TWO WEIGHTED CARLESON EMBEDDINGS ON MULTI-TREES

AND MULTI-DISK

By

Georgios Psaromiligkos

A DISSERTATION

Michigan State University

in partial fulfillment of the requirements

Submitted to

for the degree of

Mathematics — Doctor of Philosophy

2020

TWO WEIGHTED CARLESON EMBEDDINGS ON MULTI-TREES AND MULTI-DISK

ABSTRACT

By

Georgios Psaromiligkos

Given two measures µ, w on a multi-tree T n we prove a two weighted multi-parameter
dyadic embedding theorem for the Hardy operator, assuming w is a product weight and a

certain “Box” condition holds. The main result has been long proven for dimension n = 1,

however, for higher dimensions the result was not known. There was a general feeling such

an embedding was not possible under the Box condition, due to a famous counterexample by

Lennart Carleson. In this counterexample, the measure µ was the two-dimensional Lebesgue

measure, which is a product measure along with a non-product weight w. Shortly after,

A. Chang imposed a (strictly) more general condition than the Box one and showed it

is sufficient to get the same embedding in dimension n = 2. This was later used by A.

Chang and R. Fefferman to characterize the dyadic n-dimensional product BMO, denoted
by BM Odprod(Rn).

Recently, the question of embedding the Dirichlet space on the bi-disk D2 into L2(D2)

appeared. This is equivalent to proving a general measure µ is “Carleson” for the Dirichlet
space on D2.

It was shown that proving the (discrete) analogue of the embedding on a

bi-tree is enough to get the same for the bi-disc. To do this, however, we need to change the

restrictions on the measures; we will assume µ to be general and w to be a product weight.

Given these restrictions, we managed to prove the surprising result that the Box condition

is enough to imply the embedding for dimensions n = 2, 3. This is not contradictory to

Carleson’s counterexample as the weight w was non-product.

Αυτός ο άνθρωπος ακίνητος στο πλάι
μες το δωμάτιο που βλέπει στο γιαλό
είναι ο πατέρας μου που ϕεύγει αργά

από ένα κόσμο αναίτιο - μαγικό.

– Sokratis Malamas
To my father

iii

ACKNOWLEDGMENTS

My journey from a small Greek island to the U.S.A is something I could have never

imagined. Many great people have helped to make this possible. I would like to thank them,

as without them nothing would be the same.

I still remember the day I met my advisor Alexander Volberg. It was in early September

of 2016 and I was a first-year graduate student at MSU. He immediately agreed to be my

advisor and we soon started to collaborate. He introduced me to many interesting subjects

of Harmonic analysis, like weighted theory, Bellman functions and of course multi-parameter

embeddings. Sasha’s enthusiasm for math is something incredible and motivated me to study

and solve problems. His guidance and help were invaluable, and his experience made me feel

comfortable. It was a great adventure and I thank you, Sasha, for everything!

I would also like to thank Ignacio Uriarte-Tuero with whom I had several interesting

discussions about math, research, and academia in general. His teaching was one of the

highlights of my studies in MSU and I am grateful for his persistence on detail in every

proof.

I want to also thank my colleague and friend, Irina Holmes, for everything; our

math discussions, the drinks, the food and the hospitality. The great conversationalist Pavel

Mozolyako for all the Thursday night drinks at Beggars. Also, Maxim Gilula for the time

we spent over Stein’s book and for driving to Kent. Finally, a thank you is due to all the

professors who served as members of my committee; Ilya Kachkovskiy, Dapeng Zhan and

Vladimir Peller.

As a graduate student I was able to teach my own classes. This gave me the opportunity

to improve as a teacher with the help of some true leaders and excellent role-models for

anyone persuading a career in teaching math. These people are Tsveta Sendova, Andy

iv

Krause and Shiv Karunakaran. They showed me how to teach mathematics effectively,

how to ask questions to students and, more importantly, to understand a student’s way of

thinking. Thank you, guys!

I have spent four years in the U.S and it seemed like a month. This is thanks to the

many great friends I have made there. My roommate Michalis Paparizos with whom I’ve

spent many hours cooking and talking about math, life, sports, movies, literature, politics

and many more. Along with our friends Yannis and Christos we spent my first year in MSU

together and it made my transition to the new culture smoother. Our regular trips to Ann

Arbor are going to be my happiest memories of Michigan. Of course, the gatherings in East

Lansing were never complete without Ilias and Eleni who were kind and helpful since the very

first day I set foot in the U.S. Their little Maria was the icing on the cake! Ana-Maria and

Andriana, thank you for your friendship and the great birthday cake. Christina, Christiana,

Alexandros x2 and Neophytos it was great to hang out with you in A2 and EL. Lastly,

because he was always late, I thank my academic brother Dimitris Vardakis for everything,

especially for being patient when proofreading my drafts.

Cultural exchange is one of the coolest experiences when traveling or living abroad. I met

and spent lots of time discussing and eating with Gora, Arman, Armstrong, Seonghyeon,

Jihye, Jian, Zhe, Dan, Nick, Tim, Josh, Jared, Luba, Chamila, Abhishek and Hitesh. It was

a pleasure to meet and spend time with Nina, Lax, Savannah who were part of the Fall 18’

Calc 2 class, one of the best to teach.

During my undergraduate and master studies at the university of Crete I have met many

great professors. First of all, I am grateful to professor Mihalis Papadimitrakis, who has

been my biggest influence while at the university of Crete. His excellent teaching picked my

interest at my fourth year and it was that very moment I realized I wanted to work with

v

him for my Masters. I would never have thought about going to the U.S if it were not for

him and I will always be indebted to him.

I was lucky to meet Souzana Papadopoulou,

Alexandros Kouvidakis, Themis Mitsis, Nikos Frantzikinakis all of which taught me many

interesting courses and enhanced my understanding of math.

Living in a new place all by myself was challenging at the beginning, but it soon became a

great experience given the friendship of Giorgos N., Kostas R. and Manolis T. With Giorgos

T., Katerina A. and Maria K. we shared, apart from friendship, the passion for the “simple

thing that is so hard to do”. I feel very fortunate to have met you and walk side by side with

you. I spent a lot of time hanging out and having fun with people like Chrysoula A., Danai

F., Alexandros D., Argyris K., Yannis S., Giorgos G. and all of my friends from Folegandros,

especially Giannis L., Apostolis P. and Ioulia. As for Andriani V., Konstantina P. and Maria

A., I hope we could meet more often than a few days per year. Nevertheless, we do know

how to have fun even in this short time!

Going to the university would never be possible without the help my high-school profes-

sors in my hometown, Folegandros. They have spent an enormous amount of time helping,

guiding and supporting me, always with patience. First, my mathematics teachers; Stelios

Athanasiou who set the path for me and second Frantzeska Pappa and Christos Kanellos

who helped me excel in problem solving. Dimitris Traperas and Dimitris Mouratidis who

introduced me to many cool bands, apart from strenuously trying to teach me Physics and

modern Greek respectivelly. I also thank Foivi Mpaloti, Despina Christofi and Yannis Yan-

nakakis for the many hours we have spent together studying in my 3rd year of high-school.

Before junior highschool, life was so much easier and going to school was really fun.

There, I’ve met many inspiring teachers. Thank you Maria Zerefou for teaching me how

to read and write. Thank you, Eleni Lappa, for encouraging me to write more of my silly

vi

poems and for the Shakespeare book. Yannis Kotsiolis is no more with us, but I will always

remember him for the great last year of elementary school.

Finally, I would like to thank my family and closest friends. My girlfriend Anna-Maria

for her patience, support and love for the last four years. My friends Christos and Thanasis

for their friendship which has been strong for more than 20 years now. My biggest influence

to-date, my late grandfather Giorgos, to whom, even if he were to live a hundred more years,

I would never be able to show him my gratitude for teaching me how to think, how to oppose

injustice, and how to work hard. My late grandparents Kalliopi and Christos, thank you

for the memories and everything you taught me. My uncle Nikos and my aunt Evgenia for

being close to me whenever I needed them. My cousins Irini, Giorgos, Popi and Kyriakos

for growing up together. I cannot be more grateful to my sister Polina for always paving

the way for me and for bringing her children Alkis and Myrto, to my life. My Grandma

Polina for the coffee, cookies, cakes, and the stories of people long gone. Finally, to my late

father Iakovos and my mother Evangelia for bringing me into this world, raising me and

being supportive to me all these years.

vii

TABLE OF CONTENTS

KEY TO SYMBOLS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

ix

Chapter 1

1.1 The one dimensional case
1.2 Higher dimensions

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Chapter 2 Definitions of different conditions . . . . . . . . . . . . . . . . . .

Chapter 3 The proof of Theorem 1.1

. . . . . . . . . . . . . . . . . . . . . .
3.1 The known case of n = 1, a new proof
. . . . . . . . . . . . . . . . . . . . .
3.2 The case of higher dimensions - Potential theory on n-tree . . . . . . . . . .
3.2.1 Box condition implies Hereditary Carleson for n = 2, 3 . . . . . . . .
3.2.2 Hereditary Carleson implies embedding . . . . . . . . . . . . . . . . .
3.2.3 Proof of Theorem 3.5.
. . . . . . . . . . . . . . . . . . . . . . . . . .
Improvement in the case of n = 2 . . . . . . . . . . . . . . . . . . . . . . . .

3.3

Chapter 4 Capacitary conditions and Embedding
. . . . . . . . . . . . . .
4.1 General Capacitary condition implies embedding
. . . . . . . . . . . . . . .
4.2 Box condition implies Box Capacitary Condition . . . . . . . . . . . . . . . .
4.3 Box capacitary does not imply general capacitary condition . . . . . . . . .

Chapter 5 Connections with known results and applications . . . . . . . .
5.1 Maximal function and embedding . . . . . . . . . . . . . . . . . . . . . . . .
5.2 A comparison with a result of E. Sawyer . . . . . . . . . . . . . . . . . . . .

Chapter 6 Necessary and sufficient conditions . . . . . . . . . . . . . . . . .
6.1 A product weight is not necessary . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Box condition does not imply Carleson condition for general weight w . . . .
6.3 Carleson condition does not imply REC . . . . . . . . . . . . . . . . . . . .
6.4 The lack of maximal principle matters
. . . . . . . . . . . . . . . . . . . . .
6.5 REC does not imply embedding . . . . . . . . . . . . . . . . . . . . . . . . .

Chapter 7 Open questions

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.1 Failure of Schur’s test for n (cid:21) 2 . . . . . . . . . . . . . . . . . . . . . . . . .
7.1.1 Estimates for A1, A2 . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 Potential theory and the hurdles for n (cid:21) 4 . . . . . . . . . . . . . . . . . . .
7.3 Counterexample for n (cid:21) 4?
. . . . . . . . . . . . . . . . . . . . . . . . . . .
7.4 Embedding for p 6= 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

viii

1
3
3

10

13
13
18
20
29
31
40

42
42
46
50

53
53
56

59
59
60
61
62
65

68
68
72
73
76
77

79

KEY TO SYMBOLS

The following is a list of some of the notation used throughout this paper.

• Rn the real coordinate space of dimension n
• Dn the n-disk, i.e. the Cartesian product of n unit disks in the complex plane
• T a tree of finite depth (for example, a dyadic tree)
• T n is the Cartesian product of n trees
• I, I, I the one, two and three-dimensional Hardy operator respectively
• I
• A . B if there exists a universal constant C such that A (cid:20) CB
• A (cid:24) B when A . B and B . A
• D(D2) the Dirichlet space over the bi-disk
• Cap(E) the discrete bi-logarithmic capacity of a set E (cid:18) T n or the ( 1

(cid:3), I(cid:3), I(cid:3) the adjoints of the operators above

capacity of a closed set E (cid:18) (∂D)n, according to context

2 , .., 1

2)-Bessel

• Vµ the potential of measure µ
• E[µ] the total energy of measure µ
• Vµ
δ the cut-off of the potential at level δ
• Eδ[µ] the cut-off of the energy at level δ
• (cid:20),(cid:21) the ordering on the n-tree T n
• ch(α) the set of maximal elements strictly smaller than α
• jΩj the one-dimensional Lebesgue measure of the set Ω
• m2(Ω) the two-dimensional Lebesgue measure of the set Ω
• jµj the total mass of a measure µ
• [w, µ]Box the Box condition constant
• [w, µ]C the Carleson condition constant
• [w, µ]HC the hereditary Carleson (or restricted energy condition) constant
• [w, µ]CE the Carleson embedding constant

ix

Chapter 1

Introduction

This thesis deals with the boundedness of a specific dyadic paraproduct, under a certain

condition. Dyadic paraproducts are a special form of paraproducts which are arguably one of

the most important classes of operators in harmonic analysis: their boundedness properties

are at the core of many problems. Initially, they appeared through PDE questions, such as

the Leibniz rule for fractional derivatives (see below). A typical example of a paraproduct

operator appears when one deals with the classical T 1 theorem of David and Journé: For a
function b 2 BM O(Rn) we define

∫ 1

Lf =

0

ψt (cid:3)(

) dt
(ψt (cid:3) b)(ϕt (cid:3) f )

t

(for formal details see [12]). The operator L is a paraproduct and can be shown to be a

bounded Calderón-Zygmund operator on L2 with L1 = b.

However, in the literature paraproducts usually appear as bi-linear operators and one

needs to prove the next bound for paraproducts:

kΠ(f, g)kr . kfkpkgkq,

1 < p, q (cid:20) 1,

1
r

=

1
p

+

1
q

,

0 < r < 1.

(1.1)

where Π is a paraproduct operator. For example, consider the fractional derivative of order
α: f 2 S, ˆDαf (ξ) = jξjα ˆf (ξ), where S is the class of Schwartz functions. Then, the Leibniz

1

rule is:

kDα(f g)kr . kDαfkpkgkq + kfkpkDαgkq

(1.2)

for r, p, q as above. This rule appears when one works on regularity questions for linear and

non-linear PDE (see [17], [18]). In order to prove (1.2) one actually needs paraproducts and

bounds as in (1.1). Such estimates are proven in the works [15], [18] and [13].

The need to study more complicated operators, namely multi-parameter paraproducts,

appears naturally — again primarily from PDE. These are easy to describe, but difficult to

work with. For n = 2, a bi-parameter paraproduct is an operator given by the coordinate-

wise tensor product of two one-dimensional paraproduct operators. Proving the boundedness

as in (1.1) of those operators is much harder, but still feasible (see [22] or [23]), and it was

used by Kenig in [16] to treat the well-posedness of the Kadomtsev–Petviashvili equation

describing non-linear wave motions. For this, another type of Leibniz rule was required,

which was again reduced to the boundedness of paraproducts; this time of the aforementioned

multi-parameter (tensor) type.

It has been noticed that the right model for studying paraproducts are the so-called dyadic
paraproducts. For a dyadic n-rectangle R = I1 (cid:2) (cid:1)(cid:1)(cid:1) (cid:2) In (cid:18) [0, 1)n let hR(x1, . . . , xn) :=
is the Haar function in R associated with the dyadic interval

hI1
Ii. Then, a simple example of a dyadic multi-parameter paraproduct, is the operator

(x1) (cid:1) (cid:1) (cid:1) hIn(xn) where hIi

∑

hφiR (cid:1) βR (cid:1) hR

Πb φ :=

R

where hφiR is the average of φ over R and βR := hb, hRi
L2 are the Haar coefficients of the
function b. The boundedness of the operator Πb has been studied extensively in the last 60
years.

2

1.1 The one dimensional case

If we consider the case n = 1 and a dyadic lattice D of intervals in R (or [0, 1) for simplicity),
then the operator Πb is bounded in L2 if one assumes the following “Carleson Box” condition:

∑

Q2D, Q(cid:26)P

(cid:20) C jPj,

β2
Q

8P 2 D

(1.3)

The boundedness of Πb is additionally equivalent with b 2 BM Od(Rn), where d stands for
dyadic.

The one dimensional case is well-known and exhausted in every possible way. First, it

appeared in the work of L. Carleson (see [8]) in the 60’s and was used in complex interpolation

and corona results. The underlying measure was the Lebesgue one. Much later, another proof

emerged using the Bellman function method (see [25]) and in [2] an equivalent formulation

of the problem was proven on a dyadic tree. Finally, in [26] Sawyer considered the weighted

situation (with general measure ν) and used it to deal with weighted Calderón–Zygmund

operators.

1.2 Higher dimensions

By a simple observation, the arguments above are verbatim the same in the case of Rn, with

n > 1, if we replace dyadic intervals with dyadic cubes. A cube is a Cartesian product of

n dyadic intervals of the same length. The dyadic lattice produced by these cubes enjoys

similar properties to the one-dimensional case; each cube has a single parent and it is covered

only once by its 2n descendants. A question appears here; What happens if we replace cubes

by rectangles? That is, Cartesian products of dyadic intervals but not necessarily of the same

3

length. In this case, there is much more overlap as each rectangle is covered 2n (cid:0) 1 times by
its closest descendants. Another notable difference is this; when n = 1 and for I, J dyadic
intervals with I \ J 6= ?, we have I (cid:18) J or J (cid:18) I. However, it is obvious this property
is not necessarily true if n > 2 and I, J are dyadic n-rectangles. This is the setting of the

multi-parameter theory.

In the case of higher dimensions one would ask whether the condition (1.3) is enough to

imply the boundedness of the operator Πb for n = 2 with the Lebesgue measure. It turns
out, this is not true; a counterexample constructed by Carleson in [7] (see also [30]). Hence,

the need for a new condition appeared: S.-Y. A. Chang in [10] found the necessary and

sufficient condition for (1.1) to be valid:

∑

R(cid:18)Ω,R2D2

(cid:20) C m2(Ω)

β2
R

8Ω dyadic open set,

(1.4)

where a dyadic open set, Ω, is any finite union of dyadic rectangles.

Of course, the Chang-Carleson condition (1.4) is a strictly stronger requirement than the

Carleson box condition (1.3) as shown by Carleson’s counterexample. The same criterion

of Chang works for n > 2, but again only for Lebesgue measure. Chang’s criterion led to

the understanding that multi-parameter BM O space, studied by Chang and R. Fefferman,

is a much more subtle object than the “usual" BM O. The reader may guess that product

BM Od of Chang-Fefferman (see [11]) consists of the functions b whose Haar coefficients, βR,
satisfy (1.4).

Most of these developments happened in the 80’s, and apart from Sawyer’s work [27],

there were no developments in weighted multi-parameter theory until recently. In particu-

lar, the case where Lebesgue measure is replaced by an arbitrary measure was left totally

4

unsettled. The need to find a criterion for the boundedness of Πb on L2 for n (cid:21) 2 and gen-
eral measure was initiated by natural questions from several complex variables theory and

especially questions about Carleson measures for a certain scale of Hilbert spaces of analytic

functions on the bi-disc.

Namely, given s = (s1, s2) 2 R2 consider the space Hs of analytic functions f on the

unit bi-disc for which the norm

∑

n1,n2(cid:21)0

jjfjj2

s :=

(n1 + 1)s1(n2 + 1)s2j ˆf (n1, n2)j2

is finite. One then can ask for which measures µ the embedding Hs ! L2(D2, µ) is bounded.
The Hardy space on the bi-disc, H
(0,0), corresponds to the Carleson-Chang-Fefferman case.
It turns out that the special form of the second case, with arbitrary measure but without
weight, describes the embedding of the Dirichlet space H
(1,1) on the bi-disc (see [4] and [6]).
It is worth mentioning that this latter issue is always present when attempting to solve the

corona problem in several variables.

The measure µ in [4] was general, as the question there is whether µ is a Carleson measure
for the Dirichlet space D(D2). For this reason, we change the constraints; instead of the
Lebesgue measure and arbitrary weight, we consider an arbitrary measure and a weight w

of product form. Such weight is the tensor product (coordinate-wise) of n positive functions
in R. We examine whether the analogue of the implication (1.3) =) (1.1) (for p = q = 2),
still holds true:

8f 2 L2([0, 1)n, ν)

(1.5)

∑

(∫

wR

R2Dn

f dν

R

∫

)
2 (cid:20) c

f 2dν,

[0,1)n

5

whenever

∑

P(cid:18)R,P2Dn

wP ν(P )2 (cid:20) ν(R),

8R 2 Dn

(1.6)

where Dn, as before, is the Cartesian product of n dyadic intervals which we assume to be
subsets of [0, 1) (without loss of generality). Note that on the RHS of (1.6) we could have a

constant c, which we re-normalize by considering the measure ˜ν = ν/c, for simplicity.

We now explain why this condition is the correct candidate. For n = 2 and wP (cid:17) 1 on
D2, the authors in [4] were able to prove an equivalent relation to (1.5) on Bi-disc, if they
assume a “General Capacitary condition”, which we will describe later. However, one can

avoid the usage of capacity, if another condition is being assumed (see [3]):

∑

P(cid:18)Ω,P2D2

wP ν(P )2 (cid:20) ν(Ω),

8Ω dyadic open set

(1.7)

This condition is analogous to the one of S.-Y. A. Chang in [10].

Obviously, (1.7) is stronger than (1.6). Actually, without any restrictions on w, ν it is

strictly stronger; L. Carleson in [7] for ν the 2-dimensional Lebesgue measure, constructed

a particular sequence wP such that the condition (1.6) holds, but for any constant C there
exists a specific f satisfying opposite inequality of (1.5). The objects in this construction

were dyadic 2-rectangles (not of the same side length necessarily). However, Carleson’s
sequence wP was equal to
m2(P ) for certain rectangles and 0 otherwise. In [14] the authors
tried to construct an analogous counterexample on bi-trees with wP (cid:17) 1, however the only
managed to construct a weight w with wP 2 f0, 1g.

1

Nevertheless, as surprising as it seems, such a weight cannot exist. The main result
presented in this thesis is this; if wP (cid:17) 1 then, it is a product weight and for such weights

6

the condition (1.6) implies (1.5). This was proven for n = 2 in [5] and for n = 3 in [19]. For

higher dimensions the question is still open, as we highlight in section 7.

Note that in Carleson’s counterexample the weight was not of product form. Indeed, the

absence of a product weight can trigger several other counterexamples, in similar manner as

in [7], see [21].

Next, we give an overview of how we approach the proof of the implication (1.6) =)
(1.5) for n = 2, 3. Thanks to an approximation argument we can transfer the implication
(1.6) =) (1.5) to an equivalent implication on finite multi-trees (also called n-trees).
Such objects are the Cartesian product of n simple trees of the same finite depth. A finite
simple tree T is a partially ordered set with the following property: For any ω 2 ∂T the
set fα : α (cid:21) ωg is totally ordered, where ∂T is the set of minimal elements of T . Any
element α 2 T is called a node. For simplicity we consider dyadic trees, i.e. simple trees
with their nodes being dyadic intervals, but the results below can be proven for any partially
ordered set with this property. If T is a finite dyadic tree of depth N, then ∂T consists of
all intervals of side length 2

(cid:0)N.

We continue by defining an n-tree. For n 2 N, an n-tree, denoted by T n is Cartesian
product of n simple trees T of depth N. All the constants below are independent of N. A
measure µ on T n is a positive function with domain ∂T n = (∂T )n (Cartesian products of
intervals of side length 2

(cid:0)N). The Hardy operator on T n is defined for f : T n ! R as

∑

f (α)

α(cid:21)β

∑

f (α)

α(cid:20)β

7

and its adjoint by

If (β) =

I(cid:3)

f (β) =

(1.8)

(1.9)

Our investigation revolves around the following. Let any f 2 ℓ2(T n, µ). We want to

check if

∑

(

I(cid:3)

)

(f µ)(β)

2 .

∫

f 2dµ

T n

w(β)

β2T n

under the assumption

∑

α(cid:20)β

w(α)I(cid:3)

µ(α)2 . I(cid:3)

µ(β),

8β 2 T n

(1.10)

(1.11)

The smallest constant such the above inequalities hold are denoted by [w, µ]CE and [w, µ]Box
respectively. The weight w is said to be of product form if for any α = α1 (cid:2) (cid:1)(cid:1)(cid:1) (cid:2) αn 2 T n
we have w(α) = w1(α1)(cid:1)(cid:1)(cid:1) wn(αn) where wi are positive functions on the simple tree T .
The main theorem is this:

Theorem 1.1. Let n = 2, 3 and µ be a measure on T n. Assume w is a product weight on
T n. Then [w, µ]CE . [w, µ]Box

Before proceeding with our investigation let us make a comment. As we said, the constant

in (1.10) is independent of the depth N of the n-tree. Hence, we can easily show the
implication (1.11) =) (1.10) gives the implication (1.6) =) (1.5). Suppose we have
the condition (1.6) and let a positive step function f 2 L2([0, 1)n, ν). Then, there is N0
large enough (which depends on f) such that for any fixed N (cid:21) N0, f is constant in each
(cid:0)N. These squares form the boundary of an n-tree of depth
n-dimensional square of side 2
∫
N. Thus, we can think of f as a function on T n with domain ∂T n. Also we define a measure
µ on ∂T n by I(cid:3)
β f dν for β 2 T n.
Hence, as the assumption (1.6) trivially implies (1.11) for the measure µ, and if we assume

µ(α) := ν(α) for α 2 T n. Additionally, I(cid:3)

(f µ)(β) =

8

momentarily that Theorem 1.1 is true, we get (1.10). This becomes, given the definitions

above:

∑

(∫

)

w(β)

f dν

β

β2T n

∑

ω2∂T n

f (ω)2ν(ω)

2 (cid:20) C
∫

and the RHS of this inequality is equal than

[0,1)n f 2dν. Note on the left-hand we have a
positive, increasing sequence (on N) which is also bounded. As the constant C is independent
of N and this is true for any N (cid:21) N0, we get (1.5) for any positive step function. By a
limiting argument we can extend this for any f 2 L2([0, 1)n, ν). Therefore the following
equivalence holds:

(
(1.6) =) (1.5)

) () (

(1.11) =) (1.10)

)

9

Chapter 2

Definitions of different conditions

Apart from the box condition (1.11) other conditions play an important role. First, we have

the “Carleson” condition, related to the work of Chang mentioned in the introduction. For
any D (cid:18) T n with D a down-set (a set of maximal nodes along with all of their descendants)

∑

α2D

∑

w(α)I(cid:3)

µ(α)2 . µ(D) .

(2.1)

where µ(D) =
denoted by [w, µ]C. The Restricted energy (or Hereditary Carleson) condition holds if

µ(ω). The smallest constant such the above inequality holds is

ω2D\∂T n

∑

(

I(cid:3)

w(α)

α2T n

(µ1E)(α)

)
2 . jµ1Ej

(2.2)

where the latter is the total mass of the restriction of µ on E. The smallest constant such

the above inequality holds is denoted by [w, µ]HC. In other words, a measure µ satisfies the
Hereditary Carleson condition if for any set E (cid:18) T n the measure µ1E satisfies the Carleson
condition (2.1). The reader might notice the following inequalities are obvious:

[w, µ]Box (cid:20) [w, µ]C (cid:20) [w, µ]HC (cid:20) [w, µ]CE

(2.3)

As noted in the introduction, for a product weight w we also prove [w, µ]CE . [w, µ]Box
which makes the conditions above equivalent. More precisely we have the equivalence be-

10

tween the constants [w, µ]C and [w, µ]Box, which even in the case w (cid:17) 1 is surprising: In
the definition (2.1), the set D could be “wild” in the sense that any covering of D by dyadic
rectangles can have huge overlap, similar to what happens in Carleson’s counterexample.

Even though this is true, still conditions (1.11) and (2.1) turned out to be equivalent.

The equivalence (2.3) was proven in [5] for n = 2 and in [19] for n = 3. Before this, for
n = 2, the inequality [w, µ]CE . [w, µ]C was proven in [3]. The latter work tried to avoid
the usage of capacity. For a set E (cid:18) T n, the capacity Cap(E) is

{
kφk2

Cap(E) = inf
φ

ℓ2(T n)

: Iφ(α) (cid:21) 1,8α 2 E

}

(2.4)

(2.5)

For n = 2 is was proven in [4] that the embedding (1.5) holds under the assumption

ν(E) . Cap(E)

where E is any subset of ∂T n. Almost trivially it follows that the embedding implies (2.5).
In subsection 4.1 we show the same is true for n = 3.

(

(log2

1jI1j +1)(cid:1)(cid:1)(cid:1)(log2

1jInj +1)

)

I(cid:3)

µE

If in the definition above we take E = fRg with R = I1 (cid:2) (cid:1)(cid:1)(cid:1) (cid:2) In a dyadic n-rectangle
. To see this, as noted in Section 2.3 of [4] the min-

then Cap(R) =

1

2. By Lemma 5.6 of [4], for any other measure µ satisfying Vµ := II(cid:3)

∫
imizer in (2.4) is the unique equilibrium measure µE of E. This measure satisfies Cap(E) =
µ (cid:20) 1 inside
T n
E, we have µ(E) (cid:20) Cap(E). Let us construct such a measure µ and show the opposite
inequality as well, in the case E = fRg. Suppose R := I1 (cid:2) (cid:1)(cid:1)(cid:1) (cid:2) In and ki are such that
jIij = 2
2N(cid:0)k1(cid:1)(cid:1)(cid:1)2N(cid:0)kn
and µ(ω) = 0 if ω 6(cid:20) R. For any α (cid:21) R we set φ(α) := I(cid:3)
µ(α) and φ(α) = 0 for
α 6(cid:21) R. Hence, for any α (cid:21) R we have φ(α) =
(k1+1)(cid:1)(cid:1)(cid:1)(kn+1) and so Iφ(α) = 1, which

(cid:0)ki. Then for each ω (cid:20) R, ω 2 ∂T n we define µ(ω) =

(k1+1)(cid:1)(cid:1)(cid:1)(kn+1)

1

1

1

(cid:1)

11

Cap(E) (cid:20) ∫

also means Vµ(α) = 1. Using the result above, we get µ(E) (cid:20) Cap(E). Finally, we get
= µ(E). Therefore the
measure µ is the minimizer of the set fRg. Our final condition, is called the “box-capacitary
condition”:

(k1+1)(cid:1)(cid:1)(cid:1)(kn+1) =

T n φ2 =

1

(log2

1

1jI1j +1)(cid:1)(cid:1)(cid:1)(log2

1jInj +1)

ν(R) .

1

1jI1j + 1)(cid:1)(cid:1)(cid:1) (log2

(log2

1jInj + 1)

(2.6)

As we just proved, the RHS is equal to Cap(R). This condition is the subject of subsection

4.2.

12

Chapter 3

The proof of Theorem 1.1

Before proving Theorem 1.1 is true we start with the known case of n = 1. We give a new

proof of the equivalence between (1.5) and (1.11) for the case n = 1 , using Schur’s test. For

this idea we are grateful to M. Christ. As we mentioned in the introduction, this result is

long-known, proven first in [8], then in [27] and later in [25] and [2] (in the case of a simple

tree). The sequence wI can be arbitrary but as we stated in Theorem 1.1, for n = 2, 3
our weight must be of product form. Later, we inspect non-product weights for which the

embedding holds (subsection 6.1) and does not hold (section 6).

3.1 The known case of n = 1, a new proof

For this section we change our notation a little. Let ν be any measure in [0, 1) and w be any
positive weight with domain D, the collection of all dyadic intervals in [0, 1). Also, let T be
a simple tree of finite depth N. For a function f 2 L2
and a β 2 T we define the
operator J as

[0, 1), ν

(

)

∫

Jf (β) =

f dν

β

As we said before, we are interested to see whether J is bounded from L2

(

)

[0, 1), ν

to ℓ2(T , w),

13

with a norm independent of N, under the (normalized) assumption:

∑

α(cid:20)β

w(α) (cid:1) ν(α)2 (cid:20) ν(β),

8β 2 T

(3.1)

Note that if we show this for any finite simple tree T then the implication (1.6) =) (1.5)
is also true by letting N ! 1. To draw a direct connection with the other setting, note
µ on T . For a
that kJk2
function g on T and x 2 [0, 1), its formal adjoint operator is given by

= [w, µ]CE where µ is defined such that ν = I(cid:3)

L2([0,1),ν)!ℓ2(T ,w)

∑

β2T

J(cid:3)

(g)(x) =

(wg)(β) (cid:1) 1β(x)

We will show J : L2([0, 1), ν) ! ℓ2(T , w) is bounded by using Schur’s test:

∫

Theorem 3.1. Schur’s Test. Let 1 < p < 1 and µ, σ be σ-finite measures on the
measurable spaces X, Y resp. Suppose K is an operator of the form

Y k(x, y)f (y) dµ(y), where k : X (cid:2) Y ! R is measurable and non-negative.

Kf (x) =
Then K : Lp(Y, µ) ! Lp(X, σ) is bounded if there are functions u, v with
0 < u < 1 for a.e. x 2 X and 0 < v < 1 for a.e. y 2 Y such that

0

K(vp

)(x) . up

0

(x)

a.e. x 2 X

and

(cid:3)

K

(up)(y) . vp(y)

a.e. y 2 Y

Where K

(cid:3) is the formal adjoint of K, i.e. the operator K

(cid:3)

g(y) =

14

∫

X k(x, y)g(x) dσ(x).

We will use this theorem for p = 2. We take K = J, X = T , Y = [0, 1) and µ = ν and
σ = w (weighted counting measure on T ). Note that k(β, x) = 1β(x). For simplicity let
S := JJ(cid:3) and note that for β 2 T :

∑

(wg)(α) (cid:1) ν(α ^ β)

S(g)(β) =

α2T

Our goal is to construct an auxiliary function F such that J(J(cid:3)
take u2 := F and v2 := J(cid:3)
with the function f0(β) = ν(β) for β 2 T . Then we have:

(F ). To construct such F we use a recursive argument, starting

(F )) . F. Then we will

∑
∑
(∑

α2T

α2T

+

α>β

S(f0)(β) =

=

=

(wf0)(α)ν(α ^ β)

(wν)(α) (cid:1) ν(α ^ β)

)
(wν)(α) (cid:1) ν(α ^ β)

∑
∑

α(cid:20)β

w(α) (cid:1) ν(α)2

:= f1(β) +

α(cid:20)β
(cid:20) f1(β) + f0(β)

using our assumption. Moreover, for any i 2 f1, .., mg, m to be specified later, we define
the functions fi by the formula

∑

fi(β) := ν(β) (cid:1)

(wfi(cid:0)1)(α)

(3.2)

and we will prove

α>β

15

the following recursive formula

S(fi)(β) (cid:20) fi+1(β) + S(fi(cid:0)1)(β)

(3.3)

To see this we have (as before)

S(fi)(β) = fi+1(β) +

and the second term is estimated as follows:

∑

γ(cid:20)β

(wfi)(γ) (cid:1) ν(γ)

γ(cid:20)β

∑
(wfi)(γ) (cid:1) ν(γ)
∑
∑
∑
∑

γ(cid:20)β

γ(cid:20)β

=

=

=

+

γ(cid:20)β

γ<α(cid:20)β

α>β

α>γ

(wν)(γ) (cid:1)
∑
w(γ) (cid:1) ν(γ)2 (cid:1) (wfi(cid:0)1)(α)
( ∑

(wfi(cid:0)1)(α) (cid:1) ν(γ)
)
w(γ) (cid:1) ν(γ)2 (cid:1) (wfi(cid:0)1)(α)

∑

α>γ

Now we change the order of summation for both sums and we use our assumption to get:

)
w(γ) (cid:1) ν(γ)2 (cid:1) (wfi(cid:0)1)(α)
)
∑

w(γ) (cid:1) ν(γ)2 (cid:1) (wfi(cid:0)1)(α)
∑

ν(β) (cid:1) (wfi(cid:0)1)(α)

∑
∑

α>β

+

+

( ∑
∑
(∑
∑
∑

γ<α(cid:20)β

γ(cid:20)β

=

(cid:20)

α>β

γ<α

γ(cid:20)β

α(cid:20)β
ν(α) (cid:1) (wfi(cid:0)1)(α) +

α(cid:20)β

α>β

16

which is equal to

∑

α(cid:20)β

=

ν(α ^ β) (cid:1) (wfi(cid:0)1)(α) +

∑

α>β

ν(α ^ β) (cid:1) (wfi(cid:0)1)(α)

= S(fi(cid:0)1)(β)

which proves (3.3). Now recall that S(f0) (cid:20) f1(β) + f0(β) and by using (3.3) recursively,
we get for every m

S(fm)(β) (cid:20) m+1∑

i=0

fi(β)

(3.4)

Now we look at the definition of fi in (3.2). It is defined as a succession of sums, and each
such sum is over nodes which are strictly bigger than those of the previous sum. Since the
tree T has finite depth N, there is some M (cid:20) N such that fm (cid:17) 0 for m > M. For
minimal such M we define

M∑

fm(β)

2m

(3.5)

Lets prove that S(F )(β) . F (β) using (3.4):

F (β) :=

m=0

S(F )(β) =

m=0

M∑
(cid:20) M∑
M∑
M∑

m=0

i=0

=

(cid:20) 4

i=0

17

Sfm(β)

2m

m+1∑
M∑

i=0

fi(β)
2m

fi(β)
2m

m=i(cid:0)1
fi(β)
2i = 4F (β)

Remark 3.2. Here we proved the one-dimensional result for p = 2. In addition, the result is
true for general 1 < p < 1, see for example [25]. The proof there uses the Bellman
function method and although the authors take p = 2, the same is true for any 1 < p < 1
with a slight modification. By generalizing this Bellman function appropriately, [9] proves

the same result on dyadic trees.

Remark 3.3. From the proof of Schur’s test we can see that kJk2
(cid:20) 4. As
it is known, this is the sharp constant, see [25]. It is also known that the squared norm of

L2([0,1),ν)!ℓ2(T ,w)

the Maximal function operator is at most 4. For a connection between the Carleson

embedding theorem and the Maximal function, see subsection 5.1.

Remark 3.4. Using the inequality S(F ) . F on T and modifying Schur’s test
appropriately, we see that S : ℓ2(T, w) ! ℓ2(T, w) is bounded. However, since ℓ2(T, w) is a
Hilbert space, the boundedness of S is equivalent with the boundedness of
J : L2([0, 1), ν) ! ℓ2(T, w) and moreover, kSk

ℓ2(T ,w)!ℓ2(T ,w) = kJk2

L2([0,1),ν)!ℓ2(T ,w)

.

3.2 The case of higher dimensions - Potential theory

on n-tree

For this section we deploy the techniques first presented in [4]. The main tool there is

capacity on a bi-tree. A capacity-free theory is developed in [3] and later evolved even

more in [5] and [19]. Our presentation is based on the latter.
Recall the operators I, I(cid:3) from the beginning of the section. This will be the notation for
n = 2 and for general n too. Specifically, for n = 3 we use the symbols I and I(cid:3). The
corresponding operators for dimension n = 1 are denoted by I, I

(cid:3). Note that for n = 2 the

18

operator I can be written as I = I1I2 and for n = 3 we have I = I1I2I3. We fix positive a
product weight w on T n and we define the potential Vµ as

Vµ := I(wI(cid:3)

µ)

(the symbol for the potentials is the same for any dimension) and the energy E[µ] as

∑

α2T n

E[µ] :=

w(α)I(cid:3)

µ(α)2 =

Vµdµ

T n

∫

Moreover, for δ > 0 we define the truncated potential and energy

∫

∑

δ := I(1Vµ(cid:20)δwI(cid:3)
Vµ

µ),

Eδ[µ] :=

Vµ

δ dµ =

T n

α2fVµ(cid:20)δg

w(α)I(cid:3)

µ(α)2

The following is the main result of this section. As we already mentioned the maximum

principle fails. However, this quasi- maximum principle holds:

Theorem 3.5. Let n = 1, 2, 3. Let T n be an n-tree and µ, ρ positive measures on T n.
Then the following is true for r = 1
n :

∫

(

(

δjρj)

r

δ dρ .
Vµ

Eδ[µ]E[ρ]

) 1(cid:0)r

2

T n

By taking µ = ρ we get

Corollary 3.6. Let n = 1, 2, 3. Let T n be an n-tree and µ positive measure on T n. Then
the following is true for r = 1
n :

∫

Eδ[µ] =

T n

δ dµ .
Vµ

19

(
δjµj) 2r

1+rE[µ]

1(cid:0)r
1+r

For n = 1 the proof of Theorem 3.5 is immediate, as on a simple tree we have Vµ
δ

(cid:20) δ
(maximum principle). By using Cauchy-Schwarz inequality we can prove for a positive
measure ρ on T and every r 2 [0, 1] that

∫

T

δ dρ (cid:20) δrjρjrEδ[µ]
Vµ

1(cid:0)r
2 E[ρ]

1(cid:0)r
2

where jρj is the total mass of ρ on the simple tree T . However, the maximum principle fails
on the n-tree, for n (cid:21) 2 (see section 6.4). Before we give the proof of this theorem, lets see
how we can use it in order to get the desired embedding. This part of the proof is purely of

combinatorial nature.

3.2.1 Box condition implies Hereditary Carleson for n = 2, 3

We start with two definitions

Vµ

α(ω) :=

Vµ
ϵ0,good

(ω) :=

∑
∑

β:ω(cid:20)β(cid:20)α

w(β)I(cid:3)

µ(β),

w(α)I(cid:3)

µ(α).

α(cid:21)ω:Vα(ω)>ϵ0

(3.6)

(3.7)

The next results work for any n (cid:21) 2, as long as Theorem 3.5 is true. However, we are only
able to prove it in the case n = 2, 3. Hence, one needs to prove 3.5 for any n (cid:21) 4 and this
would be sufficient for settling the embedding (more in section 7).

Lemma 3.7. Let n (cid:21) 2 and µ : T n ! [0,1). Let w : T n ! [0,1) be a product weight and
assume Theorem 3.5 holds for this n. Suppose that E[µ] (cid:20) jµj and

Vµ (cid:21) 1/3 on supp µ.

20

(3.8)

Then, if ϵ

0 is small enough, we have∫

Vµ
ϵ0,good

dµ & jµj.

Proof. It suffices to show that, for some ϵ

0 and ϵn(cid:0)1, we have

µfω 2 T n j Vµ

ϵ0,good

(ω) (cid:21) ϵn(cid:0)1g (cid:21) jµj/2.

Let ϵ > 0 be chosen later and define

ϵ1 := ϵ,

1/κ
ϵ2 := ϵϵ
1

,

1/κ
ϵ3 := ϵϵ
2

, . . .

where κ = 2r

1+r, r = 1/n. By corollary 3.6, we have

∫

ϵj dµ . ϵκ
Vµ

j

jµjκE[µ]1(cid:0)κ . ϵκ

j

∫

dµ

By Chebyshov’s inequality, it follows that

ϵj (ω) (cid:20) (ϵj/ϵ)κ/10
Vµ

(3.9)

for a µ- proportion (cid:21) (1 (cid:0) Cϵκ) of ω’s. So we only consider ω’s for which (3.9) holds for all
j = 1, . . . , n (cid:0) 1. Similarly, we may restrict to those ω’s for which Vµ(ω) . 1.

Let

0
ϵ

:= ϵ (cid:1) ϵ1 (cid:1)(cid:1)(cid:1) ϵn(cid:0)1.

21

For a fixed ω, let

and

U := fα (cid:21) ω j Vµ

α(ω) > ϵ

0g

Wj := fα (cid:21) ω j Vµ(α) (cid:20) ϵjg,

1 (cid:20) j (cid:20) n (cid:0) 1.

Note these sets are decreasing in j. For p 2 T n, write

For p 2" ω, let

" p := fα 2 T n j α (cid:21) pg.

# p := fα 2 T n j ω (cid:20) α (cid:20) pg.

(3.10)

(3.11)

If U 6(cid:18) Wn(cid:0)1, then this means that there exists p 62 Wn(cid:0)1 with " p (cid:18) U. Hence,

Vµ
ϵ0,good

(ω) (cid:21)

(wI(cid:3)

µ)(p

0

) = Vµ(p) (cid:21) ϵn(cid:0)1.

∑

p02"p

Assume now that U (cid:18) Wn(cid:0)1. In this case, we will cover " ω n W1 by boundedly many sets
of the form # q with q 2" ω n U. This will lead to a contradiction with (3.8), since, by (3.9)
and (3.10), the integral of

f := wI(cid:3)

µ

is small on W1 and on each such set # q.

For a set of coordinates J (cid:18) f1, . . . , ng and a point p 2 T n, let

"J p := fq 2 T n j qj (cid:21) pj for j 2 J, qj = pj for j 62 Jg.

22

Given J (cid:18) f1, . . . , ng with J 6= ; and p 2 T n, we define a set QJ (p) (cid:26) T n as follows. If
jJj = 1, then QJ (p) consists of the (unique) maximal element of "J p n U, if the latter set is
nonempty, and is empty otherwise. If jJj (cid:21) 2, then QJ (p) is a maximal set of maximal
elements of "J p n W
n(cid:0)jJj+2 are pairwise disjoint for
q 2 QJ (p).

n(cid:0)jJj+1 such that the sets "J q n W

Then, recursively, let R;(p) := fpg,

RJ (p) := [

J0(cid:26)J

[
p02Q

J (p)

0

R
J0(p

),

where the first union runs ovel all subsets of J with cardinality jJ

0j = jJj (cid:0) 1.

We claim that, for every p 2" ω and every J (cid:18) f1, . . . , ng with J 6= ;, we have

∪

p02R

J (p)

# p

0 (cid:19)"J p n W

n(cid:0)jJj+1,

(3.12)

where we set Wn := U to simplify notation. Relation (3.12) is our main combinatorial
statement and we prove it by induction on jJj. For jJj = 1, the claim (3.12) obviously
holds. Let now J with jJj (cid:21) 2 be given, and suppose that (3.12) is known for all proper
subsets of J. Let

0

# p

, P :="J p n W

n(cid:0)jJj+1.

∪

p02R

J (p)

D :=

By the inductive hypothesis,

D (cid:19)"

J0 p

0 n W

n(cid:0)jJj+2

23

(3.13)

for every p

0 2 QJ (p) and every J

0 ( J. Suppose that

D 6(cid:19) P.

(3.14)

Choose a maximal q 2 P n D. Since D is a down-set, q is also a maximal element of P. We
claim that

("J q\ "J p

0

) n W

n(cid:0)jJj+2 = ; for all p

0 2 QJ (p).

(3.15)

q

Indeed, suppose for a contradiction that there exists q
0 be minimal with this property. Since W
of "J q\ "J p
Since q and p

0. Since q, p
0 are distinct maximal elements of P, in fact q

0 2"J p then q

n(cid:0)jJj+2 is an up-set, q

0 is in fact the coordinatewise maximum of q, p

0 2"
J0 p
one coordinate, so q
0 (cid:21) q, also q 2 D, a contradiction.
D is a down-set and q

0 for some J

0 ( J. Now, (3.13) implies that q

0 2 ("J q\ "J p

0

) n W

n(cid:0)jJj+2, and let
0 is also a minimal element

0 coincides with p

0.
0 in at least
0 2 D, and, since

Therefore, (3.15) holds. But this contradicts the maximality of QJ (p). Thus the
assumption (3.14) is false, and we obtain (3.12).

Let p (cid:21) ω. For 2 (cid:20) jJj (cid:20) n, we have

∫

1 & Vµ(ω)
∑
(cid:21) Vµ(p)
(cid:21)
∑

q2Q

(cid:21)

J (p)

q2Q

J (p)

f

"

J qnW

n(cid:0)jJj+2
(If (q) (cid:0) I(f 1W

n(cid:0)jJj+2

)(ω))

24

by definition, as q 62 W

n(cid:0)jJj+1 and by (3.9),

∑

(cid:21)

(ϵn(cid:0)jJj+1

(cid:0) (ϵn(cid:0)jJj+2/ϵ)κ/10)

J (p)

q2Q
& jQJ (p)jϵn(cid:0)jJj+1.

We multiply those inequalities to get:

ϵ1 (cid:1)(cid:1)(cid:1) ϵn(cid:0)1jRf1,...,ng(ω)j . 1.

Hence, by (3.12),

Vµ(ω) (cid:0) Vµ

ϵ1(ω) =
(cid:20)

∫

∫
#p0 f

Vµ
p0(ω)

"ωnW1

f

∑
∑

p02Rf1,...,ng(ω)

=

p02Rf1,...,ng(ω)
0jRf1,...,ng(ω)j
(cid:20) ϵ
.
ϵ1 (cid:1)(cid:1)(cid:1) ϵn(cid:0)1

0
ϵ

= ϵ.

Therefore, by (3.9),

1/3 (cid:20) Vµ(ω) = (Vµ(ω) (cid:0) Vµ

ϵ1(ω)) + Vµ

ϵ1(ω) (cid:20) Cϵ + 1/10.

This inequality is false if ϵ is sufficiently small, contradicting the assumption
U (cid:18) Wn(cid:0)1.

25

The following lemma [3, Lemma 3.1] is in place.
Lemma 3.8 (Balancing lemma). Let µ : T n ! [0,1) with

∫

E[µ] =

Vµdµ (cid:21) Ajµj.

Then there exists a down-set ˜E (cid:26) T n such that for the measure ˜µ := µ1 ˜E we have

and

V˜µ (cid:21) A
3

on ˜E,

E[˜µ] (cid:21) 1
3

E[µ].

With this at hand we complete the proof the implication “Box to Hereditary Carleson”.
Theorem 3.9. Let n (cid:21) 2 and µ be a measure on T n. Let w : T n ! [0,1) be a product
weight and assume Theorem 3.5 holds for this n. Then,

[w, µ]HC . [w, µ]Box.

Proof. By scaling, we may assume [w, µ]Box = 1 without loss of generality. Let
A := [w, µ]HC. We will show A is bounded by an absolute constant. We start with
E (cid:26) T n be a subset such that µE = µ1E 6= 0 and E[µE] = AjµEj (such a subset exists
because we assume that T n is finite). By Lemma 3.8, there exists a further subset ˜E (cid:26) T n
such that ˜µE := µE1 ˜E satisfies

V˜µ (cid:21) A
3

on ˜E

and ˜µE 6= 0. Thus, replacing µE by ˜µE, we may assume VµE (cid:21) A/3 on supp µE.

26

By Lemma 3.7 applied with µE
A

in place of µ, for sufficiently small ϵ, θ > 0, we have

ϵA,gooddµE (cid:21) 2θE[µE].
VµE

(3.16)

We claim that, with these values of ϵ and θ, we have

∑

E[µE] (cid:20) θ
1 (cid:0) θ

α:θϵAI(cid:3)µE (α)(cid:20)Eα[µE ]

w(α)(I(cid:3)

µE(α))2.

(3.17)

∫

∑

ω(cid:20)α

Indeed, suppose that α is such that

θϵAI(cid:3)

µE(α) > Eα[µE] =

µE(ω)VµE

α (ω), VµE

α (ω) =

where the latter definition is from (3.6). Then we have

∑

µE(ω) = I(cid:3)

µE(α) (cid:0)

ω(cid:20)α:VµE

α (ω)(cid:20)ϵA

ω(cid:20)α:VµE

α (ω)>ϵA

∑
∑

ω(cid:20)α

(cid:21) I(cid:3)

µE(α) (cid:0) 1

ϵA

(cid:21) (1 (cid:0) θ)I(cid:3)

µE(α).

VµE

α (ω)µE(ω)

∑

β:ω(cid:20)β(cid:20)α

w(β)(I(cid:3)

µE)(β),

µE(ω)

It follows that

∑

α:θϵAI(cid:3)µE (α)>Eα[µE ]

w(α)(I(cid:3)

µE(α))2 (cid:20)

=

∑

α

∑

µE(α)

1
1 (cid:0) θ

w(α)I(cid:3)
∑

1
1 (cid:0) θ

µE(ω)

ω

ω(cid:20)α:VµE

∑

α (ω)(cid:20)ϵA
w(α)I(cid:3)

µE(ω)

µE(α)

α(cid:21)ω:VµE

α (ω)(cid:20)ϵA

27

which is equal to

=

∑
1
1 (cid:0) θ
(cid:20) 1 (cid:0) 2θ
1 (cid:0) θ

ω

E[µE].

µE(ω)(VµE (cid:0) VµE

good,ϵA)(ω)

This implies the claim (3.17).

By corollary 3.6 again, and since VµE (cid:21) A/4 on supp µE, we also have

E
c0A[µE] . (c

0

A)κjµEjκE[µE]1(cid:0)κ . (c

0

)κE[µE].

0 sufficiently small and combining (3.18) with (3.17), we obtain
∑

w(α)(I(cid:3)

µE(α))2, R := fα 2 T n j θϵAI(cid:3)

µE(α) (cid:20) Eα[µE], VµE (α) (cid:21) c

Taking c

E[µE] .

α2R

(3.18)

0

Ag.

For each α 2 R, we have

θϵAI(cid:3)

µE(α) (cid:20) Eα[µE] (cid:20) Eα[ν] (cid:20) [w, ν]BoxI(cid:3)

ν(α) = I(cid:3)

σ(α),

where σ := ν1F , F := fβ 2 T n j 9α 2 R, α (cid:21) βg. It follows that

A2E[µE] . E[σ].

(3.19)

On the other hand, using the definition of A, the fact that VµE & A on supp σ, and the

28

Cauchy–Schwarz inequality, we obtain

E[σ] (cid:20) Ajσj .

∫

VµE dσ (cid:20) E[µE]1/2E[σ]1/2.

(3.20)

From (3.20), we obtain E[σ] . E[µE], and inserting this into (3.19) gives A . 1.

3.2.2 Hereditary Carleson implies embedding

Theorem 3.10. Let n = 1, 2, 3 and w : T n ! [0,1) be a positive product weight. Let µ, ρ
be positive measures on T n with

[w, µ]HC (cid:20) 1,

[w, ρ]HC (cid:20) 1.

Then, for some 0 < κ (cid:20) 1

4, we have∫

T n

Vµdρ . jµj1/2(cid:0)κjρj1/2+κ.

(3.21)

(3.22)

Proof. Let δ > 0 be chosen later and consider the set E := fVµ > δg (cid:26) T n. As E is a
down-set we have 1EI(cid:3)
have

(µ1E). Thus, by the Hereditary Carleson condition (2.2), we

(Vµ (cid:0) Vµ

δ )dρ =

w1EI(cid:3)

µI(cid:3)

ρ (cid:20) E[µ1E]1/2E[ρ]1/2 (cid:20) jµ1Ej1/2E[ρ]1/2

∫

T n

∫

µ (cid:20) I(cid:3)
∑

T n

∑

T n

Again, using the Hereditary Carleson condition

δjµ1Ej (cid:20)

Vµdµ =

E

wI(cid:3)

(µ1E)I(cid:3)

µ (cid:20) E[µ]1/2E[µ1E]1/2 (cid:20) E[µ]1/2jµ1Ej1/2

(3.23)

29

Therefore,

which implies

jµ1Ej1/2 (cid:20) δ

(cid:0)1E[µ]1/2,

∫

T n

(Vµ (cid:0) Vµ

δ )dρ (cid:20) δ

(cid:0)1E[ρ]1/2E[µ]1/2.

Next, by Theorem 3.5, Corollary 3.6 and (3.21), we obtain

∫

T n

δ dρ . δrjµj(1(cid:0)r)/2jρj(1+r)/2.
Vµ

and thus

∫

T n

Vµdρ (cid:20) Cδrjµj(1(cid:0)r)/2jρj(1+r)/2 + δ

(cid:0)1jρj1/2jµj1/2.

We choose δ which makes the two terms equal, to obtain

∫

∑

Vµdρ . jµj 1/2

1+rjρj 1/2+r
1+r .

∫

f 2dµ.

T n

w(I(cid:3)

(f µ))2 .

30

and we take κ := r

2(r+1)

(cid:20) 1
4.

Theorem 3.11. Let n = 1, 2, 3. Let µ, w : T n ! [0,1). Assume that w is a positive
product weight and that the Hereditary Carleson condition (2.2) holds. Then

T n

for any f 2 ℓ2(T n, µ).

The argument below is similar to the proof of [1, Theorem 7.1.1].

Proof. Without loss of generality [w, µ]HC = 1. Let f : T n ! [0,1) and consider

∫ 1

f µ =

0

µtdt, µt := µ1ff >tg.

Then [w, µt]HC (cid:20) [w, µ]HC = 1 for every 0 < t < 1. Using symmetry, we obtain

∑

T n

w(I(cid:3)

(f µ))2 = 2

by Theorem (3.10) . 2
s 7! rt and Fubini = 2

= 2
by Hölder (cid:20) 2

0

t

t

0

0

∫

µt)dsdt

w(I(cid:3)

µs)(I(cid:3)

∫
∫ 1
∫ 1
∫
T n
∫ 1
∫
jµsj1/2(cid:0)κjµtj1/2+κdsdt
∫
∫ 1
tjµrtj1/2(cid:0)κjµtj1/2+κdtdr
∫
(∫ 1
)
)
(r2tjµrtj)1/2(cid:0)κ(tjµtj)1/2+κdtdr
∫ 1
(∫
)
)(
r2tjµrtjdt
tjµtjdt
∫ 1
(cid:0)(1(cid:0)2κ)dr
tjµtjdt
∫
tjµtjdt

(∫ 1

0
(cid:0)(1(cid:0)2κ)

(cid:0)(1(cid:0)2κ)

1/2(cid:0)κ

r

r

0
1

0

1

0

1

0

1

0

0

2

r

0

0

0

0

=

. 2

1/2+κ

dr

=

T n

f 2dµ.

3.2.3 Proof of Theorem 3.5.

Some more definitions are in order.

Definition 3.12. Given a simple tree T , the set of children of a vertex β 2 T consists of
the maximal elements of T that are strictly smaller than β:

ch(β) := maxfβ

0 2 T j β

0

< βg

31

A function g : T ! R is called superadditive if for every β 2 T we have

∑

g(β) (cid:21)

0

)

g(β

β02ch(β)

The difference operator is defined by

∆g(β) := g(β) (cid:0)

∑

β02ch(β)

0

)

g(β

Thus, a super-additive function g satisfies ∆g (cid:21) 0 on T .

Lets see our first lemma
Lemma 3.13. Let T be a simple tree and f, g : T ! [0,1) be any functions. Then

(If )(Ig) (cid:20) I(f (cid:1) Ig) + I(If (cid:1) g)

Proof.

If (α)Ig(α) =

=

=

α0(cid:21)α

∑
∑
∑

α0(cid:21)α

α0(cid:21)α

∑
∑

α00(cid:21)α

α00(cid:21)α0
0

f (α

0

f (α

00

) =

)g(α

0

f (α

)g(α

0

) +

)Ig(α

00

) +

∑

α00(cid:21)α

∑
∑

α0(cid:21)α

( ∑
∑

α00(cid:21)α0

+

)

∑

α(cid:20)α00<α0
00

0

f (α

)g(α

)

α(cid:20)α00<α0

α0(cid:21)α
(If (cid:0) f )(α

00

00

)

)g(α

0

f (α

00

)

)g(α

= I(f (cid:1) Ig)(α) + I(If (cid:1) g)(α) (cid:0) I(f g)(α)

We have the next lemma.
Lemma 3.14 (Partial summation). Let T be a simple tree. For any functions

32

f, g : T ! R, we have

∑

α2T

∑

α2T

f (α)g(α) =

∆f (α)Ig(α)

Proof. By induction on the size of the tree, one can show

(cid:3) is the inverse of ∆ (as ∆I

(cid:3)

= id, trivially). It follows that

0

)

∆f (α

f (α) =

α0(cid:20)α

∑

∑

α02T

which means I

∑

∑

∑

f (α)g(α) =

α2T

α2T

α0(cid:20)α

0

∆f (α

)g(α) =

0

)

∆f (α

0

0

)

)Ig(α

∆f (α

∑

∑

g(α) =

α(cid:21)α0

α02T

Corollary 3.15. Let T be a simple tree and f, g : T ! [0,1) with f superadditive. Then

(cid:3)

(f g) (cid:20) I

(cid:3)

(∆f (cid:1) Ig)

I

Proof.

(cid:3)

I

(f g)(β) =

∑

α(cid:20)β

∑

α2T

f (α)

(

g1fγ:γ(cid:20)βg

)

(α) =

∑

α2T

f (α)g(α) =

∆f (α) (cid:1) I(g1fγ:γ(cid:20)βg)(α)

Where we used the previous lemma on the third equality.
Note f is superadditive on T and so ∆f (cid:21) 0. Finally, for each α 2 T , we have
I(g1fγ:γ(cid:20)βg)(α) (cid:20) Ig(α) (cid:1) 1fγ:γ(cid:20)βg(α) and so the desired inequality follows.

Next we have a two-dimensional analogue of Lemma (3.13).

33

Lemma 3.16. Let T 2 be a bi-tree and f, g : T 2 ! [0,1) Then

(If )(Ig) (cid:20) I(If (cid:1) g + I1f (cid:1) I2g + I2f (cid:1) I1g + f (cid:1) Ig)

Proof. Recall that I = I1I2. We apply Lemma (3.13) first to I1 and then to I2 (note I1, I2
commute) to get the desired result.

The following estimate is key for the proof of Theorem 3.5 for n = 2.

Lemma 3.17. Let T 2 be a bi-tree and f : T 2 ! [0,1) a function that is superadditive in
each parameter separately. Let w be a positive product weight. Suppose that
supp f (cid:18) fI(wf ) (cid:20) δg. Then
∑

∑

w(I1(w1f ))2(I2(w2f ))2 (cid:20) 4δ2

T 2

wf 2

T 2

Proof. By Lemma 3.13 and commutativity of operations in different coordinates,

∑

T 2

w(I1(w1f ))2(I2(w2f ))2 (cid:20) 4

= 4

= 4

T 2

∑
∑
∑
∑

T 2

T 2

= 4

T 2

By Corollary 3.15, we have

wI1(w1f (cid:1) I1(w1f )) (cid:1) I2(w2f (cid:1) I2(w2f ))

I1(w1f (cid:1) I1(wf )) (cid:1) I2(w2f (cid:1) I2(wf ))

(cid:3)
2 (w1f (cid:1) I1(wf )) (cid:1) I

(cid:3)
1 (w2f (cid:1) I2(wf ))

I

(cid:3)
2 (f (cid:1) I1(wf )) (cid:1) I

(cid:3)
1 (f (cid:1) I2(wf ))

wI

I

(cid:3)
(cid:3)
1 (∆1f (cid:1) I1I2(wf )).
1 (f (cid:1) I2(wf )) (cid:20) I
34

Since f is superadditive, ∆1f (cid:21) 0. Hence,

(cid:3)
1 (f (cid:1) I2(wf )) (cid:20) I

(cid:3)
1 (∆1f (cid:1) I(wf )) (cid:20) I

(cid:3)
1 (∆1f (cid:1) δ) = δf.

I

(3.24)

where we used ∆1f is supported on fI(wf ) (cid:20) δg (If ∆1f (α) > 0 then f (α) 6= 0 for
otherwise ∆1f (α) (cid:20) 0). Arguing similarly for the other term and inserting these in the last
estimate above, we obtain the claim.

Finally we prove theorem 3.5 for n = 2.

Lemma 3.18. Let µ, ρ be positive measures on T 2 and δ > 0. Let w be a positive product
weight. Then

∫

δ dρ . δ
Vµ

1

2jρj 1

T 2

1

4E[ρ]

1
4

(3.25)

µ. Note that f is superadditive in each parameter. Also

I(cid:3)

Proof. Let f := 1Vµ(cid:20)δ
I(wf ) = Vµ
δ

(cid:20) Vµ (cid:20) δ on suppf. Finally, Eδ[µ] =
∫

∫

wf 2. Then

(∫
(
(∫

2

Vµ

I(wf )dρ

δ dρ =

T 2

T 2

∫

T 2
(cid:20) jρj1/2
by Lemma 3.16 (cid:20) jρj1/2
. jρj1/2
T 2
. jρj1/2E[ρ]1/4

)

1/2

1/2

(I(wf ))2dρ
)
I(I1(wf ) (cid:1) I2(wf ) + (wf ) (cid:1) I(wf ))dρ
T 2
(∑
w(I1(w1f ) (cid:1) I2(w2f ) + f (cid:1) I(wf ))I(cid:3)

(
)
I1(w1f ) (cid:1) I2(w2f ) + f (cid:1) I(wf )

ρ

1/2

)

1/4

2

2Eδ[µ]
∑

T 2

)

w

T 2

35

using the inequality (a + b)2 (cid:20) 2(a2 + b2) and Lemma 3.17
)

(

∑

1/4

. jρj1/2E[ρ]1/4

10δ2

wf 2

T 2

. jρj1/2E[ρ]1/4δ1/2Eδ[µ]1/4

Next we develop further our techniques for dimension n = 3. The next lemma is an

analogue of Lemma 3.16.

Lemma 3.19. Let T 3 be a tri-tree and f, g : T 3 ! [0,1). Then
)

( ∑

(If )(Ig) (cid:20) I

IAf (cid:1) IAcg

,

A(cid:18)f1,2,3g

∏

i2A Ii.

where IA =

Proof. Note that I = I1I2I3. We use Lemma (3.13) three times along with the
commutativity of I1, I2, I3.

Corollary 3.20. Let 0 < δ (cid:20) λ/4. Let f : T 3 ! [0,1) with supp f (cid:18) fIf (cid:20) δg. Then

(If )1λ(cid:20)If(cid:20)2λ (cid:20) 4λ

(cid:0)1I

Iif (cid:1) I(i)f (cid:1) 1If(cid:20)2λ

,

( ∑

i2f1,2,3g

∏

where I(i) :=

j6=i Ij.

)

)

Proof. Substituting f = g, Lemma 3.19 implies that

(If )2 (cid:20) I

Iif (cid:1) I(i)f + 2f (cid:1) If

(

∑

2

i2f1,2,3g

36

Using the support condition, this implies

(If )1λ(cid:20)If(cid:20)2λ (cid:20) λ
(cid:20) λ
[

(
∑
(cid:0)1(If )21λ(cid:20)If(cid:20)2λ
(cid:0)1I
(
∑

i2f1,2,3g

2

(cid:20)

(cid:0)1I
λ

2

i2f1,2,3g

)

Iif (cid:1) I(i)f + 2δf
)

Iif (cid:1) I(i)f

1λ(cid:20)If(cid:20)2λ

]

(cid:0)1If

+ 2δλ

1λ(cid:20)If(cid:20)2λ

Since 2δλ

(cid:0)1 (cid:20) 1/2, this implies

(If )1λ(cid:20)If(cid:20)2λ (cid:20) 4λ

(cid:0)1I

(cid:20) 4λ

(cid:0)1I

)

Iif (cid:1) I(i)f

1λ(cid:20)If(cid:20)2λ

)

Iif (cid:1) I(i)f (cid:1) 1If(cid:20)2λ

i2f1,2,3g

( ∑
( ∑
∏

i2f1,2,3g

Now we need one more lemma. Denote w(i) :=
Lemma 3.21. Let f : T 3 ! [0,1) be superadditive on each parameter separately. Let w be
a positive product weight. Suppose that supp f (cid:18) fI(wf ) (cid:20) δg. Then for every i 2 f1, 2, 3g

j6=i wj and so w = wi (cid:1) w(i).

(
)
Ii(wif ) (cid:1) I(i)(w(i)f )

∑

w

T 3

21I(wf )(cid:20)λ

(cid:20) 2δλ

wf 2

∑

T 3

Proof. By Lemma 3.13 for Ii we have

∫

w

(

)
∫
Ii(wf ) (cid:1) I(i)(w(i)f )
∫
(cid:20) 2
∫

= 2

= 2

21I(wf )(cid:20)λ

37

wIi(wif (cid:1) Ii(wif )) (cid:1) (I(i)(w(i)f ))21I(wf )(cid:20)λ
Ii(wif (cid:1) Ii(wf )) (cid:1) (I(i)(w(i)f )) (cid:1) (I(i)(wf ))1I(wf )(cid:20)λ
(I(i)(w(i)f )) (cid:1) (I(i)(wf ))1I(wf )(cid:20)λ
wif (cid:1) Ii(wf ) (cid:1) I

(

(cid:3)
i

(3.26)

)

By Corollary 3.15, we have

(

(cid:3)
I
i

(I(i)(w(i)f )) (cid:1) (I(i)(wf )) (cid:1) 1I(wf )(cid:20)λ

)

(

(cid:20) I

(cid:3)
i

∆i(1I(wf )(cid:20)λ

)
(cid:1) I(i)(w(i)f )) (cid:1) Ii(I(i)(wf )

As f is superadditive in the i-th coordinate, the function ˜f := 1I(wf )(cid:20)λ
˜f (cid:21) 0. Moreover, if ∆i
superadditive in the i-th coordinate. Hence ∆i
˜f (α) 6= 0 which means α 2 fI(wf ) (cid:20) λg and so supp ∆i
)
˜f (cid:1) λ

))

(

(

(

(cid:3)
I
i

˜f (cid:1) Ii

∆i

= λ1I(wf )(cid:20)λ

˜f (α) > 0 then
˜f (cid:18) fI(wf ) (cid:20) λg. Hence,

I(i)(wf )

(cid:20) I

(cid:3)
i

∆i

(cid:1) I(i)(w(i)f )

(cid:1) I(i)(w(i)f ) is also

Using this bound, we obtain

(3.26) (cid:20) 2λ

= 2λ

= 2λ

∫
∫
∫

wif (cid:1) Ii(wf ) (cid:1) I(i)(w(i)f )
f (cid:1) Ii(wf ) (cid:1) I(i)(wf )
(cid:3)
i (f (cid:1) I(i)(wf ))
wf (cid:1) I

As in (6.10), we see that

(cid:3)
i (f (cid:1) I(i)(wf )) (cid:20) δf.

I

This implies the conclusion of the lemma.

The next result is a key tool for dimension 3.

Lemma 3.22 (Small energy majorization on tri-tree). Let T 3 be a tri-tree and
f : T 3 ! [0,1) a function that is superadditive in each parameter separately. Let w be a
positive product weight. Suppose that supp f (cid:18) fI(wf ) (cid:20) δg. Let λ (cid:21) 4δ. Then there exists

38

φ : T 3 ! [0,1) such that

a) I(wφ) (cid:21) I(wf ), where I(wf ) 2 [λ, 2λ],

∑

∑

T 3

wφ2 (cid:20) C

δ
λ

wf 2,

b)

T 3

where C is an absolute constant.

Proof. Put

(cid:0)1

φ := 4λ

∑

i2f1,2,3g

Iiwif (cid:1) I(i)w(i)f (cid:1) 1I(wf )(cid:20)2λ

For part a) we use Corollary 3.20 with f replaced by wf. Note supp(wf ) (cid:18) supp(f ) and
thus

I(wf )1λ(cid:20)I(wf )(cid:20)2λ

(cid:20) 4λ

(cid:0)1I

Ii(wf ) (cid:1) I(i)(wf ) (cid:1) 1I(wf )(cid:20)2λ

)

( ∑

i2f1,2,3g

Ii(wif ) (cid:1) I(i)(w(i)f ) (cid:1) 1I(wf )(cid:20)2λ

)

= I(wφ)

(

∑

(cid:0)1I

4λ

w

i2f1,2,3g
For part b) we apply Lemma 3.21.

Lets prove Theorem 3.5 for n = 3

Lemma 3.23. Let µ, ρ be positive measures on T 3 and δ > 0. Let w be a positive product
weight. Then

∫

δ dρ .
Vµ

T 3

(

) 1
δEδ[µ]E[ρ]jρj

3

(3.27)

Proof. Without loss of generality, Eδ[µ] 6= 0 and ρ 6(cid:17) 0. Let λ > 0 be chosen later.
Let f := I(cid:3)
supp f, and Eδ[µ] =

µ (cid:1) 1Vµ(cid:20)δ(α). This function is superadditive. Also, I(wf ) = Vµ

wf 2.

∫

δ

(cid:20) Vµ (cid:20) δ on

39

For m = 0, 1, . . . we use Lemma 3.22 with data (w, f, δ, 2mλ) to get

and

Hence,

∫

T 3

I(wf ) (cid:1) 12mλ<I(wf )(cid:20)2m+1λ
∑

∑

wϕ2

m . δ
2mλ

T 3

(cid:20) I(wϕm),

wf 2.

T 3

∫

Vµ

δ dρ =

Vµ

δ dρ +

(cid:20)λg

fVµ
δ
(cid:20) λjρj +

∫

1∑

f2mλ<Vµ
δ

(cid:20)2m+1λg

m=0

Vµ

δ dρ

T 3

∫
∫
(∑

T 3

m=0

1∑
1∑
1∑
1∑

m=0

= λjρj +

(cid:20) λjρj +

(cid:20) λjρj +

I(wϕm)dρ

dρ

wϕmI(cid:3)
)
1/2E[ρ]1/2

wϕ2
m

m=0

T 3
C(δ/(2mλ))1/2Eδ[µ]1/2E[ρ]1/2.

m=0

(cid:20) λjρj + C(δ/λ)1/2Eδ[µ]1/2E[ρ]1/2

Recall we need λ (cid:21) 4δ. Let λ := (δEδ[µ]E[ρ])1/3jρj(cid:0)2/3. If λ (cid:21) 4δ then we obtain (3.27) by
substituting it in the last line above. Otherwise, if λ < 4δ, then we automatically get (3.27)

without using any lemmata.

3.3 Improvement in the case of n = 2

We state here without a proof an improvement of the energy estimate of Lemma 3.18.

40

Theorem 3.24. Let µ be a measure on T 2 and w (cid:17) 1. Then, for every τ 2 (0, 1):

Eδ[µ] .τ (δjµj)1(cid:0)τE[µ]τ

with the implicit constant depending on τ.

41

Chapter 4

Capacitary conditions and Embedding

4.1 General Capacitary condition implies embedding

As we said this result is known for n = 2, first proven in [4]. We prove it here for n = 3

following closely the proof of n = 2. What would be useful here is Lemma 3.22 which first

appeared in [19]. Hence, as we have established the equivalence between the Box condition

and the embedding and since we show the General capacitary condition (2.5) is also

equivalent to the embedding, then the Box condition implies the general capacitary

condition. As we discuss in subsection 4.2 the Box condition is enough to prove the much

weaker Box capacitary condition, but here, indirectly we show something much stronger.
Here we take w (cid:17) 1.
To begin, suppose we have a measure µ on T 3 and for any E (cid:18) ∂T 3 the following
condition holds

µ(E) . Cap(E)

which is condition (2.5). This condition implies for any f 2 ℓ2(T 3)
∫
∑

∫ 1

λ µfIf (cid:21) λg .

22kµfIf (cid:21) 2kg .

(If )2dµ (cid:24)

T 3

0

∑

k2Z

42

22k CapfIf (cid:21) 2kg

(4.1)

k2Z

Recall the measure µ is non-zero only on the boundary of T 3 and thus we are interested to
estimate the capacity of the sets Ek := fIf (cid:21) 2kg \ ∂T 3. More specifically we wish to show

∑

k2Z

22k Cap(Ek) . kfk2

ℓ2(T 3)

(4.2)

and this would finish the inequality in (4.1). By duality we conclude (1.10) holds. As it is
mentioned in [1], for a set E, the minimizer in (2.4) is a function of the form I(cid:3)
µE is measure on T 3 which is called the equilibrium measure of E. The measure µE
satisfies VµE (cid:17) 1 on E and also, since E is compact we also have supp(µE) (cid:18) E. This
implies Cap(E) =

VµE dµE = µE(E). For every k 2 Z we get such a

µE where

µE)2 =

(I(cid:3)

∫

∫

measure µk and by arguments found in the Section 3.1 of [4] we deduce (4.2) as long as the
following is proven:

∫

∑

∑

k2Z

j(cid:20)k

∫

∑

k2Z

2j+k

T 3

Vµk dµj .

22k

Vµk dµk

T 3

(4.3)

In turn (4.3) is proven exactly in the same manner as in this paper, if the next Lemma is

true;

Lemma 4.1 (Lemma 3.2 of [4]). Let F, E (cid:18) ∂T 3 be two sets such that Cap(F ) (cid:20) Cap(E)
and let µF , µE be their equilibrium measures. Then, we have

∫

T 3

VµE dµF . jµEj1/3 (cid:1) jµFj2/3

Proof. We have seen above that Cap(F ) = jµFj. If this number is 0 we have nothing to
prove.

43

Otherwise, we set λ =

∫

T 3 VµE dµF

jµFj

and for k 2 N [ f0g we define the sets

F(cid:0)1 = fω 2 F : VµE (cid:20) 1g

Fk = fω 2 F : 2k < VµE (cid:20) 2k+1g

As T 3 is finite, then there exists some m 2 N such that for any k (cid:21) m, Fk = ?. Now, by
using Proposition 4.2 we see that

Cap(Fk) . jµEj

23k

(4.4)

Next, let j such that 2j < λ (cid:20) 2j+1 and we write

VµE dµF =

∑

∫
2k+1 (cid:1) µF (Fk) (cid:20) j(cid:0)2∑

k(cid:21)(cid:0)1

Fk

∫
∑

T 3

k(cid:21)(cid:0)1

VµE dµF (cid:20)

2k+1 (cid:1) µF (Fk) +

k=(cid:0)1

λ (cid:1) jµFj =

(cid:20)

∑

k(cid:21)j(cid:0)1

2k+1 (cid:1) Cap(Fk)

(4.5)

and the last inequality follows by Lemma 5.6 of [4] : As VµF = 1 on Fk then
µF (Fk) (cid:20) Cap(Fk). Using (4.4) and that the sets Fk are disjoint, we get

∑

k(cid:21)j(cid:0)1

2k+1 1
23k

(cid:20)

(4.5) (cid:20) 2j(cid:0)1 (cid:1) µF (F ) + C (cid:1) jµEj (cid:1)

(cid:20) λ
2

(cid:1) jµFj + C (cid:1) jµEj

λ2

44

which implies

jµFj . jµEj

λ3

and this finishes the proof given the definition of λ.

Proposition 4.2. Let µ be a measure on T 3 with Vµ (cid:20) 1 on supp µ. Then, for any λ (cid:21) 1
we have

Cap(fVµ > λg) . E[µ]

λ3

Proof of Proposition 4.2. Consider f = I(cid:3)
that β 2 supp µ. But then by assumption If (β) = II(cid:3)
of I we have that If (α) (cid:20) 1. Hence

µ, δ = 1. If f (α) 6= 0 then there is β (cid:20) α such

µ(β) = Vµ(β) (cid:20) 1. By monotonicity

supp f (cid:26) fIf (cid:20) δ = 1g,

and we are in the assumptions of small energy majorization Lemma on tri-tree 3.22. We

apply it with data (f, δ = 1, λ := 2mλ) to get functions φm, m = 0, 1, . . . such that

Iφm (cid:21) If = Vµ, where Vµ 2 [2mλ, 2m+1λ],

(cid:0)mλ

2

(cid:0)1Iφm (cid:21) 1, where Vµ 2 [2mλ, 2m+1λ],

(cid:0)mλ

(cid:0)1φm and hence,

m 2

which means that

∑

Now define φ :=

Iφ (cid:21) 1, where Vµ 2 [λ,1),

45

Now, by Minkowski’s integral inequality we get:

∫

(
(
φ2 (cid:20)
1∑

(cid:0)1

λ

T 3

.

(cid:0)1
λ

1∑

m=0

(cid:0)m
2

(∫
(∫

T 3

(cid:0)1/22

(cid:0) 3m
2

λ

)
)

)
)
2 . λ

2 .

1/2

φ2
m

∫

f 2

T 3

(cid:0)3

m=0

which finishes the proof, since f = I(cid:3)

µ and so

1/2

f 2

T 3

∫
T 3 f 2 = E[µ].

4.2 Box condition implies Box Capacitary Condition

Let ν be a finite positive measure in [0, 1)n with jνj (cid:20) 1. In this section we show that the
box condition (1.6) on n-tree implies box-capacitary condition (2.6) on n-tree of depth N,
for all n (cid:21) 2. In other words, if
∑

wP ν(P )2 (cid:20) ν(R),

8R 2 T n

P(cid:18)R

then

ν(R) . Cap(R),

[

8R 2 T n
)

, 0 (cid:20) k (cid:20) N, 0 (cid:20) m (cid:20) 2N(cid:0)k (cid:0) 1 and
Recall the nodes of T n have the form
we call these dyadic n-rectangles (Cartesian product of dyadic intervals). By construction

2N(cid:0)k , m+1
2N(cid:0)k

m

these are inside the unit n-cube [0, 1)n. Recall that

Cap(R) (cid:25)

1

1jI2j (cid:1) .. (cid:1) log2

1jInj

log2

1jI1j (cid:1) log2
46

where R = I1 (cid:2) .. (cid:2) In and if jIij = 1 we just replace this term by 1 in the denominator.
We will prove our result for a hooked dyadic n-rectangle, i.e. a dyadic n-rectangle which
has a vertex at (0, 0, .., 0) 2 Zn. This does not change at all the arguments although it
simplifies the calculations.

Let R be a hooked dyadic n-rectangle of the form

[0, 2

2

(cid:0)N +m2] (cid:2) .. (cid:2) [0, 2
(cid:0)ns+n. If we define a := 2

(cid:0)N +mn]. Then there is an s 2 N such that
(cid:0)s then we have an < ν(R) (cid:20) 2nan.

(cid:0)N +m1] (cid:2) [0, 2
(cid:0)ns < ν(R) (cid:20) 2
We connect the points (m1, m2, .., mn) and (N, N, .., N ) 2 Zn by a line L. In parametric
form this line is L : (x1, x2, .., xn) = (m1, m2, .., mn) + t(N (cid:0) m1, N (cid:0) m2, .., N (cid:0) mn)
where 0 (cid:20) t (cid:20) 1. We are going to consider points on that line which satisfy some
conditions. The points are defined recursively by

Pk = (m1 + m1,1 + .. + m1,k, m2 + m2,1 + .. + m2,k, .., mn + mn,1 + .. + mn,k) where
mj.k 2 Z+ with j 2 f1, .., ng, k 2 f1, .., ig and i 2 N is fixed (to be specified). Also,
P0 = (m1, .., mn).
For every k 2 f1, .., ig, the first condition we impose is Pk 2 L and the second condition is

m1,k (cid:1) m2,k (cid:1) .. (cid:1) mn,k =

2n
k(cid:0)1
an

where a0 := a and ak = 2ak(cid:0)1 with k 2 f1, .., ig. Working recursively from k = 1, we see
that for every j, l 2 f1, .., ng we get

mj,k
N (cid:0) mj

=

ml,k
N (cid:0) ml

for all k 2 f1, .., ig

(4.6)

47

Using these equalities and m1,k (cid:1) m2,k (cid:1) .. (cid:1) mn,k = 2n
k(cid:0)1
an
every j 2 f1, .., ng

we deduce the next formula for

mj,k =

(cid:1)

2
ak(cid:0)1

∏
(N (cid:0) mj)

n(cid:0)1
n
(N (cid:0) ml)

l6=j

l2f1,..,ng

1
n

(4.7)

be the dyadic n-rectangle with fixed vertex at (0, 0, .., 0) 2 Zn and of size
0
0
0
Let Qm
2,..,m
1,m
n
0
0
0
1 (cid:2) 2N(cid:0)m
2N(cid:0)m
2 (cid:2) .. (cid:2) 2N(cid:0)m
n. For simplicity, let ν(m
).
0
0
Then, for (m
1, m
2, .., m
m1,1 (cid:1) m2,1 (cid:1) .. (cid:1) mn,1 other such dyadic n-rectangles which contain R. Now we start using
(1.6) for the first time (we normalize the constant to be 1)

0
n) = (m1 + m1,1, m2 + m2,1, .., mn + mn,1) we have at least

0
0
n) := ν(Qm
1,m

0
0
2,..,m
n

0
1, m

0
2, .., m

ν(m1 + m1,1, m2 + m2,1, .., mn + mn,1) (cid:21) m1,1 (cid:1) m2,1 (cid:1) .. (cid:1) mn,1 (cid:1) ν(R)2 (cid:21)

m1,1 (cid:1) m2,1 (cid:1) .. (cid:1) mn,1 (cid:1) a2n = 2n (cid:1) an

0 = an
1

using that m1,1 (cid:1) m2,1 (cid:1) .. (cid:1) mn,1 = 2n
an
0
and after k steps we have

and the definition of a0, a1. We continue recursively

ν(m1 + m1,1 + .. + m1,k, m2 + m2,1 + .. + m2,k, .., mn + mn,1 + .. + mn,k) (cid:21) an

k

We stop the recursion (and choose i = k) at the minimal k such that

m1 + m1,1 + .. + m1,k > N (by (4.6) this is equivalent to mj + mj,1 + .. + mj,k > N for any
j 2 f1, .., ng) or when k = s.

48

If the first case happens, then by (4.7) we have

N (cid:0) m1 < m1,1 + m1,2 + .. + m1,i =

∏
(N (cid:0) m1)
l6=1

n(cid:0)1
n
(N (cid:0) ml)

l2f1,..,ng

(

(cid:1)

1
n

2
a0

+

2
a1

+ .. +

2
ai(cid:0)1

)

(cid:20) (N (cid:0) m1)

n(cid:0)1
n
(N (cid:0) ml)

(cid:1) 4
a

1
n

∏

l6=1

l2f1,..,ng

as the sum is a geometric series by the definition of ai. If we rearrange the two ends of the
above inequalities we get

an .

(N (cid:0) m1) (cid:1) (N (cid:0) m2) (cid:1) .. (cid:1) (N (cid:0) mn)

1

= C (cid:1) cap(R)

Keeping in mind that ν(R) (cid:20) 2nan we have the desired inequality. If the second case
happens, namely i = s and since as = 1 we see that

ν(m1 + m1,1 + .. + m1,s, m2 + m2,1 + .. + m2,s, .., mn + mn,1 + .. + mn,s) (cid:21) 1

Our measure is normalized such that jνj (cid:20) 1. Intuitively this means there is no much room
left. If m1 + m1,1 + .. + m1,k > N
inequality. If m1 + m1,1 + .. + m1,k (cid:20) N
recursion to get

2 then by reasoning as above we get the same desired

. We do one more step in the

⌊

2 then let M =

N
2

⌋

ν(m1 + m1,1 + .. + m1,s + M, m2 + m2,1 + .. + m2,s, .., mn + mn,1 + .. + mn,s) (cid:21)

M (cid:1) ν(m1 + m1,1 + .. + m1,s, m2 + m2,1 + .. + m2,s, .., mn + mn,1 + .. + mn,s) (cid:21) M

49

which is of course a contradiction as N is large. Therefore the desired inequality has been

established.

4.3 Box capacitary does not imply general capacitary

condition

One could ask: Since Box condition (1.6) is able to imply the embedding (1.5), is it also

true that Box capacitary condition (2.6) also implies the same embedding? The answer is

no. This was proven by Stegenga in [28] for n = 1. We now give a counterexample (due to

P. Mozolyako) in the case of a simple tree.
Consider a simple tree T of depth 2N for N 2 N. Let us describe the construction of a
“Cantor” type set with small capacity and of full-measure. For simplicity we assume the

root of the tree is of generation 1. We choose the left-most and right-most boundary points

of the tree. Then we move to the 2nd generation and for each of these two nodes we

consider the sub-graphs with these as a root. We choose the left-most and right-most

boundary points of each sub-graph (we count duplicate nodes just once). Then we go to

the 4th generation and we consider the sub-graphs with roots the left-most and right-most

descendants of the 2nd generation (hence we look at 4 sub-graphs in total). As before, for

each of these sub-graphs we choose the left-most and right-most boundary points. We
continue doing this for all generations of order 2k with 0 (cid:20) k (cid:20) N (cid:0) 1. The total number of
the boundary points we choose is 2N out of a total 22N(cid:0)1 boundary points. Their selection
is uniform and we now give the exact positions of these boundary points.

Let us look at the boundary of the simple tree from the left to the right and consider the
set E which is comprised of the boundary nodes in positions m (cid:1) 22N(cid:0)N , n (cid:1) 22N(cid:0)N + 1 for

50

m = 1, .., 2N(cid:0)1 and n = 0, .., 2N(cid:0)1 (cid:0) 1. For n = 0 we get the left-most and for m = 2N(cid:0)1
we get the right-most boundary point. The measure µ is constructed as follows: to each

2N . Therefore, jµj = 1. We also set ν = I(cid:3)

node ω of E we set µ(ω) = 1
jνj = ν(E) = 1. For all I 2 T we show ν(I) . Cap(I), and also Cap(E) (cid:20) 2
box capacitary condition holds with bounded constant while the general capacitary

µ and so

N . Hence, the

condition can only hold with a constant which depends on N. We now prove these two

facts.
First, let any I 2 T . Let ℓ be the generation of I from the top, with 1 (cid:20) ℓ (cid:20) 2N. Then,
there is some k, 0 (cid:20) k (cid:20) N such that 2k (cid:20) ℓ < 2k+1. By our construction, the amount of
boundary points which are descendants of I and have µ-mass is 2N(cid:0)k. Therefore,
ν(I) = 2N(cid:0)k 1
2N
>

1
as each such boundary node has mass 1
. On the other hand,
2N
2k
and thus ν(I) (cid:20) 2 (cid:1) Cap(I). Hence ν satisfies the Box capacitary

Cap(I) =

=

1

1
ℓ

2k+1

(cid:1)∫

N (cid:20) 2. Using this and

condition.
Second, for any ω 2 ∂T we show N (cid:20) Vµ(ω) (cid:20) 2N and so 1 (cid:20) V µ
N dµ (cid:20) 2jµj
the Cauchy-Schwarz inequality we get Cap(E) (cid:20) 1
V µ
which gives the
second estimate as jµj = 1. To prove this, fix ω 2 ∂T and let any α (cid:21) ω. As before, we
have I(cid:3)
Next, let ω =: α
of the tree T . Then we have

where k is such that, 2k (cid:20) ℓ < 2k+1 and ℓ is the generation of α.
2N (cid:20) .. (cid:20) α1 be an enumeration of the ancestors of ω where α1 is the root

µ(α) = ν(α) =

1
2k

N

N

2N∑

ℓ=1

Vµ(ω) =

I(cid:3)

µ(αℓ) = I(cid:3)

µ(ω) +

N(cid:0)1∑

∑

k=0

2k(cid:20)ℓ<2k+1

I(cid:3)

µ(αℓ) = 1 +

N(cid:0)1∑
(2k+1 (cid:0) 2k)

k=0

1
2k

= N + 1

Finally, for any n 2 N with n (cid:21) 2 and an n-tree T n, consider a measure µ on ∂T n which is

51

a product of measures constructed above. Hence, as capacity is of product nature, the

above counterexample gives rise to a counterexample for higher dimensions.

52

Chapter 5

Connections with known results and

applications

5.1 Maximal function and embedding

In this section we consider a set P which is any finite n-tree with n 2 N. More generally
one could think of P as any partially ordered set. Given a measure µ on P we define the
corresponding maximal operator by

Mµψ(ω) := sup
α(cid:21)ω

hjψjiµ(α),

hψiµ(α) :=

I(cid:3)
(ψµ)(α)
I(cid:3)µ(α)

,

(5.1)

with the convention 0/0 = 0. This definition recovers the usual dyadic maximal operator

on the tree and the bi-parameter maximal operator on the bi-tree. We have the following

connection between these constants:

Proposition 5.1. Let µ be a measure on a set P, with P as above. Then

[w, µ]CE = kMµk2

L2(µ)!L2(µ)

sup
w:[w,µ]C

(cid:20)1

.

(5.2)

Example 5.2. If P = T is a usual tree, then the maximal function Mµ is essentially the
martingale maximal function, and it is well-known that it is bounded on Lp(µ) with norm

53

at most p

0. In particular the right-hand side of (5.2) equals 4. This is the sharp constant in

the Carleson embedding theorem on the tree [25].
Example 5.3. If P = T n = T1 (cid:2) .. (cid:2) Tn is an n-tree and µ = µ1 (cid:2) .. (cid:2) µn is a product
measure, then the n-parameter maximal operator (5.1) can be majorized by the

composition of n one-parameter maximal operators

Mµψ (cid:20) M1,µ1

(cid:14) .. (cid:14) Mn,µnψ,

which are also defined by (5.1) but on the simple trees Ti. Using L2 bounds for the
one-parameter maximal operators we see that the right-hand side of (5.2) is bounded by

4n. Hence for product measures µ and arbitrary weights w Proposition 5.1 gives the
implication (2.1) =) (1.10). As the n-dimensional Lebesgue measure is product, this is
connected to the results of Chang in the case of the bi-disc (see [10]).

Remark 5.4. As we said in the first example, it is known for n = 1 the boundedness of

Maximal function is equivalent to the Carleson Embedding theorem. Additionally, by
Proposition 5.1 and our results we see the rectangular strong Maximal function on T n for
n = 2, 3 is bounded as soon as the Box condition is true and the weight w is of product
form: recall that [w, µ]Box (cid:20) [w, µ]C . [w, µ]Box.
Proof of Proposition 5.1. We begin with the inequality (cid:20) in (5.2). Let ψ : P ! [0,1) be a
non-negative function with kψk

L2(µ) = 1. Then

∑

α2P

w(α)I(cid:3)

(ψµ)(α)2 =

∫ hψiµ(α)

2sds

0

w(α)I(cid:3)

µ(α)2ds

∑
∫ 1

α2P

µ(α)2

w(α)I(cid:3)
∑

=

2s

0

α:hψiµ(α)>s

54

w(α)I(cid:3)

µ(α)2ds

2sµfα : Mµψ(α) > sgds

∫ 1

2s

0

(cid:20)

∑
∫ 1

α:Mµψ(α)>s

0

(cid:20) [w, µ]C
= [w, µ]CkMµ(ψ)k2
(cid:20) [w, µ]CkMµk2

L2(µ)

L2(µ)!L2(µ)

.

Here we have used that the superlevel sets fα : Mµψ(α) > sg are down-sets. Taking
supremum over all weight w such that [w, µ]C (cid:20) 1 we obtain the inequality (cid:20) in (5.2).

Now we will show the inequality (cid:21) in (5.2). Let ψ : P ! [0,1) be a non-negative function
such that kψk
L2(µ) = 1. The set ∂P consists of the minimal elements of P. Recall that µ is
non-zero only on ∂P. For each α 2 P with I(cid:3)

(ψµ)(α) 6= 0 let

0

A

(α) := fω 2 ∂P, ω (cid:20) α j Mµψ(ω) = hψiµ(α)g,

0
and let A

(α) := ; otherwise. Enumerate P = fα1, α2, . . .g and set

0
A(αj) := A

(αj) n

0
A

(αj0).

∪

j0<j

Then, as the sets A(α) are pairwise disjoint:

∑

ω2P

(Mµψ)2(ω)µ(ω) =

=

∑

ω2A(α)
w(α)(I(cid:3)

hψiµ(α)2µ(ω)

(ψµ)(α))2,

∑
∑

α2P

α2P

55

where

w(α) := (I(cid:3)

(cid:0)2

µ(α))

∑

µ(ω)

ω2A(α)

with the convention w(α) = 0 if A(α) = ;. Since the sets A(α) are disjoint and consist of
minimal elements then for every down-set D (cid:18) P we have

∑

α2D

w(α)(I(cid:3)

(µ)(α))2 =

µ(ω) (cid:20)

µ(ω) (cid:20) µ(D),

∑

∑

∑

α2D

ω2A(α)

ω2D\∂P

and thus [w, µ]C (cid:20) 1. Also, by the calculations above

[w, µ]CE (cid:21) kMµψk2

L2(µ)

.

By first taking the supremum over w with [w, µ]C (cid:20) 1 and then over ψ we obtain the
inequality (cid:21) in (5.2).

Similarly we can also prove:

Proposition 5.5. Let µ be a measure on ∂P. Then

sup
w:[w,µ]C

[w, µ]HC = sup
E(cid:18)P

(cid:20)1

1jµ1EjkMµ(1E)k2

L2(µ)

.

5.2 A comparison with a result of E. Sawyer

As we mentioned before, two weighted embeddings have enjoyed much attention in the last

50 years. In a continuous setting the paper [26] of E. Sawyer gives a characterization of the

2-dimensional embedding in terms of three “Box Conditions”. Lets translate his results to

56

the case of bi-trees. Suppose the weight w is supported on a set of the form:

supp w (cid:26) fα 2 T 2 j α (cid:21) ω0g.

(5.3)

where ω is a fixed point in ∂T 2. Then, in [26] we consider the discrete measure on a bi-tree
instead of the Lebesgue measure. The result becomes:

Theorem 5.6 (cf. [26, Theorem 1(A)]). Suppose that the weight w satisfies the support
condition (5.3). Then [w, µ]CE is finite if and only for some A < 1 and every β 2 T 2 with
β (cid:21) ω0 the following conditions hold:
∑
∑

I(cid:3)
µ(β)Iw(β) (cid:20) A2,
µ(α)Iw(α)2 (cid:20) A2Iw(β),

α(cid:21)β(cid:21)ω0

(5.4b)

w(α)I(cid:3)

µ(α)2 (cid:20) A2I(cid:3)

µ(β).

(5.4a)

(5.4c)

ω0(cid:20)α(cid:20)β

No two of these conditions suffice to ensure [w, µ]CE < 1.

Remark 5.7. The last condition (5.4c) is just the box condition (1.11). In other words, if

we restrict the weight to be supported only on the hooked rectangles, but drop the

requirement that it has a product structure, we see that the single box test (1.11) is getting

replaced by three single box tests for the pair w, µ.

Remark 5.8. The second condition (5.4b) is not needed in our setting. Recall that µ is

non-zero only on the boundary of the bi-tree and hence (5.4b) is implied by the first one.

Remark 5.9. A careful reading of our proof in section 4.2 shows that for a measure µ on
the boundary of the bi-tree, a weight w and for every α 2 T 2, we have

57

the following condition:

w(α) (cid:1) I(cid:3)

µ(α) . Cap(α)

as long as the box condition (1.11) holds. Therefore, if the weight w satisfies the support

condition (5.3) and moreover is non-increasing, the inequality above easily implies (5.4a).

To summarize: if w is non-increasing and satisfies the support condition (5.3), then the

Box condition (5.4c) implies the conditions (5.4a) and (5.4b) and hence the embedding
(1.10) holds as well. For an application of this, consider the weight w which satisfies w (cid:17) 1
on the set fα 2 T 2 j α (cid:21) ω0g and 0 otherwise. However, this example is the connecting
point between our results and the result in [26]: as this particular weight w is a product

weight, we already know the single box condition (1.11) is sufficient to imply (1.10).

58

Chapter 6

Necessary and sufficient conditions

6.1 A product weight is not necessary

As we know, for n (cid:21) 2 the n-tree does not have a nice geometric structure. In the case
n = 1, for any ω 2 ∂T the set fα : α (cid:21) ωg is totally-ordered. However for n (cid:21) 2 this is not
true for any ω 2 ∂T 2. Albeit, one could consider a sub-graph of the n-tree which has this
property; We start with [0, 1)n and we keep all its descendants which are cubes, namely

Cartesian products of n dyadic sub-intervals of [0, 1), of generation at most N and of the

same length. Thus, if we modify the weight w to be non-zero only on such cubes, then we

have a structure similar to a simple tree. Note that such a weight is non-product and that
the sum on the LHS of (1.5) is over a partially ordered set P with the following property;
for any ω 2 ∂P the set fα : α (cid:21) ωg is totally-ordered. As we discussed in the introduction,
for a set P with this property, the same proof as in the case n = 1 holds. Therefore, for
such a weight w the Box condition (1.6) implies (1.5) which implies a product weight is not
a necessary condition for any n (cid:21) 2 (although it is sufficient for n = 2, 3, as we already
proved).

In the last section we see another proof of this fact emerging from our try to push the idea

of Shur’s test to work for higher dimensions. However such proof might not be possible: As

we have seen in the case n = 1, the weight and its structure played no role. But the weight

59

w cannot be general for higher dimensions, given Carleson’s counterexample. This might

be an issue if one tries to use Schur’s test for higher dimensions (more in section 7).

6.2 Box condition does not imply Carleson condition

for general weight w

Let m2 to be the planar Lebesgue measure. As mentioned in the introduction, L. Carleson
constructed in [7] families Ri of dyadic sub-rectangles of Q = [0, 1)2 having the following
two properties:

∑

R(cid:18)R0
R2R
i

m2(R) (cid:20) m2(R0),
∑

8R0 (cid:18) [0, 1)2
∪

R2R
i

m2(R) = 1 >> m2(

R2R
i

R)

(6.1)

(6.2)

but

with the latter area a small as one wishes. By taking a finite bi-tree T 2
contain the family Ri, and choosing µ s.t. I(cid:3)

µ = m2 and

i , large enough to

 1

m2(R) , R 2 Ri,
otherwise

0,

wi(R) :=

we can identify the left-hand sides of (6.1) and (6.2) with the left-hand sides of (1.11) and
(2.1), respectively. The measure µ is fixed, and for any M > 0 there is i 2 N such that

sup

w:[w,µ]Box

[w, µ]C (cid:21) [wi, µ]C (cid:21) M

(cid:20)1

60

6.3 Carleson condition does not imply REC

Our aim here is to show that for general w, µ the Carleson condition (2.1) is no longer

sufficient for the Restricted Energy Condition (2.2). This example is quite simple and is

inspired by the counterexample of R. Fefferman in [24] for the Lp-boundedness of the
bi-parameter maximal function for arbitrary measure µ. Our examples are given in T 2 but
we can easily extend these to higher dimensions. Our weight w will not be product,

otherwise we know that Hereditary Carleson condition and Carleson condition are

equivalent. It will be supported on a very small subset of the bi-tree, which differs greatly

from the original graph.
For any N 2 N we construct a measure µ and a weight w such that [w, µ]C (cid:20) 4 but
[w, µ]HC (cid:21) N. Therefore, for any C > 0 there is an N 2 N such that

[w, µ]HC (cid:21) [w, µ]HC (cid:21) C

sup

w,µ:[w,µ]C

(cid:20)1

(cid:0)i+1) (cid:2) [0, 2

We start by letting Qi = [0, 2
the measure µ satisfy µ(Q0) = I(cid:3)
Qi, and µ (cid:17) 0 everywhere else. We also define the weight w to be:

µ(Q++

(cid:0)N +i) for i = 1, .., N and Q0 = [0, 2
) = 1 where Q++

(cid:0)N )2. Let
is the upper right quadrant of

i

i

1 if R = Qj for some j 2 f0, .., Ng,

0 otherwise.

w(R) :=

So we have N + 1 nodes α where w(a) is equal to 1. For the node Q0 we have

(

I(cid:3)(

)

µ1Q0

(α)

N∑

(

)

2

=

∑

w(α)

α2T 2
N

I(cid:3)

µ(Q0 \ Qi)

i=0

61

)
2 = (N + 1) (cid:1) 12 = (N + 1)jµ1Q0

j

which means [w, µ]HC (cid:21) N. Now, for an arbitrary down-set D (cid:18) T 2

N we have

w(α)I(cid:3)

µ(α)2 =

I(cid:3)

µ(α)2 =

I(cid:3)

µ(Qj)2

∑

2D

j:Qj

∑

α2D

∑

α2D, w(α)6=0

Then since Q++

i

\ Qj = ; unless i = 0, j, we have
∑

∑

I(cid:3)

µ(Qj)2 (cid:20)

2D

j:Qj

2D

j:Qj

22 (cid:20) 4µ(D)

which finishes the counterexample.

6.4 The lack of maximal principle matters

In this section we construct a measure µ which gives an example of the following fact about

the potentials on a bi-tree: The maximal principle fails. Another example of such a
measure µ can be found in [4] (Proposition 5.2). The weight there satisfies w (cid:17) 1, but here
it takes the values 0, 1. For s 2 N and N = 2s we construct a measure µ such that

but

Vµ . 1 on supp µ,

max Vµ (cid:21) Vµ(ω0) & s.

(6.3)

(6.4)

(cid:0)N )2.

where ω0 = [0, 2
We define a collection of rectangles

(cid:0)2j

Qj := [0, 2

] (cid:2) [0, 2

(cid:0)2

(cid:0)j+s

62

],

j = 1, .., s

(6.5)

and we define

R := fR : Qj (cid:26) R for some j = 1, .., sg
wQ := 1R(Q)

s∑

µ(ω) :=

1
N

1

#Q++

j

j=1

1
Q++

j

(ω).

(6.6)

where #Q denotes the total amount of ω 2 ∂T 2
measure is basically a uniform distribution of the mass 1

N with ω (cid:20) Q. Observe that on Qj the
N over the upper right quarter

Q++

j

of the rectangle Qj (and these quadrants are disjoint).

To prove (6.3) we fix ω 2 Q++
resp.)

j

and we split: (Qu

j , Qr

j are the upper and right half of Qj

Vµ(ω) = Vµ

j ) + µ(Qr

j) + Vµ(Qj),

(ω) + µ(Qu

j

Q++
µ(α) for α with ω (cid:20) α (cid:20) Q++

j

. It is easy to see that

where the first term sums up I(cid:3)
Vµ

(ω) . 1

Q++
j
µ(Qu

j ) + µ(Qp

N (the left-hand side is a double geometric sum). Trivially
j ) (cid:20) 2

N . The non-trivial part is the estimate

Vµ(Qj) . 1 .

For each dyadic rectangle R (cid:21) ω0 and each j

0 we have

either Qj0 (cid:18) R, or Q++

j0 \ R = ;.

(6.7)

(6.8)

Moreover, since the sides of rectangles Qj are nested, the set fj

0

: Qj0 (cid:18) Rg is an interval

63

that contains j. For an interval of integers [m, m + k] let

C[m,m+k] :=

R (cid:21) ω0 j fj

0

: Qj0 (cid:18) Rg = [m, m + k]

{

}

Since each rectangle in C[m,m+k] contains [0, 2

(cid:0)2m

] (cid:2) [0, 2

(cid:0)2

(cid:0)m(cid:0)k+s

], we have

#C[m,m+k] (cid:20) (2m + 1)(2

(cid:0)m(cid:0)k+s + 1) . 2

(cid:0)k+s

(6.9)

It follows that

Vµ(Qj) =

∑

[m,m+k]3j

(#C[m,m+k])(k + 1)

.

1
2s

This shows (6.7), and hence (6.3) is also proved.

∑

k(cid:21)0

(k + 1)22

(cid:0)k+s 1
2s

. 1.

(6.10)

Now we will estimate Vµ(ω0) from below. To this end we need a more careful lower bound
fjg contains all rectangles R that contain Qj and are contained
on #C[m,m+k]. The set C
(cid:0)2j(cid:0)1(cid:0)1] (cid:2) [0, 2
(cid:0)j(cid:0)s(cid:0)1(cid:0)1], so
(cid:0)2

in [0, 2

Hence

fjg (cid:21) 2j(cid:0)1 (cid:1) 2

(cid:0)j(cid:0)s(cid:0)1 & 2s.

#C

Vµ(ω0) (cid:21) s∑

j=1

(#C

fjg

)

1
2s

& s.

64

(6.11)

(6.12)

6.5 REC does not imply embedding

In this section we use the same dyadic rectangles fQjg as above. We also keep the same
measure µ to which we add an extra piece of measure. We have a few more definitions:

Q0,j := Qj, µ0 := µ from the previous section .

Next we continue with defining a sequence of collections Qk, k = 0, . . . , K (cid:25) log s of dyadic
rectangles as follows

Qk,j :=

j+2k(cid:0)1∩

i=j

 , k = 1, . . . , K.

Qk :=

Q0,i, j = 1, . . . , s (cid:0) 2k

(6.13)

In other words, Qk consists of the intersections of 2k consecutive elements of the basic
collection Q0. The total amount of rectangles in Qk is denoted by sk = s (cid:0) 2k + 1.

For k = 1, . . . , K let

sk∑

1

#Q++
k,j

j=1

(cid:0)2k
2
N

µk(ω) :=

(ω), ω 2 ∂T 2,

1
Q++
k,j

and define

K∑

µ := µ0 +

µk.

k=1

By duality the inequality (1.10) is equivalent to the Carleson embedding inequality

∫

T 2

(I(f w))2dµ (cid:20) [w, µ]CE

65

∫

T 2

f 2 (cid:1) w.

(6.14)

We test the inequality (6.14) with the function

f (R) := I(cid:3)

µ0(R).

Using (6.3) we obtain

∫

T 2

f 2 (cid:1) w =

∫

T 2

Vµ0dµ0 . jµ0j =

s
N

.

On the other hand, by definition (6.13) and replacing s by 2k in (6.12) we obtain

Vµ0(Qk,j) & 2k.

(6.15)

(6.16)

It follows that

∫

∫

(I(f w))2dµ =

T 2

(Vµ0)2dµ =

T 2

∫

K∑

k=1

K∑

k=1

(Vµ0)2dµk &

22kjµkj (cid:24) s

N

log s.

(6.17)

Substituting (6.15) and (6.17) in (6.14) we obtain [w, µ]CE & log s.

We claim that [w, µ]HC . 1. This means that for any collection A of dyadic rectangles,
setting A := [R2AR, we have

E[µjA] . jµ1Aj.

(6.18)

To show (6.18) let νk := µkjA, k = 0, . . . , K. Then
∫
∑

∑

∫

E[µjA] =

Vνnνk (cid:20) 2

Vνnνk (cid:20) 2

n,k

n(cid:21)k

66

∫

∑

n(cid:21)k

Vµnνk.

Since supp νk (cid:18) supp µk it suffices to show

∑

n(cid:21)k

Vµn . 1 on

supp µk.

(6.19)

The claim (6.19) has the advantage that it does not depend on A any more.
For every R 2 R we have

µn(R) = 2

(cid:0)2n#fQn,j (cid:18) Rg (cid:20) 2

(cid:0)2n(#fQ0,j (cid:18) Rg + 2n)

(cid:20) 2

(cid:0)n(#fQ0,j (cid:18) Rg + 1) (cid:20) 2 (cid:1) 2

(cid:0)nµ0(R).

It follows that

Vµn(Qk,j) . 2

(cid:0)nVµ0(Qk,j) (cid:20) 2

(cid:0)n

j+2k(cid:0)1∑

i=j

Vµ0(Q0,i) . 2k(cid:0)n,

where the last inequality follows from (6.3). This implies (6.19) and therefore (6.18).
Finally, for any C > 0 there is an s 2 N with logs (cid:21) C and so we get that

sup

w,µ:[w,µ]HC

[w, µ]CE (cid:21) C

(cid:20)1

67

Chapter 7

Open questions

7.1 Failure of Schur’s test for n (cid:21) 2
Now let T n be a finite n-tree. In analogy to subsection 3.1 we define the operator S as
follows: for β 2 T n and a function f 2 ℓ2(T n, w)

∑

S(f )(β) =

α2T n

(wf )(α) (cid:1) ν(α ^ β)

We want to show S : ℓ2(T n, w) ! ℓ2(T n, w) is bounded if the following “box” condition
holds

∑

α(cid:20)β

w(α) (cid:1) ν(α)2 (cid:20) ν(β),

8β 2 T n

For dimension n = 1 we used Schur’s Test successfully in subsection 3.1. Hence, we try to

follow the same scheme for higher dimensions, although we hit an obstacle as we are not

able to estimate certain terms which appear only on poly-trees. It would become obvious

the problem is the same for any dimension bigger than 2, hence we use n = 2 for simplicity.

By α>1β we mean inclusion in the first coordinate and by α>2β in the second.
As in the case n = 1 we start with a function f0 defined as f0(β) = ν(β).

68

Then we write

S(f0)(β) =

=

=

∑
∑
(∑

α2T n

α2T n

α>β

(wf0)(α) (cid:1) ν(α ^ β)
(wν)(α) (cid:1) ν(α ^ β)
∑
∑
∑
∑

α(cid:20)β

+

+

+

α>1β
α(cid:20)2β

α(cid:20)1β
α>2β
w(α) (cid:1) ν(α)2 + g1(β)

)
(wν)(α) (cid:1) ν(α ^ β)

:= f1(β) +

α(cid:20)β

where

and

f1(β) = ν(β) (cid:1)

∑

α>β

(wf0)(α)

( ∑

)

∑

+

α(cid:20)1β
α>2β

α>1β
α(cid:20)2β

g1(β) =

(wf0)(α) (cid:1) ν(α ^ β)

The subscripts 1, 2 mean we sum with respect to the 1st and 2nd coordinate respectively.

As before, the middle term is estimated using our assumption and we thus get

S(f0)(β) (cid:20) f1(β) + f0(β) + g1(β)

We define recursively the functions fi, gi by the formula

fi(β) = ν(β) (cid:1)

(wfi(cid:0)1)(α)

∑

α>β

69

and

Then, we see

( ∑

)

∑

+

α(cid:20)1β
α>2β

α>1β
α(cid:20)2β

gi(β) =

(wfi(cid:0)1)(α) (cid:1) ν(α ^ β)

∑

γ(cid:20)β

(wfi)(γ) (cid:1) ν(γ) + gi+1(β)

S(fi)(β) = fi+1(β) +

For the term in the middle we have

∑

γ(cid:20)β

(wfi)(γ) (cid:1) ν(γ) =

=

∑
∑

γ(cid:20)β

γ(cid:20)β

∑

α>γ

(wfi(cid:0)1)(α)

w(γ) (cid:1) ν(γ) (cid:1) ν(γ) (cid:1)
∑

w(γ) (cid:1) ν(γ)2 (cid:1) (wfi(cid:0)1)(α)

α>γ

:= A1(β) + A2(β)

Where

A1(β) =

(∑
∑
( ∑

γ(cid:20)β

γ<α(cid:20)β

A2(β) =

+

∑
∑

γ(cid:20)β

)
∑
w(γ) (cid:1) ν(γ)2 (cid:1) (wfi(cid:0)1)(α)
)
w(γ) (cid:1) ν(γ)2 (cid:1) (wfi(cid:0)1)(α)

α>β

+

α>1β
γ<2α(cid:20)2β

α>2β
γ<1α(cid:20)1β

70

Below we show that

which imply that

and as

by recursion we get

A1(β) (cid:20) S(fi(cid:0)1)(β)
A2(β) (cid:20) gi(β)

S(fi)(β) (cid:20) fi+1(β) + gi+1(β) + S(fi(cid:0)1)(β) + gi(β)

S(f0)(β) (cid:20) f1(β) + f0(β) + g1(β)

S(fm)(β) (cid:20) m+1∑

i=0

hi(β)

(7.1)

where h0 := f0 and for i (cid:21) 1, hi = fi + gi.

Remark 7.1. What is missing is an estimate for S(gi). Given the discussion in the
beginning of chapter 6 we believe its impossible to get such an estimate, if w is general.

However, as the functions gi collect all the terms we can not estimate, we might try to
“cancel” their contribution by considering a weight w which is non-zero only on

sub-squares of [0, 1)n. This results to a structure similar to a simple tree: for non-trivially
intersecting α, β 2 T n we either have w(α) = 0 or w(β) = 0. This implies that gi (cid:17) 0 for
any i 2 N and so we have hi = fi. Then, by taking F as in (3.5) and using (7.1) we get S

71

is bounded from ℓ2(T n, w) ! ℓ2(T n, w). Of course we saw a much simpler proof of this
using the one-dimensional case in subsection 6.1.

7.1.1 Estimates for A1, A2

For A1 we change the order of summation for both double sums and we use our assumption
to get:

A1(β) =

=

=

(cid:20)

=

γ(cid:20)β

(∑
(∑
∑
∑

α(cid:20)β

α2T 2

∑
∑

γ<α(cid:20)β

+

+

)
w(γ) (cid:1) ν(γ)2 (cid:1) (wfi(cid:0)1)(α)

∑
∑
)
∑
∑
w(γ) (cid:1) ν(γ)2 (cid:1) (wfi(cid:0)1)(α)
∑

γ(cid:20)β

α>β

ν(β) (cid:1) (wfi(cid:0)1)(α)

α>β

γ<α

γ(cid:20)β

α(cid:20)β
ν(α) (cid:1) (wfi(cid:0)1)(α) +

α>β

w(α) (cid:1) ν(α ^ β) (cid:1) fi(cid:0)1(α)

Next, recall A2 is a sum of two double sums:

= S(fi(cid:0)1)(β)

∑

( ∑

∑

)

w(γ) (cid:1) ν(γ)2 (cid:1) (wfi(cid:0)1)(α)

A2(β) =

γ(cid:20)β

+

α>1β
γ<2α(cid:20)2β

α>2β
γ<1α(cid:20)1β

We will take care of the first double sum only as the second follows similarly. By carefully

writing the double sum as four sums (each one on the 1-tree) and by changing the order of

summation for the second and third sum we have

72

the following:

∑
∑

γ(cid:20)1β

∑
∑

γ(cid:20)2β

γ(cid:20)1β

α(cid:20)2β

A2(β) =

=

∑
∑

γ<2α(cid:20)2β

∑

γ<2α

α>1β

∑

w(γ) (cid:1) ν(γ)2 (cid:1) (wfi(cid:0)1)(α)

α>1β
w(γ) (cid:1) ν(γ)2 (cid:1) (wfi(cid:0)1)(α)

Notice the restrictions above would imply γ (cid:20) α ^ β. By changing the summation again
and use our assumption we see

w(γ) (cid:1) ν(γ)2 (cid:1) (wfi(cid:0)1)(α)

∑
∑

γ<2α

α>1β
(cid:20)

(cid:20)

α(cid:20)2β

γ(cid:20)1β

∑
∑
∑
∑
∑
∑
∑
∑
∑

α(cid:20)2β

α(cid:20)2β

α>1β

α>1β

=

α>1β

α(cid:20)2β

w(γ) (cid:1) ν(γ)2 (cid:1) (wfi(cid:0)1)(α)

γ(cid:20)α^β
ν(α ^ β) (cid:1) (wfi(cid:0)1)(α)

w(γ) (cid:1) ν(γ)2 (cid:1) (wfi(cid:0)1)(α)

The symmetric of this last sum is obtained by manipulating accordingly the second double
sum, which means A2(β) (cid:20) gi(β) as claimed.

7.2 Potential theory and the hurdles for n (cid:21) 4
The main problem with pushing the results to n-trees, n (cid:21) 4, lies with the analogue of
Lemma 3.22.

73

For exampe for dimension n = 4 the analogous candidate φ would be

(

2
λ

φ =

I1f (cid:1) I234f + I2f (cid:1) I134f + I3f (cid:1) I124f + I4f (cid:1) I123f +

I12f (cid:1) I34f + I13f (cid:1) I24f + I14f (cid:1) I23f + f If

)

.

(7.2)

where I12 = I1I2, et cetera and I = I1234. We can show part a) in Lemma 3.21
analogously. But for part b) we need to estimate the energy of each one of these terms. For

the first four terms we can do it successfully exactly as in Lemma 3.21. Also, the last term
satisfies If (cid:20) δ on supp(f ). However, we can not prove the analogue of Lemma 3.21 for the
other terms and we have the next question.

Question 7.2. Let f : T 4 ! [0,1) be superadditive on each parameter separately. Let w
be a positive product weight. Suppose that supp f (cid:18) fI(wf ) (cid:20) δg. Then
∑

)
I1I2(w1w2f ) (cid:1) I3I4(w3w4f )

∑

wf 2

21I(wf )(cid:20)λ

. δκλr

(

w

T 4

T 4

for some appropriate powers κ, r. The problem here is we can not use the same proof as we

now have the product of two two-dimensional Hardy operators.

One might ask: can we do it differently? The method above first appeared in [19]. In [5]

there is a slightly different majorization which is based on the following lemma:

Lemma 7.3. Let supp f (cid:18) fIg (cid:20) δg. Let g be a superadditive function. There exists
φ : T ! [0,1) such that

a) Iφ(ω) (cid:21) If (ω) 8ω 2 ∂T : Ig(ω) 2 [λ, 2λ]

(7.3)

74

∫
T φ2 (cid:20) C

δ
λ

∫
T f 2.

b)

(cid:0)1If (cid:1) g (cid:1) 1Ig(cid:20)4λ ,

φ = λ

(7.4)

Proof. Put

and see [5].

Using this lemma, one is able to prove the corresponding of Lemma 3.22 on a bi-tree.

However if we try to prove a 2-dimensional analogue of Lemma 7.3 as a way to get Lemma

3.22 then we are going to fail. The analogous of the function φ in Lemma 7.3 is not a good

candidate anymore as, according to the methods found in [5], it is impossible to get

(

∫

T 2

If (cid:1) g)21Ig(cid:20)λ . δκλr

f 2

∑

T 2

for some appropriate powers κ, r. Hence, we are, again, in need of different function φ. We

have the following question.

Question 7.4. Suppose supp f (cid:18) fIg (cid:20) δg and g be a function which is superadditive for
each parameter separately. Then, there exists φ : T 2 ! [0,1) such that

a) Iφ(ω) (cid:21) If (ω) 8ω 2 ∂T 2 : Ig(ω) 2 [λ, 2λ]

∫

∫

)

τ

(

δ
λ

f 2

T 2

φ2 (cid:20) C

b)

T 2

with some positive τ.

(7.5)

(7.6)

The possible positive answer of this question could finish the argument for n = 3 (which we

already know) but could also open up the road for higher dimensions. Therefore, if we

75

could prove Question 7.4 for n (cid:0) 1 (instead of 2) then following Theorem 3.1 of [5] we could
get the analogue of Lemma 3.21 on the n-tree.

One of the main issues we encounter is that a function g which is super-additive on each
variable satisfies ∆ig (cid:21) 0 for each coordinate i, but the same is not true if we consider
compositions of these operators. For example, our assumptions can not force ∆1∆2g to be
non-negative. Hence our methods fail to be generalized in higher dimensions. Therefore,

one may wonder if these methods are limited or if the developed potential theory can not
answer the question for dimensions n (cid:21) 4 and we are again in need of new methods.
Finally, a natural question arises: what if its impossible to prove the embedding for
dimension n (cid:21) 4?

7.3 Counterexample for n (cid:21) 4?

It is natural to ask this question under these circumstances. One might think there is a lot

of freedom in four dimensions which might trigger a counterexample. Hence, the following

question is also possible:
Question 7.5. Let n (cid:21) 4 and Dn be the collection of all dyadic n rectangles in [0, 1)n.
There is a measure µ on [0, 1)n such that for any dyadic n-rectangle R0:

∑

R(cid:18)R0
R2Dn

µ(R)2 (cid:20) µ(R0)

but for any C > 0 there is some ψ 2 L2([0, 1)n, µ) with
∫

∑

(∫

)

R2Dn

ψdµ

R

2 > C

ψ2dµ

[0,1)n

76

Similar questions could be posed by substituting the Box condition with any of the other

conditions (Carleson, Hereditary Carleson or even the capacity condition of [4]). So far, we
do not know if any of these imply the embedding for n (cid:21) 4. However, the methods we used
for n = 2, 3 reveal that as soon as one proves the analogue of Theorem 3.5 for higher

dimensions then we get “Box condition to Embedding” and “General Capacitary to

Embedding”.

7.4 Embedding for p 6= 2

So far our investigation is around the case p = 2. As noted in the second chapter the
Carleson embedding theorem is true for any 1 < p < 1 and a proof can be found in [25].
Although they proved it for p = 2 the exact same proof works for general p. In the case of

a simple tree and p = 2 the proof (which used the same Bellman function) can be found in
[2] and for general 1 < p < 1 in a recent preprint, see [9].
Hence, as the embedding theorem is true for dimension n = 2, 3 a natural question arises:
Is it true for any 1 < p < 1 other than 2? We do not know the answer. It could be true
but our potential theoretic methods lose some important properties. For example, the Box
condition becomes, for β 2 T n

α(cid:20)β

µ(α)p . I(cid:3)
(
(I(cid:3)
and so one should define the potential to be Vµ = I
is not linear anymore and moreover the function g = (I(cid:3)
variable for p > 2, but the energy estimates do not hold with exponent p

µ(β)

)

0 (the conjugate of
p). One can construct a superadditive f which gives a counterexample of the corresponding

∑

I(cid:3)

77

. As we can see our potential

µ)p(cid:0)1
µ)p(cid:0)1 is superadditive in each

of Lemma 3.21 on a simple tree T and then consider a product of such functions in the
case of bi-tree or tri-tree.

0 and some
On the other hand, if 1 < p < 2 then the energy estimates hold with power p
appropriate powers for δ, λ (by interpolating between 2 and 1) but g is not superadditive
anymore, as the opposite inequality holds (i.e. g is subadditive). Hence p = 2 is the only p

such the conditional (on p) statements “g superadditive” and “energy estimates hold” are

both true. Therefore one might need to find new techniques in order to prove the

embedding for general p. These possible new techniques might have the advantage of
resolving the question for dimensions n (cid:21) 4 as well. Until then, all these interesting
questions remain open.

78

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