoint for all T ∈ T selectedk . This
and the fact that |G| ≤ 1 proves (4.70). To argue for (4.69), we see
|RT| . δ−12−10k/κ
T∈T selected
(2kRT) ∩ v−1(ωT) ∩G
. δ−12−10k/κ
(2kRT) ∩G
∣∣∣∣ .
At this point, Conjecture 1.12 enters. Observe that we can estimate
∣∣∣ . |{Mδ′,v,(σκ/10ann)−1 1F > σ1+κ/4N}|
. (δ′)−Npσ−p(1+κ/4N)|F |.
. (δ)−Np2−kσ−p(1+κ/4N)|F |.
(4.76)
Here, δ′ = 220k/κ
2−1δ, the choice of δ′ permitted to us by (4.75), and
we have used (4.73) in the first line, to pass to the Lipschitz Kakeya
Maximal Function.
Hence,
|sh(T)| .
. δ−1
2−10k/κ
(2kRT) ∩G
. δ−1
k : 1≤2k≤σ−κ/10
(2kRT)
+ δ−1
k : 2k>σ−κ/10
2−10k/κ
2 |G| .
On the first sum in the last line, we use (4.76), and on the second, we
just sum the geometric series, and recall that |G| = 1.
PROOFS OF LEMMATA 53
We can now begin the principal line of reasoning for the proof of
Lemma 4.50.
The Construction of S1large. We use an orthogonality, or TT
∗ argu-
ment that has been used many times before, especially in [20] and [15].
(There is a feature of the current application of the argument that is
present due to the fact that we are working on the plane, and it is
detailed by Lacey and Li [15].)
We may assume that all intervals ωs are contained in the upper
half of the unit circle in the plane. Fix S ⊂ Av, and σ = size(S).
We construct a collection of trees T 1large for the collection S1, and
a corresponding collection of 1–trees T 1,1large, with particular properties.
We begin the recursion by initializing
T 1large ← ∅, T
large ← ∅,
S1large ← ∅, Sstock ← S1.
In the recursive step, if size(Sstock) < 1
σ1+κ/100, then this recursion
stops. Otherwise, we select a tree T ⊂ Sstock such that three conditions
are met.
a: The top of the tree s(T) (which need not be in the tree)
satisfies dense(s(T)) ≥ δ/4.
b: T contains a 1–tree T1 with
(4.77) −
∆(T1) dx ≥ 1
σ1+κ/100 .
c: And that ωT is in the first place minimal and and in the
second most clockwise among all possible choices of T. (Since
all ωs are in the upper half of the unit circle, this condition
can be fulfilled.)
We take T to be the maximal tree in Sstock which satisfies these condi-
tions.
We then update
T 1large ← {T} ∪ Tlarge, T
large ← {T1} ∪ T
large,
S1large ← T ∪ S1large Sstock ← Sstock −T.
The recursion then repeats. Once the recursion stops, we update
S1 ← Sstock
It is this collection that we analyze in the next subsection.
Note that it is a consequence of the recursion, and Remark 4.44,
that the tops of the trees {s(T) | T ∈ T 1large} are pairwise incomparable
under ..
54 4. L
ESTIMATE FOR Hv
The bottom estimate of (4.53) is then immediate from the construc-
tion and (4.70).
First, we turn to the deduction of the first estimate of (4.53). Let
T 1,(1)large be the set
T 1,(1)large =
T ∈ T 1large :
|〈Sann 1F , βs〉|2 < 116σ
2+κ/50|RT|
And let T 1,(2)large be the set
T 1,(2)large =
T ∈ T 1large :
|〈Sann 1F , βs〉|2 ≥ 116σ
2+κ/50|RT|
In the inner products, we are taking βs, which is supported off of γsRs.
Since T ∈ T 1large satisfies
(4.78) −
∆(T) dx ≥ 1
σ1+κ/100 ,
we have ∑
|〈Sann 1F , αs〉|2 ≥ 14σ
2+κ/50|RT| .
Thus, if T ∈ T 1,(1)large , we have
|〈1F , ϕs〉|2 ≥ 18σ
2+κ/50|RT| .
The replacement of αs by ϕs in the inequality above is an important
point for us. That we can then drop the Sann is immediate.
With this construction and observation, we claim that
(4.79)
T∈T 1,(1)
large
|RT| . (log 1/σ)2σ−2−κ/50|F |.
Proof of (4.79). This is a variant of the the argument for the
‘Size Lemma’ in [15], and so we will not present all details. Begin by
making a further decomposition of the trees T ∈ T 1,(1)large . To each such
tree, we have a 1-tree T1 ⊂ T which satisfies (4.77). We decompose
T1. Set
T1(0) =
s ∈ T1 | |〈f, ϕs〉|√
< σ1+κ/100
T1(j) =
s ∈ T1 | 4j−1σ1+κ/100 ≤ |〈f, ϕs〉|√
< 4jσ1+κ/100
1 ≤ j ≤ j0 = C log 1/σ .
PROOFS OF LEMMATA 55
Now, set T (j) to be those T ∈ T 1,(1)large for which
(4.80)
s∈T1(j)
|〈f, ϕs〉|2 ≥ (2j0)−1σ2+κ/50|RT| , 0 ≤ j ≤ j0 .
It is the case that each T ∈ T 1,(1)large is in some T (j), for 0 ≤ j ≤ j0.
The central case is that of j = 0. We can apply the ‘Size Lemma’
of [15] to deduce that
T∈T (0)
|RT| ≤ (2j0)σ−2−κ/50
T∈T (0)
s∈T1(0)
|〈f, ϕs〉|2
. (log 1/σ)σ−2−κ/50|F | .
The point here is that to apply the argument in the ‘Size Lemma’ one
needs an average case estimate, namely (4.80), as well as a uniform
control, namely the condition defining T1(0). This proves (4.79) in
this case.
For 1 ≤ j ≤ j0, we can apply the ‘Size Lemma’ argument to the
individual tiles in the collection
{T1(j) | T ∈ T (j)} .
The individual tiles satisfy the definition of a 1-tree. And the defining
condition of T1(j) is both the average case estimate, and the uniform
control needed to run that argument. In this case we conclude that
T∈T (j)
s∈T1(j)
|〈f, ϕs〉|2 . |F | .
Thus, we can estimate
T∈T (j)
|RT| . (log 1/σ)σ−1−κ/50|F | .
This summed over 1 ≤ j ≤ j0 = C log 1/σ proves (4.79). �
For T 1,(2)large , we have
T∈T 1,(2)
large
|RT| . σ−2−κ/50
scl≥κ−1‖v‖Lip
s:scl(s)=scl
|〈Sann 1F , βs〉|2
. σ−2−κ/50|F |
scl≥κ−1‖v‖Lip
‖v‖Lip
. σ−2−κ/50|F | ,
56 4. L
ESTIMATE FOR Hv
since βs has fast decay. The Bessel inequality in the last display can
be obtained by using the same argument in the proof of Lemma 4.34.
Hence we get
(4.81)
T∈T 1,(2)
large
|RT| . σ−2−κ/50|F |.
Combining (4.79) and (4.81), we obtain the first estimate of (4.53).
Second, we turn to the deduction of the middle estimate of (4.53),
which relies upon the Lipschitz Kakeya Maximal Function. Let T 1,goodlarge
be the set
T ∈ T 1large : γT ≥ κ−1/2σ−κ/5N
And let T 1,badlarge be the set
T ∈ T 1large : γT < κ−1/2σ−κ/5N
The ‘good’ collection can be controlled by facts which we have already
marshaled together. In particular, we have been careful to arrange the
construction so that Lemma 4.67 applies. By the main conclusion of
that Lemma, (4.69), we have
(4.82)
T∈T 1,good
large
|RT| . δ−Mσ−1−3κ/4|F |+ σ1/κδ−1 .
Here, M is a large constant that only depends upon N in Conjec-
ture 1.14.
For T ∈ T 1,badlarge , there are at most K = O(log(σ−κ)) many possible
scales for scl(ωT × RT). Let scl(T) = scl(ωT × RT). Thus we have
T∈T 1,bad
large
|RT| .
T:scl(T)=2mκ−1‖v‖Lip
|RT| .
Since T satisfies (4.78), we have
|F ∩ γTRT| & σ1+κ/2|RT| .
Thus, we get
T∈T 1,bad
large
|RT| . σ−1−κ/2
T:scl(T)=2mκ−1‖v‖Lip
1σ−κRT(x)dx .
PROOFS OF LEMMATA 57
For the tiles with a fixed scale, we have the following inequality, which
is a consequence of Lemma 4.25.
T:scl(T)=2mκ−1‖v‖Lip
1σ−κRT
. σ−κ/5δ−1 .
Hence we obtain
(4.83)
T∈T 1,bad
large
|RT| . δ−1σ−1−3κ/4|F | .
Combining (4.82) and (4.83), we obtain the middle estimate of
(4.53). Therefore, we complete the proof of (4.53).
The Construction of S2large. It is important to keep in mind that we
have only removed trees of nearly maximal size, with tops of a given
density. In the collection of tiles that remain, there can be trees of
large size, but they cannot have a top with nearly maximal density.
We repeat the TT∗ construction of the previous step in the proof, with
two significant changes.
We construct a collection of trees T 2large from the collection S1, and
a corresponding collection of 1–trees T 2,1large, with particular properties.
We begin the recursion by initializing
T 2large ← ∅ , T
large ← ∅ ,
S2large ← ∅ , Sstock ← S1 .
In the recursive step, if size(Sstock) < σ/2, then this recursion stops.
Otherwise, we select a tree T ⊂ Sstock such that two conditions are
a: T satisfies ‖∆(T)‖2 ≥ σ2 |RT|
b: ωT is both minimal and most clockwise among all possible
choices of T.
We take T to be the maximal tree in Sstock which satisfies these condi-
tions. We take T1 ⊂ T to be a 1–tree so that
(4.84) −
∆(T1) dx ≥ κσ .
This last inequality must hold by Lemma 4.49.
We then update
T 2large ← {T} ∪ Tlarge, T
large ← {T1} ∪ T
large,
Sstock ← Sstock −T.
The recursion then repeats.
58 4. L
ESTIMATE FOR Hv
Once the recursion stops, it is clear that the size of Sstock is at most
σ/2, and so we take Ssmall := Sstock.
The estimate ∑
T∈T 2
large
|RT| . σ−2|F |
then is a consequence of the TT ∗ method, as indicated in the previous
step of the proof. That is the first estimate claimed in (4.54).
What is significant is the second estimate of (4.54), which involves
the density. The point to observe is this. Consider any tile s of density
at least δ/2. Let Ts be those trees T ∈ T 2large with top ωs(T) ⊃ ωs and
Rs(T) ⊂ KRs. By the construction of S1large, we must have
∆(T1) dx ≤ σ1+κ/100 ,
for the maximal 1–tree T 1 contained in
T∈Ts T. But, in addition, the
tops of the trees in T 2large are pairwise incomparable with respect to the
order relation ‘.,’ hence we conclude that
|RT| . σ2+κ/50|Rs|.
Moreover, by the construction of Slight, for each T ∈ T 2large we must be
able to select some tile s with density at least δ/2 and ωs(T) ⊃ ωs and
Rs(T) ⊂ KRs.
Thus, we let S∗ be the maximal tiles of density at least δ/2. Then,
the inequality (4.70) applies to this collection. And, therefore,
T∈T 2
large
|RT| ≤ σκ/50
|Rs| . σκ/50δ−1.
This completes the proof of second estimate of (4.54). �
The Estimates For a Single Tree.
The Proof of Lemma 4.55. It is a routine matter to check that for
any 1–tree we have ∑
|〈f, ϕs〉|2 . ‖f‖22.
Indeed, there is a strengthening of this estimate relevant to our concerns
here. Recalling the notation (4.40), we have
(4.85)
|〈f, ϕs〉|2
]1/2∥∥∥
. ‖χ(∞)RT f‖p , 1 < p <∞ .
PROOFS OF LEMMATA 59
This is variant of the Littlewood-Paley inequalities, with some addi-
tional spatial localization in the estimate.
Using this inequality for p = 1 + κ/100 and the assumption of the
Lemma, we have
σ1+κ/100 ≤
∆T dx
]1+κ/100
1+κ/100
≤ |RT|−1
|〈f, ϕs〉|2
]1/2∥∥∥
1+κ/100
1+κ/100
. |RT|−1
dx.(4.86)
This inequality can only hold if |F ∩ σ−κRT| ≥ σ1+κ|RT|. �
The Proof of Lemma 4.57. This Lemma is closely related to the
Tree Lemma of [15]. Let us recall that result in a form that we need
it. We need analogs of the definitions of density and size that do not
incorporate truncations of the various functions involved. Define
dense(s) :=
G∩v−1(ωs)
(x) dx.
(Recall the notation from (4.40).)
dense(T) := sup
dense(s).
Likewise define
size(T) := sup
′ is a 1–tree
|RT′|−1
|〈1F , ϕs〉|2
Then, the proof of the Tree Lemma of [15] will give us this inequality:
For T a tree,
(4.87)
|〈Sann 1F , ϕs〉〈φs, 1G〉| . dense(T) size(T)|RT|.
Now, consider a tree T with dense(T) = δ, and size(T) = σ, where
we insist upon using the original definitions of density and size. If
in addition, γs ≥ K(σδ)−1 for all s ∈ T, we would then have the
inequalities
dense(T) . δ,
size(T) . σ,
60 4. L
ESTIMATE FOR Hv
This places (4.87) at our disposal, but this is not quite the estimate we
need, as the functions ϕs and φs that occur in (4.87) are not truncated
in the appropriate way, and it is this matter that we turn to next.
Recall that
ϕs = αs + βs ,
αs(x− yv(x))ψs(y) dy = αs−(x) + αs+(y) .
One should recall the displays (4.30), (4.31), and (4.38).
As an immediate consequence of the definition of βs, we have∫
|βs(x)| dx . γ−2s
|Rs|.
Hence, if we replace ϕs by βs, we have
|〈Sann 1F , βs〉〈φs, 1G〉| .
|Rs||〈φs, 1G〉|
γ−1s |Rs|
. σδ|RT|.
And by a very similar argument, one sees corresponding bounds, in
which we replace the φs by different functions. Namely, recalling the
definitions of as± in (4.31) and estimate (4.38), we have
|Rs||〈as+, 1G〉| . σ
(‖v‖Lip
scl(s)
(x) dx(4.88)
(‖v‖Lip
scl(s)
. σδ|RT| .
Similarly, we have
|Rs||〈φs − as+ − as−, 1G〉| . σδ|RT|,
Putting these estimates together proves our Lemma, in particular (4.58),
under the assumption that γs ≥ K(σδ)−1 for all s ∈ T.
Assume that T is a tree with scl(s) = scl(s′) for all s, s′ ∈ T. That
is, the scale of the tiles in the tree is fixed. Then, T is in particular a 1–
tree, so that by an application of the definitions and Cauchy–Schwartz,
|〈Sann 1F , αs〉〈as−, 1G〉| ≤ δ
|〈Sann 1F , αs〉|
≤ δσ|RT|.
PROOFS OF LEMMATA 61
But, γs ≥ 1 increases as does
scl(s). Thus, any tree T with γs ≤
K(σδ)−1 for all s ∈ T, is a union of O(|log δσ|) trees for which the last
estimate holds. �
CHAPTER 5
Almost Orthogonality Between Annuli
Application of the Fourier Localization Lemma
We are to prove Lemma 4.23, and in doing so rely upon a technical
lemma on Fourier localization, Lemma 5.56 below. We can take a
choice of 1 < α < 9
, and assume, after a dilation, that ‖v‖Cα = 1.
The first inequality we establish is this.
Lemma 5.1. Using the notation of of Lemma 4.23, and assuming that
‖v‖Cα . 1, we have the estimate ‖C‖2 . 1, where
ann≥1
where the Cann are defined in (4.21).
We have already established Lemma 4.22, and so in particular know
that ‖Cann‖2 . 1. Due to the imposition of the Fourier restriction in
the definition of these operators, it is immediate that CannC∗ann′ ≡ 0 for
ann 6= ann′. We establish that
Cann′‖2 . max(ann, ann′)−δ ,
δ = 1
(α− 1) , |log ann(ann′)−1| > 3 .
(5.2)
Then, it is entirely elementary to see that C is a bounded operator. Let
Pann be the Fourier projection of f onto the frequencies ann < |ξ| <
2ann. Observe,
‖Cf‖22 =
ann≥1
Cann Pann f
ann≥1
ann′>1
〈Cann Pann f, Cann′ Pann′ f〉
≤ 2‖f‖2
ann≥1
ann′>1
Cann′ Pann′ f‖2
. ‖f‖22
ann≥1
ann′>1
max(ann, ann′)−δ
. ‖f‖22.
64 5. ALMOST ORTHOGONALITY
There are only O(log ann) possible values of scl that contribute to
Cann, and likewise for Cann′. Thus, if we define
(5.3) Cann,sclf =
s∈AT (ann)
scl(s)=scl
〈f, ϕs〉φs ,
it suffices to prove
Lemma 5.4. Using the notation of of Lemma 4.23, and assuming that
‖v‖Cα . 1, we have
ann,sclCann′,scl′‖2 . (max(ann, ann′))−δ .
Here, we can take δ′ = 1
(α − 1), and the inequality holds for all
|log ann(ann′)−1| > 3, 1 < scl ≤ ann and 1 < scl′ ≤ ann′.
Proof of Lemma 4.23. In this proof, we assume that Lemma 5.1
and Lemma 5.4 are established. The first Lemma clearly establishes
the first (and more important) claim of the Lemma.
Let us prove the inequality (4.24). Using the notation of this sec-
tion, this inequality is as follows.
(5.5)
ann=−∞
Cann,scl
. (1 + log(1 + scl−1‖v‖Cα)).
This inequality holds for all choices of Cα vector fields v.
Note that Lemma 5.4 implies immediately
ann=3
Cann,scl
. 1 , ‖v‖Cα = 1 .
We are however in a scale invariant situation, so that this inequality
implies this equivalent form, independent of assumption on the norm
of the vector field.
(5.6)
ann≥8‖v‖Cα
Cann,scl
. 1 .
On the other hand, Lemma 4.25, implies that independent of any
assumption other than measurability, we have have the inequality
‖Cann,scl‖2 . 1 .
To prove (5.5), use the inequality (5.6), and this last inequality together
with the simple fact that for a fixed value of scl, there are at most
. 1 + log(1 + scl−1‖v‖Cα)) values of ann with scl ≤ ann ≤ 8‖v‖Cα.
APPLICATION OF THE FOURIER LOCALIZATION LEMMA 65
We use the notation
AT (ann, scl) := {s ∈ AT (ann) : scl(s) = scl} ,
Observe that as the scale is fixed, we have a Bessel inequality for the
functions {ϕs | s ∈ AT (ann, scl)}. Thus,
ann,sclCann′,scl′f‖22 =
s∈AT (ann,scl)
s∈AT (ann′,scl′)
〈φs, φs′〉〈ϕs′, f〉ϕs
s∈AT (ann,scl)
s∈AT (ann′,scl′)
〈φs, φs′〉〈ϕs′, f〉
At this point, the Schur test suggests itself, and indeed, we need a
quantitative version of the test, which we state here.
Proposition 5.7. Let A = {ai,j} be a matrix acting on ℓ2(N) by
ai,jxj
Then, we have the following bound on the operator norm of A.
‖A‖2 . sup
|ai,j| · sup
|ai,j|
We assume that 1 ≤ ann < 1
′. For a subset S ⊂ AT (ann, scl)×
AT (ann′, scl′) Consider the operator and definitions below.
AS f =
(s,s′)∈S
〈φs, φs′〉〈ϕs′, f〉ϕs ,
FL(s,S) =
s′∈AT (ann′,scl′)
|〈φs, φs′〉| ,
FL(S) = sup
FL(s,S) .
Here ‘FL’ is for ‘Fourier Localization’ as this term is to be controlled
by Lemma 5.56. We will use the notations FL(s′,S), and FL′(S),
which are defined similarly, with the roles of s and s′ reversed. By
Proposition 5.7, we have the inequality
(5.8) ‖AS‖22 . FL(S) · FL′(S) .
We shall see that typically FL(S) will be somewhat large, but is bal-
anced out by FL′(S).
We partition AT (ann, scl) × AT (ann′, scl′) into three disjoint sub-
collections Su, u = 1, 2, 3, defined as follows. In this display, (s, s′) ∈
66 5. ALMOST ORTHOGONALITY
AT (ann, scl)×AT (ann′, scl′).
(s, s′) | scl
≥ scl
,(5.9)
(s, s′) | scl
, scl < scl′
,(5.10)
(s, s′) | scl
, scl′ < scl
.(5.11)
A further modification to these collections must be made, but it
is not of an essential nature. For an integer j ≥ 1, and (s, s′) ∈ Su,
for u = 1, 2, 3, write (s, s′) ∈ Su,j if j is the smallest integer such that
2j+2Rs ∩ 2j+2Rs′ 6= ∅.
We apply the inequalities (5.8) to the collections Su,j, to prove the
inequalities
(5.12) ‖ASu,j‖2 . 2−j(ann′)−δ
where δ′ = 1
(α − 1). This proves Lemma 5.4, and so completes the
proof of Lemma 5.1.
In applying (5.8) it will be very easy to estimate FL(s,S), with
a term that decreases like say 2−10j. The difficult part is to estimate
either FL(s,S) or FL′(S) by a term with decreases faster than a small
power of (ann′)−1. for which we use Lemma 5.56.
Considering a term 〈φs, φs′〉, the inner product is trivially zero if
ωs ∩ ωs′ = ∅. We assume that this is not the case below. To apply
Lemma 5.56, fix e ∈ ω′s ∩ωs. Let α be a Schwartz function on R with
α̂ supported on [ann′, 2ann′], and identically one on 3
[ann′, 2ann′]. Set
β̂(θ) := α̂(θ − 3
′). We will convolve φs with β in the direction e,
and φs′ with α also in the direction e, thereby obtaining orthogonal
functions.
Define
Ie g(x) =
g(x− ye)β(y) dy,(5.13)
∆s = φs − Ie φs
∆s′ = φs′ − Ie φs′
(5.14)
By construction, we have
〈φs, φs′〉 = 〈Ie φs +∆s, Ie φs′ +∆s′〉
= 〈Ie φs,∆s′〉+ 〈∆s, Ie φs′〉+ 〈∆s,∆s′〉 .
APPLICATION OF THE FOURIER LOCALIZATION LEMMA 67
It falls to us to estimate terms like
s∈Sℓ,j
|〈∆s, Ie φs′〉|,(5.15)
s∈Sℓ,j
|〈Ie φs,∆s′〉|,(5.16)
s∈Sℓ,j
|〈∆s,∆s′〉|.(5.17)
as well as the dual expressions, with the roles of s and s′ reversed.
The differences ∆s and ∆s′ are frequently controlled by Lemma 5.56.
Concerning application of this Lemma to ∆s, observe that
Mod−c(ωs)∆s = Mod−c(ωs) φs −
[Mod−c(ωs) φs(x− ye)]β̃(y) dy
where β̃(y) = e(c(ωs)·e)y β(y). Now the Fourier transform of β is identi-
cally one in a neighborhood of the origin of width comparable to ann′,
where as |c(ωs) · e| is comparable to ann. Since we can assume that
′ > ann+3, say, the function β̃ meets the hypotheses of Lemma 5.56,
namely it is Schwarz function with Fourier transform identically one
in a neighborhood of the origin, and the width of that neighborhood
is comparable to ann′. And so ∆s is bounded by the bounded by
the three terms in (5.57)—(5.59) below. In these estimates, we take
2k ≃ ann′ > 1. By a similar argument, one sees that Lemma 5.56 also
applies to ∆s′.
We will let ∆s,m, for m = 1, 2, 3, denote the terms that come from
(5.57), (5.58), and (5.59) respectively. We use the corresponding no-
tation for ∆s,m, for m = 1, 2, 3. A nice feature of these estimates, is
that while ∆s and ∆s′ depend upon the choice of e ∈ ωs′ ∩ ωs, the
upper bounds in the first two estimates do not depend upon the choice
of e. While the third estimate does, the dependence of the set Fs on
the choice of e is rather weak.
In application of (5.58), the functions ∆s,2 will be very small, due
to the term (ann′)−10 which is on the right in (5.58). This term is so
much smaller than all other terms involved in this argument that these
terms are very easy to control. So we do not explicitly discuss the case
of ∆s,2, or ∆s′,2 below.
In the analysis of the terms (5.15) and (5.16), we frequently only
need to use an inequality such as |Ie φs′| . χ(2)Rs′ . When it comes to the
analysis of (5.17), the function ∆s′ obeys the same inequality, so that
68 5. ALMOST ORTHOGONALITY
these sums can be controlled by the same analysis that controls (5.15),
or (5.16). So we will explicitly discuss these cases below.
In order for 〈φs, φs′〉 6= ∅, we must necessarily have ωs ∩ ωs′ 6= ∅.
Thus, we update all Sℓ,j as follows.
Sℓ,j ← {(s, s′) ∈ Sℓ,j | ωs ∩ ωs′ 6= ∅} .
The Proof of (5.12) for S1,j, j ≥ 1. Recall the definition of S1,j
from (5.9). In particular, for (s, s′) ∈ S1,j , we must have ωs ⊂ ωs′.
We will use the inequality (5.8), and show that for 0 < ǫ < 1,
FL(S1,j) . 2−10j(ann′)−eα
′ · ann′
scl · ann(5.18)
FL′(S1,j) . 22j(ann′)ǫ ·
scl · ann
′ · ann′ .(5.19)
Notice that in the second estimate, we permit some slow increase in
the estimates as a function of 2j and ann′. But, due to the form of the
estimate of the Schur test in (5.8), this slow growth is acceptable.
The terms inside the square root in these two estimates cancel out.
These inequalities conclude the proof of the inequality (5.12) for the
collection S1,j , j ≥ 1.
We prove (5.18). For this, we use Lemma 5.56. That is, we should
bound the several terms
s′ : (s,s′)∈S1,j
|〈∆s, Ie φs′〉| ,(5.20)
s′ : (s,s′)∈S1,j
|〈Ie φs,∆s′〉| ,(5.21)
s′ : (s,s′)∈S1,j
|〈∆s,∆s′〉| .(5.22)
Here ∆s and ∆s′ are as in (5.14). And, Ie is defined as in (5.13). We
can regard the tile s as fixed, and so fix a choice of e ∈ ωs. In the next
two cases, we will need to estimate the same expressions as above. In
all three cases, Lemma 5.56 is applied with 2k ≃ ann′, and we can take
ǫ in this Lemma to be ǫ = 1
(α− 1). For ease of notation, we set
(5.23) α̃ = (α− 1)(1− ǫ)2 − ǫ > 0
As we have already mentioned, we do not explicitly discuss the
upper bound on the estimate for (5.22).
APPLICATION OF THE FOURIER LOCALIZATION LEMMA 69
The Upper Bound on (5.20). We write ∆s = ∆s,1 + ∆s,2 + ∆s,3,
where these three terms are those on the right in (5.57)—(5.59) respec-
tively. Note that
(5.24) |Ie φs′| . χ(2)Rs′ ,
since Ie is convolution in the long direction of Rs′ , at the scale of
(ann′)−1, which is much smaller than the length of Rs′ in the direc-
tion e. Therefore, we can estimate the term in (5.20) by
s′ : (s,s′)∈S1,j
|〈∆s,1, Ie φs′〉| . (ann′)−eα2−10j
s′ : (s,s′)∈S1,j
. (ann′)−eα2−10j
′ · ann′
scl · ann .(5.25)
This is as required to prove (5.18) for these sums.
For the terms associated with ∆s,3, we have
s′ : (s,s′)∈S1,j
|〈∆s,3, Ie φs′〉| .
s′ : (s,s′)∈S1,j
|Rs|−1/2 · χ(2)R′s dx
. 2−10j|Fs|
ann′ · scl′ · ann · scl
. 2−10j(ann′)−α+ǫ
′ · ann′
scl · ann .
That is, we only rely upon the estimate (5.60). This completes the
analysis of (5.20). (As we have commented above, we do not explicitly
discuss the case of ∆s,2.)
The Upper Bound for (5.21). Since ωs ⊂ ωs′ , the only facts about
∆s′ we need are
(ann′)ǫR′s
|∆s′| dx . (ann′)−eα+ǫ
′ · ann′ ,
|∆s′(x)| . (ann′)−eαχ(2)Rs′ (x) , x 6∈ (ann
′)ǫRs′ .
(5.26)
Indeed, this estimate is a straightforward consequence of the various
conclusions of Lemma 5.56. (We will return to this estimate in other
cases below.)
70 5. ALMOST ORTHOGONALITY
These inequalities, with |Ie φs| . χ(2)Rs , permit us to estimate
(5.21) . 2−20j |Rs|−1/2
s′ : (s,s′)∈S1,j
(ann′)ǫR′s
|∆s′| dx
. 2−20j(ann′)−eα
scl · ann
′ · ann′ × ♯{s
′ : (s, s′) ∈ S1,j}
. 2−20j(ann′)−eα
′ · ann′
scl · ann .
which is the required estimate. Here of course we use the estimate
♯{s′ : (s, s′) ∈ S1,j} . 22j
′ · ann′
scl · ann .
We now turn to the proof of (5.19), where it is important that we
justify the small term √
scl · ann
′ · ann′
on the right in (5.19). We estimate the terms dual to (5.20)—(5.22),
namely
s : (s,s′)∈S1,j
|〈∆s, Ies φs′〉| ,(5.27)
s : (s,s′)∈S1,j
|〈Ies φs,∆s′〉| ,(5.28)
s : (s,s′)∈S1,j
|〈∆s,∆s′〉| .(5.29)
Here, for each choice of tile s, we make a choice of es ∈ ωs ⊂ ωs′.
The Upper Bound on (5.27). We have an inequality analogous to
(5.24).
(5.30) |Ies φs′| . χ
Note that as we can view s′ as fixed, all the tiles {s : (s, s′) ∈ S1,j}
have the same approximate spatial location. Let us single out a tile s0
in this collection. Then, for all s, we have Rs ⊂ 2j+2Rs0 .
Recalling the specific information about the support of the functions
of ∆s from (5.57), (5.59) and (5.61), it follows that
s : (s,s′)∈S1,j
|∆s| . 22j(ann′)ǫχ(2)2j+2Rs0 .
APPLICATION OF THE FOURIER LOCALIZATION LEMMA 71
In particular, we do not claim any decay in ann′ in this estimate. (The
small growth of (ann′)ǫ above arises from the overlapping supports of
the functions ∆s, as detailed in Lemma 5.56.) Therefore, we can esti-
s : (s,s′)∈S1,j
|〈∆s, Ies φs′〉| . 22j(ann′)ǫ
2jRs0
. 2−10j(ann′)ǫ
scl · ann
′ · ann′ .
This is as required in (5.19).
Remark 5.31. It is the analysis of the term
s : (s,s′)∈S1,j
|〈∆s,3, Ies φs′〉|
which prevents us from obtaining a decay in ann′, at least in some
choices of the parameters scl , ann , scl′, and ann′.
The Upper Bound on (5.28). The fact about ∆s′ we need is the
simple inequality |∆s′| . χ(2)Rs′ .
As in the previous case, we turn to the fact that all the tile {s :
(s, s′) ∈ S1,j} have the same approximate spatial location. Single out
a tile s0 in this collection, so that Rs ⊂ 2j+2Rs0 for all such s.
Our claim is that
(5.32)
s : (s,s′)∈S1,j
|Ies φs| . 22jχ
22jRs
(We will have need of related inequalities below.) Suppose that s ∈
{s : (s, s′) ∈ S1,j}. These intervals all have the same length, namely
scl/ann. And x 6∈ supp(φs) implies v(x) 6∈ ωs, so that by the Lipschitz
assumption on the vector field
dist(x, supp(φs)) & dist(v(x),ωs) .
This means that
(5.33) |Ies φs(x)| . χ
1 + ann′ · dist(v(x),ωs)
Here, we recall that the operator Ie is dominated by the operator which
averages on spatial scale (ann′)−1 in the direction e. Moreover, we have
(5.34) ann′ · dist(ωs,ω) & scl .
Here, we partition the unit circle into disjoint intervals ω ∈ Ω of length
|ω| ≃ scl/ann, so that for all s ∈ {s : (s, s′) ∈ S1,j}, we have ωs ∈ Ω.
72 5. ALMOST ORTHOGONALITY
Figure 5.1. The relative positions of Rs and Rs′ in for
pairs (s, s′) ∈ Sℓ, for ℓ = 2 and ℓ = 3 respectively.
In fact, the term on the left in (5.34) can be taken to be integer
multiples of scl. Combining these observations proves (5.32). Indeed,
we can estimate the term in (5.32) as follows. For x, fix ω ∈ Ω with
v(x) ∈ ω. Then,
s : (s,s′)∈S1,j
|Ies φs| .
s : (s,s′)∈S1,j
1 + ann′ · dist(ω,ωs)
The important point is that the term involving the distance allows us
to sum over the possible values of ωs ⊂ ωs′ to conclude (5.32).
To finish this case, we can estimate
s : (s,s′)∈S1,j
|〈Ies φs,∆s′〉| . 2−10j
scl · ann
′ · ann′ .
This completes the upper bound on (5.28).
The Proof (5.12) for S2,j, j ≥ 1. In this case, note that the
assumptions imply that we can assume that ωs′ ⊂ ωs, and that di-
mensions of the rectangle Rs′ are smaller than those for Rs in both
directions. See Figure 5.
We should show these two inequalities, in analogy to (5.18) and
(5.19).
FL(S2,j) . 2−10j(ann′)−eα
′ · ann′
scl · ann(5.35)
FL′(S2,j) . 2−10j(ann′)−eα ·
scl · ann
′ · ann′ .(5.36)
Here, α̃ is as in (5.23).
APPLICATION OF THE FOURIER LOCALIZATION LEMMA 73
For the proof of (5.35), we should analyze the sums
s′ : (s,s′)∈S2,j
|〈∆s, Ies′ φs′〉| ,(5.37)
s′ : (s,s′)∈S2,j
|〈Ies′ φs,∆s′〉| ,(5.38)
s′ : (s,s′)∈S2,j
|〈∆s,∆s′〉| .(5.39)
These inequalities are in analogy to (5.20)—(5.22), and es′ ∈ ωs′ ⊂ ωs.
The Upper Bound on (5.37). Fix the tile s. Fix a translate Rs of
Rs with 2
jRs ∩ 2jRs = ∅, but 2j+1Rs ∩ 2j+2Rs 6= ∅. Let us consider
(5.40) S2,j = {(s, s′) ∈ S2,j | Rs′ ⊂ Rs}
and we restrict the the sum in (5.37) to this collection of tiles. Note
that with . 22j choices of Rs, we can exhaust the collection S2,j . So
we will prove a slightly stronger estimate in the parameter 2j for the
restricted collection S2,j.
The point of this restriction is that we can appeal to an inequality
similar to (5.32). Namely,
(5.41)
s′ : (s,s′)∈S2,j
|Ie′s φs′| .
′ · ann′
scl · ann χ
Note that the term in the square root takes care of the differing L2
normalizations of φs′ and χ
. Indeed, the proof of (5.32) is easily
modified to give this inequality.
Next, we observe that the analog of (5.26) holds for ∆s. Just replace
s′ in (5.26) with s. It is a consequence that we have
s′ : (s,s′)∈S2,j
|〈∆s, Ies′ φs′〉| . 2
−12j(ann′)−eα
′ · ann′
scl · ann .
This is enough to finish this case.
The Upper Bound on (5.38). Let us again appeal to the notations
Rs and S2,j as in (5.40).
We have the estimates
|Ies′ φs| . χ
74 5. ALMOST ORTHOGONALITY
As for the sum over ∆s′ , we have an analog of the estimates (5.26).
Namely,
s′ : (s,s′)∈S2,j
(ann′)ǫRs
|∆s′,1| dx . (ann′)−eα+ǫ
′ · ann′
scl · ann
s′ : (s,s′)∈S2,j
|∆s′,1| .
′ · ann′
scl · ann χ
, x 6∈ (ann′)ǫRs .
Note that we again have to be careful to accommodate the different
normalizations here. The proof of (5.26) can be modified to prove this
estimate.
Putting these two estimates together clearly proves that
s′ : (s,s′)∈S2,j
|〈Ies′ φs,∆s′〉| . 2
−10j(ann′)−eα
′ · ann′
scl · ann ,
as is required.
We now turn to the proof of the inequality (5.36), which will follow
from appropriate upper bounds on the sums below.
s : (s,s′)∈S2,j
|〈∆s, Ie φs′〉| ,(5.42)
s : (s,s′)∈S2,j
|〈Ie φs,∆s′〉| ,(5.43)
s : (s,s′)∈S2,j
|〈∆s,∆s′〉| .(5.44)
Here, we can regard s′ as a fixed tile, and e ∈ ωs′ ⊂ ωs. In this case,
observe that we have the inequality
(5.45) ♯{s : (s, s′) ∈ S1,j} . 22j .
This is so since Rs has larger dimensions in both directions than does
The Upper Bound on (5.42). We use the decomposition of ∆s =
∆s,1 +∆s,2 +∆s,3. In the first case, we can estimate
s : (s,s′)∈S2,j
|〈∆s,1, Ie φs′〉| . 22j sup|〈∆s,1, Ie φs′〉|
. 2−10j(ann′)−eα
scl · ann
′ · ann′ .
APPLICATION OF THE FOURIER LOCALIZATION LEMMA 75
For the last case, of ∆s,3, we estimate
s : (s,s′)∈S2,j
|〈∆s,3, Ie φs′〉| . 22j sup|〈∆s,3, Ie φs′〉|
. 22j min
|Fs| ·
scl · ann · scl′ · ann′ ,
2−30j
scl · ann
′ · ann′
Examining the two terms of the minimum, note that by (5.61),
|Fs| ·
scl · ann · scl′ · ann′ . (ann′)−α+ǫ
′ · ann′
scl · ann
. (ann′)−α+1+ǫ
′ · ann′ · scl · ann
. (ann′)−α+1+ǫ
scl · ann
′ · ann′ .
Here it is essential that we have the estimate (5.60) as stated, with
|Fs| . (ann′)−α+ǫ|Rs|. This is an estimate of the desired form, but
without any decay in the parameter j. The second term in the min-
imum does have the decay in j, but does not have the decay in ann′.
Taking the geometric mean of these two terms finishes the proof, pro-
vided (α−ǫ)/2 > α̃, which we can assume by taking α sufficiently close
to one.
The Upper Bound on (5.43). Using the inequality |Ie φs| . χ(2)Rs ,
and the inequalities (5.26) and (5.45), it is easy to see that
s : (s,s′)∈S2,j
|〈Ie φs,∆s′〉| . 2−12j(ann′)−eα
scl · ann
′ · ann′ .
This is the required estimate.
The Proof of (5.12) for S3,j, j ≥ 1. In this case, we have that
the length of the rectangles Rs′ are greater than those of the rectangles
Rs, as depicted in Figure 5. We show that
FL(S3,j) . (ann′)ǫ
′ · ann′
scl · ann(5.46)
FL′(S3,j) . 2−10j(ann′)−eα
scl · ann
′ · ann′ .(5.47)
In particular, we do not claim any decay in the term FL(S3,j), in fact
permitting a small increase in the parameter ann′. Recall that 0 < ǫ < 1
is a small quantity. See (5.23). But due to the form of the estimate in
76 5. ALMOST ORTHOGONALITY
Proposition 5.7, with the decay in 2j and ann′ in the estimate (5.47),
these two estimates still prove (5.12) for S3,j .
For the proof of (5.46), we analyze the sums
s′ : (s,s′)∈S3,j
|〈∆s, Ies′ φs′〉| ,(5.48)
s′ : (s,s′)∈S3,j
|〈Ies′ φs,∆s′〉| ,(5.49)
s′ : (s,s′)∈S3,j
|〈∆s,∆s′〉| .(5.50)
Here, es′ ∈ ωs′ ⊂ ωs.
The Upper Bound on (5.48). Regard s as fixed. We employ a vari-
ant of the notation established in (5.40). Let R̃s be a rectangle with
in the same coordinates axes as Rs. In the direction es, let it have
length 1/scl′, that is the (longer) length of the rectangles Rs′, and let
it have the same width of Rs. Further assume that 2
jRs ∩ R̃s = ∅ but
2j+4Rs ∩ R̃s 6= ∅. (There is an obvious change in these requirements
for j = 1.) Then, define
S̃3,j = {(s, s′) ∈ S3,j | Rs′ ⊂ R̃s} .
With . 22j choices of R̃s, we can exhaust the collection S3,j . Thus, we
prove a slightly stronger estimate in the parameter 2j for the collection
S̃3,j.
The main point here is that we have an analog of the estimate
(5.32):
s′ : (s,s′)∈ eS3,j
|Ie φs′| .
The term in the square root takes into account the differing L2 normal-
izations between the φs′ and χ
. The proof of (5.32) can be modified
to prove the estimate above.
We also have the analogs of the estimate (5.26). Putting these two
together proves that
s′ : (s,s′)∈S3,j
|〈∆s, Ies′ φs′〉| . 2
−10j(ann′)−eα
. 2−10j(ann′)−eα
′ · ann′
scl · ann .
APPLICATION OF THE FOURIER LOCALIZATION LEMMA 77
That is, we get the estimate we want with decay in ann′, we do not
claim in general.
The Upper Bound on (5.49). We use the inequality
|Ies′ φs| . χ
And we use the decomposition ∆s′ = ∆s′,1 +∆s′,2 +∆s′,3.
For the case of ∆s′,1, we have ωs′ ⊂ ωs. And the supports of the
functions ∆s′ are well localized with respect to the vector field. See
(5.57). Thus, in particular we have
s′ : (s,s′)∈S3,j
|∆s′| . (ann′)ǫ
′ · ann′ .
Hence, we have
s′ : (s,s′)∈S3,j
|〈χ(2)Rs ,∆s′〉| . (ann
′ · ann′
scl · ann
which is the desired estimate.
Remark 5.51. It is the analysis of the sum
s′ : (s,s′)∈S3,j
|〈Ie φs,∆s′,3〉|
that prevents us from obtaining decay in the parameter ann′ for cer-
tain choices of parameters scl , ann , scl′ and ann′. This is why we have
formulated (5.46) the way we have.
For the proof of (5.47), we analyze the sums
s : (s,s′)∈S3,j
|〈∆s, Ie φs′〉| ,(5.52)
s : (s,s′)∈S3,j
|〈Ie φs,∆s′〉| ,(5.53)
s : (s,s′)∈S3,j
|〈∆s,∆s′〉| .(5.54)
Here es′ ∈ ωs′ ⊂ ωs, and one can regard the interval ωs′ as fixed. It is
essential that we obtain the decay in 2j and ann′ in these cases.
Indeed, these cases are easier, as the sum is over s. For fixed s′,
there is a unique choice of interval ωs ⊃ ωs′. And the rectangles Rs
are shorter than Rs′, but wider. Hence,
(5.55) ♯{s : (s, s′) ∈ S3,j} . 22j
78 5. ALMOST ORTHOGONALITY
The Upper Bound on (5.52). We use the decomposition ∆s = ∆s,1+
∆s,2 +∆s,3, and the inequality |Ies′ φs′, | . χ
For the sum associated with ∆s,1, we have
s : (s,s′)∈S3,j
|〈∆s,1, Ies′ φs′〉| . (ann
′)−eα
s : (s,s′)∈S3,j
〈χ(2)Rs , χ
. 2−12j(ann′)−eα
′ · ann
scl · ann′
× ♯{s : (s, s′) ∈ S3,j}
. 2−10j(ann′)−eα
′ · ann′
scl · ann .
This is the required estimate.
For the sum associated with ∆s,3, the critical properties are those
of the corresponding sets Fs, described in (5.60) and (5.61). Note that
the sets ∑
s : (s,s′)∈S3,j
1Fs . (ann
′)2ǫ .
On the other hand,
s : (s,s′)∈S3,j
|Fs| . 22j
′ sup
s : (s,s′)∈S3,j
. 22j(ann′)−α+ǫ
′ · ann .
Here, we have used the estimate (5.55).
This permits us to estimate
s : (s,s′)∈S3,j
|〈∆s,3, Ies′ φs′〉| . 2
−10j(ann′)−α+3ǫ
scl · ann′
′ · ann
Note that the parity between the ‘primes’ is broken in this estimate.
By inspection, one sees that this last term is at most
. 2−10j(ann′)−eα
′ · ann′
scl · ann .
Indeed, the claimed inequality amounts to
(ann′)−α+3ǫscl . (ann′)−eαscl′ .
We have to permit scl′ to be as small as 1, whereas scl can be as big
as ann. But α > 1, and ann < ann′, so the inequality above is trivially
true. This completes the analysis of (5.52).
THE FOURIER LOCALIZATION ESTIMATE 79
The Upper Bound on (5.53). We only need to use the inequality
|Ies′ φs| . χ
, and the inequalities (5.26). It follows that
s : (s,s′)∈S3,j
|〈Ies′ φs,∆s′〉| .
s : (s,s′)∈S3,j
〈χ(2)Rs , |∆s′|〉
. 2−10j(ann′)−eα
′ · ann′
scl · ann .
The Fourier Localization Estimate
The precise form of the inequalities quantifying the Fourier local-
ization effect follows.
Fourier Localization Lemma 5.56. Let 1 < α < 2, ǫ < (α− 1)/20,
and v be a vector field with ‖v‖Cα ≤ 1. Let s be a tile with
1 < scl(s) = scl ≤ ann(s) = ann < 1
fs = Mod−c(ωs) φs
Let ζ be a smooth function on R, with 1(−2,2) ≤ ζ̂ ≤ 1(−3,3) and set
ζ2k(y) = 2
kζ(y2k). We have this inequality valid for all unit vectors e
with |e− es| ≤ |ωs|.
∣∣∣fs(x)−
fs(x− ye)ζ2k(y) dy
.(scl2(α−1)k)−1+ǫχ
(x)1fωs(v(x))(5.57)
+ (2kscl)−10χ
(x)(5.58)
+ |Rs|−1/21Fs(x) ,(5.59)
where ω̃s is a sub arc of the unit circle, with ω̃s = λωs, and 1 < λ <
2ǫk. Moreover, the sets Fs ⊂ R2 satisfy
|Fs| . 2−(α−ǫ)k(1 + scl−1)α−1|Rs|,(5.60)
Fs ⊂ 2ǫkRs ∩ v−1(ω̃s) ∩
∂(v ·es⊥)
∣∣∣ > 2(1−ǫ)k
.(5.61)
The appearance of the set Fs is explained in part because the only
way for the function φs to oscillate quickly along the direction es is
that the vector field moves back and forth across the interval ωs very
quickly. This sort of behavior, as it turns out, is the only obstacle to
the frequency localization described in this Lemma.
Note that the degree of localization improves in k. In (5.57), it
is important that we have the localization in terms of the directions
80 5. ALMOST ORTHOGONALITY
of the vector field. The terms in (5.58) will be very small in all the
instances that we apply this lemma. The third estimate (5.59) is the
most complicated, as it depends upon the exceptional set. The form of
the exceptional set in (5.61) is not so important, but the size estimate,
as a function of α > 1, in (5.60) is.
Proof. We collect some elementary estimates. Throughout this
argument, ~y := y e ∈ R2.
(5.62)
|y|>t2−k
|y2k||ζ2k(y)| dy . t−N , t > 1.
This estimate holds for all N > 1. Likewise,
(5.63)
|u|>tscl−1
|uscl||sclψ(scl u)| du . t−N , t > 1.
More significantly, we have for all x ∈ R2,
(5.64)
eiξ0y ϕ
(x− ~y)ζ2k(y) dy = ϕ
(x) − 2k < ξ0 < 2k,
where ϕ
= Tc(Rs)D
ϕ. This is seen by taking the Fourier transform.
Likewise, by (4.17), for vectors v0 of unit length,∫
e−2πiuλ0 ϕ
(x− uv0)sclψ(scl u) du 6= 0
implies that
(5.65) scl ≤ λ0 + ξ · v0 ≤ 98scl, for some ξ ∈ supp(ϕ̂
At this point, it is useful to recall that we have specified the fre-
quency support of ϕ to be in a small ball of radius κ in (4.16). This
has the implication that
(5.66) |ξ · es| ≤ κscl, |ξ · es⊥| ≤ κann ξ ∈ supp(ϕ̂(2)Rs )
We begin the main line of the argument, which comes in two stages.
In the first stage, we address the issue of the derivative below exceeding
a ‘large’ threshold.
e ·Dv(x) · es⊥ =
∂v · es⊥
We shall find that this happens on a relatively small set, the set Fs
of the Lemma. Notice that due to the eccentricity of the rectangle
Rs, we can only hope to have some control over the derivative in the
long direction of the rectangle, and e essentially points in the long
direction. We are interested in derivative in the direction es⊥ as that is
THE FOURIER LOCALIZATION ESTIMATE 81
the direction that v must move to cross the interval ωs2. A substantial
portion of the technicalities below are forced upon us due to the few
choices of scales 1 ≤ scl ≤ 2εk, for some small positive ε.1
Let 0 < ε1, ε2 < ǫ to be specified in the argument below. In partic-
ular, we take
0 < ε1 ≤ min
, κα−1
, 0 < ε2 <
(α− 1).
We have the estimate
|fs(x)|+
fs(x− ~y)ζ2k(y) dy
∣∣∣ . 2−10kχ(2)Rs (x), x 6∈ 2
ε1kRs.
This follows from (5.62) and the fact that the direction e differs from
es by an no more than the measure of the angle of uncertainty for Rs.
This is as claimed in (5.58). We need only consider x ∈ 2ε1kRs.
Let us define the sets Fs, as in (5.59). Define
λs :=
2ε1k scl
< 2−2ε1k
8 otherwise
Let λωs denote the interval on the unit circle with length λ|ωs|, and
the same center as ωs.
2 This is our ω̃ of the Lemma; the set Fs of the
Lemma is
(5.67) Fs := 2
ε1kRs ∩ v−1(λsωs) ∩
∂(v ·es⊥)
∣∣∣ > 2(1−ε2)k
And so to satisfy (5.61), we should take ε1 < 1/1200.
Let us argue that the measure of Fs satisfies (5.60). Fix a line ℓ in
the direction of e. We should see that the one dimensional measure
(5.68) |ℓ ∩ Fs| . 2−k(α−ǫ)(1 + scl−1)α−1scl−1.
For we can then integrate over the choices of ℓ to get the estimate in
(5.60).
The set ℓ ∩ Fs is viewed as a subset of R. It consists of open
intervals An = (an, bn), 1 ≤ n ≤ N . List them so that bn < an+1
for all n. Partition the integers {1, 2, . . . , N} into sets of consecutive
integers Iσ = [mσ, nσ]∩N so that for all points x between the left-hand
endpoint of Amσ and the right-hand endpoint of Anσ , the derivative
∂(v · es⊥)/∂e has the same sign. Take the intervals of integers Iσ to be
maximal with respect to this property.
1The scales of approximate length one are where the smooth character of the
vector field helps the least. The argument becomes especially easy in the case that√
ann ≤ scl, as in the case, |ωs| & scl−1.
2We have defined λs this way so that λsωs makes sense.
82 5. ALMOST ORTHOGONALITY
For x ∈ Fs, the partial derivative of v, in the direction that is
transverse to λsωs, is large with respect to the length of λsωs. Hence,
v must pass across λsωs in a small amount of time:
|Am| . 2−(1−ε1−ε2)k for all σ.
Now consider intervals Anσ and A1+nσ = Amσ+1. By definition,
there must be a change of sign of ∂v(x) · es⊥/∂e between these two
intervals. And so there is a change in this derivative that is at least as
big as 2(1−ε2)k scl
. The partial derivative is also Hölder continuous of
index α − 1, which implies that Anσ and Amσ+1 cannot be very close,
specifically
dist(Anσ , Amσ+1) ≥
2(1−ε2)k
As all of the intervals An lie in an interval of length 2
ε1kscl
−1, it follows
that there can be at most
1 ≤ σ . 2ε1kscl−1
2(1−ε2)k
)−α+1
intervals Iσ. Consequently,
|ℓ ∩ Fs| . 2−(1−2ε1−ε2+(1−ε2)(α−1))kscl−1
. 2−(α−2ε1−2ε2)kscl−α+1scl−1
We have already required 0 < ε1 <
and taking 0 < ε2 <
achieve the estimate (5.68). This completes the proof of (5.60).
The second stage of the proof begins, in which we make a detailed
estimate of the difference in question, seeking to take full advantage of
the Fourier properties (5.62)—(5.65), as well as the derivative informa-
tion encoded into the set Fs.
We consider the difference in (5.57) in the case of x ∈ 2ε1kRs −
v−1(λsωs). In particular, x is not in the support of fs, and due to the
smoothness of the vector field, the distance of x to the support of fs is
at least
& 2ε1k
so that by (5.62), we can estimate
∣∣∣fs(x)−
fs(x− ~y)ζ2k(y) dy
∣∣∣ . (2ε1kscl)−N |Rs|−1/2
which is the estimate (5.58).
THE FOURIER LOCALIZATION ESTIMATE 83
We turn to the proof of (5.57). For x ∈ 2ε1kRs ∩ v−1(λsωs), we
always have the bound
∣∣∣fs(x)−
fs(x− ~y)ζ2k(y) dy
∣∣∣ . 210ε1k/κχ(2)Rs (x)
101λsωs(x).
It is essential that we have |e − es| ≤ |ωs| for this to be true, and κ
enters in on the right hand side through the definition (4.40).
We establish the bound
∣∣∣fs(x)−
fs(x− ~y)ζ2k(y) dy
∣∣∣ . (scl2(α−1)k)−1|Rs|−1/2,
x ∈ 2ε1kRs ∩ v−1(λsωs) ∩ F cs .
(5.69)
We take the geometric mean of these two estimates, and specify that
0 < ε1 < κ
to conclude (5.57).
It remains to consider x ∈ 2ε1kRs ∩ v−1(λsωs) ∩ F cs , and now some
detailed calculations are needed. To ease the burden of notation, we
exp(x) := e−2πiuc(ωs)·v(x), Φ(x, x′) = ϕ
(x− uv(x′)),
with the dependency on u being suppressed, and define
w(du, d~y) := sclψ(scl u)ζ2k(y) du d~y.
In this notation, note that
fs = Mod−c(ωs) φs
ec(ωs)(x−uv(x)−c(ωs))x ϕ
(x− uv(x)) sclψ(sclu) du
exp(x)Φ(x, x, )sclψ(sclu) du
exp(x)Φ(x, x, )w(du, d~y) ,
since ζ has integral on R2. In addition, we have
fs(x− ~ye)ζ2k(~y) d~y =
ec(ωs)(x−uv(x−~y)−c(ωs))x
× ϕ(2)Rs (x− uv(x− ~y)) sclψ(sclu) du d~y
exp(x− ~y)Φ(x− ~y, x− ~y)w(du, d~y) .
84 5. ALMOST ORTHOGONALITY
We are to estimate the difference between these two expressions, which
is the difference of
Diff1(x) :=
exp(x)Φ(x, x)− exp(x− ~y)Φ(x− ~y, x)w(du, d~y)
Diff2(x) :=
exp(x− ~y){Φ(x− ~y, x− ~y)− Φ(x− ~y, x)}w(du, d~y)
The analysis of both terms is quite similar. We begin with the first
term.
Note that by (5.64), we have
Diff1(x) =
{exp(x)− exp(x− ~y)}Φ(x− ~y, x)w(du, d~y).
We make a first order approximation to the difference above. Observe
exp(x)− exp(x− ~y) = exp(x){1− exp(x− ~y)exp(x)}
= exp(x){1− e−2πiu[c(ωs)·Dv(x)·e]~y}(5.70)
+O(|u|ann|~y|α).
In the Big–Oh term, |u| is typically of the order scl−1, and |~y| is of the
order 2−k. Hence, direct integration leads to the estimate of this term
|u|ann|y|α|Φ(x− ~y, x)|·|w(du, d~y)|
.|Rs|−1/2
.|Rs|−1/2(scl2(α−1)k)−1.
This is (5.69).
The term left to estimate is
Diff ′1(x) :=
exp(x)(1−e−2πiu[c(ωs)·Dv(x)·e]~y)
Φ(x− ~y, x)w(du, d~y) .
Observe that by (5.64), the integral in y is zero if
|u[c(ωs) ·Dv(x) · e]| ≤ 2k.
Here we recall that c(ωs) =
ann es⊥. By the definition of Fs, the
partial derivative is small, namely
|es⊥ ·Dv(x) · e| . 2(1−ε1)k
THE FOURIER LOCALIZATION ESTIMATE 85
Hence, the integral in y in Diff ′1(x) can be non–zero only for
scl|u| & 2ε1k.
By (5.63), it follows that in this case we have the estimate
|Diff ′1(x)| . 2−2k|Rs|−1/2
This estimate holds for x ∈ 2ε1kRs∩v−1(λsωs)∩F cs and this completes
the proof of the upper bound (5.69) for the first difference.
We consider the second difference Diff2. The term v(x− ~y) occurs
twice in this term, in exp(x − ~y), and in Φ(x − ~y, x − ~y). We will use
the approximation (5.70), and similarly,
Φ(x− ~y, x−~y)− Φ(x− ~y, x)
(x− ~y − uv(x− ~y))− ϕ(2)Rs (x− ~y − uv(x))
(x− ~y − uv(x)− uDv(x)~y)
− ϕ(2)Rs (x− ~y − uv(x)) +O(ann |u||y|
= ∆Φ(x, ~y) +O(ann |u||~y|α)
The Big–Oh term gives us, upon integration in u and ~y, a term that is
no more than
. |Rs|−1/2
2−αk . |Rs|−1/2(scl2(α−1)k)−1.
This is as required by (5.69).
We are left with estimating
Diff ′2(x) :=
e−2πiuc(ωs)·(v(x)−Dv(x)·~y)∆Φ(x, ~y)w(du, d~y).
By (5.64), the integral in y is zero if both of these conditions hold.
|uc(ωs)Dv(x) · e| < 2k,
|[uc(ωs)Dv(x)− ξ − uξDv(x)] · e| < 2k, ξ ∈ supp(ϕ̂(2)Rs )
Both of these conditions are phrased in terms of the derivative which
is controlled as x 6∈ Fs. In fact, the first condition already occurred in
the first case, and it is satisfied if
scl|u| . 2ε1k.
Recalling the conditions (5.66), the second condition is also satisfied
for the same set of values for u. The application of (5.63) then yields a
very small bound after integrating |u| & 2ε1kscl−1. This completes the
proof our technical Lemma. �
REFERENCES 87
References
[1] Angeles Alfonseca, Fernando Soria, and Ana Vargas, A remark on maximal
operators along directions in R2, Math. Res. Lett. 10 (2003), no. 1, 41–
49.MR1960122 (2004j:42010) ↑8
[2] Angeles Alfonseca, Strong type inequalities and an almost-orthogonality princi-
ple for families of maximal operators along directions in R2, J. London Math.
Soc. (2) 67 (2003), no. 1, 208–218.MR1942421 (2003j:42015) ↑8
[3] J. Bourgain, A remark on the maximal function associated to an analytic vec-
tor field, Analysis at Urbana, Vol. I (Urbana, IL, 1986–1987), 1989, pp. 111–
132.MR 90h:42028 ↑4, 6, 23, 24
[4] Camil Muscalu, Terence Tao, and Christoph Thiele, Uniform estimates on
paraproducts, J. Anal. Math. 87 (2002), 369–384. Dedicated to the memory of
Thomas H. Wolff. MR 1945289 (2004a:42023) ↑42
[5] Anthony Carbery, Andreas Seeger, Stephen Wainger, and James Wright,
Classes of singular integral operators along variable lines, J. Geom. Anal. 9
(1999), no. 4, 583–605.MR 2001g:42026 ↑6, 28
[6] Lennart Carleson, On convergence and growth of partial sumas of Fourier se-
ries, Acta Math. 116 (1966), 135–157.MR 33 #7774 ↑vii, 2
[7] Michael Christ, Alexander Nagel, Elias M. Stein, and Stephen Wainger, Sin-
gular and maximal Radon transforms: analysis and geometry, Ann. of Math.
(2) 150 (1999), no. 2, 489–577.MR 2000j:42023 ↑6
[8] A. Córdoba and R. Fefferman, On differentiation of integrals, Proc. Nat. Acad.
Sci. U.S.A. 74 (1977), no. 6, 2211–2213.MR0476977 (57 #16522) ↑7, 11
[9] Charles Fefferman, Pointwise convergence of Fourier series, Ann. of Math. (2)
98 (1973), 551–571.MR 49 #5676 ↑40
[10] , The multiplier problem for the ball, Ann. of Math. (2) 94 (1971), 330–
336.MR 45 #5661 ↑vii
[11] Loukas Grafakos and Xiaochun Li, Uniform bounds for the bilinear Hilbert
transform, I, Ann. of Math. 159 (2004), 889–933. ↑31, 42
[12] Nets Hawk Katz, Maximal operators over arbitrary sets of directions, Duke
Math. J. 97 (1999), no. 1, 67–79.MR 2000a:42036 ↑8
[13] , A partial result on Lipschitz differentiation, Harmonic analysis at
Mount Holyoke (South Hadley, MA, 2001), 2003, pp. 217–224.1 979 942 ↑6
[14] Joonil Kim, Maximal Average Along Variable Lines (2006). ↑28
[15] Michael T. Lacey and Xiaochun Li, Maximal theorems for the directional
Hilbert transform on the plane, Trans. Amer. Math. Soc. 358 (2006), no. 9,
4099–4117 (electronic). MR 2219012 ↑2, 4, 10, 31, 35, 37, 40, 53, 54, 55, 59
[16] Michael T. Lacey and Christoph Thiele, Lp estimates on the bilinear Hilbert
transform for 2 < p < ∞, Ann. of Math. (2) 146 (1997), no. 3, 693–
724.MR1491450 (99b:42014) ↑2
[17] , On Calderón’s conjecture for the bilinear Hilbert transform, Proc.
Natl. Acad. Sci. USA 95 (1998), no. 9, 4828–4830 (electronic). MR 1619285
(99e:42013) ↑2
[18] Michael Lacey and Christoph Thiele, On Calderón’s conjecture, Ann. of Math.
(2) 149 (1999), no. 2, 475–496. MR 1689336 (2000d:42003) ↑2
[19] Michael T. Lacey and Christoph Thiele, Lp estimates for the bilinear Hilbert
transform, Proc. Nat. Acad. Sci. U.S.A. 94 (1997), no. 1, 33–35. MR 1425870
(98e:44001) ↑2
88 5. ALMOST ORTHOGONALITY
[20] Michael Lacey and Christoph Thiele, A proof of boundedness of the Carleson
operator, Math. Res. Lett. 7 (2000), no. 4, 361–370.MR 2001m:42009 ↑2, 10,
40, 43, 53
[21] Camil Muscalu, Terence Tao, and Christoph Thiele, Multi-linear operators
given by singular multipliers, J. Amer. Math. Soc. 15 (2002), no. 2, 469–496
(electronic).MR 2003b:42017 ↑5, 48
[22] Alexander Nagel, Elias M. Stein, and Stephen Wainger, Hilbert transforms
and maximal functions related to variable curves, Harmonic analysis in Eu-
clidean spaces (Proc. Sympos. Pure Math., Williams Coll., Williamstown,
Mass., 1978), Part 1, 1979, pp. 95–98.MR 81a:42027 ↑6, 23
[23] D. H. Phong and Elias M. Stein, Hilbert integrals, singular integrals, and Radon
transforms. II, Invent. Math. 86 (1986), no. 1, 75–113.MR 88i:42028b ↑6
[24] , Hilbert integrals, singular integrals, and Radon transforms. I, Acta
Math. 157 (1986), no. 1-2, 99–157.MR 88i:42028a ↑6
[25] Elias M. Stein, Problems in harmonic analysis related to curvature and oscil-
latory integrals, Proceedings of the International Congress of Mathematicians,
Vol. 1, 2 (Berkeley, Calif., 1986), 1987, pp. 196–221.MR 89d:42028 ↑2, 6
[26] E. M. Stein, Harmonic analysis: real-variable methods, orthogonality, and os-
cillatory integrals, Princeton Mathematical Series, vol. 43, Princeton Univer-
sity Press, Princeton, NJ, 1993. With the assistance of Timothy S. Murphy;
Monographs in Harmonic Analysis, III.MR 95c:42002 ↑vii
[27] Jan-Olov Strömberg,Maximal functions associated to rectangles with uniformly
distributed directions, Ann. Math. (2) 107 (1978), no. 2, 399–402.MR0481883
(58 #1978) ↑2, 8, 19
[28] , Weak estimates on maximal functions with rectangles in certain di-
rections, Ark. Mat. 15 (1977), no. 2, 229–240.MR0487260 (58 #6911) ↑2, 8
	Preface
	Chapter 1. Overview of Principal Results
	Chapter 2. Connections to Besicovitch Set and Carleson's Theorem
	Besicovitch Set
	The Kakeya Maximal Function
	Carleson's Theorem
	The Weak  L 2  Estimate in Theorem 1.15 is Sharp
	Chapter 3. The Lipschitz Kakeya Maximal Function
	The Weak  L 2  Estimate
	An Obstacle to an  Lp estimate, for  1<p<2
	Bourgain's Geometric Condition
	Vector Fields that are a Function of One Variable
	Chapter 4. The  L2 Estimate for Hilbert Transform on Lipschitz Vector Fields
	Definitions and Principle Lemmas
	Truncation and an Alternate Model Sum 
	Proofs of Lemmata 
	Chapter 5. Almost Orthogonality Between Annuli
	Application of the Fourier Localization Lemma
	The Fourier Localization Estimate
	References

  Let $ v$ be a smooth vector field on the plane, that is a map from the plane
to the unit circle. We study sufficient conditions for the boundedness of the
Hilbert transform
  \operatorname H_{v, \epsilon}f(x) := \text{p.v.}\int_{-\epsilon}^ \epsilon
f(x-yv(x)) \frac{dy}y where $ \epsilon $ is a suitably chosen parameter,
determined by the smoothness properties of the vector field. It is a
conjecture, due to E.\thinspace M.\thinspace Stein, that if $ v$ is Lipschitz,
there is a positive $ \epsilon $ for which the transform above is bounded on $
L ^{2}$. Our principal result gives a sufficient condition in terms of the
boundedness of a maximal function associated to $ v$. This sufficient condition
is that this new maximal function be bounded on some $ L ^{p}$, for some $
1<p<2$. We show that the maximal function is bounded from $ L ^{2}$ to weak $ L
^{2}$ for all Lipschitz maximal function. The relationship between our results
and other known sufficient conditions is explored.

On a Conjecture of E. M. Stein on
the Hilbert Transform on Vector
Fields
Michael Lacey and Xiaochun Li
Michael Lacey, School of Mathematics, Georgia Insti-
tute of Technology, Atlanta GA 30332
Xiaochun Li, Department of Mathematics, University
of Illinois, Urbana IL 61801
E-mail address : xcli@math.uiuc.edu
http://arxiv.org/abs/0704.0808v3
1991 Mathematics Subject Classification. Primary 42A50, 42B25
The authors are supported in part by NSF grants. M.L. was
supported in part by the Guggenheim Foundation.
Abstract. Let v be a smooth vector field on the plane, that is a
map from the plane to the unit circle. We study sufficient condi-
tions for the boundedness of the Hilbert transform
Hv,ǫ f(x) := p.v.
f(x− yv(x)) dy
where ǫ is a suitably chosen parameter, determined by the smooth-
ness properties of the vector field. It is a conjecture, due to
E.M. Stein, that if v is Lipschitz, there is a positive ǫ for which
the transform above is bounded on L2. Our principal result gives
a sufficient condition in terms of the boundedness of a maximal
function associated to v, namely that this new maximal function
be bounded on some Lp, for some 1 < p < 2. We show that
the maximal function is bounded from L2 to weak L2 for all Lips-
chitz vector fields. The relationship between our results and other
known sufficient conditions is explored.
Contents
Preface vii
Chapter 1. Overview of Principal Results 1
Chapter 2. Connections to Besicovitch Set and Carleson’s
Theorem 7
Besicovitch Set 7
The Kakeya Maximal Function 7
Carleson’s Theorem 8
The Weak L2 Estimate in Theorem 1.15 is Sharp 10
Chapter 3. The Lipschitz Kakeya Maximal Function 11
The Weak L2 Estimate 11
An Obstacle to an Lp estimate, for 1 < p < 2 22
Bourgain’s Geometric Condition 23
Vector Fields that are a Function of One Variable 27
Chapter 4. The L2 Estimate for Hilbert Transform on Lipschitz
Vector Fields 31
Definitions and Principle Lemmas 31
Truncation and an Alternate Model Sum 36
Proofs of Lemmata 39
Chapter 5. Almost Orthogonality Between Annuli 63
Application of the Fourier Localization Lemma 63
The Fourier Localization Estimate 79
References 87
Preface
This memoir is devoted to a question in planar Harmonic Analysis,
a subject which is a circle of problems all related to the Besicovitch set.
This anomalous set has zero Lebesgue measure, yet contains a line seg-
ment of unit length in each direction of the plane. It is a known, since
the 1970’s, that such sets must necessarily have full Hausdorff dimen-
sion. The existence of these sets, and the full Hausdorff dimension, are
intimately related to other, independently interesting issues [26]. An
important tool to study these questions is the so-called Kakeya Maxi-
mal Function, in which one computes the maximal average of a function
over rectangles of a fixed eccentricity and arbitrary orientation.
Most famously, Charles Fefferman showed [10] that the Besicovitch
set is the obstacle to the boundedness of the disc multiplier in the
plane. But as well, this set is intimately related to finer questions of
Bochner-Riesz summability of Fourier series in higher dimensions and
space-time regularity of solutions of the wave equation.
This memoir concerns one of the finer questions which center around
the Besicovitch set in the plane. (There are not so many of these
questions, but our purpose here is not to catalog them!) It concerns
a certain degenerate Radon transform. Given a vector field v on R2,
one considers a Hilbert transform computed in the one dimensional line
segment determined by v, namely the Hilbert transform of a function
on the plane computed on the line segment {x+ tv(x) | |t| ≤ 1}.
The Besicovitch set itself says that choice of v cannot be just mea-
surable, for you can choose the vector field to always point into the set.
Finer constructions show that one cannot take it to be Hölder continu-
ous of any index strictly less than one. Is the sharp condition of Hölder
continuity of index one enough? This is the question of E. M. Stein,
motivated by an earlier question of A. Zygmund, who asked the same
for the question of differentiation of integrals.
The answer is not known under any condition of just smoothness
of the vector field. Indeed, as is known, and we explain, a positive
answer would necessarily imply Carleson’s famous theorem on the con-
vergence of Fourier series, [6]. This memoir is concerned with reversing
viii PREFACE
this implication: Given the striking recent successes related to Car-
leson’s Theorem, what can one say about Stein’s Conjecture? In this
direction, we introduce a new object into the study, a Lipschitz Kakeya
Maximal Function, which is a variant of the more familiar Kakeya Max-
imal Function, which links the vector field v to the ‘Besicovitch sets’
associated to the vector field. One averages a function over rectan-
gles of arbitrary orientation and—in contrast to the classical setting—
arbitrary eccentricity. But, the rectangle must suitably localize the
directions in which the vector field points. This Maximal Function ad-
mits a favorable estimate on L2, and this is one of the main results of
the Memoir.
On Stein’s Conjecture, we prove a conditional result: If the Lips-
chitz Kakeya Maximal Function associated with v maps is an estimate
a little better than our L2 estimate, then the associated Hilbert trans-
form is indeed bounded. Thus, the main question left open concerns
the behavior of these novel Maximal Functions.
While the main result is conditional, it does contain many of the
prior results on the subject, and greatly narrows the possible avenues
of a resolution of this conjecture.
The principal results and conjectures are stated in the Chapter 1;
following that we collect some of the background material for this sub-
ject, and prove some of the folklore results known about the subject.
The remainder of the Memoir is taken up with the proofs of the The-
orems stated in the Chapter 1.
Acknowledgment. The efforts of a strikingly generous referee has
resulted in corrections of arguments, and improvements in presentation
throughout this manuscript. We are indebted to that person.
Michael T. Lacey and Xiaochun Li
CHAPTER 1
Overview of Principal Results
We are interested in singular integral operators on functions of two
variables, which act by performing a one dimensional transform along
a particular line in the plane. The choice of lines is to be variable.
Thus, for a measurable map, v from R2 to the unit circle in the plane,
that is a vector field, and a Schwartz function f on R2, define
Hv,ǫ f(x) := p.v.
f(x− yv(x)) dy
This is a truncated Hilbert transform performed on the line segment
{x+ tv(x) : |t| < 1}. We stress the limit of the truncation in the defi-
nition above as it is important to different scale invariant formulations
of our questions of interest. This is an example of a Radon transform,
one that is degenerate in the sense that we seek results independent of
geometric assumptions on the vector field. We are primarily interested
in assumptions of smoothness on the vector field.
Also relevant is the corresponding maximal function
(1.1) Mv,ǫ f := sup
0<t≤ǫ
(2t)−1
|f(x− sv(x))| ds
The principal conjectures here concern Lipschitz vector fields.
Zygmund Conjecture 1.2. Suppose that v is Lipschitz. Then, for
all f ∈ L2(R2) we have the pointwise convergence
(1.3) lim
(2t)−1
f(x− sv(x)) ds = f(x) a. e.
More particularly, there is an absolute constant K > 0 so that if ǫ−1 =
K‖v‖Lip, we have the weak type estimate
(1.4) sup
λ|{Mv,ǫ f > λ}|1/2 . ‖f‖2 .
The origins of this question go back to the discovery of the Besicov-
itch set in the 1920s, and in particular, constructions of this set show
that the Conjecture is false under the assumption that v is Hölder con-
tinuous for any index strictly less than 1. These constructions, known
2 1. OVERVIEW OF PRINCIPAL RESULTS
since the 1920’s, were the inspiration for A. Zygmund to ask if integrals
of, say, L2(R2) functions could be differentiated in a Lipschitz choice
of directions. That is, for Lipschitz v, and f ∈ L2, is it the case that
(2ǫ)−1
f(x− yv(x)) dy = f(x) a.e.(x)
These and other matters are reviewed in the next chapter.
Much later, E. M. Stein [25] raised the singular integral variant of
this conjecture.
E.M. Stein Conjecture 1.5. There is an absolute constant K > 0
so that if ǫ−1 = K‖v‖Lip, we have the weak type estimate
(1.6) sup
λ|{|Hv,ǫ f | > λ}|1/2 . ‖f‖2 .
These are very difficult conjectures. Indeed, it is known that if
the Stein Conjecture holds for, say, C2 vector fields, then Carleson’s
Theorem on the pointwise convergence of Fourier series [6] would follow.
This folklore result is recalled in the next Chapter.
We will study these questions using modifications of the phase plane
analysis associated with Carleson’s Theorem [15–20] and a new tool,
which we term a Lipschitz Kakeya Maximal Function.
Associated with the Besicovitch set is the Kakeya Maximal Func-
tion, a maximal function over all rectangles of a given eccentricity.
A key estimate is that the L2 −→ L2,∞ norm of this operator grows
logarithmically in the eccentricity, [27, 28].
Associated with a Lipschitz vector field, we define a class of maxi-
mal functions taken over rectangles of arbitrary eccentricity, but these
rectangles are approximate level sets of the vector field. Perhaps sur-
prisingly, these maximal functions admit an L2 bound that is indepen-
dent of eccentricity. Let us explain.
A rectangle is determined as follows. Fix a choice of unit vectors in
the plane (e, e⊥), with e⊥ being the vector e rotated by π/2. Using these
vectors as coordinate axes, a rectangle is a product of two intervals
R = I × J . We will insist that |I| ≥ |J |, and use the notations
(1.7) L(R) = |I|, W(R) = |J |
for the length and width respectively of R.
The interval of uncertainty of R is the subarc EX(R) of the unit
circle in the plane, centered at e, and of length W(R)/ L(R). See Fig-
ure 1.1.
1. OVERVIEW OF PRINCIPAL RESULTS 3
0 EX(R) R
Figure 1.1. An example eccentricity interval EX(R).
The circle on the left has radius one.
We now fix a Lipschitz map v of the plane into the unit circle. We
only consider rectangles R with
(1.8) L(R) ≤ (100‖v‖Lip)−1 .
For such a rectangle R, set V(R) = R ∩ v−1(EX(R)). It is essential to
impose a restriction of this type on the length of the rectangles, for with
out it, one can modify constructions of the Besicovitch set to provide
examples which would contradict the main results and conjectures of
this work.
For 0 < δ < 1, we consider the maximal functions
(1.9) Mv,δ f(x)
= sup
|V(R)|≥δ|R|
1R(x)
|f(y)| dy.
That is we only form the supremum over rectangles for which the vector
field lies in the interval of uncertainty for a fixed positive proportion δ
of the rectangle, see Figure 1.2.
Weak L2 estimate for the Lipschitz Kakeya Maximal Function
1.10. The maximal function Mδ,v is bounded from L
2(R2) to L2,∞(R2)
with norm at most . δ−1/2. That is, for any λ > 0, and f ∈ L2(R2),
this inequality holds:
(1.11) λ2|{x ∈ R2 : Mδ,v f(x) > λ}| . δ−1‖f‖22 .
The norm estimate in particular is independent of the Lipschitz vector
field v.
A principal Conjecture of this work is:
Conjecture 1.12. For some 1 < p < 2, and some finite N and all
0 < δ < 1 and all Lipschitz vector fields v, the maximal function Mδ,v
is bounded from Lp(R2) to Lp,∞(R2) with norm at most . δ−N .
4 1. OVERVIEW OF PRINCIPAL RESULTS
Figure 1.2. A rectangle, with the vector field pointing
in the long direction of the rectangle at three points.
We cannot verify this Conjecture, only establishing that the norm
of the operator can be controlled by a slowly growing function of ec-
centricity.
In fact, this conjecture is stronger than what is needed below. Let us
modify the definition of the Lipschitz Kakeya Maximal Function, by re-
stricting the rectangles that enter into the definition to have an approx-
imately fixed width. For 0 < δ < 1, and choice of 0 < w < 1
‖v‖Lip,
parameterizing the width of the rectangles we consider, define
(1.13) Mv,δ,w f(x)
= sup
|V(R)|≥δ|R|
w≤W(R)≤2w
1R(x)
|f(y)| dy.
We can restrict attention to this case as the primary interest below
is the Hilbert transform on vector fields applied to functions with fre-
quency support in a fixed annulus. By Fourier uncertainty, the width
of the fixed annulus is the inverse of the parameter w above.
Conjecture 1.14. For some 1 < p < 2, and some finite N and all
0 < δ < 1, all Lipschitz vector fields v and 0 < w < 1
‖v‖Lip the
maximal function Mδ,v,w is bounded from L
p(R2) to Lp,∞(R2) with norm
at most . δ−N .
These conjectures are stated as to be universal over Lipschitz vec-
tor fields. On the other hand, we will state conditional results below
in which we assume that a given vector field satisfies the Conjecture
above, and then derive consequences for the Hilbert transform on vec-
tor fields. We also show that e. g. real-analytic vector fields [3] satisfy
these conjectures.
We turn to the Hilbert transform on vector fields. As it turns out,
it is useful to restrict functions in frequency variables to an annulus.
Such operators are given by
St f(x) =
1/t≤|ξ|≤2/t
f̂(ξ) eiξ·x dξ .
The relevance in part is explained in part by this result of the authors
[15], valid for measurable vector fields.
1. OVERVIEW OF PRINCIPAL RESULTS 5
Theorem 1.15. For any measurable vector field v we have the L2 into
λ|{|Hv,∞ ◦ St f | > λ}|1/2 . ‖f‖2 .
The inequality holds uniformly in t > 0.
It is critical that the Fourier restriction St enters in, for otherwise
the Besicovitch set would provide a counterexample, as we indicate in
the first section of Chapter 2. This is one point at which the differ-
ence between the maximal function and the Hilbert transform is strik-
ing. The maximal function variant of the estimate above holds, and
is relatively easy to prove, yet the Theorem above contains Carleson’s
Theorem on the pointwise convergence of Fourier series as a Corollary.
The weak L2 estimate is sharp for measurable vector fields, and so
we raise the conjecture
Conjecture 1.16. There is a universal constant K for which we have
the inequalities
(1.17) sup
0<t<‖v‖Lip
‖Hv,ǫ ◦ St‖2→2 <∞ ,
where ǫ = ‖v‖Lip/K.
Modern proofs of the pointwise convergence of Fourier series use the
so-called restricted weak type approach, invented by Muscalu, Tao and
Thiele in [21]. This method uses refinements of the weak L2 estimates,
together with appropriate maximal function estimates, to derive Lp
inequalities, for 1 < p < 2. In the case of Theorem 1.15—for which
this approach can not possibly work—the appropriate maximal func-
tion is the maximal function over all possible line segments. This is
the unbounded Kakeya Maximal function for rectangles with zero ec-
centricity. One might suspect that in the Lipschitz case, there is a
bounded maximal function. This is another motivation for our Lips-
chitz Kakeya Maximal Function, and our main Conjecture 1.12. We
illustrate how these issues play out in our current setting, with this
conditional result, one of the main results of this memoir.
Theorem 1.18. Assume that Conjecture 1.14 holds for a choice of
Lipschitz vector field v. Then we have the inequalities
(1.19) ‖Hv,ǫ ◦ St‖2 . 1 , 0 < t < ‖v‖Lip .
Here, ǫ is as in (1.17). Moreover, if the vector field as 1+η derivatives,
we have the estimate
(1.20) ‖Hv,ǫ‖2 . (1 + log‖v‖C1+η)2 .
In this case, ǫ = K/‖v‖C1+η and η > 0.
6 1. OVERVIEW OF PRINCIPAL RESULTS
While this is a conditional result, we shall see that it sheds new
light on prior results, such as one of Bourgain [3] on real analytic vector
fields. See Proposition 3.30, and the discussion of that Proposition.
The authors are not aware of any conceptual obstacles to the fol-
lowing extension of the Theorem above to be true, namely that one can
establish Lp estimates, for all p > 2. As our argument currently stands,
we could only prove this result for p sufficiently close to 2, because of
our currently crude understanding of the underlying orthogonality ar-
guments.
Conjecture 1.21. Assume that Conjecture 1.14 holds for a choice of
vector field v with 1 + η > 1 derivatives, then we have the inequalities
below
(1.22) ‖Hv,ǫ‖p . (1 + log‖v‖C1+η)2 , 2 < p <∞ .
In this case, ǫ = K/‖v‖C1+η .
For a brief remark on what is required to prove this conjecture, see
Remark 4.65.
The results of Christ, Nagel, Stein and Wainger [7] apply to certain
vector fields v. This work is a beautiful culmination of the ‘geometric’
approach to questions concerning the boundedness of Radon trans-
forms. Earlier, a positive result for analytic vector fields followed from
Nagel, Stein and Wainger [22]. E.M. Stein [25] specifically raised the
question of the boundedness of Hv for smooth vector fields v. And
the results of D. Phong and Stein [23, 24] also give results about Hv.
J. Bourgain [3] considered real–analytic vector field. N. H. Katz [13]
has made an interesting contribution to maximal function question.
Also see the partial results of Carbery, Seeger, Wainger and Wright [5].
CHAPTER 2
Connections to Besicovitch Set and Carleson’s
Theorem
Besicovitch Set
The Besicovitch set is a compact set that contains a line segment of
unit length in each direction in the plane. Anomalous constructions of
such sets show that they can have very small measure. Indeed, given
ǫ > 0 one can select rectangles R1, . . . , Rn, with disjoint eccentricities,
|EX(R)| ≃ n−1, and of unit length, so that |B| ≤ ǫ for B :=
n=1Rj.
On the other hand, letting ej ∈ EX(Rj), one has that the rectangles
Rj + ej are essentially disjoint. See Figure 2.1. Call the ‘reach’ of the
Besicovitch set
Reach :=
Rj + ej .
This set has measure about one. On the Reach, one can define a vector
field with points to a line segment contained in the Besicovitch set.
Clearly, one has
|Hv 1B(x)| ≃ 1 , x ∈ Reach .
Further, constructions of this set permit one to take the vector field
to be Lipschitz continuous of any index strictly less than one. And
conversely, if one considers a Besicovitch set associated to a vector field
of sufficiently small Lipschitz norm, of index one, the corresponding
Besicovitch set must have large measure. Thus, Lipschitz estimates
are critical.
The Kakeya Maximal Function
The Kakeya maximal function is typically defined as
(2.1) MK,ǫ f(x) := sup
|EX(R)|≥ǫ
1R(x)
|f(y)| dy , ǫ > 0 .
One is forced to take ǫ > 0 due to the existence of the Besicovitch set.
It is a critical fact that the norm of this operator admits a norm bound
on L2 that is logarithmic in ǫ. See Córdoba and Fefferman [8], and
8 2. CONNECTIONS
Figure 2.1. A Besicovitch Set on the left, and it’s
Reach on the right.
Strömberg [27, 28]. Subsequently, there have been several refinements
of this observation, we cite only Nets H. Katz [12], Alfonseca, Soria
and Vargas [1], and Alfonseca [2]. These papers contain additional
references. For the L2 norm, the following is the sharp result.
Theorem 2.2. We have the estimate below valid for all 0 < ǫ < 1.
‖MK,ǫ‖2→2 . 1 + log 1/ǫ .
The standard example of taking f to be the indicator of a small
disk show that the estimate above is sharp, and that the norm grows
as an inverse power of ǫ for 1 < p < 2.
Carleson’s Theorem
We explain the connection between the Hilbert transform on vector
fields and Carleson’s Theorem on the pointwise convergence of Fourier
series. Since smooth functions have a convergent Fourier expansion,
the main point of Carleson’s Theorem is to provide for the control of
an appropriate maximal function. We recall that maximal function in
this Theorem.
Carleson’s Theorem 2.3. For all measurable functions N : R −→
R, the operator below maps L2 into itself.
CNf(x) := p.v.
eiN(x)y f(x− y)dy
The implied operator norm is independent of the choice of measurable
N(x).
CARLESON’S THEOREM 9
v(x1)
N(x1)
ξ2 = J
σ(ξ1 −N(x1))
Figure 2.2. Deducing Carleson’s Theorem from Stein’s Conjecture.
For fixed function f , an appropriate choice of N will give us
∣∣∣p.v.
eiNy f(x− y)dy
∣∣∣ . |CNf(x)| .
Thus, in the Theorem above we have simply linearized the supremum.
Also, we have stated the Theorem with the un-truncated integral. The
content of the Theorem is unchanged if we make a truncation of the
integral, which we will do below.
Let us now show how to deduce this Theorem from an appropriate
bound on certain bound on Hilbert transforms on vector fields. (This
observation is apparently due to R.Coifman from the 1970’s.)
Proposition 2.4. Assume that we have, say, the bound
‖Hv,1‖2→2 . 1 ,
assuming that ‖v‖C2 ≤ 1. It follows that the Carleson maximal operator
is bounded on L2(R).
Proof. The Proposition and the proof are only given in their most
obvious formulation. Set σ(ξ) =
iξy dy
. For a C2 function N :
R −→ R we deduce that the operator with symbol σ(ξ − N(x)) maps
L2(R) into itself with norm that is independent of the C2 norm of
the function N(x). A standard limiting argument then permits one
to conclude the same bound for all measurable choices of N(x), as is
required for the deduction of Carleson’s inequality.
This argument is indicated in Figure 2.2. Take the vector field to
be v(x1, x2) = (1,−N(x1)/n) where n is chosen much larger than the
10 2. CONNECTIONS
C2 norm of the function N(x1). Then, Hv,1 is bounded on L
2(R2) with
norm bounded by an absolute constant. The symbol of Hv,1 is
σ(ξ1, ξ2) = σ(ξ1 − ξ2N(x1)/n) .
The trace of this symbol along the line ξ2 = J defines a symbol of a
bounded operator on L2(R). Taking J very large, we obtain a very good
approximation to symbol σ(ξ1 − ξ2N(x1)/n), deducing that it maps
L2(R) into itself with a bounded constant. Our proof is complete. �
The Weak L2 Estimate in Theorem 1.15 is Sharp
An example shows that under the assumption that the vector field
is measurable, the sharp conclusion is that Hv ◦ S1 maps L2 into L2,∞.
And a variant of the approach to Carleson’s theorem by Lacey and
Thiele [20] will prove this norm inequality. This method will also show,
under only the measurability assumption, that Hv S1 maps L
p into itself
for p > 2, as is shown by the current authors [15]. The results and
techniques of that paper are critical to this one.
CHAPTER 3
The Lipschitz Kakeya Maximal Function
The Weak L2 Estimate
We prove Theorem 1.10, the weak L2 estimate for the maximal
function defined in (1.9), by suitably adapting classical covering lemma
arguments.
The Covering Lemma Conditions. We adopt the covering lemma
approach of Córdoba and R. Fefferman [8]. To this end, we regard the
choice of vector field v and 0 < δ < 1 as fixed. Let R be any finite
collection of rectangles obeying the conditions (1.8) and |V(R)| ≥ δ|R|.
We show that R has a decomposition into disjoint collections R′ and
R′′ for which these estimates hold.
. δ−1
,(3.1)
R∈R′′
∣∣∣ .
(3.2)
The first of these conditions is the stronger one, as it bounds the L2
norm squared by the L1 norm; the verification of it will occupy most
of the proof.
Let us see how to deduce Theorem 1.10. Take λ > 0 and f ∈ L2
which is non negative and of norm one. Set R to be all the rectangles
R of prescribed maximum length as given in (1.8), density with respect
to the vector field, namely |V(R)| ≥ δ|R|, and
f(y) dy ≥ λ|R| .
We should verify the weak type inequality
(3.3) λ
. δ−1/2 .
12 3. LIPSCHITZ KAKEYA
Apply the decomposition to R. Observe that
. δ−1/2
Here of course we have used (3.1). This implies that
. δ−1/2.
Therefore clearly (3.3) holds for the collection R′.
Concerning the collection R′′, apply (3.2) to see that
R∈R′′
. δ−1/2 .
This completes our proof of (3.3).
The remainder of the proof is devoted to the proof of (3.1) and
(3.2).
The Covering Lemma Estimates.
Construction of R′ and R′′. In the course of the proof, we will need
several recursive procedures. The first of these occurs in the selection
of R′ and R′′.
We will have need of one large constant κ, of the order of say 100,
but whose exact value does not concern us. Using this notation hides
distracting terms.
Let Mκ be a maximal function given as
Mκ f(x) = sup
|f(y)| dy , sup
|f(x+ σω)| dσ
Here, Q is the unit square in plane, and Ω is a set of uniformly dis-
tributed points on the unit circle of cardinality equal to κ. It follows
from the usual weak type bounds that this operator maps L1(R2) into
weak L1(R2).
To initialize the recursive procedure, set
R′ ← ∅ ,
STOCK← R .
THE WEAK L
ESTIMATE 13
R′ κR
Figure 3.1. The rectangle R′ would have been removed
from STOCK upon the selection of R as a member of R′.
The main step is this while loop. While STOCK is not empty, select
R ∈ STOCK subject to the criteria that first it have a maximal length
L(R), and second that it have minimal value of |EX(R)|. Update
R′ ←R′ ∪ {R}.
Remove R from STOCK. As well, remove any rectangle R′ ∈ STOCK
which is also contained in
1κR ≥ κ−1
As the collection R is finite, the while loop will terminate, and at
this point we set R′′ def= R−R′. In the course of the argument below,
we will refer the order in which rectangles were added to R′.
With this construction, it is obvious that (3.3) holds, with a bound
that is a function of κ. Yet, κ is an absolute constant, so this depen-
dence does not concern us. And so the rest of the proof is devoted to
the verification of (3.1).
An important aspect of the qualitative nature of the interval of
eccentricity is encoded into this algorithm. We will choose κ so large
that this is true: Consider two rectangles R and R′ with R ∩ R′ 6= ∅,
L(R) ≥ L(R′), W(R) ≥ W(R′), |EX(R)| ≤ |EX(R′)| and EX(R) ⊂
10EX(R′) then we have
(3.4) R′ ⊂ κR .
See Figure 3.1.
Uniform Estimates. We estimate the left hand side of (3.1). In
so doing we expand the square, and seek certain uniform estimates.
14 3. LIPSCHITZ KAKEYA
Expanding the square on the left hand side of (3.1), we can estimate
l.h.s. of (3.1) ≤
|R|+ 2
(ρ,R)∈P
|ρ ∩R|
where P consists of all pairs (ρ, R) ∈ R′×R′ such that ρ∩R 6= ∅, and
ρ was selected to be a member of R′ before R was. It is then automatic
that L(R) ≤ L(ρ). And since the density of all tiles is positive, it follows
that dist(EX(ρ),EX(R)) ≤ 2‖v‖LipL(ρ) < 150 .
We will split up the collection P into sub-collections {SR : R ∈ R′}
and {Tρ : ρ ∈ R′}.
For a rectangle R ∈ R′, we take SR to consist of all rectangles ρ
such that (a) (ρ, R) ∈ P; and (b) EX(ρ) ⊂ 10EX(R). We assert that
(3.5)
|R ∩ ρ| ≤ |R|, R ∈ R.
This estimate is in fact easily available to us. Since the rectangles
ρ ∈ SR were selected to be in R′ before R was, we cannot have the
inclusion
(3.6) R ⊂
1κρ > κ
Now the rectangle ρ are also longer. Thus, if (3.5) does not hold, we
would compute the maximal function of
in a direction which is close, within an error of 2π/κ, of being orthog-
onal to the long direction of R. In this way, we will contradict (3.6).
The second uniform estimate that we need is as follows. For fixed
ρ, set Tρ to be the set of all rectangles R such that (a) (ρ, R) ∈ P and
(b) EX(ρ) 6⊂ 10EX(R). We assert that
(3.7)
|R ∩ ρ| . δ−1|ρ|, ρ ∈ R′.
This proof of this inequality is more involved, and taken up in the next
subsection.
Remark 3.8. In the proof of (3.7), it is not necessary that ρ ∈ R′.
Writing ρ = Iρ × Jρ, in the coordinate basis e and e⊥, we could take
any rectangle of the form I × Jρ.
THE WEAK L
ESTIMATE 15
. . .
Figure 3.2. Proof of Lemma 3.9
These two estimates conclude the proof of (3.1). For any two dis-
tinct rectangles ρ, R ∈ P, we will have either ρ ∈ SR or R ∈ Tρ. Thus
(3.1) follows by summing (3.5) on R and (3.7) on ρ.
The Proof of (3.7). We do not need this Lemma for the proof of
(3.7), but this is the most convenient place to prove it.
Lemma 3.9. Let S be any finite collection of rectangles with L(R) ≤
2 L(R′), and with |V(R)| ≥ δ|R| for all R,R′ ∈ S. Then it is the case
(3.10)
≤ 2δ−1 .
Proof. Fix a point x at which we give an upper bound on the
sum above. Let C(x) be any circle centered at x. We shall show that
there exists at most one R in S such that V(R) ∩ C(x) 6= ∅. By the
assumption |V(R)| ≥ δ|R| this proves the Lemma.
We prove this last claim by contradiction of the Lipschitz assump-
tion on the vector field v. Assume that there exist at least two rectan-
gles R,R′ ∈ S for which the sets V(R) and V(R′) intersect C(x). Thus
there exist y and y′ in C(x) such that v(y) ∈ EX(R) and v(y′) ∈ EX(R′).
Since v is Lipschitz, we have
|v(y)− v(y′)| ≤ ‖v‖Lip|y − y′|
≤ 4‖v‖Lip L(R)|v(y)− v(y′)| ,
but this is a contradiction to our assumption (1.8). See Figure 3.2. �
We fix ρ, and begin by making a decomposition of the collection Tρ.
Suppose that the coordinate axes for ρ are given by eρ, associated with
the long side of R, and e⊥ρ , with the short side. Write the rectangle
as a product of intervals Iρ × J , where |Iρ| = L(ρ). Denote one of the
endpoints of J as α. See Figure 3.3.
16 3. LIPSCHITZ KAKEYA
Figure 3.3. Notation for the proof of (3.7).
For rectangles R ∈ Tρ, let IR denote the orthogonal projection R
onto the line segment 2Iρ×{α}. Subsequently, we will consider different
subsets of this line segment. The first of these is as follows. For R ∈ Tρ,
let VR be the projection of the set V(R) onto 2Iρ × {α}. The angle θ
between ρ and R is at most |θ| ≤ 2‖v‖Lip L(ρ) ≤ 150 . It follows that
(3.11) 1
L(R) ≤ |IR| ≤ 2 L(R), and δ L(R) . |VR|.
A recursive mechanism is used to decompose Tρ. Initialize
STOCK← Tρ ,
U ← ∅ .
While STOCK 6= ∅ select R ∈ STOCK of maximal length. Update
U ← U ∪ {R},
U(R)← {R′ ∈ STOCK : VR ∩ VR′ 6= ∅}.
STOCK← STOCK− U(R).
(3.12)
When this while loop stops, it is the case that Tρ =
R∈U U(R).
With this construction, the sets {VR : R ∈ U} are disjoint. By
(3.11), we have
(3.13)
L(R) . δ−1 L(ρ) .
The main point, is then to verify the uniform estimate
(3.14)
R′∈U(R)
|R′ ∩ ρ| . L(R) ·W(ρ) , R ∈ U .
Note that both estimates immediately imply (3.7).
THE WEAK L
ESTIMATE 17
Figure 3.4. Proof of Lemma 3.15: The rectangles
R,R′ ∈ U(ρ), and so the angles R and R′ form with
ρ are nearly the same.
Proof of (3.14). There are three important, and more technical,
facts to observe about the collections U(R).
For any rectangle R′ ∈ U(R), denote its coordinate axes as eR′ and
e⊥R′ , associated to the long and short sides of R
′ respectively.
Lemma 3.15. For any rectangle R′ ∈ U(R) we have
|eR′ − eR| ≤ 12 |eρ− eR|
Proof. There are by construction, points x ∈ V(R) and x′ ∈ V(R′)
which get projected to the same point on the line segment Iρ × {α}.
See Figure 3.4. Observe that
|eR′ − eR| ≤ |EX(R′)|+ |EX(R)|+ |v(x′)− v(x)|
≤ |EX(R′)|+ |EX(R)|+ ‖v‖Lip · L(R) · |eρ− eR|
≤ |EX(R′)|+ |EX(R)|+ 1
|eρ− eR|
Now, |EX(R)| ≤ 1
|eρ− eR|, else we would have ρ ∈ SR. Likewise,
|EX(R′)| ≤ 1
|eR′ − eR|. And this proves the desired inequality.
Lemma 3.16. Suppose that there is an interval I ⊂ Iρ such that
(3.17)
R′∈U(R)
L(R′)≥8|I|
|R′ ∩ I × J | ≥ |I × J | .
Then there is no R′′ ∈ U(R) such that L(R′′) < |I| and R′′∩4I×J 6= ∅.
Proof. There is a natural angle θ between the rectangles ρ and R,
which we can assume is positive, and is given by |eρ− eR|. Notice that
we have θ ≥ 10|EX(R)|, else we would have ρ ∈ SR, which contradicts
our construction.
18 3. LIPSCHITZ KAKEYA
4I × J
I × J
Figure 3.5.
Moreover, there is an important consequence of Lemma 3.15: For
any R′ ∈ U(R), there is a natural angle θ′ between R′ and ρ. These
two angles are close. For our purposes below, these two angles can be
regarded as the same.
For any R′ ∈ U(R), we will have
|κR′ ∩ ρ|
|I × J | ≃ κ
W(R′) ·W(ρ)
θ|I|W(ρ)
W(R′)
θ · |I| .
Recall Mκ is larger than the maximal function over κ uniformly
distributed directions. Choose a direction e′ from this set of κ directions
that is closest to e⊥ρ . Take a line segment Λ in direction e
′ of length
κθ|I|, and the center of Λ is in 4I × J . See Figure 3.5. Then we have
|κR′ ∩ Λ|
|Λ| ≥
W(R′)
θ · |I|
Thus by our assumption (3.17),
R′∈U(R)
|κR′ ∩ Λ| & 1 .
That is, any of the lines Λ are contained in the set
1κR′ > κ
Clearly our construction does not permit any rectangle R′′ ∈ U(R)
contained in this set. To conclude the proof of our Lemma, we seek a
contradiction. Suppose that there is an R′′ ∈ U(R) with L(R′′) < |I|
and R′′ intersects 2I × J . The range of line segments Λ we can permit
THE WEAK L
ESTIMATE 19
4I × J
Figure 3.6. The proof of Lemma 3.18
is however quite broad. The only possibility permitted to us is that
the rectangle R′′ is quite wide. We must have
W(R′′) ≥ 1
|Λ| = κ
· θ · |I|.
This however forces us to have |EX(R′′)| ≥ κ
θ. And this implies that
ρ ∈ SR′′ , as in (3.5). This is the desired contradiction.
Our third and final fact about the collection U(R) is a consequence
of Lemma 3.15 and a geometric observation of J.-O. Stromberg [27,
Lemma 2, p. 400].
Lemma 3.18. For any interval I ⊂ IR we have the inequality
(3.19)
R′∈U(R)
L(R′)≤|I|≤
κL(R′)
|R′ ∩ I × J | ≤ 5|I| ·W(ρ) .
Proof. For each point x ∈ 4I × J , consider the square S centered
at x of side length equal to
κ · |I| · |eR− eρ|. See Figure 3.6. It is
Stromberg’s observation that for R′ ∈ U(R) we have
|κR′ ∩ I × J |
|I × J | ≃
|S ∩ κR′|
with the implied constant independent of κ. Indeed, by Lemma 3.15,
we have that
|κR′ ∩ I × J |
|I × J | ≃
κW(R′)
|eR− eρ| · |I|
≃ κW(R
′) · |I| · |eR− eρ|
(|eR− eρ| · |I|)2
≃ |S ∩ κR
|S| ,
20 3. LIPSCHITZ KAKEYA
as claimed.
Now, assume that (3.19) does not hold and seek a contradiction.
Let U ′ ⊂ U(R) denote the collection of rectangles R′ over which the
sum is made in (3.19). The rectangles in U ′ were added in some order
to the collection R′, and in particular there is a rectangle R0 ∈ U ′ that
was the last to be added to U ′. Let U ′′ be the collection U ′−{R0}. We
certainly have ∑
R′∈U ′′
|R′′ ∩ I × J | ≥ 4|I × J |.
Since we cannot have ρ ∈ SR0 , Stromberg’s observation implies that
R′∈U ′′
1κR′ > κ
Here, we rely upon the fact that the maximal function Mκ is larger than
the usual maximal function over squares. But this is a contradiction
to our construction, and so the proof is complete. �
The principal line of reasoning to prove (3.14) can now begin with
it’s initial recursive procedure. Initialize
C(R′)← R′ ∩ ρ .
We are to bound the sum
R′∈U(R)|C(R′)|. Initialize a collection of
subintervals of IR to be
I ← ∅
WHILE there is an interval I ⊂ IR satisfying∑
R′∈V(I)
|C(R′) ∩ I × J | ≥ 40|I| ·W(ρ) ,(3.20)
V(I) = {R′ ∈ U(R) | |C(R′) ∩ I × J | 6= ∅ , L(R′) ≥ 8|I|} ,(3.21)
we take I to be an interval of maximal length with this property, and
update
I ← I ∪ {I} ;
C(R′, I) = C(R′) ∩ I × J , R′ ∈ V(I);
C(R′)← C(R′)− I × J, R′ ∈ V(I) .
[We remark that this last updating is not needed in the most important
special case when all rectangles have the same width. But the case we
are considering, rectangles can have variable widths, so that |C(R)| can
be much larger than any |I| · |J | that would arise from this algorithm.]
Once the WHILE loop stops, we have
R′ ∩ ρ = C(R′) ∪
{C(R′, I) | I ∈ I , R′ ∈ V(I)} .
THE WEAK L
ESTIMATE 21
Here the union is over pairwise disjoint sets.
We first consider the collection of sets {C(R′) | R′ ∈ U(R)} that
remain after the WHILE loop has finished. Since we must not have
R′ ⊂ 1/4κ · ρ, it follows that the minimum value of L(R′) is 1
W(ρ).
Thus, if in (3.20), we consider an interval I of length 1
W(ρ), the
condition L(R′) ≥ 8|I| in the definition of V(I) in (3.21) is vacuous.
Thus, we necessarily have
R′∈V(I)
|C(R′) ∩ I × J | ≤ 40|I| ·W(ρ) .
For if this inequality failed, the WHILE loop would not have stopped.
We can partition IR by intervals of length close to
W(ρ), showing
that we have ∑
R′∈U(R)
|C(R′)| . |IR| ·W(ρ) .
Turning to the central components of the argument, namely the
bound for the terms associated with the intervals in I, consider I ∈ I.
The inequality (3.20) and Lemma 3.18 implies that each I ∈ I must
have length |I| ≤ κ−1/2|Iρ|. But we choose intervals in I to be of
maximal length. Thus,
R′∈V(I)
|C(R′, I)| ≤ 100 · |I| ·W(ρ) .
(3.22)
Indeed, suppose this last inequality fails. Let I ⊂ Ĩ ⊂ Iρ be an interval
twice as long as I. By Lemma 3.18, we conclude that
R′∈V(I)
L(R′)≤8|eI|
|R′ ∩ Ĩ × J | ≤ 10|I| ·W(ρ) .
Notice that we are restricting the sum on the left by the length of |Ĩ|.
Therefore, we have the inequalities
R′∈V(I)
L(R′)>8|eI|
|C(R′, I)| ≥ 90 · |I| ·W(ρ) > 40 · |Ĩ| ·W(ρ) .
That is, Ĩ would have been selected, contradicting our construction.
Lemmas 3.16 and 3.18 place significant restrictions on the collection
of intervals I. If we have I 6= I ′ ∈ I with 3
I ∩ 3
I ′ 6= ∅, then we must
have e.g.
κ|I ′| < |I|, as follows from Lemma 3.18. Moreover, V(I ′)
must contain a rectangle R′ with L(R′) < |I|. But this contradicts
Lemma 3.16.
22 3. LIPSCHITZ KAKEYA
Therefore, we must have
|I| . |IR| . L(R).
With (3.22), this completes the proof of (3.14).
An Obstacle to an Lp estimate, for 1 < p < 2
We address one of the main conjectures of this memoir, namely
Conjecture 1.12. Let us first observe
Proposition 3.23. We have the estimate below valid for all 0 < w <
‖v‖Lip.
λ|{Mv,w f > λ}|2/3 . δ−1/3(1 + logw−1‖v‖Lip)1/3‖f‖3/2
Proof. Let ‖v‖Lip = 1. This just relies upon the fact that with
0 < w < 1
fixed, there are only about log 1/w possible values of L(R).
This leads very easily to the following two estimates. Following the
earlier argument, consider an arbitrary collection of rectangles R with
each R ∈ R satisfying (1.8) and |V(R)| ≥ δ|R|. We can then decompose
R into disjoint collections R′ and R′′ for which these estimates hold.
. δ−1(log 1/w)
,(3.24)
R∈R′′
∣∣∣ .
(3.25)
Compare to (3.1) and (3.2). Following the same line of reasoning that
was used to prove (3.3), we prove our Proposition.
We can devise proofs of smaller bounds on the norm of the maximal
function than that given by this proposition. But no argument that we
can find avoids the some logarithmic term in the width of the rectangle.
Let us illustrate the difficulty in the estimate with an object pointed
out to us by Ciprian Demeter. We term it a pocketknife, and it is
pictured in Figure 3.7.
A pocketknife comes with a handle, namely a rectangle Rhandle that
is longer than any other rectangle in the pocketknife. We call a collec-
tion of rectangles B a set of blades if these two conditions are met. In
the first place,
(3.26) Rhandle ∩
R 6= ∅ .
BOURGAIN’S GEOMETRIC CONDITION 23
handle
blades
. . .
hinges
Figure 3.7. A pocketknife.
In the second place, we have
angle(R,Rhandle) ≃ angle(R′, Rhandle) , R, R′ ∈ B .
Let θ(B) denote the angle between Rhandle and the rectangles in the
blade B. We refer to as a hinge a rectangle of dimensions w/θ(B) by w,
in the same coordinate system of Rhandle that contains the intersection
in (3.26).
Now, let B be a collection of blades for the handle Rhandle. Our proof
of the weak L2 estimate for the Lipschitz Kakeya Maximal function
shows that we can assume
♯B · w2 · θ(B)−1 . |Rhandle| .
This is essentially the estimate (3.5).
But, to follow the covering lemma approach to the L3/2 estimate
for the maximal function, we need to control
(♯B)2 · w2 · θ(B)−1 .
We can only find control of expressions of this type in terms of some
slowly varying function of w−1.
Bourgain’s Geometric Condition
Jean Bourgain [3] gives a geometric condition on the Lipschitz vec-
tor field that is sufficient for the L2 boundedness of the maximal func-
tion associated with v. We describe the condition, and show how it
immediately proves that the corresponding Lipschitz Kakeya maximal
function admits a weak type bound on L1. In particular our Conjec-
ture 1.12 holds for these vector fields.
To motivate Bourgain’s condition, let us recall the earlier condition
considered by Nagel, Stein and Wainger in [22]. This condition imposes
24 3. LIPSCHITZ KAKEYA
a restriction on the maximum and minimum curvatures of the integral
curves of the vector field through the assumption that
supx∈Ω det[∇v(x)v(x), v(x)]
infx∈Ω det[∇v(x)v(x), v(x)]
Here, Ω is a domain in R2, and one can achieve an upper bound on the
norm of the maximal function associated to v, appropriately restricted
to Ω, in terms of this ratio.
Bourgain’s condition permits the vector field to have integral curves
which are flat. Suppose that v is defined on all of R2. Define
(3.27) ω(x; t) := |det[v(x+ tv(x)), v(x)]| , |t| ≤ 1
‖v‖Lip .
Assume a uniform estimate of the following type: For absolute con-
stants 0 < c, C <∞ and 0 < ǫ0 < 12‖v‖Lip,
(3.28) |{|t| ≤ ǫ | ω(x; t) < τ sup
|s|≤ǫ
ω(x, s)}| ≤ Cτ cǫ ,
this condition holding for all x ∈ R2, 0 < τ < 1 and 0 < ǫ < ǫ0.
The interest in this condition stems from the fact [3] that real-
analytic vector fields satisfy it. Also see Remark 3.35. Bourgain proved:
Theorem 3.29. Assume that (3.28) holds. Then, the maximal opera-
tor Mv,ǫ0 defined in (1.1) maps L
2 into itself.
This paper claims that the same methods would prove the bounds
‖Mv,ǫ0‖p . ‖f‖p , 1 < p <∞ .
And suggests that similar methods would apply to the localized Hilbert
transform with respect to these vector fields.
Here, we prove
Proposition 3.30. Assume that (3.28) holds. Then, the Lipschitz
Kakeya Maximal Functions
Mv,δ,w , 0 < δ < 1 , 0 < w < ǫ0
defined in (1.13) satisfy the weak L1 estimate
λ|{Mv,δ,w f > λ}| . δ−1(1 + log 1/δ)‖f‖1 .
The implied constants depend upon the constants in (3.28).
That is, these vector fields easily fall within the scope of our anal-
ysis. As a corollary to Theorem 1.18, we see that Hv maps L
2 into
itself.
BOURGAIN’S GEOMETRIC CONDITION 25
Figure 3.8. Proof of (3.31).
Proof. Let us assume that ‖v‖Lip = 1. Fix δ > 0 and 0 < w < ǫ0.
Let R be the class of rectangles with L(R) < κ and satisfying |V(R)| ≥
δ|R|.
Say that R′ ⊂ R has scales separated by s > 3 iff for R,R′ ∈ R′ the
condition 4 L(R) < L(R′) implies that 2s L(R) < L(R′). One sees that
R can be decomposed into ≃ s sub-collections with scales separated
by s.
The fortunate observation is this: Assuming (3.28), and taking s ≃
log 1/δ, any subset R′ ⊂ R with scales separated by s further enjoys
this property: If R,R′ ∈ R′ with C EX(R) ∩ C EX(R′) = ∅, with C a
fixed constant, then
(3.31) L(R) ≃ L(R′) or R ∩R′ = ∅ .
Let us see why this is true, arguing by contradiction. Thus we
assume that L(R′) ≤ 2−s L(R), R∩R′ 6= ∅ and C EX(R)∩C EX(R′) = ∅.
Since the rectangles have an essentially fixed width, it follows that
2|EX(R′)| ≥ |EX(R)|. Fix a line ℓ in the long direction of 2R with
|{x ∈ ℓ | v(x) ∈ V(R)}| ≥ δ
|ℓ| = δ
L(R) .
Let x0 be in the set above, x
0 ∈ V (R′) and x′0 is the projection of x′′0
onto the line ℓ. See Figure 3.8. Observe that we can estimate
|v(x′′0)− v(x′0)| ≤ 2|v(x0)− v(x′′0)| L(R′)(3.32)
Therefore, for C sufficiently large, we have
|v(x0)− v(x′0)| ≥
∣∣|v(x′0)− v(x′′0)| − |v(x′′0)− v(x0)|
≥ |v(x′′0)− v(x0)|(1− 2 L(R′))
≥ |EX(R′)|
provided C is large enough.
Now, after a moments thought, one sees that
|det[v(x0), v(x′0)]| ≃ angle(v(x0), v(x′0)) .
26 3. LIPSCHITZ KAKEYA
Therefore, for any x ∈ ℓ
s≤L(R)
ω(x; s) & |EX(R′)| .
But the vector field satisfies (3.28), which we will apply with
τ ≃ EX(R)
EX(R′)
≃ L(R
It follows that
L(R) ≤ |{x ∈ ℓ | ω(x; s) ≤ cτ |EX(R′)|}|
≤ |{x ∈ ℓ | ω(x; s) ≤ τ sup
|s|≤ǫ
ω(x, s)}|
. τ c L(R) .
Therefore, we see
(δ/2)1/c .
L(R′)
which is a contradiction to R′ have scales separated by s, and s ≃
1 + log 1/δ.
Let us see how to prove the Proposition now that we have proved
(3.31). Take s ≃ log 1/δ, and a finite sub-collection R′ ⊂ R of rectan-
gles with scales separated by s. We may take a further subset R′′ ⊂ R′
such that ∥∥∥
R′′∈R′′
. δ−1 ,(3.33)
R∈R′−R′′
∣∣∣ .
R∈R′′
|R′′| .(3.34)
These are precisely the covering estimates needed to prove the weak
L1 estimate claimed in the proposition.
But, in choosing R′′ to satisfy (3.33), it is clear that we need only
be concerned about rectangles with a fixed length, and the separation
in scales are (3.31) will control rectangles of distinct lengths.
The procedure that we apply to select R′′ is inductive. Set
R′′ ← ∅ ,
S ← ∅ ,
STOCK←R′ .
WHILE STOCK 6= ∅, select R ∈ STOCK with maximal length, and
update R′′ ← R′′ ∪ {R}, as well as STOCK ← STOCK − {R}. In
addition, for any R′ ∈ STOCK with R′ ⊂ 4CR, where C ≥ 1 is the
VECTOR FIELDS THAT ARE A FUNCTION OF ONE VARIABLE 27
constant that insures that (3.31) holds, remove these rectangles from
STOCK and add them to S.
Once the WHILE loop stops, we will have STOCK = ∅ and we have
our decomposition of R′. By construction, it is clear that (3.34) holds.
We need only check that (3.33) holds. Now, consider R,R′ ∈ R′, with
two rectangles have their scales separated, thus 2s L(R′) < L(R). If it
is the case that R∩R′ 6= ∅ and C EX(R)∩C EX(R′) 6= ∅, then R would
been selected to be in R′ first, whence R′ would have been placed in
Therefore, C EX(R) ∩ C EX(R′) = ∅, but then (3.31) implies that
R ∩ R′ = ∅. Thus, the only contribution to the L∞ norm in (3.33)
can come from rectangles of about the same length. But Lemma 3.9
then implies that such rectangles can overlap only about δ−1 times.
Our proof is complete. (As the interest in (3.28) is in small values of
c, it will be more efficient to use Lemma 3.9 to handle the case of the
rectangles having approximately the same length.) �
Remark 3.35. To conclude that the Hilbert transform on vector fields
is bounded, one could weaken Bourgain’s condition (3.28) to
|{|t| ≤ ǫ | ω(x; t) < τ sup
|s|≤ǫ
ω(x, s)}| ≤ C exp(−(log 1/τ)c)ǫ .
This inequality is to hold universally in x ∈ R2, 0 < τ < 1, and
0 < ǫ < ‖v‖Lip. This is of interest for 0 < c < 1. The proof above
can be modified to show that the maximal functions Mv,δ,w satisfy the
weak L1 inequality, with constant at most . δ−1−1/c.
Vector Fields that are a Function of One Variable
We specialize to the vector fields that are a function of just one real
variable. Assume that the vector field v is of the form
(3.36) v(x1, x2) = (v1(x2), v2(x2)) ,
and for the moment we do not impose the condition that the vector
field take values in the unit circle. The point is simply this: If we are
interested in transforms where the kernel is not localized, the restriction
on the vector field is immaterial. Namely, for any vector field v
Hv,∞ f(x) = p.v.
f(x− yv(x)) dy
= p.v.
f(x− yṽ(x)) dy
, ṽ(x) =
|v(x)| .
We return to a theme implicit in the proof of Proposition 2.4. This
proof only relies upon vector fields that are only a function of one
28 3. LIPSCHITZ KAKEYA
variable. Thus, it is a significant subcase of the Stein Conjecture to
verify it for Lipschitz vector fields of just one variable. Indeed, the
situation is this.
Proposition 3.37. Suppose that a choice of vector field v(x1, x2) =
(1, v1(x1)) is just a function of, say, the first coordinate. Then, Hv,∞
maps L2(R2) into itself.
Proof. The symbol of Hv,∞ is
sgn(ξ1 + ξ2v1(x1)) .
For each fixed ξ2, this is a bounded symbol. And in the special case
of the L2 estimate, this is enough to conclude the boundedness of the
operator. �
It is of interest to extend this Theorem in any Lp, for p 6= 2, for
some reasonable choice of vector fields.
The corresponding questions for the maximal function are also of
interest, and here the subject is much more developed. The paper [5]
studies the maximal function Mv,∞. They proved the boundedness of
this maximal function on Lp, p > 1, assuming that the vector field was
of the form v(x) = (1, v2(x)), that D v2 was positive, and increasing,
and satisfied a third more technical condition. More recently, [14] has
showed that the third condition is not needed. Namely the following is
true.
Theorem 3.38. Assume that v(x) = (1, v2(x)), and moreover that
D v2 ≥ 0 and is monotonically increasing. Then, Mv,∞ is bounded on
Lp, for 1 < p <∞.
These vector fields present far fewer technical difficulties than a gen-
eral Lipschitz vector field, and there are a richer set of proof techniques
that one can bring to bear on them, as indicated in part in the proof
of Proposition 3.37. The papers [5, 14] cleverly exploit the Plancherel
identity (in the independent variable), and other orthogonality consid-
erations to prove their results.
These considerations are not completely consistent with the domi-
nant theme of this monograph, in which the transforms are localized.
Nevertheless, it would be interesting to explore methods, possibly mod-
ifications of this memoir, that could provide an extension of Proposi-
tion 3.37.
In this direction, let us state a possible direction of study. The
definition of the the sets V(R) for vector fields of magnitude 1 is given
as V(R) = R ∩ v−1(EX(R)). For vector fields of arbitrary magnitude,
VECTOR FIELDS THAT ARE A FUNCTION OF ONE VARIABLE 29
we define these sets to be
V(R) = {x ∈ R | v(x)|v(x)| ∈ EX(R)} .
Define a maximal function—an extension of our Lipschitz Kakeya Max-
imal Function—by
(3.39) M̃v,δf(x) = sup
|V(R)|≥δ|R|
|R|−1
f(y) dy .
In this definition, we require the rectangles to have density δ, but do
not restrict their eccentricities, or lengths.
Conjecture 3.40. Assume that the vector field is of the form v(x) =
(1, v2(x2)), and the derivative D v ≥ 0 and is monotone. Then for all
0 < δ < 1, we have the estimate
‖M̃v,δ‖p . δ−1 , 1 < p <∞ .
One can construct examples which show that the L1 to weak L1
norm of the maximal function is not bounded in terms of δ. Indeed,
recalling the ‘pocketknife’ examples of Figure 3.7, we comment that one
can construct examples of vector fields with these properties, which we
describe with the terminology associated with the pocketknife exam-
ples.
• The width of all rectangles are fixed. And all rectangles have
density δ.
• The ‘handle’ of the pocketknife has positive angle θ with the
x1 axis.
• There is ‘hinge’ whose blades have angles which are positive,
and greater than θ. The number of blades can be unbounded,
as the width of the rectangles decreases to zero.
The assumption that the vector field is only a function of x2 then
greatly restricts, but does not completely forbid, the existence of addi-
tional hinges. So the combinatorics of these vector fields, as expressed
in the Lipschitz Kakeya Maximal Function, are not so simple.
CHAPTER 4
The L2 Estimate for Hilbert Transform on
Lipschitz Vector Fields
We prove one of our main conditional results about the Hilbert
transform on Lipschitz vector fields, the inequality (1.19) which is the
estimate at L2, for functions with frequency support in an annulus,
assuming an appropriate estimate for the Lipschitz Kakeya Maximal
Function.
We begin the proof by setting notation appropriate for phase plane
analysis for functions f on the plane supported on an annulus. With
this notation, we can define appropriate discrete analogs of the Hilbert
transform on vector fields. The Lemmas 4.22 and 4.23 are the combi-
natorial analogs of our Theorem 1.15. We then take up the proofs. The
main step in the proof is Lemma 4.50 which combines the (standard)
orthogonality considerations with the conjectures about the Lipschitz
Kakeya Maximal Functions.
Definitions and Principle Lemmas
Throughout this chapter, κ will denote a fixed small positive con-
stant, whose exact value need not concern us. κ of the order of 10−3
would suffice. The following definitions are as in the authors’ previous
paper [15].
Definition 4.1. A grid is a collection of intervals G so that for all
I, J ∈ G, we have I ∩ J ∈ {∅, I, J}. The dyadic intervals are a grid. A
grid G is central iff for all I, J ∈ G, with I ⊂6= J we have 500κ−20I ⊂ J .
The reader can find the details on how to construct such a central
grid structure in [11].
Let ρ be rotation on T by an angle of π/2. Coordinate axes for R2
are a pair of unit orthogonal vectors (e, e⊥) with ρ e = e⊥.
Definition 4.2. We say that ω ⊂ R2 is a rectangle if it is a product
of intervals with respect to a choice of axes (e, e⊥) of R
2. We will
say that ω is an annular rectangle if ω = (−2l−1, 2l−1)× (a, 2a) for an
integer l with 2l < κa, with respect to the axes (e, e⊥). The dimensions
of ω are said to be 2l × a. Notice that the face (−2l−1, 2l−1) × a is
32 4. L
ESTIMATE FOR Hv
es Rs
Figure 4.1. The two rectangles ωs and Rs whose prod-
uct is a tile. The gray rectangles are other possible loca-
tions for the rectangle Rs.
tangent to the circle |ξ| = a at the midpoint to the face, (0, a). We
say that the scale of ω is scl(ω) := 2l and that the annular parameter
of ω is ann(ω) := a. In referring to the coordinate axes of an annular
rectangle, we shall always mean (e, e⊥) as above.
Annular rectangles will decompose our functions in the frequency
variables. But our methods must be sensitive to spatial considerations;
it is this and the uncertainty principle that motivate the next definition.
Definition 4.3. Two rectangles R and R are said to be dual if they
are rectangles with respect to the same basis (e, e⊥), thus R = r1 × r2
and R = r1 × r2 for intervals ri, ri, i = 1, 2. Moreover, 1 ≤ |ri| · |ri| ≤ 4
for i = 1, 2. The product of two dual rectangles we shall refer to as a
phase rectangle. The first coordinate of a phase rectangle we think of
as a frequency component and the second as a spatial component.
We consider collections of phase rectangles AT which satisfy these
conditions. For s, s′ ∈ AT we write s = ωs × Rs, and require that
ωs is an annular rectangle,(4.4)
Rs and ωs are dual,(4.5)
The rectangles Rs are from the product of central grids.(4.6)
{1000κ−100R | ωs × R ∈ AT } covers R2, for all ωs.(4.7)
ann(ωs) = 2
j for some integer j,(4.8)
♯{ωs | scl(s) = scl, ann(s) = ann} ≥ c
,(4.9)
scl(s) ≤ κann(s).(4.10)
DEFINITIONS AND PRINCIPLE LEMMAS 33
0 ρωs1
Figure 4.2. An annular rectangular ωs, and three as-
sociated subintervals of ρωs1, ωs1, and ωs2.
We assume that there are auxiliary sets ωs,ωs1,ωs2 ⊂ T associated to
s—or more specifically ωs—which satisfy these properties.
Ω := {ωs,ωs1,ωs2 | s ∈ AT } is a grid in T,(4.11)
ωs1 ∩ ωs2 = ∅, |ωs| ≥ 32(|ωs1|+ |ωs2|+ dist(ωs1,ωs2))(4.12)
ωs1 lies clockwise from ωs2 on T,(4.13)
|ωs| ≤ K
scl(ωs)
ann(ωs)
,(4.14)
{ ξ|ξ| | ξ ∈ ωs} ⊂ ρωs1.(4.15)
In the top line, the intervals ωs1 and ωs2 are small subintervals of
the unit circle, and we can define their dilate by a factor of 2 in an
obvious way. Recall that ρ is the rotation that takes e into e⊥. Thus,
eωs ∈ ωs1. See the figures Figure 4.1 and Figure 4.2 for an illustration
of these definitions.
Note that |ωs| ≥ |ωs1| ≥ scl(ωs)/ann(ωs). Thus, eωs is in ωs1, and
ωs serves as ‘the angle of uncertainty associated to Rs.’ Let us be
more precise about the geometric information encoded into the angle
of uncertainty. Let Rs = rs × rs⊥ be as above. Choose another set of
coordinate axes (e′, e′⊥) with e
′ ∈ ωs and let R′ be the product of the
intervals rs and rs⊥ in the new coordinate axes. Then K
′ ⊂ Rs ⊂
′ for an absolute constant K0 > 1.
We say that annular tiles are collections AT satisfying the condi-
tions (4.4)—(4.15) above. We extend the definition of e⊥, eω⊥, ann(ω)
and scl(ω) to annular tiles in the obvious way, using the notation es,
es⊥, ann(s) and scl(s).
34 4. L
ESTIMATE FOR Hv
A phase rectangle will have two distinct functions associated to it.
In order to define these functions, set
Ty f(x) := f(x− y), y ∈ R2 (Translation operator)
Modξ f(x) := e
iξ·x f(x), ξ ∈ R2 (Modulation operator)
R1×R2 f(x1, x2) :=
(|R1||R2|)1/p
(Dilation operator).
In the last display, 0 < p ≤ ∞, and R1 × R2 is a rectangle, and the
coordinates (x1, x2) are those of the rectangle. Note that the definition
depends only on the side lengths of the rectangle, and not the location.
And that it preserves Lp norm.
For a function ϕ and tile s ∈ AT set
(4.16) ϕs := Modc(ωs)Tc(Rs)D
We shall consider ϕ to be a Schwartz function for which ϕ̂ ≥ 0 is
supported in a small ball, of radius κ, about the origin in R2, and is
identically 1 on another smaller ball around the origin. (Recall that κ
is a fixed small constant.)
We introduce the tool to decompose the singular integral kernels.
In so doing, we consider a class of functions ψt, t > 0, so that
Each ψt is supported in frequency in [−θ − κ,−θ + κ].(4.17)
|ψt(x)| . CN(1 + |x|)−N , N > 1 .(4.18)
In the top line, θ is a fixed positive constant so that the second half of
(4.19) is true.
Define
φs(x) :=
ϕs(x− yv(x))ψs(y) dy
(v(x))
ϕs(x− yv(x))ψs(sy) dy.
(4.19)
ψs(y) := scl(s)ψscl(s)(scl(s)y).(4.20)
An essential feature of this definition is that the support of the integral
is contained in the set {v(x) ∈ ωs2}, a fact which can be routinely
verified. That is, we can insert the indicator 1
(v(x)) without loss
of generality. The set ωs2 serves to localize the vector field, while ωs1
serves to identify the location of ϕs in the frequency coordinate.
DEFINITIONS AND PRINCIPLE LEMMAS 35
The model operator we consider acts on a Schwartz functions f ,
and it is defined by
(4.21) Cannf :=
s∈AT (ann)
scl(s)≥‖v‖Lip
〈f, ϕs〉φs.
In this display, AT (ann) := {s ∈ AT | ann(s) = ann}, and we have
deliberately formulated the operator in a dilation invariant manner.
Lemma 4.22. Assume that the vector field is Lipschitz, and satisfies
Conjecture 1.14. Then, for all ann ≥ ‖v‖−1Lip, the operator Cann extends
to a bounded map from L2 into itself, with norm bounded by an absolute
constant.
We remind the reader that for 2 < p <∞ the only condition needed
for the boundedness of Cann is the measurability of the vector field, a
principal result of Lacey and Li [15]. It is of course of great importance
to add up the Cann over ann. The method we use for doing this are
purely L2 in nature, and lead to the estimate for C :=
j=1 C2j .
Lemma 4.23. Assume that the vector field is of norm at most one
in Cα for some α > 1, and satisfies Conjecture 1.14. Then C maps
L2 into itself. In addition we have the estimate below, holding for all
values of scl.
(4.24)
ann=−∞
s∈AT (ann)
scl(s)=scl
〈f, ϕs〉φs
. (1 + log(1 + scl−1‖v‖Cα)).
Moreover, these operators are unconditionally convergent in s ∈ AT .
These are the principal steps towards the proof of Theorem 1.15.
In the course of the proof, we shall not invoke the additional notation
needed to account for the unconditional convergence, as it is entirely
notational. They can be added in by the reader.
Observe that (4.24) is only of interest when scl < ‖v‖Cα. This
inequality depends critically on the fact that the kernel sclψ(scly) has
mean zero. Without this assumption, this inequality is certainly false.
The proof of Theorem 1.15 from these two lemmas is an argument
in which one averages over translations, dilations and rotations of grids.
The specifics of the approach are very close to the corresponding argu-
ment in [15]. The details are omitted.
The operators Cann and C are constructed from a a kernel which is a
smooth analog of the truncated kernel p.v. 1
1{|t|≤1}. Nevertheless, our
36 4. L
ESTIMATE FOR Hv
main theorem follows,1 due to the observation that we can choose a
sequence of Schwartz kernels ψ(1+κ)n , for n ∈ Z, which satisfy (4.17)
and (4.18), and so that for
K(t) :=
an(1 + κ)
nψ(1+κ)n((1 + κ)
we have p.v. 1
1{|t|≤1} = K(t)−K(t). Here, for n ≥ 0 we have |an| . 1.
And for n < 0, we have |an| . (1+κ)n. The principal sum is thus over
n ≥ max(0, ‖v‖Cα), and this corresponds to the operator C. For those
n < max(0, ‖v‖Cα), we use the estimate (4.24), and the rapid decay of
the coefficient an.
Truncation and an Alternate Model Sum
There are significant obstacles to proving the boundedness of the
model sum Cann on an Lp space, for 1 < p < 2. In this section, we
rely upon some naive L2 estimates to define a new model sum which is
bounded on Lp, for some 1 < p < 2.
Our next Lemma is indicative of the estimates we need. For choices
of scl < κann, set
AT (ann, scl) := {s ∈ AT (ann) | scl(s) = scl}.
Lemma 4.25. For measurable vector fields v and all choices of ann
and scl. ∥∥∥
s∈AT (ann,scl)
〈f, ϕs〉φs
. ‖f‖2
Proof. The scale and annulus are fixed in this sum, making the
Bessel inequality
s∈AT (ann,scl)
|〈f, ϕs〉|2 . ‖f‖22
evident. For any two tiles s and s′ that contribute to this sum, if ωs 6=
ωs′, then φs and φs′ are disjointly supported. And if ωs = ωs′, then
Rs and R
s are disjoint, but share the same dimensions and orientation
in the plane. The rapid decay of the functions φs then gives us the
1In the typical circumstance, one uses a maximal function to pass back and
forth between truncated and smooth kernels. This route is forbidden to us; there
is no appropriate maximal function to appeal to.
TRUNCATION AND AN ALTERNATE MODEL SUM 37
estimate
s∈AT (ann,scl)
〈f, ϕs〉φs
s∈AT (ann,scl)
|〈f, ϕs〉|2
. ‖f‖2
Consider the variant of the operator (4.21) given by
(4.26) Φf =
s∈AT (ann)
scl(s)≥κ−1‖v‖Lip
〈f, ϕs〉φs.
As ann is fixed, we shall begin to suppress it in our notations for oper-
ators. The difference between Φ and Cann is the absence of the initial
. log(1+ ‖v‖Lip) scales in the former. The L2 bound for these missing
scales is clearly provided by Lemma 4.25, and so it remains for us to
establish
(4.27) ‖Φ‖2 . 1,
the implied constant being independent of ann, and the Lipschitz norm
of v.
It is an important fact, the main result of Lacey and Li [15], that
(4.28) ‖Φ‖p . 1, 2 < p <∞.
This holds without the Lipschitz assumption.
We are now at a point where we can be more directly engaged with
the construction of our alternate model sum. We only consider tiles
with κ−1‖v‖Lip ≤ scl(s) ≤ κann. A parameter is introduced which is
used to make a spatial truncation of the functions ϕs; it is
(4.29) γ2s := 100
−2 scl(s)
‖v‖Lip
Write ϕs = αs + βs where αs = (Tc(Rs)D
ζ)ϕs, and ζ is a smooth
Schwartz function supported on |x| < 1/2, and equal to 1 on |x| < 1/4.
Write for choices of tiles s,
(4.30) ψs(y) = ψs−(y) + ψs+(y)
where ψs−(y) is a Schwartz function on R, with
supp(ψs−) ⊂
γs(scl(s))
−1[−1, 1] ,
38 4. L
ESTIMATE FOR Hv
and equal to ψscl(s)(y) for |y| < 14γs(scl(s))
−1. Then define
(4.31) as±(x) = 1ωs2(v(x))
φs(x− yv(x))ψs±(y) dy.
Thus, φs = as− + as+. Recalling the notation Sann in Theorem 1.15,
define
(4.32) A± f :=
s∈AT (ann)
scl(s)≥κ−1‖v‖Lip
〈Sann f, αs〉as±
We will write Φ = ΦSann = A+ +A− +B, where B is an operator
defined in (4.35) below. The main fact we need concerns A−.
Lemma 4.33. There is a choice of 1 < p0 < 2 so that
‖A−‖p . 1, p0 < p <∞.
The implied constant is independent of the value of ann, and the Lips-
chitz norm of v.
The proof of this Lemma is given in the next section, modulo three
additional Lemmata stated therein. The following Lemma is important
for our approach to the previous Lemma. It is proved below.
Lemma 4.34. For each choice of κ−1‖v‖Lip < scl < κann, we have the
estimate ∑
s∈AT (ann,scl)
|〈Sann f, αs〉|2 . ‖f‖22.
Define
(4.35) B f :=
s∈AT (ann)
scl(s)≥κ−1‖v‖Lip
〈Sann f, βs〉φs
Lemma 4.36. For a Lipschitz vector field v, we have
‖B‖p . 1, 2 ≤ p <∞.
Proof. For choices of integers κ−1‖v‖Lip ≤ scl < κann, consider
the vector valued operator
Tj,k f :=
{〈Sann f, βs〉√
1{v(x)∈ωs2}Tc(Rs)D
(1 + | · |2)N )(x)
| s ∈ AT (ann, scl)
where N is a fixed large integer.
PROOFS OF LEMMATA 39
Recall that βs is supported off of
γsRs. This is bounded linear op-
erator from L∞(R2) to ℓ∞(AT (ann, scl)). It has norm. (scl/‖v‖Lip)−10.
Routine considerations will verify that
Tj,k : L
2(R2) −→ ℓ2(AT (ann, scl))
with a similarly favorable estimate on its norm. By interpolation, we
achieve the same estimate for Tj,k from L
p(R2) into ℓp(AT (ann, scl)),
2 ≤ p <∞.
It is now very easy to conclude the Lemma by summing over scales
in a brute force way, and using the methods of Lemma 4.25. �
We turn to A+, as defined in (4.32).
Lemma 4.37. We have the estimate
‖A+‖p . 1 2 ≤ p <∞.
Proof. We redefine the vector valued operator Tj,k to be
Tj,k f :=
{〈Sann f, αs〉√
1{v(x)∈ωs2}Tc(Rs)D
(1 + |x|2)N
| s ∈ AT (ann, scl)
where N is a fixed large integer. This operator is bounded from
Lp(R2) −→ ℓp(AT (ann, scl)) , 2 ≤ p <∞
Its norm is at most . 1.
But, for s ∈ AT (ann, scl), we have
(4.38) |as+| . (scl/‖v‖Lip)−10|Rs|−1/2(M1Rs)100.
Here M denotes the strong maximal function in the plane in the coordi-
nates determined by Rs. This permits one to again adapt the estimate
of Lemma 4.25 to conclude the Lemma. �
Now we conclude that ‖Φ‖2 . 1. And since Φ = A− +A++B, it
follows from the Lemmata of this section.
Proofs of Lemmata
Proof of Lemma 4.33. We have Φ = A−+A+ +B, so from
(4.28), Lemma 4.36 and Lemma 4.37, we deduce that ‖A−‖p . 1 for
all 2 < p <∞. It remains for us to verify that A− is of restricted weak
type p0 for some choice of 1 < p0 < 2. That is, we should verify that
for all sets F,G ⊂ R2 of finite measure
(4.39) |〈A− 1F , 1G〉| . |F |1/p|G|1−1/p, p0 < p < 2.
40 4. L
ESTIMATE FOR Hv
Since A− maps L
p into itself for 2 < p < ∞, it suffices to consider
the case of |F | < |G|. Since we assume only that the vector field is
Lipschitz, we can use a dilation to assume that 1 < |G| < 2, and so
this set will not explicitly enter into our estimates.
We fix the data F ⊂ R2 of finite measure, ann, and vector field v
with ‖v‖Lip ≤ κann. Take p0 = 2 − κ2. We need a set of definitions
that are inspired by the approach of Lacey and Thiele [20], and are
also used in Lacey and Li [15]. For subsets S ⊂ Av := {s ∈ AT (ann) |
κ−1‖v‖Lip ≤ scl(s) < κann}, set
〈Sann 1F , αs〉as−
Set χ(x) = (1 + |x|)−1000/κ. Define
(4.40) χ
:= χ(p)s = Tc(Rs)D
χ, 0 ≤ p ≤ ∞.
And set χ̃
s = 1γsRsχ
Remark 4.41. It is typical to define a partial order on tiles, following
an observation of C. Fefferman [9]. In this case, there doesn’t seem to
be an appropriate partial order. Begin with this assumption on the
order relation ‘<’ on tiles:
(4.42)
If ωs ×Rs ∩ ωs′ ×Rs′ 6= ∅, then s and s′ are comparable under ‘<’.
It follows from transitivity of a partial order that that one can have
tiles s1, . . . , sJ , with sj+1 < sj for 1 ≤ j < J , J ≃ log(‖v‖Lip · ann), and
yet the rectangles RsJ and Rs1 can be far apart, namely RsJ ∩ (cJ)Rs1 ,
for a positive constant c. See Figure 4.3. (We thank the referee for
directing us towards this conclusion.) Therefore, one cannot make
the order relation transitive, and maintain control of the approximate
localization of spatial variables, as one would wish. The partial order
is essential to the argument of [9], but while it is used in [20], it is not
essential to that argument.
We recall a fact about the eccentricity. There is an absolute con-
stant K ′ so that for any two tiles s, s′
(4.43) ωs ⊃ ωs′ , Rs ∩ Rs′ 6= ∅ implies Rs ⊂ K ′Rs′ .
Figure 3.1 illustrates this in the case where the two rectangles Rs and
Rs′ have different widths, which is not the case here.
We define an order relation on tiles by s . s′ iff ωs ) ωs′ and
Rs ⊂ κ−10Rs′. Thus, (4.42) holds for this order relation, and it is
certainly not transitive.
PROOFS OF LEMMATA 41
Figure 4.3. The rectangles Rs1 , . . . , RsJ of Remark 4.41.
A tree is a collection of tiles T ⊂ Av, for which there is a (non–
unique) tile ωT × RT ∈ AT (ann) with Rs ⊂ 100κ−10RT, and ωs ⊃ ωT
for all s ∈ T. Here, we deliberately use a somewhat larger constant
100κ−10 than we used in the definition of the order relation ‘..’
For j = 1, 2, call T a i–tree if the tiles for all s, s′ ∈ T, if scl(s) 6=
scl(s′), then ωsi∩ωs′i = ∅. 1–trees are especially important. A few tiles
in such a tree are depicted in Figure 4.4.
Remark 4.44. This remark about the partial order ‘.’ and trees is
useful to us below. Suppose that we have two trees T, with top s(T)
and T′ with top s(T′). Suppose in addition that s(T′) . s(T). Then,
it is the case that T ∪ T′ is a tree with top s(T). Indeed, we must
necessarily have ωT ( ωT, since the Rs are from products of a central
grid. Also, 100κ−1RT′ ⊂ 100κ−1RT. And so every tile in T′ could also
be a tile in T.
Our proof is organized around these parameters and functions as-
sociated to tiles and sets of tiles. Of particular note here are the first
definitions of ‘density,’ which have to be formulated to accommodate
the lack of transitivity in the partial order. Note that in the first defi-
nition, the supremum is taken over tiles s′ ∈ AT of the same annular
parameter as s. We choose the collection AT as it is ‘universal,’ cover-
ing all scales in a uniform way, due to different assumptions including
42 4. L
ESTIMATE FOR Hv
(4.7).
dense(S) := sup
s′∈AT
G∩v−1(ωs′ )
s′ dx | ∃ s , s′′ ∈ S :
ωs ⊃ ωs′ ⊃ ωs′′ , Rs ⊂ 100κ−10Rs′ ,
Rs′ ⊂ 100κ−10Rs′
(4.45)
∆(T)2 :=
|〈Sann 1F , αs〉|2
1Rs , T is a 1–tree,(4.46)
size(S) := sup
T is a 1–tree
∆(T) dx.(4.47)
Observe that dense(S) only really applies to ‘tree-like’ sets of tiles, and
that—and this is important—the tile s′ that appear in (4.45) are not
in S, but only assumed to be in AT . Finally, note that
dense(s) ≃
G∩v−1(ωs)
χ̃(1)s dx
with the implied constants only depending upon κ, χ, and other fixed
quantities.
Observe these points about size. First, it is computed relative to
the truncated functions αs, recall (4.29). Second, that for p > 1,
(4.48) ‖∆(T)‖p . |F |1/p ,
because of a standard Lp estimate for a Littlewood-Paley square func-
tion. Third, that size(Av(ann)) . 1. And fourth, that one has an
estimate of John-Nirenberg type.
Lemma 4.49. For a 1-tree T we have the estimate
‖∆(T)‖p . size(T)|RT|1/p , 1 < p <∞ .
Proofs of results of this type are well represented in the literature.
See [4, 11].
Given a set of tiles, say that count(S) < A iff S is a union of trees
T ∈ T for which ∑
|sh(T)| < A.
We will also use the notation count(S) . A, implying the existence of
an absolute constant K for which count(S) ≤ KA.
The principal organizational Lemma is
PROOFS OF LEMMATA 43
Figure 4.4. A few possible tiles in a 1–tree. Rectangles
ωs are on the left in different shades of gray. Possible
locations of Rs are in the same shade of gray.
Lemma 4.50. Any finite collection of tiles S ⊂ Av is a union of four
subsets
Slight, Ssmall, S
large, ℓ = 1, 2.
They satisfy these properties.
size(Ssmall) <
size(S),(4.51)
dense(Slight) <
dense(S),(4.52)
and both Sℓlarge are unions of trees T ∈ T ℓ, for which we have the
estimates
count(S1large) .
size(S)
−2−κ|F |
size(S)−p dense(S)−M |F |
+ size(S)
dense(S)−1
dense(S)−1
(4.53)
count(S2large) .
size(S)
(log 1/ size(S))3|F |
size(S)κ/50 dense(S)−1
(4.54)
What is most important here is the middle estimate in (4.53). Here,
p is as in Conjecture 1.14, and M > 0 is only a function of N in that
Conjecture.
The estimates that involve size(S)−2|F | are those that follow from
orthogonality considerations. The estimates in dense(S)−1 are those
that follow from density considerations which are less complicated.
However, in the second half of (4.54), the small positive power of size
is essential for us. All of these estimates are all variants of those in
[20].
The middle estimate of (4.53) is not of this type, and is the key
ingredient that permits us to obtain an estimate below L2. Note that
44 4. L
ESTIMATE FOR Hv
it gives the best bound for collections with moderate density and size.
For it we shall appeal to our assumed Conjecture 1.14.
Logarithms, such as those that arise in (4.54), arise from our trun-
cation arguments, associated with the parameters γs in (4.29).
For individual trees, we need two estimates.
Lemma 4.55. If T is a 1–tree with −
∆(T) ≥ σ, then we have
(4.56) |F ∩ σ−κRT| & σ1+κ|RT|.
Lemma 4.57. For trees T we have the estimate
(4.58)
|〈Sann1F , αs〉〈as−, 1G〉| . Ψ
dense(T) size(T)
|sh(T)|.
Here Ψ(x) = x|log cx|, and inside the logarithm, c is a small fixed
constant, to insure that c dense(T) · size(T) < 1
, say.
Sum(S) :=
|〈Sann1F , αs〉〈as−, 1G〉|
We want to provide the bound Sum(Av) . |F |1/p for p0 < p < 2. We
have the trivial bound
(4.59) Sum(S) . Ψ
dense(S) size(S)
count(S).
It is incumbent on us to provide a decomposition of Av into sub-
collections for which this last estimate is effective.
By inductive application of our principal organizational Lemma 4.50,
Av is the union of Sℓδ,σ, ℓ = 1, 2 for δ, σ ∈ 2 := {2n | n ∈ Z , n ≤ 0},
satisfying
dense(Sℓδ,σ) . δ,(4.60)
size(Sℓδ,σ) . σ,(4.61)
count(Sℓδ,σ) .
min(σ−2−κ|F |, δ−Mσ−p|F |+ σ1/κδ−1, δ−1) ℓ = 1,
min(σ−2(log 1/σ)3|F |, δ−1σκ/50) ℓ = 2
(4.62)
Using (4.59), we see that
Sum(S1δ,σ) . min(Ψ(δ)σ
−1−κ|F |, δ−M+1σ−p+1|F |+ σ1/κ+1, σ)
Sum(S2δ,σ) . min(Ψ(δ)σ
−1(log 1/σ)4|F |, σ1+κ/50)
(4.63)
One can check that for ℓ = 1, 2,
(4.64)
δ,σ∈2
Sum(Sℓδ,σ) . |F |1/p, p0 < p < 2.
PROOFS OF LEMMATA 45
This completes the proof of Lemma 4.33, aside from the proof of
Lemma 4.50.
Proof of (4.64). We can assume that |G| = 1, and that |F | ≤ 1,
for otherwise the result follows from the known Lp estimates, for p > 2
and measurable vector fields, see Theorem 1.15.
The case of ℓ = 2 in (4.64) is straightforward. Notice that in (4.63),
for ℓ = 2, the two terms in the minimum are roughly comparable,
ignoring logarithmic terms, for
δ|F | ≃ σ2+κ/50 .
Therefore, we set
T1 = {(δ, σ) ∈ 2× 2 | δ|F | ≤ σ2+κ/50 ≤ |F |} ,
T2 = {(δ, σ) ∈ 2× 2 | σ2+κ/50 ≤ δ|F |}
and T3 = 2× 2− T1 − T2.
We can estimate
(δ,σ)∈T1
Sum(S2δ,σ) .
(δ,σ)∈T1
Ψ(δ)σ−1(log 1/σ)4|F |
σ2+κ/50≤|F |
σ1+κ/75
. |F |1/p0 , p0 =
2 + κ/50
1 + κ/75
< 2 .
Notice that we have absorbed harmless logarithmic terms into a slightly
smaller exponent in σ above.
The second term is
(δ,σ)∈T2
Sum(S2δ,σ) .
(δ,σ)∈T2
σ1+κ/50
(δF )1/p1 , p1 =
2 + κ/50
1 + κ/50
< 2 ,
. |F |1/p1 .
46 4. L
ESTIMATE FOR Hv
The third term is
(δ,σ)∈T3
Sum(S2δ,σ) .
(δ,σ)∈T3
Ψ(δ)σ−1(log 1/σ)4|F |
σ2+κ/50≥|F |
σ−1|F |1−κ/75
. |F |1/p0 .
Here, we have again absorbed harmless logarithms into a slightly smaller
power of |F |, and p0 < 2 is as in the first term.
The novelty in this proof is the proof of (4.64) in the case of ℓ = 1.
We comment that if one uses the proof strategy just employed, that
is only relying upon the first and last estimates from the minimum in
(4.63), in the case of ℓ = 1, one will only show that |F |1/2.
In the definitions below, we will have a choice of 0 < τ < 1, where
τ = τ(M, p) ≃ M−1·(2−p) will only depend uponM and p in (4.63). (τ
enters into the definition of T4 and T5 below.) The choice of 0 < κ < τ
will be specified below.
T1 = {(δ, σ) ∈ 2× 2 | |F |
(2+κ)(1+κ) ≤ σ} ,
T2 = {(δ, σ) ∈ 2× 2 | σ < |F |
2−κ , δ ≥ σ1/κ} ,
T3 = {(δ, σ) ∈ 2× 2 | σ < |F |
2−κ , δ > σ1/κ} ,
T4 = {(δ, σ) ∈ 2× 2 | |F |
2−κ ≤ σ < |F |
(2+κ)(1+κ) , δ > στ} ,
T5 = {(δ, σ) ∈ 2× 2 | |F |
2−κ ≤ σ < |F |
(2+κ)(1+κ) , δ ≤ στ} ,
T (T ) =
(δ,σ)∈T
Sum(S1δ,σ) .
Note that for T1 we can use the first term in the minimum in (4.63).
T (T1) .
(δ,σ)∈T1
δσ−1−κ|F |
σ≥|F |
(2+κ)(1+κ)
σ−1−κ|F |
. |F |1−
2+κ .
This last exponent on |F | is strictly larger than 1
as desired. The point
of the definition of T1 is that when it comes time to use the middle term
PROOFS OF LEMMATA 47
of the minimum for ℓ = 1 in (4.63), we can restrict attention to the
δ−M+1σ−p+1|F | .
For the collection T2, use the last term in the minimum in (4.63).
T (T2) .
(δ,σ)∈T2
σ≤|F |
σ log 1/σ
. |F |
2−κ/2 .
Again, for 0 < κ < 1, the exponent on |F | above is strictly greater
than 1/2.
The term T3 can be controlled with the first term in the minimum
in (4.63).
T (T3) .
(δ,σ)∈T3
δσ−1−κ|F |
σ|F | . |F | .
The term T4 is the heart of the matter. It is here that we use the
middle term in the minimum of (4.63), and that the role of τ becomes
clear. We estimate
T (T4) .
(δ,σ)∈T4
δ−Mσ−p+1|F |
δ≥|F |τ
δ−M |F |1−
. |F |1−
−Mτ .
Recall that 1 < p < 2, so that 0 < p − 1 < 1. Therefore, for 0 < κ
sufficiently small, of the order of 2− p, we will have
1− p− 1
2− κ >
+ 2−p
Therefore, choosing τ ≃ (2 − p)/M will leave us with a power on |F |
that is strictly larger than 1
The previous term did not specify κ > 0. Instead it shows that
for 0 < κ < 1 sufficiently small, we can make a choice of τ , that is
48 4. L
ESTIMATE FOR Hv
independent of κ, for which T (T4) admits the required control. The
bound in the last term will specify a choice of κ on us. We estimate
T (T5) .
(δ,σ)∈T5
δσ−1−κ|F |
σ≥|F |
σ−1−κ|F |1+τ
. |F |1+τ+
Choosing κ = τ/6 will result in the estimate
which is as required, so our proof is finished. �
Remark 4.65. The resolution of Conjecture 1.21 would depend upon
refinements of Lemma 4.50, as well as using the restricted weak type
approach of [21].
Proof of Lemma 4.34. We only consider tiles s ∈ AT (ann, scl),
and sets ω ∈ Ω which are associated to one of these tiles. For an
element a = {as} ∈ ℓ2(AT (ann, scl)),
s :ωs=ω
asSannαs
For |ωs| = |ωs′|, note that dist(ωs,ωs′) is measured in units of scl/ann.
By a lemma of Cotlar and Stein, it suffices to provide the estimate
′‖2 . ρ−3, ρ = 1 +
dist(ω,ω′).
Now, the estimate ‖T
‖2 . 1 is obvious. For the case ω 6= ω′, by
Schur’s test, it suffices to see that
(4.66) sup
s′ :ωs′=ω
s :ωs=ω
|〈Sannαs, Sannαs′〉| . ρ−3.
For tiles s′ and s as above, recall that 〈ϕs, ϕs′〉 = 0, note that
|Rs′ ∩ Rs|
ann dist(ω,ω′)
≃ ρ−1,
and in particular, for a fixed s′, let Ss′ be those s for which ρRs∩ρRs′ 6=
∅. Clearly,
card(Ss′) .
|ρRs|
|2ρRs′ ∩ 2ρRs|
ρ ≃ ρ2
PROOFS OF LEMMATA 49
If for r > 1, rRs ∩ rRs′ = ∅, then it is routine to show that
|〈Sannαs, Sannαs′〉| . r−10
And so we may directly sum over those s 6∈ Ss′ ,
s 6∈Ss′
|〈Sannαs, Sannαs′〉| . ρ−3.
For those s ∈ Ss′, we estimate the inner product in frequency vari-
ables. Recalling the definition of αs = (Tc(Rs)D
ζ)ϕs, we have
α̂s = (Mod−c(Rs) D
ζ̂) ∗ ϕ̂s.
Recall that ζ is a smooth compactly supported Schwartz function. We
estimate the inner product
|〈Ŝannαs, Ŝannαs′〉|
without appealing to cancellation. Since we choose the function λ̂ to
be supported in an annulus 1
< |ξ| < 3
so that λ̂ann = λ̂(·/ann) is
supported in the annulus 1
ann < |ξ| < 3
ann. We can restrict our
attention to this same range of ξ. In the region |ξ| > ann/4, suppose,
without loss of generality, that ξ is closer to ωs than ωs′. Then since
ωs and ωs′ are separated by an amount & anndist(ω,ω
|α̂s(ξ)α̂s′(ξ)| . χ(2)ωs (ξ)χ
dist(ω,ω)
. χ(2)ωs (ξ)χ
(ξ)ρ−20.
Here, χ is the non–negative bump function in (4.40). Hence, we have
the estimate ∫
|λ̂ann(ξ)|2|α̂s(ξ)α̂s′(ξ)|dξ . ρ−10.
This is summed over the . ρ2 possible choices of s ∈ Ss′ , giving the
estimate ∑
s∈Ss′
|〈Sannαs, Sannαs′〉| . ρ−8 . ρ−3.
This is the proof of (4.66). And this concludes the proof of Lemma 4.34.
Proof of the Principal Organizational Lemma 4.50. Recall
that we are to decompose S into four distinct subsets satisfying the
favorable estimates of that Lemma. For the remainder of the proof set
dense(S) := δ and size(S) := σ. Take Slight to be all those s ∈ S for
which there is no tile s′ ∈ AT of density at least δ/2 for which s . s′.
It is clear that this set so constructed has density at most δ/2, that
this is a set of tiles, and that S1 := S− Slight is also .
50 4. L
ESTIMATE FOR Hv
The next Lemma and proof comment on the method we use to
obtain the middle estimate in (4.53) which depends upon the Lipschitz
Kakeya Maximal Function Conjecture 1.14. It will be used to obtain
the important inequality (4.82) below.
Lemma 4.67. Suppose we have a collection of trees T ∈ T , with these
conditions.
a: For T ∈ T there is a 1-tree T1 ⊂ T with
(4.68)
∆(T1) dx ≥ κσ ,
b: Each tree has top element s(T) := ωT×RT of density at least
c: The collections of tops {s(T) | T ∈ T } are pairwise incompa-
rable under the order relation ‘.’.
d: For all T ∈ T , γT = γωT×RT ≥ κ−1/2σ−κ/5N . Here, N is the
exponent on δ in Conjecture 1.14.
Then we have
|RT| . δ−Np−1σ−p(1+κ/4)|F |+ σ1/κδ−1.(4.69)
|RT| . δ−1.(4.70)
Concerning the role of γT, recall from the definition, (4.29), that
γs is a quantity that grows as does the ratio scl(s)/‖v‖Lip, hence there
are only . log σ−1 scales of tiles that do not satisfy the assumption d
above.
Proof. Our primary interest is in (4.69), which is a consequence of
our assumption about the Lipschitz Kakeya Maximal Functions, Con-
jecture 1.12.
s(T) := ωT × σ−κ/10NRT .
Let us begin by noting that
κ−1‖v‖Lip ≤ scl(s(T)) ≤ κann(s(T)), T ∈ T ,(4.71)
dense(s(T)) ≥ δσκ/10N , T ∈ T ,(4.72)
|F ∩Rs(T)| ≥ σ1+κ/4N |Rs(T)| .(4.73)
The conclusion (4.71) is straightforward, as is (4.72). The inequality
(4.73) follows from Lemma 4.55.
PROOFS OF LEMMATA 51
Note that the length of σ−κ/10NRT satisfies
σ−κ/10N L(RT) ≤ γT L(RT)
scl(s)
‖v‖Lip
≤ (100‖v‖Lip)−1 .
(4.74)
This is the condition (1.8) that we impose in the definition of the
Lipschitz Kakeya Maximal Functions.
Observe that we can regard ann(s(T)) ≃ σκ/10ann as a constant
independent of T.
The point of these observations is that our assumption about the
Lipschitz Kakeya Maximal Function applies to the maximal function
formed over the set of tiles {s(T) | T ∈ T }. And it will be applied
below.
Let Tk be the collection of trees so that T ∈ Tk if k ≥ 0 is the
smallest integer such that
(4.75) |(2kRT) ∩ v−1(ωT) ∩G| ≥ 220k/κ
2−1δ|RT| .
Then since the density of s(T) for every tree T ∈ T is at least δ, we
have T =
k=0 Tk. We can apply Conjecture 1.12 to these collections,
with the value of δ in that Conjecture being 220k/κ
2−1δ.
For each Tk, we decompose it by the following algorithm. Initialize
T selectedk ← ∅, T stockk ← Tk .
While T stockk 6= ∅, select T ∈ T stockk such that scl(s(T)) is minimal.
Define Tk(T) by
Tk(T) = {T′ ∈ Tk : (2kRT) ∩ (2kRT′) 6= ∅ and ωT ⊂ ωT′} .
Update
T selectedk ← T selectedk ∪ {T} , T stockk ← T stockk \Tk(T) .
Thus we decompose Tk into
T∈T selected
T′∈Tk(T)
{T′} .
And ∑
|RT| =
T∈T selected
T′∈Tk(T)
|RT′| .
Notice that RT′ ’s are disjoint for all T
′ ∈ Tk(T) and they are contained
in 5(2kRT). This is so, since the tops of the trees are assumed to be
incomparable with respect to the order relation ‘.’ on tiles.
52 4. L
ESTIMATE FOR Hv
Thus we have
|RT| .
T∈T selected
22k|RT|
. δ−12−10k/κ
T∈T selected
|(2kRT) ∩ v−1(ωT) ∩G| .
Observe that (2kRT)∩v−1(ωT)’s are disjoint for all T ∈ T selectedk . This
and the fact that |G| ≤ 1 proves (4.70). To argue for (4.69), we see
|RT| . δ−12−10k/κ
T∈T selected
(2kRT) ∩ v−1(ωT) ∩G
. δ−12−10k/κ
(2kRT) ∩G
∣∣∣∣ .
At this point, Conjecture 1.12 enters. Observe that we can estimate
∣∣∣ . |{Mδ′,v,(σκ/10ann)−1 1F > σ1+κ/4N}|
. (δ′)−Npσ−p(1+κ/4N)|F |.
. (δ)−Np2−kσ−p(1+κ/4N)|F |.
(4.76)
Here, δ′ = 220k/κ
2−1δ, the choice of δ′ permitted to us by (4.75), and
we have used (4.73) in the first line, to pass to the Lipschitz Kakeya
Maximal Function.
Hence,
|sh(T)| .
. δ−1
2−10k/κ
(2kRT) ∩G
. δ−1
k : 1≤2k≤σ−κ/10
(2kRT)
+ δ−1
k : 2k>σ−κ/10
2−10k/κ
2 |G| .
On the first sum in the last line, we use (4.76), and on the second, we
just sum the geometric series, and recall that |G| = 1.
PROOFS OF LEMMATA 53
We can now begin the principal line of reasoning for the proof of
Lemma 4.50.
The Construction of S1large. We use an orthogonality, or TT
∗ argu-
ment that has been used many times before, especially in [20] and [15].
(There is a feature of the current application of the argument that is
present due to the fact that we are working on the plane, and it is
detailed by Lacey and Li [15].)
We may assume that all intervals ωs are contained in the upper
half of the unit circle in the plane. Fix S ⊂ Av, and σ = size(S).
We construct a collection of trees T 1large for the collection S1, and
a corresponding collection of 1–trees T 1,1large, with particular properties.
We begin the recursion by initializing
T 1large ← ∅, T
large ← ∅,
S1large ← ∅, Sstock ← S1.
In the recursive step, if size(Sstock) < 1
σ1+κ/100, then this recursion
stops. Otherwise, we select a tree T ⊂ Sstock such that three conditions
are met.
a: The top of the tree s(T) (which need not be in the tree)
satisfies dense(s(T)) ≥ δ/4.
b: T contains a 1–tree T1 with
(4.77) −
∆(T1) dx ≥ 1
σ1+κ/100 .
c: And that ωT is in the first place minimal and and in the
second most clockwise among all possible choices of T. (Since
all ωs are in the upper half of the unit circle, this condition
can be fulfilled.)
We take T to be the maximal tree in Sstock which satisfies these condi-
tions.
We then update
T 1large ← {T} ∪ Tlarge, T
large ← {T1} ∪ T
large,
S1large ← T ∪ S1large Sstock ← Sstock −T.
The recursion then repeats. Once the recursion stops, we update
S1 ← Sstock
It is this collection that we analyze in the next subsection.
Note that it is a consequence of the recursion, and Remark 4.44,
that the tops of the trees {s(T) | T ∈ T 1large} are pairwise incomparable
under ..
54 4. L
ESTIMATE FOR Hv
The bottom estimate of (4.53) is then immediate from the construc-
tion and (4.70).
First, we turn to the deduction of the first estimate of (4.53). Let
T 1,(1)large be the set
T 1,(1)large =
T ∈ T 1large :
|〈Sann 1F , βs〉|2 < 116σ
2+κ/50|RT|
And let T 1,(2)large be the set
T 1,(2)large =
T ∈ T 1large :
|〈Sann 1F , βs〉|2 ≥ 116σ
2+κ/50|RT|
In the inner products, we are taking βs, which is supported off of γsRs.
Since T ∈ T 1large satisfies
(4.78) −
∆(T) dx ≥ 1
σ1+κ/100 ,
we have ∑
|〈Sann 1F , αs〉|2 ≥ 14σ
2+κ/50|RT| .
Thus, if T ∈ T 1,(1)large , we have
|〈1F , ϕs〉|2 ≥ 18σ
2+κ/50|RT| .
The replacement of αs by ϕs in the inequality above is an important
point for us. That we can then drop the Sann is immediate.
With this construction and observation, we claim that
(4.79)
T∈T 1,(1)
large
|RT| . (log 1/σ)2σ−2−κ/50|F |.
Proof of (4.79). This is a variant of the the argument for the
‘Size Lemma’ in [15], and so we will not present all details. Begin by
making a further decomposition of the trees T ∈ T 1,(1)large . To each such
tree, we have a 1-tree T1 ⊂ T which satisfies (4.77). We decompose
T1. Set
T1(0) =
s ∈ T1 | |〈f, ϕs〉|√
< σ1+κ/100
T1(j) =
s ∈ T1 | 4j−1σ1+κ/100 ≤ |〈f, ϕs〉|√
< 4jσ1+κ/100
1 ≤ j ≤ j0 = C log 1/σ .
PROOFS OF LEMMATA 55
Now, set T (j) to be those T ∈ T 1,(1)large for which
(4.80)
s∈T1(j)
|〈f, ϕs〉|2 ≥ (2j0)−1σ2+κ/50|RT| , 0 ≤ j ≤ j0 .
It is the case that each T ∈ T 1,(1)large is in some T (j), for 0 ≤ j ≤ j0.
The central case is that of j = 0. We can apply the ‘Size Lemma’
of [15] to deduce that
T∈T (0)
|RT| ≤ (2j0)σ−2−κ/50
T∈T (0)
s∈T1(0)
|〈f, ϕs〉|2
. (log 1/σ)σ−2−κ/50|F | .
The point here is that to apply the argument in the ‘Size Lemma’ one
needs an average case estimate, namely (4.80), as well as a uniform
control, namely the condition defining T1(0). This proves (4.79) in
this case.
For 1 ≤ j ≤ j0, we can apply the ‘Size Lemma’ argument to the
individual tiles in the collection
{T1(j) | T ∈ T (j)} .
The individual tiles satisfy the definition of a 1-tree. And the defining
condition of T1(j) is both the average case estimate, and the uniform
control needed to run that argument. In this case we conclude that
T∈T (j)
s∈T1(j)
|〈f, ϕs〉|2 . |F | .
Thus, we can estimate
T∈T (j)
|RT| . (log 1/σ)σ−1−κ/50|F | .
This summed over 1 ≤ j ≤ j0 = C log 1/σ proves (4.79). �
For T 1,(2)large , we have
T∈T 1,(2)
large
|RT| . σ−2−κ/50
scl≥κ−1‖v‖Lip
s:scl(s)=scl
|〈Sann 1F , βs〉|2
. σ−2−κ/50|F |
scl≥κ−1‖v‖Lip
‖v‖Lip
. σ−2−κ/50|F | ,
56 4. L
ESTIMATE FOR Hv
since βs has fast decay. The Bessel inequality in the last display can
be obtained by using the same argument in the proof of Lemma 4.34.
Hence we get
(4.81)
T∈T 1,(2)
large
|RT| . σ−2−κ/50|F |.
Combining (4.79) and (4.81), we obtain the first estimate of (4.53).
Second, we turn to the deduction of the middle estimate of (4.53),
which relies upon the Lipschitz Kakeya Maximal Function. Let T 1,goodlarge
be the set
T ∈ T 1large : γT ≥ κ−1/2σ−κ/5N
And let T 1,badlarge be the set
T ∈ T 1large : γT < κ−1/2σ−κ/5N
The ‘good’ collection can be controlled by facts which we have already
marshaled together. In particular, we have been careful to arrange the
construction so that Lemma 4.67 applies. By the main conclusion of
that Lemma, (4.69), we have
(4.82)
T∈T 1,good
large
|RT| . δ−Mσ−1−3κ/4|F |+ σ1/κδ−1 .
Here, M is a large constant that only depends upon N in Conjec-
ture 1.14.
For T ∈ T 1,badlarge , there are at most K = O(log(σ−κ)) many possible
scales for scl(ωT × RT). Let scl(T) = scl(ωT × RT). Thus we have
T∈T 1,bad
large
|RT| .
T:scl(T)=2mκ−1‖v‖Lip
|RT| .
Since T satisfies (4.78), we have
|F ∩ γTRT| & σ1+κ/2|RT| .
Thus, we get
T∈T 1,bad
large
|RT| . σ−1−κ/2
T:scl(T)=2mκ−1‖v‖Lip
1σ−κRT(x)dx .
PROOFS OF LEMMATA 57
For the tiles with a fixed scale, we have the following inequality, which
is a consequence of Lemma 4.25.
T:scl(T)=2mκ−1‖v‖Lip
1σ−κRT
. σ−κ/5δ−1 .
Hence we obtain
(4.83)
T∈T 1,bad
large
|RT| . δ−1σ−1−3κ/4|F | .
Combining (4.82) and (4.83), we obtain the middle estimate of
(4.53). Therefore, we complete the proof of (4.53).
The Construction of S2large. It is important to keep in mind that we
have only removed trees of nearly maximal size, with tops of a given
density. In the collection of tiles that remain, there can be trees of
large size, but they cannot have a top with nearly maximal density.
We repeat the TT∗ construction of the previous step in the proof, with
two significant changes.
We construct a collection of trees T 2large from the collection S1, and
a corresponding collection of 1–trees T 2,1large, with particular properties.
We begin the recursion by initializing
T 2large ← ∅ , T
large ← ∅ ,
S2large ← ∅ , Sstock ← S1 .
In the recursive step, if size(Sstock) < σ/2, then this recursion stops.
Otherwise, we select a tree T ⊂ Sstock such that two conditions are
a: T satisfies ‖∆(T)‖2 ≥ σ2 |RT|
b: ωT is both minimal and most clockwise among all possible
choices of T.
We take T to be the maximal tree in Sstock which satisfies these condi-
tions. We take T1 ⊂ T to be a 1–tree so that
(4.84) −
∆(T1) dx ≥ κσ .
This last inequality must hold by Lemma 4.49.
We then update
T 2large ← {T} ∪ Tlarge, T
large ← {T1} ∪ T
large,
Sstock ← Sstock −T.
The recursion then repeats.
58 4. L
ESTIMATE FOR Hv
Once the recursion stops, it is clear that the size of Sstock is at most
σ/2, and so we take Ssmall := Sstock.
The estimate ∑
T∈T 2
large
|RT| . σ−2|F |
then is a consequence of the TT ∗ method, as indicated in the previous
step of the proof. That is the first estimate claimed in (4.54).
What is significant is the second estimate of (4.54), which involves
the density. The point to observe is this. Consider any tile s of density
at least δ/2. Let Ts be those trees T ∈ T 2large with top ωs(T) ⊃ ωs and
Rs(T) ⊂ KRs. By the construction of S1large, we must have
∆(T1) dx ≤ σ1+κ/100 ,
for the maximal 1–tree T 1 contained in
T∈Ts T. But, in addition, the
tops of the trees in T 2large are pairwise incomparable with respect to the
order relation ‘.,’ hence we conclude that
|RT| . σ2+κ/50|Rs|.
Moreover, by the construction of Slight, for each T ∈ T 2large we must be
able to select some tile s with density at least δ/2 and ωs(T) ⊃ ωs and
Rs(T) ⊂ KRs.
Thus, we let S∗ be the maximal tiles of density at least δ/2. Then,
the inequality (4.70) applies to this collection. And, therefore,
T∈T 2
large
|RT| ≤ σκ/50
|Rs| . σκ/50δ−1.
This completes the proof of second estimate of (4.54). �
The Estimates For a Single Tree.
The Proof of Lemma 4.55. It is a routine matter to check that for
any 1–tree we have ∑
|〈f, ϕs〉|2 . ‖f‖22.
Indeed, there is a strengthening of this estimate relevant to our concerns
here. Recalling the notation (4.40), we have
(4.85)
|〈f, ϕs〉|2
]1/2∥∥∥
. ‖χ(∞)RT f‖p , 1 < p <∞ .
PROOFS OF LEMMATA 59
This is variant of the Littlewood-Paley inequalities, with some addi-
tional spatial localization in the estimate.
Using this inequality for p = 1 + κ/100 and the assumption of the
Lemma, we have
σ1+κ/100 ≤
∆T dx
]1+κ/100
1+κ/100
≤ |RT|−1
|〈f, ϕs〉|2
]1/2∥∥∥
1+κ/100
1+κ/100
. |RT|−1
dx.(4.86)
This inequality can only hold if |F ∩ σ−κRT| ≥ σ1+κ|RT|. �
The Proof of Lemma 4.57. This Lemma is closely related to the
Tree Lemma of [15]. Let us recall that result in a form that we need
it. We need analogs of the definitions of density and size that do not
incorporate truncations of the various functions involved. Define
dense(s) :=
G∩v−1(ωs)
(x) dx.
(Recall the notation from (4.40).)
dense(T) := sup
dense(s).
Likewise define
size(T) := sup
′ is a 1–tree
|RT′|−1
|〈1F , ϕs〉|2
Then, the proof of the Tree Lemma of [15] will give us this inequality:
For T a tree,
(4.87)
|〈Sann 1F , ϕs〉〈φs, 1G〉| . dense(T) size(T)|RT|.
Now, consider a tree T with dense(T) = δ, and size(T) = σ, where
we insist upon using the original definitions of density and size. If
in addition, γs ≥ K(σδ)−1 for all s ∈ T, we would then have the
inequalities
dense(T) . δ,
size(T) . σ,
60 4. L
ESTIMATE FOR Hv
This places (4.87) at our disposal, but this is not quite the estimate we
need, as the functions ϕs and φs that occur in (4.87) are not truncated
in the appropriate way, and it is this matter that we turn to next.
Recall that
ϕs = αs + βs ,
αs(x− yv(x))ψs(y) dy = αs−(x) + αs+(y) .
One should recall the displays (4.30), (4.31), and (4.38).
As an immediate consequence of the definition of βs, we have∫
|βs(x)| dx . γ−2s
|Rs|.
Hence, if we replace ϕs by βs, we have
|〈Sann 1F , βs〉〈φs, 1G〉| .
|Rs||〈φs, 1G〉|
γ−1s |Rs|
. σδ|RT|.
And by a very similar argument, one sees corresponding bounds, in
which we replace the φs by different functions. Namely, recalling the
definitions of as± in (4.31) and estimate (4.38), we have
|Rs||〈as+, 1G〉| . σ
(‖v‖Lip
scl(s)
(x) dx(4.88)
(‖v‖Lip
scl(s)
. σδ|RT| .
Similarly, we have
|Rs||〈φs − as+ − as−, 1G〉| . σδ|RT|,
Putting these estimates together proves our Lemma, in particular (4.58),
under the assumption that γs ≥ K(σδ)−1 for all s ∈ T.
Assume that T is a tree with scl(s) = scl(s′) for all s, s′ ∈ T. That
is, the scale of the tiles in the tree is fixed. Then, T is in particular a 1–
tree, so that by an application of the definitions and Cauchy–Schwartz,
|〈Sann 1F , αs〉〈as−, 1G〉| ≤ δ
|〈Sann 1F , αs〉|
≤ δσ|RT|.
PROOFS OF LEMMATA 61
But, γs ≥ 1 increases as does
scl(s). Thus, any tree T with γs ≤
K(σδ)−1 for all s ∈ T, is a union of O(|log δσ|) trees for which the last
estimate holds. �
CHAPTER 5
Almost Orthogonality Between Annuli
Application of the Fourier Localization Lemma
We are to prove Lemma 4.23, and in doing so rely upon a technical
lemma on Fourier localization, Lemma 5.56 below. We can take a
choice of 1 < α < 9
, and assume, after a dilation, that ‖v‖Cα = 1.
The first inequality we establish is this.
Lemma 5.1. Using the notation of of Lemma 4.23, and assuming that
‖v‖Cα . 1, we have the estimate ‖C‖2 . 1, where
ann≥1
where the Cann are defined in (4.21).
We have already established Lemma 4.22, and so in particular know
that ‖Cann‖2 . 1. Due to the imposition of the Fourier restriction in
the definition of these operators, it is immediate that CannC∗ann′ ≡ 0 for
ann 6= ann′. We establish that
Cann′‖2 . max(ann, ann′)−δ ,
δ = 1
(α− 1) , |log ann(ann′)−1| > 3 .
(5.2)
Then, it is entirely elementary to see that C is a bounded operator. Let
Pann be the Fourier projection of f onto the frequencies ann < |ξ| <
2ann. Observe,
‖Cf‖22 =
ann≥1
Cann Pann f
ann≥1
ann′>1
〈Cann Pann f, Cann′ Pann′ f〉
≤ 2‖f‖2
ann≥1
ann′>1
Cann′ Pann′ f‖2
. ‖f‖22
ann≥1
ann′>1
max(ann, ann′)−δ
. ‖f‖22.
64 5. ALMOST ORTHOGONALITY
There are only O(log ann) possible values of scl that contribute to
Cann, and likewise for Cann′. Thus, if we define
(5.3) Cann,sclf =
s∈AT (ann)
scl(s)=scl
〈f, ϕs〉φs ,
it suffices to prove
Lemma 5.4. Using the notation of of Lemma 4.23, and assuming that
‖v‖Cα . 1, we have
ann,sclCann′,scl′‖2 . (max(ann, ann′))−δ .
Here, we can take δ′ = 1
(α − 1), and the inequality holds for all
|log ann(ann′)−1| > 3, 1 < scl ≤ ann and 1 < scl′ ≤ ann′.
Proof of Lemma 4.23. In this proof, we assume that Lemma 5.1
and Lemma 5.4 are established. The first Lemma clearly establishes
the first (and more important) claim of the Lemma.
Let us prove the inequality (4.24). Using the notation of this sec-
tion, this inequality is as follows.
(5.5)
ann=−∞
Cann,scl
. (1 + log(1 + scl−1‖v‖Cα)).
This inequality holds for all choices of Cα vector fields v.
Note that Lemma 5.4 implies immediately
ann=3
Cann,scl
. 1 , ‖v‖Cα = 1 .
We are however in a scale invariant situation, so that this inequality
implies this equivalent form, independent of assumption on the norm
of the vector field.
(5.6)
ann≥8‖v‖Cα
Cann,scl
. 1 .
On the other hand, Lemma 4.25, implies that independent of any
assumption other than measurability, we have have the inequality
‖Cann,scl‖2 . 1 .
To prove (5.5), use the inequality (5.6), and this last inequality together
with the simple fact that for a fixed value of scl, there are at most
. 1 + log(1 + scl−1‖v‖Cα)) values of ann with scl ≤ ann ≤ 8‖v‖Cα.
APPLICATION OF THE FOURIER LOCALIZATION LEMMA 65
We use the notation
AT (ann, scl) := {s ∈ AT (ann) : scl(s) = scl} ,
Observe that as the scale is fixed, we have a Bessel inequality for the
functions {ϕs | s ∈ AT (ann, scl)}. Thus,
ann,sclCann′,scl′f‖22 =
s∈AT (ann,scl)
s∈AT (ann′,scl′)
〈φs, φs′〉〈ϕs′, f〉ϕs
s∈AT (ann,scl)
s∈AT (ann′,scl′)
〈φs, φs′〉〈ϕs′, f〉
At this point, the Schur test suggests itself, and indeed, we need a
quantitative version of the test, which we state here.
Proposition 5.7. Let A = {ai,j} be a matrix acting on ℓ2(N) by
ai,jxj
Then, we have the following bound on the operator norm of A.
‖A‖2 . sup
|ai,j| · sup
|ai,j|
We assume that 1 ≤ ann < 1
′. For a subset S ⊂ AT (ann, scl)×
AT (ann′, scl′) Consider the operator and definitions below.
AS f =
(s,s′)∈S
〈φs, φs′〉〈ϕs′, f〉ϕs ,
FL(s,S) =
s′∈AT (ann′,scl′)
|〈φs, φs′〉| ,
FL(S) = sup
FL(s,S) .
Here ‘FL’ is for ‘Fourier Localization’ as this term is to be controlled
by Lemma 5.56. We will use the notations FL(s′,S), and FL′(S),
which are defined similarly, with the roles of s and s′ reversed. By
Proposition 5.7, we have the inequality
(5.8) ‖AS‖22 . FL(S) · FL′(S) .
We shall see that typically FL(S) will be somewhat large, but is bal-
anced out by FL′(S).
We partition AT (ann, scl) × AT (ann′, scl′) into three disjoint sub-
collections Su, u = 1, 2, 3, defined as follows. In this display, (s, s′) ∈
66 5. ALMOST ORTHOGONALITY
AT (ann, scl)×AT (ann′, scl′).
(s, s′) | scl
≥ scl
,(5.9)
(s, s′) | scl
, scl < scl′
,(5.10)
(s, s′) | scl
, scl′ < scl
.(5.11)
A further modification to these collections must be made, but it
is not of an essential nature. For an integer j ≥ 1, and (s, s′) ∈ Su,
for u = 1, 2, 3, write (s, s′) ∈ Su,j if j is the smallest integer such that
2j+2Rs ∩ 2j+2Rs′ 6= ∅.
We apply the inequalities (5.8) to the collections Su,j, to prove the
inequalities
(5.12) ‖ASu,j‖2 . 2−j(ann′)−δ
where δ′ = 1
(α − 1). This proves Lemma 5.4, and so completes the
proof of Lemma 5.1.
In applying (5.8) it will be very easy to estimate FL(s,S), with
a term that decreases like say 2−10j. The difficult part is to estimate
either FL(s,S) or FL′(S) by a term with decreases faster than a small
power of (ann′)−1. for which we use Lemma 5.56.
Considering a term 〈φs, φs′〉, the inner product is trivially zero if
ωs ∩ ωs′ = ∅. We assume that this is not the case below. To apply
Lemma 5.56, fix e ∈ ω′s ∩ωs. Let α be a Schwartz function on R with
α̂ supported on [ann′, 2ann′], and identically one on 3
[ann′, 2ann′]. Set
β̂(θ) := α̂(θ − 3
′). We will convolve φs with β in the direction e,
and φs′ with α also in the direction e, thereby obtaining orthogonal
functions.
Define
Ie g(x) =
g(x− ye)β(y) dy,(5.13)
∆s = φs − Ie φs
∆s′ = φs′ − Ie φs′
(5.14)
By construction, we have
〈φs, φs′〉 = 〈Ie φs +∆s, Ie φs′ +∆s′〉
= 〈Ie φs,∆s′〉+ 〈∆s, Ie φs′〉+ 〈∆s,∆s′〉 .
APPLICATION OF THE FOURIER LOCALIZATION LEMMA 67
It falls to us to estimate terms like
s∈Sℓ,j
|〈∆s, Ie φs′〉|,(5.15)
s∈Sℓ,j
|〈Ie φs,∆s′〉|,(5.16)
s∈Sℓ,j
|〈∆s,∆s′〉|.(5.17)
as well as the dual expressions, with the roles of s and s′ reversed.
The differences ∆s and ∆s′ are frequently controlled by Lemma 5.56.
Concerning application of this Lemma to ∆s, observe that
Mod−c(ωs)∆s = Mod−c(ωs) φs −
[Mod−c(ωs) φs(x− ye)]β̃(y) dy
where β̃(y) = e(c(ωs)·e)y β(y). Now the Fourier transform of β is identi-
cally one in a neighborhood of the origin of width comparable to ann′,
where as |c(ωs) · e| is comparable to ann. Since we can assume that
′ > ann+3, say, the function β̃ meets the hypotheses of Lemma 5.56,
namely it is Schwarz function with Fourier transform identically one
in a neighborhood of the origin, and the width of that neighborhood
is comparable to ann′. And so ∆s is bounded by the bounded by
the three terms in (5.57)—(5.59) below. In these estimates, we take
2k ≃ ann′ > 1. By a similar argument, one sees that Lemma 5.56 also
applies to ∆s′.
We will let ∆s,m, for m = 1, 2, 3, denote the terms that come from
(5.57), (5.58), and (5.59) respectively. We use the corresponding no-
tation for ∆s,m, for m = 1, 2, 3. A nice feature of these estimates, is
that while ∆s and ∆s′ depend upon the choice of e ∈ ωs′ ∩ ωs, the
upper bounds in the first two estimates do not depend upon the choice
of e. While the third estimate does, the dependence of the set Fs on
the choice of e is rather weak.
In application of (5.58), the functions ∆s,2 will be very small, due
to the term (ann′)−10 which is on the right in (5.58). This term is so
much smaller than all other terms involved in this argument that these
terms are very easy to control. So we do not explicitly discuss the case
of ∆s,2, or ∆s′,2 below.
In the analysis of the terms (5.15) and (5.16), we frequently only
need to use an inequality such as |Ie φs′| . χ(2)Rs′ . When it comes to the
analysis of (5.17), the function ∆s′ obeys the same inequality, so that
68 5. ALMOST ORTHOGONALITY
these sums can be controlled by the same analysis that controls (5.15),
or (5.16). So we will explicitly discuss these cases below.
In order for 〈φs, φs′〉 6= ∅, we must necessarily have ωs ∩ ωs′ 6= ∅.
Thus, we update all Sℓ,j as follows.
Sℓ,j ← {(s, s′) ∈ Sℓ,j | ωs ∩ ωs′ 6= ∅} .
The Proof of (5.12) for S1,j, j ≥ 1. Recall the definition of S1,j
from (5.9). In particular, for (s, s′) ∈ S1,j , we must have ωs ⊂ ωs′.
We will use the inequality (5.8), and show that for 0 < ǫ < 1,
FL(S1,j) . 2−10j(ann′)−eα
′ · ann′
scl · ann(5.18)
FL′(S1,j) . 22j(ann′)ǫ ·
scl · ann
′ · ann′ .(5.19)
Notice that in the second estimate, we permit some slow increase in
the estimates as a function of 2j and ann′. But, due to the form of the
estimate of the Schur test in (5.8), this slow growth is acceptable.
The terms inside the square root in these two estimates cancel out.
These inequalities conclude the proof of the inequality (5.12) for the
collection S1,j , j ≥ 1.
We prove (5.18). For this, we use Lemma 5.56. That is, we should
bound the several terms
s′ : (s,s′)∈S1,j
|〈∆s, Ie φs′〉| ,(5.20)
s′ : (s,s′)∈S1,j
|〈Ie φs,∆s′〉| ,(5.21)
s′ : (s,s′)∈S1,j
|〈∆s,∆s′〉| .(5.22)
Here ∆s and ∆s′ are as in (5.14). And, Ie is defined as in (5.13). We
can regard the tile s as fixed, and so fix a choice of e ∈ ωs. In the next
two cases, we will need to estimate the same expressions as above. In
all three cases, Lemma 5.56 is applied with 2k ≃ ann′, and we can take
ǫ in this Lemma to be ǫ = 1
(α− 1). For ease of notation, we set
(5.23) α̃ = (α− 1)(1− ǫ)2 − ǫ > 0
As we have already mentioned, we do not explicitly discuss the
upper bound on the estimate for (5.22).
APPLICATION OF THE FOURIER LOCALIZATION LEMMA 69
The Upper Bound on (5.20). We write ∆s = ∆s,1 + ∆s,2 + ∆s,3,
where these three terms are those on the right in (5.57)—(5.59) respec-
tively. Note that
(5.24) |Ie φs′| . χ(2)Rs′ ,
since Ie is convolution in the long direction of Rs′ , at the scale of
(ann′)−1, which is much smaller than the length of Rs′ in the direc-
tion e. Therefore, we can estimate the term in (5.20) by
s′ : (s,s′)∈S1,j
|〈∆s,1, Ie φs′〉| . (ann′)−eα2−10j
s′ : (s,s′)∈S1,j
. (ann′)−eα2−10j
′ · ann′
scl · ann .(5.25)
This is as required to prove (5.18) for these sums.
For the terms associated with ∆s,3, we have
s′ : (s,s′)∈S1,j
|〈∆s,3, Ie φs′〉| .
s′ : (s,s′)∈S1,j
|Rs|−1/2 · χ(2)R′s dx
. 2−10j|Fs|
ann′ · scl′ · ann · scl
. 2−10j(ann′)−α+ǫ
′ · ann′
scl · ann .
That is, we only rely upon the estimate (5.60). This completes the
analysis of (5.20). (As we have commented above, we do not explicitly
discuss the case of ∆s,2.)
The Upper Bound for (5.21). Since ωs ⊂ ωs′ , the only facts about
∆s′ we need are
(ann′)ǫR′s
|∆s′| dx . (ann′)−eα+ǫ
′ · ann′ ,
|∆s′(x)| . (ann′)−eαχ(2)Rs′ (x) , x 6∈ (ann
′)ǫRs′ .
(5.26)
Indeed, this estimate is a straightforward consequence of the various
conclusions of Lemma 5.56. (We will return to this estimate in other
cases below.)
70 5. ALMOST ORTHOGONALITY
These inequalities, with |Ie φs| . χ(2)Rs , permit us to estimate
(5.21) . 2−20j |Rs|−1/2
s′ : (s,s′)∈S1,j
(ann′)ǫR′s
|∆s′| dx
. 2−20j(ann′)−eα
scl · ann
′ · ann′ × ♯{s
′ : (s, s′) ∈ S1,j}
. 2−20j(ann′)−eα
′ · ann′
scl · ann .
which is the required estimate. Here of course we use the estimate
♯{s′ : (s, s′) ∈ S1,j} . 22j
′ · ann′
scl · ann .
We now turn to the proof of (5.19), where it is important that we
justify the small term √
scl · ann
′ · ann′
on the right in (5.19). We estimate the terms dual to (5.20)—(5.22),
namely
s : (s,s′)∈S1,j
|〈∆s, Ies φs′〉| ,(5.27)
s : (s,s′)∈S1,j
|〈Ies φs,∆s′〉| ,(5.28)
s : (s,s′)∈S1,j
|〈∆s,∆s′〉| .(5.29)
Here, for each choice of tile s, we make a choice of es ∈ ωs ⊂ ωs′.
The Upper Bound on (5.27). We have an inequality analogous to
(5.24).
(5.30) |Ies φs′| . χ
Note that as we can view s′ as fixed, all the tiles {s : (s, s′) ∈ S1,j}
have the same approximate spatial location. Let us single out a tile s0
in this collection. Then, for all s, we have Rs ⊂ 2j+2Rs0 .
Recalling the specific information about the support of the functions
of ∆s from (5.57), (5.59) and (5.61), it follows that
s : (s,s′)∈S1,j
|∆s| . 22j(ann′)ǫχ(2)2j+2Rs0 .
APPLICATION OF THE FOURIER LOCALIZATION LEMMA 71
In particular, we do not claim any decay in ann′ in this estimate. (The
small growth of (ann′)ǫ above arises from the overlapping supports of
the functions ∆s, as detailed in Lemma 5.56.) Therefore, we can esti-
s : (s,s′)∈S1,j
|〈∆s, Ies φs′〉| . 22j(ann′)ǫ
2jRs0
. 2−10j(ann′)ǫ
scl · ann
′ · ann′ .
This is as required in (5.19).
Remark 5.31. It is the analysis of the term
s : (s,s′)∈S1,j
|〈∆s,3, Ies φs′〉|
which prevents us from obtaining a decay in ann′, at least in some
choices of the parameters scl , ann , scl′, and ann′.
The Upper Bound on (5.28). The fact about ∆s′ we need is the
simple inequality |∆s′| . χ(2)Rs′ .
As in the previous case, we turn to the fact that all the tile {s :
(s, s′) ∈ S1,j} have the same approximate spatial location. Single out
a tile s0 in this collection, so that Rs ⊂ 2j+2Rs0 for all such s.
Our claim is that
(5.32)
s : (s,s′)∈S1,j
|Ies φs| . 22jχ
22jRs
(We will have need of related inequalities below.) Suppose that s ∈
{s : (s, s′) ∈ S1,j}. These intervals all have the same length, namely
scl/ann. And x 6∈ supp(φs) implies v(x) 6∈ ωs, so that by the Lipschitz
assumption on the vector field
dist(x, supp(φs)) & dist(v(x),ωs) .
This means that
(5.33) |Ies φs(x)| . χ
1 + ann′ · dist(v(x),ωs)
Here, we recall that the operator Ie is dominated by the operator which
averages on spatial scale (ann′)−1 in the direction e. Moreover, we have
(5.34) ann′ · dist(ωs,ω) & scl .
Here, we partition the unit circle into disjoint intervals ω ∈ Ω of length
|ω| ≃ scl/ann, so that for all s ∈ {s : (s, s′) ∈ S1,j}, we have ωs ∈ Ω.
72 5. ALMOST ORTHOGONALITY
Figure 5.1. The relative positions of Rs and Rs′ in for
pairs (s, s′) ∈ Sℓ, for ℓ = 2 and ℓ = 3 respectively.
In fact, the term on the left in (5.34) can be taken to be integer
multiples of scl. Combining these observations proves (5.32). Indeed,
we can estimate the term in (5.32) as follows. For x, fix ω ∈ Ω with
v(x) ∈ ω. Then,
s : (s,s′)∈S1,j
|Ies φs| .
s : (s,s′)∈S1,j
1 + ann′ · dist(ω,ωs)
The important point is that the term involving the distance allows us
to sum over the possible values of ωs ⊂ ωs′ to conclude (5.32).
To finish this case, we can estimate
s : (s,s′)∈S1,j
|〈Ies φs,∆s′〉| . 2−10j
scl · ann
′ · ann′ .
This completes the upper bound on (5.28).
The Proof (5.12) for S2,j, j ≥ 1. In this case, note that the
assumptions imply that we can assume that ωs′ ⊂ ωs, and that di-
mensions of the rectangle Rs′ are smaller than those for Rs in both
directions. See Figure 5.
We should show these two inequalities, in analogy to (5.18) and
(5.19).
FL(S2,j) . 2−10j(ann′)−eα
′ · ann′
scl · ann(5.35)
FL′(S2,j) . 2−10j(ann′)−eα ·
scl · ann
′ · ann′ .(5.36)
Here, α̃ is as in (5.23).
APPLICATION OF THE FOURIER LOCALIZATION LEMMA 73
For the proof of (5.35), we should analyze the sums
s′ : (s,s′)∈S2,j
|〈∆s, Ies′ φs′〉| ,(5.37)
s′ : (s,s′)∈S2,j
|〈Ies′ φs,∆s′〉| ,(5.38)
s′ : (s,s′)∈S2,j
|〈∆s,∆s′〉| .(5.39)
These inequalities are in analogy to (5.20)—(5.22), and es′ ∈ ωs′ ⊂ ωs.
The Upper Bound on (5.37). Fix the tile s. Fix a translate Rs of
Rs with 2
jRs ∩ 2jRs = ∅, but 2j+1Rs ∩ 2j+2Rs 6= ∅. Let us consider
(5.40) S2,j = {(s, s′) ∈ S2,j | Rs′ ⊂ Rs}
and we restrict the the sum in (5.37) to this collection of tiles. Note
that with . 22j choices of Rs, we can exhaust the collection S2,j . So
we will prove a slightly stronger estimate in the parameter 2j for the
restricted collection S2,j.
The point of this restriction is that we can appeal to an inequality
similar to (5.32). Namely,
(5.41)
s′ : (s,s′)∈S2,j
|Ie′s φs′| .
′ · ann′
scl · ann χ
Note that the term in the square root takes care of the differing L2
normalizations of φs′ and χ
. Indeed, the proof of (5.32) is easily
modified to give this inequality.
Next, we observe that the analog of (5.26) holds for ∆s. Just replace
s′ in (5.26) with s. It is a consequence that we have
s′ : (s,s′)∈S2,j
|〈∆s, Ies′ φs′〉| . 2
−12j(ann′)−eα
′ · ann′
scl · ann .
This is enough to finish this case.
The Upper Bound on (5.38). Let us again appeal to the notations
Rs and S2,j as in (5.40).
We have the estimates
|Ies′ φs| . χ
74 5. ALMOST ORTHOGONALITY
As for the sum over ∆s′ , we have an analog of the estimates (5.26).
Namely,
s′ : (s,s′)∈S2,j
(ann′)ǫRs
|∆s′,1| dx . (ann′)−eα+ǫ
′ · ann′
scl · ann
s′ : (s,s′)∈S2,j
|∆s′,1| .
′ · ann′
scl · ann χ
, x 6∈ (ann′)ǫRs .
Note that we again have to be careful to accommodate the different
normalizations here. The proof of (5.26) can be modified to prove this
estimate.
Putting these two estimates together clearly proves that
s′ : (s,s′)∈S2,j
|〈Ies′ φs,∆s′〉| . 2
−10j(ann′)−eα
′ · ann′
scl · ann ,
as is required.
We now turn to the proof of the inequality (5.36), which will follow
from appropriate upper bounds on the sums below.
s : (s,s′)∈S2,j
|〈∆s, Ie φs′〉| ,(5.42)
s : (s,s′)∈S2,j
|〈Ie φs,∆s′〉| ,(5.43)
s : (s,s′)∈S2,j
|〈∆s,∆s′〉| .(5.44)
Here, we can regard s′ as a fixed tile, and e ∈ ωs′ ⊂ ωs. In this case,
observe that we have the inequality
(5.45) ♯{s : (s, s′) ∈ S1,j} . 22j .
This is so since Rs has larger dimensions in both directions than does
The Upper Bound on (5.42). We use the decomposition of ∆s =
∆s,1 +∆s,2 +∆s,3. In the first case, we can estimate
s : (s,s′)∈S2,j
|〈∆s,1, Ie φs′〉| . 22j sup|〈∆s,1, Ie φs′〉|
. 2−10j(ann′)−eα
scl · ann
′ · ann′ .
APPLICATION OF THE FOURIER LOCALIZATION LEMMA 75
For the last case, of ∆s,3, we estimate
s : (s,s′)∈S2,j
|〈∆s,3, Ie φs′〉| . 22j sup|〈∆s,3, Ie φs′〉|
. 22j min
|Fs| ·
scl · ann · scl′ · ann′ ,
2−30j
scl · ann
′ · ann′
Examining the two terms of the minimum, note that by (5.61),
|Fs| ·
scl · ann · scl′ · ann′ . (ann′)−α+ǫ
′ · ann′
scl · ann
. (ann′)−α+1+ǫ
′ · ann′ · scl · ann
. (ann′)−α+1+ǫ
scl · ann
′ · ann′ .
Here it is essential that we have the estimate (5.60) as stated, with
|Fs| . (ann′)−α+ǫ|Rs|. This is an estimate of the desired form, but
without any decay in the parameter j. The second term in the min-
imum does have the decay in j, but does not have the decay in ann′.
Taking the geometric mean of these two terms finishes the proof, pro-
vided (α−ǫ)/2 > α̃, which we can assume by taking α sufficiently close
to one.
The Upper Bound on (5.43). Using the inequality |Ie φs| . χ(2)Rs ,
and the inequalities (5.26) and (5.45), it is easy to see that
s : (s,s′)∈S2,j
|〈Ie φs,∆s′〉| . 2−12j(ann′)−eα
scl · ann
′ · ann′ .
This is the required estimate.
The Proof of (5.12) for S3,j, j ≥ 1. In this case, we have that
the length of the rectangles Rs′ are greater than those of the rectangles
Rs, as depicted in Figure 5. We show that
FL(S3,j) . (ann′)ǫ
′ · ann′
scl · ann(5.46)
FL′(S3,j) . 2−10j(ann′)−eα
scl · ann
′ · ann′ .(5.47)
In particular, we do not claim any decay in the term FL(S3,j), in fact
permitting a small increase in the parameter ann′. Recall that 0 < ǫ < 1
is a small quantity. See (5.23). But due to the form of the estimate in
76 5. ALMOST ORTHOGONALITY
Proposition 5.7, with the decay in 2j and ann′ in the estimate (5.47),
these two estimates still prove (5.12) for S3,j .
For the proof of (5.46), we analyze the sums
s′ : (s,s′)∈S3,j
|〈∆s, Ies′ φs′〉| ,(5.48)
s′ : (s,s′)∈S3,j
|〈Ies′ φs,∆s′〉| ,(5.49)
s′ : (s,s′)∈S3,j
|〈∆s,∆s′〉| .(5.50)
Here, es′ ∈ ωs′ ⊂ ωs.
The Upper Bound on (5.48). Regard s as fixed. We employ a vari-
ant of the notation established in (5.40). Let R̃s be a rectangle with
in the same coordinates axes as Rs. In the direction es, let it have
length 1/scl′, that is the (longer) length of the rectangles Rs′, and let
it have the same width of Rs. Further assume that 2
jRs ∩ R̃s = ∅ but
2j+4Rs ∩ R̃s 6= ∅. (There is an obvious change in these requirements
for j = 1.) Then, define
S̃3,j = {(s, s′) ∈ S3,j | Rs′ ⊂ R̃s} .
With . 22j choices of R̃s, we can exhaust the collection S3,j . Thus, we
prove a slightly stronger estimate in the parameter 2j for the collection
S̃3,j.
The main point here is that we have an analog of the estimate
(5.32):
s′ : (s,s′)∈ eS3,j
|Ie φs′| .
The term in the square root takes into account the differing L2 normal-
izations between the φs′ and χ
. The proof of (5.32) can be modified
to prove the estimate above.
We also have the analogs of the estimate (5.26). Putting these two
together proves that
s′ : (s,s′)∈S3,j
|〈∆s, Ies′ φs′〉| . 2
−10j(ann′)−eα
. 2−10j(ann′)−eα
′ · ann′
scl · ann .
APPLICATION OF THE FOURIER LOCALIZATION LEMMA 77
That is, we get the estimate we want with decay in ann′, we do not
claim in general.
The Upper Bound on (5.49). We use the inequality
|Ies′ φs| . χ
And we use the decomposition ∆s′ = ∆s′,1 +∆s′,2 +∆s′,3.
For the case of ∆s′,1, we have ωs′ ⊂ ωs. And the supports of the
functions ∆s′ are well localized with respect to the vector field. See
(5.57). Thus, in particular we have
s′ : (s,s′)∈S3,j
|∆s′| . (ann′)ǫ
′ · ann′ .
Hence, we have
s′ : (s,s′)∈S3,j
|〈χ(2)Rs ,∆s′〉| . (ann
′ · ann′
scl · ann
which is the desired estimate.
Remark 5.51. It is the analysis of the sum
s′ : (s,s′)∈S3,j
|〈Ie φs,∆s′,3〉|
that prevents us from obtaining decay in the parameter ann′ for cer-
tain choices of parameters scl , ann , scl′ and ann′. This is why we have
formulated (5.46) the way we have.
For the proof of (5.47), we analyze the sums
s : (s,s′)∈S3,j
|〈∆s, Ie φs′〉| ,(5.52)
s : (s,s′)∈S3,j
|〈Ie φs,∆s′〉| ,(5.53)
s : (s,s′)∈S3,j
|〈∆s,∆s′〉| .(5.54)
Here es′ ∈ ωs′ ⊂ ωs, and one can regard the interval ωs′ as fixed. It is
essential that we obtain the decay in 2j and ann′ in these cases.
Indeed, these cases are easier, as the sum is over s. For fixed s′,
there is a unique choice of interval ωs ⊃ ωs′. And the rectangles Rs
are shorter than Rs′, but wider. Hence,
(5.55) ♯{s : (s, s′) ∈ S3,j} . 22j
78 5. ALMOST ORTHOGONALITY
The Upper Bound on (5.52). We use the decomposition ∆s = ∆s,1+
∆s,2 +∆s,3, and the inequality |Ies′ φs′, | . χ
For the sum associated with ∆s,1, we have
s : (s,s′)∈S3,j
|〈∆s,1, Ies′ φs′〉| . (ann
′)−eα
s : (s,s′)∈S3,j
〈χ(2)Rs , χ
. 2−12j(ann′)−eα
′ · ann
scl · ann′
× ♯{s : (s, s′) ∈ S3,j}
. 2−10j(ann′)−eα
′ · ann′
scl · ann .
This is the required estimate.
For the sum associated with ∆s,3, the critical properties are those
of the corresponding sets Fs, described in (5.60) and (5.61). Note that
the sets ∑
s : (s,s′)∈S3,j
1Fs . (ann
′)2ǫ .
On the other hand,
s : (s,s′)∈S3,j
|Fs| . 22j
′ sup
s : (s,s′)∈S3,j
. 22j(ann′)−α+ǫ
′ · ann .
Here, we have used the estimate (5.55).
This permits us to estimate
s : (s,s′)∈S3,j
|〈∆s,3, Ies′ φs′〉| . 2
−10j(ann′)−α+3ǫ
scl · ann′
′ · ann
Note that the parity between the ‘primes’ is broken in this estimate.
By inspection, one sees that this last term is at most
. 2−10j(ann′)−eα
′ · ann′
scl · ann .
Indeed, the claimed inequality amounts to
(ann′)−α+3ǫscl . (ann′)−eαscl′ .
We have to permit scl′ to be as small as 1, whereas scl can be as big
as ann. But α > 1, and ann < ann′, so the inequality above is trivially
true. This completes the analysis of (5.52).
THE FOURIER LOCALIZATION ESTIMATE 79
The Upper Bound on (5.53). We only need to use the inequality
|Ies′ φs| . χ
, and the inequalities (5.26). It follows that
s : (s,s′)∈S3,j
|〈Ies′ φs,∆s′〉| .
s : (s,s′)∈S3,j
〈χ(2)Rs , |∆s′|〉
. 2−10j(ann′)−eα
′ · ann′
scl · ann .
The Fourier Localization Estimate
The precise form of the inequalities quantifying the Fourier local-
ization effect follows.
Fourier Localization Lemma 5.56. Let 1 < α < 2, ǫ < (α− 1)/20,
and v be a vector field with ‖v‖Cα ≤ 1. Let s be a tile with
1 < scl(s) = scl ≤ ann(s) = ann < 1
fs = Mod−c(ωs) φs
Let ζ be a smooth function on R, with 1(−2,2) ≤ ζ̂ ≤ 1(−3,3) and set
ζ2k(y) = 2
kζ(y2k). We have this inequality valid for all unit vectors e
with |e− es| ≤ |ωs|.
∣∣∣fs(x)−
fs(x− ye)ζ2k(y) dy
.(scl2(α−1)k)−1+ǫχ
(x)1fωs(v(x))(5.57)
+ (2kscl)−10χ
(x)(5.58)
+ |Rs|−1/21Fs(x) ,(5.59)
where ω̃s is a sub arc of the unit circle, with ω̃s = λωs, and 1 < λ <
2ǫk. Moreover, the sets Fs ⊂ R2 satisfy
|Fs| . 2−(α−ǫ)k(1 + scl−1)α−1|Rs|,(5.60)
Fs ⊂ 2ǫkRs ∩ v−1(ω̃s) ∩
∂(v ·es⊥)
∣∣∣ > 2(1−ǫ)k
.(5.61)
The appearance of the set Fs is explained in part because the only
way for the function φs to oscillate quickly along the direction es is
that the vector field moves back and forth across the interval ωs very
quickly. This sort of behavior, as it turns out, is the only obstacle to
the frequency localization described in this Lemma.
Note that the degree of localization improves in k. In (5.57), it
is important that we have the localization in terms of the directions
80 5. ALMOST ORTHOGONALITY
of the vector field. The terms in (5.58) will be very small in all the
instances that we apply this lemma. The third estimate (5.59) is the
most complicated, as it depends upon the exceptional set. The form of
the exceptional set in (5.61) is not so important, but the size estimate,
as a function of α > 1, in (5.60) is.
Proof. We collect some elementary estimates. Throughout this
argument, ~y := y e ∈ R2.
(5.62)
|y|>t2−k
|y2k||ζ2k(y)| dy . t−N , t > 1.
This estimate holds for all N > 1. Likewise,
(5.63)
|u|>tscl−1
|uscl||sclψ(scl u)| du . t−N , t > 1.
More significantly, we have for all x ∈ R2,
(5.64)
eiξ0y ϕ
(x− ~y)ζ2k(y) dy = ϕ
(x) − 2k < ξ0 < 2k,
where ϕ
= Tc(Rs)D
ϕ. This is seen by taking the Fourier transform.
Likewise, by (4.17), for vectors v0 of unit length,∫
e−2πiuλ0 ϕ
(x− uv0)sclψ(scl u) du 6= 0
implies that
(5.65) scl ≤ λ0 + ξ · v0 ≤ 98scl, for some ξ ∈ supp(ϕ̂
At this point, it is useful to recall that we have specified the fre-
quency support of ϕ to be in a small ball of radius κ in (4.16). This
has the implication that
(5.66) |ξ · es| ≤ κscl, |ξ · es⊥| ≤ κann ξ ∈ supp(ϕ̂(2)Rs )
We begin the main line of the argument, which comes in two stages.
In the first stage, we address the issue of the derivative below exceeding
a ‘large’ threshold.
e ·Dv(x) · es⊥ =
∂v · es⊥
We shall find that this happens on a relatively small set, the set Fs
of the Lemma. Notice that due to the eccentricity of the rectangle
Rs, we can only hope to have some control over the derivative in the
long direction of the rectangle, and e essentially points in the long
direction. We are interested in derivative in the direction es⊥ as that is
THE FOURIER LOCALIZATION ESTIMATE 81
the direction that v must move to cross the interval ωs2. A substantial
portion of the technicalities below are forced upon us due to the few
choices of scales 1 ≤ scl ≤ 2εk, for some small positive ε.1
Let 0 < ε1, ε2 < ǫ to be specified in the argument below. In partic-
ular, we take
0 < ε1 ≤ min
, κα−1
, 0 < ε2 <
(α− 1).
We have the estimate
|fs(x)|+
fs(x− ~y)ζ2k(y) dy
∣∣∣ . 2−10kχ(2)Rs (x), x 6∈ 2
ε1kRs.
This follows from (5.62) and the fact that the direction e differs from
es by an no more than the measure of the angle of uncertainty for Rs.
This is as claimed in (5.58). We need only consider x ∈ 2ε1kRs.
Let us define the sets Fs, as in (5.59). Define
λs :=
2ε1k scl
< 2−2ε1k
8 otherwise
Let λωs denote the interval on the unit circle with length λ|ωs|, and
the same center as ωs.
2 This is our ω̃ of the Lemma; the set Fs of the
Lemma is
(5.67) Fs := 2
ε1kRs ∩ v−1(λsωs) ∩
∂(v ·es⊥)
∣∣∣ > 2(1−ε2)k
And so to satisfy (5.61), we should take ε1 < 1/1200.
Let us argue that the measure of Fs satisfies (5.60). Fix a line ℓ in
the direction of e. We should see that the one dimensional measure
(5.68) |ℓ ∩ Fs| . 2−k(α−ǫ)(1 + scl−1)α−1scl−1.
For we can then integrate over the choices of ℓ to get the estimate in
(5.60).
The set ℓ ∩ Fs is viewed as a subset of R. It consists of open
intervals An = (an, bn), 1 ≤ n ≤ N . List them so that bn < an+1
for all n. Partition the integers {1, 2, . . . , N} into sets of consecutive
integers Iσ = [mσ, nσ]∩N so that for all points x between the left-hand
endpoint of Amσ and the right-hand endpoint of Anσ , the derivative
∂(v · es⊥)/∂e has the same sign. Take the intervals of integers Iσ to be
maximal with respect to this property.
1The scales of approximate length one are where the smooth character of the
vector field helps the least. The argument becomes especially easy in the case that√
ann ≤ scl, as in the case, |ωs| & scl−1.
2We have defined λs this way so that λsωs makes sense.
82 5. ALMOST ORTHOGONALITY
For x ∈ Fs, the partial derivative of v, in the direction that is
transverse to λsωs, is large with respect to the length of λsωs. Hence,
v must pass across λsωs in a small amount of time:
|Am| . 2−(1−ε1−ε2)k for all σ.
Now consider intervals Anσ and A1+nσ = Amσ+1. By definition,
there must be a change of sign of ∂v(x) · es⊥/∂e between these two
intervals. And so there is a change in this derivative that is at least as
big as 2(1−ε2)k scl
. The partial derivative is also Hölder continuous of
index α − 1, which implies that Anσ and Amσ+1 cannot be very close,
specifically
dist(Anσ , Amσ+1) ≥
2(1−ε2)k
As all of the intervals An lie in an interval of length 2
ε1kscl
−1, it follows
that there can be at most
1 ≤ σ . 2ε1kscl−1
2(1−ε2)k
)−α+1
intervals Iσ. Consequently,
|ℓ ∩ Fs| . 2−(1−2ε1−ε2+(1−ε2)(α−1))kscl−1
. 2−(α−2ε1−2ε2)kscl−α+1scl−1
We have already required 0 < ε1 <
and taking 0 < ε2 <
achieve the estimate (5.68). This completes the proof of (5.60).
The second stage of the proof begins, in which we make a detailed
estimate of the difference in question, seeking to take full advantage of
the Fourier properties (5.62)—(5.65), as well as the derivative informa-
tion encoded into the set Fs.
We consider the difference in (5.57) in the case of x ∈ 2ε1kRs −
v−1(λsωs). In particular, x is not in the support of fs, and due to the
smoothness of the vector field, the distance of x to the support of fs is
at least
& 2ε1k
so that by (5.62), we can estimate
∣∣∣fs(x)−
fs(x− ~y)ζ2k(y) dy
∣∣∣ . (2ε1kscl)−N |Rs|−1/2
which is the estimate (5.58).
THE FOURIER LOCALIZATION ESTIMATE 83
We turn to the proof of (5.57). For x ∈ 2ε1kRs ∩ v−1(λsωs), we
always have the bound
∣∣∣fs(x)−
fs(x− ~y)ζ2k(y) dy
∣∣∣ . 210ε1k/κχ(2)Rs (x)
101λsωs(x).
It is essential that we have |e − es| ≤ |ωs| for this to be true, and κ
enters in on the right hand side through the definition (4.40).
We establish the bound
∣∣∣fs(x)−
fs(x− ~y)ζ2k(y) dy
∣∣∣ . (scl2(α−1)k)−1|Rs|−1/2,
x ∈ 2ε1kRs ∩ v−1(λsωs) ∩ F cs .
(5.69)
We take the geometric mean of these two estimates, and specify that
0 < ε1 < κ
to conclude (5.57).
It remains to consider x ∈ 2ε1kRs ∩ v−1(λsωs) ∩ F cs , and now some
detailed calculations are needed. To ease the burden of notation, we
exp(x) := e−2πiuc(ωs)·v(x), Φ(x, x′) = ϕ
(x− uv(x′)),
with the dependency on u being suppressed, and define
w(du, d~y) := sclψ(scl u)ζ2k(y) du d~y.
In this notation, note that
fs = Mod−c(ωs) φs
ec(ωs)(x−uv(x)−c(ωs))x ϕ
(x− uv(x)) sclψ(sclu) du
exp(x)Φ(x, x, )sclψ(sclu) du
exp(x)Φ(x, x, )w(du, d~y) ,
since ζ has integral on R2. In addition, we have
fs(x− ~ye)ζ2k(~y) d~y =
ec(ωs)(x−uv(x−~y)−c(ωs))x
× ϕ(2)Rs (x− uv(x− ~y)) sclψ(sclu) du d~y
exp(x− ~y)Φ(x− ~y, x− ~y)w(du, d~y) .
84 5. ALMOST ORTHOGONALITY
We are to estimate the difference between these two expressions, which
is the difference of
Diff1(x) :=
exp(x)Φ(x, x)− exp(x− ~y)Φ(x− ~y, x)w(du, d~y)
Diff2(x) :=
exp(x− ~y){Φ(x− ~y, x− ~y)− Φ(x− ~y, x)}w(du, d~y)
The analysis of both terms is quite similar. We begin with the first
term.
Note that by (5.64), we have
Diff1(x) =
{exp(x)− exp(x− ~y)}Φ(x− ~y, x)w(du, d~y).
We make a first order approximation to the difference above. Observe
exp(x)− exp(x− ~y) = exp(x){1− exp(x− ~y)exp(x)}
= exp(x){1− e−2πiu[c(ωs)·Dv(x)·e]~y}(5.70)
+O(|u|ann|~y|α).
In the Big–Oh term, |u| is typically of the order scl−1, and |~y| is of the
order 2−k. Hence, direct integration leads to the estimate of this term
|u|ann|y|α|Φ(x− ~y, x)|·|w(du, d~y)|
.|Rs|−1/2
.|Rs|−1/2(scl2(α−1)k)−1.
This is (5.69).
The term left to estimate is
Diff ′1(x) :=
exp(x)(1−e−2πiu[c(ωs)·Dv(x)·e]~y)
Φ(x− ~y, x)w(du, d~y) .
Observe that by (5.64), the integral in y is zero if
|u[c(ωs) ·Dv(x) · e]| ≤ 2k.
Here we recall that c(ωs) =
ann es⊥. By the definition of Fs, the
partial derivative is small, namely
|es⊥ ·Dv(x) · e| . 2(1−ε1)k
THE FOURIER LOCALIZATION ESTIMATE 85
Hence, the integral in y in Diff ′1(x) can be non–zero only for
scl|u| & 2ε1k.
By (5.63), it follows that in this case we have the estimate
|Diff ′1(x)| . 2−2k|Rs|−1/2
This estimate holds for x ∈ 2ε1kRs∩v−1(λsωs)∩F cs and this completes
the proof of the upper bound (5.69) for the first difference.
We consider the second difference Diff2. The term v(x− ~y) occurs
twice in this term, in exp(x − ~y), and in Φ(x − ~y, x − ~y). We will use
the approximation (5.70), and similarly,
Φ(x− ~y, x−~y)− Φ(x− ~y, x)
(x− ~y − uv(x− ~y))− ϕ(2)Rs (x− ~y − uv(x))
(x− ~y − uv(x)− uDv(x)~y)
− ϕ(2)Rs (x− ~y − uv(x)) +O(ann |u||y|
= ∆Φ(x, ~y) +O(ann |u||~y|α)
The Big–Oh term gives us, upon integration in u and ~y, a term that is
no more than
. |Rs|−1/2
2−αk . |Rs|−1/2(scl2(α−1)k)−1.
This is as required by (5.69).
We are left with estimating
Diff ′2(x) :=
e−2πiuc(ωs)·(v(x)−Dv(x)·~y)∆Φ(x, ~y)w(du, d~y).
By (5.64), the integral in y is zero if both of these conditions hold.
|uc(ωs)Dv(x) · e| < 2k,
|[uc(ωs)Dv(x)− ξ − uξDv(x)] · e| < 2k, ξ ∈ supp(ϕ̂(2)Rs )
Both of these conditions are phrased in terms of the derivative which
is controlled as x 6∈ Fs. In fact, the first condition already occurred in
the first case, and it is satisfied if
scl|u| . 2ε1k.
Recalling the conditions (5.66), the second condition is also satisfied
for the same set of values for u. The application of (5.63) then yields a
very small bound after integrating |u| & 2ε1kscl−1. This completes the
proof our technical Lemma. �
REFERENCES 87
References
[1] Angeles Alfonseca, Fernando Soria, and Ana Vargas, A remark on maximal
operators along directions in R2, Math. Res. Lett. 10 (2003), no. 1, 41–
49.MR1960122 (2004j:42010) ↑8
[2] Angeles Alfonseca, Strong type inequalities and an almost-orthogonality princi-
ple for families of maximal operators along directions in R2, J. London Math.
Soc. (2) 67 (2003), no. 1, 208–218.MR1942421 (2003j:42015) ↑8
[3] J. Bourgain, A remark on the maximal function associated to an analytic vec-
tor field, Analysis at Urbana, Vol. I (Urbana, IL, 1986–1987), 1989, pp. 111–
132.MR 90h:42028 ↑4, 6, 23, 24
[4] Camil Muscalu, Terence Tao, and Christoph Thiele, Uniform estimates on
paraproducts, J. Anal. Math. 87 (2002), 369–384. Dedicated to the memory of
Thomas H. Wolff. MR 1945289 (2004a:42023) ↑42
[5] Anthony Carbery, Andreas Seeger, Stephen Wainger, and James Wright,
Classes of singular integral operators along variable lines, J. Geom. Anal. 9
(1999), no. 4, 583–605.MR 2001g:42026 ↑6, 28
[6] Lennart Carleson, On convergence and growth of partial sumas of Fourier se-
ries, Acta Math. 116 (1966), 135–157.MR 33 #7774 ↑vii, 2
[7] Michael Christ, Alexander Nagel, Elias M. Stein, and Stephen Wainger, Sin-
gular and maximal Radon transforms: analysis and geometry, Ann. of Math.
(2) 150 (1999), no. 2, 489–577.MR 2000j:42023 ↑6
[8] A. Córdoba and R. Fefferman, On differentiation of integrals, Proc. Nat. Acad.
Sci. U.S.A. 74 (1977), no. 6, 2211–2213.MR0476977 (57 #16522) ↑7, 11
[9] Charles Fefferman, Pointwise convergence of Fourier series, Ann. of Math. (2)
98 (1973), 551–571.MR 49 #5676 ↑40
[10] , The multiplier problem for the ball, Ann. of Math. (2) 94 (1971), 330–
336.MR 45 #5661 ↑vii
[11] Loukas Grafakos and Xiaochun Li, Uniform bounds for the bilinear Hilbert
transform, I, Ann. of Math. 159 (2004), 889–933. ↑31, 42
[12] Nets Hawk Katz, Maximal operators over arbitrary sets of directions, Duke
Math. J. 97 (1999), no. 1, 67–79.MR 2000a:42036 ↑8
[13] , A partial result on Lipschitz differentiation, Harmonic analysis at
Mount Holyoke (South Hadley, MA, 2001), 2003, pp. 217–224.1 979 942 ↑6
[14] Joonil Kim, Maximal Average Along Variable Lines (2006). ↑28
[15] Michael T. Lacey and Xiaochun Li, Maximal theorems for the directional
Hilbert transform on the plane, Trans. Amer. Math. Soc. 358 (2006), no. 9,
4099–4117 (electronic). MR 2219012 ↑2, 4, 10, 31, 35, 37, 40, 53, 54, 55, 59
[16] Michael T. Lacey and Christoph Thiele, Lp estimates on the bilinear Hilbert
transform for 2 < p < ∞, Ann. of Math. (2) 146 (1997), no. 3, 693–
724.MR1491450 (99b:42014) ↑2
[17] , On Calderón’s conjecture for the bilinear Hilbert transform, Proc.
Natl. Acad. Sci. USA 95 (1998), no. 9, 4828–4830 (electronic). MR 1619285
(99e:42013) ↑2
[18] Michael Lacey and Christoph Thiele, On Calderón’s conjecture, Ann. of Math.
(2) 149 (1999), no. 2, 475–496. MR 1689336 (2000d:42003) ↑2
[19] Michael T. Lacey and Christoph Thiele, Lp estimates for the bilinear Hilbert
transform, Proc. Nat. Acad. Sci. U.S.A. 94 (1997), no. 1, 33–35. MR 1425870
(98e:44001) ↑2
88 5. ALMOST ORTHOGONALITY
[20] Michael Lacey and Christoph Thiele, A proof of boundedness of the Carleson
operator, Math. Res. Lett. 7 (2000), no. 4, 361–370.MR 2001m:42009 ↑2, 10,
40, 43, 53
[21] Camil Muscalu, Terence Tao, and Christoph Thiele, Multi-linear operators
given by singular multipliers, J. Amer. Math. Soc. 15 (2002), no. 2, 469–496
(electronic).MR 2003b:42017 ↑5, 48
[22] Alexander Nagel, Elias M. Stein, and Stephen Wainger, Hilbert transforms
and maximal functions related to variable curves, Harmonic analysis in Eu-
clidean spaces (Proc. Sympos. Pure Math., Williams Coll., Williamstown,
Mass., 1978), Part 1, 1979, pp. 95–98.MR 81a:42027 ↑6, 23
[23] D. H. Phong and Elias M. Stein, Hilbert integrals, singular integrals, and Radon
transforms. II, Invent. Math. 86 (1986), no. 1, 75–113.MR 88i:42028b ↑6
[24] , Hilbert integrals, singular integrals, and Radon transforms. I, Acta
Math. 157 (1986), no. 1-2, 99–157.MR 88i:42028a ↑6
[25] Elias M. Stein, Problems in harmonic analysis related to curvature and oscil-
latory integrals, Proceedings of the International Congress of Mathematicians,
Vol. 1, 2 (Berkeley, Calif., 1986), 1987, pp. 196–221.MR 89d:42028 ↑2, 6
[26] E. M. Stein, Harmonic analysis: real-variable methods, orthogonality, and os-
cillatory integrals, Princeton Mathematical Series, vol. 43, Princeton Univer-
sity Press, Princeton, NJ, 1993. With the assistance of Timothy S. Murphy;
Monographs in Harmonic Analysis, III.MR 95c:42002 ↑vii
[27] Jan-Olov Strömberg,Maximal functions associated to rectangles with uniformly
distributed directions, Ann. Math. (2) 107 (1978), no. 2, 399–402.MR0481883
(58 #1978) ↑2, 8, 19
[28] , Weak estimates on maximal functions with rectangles in certain di-
rections, Ark. Mat. 15 (1977), no. 2, 229–240.MR0487260 (58 #6911) ↑2, 8
	Preface
	Chapter 1. Overview of Principal Results
	Chapter 2. Connections to Besicovitch Set and Carleson's Theorem
	Besicovitch Set
	The Kakeya Maximal Function
	Carleson's Theorem
	The Weak  L 2  Estimate in Theorem 1.15 is Sharp
	Chapter 3. The Lipschitz Kakeya Maximal Function
	The Weak  L 2  Estimate
	An Obstacle to an  Lp estimate, for  1<p<2
	Bourgain's Geometric Condition
	Vector Fields that are a Function of One Variable
	Chapter 4. The  L2 Estimate for Hilbert Transform on Lipschitz Vector Fields
	Definitions and Principle Lemmas
	Truncation and an Alternate Model Sum 
	Proofs of Lemmata 
	Chapter 5. Almost Orthogonality Between Annuli
	Application of the Fourier Localization Lemma
	The Fourier Localization Estimate
	References
