THE LP -OPERATOR NORM OF A QUADRATIC PERTURBATION OF THE REAL
PART OF THE AHLFORS–BEURLING OPERATOR
By
Nicholas Boros

A DISSERTATION
Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of
DOCTOR OF PHILOSOPHY
Mathematics
2012

ABSTRACT
THE LP -OPERATOR NORM OF A QUADRATIC PERTURBATION OF
THE REAL PART OF THE AHLFORS–BEURLING OPERATOR
By
Nicholas Boros

Given a sequence of martingale differences, Burkholder found the sharp constant for the
Lp -norm of the corresponding martingale transform. We are able to determine the sharp Lp norm of small “quadratic perturbations” of the martingale transform in Lp . By “quadratic
perturbation” of the martingale transform we mean the Lp norm of the square root of the
squares of the martingale transform and the original martingale (with small constant). The
problem of perturbation of martingale transform appears naturally if one wants to estimate
the linear combination of Riesz transforms (as, for example, in the case of Ahlfors–Beurling
operator). Let {dk }k≥0 be a complex martingale difference in Lp [0, 1], where 1 < p < ∞,
and {εk }k≥0 a sequence of signs. We obtain the following generalization of Burkholder’s
famous result. If n ∈ Z+ , then we have the following estimate
n
k=0

1
εk , τ dk p
≤ (p∗ − 1)2 + τ 2 2
L ([0,1],C2 )

n
dk
k=0

Lp ([0,1],C)

,

1
1
with (p∗ −1)2 +τ 2 2 being the sharp constant in the estimate for 1 < p < 2 and τ 2 ≤ 2p−1
1
or 2 ≤ p < ∞ and τ ∈ R, where p∗ − 1 = max{p − 1, p−1 }. This result is significant not
only because it is a generalization of Burkholder’s famous result from the 1980’s, but also
because we can apply to obtaining the exact operator norm of a certain singular integral
operator.

Let R1 and R2 be the planar Riesz transforms. We compute the Lp -operator norm of a
2
2
quadratic perturbation of R1 − R2 as
1
2
2
= (p∗ − 1)2 + τ 2 2 ,
(R1 − R2 , τ I) p
L (C,C)→Lp (C,C2 )
1
for 1 < p < 2 and τ 2 ≤ 2p−1 or 2 ≤ p < ∞ and τ ∈ R. To obtain the lower bound estimate
2
2
of, what we are calling a quadratic perturbation of R1 − R2 , we discuss a new approach of
constructing laminates (a special type of probability measure on matrices) to approximate
the Riesz transforms.

To my wife Lindsey and daughter Elizabeth

iv

ACKNOWLEDGMENT

First of all, I would like to express my deepest and sincerest gratitude to my dissertation
advisor, Dr. Alexander Volberg for his invaluable support, guidance and personal care. It is
beyond my words of appreciation that he is a far-sighted person with a broad perspective,
and that he has a unique creative insight which often sheds light on undiscovered potentialities of a subject. Throughout these years, he has been cultivating my passion in science and
academic aspirations. He has had enormous patience with me, even during the most challenging times in my Ph.D. study. It has been an enormously enjoyable experience working
with him.
My gratitude also goes to my defense committee members Dr. Fedor Nazarov, Dr.
Michael Shapiro, Dr. Yang Wang, Dr. Clifford Weil, and Dr. Dapeng Zhan for their expertise and precious time. I also would like to thank Dr. Zhengfang Zhou and Dr. Baisheng
Yan their generous support throughout my Ph.D. program, and Dr. Ignacio Uriarte-Tuero
for his guidance, support and the inspiring classroom discussions.
Moreover, I would like to thank Dr. Prabhu Janakiraman, Dr. Vasily Vasyunin, Dr.
L´zl´ Sz´kylhidi Jr., Mr. Nikolaos Pattakos and Dr. Alberto Condori for their continuing
a o e
help, encouragement and involvement in my research. My thanks also go to some of current
and former members in Dr. Volberg’s group: Dr. Leonid Slavin, Dr. Oliver Dragiˇevi´,
c c
Dr. Matthew Bond, Mr. Nikolaus Pattakos and Mr. Alexander Reznikov. This group has
been very close, much like a family and I will always treasure our friendships. I am also
grateful to Ms. Barbara Miller, Graduate Secretary in the Department of Mathematics, for
her generous help during my graduate study.
I am, very much, indebted to Ms. Sharon Griffin for all of her support, guidance and
v

mentoring which has definitely shaped me as an educator and to Dr. Jerry Caldwell and Mr.
Theodore Caldwell for the opportunities and support that they have given me. Teaching for
the Drew Science Scholars and the Diversity Programs Organization within Engineering has
completely changed the way that I teach, as well as, my career path.
Finally, I would like to thank my wife for her love, support and understanding throughout
my Ph.D. study. She has always been patient with me through this long and difficult process.
I cannot come up with the words to appropriately express my gratitude.

vi

TABLE OF CONTENTS

List of Figures

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ix

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

2 A Quadratic Perturbation of the Martingale Transform . . . . . . . . . .
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.1 Motivation of the Bellman function . . . . . . . . . . . . . . . . . . .
2.1.2 Outline of Argument to Prove Main Result . . . . . . . . . . . . . . .
2.1.3 Properties of the Bellman function . . . . . . . . . . . . . . . . . . .
2.1.4 Monge-Amp`re equation and method of “characteristics” . . . . . . .
e
2.2 Computing the Bellman function candidate from the Monge-Amp`re solution
e
2.2.1 Bellman candidate for 2 < p < ∞ . . . . . . . . . . . . . . . . . . . .
2.2.1.1 Case (12 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.1.2 Case (22 ) for 2 < p < ∞ . . . . . . . . . . . . . . . . . . . .
2.2.1.3 Gluing together partial candidates from Cases (12 ) and (22 )
2.2.2 The Bellman function candidate for 1 < p < 2 . . . . . . . . . . . . .
2.3 Monge-Amp`re solution is the Bellman function . . . . . . . . . . . . . . . .
e
2.4 Proving the main result . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5 Proof of Proposition 25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.1 Considering Case (32 ) . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.2 Case (2) for 1 < p < 2 . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6 Remaining cases and why they do not give the Bellman function candidate .
2.6.1 Case (12 ) for 1 < p < 2 and Case (32 ) for 2 < p < ∞ . . . . . . . . .
2.6.2 Case (11 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.3 Case (31 ) does not provide a Bellman function candidate . . . . . . .
2.6.4 Case(21 ) gives a partial Bellman function candidate . . . . . . . . . .
2.6.5 Case (4) may or may not yield a Bellman function candidate . . . . .

5
5
8
11
12
16
21
22
22
32
34
40
42
49
57
57
62
67
67
67
72
74
74

3 Laminates Meet Burkholder Functions
3.1 Introduction . . . . . . . . . . . . . . .
3.2 Lower Bound Estimate . . . . . . . . .
3.2.1 Laminates and gradients . . . .
3.2.2 Laminates and lower bounds . .

76
76
78
78
80

vii

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3.3

3.4

3.2.3 Proof of Theorem 73 . . . . . . . . . . . . . . . . . . . . . . . . .
Comparison with Burkholder functions . . . . . . . . . . . . . . . . . . .
3.3.1 Analyzing the Burkholder functions U and v . . . . . . . . . . . .
3.3.2 Why the laminate sequence νN worked in Theorem 73 . . . . . .
Upper Bound Estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4.1 Background information and notation . . . . . . . . . . . . . . . .
3.4.2 Extending the martingale estimate to continuous time martingales
3.4.3 Connecting the martingales to the Riesz transforms . . . . . . . .
3.4.4 Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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81
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86
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93
93
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97
98

4 Dissertation achievements and future work . . . . . . . . . . . . . . . . . . 100
4.1 Contributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
4.2 Future work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
Bibliography . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . 107

LIST OF FIGURES

Figure 2.1

Sample characteristic of solution from Case (12 ) . . . . . . . . . . .

22

Figure 2.2

Sector for characteristics in Case (12 ), when p > 2. . . . . . . . . . .

24

Figure 2.3

Sample characteristic of Monge-Amp`re solution in Case (21 ) . . . .
e

32

Figure 2.4

Sample characteristic of Monge-Amp`re solution for Case (22 ) . . .
e

32

Figure 2.5

Characteristics of Bellman candidate for 2 < p < ∞ . . . . . . . . .

35

Figure 2.6

Characteristics of Bellman candidate for 1 < p < 2.

. . . . . . . . .

40

Figure 2.7

Location of Implicit (I) and Explicit (E) part of Bτ for 2 ≤ p < ∞. .

44

Figure 2.8

Sample characteristic of Monge-Amp`re solution in Case (32 ) . . . .
e

57

Figure 2.9

Range of characteristics in Case (32 ) for 1 < p < 2. . . . . . . . . . .

59

Figure 2.10

Splitting [−1, 1] × (1, 2) in the (τ × p)–plane into three regions A, B
1
and C. Region B = τ 2 ≤ 2p−1 . . . . . . . . . . . . . . . . . . .

64

Sample characteristic for Monge-Amp`re solution in Cases (11 ) and
e
(31 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

68

Figure 2.12

Characteristic of solution in Case (41 ). . . . . . . . . . . . . . . . .

75

Figure 2.13

Characteristic for the solution from Case (42 ) . . . . . . . . . . . . .

75

Figure 3.1

Splitting between u and v in y1 × y2 -plane. . . . . . . . . . . . . . .

88

Figure 2.11

ix

Chapter 1
Introduction
Determining the exact Lp -operator norm of a singular integral operator is a difficult task to
accomplish in general. Classically, the Hilbert transform’s operator norm was determined
by Pichorides [32]. More recently, the real and imaginary parts of the Ahlfors–Beurling
operator were determined by Nazarov,Volberg [31] and Geiss, Montgomery-Smith, Saksman
[24]. Considering the full Alhfors–Beurling operator, the lower bound was determined as
p∗ − 1 by Lehto [28], or a new proof of this fact in [10]. On the other hand the upper bound
has been quite a bit more difficult. Iwaniec conjectured in [25] that the upper bound is p∗ −1.
However, attempts at getting the conjectured upper bound of p∗ − 1 have been unsuccessful
so far. The works of Ba˜uelos,Wang [4], Nazarov, Volberg [31], Ba˜uelos, M´ndez-Hern´ndez
n
n
e
a
[2], Dragiˇevi`, Volberg [22], Ba˜uelos, Janakiraman [3], and Borichev, Janakiraman, Volberg
c c
n
[5] have progressively gotten closer to p∗ −1 as the upper bound, but no one has yet achieved
it. We note that all of these upper bound estimates crucially rely on Burkholder’s estimates
[15] of the martingale transform. Burkholder’s estimates of the martingale transform even
play a crucial part in determining the sharp lower bounds of the real and imaginary parts of
1

the Alhfors-Beurling operator as seen in Geiss, Montgomery-Smith, Saksman [24]. Also, in
[24] it becomes clear that if one can determine the Lp -operator norm of some perturbation
of the martingale transform then one can use it to determine the Lp -operator norm of the
same perturbation of the real or imaginary part of the Alhfors-Beurling operator. The
focus of Chapter 2 is determining the Lp -operator norm for a “quadratic perturbation” of
the martingale transform using the Bellman function technique, which is similar to how
Burkholder originally did, for τ = 0, in [15]. By “quadratic perturbation”, we are referring
1
to the quantity (Y 2 + τ 2 X 2 ) 2 , where τ ∈ R is small, X is a martingale and Y is the
corresponding martingale transform.
We claim that our operator

2
2
(R1 − R2 , τ I) p
,
L (C,C)→Lp (C,C2 )

represents a simpler model of the difficulties encountered by treating the Ahlfors–Beurling
transform. An extra interesting feature of this operator is that there it breaks the symmetry
between p ∈ (1, 2) and p ∈ (2, ∞). Another interesting feature is that it is simpler to treat
2
2
than a seemingly simpler perturbation R1 − R2 + τ I, whose norm in Lp is horrendously
difficult to find. The last assertion deserves a small elaboration. Burkholder found the
norm in Lp of the martingale transform, where the family of transforming multipliers I
run over [−1, 1]. It is not known how (and it seems very difficult) to find the norm of the
martingale transform, where the family of transforming multipliers I run over [−0.9, 1.1].
2
2
2
2
That would be the norm in Lp of R1 − R2 + 0.1 I = 1.1R1 − 0.9R2 . However, if one writes
the perturbation in the form we did this above, the problem immediately becomes much
more treatable, and we manage to find the precise formula for the norm for a wide range of
2

p’s and τ ’s.
The method of Bourgain [11], for the Hilbert transform, which was later generalized for
a large class of Fourier multiplier operators by Geiss, Montgomery-Smith, Saksman [24], is
to discretize the operator and generalize it to a higher dimensional setting. This operator in
the higher dimensional setting will turn out to have the same operator norm and it naturally
connects with discrete martingales, if done in a careful and clever way. At the end, one
has the operator norm of the singular integral bounded below by the operator norm of
the martingale transform, which Burkholder found in [15]. This approach can be used for
2
2
estimating (R1 − R2 , τ I) p
from below as well, see [10]. However, we will present
L →Lp
an entirely different approach to the problem.
Rather than working with estimates on the martingale transform, we only need to consider
the “Burkholder” functions that were used to find those sharp estimates on the martingale
transform. More specifically, we analyze the behavior of the Burkholder functions, U and
v found in Chapter 2, associated with determining the Lp -operator norm of the quadratic
perturbation of the martingale transform. Using the fact that U is the least bi-concave
majorant of v (in the appropriately chosen coordinates), in addition to some of the ways
in which the two functions interact will allow us to construct an appropriate sequence of
laminates, which approximate the push forward of the 2-dimensional Lebesgue measure by
the Hessian of a smooth function with compact support. Once the appropriate sequence
of laminates is constructed, we are finished since the norm of the Riesz transforms can be
approximated by certain fractions of the partial derivatives of smooth functions. The beauty
of this method is that it quickly gets us the sharp lower bound constant with very easy
calculations. This lower bound argument is discussed in Section 3.2.
3

The use of laminates for obtaining lower bounds on Lp -estimates has been first recognized
by D. Faraco. In [23] he introduced the so-called staircase laminates. These have also been
used in refined versions of convex integration in [1] in order to construct special quasiregular
mappings with extremal integrability properties. Staircase laminates also proved useful in
several other problems, see for instance [21].
In this text we will use a continuous, rather than a discrete laminate. More importantly,
in contrast to the techniques used in [23, 1, 21] we will construct the laminate indirectly
using duality. The advantage is essentially computational, we very quickly obtain the sharp
lower bounds. Indeed, the lower bound follows from the following inequality

f (1, 1) ≤

∞
dt
1
[f (kt, t) + f (t, kt)]t−pk ,
(1 − k) 1
t

where k ∈ (−1, 1) and pk = 2 , and valid for all biconvex functions f : R2 → R with
1−k
f (x) = o(|x|pk ) as |x| → ∞ (see Section 3.2.3).
The “Burkholder” functions U and v also play a crucial role in obtaining the sharp
upper bound estimate as well. With the Burkholder functions we are able to extend sharp
1
estimates of (Y 2 + τ 2 X 2 ) 2 , obtained in Chapter 2, from the discrete martingale setting to
the continuous martingale setting. The use of “heat martingales”, as in [2] and [3], will allow
us to connect the Riesz transforms to the continuous martingales estimate, without picking
up any additional constants. This upper bound argument is presented in Section 3.4.

4

Chapter 2
A Quadratic Perturbation of the
Martingale Transform

2.1

Introduction

In a series of papers, [12] to [19], Burkholder was able to compute the Lp operator norm
of the martingale transform, which we will denote as M T. This was quite a revolutionary
result, not only because of the result itself but because of the method for approaching the
problem. Burkholder’s method in these early papers was inspiration for the Bellman function
technique, which has been a very useful tool in approaching modern and classical problems
in harmonic analysis (this chapter will demonstrate the Bellman function technique as well).
But, the result itself has many applications. One particular application of his result is for
obtaining sharp estimates for singular integrals. Consider the Ahlfors-Beurling operator,
which we will denote as T. Lehto, [28], showed in 1965 that T p := T p→p ≥ (p∗ − 1) =
1
max p − 1, p−1 . Iwaniec conjectured in 1982, [25], that T p = p∗ −1. The only progress
5

toward proving that conjecture has been using Burkholder’s result, see [31], [2] and [3] for
the major results toward proving the conjecture. However, Burkholder’s estimates have
been useful for lower bound estimates as well. For example, Geiss, Montgomery-Smith and
Saksman, [24], were able to show that

T p,

T p ≥ p∗ − 1, by using Burkholder’s

estimates. The upper bound for these two operators were determined as p∗ − 1 by Nazarov,
Volberg, [31] and Ba˜uelos, M`ndez-Hern`ndez [2], so we now have
n
e
a
p∗ − 1. Note that

T p =

T p =

T the difference of the squares of the planar Riesz transforms, i.e.

2
2
T = R1 − R2 .
A recent result of Geiss, Montomery-Smith and Saksman, [24] points to the following
observation, though not immediately. We can estimate linear combinations of squares of
Riesz transforms if we know the corresponding estimate for a linear combination of the
martingale transform and the identity operator. In other words, one can get at estimates
2
2
of the norm of (R1 − R2 ) + τ · I, by knowing the estimates of the norm of M T + τ · I.
M T +τ ·I p has only been computed for either τ = 0 by Burkholder [15] or τ = ±1 by Choi
[20]. The problem is still open for all other τ -values and seems to be very difficult, though
we have had some progress. But, if we consider “quadratic” rather than linear perturbations
then things become more manageable. This brings us to the focus of this paper, which is
determining estimates for quadratic perturbations of the martingale transform, which will
have connections to quadratic combinations of squares of Riesz transforms.
To prove our main result we are going to take a slightly indirect approach. Burkholder
(see [15]) defined the martingale transform, M Tε , as
n
M Tε

n
dk :=

k=1

εk d k .
k=1

6

Then the main result can be stated as

sup
ε

M Tε , τ I

n
1
k=1 εk , τ dk p
= ((p∗ − 1)2 + τ 2 ) 2 ,
n d
k=1 k p

= sup
Lp (C)→Lp (C2 )
ε

where I is the identity transformation and τ is “small”. However, rather than working
with this martingale transform in terms of the martingale differences, in a probabilistic
setting, we will define another martingale transform in terms of the Haar expansion of
Lp [0, 1] functions and set up a Bellman function in that context. Burkholder showed, in
[19], that these two different martingale transforms have the same Lp operator norm, for
τ = 0, so we expected a perturbation of these to act similarly and it turns out that they
do. For convenience, we will work with the martingale transform in the Haar setting. Using
the Bellman function technique will turn the problem of finding the sharp constant of the
above estimate into solving a second order partial differential equation. The beauty of this
approach is that it gets right to the heart of the problem with very little advanced techniques
needed in the process. In fact, the only background material that is needed for the Bellman
function technique approach, is some basic knowledge of partial differential equations and
some elementary analysis.
Observe that for 2 ≤ p < ∞, the estimate from above is just an application of Minkowski’s
p
inequality on L 2 and Burkholder’s original result. But, this argument does not address
sharpness, even though the constant obtained turns out to be the sharp constant for all τ .
2
For 1 < p < 2, Minkowski’s inequality (in l p ) also plays a role, but to a lesser extent and
cannot give the sharp constant, as we will see Proposition 33. It is, indeed, very strange that
such sloppy estimation could give the estimate with sharp constant for 1 ≤ p < ∞. We will
now rigorously develop some background ideas needed to set up the Bellman function.
7

In our calculations we follow the scheme of [37], but our “Dirichlet problem” for MongeAmp`re is different. For small τ the scheme works. For large τ and 1 < p < 2 it definitely
e
must be changed, as will be shown later. The amazing feature is the “splitting” of the result
to two quite different cases: 1 < p < 2 and 2 ≤ p < ∞, where in the former case we know
the result only for small τ, but in the latter one τ is unrestricted.

2.1.1

Motivation of the Bellman function

Let I be an interval and α± ∈ R+ such that α+ +α− = 1. These α± generate two subintervals I ± such that |I ± | = α± |I| and I = I − ∪I + . We can continue this decomposition indefinitely as follows. For any sequence {αn,m : 0 < αn,m < 1, 0 ≤ m < 2n , 0 < n < ∞, αn,2k +
αn,2k+1 = 1}, we can generate the sequence I := {In,m : 0 ≤ m < 2n , 0 < n < ∞}, where
−
+
In,m = In,m ∪ In,m = In+1,2m+1 ∪ In+1,2m+1 and α− = αn+1,2m , α+ = αn+1,2m+1 .
Note that I0,0 = I.
For any J ∈ I we define the Haar function hJ := −

α+ χ
+
α− |J| J −

α− χ . If
α+ |J| J +

max{|In,m | : 0 ≤ m < 2n } → 0 as n → ∞, then {hJ }J∈I is an orthonormal basis for
L2 (I) := {f ∈ L2 (I) : I f = 0}. However, if we add one extra function, then Haar functions
0
form an orthonormal basis in L2 [0, 1] function. We will show that this is true in the case
I0 = [0, 1] and I = D as the dyadic subintervals of I0 . The Bellman function will also be set
up under this specific choice of I = D for clarity. As it will turn out the Bellman function
will be the same under the the choice of I, so the fact that we set up the Bellman function
under this specific choice of I will not matter.
Let Dj = {I ∈ D : |I| = 2−j }. We use the notation f J to represent the average
integral of f over the interval J ∈ D and ∆j f = f I
− f I , where Ij ∈ Dj and
j+1
j
8

Ij+1 ∈ Dj+1 . For any f ∈ L1 (I0 ) we have ∆j f =

I∈Dj (f, hI )hI . Then

∞
∆j f =
j=0

lim f I
− f I
0
N +1
N →∞

(2.1)

By Lebesgue differentiation, the limit in (2.1) converges to f almost everywhere as N → ∞.
So any f ∈ Lp (I0 ) ⊂ L1 (I0 ) can be decomposed in terms of the Haar system as

f = f I χI +
0 0

(f, hI )hI .
I∈D

In terms of the expansion in the Haar system we define the martingale transform, g of f, as

g := g I χI +
0 0

εI (f, hI )hI ,
I∈D

where εI ∈ {±1}. Requiring that |(g, hJ )| = |(f, hJ )|, for all J ∈ D, is equivalent to g being
the martingale transform of f, for f, g ∈ Lp (I0 ).

Now we define the Bellman function as B(x1 , x2 , x3 ) :=

sup
f,g

p
(g 2 + τ 2 f 2 ) 2 I : x1 = f I , x2 = g I , x3 = |f |p I , |(f, hJ )| = |(g, hJ )|, ∀J ∈ D

on the domain Ω = {x ∈ R3 : x3 ≥ 0, |x1 |p ≤ x3 }. The Bellman function is defined in this
way, since we would like to know the value of the supremum of (g, τ f ) p , where g is the
martingale transform of f . Note that |x1 |p ≤ x3 is just H¨lder’s inequality. Even though
o
the Bellman function is only being defined for real-valued functions, we can “vectorize” it to
work for complex-valued (and even Hilbert-valued) functions, as we will later demonstrate.
9

Finding the Bellman function will make proving the following main result quite easy. We
p 1
2 + τ 2 f 2 ) 2 p the “quadratic perturbation” of the martingale transform’s norm
will call (g
I
1
p
|g|p I .
Theorem 1. Let {dk }k≥1 be a complex martingale difference in Lp [0, 1], where 1 < p < ∞,
and {εk }k≥1 a sequence in {±1}. If n ∈ Z+ , then
n
k=1

1
≤ (p∗ − 1)2 + τ 2 2
εk , τ dk p
L ([0,1],C2 )

n
k=1

,
dk p
L ([0,1],C)

p
1
with (p∗ − 1)2 + τ 2 2 as the sharp constant for τ 2 ≤ 2p−1 and 1 < p < 2 or τ ∈ R and
2 ≤ p < ∞.

Note that when τ = 0 we get Burkholder’s famous result [15].
Now that we have the problem formalized, notice that B is independent of the initial
choice of I0 (which we will just denote I from now on) and {αn,m }n,m , so we return to
having them arbitrary. Finding B when p = 2 is easy, so we will do this first.
Proposition 2. If p = 2 then B(x) = x2 − x2 + (1 + τ 2 )x3 .
2
1
Proof. Since f ∈ L2 (I) then f = f I χI +

J∈D (f, hJ )hJ implies

1
|f |2
|f |2 I =
|I| I
= f 2+2 f I
I

(f, hJ )
J∈D

1
= f 2+
I |I|

1
1
hJ +
(f, hJ )(f, hK )hJ hK
|I| I
|I| I
J,K∈D

|(f, hJ )|2 .
J∈D
10

So f 2 = |I|x3 = |I|x2 +
1
2

2
J∈D |(f, hJ )| and similarly

|(g, hJ )|2 = |I|x2 +
2

g 2 = |I|x2 +
2
2

|(f, hJ )|2 .
J∈D

J∈D
Now we can compute B explicitly, (p = 2)

p
1
(g 2 + τ 2 f 2 ) 2 I = |g|2 I + τ 2 |f |2 I = x2 + τ 2 x2 + (1 + τ 2 )
1
2
|I|

|(f, hJ )|2
J∈D

= x2 + τ 2 x2 + (1 + τ 2 )(x3 − x2 ).
1
1
2

Remark 3. The fact that τ remains restricted when p approaches 2 from the left is by-product
of a certain inefficiency in the proof. But for the time being we do not know how to lighten
the restriction on τ when p approaches 2 from the left.

2.1.2

Outline of Argument to Prove Main Result

Computing the Bellman function, B, for p = 2, is much more difficult, so more machinery is
needed. In Section 2.1.3 we will derive properties of the Bellman function, the most notable
of which is concavity under certain conditions. Finding a B to satisfy the concavity will
amount to solving a partial differential equation, after adding an assumption. This PDE
has a solution on characteristics that is well known, so we just need to find an explicit
solution from this, using the Bellman function properties. How the characteristics behave
in the domain of definition for the Bellman function will give us several cases to consider.
In Section 2.2 we will get a Bellman function candidate for 1 < p < ∞ by putting together
several cases. Once we have what we think is the Bellman function, we need to show that
11

it has the necessary smoothness and that Assumption 8 was not too restrictive to give us
the Bellman function. This is covered in Section 2.3. Finally the main result is shown in
Section 2.4. In Section 2.6, we show why several cases did not lead to a Bellman function
candidate and why the value of τ was restricted for the Bellman function candidate.

2.1.3

Properties of the Bellman function

One of the properties we nearly always have (or impose) for any Bellman function, is concavity (or convexity). It is not true that B is globally concave, on all of Ω, but under certain
conditions it is concave. The needed condition is that g is the martingale transform of f, or
|x+ − x− | = |x+ − x− | in terms of the variables in Ω.
2
2
1
1
Definition 4. We say that the function B on Ω has restrictive concavity if for all x± ∈ Ω
such that x = α+ x+ + α− x− , α+ + α− = 1 and |x+ − x− | = |x+ − x− |, then B(x) ≥
2
2
1
1
α+ B(x+ ) + α− B(x− ).
Proposition 5. The Bellman function B is restrictively concave in the x-variables.
Proof. Let ε > 0 be given and x± ∈ Ω. By the definition of B, there exists f ± , g ± on I ±
such that f

J±

= x± , g
1

J±

= x± , |f ± |p ± = x± and
2
3
I

p
B(x± ) − [(g ± )2 + τ 2 (f ± )2 ] 2

I±

≤ ε.

On I = I + ∪ I − we define f and g as f := f + χ + + f − χ − , g := g + χ + + g − χ − . So,
I
I
I
I
|x+ − x− | = | f
1
1
=

1
1
f− −
f
+ − f I−| = +
I
|I | |I − |
|I | I −

1
1
1
+ |I| |I − | f − α− |I| I − f = |I|
α
12

f

1 χ
1
+ I + − α− χI −
α

=

|I|
α+ α−

f hI =:

|I|
(f, hI ) .
α+ α−

|I|
−
−
+
+
+ α− (g, hI ) . So our assumption |x1 − x1 | = |x2 − x2 | is
α
equivalent to |(f, hI )| = |(g, hI )|. Since x1 = f I , x2 = g I and x3 = |f |p I , f and g are

Similarly,|x+ − x− | =
2
2

test functions and so
p
B(x) ≥ (g 2 + τ 2 f 2 ) 2 I
p
= α+ [(g + )2 + τ 2 (f + )2 ] 2

p
+ α− [(g − )2 + τ 2 (f − )2 ] 2 −
I+
I

≥ α+ B(x+ ) + α− B(x− ) − ε.

At this point we do not quite have concavity of B on Ω since there is the restriction
|x+ − x− | = |x+ − x− | needed. To make this condition more manageable, we will make a
1
1
2
2
x +x
x −x
change of coordinates. Let y1 := 2 2 1 , y2 := 2 2 1 and y3 := x3 . We will also change
notation for the Bellman function and corresponding domain in the new variable y. Let
M(y1 , y2 , y3 ) := B(x1 , x2 , x3 ) = B(y1 − y2 , y1 + y2 , y3 ). Then the domain of definition for
M will be Ξ := {y ∈ R3 : y3 ≥ 0, |y1 − y2 |p ≤ y3 }.
If we consider x± ∈ Ω such that |x+ − x− | = |x+ − x− |, then the corresponding points
1
1
2
2
−
−
−
+
+
+
y ± ∈ Ξ satisfy either y1 = y1 or y2 = y2 . This implies that fixing y1 as y1 = y1 or y2
−
+
as y2 = y2 will make M concave with respect to y2 , y3 under fixed y1 and with respect to
y1 , y3 under y2 fixed.
Rather than using Proposition 5 to check the concavity of the Bellman function we can
just check it in the following way, assuming M is C 2 . Let j = i ∈ {1, 2} and fix yi as
13

−
+
yi = yi . Then M as a function of yj , y3 is concave if




 My j y j My j y 3 

 ≤ 0,
My 3 y j My 3 y 3
which is equivalent to

Myj yj ≤ 0, My3 y3 ≤ 0, Dj = Myj yj My3 y3 − My3 yj Myj y3 ≥ 0.

Proposition 6. (Restrictive Concavity in y-variables) Let j = i ∈ {1, 2} and fix yi as
−
+
yi = yi . If Myj yj ≤ 0, My3 y3 ≤ 0 and Dj = Myj yj My3 y3 − (Myj y3 )2 ≥ 0 for j = 1
and j = 2, then M is Restrictively concave.
The Bellman function, as it turns out, has many other nice properties.
Proposition 7. Suppose that M is C 1 (Ξ), then M has the following properties.
(i) Symmetry: M(y1 , y2 , y3 ) = M(y2 , y1 , y3 ) = M(−y1 , −y2 , y3 )
p
(ii) Dirichlet boundary data: M(y1 , y2 , (y1 − y2 )p ) = ((y1 + y2 )2 + τ 2 (y1 − y2 )2 ) 2
(iii) Neumann conditions: My1 = My2 on y1 = y2 and My1 = −My2 on y1 = −y2
(iv) Homogeneity: M(ry1 , ry2 , rp y3 ) = rp M(y1 , y2 , y3 ), ∀r > 0
(v) Homogeneity relation: y1 My1 + y2 My2 + py3 My3 = pM
Proof.

(i) Note that we get B(x1 , x2 , x3 ) = B(−x1 , x2 , x3 ) = B(x1 , −x2 , x3 ) by consider-

ing test functions f = −f and g = −g. Change coordinates from x to y and the result
follows.
14

(ii) On the boundary {x3 = |x1 |p } of Ω we see that
p
1
1
|f |p = |f |p I = x3 = |x1 |p = | f I |p =
f
|I| I
|I| I
is only possible if f ≡ const. (i.e. f = x1 ). But, |(f, hJ )| = |(g, hJ )| for all J ∈ I,
p
which implies that g ≡ const. (i.e. g = x2 ). Then B(x1 , x2 , |x1 |p ) = (g 2 +τ 2 f 2 ) 2 I =
p
(x2 +τ 2 x2 ) 2 . Changing coordinates gives the result. If we are not on the boundary, but
2
1
close to it, then any test function has the p-th power of its average close to the average
of its p-th power. This means that the function itself is close to its average in Lp norm. We know a priori that the martingale transform is bounded in Lp , 1 < p < ∞.
Hence the value of B for the data close to the boundary will be close to the martingale
transform of a constant function considered above. We are done.
(iii) This follows from from (i).
(iv) Consider the test functions f = rf, g = rg
(v) Differentiate (iv) with respect to r and evaluate it at r = 1.
Now that we have all of the properties of the Bellman function we will turn our attention
to actually finding it. Proposition 6 gives us two partial differential inequalities to solve,
D1 ≥ 0, D2 ≥ 0, that the Bellman function must satisfy. Since the Bellman function is the
supremum of the left-hand side of our estimate under the condition that g is the martingale
transform of f and must also satisfy the estimates in Proposition 6, it seems reasonable that
the Bellman function (being the optimal such function) may satisfy the following, for either
j = 1 or j = 2:
Dj = Myj yj My3 y3 − (My3 yj )2 = 0.
15

The PDE that we now have is the well known Monge-Amp`re equation which has a solution.
e
Let us make it clear that we have added an assumption.

Assumption 8. Dj = Myj yj My3 y3 − (My3 yj )2 = 0, for either j = 1 or j = 2.

Adding this assumption comes with a price. Any function that we construct, satisfying
all properties of the Bellman function, must somehow be shown to be the Bellman function.
We will refer to any function satisfying some, or all Bellman function properties as a Bellman
function candidate.
In Subsection 2.1.4 we use [35] to explain what consequences the assumption 8 will have
on our search of Bellman function candidate.

2.1.4

Monge-Amp`re equation and method of “characteristics”
e

Let v = v(x1 , ..., xn ) is a smooth function satisfying the following Monge-Amp`re equation
e
in some domain Ω





 v x1 x1 · · · v x1 xn 




det d2 v = det . . . . . . . . . . . . . . . . . . . . = 0 ,




vxn x1 · · · vxn xn

∀x = (x1 , . . . , xn ) ∈ Ω ,

(2.2)

and suppose that this matrix has rank n − 1; i.e., all smaller minors of d2 v are non-zero.
Then there are functions ti (x1 , . . . , xn ), i = 0, 1, . . . , n, such that

v(x) = t0 + t1 x1 + t2 x2 + · · · + tn xn
16

(2.3)

and the following n − 1 linear equations hold:

dt0 + x1 dt1 + x2 dt2 + · · · + xn−1 dtn−1 + xn dtn = 0 .

(2.4)

Let us explain why this is n−1 equations and why they are linear. One needs to read (2.4)
as follows: we think that, say, t1 , . . . , tn−1 are n − 1 independent variables and tn , t0 are
functions of them. Then (2.4) can be rewritten as

∂t0
∂t0
∂tn
∂tn
dt1 + · · · +
dtn−1 = 0 ,
+ x1 + xn
+ xn−1 + xn
∂t1
∂t1
∂tn−1
∂tn−1
whence
xi + x n

∂tn ∂t0
+
= 0,
∂ti
∂ti

i = 1, . . . , n − 1 .

So we get n − 1 equations.

Remark. In general we can choose any n − 1 variables as independent, of course. Since the
order of variables is arbitrary, sometimes the first n − 1 is not the most convenient choice.

Now why are these linear equations? We think that t1 , . . . , tn−1 is fixed. Then the n − 1
equations give us linear relationships in x1 , . . . , xn , so n − 1 hyperplanes.
Therefore, (2.4) gives the intersection of n − 1 hyperplanes, so gives us a line. We can
call it Lt ,...,t
. These lines foliate domain Ω and (2.3) shows that v is a linear function
1
n−1
on each such line.
Let us prove all these propositions. Matrix d2 v annihilates one vector Θ(x) at every x = (x1 , ..., xn ) ∈ Ω. So we get a vector field Θ. Consider its integral curve x1 =
17

x1 (xn ), . . . , xn−1 = xn−1 (xn ). Vector Θ(x) is a tangent vector to that curve; i.e.,




 x1 




 x 
 2 




Θ = Θn  · · ·  .






xn−1 




1

(2.5)

Consider a new function g(xn ) = v(x1 (xn ), . . . , xn−1 (xn ), xn ). Due to (2.5) its second
derivative is


 



 x1   x1 

 


 

 x   x 
 2   2 

 


 
2v 
g = d  · · ·  ,  · · ·  +vx1 x1 +· · ·+vxn−1 xn−1 = vx1 x1 +· · ·+vxn−1 xn−1 .

 


 



 
xn−1  xn−1 

 


 

1
1
(2.6)
Now, let us also show that vx1 , . . . , vxn−1 are constants on this integral curve. Suppose we are standing on the integral curve. The surface vx1 = t1 = const. has normal
(vx1 x1 , . . . , vx1 xn ), that is the first row of matrix d2 v, which is orthogonal to the directional vector Θ of the integral curve. Hence Θ is in the tangent hyperplane to the surface
vx1 = t1 . The same is true for the surfaces vxi = ti = const., i = 2, . . . , n − 1. Intersection of these surfaces gives us our integral curves, because Θ is in the intersection of all
tangent planes to these surfaces. Therefore the curves Ct ,...,t
enumerated by constants
1
n−1
t1 , . . . , tn are just the integral curves of the tangent bundle Θ. Thus, (2.6) can be rewritten
18

as
d2
g − (t1 x1 + · · · + tn xn−1 ) = 0.
dx2
n

(2.7)

We obtain that the second derivative of a function g(xn ) = v(x1 (xn ), . . . , xn−1 (xn ), xn )
in (2.7) is zero. So function this function is linear in xn , that is g(xn ) = tn xn + t0 , where
the constants tn , t0 depend only on the curve Ct ,...,t
, that is
1
n−1

tn = tn (t1 , . . . , tn−1 ) ,

t0 = t0 (t1 , . . . , tn−1 ) .

Looking at the definition of g(xn ) we see that we obtained on Ct ,....,t
the following
1
n−1

v(x1 (xn ), . . . , xn−1 (xn ), xn ) = t0 + x1 t1 + · · · + xn−1 tn−1 + tn xn .

Since we assumed our vector field to be smooth and its integral curves foliate the whole
domain, varying parameters t1 , . . . , tn we get (2.3). To check (2.4) take a full differential
in (2.3). Then

[vx1 dx1 + · · · + vxn dxn ] = dv = [dt0 + t1 dx1 + · · · + tn dxn ] + x1 dt1 + · · · + xn dtn . (2.8)

and so vxi = ti , i = 1, . . . , n − 1 as we established already. But it is
We are on Ct ,...,t
1
n−1
also easy to see that
vxn = tn (t1 , . . . , tn−1 )
on Ct ,...,t
. In fact, all vxi and all ti are symmetric. We could have chosen to represent
1
n−1
the integral curve of Θ not as xi = xi (xn ), i = 1, . . . , n − 1 but as xj = xj (x1 ), j =
2, . . . , n. Now we see that two expression in brackets in (2.8) are equal. Then we obtain
19

x1 dt1 + · · · + xn dtn + dt0 = 0, which is the desired n − 1 linear relationships (2.4).

From the consideration above we can see that the following proposition holds.

Proposition 9. For j = 1 or 2, if Myj yj My3 y3 − (My3 yj )2 = 0 has a smooth solution,
then it should have a form M (y) = yj tj + y3 t3 + t0 . Here ti = ti (yj , y3 ), i = 0, 1, 2 are
functions of yj , y3 (and parameter y3−j ). Furthermore, t0 , tj , t3 have common level sets
called characteristics, and the differentials of these functions satisfy yj dtj + y3 dt3 + dt0 = 0.
This shows that these common level sets (characteristics) are straight lines in the yj × y3
plane. Moreover, one can choose functions tj , t3 as follows: tj = Myj , t3 = My3 .

This is a result of Pogorelov, see [33]. We explained it above in Subsection 8. Now that
we have a solution M to the Monge-Amp`re, we need get rid of t0 , tj , t3 so that we have
e
an explicit form of M, without the characteristics. We note that a solution to the MongeAmp`re is not necessarily the Bellman function. It must satisfy the restrictive concavity of
e
Proposition 6, be C 1 -smooth, and satisfy the properties of Proposition 7. The restrictive
concavity property is one of the key deciding factors of whether or not we have a Bellman
function in many cases. Even if the Monge-Amp`re solution satisfies all of those conditions,
e
it must still be shown to be equal to the Bellman function, because we added an additional
assumption (Assumption 8) to get the Monge-Amp`re solution as a starting point. This
e
will be considered rigorously in Section 2.3, after we obtain a solution to the Monge-Amp`re
e
equation, with the appropriate Bellman function properties. So from this point on we will
use M and B to denote solutions to the Monge-Amp`re equation; i.e. Bellman function
e
candidates, and M and B to denote the true Bellman function.
20

2.2

Computing the Bellman function candidate from
the Monge-Amp`re solution
e

Due to the symmetry property of M, from Proposition 7, we only need to consider a portion
of the domain Ξ, which we will denote as, Ξ+ := {y : −y1 ≤ y2 ≤ y1 , y3 ≥ 0, (y1 −
y2 )p ≤ y3 }. Since the characteristics are straight lines, one end of each line must be on
the boundary. Let U denote the point at which the characteristic touches the part of the
boundary {y : (y1 −y2 )p = y3 } (if it does). Furthermore, the characteristics can only behave
in one of the following four ways, since they are straight lines in the plane:
(1) The characteristic goes from U to {y : y1 = y2 }
(2) The characteristic goes from U to to infinity, running parallel to the y3 -axis
(3) The characteristic goes from U to {y : y1 = −y2 }
(4) The characteristic goes from U to {y : (y1 − y2 )p = y3 }
(5) The characteristic goes from the “wall” y2 = −y1 to the “wall” y2 = y1
To find a Bellman function candidate we must first fix a variable (y1 or y2 ) and a
case for the characteristics. Then we use the Bellman function properties to get rid of the
characteristics. If the Monge-Amp`re solution satisfies restrictive concavity, then it is a
e
Bellman function candidate. However, checking the restrictive concavity is quite difficult in
many of the cases, since it amounts to doing second derivative estimates for an implicitly
defined function. Let us now find our Bellman function candidate.
Remark 10. Since we will have either y1 or y2 fixed in each case, there will be eight cases in
all. Let (1j ), (2j ), (3j ), (4j ), (5j ) denote the case when Myj yj My3 y3 − (My3 yj )2 = 0 and
21

yi is fixed, where i = j. Also, we will denote G(z1 , z2 ) := (z1 + z2 )p−1 [z1 − (p − 1)z2 ] and
1
M(y) p
ω :=
from this point on.
y3
Remark 11. Characteristics foliate the domain. Therefore the case of the existence of characteristic of the type (5j ) automatically implies that there exists a characteristic of type
(4j ). Therefore we are not looking at the case (5j ) at all in what follows.

2.2.1

Bellman candidate for 2 < p < ∞

The solution to the Monge-Amp`re equation when 2 < p < ∞, is only partially valid on the
e
domain in two cases, due to restrictive concavity. Case (12 ), will give us an implicit solution
that is valid on part of Ξ+ and Case (22 ) will give us an explicit solution for the remaining
part of Ξ+ . First, we deal with Case (12 ).

2.2.1.1

Case (12 )

Since we are considering Case (12 ), y1 ≥ 0 is fixed until the point that we have the implicit
solution independent of the characteristics satisfying all of the Bellman function properties.

T3
y

U
−y1

y1

E

y2

Figure 2.1: Sample characteristic of solution from Case (12 )

22

p−2
Proposition 12. For 1 < p < ∞ and p y1 < y2 < y1 , M is given implicitly by the relation
G(y1 + y2 , y1 − y2 ) = y3 G( ω 2 − τ 2 , 1), where G(z1 , z2 ) := (z1 + z2 )p−1 [z1 − (p − 1)z2 ]
1
M (y) p
on z1 + z2 ≥ 0 and ω :=
.
y3
This is proven through a series of Lemmas.
Lemma 13. M (y) = t2 y2 + t3 y3 + t0 on the characteristic y2 dt2 + y3 dt3 + dt0 = 0 can be

p
(y1 +u)2 +τ 2 (y1 −u)2
 y , where u is the unique solution to the
simplified to M (y) = 
3
y1 −u
equation

2
2
y2 +( p −1)y1
u+( p −1)y1
p−2
=
and p y1 < y2 < y1
y3
(y1 −u)p

Proof. A characteristic in Case (12 ) is from U = (y1 , u, (y1 − u)p ) to W = (y1 , y1 , w).
Throughout the proof we will use the properties of the Bellman function from Proposition 7.
Using the Neumann property and the property from Proposition 9 we get My1 = My2 = t2
at W. By homogeneity at W we get

py2 t2 + pwt3 + pt0 = pM (W ) = y1 My1 + y2 My2 + py3 My3 = 2y1 t2 + pwt3 .
2
2
Then t0 = ( p − 1)y1 t2 and dt0 = ( p − 1)y1 dt2 , since y1 is fixed. So M (y) = [y2 +
2
2
( p − 1)y1 ]t2 + y3 t3 on [y2 + ( p − 1)y1 ]dt2 + y3 dt3 = 0. By substitution we get, M (y) =
dt
dt
y3 [t3 − t2 3 ] on characteristics. But, t2 , t3 , 3 are constant on characteristics, which gives
dt2
dt2
M (y)
that y
≡ const. as well. We can calculate the value of the constant by using the Dirichlet
3

p
(y1 +u)2 +τ 2 (y1 −u)2
 y , where u is
boundary data for M at U. Therefore, M (y) = 
3
y1 −u
the solution to the equation
2
y2 + ( p − 1)y1
y3

2
u + ( p − 1)y1
=
.
(y1 − u)p
23

(2.9)

2
2
Now fix u = −( p − 1)y1 . Then we see that y2 = −( p − 1)y1 = u is also fixed by (2.9).
This means that the characteristics are limited to part of the domain, as shown in Figure 2.2,
since they start at U and end at W ∈ {y1 = y2 }. All that remains is verifying the equation
y2

y = y1

  2
 
 
y
 
 2

 
 

d
d
d
d
d y2
d
T

=

p−2
p y1
Ey

1

= −y1

Figure 2.2: Sector for characteristics in Case (12 ), when p > 2.

(2.9) has exactly one solution u = u(y1 , y2 , y3 ) in the sector

p−2
p y1 < y2 < y1 . Indeed, the

function
f (u) := y3 u +

2
− 1 y1 − (y1 − u)p y2 +
p

2
− 1 y1
p

2
2
2
is monotone increasing for u < y1 , f (−( p − 1)y1 ) = − p y1 )p y2 + p − 1 y1 < 0 and
2
f (y1 ) = p y1 y3 > 0. Therefore, we do get a unique solution, u, in the sector.

p
2 +τ 2 (y −u)2
(y1 +u)
1
 y can be rewritten as G(y + y , y −
Lemma 14. M (y) = 
3
1
2 1
y1 −u
y2 ) = y3 G( ω 2 − τ 2 , 1) for

p−2
p y1 < y2 < y1 .

1
M (y) p
=
y3

(y1 +u)2 +τ 2 (y1 −u)2
|y −u|
Proof. ω =
≥ |τ | y1 −u = |τ |.
y1 −u
1
2 −τ 2 −1
Since y1 ± u ≥ 0 and ω 2 − τ 2 ≥ 0, u = ω
y1 by inversion. Substituting this
ω 2 −τ 2 +1
2
2
y2 +( p −1)y1
u+( p −1)y1
into
=
gives
y3
(y1 −u)p
p−1
2p−1 y1 [py2 − (p − 2)y1 ] = y3

ω2 − τ 2 + 1
24

p−1

ω 2 − τ 2 − (p − 1)

or (x1 + x2 )p−1 [x2 − (p − 1)x1 ]


=


p−1 
1
1
2

p − τ x p  + x p 

B

3
3
2





2



1
1
2
p − τ x p  − (p − 1)x p  .
B
3
3

Thus,

2
1
1
2
p − τ x p  , x p  or G(y + y , y − y ) = y G( ω 2 − τ 2 , 1).
B
1
2 1
2
3
3
3






G(x2 , x1 ) = G 


This proves Proposition 12. We have constructed a partial Bellman function candidate
from the Monge-Amp`re solution in Case(12 ), so y1 no longer needs to be fixed. All of the
e
properties of the Bellman function were used to derive this partial Bellman candidate, but
the restrictive concavity from Proposition 6 still needs to be verified. To verify restrictive
concavity, we need that My2 y2 ≤ 0, My3 y3 ≤ 0, D2 ≥ 0 and My1 y1 ≤ 0, D1 ≥ 0. By
assumption D2 = 0, so we need not worry about that estimate. The remaining estimates
will be verified in a series of Lemmas. The first Lemma is an idea taken from Burkholder [13]
to make the calculations for computing mixed partials shorter. In the Lemma, we compute
the partials of arbitrary functions which we will choose specifically later, although it is not
hard to see what the appropriate choices should be.

Lemma 15. Let H = H(y1 , y2 ), Φ(ω) =

H(y1 ,y2 )
, R1 = R1 (ω) := 1 and R2 = R2 (ω) :=
y3
Φ

R1 = − Φ 2 . Then
Φ

My y =
3 3

pω p−2 R1 H 2
[ωR2 + (p − 1)R1 ]
3
y3
25

My y
3 i
My y
i i
Di

where ω =

pω p−2 R1 HH
=−
[ωR2 + (p − 1)R1 ]
2
y3
pω p−2 R1
=
[ωR2 + (p − 1)R1 ](H )2 + ωy3 H
y3
2
p2 ω 2p−3 R1 H 2 H
2
[ωR2 + (p − 1)R1 ],
= My y My y − My y =
3
3 3
i i
3 i
y3

M (y)
y3 .

Proof. First of all we calculate the partial derivatives of ω:

H
Φ ωy = −
2
3
y3
Hy
i
Φ ωy =
i
y3

=⇒
=⇒

R H
ωy = − 1 ,
2
3
y3
R1 Hy
i = R1 H ,
ωy =
i
y3
y3

i = 1, 2 .

Here and further we shall use notation H for any partial derivative Hyi , i = 1, 2. This
cannot cause any confusion since only one i ∈ {1, 2} participates in the calculation of Di .
R2 ωy H
3 + 2 R1 H = R1 H (R H + 2y ) ,
2
3
2
3
4
y3
y3
y3
R2 ωy H R H
R H
i − 1
= − 1 (R2 H + y3 ) ,
=−
2
2
3
y3
y3
y3
R2 ωy H R H
R
i + 1
=
= 1 (R2 (H )2 + y3 H ) .
2
y3
y3
y3

ωy y = −
3 3
ωy y
3 i
ωy y
i i

Now we pass to the calculation of derivatives of M = y3 ω p :

My = py3 ω p−1 ωy + ω p ,
3
3
My = py3 ω p−1 ωy ;
i
i
26

2
My y = py3 ω p−1 ωy y + 2pω p−1 ωy + p(p − 1)y3 ω p−2 ωy
3 3
3 3
3
3
p−2 R H 2
pω
1
[ωR2 + (p − 1)R1 ] ,
=
3
y3

(2.10)

My y = py3 ω p−1 ωy y + pω p−1 ωy + p(p − 1)y3 ω p−2 ωy ωy
3 i
3 i
3 i
i
=−

pω p−2 R1 HH
[ωR2 + (p − 1)R1 ] ,
2
y3

2
My y = py3 ω p−1 ωy y + p(p − 1)y3 ω p−2 ωy
i i
i i
i
p−2 R
pω
1 [ωR + (p − 1)R ](H )2 + ωy H
=
2
1
3
y3

.

(2.11)

This yields
2
p2 ω 2p−3 R1 H 2 H
2
Di = My y My y − My y =
[ωR2 + (p − 1)R1 ] .
3
3 3
i i
3 i
y3

Lemma 16. If αi , βi ∈ {±1} and H(y1 , y2 ) = G(α1 y1 + α2 y2 , β1 y1 + β2 y2 ), then



4G
 z z , αj = βj
1 2
H =


0,

αj = −βj .
Consequently, in Case (12 ), sign H = − sign(p − 2).
Proof.

H =

∂2
G(α1 y1 + α2 y2 , β1 y1 + β2 y2 )
2
∂yi

2
2
= αi Gz z + 2αi βi Gz z + βi Gz z
11
12
22
= Gz z + Gz z ± 2Gz z ,
11
22
12
27

(2.12)

where the “+” sign has to be taken if the coefficients in front of yi are equal and the “−”
sign in the opposite case.
The derivatives of G are simple:

Gz1 = p(z1 + z2 )p−2 z1 − (p − 2)z2 ,
Gz2 = −p(p − 1)z2 (z1 + z2 )p−2 ;

Gz z = p(p − 1)(z1 + z2 )p−3 z1 − (p − 3)z2 ,
12
Gz z = −p(p − 1)(p − 2)z2 (z1 + z2 )p−3 ,
12
Gz z = −p(p − 1)(z1 + z2 )p−3 z1 + (p − 1)z2 .
22

Note that Gz z + Gz z = 2Gz z , and therefore,
11
22
12



4G
 z z ,
1 2
H =


0, α = −β

j
j

α j = βj

Now in Case (12 ), we must choose α1 = 1, α2 = 1, β1 = 1 and β2 = −1 for H
to match how the implicit solution was defined in terms of G in Proposition 12. Then
Gz z = −p(p − 1)(p − 2)(y1 − y2 )(2y1 )p−3 .
12

ω 2 − τ 2 from this point on. In Case (12 ), β > p − 1 in the sector
p
p−2
2 + (p − 1)2 ) 2 in p−2 y < y < y .
2
1
p y1 < y2 < y1 . Equivalently, B > (τ
p 1

Remark 17. Let β :=

28

This is an easy application of Proposition 12:

(β + 1)p−1 [β − p + 1] = G(β, 1) =

1
G(y1 + y2 , y1 − y2 )
y3

= (2y1 )p−1 [−(p − 2)y1 + py2 ] > 0.

Before we can compute the signs of My1 y1 , My2 y2 , My3 y3 and D1 we need a technical
lemma.

Lemma 18. If 1 < p < ∞ and τ ∈ R, then

g(β) := −p(p − 2)ωβ −3 (β + 1)p−3 [(τ 2 + p − 1)β 2 − τ 2 (p − 3)β + τ 2 ]

satisfies sign g(β) = − sign(p − 2) in Case (12 ).
Proof. The only terms controlling the sign in g are (p − 2) and the quadratic part, which we
will denote q(β). So all that is needed is to simply determine the sign of q. The discriminant
of q is τ 2 (p − 1)[τ 2 (p − 5) − 4].
If p ≤ 5, then the discriminant of q is negative and so q(β) > 0. If p > 5 and τ 2 (p−5)−4 <
0, then q(β) > 0 once again.
The only case left to consider is if p > 5 and τ 2 (p − 5) − 4 ≥ 0. The zeros of q are given
√
τ 2 (p−3)±|τ | p−1 τ 2 (p−5)−4
. Let β1 , β2 be the zeros such that β2 ≥ β1 . We
by β =
2(τ 2 +p−1)
claim that max{p − 1, β2 } = p − 1. Indeed, p − 1 − β2 > 0

⇐⇒ (p + 1)τ 2 + 2(p − 1)2 > |τ | p − 1 τ 2 (p − 5) − 4
⇐⇒ 4(p − 1)4 + 4τ 2 (p + 1)(p − 1)2 + τ 4 (p + 1)2 > τ 2 (p − 1)(τ 2 (p − 5) − 4)
29

⇐⇒ (p − 1)4 + τ 2 p2 (p − 1) + τ 4 (2p − 1) > 0,

which is obviously true for all τ ∈ R. Now that we have proven the claim, recall that β > p−1,
as shown in Remark 17. Therefore, β > β2 , so q(β) > 0 in this case.
Lemma 19. D1 > 0 in Case (12 ) for all τ ∈ R.
Proof. We use the partial derivatives of G computed in the proof of Lemma 16 to make the
computations of Φ and Φ easier.

Φ(ω) = G(β, 1)
Φ (ω) = pω[β + 1]p−2 [1 − (p − 2)β −1 ]

(2.13)

Φ (ω) = p(β + 1)p−2 [1 − (p − 2)β −1 ] + p(p − 2)ω 2 β −1 [β + 1]p−3 [1 − (p − 2)β −1 ]
+ p(p − 2)ω 2 β −3 [β − 1]p−2
Λ = (p − 1)Φ − ωΦ
= p(p − 2)ωβ −1 (β + 1)p−2 [1 − (p − 2)β −1 ]
− p(p − 2)ω 3 β −3 (β + 1)p−3 [β(β − p + 2) + β + 1]
= p(p − 2)ωβ −2 (β + 1)p−3 [β − p + 2]{β(β + 1) − ω 2 } − p(p − 2)ω 3 β −3 (β + 1)p−2
= p(p − 2)ωβ −3 [β(β − p + 2)(β − τ 2 ) − ω 2 (β + 1)]
= −p(p − 2)ωβ −3 (β + 1)p−3 [(τ 2 + p − 1)β 2 − τ 2 (p − 3)β + τ 2 ]

So we can see that sign Λ = sign g(β) = − sign(p − 2), by Lemma 18. Therefore, sign D1 =
sign H sign Λ = [− sign(p − 2)]2 by (2.12) and Lemma 16.
Since D1 > 0, then all that remains to be checked, for the restrictive concavity of M, is
30

that Myi yi (for i = 1, 2) and My3 y3 have the appropriate signs. But, it turns out that only
for 2 < p < ∞, will these have the appropriate signs.

Lemma 20. sign My1 y1 = sign My2 y2 = sign My3 y3 = − sign(p − 2) in Case (12 ) for all
τ ∈ R. Therefore, M is a partial Bellman function candidate for 2 < p < ∞ but not for
1 < p < 2, since it does not satisfy the required restrictive concavity.

Proof. By (2.10),
My3 y3 =

2
pω p−2 R1 H 2 Λ
3
Φ
y3

Remark 17 gives Φ > 0. From Lemma 18, sign My3 y3 = sign Λ = sign g(β) = − sign(p − 2).
By (2.11), for i = 1or 2,
pω p−2 R1
(ωR2 + (p − 1)R1 )(H )2 + ωy3 H
y3
pω p−2
=
Λ(H )2 + ωy3 H (Φ )2 ,
3
y3 (Φ )

Myi yi =

giving sign My2 y2 = − sign(p − 2).

The previous two lemmas established that the partial Bellman function candidate, from
Case (12 ) satisfies the restrictive concavity property, for 2 < p < ∞. The candidate was constructed using the remaining Bellman function properties, so it is in fact a partial candidate.
Now we will turn our attention to Case (2). As it turns out, Case (22 ) also gives a partial
Bellman function candidate, which, as luck would have it, is the missing half of the partial
Bellman candidate just constructed.
31

2.2.1.2

Case (22 ) for 2 < p < ∞

We can obtain a Bellman candidate from Case (2) without having to separately fix y1 or y2 .
Let us compute the solution in this case.

T3
y

T3
y

U
U
E

y2

y1

E

−y2

y1

Figure 2.3: Sample characteristic of Monge-Amp`re solution in Case (21 )
e

T3
y

U
−y1

y1

E

y2

Figure 2.4: Sample characteristic of Monge-Amp`re solution for Case (22 )
e

Lemma 21. In Case (2) we obtain
p
p
2 ) 2 [y 2 + 2γy y + y 2 ] 2 + c[y − (y − y )p ]
M (y) = (1 + τ
1 2
3
1
2
1
2
2
as a Bellman function candidate, where c > 0 is some constant and γ = 1−τ 2 .
1+τ
32

(2.14)

Proof. In Case (2), on the characteristic yi dti + y3 dt3 + dt0 = 0, y1 and y2 are fixed.
Furthermore, on the characteristic, t0 , ti , t3 are fixed, so we have

M (y) = yi ti + y3 t3 + t0
= (yi ti + t0 ) + y3 t3
= c1 (y1 , y2 ) + c2 (y1 , y2 )y3

Then My3 y3 = 0 and My3 yi = ∂yi c2 . Recall that Di ≥ 0 by Remark 6, so ∂yi c2 (y1 , y2 ) = 0.
This implies that c2 is a constant. Using the boundary data from Proposition 7 gives
p
((y1 +y2 )2 +τ 2 (y1 −y2 )2 ) 2 = M (y1 , y2 , (y1 −y2 )p ) = c1 (y1 , y2 )+c2 (y1 −y2 )p . Solving for
c1 (y1 , y2 ) gives the result. To see that c2 > 0, just notice that as y3 → ∞, M (y) → ∞.
It is not possible to determine if this Bellman function candidate satisfies restrictive
concavity, unless we know the value of the constant c in Lemma 21. This constant can be
computed by using the fact that (2.14) must agree with the partial candidate in Case (12 )
p−2
at y2 = p y1 , if (2.14) is in fact a candidate itself.
p
2 + τ 2) 2
Lemma 22. In Case (22 ), the value of the constant in Lemma 21 is c = ((p − 1)
for 2 < p < ∞.
p
p
2
2
2
Proof. If M (y) = (1 + τ 2 ) 2 [y1 + 2γy1 y2 + y2 ] 2 + c[y3 − (y1 − y2 )p ] (where γ = 1−τ 2 ) is to
1+τ
p−2
be a candidate, or partial candidate, then it must agree at y2 = p y1 , with the solution M
given implicitly by the relation G(y1 + y2 , y1 − y2 ) = y3 G(
p−2
12. At y2 = p y1 ,

ω2 − τ 2 + 1

p−1

ω2 − τ 2 − p + 1 = G
33

ω 2 − τ 2 , 1), from Proposition

ω2 − τ 2, 1

=

Since

ω 2 − τ 2 + 1 = 0,

1
(2y )p−1 [−(p − 2)y1 + (p − 2)y1 ] = 0.
y3 1

1
ω 2 − τ 2 = p − 1, which implies ω = ((p − 1)2 + τ 2 ) 2 . So,

p
((p − 1)2 + τ 2 ) 2 y3 = ω p y3
= M (y1 ,
=

p−2
y ,y )
p 1 3

p−1
2
y
p 1

2

+ τ2

2
y
p 1

p
2 2

+ c y3 −

p
2
y
.
p 1

Now just solve for c.

2.2.1.3

Gluing together partial candidates from Cases (12 ) and (22 )

It turns out that the Bellman function candidate obtained from Case (22 ) is only valid on
part of the domain Ξ+ , since it does not remain concave throughout (for example at or
near (y1 , y1 , y3 )). As luck would have it, the partial candidate has the necessary restrictive
concavity on the part of the domain where the candidate from Case (12 ) left off; i.e. in
p−2
−y1 < y2 < p y1 . This means that we can glue together the partial candidate from Cases
(12 ) and (22 ) to get a candidate on Ξ+ for 2 < p < ∞. The characteristics for this solution
can be seen in Figure 2.5.

Proposition 23. For 2 < p < ∞, γ =

1 − τ2
and τ ∈ R, the solution to the Monge-Amp`re
e
1 + τ2

equation is given by
p
p
p
2 ) 2 [y 2 + 2γy y + y 2 ] 2 + ((p − 1)2 + τ 2 ) 2 [y − (y − y )p ]
M (y) = (1 + τ
1 2
3
1
2
1
2
34

T3
y

¥¥
¥
¥
¥
¥
¥

p−2
y
p y1 1

−y1

E

y2

Figure 2.5: Characteristics of Bellman candidate for 2 < p < ∞
when −y1 < y2 ≤

p−2
p y1 and is given implicitly by

p−2
ω 2 − τ 2 , 1) when p y1 ≤ y2 < y1 , where G(z1 , z2 ) =
1
M (y) p
(z1 + z2 )p−1 [z1 − (p − 1)z2 ] and ω =
. This solution satisfies all properties of
y3
G(y1 + y2 , y1 − y2 ) = y3 G(

the Bellman function.
We already know that the implicit part of the solution has the correct restrictive concavity
property of the Bellman function, as shown in Section 2.2.1.1. However, the restrictive
concavity still needs to be verified for the explicit part. Since the explicit part of the solution
satisfies My3 yi = My3 y3 = 0, Di = 0 for i = 1, 2. So all that remains to be verified for the
restrictive concavity of the explicit part is checking the sign of Myi yi , for i = 1, 2. Observe
that the explicit part can be written as
p
M (y) = [(y1 + y2 )2 + τ 2 (y1 − y2 )2 ] 2 + Cp,τ [y3 − (y1 − y2 )p ].

It is easy to check that My2 y2 ≤ My1 y1 on −y1 < y2 ≤

p−2
p y1 for 2 < p < ∞. So we only

need to find the largest range of τ ’s such that My1 y1 ≤ 0.
Lemma 24. In Case (22 ), My1 y1 ≤ 0 on −y1 < y2 ≤
35

(2.15)

p−2
p y1 for all τ ∈ R.

Proof. Changing coordinates back to x will make the estimates much easier. So we would
like to show that, on 0 ≤ x2 ≤ (p − 1)x1 , we have,

My1 y1 ≤ 0,

(2.16)

p
1
where Cp,τ = ((p − 1)2 + τ 2 ) 2 and p My1 y1 =
p−4
p−2
p−2
(p − 2)(x2 + τ 2 x2 ) 2 (x2 + τ 2 x1 )2 + (1 + τ 2 )(x2 + τ 2 x2 ) 2 − (p − 1)Cp,τ x1 .
1
1
2
2

First, consider 4 ≤ p < ∞. If p = 4, then showing (2.16) is equivalent to

(p − 2)(p − 1 + τ 2 )2 + (1 + τ 2 )((p − 1)2 + τ 2 ) − (p − 1)((p − 1)2 + τ 2 )2 ≤ 0.

Using the fact that (p − 1 + τ 2 )2 ≤ ((p − 1)2 + τ 2 )2 , it suffices to show

(p − 2)((p − 1)2 + τ 2 ) + 1 + τ 2 ≤ (p − 1)((p − 1)2 + τ 2 ) ⇐⇒ 1 ≤ (p − 1)2 .

x
4So we have verified that the estimate is true for all τ. Let s = x2 , then (2.16) simplifies to
1
showing,
4−p
F (s) = (p − 2)(s + τ 2 )2 + (1 + τ 2 )(s2 + τ 2 ) − Cp,τ (p − 1)(s2 + τ 2 ) 2 ≤ 0,

where 0 ≤ s ≤ p−1. For p = 4, F is a quadratic function that is increasing on (

−2τ 2
, p−1).
τ2 + 3

Since F (3) ≤ 0, F (s) ≤ 0 on (0, 3).
Now we will consider 2 ≤ p < 4. Note that F (s) = 0 at p = 2, so we can assume that
36

p = 2. Breaking up the domain of F will make things easier. For s ∈ (1, p − 1), we have the
following estimate, (s + τ 2 )2 ≤ (s2 + τ 2 )2 . Let t = s2 + τ 2 . Then
2−p
1
F (s) ≤ (p − 2)t + 1 + τ 2 − Cp,τ (p − 1)t 2 := g1 (t).
t
Observe that g1 is increasing on 1 + τ 2 ≤ t ≤ (p − 1)2 + τ 2 and g1 ((p − 1)2 + τ 2 ) ≤ 0.
Therefore, F (s) ≤ 0 on (1, p − 1).
Now we will show that F (s) ≤ 0, for s ∈ (0, 1). Let r = τ 2 > 0. Since (s + r)2 ≤ (1 + r)2 ,
it suffices to show

(p − 2)(1 + r)2 + (1 + r)(s2 + r) ≤ (p − 1)((p − 1)2 + r)p/2 (s2 + r)(4−p)/2
⇐⇒ (1 + r)[s2 + (p − 1)r + p − 2] ≤ (p − 1)((p − 1)2 + r)p/2 (s2 + r)(4−p)/2
⇐⇒ H(s) := log(1 + r) + log[s2 + (p − 1)r + p − 2] − log(p − 1)
−

4−p
p
log((p − 1)2 + r) −
log(s2 + r) ≤ 0
2
2

H (s) ≥ 0 and H(1) ≤ 0 will imply H ≤ 0 on (0, 1). H (s) ≥ 0 ⇐⇒ As2 + B ≥ 0, where
A = p − 2 and B = 2r − (4 − p)[(p − 1)r + p − 2]. For this we simply need to check the
conditions under which B ≥ 0. It is easy to see that

B ≥ 0 ⇐⇒ r ≥

(4 − p)(p − 2)
,
2 − (4 − p)(p − 1)

which is satisfied for all r ≥ 0, if p ∈ [2, 3]. Checking that H(1) ≤ 0 is easy:

H(1) =

p
log
2

1+r
(p − 1)2 + r
37

≤0

⇐⇒ 1 ≤ (p − 1)2 ,

which is obviously true for all 2 ≤ p ≤ 3 (or better yet for all p ≥ 2, but we don’t have
the derivative estimate for that large range of p’s). Therefore, F (s) ≤ 0 for s ∈ (0, 1) and
2 ≤ p ≤ 3.

Let us now show that F (s) ≤ 0, for s ∈ (0, 1) and 3 < p < 4.

F (s) ≤ (p − 2)(1 + r)2 + (1 + r)(1 + r) − (p − 1)(4 + r)p/2 r(4−p)/2 ≤ 0
⇐⇒ (1 + r)2 ≤ (4 + r)p/2 r(4−p)/2
⇐⇒ 4 log(1 + r) − 4 log r ≤ p[log(4 + r) − log r]
⇐⇒ p ≥

4 log(1/r + 1)
:= K(r)
log(4/r + 1)

K(r) = 4 and that K decreases rapidly to limr→∞ K(r) = 1. Let
Observe that lim
r→0+
us find where K(r) = 3. This reduces to the equation

8r3 + 42r2 + 60r − 1 = 0,

which has a positive zero of r ≈ 0.0165. Therefore, K(r) ≤ p, for all r ≥ 0.02. This gives us
F (s) ≤ 0, for s ∈ (0, 1), 3 < p < 4 and τ 2 ≥ 0.02.

The proof is finished except that we still need to show that F (s) ≤ 0, for s ∈ (0, 1), with
1 − τ2
1 − τ2
τ 2 ≤ 0.02 and p ∈ (3, 4). We will now proceed to break (0, 1) into (0,
)∪(
, 1)
2
2
and show that F (s) ≤ 0 on each piece separately. Though we only need the estimate for
τ 2 ≤ 0.02, the estimates below work for |τ | ≤ 1.
38

For s ∈ (0,

1 − τ2
), we have the estimate (s + τ 2 )2 ≤ s2 + τ 2 . Let t = s2 + τ 2 . Then
2
2−p
1
2 − C (p − 1)t 2 := g (t).
F (s) ≤ p − 1 + τ
p,τ
2
t

Since g2 is increasing on τ 2 ,
1 − τ2
.
on 0,
2

(1 − τ 2
4

2
+ τ 2 ) and since g2

All that remains is to show that F (s) ≤ 0 on (

(1 − τ 2 )2
+ τ 2 ≤ 0, F (s) ≤ 0
4

1 − τ2
, 1), for p ∈ (3, 4).
2

p
1
F (s) ≤ (1 + τ 2 )2 − ((p − 1)2 + τ 2 ) 2
p−1

(1 − τ 2 )2
4

4−p
2
+ τ2
≤0

4−p
p
4−p
⇐⇒ 4 2 (1 + τ 2 )2 ≤ ((p − 1)2 + τ 2 ) 2 ((1 − τ 2 )2 + 4τ 2 ) 2
p
⇐⇒ 24−p (1 + τ 2 )p−2 ≤ ((p − 1)2 + τ 2 ) 2

Now we perform the second order Taylor expansion of the two terms involving p and τ and
obtain

1
p−4
24−p [1 + (p − 2)τ 2 + (p − 2)(p − 3)ξ2 τ 4 ]
2
p−4
p
2 + τ 2 − 1) + p (p − 2)((p − 1)2 + τ 2 − 1)2 ξ 2 ,
≤ 1 + ((p − 1)
1
2
8

where 1 ≤ ξ2 ≤ 2 and 4 ≤ ξ1 ≤ 10. Using these estimates of ξ1 and ξ2 and considering
p ∈ [3, 4], it suffices to show

3
3
3
2 + 4τ 2 + τ 4 ≤ 1 + (3 + τ 2 ) + (3 + τ 2 )2 ⇐⇒ −71τ 4 − 146τ 2 + 361 ≥ 0,
2
8
10
39

which is obviously true for all |τ | ≤ 1 and p ∈ [3, 4]. So we have shown that F (s) ≤ 0 on
1 − τ2
(
, 1) for p ∈ [3, 4] and |τ | ≤ 1. Thus, F (s) ≤ 0, for s ∈ [0, p − 1] and 2 < p < 4, which
2
completes the proof.
We have now verified that the explicit part of the Bellman function candidate, from
Case (22 ), has the appropriate restrictive concavity. So we have proven Proposition 23, by
Lemmas 19, 20 and 24. Now that we have a Bellman candidate for 2 < p < ∞, we will turn
our attention to p-values in the dual range 1 < p < 2.

2.2.2

The Bellman function candidate for 1 < p < 2

In order to get a Bellman function candidate for 1 < p < 2 we just need to glue together
candidates from Cases (22 ) and (32 ) in almost the same way as we did for 2 < p < ∞ in
Section 2.2.1. Refer to Addendum 1 (Section 2.5) for full details.

T3
y

−y1

2−p
p y1

y1

E

y2

Figure 2.6: Characteristics of Bellman candidate for 1 < p < 2.

Proposition 25. Let 1 < p < 2 and γ =
Monge-Amp`re equation is given by
e
p
p
2
2
M (y) = (1 + τ 2 ) 2 [y1 + 2γy1 y2 + y2 ] 2 +

1 − τ2
1
. If τ 2 ≤ 2p−1 , then a solution to the
2
1+τ
p
2
1
+ τ2
[y3 − (y1 − y2 )p ]
(p−1)2
40

when

2−p
p y1 ≤ y2 < y1

and if τ 2 ≤ p∗ − 1, then the solution is given implicitly by
G(y1 −y2 , y1 +y2 ) = y3 G(1,

2−p
ω 2 − τ 2 ) when −y1 < y2 ≤ p y1 , where ω =

1
M (y) p
.
y3

This solution satisfies all of the properties of the Bellman function (once we restrict the τ values to the smaller set between the two restrictions above).

Most of the remaining cases do not yield a Bellman function candidate. If we fix y2 , then
the Monge-Amp`re solution from Cases (1) and (3) do not satisfy the restrictive concavity
e
needed to be a Bellman function candidate. Case (2) yields the same partial solution if we
first fix y1 or y2 , since restrictive concavity is only valid on part of the domain. So, all
that remains is Case (4). However, we do not know whether or not Case (4) gives a Bellman
function candidate. For τ = 0, it was shown in [37] that Case (4) does not produce a Bellman
function candidate, since some simple extremal functions give a contradiction to linearity of
the Monge-Amp`re solution on characteristics. However, for τ = 0 these extremal functions
e
only work as a counterexample for some p-values and some signs of the martingale transform.
Case (4) could give a solution throughout Ξ+ or could yield a partial solution that would
work well with the characteristics from Case (21 ). Since Case (4) does not provide a Bellman
candidate for τ = 0, we expect the same for small τ. The picture probably changes most
drastically for large τ. But it does not matter, since we will now show that our Bellman
candidate is actually the Bellman function (which we would have to check anyways because
of the added assumption). The details for the remaining cases that do not yield a Bellman
function candidate are in Section 2.6.
41

2.3

Monge-Amp`re solution is the Bellman function
e

We will now show that the Monge-Amp`re solution obtained in Proposition 23 and 25 is
e
actually the Bellman function. To this end, let us revert back to the x-variables. We
will denote the Bellman function candidate as Bτ and use Bτ to denote the true Bellman
function. Extending the function G on part of Ω+ to Uτ on all of Ω, appropriately, makes
it possible to define the solution in terms of a single relation.
p
p
Definition 26. Let v(x1 , x2 ) := vp,τ (x1 , x2 ) = (τ 2 |x1 |2 +|x2 |2 ) 2 −((p∗ −1)2 +τ 2 ) 2 |x1 |p ,
p−2
2
2
u(x1 , x2 ) := up,τ (x1 , x2 ) = p(1− 1 )p−1 1 + ∗τ 2
(|x1 |+|x2 |)p−1 [|x2 |−(p∗ −
∗
p
(p −1)
1)|x1 |] and

U (x1 , x2 ) := Up,τ (x1 , x2 ) =





v(x1 , x2 )

: |x2 | ≥ (p∗ − 1)|x1 |



u(x , x ) : |x | ≤ (p∗ − 1)|x |.
 1 2
2
1
for 1 < p < 2. For 2 < p < ∞ we interchange the two pieces in U.

Remark 27. Wherever B, U, u and v are considered, we are assuming that if 1 < p < 2, then
τ2 ≤

1
and if 2 ≤ p < ∞, then τ ∈ R.
2p − 1

Proposition 28. For 1 < p < 2 and τ 2 ≤

1
or 2 < p < ∞ and τ ∈ R the Bellman
2p − 1

function candidate is the unique positive solution given by
1
p
U (x1 , x2 ) = U x3 ,
Furthermore, U is C 1 -smooth on Ω.
42

2
2
p
2x p
Bτ − τ 3

.

Proof. First consider 2 ≤ p < ∞. It is clear that
1
p
U (x1 , x2 ) = U x3 ,

2
2
p
2x p
Bτ − τ 3

,

(2.17)

by comparing the solution obtained in Proposition 23 and using the symmetry property in
p−2
1 p−1
τ2
2
Proposition 7. The constant αp,τ = p 1 − ∗
1+
was determined so
p
(p∗ − 1)2
that Ux = Uy at |y| = (p∗ − 1)|x|. The partial derivatives are given by,

ux1 = αp,τ [(p − 1)x1 (|x1 | + |x2 |)p−2 (|x2 | − (p∗ − 1)|x1 |) − (p∗ − 1)x1 (|x1 | + |x2 |)p−1 ],
p
p−2
2 x (τ 2 |x |2 + |x |2 ) 2 − px ((p∗ − 1)2 + τ 2 ) 2 |x |p−1 ,
vx1 = pτ 1
1
1
2
1
ux2 = αp,τ (p − 1)x2 (|x1 | + |x2 |)p−2 (|x2 | − (p∗ − 1)|x1 |) + αp,τ x2 (|x1 | + |x2 |)p−1 ,
p−2
vx2 = px2 (τ 2 |x1 |2 + |x2 |2 ) 2 ,
x
x
where x1 = 1 and x2 = 2 . U is C 1 -smooth, except possibly at gluing and symmetry
|x1 |
|x2 |
lines. It is easy to verify that ux is continuous at {x1 = 0}, Ux1 and Ux2 are continuous at
{|x2 | = (p∗ − 1)|x1 |} and vx2 is continuous at {x2 = 0}. This proves that U is C 1 -smooth
on Ω.

Observe that Ux2 > 0 for x2 = 0 and Ux1 < 0 for x1 = 0. This is enough to show that
1
p
Bτ is the unique positive solution to (2.17). Indeed, if x ∈ Ω such that |x1 | = x3 , then
2
2
p
2 x p = |x | by the Dirichlet boundary conditions. This gives us (2.17) uniquely at
Bτ − τ 3
2




1
1
2
2
2
2
 p

p
p
p 
p
p 
Bτ (x). Fix x1 , such that |x1 | < x3 . Then U x3 , Bτ − τ 2 x3  < U x1 , Bτ − τ 2 x3  .
43

Since x1 is fixed,

U (x1 , x2 ), as


2
2

p
p
Bτ − τ 2 x3 > |x2 |, so U x1 ,


2
2
p
p 
Bτ − τ 2 x3  strictly decreases to

2
2
p
2 x p decreases to |x |, giving us a unique Bτ (x) for which (2.17)
Bτ − τ 3
2

holds.
Now consider 1 < p < 2. U is C 1 -smooth on Ω, since vx1 is continuous at {x1 = 0}, ux2
is continuous at {x2 = 0} and Ux1 and Ux2 are continuous at {|x2 | = (p∗ − 1)|x1 |}. This is
easily verified since the partial derivatives are computed above (just switch the two pieces of
each function). Observe that for x1 = 0 and x2 = 0, Ux1 < 0 and for x2 = 0, Ux2 > 0. Then


1
2
2
 p
p
p 
the argument above showing U (x1 , x2 ) = U x3 , Bτ − τ 2 x3  uniquely determines
Bτ also holds for this range of p-values as well, except maybe at x1 = x2 = 0. Suppose


1
2
2
p
 p
p
p 
U (0, 0) = U x3 , Bτ − τ 2 x3  . Then Bτ (x) = ((p∗ − 1)2 + τ 2 ) 2 x3 . So Bτ (x) is
uniquely determined by the fixed x-value.
1
Corollary 29. Bτ is continuous in Ω, for τ 2 ≤ 2p−1 and 1 < p < 2 or τ ∈ R and
2 ≤ p < ∞.
x2 = (p∗ − 1)x1

x2
T

d
d

I

d
d

E d

 
 
 

I
 
 

E

d 
Ex
1
 d
E  
d E
Id
  I
 
d
 
d x2
 
d

= −(p∗ − 1)x1

Figure 2.7: Location of Implicit (I) and Explicit (E) part of Bτ for 2 ≤ p < ∞.

Proof. In this proof only we will revert back to the notation Up,τ , rather than U, to make
44

clear the distinction when τ = 0 or τ = 0. We only consider 2 < p < ∞ as the dual range
is handled identically. By Proposition 28, we have that Bτ is the unique positive solution
2
2 p
p
2 x p 2 on |x | ≥ (p∗ − 1)|x |,
to (2.17). Since this is true for all τ ∈ R, B0 = Bτ − τ 3
2
1
p−2
2
2
because Up,τ = 1 + ∗τ 2
Up,0 . Equivalently, we have
(p −1)
p
2
2 2
p
p
Bτ = B0 + τ 2 x3  .


(2.18)

Since B0 was shown to be continuous in [37] (pg. 26), Bτ is also continuous on |x2 | ≥
(p∗ − 1)|x1 |, using the relation. This takes care of the implicit part of Bτ . The explicit part
of Bτ is clearly continuous on |x2 | ≤ (p∗ − 1)|x1 |.
1
Lemma 30. Let τ 2 ≤ 2p−1 and 1 < p < 2 or τ ∈ R and 2 ≤ p < ∞. Then, Bτ L is
C 1 -smooth on Ω, where L is any line in Ω.
Proof. Since Bτ L is C 2 -smooth on Ω+ , all that remains to be checked is the smoothness
at the gluing and symmetry lines; i.e., at {x1 = 0}, {x2 = 0} and {|x2 | = (p∗ − 1)|x1 |}.
Let L = L(t), t ∈ R, be any line in Ω passing through any of the planes in question, such
that L(0) is on the plane. Now plug L(t) into (2.17) and differentiate with respect to t. Let
t → 0+ and t → 0− and equate the two relations. This gives

d
d
Bτ (L(t))
− = dt Bτ (L(t)) t=0+ .
t=0
dt
1
Proposition 31. (Restrictive Concavity) Let 1 < p < 2 and τ 2 ≤ 2p−1 or 2 ≤ p < ∞
and τ ∈ R. Suppose x± ∈ Ω such that x = α+ x+ + α− x− , α+ + α− = 1. If |x+ − x− | =
1
1
|x+ − x− |, then Bτ (x) ≥ α+ Bτ (x+ ) + α− Bτ (x− ).
2
2
45

Proof. Recall that Propositions 23 and 25, together with the symmetry property of Bτ ,
establish this result everywhere, except at {x1 = 0}, {x2 = 0} and {|x2 | = (p∗ − 1)|x1 |}.
Let f (t) = Bτ L(t) , where L is any line in Ω, such that L(0) ∈ {x1 = 0}, {x2 = 0} or
{|x2 | = (p∗ − 1)|x1 |}. Since f < 0 for t < 0 and t > 0 and f is C 1 -smooth (by Lemma 30),
f is concave.
Proposition 32. Let 1 < p < ∞. If a function B has restrictive concavity and satisfies
p
the Dirichlet estimate Bτ (x1 , x2 , |x1 |p ) ≥ (τ 2 x2 + x2 ) 2 , then Bτ ≥ Bτ . In particular,
2
1
1
Bτ ≥ Bτ , for τ 2 ≤ 2p−1 and 1 < p < 2 or τ ∈ R and 2 ≤ p < ∞.
Proof. This was proven in [37] for B0 (Lemma 2 on page 29). The same proof will apply
here to Bτ .
1
Proposition 33. For τ 2 ≤ 2p−1 and 1 < p < 2 or τ ∈ R and 2 ≤ p < ∞, Bτ ≤ Bτ .
Proof. For 1 < p < 2 there is a direct proof, which will be discussed first. By (2.18) we
2 p
2
2 p
2
p
p
p
p
know that B0 = Bτ − τ 2 x3 2 on {|x2 | ≤ (p∗ − 1)|x1 |}. Consider, B0 = Bτ − τ 2 x3 2 .
It suffices to show that B0 ≤ B0 . But, B0 = B0 (as Burkholder showed), so without the
supremum’s we can reduce to simply showing
2
2
p
p p + τ 2 |f |p p ≤ (τ 2 |f |2 + |g|2 ) 2 .
|g| I
I
I

I (A, C) 2 ≤ I (A, C)
lp
p proves the result. So we have shown that B ≤ B
C = |τ f |
τ
τ
Apply Minkowski’s inequality:

p
2 . Choosing A = |g| and
lp
on {|x2 | ≤ (p∗ − 1)|x1 |}.

Now we would like to show that Bτ ≤ Bτ on {|x2 | ≥ (p∗ − 1)|x1 |}. Let H1 (x1 , x2 , x3 ) =
Bτ (x1 , x2 , x3 )−Bτ (0, 0, 1)x3 . Lemma 38, in the next section, proves that H1 (x1 , x2 , ·) is an
increasing function starting at H1 (x1 , x2 , |x1 |p ) = vτ (x1 , x2 ) and increasing to Up,τ (x, y) :=
46

supt≥|x|p {Bτ (x, y, t) − Bτ (0, 0, 1)t}. The same proof works for

H2 (x1 , x2 , x3 ) = Bτ (x1 , x2 , x3 ) − Bτ (0, 0, 1)x3 .

So
H2 (x1 , x2 , x3 ) ≥ vτ (x1 , x2 ) = Bτ (x1 , x2 , x3 ) − Bτ (0, 0, 1)x3 .
Since Bτ (0, 0, 1) ≤ Bτ (0, 0, 1), Bτ ≤ Bτ on {|x2 | ≥ (p∗ − 1)|x1 |}.
Now we consider 2 < p < ∞. Let ε > 0 be arbitrarily small and consider the following
extremal functions



 −c




t−ε
f (x) =
 γf 1−2ε




 c


 d
 −



t−ε
g(x) =
 γg 1−2ε




 d
+

:1<x<ε
:ε<x<1−ε
: 1 − ε < x < 1,

:1<x<ε
:ε<x<1−ε
: 1 − ε < x < 1,

where c, d± and γ are defined so that f and g are a pair of test functions at (0, x2 , x3 ). We
can use f and g to show, just as in [37] (Lemma 3, pg. 30), that

Bτ (0, x2 , x3 ) ≤ Bτ (0, x2 , x3 ).

(2.19)

Now we need to take care of the estimate when x1 = 0. Making a change of coordinates from x to y we only need to consider y ∈ Ξ+ , by the symmetry property of the
Bellman function and Bellman function candidate. So far we have that Mτ (y1 , y1 , y3 ) ≤
47

Mτ (y1 , y1 , y3 ) by (2.19). The Dirichlet boundary conditions give that M (y1 , y2 , (y1 −
p−2
y2 )p ) = M(y1 , y2 , (y1 −y2 )p ). On any characteristic in { p y1 ≤ y2 ≤ y1 }, see Figure 2.5,
Mτ is linear (since it is the Monge-Amp`re solution) and Mτ is concave (by Proposition 5).
e
p−2
Therefore, Mτ (y1 , y2 , y3 ) ≤ Mτ (y1 , y2 , y3 ) on { p y1 ≤ y2 ≤ y1 }. For the remaining part
of Ξ+ , we can use the same proof as for 1 < p < 2 to get Mτ (y1 , y2 , y3 ) ≤ Mτ (y1 , y2 , y3 )
on {−y1 ≤!y2 ≤

p−2
p y1 }.

Now that we have proven B = B, we will mention another surprising fact.

Definition 34. We define B l = B l (x1 , x2 , x3 ) as the least restrictively concave majorant of
p
(x2 + τ 2 x2 ) 2 in Ω.
2
1

1
Proposition 35. For 1 < p < 2 and τ 2 ≤ 2p−1 or 2 ≤ p < ∞ and τ ∈ R we have
B = B = Bl .

Proof. First we will show B ≥ B l . By Minkowski’s inequality,

B(x1 , x2 , x3 ) = B(x1 , x2 , x3 ) ≥

p
(|g|2 + τ 2 |f |2 ) 2 =

p
(g, τ f ) 2 ≥
l

p
(g, τ f )

p
p
2 + τ 2 | f |2 ) 2 = (x2 + τ 2 x2 ) 2 ,
= (| g |
1
2

l2

proving the estimate.
Conversely, one can show that B ≤ B l just as B ≤ B was shown in Proposition 32 (simply
apply the same argument to B l using the restrictive concavity of this function).
48

2.4

Proving the main result

Now that we have the Bellman function, the main result can be proven without too much
difficulty. But first, we will find another relationship between U and v. Quite surprisingly,
U is the least zigzag-biconcave majorant of v.
Definition 36. A function of (x, y) that is biconcave in (x+y, x−y) we call zigzag-biconcave.
Remark 37. For this section, while proving the main result, we will stop using (x1 , x2 , x3 )
and (y1 , y2 , y3 ) coordinates to denote a rotation by π . Definition 36 makes the rotation
4
clear and we can simply use (x, y, t) as the coordinates and relax the use of subscripts.
Lemma 38. Let 1 < p < ∞ and U (x, y) = supt≥|x|p {Bτ (x, y, t) − Bτ (0, 0, 1)t}. Fix (x, y).
The function H(x, y, t) = Bτ (x, y, t) − Bτ (0, 0, 1)t is increasing in t from H(x, y, |x|p ) =
p
p
v(x, y) := (τ 2 |x|2 + |y|2 ) 2 − ((p∗ − 1)2 + τ 2 ) 2 |x|p to Up,τ (x, y).
Proof. Recall that Bτ is continuous in Ω and for (x, y) fixed, Bτ (x, y, ·) is concave. Then
H(x, y, ·) is also concave. Since Up,τ (x, y) = supt≥|x|p {Bτ (x, y, t) − Bτ (0, 0, 1)t}, it either
increases to U (x, y), or there exists t0 such that H(x, y, t0 ) = U (x, y) and H is decreasing
for t > t0 . If H is decreasing for t > t0 , then H → −∞ as t → ∞ by concavity. Then there
exists ε > 0 and t > t0 such that H(x, y, t ) < εt . So we have, lim supt→∞

H(x,y,t)
< −ε.
t

But,

lim
t→∞







H(x, y, t)
x y
= lim Bτ  , , 1 − Bτ (0, 0, 1) = 0,
1 1
t
t→∞
tp tp

by continuity of Bτ at (0, 0, 1). This gives us a contradiction. Therefore, H(x, y, t) ≥ −εt,
for all t and all ε > 0; i.e., H is non-negative concave function on [|x|p , ∞). So H(x, y, ·)
is increasing and H(x, y, |x|p ) = vp,τ (x, y) by the Dirichlet boundary conditions of Bτ in
Proposition 7.
49

1
Proposition 39. For 1 < p < 2 and τ 2 ≤ 2p−1 or 2 ≤ p < ∞ and τ ∈ R, Up,τ (x, y) =
Up,τ (x, y).

Proof. Suppose 2 ≤ p < ∞ and |y| ≥ (p − 1)|x|. Then

U0 (x, y) = lim (B0 (x, y, t) − B0 (0, 0, 1)t)
t→∞
x , y , 1 − B (0, 0, 1)
B0
0
1/p t1/p
t
= lim
1/t
t→∞
=

d
B (u1/p x, u1/p y, 1)
.
du 0
u=0

Now we repeat the same steps and obtain

Uτ (x, y) = lim (Bτ (x, y, t) − Bτ (0, 0, 1)t)
t→∞
p/2
d
2/p
B0 (u1/p x, u1/p y, 1) + τ 2
du
u=0



p−2
2−p 1
2
2
1
1
1
1
 p 1

d
p
= B0 (u p x, u p y, 1) + τ 2
B0
(u p x, u p y, 1) B0 (u p x, u p y, 1)


du

=

=

τ2
1+
(p − 1)2

p−2
2
U0 (x, y) =

τ2
1+
(p − 1)2

u=0

p−2
2
U0 (x, y),

where the last equality is by [15]. Therefore, Uτ (x, y) = Uτ (x, y).
Now suppose |y| ≤ (p − 1)|x|. Looking at the explicit form of Bτ in the region, note that
Bτ (x, y, ·) is linear. So

Uτ (x, y) = sup {Bτ (x, y, t) − Bτ (0, 0, 1)t}
t≥|x|p
50

= sup {Bτ (x, y, 0)} = vτ (x, y) = Uτ (x, y).
t≥|x|p
We can apply the same proof to show that Uτ (x, y) = Uτ (x, y) for 1 < p < 2.
1
Proposition 40. For τ 2 ≤ 2p−1 and 1 < p < 2 or τ ∈ R and 2 ≤ p < ∞, U is the least
p
p
zigzag-biconcave majorant of v(x, y) = (y 2 + τ 2 x2 ) 2 − ((p∗ − 1)2 + τ 2 ) 2 |x|p .
Proof. Recall the following facts just proved in Lemma 38 and Proposition 39:

U (x, y) =

p
sup
{B(x, y, t) − B(0, 0, 1)t} ≥ v(x, y) = (y 2 + τ 2 x2 ) 2 − Cp t,
t:(x,y,t)∈Ω

p
where Cp = ((p∗ − 1)2 + τ 2 ) 2 .
Suppose that w is a zigzag-biconcave function such that v ≤ w ≤ U. Then W (x, y, t) :=
p
2 + τ 2 x2 ) 2 .
w(x, y) + Cp t has restrictive concavity and W (x, y, t) ≥ v(x, y) + Cp t = (y
Therefore, by Proposition 32, we have W ≥ B. So we have

w(x, y) =
≥

sup
{W (x, y, t) − Cp t}
t:(x,y,t)∈Ω
sup
{B(x, y, t) − Cp t} = U (x, y).
t:(x,y,t)∈Ω

We now have enough machinery to easily prove the main result, in terms of the Haar
expansion of a R-valued Lp function.
1
Theorem 41. Let 1 < p < 2 and τ 2 ≤ 2p−1 or 2 ≤ p < ∞ and τ ∈ R. Let f, g :
[0, 1] → R. If | g [0,1] | ≤ (p∗ − 1)| f [0,1] | and |(f, hJ )| = |(g, hJ )| for all J ∈ D, then
p
p
(τ 2 |f |2 + |g|2 ) 2 [0,1] ≤ (p∗ − 1)2 + τ 2 2 |f |p [0,1] , where ((p∗ − 1)2 + τ 2 ) is the sharp
1
constant and p∗ − 1 = max p − 1, p−1 .
51

Proof. Suppose that 2 ≤ p < ∞ and τ ∈ R. The proof relies on the fact that B = B
(Propositions 39 and 33) and U (x, y) = supt≥|x|p {B(x, y, t) − B(0, 0, 1)t} (Lemma 39 and
Proposition 38).
Since |y| ≤ (p∗ − 1)|x| on Ω,
p
p
U (x, y) = v(x, y) = (|y|2 + τ 2 |x|2 ) 2 − ((p∗ − 1)2 + τ 2 ) 2 |x|p ≤ 0.

Then,
sup
{B(x, y, t) − B(0, 0, 1)t} ≤ 0.
p
t>|x|
|y|≤(p∗ −1)|x|
But, U (0, 0) = 0. Therefore

sup
t>|x|p
|y|≤(p∗ −1)|x|

p
B(x, y, t)
= B(0, 0, 1) = ((p∗ − 1)2 + τ 2 ) 2 .
t

(2.20)

Observing the relationship B = B, gives the desired result.
1
For 1 < p < 2, τ 2 ≤ 2p−1 and |y| ≤ (p∗ − 1)|x|,

1
U (x, y) = p 1 − ∗
p

1+

τ2

p−2
2

(p∗ − 1)2

(|x| + |y|)p−1 [|y| − (p∗ − 1)|x|] ≤ 0,

so we have (2.20) by the same reasoning as for 2 ≤ p < ∞.

Remark 42. Note that Minkowski’s inequality together with Burkholder’s original result
gives the same upper estimate for 2 ≤ p < ∞.
Indeed, if f ∈ Lp [0, 1] and g is the corresponding martingale transform, then Minkowski’s
52

inequality gives,
p
2 + τ 2f 2 2
g

p
p
≤ ( g2 p + τ 2f 2 p ) 2 = ( g 2 p + τ f 2 p ) 2
p
L
L
L2
L2
L2
p
p
≤ f p ((p∗ − 1)2 + τ 2 ) 2 .
L

p
This is very surprising in the sense that the “trivial” constant (p∗ − 1)2 + τ 2 2 is
actually the sharp constant.
Now we will prove the main result for Hilbert-valued martingales. The same ideas can
be used to extend the previous result to Hilbert-valued Lp -functions as well. Let H be a
separable Hilbert space with

· H as the induced norm.

Theorem 43. Let 1 < p < ∞, (W, F, P) be a probability space and {fk }
, {g }
:
k∈Z+ k k∈Z+
W → H be two H-valued martingales with the same filtration {Fk }
. Denote dk =
k∈Z+
fk − fk−1 , d0 = f0 , ek = gk − gk−1 , e0 = g0 as the associated martingale differences. If
ek (ω) H ≤ dk (ω) H , for all ω ∈ W and all k ≥ 0, then we have the following estimate
n

n
ek , τ

k=0

dk
k=0

Lp (W,H2 )

p
∗ − 1)2 + τ 2 2
≤ (p

n
dk
k=0

Lp (W,H)

,

p
∗ − 1)2 + τ 2 2 as the best possible constant for and τ 2 ≤ 1 and 1 < p < 2 or
with (p
2p−1
1
τ ∈ R and 2 ≤ p < ∞, where p∗ − 1 = max{p − 1, p−1 }.

1
In the theorem, “best possible” constant means that if Cp,τ < ((p∗ − 1)2 + τ 2 ) 2 , then
for some probability space (W, G, P ) and a filtration F, there exists H-valued martingales
53

{f }k and {g}k , such that

gk , τ f k

Lp ([0,1],H2 )

> Cp,τ fk Lp ([0,1],H) .

Proof. We will prove the result for 2 ≤ p < ∞, since the result for 1 < p < 2 is similar.
Replace | · | with

· H , in Up,τ . Let fn =

n d and g =
n
k=0 k

n e . Recall that
k=0 k

U := Up,τ is the least zigzag-biconcave majorant of v := vp,τ . As in [16] (pages 77-79),

Up,τ (x + h, y + k) ≤ Up,τ (x, y) + (∂x Up,τ , h) + (∂y Up,τ , k),

(2.21)

for all x, y, h, k ∈ H, such that |k| ≤ |h| and x + ht H x + kt H > 0. The result in
(2.21) follows from the zigzag-biconcavity and implies that E[U (fk , gk )] is a supermartingale.
Lemma 38 gives that v(fn , gn ) ≤ U (fn , gn ). Therefore,

E[v(fn , gn )] ≤ E[U (fn , gn )] ≤ E[U (fn−1 , gn−1 )] ≤ · · · ≤ E[U (d0 , e0 )].

But, E[U (d0 , e0 )] ≤ 0 in both pieces of Uτ since 2 − p∗ ≤ 0 and e0 H ≤ d0 H . Thus,
E[vτ (fn , gn )] ≤ 0. The constant, in the estimate, is best possible, since it was attained in
Theorem 41.

The following construction originated from a conversation with Fedja Nazarov.
1
Remark 44. For 1 < p < 2 and τ 2 > 2p−1 , chosen sufficiently large, the “trivial” constant,
p
(p∗ − 1)2 + τ 2 2 , in the main result is no longer sharp because of a “phase transition”. To
give a sense of why this is true one can show that for 1 < p < 2 fixed, the constant is no
longer sharp for τ sufficiently large.
54

Let us construct such a function f to do this. First of all, fn ∈ Lp [0, 1] will be chosen so
that fn = 0 a.e. Let Cp = (p∗ − 1). Note that

p
(|gn |2 + τ 2 |fn |2 ) 2 = |τ |p

|fn |p

= |τ |p

1 |gn |2
1+
|τ |2 |fn |2

|fn |p +

=: |τ |p

|gn |4
|fn |4−p

p
2− 2
1
1 + θn (x, τ )

|fn |p + A + B

|fn |p = |τ |p

2
Cp
1+
|τ |2

= |τ |p

p
((p∗ − 1)2 + τ 2 ) 2

|gn |2
|fn |2−p

p
2|τ |2−p

p p
1
+
−1
2 2
2|τ |4−p

p
2

|fn |p +

p
2

|fn |p

2
pCp
2|τ |2−p

4
Cp
p p
−1
+
2 2
2|τ |4−p
=: |τ |p

|fn |p
|fn |p

p
2− 2
1
˜
1 + θn

|fn |p + C + D,

˜
where θn (x, τ ), θn ≥ 0.

Choose fn = χ 1 1
−χ 3 1
− εn χ 1
+ εn χ 1 3
,
5
3
,4 ∪ 8,3
, 2 ∪ 7 ,1
0, 8 ∪ 1 , 5
,8 ∪ 4,7
8
4
8
8
2 8
4
8
where εn > 0 is small. On the sets where |fn | = εn , we can choose the martingale transform
gn of fn not small. Indeed, without loss of generality choose x ∈ 0, 1 and denote J1 =
8
0, 1 . Then
4
(f, hI )hI (x) = −εn .

fn (x) =
I:I⊃J1

55

Define a martingale transform gn as

gn (x) =
I:I J1

(f, hI )hI (x) − (f, hJ )hJ (x).
1
1

Then
fn (x) − gn (x) = 2(fn , hJ )hJ (x) = (2 − εn ) |J1 | −
1
1

1
.
|J1 |

Therefore, fn (x) = −εn , yet its martingale transform gn (x) = 2−2εn , for x ∈ J1 . The same
can be done for other intervals of smallness of fn . Note that

|gn |2 =

|fn |2 →

1
|f |2 = 2 ,

1
|gn |2 ≈ 2 , for n sufficiently large.
p
p
(g 2 + τ 2 f 2 ) 2 > ((p∗ −1)2 +τ 2 ) 2 |f |p it suffices to show A+B > C +D.

if we choose εn → 0. So
To show that

As D ≤ 0 it is enough to prove A+B −C > 0. Let A = τ 2−p A, C = τ 2−p C, B = τ 4−p B.
Regardless of the choice of τ we have A > 2C if n is chosen sufficiently large. In fact looking
at A we see that it is bigger than the integral, where the integrand has numerator close to
2 and denominator equal εn . On the other hand C involves just an integral with uniformly
(in n) bounded integrand. Then we fix n, of course |B | is very large, but we notice that
choosing τ to be very large makes the following inequality true:

A+B−C =

1
1
[(A − C ) −
|B |] > 0 .
2−p
|τ |2
|τ |

This completes the example.

Remark 45. We would like to point out that that as p approaches 2 from the left that the τ
values for the main result do not improve to infinity, as we expect. So either there can be
some improvement in the range of τ -values for which we have the Bellman function, or the
56

operator which we are studying in this chapter behaves in a very nonintuitive way.

2.5

Proof of Proposition 25

Throughout this section the arguments may seem brief in comparison to Section 2.2.1. The
reason for this is because we cover the exact same argument as in Section 2.2.1, only with
slightly different cases. So if any arguments are unclear, then returning to Section 2.2.1
should help to clear up any difficulties. We will first consider Case (32 ) to get a partial
Bellman function candidate.

2.5.1

Considering Case (32 )

T3
y

U
−y1

y1

E

y2

Figure 2.8: Sample characteristic of Monge-Amp`re solution in Case (32 )
e

Proposition 46. For 1 < p < ∞ and −y1 < y2 <
relation G(y1 − y2 , y1 + y2 ) = y3 G(1,

ω 2 − τ 2 ).

This is proven through a series of Lemmas.

57

2−p
p y1 , M is given implicitly by the

Lemma 47. M (y) = t2 y2 + t3 y3 + t0 on the characteristic y2 dt2 + y3 dt3 + dt0 = 0 can be
(y1 +u)2 +τ 2 (y1 −u)2 p
simplified to M (y) =
y3 , where u is the unique solution to the
y1 −u
2
2
y2 +(1− p )y1
u+(1− p )y1
2−p
equation
and −y1 < y2 < p y1 .
=
y3
(y1 −u)p

Proof. Any characteristic, in Case(32 ), goes from U = y1 , u, (y1 −u)p to W = (y1 , −y1 , w).
Recall the properties of the Bellman function we derived in Proposition 7, as we will be using
them throughout the proof. Using the Neumann property and the property from Proposition
9, we get My1 = −My2 = −t2 at W. By homogeneity at W we get

−py1 t2 + pwt3 + pt0 = pM (W ) = y1 My1 + y2 My2 + py3 My3 = −2y1 t2 + pwt3 .

Now we follow the same idea as in Lemma 13, to get

M (y) =

(y1 + u)2 + τ 2 (y1 − u)2 p
y3 ,
y1 − u

where u = u(y1 , y2 , y3 ) is the solution to the equation
2
2
y2 + (1 − p )y1
u + (1 − p )y1
=
.
y3
(y1 − u)p

(2.22)

2
2
Fix u = −(1 − p )y1 . Then we see that y2 = −( p − 1)y1 = u is also fixed by (2.22). This
means that the characteristics must lie in the sector shown in Figure 2.9, since they go from
U to W ∈ {y2 = −y1 }.

The same argument as in Lemma 13 can be used to verify that

equation (2.22) has a unique solution in the sector −y1 < y2 <
58

2−p
p y1 .

y2

y = y1

  2
 
 
y
 
 2
 
 

d
d
d
d
d y2
d
T

2−p
p y1

=

Ey

1

= −y1

Figure 2.9: Range of characteristics in Case (32 ) for 1 < p < 2.

(y1 +u)2 +τ 2 (y1 −u)2 p
y3 can be rewritten as
y1 −u

Lemma 48. M (y) =

G(y1 − y2 , y1 + y2 ) = y3 G(1,

ω 2 − τ 2 ), for − y1 < y2 <

2−p
y .
p 1

1
M (y) p
=
y3

(y1 +u)2 +τ 2 (y1 −u)2
Proof. ω =
≥ |τ |
y1 −u
2 2
Since y1 ± u ≥ 0 and sinceω 2 − τ 2 ≥ 0, u = ω −τ −1 y1 by inversion. Substituting
ω 2 −τ 2 +1
2 )y
2 )y
y2 +(1− p 1
u+(1− p 1
gives
u into
=
y3
(y1 −u)p
p−1
2p−1 y1 [py2 + (p − 2)y1 ] = y3 (

ω 2 − τ 2 + 1)p−1 [ ω 2 − τ 2 − (p − 1)]

or (x1 + x2 )p−1 [(p − 1)x2 − x1 ]

=

1/p
1/p p−1
1/p
1/p
B 2/p − τ x3 2 + x3
(p − 1) B 2/p − τ x3 2 − x3 .

1/p
Thus, G(x1 , x2 ) = G x3 ,

1/p
B 2/p − τ x3 2 or equivalently

G(y1 − y2 , y1 + y2 ) = y3 G(1,
59

ω 2 − τ 2 ).

As before, we must verify that this partial Bellman function candidate has the restrictive
concavity property, so y1 is no longer fixed. To check restrictive concavity, we must show
that My1 y1 ≤ 0, My2 y2 ≤ 0, My3 y3 ≤ 0 and D1 ≥ 0 (note that D2 = 0 by assumption).
These estimates are verified in the following series of lemmas.
Lemma 49. In Case (32 ) we choose H(y1 , y2 ) = G(y1 − y2 , y1 + y2 ) because of how the
implicit solution is defined and obtain sign H = − sign(p − 2).
Proof. We already computed



4G
 z z , αj = βj
1 2
H =


0,

αj = −βj
in Lemma 16. Since, α1 = 1, α2 = −1, β1 = 1 and β2 = 1 then Gz z = −p(p − 1)(p −
12
2)(y1 + y2 )(2y1 )p−3 .
2−p
1
Remark 50. In Case (32 ), β > p−1 in the sector −y1 < y2 < p y1 , where β :=
p
2−p
Equivalently, B(x1 , x2 , x3 ) ≥ ((p∗ − 1)2 + τ 2 ) 2 x3 in −y1 < y2 < p y1 .

ω2 − τ 2.

This is trivial since

(β + 1)p−1 [1 − (p − 1)β] = G(1, β) =

1
G(y1 − y2 , y1 + y2 )
y3

= (2y1 )p−1 [(p − 2)y1 + py2 ] < 0.

Now we have enough information to check the sign of D1 . We will start limiting the
values of τ, since it will be essential for having the restrictive concavity of the partial Bellman
60

candidate from Case (22 ).
1
Lemma 51. D1 > 0 in Case (32 ) for all τ -values such that p−1 ≥ τ 2 .
Proof. We use the partial derivatives of G computed in the proof of Lemma 16 to make the
computations of Φ and Φ easier.

Φ(ω) = G(1, β)
Φ (ω) = −p(p − 1)ω[β + 1]p−2
Φ (ω) = −

(2.23)

p(p − 1)(1 + β)p−3
β(1 + β) + (p − 2)ω 2
β

Λ = (p − 1)Φ − ωΦ
p(p − 1)ω(1 + β)p−3
β(1 + β) + (p − 2)ω 2
= −p(p − 1)2 ω(β + 1)p−2 +
β
p(p − 1)ω(1 + β)p−3
−(p − 1)(1 + β)β + β(1 + β) + (p − 2)ω 2
=
β
p(p − 1)(p − 2)ω(1 + β)p−3 [β − τ 2 ]
=−
(2.24)
β
Thus, sign D1 = sign H sign Λ = [− sign(p − 2)]2 sign(β − τ 2 ) by (2.12) and Lemma 49.
1
In order to have D1 > 0, we must have that β > τ 2 . By Remark 50, β > p−1 . So if we
1
impose that p−1 ≥ τ 2 then we will have β > τ 2 and therefore D1 > 0.
The following lemma restricts the p-values for which our solution is a Bellman function
candidate to 1 < p < 2.
Lemma 52. sign My1 y1 = sign My2 y2 = sign My3 y3 = sign(p − 2) in Case (32 ) for all τ
such that τ 2 ≤ p∗ − 1. Consequently, M is a Bellman function candidate for 1 < p < 2 but
not for 2 < p < ∞, since it would not satisfy the restrictive concavity needed.
61

Proof. By (2.10), (2.23), (2.24)),
2
pω p−2 R1 H 2 Λ
My3 y3 =
,
3
Φ
y3
giving sign My3 y3 = (−1)[− sign(p − 2)]. By (2.11), for i = 1, 2,
pω p−2 R1
(ωR2 + (p − 1)R1 )(H )2 + ωy3 H
y3
pω p−2
=
Λ(H )2 + ωy3 H (Φ )2 ,
y3 (Φ )3

Myi yi =

giving sign Myi yi = (−1)[− sign(p − 2)], since Φ < 0.
Now that we have a partial Bellman function candidate for 1 < p < 2, from Case (32 ),
satisfying all of the properties of the Bellman function, including restrictive concavity, we
can turn our attention to Case (22 ). From Case (22 ) we will get a Bellman candidate on all
Ξ+ , or part of it, depending on the τ - and p-values. The partial Bellman candidate, from
Case (22 ), turns out to be the missing half for Case (32 ). We already have the solution for
Case (2) from Lemma 21, but the value of the constant is needed before we can progress
further.

2.5.2

Case (2) for 1 < p < 2

Lemma 53. If 1 < p < 2, then in Case (22 ), the value of the constant in Lemma 21 is
p/2
1
+ τ2
c=
= (p∗ − 1)2 + τ 2 p/2 .
2
(p−1)
p
p
2
2 ) 2 [y 2 + 2γy y + y 2 ] 2 + c[y − (y − y )p ] (where γ = 1 − τ )
Proof. If M (y) = (1 + τ
1 2
3
1
2
1
2
1 + τ2
2−p
is to be a candidate or partial candidate, then it must agree, at y2 = p y1 , with the
62

solution M given implicitly by the relation G(y1 − y2 , y1 + y2 ) = y3 G(1,
2−p
Proposition 46. At y2 = p y1 ,

( ω 2 − τ 2 + 1)p−1 [1 − (p − 1) ω 2 − τ 2 ] = G(1,
=

Since

ω 2 − τ 2 + 1 = 0,

ω 2 − τ 2 ), from

ω2 − τ 2)

1
(2y )p−1 [(2 − p)y1 + (p − 2)y1 ] = 0.
y3 1

1
ω 2 − τ 2 = p−1 , which implies ω =

1
1
2 2 . So,
+τ
(p−1)2

p
1
2
y3 = ω p y3
+ τ2
2
(p − 1)
= M (y1 ,
=

2
y
p 1

2−p
y ,y )
p 1 3
2

+ τ2

2(p − 1)
y1
p

2
2 p
+ c y3 −

p
2(p − 1)
y1
.
p

Now just solve for c.

1
In the following Lemma the value of τ has to be restricted to τ 2 ≤ 2p−1 , so that
restrictive concavity is satisfied for our Bellman candidate. Actually, the τ -values play an
even bigger role. Depending on the value of (τ, p) ∈ [−1, 1] × (1, 2), there is either one
or two Bellman function candidates. For (τ, p) ∈ B, from Figure 2.10, there is a partial
Bellman candidate arising from Case (22 ). So we can glue this together with the other
partial candidate obtained in Case (32 ). This gives a Bellman candidate, as before, having
characteristics as in Figure 2.6. For (τ, p) ∈ A ∪ C the candidate obtained from Case (22 )
maintains restrictive concavity throughout Ξ+ and is therefore requires no gluing. To avoid
the difficulty of determining which candidate to choose and how to determine the optimal
63

constant from Case (2), we restrict (τ, p) to region B.
p
T

2

A

1.5
B

−1

C

B

−0.6

0.6

1

E

τ

Figure 2.10: Splitting [−1, 1] × (1, 2) in the (τ × p)–plane into three regions A, B and C.
1
Region B = τ 2 ≤ 2p−1 .

Recall that the partial Bellman candidate, M, obtained from Case (22 ), for 1 < p < 2,
satisfies Myi y3 = My3 y3 = 0 and hence Di = 0, for i = 1, 2. So all that still needs to be
checked for restrictive concavity is the sign of My1 y1 and My2 y2 . Since My1 y1 ≤ My2 y2 ,
2−p
we just need to show that My2 y2 ≤ 0 on p y1 ≤ y2 ≤ y1 in Ξ+ . This is considered in the
following Lemmas.
2−p
1
Lemma 54. In Case (22 ), My2 y2 (y1 , p y1 , y3 ) ≤ 0 for τ 2 ≤ 2p−1 and 1 < p < 2.
Proof. The solution M that we get from (22 ), when 1 < p < 2, is obtained from Lemmas 21
1 − τ2
2
2
and 53. Let γ =
, f (y) = y1 + y2 + 2γy1 y2 , f2 (y) = (p − 2)(y2 + γy1 )2 + f1 (y) and
2 1
1+τ
f3 (y) = y1 − y2 . Then
p p−4
2 ) 2 f 2 f − p(p − 1)
My2 y2 = p(1 + τ
2
1

p
1
2 p−2
+ τ2
f3 .
2
(p − 1)

2−p
To show that My2 y2 (y1 , p y1 , y3 ) ≤ 0 we change to x-variables, make the substitution
4−p
x
s = x2 and multiply both sides of the inequality by (s2 + τ 2 ) 2 . Denote F (s) as the left
1
64

side of the new inequality. So we just need to show that
p
4−p
F (s) = (p − 2)(s − τ 2 )2 + (1 + τ 2 )(s2 + τ 2 ) − (p − 1)((p∗ − 1)2 + τ 2 ) 2 (s2 + τ 2 ) 2 ≤ 0,

at s = p∗ − 1. Or equivalently we need to show that

τ 2 ((p∗ − 1)2 + τ 2 ) ≤ (p∗ − 1 − τ 2 )2 .

(2.25)

2−p
1
Note that (2.25) reduces to τ 2 ≤ 2p−1 . Therefore, My2 y2 (y1 , p y1 , y3 ) ≤ 0 for τ 2 ≤
1
2p−1 and 1 < p < 2.

Lemma 55. In Case (22 ), My2 y2 (y1 , cy1 , y3 ) ≤ 0, for all c ∈

2−p
1
2
p , 1 , for τ ≤ 2p−1

and 1 < p < 2.

Proof. Using My2 y2 from Lemma 54 we see that My2 y2 ≤ 0 is equivalent to
p−4
p
2 ) 2 f 2−p f f 2 − p(p − 1)
(1 + τ
2 1
3

p
1
2 2 ≤ 0.
+τ
(p − 1)2

4−p
Observe that the function f2 /(f1 2 ) is strictly positive, has a horizontal asymptote at the
2−p
y2 -axis, increases on (−∞, −γ), and decreases on (−γ, ∞). As y2 increases from p to 1,
p−4
2−p
2−p
f3
and f2 f1 2 both decrease. Since My2 y2 (y1 , p y1 , y3 ) ≤ 0 (as shown in Lemma
54), the result follows here as well.

Lemma 56. The Monge-Amp`re solution in Case (22 ) yields the following results for 1 <
e
p < 2. My2 y2 (y1 , y1 , y3 ) < 0 for |τ | ≤ 1 and My2 y2 (y1 , −y1 , y3 ) > 0 for |τ | ≤ 1
2
65

Proof. Let f1 , f2 and f3 be as in Lemma 54 and
p 4−p
p
1
2 ) 2 f 2−p f − (p − 1)
2 2f 2 .
g = (1 + τ
+τ
2
1
3
(p − 1)2
Note that My2 y2 and g have the same signs. It is clear that g(y1 , y1 , y3 ) < 0, proving the
1
first inequality. One can now verify that g(y1 , −y1 , y3 ) > 0 for |τ | ≤ 2 which proves the
second inequality.

2−p
Remark 57. One can see in the graph of y1 g(y1 , y1 , y3 ) that g(y1 , y1 , y3 ) < 0, in regions
A and C, (see Figure 2.10). This tells us that the Bellman candidate from Case (22 ) will
maintain restrictive concavity throughout the domain in for (τ, p) ∈ A ∪ C. Furthermore,
p
there will be an improvement in the constant (p∗ − 1)2 + τ 2 2 that can still be used to
still maintain restrictive concavity in A ∪ C.

By Lemmas 55 and 56 we obtain a partial Bellman candidate from Case (22 ), when
1
1 < p < 2 and τ satisfying τ 2 ≤ 2p−1 . As before, we will glue this partial candidate
from Case (22 ) to the partial candidate in Case (32 ) to obtain the Bellman candidate for
1 < p < 2. This glued function will be a Bellman candidate since it satisfies all Bellman
function properties, including the estimates needed for restrictive concavity. However, we
must choose the smaller set of τ restrictions between the implicit and explicit part of the
candidate, so that restrictive concavity will be satisfied. In particular we must restrict τ
1
such that τ 2 ≤ 2p−1 .
66

2.6

Remaining cases and why they do not give the Bellman function candidate

Now that we have particular cases in which the Monge-Amp`re solution gives a Bellman
e
function candidate, we would like to discuss the remaining cases. It can be shown that all
remaining cases do not yield a Bellman function candidate, except for Case (4) which is still
not determined.

2.6.1

Case (12 ) for 1 < p < 2 and Case (32 ) for 2 < p < ∞

It was shown in Lemmas 20, 52 that the Monge-Amp`re solution obtained in each case does
e
not have the appropriate restrictive concavity property to be a Bellman function candidate.
We mention this here again simply for clarity.

2.6.2

Case (11 )

We can consider Cases (11 ) and (31 ) simultaneously, for part of the calculation, since the
same argument will work in both cases. In both cases, y2 is fixed and the Monge-Amp`re
e
solution is given by M (y) = t1 y1 + t3 y3 + t0 on the characteristics dt1 y1 + dt3 y3 + dt0 = 0.
As shown in Figure 2.11, y2 ≥ 0 in case (12 ) and y2 ≤ 0 in Case (32 ), since if not then the
characteristics go outside of the domain Ξ+ .
Lemma 58. In Cases (11 ) and (31 ), the solution to the Monge-Amp`re can be written as,
e

M (y) = 

(u + y2 )2 + τ 2 (u − y2 )2
u − y2
67

p
 y
3

T3
y

T3
y

d
d
d
d
d
d

U

U
E

y2

y1

E

−y2

y1

Figure 2.11: Sample characteristic for Monge-Amp`re solution in Cases (11 ) and (31 )
e
2
2
y1 + p −1 |y2 |
u+ p −1 |y2 |
where u = u(y1 , y2 , y3 ) is the solution to the equation
=
.
y3
(u−y2 )p
Proof. Any characteristic, in Cases (11 ) and (31 ), go from U = (u, y2 , (u − y2 )p ) to W =
(|y2 |, y2 , w). Throughout the proof we will use the properties of the Bellman function derived
in Proposition 7. Using the Neumann property and the property from Proposition 9 we get
y2 My2 = y1 My1 = |y2 |t1 at W. By homogeneity at W we get

p|y2 |t1 + pwt3 + pt0 = pM (W ) = y1 My1 + y2 My2 + py3 My3 = 2t1 |y2 | + pwt3 .

Following the same argument as in Lemma 13, gives

M (y) = 

(u + y2 )2 + τ 2 (u − y2 )2
u − y2

p
 y ,
3

where u = u(y1 , y2 , y3 ) is the solution to the equation
2
2
y1 + p − 1 |y2 |
u + p − 1 |y2 |
=
.
y3
(u − y2 )p

(2.26)

Since the solution, M, does not satisfy the restrictive concavity property necessary to be
68

the Bellman function (as we will soon show), we are not concerned about existence of the
solution u in equation (2.26).
Lemma 59. If ω =

1
M (y) p
, then in Cases (11 ) and (31 ), the solution u to equation
y3

(2.26) can be expressed as u =

ω 2 −τ 2 +1 y and equation (2.26) can be rewritten as
2
ω 2 −τ 2 −1

2p |y2 |p−1 [py1 + (2 − p)|y2 |] = y3 |β − 1|p−1 [p(β + 1) + (2 − p)|β − 1|],

where β =

(2.27)

ω 2 − τ 2 . Furthermore, sign y2 = sign(β − 1).

Proof. Let us show that u =

ω 2 −τ 2 +1 y first. This follows from inverting
2
ω 2 −τ 2 −1

om =

(u + y2 )2 + τ 2 (u − y2 )2
u − y2

,

ω 2 −τ 2 +1 y , we
2
ω 2 −τ 2 −1
ω 2 − τ 2 ≥ 0, which implies that

and using the properties ω ≥ |τ | and u ± y2 ≥ 0. Now that we have u =
can use it to get the next result. Note that u ≥ 0 and

sign y2 = sign( ω 2 − τ 2 − 1). To get (2.27), simply plug u =

ω 2 −τ 2 +1 y in equation
2
ω 2 −τ 2 −1

(2.26).
We can no longer discuss Cases (11 ) and (31 ) together, so for the remainder of the
Subsection the focus will be on Case (11 ) only.
Lemma 60. In Case (11 ), the solution M from Lemma 58 can be rewritten in the implicit
form
G(y2 + y1 , y2 − y1 ) = y3 G( ω 2 − τ 2 , −1),
where G(z1 , z2 ) = (z1 + z2 )p−1 [z1 − (p − 1)z2 ].
69

Proof. Recall that for Case (11 ) we have y2 > 0.
1
y2 = (x2 − x1 ) > 0
2
sign( ω 2 − τ 2 − 1) = sign y2 > 0

=⇒

=⇒

x2 > x1

ω2 − τ 2 > 1

=⇒

ω>

τ2 + 1

p
So, B(x) = M (y) > y3 (τ 2 + 1) 2 . Now (2.27) can be rewritten as

(x2 − x1 )p−1 [(p − 1)x1 + x2 ] =
2/p
1/p p−1
B 2/p − τ 2 x3 − x3

2/p
1/p
B 2/p − τ 2 x3 + (p − 1)x3
.

Therefore,
G(x2 , −x1 ) = G

2/p
2/p
B 2/p − τ 2 x3 , −x3

1
p
or by factoring out x3 on the right side we get

G(y2 + y1 , y2 − y1 ) = y3 G( ω 2 − τ 2 , −1).

Recall that the Monge-Amp`re solution must satisfy the restrictive concavity conditions
e
in Proposition 6 to be a Bellman function candidate. We will show that the Monge-Amp`re
e
solution obtained in Case (11 ) has D1 < 0 and therefore cannot be a Bellman candidate.

Lemma 61. In Case (11 ) we choose H(y1 , y2 ) = G(y1 + y2 , −y1 + y2 ) because of how the
implicit solution is defined and obtain sign H = sign(p − 2)
70

Proof. We already computed


 4Gz z , α = β
j
j
1 2
H =

 0,
αj = −βj
in Lemma 16.
Since, α1 = 1 since α2 = 1 since β1 = −1 and since β2 = 1,

Gz z = p(p − 1)(p − 2)(y1 − y2 )(2y2 )p−3 .
12

Lemma 62. If p = 2 then D2 < 0 in Case (11 ) for all τ.
Proof. We use the partial derivatives of G from the proof of Lemma 16 to make the compup(p−1)ω(β−1)p−3
and β = ω 2 − τ 2 .
tations of Φ and Φ easier. Let αp =
β3

Φ(ω) = G(β, 1)
Φ (ω) = p[β − 1]p−2 [ω + (p − 2)ωβ −1 ]
Φ (ω) = Gz1 z2 (β, −1)β −2 ω 2 + Gz1 (β, −1)[−ωβ −3 + β −1 ]
ω2
τ2
= p(p − 1)[β − 1]p−3 [β + p − 3]
− p [β − 1]p−2 [β + p − 2]
β2
β3
Λ = (p − 1)Φ − ωΦ
= αp [(β − 1)β 3 (β −1 (p − 2) + 1) − ω 2 β(β + p − 3) + τ 2 (β − 1)(β + p − 2)]
= αp [(β 2 + τ 2 )(β − 1)(β + p − 2) − ω 2 β(β + p − 3)]
= αp ω 2 [β 2 + β(p − 2) − β − (p − 2) − β 2 − β(p − 3)]
=−

p(p − 1)(p − 2)ω 3 ( ω 2 − τ 2 − 1)p−3
( ω 2 − τ 2 )3
71

From Lemma 59, sign(β − 1) = sign y2 > 0 and ω 2 > τ 2 > 0. Therefore, by Lemma 61 and
(2.12) sign D2 = sign H sign Λ = −(sign(p − 2))2 < 0.
Since D2 < 0 in Case (11 ), we get the following result.
Proposition 63. Case (11 ) does not give a Bellman function candidate.

2.6.3

Case (31 ) does not provide a Bellman function candidate

Much of the work needed to show that the Monge-Amp`re solution cannot be the Bellman
e
function, in Case (31 ), has already been started in Section 2.6.2. Let us finish the argument.
Lemma 64. In Case (31 ), the solution M from Lemma 58 can be rewritten in the implicit
form G(y2 − y1 , −y1 − y2 ) = y3 G(1, −

ω 2 − τ 2 ), where G(z1 , z2 ) = (z1 + z2 )p−1 [z1 −

(p − 1)z2 ].
Proof. Recall that in Case (32 ) we have that y2 < 0.
1
y2 = (x2 − x1 ) < 0
2
sign( ω 2 − τ 2 − 1) = sign y2 < 0

=⇒

=⇒

x2 < x1

ω2 − τ 2 < 1

=⇒

ω<

τ 2 + 1.

p
So, B(x) = M (y) < y3 (τ 2 + 1) 2 . Now (2.27) can be rewritten as
1/p
(x1 −x2 )p−1 [x1 +(p−1)x2 ] = x3 −

2/p
2/p 1/p
B 2/p − τ 2 x3
(p−1) B 2/p − τ 2 x3 +x3 .

Therefore,
1/p
G(x1 , −x2 ) = G x3 , −
72

2/p
,
B 2/p − τ 2 x3

1/p
or by factoring out x3 on the right side we get

G(y1 − y2 , −y1 − y2 ) = y3 G(1, − ω 2 − τ 2 ).

Since y2 is fixed, D2 ≥ 0 must be true in order that the Monge-Amp`re solution from
e
Case (31 ) is the Bellman function (see Proposition 6). However, the contrary is true: D2 < 0.
Lemma 65. In Case (31 ) we choose H(y1 , y2 ) = G(y1 − y2 , −y1 + y2 ) because of how the
implicit solution is defined and obtain sign H = sign(p − 2)
Proof. We already computed


 4Gz z , α = β
j
j
1 2
H =

 0,
αj = −βj
in Lemma 16.
Since, α1 = 1, α2 = −1, β1 = −1 and β2 = 1, it follows that

Gz z = p(p − 1)(p − 2)(y1 − y2 )(2y2 )p−3 .
12

Lemma 66. If p = 2, then D2 < 0 in Case (31 ) for all τ.
Proof. We use the partial derivatives of G computed in the proof of Lemma 16 to make the
following computations of Φ and Φ easier. Let β =

Φ(ω) = G(1, −β)
Φ (ω) = −p(p − 1)ω(1 − β)p−2
73

ω2 − τ 2.

Φ (ω) = −p(p − 1)[(1 − β)p−2 − (p − 2)ω 2 β −1 ]
Λ = (p − 1)Φ − ωΦ
= p(p − 1)ω(1 − β)p−3 [−(p − 1)(1 − β) + (1 − β) − (p − 2)ω 2 β −1 ]
= −p(p − 1)ω(1 − β)p−3 (p − 2)[1 − β + ω 2 β −1 ]
= −p(p − 1)(p − 2)ω(1 − β)p−3 1 +

τ2
β

.

From Lemma 59, 1 − β > 0 and ω 2 > τ 2 > 0. Therefore, by Lemma 65 and (2.12) sign D2 =
sign H sign Λ = −(sign(p − 2))2 < 0.
Having shown that D2 < 0 in Case (31 ) implies that the Monge-Amp`re solution in that
e
case cannot be the Bellman function.
Proposition 67. Case (31 ) does not give a Bellman function candidate.

2.6.4

Case(21 ) gives a partial Bellman function candidate

Case (2) was considered without having to fix either y1 or y2 first, so there is nothing new
to do here. Refer to Sections 2.5.2 and 2.2.1.3 for more details.

2.6.5

Case (4) may or may not yield a Bellman function candidate

For τ = 0, it was shown in [36] that Case (4) does not produce a Bellman function candidate,
since some simple extremal functions give a contradiction to linearity of the Monge-Amp`re
e
solution on characteristics. However, for τ = 0 it is much more difficult to show this. Those
same extremal functions do contradict linearity for some p-values and some signs of the
Martingale transform. For the sign of the Martingale transform where we do not have a
74

T3
y

y3

T

U
U
E

y2

y1

E

−y2

y1

Figure 2.12: Characteristic of solution in Case (41 ).
T3
y

U
E

−y1

y1

y2

Figure 2.13: Characteristic for the solution from Case (42 )
contradiction, a new set of test of extremal functions would have to be found. Since the
Bellman function has already been constructed from other cases, this case has not been
investigated any further than just described. So, for p and τ values not mentioned in the
main result, Case (4) could give a Bellman candidate throughout Ξ+ or we could get a
partial Bellman candidate that may work well with the characteristics from Case (21 ).

75

Chapter 3
Laminates Meet Burkholder
Functions

3.1

Introduction

In Chapter 2 the Lp -operator norm was computed for a “quadratic perturbation” of the martingale transform using the Bellman function technique, which is similar to how Burkholder
originally did, for τ = 0, in [15]. By “quadratic perturbation”, we are referring to the quan1
tity (Y 2 + τ 2 X 2 ) 2 , where τ ∈ R is small, X is a martingale and Y is the corresponding
martingale transform.
The method of Bourgain [11], for the Hilbert transform, which was later generalized for
a large class of Fourier multiplier operators by Geiss, Montgomery-Smith, Saksman [24], is
to discretize the operator and generalize it to a higher dimensional setting. This operator in
the higher dimensional setting will turn out to have the same operator norm and it naturally
connects with discrete martingales, if done in a careful and clever way. In the end, one
76

has the operator norm of the singular integral bounded below by the operator norm of
the martingale transform, which Burkholder found in [15]. This approach can be used for
estimating

2
2
R1 − R2 , τ I

from below as well, see [10]. However, we will present
Lp →Lp

an entirely different approach to the problem.
Rather than working with estimates on the martingale transform, we only need to consider
the “Burkholder” functions that were used to find those sharp estimates on the martingale
transform. More specifically, we analyze the behavior of the “Burkholder” functions, U
and v found in Chapter 2 (and reiterated in Definition 76 and Theorem 78), associated
with determining the Lp -operator norm of the quadratic perturbation of the martingale
transform. Using the fact that U is the least bi-concave majorant of v (in the appropriately
chosen coordinates), in addition to some of the ways in which the two functions interact will
allow us to construct an appropriate sequence of laminates, which approximate the push
forward of the 2-dimensional Lebesgue measure by the Hessian of a smooth function with
compact support. Once the appropriate sequence of laminates is constructed, we are finished
since Riesz transforms can be written as fractional derivatives of some smooth function. The
beauty of this method is that it quickly gets us the sharp lower bound constant with very
easy calculations. This lower bound argument is discussed in Section 3.2.
The Burkholder functions U and v also play a crucial role in obtaining the sharp upper
bound estimate as well. With the Burkholder functions we are able to extend sharp esti1
mates of (Y 2 + τ 2 X 2 ) 2 , obtained in Chapter 2, from the discrete martingale setting to the
continuous martingale setting. The use of “heat martingales”, as in [2] and [3], will allow us
to connect the Riesz transforms to the continuous martingales estimate, without picking up
any additional constants. This upper bound argument is presented in Section 3.4.
77

3.2
3.2.1

Lower Bound Estimate
Laminates and gradients

We denote by Rm×n the space of real m × n matrices and by Rn×n the set of real n × n
sym
symmetric matrices.
Definition 68. We say that a function f : Rm×n → R is rank-one convex, if t → f (A + tB)
is convex for all A, B ∈ Rm×n with rank B = 1.
Let P(Rm×n ) denote the set of all compactly supported probability measures on Rm×n .
For ν ∈ P we denote by ν =

Rm×n

X dν(X) the center of mass or barycenter of ν.

Definition 69. A measure ν ∈ P is called a laminate, denoted ν ∈ L, if

f (ν) ≤

Rm×n

f dν

(3.1)

for all rank-one convex functions f . The set of laminates with barycenter 0 is denoted by
L0 (Rm×n ).
Laminates play an important role in several landmark applications of convex integration
for producing unexpected counterexamples, see for instance [30, 27, 1, 34, 21]. For our
purposes the case of 2 × 2 symmetric matrices is of relevance, therefore in the following we
restrict attention to this case. The key point is that laminates can be viewed as probability
measures recording the gradient distribution of maps, see Theorem 72 below. This is by now
a very standard technique. Refer to [9] for the main steps of the argument. Detailed proofs
of these statements can be found for example in [30, 26, 34].
78

Definition 70. Given a set U ⊂ R2×2 we call PL(U ) the set of prelaminates generated in
U . This is the smallest class of probability measures on R2×2 which
• contains all measures of the form λδA + (1 − λ)δB with λ ∈ [0, 1] and rank(A − B) = 1;
• is closed under splitting in the following sense if λδA + (1 − λ)˜ belongs to PL(U ) for
ν
some ν ∈ P(R2×2 ) and µ also belongs to PL(U ) with µ = A, then also λµ + (1 − λ)˜
˜
ν
belongs to PL(U ).
The order of a prelaminate denotes the number of splittings required to obtain the measure from a Dirac measure.
Example 71. The measure

1
1
1
δdiag(1,1) + δdiag(−1,1) + δdiag(0,−1) ,
4
4
2

where




x 0 
diag(x, y) := 
,
0 y
is a second order prelaminate with barycenter 0.
It is clear from the definition that PL(U ) consists of atomic measures. Also, from a
repeated application of Jensen’s inequality it follows that PL ⊂ L. The following statement,
links laminates supported on symmetric matrices with second derivatives of functions.
∞
Theorem 72. Let ν ∈ L0 (R2×2 ). Then there exists a sequence uj ∈ Cc (B1 (0)) with
sym
uniformly bounded second derivatives, such that

B1 (0)

φ(D2 uj (x)) dx →
79

φ dν
R2×2
sym

for all continuous φ : R2×2 → R.
sym

3.2.2

Laminates and lower bounds

Let τ ∈ R be fixed. Our goal is to find

sup
ϕ∈S(R2 )

2
2
2
2
(R1 ϕ − R2 ϕ)2 + τ 2 (R1 ϕ + R2 ϕ)2 1/2
p
,
2
2
R1 ϕ + R2 ϕ p

(3.2)

for the planar Riesz transforms R1 and R2 , where S(R2 ) is the Schwartz class. We can
rework the Riesz transforms acting on ϕ into the second derivative of a function u ∈ S(R2 )
in the following way.

2
Ri ϕ =

ξ2
− i ϕ
|ξ|2

∨
2
= ∂i u,

where “ ” denotes the Fourier transform, “∨” denotes the inverse Fourier transform and
−∆u = ϕ. So (3.2) is equivalent to

sup
u∈S(R2 )

p
2
2
2
2
|(∂11 u − ∂22 u)2 + τ 2 (∂11 u + ∂22 u)2 | 2
.
2
2
|∂11 u + ∂22 u|p

(3.3)

Let Aij denote, as usual, the ij-entry of a matrix A and put
p
φ1 (A) = |(A11 − A22 )2 + τ 2 (A11 + A22 )2 | 2 ,
φ2 (A) = |A11 + A22 |p .
80

(3.4)

∞
Using a standard cut-off argument we can write replace S(R2 ) with Cc (R2 ) and write (3.3)
as
sup
∞
u∈Cc (R2 )

φ1 (D2 u) dx
.
φ2 (D2 u) dx

(3.5)

From Corollary 72 we deduce that

sup
∞
u∈Cc (R2 )

φ1 (D2 u) dx
φ2 (D2 u) dx

≥

sup
ν∈L0 (R2×2 )
sym

φ1 dν
.
φ2 dν

(3.6)

Our goal is therefore to prove the following
Theorem 73. For any 1 < p < ∞ and τ ∈ R there exists a sequence νN ∈ L0 (R2×2 ) such
sym
that
p
φ1 dνN
→ ((p∗ − 1)2 + τ 2 ) 2 .
φ2 dνN

3.2.3

Proof of Theorem 73

A function f (x, y) of two variables is said to be biconvex if the functions x → f (x, y)
and y → f (x, y) are convex for all x, y. We start with the following inequality for biconvex
functions in the plane.
Lemma 74. Let k ∈ (−1, 1) and N > 1. For every f ∈ C(R2 ) biconvex we have
N
− 2 dt
− 2
1
1−k
f (1, 1) ≤
f (kt, t) + f (t, kt) t
+ f (N, N )N 1−k .
1−k 1
t

(3.7)

Proof. By a standard regularization argument it suffices to show the inequality for f ∈
81

C 1 (R2 ) biconvex. The biconvexity implies the following elementary inequalities:

f (t, t) ≤ λ f (t, t + ) + (1 − λ )f (t, kt),

(3.8)

f (t, t + ) ≤ µ f (t + , t + ) + (1 − µ )f (k(t + ), t + ),

(3.9)

where

λ =1−
µ =1−

t(1 − k) +
t(1 − k) + (1 − k)

Combining (3.8) and (3.9) and observing that λ , µ = 1 −

f t + ,t +

− f (t, t)

−

.

t(1−k)

+ o( ), we obtain

2
f (t + , t + ) ≥
t(1 − k)

1
−
f (k(t + ), t + ) + f (t, kt) + o(1).
t(1 − k)

(3.10)

Letting → 0+ this yields

−

2
1
∂
f (t, t) +
f (t, t) ≤
f (kt, t) + f (t, kt)
∂t
t(1 − k)
t(1 − k)

− 2
Multiplying both sides by t 1−k and integrating, we obtain (3.7) as required.

The method of obtaining continuous laminates by integrating a differential inequality as
above is due to Kirchheim, and appeared first in the context of separate convexity in R3 in
[27].
82

Next, for 1 < p < ∞ let
2
k =1− ,
p
so that p =

2 . We need to differentiate between the cases 1 < p ≤ 2 and 2 < p < ∞.
1−k

The case 1 < p ≤ 2.

Let µN ∈ P(R2×2 ) be defined by the RHS of (3.7), more precisely

f dµN :=

N
f diag(N, N )
dt
1
f diag(kt, t) + f diag(t, kt) t−p
+
1−k 1
t
Np

for f ∈ C(R2×2 ). Then µN is a probability with barycenter µN = diag(1, 1). Moreover,
observe that if f is rank-one convex, then (x, y) → f (diag(x, y)) is biconvex. Therefore,
using Lemma 74 we see that µN is a laminate. Then, combining with the measure from
Example 71 (c.f. splitting procedure from Definition 70) we conclude that the measure

1
1
1
νN := µN + δdiag(−1,1) + δdiag(0,−1)
4
4
2

is a laminate with barycenter ν N = 0. We claim that this sequence of laminates has the
desired properties for Theorem 73. To this end we calculate

φ1 dµN = p|(1 − k)2 + τ 2 (1 + k)2 |p/2 log N + 2p ,
φ2 dµN = p(1 + k)p log N.
83

In particular we see that as N → ∞
φ1 dνN
φ2 dνN

→

|(1 − k)2 + τ 2 (1 + k)2 |p/2
(1 + k)p
p
1−k 2
=
+ τ2
1+k
2
p
1
+ τ 2 = [(p∗ − 1)2 + τ 2 ]p .
=
p−1

The case 2 < p < ∞.

Let µN ∈ P(R2×2 ) be defined by
˜

f d˜N :=
µ

N
f diag(−N, N )
dt
1
f diag(−kt, t) + f diag(−t, kt) t−p
+
1−k 1
t
Np

˜
for f ∈ C(R2×2 ). Then µN is a probability with barycenter µN = diag(−1, 1). Moreover,
˜
as before, we see that if f is rank-one convex, then (x, y) → f (diag(−x, y)) is biconvex.
Therefore µN is again a laminate, hence also
˜
1
1
1
νN := µN + δdiag(1,1) + δdiag(0,−1)
˜
˜
4
4
2

is a laminate with barycenter 0. Repeating the calculations above, we obtain
φ1 d˜N
ν
φ2 d˜N
ν

−→
N →∞

|(1 + k)2 + τ 2 (1 − k)2 |p/2
(1 − k)p
p
1+k 2
=
+ τ2
1−k
= [(p − 1)2 + τ 2 ]p = [(p∗ − 1)2 + τ 2 ]p .
84

3.3

Comparison with Burkholder functions

Now we will discuss the Burkholder functions introduced Chapter 2. Let p∗ − 1 :=
1
max p − 1, p−1

and x := (x1 , x2 ) denote a point in R2 . We will denote the coordinates

y := (y1 , y2 ) ∈ R2 , as the rotation of x by π , that is
4
x + x2
x − x2
y1 = 1
, y2 = 1
.
2
2

Definition 75. We say that a function f := f (x1 , x2 ) is zigzag concave if it is bi-concave
in the y-variables.
p
p
Definition 76. Let v(x1 , x2 ) := vp,τ (x1 , x2 ) = (τ 2 |x1 |2 + |x2 |2 ) 2 − ((p∗ − 1)2 + τ 2 ) 2 |x1 |p
and u(x1 , x2 ) := up,τ (x1 , x2 ) = αp (|x1 | + |x2 |)p−1 [|x2 | − (p∗ − 1)|x1 |], where
p−2
2
1 )p−1 1 +
τ2
αp = p(1 − ∗
. For 1 < p < 2, we define
p
(p∗ −1)2


 v(x , x ) : |x | ≥ (p∗ − 1)|x |
1 2
2
1
U (x1 , x2 ) := Up,τ (x1 , x2 ) =

 u(x , x ) : |x | ≤ (p∗ − 1)|x |,
1 2
2
1
and for 2 < p < ∞,


 u(x , x ) : |x | ≥ (p∗ − 1)|x |
1 2
2
1
U (x1 , x2 ) := Up,τ (x1 , x2 ) =

 v(x , x ) : |x | ≤ (p∗ − 1)|x |.
1 2
2
1
Definition 77. Denote

cM := inf {c : vc has a zigzag concave majorant and U is such that U (0, 0) = 0} .
85

Now we will see the key relationship between the Burkholder functions U and v.
p
Theorem 78. 1) cM ≥ ((p∗ − 1)2 + τ 2 ) 2 , for all 1 < p < ∞ and all τ ∈ R.
p
1
2) cM = ((p∗ − 1)2 + τ 2 ) 2 , for 1 < p < 2 and τ 2 ≤ 2p−1 or 2 ≤ p < ∞ and τ ∈ R.
p
3) If 1 < p < 2 and τ is sufficiently large then cM > ((p∗ − 1)2 + τ 2 ) 2 .
p
Proof. 1) By way of contradiction, suppose that there is such a c ∈ [0, ((p∗ − 1)2 + τ 2 ) 2 )
that vc has a zigzag concave majorant. Then following the upper bound estimate in Section
3.4, Theorem 93 would have c as the upper bound of our quadratic perturbation. However,
this is impossible because of Theorem 73.
1
2) In Proposition 40 it was shown for 1 < p < 2 and τ 2 ≤ 2p−1 or 2 ≤ p < ∞ and
p
τ ∈ R that for c = ((p∗ − 1)2 + τ 2 ) 2 , vc has a zigzag concave majorant. This proves that
p
cM ≤ ((p∗ − 1)2 + τ 2 ) 2 . Combining with 1) we get equality.
3) For 1 < p < 2 and τ ∈ R sufficiently large, U is no longer zigzag concave, while
p
still being a majorant of vc , with c = ((p∗ − 1)2 + τ 2 ) 2 . We know that if τ is sufficiently
large and 1 < p < 2, the least c0 for which vc0 has a zigzag concave majorant must satisfy
p
1
c0 > ((p∗ − 1)2 + τ 2 ) 2 . See [8], Remark 27. The condition τ 2 ≤ 2p−1 is a sufficient
condition for U to be the zigzag concave majorant, but not necessary.

3.3.1

Analyzing the Burkholder functions U and v

We will use the y-coordinates, unless otherwise stated, from this point on.
coordinates,
U (y1 , y2 ) := U (y1 − y2 , y1 + y2 ) = U (x1 , x2 ),
86

In the y-

and it takes the following form. For 2 ≤ p < ∞,


 u(y), p−2 y ≤ y ≤ p y
2 p−2 1
p 1
U (y) =

 v(y), otherwise,
and for 1 < p < 2,


 v(y), 2−p y ≤ y ≤ p y
2 2−p 1
p 1
U (y) =

 u(y), otherwise,

(3.11)

where

u(y1 , y2 ) := u(y1 − y2 , y1 + y2 ) = u(x1 , x2 ), v(y1 , y2 ) := v(y1 − y2 , y1 + y2 ) = v(x1 , x2 ).

We will fix 2 ≤ p < ∞, as the dual range of p values is handled similarly. Denote

k :=

p
.
p−2

(3.12)

Then p = 2k and p − 1 = k+1 . Also denote
k−1
k−1
1
Lk := {y2 = ky1 }and L 1 := y2 = y1 .
k
k
Observe that in the cone
C1 = {y1 ≤ y2 ≤ ky1 },
U is linear if we fix y2 . Also, U is linear if we fix y1 in the cone

C2 =

1
y ≤ y2 ≤ y1 .
k 1
87

(3.13)

Consequently, U is almost linear in the “T-shape” graph, which we will denote as T , with
vertices
1
1
( (y1 + h), y1 + h), (y1 , y1 + h), (y1 + h, y1 + h), (y1 , y1 ) .
k
k
The only portion where U is not linear is on the segment from (y1 , y1 ) to (y1 , y1 + h). It is
very small in comparison with the graph T.
y2
T

y2

y2 = y1
Lk
¡
¡

 
 
 
 

¡ C1  
¡
 
L
¡ 1   C2
¨ 1/k
¡
¨¨
 
2 ¨¨
¡  
¡   ¨¨¨
¡   ¨
¡ ¨¨
¡ ¨
¨
 
E

y1 + h
y1

y2 = y1

Lk

T

 
 

¡
¡
¡
¡
¡
¡
¡

 
 
 
 

L

¨ 1/k
¡
¨¨
¨
1y
¡  
¨
k 1 ¡  
¨¨
¨
¡   ¨
¡ ¨¨
¨¨
¡
 
Ey

y1

 
 

y1 +h y1 y1 + h
k

1

Figure 3.1: Splitting between u and v in y1 × y2 -plane.
By Theorem 78, U ≥ v in R2 . But, on Lk and L 1 , U = v. Also, observe that U (0, 0) =
k
v(0, 0) and v ≥ 0 on C1 and C2 . (This is easy to see in x-coordinates.) We will summarize
these important facts, so that we can later refer to them.
Proposition 79. 1) v ≥ 0 on C1 and C2
2) v(0, 0) = U (0, 0) = 0
3) U = v on Lk and L 1 .
k
4) U is nearly linear on T.

3.3.2

Why the laminate sequence νN worked in Theorem 73

88

p
Let φ1 (y1 , y2 ) := (|y1 − y2 |2 + τ 2 |y1 + y2 |2 ) 2 , φ2 (y1 , y2 ) := |y1 + y2 |p .
Definition 80. Let cL := sup

φ1 dν
: ν ∈ L0 .
φ2 dν

Theorem 81. cM ≥ cL .
Proof. By the definition of cL , there exists a laminate ν ∈ L with barycenter 0, such that
φ1 dν
> cL − ε. This is equivalent to
φ2 dν
[φ1 − (cL − ε)φ2 ]dν > 0.

(3.14)

We will now show that φ1 − (cL − ε)φ2 does not have a biconcave majorant. By changing
the variables back to x1 , x2 we see that this means that vc −ε (x1 , x2 ) would not have a
L
zigzag concave majorant, thus proving that cL − ε ≤ cM . By way of contradiction, suppose
that b := cL − ε, and that φ1 − bφ2 ≤ U, which is biconcave. Then by (3.14),

0<

(φ1 − bφ2 )dν ≤

U dν ≤ U (ν) = U (0, 0) = 0.

This gives a contradiction and we are finished with the proof.
1
Definition 82. We denote (p, τ ) ∈ T , if 1 < p < 2 and τ 2 ≤ 2p−1 or 2 ≤ p < ∞ and
τ ∈ R.
Let us compose vc (x1 , x2 ) with the change of variables

x1 = y 1 + y 2
x2 = y 1 − y 2 .
89

We get the function called
p
vc (y1 , y2 ) := (|y1 − y2 |2 + τ 2 |y1 + y2 |2 ) 2 − c|y1 + y2 |p .

Recall that similarly U (y1 , y2 ) := U (y1 + y2 , y1 − y2 ) = U (x1 , x2 ).
Let us introduce the following notation.
p
Definition 83. cB := (p∗ − 1)2 + τ 2 2 .
Here B stands for Burkholder. Recall Definition 77 and let us also denote
Definition 84. cN :=

2
2
R1 − R2 , τ I

.
Lp (C,C)→Lp (C,C2 )

Remark 85. For (p, τ ) ∈ T , cL = cM = cB .
This follows immediately from Theorems 73, 78 and 81.
Moreover, we saw that for all p, τ

cL ≤ cN .

This is Corollary 72 essentially.
In Section 3.4 we are proving that

cN ≤ cM .

We wish to discuss the set Ω of pairs (p, τ ), for which cL = cM . This set of pairs contains
T introduced above, but we do not know exactly the whole Ω. By what we have just said
if (p, τ ) ∈ Ω, then the norm of our operator is cL , and we also know that for such pairs the
90

sharp estimate from above for the norm is obtainable by means of finding the least c for
which vc has a zigzag concave majorant.
Now we want to see what kind of restriction the equality cL = cM imposes a priori.
Conjecture 86. If p, τ are such that cL = cM , then there exists a sequence of laminates
{νN } with barycenter 0, such that

UcM dνN increases to 0.

1
1
Remark 87. In fact, this is exactly what happens on T . Namely, if νN = 4 µN + 4 δ(−1,1) +
1δ
2 (0,−1) and (p, τ ) ∈ T , then

−O(1) ≤

UcM dνN ≤ 0.

If (p, τ ) is not in T , but cL = cM , then we get something interesting. By the definition
φ1 dνN
1
of cL , there exists some νN with barycenter 0, and such that
≥ cM − N . Then
φ2 dνN
we get
−

1
≤
N

(φ1 − cM φ2 )dνN
≤
φ2 dνN

UcM dνN
≤ 0.
φ2 dνN

Therefore,
UcM dνN = o

φ2 dνN .

For (3.15) it would be sufficient to have

UcM dνN = O(1).

Remembering that

UcM dνN ≤ 0 we can write this as

−c ≤

UcM dνN ≤ 0.
91

(3.15)

1
This was exactly the case for (p, τ ) ∈ T with νN = 1 µN + 4 δ(−1,1) + 1 δ(0,−1) . The reason
4
2
for that was because

µN = (1, 1) and

UcM dµN = UcM (1, 1)

For any w biconcave and µ with barycenter (1, 1) we have

(3.16)

w dµ ≤ w(1, 1). But in (3.16)

we have the case when the equality is attained. To understand better the case when the
equality can be attained when integrating a biconcave function against a laminate, let us
consider first a simpler question when equality is attained in integrating the usual concave
function against a usual probability measure.
If w is concave, then

w dµ ≤ w(1, 1) is true for any probability measure µ. There are

only two ways to get equality (i.e.,

w dµ = w(1, 1)): 1) if µ is a delta measure at (1, 1) or

2) if w is linear on the convex hull of the support of measure µ (degenerate concave).
Coming back to the attained equality in (3.16), for biconcave UcM , we see that (3.16)
happened also exactly because the Burkholder function UcM , is not only biconcave on the
cones C1 ∪ C2 , but degenerate biconcave, meaning that C1 ∪ C2 is foliated by curves on
which one of the concavities degenerates into linearity. We may conjecture that the same
geometric picture happens for those (p, τ ) outside of T , for which cL = cM , but we do not
know how to prove this.
To summarize, we have the following.
p
• For all p and τ, cM ≥ cN ≥ cL ≥ (p∗ − 1)2 + τ 2 2 =: cB (p, τ ). All four constants
coincide at least for (p, τ ) ∈ T .

• For all p ∈ (1, 2), there exists a τ0 such that for all |τ | > τ0 , cM > cB (p, τ ) (by [7]
92

Remark 27).

• (p, τ ) such that cL = cM holds for all (p, τ ) ∈ T , but may also be true outside of T .
• By a modification of [7], Remark 27, one can prove that for all p ∈ (1, 2), there exits
τ0 such that, for all |τ | ≥ τ0 , cN > cB (p, τ ).
• We of course expect that always cN = cL .

3.4
3.4.1

Upper Bound Estimate
Background information and notation

We will use similar notation, estimates and reasoning developed in [2] and [3]. Let Bt =
(Zt , T − t) denote space-time Brownian motion starting at (0, T ) ∈ R3 := R2 × (0, ∞),
+
where Zt is standard Brownian motion in the plane. There is a pseudo-probability measure
P T associated with the process and we will denote ET as the corresponding expectation.
∞
For φ ∈ Cc (C), we denote Uφ (z, t), as the heat extension to the upper half space, in
other words Uφ is the solution to


 ∂t U − 1 ∆U , R3
φ 2
φ
+

 U = φ,
R2 .
φ
By Itˆ’s formula we get the relation,
o
t
Uφ (Bt ) − Uφ (B0 ) =
93

0

Uφ (Bs ) · dZs ,

(3.17)

which is a martingale. For a 2 × 2 matrix A we denote
t
(A ∗ Uφ )t :=

0

A Uφ (Bs ) · dZs

as a martingale transform. Throughout Section 3.4 we will refer to the matrices






1 0
0 0 
1 0 
A1 = 
 , A2 = 
, I = 

0 0
0 1
0 1

If we rewrite (3.17) in the form
t

t
t
Xt = X1 + iX2 = (I ∗ φ)t =

0

Uφ (Bs ) · dZs ,

then its martingale transform will be denoted as

t
t
Yt = Y1 + iY2 = ((A1 − A2 ) ∗ φ)t =

t
0

(A1 − A2 ) Uφ (Bs ) · dZs .

The quadratic variation of Xi and Yi are




t  ∂x Uφ (Bs )  2
2 ds =
i
Xi t =
| Uφ (Bs )|

 ds = Yi t , for i = 1, 2.
i
0
0
−∂y Uφ (Bs )
i
t

Then, X t = X1 t + X2 t = Y1 t + Y2 t = Y t .
Definition 88. A process H is called differentially subordinate to a process K, if d H t ≤
dt
d K .
t
dt
We have computed that Y is differentially subordinate to X. Note that Y is the contin94

uous version of the martingale transform (the discrete version of Burkholder’s martingale
transform is

n d →
k=1 k

n ε d , where ε ∈ {±1} and {d } is a martingale differk
k k
k=1 k k

ence sequence and n ∈ Z+ ), since Y is differentially subordinate to X.

3.4.2

Extending the martingale estimate to continuous time martingales

Theorem 89. Let X and Y be two complex-valued martingales, such that Y is the martingale
transform of X (in other words d X t ≤ d Y t ). Then
dx
dx
1
τ 2 |X|2 + |Y |2 p ≤ (p∗ − 1)2 + τ 2 2 X p ,

1
with the best possible constant for 1 < p < 2 and τ 2 ≤ 2p−1 or 2 ≤ p < ∞ and τ ∈ R.

This was basically shown in Chapter 2, but we will give the idea of the proof. The proof
here only requires the same modification to continuous time martingales as was done in [3],
for τ = 0. Let
p−2
τ2
1 p−1
u(x, y) = p 1 − ∗
1+
(|y| − (p∗ − 1)|x|)(|x| + |y|)p−1 and
∗ − 1)2
p
(p
p
1
v(x, y) = (τ 2 |x|2 + |y|2 ) 2 − (p∗ − 1)2 + τ 2 2 |x|p .

It was shown in Theorem 78 that v ≤ u. The key properties of u and v, that will be used,
are:
(1) v(x, y) ≤ u(x, y)
95

(2) For all x, y, h, k ∈ C, if |x||y| = 0, then

huxx (x, y), h + 2 huxy (x, y), k + kuyy (x, y), k = −cp,τ (A + B + C),

where cp,τ > 0 is a constant only depending on τ and p and
p−1
A = p(p − 1)(|h|2 − |k|2 )(|x| + |y|)p−2 , B = p(p − 2)[|k|2 − (y , k)2 ]|y|−1(|x|+|y|)
,
C = p(p − 1)(p − 2)[(x , h) + (y , k)]2 |x|(|x| + |y|)p−3 ,

where x = x/|x|, y = y/|y|.
(3) u(x, y) ≤ 0 if |y| ≤ |x|.
Since u here only differs from the one in [3] by a multiple of 1 +

p−2
τ2
, the rest of
∗ −1)2
(p

the argument follows in an identical way which we briefly outline.
By Itˆ’s formula,
o
t
u(Xt , Yt ) = u(X0 , Y0 ) +

0

t

ux (Xs , Ys ), dXs +

I
uy (Xs , Ys ), dYs + t ,
2
0

where It contains the second order terms. We can assume, without loss of generality that
|Y0 | ≤ |X0 |, so that when we take expectation of u(Xt , Yt ), we obtain Eu(Xt , Yt ) ≤ E(It /2).
Using property (2) above, in the martingale setting, one can obtain
t
It ≤ −cp,τ

0

(|Xs | + |Ys |)p−2 d( X s − Y s ) ≤ 0,

since B, C ≥ 0 and using the differential subordinate assumption. Therefore, Ev(Xt , Yt ) ≤ 0
by property (1) above.
96

3.4.3

Connecting the martingales to the Riesz transforms

Now we choose Xt := (I ∗ Uφ )t and Yt := ((A1 − A2 ) ∗ Uφ )t to obtain the following
corollary of Theorem 89.

1
Corollary 90. If 1 < p < 2 and τ 2 ≤ 2p−1 or 2 ≤ p < ∞ and τ ∈ R, then
1
τ 2 |(I ∗ Uφ )t |2 + |((A1 − A2 ) ∗ Uφ )t |2 p ≤ ((p∗ − 1)2 + τ 2 ) 2 (I ∗ Uφ )t p .

∞
Proposition 91. For all φ ∈ Cc and all p ∈ (1, ∞), limT →∞ (I ∗ Uφ )T p ≤ φ p .

This result was proven in [2].
Now we will connect the martingales Xt and Yt with the planar Riesz transforms,R1 and
R2 , in the following way.
∞
Proposition 92. For all φ ∈ Cc (C),
p
lim
|ET (YT |BT = z)|2 + τ 2 |ET (XT |BT = z)|2 2 dz
T →∞ C
p
2 )φ|2 + τ 2 |(R + R2 )φ|2 2 dz.
=
|(R1 − R2
1
2
C
∞
This result follows almost immediately from the fact that, for all ψ, φ ∈ Cc (C),

2
2
lim
ψET [YT |BT = z] dz =
ψ(R1 − R2 )φ dz and
T →∞ C
C

(3.18)

2
2
lim
ψET [XT |BT = z] dz =
ψ(R1 + R2 )φ dz,
T →∞ C
C

(3.19)

97

∞
by [2]. By (3.18) and (3.19) we obtain that for all ψ, φ ∈ Cc (C),
1
lim
ψ[|ET (YT |BT = z)|2 + τ 2 |ET (XT |BT = z)|2 ] 2 dz
T →∞ C
1
2
2
=
ψ[|(R1 − R2 )φ|2 + τ 2 |(R1 + R2 )φ|2 ] 2 dz.
C

3.4.4

Main Result

1
Theorem 93. For 1 < p < 2 and τ 2 ≤ 2p−1 or 2 ≤ p < ∞ and τ ∈ R we have the
following estimate.
1
1
2
2
[|(R1 − R2 )f |2 + τ 2 |f |2 ] 2 p ≤ ((p∗ − 1)2 + τ 2 ) 2 f p

Let Ez0 ,T correspond to Brownian motion starting at (z0 , T ) ∈ R3 . Let φ ≥ 0 and
+
1 1
p + q = 1. Then

C

1
2
2
(|(R1 − R2 )φ|2 + τ 2 |φ|2 ) 2 ψ(z)dz

1
lim
(|E(z0 ,T ) (YT |BT = z)|2 + τ 2 |E(z0 ,T ) (XT |BT = z)|2 ) 2 dz0 ψ(z)dz
C T →∞ C


(z0 ,T ) (Y ψ(Z )|B = z)
T
T
T
E

= lim

 dz dz0
T →∞ C C
(z0 ,T ) (X ψ(Z )|B = z)
τE
T
T
T


 YT ψ(ZT ) 
≤ lim
E(z0 ,T ) 
 dz0
T →∞ C
τ XT ψ(ZT )


1
1
p
Y
(z0 ,T )  T  dz p
(z0 ,T ) |φ(Z )|q dz q
≤ lim
E
lim
E


0
0
T
T →∞ C
T →∞ C
τ XT
=

98





1
p
Y
(z0 ,T )  T  dz p ψ q ,
= lim
E


0
L
T →∞ C
τ XT
where the last equality is by Proposition 91. By linearity we have this result for any ψ ∈ Lq .
Therefore, by duality

C

p 1
2 − R2 )φ|2 + τ 2 |φ|2 ) 2 p ≤
(|(R1
2





1
p
Y
(z0 ,T )  T  dz p
lim
E


0
T →∞ C
τ XT

p 1
1
1
T [(|Y | + τ 2 |X |2 ) 2 ]) p ≤ ((p∗ − 1)2 + τ 2 ) 2 lim (E|X |p ) p
=( lim E
T
T
T
T →∞
T →∞
1
=((p∗ − 1)2 + τ 2 ) 2 φ Lp ,

where the last inequality is due to Theorem 89 and the last equality is by Proposition 92.
1
Corollary 94. For 1 < p < 2 and τ 2 ≤ 2p−1 or 2 ≤ p < ∞ and τ ∈ R,

(2R1 R2 , τ I) p
=
L (C,C)→Lp (C,C2 )

2
2
R1 − R2 , τ I

Lp (C,C)→Lp (C,C2 )

1
= (p∗ − 1)2 + τ 2 2 .

2
2
Since R1 −R2 and 2R1 R2 are just rotations of one another by π/4, we have the equality of
1
the two operator norms. The lower bound was computed as (p∗ −1)2 +τ 2 2 in Theorem 73
(or by another technique in [10]). The upper bound was just computed as the same, giving
the desired result.

99

Chapter 4

Dissertation achievements and future
work

4.1

Contributions

A large class of Fourier multiplier operators, can be built out of linear combinations of
the product of two Riesz transforms. The Ahlfors–Beurling operator is one such example,
which has been discussed extensively in this dissertation. Fourier multiplier operators are
also referred to as singular integrals, a designation which becomes clear by looking at the
definition in (4.1). Let us now define what the Riesz transforms are.
The jth Riesz transform operator on Rd is defined as
(xj − tj )f (t)
dt
Rj f (x) := cd lim
ε→0 Rd \B(x,ε) |x − t|d+1
100

(4.1)

and is a well defined operator that takes Lp functions to Lp functions, where

cd :=

Γ((d + 1)/2)
π (d+1)/2

is a constant which was suitably chosen so that (4.2) does not have any extra constants (and
Γ is the well known “gamma function”). One of the key properties of this operator is that
it is also a Fourier multiplier operator; in other words, if “ ” denotes the Fourier transform,
then
ξj
Rj f (ξ) = i f (ξ),
|ξ|

(4.2)

ξj
being the associated Fourier multiplier. (Throughout g denotes the
with mj (ξ) = i
|ξ|
Fourier transform of g and g ∨ denotes the inverse Fourier transform.)
The Riesz transforms arise naturally in partial differential equations in the following way.
If f is a rapidly decreasing function on Rd , and u is a solution to the Laplace equation,

∆u = f,

then taking the Fourier transform of the equation twice gives the well known formula u =
1 f . So we have
|ξ|2

Rj Rk f =

∨
ξj ξk
−
f
= (−ξj ξk u)∨ = −uyj yk .
2
|ξ|

(4.3)

Note that, we are only looking at the product (composition) of two Riesz transforms, since
that will be the focus here, but there is a similar formula for just one Riesz transform. So,
these Riesz transforms naturally arise by taking partial derivatives of solutions of partial
101

differential equations. A. Calder´n and A. Zygmund showed in the 1950’s that a large class
o
of singular integrals, including the Fourier multiplier operators (as long as the corresponding
multiplier is bounded), are bounded in Lp (meaning that their operator norm from Lp to Lp
is finite). Knowing that a singular integral is bounded can help gain more information about
the associated partial differential equation and its solution. But, being able to compute its
exact operator norm can give even more information. One such singular integral whose Lp
2
2
operator norm we computed exactly is a quadratic perturbation of the R1 − R2 (the real
part of the Ahlfors–Beurling operator).
2
2
The approach of determining the operator norm of the perturbation of R1 − R2 , was
to first determine the operator norm of the same perturbation of the martingale transform.
In Chapter 2, the exact Lp - operator norm was found for the quadratic perturbation of the
martingale transform. Furthermore, we found the exact Bellman function and the Burkholder
functions associated with the problem. Therefore, the celebrated result of Donald Burkholder
proven in the series of papers, [12] to [19], has been generalized.
From the results of Chapter 2, there are several different approaches for determining
the sharp lower bound estimate of the following operator norm:

2
2
R1 − R2 , τ I

Lp →Lp

,

where R1 and R2 are the two components of the Riesz transform in the plane. One approach
uses a similar technique developed by J. Bourgain [11] which has then been generalized by
Geiss, Montgomery-Smith, Saksman [24]. The sharp lower bound estimate can be found
using this approach (in N. Boros, A. Volberg [10]), however we will instead discuss a new
approach described in Chapter 3. This new approach does not use the estimates obtained
in Theorem 43 of Chapter 2, but relies on the interaction between the Burkholder functions
U and v used to obtain the sharp estimate. This is an entirely new approach of estimating
102

singular integral operators from below, in which we constructed an almost extremal sequence
for the operator, by means of laminates. The appropriate laminate sequence was constructed
by carefully studying the relationships of the Burkholder functions.
It should also be noted that there are very few singular integral operators whose exact
operator norm is known, as it is quite difficult to calculate this quantity for many singular
integral operators. The astute reader of Chapters 2 and 3, can see that this truly is the case
2
2
for R1 − R2 , τ I .
Most of the materials of this dissertation are adopted from the following publications and
preprints:
• N. Boros, L. Sz`kelyhidi, Jr., A. Volberg, “Laminates Meet Burkholder Functions.”
e
submitted to the Duke Mathematical Journal, (2011).
• N. Boros, P. Janakiraman, A. Volberg, “Perturbation of Burkholder’s martingale
transform and Monge-Amp`re equation”, submitted to Advances in Mathematics Joure
nal, (2011).
• N. Boros, P. Janakiraman, A. Volberg, “Sharp Lp -bounds for a small perturbation
of Burkholder’s martingale transform”, to appear in Indiana University Mathematics
Journal, (2011).
• N. Boros, P. Janakiraman, A. Volberg, “Sharp Lp -bounds for a perturbation of
Burkholders Martingale Transform”, Ser. I, C. R. Acad. Sci. Paris, Ser. I, 349:
303–307, (2011).
• N. Boros, A. Volberg, “Sharp Lower bound estimates for vector-valued and matrixvalued multipliers in Lp ”, arXiv:1110.5405v1, (2010).
103

4.2

Future work

Let B denote the Ahlfors-Beurling operator and B p denote the operator norm from Lp to
Lp . It was shown in 1965, by O. Lehto [28], that B p→p ≥ p∗ − 1. T. Iwaniec conjectured
in 1982, see [25], that B p→p = p∗ − 1. There have been many attempts at proving this
conjecture: Ba˜uelos, Wang [4], Nazarov, Volberg [31], Ba˜uelos, M`ndez-Hern`ndez [2],
n
n
e
a
Dragiˇevi´, Volberg [22], Ba˜uelos, Janakiraman [3] is the current list of attempts, which
c c
n
have slowly gotten closer to the result but have still not attained it. The implications
of the validity of this nearly 30 year old conjecture are broad reaching into several fields,
such as harmonic analysis, partial differential equations and in the study of quasi-conformal
mappings (for more details on the implications see [22]).
There are two difficulties in estimating B. One is that B is a complex operator, in
fact there are very few complex operators for which we know the exact Lp operator norm.
Secondly, B is a linear combination of squares of Riesz transforms and a product of two Riesz
transforms. (Recall that

2
2
B = R1 − R2 and

B = 2R1 R2 .) The method developed and

implemented for attaining the result in Corollary 94 can be applied to computing the exact
Lp norm of many other perturbation operators. In fact, two other such perturbations are a
linear perturbation already mentioned,

B + τ · I and a complex perturbation

Since

2
2 a − b (R2 − R2 ) + a + b (R2 + R2 )
aR1 + bR2 =
2
2
1
1
2
2
a−b
a+b 2
2
2
2
=
(R1 − R2 ) +
(R + R2 )
2
a−b 1
a−b
a+b
2
2
=
(R1 − R2 ) +
I ,
2
a−b
104

B + τ · iI.

2
2
the linear perturbation is just a constant multiple of aR1 + bR2 . So we now have a way for
approaching the Lp operator norm of a linear combination of squares of Riesz transforms as
well as for a complex operator. The two results would be interesting not just in obtaining
the exact Lp of the perturbation of

B, but along the way two other very interesting

generalizations of Burkholder’s famous result would be found as well. These two problems
are joint works with P. Janakiraman and A. Volberg.

105

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106

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