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BELLMAN FUNCTION AND BMO
By

Leonid Slavin

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY
Department of Mathematics
2004

ABSTRACT
BELLMAN FUNCTION AND BMO
By

Leonid Slavin

The Bellman function method is applied to three different problems in harmonic anal-
ysis. The first, introductory chapter outlines the specifics of the method, addresses its
stochastic control origins, and gives a harmonic analysis perspective. A brief descrip-
tion of the main function space under consideration, BM 0, is also provided. In the
second chapter, the integral form of the John-Nirenberg inequality for BM 0 func-
tions is examined, the corresponding Bellman function explicitly found, and the sharp
constants in the inequality as well as the exact bounds on the region of its validity
established. Two cases, those of the continuous and dyadic BM 0, are treated and
the results differ significantly between the cases. In the third chapter, the dyadic ver-
sion of the Chang-Wilson—Wollf theorem for functions whose s-function is uniformly
bounded is proved using a Bellman-type argument. Furthermore, a local version
of the theorem is established, whereby the s-function is assumed to be bounded
on a measurable subset E of [0,1]. Consequently, the exponential summability of
the second order over E is derived. In the fourth chapter, the famous question of
H 1 — BM 0 duality is considered. Two cases, the continuous and dyadic ones, are
treated and the same key lemma, based on a Bellman-type argument, is used in both
to establish the embedding of BM 0 in the corresponding dual space. Moreover, in

the dyadic case, an explicit estimate for the norm of the embedding is found.

To the memory of my father

iii

ACKNOWLEDGMENTS

I would like to express my deepest and sincerest gratitude to my advisor, Professor
Alexander Volberg. He has taught me all the skills of the trade I have had the ability
to grasp and he is responsible for my reaching this stage in my career. I am very
grateful for his patience and generosity with his time, ideas, and expertise. I am also
very happy to be working in this field and credit him solely for the opportunity to
be on the cutting edge of research. Whatever mathematical accomplishments I dare
count to my name would not have existed, were it not for his help and encouragement.

The time I have spent as his apprentice will forever be appreciated.

I would also like to thank my mentor and coauthor, Professor Vasily Vasyunin. He
has opened my eyes to many a subtle mathematical phenomenon in the process of
our joint work and taken time to educate me on not so subtle matters. His friendship,
truly infinite patience, and preparedness to give the youngling an Opportunity after

opportunity to establish himself have helped me immensely.

I am very thankful to Professor Michael Frazier for the many useful discussions I have
had the pleasure and advantage of having with him. He has helped me prepare for my
talks and his insight has given me a broader perspective on certain areas of research.

I am also very thankful for his help in preparation for my job interviews.

I would like to express my appreciation to Professors Alexander Volberg, Clifford
Weil, Vladimir Peller, Joel Shapiro, and Kirill Vaninsky for agreeing to be on my
committee, to have their patience tried with delays, and their kindness tested with

this imperfect text.

Last but not least, I would like to thank my best friends Leonid Freidovich and Zeynep

Altinsel for being there for me all these years.

iv

Contents

List of Figures vi
1 Preliminaries 1
1.1 Introduction ................................ 1
1.2 Stochastic control Bellman function ................... 4
1.3 Harmonic analysis Bellman function ................... 7
1.4 The space BMO ............................. 9
1.4.1 John-Nirenberg inequality .................... 9

1.4.2 Hl — BMO duality ....................... 11

1.5 The Bellman equation for the John-Nirenberg inequality ....... 12
1.6 Bellman-function-type proofs ...................... 13

2 Sharp constants and bounds in the J ohn-N irenberg inequality 16
2.1 Introduction ................................ 16
2.2 Main results ................................ 18
2.3 The continuous case ........................... 20
2.3.1 The key lemmas .......................... 20

2.3.2 How to find the Bellman function ................ 26

2.3.3 How to find the extremal function ................ 29

2.4 The dyadic case .............................. 32
2.4.1 Preliminary considerations .................... 32

2.4.2 Detailed considerations ......................
2.4.3 Stage 1 ...............................
2.4.4 Stage 2 ...............................

3 Bellman-function-type proof of a local Chang-Wilson—Wolfl' theorem

and related results
3. 1 Introduction ................................
3.2 Main results ................................

3.3 Bellman function considerations .....................

4 Bellman function and the H1 — BM 0 duality
4.1 Introduction ................................
4.2 The dyadic case ..............................
4.3 The continuous case ...........................
4.3.1 Multi-dimensional setting ....................
4.4 The proof of the key lemma .......................

4.5 A sample Bellman function ........................
Research Prospects

Bibliography

vi

37
40
52

54
54
56
61

64
64
65
70
74
75
77

78

81

List of Figures

2.1
2.2
2.3
2.4

4.1
4.2

The initial splitting: a- = 0+ = = .r’“.

[Olr—
I

The stopping time: [.L“,£] is tangent to the parabola I2 = If + 62.

The counterexample to the conjecture 53 = 1. .............

The worst case scenario: the largest. portion of [1", .z'+] is outside 95.

The region II, (6”). ...........................

The clecomposition Q; = U TQ1. .................
IEDJCJ

25
33
35

71

Chapter 1

Preliminaries

1. 1 Introduction

This work is a compilation of three results united by a common theme. In all three,
a relatively new harmonic analysis technique is used to establish the crucial segments
of the argument: the Bellman function method. With its origins in stochastic control,
its timeline in analysis both short and rich, and its scope seemingly unlimited, the
method is as novel as it is powerful. Every successful application, therefore, serves
to further the method’s legacy as well as explore and develop its subtle and intricate
aspects that only come to light in a particular problem. That is why applying the
Bellman function technique to problems that have been solved by other methods
merits a researcher’s time. Due to its nature, this technique often streamlines the
proofs and / or reveals new properties of the function spaces under consideration. What
is more, it has been applied to obtain sharp and dimensionless results in existing
estimates and inequalities, establish their regions of validity, and, sometimes, disprove

intriguing conjectures.

The other theme linking the results presented is, as the title suggests, the function

space for whose elements they are valid. It is BM 0, the space of functions of

bounded mean oscillation (see below for detailed description). The first and the last
results establish certain properties of B M 0 functions explicitly, whereas the second
one deals with a space that is better (smaller) than BM 0, and whose elements
thus possess properties that are similar in nature to those of BM 0 functions, but
stronger; that is to say it proves exponential integrability but of higher order than
that of BM 0.

Chapter 2 is on the John-Nirenberg inequality for BM 0, perhaps the most fun-
damental property of BM 0 functions. It establishes sharp constants and bounds
previously unknown. This is a joint result with Professor Vasily Vasyunin of St.
Petersburg Department of Steklov Mathematical Institute and Mathematics Depart-
ment of St. Petersburg State University. This work was inspired by that of Professor
Alexander Volberg of Michigan State University, the mathematician who is, perhaps,
the most responsible for establishing and promoting the Bellman function method
as a tool in harmonic analysis. It is his presentation of the stochastic control Bell-
man set-up for the John-Nirenberg inequality in [15] and an earlier version of [24]
that, at about the same time but at vastly distant venues, caught our attention and
spurred further research. We then independently found the Bellman function for the
problem, after which we joined the efforts to solve the problem completely. Many
useful discussions with A. Volberg have allowed us to use his insight and expertise.
Two formulations, the continuous and dyadic, are considered. Notably, the results

are significantly different in the two cases.

Chapter 3 is on a joint result with A. Volberg. It gives a Bellman-function—type
proof of a famous result of Chang, Wilson, and Wolff for functions whose square
function is bounded and uses that result to establish that the square of such a function
is exponentially summable. More precisely, for any cp from the unit ball in the
corresponding space (with the usual factorization over constant functions and the s-

. . 2 . . . .
function as the semi-norm), e0“? 18 integrable. Furthermore, we give a local versmn

of the theorem, whereas the s-function is assumed to be from L°°(E), as opposed to
L°°([0, 1]), for a measurable subset E of [0, 1]. The result is a weak-form estimate,
similar to that in the Chang, Wilson, Wolff’s original paper [2] and a uniform bound
on the integral f3 emf”! for some a > 0. The article [2] does not give the local
version, although it seems that the authors’ reasoning can be modified to make it
work for an arbitrary E. Our proof is short, straightforward, and constructive —
the usual advantages of the method used. The formulation of an extremal problem is
discussed, although the “Bellman function” used in the proof is a supersolution of the
optimization problem, i.e. a majorate of the “true” Bellman function. The Bellman
approach to this problem has very intriguing implications, as discussed in the section

on research prospects.

Chapter 4 is on a joint result with A.Volberg. It examines the old question of
H1 — BM 0 duality, first addressed by Fefferman in [5]. We use a Bellman-type
reasoning to establish a key lemma, which then works to prove two duality results:
the continuous one and its dyadic analog. No optimization problem is set up; thus
we use a hands-on, heuristic approach to find a “Bellman function,” whose properties
are dictated by the differential estimates in the lemma. As is often the case with this
technique, we are able to obtain an explicit embedding constant in the dyadic case,
although we do not discuss whether it is sharp. Moreover, the same proof seems to
work in a multidimensional setting and, with the types of H1 and BM 0 norms

used, the constants of embedding are dimensionless.

One common feature, the hallmark of a Bellman function or Bellman-function-type
proof, that is present in all three results is the unwrapping of a certain integral sum,
undoubtedly the main connection between stochastic control and harmonic analysis.

We expand on this connection in the next two sections.

1.2 Stochastic control Bellman function

We will formally derive the Bellman equation for a controlled stochastic process fol-

lowing the exposition in [15, 24, 10].

Let :r‘ be an d-dimensional stochastic process, satisfying the stochastic differential

equation

t t
33‘ = a: +/ o(a",:r’) dw’ +/ b(a’,:r’) ds. (1.1)
o 0

Here t is the time, w‘ is a d1 —dimensional Wiener process, o(a, y) is a d x d1
matrix, and b is a d-dimensional vector. Different choices of the control, a’, which
is a d-dimensional stochastic process, give us different trajectories, i.e. different
solutions of (1.1). The derivation we give will be entirely formal; thus we do not

address questions of existence or uniqueness of solutions.

Equation (1.1) is a part of an optimal control problem. Namely, given a profit function

f“, on the trajectory x‘, for the interval [t,t + At], the profit is

fat(a:‘)At + 0(At).
Therefore, on the whole trajectory we earn

/ fa‘ (17‘) dt.
0
We want to choose the control a = {as} to maximize the average profit
210(33): IE / fa‘(xt)dt+;1i‘ffia(r(xt)), (1.2)
0 —ioo

for the process starting at :5. Here F 2 0 is the bonus function — one gets it when

one retires. The Bellman function for the process (1.1) is the optimal average gain,

v(:r:) = sup v"(x), (1.3)

06.4

where A is the set of admissible controls; 1) satisfies the well-known Bellman (par-
tial) differential equation, which is based on two ingredients: Bellman’s principle and

Ito’s formula.

Bellman’s principle states that

v(-:c) —— suplE [A( f"‘3 x)ds + v(:r‘) . (1.4)

06A

To explain it, we fix t > 0 and consider an individual trajectory. The profit for the

[SM

Suppose the trajectory has reached the point y at the moment t. The maximal

interval [0, t] is given by

average profit we can make starting at t and at the point y is precisely v(y).
Indeed, since the increments of w‘ for s 2 t do not depend on wT,T < t, and
equation (1.1) is time-invariant, there is no difference between starting at time 0 or
at time t. Applying the full probability formula to take into account all possible

endpoints y = x‘, we obtain (1.4).

Let us now explain the version of Ito’s formula that we need. Fix a moment of time
s and a small increment As. We want to estimate the difference v(:r’+A") — v(3:’).
Let Aw’ = w’+A‘ -— w’. Assuming enough smoothness, we can use Taylor’s formula.

Among others, we will have the term

d1

d
okj(a’, x”) )Awf-l- +2; a—(f’ )bk( (0 ,x’)As.
_1 k:

HQJIQD

623;.

J:

After taking the expectation, the first term will vanish, since each Aw}: is indepen-

dent of :c’ and has zero mean. The second term can be rewritten as
IE (£?s(:c’)v) (x’)As,
with the first order differential operator .6? given by

“ a
1 : gbk(a, S) a.

The next term in the Taylor formula will be

 

2
'1': 6 v 2 UjkAwZ + M018, $3)AS Z UikAwi + b.-(a’, x")As .
2 k,j 611.1323]; 1: k

Averaging over probability, taking into account the-fact that lEszAwfn = As if

k = m and 0 otherwise, and omitting the terms with (As)2, we get
IE (£38 (x’)v) ($’)As,

where the second-order differential operator CS is given by

 

d . a2 1 “1
£3 = Z av<a,x> 33.3.; arm) = ,- ganammm).
i,j=l =1

Gathering all the terms together and omitting the ones with powers of As greater

than one, we obtain

lE('U(iL‘t)) = v(:1:) + EA £08053) v(:c’) d3, (1.5)

where L“ = Li" + £3. That is the application of Ito’s formula we need. Putting (1.5)

into Bellman’s principle (1.4), we get

0 = sup [1: fa’w) + [at castes) v(:1:") d3] .

06A

Dividing by t an taking the limit as t —» 0 (assuming it is justified), we get Bellman’s

partial differential equation supplemented by the obstacle condition 1} 2 F

:13 [£°‘(x)v(:r) + f°’($)] = 0, x E Q

v(:1:) 2 F(m), a: E 9.

1.3 Harmonic analysis Bellman function

The scope of applications of the Bellman function method has been exceptionally
broad. While the method seems well suited for proving weighted norm inequalities
(see the early 1995 version of paper [16], the ground-breaking proof of the matrix
Hunt-Muckenhoupt-Wheeden theorem in [13], or a more recent result in [20]; alter-
natively, see [12] for a two-weight negative result), it has found use in areas far from
its origins. The (much needed) anthology of the existing Bellman function results is
far beyond the scope of this chapter. An incomplete list of references, besides those
already named, includes [7, 14, 17, 18, 19] and, perhaps the closest in spirit to the

current work, [23].

This panoply of results seemed to necessitate the development of a uniform founda-
tion. Such a foundation has been found in the very field from which the notion of a
Bellman function first arose — that of stochastic control. The work on developing
the stochastic control framework for harmonic analysis problems was begun in [15]
and continued in [24]. It is in paper [15] that the famous result of Burkholder for
martingale transforms [3] was interpreted from the stochastic perspective. We refer

the reader to these articles for a thorough exposition and many interesting examples.

Here, we would like to indicate the general principles of the stochastic control frame
work. We will use them to develop the Bellman equation for the John-Nirenberg

inequality after we introduce the space BM 0 in the next section.

The problems that can be treated using the Bellman function from stochastic control
and thus the machinery of the previous section are always dyadic. It is often possible
to pass from a dyadic problem to the corresponding problem with analytic or harmonic
function using some kind of Green’s formula. (That is exactly what is done in Chapter
4 in the continuous case.) It is, therefore, very interesting and unexpected that in the
John-Nirenberg setting of Chapter 2, we, motivated by the stochastic control Bellman
function formally derived for the inequality, first solve the continuous problem and

only then use those results to return to the dyadic case.

First of all, we make a choice of variables so that the function space under con-
sideration is mapped onto a Euclidean domain. Namely, assume that we want to
prove a certain inequality for all functions from the 6-ball, F5, of a space F. Then,
with every pair (tp, J), where cp E F; and J is a dyadic interval, we associate a
d-dimensional vector :1: = (1151,15,. . . ,xd) in a domain 05 whose geometry is de-
termined by the space F. Very often, the coordinates x,- will have a martingale
structure; we often see 1:10p, J) = (cp)J. (We make a choice like this when dealing
with the John-Nirenberg inequality below.) The vector x then is the state vector of
our system, i.e. a solution of (1.1). Thus, given a function (,0 6 F5, we know the state
of the system on every dyadic level (for every generation of dyadic intervals), that is
to say, at every moment t. Naturally, the time is now discrete and is equivalent to

the order of the current generation. Instead of (1.1), we now have a discrete model

10”“ = :12” + 0(0", :13”)A"w + b(a”, :13"). (1.7)

Most of the time, we have a = diag{a1,a2, . . . ,ad}. The exact form of the matrix

o and the vector b is determined by the difference 23"“ — x”. The Wiener process
is very often just a series of coin tosses: we either move to the left or the right half

of the interval I. Thus the equation (1.7) often becomes
xvi-+1: 1'" + anén + b(an,$n),

where 6" is either 1 or —1. If we are maximizing a sum of the form

that sum can be interpreted as the expectation of the cumulative gain f0°° f as (:r’) ds,
allowing us to choose the profit function correctly. On the other hand, if we have a
term of the form (g(<p)), to maximize, and $1 = (cp), then the bonus (obstacle)

function is g(x1).

We have described the method very empirically but just enough to be able to for-
mulate the Bellman optimization problem for the John-Nirenberg inequality. First,
however, we need to introduce the space in question, BM 0, in order to understand

the geometry of the state domain (I.

1.4 The space BMO

1.4.1 J ohn-Nirenberg inequality

The space of functions of bounded mean oscillation, or BM 0, was introduced by
John and Nirenberg in [8] in their work on partial differential equations and quickly
began to play a very prominent role in harmonic analysis. An excellent reference on
the complex-variable approach to BM 0 is [6], whereas [21] describes this space in

the real-variable and multi-dimensional setting in great detail.

The original definition follows. Let I be an interval. Then

BMO(I) = {p E L1(I): sltépl—‘lI—l/Jlflt) — (¢)J| dt < 00}. (1.8)

Here (cp), = TIT f1 <p(t) dt, and J is a subinterval of I. If we factorize the space

( 1.8) over constant functions, it becomae a Banach space with the norm

1
ll<P||BMom = sup — / Mt) — «an dt. (19)
JcI |J| J

The most fundamental result for BM 0 is the weak-form John-Nirenberg theorem,
first proved in [8], that states that for every (,0 E BM 0(I ) one has

m ({23 € 1: MI) - <<p>1| > M) S cle—C2AIIMIBMO(I)- (1.10)

Finding the sharp constants in the inequality (1.10) is a natural goal. In an important

development, Korenovskii found the sharp constant 02 = 2 / e in [9].

A remarkable consequence of the John-Nirenberg inequality is that every BM 0 func-

tion (,0 is in LP(I),l£p<oo and

(supl—j—l [J W) — <90)le at)” (1.11)

.19

defines an equivalent norm on B M 0(1 ) We are particularly interested in using the

L2 -based norm. Let

1 . 1/2
llwIIBMoum = (sup— / |<p(t)—<so>1|2dt) . (1.12)
JcI IJI .1
We can — and this is the main reason for using p = 2 — rewrite (1.12) as
1/2
2
”WIIBMO(1),2 = (93211) “902% - Will) . (1-13)

10

Phrthermore, we have the corresponding weak-form John-Nirenberg inequality
m ({x e I : we) — «am > A» s ale-:2*/”r"BMO<I>i2. (1.14)

with constants c1 and C2 different from those in ( 1.10). One can rewrite (1.14) in
the integral form, as follows. There exists so > 0 such that for every s < so and

every 90 E BMO(I) 31101] that "‘pllBMO(I),2 _<. 5, we have
or). s C(aeW (1.15)

for some constant C (s) The inequality (1.15) is the reverse Jensen inequality. Find-
ing the sharp value for so and the sharp expression for C(s) is highly desirable.

Chapter 2 describes how it is done in both continuous and dyadic settings.

1.4.2 Hl—BMO duality

A major reason BM 0 gained prominence is paper [5] in which Fefferman established
it as dual to H1. This is the result we take up in Chapter 4. Both spaces are
considered on the unit circle T. Again, we treat two cases, the continuous and

dyadic. In the dyadic case, we use the following BM 0 norm

||<Pl| = sup i 2: (tot, — <w>1_)2lll. (1.16)

JeDlJl 19

where D is the dyadic lattice rooted in T and 1., 1+ are the left and right halves
of the dyadic arc I, correspondingly. Here, the advantage of the L2-formulation for

BM 0 is evident, since

fi 2 (M1, — <90)1_)2 III = 4 («02), — M3)-

11

(See Chapter 4 for a detailed explanation.) Thus we are using virtually the same

BMO norm as in (1.12).

In the continuous case, the equivalence of the norm (1.12) and

1 I 2 __
llcpll— sup m [01 moi (1 |€I)dA(€), (1.17)

are ICT

where 90(2) is the harmonic extension of (,0 into the disk and Q; is the Carleson

square based on the arc I, can be found, for instance, in [21].

1.5 The Bellman equation for the J ohn—N irenberg
inequality

To preserve history, we will develop the equation in the way we first encountered
it, although the attentive reader will notice that we use a slightly different choice
of variables in Chapter 2. According to the previous section, we have the follow-
ing underlying dyadic problem: Given that ((90 — (90) J)2)J S 62, for every dyadic

subinterval J of I , prove that
(e‘P), S C(6)e("°>1.

For every J, let $1 = (90)] , 3:2 = ((cp — (<p)J)2)J. The Cauchy inequality and the

assumption of the theorem give {25 = {($1,222) : m1 6 IR, 0 S 332 S 6}.

If we take conditional expectations IE( |a:") of (1.7), we get

11 — n +
b1(a.:r") = 1E(xi‘“lx?) — x? = “’1’ 3“”) -:c'; = o

12

and

n - In + xn — _ 1.11 + 2
b2(a,$")=lE($3+1|$3)—:r3= (£132) :( 2) —$3= (( 2) 2( 2) ) =(a?)2.

 

 

Furthermore, according to the reasoning of section 1.3, the profit function is 0 and

the obstacle function is F (SE) = e11. We, therefore, solve the optimization problem
1 8v

sup -2- (d221, v) — 5;?- a?] = 0

a=(al,02)

(1.18)

v(:r) Z e’1,x E 05.

This is the formulation that led us (independently) to the corresponding family of

solutions
___M e$1+v52—I2-5
1 — 6 '

125(27) =
As it turned out, this function was NOT the dyadic Bellman function for the domain
95. It was not until the crucial splitting tool (Lemma 4 in Chapter 2) was developed in
[23] and the realization that the continuous case was the one to consider ensued, that
the result started to develop further. The continuous Bellman function was first dis-
covered and after much effort the dyadic one was found in the same family, although,

somewhat bafllingly, not on the same level (to clarify these cryptic statements, the

reader is encouraged to refer to Chapter 2 for a complete presentation).

1.6 Bellman-function-type proofs

We now attempt to address the type of Bellman function argument in which no
explicit Bellman function is found. We can roughly consider two categories of proofs
of this sort. First, we may have an extremal problem posed and, instead of finding

the Bellman function itself, we find its majorate. Provided that is bounded, so is the

13

Bellman function and the result follows. The influential papers [15, 24] describe the
procedure for finding such majorates, provided the corresponding Bellman equation
has been obtained. Chapter 3 details a result of this kind. However, we do not
even attempt to obtain the Bellman equation there. The reason is that the most
important property of the formal stochastic-control—like development in the previous
sections was that the Bellman function, defined as the supremum (taken over all
elements of a function space) of an integral or a dyadic sum over an interval, would
not depend on the interval. For instance, in the John-Nirenberg setting of Chapter
2 (slightly different from that of the previous section), the Bellman function, defined
by

Beta) = sup {<e“’>1= (e): = $1, (902); = $2},

ll‘PllBMO(I)SE
does not depend on the interval I . This allows for effective modeling using the
stochastic control technique. In the case of the local Chang-Wilson-Wolff theorem,

given a measurable subset E of [0, l], the corresponding Bellman function

Bf(x) = sup {/ ems) ds : (90)“),1] = 331,561] = 1:2}
E

‘PEFQ

is very E-specific. This is, of course, due to the fact that E does not scale as well
as [0,1] itself does, i.e. dividing an interval in half, we will not necessarily divide
that interval’s portion of E in half. Of equal importance is the fact that the variable
S; defined in the chapter does not have a martingale character to it; thus it does not

scale correctly either.

A reasonable question then is whether one can obtain a more manageable Bellman
function set-up in the case E = [0,1] (the original Chang-Wilson—Wolff case), if,
instead of 5;, one uses the s-function itself as a variable. This remains subject of

further investigation.

The second category of Bellman-function-type arguments are the ones where the

14

extremal problem does not even enter the consideration. Chapter 4 deals with a result
of this kind, the H1 — B M 0 duality. Although it is possible to think of an underlying
optimization problem, the Bellman function involved appears simply as a means to
carry out a clever segment of the proof. It is a very utilitarian approach — one
attempts to unwrap the corresponding sum over dyadic intervals with a nonexistent
(as of yet) function B and imposes such differential and quantitative properties on
the function as are required so that the estimates work out the right way, the constants
are not too big, etc. After that, one attempts to find the function satisfying the
prOperties so determined and, perhaps, Optimize its parameters or tweak it otherwise.
We daresay this is the way many explicit Bellman functions as well as their majorates
are constructed. Admittedly, this is how the function in Chapter 3 was constructed
(The function in Chapter 2, however, stands out as a pure product of stochastic

control formalism.)

What, then, distinguishes a Bellman-type proof from any other? Precisely the un-
wrapping of an integral/dyadic sum, mentioned in the preceding paragraph. The
three key lemmas of the three chapters that follow this introduction, the reader will

notice, accomplish just that.

We are now in a position to present the main results of this thesis. It is our hope
that the reader will appreciate the unifying ideas behind the proofs and the flexibility
of the method in dealing with subtleties of each particular case. Some prospects for

future research are detailed in the last section.

15

Chapter 2

Sharp constants and bounds in the

J ohn—N irenb erg inequality

2. 1 Introduction

For any interval I C R and a function cp 6 DH), we denote by (<0) I the average
of cp over I, (90), = l—i|f190(t)dt' We define the space BMO(I) as

BMO(I) = {90 E L2(I) :/|<p(t)—(g0)1|2dts C2|J|,V interval J C I} (2.1)
J

with the best such C being the corresponding norm of cp. This definition can be

rewritten in a more useful form:
BMO(I) = {<0 6 L2(I) : (o2), — m3 g 0%.] c I} (2.2)
with the norm

2 1/2
”SOHBMOU) = (31?!) {(902), — (spiJi) - (2-3)

16

We also introduce BM 0“(I ), the dyadic analog of BM 0(I ), whose definition is
identical to that of BM 0(I ), except that the intervals J over which the supre-
mum is taken are members of the dyadic lattice based on I. Finally, by BM 05(1)
and BM 03(1 ) we denote the s-ball (the ball of radius s centered at 0) in the

corresponding space.
The following result is well known. (This is the integral form of the John-Nirenberg

theorem.)

Theorem. There exists so > 0 such that for every 0 S s < so there is C(s) > 0

such that for any function <0 6 BM 05(1 ),
(er), s C(s)e<‘p)1. (2.4)

We are interested in determining the sharp bound so and the exact expression for
C (s) We will do that in the case of both, the conventional (continuous) BM 0 and

the dyadic BM 0.

In accordance with the ideology of the Bellman function method, with every ball
BMO,-(I) (BM 03(1 )) and the set of all subintervals J C I we associate the

domain 05 = {x = (2:1,:r2) : :31 E Rm? 3 :52 g at? + 82} as follows

(so, J)*—* (<¢>,. ($51) (25)

This map is well-defined because (cp)3 S (902)] (Cauchy inequality) and «p E
BM 05(1 ) (BM 021(1 )) On 95, we define the following Bellman functions

B.(x)= sup {<er>,= (¢)1=w1.(902),=$2}, (2.6)
‘PEBMOsUI

17

Bg(.r) = sup {(e‘p), : (<0), = x1,((02)1 = 51:2}. (2.7)
wEBMO‘gU)

Observe that these functions do not depend on I . Finding them explicitly would
provide us with the complete solution of the John-Nirenberg problem. That is exactly
what we are able to accomplish. Remarkably and unlike many other Bellman function
problems, the results for the continuous and dyadic cases differ significantly. Which
is more, we first solve the continuous problem and only then use the corresponding

Bellman function family to solve the dyadic case.

2.2 Main results

Theorem 1. Let so = 1. For every 0 g s < so, let

 

C(s) = . (2.8)
Then, for any <0 6 BMOE(I),
(e‘p), S C(s)e(‘p)1. (2.9)

Moreover, so and C(s) are sharp.

Theorem 2. Let s3 = x/2—log 2. For every 0 S s < s3, let
01(5) = 0(6), (2.10)
where C(6) is defined by (2.8) and 6 = 6(5) is the unique solution of the equation

(1 — «52 — €2)e .2-.2 [2 — 8W5] — (1 — (net-W5 = 0. (2.11)

18

Then, for any <0 6 BMO,?(I),
(er), s C“(s)e("°>1. (2.12)

Moreover, s3 and C“(s) are sharp.

Theorems 1 and 2 are immediate consequences of the following results for the Bellman

functions (2.6) and (2.7). Let

_ 1/ 2 5-
85(13): 1 61 1.31:1 x2 exp (2:1 + V62 + :13? - x2 — 6). (2.13)

 

Theorem 3. If 0 S s < 1, then
BE(.v) = Bs(:r); (2.14)
if s 2 1, then

e‘cl if :02 = 113%

+00 ifxg > in?

Theorem 4. If 0 S s < x/210g 2, then

I323(3) = Ba(.)(:r); (2-15)

if s 2 filog 2, then

eI1 if 2:2 = at?

+00 ifxo > :10?

Indeed, since the function t H (1 — t)e‘ is decreasing for positive t, B assumes its

19

maximum when :32 = a3? + 52, i.e.

—€
e3E1 ,

e
<
B(x)__ 1-s

 

giving (2.9) and (2.12) with the sharp constant (2.8).

2.3 The continuous case

2.3.1 The key lemmas

We first consider the continuous case and prove Theorem 3. One can observe that
the proof does not work in the dyadic case. We split the proof of the identity (2.14)

into two parts.

Lemma 1. For every 1: E 05,

B5(x) 2 85(3).

Proof. We prove this inequality by explicitly finding a function (0 for every point
x E 9. such that (<¢)1.<<p2)1)= ($1,322) and

(aw)! = 3431,32)-

Since 1:2 = 2:2 occurs if and only if <0 = :01 = const, it is clear that Bo x =
1

Bo(a:) = e31. So we only need to consider 5 > 0.

Take I = [0,1], a 6 (0,1], b 6 IR, 7 E IR\{O}. Let

ylog%+b for0_<_t§a
‘pa.bn(t) “——
b foragtg 1.

20

Direct calculation shows that <00;w E BM 0|,|(I ), (<pa,bn)1 = 7a + b, ($312.7); =

272a + 2yab + b2, and

l—1+al -
(epa'bn)l = 1—7 If 7 < 1

oo if'yZl.

Since BE $1,202 = B,E 2:1,2‘2 = ell for all s, we only need to consider the points
1 1

 

a: E 95 with x2 > 32%. Then wecan set a: 1— fl2+x§—$2 and b=x1—7a,
which yields (moon), = x1,(<0§,bfl)1 = 432. Now, if we put 7 = s 2 1, we get

B£(:c) = 00. For '7 = s 6 (0,1), we get

 

 

1—\/s2+:c"1’—;r2
l—s

 

BE(.r) _>_ (yam)! = exp (271+ s2 + :13? — 1:2 — s) = BE(:r). Cl

Lemma 2. For every x 6 fig,
BE(:c) S Bs(:r) (2.16)

Proof. To establish (2.16), we first prove that B€(:r) _<_ 851(3), Vsl > s,Va: E 95,
and take the limit as sl —+ s. (Observe that BE is continuous in s from above.)

We need the following two results:

Lemma 3. The function 8.; is concave in $25, i.e.
Bs(oz_:r' + a+x+) 2 a_B€(:r’) + a+B€(;r+) (2.17)

for any straight-line segment with the endpoints 23* that lies entirely in DE and an
I!

pair of nonnegative numbers at such that a- + 0+ = 1.

Lemma 4. Fix s. Take any s1 > 5. Then for every interval I and every <0 6

BMOEU), there exists such a splitting I = I_ U 1+ that the whole straight—line

21

segment with the endpoints xi = ((<0)Ii,(<02)1i) is inside 95,. Moreover, the
splitting parameter 0+ = [1+] / II [ can be chosen uniformly {with respect to <0 and

I ) separated from 0 and 1.

Assuming these lemmas for the moment, take <0 6 BM 05(1 ) Take any 51 > s.
Observe that <0 6 BM 06(J ) for any subinterval J of I. Split I according to the
rule from Lemma 4. Let 10'0 = I, 1110 = I_, 11'1 = 1+. Now split I. and 1+
according to the rule of Lemma 4 and continue this splitting. By I "I'" we denote
the intervals of the n-th generation, as follows: I ”'2’“ = If-l’k and I "12““ = If”,
so the second index runs from O to 2" - 1. The corresponding points given by (2.5)
are indexed by the same pair of indices. Also, let awn = |I"I'"|/|I|. Since Lemma 4
provides for the value of (1+ uniformly separated from O and 1 on every step, we
have

max {|I”’k|} -—> 0 as n —> oo.
k=0,1,...,2"—1

Then, using Lemma 3 repeatedly, we have

II‘IOI II“!

 

 

 

0, , 1,
361(5): 0) Z EEO—[851(x10)+I-jO—,O-IBEI(SB 1)
[Il’ol [[2,0] 2,0 l__Il’0l l___’_121lB 2,1
2 0,0 1,0 Bel(x )+— 0,0 1,0 Bel“ )
II III | II III I
”I’ll [122' 2,2 [Il’ll [12’3" 2,3
+I1‘09II1—IIIIB‘1“: H IIIIII I1—1»IIB€1“‘ ’ (2'18)
[I2’Ol 2,0 [12,1] 2,1 [12'2l 2,2 _l___12’3lB 23
= l10,0|B€1($ )+ l10,0|B€1($ )+ [10,0lB€1($ )+— [[0,0] 851(55 )
2"—1 2"-—1 1
2 II'”"IB — —- /B”""’ds,
,,.——., Z; III ,

                 

where we have used the fact that 851(3) 2 erl and <0,, is the step function,

22

<0,,(s) = 2:?” for s E 1”“. Since <0” converges almost everywhere to 90, Fatou’s

Lemma yields

1 1 I
B51(:r) 2 —lim sup/e‘PMsM’ 2 — lim sup elm“) ds = — /e"’(3) ds = (6”)1.
III I [I] I III 1

Taking supremum over all <0 with ((0), = x?” = x1 and (<02), = x310 = :02, we

obtain the inequality
Bil (x) Z B€($)7

thus proving the lemma. E]

To finish the proof of Theorem 3, we need to prove Lemmas 3 and 4. In the section
“How to find the Bellman function” below, the function BE is explicitly constructed
to be concave in $25, which is what Lemma 3 states. Thus Theorem 3 is contingent

on Lemma 4.

Proof of Lemma 4. We fix an interval I and a function <0 6 BM 05(1 ) We now
explicitly construct an algorithm to find the splitting I = I. U 1., i.e. choose the

splitting parameters ai = |Ii|/|I|. As before, 501* = (<0) 2:; = (<02) 11' Also,

15:,

put at? = (<0) I and $3 = (<02),. Lastly, by [s,t] we will denote the straight-line
segment connecting two points 3 and t in the plane.

1

First, we take a- = (1+ = . If the whole segment [x‘,x+] is in 05,, we fix this

[0|

splitting. Assuming it is not the case, there exists a point a: on this segment with
2:2 — 2:? > sf. Observe that only one of the segments [:r",:1:°] and [23+, :00] contains
such points. Call the corresponding endpoint ( :13“ or 23+ ) 6.

Its position is completely defined by the choice of 0+. Define the function 0 by:
p(a+) = maxxelmoflxo —- 10%}. By assumption, 0 G) > sf. We will now change 01+

0

so that 6 approaches a: , i.e. we will increase a+ if .5 = x+ and deCrease it if

E = :r‘. We stop when p(a+) = sf and fix that splitting. It remains to check that

23

 

 

 

+

Figure 2.1: The initial splitting: a- = (1+ = , 5 = 2: .

NIH

such a moment occurs at all and that the corresponding 0+ is separated from 0 and
1. Without loss of generality, assume that 5 = 22". Let I = [a, b]. Since <0 6 L2(I),

the functions 5101+) = 31: fbb—lllm <0(w)dw and {2(a+) = a—1+- fbb-llla+ <02(w)dw are

continuous on the interval (0,1] and 5 (1) = 23°

. Therefore, 0 is continuous on
(0,1]. Since 0(%) > sf and 0(1) 3 52 < sf (recall, 22° E (25 ), we conclude that

there is a point m, E [%, 1] with 0(a+) = sf.

Having just proved that the desired point exists, we need to check that the cor-

responding a+ is not too close to 0 or 1. If 5 = 2”“, we have 0+ > and

l
2
{1 — 2:? = 2:? — 2:? = a.(2:1+ — 20f). Analogously, if 6 = 23‘, we have a- > % and
£1— 2:? = 2:1- — 23(1): a+(:rl_ - 231+). Thus [£1 — $9] = min{ai}]2:1‘ — 25?].

For the stopping value of 0+, the straight line through the points 27,2:+ and 2:0

is tangent to the parabola 2:2 = 20? + sf at some point y. The equation of this line

24

 

 

 

Figure 2.2: The stopping time: [2:‘, 5] is tangent to the parabola 232 = 2:? + 62.

is, therefore, 22 = 2231311 — y? + sf. The line intersects the graph of 222 = 22% + s2 at

2:(s)i =(y1 :1: Vs? — s2,y2 :l: 2y1(/s¥ — s2)

and the graph of :02 = 2:? at the points

the points

917(0)’c = (w i 61.112 i 21/161)-

We then have

lw(€)‘.x(€)+l C [300.6] C [$193+] C lx(0)‘,$(0)+l

25

and, therefore,

2 6i - 52 = |£I(€)f - $(€)f| S I23? - {II = min{ai}|$l - III

S min{ai}|x(0)'f — 23(0)l"| = min{ai}2sl,

which implies

As promised, this estimate does not depend on <0 or I. E]

2.3.2 How to find the Bellman function

We first observe that the Bellman function B must be of the form
BE(2:) = exp {2:1 + w€(2:2 — 20%)} (2.19)

for some positive function w on [0, 52] such that w€(0) = 0.

Indeed, fix an interval I. Then <0 6 BMO,;(I) if and only if <0 + c E BMO,-(I),
where c is an arbitrary constant. Let <0 = <0 + c. We have (all averages are over I )

(<0) = (<0) + c, (<02) = (902) + 2c (<0) + (:2, and (e‘f’) = e‘ (e‘P) . Then

sup {(.¢).(,.)=.,,(,.2)=.,}-.c sup {<er>=<¢>=x.,<cp2)=x.}
063M050) I0€BMOE(I)

OI‘

sup {(e‘f’): ((0) = 21+ c, (<02) = 202 + 202:1 + c2}
¢€BM05(I)

= ‘36 SUP {(5‘)}: (if?) = $1, ($2) = 332}
IpEBMOgU)

26

OI'

B€(2:1 + c, 232 + 2c271 + c2) = e°B€(2:1,272).

Setting c = —2:1, and omitting the index s we get

B(0,:r2 — 2:?) = e'11B(2:1,2:2).

By the Jensen inequality ( (e‘P) 2 em ), we get that B(0, 2:2 — 2:?) Z 1. Hence, there
exists a positive function w = log B(0, ) defined on the interval [0,s2] such that
(2.19) holds. Furthermore, 2:2 = 2:1 = 0 if and only if (,0 = 0. Thus B(0,0) = 1
and w(0) = 0.

To use the machinery of Lemma 2, we need the Bellman function candidate B to be

a concave function. We thus want

 

 

623
—82:-82:- (220)
I J
to be a nonnegative matrix.
Using (2.19), we get
32:3; = (1— 22:1w')B,
BB
_ = 'B
6172 w ,
2B
6;? = ((1 — 2271211')2 — 2w' + 42:?w”) B,
1
2B
a: 62: = (w'(1— 22:1w') - 22:1w") B,
1 2
B2B ,
8—2122— : ((11),)2 + w') B.
2

27

The matrix (2.20) turns into

028 (328

_ $2 033;? =-23 1 ”‘ 11 1 U I22u
[C ., _ . l) I —2.l'1 1
02:10.1'3 0.12:;
where
1 — 2111’ ll
1? = (2.22)
u" (u ')2 + 11'"

For the extremal function (if any) we must have equality at every step in (2.18) from
Lemma 2. so the matrix (2.20) has to he degenerate. Because of the representation

(2.21) and (2.22). this translates into

(1 — 2111') ((u")2 + III") 2 (II/)2. (2.23)

[0

u." — 1 Z 0. ( .

[Q
to
.L.
V

we solve equation (2.23) by

 

 

ll

1‘ + (,‘(Ilisf

_ (1— _) = i + no.1
211"

This iIIIplies that the mustaut has to he urinpnsitive. \Ve parametrize the futility (If

n)\‘v

—R

possible solutions by a positive parameter 6 setting const = —62. Then we continue

and

1— 1 =i\/62—t. (2.25)

210’

 

Condition (2.24) requires that the square root be strictly less than 1. Therefore, the
only feasible solutions for w are those for 6 < 1 and such a solution is defined on

the interval [0,62]. Equation (2.25) gives

,_ 1
w — 2(1— «62' ‘— t)

 

and, taking into account that w(O) = 0, we obtain

1 ‘ 1 1— 62—t
:— — : __ 2- _
w(t) 2./01_\/67__Sds log 1—6 +V6 6,

which, together with (2.19), gives formula (2.13)

_ 2 2—
35(1)) = 1 \/61 +3171 $2 exp (131 + V62 +113? — I122 -‘ 6) .

2.3.3 How to find the extremal function

 

 

We now show how to find the extremal function that appeared without any explana-
tion in the proof of Lemma 1. As mentioned in the previous section, for the extremal
function there is equality at every step in the chain of inequalities (2.18). Thus in the

splitting process we only proceed along the vector field defined by the kernel vectors

29

of the matrix (2.20). Using (2.13), we obtain the quadratic form of that matrix

 

(2.26)

 

AH
A

 

 

2
2:1+\/62+2:? —:ro) A1 — %A2)
= exp{x1+(/52+2:?—2:2—6}.
\/62+2:?—:02(1—6)

Hence, the trajectories along which B is a linear function are given by

1
(2:1 + V62 + 20? — 2:2) (12:1 = -2-d:rg. (2.27)

Introducing the variable t = V62 + x? — 2:2, we have t2 = 62 + 2:? — 2:2 and 2t dt =
22:1 d2:1 — d2:2. Replacing %d$2 in (2.27) by 2:1 d2:1 - tdt, we get td2:1 = —t dt, i.e.
t = c — $1 and

32:62+$?—t2=26$1+62—C2.

The corresponding trajectories are the family of the straight lines tangent to the upper
bound 2:2 = 2:? + 62 of 95 at the point 2: = (c,c2 + 62). Consider the following

family of straight-line segments
025(0) = {2: = (2:1,2c2:1+62 —c2) : c—6 _<_ 2:1 S c}.
It covers the whole domain, i.e.

95 = U 0J5(C).

061R
Furthermore, B is a linear function on each segment wo(c). Indeed, since

V62 + m? - 2:2 = [2:1— c] on the line 2:2 = 2c2:1+ 62 —- 02, we have

1+$1—

Bo(x1,2c2:1+ 62 — c2) = 1 6 Get"5 for c — 6 S :01 S c.

30

Therefore, if the points 2* are on such a segment, we have equality in (2.17) (with

6:5)
Note that we have one more “acceptable trajectory,” the envelope of the segments
wo(c), the parabola 202 = :r? + 62.

We now write what it means to be on one of our trajectories in terms of the function

0 0

<0. First, consider the case when the point 2: is on the boundary, i.e. 2:1 is arbitrary
and 2:? = (2:?)2 -i- 62. Let I = [0,1]. Split it at point a : I- = [0, a]; 1+ = [a, 1].
We can choose which point is to the right of 23°. Assume it is 2.". We try to place

it on the trajectory 2:2 = 2:? + 62. Then for all a 6 [0,1] we have

62 = x,- — (22,—)2 = if: <02(t) dt — G [on <0(t) (2)2. (2.28)

Introducing the function w(a) = % f0“ <0(t) dt, we have <,0 = 610’ and, after multipli-

cation by a, (2.28) turns into

A“ WU? dt: a+§102(a)
2 a a ' a
¢’(a)2 = 21% (a+ 22/1291)) =1_ 1I[’()+ 2‘“ It“)

(Ii/(a) — M)” = 1

a

0(a) - 31—292 = i1

<—)

w(a) = :l:a (log; + const) .

 

D l

Finally, we have (since we seek to maximize e‘t", we choose the plus sign)

<0(t) = 6 (log% — 1) + 20?, (2.29)

31

where the constant has been chosen so that 2:? is the mean value of <0.

Now, let 2:0 be an arbitrary point inside {25. Then we make the splitting so that 2:"
is on the boundary 2:2 = 2:? + 62 and the segment w(xl') passes through the point

20°. Every point on that segment satisfies the equation

2.2 = 2:1:fo + 62 — (ml—)2,

 

so 2:1“ = 2:? + \/62 + (2:?)2 — 2:3. We choose the second endpoint 23+ to be the point
of intersection of (05(231’) and the lower boundary of {25, 2:2 = 2:?. This is equivalent
to letting <0 be constant on 1+. Then 2:; = (2:?)2 = 2:01"ch +6?--(2:1‘)2 and, hence,

2:? = 2:1“ — 6. Then for the splitting parameter a = a- we have the equation
20? = (1-2:; + n+1? = axf +(1— a)2:fL = 2:?r + a6,

giving us
0 + — 0

_ _ _ 1 0 0
a— 5 —1——6——l-3\/62+(2:1)2—2:2.

 

 

Since 2:2; = (2:?)2, <0 is constant on 1., <0|1+ = 23?, and for the interval I.
we use the procedure outlined above, since the point 2:“ is on the top boundary
curve. We only need to renormalize the function from (2.29) to this interval: <Plr_ =

6 (log?- - l) + 2:17. .This yields the function we used to prove Lemma 1.

2.4 The dyadic case

2.4.1 Preliminary considerations

We now explicitly construct BS, proving Theorem 4 and thus Theorem 2. It is
apparent that the proofs given above for the continuous case do not go through in

the dyadic case. In particular, Lemma 4 becomes irrelevant, since in the dyadic case

32

one cannot choose the splitting of an interval because it is always split in half, i.e., in
the language of Lemma 3 and Lemma 4, a- = a. = in This seemingly makes the
proof of Lemma 2 impossible to use. We, however, are still able to use the chain of

inequalities in Lemma 2 to establish the results in the dyadic case.

A fundamental question that comes to mind first is whether the dyadic Bellman
function is different from the continuous one and if so, whether the size of the BM 0“
ball in which the John-Nirenberg inequality is valid is the same in both cases (namely,
if it is still the unit ball). The following (counter)example answers the former question

in the positive and the latter, in the negative.

Fix s > 0. Let the function <0 be defined on I = (0,1] by

90l(2-(k+l)’2—k] = (k — 1)a, k = 0,1,. . . , (2.30)

with the constant a to be determined later. We have the following picture for <0

5a“
4a“
3a“
2a,_

‘7

 

Q

v—t '0

cal“

(Jolt--W

.c-II-I

NIH
H r

 

 

Figure 2.3: The counterexample to the conjecture s3 = 1.

We now calculate the BM 0" norm of <0 and choose a so that ||<0||BMOd = s. The

only dyadic intervals on which <0 is not constant and, hence, (<0?) - (<0)2 75 0 are

33

the ones with 0 as their left endpoint. Let 1,. = (0, 2‘"] . Then
”2“ 0° ha a 1 "-2
= 2" d = 2" _ = -2" _ =
(80)!“ [0 ‘P(3) 3 k§12k+2 4 (2) n an

and

V2" 0° We2 a2 l "—2
($02)“ = 2"] 902(3) ds = 2" 2: 5,23 = 42“ (2) (n2 + 2) = a2(n2 + 2),

0 k=n—-l

where we have used the identities
oo 1 k 1 N—2 oo 1 k 1 N—2
2 .(-) = t) I Z .2 <-) = (I (N2 + I
k=N-1 2 2 k=N—1 2 2

Then

2 _ 2 _ 2
llfpllgMod _' styldgcl {<fp >J (99>J}

= sup {(<02),n — (<0)?n} = sup {a2(n2 + 2) - a2n2} = 2a2.

Setting ||<0||BMOd = s, we get a = s/\/2. Now,

 

<0 —°° eke—m1 eak
<e>1_ 2k+2—ZZ§ '

The latter sum converges if and only if e“ < 2, i.e. a < log 2. In terms of 53 from

Theorem 2, we obtain the crucial estimate
53 S x/210g 2. (2.31)

Informally, we have just shown that the space BM 0‘1 is substantially worse than
BM 0. However, having developed the Bellman function family (2.13) for the con-

tinuous BM 0, we would like to look for the dyadic Bellman function within that

34

family. From the example (2.30) above, it is clear that, should the dyadic Bellman

function be in fact found in that family, we would have to have
B: = 36(5) (2.32)

for some 6(5) > 5. One straightforward approach would be to choose 6(5) large
enough so that any straight-line segment [1", x+] with 23-, 17+ 6 05 would fit entirely
inside 95(5). Then we would be able to use the chain of inequalities in Lemma 2
without the help of Lemma 4. Let us investigate how large the 6(5) so chosen would

be with regard to 5.

Consider the situation when the segment [x‘, 3*] “sticks out” the most. This hap-
pens when the middle point x0 = 11,-(15’ + 3:“) as well as one of the endpoints, say
23+, are on the top boundary, m2 = 1:? + 52, and the other endpoint, 2:", is on the

bottom boundary, x2 = mg.

2:2 2 x? + 62(5)

$2=x§+52

 

 

Figure 2.4: The worst case scenario: the largest portion of [1:", 1+] is outside (25.

We have

:1:(2’=(:r(1))2 + 52, 2:; =(:171+)2 + 52, 11:3 = ($92 + 52,

35

and

0 xf+srf 0 LUZ-+33;
:cl=—, x2=———.
2 2

Eliminating :r‘ and parameterizing everything by $9, we get

0
xf=$1+

72'“
The equation of the line segment [2:0, x+] is then

332: (T;_—2-+2:r(1’) x1—::_-2—x1"—(a:(1))2+52, $93$13x?+%

and the distance between this segment and the top boundary curve is

E 5

5
dCL‘ —-$2+( +2x°):c——:r°— $02, x°<$ <xO+—,
(1) 1 fl 1 1 fil (1) 1 1 1 fl

with its maximum d (at? + 555) = £83. Therefore, any segment [x‘,x+] with

x‘,a:+ E 95 will lie inside 973-? Thus, for any 8 < 243g we can run the machine
2 2

of Lemma 2 to establish that

d
B72435) 2 BE(:1:), Va: 6 95. (2.33)

2

This gives us the following estimates:

2 2
—\3/- S 53' S \/2_log2 (2-34)

and, provided the dyadic Bellman function can indeed be found inside the family Ba,

5(6) (2.35)

3
< — 5.
“ N?
In what follows, we are able to prove that the example (2.30) does indeed produce

36

the sharp constant and that the dyadic Bellman function is a member of the family

8.5 . Clearly, we need to employ more subtle reasoning than the one above.

2.4.2 Detailed considerations

The following simple lemma shows that the dyadic Bellman function is concave, some-

thing that could not be shown directly in the continuous case.

Lemma. For any three points 17', 15+,a: E Q; such that :c = %(:r’ + 3:”) we have

Bite) 2 ngm + $3.1m. (236)

Proof. Take a sequence {cpn} E BMOSU-) U BMO§(I+) such that

(6”), —* Bflxi) as 71—» 00.

:l:

We need to check that cpn e BMO,?(I). But BMO,?(I) = BMO§(I_)UBMO;’(I+)U
{cp : ($2), — (m)? S 52} . Since, by assumption, :1: E {25, we have ($2),— (cp)? S 52.

Then we can pass to the limit in the identity

1 1
(e‘Pfl)! = E (e‘pn)[_ + 5 (e‘Pn)I+
to get
. 1 _ 1
Base) 211m<ern>1 = glaze )+ 513%),
which completes the proof. Cl

Observe that in the continuous case BM05(I) # BMO,;(L) U BMOE(I+) U
{cp : (902), - (90)? S 52} , since there are other intervals to consider, those with the

left endpoint in I- and the right one, in 1+.

37

We have just proved that B? is concave in $25. Phrthermore, the reasoning of (2.19)

still works and we conclude that
Bg(a:) = exp {1:1 + 10({132 — 1%)} (2.37)

for a nonnegative function U) such that w(O) = 0. What is more, we expect the
corresponding matrix —-d2Bg (assuming sufficient smoothness) to be degenerate, in
order for the supremum to be attained for an extremal function. But we have already
described all functions with these properties. They are the functions B; from (2.13)
with 6 Z 5. This, somewhat heuristic, argument supports our believe that the dyadic

Bellman function is a member of that family.

In our desire to use Lemma 2, we have been trying to ensure that the segment [:r‘ , 33+]
lies inside the domain of concavity of a certain function B, so that we can conclude

that

1 1
and proceed with the unwrapping of the integral sum (2.18). Now, we try to enforce

the condition (2.38) directly instead. Since we are searching for 6(6) such that

B: = 85(6), we attempt to solve the extremal problem

(5(5) = min {62 85(3)) 2 185017) + l35(174'),
s<6<min{fi55,l} 2 2
(2.39)

 

- +
Vx',:r+ 6 525 such that :r = :1: :3 E (25},

where we have used (2.35). Recalling the definition (2.13) of B5, we write the

38

“concavity” condition as

_FT
1 (ii-:1 x2exp<x1+(/62+x§—x2—6)

 

 

 

 

Z

11—\/62+(:r1')2—:r2_ _ _ _
5 1_5 exp 171 + fi2+(:r1)2 -552 ‘6 (2'40)

 

 

11-\/62+(x'{)2-x:;
2 1—5

 

exp (sf + \/62 + (at?)2 -a:'2* — 6).

Let

 

l
a=(/62+:rf—:1:2, at = \/(52+ (milky—xi 6: 5(131“ —:rf').

Using this notation and multiplying (2.40) by 2(1 - (De-’1”, we can rewrite it as
2(1— a)e°' 2 (1 — a_)eo+°" + (1 — a+)e‘o+“+.

A straightforward calculation shows that 03 + a: = 20:2 + 202. The condition

x,:1:_,:1:+ E (25 can be rewritten as a,a_,a+ 6 [V62 — 52,6]. Finally, let
f(a,a_,a+,6) = 2(1- a)ea — (1 - a_)eg+°- -(1— a+)e'0+°‘+ (2.41)

We will solve the extremal problem (2.39) as the following two-stage problem:

- 3
For O<5S6<m1n{m5,l}, let

35,5 = {(a,b, c, d) E R4: a,b,c 6 [V62 -— 52,6]; b2 + c2 = 2a2 + 2d2} (2 42)
9(6i5) = min {f(a)a—1a+30) Z (a7a-7a‘l'30) 6 56,6}:
6(5) = min{6 : g(6,e) 2 0}. (2.43)

While it is not a priori obvious that the set {6 : g(6,€) Z 0} is not empty, in

39

 

what follows we show that the problem (2.42), (2.43) has a solution (unique, by our

formulation).

2.4.3 Stage 1

We will often refer to finding the solution of (2.42) as optimization over the cube
[Wt-7:7, 6]3, even though 35,5 is in 4-space. We will use Lagrange multipliers in
the interior and on the faces of the cube and straightforward one-variable optimization
on the edges. We make extensive use of the inherent symmetry of the problem. In
addition, since we are only interested in the possible negative values of f, we will

disregard any and all other possible minimums.

Interior of the cube

Let

H(a, a_, 01+, 6’, A) = 2(1—a)e°‘—(1—a_)eg+°‘" —(1—a+)e—‘9+‘"+ —A(a2_+ai—2a2—202).

From VH = 0 we get

—2ae" + 4a). = 0
01-6“"- — 2(1_/\ = 0
a+e‘9+°‘+ — 2a+/\ = 0

—(1 — a_)e9+°'— +(1— 01+)e“9+"+ + 49A = O

a2 +ai—2a2—262=0.

Since no a can be zero, we get

e“ 2 6"“ = e0“+ = 2A, so a- +a+ = 2a,

40

and so at any possible point of minimum in the interior
f = 2(1- a)e°‘ — (1 — a_)e" — (1 — a+)e°‘ = e“[—201 + a- + 01+] = 0.

Faces

There are six faces. We make the following observations:

— if a = 6, then a- = (1+ = 6 and 0 = 0. Hence, the intersection of the domain

35,5 and the interior of this face is empty;

— because of the symmetry in a_,a+ (0 can be S O or 2 0 ), it suffices to
consider only the faces 0+ = 6 and 01+ = V62 — 52. The extremum points (if
any) will have their “twins” on the faces a_ = 6 and a- = V62 - 52.

Thus the only faces we need to consider are 01 = V62 — 52, (1+ = 6, and (1+ =

V62 — 62.

1. Face (1 = V62 — 52
Let

h1(a”a+’67A) = H( V 62 - €2ia—va+76>/\) = 2(1- “Fifi-236 V 62—52

—(1 — a_)e9+°‘- — (1 — a+)e‘9+a+ — Ma": + a: — 2(62 — 52) — 292).

From Vhl = O we get

a_e6+°‘- — 2a_A = 0
a+e'9+°‘+ — 204A = O
——(1 — a-)e9+°- + (1 — a+)e-0+°‘+ + 40)\ = 0

(13+ a1 = 2(62 — 52) + 262.

Therefore, e9+°- = e“9+"+ = 2). and (1+ = a_ + 26. Plugging this into the last

41

equation, we get
(a- +6)2 = 62—52, a- = —6+\/62 —e2, a+'= 0+\/62 —e'~’,

where the plus sign is chosen in the square root to ensure that each a is positive.

Lastly, we obtain at any possible extremum point,

h1 = 2(1— «62 — €2)eV62“€2 — 2(1— «62 — e2)e 52-52 = 0.

2. Face (1+ = 6

Let

h2(a,a-,0,x\) = H(a,a_,6,0,)\)
= 2(1— a)e°‘ — (1 — a_)e9+°‘- — (1 — 6)e“9+‘S — Mari + 62 — 2a2 — 262).

From th = 0 we get

6“ = e9+°'- = 2A
—(1 — a_)eg+°" +(1— 6)e"“HHS + 40A = 0

a2_ + 62 = 2012 + 262,

from which we get a = 0 + a_, (1 -— (De—0+6 = (1 — a- — 26)e9+°‘-, and 62 =
(a- +26)2. If a- = -—20+6, then we get (1 — 6)e" = (1 +6)e‘6, which only has the
trivial solution 6 = 0. So a- = 20 + 6, a = —0 + 6 and, at any possible extremum
point,

h2 = 2(1+ 0 — 6)e_9+5 — (1+ 29 — 6)e‘9+6 — (1 — 6)e_9+6 = 0.

42

3. Face 0+ = V62 — 52

Let

h3(a,a_,6, A) = H(a,a__, V62 _ 82,0,A) = 2(1_ (1)60 _(1_ a_)ea+a_
‘ (1 - V52 - 826“” ‘52"2 - Mai + (62 — 52) — 2a2 — 292).

We observe that this case is identical with the previous one if V62 — e2 is substituted

for 6. Therefore, at any possible extremum point, h3 = O.
Edges

We have a total of 12 edges:

(1)a=\/62———€7,a+—5
(2)0: 62—52, 01+: 62—52
(3)a=\/67-—52,a_=6
(4)0: 62-52,a_= 62—52
(5)a=6,a+—-6
(6)a=6,a+=\/67_—?
(7)a=6,a_=6
(8)a=6,a_=m
(9)a+=6, (1-26
(10)a+=6,a_= 62—52
(11)a+= 62—52,a_=6

)

Edges (5) through (8) have a = 6, which implies a- = (1+ = 6, t9 = O. The
pairs (1)-(3), (2)-(4), and (10)—(11) are symmetric. On edge (12), using the fact that
ag+afi =2a2+202, we get a2 =52—82—92, giving 0:0, 0:0- 201+ and

f = 0. This leaves us with the following four (renumbered) edges to consider (out of

43

the symmetric pairs we chose the edges sharing a vertex):

(1) a+=6, a_=6

(2) a+=\/'6—2-——e2, a_=6

(3) a=\/6"’——8_2, 01.26

(4) a=m, a+=m

1. Edge a+ = 6, a- = 6
Since a“: + 0?, = 20:2 + 202, we have a = V62 — 62, —e S 0 S e. The function to

minimize is, therefore,
17(0) = 2(1 — J52 — gays/62*? — (1 — 6)e6(ee + e-O).

We only need to show that F (00) 2 0 at any possible extremum point 90 E (-—e, e).

We will, however, show more; namely, that
F(6) Z 0, V0 6 {—6, 6], V6 E [0,1].

Let u = 6 + 0, v =6 — 0. Then, slightly abusing notation, we have

 

F(u,v)=2(1—\/uv)e‘/fi— (1_u~2l-v) (eu-l-e”), u+v=26, 0Su,vS26.

We will accomplish our goal if we show that F (u, v) 2 0, 0 S u S 2,

OSvS2—u.

 

44

Every point (11,12) of minimum inside the region in the picture would have to satisfy

the equations

1
Fuz—vem+§(e“+e”)— 1_u-;-v e“=0

1
Fv=—ue‘/fi+-2-(e“+e”)- 1_u-2l-v e”=0.

 

 

Adding the two equations, we get

 

e + e = em. (2.44)
Subtracting the second equation from the first and using (2.44), we get
e”(1 — v) = e“(1 — 11.),

which implies (since u, v 2 O ) that u = v. The value of F on any possible extremum
point in the interior is then F (u, u) = 0.

On the boundary 1) = 0, 0 S u S 2, we have F(u,0) = 2 — (1 — g) (e“ +1). The
equation [F(u,0)]' = 0 gives e"(1 — u) = 1. Since 11 Z 0, we conclude that u = 0.
Then F(0, 0) = 0. The case u = O is identical.

If u + v = 2, we have F(u,v) = F(u,2 — u) = 2(1 — \/u(2 — u))eV“(2'“). Since
0 S u(2 — u) = l — (u - 1)2 S 1, and the function 3 H e’(1 - s) is nonnegative for
s 6 [0,1], we conclude that F 2 0 on this piece of the boundary. This completes

the consideration of this edge.

2. Edge a+ = V62 — 52, a- = 6

Since oz?’_+oz_2+ = 2024-202, we have a: 62———l92,—E SOS %. The

45

function to minimize is, therefore,

 

F(6)=2(1— \/62——2——02)e eV 62-27-92 ——(1 6).: “9— —1( —\/62——?)em-9.

Assume that F has a local minimum for some 60 E (-—}v, 3‘5) . Then F’ (60) = O,

2

i.e.

62 5—2 —92

290a ’ 2 ° -— (1 — 6)e"+"0 + (1 — 62 — e2) eV62'52’90 = 0.

Expressing (1 -— V62 — 52) 6" 62—52490 from (2.45), we get

(2.45)

F(60) = 2.990 [(1- (M62 — é — 03 — 60)) eV62—%_93_90 — (1 — (5)125] . (2.46)

Observe that (1 - :1:)e3c - (1 — y)ey 2 0 if y 2 0 and —y S a: S y. Together with

(2.46), this consideration implies that if

 

2
—6g\/62—52——ag—aoga,

then F (60) Z 0. We now solve the inequality (2.47).

First, since |60| S €/\/2 S 6, we have that

 

((52-5493 —6> —,6V06[—-£- 5—]
2 O 0 fi,\/§ '

Secondly, solving the (right-hand) inequality (2.47), we obtain that

 

52 2
62—3—90— —60S6

if and only if

_ 2__2 __ 2_2
602 6+\/26 e or 003 6 6 e.

 

 

46

(2.47)

_ _ 2- 2 . . . . .
We must check whether —% S $42. This inequality is equivalent to
V62—52S \/2€—6.

Since €> >2—‘Q6 62 76, we can square the last inequality, getting as a result

2f

5>—6

again, which is true by the formulation of our extremal problem (2.42)—(2.43). Finally,
we conclude that (2.47) holds if and only if

_ _ 2_ 2
$003 6 5 5. (2.48)

_ 2_ 2
6+\/6 e $903-$—
2 J2

 

 

We now try and fill the gap in the condition (2.48). Expressing (1 — 6)e‘5+90 from
(2.45), we get

‘/ __2_ 2
F(90)=2€-9° [(1- (y/52-E23—03+00))e 62 90”“

 

(2.49)
— (1 - V62 — 52) eV52‘52].
Similarly to (2.47), if
2
—x/62—52S (/62—52——63+90SV62—52, (2.50)

then F (60) _>_ 0. Carefully solving this inequality, we obtain that (2.50) holds if and

only if

 

 

—\/62-€2 —6+\/62—€2
$903 2 -

2 (2.51)

47

To summarize, the fact that 60 E (—§5, %) implies either (2.48) or (2.51), which, in
turn, imply (2.47 ) or (2.50), respectively. Either one of the latter inequalities implies

that F(00) Z O. This completes the consideration of this edge.

3. Edge a=\/62—52, a_=6

 

Since of“: + 01 = 2012 + 262, we have 02+ = \/62 — 282 + 262, i S |0| S E. The

function to minimize is, therefore,

 

15(0) = 2 (1 — «62 — e2) eV62-52—(1—5)e<‘+9—(1 — J52 — 252 + 292) eV52-2€2+292-9.

We seek the absolute minimum of F. First, we observe that we only need to consider

 

6 < 0. Indeed, since 0 3 J62 - 262 + 202 g 6, we have

 

(1 — 6):? g (1 — J62 — 252 + 262) #5245244"?

So, if 6 Z 0, then

 

(1 _ (5)66” + (1 - \/62 — 252 + 202) eV 62-2€2+202—9

 

g (1 _ (5)65—9 + (1 _ \/52 _ 252 + 292) e\/62—2£2+202+0.

 

 

i.e. F(0) Z F(—6). Therefore,

min F(6)= min F(6).

e e
753955 —559$—75

We now look for the minimum of F over the interval —5 S 0 g —755. We have

 

F’(6) = e9 [(1— (x/(s2 — 252 + 262 - 20)) eV52-2€2+292-29 — (1 — 6).? .

48

We claim that F’(6) 5 0, — 5 S 6 S —752-. Indeed, since 6 5 755 and 5 2 2496,

 

we have 26 S —\/25 g —§6. Thus 6+26 S O and J62 — 252 + 262 - 26 2 6. Since

the function .9 H (1 — s)e’ is decreasing for s 2 0, we conclude that F’(0) S 0.

Finally, since there are no extremum points in the interior of the interval, and F’ (6) S

0, we conclude that

5:
min F6: min F6=F --—,
isms: ( ) asses-75; ( ) ( V2)

i.e. the minimum of f over this edge is attained at a vertex.

4. Edge a 2 V62 — 52, 0+ = V62 — 52

Since 013 + (11: 2(12 + 262, we have a_ = J62 — e2 + 262, ——§§ S |6| 5 i. The

 

function to minimize is, therefore,

 

F(6)— - 2(1— V62 — 52)ev522-€ _(1_ V52 _ 52 + 292),r3 \/62-—e2+262+6
— (1— x/6—2-—52) 6 62‘52‘9.

As in the previous case, we only need to consider 6 < 0. Indeed, assume that 6 Z 0.

Since

 

(1— \/62 — 52 + 262) e 62‘€2+292 S (1 _ 52 _ 52) em

2

we have

 

(1 — V62 — e2 + 202) eV52-€2+292+9 + (1 _ m) 8 5242—9
S (1— V62 - 52 + 262) eV 62“2+292‘9 + (1 _ 52 _ 52) e\/62—-e2+6.

 

Therefore, as above F(6) 2 F (-—6) and we have

49

Assume now that there exists a local extremum 60 in this interval. Then

F’ (60) = 0. This gives

 

(1— 007:2) 8(62-62—90 = (1.. (fl? _ 52 + 203 — 200) 3(62-62+298+90

and

F(60) = 2800 [(1 __ W) e‘/¢§2_E2
__ (1— (\/62 _ 52 + 20(2) + 00)) BW+90] .

 

 

We know that if W52 — 52 + 203 + 60 2 V62 - 52, then F(60) Z 0. Solving this
inequality we obtain

)0.) 2 2052 — 52.

What to do if |6o| < 2 62 - E2? Assume that is the case. Here we have to consider

the behavior of the derivative. We have

 

F'(6) = 8-9 [(1 — M) 6/63:2 — (1 — («52 — e2 + 202 + 20)) eV62-E2+2"2+29] .

Recall that (1— :1:)eJr — (1— y)ey Z 0 if y 2 0 and —y S :1: S y. Thus if

 

—\/6"’—52 < \/62—52+262+26< V62—52,

then F’ (6) < O. The left-hand inequality translates into 0 < |6| < 2V62 — 52, while

the right-hand one, into |6| > 0.

To summarize: we have F’(6) < 0 if —2\/62 —e2 < 6 < 0, thus there are no
points of local extremum on this subinterval; if there is an extremum point 60 S
—2\/62 — 52, then F(60) 2 O. Altogether, the two “interesting” points where the

absolute minimum may be attained are 6 = 0 and 6 = f5, both corresponding to

50

some vertices of the cube, to be considered in the next subsection. This completes

the consideration of the last edge.

Vertices

We formally have a total of eight vertices, but some are non-existent and only two

give us nontrivial results.

1. As mentioned before, the set 35,5 intersects the face a = 6 at one point only,

the vertex 0 = a- = 01+ = 6 where we have f = 0.
2. The vertex a = a- 2 0+ = V62 — 52 also gives 6 = O and f = 0.

3. The vertex (1 = V62 — 52, a- = 01... = 6 was considered above as a part of the

edge a_ = (1+ = 6 on which we have f 2 0.

4. The vertex 0 = a- = \/62———€2, (2+ = 6 gives 6 = i%, Then
fl9=755 = 2(1— m) e 62-52-(1— W) 6W+s/¢2_(1_5)es_em
and
f'6=— 2 = 2 (1 - W) e ”-62—(1 — W) .m—e/«a_(,_6,e.../a

Since (1- V62 — 52) (2"‘52‘£2 2 (1 — 6)e‘5, we have

5. The vertex (1 = (1+ = V62 — 52, a- = 6 is symmetric to the previous one and

gives the same result.

We have thus solved the first part of the extremal problem (2.42)—(2.43) in the sense

that the only possible nontrivial (meaning negative) minimum the function f can

51

have on 35,5 is given by

9(05) = (1 — W) «9/35752 (2 — (BE/‘5) — (1 — 5)e6-W~". (2.52)

2.4.4 Stage 2

We are now in a position to solve the minimization problem (2.43)

6(5) 2 min {6 E (e,min {7:28, 1}) : g(6,e) Z 0}. (2.53)

Differentiating g with respect to 6, we get

96(5,€) : 6 [eé—s/fi _ e\/52_s2 (2 _ Bah/2)] .

We observe that 95 > 0. Indeed, checking if 6 — i > V62 — 52, we obtain that this

condition is equivalent to 6 < '273/"5 6, hence it is satisfied. Then

9505,13) Z 56 V 62—52 [—1 + (as/‘5] > 0.

Therefore, if the equation g(6, a) = 0 has a solution, then it is unique and solves our

extremal problem. We thus look for a solution of the equation
g(6,€) = 0 (2.54)

in the interval [5, min {237 5, 1}) . We have two cases

22
<_\£

1.
E 3

We seek the solution 6 E [5, 5—312 5) . We have

g(€,€) .—.. 2 — 6W? — (1 - gee-6N5 < 0

52

and

3E 35 5
—,5 = 1__ 2.66% _ (1___)]em >0,
g(zfi) l< 2fi)( ) 2f
which implies that there is solution 6 of (2.54) inside the interval [5, 2.3/‘2 5) .

Ni

2. —<5 <1

We seek the solution 6 6 [5,1). As above, g(5,5) < 0. On the other hand,
90,5) = (1— v1— 52) 5V 1‘52 [2 —eE/‘/§].

If 5 < s/210g 2, we have g(1,5) > O and, hence, there is a unique solution of

(2.54) in the interval [249, \/210g 2) .

Putting the two cases together, we see that, provided 5 E (0, x/2 log 2) there is
a unique solution 6(5) of equation (2.54), which also solves our extremal problem
(2.53). On the other hand, in light of example (2.30) this is the best we can hope
for, since the example implies that 33(5) = 00 if 5 2 \/210g 2. Therefore, we have
proved that

53 = \/2 log 2,

as Theorem 2 asserts. Cl

Some of the prospects and future directions of research on this topic are discussed
at the end. We are now turning to the result that deals with a property not unlike
the one this chapter has been devoted to. The space considered there, the Chang-
Wilson-Wolff space, is better than BM 0. Thus the result is stronger: instead of
summability of the exponent 5‘6 we get summability of 5°82. We deviate from the
formalism of the Bellman function method, so the proofs are Bellman-function—type

proofs.

53

Chapter 3

Bellman-function—type proof of a
local Chang-Wilson—Wolff theorem

and related results

3. 1 Introduction

Let D be the dyadic lattice on [0,1]. For each I E D and every :1: 6 [0,1], define
the dyadic cone I‘ [(513) to be

F1(:1:)={JED: JQI,J3:I:}. (3.1)

Let (p E L1([0, 1]) For every I E D, let (cp), = fif1<p(s)ds. Also, let I- and 1+

be the left and right halves of I, respectively. Let

1/2
810(x)=( Z ((<P>1+—<90)1_)2) . (3.2)

IeI‘lO,” (1:)

54

be the s-function of 1p. Let E be a measurable subset of [0, 1]. We introduce the
Chang-Wilson-Wolff space

F(E) = {90 5 L1 = 11506111005) < oo}. (32»)

Then ||S¢p(a:)||Lw(E) defines a semi-norm in this space. (In the case E = [0, 1], it is

actually a norm, with the usual factorization over constant functions).

Fix (,0 E F(E). For every I E D such that m (If) E) 96 0, let

2
SI = Z (<‘PlJ+ " (WlL) (3-4)
J6F1(:z:) L°°(E)
and 51 = 0 if m(I DE) = 0. Obviously, 5; depends on 1p but we will not
indicate this dependence when the context is unambiguous. We prove that if SIM] =

||S<p(:r)||i.,o(E) S 1, then there exist absolute constants a > 0 and C > 0 such that
2
/ e“(~°-<r>10,11) s c. (3.5)
E

The integrability of 50"“2 over the disk D under the assumption f0 |Vu(:c)|2da: S 1
has been studied in [22] and [11]. In [1], the authors study a question like ours
but for functions analytic in ID). Namely, they answer in the affirmative the ques-
tion of Beurling and Moser as to whether the fact that In | f’ (z)|2 dz S 7r implies
fr exp I f (15“9)|2 d6 S C for some absolute constant 0. Considering for an instant 11‘
instead of [0,1], we note that even in the case E = T (3.5) holds under weaker
conditions, since if 1p 6 L2('ll‘) and <p(z) is its harmonic extension into the disk,
f0 |V<,o(z)|2 dz S C implies ||S<p||Loo(T) S C with a different constant (see, for in-

stance, [4]).

Our result follows immediately from a local version of the famous Chang-Wilson-Wolff

55

theorem, which we prove first using a Bellman-type argument. In [2], the authors treat
the case E = [0,1], while we prove the theorem for arbitrary E. It has to be said
that the authors’ reasoning would seem to work for arbitrary E as well. However,

the Bellman-function—type proof we give incorporates any E effortlessly.

3.2 Main results

We split the proof into several lemmas. Lemma 1 sets up the stage for the unwrapping
of the typical Bellman-function integral sum and Lemma 2 uses it to prove a certain

integral estimate, not unlike those in Chapter 1.

Lemma 1. Assume there exists a C2 Junction B = B(:c, L) : R X R+ -—> 11%, such

that 98% > 0 and there exists 6 6 (0,1) such that

flab)

132*— < 4(1— 0) (3.6)
63 — ,
811 (C7 d)

for any (a,b), (c,d) E IR x IR+ such that b —- d S --662 and [a — cl S %6 for some
6 Z 0. Then, for every cp E F(E) and every I E D such that m(I n E) 79 0,

1 1

B «was» 2 53 (<<p>,_ .31-) + §B(<<0>1+ .81..) . (3.7)

Proof. Assume the existence of such a function B. By definition (3.4),

S, = maX{SI_1SI+}+ (0P);+ — <¢)1_)2-

56

Let L0 = maX{S[_,S]+} , L = S] = Led-62, L_ = S]__, L+ = 31+, CE = ((p)l, :1;- =

($0)1_ , 33+ =<‘P)1+, 5 = [33+ - x’l. Then

B (<10, .51) - ,l-B (<0)._ .51-) — $30.01, .31..)
= B(z,L) — é— (B(:r+,L+) + B(:c",L"))

1
2 301.52) — 5 (3501.). 30,5.»
since %-L > 0. We continue

= [B(:z:, Lo + 62) - 3(117, Lo + (952)]
+ [303,110 + 952) - 3(‘7’3 L0)]

+ B(:1:, L0) — % (B(:1:+, Lo) + B(:1:_, 110”]

BB 1 823

>— —(:z:, fi)62(1 -6)— 4 83: ——2-

2
6L —(77 L0)6

for some 6 6 [Lo + 662, L0 + 62] and some 7 between 12‘ and :19“.

3(71L0)
m0) [ 2 0’

since 7 S [as — x'l = la: — z+| = %6 and L0 — 6 S —662.

168
=10L“ 0)[4<1—6>—

58:13?

Q)

This completes the proof. E]

The choice of B

We introduce the family of functions B parametrized by t 2 0. Let
B.(a:, L) = etI‘LAtzL (3.8)
for some A 2 0. We find the best suitable A below. For now, we check that the

57

functions Bt satisfy the conditions of Lemma 1. Fix any 6 6 (0,1). Take a, b, c,d

as in the formulation of Lemma 1. Then

828, (a b) 2 2
3x2 1 t exp(at + At b) 1 2

= = — - t At b—d
92% d, At2€xp(ct+At2d) Aexp<<a C) + < ))
01. ’

 

 

 

3 exp (,0 — 2000) = fiexp (§ ((60 - 2249602))

Zexp (a— (23—210)) = 4(1— 0),

where we have used the fact that y— 2A6y2 S fig and the last equality is guaranteed

|/\

by a suitable choice of A. In Lemma 3, we use Lemma 1 with the function B defined

by (3.8) to prove the estimate

m ({‘P — <¢)[0,1] > A» S 8—62/(2Asm'11).

This suggests that we need to choose 6 so that A is the smallest possible. Let

D = 4%,; Then the equation relating 6 and A becomes

DeD/(49) = 1 — 0, (3.9)

which defines D as a function of 6 in the interval (0,1). We are seeking the

maximum of D. Differentiating (3.9) with respect to 6, we get

D D2
D/(49) ’ + _ _ _ + —

Setting D’ = 0 and solving for D gives

462
1—6'

 

D = (3.10)

58

Solving (3.9) and (3.10) simultaneously yields

2+) D W (3...)

1
4
:1+2 i)’1+2w(

where w is Lambert’s w-function, i.e. w(z) is the solution of the equation we‘” = 2:.

Therefore, for the best A we obtain
————)- z 0.5291. (3.12)

We are now in a position to prove the following lemma.

Lemma 2. For every (p 6 F(E) and every t 2 O,
/ et(‘P(3)—(‘P)[0,1])ds S eAt2S[O,1]. (313)
E

Proof. Let B; be given by (3.8) with A given by (3.12). We will apply Lemma 1
repeatedly, at every step omitting the terms corresponding to dyadic intervals whose

intersection with E has zero measure, i.e at every step we use the simple fact that

Z IIJIB<<¢>J.SJ) 2 2 (JIB(((o>.,,sJ),
”'5’" 45:53:...

which allows us to apply Lemma 1 to every term in the latter sum. Thus we have

et<¢l[o,1]+At25[o,ll = B (((p)[0 1] ,S[o,1])
1 1
-B ((<p)[0,1/2] ,S[o,1/2]) + 53 («PM/2,1] 1511/2”)

> 2 (1130.0 >,.SI)2 Z IIIB((<.o>,.S,)

(1| =12- |1|=2--n
m(IUE);£0 m(IUE)9éO

59

= 2 [Il et(1p)]+At2SI.

|1|=2~n
m(1uE)¢o

Now let n -+ 00 and observe that, since SW] < 00, we have 5'; —) 0 as [I] —> 0.

On the other hand,

2 |I|et<f>1 —*/et""(’)ds asn-—>oo,
E

III=2-"
m(IUE);éO

which completes the proof. Cl

Lemma 3. Let (,0 E F(E). Let E,\ = {x E E: w(m) — ((p)[0,1] > A}. Let S =
510,1]. Then

m(E1) s erg/(“5), (3.14)
where A is given by (3.12).
Proof. We apply Chebyshev’s inequality to (3.13) with t = A/ (AS).

m(E,\) 6A2/(AS) < / eA(SP(8)—(‘P)[0,1])/(A3)ds S €A2/(2AS),
E

concluding the proof. C]
We are now in a position to prove the main result.

Theorem. If ||(,0||p(3)S1, then
2
/ $946100!) 3 C(01) (3.15)
E

for any a < —1-, with A given by (3.12.
2A

60

Proof. Lemma 3 implies that

m (“9" ‘ (6)1011!

- 2 1
F1x any 7' > 1 such that ar < 57. Let

 

2 2.)) g 2562““).

r" r 1
Fk={z<l<P—<<P>(o,1][37}1 k=0,2,--- (3-16)
Take any a < fl. Then, using (3.16),

2
/ e“(¢"(“°>[0,ll) S m(Fk)ear2k+2/A2 S 2e—1~2k/(2A3)ec1rr'2k'l'2/A2 = 2e—r2k(1/(2A)—orr2)/A2.
Fl:

Therefore,

[60(‘p_(¢)[0,1])2 :/ (-<p (90)“) 1]) 2+Z/kea(¢— (‘P>[01])2
E {Ir—(610.1151 }

1 °‘ 00 2k
3 m ({l“? ‘ (01ml S 2}) a” + 226’ (“WM-MW,
k=0

where the last sum converges and depends only on 0 (provided the best choice of r

has been made). This completes the proof. E]

3.3 Bellman function considerations

Lemma 1 and Lemma 2 are where the Bellman function technique is used, the rest
of the proof follows using standard arguments. One can observe that no extremal
problem as such has been posed. However, in the spirit of Chapter 1, we can try and
associate with every dyadic interval I such that m(I 0 E) aé O and every function

(p E F(E) a point in a certain two-dimensional domain. First, we let Fa be the

61

“oz-ball” in F(E), Fa = {cp E F(E) : ||Scp(a:)||Lm(E) g a}. What is the domain in
this case? This depends on the choice of the variables. Lemma 1 suggests the choice
($1,152) = ((90),,31). Then the domain associated with F0 is {20 = IR x [0,0]. We

can define the Bellman function

Bf(:l:) = sup {/ ems) ds : (99)“),1] = $1,561] = 1:2} . (3.17)
E

sOEFa

If we let cfi = 90 + c, then (95) 2 (4p) + c and 397’ = .9". Furthermore,
/ em”) ds = etc / ems) ds.
E E
Taking the supremum in the last identity, we get
B?(:c1 + c, 232) = 6“ Bf"(:r1,:rg).
If we set c = —:c1, we obtain that
B?($1,$2) = €trlft($2) (3-18)

for some ft 2 0. This is one of the considerations that led us to choose the family

(3.8) of Bellman function majorates. We observe that the inequality (3.7)

New-)+éB<<¢>I+,s:+>

NIH

B(<SO>I’SI) 2.

implies some sort of concave behavior about B, although the functions (3.8) are

828

decidedly not concave and (3.7) holds because of the subtle interaction between 5—13-

and g—E. If one were to produce a concave function U of the form (3.18), which

would also be increasing with respect to the second argument, one could conclude

62

%U(<<P>1 :51-) + -;-U ((99)1+ :SI+) S U((90)1,%(SI_ + SI+)) S U(<99>1’SI)’

where the last inequality is due to the definition of SI and the fact that % Z 0.
However, a concave function of the form (3.18) does not exist, signaling the need for

more delicate considerations.

The approach used to obtain the results of this chapter may benefit from being put on
a more formal Bellman function method footing. However, as the preceding discussion
demonstrates, the questions of formulating the extremal problem exactly, the choice
of the variables, and, of course, finding the Bellman function explicitly can be very

complicated in this case. Some intriguing perspectives are discussed at the end.

We now turn to a result, in which the Bellman function does not appear in con-
nection with any extremal problem at all, although, doubtless, the corresponding
formalism may be developed. This result brings us back to the space BM 0, the
main space of interest in this work. The old and famous question of H1 — BM 0 du-
ality is examined using the Bellman-function-type approach, which proves powerful;
one Bellman-function lemma yields concise proofs in both, the dyadic and continuous

settings.

63

Chapter 4

Bellman function and the

H1 — BMO duality

4. 1 Introduction

In this chapter, we aim to demonstrate technique rather than new results. Specif-
ically, with the aid of the Bellman function method we prove one, the more tech-
nically involved, direction of the famous Fefferman duality theorem by elementary
means. Namely, we establish the fact that BM00('lI‘) C H1('Il‘)‘ (BM00('II‘) =
{cp E BMO(T),<p(0) = 0}, and as usual, <p(z) is the harmonic continuation of cp
into 1D ). The most complex technical tool we use is Green’s formula; thus the proof
serves to show further the power of the Bellman function method in harmonic anal-
ysis. An application of the same method in the dyadic case serves to establish that
BM 0“ C (1:192): (with an explicit estimate for the constant of embedding), with
the Triebel-Lizorkin space 13592 giving a convenient characterization for H 3(11‘), the
dyadic version of H1('ll‘). A simple argument demonstrates the converse inclusion.
One technical “trick” allows us to deal with both, the dyadic and continuous cases.

The key to the proof is a lemma whose hypotheses include the existence of a certain

64

function. We will call it the Bellman function slightly abusing the language, since we

make no claim as to its uniqueness.

The formulation of the key lemma. Let D 2 D10 be the dyadic lattice rooted
in an interval Io. For an interval I E D, let I- and 1+ be its left and right halves,
respectively. Consider two functions, 5' : D -—> (0,00) and M : D —) [0,1171], such

that
1
S[_ = 31+ 2 S] and M1 2 §(M1_ + M1+),VI E D. (4.1)

Lemma. Let S and M be as above. Assume there exists a 02-function B :

(0,00) x [0, M] —+ R satisfying

- 2
BB BB M 9—? < 0, @- 2 0. (4.2)

o_<_B(x,y)s2M\/i 7537?? 8:132— 6y?

Then, for any positive integer n,

 

Z |J|\/(SJ+ - SJ) (MJ - g(MJ_ + MJ+)) S g 2'" Z \/S—J- (4-3)

JED JED
lJIZ2""+1 IJI=2-"

We will prove the lemma and demonstrate our Bellman function later. For now, we

will establish the main results.

4.2 The dyadic case

d
Consider the dyadic lattice D = DT on T. Let F? be the dyadic Triebel-Lizorkin

space,

1/2
£92: feLl; /1~( Z ((f),+—(f),_)2) d6<oo (4.4)

139;I€D

65

with the norm

2 1/2
”qu [.- ( Z (<f>1+-(f)1_)) do.

196:160

The definition of the dyadic BM 0 is the same (up to a constant multiple) as the

one we used in Chapter 2:

BMOdz cpELI: sup—
JED'JI

12((cp)l+ )_)2 |1| < 00} (4.5)

IcJ

with the norm

1/2
ll‘PHBMOd: 5111;0le ((‘PM— ‘P>1_2) III) -

ICJ

To see the equivalence of the definitions (2.3) and (4.5), recall the Haar system: for

every dyadic arc J, let

1
—— on J.
(M
h] = —— on J+ . (4.6)
VI
0 elsewhere

It is easy to check that {h J} K. D form an orthonormal system in L2 = {f E L2 (T) :
f1. f (0) d0 = 0}, what is more (and well-known) is the fact that the Haar system
actually is a basis for Lg. For any function f E L1 and every J E D one can

compute the corresponding Haar coefficient,

(M )= [fl (<f>._ — m...) (4.7)

For f 6 L3 we then have f = ZJeD(f,hJ)hJ and

66

my = Zuni)? = Z 1‘—,f1(<f>._ — mg)?

JED JED

We state our main result.
(I :-
Theorem 1. BMOd 2 (F92) .
Proof. The more difficult inclusion is handled using the Bellman-function lemma

stated above.

(I t
Lemma 1. BM 0d C (F92) . More precisely, in terms of the Haar coefi'icients, for

d
every cpEBMOd and fEF‘fz,

2 law 1w.» = $2 HI I m.+ — <f>._ Hat. — (99)J_| (4.8)

JED JeD
1
— MHWHBMOd “fIIFC‘i92
d
Proof. Fix cp E BMOd, f 6 F3”. For every J E D define
ICJ
def

Then 0 < M J < M—
Define

”scum... and M.-§(MJ++MJ_) = ((99)., - (¢)J_)2-
s. = 2 (ml, — <f>,_)2.
19.1

2 2

Then SJ.=SJ_=Z (m1, — <f>,_) and SJ.—SJ = 81:51 = (<f>.;<f>._) .
131

We thus see that the conditions (4.1) of the lemma are satisfied.

67

Assuming the existence of the function B in the lemma and using (4.3), we obtain

 

2: MI I m... — <f>._ ll «0)., — (m I s 92‘- 2-"2 Z (0)., — <f>.,_)2.
JED

IJI22-n+1 |J|=2‘"
Letting n —» 00, we get the statement (4.8) of the theorem. C]
The converse inclusion follows along more conventional lines.
0' 2 "' d
Lemma 2. (F?) C BMO .

- d
Proof. We want to show that for every continuous linear functional 1 on F92 there

exists cp E BM 0" such that

”‘pllBMOd 5 CW“ (49)

and
1m = [T 90(0)f(6) do, w e 5‘32. (4.10)

d
First, we observe that L3 6 F92. Indeed, for f 6 L3,
1

|If||:402= (fr ( Z ‘((f>1+-<f)z_)2)§d9) s f.- 2 (<f>1+-<f>z_)2d9

139;IED 139;IED
=1; /r we) (<f>1+ - <f>,_) «16 = 2m (at. — <f>,_) = 4||f|lig-

16D

d *
Let I 6 (F92) . We can apply the Riesz representation theorem to 1ng and

conclude that there exists a function (,0 6 L3 such that

W) = [T 90(9)f(9)d0. w 6 L3. (4.11)

68

We test I on appropriate elements of L3 to see that «,0 E BM 0". Let a; be an
atom associated with a dyadic arc I , i.e. be supported on I with |a1| 3 fi’ ac.

and f1 a(6) d6 = 0. We have

2 1/2
llazlllga,2 =/1~( Z ((cu).1+ - (a1)J_) ) d0

J39;JED
1/2

/<<>> .

S (flJaaZJc1(<a1)J+ " (GI)J_)2d9) 1/2 (flldfi) 1/2

2
= 2||azlng III s WWII = 2,

and hence

 

 

 

T

Since this is true for any atom a], we conclude that f, lap— (99), I g 2||l|| III and thus

[0.0- («pmaI

= ”(MI 3 Illllllazll 3 2H1”-
I F

1

that Ip E BM 0“ with the norm estimate (4.9). Here we have used the equivalence of
the L1- and L2-based BM 0 norms, which is due to the John-Nirenberg inequality
(see section 1.4.1). The proof of Lemma 2 (and hence Theorem 1) thus depends on

d
proving that L3 is dense in F3”. Together with (4.11) this will yield the result.

d
Take f 6 F92. Let fn be the truncation of its Haar expansion at the n-th generation

of the dyadic lattice,
fn = Z (fth)h-I'

JeD
IJI22‘"

While {fn} may not converge in the Lg-norm, we show that it does converge

69

d
(to f ) in the F (In-norm. We have

1/2 4 1/2
Ilf—fnllggf ](Zl-‘l—Jl (f-f.,hJ)2) d6=[r( Z mum?) d6.

J30 J30
|J|<2“'"

Since f 6 F92 , the dominated convergence theorem applies, so [If - f n” «102 —> 0 as
F

1
n —+ 00. This concludes the proof of Lemma 2 and Theorem 1. Cl

4.3 The continuous case

Following [21], we define H l = H1('ll‘) using the area integral; specifically

= {f E L1 ; [T (Amie) lft(€)|2dA(§))l/2 d6 < 00} (4.12)

with the corresponding norm

Ilfllul = [r ( / .ae, award/1(a))” am. (4.13)

Here f (z) is the harmonic extension of f into 1D. Pa(e‘9) is the cone-like region

} . We will specify the angle

lei” — z| l

 

 

with vertex 6‘”: Fa(e‘9)= z E 11):,
1 - |z| sma

a a little later.

The corresponding definition of BM 00 = BM 00(T) is

arc I CT

BMOo = {w 6 L1: SUP fi/Q |¢'(€)l2(1-|€|)dA(€) < 00, w(O) = 0}, (4-14)
I

where <p(z) is the harmonic extension of cp into ID and Q; is the Carleson square

corresponding to the arc I, Q; = {z 6 ID: z/Izl E I,1—|I|S |z| <1}. The norm

70

 

Figure 4.1: The region Fa(e‘9).

in this space is then

1/2
1

||<P||BMO= SUP — |¢’(€)|2(1—|€l)d1‘1(€) -
III Q1

arc ICT
We are now in a position to state the main result.

Theorem 2. BMOO C (Hl)‘. More precisely,

|fs0(6i9)f_(6‘9)d9l S Cllcpllsmoollfllm, W E BMOoIVf 6 H1- (4-15)
'1‘

Proof. Not surprisingly, the proof starts off dyadic. For every J E D = DT define

1

M1:—
IJI QJ

|<P'(€)|2(1-|€|)d14(€).
Clearly, M J g M déf ”LPIIBMOO- We have

M. — g (M... - M1-) = fi' [TQJ mam — |6|),dA(€)-

Here TQJ is the top half of the (dyadic) square Q], TQJ = QJ\(QJ+ U Q;_).

71

 

TQJ

 

TQJ_ TQJ+

 

 

 

 

 

 

 

J- J
J +

Figure 4.2: The decomposition Q] = U TQI.
IEDJCJ

Define
_ I 2 =
SJ —/F0Jllf(€)l dA(€)

Z / meme).

IQJJED TQI
Here P3 is the dyadic cone, IV} = U12 J TQ J. Observation: For some fixed (1,
r3 c raw”), vs e J.
We have SJ_ = 51+ = Z / |f’(€)|2dA(£) and thus, SJ_ — SJ = 31+ — SJ =
Ill-MED TQI
/ | f’ (€)|2dA(§). Therefore, the conditions (4.1) of the key lemma are satisfied.
TQJ

Assuming the existence of the function B in the lemma and using (4.3), we have

1/2
1 I 2 _ I 2
2: |J|(—|J|/TQ1|99(€)I(1 KIN/1(6)) (fmJlflflldMa)

IJIZ2-n+l
M 1/2
4;?" )3 (brew/4(5)) .
PJ

|J|=2-n

1/2

Let us estimate the left-hand side as n —> 00.

1/2
. 1 I 2 _ I 2
3320 E: HI ('7, [TQJIMOI (1 KIM/1(6)) (AQJ|f(€)IdA(€))

”lag-n+1

1/2

72

1/2

1/2
— Jl/2 ’ 21— dA ' 2dA
—JEZD|| (fQ Imam |6|) (a) (fTQJIIIoI (a)

> Z W” /QJ|<p ’()| If (€)l(1-|€|)”2dA(£)

JED

>0: .2 MIN )ldA(€)
12%| I/QJIr I

I I I .1—
> C [D I<p (0| If (el log |€| dA(€)

1
(Here we have used the fact that (1- |€|)1/2 ~ HI”2 and |J| ~ log —

|€ I
addition, U TQJ = n.)
JED

ifEETQJ. In

> C" If acpgflog -—1—dA(§)|
1D

|€|

= C”

     

(wf) 10g l—éldA €()‘

where we have used the fact that 6<p6f = 65(cpf) = %A(cpf-), since cp and f are

analytic.

Recall Green’s formula.

1 .9 _ =_1__ _
5; Tm )d6 F(0) 27: [D 23mg) log KIM g).

n—ooo(

Since <p(0)— — 0, we get lim (LHS) > Cl/T<p(e’9)f( e“9 d)6’.

73

On the right-hand side we obtain, as n —I oo,

1 I 2 )1/2
— ' dA (16,
«int/JIM), f. (fgkw) If (al (e)

where Pd 6‘9 = Fd. Since each I‘d C I}, em , we have I‘d em C Fa em , and
a J J a

Jae“
thus

fr (frgkw)|f’(€)|2dA(€))l/2d6s [I (Amie)|f’(€)|2dA(§))l/2d6= ||f||H1,

Putting together the estimates for the right- and left-hand sides, we obtain the state-
ment (4.15). [:1

4.3.1 Multi-dimensional setting

Without going into detail, we note that an almost identical proof allows us to extend
the results to the multi-dimensional setting. Fix a dyadic lattice D on IR". As
before, for I E D define Q1 and TQI. Given .7: E R", introduce the “strange”

COIl€S

r: = U TQI.

199:,IED

For f E L1(IR"), let f(y,t) be its harmonic extension into R1“. Let

H1(Rn)={f E LINK"): [Rn (frg |Vf(y,t)|2dydt) 1/2 da: < 00}.

with the corresponding norm. We use a quite natural BM OUR").

BMO(R") 2 {4p E L1(IR") : /|V<p(y,t)|2t dydt S 02|I|,VI, a cube in 1R"}.
I

74

The host surh (l' is the («Irrespnmling BAH) norm. III this notation. we can state

the following thmn‘mn

Theorem. 8310(3") C (1H1(R"))i um] HII: ('Imsfu/It of (;I7[()(‘(](/l'll‘(] {/(N-‘S not III)” III/

on (II/Hensiml.

4.4 The proof of the key lemma

Proof. Fix J E D. Let S 2 SJ: SO 2 SJ_ 2 SJ+; .l[ 2 AL]: .l[_ = AIL; .\[+ =
.11.“. Then

1 ~. 1
38(5.1_..\I.)_)+ 53(5),. Mu)

.—

1 - l
: 38(50. .11-) + 313(50. 31+) — B(&,. M) + B(SO. M) — B(S,.\1)+ B(S. M)

l 013 .
2 B (so. §(.\L + Am) — B(SO. M) + —(s..-I1)(SO — S) + B(S. M)

()S
013 - 1 OB ~
=——.'..-\ .\——.\_ . —— ._—‘ ....
0.1160 I)( 1 2(1 + \I+))+OS(S \I)(€0 S)+B(S‘ \1)

for some 5 E [SSH] and .l.[ E [%(.\I-+.\I+)..ll]. Since ”2” g (l, we have

(Isl-5
"’3 '1' ‘ “1‘ " . v 03/3 . .. on . on . .
m(bdl) 2 aflbwll) and sum "FA/'- 3 (l. \u hau- —W(SO..\I) Z —W(50.,\I),

Tllt‘l't‘fI'II'O.

1 . , l ..
513m, .I1_,_ ) + 513(5)... MI.)

 

 

UB 1 1 NB . .. y N \
2 —‘ (5”. .\/I .ll — —(.ll_.. + .l/...) + , ,(50. HHS“ — 5) + 13(5. M)
(I)! ') ()5

 

 

BB (BB

2 2 -6—-M(S°’ M)6—S(SO’ M)\/(M - é(M. + M+)) (So — S) + B(S,M)

 

2 \/2_A7I\/(M -— gm. + M+)) (s0 — S) + 3(3, M).

Now,

 

_,. 1 _. 1 1
2 Z §B(SJ,MJ) : 2 Z [§B(SJ_,MJ_) + §B(SJ+)MJ+)]

JED 1‘50

|J|=2-n |J|= -2‘"+l

Z 2'": Z2[V M\/SJ+ - SJ)( MJ _ ’(MJ— + MJ+))

JED

LII: _2-n+1

+B(SJI MJ)]

1 1
= 2‘"F,, + 2""+1 Z [§B(SJ_:MJ_) + §B(SJ+v MJ+)]

JED
|Jl=2—fl+2

2 WP. + 2“"“17.-. + 2"”1 Z 3(3), MI)

JED
|J|=2—n+2

Z . . Z 22-ka + B(SIO,M[0),
12:]

where we have set

 

Fig: 2: \/—( (SJ+—SJ) (MJ—%(MJ_+MJ+)).

JED
|J|=2—k+l

Using the fact that B(SJ, M;) S 2M\/SJ,VJ E D, 8(810, M10) 2 0 and the defini-

tion of Fk, we obtain the statement of the lemma. [:1

76

4.5 A sample Bellman function
We are looking for a function B such that

2 2
OSBSCIVE, (2712330, €3>0 —a_B§_Bi>C21

for some positive constants 01,02. The proof of the key lemma using B suggests
that we want to choose these constants in order to minimize the ratio Cl/\/Cg. To

make the estimates “sharper,” we require that gay?- = 0. (We cannot require equality

for £7?- 5 0 .) This means that B is a linear function of y. Furthermore, % must
be negative. Because of the homogeneity in the way B is used in the lemma, we
only need to choose one coefficient in that linear function. Thus we seek B in the

form

Therefore,
__3£3_B_A—y>A—M
Bar By_ 2 - 2 ’

 

since y S M. We have CI = A and for the ratio to be minimized

_C_1____\/§A _
\/’C_2_\/A—M'

The minimum of this ratio is attained at A = 2M, thus producing the function
B(:v,y) = x/HQM - y),

satisfying conditions (4.3).

Some possible directions of future research on the Bellman function and H 1 - BM 0

duality are discussed in the next section.

77

 

Research prospects

In this section, we discuss some possible directions for future research on the topics
presented throughout the thesis. We do not attempt to embrace all possible research
prospects but rather concentrate on those of immediate importance and the greatest

promise.

J ohn—Nirenberg inequality

The most natural continuation of the research on this topic would be to find the

Bellman function for the weak-form John-Nirenberg inequality (1.14)
m ({a: E I : |gp(x) — (90)” > ,\}) S cle_C2A/”‘P”BMO(I),

where the L2-based BM 0 norm is used. While it is entirely possible that the
extremal functions are different for the weak and integral form of the inequality,
it would be very interesting to find out how they are related. The corrasponding

formulation would look something like this

B(s,t) = sup{|{:1: 6 [0,1] : |<p(:z:)— <cp>[o,1] I > t}|, cp E BMO,” ”SOHBMOP S 3}.

Finding B exactly would give sharp constants for the traditional John-Nirenberg

inequality. This problem is open even in the case p = 2.

78

The next order of business would be to explore the possibility of finding the Bellman
function for the integral form in the LP-based formulations with p aé 2. We observe
that the ease with which we managed to associate a plane domain with the s-ball
in BM 0 in the L2 setting will not be there. Somewhat similar formulations are
possible in the case of even, positive p, however the number of variables involved
would grow very fast, as would the number of constraints relating those variables.
The essential feature of the formulation we have used is that, after considerations of
homogeneity, the Bellman function can be sought as a function of one variable. That

is unlikely to happen if the number of the variables in the set-up increases.

Perhaps a more perspective direction of research (and, possibly, of more interest to
the general mathematical audience) would be to try to use the results obtained to
deal with BM 0(Q) with Q being a cube in IR". The John-Nirenberg inequality

holds in higher-dimensional BM 0; hence the question of best constants.

Chang-Wilson-Wolff theorem

Having successfully treated the local Chang-Wilson-Wolff theorem in the dyadic
case using the Bellman function technique, we would like to apply the method to
the continuous version of the theorem. Namely, for f E LICK"), let P7(:c) =

{(3}: t) 6 RnH- Ix - yl < 7t}. Let

1/2
A~f($) = (fr ( )lVyf(y,t)l2tl‘"dydt) -

Assume A7 f E L°°(lR"). Then, as shown in [2],

U.) ___|_2)
su exp c < C ,
1:..R..IT|/, (I'll/17f“; 2

79

where c1, C2 depend on 7 and the dimension. A successful application of the Bell-
man function technique (how to do it is far from obvious in this case) may yield
sharp constants. If the constants are independent 1of 7, one may replace A, by
the g-function, g(f)(ac) = (foo |V3f(x,t)|2t cit) /, thus solving a famous open
problem. 0

On the other hand, it would be instrumental to try and pose the extremal problem
even in the dyadic case, so that the Bellman function could be found explicitly. The

difficulties with the choice of variables and the right scaling (an impossibility in the

case E 7:4 [0,1] ) need to be addressed.

H1 - BM 0 duality

The successful formulation of an optimization problem in this case would be a signif-
icant accomplishment, since it would not only be the first of its kind, but also would
shed some light on how to proceed to find the best function exactly. If the optimiza-
tion problem is explicitly solved, the sharp constant of embedding would be produced,
in the one- and multi-dimensional cases, which, to the best of our knowledge, is an

open problem.

80

 

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