THE CARLESON HUNT THEOREM - Giải tích · PDF filevietnam national university...

VIETNAM NATIONAL UNIVERSITY

UNIVERSITY OF SCIENCE

FACULTY OF MATHEMATICS, MECHANICS AND INFORMATICS

Trieu Quang Phong

THE CARLESON HUNT THEOREM

Undergraduate Thesis

Honors Undergraduate Program in Mathematics

Hanoi - 2013

VIETNAM NATIONAL UNIVERSITY

UNIVERSITY OF SCIENCE

FACULTY OF MATHEMATICS, MECHANICS AND INFORMATICS

Trieu Quang Phong

THE CARLESON HUNT THEOREM

Undergraduate Thesis

Honors Undergraduate Program in Mathematics

Thesis advisor:Dr. Dang Anh Tuan

Hanoi - 2013

Acknowledgments

It would not have been possible to write this undergraduate thesis without the help,and support, of the kind people around me, to only some of whom it is possible to giveparticular mention here.

This thesis would not have been possible without the help, support and patienceof my advisor, Dr. Dang Anh Tuan. I am very grateful for the enthusiastic help of him.

I would like to show my gratitude to my teachers at Faculty of Mathematics, Me-chanics and Informatics, University of Sciences, VietNam National University, Hanoi,who equip me with important mathematics knowledge during first four years at theuniversity.

I would like to thank my parents for their personal support and great patience atall times. My parents have given me their unequivocal support throughout, as always,for which my mere expression of thanks likewise does not suffice.

Contents

Acknowledgments 3

Introduction 5

1 Basic step 11.1 Carleson maximal operator . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Dyadic partition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Some definitions and theorems . . . . . . . . . . . . . . . . . . . . . . . 61.4 Basic decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.5 Some comments about proof . . . . . . . . . . . . . . . . . . . . . . . . 111.6 The norm |||f |||α and choosing the partition Πα . . . . . . . . . . . . . 12

2 Carleson analysis of the function 142.1 The seven trick . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.2 The note of f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.3 The set X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.4 The set S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3 Allowed pairs 223.1 Well situated notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.2 The length of well situated notes . . . . . . . . . . . . . . . . . . . . . 273.3 Allowed pairs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.4 The exceptional set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.5 Choosing the shift m . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.6 A bound for ‖f‖α . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.7 Selecting an allowed pair . . . . . . . . . . . . . . . . . . . . . . . . . . 373.8 All together . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4 Final part of proof 524.1 Maximal operator of Fourier series . . . . . . . . . . . . . . . . . . . . 524.2 The Carleson-Hunt theorem . . . . . . . . . . . . . . . . . . . . . . . . 54

Conclusion 57

Bibliography 57

Introduction

The Carleson Hunt theorem is a fundamental result in mathematical analysis. TheTheorem shows that the almost everywhere pointwise convergence of the Fourier seriesfor every f ∈ Lp[−π, π] for 1 < p <∞.

Historically, a fundamental question about Fourier series, asked by Fourier himselfat the beginning of the 19th century, is whether the Fourier series of a continuousfunction converges pointwise to the function. In 1913, Nikolai Nikolaevich Luzin con-jectured that every function in L2[−π, π] has an almost everywhere convergent Fourierseries. In 1966, Lennart Carleson proved Luzin’s conjecture. The next year RichardHunt proved this pointwise convergence for Lp-space where 1 < p < ∞, and we knowit as the Carleson Hunt theorem.

The purpose of this thesis is to present method and rewrite in detail for some prob-lems in the proof of the Carleson Hunt theorem. The thesis consists of Introduction,Conclusion, Bibliography and four chapters.

Chapter 1 ”Basic step” introduces the Carleson integral and general points ofview on the method. Besides, this chapter reviews some basic definitions, propositionsand theorems without proof.

Chapter 2 ”Carleson analysis of the function f” gives us some properties, func-tions and sets from the expansion of f .

Chapter 3 ”Allowed pair” is the main part of the proof. I will present the detailmethod used in the proof of Carleson theorem in this chapter.

Chapter 4 ”Final part of proof” is the summary by applying following chapter tocomplete the proof of theorem.

The thesis is based on completely the book Pointwise Convergence of Fourier Seriesby Juan Arias de Reyna.

Chapter 1

Basic step

The proof of the Carleson Theorem is based on a new method of estimatingpartial sums of Fourier series. We replace the Dirichlet integral by the singular integral

C(n,I) = p.v.

π∫−π

eint(x−t)

x− tf(t)dt.

The new method consists of applying repeatedly a basic step that we are going tointroduce in this chapter.

Given a partition Π of the interval I = [−π, π], we decompose the integral as

C(n,I) = p.v.

∫I(x)

ein(x−t)

x− tf(t)dt+

∑J∈Π,J 6⊂I(x)

∫J

EΠf(t)

x− tdt (1.1)

+∑

J∈Π,J 6⊂I(x)

∫J

ein(x−t)f(t)− EΠf(t)

x− tdt,

where I(x) is an interval containing x, that is a union of some members of Π, and EΠfis the conditional expectation of ein(xt)f(t) with respect to Π.

The principal point of the basic step is the careful choice of I(x) and Π suchthat we have good control of the last two sums. The first term is an integral of thesame type that we can treat in a similar way.

1

1.1 Carleson maximal operator

Let I ⊂ R be a bounded interval and n ∈ Z. For every f ∈ L1(I), we considerthe singular integral

p.v.

∫I

e2πin(x−t)/|I|

x− tdt. (1.2)

If x is contained in the interval I/2 with the same center as I and half length we call(1.2) a Carleson integral.

Here we have an arbitrary selection: x ∈ I/2. Every condition x ∈ θI, withθ ∈ (0, 1) would be adequate. If we allow x near the extremes of I, the simplestCarleson integral would be unbounded.

The set of pairs P = (n, I) : n ∈ Z, I a bounded interval of R will assume avery central role in the following. Given a pair α = (n, I) we denote I by I(α) and nby n(α). Also we call |I| = |I(α)| the length of α, and write it as |α|. Then for everyα ∈ P and x ∈ I(α)/2 we put

Cαf(x) = p.v.

∫I(α)

e2πin(α)(x−t)/|α|

x− tdt.

Given f ∈ L1(I) and α ∈ P with I(α) = I, Cαf(x) is defined for almost everyx ∈ I(α)/2. We shall study the Carleson maximal operator

C∗I f(x) = supI(α)=I,n(α)∈Z

|Cαf(x)|.

This is a measurable function with values in [0,∞]. We shall prove that it is a boundedoperator from Lp(I) → Lp(I/2) for every 1 0 such that if x ∈ I(γ)/2,and for every γ ∈ R,

|Cγ(eiλt)(x)| ≤ A.

Next, we will define local norm ‖·‖α with α ∈ P . Before, we associate a functioneα(x) to every pair α ∈ P . eα(x) is a function in L2(R) supported by the interval I(α)

eα(x) = e2πin(α)x/|α|χI(α)(x) = e2πiλ(α)xχI(α)(x).

The function eα is a localized wave train. It is localized in the time intervalI(α), and angular frequency λ(α) = 2πn(α)/|α|. The number |α| is the durationand n(α) the total number of cycles in the wave train.

2

The functions e with I(α) = I fixed, form an orthonormal system and everyfunction f supported by I can be developed in a series of these functions, convergentin L1(R).

f =∑I(α)=I

〈f, eα〉|α|

eα.

Observe that〈f, eα〉|α|

=1

|α|

∫I(α)

f(t)e−2πiλ(α)tdt.

We define the local norm ‖f‖α as

‖f‖α =∑j∈Z

c

1 + j2| 1

|I|

∫I

f(t)e−2πi(n(α)+j/3)t/|I|dt | .

Here c is chosen so that ∑j∈Z

c

1 + j2= 1.

Hence ‖f‖α is a mean value of absolute value of f .

Proposition 1.2. Let φ ∈ C2[a, b], δ = b− a. For every x ∈ [a, b] we have

φ(x) =∑n∈Z

cne2πinx/(3δ),

where the coefficient cn satisfies (1 + n2)|cn| ≤ A(‖φ‖∞ + δ2‖φ′′‖∞), for every n ∈ Z.

Theorem 1.1. Let f ∈ L1(I), φ ∈ C2(I) and a = (n, I) ∈ P . For some absoluteconstant B we have

| 1

|α|

∫I

e2πin(x−t)/|α|f(t)φ(t)dt| ≤ B(‖φ‖∞ + |a|2‖φ′′‖∞)‖f‖α.

Proposition 1.3. There exists an absolute constant C such that for every ω ∈ R andα ∈ P we have

‖e2πiωx‖α ≤ 1, and ‖e2πiωx‖α ≤C

|bω|α|c − n(α)|.

Proposition 1.4. There exists an absolute constant B > 0 such that, for every intervalK and ω ∈ R, there exists k ∈ N such that for κ = (k,K), we have

‖e2πiωx‖κ ≥ B.

We can choose k = b|ω| · |K|c.

3

1.2 Dyadic partition

Given an interval I = [a, b], and the central point c of I. We will have twointervals with length of |I|/2 that we denote by

I0 = [a, c], and I1 = [c, b].

By similarly way, let d be the central of I0 = [a, c], we will obtain two new intervalwith length of |I0|/2, we denote by (I0)0 = I00 = [a, d] and (I0)1 = I01 = [d, c].

Let 0, 1∗ be the set of all binary sequences. Then process defines the intervalsIu for every word u ∈ 0, 1∗ with the role that if we have found Iu by the process, itscentral will divides the interval to two sub-interval, we denote the sub-interval in theleft by Iu0, and other by Iu1. We call these intervals dyadic intervals generated fromI.

Every dyadic interval Iu has two sons Iu0 and Iu1. We call that Iu0 and Iu1 arebrothers. Every dyadic interval has a brother. But, in general, every dyadic intervalhas two contiguous intervals. We call that two intervals are contiguous if they hassame length and a unique point in common. We also speak of the grandson of Iu.They are the four intervals, Iu00, Iu01, Iu10 and Iu11.

Smoothing intervals. As we know that a dyadic interval can be divided twodyadic intervals, and they are contiguous. But, in general, the union of two contiguousdyadic intervals is not a dyadic interval. Such an interval we shall call a soothinginterval. Hence, a dyadic interval is also a smoothing interval, but the inversion isnot true.

Given an interval I we shall denote by I/2 its middle half, that is I/2 = I01∪I10.And we denote by PI the set of pairs (n, J) where n ∈ Z and J is a smoothing intervalwith respect to I.

Dyadic points. The extremes of all dyadic intervals (with respect to I) arecalled dyadic points and the set of all dyadic points will be denoted by D. D is acountable set and so it is of measure 0.

Proposition 1.5. Given I and x ∈ I/2 that is not a dyadic point, for every n =0, 1, 2, ... there is only one smoothing interval In of length |I|/2n such that x ∈ In/2.We also have I = I0 ⊃ I1 ⊃ I2 ⊃ ...

Choosing I(x). We shall consider partitions Π of some smoothing interval Jwhere every member of Π is a dyadic interval generated from I, length less than |J |/4.We always consider closed intervals, and when we speak of partitions we don’t takeinto consideration the extremes of these intervals. We also speak of disjoint intervalsto mean those whose interiors are disjoint

We assume a Carleson integral Cαf(x) and a dyadic partition Π of I = I(α),where every J ∈ Π has length |J | ≤ |I|/4, to be given. I(x) will be an interval, a unionof some members of Π, such that x ∈ I(x)/2, so that the first term in decompositionof Carleson integral will be almost the Carleson integral. We must also choose I(x) in

4

order to obtain a good bound for the other members of decomposition. The followingproposition will show that there exists the interval I(x) satisfying all condition can beattained.

Proposition 1.6. Let x ∈ I/2 and let Π be a dyadic partition of the smoothing interval,with intervals of length ≤ |I|/4. Then there exists a smoothing interval I(x) such that:

(a) x ∈ I(x)/2.(b) I(x) ≤ |I|/2.(c) I(x) is the union of some intervals of Π.(d) Some of the two sons of I(x) is a member of Π.(e) For every J ∈ Π such that J 6⊂ I(x) we have d(x, J) ≥ |I|/2.(f) Each smoothing interval J with I(x) ⊂ J ⊂ I and x ∈ J/2 is a union of

intervals of the partition Π.

5

1.3 Some definitions and theorems

Definitions of operator

The second and third terms of Carleson integral will be bounded using twofunctions. The first is a modification of the Hilbert transform.

Definition 1.1. Given f ∈ L1(I), we define H∗I f(x), the maximal Hilbert dyadictransform on I, by

H∗I f(x) = supK|p.v.

∫K

f(t)

x− tdt|,

where the supremum is over the interval K = J ∪ J ′, which are the union of twocontiguous dyadic intervals such that x ∈ K/2.Proposition 1.7. Let 1 ≤ p < ∞, f ∈ Lp(R), and let I ⊂ R be a bounded interval.Then for every x ∈ I we have

H∗I f(x) ≤ 2H∗f(x) + 6Mf(x), (1.3)

where H∗f and M are the maximal Hilbert transform and Hardy-Littlewood maximalfunction of f .

Theorem 1.2. There are absolute constant A and B > 0 such that for every f inL∞(I), and every y > 0

mH∗I f(x > y)|I|

≤ Ae−By/‖f‖∞ . (1.4)

The second function will be needed to bound the third team in decompositionof Carleson integral.

Definition 1.2. Given a finite partition Π of I by intervals Jk of length δ and centertk we define the function

∆ = (Π, x) =∑k

δ2k

(x− tk)2 + δ2k

.

Theorem 1.3. There are some absolute constants A and B > 0 such that for everypartition Π of the interval J ⊂ R by interval, we have

mx ∈ J : ∆(Π, x) > y|J |

≤ Ae−By. (1.5)

In the future reasoning a pair α = (n, I), (with some other elements that noware of no consequence) will determine a dyadic partition Πα of I(α). Hence we shalldenote by ∆α(x) the corresponding function ∆(Πα, x). We shall also use H∗f(x) todenote the maximal Hilbert dyadic transform H∗I(α)Eαf(x), where

Eαf(x) =1

|J |

∫J

eiλ(α)(x−t)f(t)dt, if x ∈ J ∈ Π.

6

Notation α/β

Definition 1.3. Given α ∈ P and an interval J ⊂ I(α), we define α/J ∈ P by

α/J = (m, J), where m = bn(α)|J ||α|c.

Definition 1.4. Given α and β ∈ P such that I(β) ⊂ I(α), we define

α/β = α/I(β).

Proposition 1.8. Given α ∈ P and two intervals J ⊂ K ⊂ I(α), if all of them aresmoothing intervals, then

α/K = (α/J)/K. (1.6)

Space of functions

The functions we are considering will be defined on an interval X ⊂ R and wewill always consider the normalized Lebesgue measure on this interval, µ. Therefore µis a probability measure µ(X) = 1.

Definition 1.5. Given a measurable function f : X → R we consider its distributionfunction

µf (y) = µx ∈ X : |f(x)| > y.

Definition 1.6. Given a measurable function f : X → R, the decreasing rear-rangement of f is defined by

f ∗(t) = my > 0 : µf (y) > t.

Definition 1.7. For every measurable function f : X → C we define for t > 0 thefunction f ∗∗(t) in the following way

f ∗∗(t) =1

t

t∫0

f ∗(s)ds.

Remark. If we assume that sup0<t<1

t1pf ∗(t) = C, then

f ∗∗(t) =1

t

t∫0

f ∗(s)ds ≤ Cp

p− 1t−

1p . (1.7)

Using (16) of Proposition 1.4.5 in Classical Fourier Analysis, Springer; 2nd ed; 2008by Loukas Grafakos, we have.

7

Proposition 1.9. Given measurable function f : X → R, and 1 0y(µf (y))

1p . (1.8)

For a characteristic function χA a simple computation shows that ‖χA‖p,1 isequal to ‖χA‖p. The space Lp,1(µ) (p > 1) can be defined as the smallest Banach spacewith this property. This is the content of the following theorem.

Theorem 1.4. There is an absolute constant C such that for every f ∈ Lp,1(µ) thereexist a sequence of measurable sets (Aj)

∞j=1 and numbers (aj)

∞j=1 such that

f =∞∑j=1

ajχAj , ‖f‖p,1 ≤∞∑j=1

|aj|µ(Aj)1p ≤ C‖f‖p,1.

Theorem 1.5. (Marcinkiewicz) Let S be a sub-linear operator defined on Lp0,1(µ)+Lp1,µ where 1 < p1 < p0 <∞. Assume that there exist constants M0 and M1 such that

‖Sf‖p0,∞ ≤M0‖f‖p0,1, and ‖Sf‖p1,∞ ≤M1‖f‖p1,1.

Then, for every p ∈ (p1, p0), S : Lp(µ)→ Lp(µ) is continuous with norm

‖S‖p ≤p(p0 − p1)

(p0 − p)(p1 − p)M1−θ

0 M θ1 , where

1

p=

1− θp0

+θ

p1

.

8

1.4 Basic decomposition

The basic step in the proof of Carleson’s Theorem is a decomposition of aCarleson integral Cαf(x) into three parts associated with a dyadic partition Π of I =I(α). We assume that every J ∈ Π has a measure less than |I|/4; then Proposition 1.6gives us an interval I(x) such that x ∈ I(x)/2. The decomposition is given by

C(n,I) = p.v.

∫I(x)

ein(x−t)

x− tf(t)dt+

∫I\I(x)

Eαf(t)

x− tdt (1.9)

+

∫I\I(x)

ein(x−t)f(t)− Eαf(t)

x− tdt.

We shall transform the first term into a Carleson integral that can be consideredsimpler than Cαf(x). The second and third terms can be bounded in terms of thefunctions Hαf(x) and ∆α(x) = ∆(Πα, x).

The first integral in (1.9) is almost a Carleson integral. We need β = (m, I(x)) ∈P , such that the difference between Cβf(x) and the first integral is small. Now we willsee a theorem about the different when changing the frequency of a Carleson integral.

Theorem 1.6. (Change of frequency) Let α, β be two pairs with I(α) = I(β) = Jand x ∈ J/2. If |n(α)− n(β)| ≤M with M > 1, then

|Cαf(x)− Cβf(x)| ≤ BM3‖f‖α, (1.10)

where B is some absolute constant.

We apply this theorem to our case it follow that

|p.v.∫I(x)

eiλ(α)(x−t)

x− tf(t)dt− Cβf(x)| ≤ C‖f‖β. (1.11)

The second term has form∫I\I(x)

Eαf(t)

x− tdt = p.v.

∫I

Eαf(t)

x− tdt− p.v.

∫I(x)

Eαf(t)

x− tdt.

Now these two intervals I and I(x) are of the form that is used in Definition 1.1 of themaximal Hilbert dyadic transform. Hence we have

|∫

I\I(x)

Eαf(t)

x− tdt| ≤ 2H∗αf(x). (1.12)

The third term of (1.9) can be bounded in the form,

|∫

I\I(x)

ein(x−t)f(t)− Eαf(t)

x− tdt| ≤ D sup

Jk

‖f‖α/Jk∆α(x), (1.13)

where Jk ∈ Π ’disjoint’ from I(x).

9

Theorem 1.7. Let Cαf(x) be a Carleson integral and Π = Πα a dyadic partition ofI = I(α). Assume that every J ∈ Π has measure less than |I|/4. Let I(x) be theinterval defined on Proposition 1.6, and β = α/I(x). Then

|Cαf(x)− Cβf(x)| ≤ C‖f‖β + 2H∗αf(x) +D supJk

‖f‖α/Jk∆α(x), (1.14)

where C and D are absolute constants.

10

1.5 Some comments about proof

1. In what sense can we say that Cβf(x) is a ’simpler’ Carleson Integral? First,we can say that we pass from Cαf(x) to Cβf(x) where if n(α) > 0 then n(β) ≤ n(α).Thus we can expect to obtain n(β) = 0. We can restrict the study to real functions,because, with f real, we have

C−αf(x) = Cαf(x) (1.15)

if α = (n, I) and −α = (−n, I) with n ∈ N. Hence we can assume that n(α) is positive.

Another case in which we can consider only values n(α) > 0 is when we studyfunctions f with |f | = χA (for some measurable set A). In this case f is of the samenature and

C−αf(x) = Cαf(x).

2. We want to prove that the Carleson maximal operator is bounded fromLp(I) → Lp(I/2) (1 y ≤Ap‖f‖ppyp

.

Hence we would like, given f ∈ Lp(I) and given y > 0, to define E withm(E) < Ap‖f‖pp/yp and such that for every x ∈ I/2 \ E and α ∈ P with I(α) = I wehave |Cαf(x)| < y.

In fact, we will first construct, given f ∈ Lp(I), y > 0 and N ∈ N, a subset ENwith m(EN) < Ap‖f‖pp/yp; such that for every x ∈ I/2 \ EN and ∈ P with I(α) = I,and 0 ≤ n(α) < 2N we will have |Cαf(x)| < y. Then

C∗I f > y ⊂⋃N

x ∈ I/2 : |Cαf(x)| > y, 0 ≤ n(α) < 2N , I(α) = I,

and since AN = x ∈ I/2 : |Cαf(x)| > y, 0 ≤ n(α) < 2N , I(α) = I is an increasingsequence of sets, we will have

mC∗I f > y = limNm(AN) ≤ Ap

‖f‖ppyp

.

Technical reasons will force us to replace 2N by θ2N (θ an absolute constant) in theabove reasoning.

3. Another important point about the proof is that we shall use the basic steprepeatedly. Therefore, given a Carleson integral Cαf(x) with I(α) = I, 0 ≤ n(α) < 2N

and x /∈ EN , we shall obtain a sequence (αj)sj=1 in P with α1 = α. Then we will have

|Cαf(x)| ≤s−1∑j=1

|Cαjf(x)− Cαj+1f(x)|+ |Cαsf(x)|. (1.16)

11

1.6 The norm |||f |||α and choosing the partition Πα

From now on we will consider a Carleson integral Cαf(x) where 0 ≤ n(α) < 2N

and I(α) = J, but we will not assume that J = I. If we want to apply the basic stepto this integral, what selection of the partition Π will be good?

The intervals of the partition Π will be dyadic with respect to J = I(α) and oflength less than |J |/4. What we want, in view of (1.14) is to have control of ‖f‖α/Jkfor every Jk ∈ Π. Hence we put bj = 2 · 2−2j

, and we will assume that for the intervalsJ00, J01, J10, and J11 we have

‖f‖α/J00 , ‖f‖α/J01 , ‖f‖α/J10 , ‖f‖α/J11 < ybj−1. (1.17)

We obtain the partition Π by a process of subdivision. We start with the four grandsonsof J = I(α) of which we assume (1.17). Then at every stage of the process we takesome interval K and we subdivide it in its two sons K0 and K1, if they satisfy thecondition ‖f‖α/K0 , ‖f‖α/K1 < ybj−1. If they do not, we consider K to be one of theintervals of the partition. Since this process can be infinite, we also stop the divisionif |K| ≤ |I|/2N and consider it to be of the partition.

As we need to consider the condition (1.17), we define for every α = (n, J) ∈ P

|||f |||α = sup‖f‖α/J00 , ‖f‖α/J01 , ‖f‖α/J10 , ‖f‖α/J11.

It is important to see that this construction and the definition of I(x) in Propo-sition 1.6, imply that either |I(x)| = 2|I|/2N or ‖f‖β ≥ ybj−1.

Now we have another answer to the question 1. If we start with ybj ≤ ‖f‖α <ybj−1 we arrive at ‖f‖β ≥ ybj−1. We go from level j to a lesser level. It is true thata lesser level means here a greater norm ‖f‖β, but it also means a smaller numberof cycles n(β) ≤ n(α), and we will arrange things so that we arrive to Cαsf withn(αs) = 0. We also have a good bound of the differences |Cαjf(x) − Cαj+1

f(x)| in(1.16). As we have motivated above, given f ∈ L(I) and α = (n, J) ∈ P , we define

|||f |||α = sup‖f‖α/J00 , ‖f‖α/J01 , ‖f‖α/J10 , ‖f‖α/J11. (1.18)

We will say a Carleson integral Cαf(x) is of level j ∈ N if

ybj ≤ |||f |||α < ybj−1.

In the construction that follows we assume f ∈ L(I), a Carleson integralCαf(x), a natural number N , and a real number y > 0 to be given; where α = (n, J)with 0 ≤ n < 2N , and J being the union of two dyadic intervals with respect to I, haslength |J | > 4|I|/2N . We also assume that ‖f‖α < ybj−1 for some natural number j(not necessarily the level of Cαf(x)).

Our objective is to select a convenient dyadic partition Π of J so that we canapply Theorem 1.7.

12

We consider now the set of dyadic intervals Ju with respect to J , such that|J |/4 ≥ |Ju| ≥ |I|/2N . For every one of these intervals we determine if they satisfy thecondition

‖f‖α/Ju < ybj−1. (1.19)

Now the interval Ju is a member of the partition Π if it is of length |Ju| = |I|/2Nand it, all its ancestors and their brothers satisfy the condition (1.19), or it, all itsancestors and their brothers satisfy the condition (1.19) but one of its sons does notsatisfy this condition.

Finally, observe that according to Proposition 1.6, the interval I(x) is alwaysthe union of one of the intervals of Π and a contiguous interval of the same length.Therefore, |I(x)| = 2|I|/2N or some of the four grandsons of I(x) will not satisfy thecondition (1.19).

Now that we have a good selection of the partition Πα we can give a betterversion of the basic step. We shall need the following comparison between the twonorms ‖ · ‖α and ||| · |||α.

Proposition 1.10. There is a constant C > 0 such that for every f ∈ L2(J) andα = (n, J)

‖f‖α ≤ C|||f |||α.

Now we can formulate the basic step with all its ingredients.

Theorem 1.8. (Basic step) Let ξ ∈ PI , and x ∈ I(ξ)/2 and assume that ‖f‖ξ <ybj−1. Given N ∈ N, let Πξ and I(x) be the corresponding partition of I(ξ) and interval,defined in Proposition 1.6. Let J be a smoothing interval such that I(x) ⊂ J ⊂ I(ξ),and x ∈ J/2. Assume that |ξ| ≥ 4|I|/2N . Then we have

|Cξ/Jf(x)− Cξ/I(x)f(x)| ≤ Cybj−1 + 2H∗ξ f(x) +Dybj−1∆ξ(x), (1.20)

where C and D are absolute constants.

13

Chapter 2

Carleson analysis of the function

In this chapter we assume a function f ∈ L2(I) ∩ Lp(I) to be given, where1 ≤ p < ∞. Later we shall assume that |f | = χA is the characteristic function of ameasurable set of I, and we shall use interpolation techniques to recover the case of ageneral f ∈ Lp(I).

2.1 The seven trick

For α ∈ P , we want to consider some properties of every grandchild J of I(α).Therefore, for every given property p, we define the set

A =⋃J : J dyadic and p(J),

where p(J) denotes that J satisfies the property p.

Now, for arbitrary measurable set A, we define a new set

A∗ =⋃7J : J ⊂ A and dyadic,

where 7J denotes the interval of length 7|J | and center interval J .

Hence for every dyadic interval J ⊂ A, we add in A∗ the interval containingJ and three contiguous intervals of the same length at each side. So, A∗ have someinteresting properties. In fact if I(α) 6⊂ A∗, then J 6⊂ A, where J is a grandchild ofI(α). Indeed, if J ⊂ A, we have J is dyadic interval. Thus, using definition of 7J andA∗ sets, we have I(α) ⊂ 7J ⊂ A∗, that is a contradiction.

Moreover, two dyadic are either disjoint or one is contained in other, and alldyadic interval is subset of I. Hence, Let V be a set of all disjoint dyadic intervalJ ⊂ A, we have

A∗ =⋃J∈V

7J.

Thereforem(A∗) ≤

∑J∈V

7|J | ≤ 7m(A).

14

In general, for J is smoothing interval such that J 6⊂ A∗, then every grandchildof J is not subset of A.

15

2.2 The note of f

We want to find the pairs α such that |||f |||α is specially great. We call thesepairs that is the allowed pairs. To this purpose we carry out the Carleson analysis ofthe function f. For every function f supported by I can be developed in a series ofthese functions, convergent in L(R).

f =∑I(α)=I

〈f, eα〉|α|

· eα

=∑I(α)=I

〈f, eα〉|α|

· eλ(α)ix

=∑I(α)=I

aα · eλ(α)ix,

where aα is Fourier coefficient of f and

aα =〈f, eα〉|α|

.

First, we fix a level j (j ∈ N), and find the pairs α = (n, I) for which thecorresponding Fourier coefficient of f is greater than bjy

p2 . Next, for α = (2n, I) and

β = (n, J) such that J is son of I then we have

λ(α) = 2π · 2n

|I|= 2π · 2n

2|J |= 2π · n

|J |= λ(β).

So, given α we may find other pairs β satisfying eα = eβ and I(α) % I(β). Hence, wedevelop the rest of the function on the two sons of I. And, we find also those termswith the coefficients greater than bjy

p2 , and so on. We will call these pairs that is the

notes of f at level j. The set of allowed pairs will be modification of these.

In our construct, we only care pairs (n, J) where J is a dyadic interval. Wedefine DI as this set:

DI = α ∈ P : I(α) a dyadic interval with respect to I.

If α = (n, J) and J is a dyadic interval with respect to I, then there is a uniqueu ∈ 0, 1∗ such that J = Iu. We put u = u(α). In general, if u ∈ 0, 1∗ we call u′

his father. For example 010010′ = 01001, 0101′ = 010. Hence if α = (n, J) and J is adyadic interval, Iu′(α) is the father of J .

We will determine set of pairs Qju that denotes the sets of notes of f at levelj. These pairs contain the information about f that we shall need. We are going toconstruct a set Qju and function P j

u(x) by induction, for every level j ∈ N and everyu ∈ 0, 1∗.

We take Qj∅ = Qj0 = Qj1 = ∅ and corresponding function P j∅ = P j

0 = P j1 = 0.

In the first step of induction we construct for u of length 2, that is u ∈ 00, 01, 10, 11:

Qju = α ∈ DI : I(α) = Iu, |〈f, eα〉| ≥ bjyp2 |Iu|. (2.1)

16

We easy to see that for α ∈ Qju then the Fourier coefficient corresponding α

|aα| =|〈f, eα〉||α|

≥ bjyp2 ,

and we define

P ju(x) =

∑α∈Qju

〈f, eα〉|Iu|

· eα(x). (2.2)

Next, assuming we have defined Qju and P ju , and that v = u0, and v = u1. We define

Qjv = α ∈ DI : I(α) = Iv, |〈f − P ju , eα〉| ≥ bjy

p2 |Iv|, (2.3)

and

P ju(x) =

∑α∈Qju

〈f − P ju , eα〉|Iu|

· eα(x). (2.4)

LetQj =

⋃u∈0,1∗

Qju,

is called the set of notes of level j of f . Using that definition, we can rewrite functionP ju(x) :

P ju(x) =

∑α∈Qj ,I(α)⊃Iu

〈f − P ju′(α)〉|α|

· eα(x) =∑

α∈Qj ,I(α)⊃Iu

aαeα(x). (2.5)

By definition P jv − P j

u is orthogonal to f − P jv on Iv, where v is a son of u. Therefore∫

Iv

|f(x)− P jv (x)|2dx+

∑α∈Qju

|aα|2|α| =∫Iv

|f(x)− P ju(x)|2dx.

For any natural number n > 1, summing over all the dyadic intervals of length |v| ≤ n,∑|u|=n

∫Iu

|f(x)− P ju(x)|2dx+

∑α∈Qju,|u(α)|≤n

|aα|2|α| =∫Iu

|f(x)|2dx.

Therefore ∑α∈Qju,|u(α)|≤n

|aα|2|α| ≤∫Iu

|f(x)|2dx, ∀n > 1.

Taking n to infinity, we have ∑α∈Qju

|aα|2|α| ≤∫Iu

|f(x)|2dx. (2.6)

The definition of aα implies that for every α ∈ Qj, |aα| ≥ bjyp2 . Hence the length

of all the pairs in Qj is bounded by∑α∈Qj|α| ≤ b−2

j

‖f‖22

yp. (2.7)

This is the bound of the length of the notes of level j.

17

2.3 The set X

In this part, we define a component of the exceptional set. When Iu 6⊂ X wehave a good bound for P j

u , and also bound for the number of terms in the sum thatdefines this function. We define the new function

Aj(x) =∑a∈Qj|aα|2χI(α)(x).

By (2.6), we have∫I

Aj(x)dx ≤ ‖f‖22. For every j we define the set

Xj = x ∈ I : Aj(x) >yp

bj.

So, we have the bound

m(Xj) =

∫I

χXj(x)dx

≤∫I

Aj(x) · bjypdx

≤ bj‖f‖2

2

yp.

Now we put

X =⋃j∈N

Xj.

Therefore

m(X) ≤∑j∈N

m(Xj) ≤∑j∈N

bj‖f‖2

2

yp< C‖f‖2

2

yp.

So, we can bound measure of X∗

m(X∗) ≤ C‖f‖2

2

yp. (2.8)

Next, we will study the proposition that gives a good bound for |P ju(x)| and the

number of terms in the sum of |P ju(x)|.

Proposition 2.1. Let Iu be a dyadic interval, such that Iu 6⊂ X. Then P ju has at most

b−3j terms and

|P ju(x)| ≤


|aα| ≤yp2

b2j

. (2.9)

18

Proof. We have condition Iu 6⊂ X implies that Iu 6⊂ Xj. Hence there exists a pointx0 ∈ Iu such that x0 /∈ Xj. So

Aj(x0) =∑α∈Qj|aα|2χI(α)(x0) ≤ yp

bj.

LetN be the number of terms in P ju , and notice that |aα| ≥ bjy

p2 . Using above inequality

we deduce

Nb2jyp ≤ yp

bj,

soN ≤ b−3

j .

On the other hand

P ju(x) =


aαeα(x).

Applying bunyakovsky inequality, we have

|P ju(x)| ≤


|aα||eα(x)|

=∑

α∈Qj ,I(α)⊃Iu

|aα|

≤√N(


|aα|2)12 =√NAj(x0)

12

≤ b−32j y

p2 b−12j =

yp2

b2j

.

19

2.4 The set S

In this section, we construct a set S such that S∗ is a component of the excep-tional set EN . We want to consider S such that every pairs α with I(α) 6⊂ S∗ satisfies|||f |||α < y.

Hence, at every point in which we are interested, Cαf(x) will have well definedlevel or |||f |||α = 0 in which case Cαf(x) = 0. So we have achieved our aim to boundCαf(x).

We define

H = J : J is dyadic respect to I and1

|J |

∫J

|f |pdm ≥ yp.

AndS =

⋃J∈H

J.

Now if α = (n, J) ∈ PI and J 6⊂ S∗, we will have for every grandson Ju of J

1

|Ju|

∫Ju

|f |dm ≤ (1

|Ju|

∫Ju

|f |pdm)1p < y,

where 1 < p < +∞. Hence‖f‖α/Ju < y.

It implies that|||f |||α < y.

Now if x /∈ S∗, α = (n, J) ∈ PI and x ∈ I(α)/2, we will also have I(α) 6⊂ S∗,and |||f |||α < y. As a consequence, if I(α) 6⊂ S∗, α ∈ PI , the Carleson integral Cαf(x)have level j ∈ N, that is,

ybj ≤ |||f |||α < ybj−1,

or |||f |||α = 0, which implies that f = 0 on I(α).

Let V be set of all disjoint interval J where J ∈ H. Using the property that twodyadic interval are disjoint or one contains other, we can rewrite the set S in the form

S =⋃J∈V

J.

Hencem(S) ≤

∑J∈V

m(J).

Therefore

m(S) ≤∑J∈V

|J | ≤ 1

yp

∑J∈V

∫J

|f |pdm ≤ 1

yp‖f‖pp.

20

Hence we also have the bound

m(S∗) ≤ 7

yp‖f‖pp. (2.10)

21

Chapter 3

Allowed pairs

3.1 Well situated notes

We are going to define the set of well situated pairs. These pairs are anenlargement of the set Q =

⋃j≥1Qj of notes of f . We add pairs that in a certain sense

are near the notes of f . This will define the set R of well situated notes of pairs. Ouraim defining the set R is that given a Calerson integral Cαf(x) there exists an allowedβ near α. This set is defined as the sets of pairs that satisfy one of two condition.

The pair α ∈ DI will be an element of Rj if it satisfies one of two conditionsA(Rj) or B(Rj). Now we see the definition of two above conditions and its properties.

Condition B(Rj) Let α ∈ DI . If there is β ∈ Qj such that I(α) ⊂ I(β) and there aretwo different elements γ and δ in Qj with I(γ) ∩ I(δ) ⊃ I(β) such that

b10j ≤ |λ(γ)− λ(δ)| · |α| ≤ 32 · b−10

j and |n(α)− n(γ/α)| < b−10j .

The role of I(β) in this condition is to simplify the calculation the sums of thelength I(α) for those α satisfy the condition. We go to the following lemma.

Lemma 3.1. (Structure of the function P ju) Let α ∈ DI be such that I(α) 6⊂ X.

Assume that α /∈ Rj and put u = u(α). Then we can write P = P ju as

P (t) = ρeiλ(δ)t + P0(t) + P1(t), t ∈ I(α) (3.1)

where P1(t) consists of the terms of P (t) for which |n(α)− n(γ/α)| ≥ b−10j , and

|ρ| ≤ b−2j y

p2 , and |P0(t)| ≤ b8

jyp2 , if t ∈ I(α). (3.2)

Furthermore, if ρ 6= 0, then δ ∈ Qj and satisfies I(δ) ⊃ I(α).

Proof. Using the form of P ju :

P (t) =∑

I(γ)⊃I(α),γ∈Qjaγ.

22

LetΓj1 = γ ∈ Qj : I(γ) ⊃ I(α) and |n(α)− n(γ/α)| < b−10

j ,

Γj2 = γ ∈ Qj : I(γ) ⊃ I(α) and |n(α)− n(γ/α)| ≥ b−10j .

Hence, if we put

P1(t) =∑γ∈Γj2

aγeγ(t)

then, we get

P (t) =∑γ∈Γj1

aγeγ(t) + P1(t),

and P1(t) satisfies the conditions of the lemma.

Now if Γj1 is an empty set, we put ρ = 0 and P0(t) = 0. Easy to see that thelemma is true in this case. If the set consists only one element δ, we can put ρ = a(δ)and P0(t) = 0. Since Iu 6⊂ X, using Proposition 2.1, we obtain |ρ| ≤ b−2

j yp2 .

Finally if there are δ, γ ∈ Γj2 such that

|n(δ)− n(γ)| · |α| ≥ b10j .

By definition of Γj2, we have that δ, γ are dyadic intervals, and I(α) ⊂ (I(δ) ∩I(γ)). Thus I(δ) ⊂ I(γ) or I(δ) ⊃ I(γ). Without loss of generality, assuming thatI(γ) ⊂ I(δ). We take β = γ, then α satisfies the condition B(Rj). Indeed, we haveI(α) ⊂ I(γ) = I(β) and I(β) = I(γ) ⊂ (I(γ) ∩ I(δ)). On the other hand,

|λ(γ)− λ(δ)| · |α| = 2π|n(γ)|α||γ|− n(δ)

|α||δ||.

By definition n(γ/α) = bn(γ) |α||γ|c and by the construction of γ and δ we have

|n(α)− n(γ/α)| < b−10j and |n(α)− n(δ/α)| < b−10

j .

Hence

b10j ≤ |λ(γ)− λ(δ)| · |α| ≤ 2π(|n(γ/α)− n(δ/α)|+ 2) ≤ 2π(2b−10

j + 2) ≤ 32 · b−10j .

Therefore α satisfies the condition B(Rj), it is contradiction by α /∈ Rj.

We have proved, for every γ, δ ∈ Γj1 then

|n(γ)− n(δ)| · |α| < b10j .

We choose δ to be one element in the set Γj1. Let t0 be the central point of I(α),we have for every t ∈ I(α) (Observe that for every γ that appears in the sum belowwe have t ∈ I(γ))∑

γ∈Γj1

aγeγ(t) = (∑γ∈Γj1

aγeiλ(γ)t0)eiλ(δ)(t−t0) +

∑γ∈Γj1

aγeiλ(γ)t0 [eiλ(γ)(t−t0) − eiλ(δ)(t−t0)]

= ρeiλ(δ)t + P0(t),

23

whereρ =

∑γ∈Γj1

aγeiλ(γ)t0 ,

andP0(t) =

∑γ∈Γj1

aγeiλ(γ)t0 [eiλ(γ)(t−t0) − eiλ(δ)(t−t0)].

Applying again Proposition 2.1 give us that

|ρ| ≤∑γ∈Γj1

|aγ| ≤∑

γ∈Qj ,I(γ)⊃I(α)

|aγ| ≤ b−2j y

p2 .

On the other hand

|P0(t)| = |∑γ∈Γj1

aγeiλ(γ)t0 [eiλ(γ)(t−t0) − eiλ(δ)(t−t0)] |

= |∑γ∈Γj1

aγei[λ(γ)t0+(λ(γ)−λ(δ))(t−t0)][ei(λ(γ)−λ(δ)(t−t0) − 1] |

≤∑γ∈Γj1

|aγ| | ei(λ(γ)−λ(δ))(t−t0) − 1 | .

Since t ∈ I(α) and t0 is the central point of this interval, we have

|(λ(γ)− λ(δ))(t− t0)| ≤ |λ(γ)− λ(δ)| · |α| ≤ b10j .

This implies| ei(λ(γ)−λ(δ))(t−t0) − 1 |≤ b10

j .

Hence, since I(α) 6⊂ X,

|P0(t)| ≤ b10j

∑γ∈Γj1

|aγ| ≤ b10j b−2j y

p2 = b8

jyp2 .

The next condition in the definition of Rj is added in order to achieve, undercertain hypothesis, that the function P j

u(x) coincides on every grandson Iv of Iu withP jv (x).

Condition A(Rj) If β ∈ Qj then α ∈ R for every α ∈ DI such that

I(α) ⊂ I(β), |α| > b10j |β|,

and |n(α)− n(γ/α)| < b−10j , (3.3)

where γ ∈ Qj is such that I(γ) ⊃ I(β).

24

Easy to see that every α ∈ Qj also satisfy the condition A(Rj) by taking β andγ as same as α. Thus α ∈ Rj. This condition needs also an exceptional set that allowsthe reasoning of Lemma 3.2.

For every α ∈ Qj let Yj(α) be the union of the two intervals of length 8b3j |α|

with centers at the two extremes of I(α). Then we set

Y =∞⋃j=1

⋃α∈Qj

Yj(α). (3.4)

Hence we have, by the bound of the length of the notes of level j,

m(Y ) ≤∞∑j=1

∑α∈Qj

16b3j |α| ≤ 16(

∞∑j=1

bj)‖f‖2

2

yp≤ C

yp‖f‖2

2. (3.5)

Lemma 3.2. Let α ∈ PI such that α/L /∈ Rj for every grandchild L of I(α), andI(α) 6⊂ Y . Let P j

u and P jv be the functions respect to two grandchildren J = Iu and

K = Iv of I(α). If there is a term aγeγ(x) of P ju , such that

|n(γ/J)− n(α/J)| < b−9j ,

then P ju = P j

v . Hence the four functions associated to the grandchildren of α are thesame.

Proof. By the hypothesis, there is a term aγeγ(x), it implies that γ ∈ Qj is such thatI(γ) ⊃ J . Since α ∈ PI so α/J ∈ DI . We take β = γ and ζ = α/J . We can see that,

I(ζ) = J ⊂ I(γ) = I(β).

On the other hand, by n(ζ) = n(α/J), and

n(γ/J) = bn(γ)|γ||J |c = bn(γ)

|γ||ζ|c = n(γ/ζ),

using the hypothesis|n(γ/J)− n(α/J)| < b−9

j < b−10j .

We have |n(ζ)− n(γ/ζ)| < b−10j , where γ ∈ Qj satisfies I(γ) = I(β). So if |ζ| > b10

j |γ|,then ζ satisfies B(Rj) condition or |ζ| = |α/J | ∈ Rj, it is a contradiction. Therefore

|α/J | ≤ b10j |γ|.

Now we shall prove that I(γ) ⊂ I(α). To this purpose we must show that itcontains every grandchildren of I(α), say K. Indeed, if K 6⊂ I(γ) there is one end pointa of I(γ) contained in I(α). Yj(α) contains an interval with center at a and length

8b3j |γ| > 8b10

j |γ| ≤ 8|J | = 2|α|.

25

Hence it must be that I(α) ⊂ Yj(α) ⊂ Y. This is contradiction with I(α) 6⊂ Y .Therefore I(γ) contains every grandchild of I(α) or I(γ) ⊃ I(α).

From the above reason, we can not only deduce that I(γ) ⊃ I(α) but also thatthe distance %1, %2 to the end points of I(γ). Let % = min(%1, %2), by I(α) 6⊂ Yj(γ) wemust have that

%+ |α| ≥ 4b3j |γ|.

It follow that% ≥ 4b3

j |γ| − |α| ≥ 4(b3j − b10

j )|γ| ≤ b7j |γ|.

In order two prove P ju = P j

v . We remember

P ju(x) =

∑b∈Qj ,I(δ)⊃J

aδeδ(x), and P jv (x) =

∑b∈Qj ,I(δ)⊃K

aδeδ(x).

Thus if we have that for every δ ∈ Qj such that I(δ) ⊃ J satisfies also I(δ) ⊃ K, thenP ju = P j

v . Now we deal above problem. We assume, there is β ∈ Qj, I(β) ⊃ J andI(β) 6⊃ K. It follows that one end point b of I(β) is contained in I(α). Since Yj(β) itfollows that

8b3j |β| < 2|α|.

Now we prove that I(γ) ⊃ I(β). Indeed, I(γ) contains an interval of length |α| + 2%with the same center as I(α). On the other hand, we have

% > b7j |γ| >

1

4b−3j |α| > |β|.

By (I(β) ∩ I(α)) ⊃ J, so I(β) is contained in the interval of length |α| + 2% with thesame center as I(α). Hence I(γ) ⊃ I(β).

We can apply now condition A(Rj), with our β and γ, to prove that α/J ∈ Rj.This contradicts our hypotheses.

Therefore, for every δ ∈ Qj such that I(δ) ⊃ J satisfies also I(δ) ⊃ K orP ju = P j

v .

26

3.2 The length of well situated notes

We must bound the length of the well situated notes. In fact we bound onlywell situated pairs α such that I(α) 6⊂ X. The importance of this condition is thatassuming it we can apply Proposition 2.1. That is, we know that P j

u(α) has at most b−3j

terms.First observe that given β ∈ Qj, the number of cycles n(α) of a pair α that

satisfies condition A(Rj) with this β is less than 2b−10j times the number k of γ ∈ Qj

such that I(γ) ⊃ I(β) by using condition |n(α)−n(γ/α)| < b−10j . Since I(α) ⊂ I(β) we

have I(β) 6⊂ X. Therefore k ≤ b−3j . Independently, the intervals I(α) must be dyadic

sub-intervals of I(β) of length between b10j |β| and |β|. The length of all these intervals

is ∑I(α)⊂I(β),|α|≥b10j |β|

|α| ≤ (1 + log2 b−10j )|β|.

Therefore, by the known estimates of the length of the notes of level j of function f .We have ∑

α∈A(Rj),I(α)6⊂X

|α| ≤∑β∈Qj

Cb−3j · b−10

j (log b−1j )|β|

≤ Cb−15j (log b−1

j )‖f‖2

2

yp≤ Cb−16

j

‖f‖22

yp.

In the same way, the number of pairs γ and δ, that with a fixed β, satisfiesconditions B(Rj) is less than b−3

j · b−3j = b−6

j . As in the previous case we are interestedonly in the case I(α) 6⊂ X. With these γ and δ fixed, the number of cycles n(α)can be selected between 2b−10

j values, and the length of |α| can attain only less than

C(log b−1j ). With these observations we can conclude that

∑α∈B(Rj),I(α)6⊂X

|α| ≤ Cb−6j · b−10

j (log b−1j )b−2

j

‖f‖yp≤ Cb−19

j

‖f‖yp

.

Therefore the length of the well situated notes of leve; j with I(α) 6⊂ X isbounded in the following way: ∑

α∈Rj ,I(α)6⊂X

|α| ≤ Cb−19j

‖f‖22

yp. (3.6)

27

3.3 Allowed pairs

Recall the basic step of the proof. We have a Carleson integral Cαf(x), I(α) 6⊂S∗ and this gives us that the Carleson integral has a well defined level j ∈ N such that

ybj ≤ |||f |||α < ybj−1

We apply the process by which we obtain a partition Πα, and to obtain a good boundfor the terms of the decomposition we carefully select a smoothing interval I(x). Now,in general, one of the halves of I(x) is an interval of Πα. This implies that grandson Jof I(x) satisfies

‖f‖α/J ≤ ybj−1. (3.7)

Now we pass from Cαf(x) to Cβf(x) where β = α/I(x).

We cane assume that α was an allowed pair, but the problem is that in generalβ is not such a pair. The idea of the proof is to choose the set of allowed pairs Sj sothat we can choose γ ∈ Sj with |Cβf(x)− Cγf(x)| conveniently bounded.

We have obtained a set Rj such that δ /∈ Rj implies that P ju sound as a rest or

a pure note ξ if we consider only the sounds of comparable pitch to that of δ. By (3.7)for some grandchildren J of I(β), ‖f‖β/J ≥ ybj−1. This implies that certain coefficientsFourier of f on J with a pitch comparable to that of β are of a certain size. The hopeis that taking δ = β/J we can arrange things so that we can in fact prove that thereexists (in P j

u(δ)) the note ξ of this pitch. Then the candidate to γ will be this note

ξ/I(x).

According to the previous considerations we must define β /∈ Sj so that itimplies β/J /∈ Rj. This will not be sufficient because we can not prove the implication,‖f‖β/J ≥ ybj−1 implies that there exists one note in P j

u(β/J) of pitch near to that of β.Instead we shall prove that there exists a natural number m only depending on p, andsuch that ‖f‖β/J ≥ ybj−1 implies that there exists one note in Pm+j

u(β/J) of pitch near tothat of β.

This shift is of no consequence to the rest of the proof. In fact the naturalnumber m must be chosen great enough to attain also another objective. For thepresent we take m depending only on p and great enough.

Then we can define Sj, the set of allowed pairs:

Definition 3.1. (Allowed pairs) A pair α is in Sj if I(α) 6⊂ X∗ is a smoothinginterval, and for some grandchild J of I(α) we have α/J ∈ Rm+j.

28

3.4 The exceptional set

Here we define the exceptional set. We 1 0 and a natural number N to the given. And ourpurpose is to define a set EN ⊂ R, with m(EN) < A‖f‖py−p such that for everyx ∈ I/2 \ EN and α = (n, I) ∈ P with 0 ≤ n < 2N , we have |Cαf(x)| < Bpy.

In fact we will assume that |f | = χA is the characteristic function of a measur-able set. Therefore once we obtain a bound of |C(n,I)f(x)| for 0 ≤ n < 2N , we shallhave also a bound for 0 ≤ |n| < 2N applying the same reasoning to f .

The set EN , and other auxiliary sets that we will define later, will dependheavily on f, p, y and N . We will not mention the dependence on f, p, and y but weshall mention the dependence on p of every constant that appears. Recall that theshift m depends on p and only on p.

The set EN will be a union D ∪ S∗ ∪ TN ∪ UN ∪ V ∪X∗ ∪ Y of various sets.

We retain N , but the main point in the proof is to make the estimates theindependent of N . The role of N is now to define the partition Πα for every allowedpair α.

We have defined the sets X∗, S∗ and Y that do not depend on N and have thebounds

m(S∗) ≤ 7

yp‖f‖pp, m(X∗) ≤ C

yp‖f‖2

2, m(Y ) ≤ C

yp‖f‖2

2.

Now to define the sets TN and UN , we must consider the allowed pairs α andthe associate partition Πα. Recall from section [1.6] the needed ingredients to define Πα:

• The function f ∈ L1(I),• The natural number N ,• The positive real number y,• The pair α, where I(α) is a smoothing interval of length |α| ≥ 4|I|

2N,

• A natural number j ≥ 1 such that |||f |||α < ybj−1.

We have been given f,N and y, so we consider, for every j ∈ N , the set ofallowed pairs α ∈ Sj such that |α| ≥ 4|I|

2Nand |||f |||α < ybj−1. For every such pair

we obtain the associated partition Πα. With this partition we can define the functions∆α(x) = ∆(Πα, x) and H∗αf(x) = H∗I(α)Eαf(x) and the two sets

UN,j(α) = x ∈ I(α) : ∆α(x) > C12mb− 1

2j−1,

TN,j(α) = x ∈ I(α) : H∗α(x) > C22mb12j−1, (3.8)

where m is the shift and C1, C2 are constant that we are going to fix.

By the maximal inequalities proved for ∆α(x) and H∗αf(x), we obtain

m(UN,j(α)) ≤ A|α|e−BC12mb− 1

2j−1 ,

29

m(TN,j(α)) ≤ A|α|e−BC22mb12j−1‖Eαf‖

−1∞ .

We have the bound ‖Eαf‖∞ ≤ C supk‖f‖βk ≤ Cybj−1. Hence with a proper

choice of the constant C2 in the definition of TN,j(α) we have

m(TN,j(α)) ≤ A|α|e−BC12mb− 1

2j−1 .

We put

UN =⋃α,j

UN,j(α), TN =⋃α,j

TN,j(α),

where the summation is on all allowed pairs, with α ∈ Sj, and |||f |||α < ybj−1.

Collecting our results, and selecting adequately C1, we get

m(TN ∪ UN) ≤ 2∞∑j=1

A2−2m+8b− 1

2j−1

∑α∈Sj|α|.

For every β ∈ Rm+j there are eight pairs α such that α/β = β. In fact there aretwo smoothing intervals I(α) such that I(β) is a grandchild of I(α), and there arefour possible values of n(α). Thus, taking into account the bound of the length of wellsituated notes, we have

m(TN ∪ UN) ≤ 32∞∑j=1

A2−2m+8b− 1

2 j−1 ·∑

β∈Rm+j ,I(β)6⊂X

|β| ≤ C‖f‖22

yp

∞∑j=1

b−19m+j2

−2m+8b− 1

2j−1 .

It is easy to see that no mater what the natural number m, there is an absoluteconstant C such that

m(TN ∪ UN) ≤ C

yp‖f‖2

2. (3.9)

Another component of the exceptional set is the set D of dyadic points withrespect to the interval. This set has zero measure.

Finally we define the set V , the last component of EN . V is defined as the set

V = x : H∗I f(x) ≥ By2m. (3.10)

Assume that f ∈ Lp(R). By Proposition 1.7,

H∗I f(x) ≤ 2H∗f(x) + 6Mf(x).

The theorem about the maximal Hilbert transform and the Hardy-Littlewood maximalfunction imply that

‖H∗I f‖p ≤ Cp2

p− 1‖f‖p.

30

Therefore

m(V ) ≤ (Cp2

B2m(p− 1))p‖f‖ppyp

.

We will see in the following part that the selection of m is such that

Ap2

p− 1≤ 2m ≤ A′

p2

p− 1.

Therefore choosing conveniently the constant B in definition of set V , we get

m(V ) ≤‖f‖ppyp

. (3.11)

31

3.5 Choosing the shift m

This section contains the most difficult part of the proof of Carleson’s Theorem,that is, how to manage to pass from a Carleson integral Cαf(x) to another where wecan apply the basis step. This is accomplished mainly by changing frequency to obtainCγ/α, where γ is an allowed pair with lesser level than that of the initial Carlesonintegral. First we will obtain a nearby allowed pair β that controls the change offrequency.

In previous section, we have constructed the exceptional set EN that is unionof sets D,S∗, TN , UN , V,X

∗, Y . And we obtain, m(S∗) and m(V ) are bounded byA‖f‖pp/yp. But the bounds we have obtained for m(X∗),m(Y ),m(TN ∪ UN) are oftype A‖f‖2

2/yp. So it is difficult to find the type of bound for EN . To overcome

this difficulty we observe that if f is a measurable function such that |f | = χA is acharacteristic function, then ‖f‖2

2 = ‖f‖pp. Later we will have to deal with more generalfunctions. We shall call such a function a special function.

There is also another reason for which it is convenient to consider the case ofthese special function. In fact this will permit us to define the shift m, depend on p,connecting the level of a Carleson integral and y.

The shift m is a natural number, depending only on p. As we have said in theintroduction to the definition of allowed pair Sj, m has a role in the process of selectinga note of f near a not allowed pair. In the following proposition we give to m anotherrole in the proof of the Lp result.

Proposition 3.1. (Selection of the shift) Let 1 < p < +∞, then there exists anatural number m = m(p) such that if α ∈ PI , f is a special function, and j ∈ N aresuch that bjy ≤ |||f |||α, and I(α) 6⊂ S∗, then

yp2 ≤ b

− 14

m+jy, and bm+j < y. (3.12)

Proof. By bjy ≤ |||f |||α = sup‖f‖α/J00 , ‖f‖α/J01 , ‖f‖α/J10 , ‖f‖α/J11, so there existsa grandchild J of I(α) such that ‖f‖α/J = |||f |||α ≥ bjy. Since I(α) 6⊂ S∗, we haveJ 6⊂ S. It follows that

1

|J |

∫J

|f |pdm < yp.

Also from the definition of ‖f‖α/J and the fact that f is a special function, it impliesthat

‖f‖α/J ≤1

|J |

∫J

|f |dm =1

|J |

∫J

|f |pdm.

Therefore bjy < yp. Hence we show that yp2 ≤ b

− 14

m+jy for every m ≥ m(p), if bjy < yp.

First, we consider the case 1 y ≥ y

p2 , and y ≥ yp−1 > bj.

32

If y < 1, we choose m such that

2m − 1

4>

1− p2

p− 1.

Then we have2m − 2−j

4> (1− 2−j)

1− p2

p− 1,

or1− 2m+j

4< (1− 2j)

1− p2

p− 1.

Since bj = 2 · 2−2j , we deduce that

b14m+j = 2

1−2m+j

4 < 2(1−2j)1− p2p−1 = b

1− p2p−1

j < y1− p2 .

This is the first inequality of proposition is proved. For the second one we consider

first the case 1 bm+j for m ≥ 1. So y > bm+j.

Now, If p ≥ 2. For y ≤ 1 it suffice to take m ≥ 1, and we have

yp2−1 ≤ 1 ≤ b

− 14

m+j, bm+j < bj < yp−1 ≤ y,

so the proposition is true in this case. If y > 1; since |f | is a characteristic function,we have bjy ≤ |||f |||α ≤ 1. So

y > 1 ≥ bjy > bj ≥ bm+j.

For the second inequality in proposition, we choose m such that

p

2− 1 <

2m − 1

4.

It implies that

(1− 2−j)(p

2− 1) <

p

2− 1 <

2m − 1

4<

2m − 2−j

4.

Therefore

(2j − 1)(p

2− 1) <

2m+j − 1

4.

Henceyp2−1 ≤ b

1− p2

j < b− 1

4m+j,

oryp2 < b

− 14

m+jy.

33

Observe that, we choose m only depending on p and

• If 1 1−p/2p−1

• If p ≥ 2 then choosing m such that 2m−14

> p2− 1.

Hence, there exists a constant 0 < A < ∞ such that the shift m can be anynatural number satisfying

2m ≥ Ap2

p− 1.

So we can choose m such that

A′p2

p− 1≥ 2m,

for some absolute constant A′.

34

3.6 A bound for ‖f‖α

The following proposition is essential to obtain, from a lower bound for a localnorm: ‖f‖α > r > 0, some note of f at the scale of α.

Proposition 3.2. Let α = (n, J) ∈ P , f ∈ L2(J) and let

f(t) =∑

β=(k,J)

akeβ(t)

be its local Fourier expansion. Assume that |ak| ≤ N for every k with |k + n| < M,where 1 < M <∞, and 0 < N <∞. Then

‖f‖α ≤ B(N logM +‖f‖2

L(J)√M

), (3.13)

where B is an absolute constant.

Proof. By definition

‖f‖α =∑j∈Z

c

1 + j2| 1

|J |

∫J

f(t)e−2πi(n(α)+ j3

) t|J|dt|.

The integral can be written as

1

|J |

∫J

f(t)e−iλ(α)t · e−2πi jt3|J|dt =

∑k∈Z

a′kbk, (3.14)

where a′k and bk are the coefficients of the expansions on L2(J)

f(t)e−iλ(α)t =∑

β=(n,J)

a′n(β)eβ(t), e2πi jt3|J| =

∑β=(n,J)

bn(β)eβ(t).

Hence

a′k =1

|J |

∫J

f(t)e−2πi(n(α)+k) t|J|dt = an(α)+k.

By hypothesis |a′k| ≤ N if |k| < M. If k + j36= 0, the coefficient b−k are bounded by

|b−k| ≤ |e2πi( j

3+k) − 1

2π( j3

+ k)| ≤ 1

| j3

+ k|.

In the case k = − j3, |bk| is bounded by 1.

Now we can bound the integral in (3.14). For |j| ≤ M2

1

|J |

∫J

f(t)e−iλ(α)t · e−2πi jt3|J|dt =

∑k∈Z

a′kbk ≤∑|k|≤M

|a′kbk|+∑|k|>M

|a′kbk|

35

≤ N +N ·∑

|k|<M,k 6=−j/3

1

|k + j3|

+ (∑k

|a′k|2)12 (

∑|k|≥M

(1

|k + j3|)2)

12 .

Hence there exists a constant C such that∑k

|a′kbk| ≤ C(N logM +‖f‖L2(J)√

M).

For |j| > M2

we use only the Schwarz inequality, and obtain∑k

|a′kbk| ≤ ‖f‖L2(J).

Therefore ‖f‖α is bounded by

C(N logM +‖f‖L2(J)√

M)∑j

c

1 + j2+ ‖f‖L2(J)

∑|j|>M

2

1

1 + j2≤ B(N logM +

‖f‖L2(J)√M

).

36

3.7 Selecting an allowed pair

Every grandson α/K of a not allowed pair α /∈ Sj is not well situated (α/K /∈Rm+j). Then, we have arranged things so that the sound f in the scale of α/K is a restor a pure note. If we also assume the we are at level j, and have selected adequatelythe shift m, then f definitely sounds, so by the structure theorem Pm+j

δ/K musst be apure note. All pure notes of f are well situated, so a grandfather β of this note will bean allowed pair nearby to α. If we are considering this β as a step in our process, wemust bound the difference |Cβ/αf(x) − Cαf(x)|, for this will be convenient to give abound for |n(β/α)−n(α)|. We will be undertaking the process of selecting this nearbyallowed pair β.

Proposition 3.3. Let f be a special function, and α ∈ PI such that bjy ≤ |||f |||α <bj−1y, α /∈ Sj, and let x ∈ I/2 be a point such that x ∈ I(α)/2 and x /∈ D∪S∗∪X∗∪Y .Then there exists β ∈ Sj, with I(β) ⊃ I(α), x ∈ I(β)/2, and such that

|n(α)− n(β/α)| < A0

bj, (3.15)

and for every γ with I(γ) = I(α), and |n(α)− n(γ)| ≤ 2A0b− 3

2j we have

|Cγf(x)− Cαf(x)| ≤ B(|||f |||β/α + bjy). (3.16)

Remark. In such a situation we shall call β a nearby allowed pair to α. Itsfunction is to control the change of frequency.

Proof. There exists a grandchild K of I(α) such that ‖f‖α/K = |||f |||α. For this K, asfor every other grandchild of I(α), we have α/K /∈ Rm+j. We hope to find a β startingwith some term of the function Pm+j

u (t), where u = u(K). A general remark aboutthe proof: we care to the shift m, we have quantities at two levels ybj and ybm+j andalways ybj+m ≪ ybj.

We divide the proof into four steps.

First step: To find δ ∈ Qm+j, with I(δ) ⊃ K

The proof starts applying Proposition 3.2 to obtain, to every k ∈ Z, a boundfor ‖f − Pm+j

u ‖(k,K).By construction of Pm+j

u (t) the local Fourier coefficients of (f − Pm+ju )(t) are

less than bm+jyp2 . Also

‖f − Pm+ju ‖L2(k) ≤ ‖Pm+j

u ‖L2(k) + ‖f‖L2(k). (3.17)

Since x ∈ I(α) and x /∈ S∗, we have I(α) 6⊂ S∗, so K 6⊂ S and

‖f‖2L2(K) = ‖f‖pL2(K) < yp.

Furthermore, since I(α) 6⊂ X∗ or K 6⊂ X and by (2.9), |Pm+ju | ≤ b−2

m+jyp2 . Thus

‖f − Pm+ju ‖L2(k) ≤ y

p2 + b−2

m+jyp2 ≤ 2b−2

m+jyp2 .

37

We are in proposition to apply Proposition 3.2 with N = bm+jyp2 ,M = b−8

m+j

and the bound obtained for ‖f − Pm+ju ‖L2(K). we obtain that for every k ∈ Z

‖f − Pm+ju ‖(k,K) ≤ B(y

p2 log(b−8

m+j) +2b−2m+jy

p2

b−4m+j

)

= Byp2 (log(b−8

m+j) + 2b2m+j)

≤ 3Byp2 b

32m+j. (3.18)

The last inequality requires that we take m big enough. In fact m ≤ 5 suffice.

Now we have a lower bound of ‖Pm+ju ‖α/K . This will follow that there are term

in this function.

‖Pm+ju ‖α/K ≥ ‖f‖α/k − ‖f − Pm+j

u ‖α/K ≥ ybj − 3Byp2 bm+j

34.

It is here were are forced to choose the shift so that yp2 ≤ b−

14 or something similar.

Since f is a special function, we are able to choose the shift so that yp2 ≤ yb

− 14

m+j,therefore

‖Pm+ju ‖α/K ≥ ybj − 3Byb

12m+j >

1

2ybj. (3.19)

We assume here that m has been chosen in order to satisfy

3Bb12m+j <

bj2.

Given that bj = 2 · 2−2j , it is easy to see that this condition is equivalent to m ≥ m0

for some absolute constant m0.

Note that the same calculation proves that for every grandchild L = Iv of I(α)we have

‖f − Pm+jv ‖(k,L) ≤ 3Byb

12m+j. (3.20)

Since α/K /∈ Rm+j and K 6⊂ X, the Lemma 3.1 on the structure of Pm+ju say

that

Pm+ju (t) = ρeiλ(δ)t + P0(t) + P1(t), t ∈ K = Iu (3.21)

where

|ρ| ≤ b−2m+jy

p2 , |P0(t)| ≤ b8

m+jyp

2, (3.22)

in which δ ∈ Γm+j1 and all the term aγeγ(t) of P1(t) satisfy γ ∈ Γm+j

2 or

|n(α/K)− n(γ/K)| ≥ b−10m+j.

38

If we have ρ 6= 0, then δ ∈ Qm+j is such that I(δ) ⊃ K. In this case we canchoose β to be grandfather of δ. Thus we will prove ρ 6= 0. Indeed, we have givenadequate bounds for the local norm of exponential in Proposition 1.3. Hence taking(3.21) into account, we have

‖Pm+ju (t)‖α/K ≤ |ρeiλ(δ)t|+ |P0(t)|+ |P1(t)|

≤ |ρeiλ(δ)t|+ b8m+jy

p2 +

∑γ∈Γm+j

2

|aγ||eγ(t)|

≤ C|ρ||bn(δ) |K||δ| c)− n(α/K)|

+ b8m+jy

p2 +

∑γ∈Γm+j

2

C|aγ||bn(γ) |K||γ| c)− n(α/K)|

=C|ρ|

|n(δ/K)− n(α/K)|+ b8

m+jyp2 +

∑γ∈Γm+j

2

C|aγ||n(γ/K)− n(α/K)|

≤ C|ρ||n(δ/K)− n(α/K)|

+ b8m+jy

p2 + Cb10

m+j

∑γ∈Γm+j

2

|aγ|.

If n(α/K) = n(γ/K) the first term is reduce to |ρ|.

Since K 6⊂ X, we have a bound for the last sum

‖Pm+ju ‖α/K ≤

C|ρ||n(δ/K)− n(α/K)|

+ b8m+jy

p2 + Cb8

m+jyp2 .

We conclude from (3.19) that

1

2ybj ≤ ‖Pm+j

u ‖α/K ≤ ‖Pm+ju ‖α/K ≤ C|ρ|

|n(δ/K)− n(α/K)|+ 2b8

m+jyp2

≤ C|ρ||n(δ/K)− n(α/K)|

+ 2b7m+jy.

We assume m is such that for every j, 8b7m+j < bj, and we obtain

ybj <4C|ρ|

|n(δ/K)− n(α/K)|. (3.23)

Therefore |ρ| 6= 0. This gives us a element δ ∈ Qm+j such that I(δ) ⊃ K.

Second step: Bound for |n(δ/K)− n(α/K)|

Now we must prove that the pitch of δ is comparable to that of α. That is wellmust bound |n(δ/K)− n(α/K)|, so we assume it is not null.

From the proof of the structure theorem (of P uj ) we know that we can take

as δ any pair in Γm+j1 that gives |n(δ/K) − n(α)/K| < b10

m+j. But our hypothesesbjy ≤ |||f |||α < bj−1y and α /∈ Sj are very strong. In fact, we can prove that every one

39

of this δ’s satisfies |n(δ/K) − n(α/K)| < Cb−1j . To this end we must prove |ρ| ≤ Cy.

A first approximate is obtained from (3.23) and the known estimate |ρ| ≤ b−2m+jy

p2 ,

|n(δ/K)− n(α/K)| ≤Cb−2

m+jyp2

ybj≤ 4Cb−4

m+j < b−5m+j. (3.24)

Now we reverse the inequalities from the previous step.

The local norm ‖Pm+ju ‖(k,K) has a maximum near k = (δ/K). Hence for δ/K =

(k,K), Proposition 1.4 give us:

‖Pm+ju ‖δ/K ≥ B|ρ| − b8

m+jyp2 −

∑γ∈Γm+j

2

C|aγ||n(γ/K)− n(δ/K)|

.

And we have∑γ∈Γm+j

2

C|aγ||n(γ/K)− n(δ/K)|

≤∑

γ∈Γm+j2

C|aγ||n(γ/K)− n(α/K)| − |n(δ/K)− n(α/K)|

.

Hence, by (3.24), this sum is bounded by

∑γ∈Γm+j

2

C|aγ|b−10m+j − b−5

m+j

≤Cy

p2 b−2m+j

b−10m+j − b−5

m+j

.

Since m ≤ 2 we obtain 2b−5m+j < b−10

m+j, so that finally we are at

‖Pm+ju ‖δ/K ≥ B|ρ| − b8

m+jyp2 − 2Cy

p2 b8m+j > B|ρ| − yb6

m+j.

But, on the other hand, by (3.18)

‖Pm+ju ‖δ/K ≤ ‖f‖δ/K + ‖f − Pm+j

u ‖δ/K ≤ ‖f‖δ/K + 3Byb6m+j.

Therefore

|ρ| ≤ C(‖f‖δ/K + yb12m+j). (3.25)

Now ‖f‖δ/K ≤ ‖f‖L1(K) ≤ ‖f‖Lp(K) < y and we see at once that

|ρ| ≤ 2Cy,

which with (3.23) establishes that

|n(δ/K)− n(α/K)| < C

bj.

Third step: Definition of β

40

We are in proposition to apply Lemma 3.2 to prove that the function Pm+ju are

the same for every grandson of α. Namely is a pair such that α/L /∈ Rm+j for everygrandchild L of I(α), I(α) 6⊂ Y by hypothesis, and there is a term δ ∈ Qm+j of Pm+j

u

such that

|n(δ/K)− n(α/K)| < C

bj< b−9

m+j.

We deduce that in fact I(α) ⊂ I(δ) and that four functions Pm+jv , corresponding to

the four grandchildren Iv of I(α), coincide.

Since δ ∈ Qm+j we know that δ is a dyadic interval with |δ| ≤ I4. Then x ∈ I/2

is not a dyadic point there is only smoothing interval I(β) of length 4|δ| such thatx ∈ I(β)/2. We define β as the pair with n(β) = 4nδ, and this I(β). It is easy tosee that I(δ) is a grandchild of I(β) and β/δ = δ. Furthermore I(β) 6⊂ X∗ sinceI(α) ⊂ I(δ) ⊂ I(β) and I(α) 6⊂ X∗. Hence prove β ∈ Sj we only have to proveδ ∈ Rm+j. But δ ∈ Qm+j, we also have δ ∈ Rm+j. Therefore β ∈ Sj

We also have that

|n(α)− n(β/α)| = |n(α)− bn(β)|α||β|c|

= |b4n(α)|K||α|c − b4n(δ)

|K||δ|c|

≤ 4|n(α/K)− n(δ/K)|+ 8 ≤ 4C

bj+ 8 <

A0

bj.

To prove (3.15).

Fourth step: Bound for |Cγf(x)− Cαf(x)|.

Let γ be a pair such that I(γ) = I(α) and |n(α)−n(γ)| ≤ 2A0b− 3

2j . Let P = Pm+j

u

and recall that we have proved that it coincides with every Pm+jv if v = v(L) for some

grandchild L of I(α). Also note that (3.21) and the inequalities in (3.22) are valid nowfor every t ∈ I(α).

First we have

|Cγf(x)− Cαf(x)| ≤ |Cγ(f − P )(x)− Cα(f − P )|+ |CγP (x)− CαP (x)|.

By a changing of frequency, by the bounds of ‖f −P‖(k,L) obtained from (3.20)at step 1, and by the comparison between the two norm ‖f − P‖α and |||f − P |||α weget

|Cγ(f − P )(x)− Cα(f − P )| ≤ C(2A0b− 3

2j )3‖f‖α ≤ C(2A0b

− 32

j )3|||f |||α

≤ C(2A0b− 3

2j )33Byb

12m+j < bjy,

by the restriction m ≥ m0.

41

Second, by the structure of P

|CγP (x)− CαP (x)| ≤ |ρ| · |Cγ(eiλ(δ)t)(x)− Cα(eiλ(δ)t)(x)||CγP0(x)− CαP0(x)|+ |CγP1(x)− CαP1(x)|.

The Carleson integral of an exponential is bounded. So for the first term we have

|ρ| · |Cγ(eiλ(δ)t)(x)− Cα(eiλ(δ)t)(x)| ≤ 2A|ρ|.

Ine the second term we have, one more time, a change of frequency. Hence

|CγP0(x)− CαP0(x)| ≤ B(2A0b− 3

2j )3‖P0‖α ≤ Ab

− 92

j b7m+jy < bjy.

Again by the assumption m ≥ m0.

The third term is bounded in the following way:

|CγP1(x)− CαP1(x)| ≤∑

η∈Γm+j2

|aη| · |Cγ(eiλ(η)t)(x)− Cα(eiλ(η)t)(x)|.

By another change of frequency, we get

|CγP1(x)− CαP1(x)| ≤∑

η∈Γm+j2

|aη|b− 9

2j ‖eiλ(η)(t)‖α

≤ Cb− 9

2j

∑η∈Γm+j

2

|aη||bn(η) |α||η| − n(α)c|

≤ Cb− 9

2j

∑η∈Γm+j

2

|aη|4|n(η/K)− n(α/K)| − 8

≤ C ′b− 9

2j b10

m+jyp2 b−2m+j ≤ Cb

− 92

j b− 9

2m+jb

7m+jy ≤ bjy.

Collecting all these inequalities we have prove that

|Cγf(x)− Cαf(x)| ≤ 2A|ρ|+ 3bjy ≤ C(|ρ|+ bjy),

and by (3.25) and (β/α)/K = δ/K

|Cγf(x)− Cαf(x)| ≤ B(‖f‖δ/K + bjy) ≤ B(|||f |||β/α + bjy).

Since β is an allowed pair (β ∈ Sj) and

|Cα/βf(x)− Cαf(x)| ≤ B(|||f |||β/α + bjy)

we can think to apply the basic step to the Carleson integral Cβ/αf(x). This can beproblematic for two reason: (a) We have not a reasonable bound for |||f |||β/α. (b) We

42

shall arrive to a Carleson integral Cβ/I(x)f(x). This is not guarantee that I(x) $ I(α).There is not more simple Carleson integral, not one with more cycles. Therefore wehave not obtain our objective. What is the new level K with bky ≤ |||f |||β/α < bk−1y?,is β/α ∈ Sk? These question have not a unique answer.

Therefore what we need is a pair ξ, and a level ` such that:

(a) We are in a good position to apply the basic step, I(ξ) ⊃ I(α),

x ∈ I(ξ)/2, ξ ∈ S`, and |||f |||ξ < b`−1.

(b) We are at an adequate level

|||f |||β/α < b`−jy, ` < j.

(c) We have a controlled change of frequency

|n(ξ/α)− n(α)|2A0b− 3

2j .

(d) Finally, to be able to apply the basic step I(α) must be a union of sets ofthe partition Πξ corresponding to this pair ξ.

This pair ξ is obtained by picking, from the set of pairs (µ, `) with some ofthese properties, one with the minimum level `. Next we prove that it satisfies all ourcondition.

Proposition 3.4. Assume that f is a special function, x ∈ I/2 and let α ∈ PI , suchthat bjy ≤ |||f |||α < bj−1y, α /∈ Sj and x ∈ I(α)/2 such that x /∈ D ∪ S∗ ∪X∗ ∪ Y. Wealso assume that 0 ≤ n(α) < 2N , and |α| > 4|I|/2N .

Then there exists ξ ∈ Sk, with 1 ≤ k ≤ j such that I(ξ) ⊃ I(α), x ∈ I(ξ)/2, andalso with β being a nearby allowed pair to α, it is satisfied that

|||f |||β/α ≤ bk−1y, |||f |||ξ < bk−1y, |n(ξ/α)− n(α)| ≤ 2A0

bj. (3.26)

Furthermore, if Πξ is the partition determined on I(ξ) and I(x) is the centralinterval then I(x) $ I(α), and I(α) \ I(x) is union of intervals of Πξ.

Proof. Let∑

be the set of pair (µ, `), where µ ∈ PI and ` ∈ N satisfying the conditions:

(i) I(µ) ⊃ I(α) and x ∈ I(µ)/2.

(ii) |||f |||β/α < b`−1y, and ` ≤ j.

(iii) |n(µ/α)− n(α)| ≤ A0

j∑i=`

B−1i .

(iv) µ ∈ S`.

Where in (iii) the constant A0 is the same as that appearing in Proposition 3.3.

43

We divide the proof in 3 steps.

Step 1:∑

is nonempty.If |||f |||β/α < bj−1y then (β, j) ∈

∑which proves our claim.

If |||f |||β/α ≥ bj−1y there exists k, with 1 ≤ k ≤ j−1 such that bky ≤ |||f |||β/α <bk−1y. (Since I(β/α) 6⊂ S∗, we know that |||f |||β/α < y, and |||f |||β/α > 0since|||f |||α ≤bjy > 0.)

Now, if β/α /∈ Sk, then (β/α, k) ∈∑.

Otherwise, β/α /∈ Sk, we obtain an allowed pair β′ nearby to β/α; hence β′ ∈ Sk,with I(β′) ⊃ I(β/α) = I(α), x ∈ I(β′)/2, and

|n(β/α)− n(β′/α)| ≤ A0

bk.

Then (β′, k) ∈∑

. We see that condition (iii) in the definition of∑

is satisfied since

|n(β′/α)− n(α)| ≤ |n(β′/α)− n(β/α)|+ |n(β/α)− n(α)

≤ A0

bk+A0

bj≤ A0

j∑i=k

b−1i .

The other condition is obviously.

Step 2: Selection of (ξ, k) ∈∑

and the proof of |||f |||ξ < bk−1y.

We pick a of (ξ, k)∑

with a minimum k. We are going to prove that ξ and ksatisfy the theorem.

If it were true that |||f |||ξ ≥ bk−1y there would be an `, with 1 ≤ ` < k andb`y ≤ |||f |||ξ < b`−1y.

If ξ ∈ S`, (ξ, `) ∈∑

in contradiction to the selection of (ξ, k).Therefore ξ /∈ S`. Then by Proposition 3.3 there exists an allowed pair β′′ ∈ S`

nearby to ξ. Then I(β′′) ⊃ I(ξ), x ∈ I(β′′)/2, and |n(ξ)− n(β′′/ξ)| < A0

b`.

Since (ξ, k) ∈∑, we have |n(ξ/α) − n(α)| ≤ A0

j∑j=k

b−1i , I(ξ) ⊃ I(α) and

|||f |||β/α < bk−1y.We can check that (β′′, `) ∈

∑. Condition (i) is obvious; (ii) follows from

|||f |||β/α < bk−1y < b`−1y.To deduce (iii), observe that

|n(β′′/α)− n(α)| ≤ |n(β′′)− n(ξ/α)− n(ξ/α)|+ |n(ξ/α)− n(α)|

≤ |n(β′′/α)− n(ξ/α)|+ A0

j∑i=k

b−1k .

Now |n(β′′/α)− n(ξ/α)| ≤ |n(β′′/ξ)− n(ξ)|. This is a particular case of the inequality

|b N2r+sc − bM

2rc| ≤ |bN

2sc −M |,

44

valid when M,N, r and s are non-negative integers. This can be easily proved usingthe binary expansion of natural numbers.

Therefore we have

|n(β′′/α)− n(α)| ≤ A0

b`+ A0

j∑i=k

b−1i ≤ A0

j∑i=k

b−1i .

Since ` < k, that (β′′, `) ∈∑

is in contradiction to the definition of (ξ, k). Thiscontradiction proves the inequality |||f |||ξ < bk−1y, and finishes step 2.

Step 3: Claim on the partition Πξ.

Now let Πξ be the partition of I(ξ) obtained with the given N, y, and theinequality |||f |||ξ < bk−1y. Let also I(x) denote the central interval determined by xand Πξ.

I(x) and I(α) are two smoothing intervals which contain x in their middlehalves. By Proposition 1.5 it follows that I(α) ⊂ I(x) or I(x) $ I(α).

The first hypothesis leads to a contradiction. In fact I(x) ⊃ I(α) and |α| >4|I|/2N . By construction, one of the two halves of I(x), say L, is a member of thepartition Πξ. Since |L| > 2|I|/2N , K, one of the two sons of L, is such that ‖f‖ξ/K ≥bk−1y. Hence there exists ` with 1 ≤ ` < k such that b`y ≤ |||f |||ξ/I(x) < b`−1y.

If ξ/I(x) ∈ ∫ `, then (ξ/I(x), `) ∈∑

. This is in contradiction to the selection of(ξ, k) in

∑.

If ξ/I(x) /∈ S`, Proposition 3.3 gives a nearby allowed pair γ ∈ S`. As in thesecond step we can prove that (γ, `) ∈

∑. This is also contradictory.

We have proved I(x) $ I(α) ⊂ I(ξ). By the last condition of Proposition 1.6 inthis case I(α) and I(x) are union of intervals of the partition Πξ.

45

3.8 All together

The following proposition is formulation the basic step of the proof. We cansummarize it by saying that a Carleson integral of level j is small with respect to itslevel or can be approximated by another Carleson integral of level j′ < j. In this part,if ‖f‖α = 0, we will call that the level of the Carleson integral Cαf(x) is ∞. Thus thelevel of a Carleson integral is always defined and is a natural number or infinity. Inthis part f is a special function, |f | = χA, and we has the bound of exceptional set ENis type

‖f‖ppyp.

Proposition 3.5. Let Cαf(x) be a Carleson integral of level j, such that x /∈ EN , and|α| > 4|I|/2N . Then there exists a natural number 1 ≤ k ≤ j satisfying one of twofollowing conditions:

(A) |Cαf(x)| ≤ C2myb12k−1.

(B) There exists δ ∈ PI , such that I(δ) $ I(α), x ∈ I(δ)/2, λ(δ) ≤ (1+b12j )λ(α),

and either

|δ| ≤ 4|I|2N

or the level of Cδf(x) is j′ < k, and

|Cδf(x)− Cαf(x)| ≤ C2myb12k−1. (3.27)

Proof. If the level of Cαf(x) is∞, then Cαf(x) = 0 and condition (A) is satisfied withk = 1. thus we assume that j <∞.

Because j is the level of Cαf(x) we have ybj ≤ |||f |||α < ybj−1. We divide theproof in two case.

First case. Assume that α ∈ Sj. Given N and |||f |||α < ybj−1, we can use theprocedure to find a partition Πα of I(α) and a smoothing interval I(x) as in Proposition1.6.

Set δ = α/I(x). Then, δ satisfies the condition (B) of the Proposition. Indeed,we will check the condition (B)

λ(δ) = 2πn(δ)

|δ|= 2πbn(α)

|δ||α|c 1

|δ|≤ 2πn(α)

|δ||α|

1

|δ|= λ(α).

If |δ| > 4|I|/2N , we can apply Theorem 1.8, with ξ = α and J = I(α) andobtain the bound

|Cαf(x)− Cδf(x)| ≤ Cybj−1 + 2H∗αf(x) +Dybj−1∆α(x).

Since α ∈ Sj and x /∈ TN(α)∪UN(α) we have by the definition of the sets TN,j(α) andUN,j(α)

H∗αf(x) ≤ C22myb12j−1, ∆α(x) ≤ C12mb

− 12

j−1.

And we get

|Cαf(x)− Cδf(x)| ≤ C2myb12j−1.

46

In this case we take k = j and condition (B) is satisfied. Since |δ| > 4|I|/2N , theconstruction of I(x) also implies that the level j′ of Cδf(x) is less than k = j, andI(x) $ I(α).

Second case. Assume that α /∈ Sj. we select a nearby allowed pair β andapply Proposition 3.4, finding a level k, with 1 ≤ k ≤ j and a pair ξ ∈ Sk such thatI(ξ) ⊃ I(α), x ∈ I(ξ)/2, and |||f |||β/α ≤ ybk−1.

Let also Πξ and I(x) be the partition and central interval arising from ξ andthe inequality |||f |||ξ < ybk−1.

First subcase. If n(α) ≤ 2A0b− 3

2j , since the nearby allowed pair β controls the

change of frequency, we have

|Cγf(x)− Cαf(x)| ≤ B(‖f‖β/α + bjy) ≤ Cybk−1,

where γ = (0, I(α)). By x /∈ EN so x /∈ V

|Cγf(x)| = |p.v.∫I(α)

f(t)

x− tdt| ≤ B2my.

Thus in this subcase, we put k = 1 and obtain

|Cαf(x)| ≤ C2myb12k−1 = C2my.

Second subcase. Now we assume n(α) > 2A0b− 3

2j . Take δ = ξ/I(x), we will prove

δ satisfied condition (B).

First, we prove λ(δ) ≤ (1 + b12j )λ(α). Observe that by definition and by the

construction of ξ

n(δ) = bn(ξ)|I(x)||ξ|c, |n(ξ/α)− n(α)| ≤ 2A0

bj.

Put n′ = n(ξ/α). Since I(ξ) ⊃ I(α) % I(δ) = I(x), we get

n(δ) = n(ξ/I(x)) = n((ξ/α)/I(x)) = bn′ |δ||α|c.

Therefore

n(δ)

|δ|=

1

|δ|bn′ |δ||α|c ≤ n′

|α|≤ n(α)

|α|+

2A0

bj

1

|α|

<n(α)

|α|+ n(α)

b12j

|α|=n(α)

|α|(1 + b

12j ).

And we have proved λ(δ) ≤ (1 + b12j )λ(α).

Now assume that |δ| > 4|I|/2N . Then by the construction of Πξ and I(x). Thelevel j′ of Cδf(x) is j′ < k. As I(α)/I(x) is union of intervals of Πξ, we can apply thebasic step to obtain

|Cξ/I(α)f(x)− Cδf(x)| ≤ Cybk−1 + 2H∗ξ f(x) +Dybk−1∆ξ(x).

47

Since ξ ∈ Sk and x /∈ EN , we obtain the bounds

H∗ξ f(x) ≤ C22myb12k−1, ∆ξ(x) ≤ C12myb

− 12

k−1.

Hence|Cξ/αf(x− Cδf(x)| ≤ C2myb

12k−1.

Furthermore, β controls the change of frequency, thus

|Cξ/αf(x)− Cαf(x)| ≤ B(ybk−1 + ybj) ≤ Cybk−1,

because |n(ξ/α) − n(α)| ≤ 2A0b−1j , and |||f |||β/α < ybk−1. Combining the last two

inequalities, we get

|Cδf(x)− Cαf(x)| ≤ C2myb12k−1.

Now we go to the important theorem

Theorem 3.1. There exists an absolute constant C such that for every special functionf, y > 0 and 1 y ≤ Bpp

‖f‖ppyp

,

where Bp ≤ C p2

p−1.

Proof. Using the approximation C p2

p−1∼ 2m, so we only prove that

mx ∈ I/2 : C∗I f(x) > y ≤ C(B2m)p‖f‖ppyp

= C‖f‖pp

( yB2m

)p.

It is equivalent to

mx ∈ I/2 : C∗I f(x) > B2my ≤ C‖f‖ppyp

As we have know that

m(EN) ≤ C‖f‖ppyp

.

We are going to prove that for every x ∈ I/2 \ EN

sup0≤n<θ2N

|C(n,I)f(x)| ≤ B2my, (3.28)

where θ is an absolute constant.Because, if we have above bound, let

E = x ∈ I/2 : supn≥0|C(n,I)f(x)| > B2my.

48

It easy to see that E ⊂ lim infN

EN , hence we get

m(E) ≤ C‖f‖ppyp

.

If we apply the same reason to f we obtain an analogous inequality for the set

E ′ = x ∈ I/2 : supn≤0|C(n,I)f(x)| > B2my.

Thus we obtain

m(E ′) ≤‖f‖ppyp

.

On the other hand,

x ∈ I/2 : C∗I f(x) > B2my ⊂ (E ∪ E ′).

Therefore,

mx ∈ I/2 : C∗I f(x) > B2my ≤ m(E) +m(E ′) ≤ 2C‖f‖ppyp

.

So the theorem is proved if (3.28) is true.

Now, we are going to prove (3.28). Take α = (n, I) with 0 ≤ n < θ2N , x ∈I/2 \ EN and proceed to show that |Cαf(x)| ≤ B2my. We start an interactive processthat takes a Carleson integral Cα of level j and obtain: a number 1 ≤ k ≤ j, and abound of Cαf(x), or a second Carleson integral of level j′ < k, and besides, a goodbound of the difference |Cδf(x)−Cαf(x)|. Then the process continues with Cδf(x) instead of Cα.

We start with α = α0 = (n, I), but the same procedure is repeated with otherpairs. So, now we assume |α| > 4|I|/2N , Carleson integral Cαf(x) of level j to begiven, and x ∈ I/2 \ EN .

Observe that since x /∈ V , if n(α) = 0 then |Cαf(x)| ≤ B2my.

We apply Proposition 3.5, and obtain a natural number k, with 1 ≤ k ≤ j andone of these two possibilities:

1 If |Cαf(x)| ≤ C2myb12k−1, we stop process, for we have the bound we are

seeking.2 In this second possibility the level j is a natural number and Proposition 3.5

give us a pair δ with I(δ) $ I(α), x ∈ I(δ)/2, and λ(δ) ≤ (1 + b12j )λ(α). We divide this

case into two other case depending on the value of n(α).2.1 If n(α) = 0, we change the value of k to k = 1 and we have |Cαf(x)| ≤

C2myb12k−1 = C2my. We stop the process at α.2.2 If n(α) > 0 we divide it into two subcases.2.2.1 |δ| ≤ 4|I|/2N . In this case we stop the process. But observe that in this

case we have not attained our object, that is, to bound |Cαf(x)| or the difference

49

|Cαf(x)− Cδf(x)|.2.2.2 We have |δ| > 4|I|/2N . Since we are now in case (B) of Proposition 3.5,

the level of Cδf(x) is j′ < k, and

|Cαf(x)− Cδf(x)| ≤ C2myb12k−1.

The process starts with α = α0, of level j0 and such that 0 ≤ n(α0) < θ2N . Weobtain k0, with 1 ≤ k ≤ j0, and stop or obtain α1 = δ, such that the level of Cα1 isj1 = j′ < k, and

|Cα1f(x)− Cα0f(x)| ≤ C2myb12k0−1.

If the process is not stopped, |α1| > 4|I|/2N and we can continue with Cα1 inthe same way we have proceeded with Cα0f(x). Then we stop the process or obtain anew α2, j2 and k2.

This process must has finitely steps because the position of x implies that thelevel is always a natural number, but j0 > j1 > j2 > ... We can repeat the reasoningto obtain a sequence of pairs

α0, α1, α2, ..., αs,

and the corresponding numbers

j0 ≥ k0 > j1 ≥ k1 > ... > js ≥ ks.

Now when we stop process we shall have

|Cαsf(x)| ≤ C2myb12k−1.

Indeed, the situation depicted on 2.2.1 never happen. In fact, before that can happenwe stopped the process at the point 2.1. The reason for this connected with the

selection θ. Assume that αs exists. We have λ(αr+1) ≤ (1 + b12jr

)λ(αr). Therefore

λ(as) ≤s−1∏ι=0

(1 + b12jι

)λ(α0) ≤∞∏j=1

(1 + b12j )λ(α0) = Cλ(α0).

Take θ so that θC < 14. We have

n(αs)

|αs|≤ C

α0

|α0|.

Hence, if n(αs) 6= 0, then n(αs) ≥ 1, and we get

|αs| ≥1

C

|α0|n(α0)

≥ 1

C

|α0|θ2N

>4|I|2N

.

Therefore we arrive at n(αs) = 0 before |αs| ≤ 4|I|/2N . The Carleson integral Cαιf(x)satisfy the inequalities

|Cαι+1f(x)− Cαιf(x)| ≤ C2myb12kι−1, 0 ≤ ι ≤ s− 1.

50

Chapter 4

Final part of proof

4.1 Maximal operator of Fourier series

Let f : I → C be a measurable function such that |f | = χA. For every y > 0and 1 y ≤ (Cp2

p− 1)pm(A)

yp. (4.1)

All the considerations of this part can be applied to the operator C∗I , but we perfectto talk about the maximal operator of Fourier series

S∗f(x) = supn∈N|Sn(f, x)|.

We will prove S∗f satisfies the same restricted weak inequality.

Proposition 4.1. There exists an absolute constant C <∞ such that for every mea-surable function f : [−π, π] → R with |f | = χA, every 1 0 wehave

µx ∈ [−π, π] : S∗f(x) > y ≤ Cpp

µ(A)

yp, Cp ≤ C

p2

p− 1. (4.2)

Proof. Let f o : R → R be equal to 0 for |t| > 2π and equal to the periodic extensionof f on [−2π, 2π].

Recalling the expression of the Dirichlet kernel

Dn(t) = 2sinnt

t+ φn(t), |t| < π,

where φ is function such that ‖φ‖∞ uniformly bounded in n ∈ N. We obtain, for everyx ∈ [−π, π],

Sn(f, x) =1

π

π∫−π

f o(x− t)sinnt

tdt+

1

2π

π∫−π

f o(x− t)φn(t)dt.

52

Since sinnt/t is uniformly bounded in [π, 3π] and [−3π,−π], and ‖φn‖∞ is uniformlybounded in n ∈ N, we get

|Sn(f, x)| = 1

π

2π∫−2π

f o(t)sinn(x− t)

x− tdt+Bm(A), |x| < π.

Therefore if I denotes the interval [−2π, 2π], we obtain

S∗f(x) ≤ C∗I fo(x) +Bm(A).

By |f | = χA, and the construction of f o following f , we can see that f o = χA′ , whereA ⊂ A′ ⊂ [−2π, π] satisfying m(A′) ≤ 2m(A), for y > 2Bm(A), if S∗f(x) > y then

C∗I fo(x) +Bm(A) ≥ S∗f(x) > y >

y

2+Bm(A),

orC∗I f

o(x) ≥ y

2.

Hence, we have

x ∈ [−π, π] : S∗f(x) > y ⊂ x ∈ [−π, π] : C∗I f(x) >y

2.

Thereforex ∈ [−π, π] : S∗f(x) > y ⊂ x ∈ I/2 : C∗I f(x) >

y

2.

Using (4.1), we get

mx ∈ [−π, π] : S∗f(x) > y ≤ mx ∈ I/2 : C∗I f(x) >y

2 ≤ (2Cp)

p2m(A)

yp,

where Cp ≤ C p2

p−1.

Finally we arrive at

mS∗f(x) > y ≤ Dpp

m(A)

yp, Bp ≤ D

p2

p− 1. (4.3)

In other case, y ≤ 2Bm(A), we pick Dp = 4Bπ, then

Dpp

m(A)

yp≥ (4Bπ)p

m(A)

(2Bm(A))p=

(2π)p

m(A)p−1

≥ 2π ≥ mx ∈ [−π, π] : S∗f(x) > y.

Thus, in generality, we can choose D big enough satisfying (4.3). Finally we change tothe measure µ.

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4.2 The Carleson-Hunt theorem

Here we drive the principal result of Hunt. Observe that our notation for thenorm of Lp,1(µ) and Lp,∞ is not standard. We have put, for 1 < p <∞,

‖f‖p,1 =1

p

1∫0

t1pf ∗(t)

dt

t, ‖f‖p,∞ = sup

0<tt1pf ∗∗(t).

They are specially useful for our purpose here.

Theorem 4.1. For every p ∈ (1,∞), the operator S∗ maps Lp,1(µ)→ Lp,∞, and

‖S∗f‖p,∞ ≤ Cp3

(p− 1)2‖f‖p,1.

Proof. Using Proposition 4.1 give us, for every characteristic function

supy>0

y(µS∗χA > y)1p ≤ Cp2

p− 1‖χA‖p.

Proposition 1.9 gives us

sup0<t<1

t1pf ∗(t) = sup

y>0yµf (y)

1p .

Thus, for every characteristic function,

sup0<t<1

t1p (S∗χA)∗(t) = sup

y>0y(µS∗χA > y)

1p ≤ Cp2

p− 1‖χA‖p.

We have seen in (1.7) that

sup0<t<1

t1pf ∗∗(t) ≤ p

p− 1sup

0<t<1t1pf ∗(t).

Thus we have

‖S∗χA‖p,∞ ≤Cp3

(p− 1)2‖χ‖p.

Now, recall that Lp,1(µ) is an atomic space. Using Theorem 1.4, we obtain, each

function f ∈ Lp,1(µ) can be written as f =∞∑j=1

ajχAj , in such a way that

∞∑j=1

|aj|‖χAj‖p ≤ C‖f‖p,1.

Hence

S∗f ≤∞∑j=1

|aj|S∗(χAj).

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Therefore

‖S∗f‖p,∞ ≤∞∑j=1

|aj|‖S∗(χAj)‖p,∞ ≤ Cp3

p− 1

∞∑j=1

|aj|‖χAj‖p

≤ C ′p3

(p− 1)2‖f‖p,1.

Theorem 4.2. (Carleson-Hunt) For every 1 < p < ∞, the operator S∗ : Lp → Lpis continuous. More precisely, there is a constant C such that for every measurablefunction f

‖S∗f‖p ≤ Cp4

(p− 1)3‖f‖p.

Therefore for f ∈ Lp[−π, π]

limn→∞

Sn(f, x) = f(x), a.e. on [−π, π].

Proof. Applying the previous theorem we know that

‖S∗f‖q,∞ ≤ Cq3

(q − 1)2‖f‖q,1,

for every 1 < p <∞.

Thus we can apply Marcinkiewicz’s Theorem for every 1 < p1 < p < p0 < ∞with the operator S∗ : Lp(µ) → Lp(µ). Then we obtain that S∗ is continuous andbounded by

|S∗‖p ≤ Cp(p0 − p1)

(p0 − p1)(p− p1)(

p30

(p0 − p1)2)1−θ(

p31

(p1 − 1)2)θ, (4.4)

where

θ =p−1 − p−1

0

p−11 − p−1

0

.

Observe that if q′, q denote the conjugate exponents, that is

1

q+

1

q′= 1, or

q3

(q − 1)2= qq′2.

Thus if we take p0 and p1 conjugate exponents into (4.4), we obtain

‖S∗‖p ≤ Cp(p0 − p1)

(p0 − p)(p− p1)p1+θ

0 p2−θ1 .

We divide the Theorem in two case of the value of p:

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First case, 1 < p < 2 we choose p0 = p+1p−1

and p1 = p+12. We can calculate θ = 1+2p−p2

p(3−p) ,and get

‖S∗‖p ≤ C2θp(p+ 1)4(3− p)

4(1 + 2p− p2)(p− 1)2+θ≤ C ′

(p− 1)3.

Second case, 2 < p <∞ we choose p0 = 2p, p1 = 2p2p−1

, then θ = 12(p−1)

and we have

‖S∗‖p ≤ C2(p− 1)(2p)4

p(2p− 3)(2p− 1)2−θ ≤ C ′p.

Therefore for all value p we obtain

‖S∗‖p ≤ Cp4

(p− 1)3.

Now, we are going to deal conclusion about the pointwise convergence. Weknow that for a dense set of functions on Lp[π, π] we have pointwise convergence at allpoints.

We define an operator

Ωf(x) = lim supn|Sn(f, x)− f(x)|.

Easy to see that Ωf(x) ≤ S∗f(x)+|f(x)|. We use the property that Ω(φ) = Ω(−φ) = 0,for every periodic and differentiable function φ. Then we have

Ω(f) ≤ Ω(f − φ) + Ω(φ) = Ω(f − φ)− Ω(−φ) ≤ Ω(f).

Therefore Ω(f) = Ω(f − φ). It implies that

Ωf(x) = Ω(f − φ)(x) ≤ S∗(f − φ)(x) + |f(x)− φ(x)|.

Hence fore every periodic and differentiable function φ, and positive number α, we have

Ωf(x) > 2α ⊂ S∗(f − φ) > α ∪ |f(x)− φ(x)| > α.

Therefore

mΩf(x) > 2α ≤ mS∗(f − φ) > α+m|f(x)− φ(x)| > α

≤ 1

α(‖S∗(f − φ)‖p + ‖f − φ‖p)

≤ 1

α(C

p4

(p− 1)3+ 1)‖f − φ‖p.

It is easy to see that, if we fix p, then Ωf(x) > 2α has zero-measure. This give

limn→∞

Sn(f, x) = f(x), a.e. on [−π, π].

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Conclusion

We have presented the proof of Carleson Hunt theorem. If chapter 1 shows the ideafor the proof, then chapter 3 is the most difficult part of the proof, and it give us agood evaluation for Carleson integral. Especially, Theorem 3.1 is important result, ithelps us make the problem become easier to solve. By applying Theorem 3.1, I haverewritten in more detail Proposition 4.1 and Theorem 4.2. Those are the final resultsthat we need.

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Bibliography

[1] Loukas Grafakos, Classical Fourier Analysis, Graduate Texts in Mathematics 249 ,Springer, 2nd ed. 2008.

[2] Juan Arias de Reyna, Pointwise Convergence of Fourier Series, Lecture Notes inMathematics 1785 . Springer, 2002.

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