GEOMETRIC ASPECTS OF EXACT SOLUTIONS OF BELLMAN EQUATIONS OF
HARMONIC ANALYSIS PROBLEMS
By
Paata Ivanisvili

A DISSERTATION
Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of
Mathematics — Doctor of Philosophy
2015

ABSTRACT
GEOMETRIC ASPECTS OF EXACT SOLUTIONS OF BELLMAN
EQUATIONS OF HARMONIC ANALYSIS PROBLEMS
By
Paata Ivanisvili
In Chapter 1 we find the sharp constant C = C(τ, p, EG, EF ) of the following inequality
(G2 +τ 2 F 2 )1/2 p ≤ C F p , where G is the transform of a martingale F under a predictable
sequence ε with absolute value 1, 1 < p < 2, and τ is any real number. Thereby we solve
the open problem posted by Boros–Janakiraman–Volberg.
In Chapter 2 under some assumptions on the vectors a1 , . . . , an ∈ Rk and the function
B : Rk → R we find the sharp estimate of the expression Rk B(u1 (a1 · x), . . . , un (an · x))dx
in terms of R uj (y)dy, j = 1, . . . , n. In some particular cases (k = 1, n − 1 and n) we show
that these assumptions on B imply that there is only one Brascamp–Lieb inequality.
In Chapter 3 we find underlying PDEs on the Bellman functions B which imply inequalities such as John–Nirenberg inequality, Prekopa–Leindler inequality, Ehrhard’s inequality,
Borell’s Gaussian noise “stability”, hypercontractivity of Ornstein–Uhlenbeck semigroup,
logarithmic Sobolev inequality, Beckner–Sobolev inequality and Bobkov’s inequality. We
also describe underlying differential geometry that arises in solving these PDEs, and we
formulate some open questions.

to my fianc´e Maria.

iii

ACKNOWLEDGMENTS

I am very thankful to my advisor A. Volberg. I appreciate every time that we spent together
and everything that he has done for me. I remember once he challenged me to understand
that there are PDEs on some Bellman functions which govern the isoperimetric inequalities,
and as a result he introduced me to semigroup methods which we developed together in
Chapter 2 and Chapter 3. From the personal life side he implicitly gave me the answers on
the questions how one can develop a life of a professor of mathematics, and in what direction
I need to move. It was a great honor for me to having A. Volberg as my academic father
during my study in Michigan State University. He was the most reliable person for me in
this foreign country and this feeling already gave me a complete freedom to do research in
this beautiful place.
I would like to express my gratitude to F. Nazarov who was always inspiring me. On his
example I saw that it is better to be a mathematician rather than just to work in one specific
field - harmonic analysis. I do appreciate his unpredictability and I believe this is the reason
that allows us to cross the boundaries of our imaginations and to reach something new.
My gratitude is due to V. Vasyunin who first introduced me to Bellman function technique
and S. V. Kislyakov who was advising me during my study in Saint-Petersburg. I am thankful
to my colleagues and close friends P. Zatitskij, D. Stiolyarov and N. Osipov. This group of
people and myself, under the supervision of V. Vasyunin developed together the theory of
minimal concave functions in Chebyshev Laboratory founded by S. Smirnov. I am thankful
to S. Smirnov for supporting our group at Chebyshev Laboratory during our early very hard
times.
I am also thankful to those professors and teachers who contributed in rising young
iv

mathematician like me: V. Babich, A. Baranov, D. Chelkak, A. Florinskii, I. Ibragimov,
S. Ivanov, A. Khrabrov, V. Khavin, K. Kokhas, A. Nazarov, Ya. Nikitin V. Osmalovsky,
A. Podkarytov, A. Vershik, O. Vinogradov, and S. Zegzhda
I am grateful to my parents and my brothers for supporting me throughout my study. I
am grateful to my fian´ce Maria for what we already have created and for what we are going
to create in the future.
In my graduate years I had the company and the support of many other friends. I am
also grateful to my fellow MSU graduate students Ch. Ang, K. Bhola, T. Bongers, Z. Cang,
W. Chen, A. Chapman, M. Che, L. Chu, R. Fakhry, B. Jaye, W. Jiang S. Lee, G. Livshyts,
B. Mackey, M. Maridakis, S. Mehri, W. Park, G. Rey, A. Reznikov, A. Sharliev, B. Wang,
K. Wu, X. Yang, E. Yildiz, W. Zhou and the rest of the graduate students to whom I owe a
lot of joyful moments.

v

TABLE OF CONTENTS

LIST OF FIGURES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viii

Introduction

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Chapter 1 Inequality for Burkholder’s martingale transform . . . . .
1.1 History of the problem . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.1 Our main results . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.2 Plan of the Chapter 1 . . . . . . . . . . . . . . . . . . . . . . .
1.2 Definitions and known results . . . . . . . . . . . . . . . . . . . . . . .
1.3 Homogeneous Monge–Amp`ere equation and minimal concave functions
1.3.1 Foliation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.2 Cup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4 Construction of the Bellman function . . . . . . . . . . . . . . . . . . .
1.4.1 Reduction to the two dimensional case . . . . . . . . . . . . . .
1.4.2 Construction of a candidate for M . . . . . . . . . . . . . . . . .
1.4.3 Concavity in another direction . . . . . . . . . . . . . . . . . . .
1.5 Sharp constants via foliation . . . . . . . . . . . . . . . . . . . . . . . .
1.5.1 Main theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5.2 Case yp ≤ s0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5.3 Case yp > s0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6 Extremizers via foliation . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6.1 Case s0 ≤ yp . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6.2 Case s0 > yp . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 2

2.1

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1
4
4
6
8
9
11
11
24
35
35
39
47
55
55
58
59
63
67
71

Hessian of Bellman functions and uniqueness of Brascamp–
Lieb inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
2.0.3 Brascamp–Lieb inequality . . . . . . . . . . . . . . . . . . . . . . . . 78
2.0.4 Bellman function in Brascamp–Lieb inequality . . . . . . . . . . . . . 85
How to find the Bellman function . . . . . . . . . . . . . . . . . . . . . . . . 94
2.1.1 Case k = 1. Jointly concave and homogeneous function . . . . . . . . 96
2.1.2 Case k = n. B(y) = Const · y1 · · · yn . . . . . . . . . . . . . . . . . . 96
2.1.2.1 First proof . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
2.1.2.2 Second proof . . . . . . . . . . . . . . . . . . . . . . . . . . 97
2.1.2.3 Third proof . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
2.1.3 Case k = n − 1. Young’s function. . . . . . . . . . . . . . . . . . . . . 99
2.1.3.1 Example of necessity of the assumption Byi yj = 0 in Theorem
2.1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
2.1.3.2 Theorem 2.1.2 does not hold in the case 1 < k < n − 1 . . . 102
vi

2.1.4

Case of Young’s function . . . . . . . . . . . . . . . . . . . . . . . . . 104

Chapter 3 Harmonic analysis, PDE and differential geometry . . . . . . .
3.0.5 Short review of some harmonic analysis problems . . . . . . . . . . .
3.0.5.1 John–Nirenberg inequality . . . . . . . . . . . . . . . . . . .
3.0.5.2 Uniform convexity . . . . . . . . . . . . . . . . . . . . . . .
3.0.5.3 Brunn–Minkowski and isoperimetric inequalities . . . . . . .
3.0.5.4 Sobolev inequality . . . . . . . . . . . . . . . . . . . . . . .
3.0.5.5 Prekopa–Leindler inequality . . . . . . . . . . . . . . . . . .
3.0.5.6 Borell–Brascamp–Lieb inequality . . . . . . . . . . . . . . .
3.0.5.7 Ehrhard’s inequality . . . . . . . . . . . . . . . . . . . . . .
3.0.5.8 Borell’s Gaussian noise “stability” . . . . . . . . . . . . . .
3.0.5.9 Hypercontractivity . . . . . . . . . . . . . . . . . . . . . . .
3.0.5.10 Logarithmic Sobolev inequalities . . . . . . . . . . . . . . .
3.0.5.11 Beckner–Sobolev inequality . . . . . . . . . . . . . . . . . .
3.0.5.12 L´evy–Gromov’s isoperimetric inequlity . . . . . . . . . . . .
3.0.5.13 Bobkov’s inequality . . . . . . . . . . . . . . . . . . . . . . .
3.0.6 Relation to PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.0.6.1 Prekopa–Leindler, Ehrhard’s inequality and its underlying
PDE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.0.6.2 log-Sobolev inequality, Beckner–Sobolev inequality, Bobkov’s
inequality and its underliyng PDE . . . . . . . . . . . . . .
3.0.6.3 Brascamp–Lieb, hypercontractivity, Borell’s Gaussian noise
“stability” and its underlying PDE . . . . . . . . . . . . . .
3.0.6.4 Extremal problems on BM O, Ap classes, Gehring classes and
its underlying PDE . . . . . . . . . . . . . . . . . . . . . . .
3.0.7 Relation to differential geometry . . . . . . . . . . . . . . . . . . . . .
3.0.7.1 Developable surfaces, concavity and the torsion of the space
curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.0.7.2 Modified Monge–Amp`ere equation . . . . . . . . . . . . . .
BIBLIOGRAPHY

107
107
107
108
109
110
111
113
114
115
115
116
116
116
117
118
118
126
133
139
147
148
150

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

vii

LIST OF FIGURES

Figure 1.1

Domain Ω . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

12

Figure 1.2

Foliation Θ(J, g) . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14

Figure 1.3

˜ g) . . . . . . . . . . . . . . . . . . . . . .
Foliations Θ(J, g) and Θ(J,

21

Figure 1.4

Gluing of B− and B+ . . . . . . . . . . . . . . . . . . . . . . . . . .

21

Figure 1.5

Foliation Θcup (J, g) . . . . . . . . . . . . . . . . . . . . . . . . . . .

24

Figure 1.6

Graph v(s) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

33

Figure 1.7

Uniqueness of the cup . . . . . . . . . . . . . . . . . . . . . . . . . .

39

Figure 1.8

Segment (y) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

41

Figure 1.9

Foliations Θcup ((c, s0 ], g) and Θ([s0 , yp ), g) . . . . . . . . . . . . . .

43

Figure 1.10

1
<0 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Case u p−1

44

Figure 1.11

1
Case u p−1
≥0

. . . . . . . . . . . . . . . . . . . . . . . . . . .

52

Figure 1.12

Domain of R(s, z) . . . . . . . . . . . . . . . . . . . . . . . . . . . .

65

viii

Introduction
The current dissertation is split into 3 chapters. The first chapter solves the open problem
put forward by Boros–Janakiraman–Volberg regarding sharp estimates of the perturbation
for Burkholder’s martingale transform (see [8], [15]). The inequality stems from important
questions concerning the Lp bounds for the perturbation of Beurling–Ahlfors operator and
hence it is of interest. We refer the reader to recent works regarding martingale inequalities and estimates of Beurling–Ahlfors operator [1, 3, 4, 5, 8] and references therein. It is
worth mentioning that the problem was solved by using the theory of minimal concave functions developed jointly with N. Osipov, D. Stolyarov, P. Zatitskyj, V. Vasyunin and myself.
For example, such notions and objects as foliation, force functions, cup and torsion already
appeared in the recent works [9], [10], [11], [12] and [13]. However, since the theory was
developed in two dimensional setting, it required some additional technical work to solve
the problem in three dimensional setting. In particular it includes finding minimal concave
solution of homogeneous Monge–Amp`ere equation with Dirichlet and Neumann boundary
data, and minimality was proved by constructing optimal martingale trajectories along the
foliations. Concave solutions were found by detailed investigation of the important object:
smooth transformation of the torsion of Dirichlet boundary data (further called force functions, see (1.3.11)) which coincides up to some positive factor (depending only on the domain

1

and foliation) with the trace of the Hessian of a Bellman function (or mean curvature).
Second chapter is devoted to the inequalities of Brascamp–Lieb in the Lebesgue measure case for a general function B (see also [14]). The particular case B(x1 , . . . , xn ) =
1/p1

x1

1/pn

· · · xn

, which corresponds to the Brascamp–Lieb inequality is important for a num-

ber of reasons, including applications in analysis and convex geometry, and, for example,
includes the sharp form of Young’s convolution inequality (established in [22], and [25]). For
general function B, it turns out that B satisfies inequality of Brascamp–Lieb if under some
mild assumptions on B it also satisfies some interesting concavity condition (see property
L3 in Subsection 2.0.4, Chapter 2). Similar concavity condition was found recently independently by Ledoux (see [31]). Our second main contribution is that under the assumptions
L1-L5 (see properties L1-L5 in Subsection 2.0.4, Chapter 2) we give complete description of
such functions B in the case k = 1, k = n − 1 and k = n. In these special cases the results
below imply uniqueness of Brascamp–Lieb inequalities and it sheds light to the works of
E. Calren, E. Lieb and M. Loss (see [25]), and J. Bannett, A. Carbery, M. Christ and T. Tao
(see [26, 27]).
Third chapter mainly contains brief overview of some old and recent isoperimetric problems, hints and new ideas about their relations to PDE and differential geometry. This
chapter is not as rigorous as previous chapters because the initial purpose was to give to the
reader very short overview and an attempt of the general picture which shows underlying
PDEs and PDIs (partial differential inequalities), which govern these classical isoperimetric
inequalities of analysis. Even though the chapter is short, it requires very careful reading
since the objects (as they are written down) only make sense under some extra assumptions
on the functions. For example, If there are no a priori assumptions on the functions B and f ,
and the reader sees the expressions of the form R B(f (x)), the reader should automatically
2

assume that we deal with those B and f such that the composition B(f (x)) is measurable,
moreover it is integrable on the real line. The material is still under the preparation, therefore, for now we decided to avoid finding the best conditions under which the theorems of
the chapter are fulfilled. However, we believe that the reader can easily find for her/himself
at at least some sufficient conditions under which all the computations are justified (or one
can extract such conditions them from the applications given after each theorem).
Our goal is to try to find underlying PDEs and PDIs for the following inequalities:
John–Nirenbeg inequality, sharp inequalities on BM O, Ap , Reverse H¨older, Gehring and
the classes of functions with bounded oscillation (see [9, 10, 11, 12] and references therein),
uniform convexity (see [16, 13, 28, 29, 30] and references therein). Isoperimetric, Prekopa–
Leindler and Ehrhard’s inequality (see [39, 38, 37, 33, 32, 34, 35, 36] and references therein ).
Borell’s Gaussian noise “stability” and hypercontractivity for Ornstein–Uhlenbeck semigroup
(see [31, 40, 42, 43, 44, 45, 46, 47] and references therein). Log-Sobolev, Beckner–Sobolev
and Bobkov’s inequality (see [48, 49, 50, 51, 52, 53, 54, 55] and references therein). Plan
of the chapter is simple. It is divided in 3 parts. First part briefly formulates these inequalities. Second part finds underlying PDEs and PDIs for these inequalities, and the third
part mentions the relation to differential geometry and in it we also formulate some open
questions.

3

Chapter 1
Inequality for Burkholder’s
martingale transform
1.1 History of the problem
Let I be an interval of the real line R, and let |I| be its Lebesgue length. By symbol B we
denote the σ-algebra of Borel subsets of I. Let {Fn }∞
n=0 be a martingale on the probability
space (I, B, dx/|I|) with a filtration {I, ∅} = F0 ⊂ F1 ⊂ ... ⊂ F. Consider any sequence of
functions {εn }∞
n=1 such that for each n ≥ 1, εn is Fn−1 measurable and |εn | ≤ 1. Let G0
be a constant function on I; for any n ≥ 1, let Gn denote
n

G0 +
k=1

εk (Fk − Fk−1 ).

∞
The sequence {Gn }∞
n=0 is called the martingale transform of {Fn }. Obviously {Gn }n=0

is a martingale with the same filtration {Fn }∞
n=0 . Note that since {Fn } and {Gn } are
martingales, we have F0 = EFn and G0 = EGn for any n ≥ 0.
In [17] Burkholder proved that if |G0 | ≤ |F0 |, 1 < p < ∞, then we have the sharp
estimate

Gn Lp ≤ (p∗ − 1) Fn Lp

4

for all n ≥ 0,

(1.1.1)

1 }. Burkholder showed that it is sufficient to prove inequality
where p∗ − 1 = max{p − 1, p−1

(1.1.1) for the sequences of numbers {εn } such that εn = ±1 for all n ≥ 1. It was also
mentioned that such an estimate as (1.1.1) does not depend on the choice of filtration {Fn }.
For example, one can consider only the dyadic filtration. For more information on the
estimate (1.1.1) we refer the reader to [17], [18].
In [20] the result was slightly generalized by Bellman function technique and Monge–
Amp`ere equation, i.e., the estimate (1.1.1) holds if and only if

|G0 | ≤ (p∗ − 1)|F0 |.

(1.1.2)

In what follows we assume that {εn } is a predictable sequence of functions such that
|εn | = 1.
In [8], a perturbation of the martingale transform was investigated. Namely, under the
same assumptions as (1.1.2) it was proved that for 2 ≤ p < ∞, τ ∈ R, we have the sharp
estimate

(G2n + τ 2 Fn2 )1/2 Lp ≤ ((p∗ − 1)2 + τ 2 )1/2 Fn Lp ,

for all n ≥ 0.

(1.1.3)

It was also claimed to be proven that the same sharp estimate holds for 1 < p < 2, |τ | ≤ 0.5,
and the case 1 < p < 2, |τ | > 0.5 was left open.
The inequality (1.1.3) stems from important questions concerning the Lp bounds for the
perturbation of Beurling–Ahlfors operator and hence it is of interest. We refer the reader to
recent works regarding martingale inequalities and estimates of Beurling–Ahlfors operator
[1, 3, 4, 5, 8] and references therein.

5

We should mention that Burkholder’s method [17] and the Bellman function approach [20], [8]
have similar traces in the sense that both of them reduce the required estimate to finding
a certain minimal diagonally concave function with prescribed boundary conditions. However, the methods of construction of such a function are different. Unlike Burkholder’s
method [17], in [20] and [8] the construction of the function is based on the Monge–Amp`ere
equation.

1.1.1 Our main results
Firstly, we should mention that the proof of (1.1.3) presented in [8] has a gap in the case
1 < p < 2, 0 < |τ | ≤ 0.5 (the constructed function does not satisfy necessary concavity
condition).
In the present paper we obtain the sharp Lp estimate of the perturbed martingale transform for the remaining case 1 < p < 2 and for all τ ∈ R. Moreover, we do not require
condition (1.1.2).
We define

def

u(z) = τ p (p − 1) τ 2 + z 2

(2−p)/2

− τ 2 (p − 1) + (1 + z)2−p − z(2 − p) − 1.

∞
Theorem 1.1.1. Let 1 < p < 2, and let {Gn }∞
n=0 be a martingale transform of {Fn }n=0 .
|G |−|F |

Set β = |G0 |+|F0 | . The following estimates are sharp:
0
0
1
1. If u p−1
≤ 0 then

(τ 2 Fn2 + G2n )1/2 Lp ≤

τ 2 + max

G0
1
,
F0 p − 1
6

2

1
2

F n Lp ,

for all

n ≥ 0.

1
> 0 then
2. If u p−1

p

p

(τ 2 Fn2 + G2n )1/2 Lp ≤ C(β) Fn Lp ,

for all n ≥ 0,

where C(β) is continuous nondecreasing, and it is defined as follows:

p/2


|G0 |2
2

τ
+
,


|F0 |2





τp
def
,
C(β) =
22−p (1−s0 )p−1


1
−

(τ 2 +1)(p−1)(1−s0 )+2(2−p)






C(β),

β ≥ s0 ;
β ≤ −1 + p2 ;
β ∈ (−1 + 2/p, s0 );

0
where s0 ∈ (−1 + 2/p, 1) is the solution of the equation u 1+s
1−s0

= 0.

Explicit expression for the function C(β) on the interval (−1+2/p, s0 ) was hard to present
in a simple way. The reader can find the value of the function C(β) in Theorem 1.5.1, part
(ii).
1
Remark 1. The condition u p−1

≤ 0 holds when |τ | ≤ 0.822. So we also obtain

Burkholder’s result in the limit case when τ = 0. It is worth mentioning that although
the proof of the estimate (1.1.3) has a gap in [8], the claimed result in the case 1 < p < 2,
|τ | < 0.5 remains true as a result of Theorem 1.1.1.
One of the important results is that we find the function (1.2.2), and the above estimates
are corollaries of this result. We would like to mention that unlike [20] and [8] the argument
exploited in the current paper is different. Instead of writing a lot of technical computations
and checking which case is valid, we present some pure geometrical facts regarding minimal
concave functions with prescribed boundary conditions, and by this way we avoid compu7

tations. Moreover, we explain to the reader how we construct our Bellman function (1.2.2)
based on these geometrical facts derived in Section 1.3.

1.1.2 Plan of the Chapter 1
In Section 1.2 we formulate results about how to reduce the estimate (1.1.3) to finding of a
certain function with required properties. These results are well-known and can be found in
[8]. A slightly different function was investigated in [20], however, it possesses almost the
same properties and the proof works exactly in the same way. We only mention these results
and the fact that we look for a minimal continuous diagonally concave function H(x1 , x2 , x3 )
(see Definition 3) in the domain Ω = {(x1 , x2 , x3 ) ∈ R3 : |x1 |p ≤ x3 } with the boundary
condition H(x1 , x2 , |x1 |p ) = (x22 + τ 2 x21 )p/2 .
Section 1.3 is devoted to the investigation of the minimal concave functions in two variables. It is worth mentioning that the first crucial steps in this direction for some special
cases were made in [9] (see also [10, 11]). In Section 1.3 we develop this theory for a slightly
more general case. We investigate some special foliation called the cup and another useful
object, called force functions.
We should note that the theory of minimal concave functions in two variables does not
include the minimal diagonally concave functions in three variables. Nevertheless, this knowledge allows us to construct the candidate for H in Section 1.4, but with some additional
technical work not mentioned in Section 1.3.
In section 1.5 we find the good estimates for the perturbed martingale transform. In
Section 1.6 we prove that the candidate for H constructed in Section 1.4 coincides with H,
and as a corollary we show the sharpness of the estimates found for the perturbed martingale
transform in Section 1.5.
8

In conclusion, the reader can note that the hard technical lies in the construction of the
minimal diagonally concave function in three variables with the given boundary condition.

1.2 Definitions and known results
def

Let EF = F

I

where

F

def
J

=

1
F (t)dt
|J| J

for any interval J of the real line. Let F and G be real valued integrable functions. Let
Gn = E(G|Mn ) and Fn = E(F |Mn ) for n ≥ 0, where {Mn } is a dyadic filtration (see [8]).
Definition 1. If the martingale {Gn } satisfies |Gn+1 − Gn | = |Fn+1 − Fn | for each n ≥ 0,
then G is called the martingale transform of F .
Recall that we are interested in the estimate

(G2 + τ 2 F 2 )1/2 Lp ≤ C F Lp .

(1.2.1)

We introduce the Bellman function

def

H(x) = sup{EB(ϕ(F, G)), Eϕ(F, G) = x, |Gn+1 − Gn | = |Fn+1 − Fn |, n ≥ 0}.
F,G

where ϕ(x1 , x2 ) = (x1 , x2 , |x1 |p ), B(ϕ(x1 , x2 )) = (x22 + τ 2 x21 )p/2 , x = (x1 , x2 , x3 ).
Remark 2. In what follows bold lowercase letters denote points in R3 .

9

(1.2.2)

Then we see that the estimate (1.2.1) can be rewritten as follows:

H(x1 , x2 , x3 ) ≤ C p x3 .
We mention that the Bellman function H does not depend on the choice of the interval
I. Without loss of generality we may assume that I = [0, 1].
Definition 2. Given a point x ∈ R3 , a pair (F, G) is said to be admissible for x if G is the
martingale transform of F and E(F, G, |F |p ) = x.
Proposition 1. The domain of H(x) is Ω = {(x1 , x2 , x3 ) ∈ R3 : |x1 |p ≤ x3 }, and H
satisfies the boundary condition

H(x1 , x2 , |x1 |p ) = (x22 + τ 2 x21 )p/2 .

(1.2.3)

Definition 3. A function U is said to be diagonally concave in Ω, if it is concave in both
Ω ∩ {(x1 , x2 , x3 ) : x1 + x2 = A} and Ω ∩ {(x1 , x2 , x3 ) : x1 − x2 = A} for every constant
A ∈ R.
Proposition 2. H(x) is a diagonally concave function in Ω.
Proposition 3. If U is a continuous diagonally concave function in Ω with boundary condition U (x1 , x2 , |x1 |p ) ≥ (x22 + τ 2 x21 )p/2 , then U ≥ H in Ω.
We explain our strategy of finding the Bellman function H. We are going to find a minimal
candidate B, that is continuous, diagonally concave, with the fixed boundary condition
B|∂Ω = (y 2 + τ 2 x2 )p/2 . We warn the reader that the symbol B denoted boundary data
previously, however, in Section 1.6 we are going to use symbol B as the candidate for the
10

minimal diagonally concave function. Obviously B ≥ H by Proposition 3. We will also
see that given x ∈ Ω and any ε > 0, we can construct an admissible pair (F, G) such that
B(x) < E(F 2 + τ 2 G2 )p/2 + ε. This will show that B ≤ H and hence B = H.
In order to construct the minimal candidate B, we have to elaborate few preliminary
concepts from differential geometry. We introduce notion of foliation and force functions.

1.3 Homogeneous Monge–Amp`ere equation and minimal concave
functions
1.3.1 Foliation
Let g(s) ∈ C 3 (I) be such that g > 0, and let Ω be a convex domain which is bounded
by the curve (s, g(s)) and the tangents that pass through the end-points of the curve (see
Figure 1.1). Fix some function f (s) ∈ C 3 (I). The first question we ask is the following: how
the minimal concave function B(x1 , x2 ) with boundary data B(s, g(s))) = f (s) looks locally
in a subdomain of Ω. In other words, take a convex hull of the curve (s, g(s), f (s)), s ∈ I,
then the question is how the boundary of this convex hull looks like.
We recall that the concavity is equivalent to the following inequalities:

det(d2 B) ≥ 0,

(1.3.1)

Bx1 x1 + Bx2 x2 ≤ 0.

(1.3.2)

The expression (1.3.1) is the Gaussian curvature of the surface (x1 , x2 , B(x1 , x2 )) up to a
positive factor (1 + (Bx1 )2 + (Bx2 )2 )2 . So in order to minimize the function B(x1 , x2 ), it is
reasonable to minimize the Gaussian curvature. Therefore, we will look for a surface with
11

zero Gaussian curvature. Here the homogeneous Monge–Amp`ere equation arises. These
surfaces are known as developable surfaces i.e., such a surface can be constructed by bending
a plane region. The important property of such surfaces is that they consist of line segments,
i.e., the function B satisfying homogeneous Monge–Amp`ere equation det(d2 B) = 0 is linear
along some family of segments. These considerations lead us to investigate such functions
B. Firstly, we define a foliation. For any segment
denote an open segment i.e.,

in the Euclidean space by symbol ◦ we

without endpoints.

y

Ω

I

s

Figure 1.1 Domain Ω

Fix any subinterval J ⊆ I. By symbol Θ(J, g) we denote an arbitrary set of nontrivial
segments (i.e. single points are excluded) in R2 with the following requirements:
1. For any

∈ Θ(J, g) we have ◦ ∈ Ω.

2. For any 1 , 2 ∈ Θ(J, g) we have 1 ∩ 2 = ∅.
3. For any

∈ Θ(J, g) there exists only one point s ∈ J such that (s, g(s)) is one of the

end-points of the segment and vice versa, for any point s ∈ J there exists ∈ Θ(J, g)
12

such that (s, g(s)) is one of the end-points of the segment .
4. There exists C 1 smooth argument function θ(s).
We explain the meaning of the requirement 4. To each point s ∈ J there corresponds
only one segment

∈ Θ(J, g) with an endpoint (s, g(s)). Take a nonzero vector with initial

point (s, g(s)), parallel to the segment

and having an endpoint in Ω. We define the value

of θ(s) to be an argument of this vector. Surely argument is defined up to additive number
2πk where k ∈ Z. Nevertheless, we take any representative from these angles. We do the
same for all other points s ∈ I. In this way we get a family of functions θ(s). If there exists
C 1 (J) smooth function θ(s) from this family then the requirement 4 is satisfied.
Remark 3. It is clear that if θ(s) is C 1 (J) smooth argument function, then for any k ∈
Z, θ(s) + 2πk is also C 1 (J) smooth argument function. Any two C 1 (J) smooth argument
functions differ by constant 2πn for some n ∈ Z.
This remark is the consequence of the fact that the quantity θ (s) is well defined regardless of the choices of θ(s). Next, we define Ω(Θ(J, g)) = ∪ ∈Θ(J,g) ◦ . Given a point
x ∈ Ω(Θ(J, g)) we denote by (x) a segment (x) ∈ Θ(J, g) which passes through the point
x. If x = (s, g(s)) then instead of ((s, g(s))) we just write (s). Surely such a segment
exists, and it is unique. We denote by s(x) a point s(x) ∈ J such that (s(x), g(s(x))) is
one of the end points of the segment (x). Moreover, in a natural way we set s(x) = s if
x = (s, g(s)). It is clear that such s(x) exists, and it is unique. We introduce a function

K(s) = g (s) cos θ(s) − sin θ(s),

s ∈ J.

(1.3.3)

Note that that K < 0. This inequality becomes obvious if we rewrite g (s) cos θ(s) −
13

sin θ(s) = (1, g ), (− sin θ, cos θ) and take into account the requirement 1 of Θ(J, g). Note
that ·, · means scalar product in Euclidean space. We need few more requirements on
Θ(J, g).

g

J
Figure 1.2 Foliation Θ(J, g)

5. For any x = (x1 , x2 ) ∈ Ω(Θ(J, g)) we have an inequality K(s(x)) + θ (s(x)) (x1 −
s(x), x2 − g(s(x))) < 0.
6. The function s(x) is continuous in Ω(Θ(J, g)) ∪ Γ(J) where Γ(J) = {(s, g(s)) : s ∈ J}.
Note that if θ (s) ≤ 0 (which happens in most of the cases) then the requirement 5 holds.
If we know the endpoints of the segments Θ(J, g), then in order to verify the requirement 5
it is enough to check at those points x = (x1 , x2 ), where x is the another endpoint of the
segment other than (s, g(s)). Roughly speaking the requirement 5 means the segments of
Θ(J, g) do not rotate rapidly counterclockwise.
Definition 4. A set of segments Θ(J, g) with the requirements mentioned above is called
foliation. The set Ω(Θ(J, g)) is called domain of foliation.
A typical example of a foliation is given in Figure 1.2.

14

Lemma 1. The function s(x) belongs to C 1 (Ω(Θ(J, g))). Moreover

(sx1 , sx2 ) =

(sin θ, − cos θ)
.
−K(s) − θ · (x1 − s, x2 − g(s))

(1.3.4)

Proof. Definition of the function s(x) implies that

−(x1 − s) sin θ(s) + (x2 − g(s)) cos θ(s) = 0.
Therefore the lemma is an immediate consequence of the implicit function theorem.
Let J = [s1 , s2 ] ⊆ I, and let (s, g(s), f (s)) ∈ C 3 (I) be such that g > 0 on I. Consider an
arbitrary foliation Θ(J, g) with an arbitrary C 1 ([s1 , s2 ]) smooth argument function θ(s). We
need the following technical lemma which describes behavior of the gradient of the function
B which satisfies homogeneous Monge–Amp`ere equation.
Lemma 2. The solutions of the system of equations

t1 (s) cos θ(s) + t2 (s) sin θ(s) = 0,
t1 (s) + t2 (s)g (s) = f (s),

s∈J

are the following functions
s

t1 (s) =

s1

f (r)
g (r)
sin θ(r) · t2 (r) −
sin θ(r) dr + f (s1 ) − t2 (s1 )g (s1 ),
K(r)
K(r)
s

g (r)
cos θ(r)dr +
s1 K(r)
s f (y)
s g (r)
exp −
cos θ(r)dr cos θ(y)dy,
s1 K(y)
y K(r)

t2 (s) = t2 (s1 ) exp −

15

s∈J

(1.3.5)
(1.3.6)

where t2 (s1 ) is an arbitrary real number.
Proof. We differentiate (1.3.6) and combine it with (1.3.5) to obtain the system


  
   
cos θ sin θ t1  0 0  t1   0 

  = 
  +  .
1
g
t2
0 −g
t2
f
This implies that

 


 
t1  g 0 sin θ  t1  f − sin θ

  +

.
 =
K
K
0 − cos θ
t2
cos θ
t2

(1.3.7)

By solving this system of differential equations and using the fact that t1 (s1 )+g (s1 )t2 (s1 ) =
f (s1 ) we get the desired result.
Remark 4. Integration by parts allows us to rewrite the expression for t2 (s) as follows
s

t2 (s) = exp −
s

−

s1

f (y)
g (y)

g (r)
f (s1 )
cos θ(r)dr
t2 (s1 ) −
g (s1 )
s1 K(r)
s g (r)
exp −
cos θ(r)dr dy.
y K(r)

+

f (s)
−
g (s)

Definition 5. We say that a function B has a foliation Θ(J, g) if it is continuous on
Ω(Θ(J, g)), and it is linear on each segment of Θ(J, g).
The following lemma describes how to construct a function B with a given foliation
Θ(J, g), boundary condition B(s, g(s)) = f (s), such that B satisfies the homogeneous
Monge–Amp`ere equation.

16

Consider a function B defined as follows

B(x) = f (s) + t(s), x − (s, g(s)) ,

x = (x1 , x2 ) ∈ Ω(Θ(J, g))

(1.3.8)

where s = s(x), and t(s) = (t1 (s), t2 (s)) satisfies the system of the equations (1.3.5), (1.3.6)
with an arbitrary t2 (s1 ).
Lemma 3. The function B defined by (1.3.8) satisfies the following properties:
1. B ∈ C 2 (Ω(Θ(J, g))) ∩ C 1 (Ω(Θ(J, g)) ∪ Γ), B has the foliation Θ(J, g) and
B(s, g(s)) = f (s) for all

s ∈ [s1 , s2 ].

(1.3.9)

2. ∇B(x) = t(s), where s = s(x), moreover B satisfies the homogeneous Monge–Amp`ere
equation.
Proof. The fact that B has the foliation Θ(J, g), and it satisfies the equality (1.3.9) immediately follows from the definition of the function B. We check the condition of smoothness.
By Lemma 1 and Lemma 2 we have s(x) ∈ C 2 (Ω(Θ(J, g))) and t1 , t2 ∈ C 1 (J), therefore the
right-hand side of (1.3.8) is differentiable with respect to x. So after differentiation of (1.3.8)
we get

∇B(x) = f (s) − t(s), (1, g (s)) (sx1 , sx2 ) + t(s) + t (s), x − (s, g(s)) (sx1 , sx2 ).
(1.3.10)
Using (1.3.5) and (1.3.6) we obtain ∇B(x) = t(s). Taking derivative with respect to x the

17

second time we get
∂ 2B
= t1 (s)sx1 ,
∂x21

∂ 2B
= t1 (s)sx2 ,
∂x2 ∂x1

∂ 2B
= t2 (s)sx1 ,
∂x1 ∂x2

∂ 2B
= t2 (s)sx2 .
∂x22

Using (1.3.5) we get that t1 (s)sx2 = t2 (s)sx1 , therefore B ∈ C 2 (Ω(Θ(J, g))). Finally, we
check that B satisfies the homogeneous Monge–Amp`ere equation. Indeed,

det(d2 B) =

∂ 2B ∂ 2B
∂ 2B
∂ 2B
·
−
·
= t1 (s)sx1 · t2 (s)sx1 − t1 (s)sx2 · t2 (s)sx1 = 0.
∂x2 ∂x1 ∂x1 ∂x2
∂x21 ∂x22

Definition 6. The function t(s) = (t1 (s), t2 (s)) = ∇B(x), s = s(x), is called gradient
function corresponding to B.
The following lemma investigates the concavity of the function B defined by (1.3.8). Let
˜(x) = (s(x) − x1 , g(s(x)) − x2 ) , where x = (x1 , x2 ) ∈ Ω(Θ(J, g)).
Lemma 4. The following equalities hold
g
∂ 2B ∂ 2B
f
+
=
−t2 +
=
2
2
˜
g
∂x1
∂x2
K(K + θ (x) )
s g (r)
g
× − exp −
cos θ(r)dr
K(K + θ ˜(x) )
s1 K(r)
s

+
s1

f (y)
g (y)

s

exp −

g (r)
cos θ(r)dr dy .
y K(r)

Proof. Note that
∂ 2B ∂ 2B
+
= t1 (s)s1 + t2 (s)s2 .
∂x21
∂x22
18

t2 (s1 ) −

f (s1 )
g (s1 )

Therefore the lemma is a direct computation and application of Equalities (1.3.4), (1.3.5),
(1.3.6) and Remark 4.
Finally, we get the following important statement.
Corollary 1. The function B is concave in Ω(Θ(J, g)) if and only if F(s) ≤ 0, where
s

g (r)
f (s1 )
cos θ(r)dr
t2 (s1 ) −
g (s1 )
s1 K(r)
s g (r)
f (s)
cos θ(r)dr dy =
− t2 (s).
exp −
g (s)
y K(r)

F(s) = − exp −
s

+
s1

f (y)
g (y)

(1.3.11)

Proof. B satisfies the homogeneous Monge–Amp`ere equation. Therefore B is concave if and
only if
∂ 2B ∂ 2B
+
≤ 0.
∂x21
∂x22

(1.3.12)

Note that

g
> 0.
K(K + θ ˜(x) )
Hence, according to Lemma 4, the inequality (1.3.12) holds if and only if F(s) ≤ 0.
Furthermore, the function F will be called force function.
Remark 5. The fact t2 (s) = f /g − F together with (1.3.7) imply that the force function

19

F satisfies the following differential equation
F +F ·

f
cos θ
g −
K
g

F(s1 ) =

f (s1 )
− t2 (s1 ).
g (s1 )

= 0,

s∈J

(1.3.13)

We remind the reader that for an arbitrary smooth curve γ = (s, g(s), f (s)), the torsion
has the following expression
f g −g f
(g )2
f
det(γ , γ , γ )
=
=
·
2
2
2
g
γ ×γ
γ ×γ
γ ×γ

.

Corollary 2. If F(s1 ) ≤ 0 and the torsion of a curve (s, g(s), f (s)), s ∈ J is negative, then
the function B defined by (1.3.8) is concave.
Proof. The corollary is an immediate consequence of (1.3.11).
Thus, we see that the torsion of the boundary data plays a crucial role in the concavity of
a surface with zero Gaussian curvature. More detailed investigations about how we choose
the constant t2 (s1 ) will be given in Subsection 1.3.2.
˜
˜ g) be foliations with some argument functions θ(s) and θ(s)
Let Θ(J, g) and Θ(J,
re˜ be the corresponding functions defined by (1.3.8), and let F, F˜
spectively. Let B and B
˜
be the corresponding force functions. Note that F(s) = F(s)
is equivalent to the equality t(s) = t˜(s) where t(s) = (t1 (s), t2 (s)) and t˜(s) = (t˜1 (s), t2 (s)) are the corresponding
˜ (see (1.3.6) and Corollary 1).
gradients of B and B
˜ are concave functions.
Assume that the functions B and B
˜ ≤ B on Ω(Θ(J, g)) ∩
˜ 1 ), then B
Lemma 5. If sin(θ˜ − θ) ≥ 0 for all s ∈ J, and F(s1 ) = F(s
˜
Ω(Θ(J,
g)).
20

˜

(x)

˜
Θ(J,g)
Θ(J,g)

g

x

(s(x),g(s(x)))

˜ g)
Figure 1.3 Foliations Θ(J, g) and Θ(J,

In other words, the lemma says that if at initial point (s1 , g(s1 )) gradients of the functions
˜ and B coincide, and the foliation Θ(J,
˜ g) is “to the left of” the foliation Θ(J, g) (see
B
˜ ≤ B provided B and B
˜ are concave.
Figure 1.3) then B
˜ defined by (1.3.3). The
˜ be the corresponding functions of B and B
Proof. Let K and K
˜ < 0 implies that the inequality sin(θ˜ − θ) ≥ 0 is equivalent to the inequality
condition K, K
cos θ˜ cos θ
≥
˜
K
K

for s ∈ J.

(1.3.14)

˜ cos θ then this simplifies to − sin θ cos θ˜ ≥
Indeed, if we rewrite (1.3.14) as K cos θ˜ ≥ K
− sin θ˜ cos θ, so the result follows. The force functions F, F˜ satisfy the differential equation
Θ−

−

Θ+

+

Ang(s2 )
g

(s2 ,g(s2 ))
J−

J+

Figure 1.4 Gluing of B− and B+

21

˜ 1 ). Then by (1.3.14) and by compar(1.3.13) with the same boundary condition F(s1 ) = F(s
ison theorems we get F˜ ≥ F on J. This and (1.3.11) imply that t˜2 ≤ t2 on J. Pick any point
˜
x ∈ Ω(Θ(J, g)) ∩ Ω(Θ(J,
g)). Then there exists a segment (x) ∈ Θ(J, g). Let (s(x), g(s(x)))
˜ g) which
be the corresponding endpoint of this segment. There exists a segment ˜ ∈ Θ(J,
has (s(x), g(s(x))) as an endpoint (see Figure 1.3).
˜ at point (s(x), g(s(x))). The fact that the
Consider a tangent plane L(x) to (x1 , x2 , B)
˜ is constant on ˜, implies that L is tangent to (x1 , x2 , B)
˜ on ˜. Therefore
gradient of B

L(x) = f (s) + (t˜1 (s), t˜2 (s)), (x1 − s, x2 − g(s)) ,
˜ implies that a value of the function B
˜
where x = (x1 , x2 ) and s = s(x). Concavity of B
˜
at point y seen from the point (s(x), g(s(x))) is less than L(y). In particular B(x)
≤ L(x).
Now it is enough to prove that L(x) ≤ B(x). By (1.3.8) we have
B(x) = f (s) + (t1 (s), t2 (s)), (x1 − s(x), x2 − g(s)) .
Therefore using (1.3.6), (−g , 1), (x1 − s, x2 − g(s)) ≥ 0 and the fact that t˜2 ≤ t2 we get
the desired result.
Let J − = [s1 , s2 ] and J + = [s2 , s3 ] where J − , J + ⊂ I. Consider arbitrary foliations
Θ− = Θ− (J − , g) and Θ+ = Θ+ (J + , g) such that Ω(Θ− ) ∩ Ω(Θ+ ) = ∅, and let θ− and θ+
be the corresponding argument functions. Let B− and B+ be the corresponding functions
−
+ +
+
defined by (1.3.8), and let t− = (t−
1 , t2 ), t = (t1 , t2 ) be the corresponding gradient

functions. Set Ang(s2 ) to be a convex hull of − (s2 ) and + (s2 ) where − (s2 ) ∈ Θ− ,
+ (s

2)

∈ Θ+ are the segments with the endpoint (s2 , g(s2 )) (see Figure 1.4). We require
22

that Ang(s2 ) ∩ Ω(Θ− ) = − and Ang(s2 ) ∩ Ω(Θ+ ) = + .
Let F − , F + be the corresponding forces, and let BAng be the function defined linearly
on Ang(s2 ) via the values of B− and B+ on − , + respectively.
+
Lemma 6. If t−
2 (s2 ) = t2 (s2 ), then the function B defined as follows






B− (x),
x ∈ Ω(Θ(J − , g)),





B(x) = B
Ang (x), x ∈ Ang(s2 ),








B+ (x),
x ∈ Ω(Θ(J + , g)),
belongs to the class C 1 (Ω(Θ− ) ∪ Ang(s2 ) ∪ Ω(Θ+ ) ∪ Γ(J − ∪ J + )).
+
−
Proof. By (1.3.6) the condition t−
2 (s2 ) = t2 (s2 ) is equivalent to the condition t (s2 ) =

t+ (s2 ). We recall that the gradient of B− is constant on − (s2 ), and the gradient of
B+ is constant on + (s2 ), therefore the lemma follows immediately from the fact that
B− (s2 , g(s2 )) = B+ (s2 , g(s2 )).
Remark 6. The fact B ∈ C 1 implies that its gradient function t(s) = ∇B is well defined, and
it is continuous. Unfortunately, it is not necessarily true that t(s) ∈ C 1 ([s1 , s3 ]). However,
it is clear that t(s) ∈ C 1 ([s1 , s2 ]), and t(s) ∈ C 1 ([s2 , s3 ]).
Finally we finish this section with the following important corollary about concave extension of the functions with zero gaussian curvature.
+
Let B− and B+ be defined as above (see Figure 1.4). Assume that t−
2 (s2 ) = t2 (s2 ).

Corollary 3. If B− is concave in Ω(Θ− ) and the torsion of the curve (s, g(s), f (s)) is
nonnegative on J + = [s2 , s3 ] then the function B defined in Lemma 6 is concave in the
domain Ω(Θ− ) ∪ Ang(s2 ) ∪ Ω(Θ+ ).
23

In other words the corollary tells us that if we have constructed concave function B−
which satisfies homogeneous Monge–Amp`ere equation, and we glued B− smoothly with B+
(which also satisfies homogeneous Monge–Amp`ere equation), then the result B is concave
function provided that the space curve (s, g(s), f (s)) has nonnegative torsion on the interval
J +.
Proof. By Lemma 1 concavity of B− implies F − (s2 ) ≤ 0. By (1.3.11) the condition t−
2 (s2 ) =
−
+
+
t+
2 (s2 ) is equivalent to F (s2 ) = F (s2 ). By Corollary 2 we get that B is concave. Thus,

concavity of B follows from Lemma 6.

1.3.2 Cup

(s1 ,g(s1 ))

g

J
a(s1 )

a(s0 )

I

s0

s1

Figure 1.5 Foliation Θcup (J, g)

In this subsection we are going to consider a special type of foliation which is called Cup.
Fix an interval I and consider an arbitrary curve (s, g(s), f (s)) ∈ C 3 (I). We suppose that
g > 0 on I. Let a(s) ∈ C 1 (J) be a function such that a (s) < 0 on J, where J = [s0 , s1 ]
is a subinterval of I. Assume that a(s0 ) < s0 and [a(s1 ), a(s0 )] ⊂ I. Consider a set of open
segments Θcup (J, g) consisting of those segments (s, g(s)), s ∈ J such that (s, g(s)) is a
segment in the plane joining the points (s, g(s)) and (a(s), g(a(s))) (see Figure 1.5).
24

Lemma 7. The set of segments Θcup (J, g) described above forms a foliation.
Proof. We need to check the 6 requirements for a set to be the foliation. Most of them are
trivial except for 4 and 5. We know the endpoints of each segment therefore we can consider
the following argument function

θ(s) = π + arctan

g(s) − g(a(s))
s − a(s)

.

Surely θ(s) ∈ C 1 (J), so requirement 4 is satisfied. We check requirement 5. It is clear that
it is enough to check this requirement for x = (a(s), g(a(s)). Let s = s(x), then

K(s) + θ (s) (a(s) − s, g(a(s)) − g(s)) =

(1, g ), (g − g(a), a − s)
+
(g(a) − g, s − a)

a · (1, g (a)), (g − g(a), a − s)
(g − a g (a))(s − a) − (1 − a )(g − g(a))
=
(g(a) − g, s − a)
(g(a) − g, s − a)
which is strictly negative.
Let γ(t) = (t, g(t), f (t)) ∈ C 3 ([a0 , b0 ]) be an arbitrary curve such that g > 0 on [a0 , b0 ].
Assume that the torsion of γ is positive on I − = (a0 , c), and it is negative on I + = (c, b0 )
for some c ∈ (a0 , b0 ).
Lemma 8. For all P such that 0 < P < min{c − a0 , b0 − c} there exist a ∈ I − , b ∈ I + such
that b − a = P and
1

a−b

1

g (a) g (b) g(a) − g(b) = 0.
f (a) f (b) f (a) − f (b)

25

(1.3.15)

Proof. Pick a number a ∈ (a0 , b0 ) so that b = a + P ∈ (a0 , b0 ). We denote
M(a, b) = (a − b)(g (b) − g (a))

g(a) − g(b)
− g (a) .
a−b

Note that the conditions a = b and g > 0 imply M(a, b) = 0. Then
1

1

a−b

f (a) − f (b) − f (a)(a − b) f (b) − f (a)
g (a) g (b) g(a) − g(b) = M(a, b) g(a) − g(b) − g (a)(a − b) − g (b) − g (a) .
f (a) f (b) f (a) − f (b)
Thus our equation (1.3.15) turns into

f (a) − f (b) − f (a)(a − b) f (b) − f (a)
−
= 0.
g(a) − g(b) − g (a)(a − b)
g (b) − g (a)

(1.3.16)

We consider the following functions V (x) = f (x) − f (a)x and U (x) = g(x) − g (a)x. Note
that U (a) = U (b) and U = 0 on (a, b). Therefore by Cauchy’s mean value theorem there
exists a point ξ = ξ(a, b) ∈ (a, b) such that
f (a) − f (b) − f (a)(a − b)
V (a) − V (b)
V (ξ)
f (ξ) − f (a)
=
=
=
.
g(a) − g(b) − g (a)(a − b)
U (a) − U (b)
U (ξ)
g (ξ) − g (a)
Now we define

def

Wa (z) =

f (z) − f (a)
,
g (z) − g (a)

z ∈ (a, b].

So the left hand side of (1.3.16) takes the form Wa (ξ) − Wa (b) = 0 for some ξ(a, P ) ∈ (a, b).
26

We consider the curve v(s) = (g (s), f (s)) which is a graph on [a0 , b0 ]. The fact that the
torsion of the curve γ(s) = (s, g(s), f (s)) changes sign from + to − at the point c ∈ (a0 , b0 )
means that the curve v(s) is strictly convex on the interval (a0 , c), and it is strictly concave
on the interval (c, b0 ). We consider a function obtained from (1.3.16)

def

D(z) =

f (z) − f (z + P ) + f (z)P
f (z + P ) − f (z)
−
,
g(z) − g(z + P ) + g (z)P
g (z + P ) − g (z)

z ∈ [a0 , c].

(1.3.17)

Note that D(a0 ) = Wa0 (ζ)−Wa0 (a0 +P ) for some ζ = ζ(a0 , P ) ∈ (a0 , a0 +P ). We know that
v(s) is strictly convex on the interval (a0 , a0 +P ). This implies that Wa0 (z)−Wa0 (a0 +P ) < 0
for all z ∈ (a0 , a0 + P ). In particular D(a0 ) < 0. Similarly, concavity of v(s) on (c, c + P )
implies that D(c) > 0. Hence, there exists a ∈ (a0 , c) such that D(a) = 0.
Let a1 and b1 be some solutions of (1.3.15) obtained by Lemma 8.
Lemma 9. There exists a function a(s) ∈ C 1 ((c, b1 ]) ∩ C([c, b1 ]) such that a(b1 ) = a1 ,
a(c) = c, a (s) < 0, and the pair (a(s), s) solves the equation (1.3.15) for all s ∈ [c, b1 ].
Proof. The proof of the lemma is a consequence of the implicit function theorem. Let a < b,
and consider the function

1

1

def

a−b

Φ(a, b) = g (a) g (b) g(a) − g(b) .
f (a) f (b) f (a) − f (b)
We are going to find the signs of the partial derivatives of Φ(a, b) at the point (a, b) = (a1 , b1 ).
We present the calculation only for ∂Φ/∂b. The case for ∂Φ/∂a is similar.

27

1

0

a−b

∂Φ(a, b)
= g (a) g (b) g(a) − g(b) =
∂b
f (a) f (b) f (a) − f (b)
= (a − b)g (b)

g(a) − g(b)
− g (a)
a−b

f (a) − f (b) − f (a)(a − b) f (b)
−
.
g(a) − g(b) − g (a)(a − b)
g (b)

Note that

(a − b)g (b)

g(a) − g(b)
− g (a)
a−b

< 0,

therefore we see that the sign of ∂Φ/∂b depends only on the sign of the expression

f (a) − f (b) − f (a)(a − b) f (b)
−
.
g(a) − g(b) − g (a)(a − b)
g (b)

(1.3.18)

We use the cup equation (1.3.16), and we obtain that the expression (1.3.18) at the point
(a, b) = (a1 , b1 ) takes the following form:

f (b) − f (a) f (b)
−
.
g (b) − g (a)
g (b)

(1.3.19)

The above expression has the following geometric meaning. We consider the curve v(s) =
(g (s), f (s)), and we draw a segment which connects the points v(a) and v(b). The above
expression is the difference between the slope of the line which passes through the segment

28

[v(a), v(b)] and the slope of the tangent line of the curve v(s) at the point b. In the case as
it is shown on Figure 1.6, this difference is positive. Recall that v(s) is strictly convex on
(a1 , c), and it is strictly concave on (c, b1 ). Therefore, one can easily note that this expression
(1.3.19) is always positive if the segment [v(a), v(b)] also intersects the curve v(s) at a point
ξ such that a < ξ < b. This always happens in our case because equation (1.3.16) means
that the points v(a), v(ξ), v(b) lie on the same line, where ξ was determined from Cauchy’s
mean value theorem. Thus
f (b) − f (a) f (b)
−
> 0.
g (b) − g (a)
g (b)

(1.3.20)

Similarly, we can obtain that ∂Φ
∂a < 0, because this is the same as to show that
f (b) − f (a) f (a)
−
> 0.
g (b) − g (a)
g (a)

(1.3.21)

Thus, by the implicit function theorem there exists a C 1 function a(s) in some neighborhood
Φ

of b1 such that a (s) = − Φ b < 0, and the pair (a(s), s) solves (1.3.15).
a

Now we want to explain that the function a(s) can be defined on (c, b1 ], and, moreover,
lims→c+0 a(s) = c. Indeed, whenever a(s) ∈ (a1 , c) and s ∈ (c, b1 ) we can use the implicit
function theorem, and we can extend the function a(s). It is clear that for each s we have
a(s) ∈ [a1 , c) and s ∈ (c, b1 ). Indeed, if a(s), s ∈ (a1 , c], or a(s), s ∈ [c, b1 ) then (1.3.15) has
a definite sign (see (1.3.17)). It follows that α(s) ∈ C 1 ((c, b1 ]), and the condition a (s) < 0
implies lims→c+0 a(s) = c. Hence a(s) ∈ C([c, b1 ]).
It is worth mentioning that we did not use the fact that the torsion of (s, g(s), f (s))
changes sign from + to −. The only thing we needed was that the torsion changes sign.
29

Let a1 and b1 be any solutions of equation (1.3.15) from Lemma 8, and let a(s) be
Fix an arbitrary s1 ∈ (c, b1 ) and consider the foliation

any function from Lemma 9.

Θcup ([s1 , b1 ], g) constructed by a(s) (see Lemma 7). Let B be a function defined by (1.3.8),
where

t2 (s1 ) =

f (s1 ) − f (a(s1 ))
.
g (s1 ) − g (a(s1 ))

(1.3.22)

Set Ωcup = Ω(Θcup ([s1 , b1 ], g)), and let Ωcup be the closure of Ωcup .
Lemma 10. The function B satisfies the following properties
1. B ∈ C 2 (Ωcup ) ∩ C 1 (Ωcup ).
2. B(a(s), g(a(s))) = f (a(s)) for all s ∈ [s1 , b1 ].
3. B is a concave function in Ωcup .
Proof. The first property follows from Lemma 3 and the fact that ∇B(x) = t(s) for s = s(x),
where s(x) is a continuous function in Ωcup .
We are going to check the second property. We recall (see (1.3.6)) that t1 (s) = f (s) −
t2 (s)g (s). Condition (1.3.22) implies that

t1 (s1 ) + t2 (s1 )g (a(s1 )) = f (a(s1 )).

(1.3.23)

Let B(a(s), g(a(s))) = f˜(a(s)). After differentiation of this equality we get t1 (s1 )+t2 (s1 )g (a(s1 )) =

30

f˜ (a(s1 )). Hence, (1.3.23) implies that f (a(s1 )) = f˜ (a(s1 )). It is clear that

t1 (s) + t2 (s)g (s) = f (s),
t1 (s) + t2 (s)g (a(s)) = f˜ (a(s)),
t1 (s)(s − a(s)) + t2 (s)(g(s) − g(a(s))) = f (s) − f˜(a(s)),
which implies

1

s − a(s)

1

g (s) g (a(s)) g(s) − g(a(s)) = 0.
f (s) f˜ (a(s)) f (s) − f˜(a(s))
This equality can be rewritten as follows:

f ·

1

s − a(s)

g (a(s)) g(s) − g(a(s))

− f˜ (a)

1

s − a(s)

g

g(s) − g(a(s))

+ (f − f˜(a))(g (a(s)) − g (s)) = 0.

By virtue of Lemma 9 we have the same equality as above except f˜ is replaced by f . We
subtract one from another one:

[f (a(s)) − f˜(a(s))] + [f (a(s)) − f˜ (a(s))] ·

31

1

s − a(s)

g

g(s) − g(a(s))
g (a(s)) − g (s)

= 0.

Note that

1

s − a(s)

g

g(s) − g(a(s))
g (a(s)) − g (s)

<0

and a(s) is invertible. Therefore we get the differential equation z(u)B(u) + z (u) = 0 where
B ∈ C 1 ([a(b1 ), a(s1 )]), z(u) = f (u) − f˜(u) and B < 0. The condition z (a(s1 )) = 0 implies
z(a(s1 )) = 0. Note that z = 0 is a trivial solution. Therefore, by uniqueness of solutions to
ODEs we get z = 0.
We are going to check the concavity of B. Let F be the force function corresponding to
B. By Corollary 2 we only need to check that F(s1 ) ≤ 0. Note that (1.3.11) and (1.3.22)
imply

F(s1 ) =

f (s1 ) f (s1 ) − f (a(s1 ))
f (s1 )
− t2 (s1 ) =
−
,
g (s1 )
g (s1 )
g (s1 ) − g (a(s1 ))

which is negative by (1.3.20).
Remark 7. The above lemma is true for all choices s1 ∈ (c, b1 ). If we send s1 to c then one
can easily see that lims1 →c+ t2 (s1 ) = 0, therefore the force function F takes the following
form
s

F(s) =

c

f (y)
g (y)

s

exp −

g (r)
cos θ(r)dr dy.
y K(r)

This is another way to show that the force function is nonpositive.
The next lemma shows that the regardless of the choices of initial solution (a1 , b1 ) of
32

v(s)

g (a)

g (ξ) g (c)

g (b)

Figure 1.6 Graph v(s)

(1.3.15), the constructed function a(s) by Lemma 9 is unique (i.e. it does not depend on the
pair (a1 , b1 )).
Lemma 11. Let pairs (a1 , b1 ), (˜
a1 , ˜b1 ) solve (1.3.15), and let a(s), a
˜(s) be the corresponding
functions obtained by Lemma 9. Then a(s) = a
˜(s) on [c, min{b1 , ˜b1 }].
Proof. By the uniqueness result of the implicit function theorem we only need to show
existence of s1 ∈ (c, min{b1 , ˜b1 }) such that a(s1 ) = a
˜(s1 ). Without loss of generality assume
that ˜b1 = b1 = s2 . We can also assume that a
˜(s2 ) > a(s2 ), because other cases can be solved
in a similar way.
˜ =Θ
˜ cup ([c, s2 ], g) be the foliations corresponding to the
Let Θ = Θcup ([c, s2 ], g) and Θ
˜ be the functions corresponding to these foliations
functions a(s) and a
˜(s). Let B and B
from Lemma 10. We consider a chord T in R3 joining the points (a(s1 ), g(a(s1 )), f (a(s1 )))
and (s1 , g(s1 ), f (s1 )) (see Figure 1.7). We want to show that the chord T belongs to the
˜ Indeed, concavity of B
˜ (see Lemma 10) implies that the chord T lies below
graph of B.
˜ 1 , x2 ), where (x1 , x2 ) ∈ Ω(Θ).
˜ Moreover, concavity of B, Ω(Θ)
˜ ⊂ Ω(Θ)
the graph of B(x
˜ consists of chords joining the points of the curve (t, g(t), f (t))
and the fact that the graph B

33

˜ In particular the chord T , belonging to the
imply that the graph B lies above the graph B.
˜ This can happen if and only if T belongs to the graph
graph B, lies above the graph B.
˜ Now we show that if s1 < s2 , then the torsion of the curve (s, g(s), f (s)) is zero for
B.
s ∈ [s1 , s2 ]. Indeed, let T˜ be a chord in R3 which joins the points (a(s1 ), g(a(s1 )), f (a(s1 )))
˜ at the point
and (s2 , g(s2 ), f (s2 )). We consider the tangent plane L(x) to the graph B
(x1 , x2 ) = (a(s1 ), g(a(s1 ))). This tangent plane must contain both chords T and T˜, and it
˜ implies that the tangent
must be tangent to the surface at these chords. Concavity of B
˜ at points belonging to the triangle, which is the convex hull of
plane L coincides with B
the points (a(s1 ), g(a(s1 ))), (s1 , g(s1 )) and (s2 , g(s2 )). Therefore, it is clear that the tangent
˜ on the segments ∈ Θ
˜ with the endpoint at (s, g(s)) for s ∈ [s1 , s2 ].
plane L coincides with B
˜
Thus L((s, g(s))) = B((s,
g(s))) for any s ∈ [s1 , s2 ]. This means that the torsion of the curve
(s, g(s), f (s)) is zero on s ∈ [s1 , s2 ] which contradicts our assumption about the torsion.
Therefore s1 = s2 .
Corollary 4. In the conditions of Lemma 8, for all 0 < P < min{c − a0 , b0 − c} there exists
a unique pair (a1 , b1 ) which solves (1.3.15) such that b1 − a1 = P .
The above corollary implies that if the pairs (a1 , b1 ) and (˜
a1 , ˜b1 ) solve (1.3.15), then
a1 = a
˜1 and b1 = ˜b1 , and one of the following conditions holds: (a1 , b1 ) ⊂ (˜
a1 , ˜b1 ), or
(˜
a1 , ˜b1 ) ⊂ (a1 , b1 ).
Remark 8. The function a(s) is defined on the right of the point c. We extend naturally its
def
definition on the left of the interval by a(s) = a−1 (s).

34

1.4 Construction of the Bellman function
1.4.1 Reduction to the two dimensional case
We are going to construct the Bellman function for the case p < 2. The case p = 2 is trivial,
and the case p > 2 was solved in [8]. From the definition of H it follows that

H(x1 , x2 , x3 ) = H(|x1 |, |x2 |, x3 ) for all (x1 , x2 , x3 ) ∈ Ω.

(1.4.1)

Also note the homogeneity condition

H(λx1 , λx2 , λp x3 ) = λp H(x1 , x2 , x3 ) for all λ ≥ 0.

(1.4.2)

These two conditions (1.4.1), (2.1.7), which follow from the nature of the boundary data
(x2 + τ 2 y 2 )2/p , make the construction of H easier. However, in order to construct the
function H, this information is not necessary. Further, we assume that H is C 1 (Ω) smooth.
Then from the symmetry (1.4.1) it follows that

∂H
= 0 on xj = 0 for j = 1, 2.
∂xj

(1.4.3)

For convenience, as in [8], we rotate the system of coordinates (x1 , x2 , x3 ). Namely, let

def

y1 =

x 1 + x2
,
2

def

y2 =

35

x2 − x1
,
2

def

y3 = x3 .

(1.4.4)

We define

def

N (y1 , y2 , y3 ) = H(y1 − y2 , y1 + y2 , y3 ) on Ω1 ,
where Ω1 = {(y1 , y2 , y3 ) : y3 ≥ 0, |y1 − y2 |p ≤ y3 }. It is clear that for fixed y1 , the function
N is concave in variables y2 and y3 ; moreover, for fixed y2 the function N is concave with
respect to the rest of variables. The symmetry (1.4.1) for N turns into the following condition

N (y1 , y2 , y3 ) = N (y2 , y1 , y3 ) = N (−y1 , −y2 , y3 ).

(1.4.5)

Thus it is sufficient to construct the function N on the domain

def

Ω2 = {(y1 , y2 , y3 ) : y1 ≥ 0, −y1 ≤ y2 ≤ y1 , (y1 − y2 )p ≤ y3 }.
Condition (1.4.3) turns into

∂N
∂N
=
on the hyperplane y2 = y1 ,
∂y1
∂y2
∂N
∂N
=−
on the hyperplane y2 = −y1 .
∂y1
∂y2

(1.4.6)
(1.4.7)

The boundary condition (1.2.3) becomes

N (y1 , y2 , |y1 − y2 |p ) = ((y1 + y2 )2 + τ 2 (y1 − y2 )2 )p/2 .

36

(1.4.8)

The homogeneity condition (2.1.7) implies that N (λy1 , λy2 , λp y3 ) = λp N (y1 , y2 , y3 ) for λ ≥
0. We choose λ = 1/y1 , and we obtain that

p

N (y1 , y2 , y3 ) = y1 N

y y
1, 2 , 3p
y1 y1

(1.4.9)

def

Suppose we are able to construct the function M (y2 , y3 ) = N (1, y2 , y3 ) on

def

Ω3 = {(y2 , y3 ) : −1 ≤ y2 ≤ 1, (1 − y2 )p ≤ y3 }
with the following conditions:
1. M is concave in Ω3
2. M satisfies (1.4.8) for y1 = 1.
3. The extension of M onto Ω1 via formulas (1.4.9) and (1.4.5) is a function with the
properties of N (see (1.4.6), (1.4.7), and concavity of N ).
4. M is minimal among those who satisfy the conditions 1,2,3.
Then the extended function M should be N . So we are going to construct M on Ω3 . We
denote

def

g(t) = (1 − t)p ,

t ∈ [−1, 1],

def

f (t) = ((1 + t)2 + τ 2 (1 − t)2 )p/2 ,

37

(1.4.10)
t ∈ [−1, 1].

(1.4.11)

Then we have the boundary condition

M (t, g(t)) = f (t),

t ∈ [−1, 1].

(1.4.12)

We differentiate the condition (1.4.9) with respect to y1 at the point (y1 , y2 , y3 ) =
(1, −1, y3 ) and we obtain that
∂N
∂N
∂N
(1, −1, y3 ) = pN (1, −1, y3 ) +
(1, −1, y3 ) − py3
,
∂y1
∂y2
∂y3

y3 ≥ 0.

Now we use (1.4.7), so we obtain another requirement for M (y2 , y3 ):

0 = pM (−1, y3 ) + 2

∂M
∂M
(−1, y3 ) − py3
(−1, y3 ),
∂y2
∂y3

for y3 ≥ 0.

(1.4.13)

Similarly, we differentiate (1.4.9) with respect to y1 at point (y1 , y2 , y3 ) = (1, 1, y3 ) and use
(1.4.6), so we obtain

0 = pM (1, y3 ) − 2

∂M
∂M
(1, y3 ) − py3
(1, y3 ),
∂y2
∂y3

for y3 ≥ 0.

(1.4.14)

So in order to satisfy conditions (1.4.6) and (1.4.7), the requirements (1.4.13) and (1.4.14)
are necessary. It is easy to see that these requirements are also sufficient in order to satisfy
these conditions.
The minimum between two concave functions with fixed boundary data is a concave
function with the same boundary data. Note also that the conditions (1.4.13) and (1.4.14)
still fulfilled after taking the minimum. Thus it is quite reasonable to construct a candidate
for M (y2 , y3 ) as a minimal concave function on Ω3 with the boundary conditions (1.4.12),

38

a(s2 )

a(s1 )=˜
a(s1 )

c

s1

s2

Figure 1.7 Uniqueness of the cup

(1.4.13) and (1.4.14). We remind that we should also have the concavity of the extended
function N (y1 , y2 , y3 ) with respect to variables y1 , y3 for each fixed y2 . This condition can
be verified after the construction of the function M (y2 , y3 ).

1.4.2 Construction of a candidate for M
We are going to construct a candidate B for M . Firstly, we show that for τ > 0, the torsion
def

τγ of the boundary curve γ(t) = (t, g(t), f (t)) on t ∈ (−1, 1), where f, g are defined by
(1.4.10) and (1.4.11), changes sign once from + to −. We call this point the root of a cup.
We construct the cup around this point. Note that g < 0, g > 0 on [−1, 1). Therefore

sign τγ = sign f

−

g
f
g

= sign f

−

2−p
f
1−t

= sign(v(t)),

where

def

v(t) = −(1 + τ 2 )2 (p − 1)t3 + (1 + τ 2 )(3τ 2 + τ 2 p + 3 − 3p)t2 +
(2τ 2 p − 9τ 4 + τ 4 p + 3 − 3p − 6τ 2 )t − p + 5τ 4 + 2τ 2 p − τ 4 p − 10τ 2 + 1.
39

Note that v(−1) = 16τ 4 > 0 and v(1) = −8((p − 1) + τ 2 ) < 0. So the function v(t) changes
3(p−1)
sign from + to − at least once. Now, we show that v(t) has only one root. For τ 2 < 3−p ,

note that the linear function v (t) is nonnegative i.e. v (−1) = 8τ 2 p(1 + τ 2 ) > 0, v (1) =
−4(1 + τ 2 )(τ 2 p − 3τ 2 + 3p − 3) ≥ 0. Therefore, the convexity of v(t) implies the uniqueness
of the root v(t) on [−1, 1].
3(p−1)
Suppose τ 2 < 3−p ; we will show that v ≤ 0 on [−1, 1]. Indeed, the discriminant of

the quadratic function v (x) has the expression

D = 16τ 2 (τ 2 + 1)2 ((3 − p)2 τ 2 − 9(p − 1)),
3(p−1)

which is negative for 0 < τ 2 < 3−p . Moreover, v (−1) = −4τ 2 (τ 2 p + 3τ 2 + 3) < 0. Thus
we obtain that v is negative.
We denote the root of v by c. It is an appropriate time to make the following remark.
Remark 9. Note that v(−1 + 2/p) < 0. Indeed,

v(−1 + 2/p) =

(3p − 2)(p2 − 2p − 4)τ 4 + (16 + 5p3 − 8p2 − 16p)τ 2 + 8(1 − p)
,
p3

which is negative because coefficients of τ 4 , τ 2 , τ 0 are negative. Therefore, this inequality
implies that c < −1 + 2/p.
Consider a = −1 and b = 1; the left side of (1.3.15) takes the positive value −22p−1 p(1 −
p). However, if we consider a = −1 and b = c, then the proof of Lemma 8 (see (1.3.17))
implies that the left side of (1.3.15) is negative. Therefore, there exists a unique s0 ∈ (c, 1)
such that the pair (−1, s0 ) solves (1.3.15). Uniqueness follows from Corollary 4. The equation

40

0
(1.3.15) for the pair (−1, s0 ) is equivalent to the equation u 1+s
1−s0

def

u(z) = τ p (p − 1) τ 2 + z 2

(2−p)/2

= 0, where

− τ 2 (p − 1) + (1 + z)2−p − z(2 − p) − 1.

(1.4.15)

Lemma 9 gives the function a(s), and Lemma 10 gives the concave function B(y2 , y3 ) for
s1 = c with the foliation Θcup ((c, s0 ], g) in the domain Ω(Θcup ((c, s0 ], g)).
The above explanation implies the following corollary.
Corollary 5. Pick any point y˜2 ∈ (−1, 1). The inequalities s0 < y˜2 , s0 = y˜2 and y˜2 > s0
1+˜
y2
are equivalent to the following inequalities respectively: u 1−˜
y2
y2
u 1+˜
1−˜
y

2

1+˜
y2
< 0, u 1−˜
y2

= 0 and

> 0.
y3

∂M
∂M
=− ∂y
∂y2
3

∂M
∂M
= ∂y
∂y2
3

h(s)
y=(y2 ,y3 )
(y)

(t,g(t))
−1

1

s=s(y)

y2

Figure 1.8 Segment (y)

Now we are going to extend C 1 smoothly the function B on the upper part of the cup.
Recall that we are looking for a minimal concave function. If we construct a function with a
foliation Θ([s0 , y˜2 ], g) where y˜2 ∈ (s0 , 1) then the best thing we can do according to Lemma 6
and Lemma 5 is to minimize sin(θcup (s0 ) − θ(s0 )) where θcup (s) is an argument function of
Θcup ((c, s0 ], g) and θ(s) is an argument function of Θ([s0 , y˜2 ], g). In other words we need to
41

choose segments from Θ([s0 , y˜2 ], g) close enough to the segments of Θcup ((c, s0 ], g).
Thus, we are going to try to construct the set of segments Θ([s0 , y˜2 ]) so that they start
from (s, g(s), f (s)), s ∈ [s0 , y˜2 ], and they go to the boundary y2 = −1 of Ω3 .
We explain how the conditions (1.4.13) and (1.4.14) allow us to construct such type of
foliation Θ([s0 , y˜2 ], g) in a unique way. Let (y) be the segment with the endpoints (s, g(s))
where s ∈ (s0 , y˜2 ) and (−1, h(s)) (see Figure 1.8).
Let t(s) = (t1 (s), t2 (s)) = ∇B(y) where s = s(y) is the corresponding gradient function.
Then (1.4.13) takes the form

0 = pB(−1, h(s)) + 2t1 (s) − ph(s)t2 (s).

(1.4.16)

We differentiate this expression with respect to s, and we obtain

2t1 (s) − ph(s)t2 (s) = 0.

(1.4.17)

Then according to (1.3.5) we find the function tan θ(s), and, hence, we find the quantity
h(s)

tan θ(s) = −

ph(s)
2

⇔

h(s) − g(s)
ph(s)
=
.
s+1
2

Therefore,

h(s) =

2g(s)
p

1
yp − s

2
def
where yp = −1 + .
p

(1.4.18)

We see that the function h(s) is well defined, it increases, and it is differentiable on −1 ≤
42

s < yp . So we conclude that if s0 < yp then we are able to construct the set of segments
Θ([s0 , yp ), g) that pass through the points (s, g(s)) , where s ∈ [s0 , yp ) and through the
boundary y2 = −1 (see Figure 1.9).
y3

∂M
∂M
=− ∂y
∂y2
3

∂M
∂M
= ∂y
∂y2
3

Θ([s0 ,yp ),g)

Ang(s0 )

h(s0 )

Θcup ((c,s0 ],g)

−1

c

s0

yp

1

y2

Figure 1.9 Foliations Θcup ((c, s0 ], g) and Θ([s0 , yp ), g)

It is easy to check that Θ([s0 , yp ), g) is a foliation. So choosing the value t2 (s0 ) of B
on Ω(Θ([s0 , yp ), g)) according to Lemma 6, then by Corollary 3 we have constructed the
concave function B in the domain Ω(Θcup ((c, s0 ], g)) ∪ Ang(s0 ) ∪ Ω(Θ([s0 , yp ], g)).
1+y

1 .
It is clear that the foliation Θ([s0 , yp ), g) exists as long as s0 < yp . Note that 1−ypp = p−1

Therefore, Corollary 5 implies the following remark.
Remark 10. The inequalities s0 < yp , s0 = yp and s0 > yp are equivalent to the following
1
1
1
< 0, u p−1
= 0 and u p−1
> 0.
inequalities respectively: u p−1

At the point yp the segments from Θ([s0 , yp ), g) become vertical. After the point (yp , g(yp ))
we should consider vertical segments Θ([yp , 1], g) (see Figure 1.10), because by Lemma 5 this
corresponds to the minimal function. Surely Θ([yp , 1], g) is the foliation. Again, choosing
the value t2 (yp ) of B on Ω(Θ([yp , 1], g)) according to Lemma 6, then by Corollary 3 we have
43

y3

∂M
∂M
=− ∂y
∂y2
3

∂M
∂M
= ∂y
∂y2
3

Θ([yp ,1],g)

Θ([s0 ,yp ),g)

Ang(s0 )

h(s0 )

Θcup ((c,s0 ],g)

−1

c

s0

yp

1

y2

1
Figure 1.10 Case u p−1
<0

constructed the concave function B on Ω3 . Note that if s0 ≥ yp (which corresponds to the
1
inequality u p−1

> 0) then we do not have the foliation Θ([s0 , yp ), g). In this case we

consider only vertical segments Θ([s0 , 1], g) (see Figure 1.11), and again choosing the value
t2 (s0 ) of B on Ω(Θ([s0 , 1], g)) according to Lemma 6 then by Corollary 3 we construct a
concave function B on Ω3 . We believe that B = M .
We still have to check the requirements (1.4.13) and (1.4.14). The crucial role is played
by symmetry of the boundary data of N . Further, the given proofs work for both of the
cases yp < s0 and yp ≥ s0 . Therefore, we do not consider them separately.
The requirement (1.4.14) follows immediately. Indeed, the condition (1.3.8) at the point
y = (1, y3 ) (note that in (1.3.8) instead of x = (x1 , x2 ) we consider y = (y2 , y3 )) implies
that B(1, y3 ) = f (1) + t2 (1)(y3 − g(1)). Therefore, the requirement (1.4.14) takes the form
0 = pf (1) − 2t1 (1). Using (1.3.6), we obtain that t1 (1) = f (1). Therefore, we see that
pf (1) − 2t1 (1) = pf (1) − 2f (1) = 0.
Now, we are going to obtain the requirement (1.4.13) which is the same as (1.4.16). The
quantities t1 , t2 of B with the foliation Θ([s0 , yp ), g) satisfy the condition (1.4.17) which was
44

obtained by differentiation of (1.4.16). So we only need to check the condition (1.4.16) at
the initial point s = s0 . If we substitute the expression of B from (1.3.8) into (1.4.16), then
(1.4.16) turns into the following equivalent condition:

t1 (s)(s − yp ) + t2 (s)g(s) = f (s).

(1.4.19)

Note that (1.3.6) allows us to rewrite (1.4.19) into the equivalent condition

t2 (s) =

f (s) − (s − yp )f (s)
.
g(s) − (s − yp )g (s)

(1.4.20)

And as it was mentioned above we only need to check condition (1.4.20) at the point s = s0 ,
i.e.

t2 (s0 ) =

f (s0 ) − (s0 − yp )f (s0 )
.
g(s0 ) − (s0 − yp )g (s0 )

(1.4.21)

On the other hand, if we differentiate the boundary condition B(s, g(s)) = f (s) at the
points s = s0 , −1, then we obtain
t1 (s0 ) + t2 (s0 )g (−1) = f (−1),
t1 (s0 ) + t2 (s0 )g (s0 ) = f (s0 ).

Thus we can find the value of t2 (s0 ):

t2 (s0 ) =

f (−1) − f (s0 )
.
g (−1) − g (s0 )
45

(1.4.22)

So these two values (1.4.22) and (1.4.21) must coincide. In other words we need to show
f (s0 ) − (s0 − yp )f (s0 )
f (−1) − f (s0 )
=
.
g(s0 ) − (s0 − yp )g (s0 )
g (−1) − g (s0 )

(1.4.23)

It will be convenient for us to work with the following notations for the rest of the current
subsection. We denote g(−1) = g− , g (−1) = g− , f (−1) = f− , f (−1) = f− g(s0 ) =
g, g (s0 ) = g , f (s0 ) = f, f (s0 ) = f . The condition (1.4.23) is equivalent to

s0 =
=

f g− + f g − f g − gf−
+ yp =
f g− − g f−

(1.4.24)

f g− + f g − f g − gf−
2
−1 + .
p
f g− − g f −

On the other hand, from (1.3.15) for the pair (−1, s0 ) we obtain that

s0 =

f g− + g− f− − g f− − f− g−
f g− + f g − f g − gf−
−1 +
.
f g− − g f −
g f− − f g−

So, from (1.4.24) we see that it suffices to show that

f g− + g− f− − g f− − f− g−
2
= .
p
g f − − f g−
We note that g− = −(p/2)g− , f− = −(p/2)f− , hence g− f− = f− g− . Therefore, we have
f g− + g− f− − g f− − f− g−
f g− − g f−
2
=
= .
p
g f− − f g−
g f− − f g−

46

1.4.3 Concavity in another direction
We are going to check the concavity of the extended function N via B in another direction.
It is worth mentioning that the both of the cases yp < s0 , yp ≥ s0 do not play any role in
the following computations, therefore we consider them together. We define a candidate for
N as

def

p

p

N (y1 , y2 , y3 ) = y1 B(1, y2 /y1 , y3 /y1 ) for

y2 y3
,
y1 y1p

∈ Ω3 ,

(1.4.25)

and we extend N to the Ω1 by (1.4.5). Then, as it was already discussed, N ∈ C 1 (Ω1 ). We
need the following technical lemma:
Lemma 12.

p−2

Ny1 y1 Ny3 y3 − (Ny1 y3 )2 = −t2 sy3 p(p − 1)y1

where s = s

y2 y3
y1 , y p
1

and

y2 y3
y1 , y p
1

y
st1 + gt2 − f + 2 t1 ·
y1

2
−1
p

∈ int(Ω3 ) \ Ang(s0 ).

As it was mentioned in Remark 6, the gradient function t(s) is not necessarily differentiable at point s0 , this is the reason of the requirement

y2
y1 ,

y3
y1p

∈ int(Ω3 ) \ Ang(s0 )

in the lemma. However, from the proof of the lemma, the reader can easily see that
Ny1 y1 Ny3 y3 − (Ny1 y3 )2 = 0 whenever the points

y2 y3
y1 , y p
1

belong to the interior of the

domain Ang(s0 ).
Proof. Definition of the candidate N (see (1.4.25)) implies Ny3 y3 = t2 (s)sy3 , Ny3 y1 = t2 sy1 ,

47

p−1

Ny1 = y1

pB

y2 y3
,
y1 y1p

y
y
− t1 2 − pt2 3p
y1
y1

.

(1.4.26)

Condition (1.3.8) implies

B

y2 y3
,
y1 y1p

= f (s) + t1 ·

We substitute this expression for B

p−1

Ny1 = y1

Condition

y2 y3
y1 , y p
1

y2 y3
y1 , y p
1

y2
− s + t2 ·
y1

y3
p − g(s) .
y1

into (1.4.26), and we obtain:

y
pf + 2 t1 (p − 1) − pst1 − pgt2 .
y1

(1.4.27)

∈ int(Ω3 ) \ Ang(s0 ) implies the equality Ny1 y3 = Ny3 y1 which in

turn gives

p−1

t2 sy1 = y1

y
pf + 2 t1 (p − 1) − (pst1 + pgt2 )s sy3 .
y1

Hence

p−1

t2 · (sy1 )2 = y1

y
pf + 2 t1 (p − 1) − (pst1 + pgt2 )s sy3 sy1 .
y1

We keep in mind this identity and continue our calculations

48

(1.4.28)

p−2

Ny1 y1 = (p − 1)y1

y
pf + 2 t1 (p − 2) − pst1 − pgt2 +
y1

y
pf + 2 t1 (p − 1) − (pst1 + pgt2 )s sy1 .
y1

p−1

y1

So, finally we obtain

Ny1 y1 Ny3 y3 − (Ny1 y3 )2 = t2 Ny1 y1 sy3 − t2 (sy1 )2 .
Now we use the identity (1.4.28), and we substitute the expression t2 (sy1 )2 :

y
pf + 2 t1 (p − 1) − (pst1 + pgt2 )s sy1
y1

p−1

Ny1 y1 Ny3 y3 − (Ny1 y3 )2 = t2 sy3 Ny1 y1 − y1

y
pf + 2 t1 (p − 2) − pst1 − pgt2
y1

p−2

t2 sy3 (p − 1)y1

p−2

− t2 sy3 p(p − 1)y1

y
st1 + gt2 − f + 2 t1 ·
y1

=

=

2
−1
p

.

p

Now we are going to consider several cases when the points (y2 /y1 , y3 /y1 ) belong to the
different subdomains in Ω3 . Note that we always have Ny3 y3 ≤ 0, because of the fact that B
is concave in Ω3 and (1.4.25). So we only have to check that the determinant of the Hessian
N is negative. If the determinant of the Hessian is zero, then it is sufficient to ensure that
Ny3 y3 is strictly negative, and if Ny3 y3 is also zero, then we need to ensure that Ny1 ,y1 is
nonpositive.
Domain Ω(Θ[s0 , yp ]).
49

In this case we can use the equality (1.4.19), and we obtain that

st1 + gt2 − f = yp t1 .
Therefore

p−2

Ny1 y1 Ny3 y3 − (Ny1 y3 )2 = −t2 sy3 p(p − 1)y1

y
t1 yp 1 + 2
y1

≥ 0.

because t1 ≥ 0. Indeed, t1 (s) is continuous on [c, 1], where c is the root of the cup and
By2 y2 = t1 sy2 ≤ 0, therefore, because of the fact sy2 > 0, it suffices to check that t1 (1) ≥ 0
which follows from the following inequality

t1 (1) = f (1) − t2 (1)g (1) = f (1) > 0.
Domain of linearity Ang(s0 ).
This is the domain which is obtained by the triangle ABC, where A = (−1, g(−1)),
B = (s0 , g(s0 )), and C = (−1, h(s0 )) if s0 < yp and by the infinity domain of linearity, which
is rectangular type, and which lies between the chords AB, BC , where C = (s0 , +∞) and
AC , where C = (−1, +∞) (see Figure 1.11).
p

Suppose the points y2 /y1 , y3 /y1 belong to the interior of Ang(s0 ). Then the gradient
function t(s) of B is constant, and moreover s

y2
y1 ,

y3
y1p

is constant. The fact that the

determinant of the Hessian is zero in the domain of linearity (note that sy3 = 0) implies that

50

we only need to check Ny1 y1 < 0. Equality (1.4.27) implies

p−2

Ny1 y1 = (p − 1)y1
p−2

(p − 1)y1

y
pf + 2 t1 (p − 2) − ps0 t1 − pgt2
y1

≤

(pf − ps0 t1 − pgt2 − t1 (p − 2)) = 0.

The last equality follows from (1.4.19). The above inequality turns into the equality if and
only if yy21 = s0 , this is the boundary point of Ang(s0 ).
Domain of vertical segments.
On the vertical segments determinant of the Hessian is zero (for example, because the
vertical segment is vertical segment in all directions) and By3 y3 = 0, therefore, we must
check that Ny1 y1 ≤ 0. We note that s(y2 , y3 ) = y2 , therefore,

p−2

Ny1 y1 = y1

× (p − 1) (pf + st1 (p − 2) − pst1 − pgt2 ) − s pf − t1 s − t1 p − pg t2

.

However, from (1.3.6) we have pf − t1 p − pg t2 = 0, therefore,
p−2

Ny1 y1 = y1

× (p − 1) (pf − 2st1 − pgt2 ) + s2 t1 .

Condition t1 ≤ 0 implies that it is sufficient to show pf − 2st1 − pgt2 ≤ 0. We use (1.3.6),
and we find t1 = f − g t2 . Hence,
pf − 2st1 − pgt2 = pf − gpt2 − 2s(f − g t2 ) = pf − 2sf − t2 (gp − 2sg ).
Note that gp − 2sg ≥ 0 (because s ≥ 0 and g ≤ 0), and we recall that from (1.3.6) and
51

the fact that on the vertical segments t2 is constant, since we have cos θ(s) = 0 (see the
expression of t2 from Lemma 2), so t2 is constant and hence 0 ≥ t1 = f − g t2 , therefore,
we have t2 ≥ f /g . Therefore,
pf − 2sf − t2 (gp − 2sg ) ≤ pf − 2sf −

f
(gp − 2sg ).
g

Now we recall the values (1.4.12), (1.4.11), and after direct calculations we obtain

pf − 2sf −

f
f (1 − s2 )p(p − 2)(τ 2 (1 + s)2 + (1 − s)2 + 2τ 2 (1 − s2 ))
(gp − 2sg ) =
≤ 0.
g
(p − 1)((1 + s)2 + τ 2 (1 − s)2 )2

Domain of the cup Ω(Θcup ((c, s0 ], g)).

y3

∂M
∂M
=− ∂y
∂y2
3

∂M
∂M
= ∂y
∂y2
3

Θ([s0 ,1],g)

Ang(s0 )

Θcup ((c,s0 ],g)

−1

c

yp s 0

1

y2

1
Figure 1.11 Case u p−1
≥0

The condition that Ny3 y3 is strictly negative in the cup implies that we only need to show
p

p

st2 +gt3 −f + yy21 t1 ( p2 −1) ≥ 0, where s = s(y2 /y1 , y3 /y1 ) and the points y = (y2 /y1 , y3 /y1 ) lie
p

p

in the cup. We can think that y2 /y1 → y2 and y3 /y1 → y3 and s(y2 /y1 , y3 /y1 ) → s(y2 , y3 ),
and we can think that the points (y2 , y3 ) lie in the cup. Therefore it suffices to show that
52

st2 + gt3 − f + y2 t1 ( p2 − 1) ≥ 0, where y = (y2 , y3 ) ∈ Ω(Θcup ((c, s0 ], g)). On a segment with
the fixed endpoint (s, g(s)) the expressions s, t1 , g(s), t2 , f (s) are constant, except of y2 , so
the expression st1 + gt2 − f + y2 t1 ( p2 − 1) is linear with respect to the y2 on the each segment
of the cup. Therefore, the worst case appears when y2 = a(s) (a(s) − is the left end (it is
abscissa) of the given segment). This is true because t1 ≥ 0 (as it was already shown) and
( p2 − 1) ≥ 0. So, as the result, we derive that it is sufficient to prove the inequality
st1 + gt2 − f + a(s)t1 ·

2
−1
p

= t1 (s − a(s)) + gt2 − f +

2a(s)
t ≥ 0.
p 1

(1.4.29)

We use the identity (1.3.8) at the point y = (a(s), g(a(s))), and we find that

t1 (s − a(s)) + gt2 − f = g(a(s))t2 − f (a(s)).
We substitute this expression into (1.4.29) then we will get that it suffices to prove the
inequality:

g(a(s))t2 − f (a(s)) +

2a(s)
t ≥ 0.
p 1

(1.4.30)

We differentiate condition B(a(s), g(a(s))) = f (s) with respect to s. Then we find
the expression for t1 (s), namely t1 (s) = f (a(s)) − t2 (s)g (a(s)). After substituting this
expression into (1.4.30) we obtain that:

g(a(s))t2 − f (a(s)) +

2a(s)
1+z
t1 =
p
g (z)

(z − 1)(τ 2 + 1)f (z)
− t2 (s) ,
((1 + z)2 + τ 2 (1 − z)2 )g (z)

53

where z = a(s). So it suffices to show that
(z − 1)(τ 2 + 1)f (z)
− t2 (s) ≤ 0
((1 + z)2 + τ 2 (1 − z)2 )g (z)

(1.4.31)

because g is negative. We are going to show that the condition (1.4.31) is sufficient to check
at the point z = −1. Indeed, note that (t2 )z ≥ 0 on [−1, c], where c is the root of the cup,
and also note that
(z − 1)(τ 2 + 1)f
((1 + z)2 + τ 2 (1 − z)2 )g

=
z

τ2 + 1
(p − 2)(1 − z)−(p−1) [(1 + z)2 + τ 2 (1 − z)2 ]p/2−2 2(1 + z) ≤ 0.
p
The condition (1.4.31) at the point z = −1 turns into the following condition
t2 (s0 ) −

τ p−2 (τ 2 + 1)
≥ 0.
p

Now we recall (1.3.21) and t2 (s0 ) = (f (−1) − f (s0 )/(g (−1) − g (s0 )), therefore we have
t2 (s0 ) −

τ p−2 (τ 2 + 1)
f (−1) τ p−2 (τ 2 + 1)
τ p (p − 1)2 + τ p−2
≥
−
=
> 0.
p
g (−1)
p
p(p − 1)

Thus we finish this section by the following remark.
Remark 11. We still have to check the cases when the points

y2
y1 ,

y3
p
y1

belong to the bound-

ary of Ang(s0 ) and the vertical rays y2 = ±1 in Ω3 . The reader can easily see that in this
case concavity of N follows from the observation that N ∈ C 1 (Ω1 ). Symmetry of N covers
the rest of the cases when

y2 y3
y1 , y p
1

∈
/ Ω3 .

54

Thus we have constructed the candidate N .

1.5 Sharp constants via foliation
1.5.1 Main theorem
We remind the reader the definition of the functions u(z), g(s), f (s) (see (1.4.15), (1.4.10),
(1.4.11)), the value yp = −1 + 2/p and the definition of the function a(s) (see Lemma 9,
Lemma 11 and Remark 8).
Theorem 1.5.1. Let 1 < p < 2, and let G be the martingale transform of F and let
β−1
|EG| ≤ β|EF |. Set β = β+1
.
1
(i) If u p−1
≤ 0 then

E(τ 2 F 2 + G2 )p/2 ≤

2
1
τ 2 + max β,
p−1

1
> 0 then
(ii) If u p−1

E(τ 2 F 2 + G2 )p/2 ≤ C(β )E|F |p ,

55

p
2

E|F |p .

(1.5.1)

where C(β ) is continuous, nondecreasing, and it is defined by the following way:



p/2


,
β ≥ s∗ ;
τ 2 + β2
















τp
def
, β ≤ −1 + p2 ;
C(β ) =
2−p (1−s )p−1
2
0

1 − (τ 2 +1)(p−1)(1−s


0 )+2(2−p)














f (s1 ) − f (a(s1 ))


,
R(s1 , β ) = 0 for

g (s1 ) − g (a(s1 ))
1+s0
where s0 ∈ (−1 + 2/p, 1) is the solution of the equation u 1−s
0

s1 ∈ (β , s0 );

= 0, and the function

R(s, z) is defined as follows

def

R(s, z) = −f (s) −

f (a(s))g (s) − f (s)g (a(s))
(z − s) +
g (s) − g (a(s))

f (s) − f (a(s))
g(s) = 0,
g (s) − g (a(s))

z ∈ [−1 + 2/p, s∗ ],

s ∈ [z, s0 ].

The value s∗ ∈ [−1 + 2/p, s0 ] is the solution of the equation
f (s∗ ) − f (a(s∗ ))
f (s∗ )
=
.
g (s∗ ) − g (a(s∗ ))
g(s∗ )

(1.5.2)

Proof. Before we investigate some of the cases mentioned in the theorem, we should make
the following observation. The inequality of the type (1.5.1) can be restated as follows

H(x1 , x2 , x3 ) ≤ Cx3 ,

(1.5.3)

where H is defined by (1.2.2) and x1 = EF, x2 = EG, x3 = E|F |p . In order to derive the
56

estimate (1.5.1) we have to find the sharp C in (1.5.3). Because of the property (1.4.1) we
can assume that both of the values x1 , x2 are nonnegative. So non-negativity of x1 , x2 and
the condition |EG| ≤ β|EF | can be reformulated as follows
x − x1
x + x2
≤ 2
≤
− 1
2
2

β−1
β+1

x1 + x 2
2

.

(1.5.4)

The condition (1.5.4) with (1.5.3) in terms of the function N and the variables y1 , y2 , y3
means that we have to find the sharp C such that

N (y1 , y2 , y3 ) ≤ Cy3

for

− y1 ≤ y2 ≤

β−1
β+1

y1 ,

y ∈ Ω2 .

Because of (1.4.9) the above condition can be reformulated as follows

B(y2 , y3 ) ≤ Cy3

for

− 1 ≤ y2 ≤

β−1
β+1

,

y3 ≥ g(y2 ),

(1.5.5)

where B(y2 , y3 ) = N (1, y2 , y3 ). So our aim is to find the sharp C, in other words the
value supy1 ,y2 B/y3 where the supremum is taken from the domain mentioned in (1.5.5).
Note that the quantity B(y2 , y3 )/y3 increases with respect to the variable y2 . Indeed,
(B(y2 , y3 )/y3 )y2 = t1 (s(y))/y3 , where the function t1 (s) is nonnegative on [c0 , 1] (see the
end of the proof of the concavity condition in the domain Ω(Θ[s0 , yp ])). Note that as we
increase the value y2 then the range of y3 also increases. This means that the supremum of
the expression B/y3 is attained on the subdomain where y2 = (β − 1)/(β + 1). It is worth
mentioning that since the quantity (β − 1)/(β + 1) ∈ [−1, 1] increases as β increases and
because of the observation made above we see that the value C increases as the β increases.

57

1.5.2 Case yp ≤ s0 .
We are going to investigate the simple case (i). Remark 10 implies that s0 ≤ yp , in other
words, the foliation of vertical segments is Θ([yp , 1], g) where the value θ(s) on [yp , 1] is
equal to π/2. This means that t2 (s) is constant on [yp , 1] (see Lemma 2), and it is equal to
1 )p/2 (see (1.4.20)).
f (yp )/g(yp ) = (τ 2 + (p−1)
2
1
If β−1
β+1 ≥ yp , or equivalently β ≥ p−1 , then the function B on the vertical segment with

the endpoint (β , g(β )) where β−1
β+1 = β ∈ [yp , 1) has the expression (see (1.3.8))
B(β , y3 ) = f (β ) +

f (yp )
(y − g(β )),
g(yp ) 3

y3 ≥ g(β ).

Therefore,
f (yp ) g(β )
B(β , y3 )
=
+
y3
g(yp )
y3

f (β ) f (yp )
−
g(β )
g(yp )

,

y3 ≥ g(β ).

(1.5.6)

The expression f (s)/g(s) is strictly increasing on (−1, 1), therefore, the expression (1.5.6)
attains its maximal value at the point y3 = g(β ). Thus, we have
p/2
B(y2 , y3 )
B(β , y3 )
B(β , g(β ))
f (β )
≤
≤
=
= τ 2 + β2
y3
y3
g(β )
g(β )

for

− 1 ≤ y2 ≤ β , y3 ≥ g(y2 ).

β−1
1 , then we can achieve such value for C which was
< yp , or equivalently β < p−1
If β+1
1 , and since the function C = C(β ) increases as β increases
achieved at the moment β = p−1

this value will be the best. Indeed, it suffices to look at the foliation (see Figure 1.10). For

58

any fixed y2 we send y3 to +∞, and we obtain that
f (s) + t1 (s)(y2 − s) + t2 (s)(y3 − g(s))
B(y2 , y3 )
= lim
=
y3 →∞
y3 →∞
y3
y3
lim

τ2 +

lim t2 (s(y)) = t2 (yp ) =

y3 →∞

p/2
1
.
(p − 1)2

1.5.3 Case yp > s0 .
As it was already mentioned, the condition in the case (ii) is equivalent to the inequality
s0 > yp (see Remark 10). This means that that the foliation of the vertical segments is
Θ([s0 , 1], g) (see Figure 1.11). We know that C(β ) is increasing. We remind that we are
going to maximize the function

B(y2 ,y3 )
y3

mentioned that we can require y2 =

on the domain mentioned in (1.5.5). It was already

β−1
β+1

= β . For such fixed y2 = β ∈ [−1, 1] we are

going to investigate the monotonicity of the function

B(β ,y3 )
.
y3

We consider several cases.

Let β ≥ s0 . We differentiate the function B(β , y3 )/y3 with respect to the variable y3 , and
we use the expression (1.3.8) for B, so we obtain that

∂
∂y3

B(β , y3 )
y3

t (β )y3 − B(β , y3 )
−f (β ) + t2 (β )g(β )
= 2
=
.
2
y3
y32

Recall that t2 (s) = t2 (s0 ) for s ∈ [s0 , 1], therefore, direct calculations imply

t2 (β ) =

f (s0 ) − (s0 − yp )f (s0 )
f (s0 )
f (β )
<
≤
,
g(s0 ) − (s0 − yp )g (s0 )
g(s0 )
g(β )

59

β ≥ s0 .

This implies that

C(β ) =

B(β , y3 )
B(β , y3 )
f (β )
=
= (τ 2 + β 2 )p/2 .
=
y
y
g(β
)
3
3
y3 ≥g(β )
y3 =g(β )
sup

Now we consider the case β < s0 .
For each point y = (β , y3 ) that belongs to the line y2 = β there exists a segment
(y) ∈ Θ((c, s0 ], g) with the endpoint (s, g(s)) where s ∈ [max{β , a(β )}, s0 ]. If the point y
belongs to the domain of linearity Ang(s0 ), then we can choose the value s0 , and consider
any segment with the endpoints y and (s0 , g(s0 )) which surely belongs to the domain of
linearity. The reader can easily see that as we increase the value y3 the value s increases as
well. So,

∂
∂y3

B(β , y3 )
y3

t (s)y3 − B(β , y3 )
−f (s) − t1 (s)(β − s) + t2 (s)g(s)
=
.
= 2
y32
y32

Our aim is to investigate the sign of the expression −f (s) − t1 (s)(β − s) + t2 (s)g(s) as
we variate the value y3 ∈ [g(β ), +∞). Without loss of generality we can forget about the
variable y3 , and we can variate only the value s on the interval [max{α(β ), β }, s0 ].
def

We consider the function R(s, z) = −f (s) − t1 (s)(z − s) + t2 (s)g(s) with the following
domain −1 ≤ z ≤ s0 and s ∈ [max{α(z), z}, s0 ] (see Figure 1.12). As we already have seen
R(s0 , s0 ) < 0. Note that R(s0 , −1) > 0. Indeed, note that R(s0 , −1) = t2 (s0 )g(−1)−f (−1).
This equality follows from the fact that

f (s0 ) − f (−1) = t1 (s0 )(s0 + 1) + t2 (s0 )(g(s0 ) − g(−1)),

60

which is consequence of Lemma 10. So, (1.4.22) and (1.3.21) imply

t2 (s0 ) =

f (−1)
f (−1)
f (−1) − f (s0 )
>
≥
.
g (−1) − g (s0 )
g (−1)
g(−1)

Note that the function R(z, s0 ) is linear with respect to z. So on the interval [−1, s0 ] it has
the root yp = −1 + 2/p. Indeed,
−f (s0 ) + t2 (s0 )g(s0 ) + t1 (s0 )s0
= yp .
t1 (s0 )
The last equality follows from (1.4.22), (1.4.24) and (1.3.6). We need few more properties
of the function R(s, z). Note that for each fixed z, the function R(s, z) is nonincreasing on
[max{α(z), z}, s0 ]. Indeed

Rs (s, z) = −f (s) − t1 (s)(z − s) + t1 (s) + t2 (s)g(s) + t2 (s)g(s).

(1.5.7)

We take into account the condition (1.3.6), so the expression (1.5.7) simplifies as follows

Rs (s, z) = t2 (s)g(s) + t1 (s)(s − z).
We remind the reader equality (1.3.5) and the fact that t2 (s) ≤ 0. Therefore we have
Rs (s, z) = y3 t2 (s) where y3 = y3 (s) > 0. Thus we see that R(s, β ) ≥ 0 for β ≤ yp .
So if the function R(·, z) at the right end on its domain [max{α(z), z}, s0 ] is positive, this
will mean that the function B/y3 is increasing, hence, the constant C(β ) will be equal to

C(β ) = lim

y3 →∞

f (−1) − f (s0 )
B(z, y3 )
= t2 (s0 ) =
y3
g (−1) − g (s0 )
61

0
(this follows from (1.4.22) and the structure of the foliation). Since u 1+s
1−s0

= 0 and

(1.4.23) direct computations show that
τp
f (−1) − f (s0 )
.
=
22−p (1−s0 )p−1
g (−1) − g (s0 )
1 − (τ 2 +1)(p−1)(1−s )+2(2−p)
0

(1.5.8)

So it follows that if β ≤ yp then (1.5.8) is the value of C(β ).
If the function R(·, z) on the left end of its domain is nonpositive this will mean that
the function B/y3 is decreasing, so the sharp constant will be the value of the function
B(z, y3 )/y3 at the left end of its domain

C(β ) =

B(z, y3 )
f (z)
=
= (τ 2 + β 2 )p/2 .
y3
g(z)
y3 =g(z)

(1.5.9)

We recall that c is the root of the cup and c < yp (see Remark 9). We will show that the
function R(z, s) is decreasing on the boundary s = z for s ∈ (yp , s0 ]. Indeed, (1.3.6) implies
(R(s, s))s = −f (s) + t2 (s)g(s) + t2 (s)g (s) = −t1 (s) + t2 (s)g(s) < 0.
The last inequality follows from the fact that t2 (s) ≤ 0 and t1 (s) > 0 on (c, 1]. Surely
R(yp , yp ) > R(s0 , yp ) = 0, and we recall that R(s0 , s0 ) < 0, so there exists unique s∗ ∈
[yp , s0 ] such that R(s∗ , s∗ ) = 0. This is equivalent to (1.5.2). So it is clear that R(z, z) ≤ 0
for z ∈ [s∗ , s0 ]. Therefore C(β ) has the value (1.5.9) for β ≥ s∗ .
The only case remains is when β ∈ [yp , s∗ ]. We know that R(z, z) ≥ 0 for z ∈ [yp , s∗ ]
and R(s0 , z) ≤ 0 for z ∈ [yp , s∗ ]. The fact that for each fixed z the function R(s, z) is
decreasing implies the following: for each z ∈ [yp , s∗ ] there exists unique s1 (z) ∈ [z, s0 ] such

62

that R(z, s1 (z)) = 0. Therefore, for β ∈ [yp , s∗ ] we have
C(β ) =

B(β , y3 (s1 (β )))
,
y3 (s1 (β ))

(1.5.10)

where the value s1 (β ) is the root of the equation R(s1 (β ), β ) = 0. Recall that

R(s1 (β ), β ) = t2 (s1 )y3 (s1 ) − B(β , y3 (s1 )) = −f (s1 ) − t1 (s1 )(β − s1 ) + t2 (s1 )g(s1 ).
(1.5.11)
So the expression (1.5.10) takes the form

C(β ) = t2 (s1 ) =

f (s1 ) − f (a(s1 ))
.
g (s1 ) − g (a(s1 ))

Finally, we remind the reader that

t2 (s) =
t1 (s) =

f (s) − f (a(s))
,
g (s) − g (a(s))

f (a(s))g (s) − f (s)g (a(s))
.
g (s) − g (a(s))

for s ∈ (c, s0 ], and we finish the proof of the theorem.

1.6 Extremizers via foliation
We set Ψ(F, G) = E(G2 + τ 2 F 2 )2/p . Let N be the candidate that we have constructed
in Section 1.4 (see (1.4.25)). We define the candidate B for the Bellman function H (see

63

(1.2.2)) as follows

B(x1 , x2 , x3 ) = N

x1 + x2 x 2 − x1
,
, x3 ,
2
2

(x1 , x2 , x3 ) ∈ Ω.

We want to show that B = H. We already know that B ≥ H (see Lemma 3). So, it remains
to show that B ≤ H. We are going to do this as follows: for each point x ∈ Ω and any ε > 0
we are going to find an admissible pair (F, G) such that

Ψ(F, G) > B(x) − ε.

(1.6.1)

Up to the end of the current section we are going to work with the coordinates (y1 , y2 , y3 )
(see (1.4.4)). It will be convenient for us to redefine the notion of admissibility of the pair.
Definition 7. We say that a pair (F, G) is admissible for the point (y1 , y2 , y3 ) ∈ Ω1 , if G
is the martingale transform of F and E(F, G, |F |p ) = (y1 − y2 , y1 + y2 , y3 ).
So in this case condition (1.6.1) in terms of the function N takes the following form: for
any point y ∈ Ω1 and for any ε > 0 we are going to find an admissible pair (F, G) for the
point y such that

Ψ(F, G) > N (y) − ε.

(1.6.2)

We formulate the following obvious observations.
Lemma 13. The following statements hold:
˜ =
1. A pair (F, G) is admissible for the point y = (y1 , y2 , y3 ) if and only if (F˜ , G)
˜ = Ψ(F, G).
˜ = (∓y2 , ∓y1 , y3 ); moreover, Ψ(F˜ , G)
(±F, ∓G) is admissible for the point y
64

s

R>0

s0

R<0

a(z)

−1

0

c y p s∗

s0

1 z

Figure 1.12 Domain of R(s, z)

˜ =
2. A pair (F, G) is admissible for the point y = (y1 , y2 , y3 ), if and only if (F˜ , G)
˜ = (λy1 , λy2 , |λ|p y3 ); moreover,
(λF, λG) (where λ = 0) is admissible for the point y
˜ = |λ|p Ψ(F, G).
Ψ(F˜ , G)
Definition 8. The pair of functions (F, G) is called an ε-extremizer for the point y ∈ Ω1 if
(F, G) is admissible for the point y and Ψ(F, G) > N (y) − ε.
Lemma 13, homogeneity, and symmetry of N imply that we only need to check (1.6.2)
for the points y ∈ Ω1 where y1 = 1 (y2 , y3 ) ∈ Ω3 . In other words, we show that Ψ(F, G) >
B(y2 , y3 )−ε for some admissible (F, G) for the point (1, y2 , y3 ) where (y2 , y3 ) ∈ Ω3 . Further,
instead of saying that (F, G) is an admissible pair (or ε-extremizer) for the point (1, y2 , y3 )
we just say that it is an admissible pair (or an ε-extremizer) for the point (y2 , y3 ). So we
only have to construct ε-extremizers in the domain Ω3 .
It is worth mentioning that we construct ε-extremizers (F, G) such that G will be the
martingale transform of F with respect to some filtration other than dyadic. A detailed
explanation on how to pass from one filtration to another the reader can find in [19].
65

We need a few more observations. For α ∈ (0, 1) we define the α− concatenation of the
˜ as follows
pairs (F, G) and (F˜ , G)

˜ α (x) =
(F • F˜ , G • G)





(F, G)(x/α)

x ∈ [0, α],




˜
(F˜ , G)((x
− α)/(1 − α)) x ∈ [α, 1].

˜ α (x)) = αΨ(F, G) + (1 − α)Ψ(F˜ , G).
˜
Clearly Ψ((F • F˜ , G • G)
Definition 9. Any domain of the type Ω1 ∩ {y1 = A} where A is some real number is said to
be a positive domain. Any domain of the type Ω1 ∩ {y2 = B} where B is some real number
is said to be a negative domain.
The following lemma is obvious.
˜ is an admissible pair
Lemma 14. If (F, G) is an admissible pair for a point y and (F˜ , G)
˜ such that either of the following is true: y, y
˜ belong to a positive domain, or
for a point y
˜ α is an admissible pair for the point
˜ belong to a negative domain, then (F • F˜ , G • G)
y, y
αy + (1 − α)˜
y.
˜ be an admissible pair for
Let (F, G) be an admissible pair for a point y, and let (F˜ , G)
˜ . Let y
ˆ be a point which belongs to the chord joining the points y and y
˜.
a point y
Remark 12. It is clear that if (F + , G+ ) is admissible for a point (y2+ , y3+ ) and (F − , G− )
is admissible for a point (y2− , y3− ) then an α− concatenation of these pairs is admissible for
the point (y2 , y3 ) = α · (y2+ , y3+ ) + (1 − α) · (y2− , y3− ).
Now we are ready to construct ε-extremizers in Ω3 . The main idea is that these functions
Ψ and B are very similar: they obey almost the same properties. Moreover, foliation plays
crucial role in the contraction of ε− extremizers.
66

1.6.1 Case s0 ≤ yp .
We want to find ε-extremizers for the points in Ω3 .
Extremizers in the domain Ω(Θcup ((c, s0 ], g)).
Pick any y = (y2 , y3 ) ∈ Ω(Θcup ((c, s0 ], g)). Then there exists a segment (y) ∈ Θcup ((c, s0 ], g).
Let y + = (s, g(s)) and y − = (a(s), g(a(s)) be the endpoints of (y) in Ω3 . We know
ε-extremizers at these points y + , y − .

Indeed, we can take the following ε-extremizers

(F + , G+ ) = (1 − s, 1 + s) and (F − , G− ) = (1 − a(s), 1 + a(s)) (i.e.

constant func-

tions). Consider an α−concatenation (F + • F − , G+ • G− )α , where α is chosen so that
y = αy + + (1 − α)y − . We have
Ψ[(F + • F − , G+ • G− )α ] = αΨ(F + , G+ ) + (1 − α)Ψ(F −1 , G− ) >
αB(y + ) + (1 − α)B(y − ) − ε = B(y) − ε.
The last equality follows from the linearity of B on (y).
Extremizers on the vertical line (−1, y3 ), y3 ≥ h(s0 ).
Now we are going to find ε-extremizers for the points (−1, y3 ) where y3 ≥ h(s0 ). We
use a similar idea mentioned in [20] (see proof of Lemma 3). We define the functions (F, G)
recursively:






−w





G(t) = γ · g t−ε
1−2ε








w

67

0 ≤ t < ε;
ε ≤ t ≤ 1 − ε;
1 − ε < t ≤ 1;






d−





F (t) = γ · f t−ε
1−2ε








d +

0 ≤ t < ε;
ε ≤ t ≤ 1 − ε;
1 − ε < t ≤ 1;

where the nonnegative constants w, d− , d+ , γ will be obtained from the requirement E(F, G, |F |p ) =
(2, 0, y3 ) and the fact that G is the martingale transform of F . Surely G
F

[0,1]

[0,1]

= 0. Condition

= 2 means that

(d− + d+ )ε + 2γ(1 − 2ε) = 2.
Condition |F |p

[0,1]

(1.6.3)

= y3 implies that
p

p

ε(d+ + d− )
.
y3 =
1 − (1 − 2ε)γ p

(1.6.4)

Now we use the condition |F0 − F1 | = |G0 − G1 |. In the first step we split the interval [0, 1]
at the point ε with the requirement F0 − F1 = G0 − G1 , from which obtain w = 2 − d− .
In the second step we split at the point 1 − ε with the requirement F1 − F2 = G2 − G1 ,
obtaining w = 2γ − d+ . From these two conditions we obtain d− + d+ = 2(1 + γ) − 2w, and
by substituting in (1.6.3) we find the γ

γ =1+

εw
.
1−ε

Now we investigate what happens as ε tends to zero. Our aim will be to focus on the limit

68

value limε→0 w = w0 . We have 1 − (1 − 2ε)γ p ≈ ε(2 − wp). So (1.6.4) becomes
p

p

ε(d+ + d− )
2(2 − w0 )p
.
y3 =
→
1 − (1 − 2ε)γ p
2 − w0 p

(1.6.5)

Note that for w0 = 1 + s equation (1.6.5) is the same as (1.4.18). By direct calculations we
see that as ε → 0 we have
(G2 + τ 2 F 2 )p/2

[0,1]

=

ε[(w2 + τ 2 d2− )p/2 + (w2 + τ 2 d2+ )p/2 ]
2f (w0 − 1)
.
→
p
1 − (1 − 2ε)γ
2 − w0 p

Now we are going to calculate the value B(−1, h(s)) where h(s) = y3 . From (1.4.16) we have
2
B(−1, h(s)) = h(s)t2 (s) − t1 (s).
p
By using (1.3.6) we express t1 via t2 , also because of (1.4.18) and (1.4.21) we have
2
B(−1, h(s)) = h(s)t2 (s) − t1 (s) = h(s)t2 −
p
f (s) − (s − yp )f (s)
2
2
t2 (h(s) + g ) − f =
p
p
g(s) − (s − yp )g (s)

2
(f − t2 g ) =
p
2g
2
+ g
p(yp − s) p

2 f (s)
2
2(2 − w0 )p
.
−f =
=
p
p yp − s
2 − w0 p

Thus we obtain the desired result

(G2 + τ 2 F 2 )p/2

[0,1]

→ B(−1, y3 ) as ε → 0.

Extremizers in the domain Ω(Θ([s0 , yp ), g)).
Pick any point y = (y2 , y3 ) ∈ Ω(Θ([s0 , yp ], g)). Then there exists a segment (y) ∈
Θ([s0 , yp ], g). Let y + and y − be the endpoints of this segment such that y + = (−1, y˜3 ) for

69

some y˜3 ≥ h(s0 ) and y − = (˜
s, g(˜
s)) for some s˜ ∈ [yp , s0 ). We remind the reader that we
know ε-extremizers for the points (s, g(s)) where s ∈ [s0 , 1], and we know ε-extremizers on
the vertical line (−1, y3 ) where y3 ≥ h(s0 ). Therefore, as in the case of a cup, taking the
appropriate α−concatenation of these ε-extremizers and using the fact that B is linear on
(y), we find an ε-extremizer at point y.
Extremizers in the domain Ang(s0 ).
Pick any y = (y1 , y2 ) ∈ Ang(s0 ). There exist the points y + ∈ + , y − ∈ − , where
+

= + (s0 , g(s0 )) ∈ Θ([s0 , yp ), g) and − = − (s0 , g(s0 )) ∈ Θ((c, s0 ], g), such that y =

αy + + (1 − α)y − for some α ∈ [0, 1]. We know ε-extremizers at the points y + and y − . Then
by taking an α−concatenation of these extremizers and using the linearity of B on Ang(s0 )
we can obtain an ε-extremizer at the point y.
Extremizers in the domain Ω(Θ([yp , 1], g)).
Finally, we consider the domain of vertical segments Ω(Θ[yp , 1], g). Pick any point y =
(y2 , y3 ) ∈ Ω(Θ[yp , 1]). Take an arbitrary point y + = (−1, y3+ ) where y3+ is sufficiently large
such that y = αy + + (1 − α)y − for some α ∈ (0, 1) and some y − = (y2− , y3− ) such that
(1, y2− , y3− ) ∈ ∂Ω1 . Surely, y + , y − belong to a positive domain. Condition (1, y2− , y3− ) ∈
∂Ω1 implies that we know an ε-extremizer (F − , G− ) at the point y − (these are constant
functions). We also know an ε-extremizer at the point y + . Let (F + • F − , G+ • G− )α be an
α−concatenation of these extremizers. Then

Ψ[(F + • F − , G+ • G− )α ] > αB(y + ) + (1 − α)B(y − ) − ε.

70

Note that the condition y = αy + + (1 − α)y − implies that

α=

y3 − y−2 y3−
y2

y3+ +

y3−
y2−

.

Recall that B(y2 , g(y2 )) = f (y2 ) and B(y + ) = f (s) + t1 (s)(−1 − s) + t2 (s)(y3+ − g(s)), where
s ∈ [s0 , yp ] is such that a segment (s, g(s)) ∈ Θ([s0 , yp ), g) has an endpoint y + .
Note that as y3+ → ∞ all terms remain bounded; moreover, α → 0, y − → (y2 , g(y2 )) and
s → yp . This means that
lim αB(y + ) + (1 − α)B(y − ) − ε =

y3+ →∞





y2 −
y
y2− 3 


 y3 −
lim t2 (s)y3+ 
 + y−  + f (y2 ) − ε = t2 (yp )(y3 − g(y2 )) + f (y2 ) − ε.
y3+ →∞
y3 + 3−
y2

We recall that t2 (s) = t2 (yp ) for s ∈ [yp , 1]. Then
B(y) = f (y2 ) + t2 (s(y))(y3 − g(y2 )) = f (y2 ) + t2 (yp )(y3 − g(y2 )).
Thus, if we choose y3+ sufficiently large then we can obtain a 2ε-extremizer for the point y.

1.6.2 Case s0 > yp .
In this case we have s0 ≥ yp (see Figure 1.11). This case is a little bit more complicated than
the previous one. Construction of ε-extremizers (F, G) will be similar to the one presented
in [21].
We need a few more definitions.

71

Definition 10. Let (F, G) be an arbitrary pair of functions. Let (y2 , g(y2 )) ∈ Ω3 and let J
˜ as follows:
be a subinterval of [0, 1]. We define a new pair (F˜ , G)

˜
(F˜ , G)(x)
=





(F, G)(x)

x ∈ [0, 1] \ J




(1 − y2 , 1 + y2 )

x ∈ J.

˜ as putting the constant (y2 , g(y2 )) on the interval J
We will refer to the new pair (F˜ , G)
for the pair (F, G)
˜ we will denote by the same
It is worth mentioning that sometimes the new pair (F˜ , G)
symbol (F, G).
Definition 11. We say that the pairs (Fα , Gα ), (F1−α , G1−α ) are obtained from the pair
(F, G) by splitting at the point α ∈ (0, 1) if
(Fα , Gα ) = (F, G)(x · α) x ∈ [0, 1];
(F1−α , G1−α ) = (F, G)(x · (1 − α) + α) x ∈ [0, 1];
It is clear that Ψ(F, G) = αΨ(Fα , Gα ) + (1 − α)Ψ(F1−α , G1−α ). Also note that if
(Fα , Gα ), (F1−α , G1−α ) are obtained from the pair (F, G) by splitting at the point α ∈
(0, 1), then (F, G) is an α−concatenation of the pairs (Fα , Gα ), (F1−α , G1−α ). Thus, such
operations as splitting and concatenation are opposite operations.
Instead of explicitly presenting an admissible pair (F, G) and showing that it is an εextremizer, we present an algorithm which constructs the admissible pair, and we show that
the result is an ε-extremizer.
By the same explanations as in the case s0 ≤ yp , it is enough to construct an ε-extremizer
72

(F, G) on the vertical line y2 = −1 of the domain Ω3 . Moreover, linearity of B implies that
for any A > 0, it is enough to construct ε-extremizers for the points (−1, y3 ), where y3 ≥ A.
(0)

Pick any point (−1, y3 ) where y3 = y3

> g(−1). Linearity of B on Ang(s0 ) and direct

calculations (see (1.3.8), (1.4.22)) show that

B(−1, y3 ) = f (−1) + t3 (s0 )(y3 − g(−1)) = f (−1) + (y3 − g(−1))

f (−1) − f (s0 )
. (1.6.6)
g (−1) − g (s0 )

We describe the first iteration. Let (F, G) be an admissible pair for the point (−1, y3 ),
whose explicit expression will be described during the algorithm. For a pair (F, G) we put a
constant (s0 , g(s0 )) on an interval [0, ε] where the value ε ∈ (0, 1) will be given later. Thus
we obtain a new pair (F, G) which we denote by the same symbol. We want (F, G) to be an
admissible pair for the point (−1, y3 ). Let the pairs (Fε , Gε ), (F1−ε , G1−ε ) be obtained from
the pair (F, G) by splitting at point ε. It is clear that (Fε , Gε ) is an admissible pair for the
point (s0 , g(s0 )). We want (F1−ε , G1−ε ) to be an admissible pair for the point P = (˜
y2 , y˜3 )
so that

(−1, y3 ) = ε(s0 , g(s0 )) + (1 − ε)P.

(1.6.7)

Therefore we require

P =

−1 − εs0 y3 − εg(s0 )
,
1−ε
1−ε

.

(1.6.8)

So we make the following simple observation: if (F1−ε , G1−ε ) were an admissible pair for
the point P , then (F, G) (which is an ε−concatenation of the pairs (1 − s0 , 1 + s0 ) and
(F1−ε , G1−ε )) would be an admissible pair for the point (−1, y3 ). Explanation of this obser73

vation is simple: note that these pairs (F1−ε , G1−ε ) and (1 − s0 , 1 + s0 ) are admissible pairs
for the points P and (s0 , g(s0 )) which belong to a positive domain (see (1.6.7)); therefore,
the rest immediately follows from Lemma 14. So we want to construct the admissible pair
(F1−ε , G1−ε ) for the point (1.6.8).
We recall Lemma 13 which implies that the pair (F1−ε , G1−ε ) is admissible for the point
0
1, −1−εs
1−ε ,

y3 −εg(s0 )
1−ε

˜ where
if and only if the pair (F˜ , G)

(F1−ε , −G1−ε ) =

1 + εs0 ˜ ˜
(F , G)
1−ε

(y −εg(s ))

ε−1 , 3
0
is admissible for a point W = 1, 1+εs
· (1 − ε)p−1 . So, if we find the admis(1+εs0 )p
0

˜ then we automatically find the admissible pair (F, G).
sible pair (F˜ , G)
Choose ε small enough so that

ε−1 (y3 −εg(s0 ))
1+εs0 , (1+εs0 )p

ε − 1 (y3 − εg(s0 ))
· (1 − ε)p−1
,
1 + εs0 (1 + εs0 )p
(1)

for some δ ∈ (0, 1) and y3
δ=
(1)
y3

· (1 − ε)p−1 ∈ Ω3 and
(1)

= δ(s0 , g(s0 )) + (1 − δ)(−1, y3 )

≥ g(−1). Then

ε
= ε + O(ε2 )
1 + εs0
=

(y3 −εg(s0 ))
(1+εs0 )p

ε g(s )
· (1 − ε)p−1 − 1+εs
0
0
= y3 (1 − ε(p + ps0 − 2)) − 2εg(s0 ) + O(ε2 ).
ε
1 − 1+εs0

(1.6.9)
˜ we put a constant (s0 , g(s0 )) on the interval [0, δ]. We split the new
For the pair (F˜ , G)
˜ at point δ so we get the pairs (F˜δ , G
˜ δ ) and (F˜1−δ , G
˜ 1−δ ). We make a similar
pair (F˜ , G)
˜ 1−δ ) for the point
observation as above. It is clear that if we know the admissible pair (F˜1−δ , G

74

(1)
˜ for the point ε−1 , (y3 −εg(s0p)) · (1 − ε)p−1 .
(−1, y3 ) then we can obtain an admissible pair (F˜ , G)
1+εs0 (1+εs0 )

˜ is a δ−concatenation of the pairs (1 − s0 , 1 + s0 ) and (F˜1−δ , G
˜ 1−δ ).
Surely (F˜ , G)
We summarize the first iteration. We took ε ∈ (0, 1), and we started from the pair
˜ 1−δ ).
(F (0) , G(0) ) = (F, G), and after one iteration we came to the pair (F (1) , G(1) ) = (F˜1−δ , G
(1)

We showed that if (F (1) , G(1) ) is an admissible pair for the point (1, y3 ), then the pair
(F (0) , G(0) ) can be obtained from the pair (F (1) , G(1) ); moreover, it is admissible for the
(0)

point (1, y3 ).
(j)

Continuing these iterations, we obtain the sequence of numbers {y3 }N
j=0 and the se(N )

quence of pairs {(F (j) , G(j) )}N
j=0 . Let N be such that y3
(N )

(F (N ) , G(N ) ) is an admissible pair for the point (−1, y3

≥ g(−1). It is clear that if

−1
) then the pairs {(F (j) , G(j) )}N
j=0

can be determined uniquely, and, moreover, (F (j) , G(j) ) is an admissible pair for the point
(j)

(−1, y3 ) for all j = 0, .., N − 1.
Note that we can choose sufficiently small ε ∈ (0, 1), and we can find N = N (ε) such
(N )

that y3

= g(−1) (see (1.6.9), and recall that s0 > yp ). In this case the admissible
(N )

pair (F (N ) , G(N ) ) for the point (−1, y3

) = (−1, g(−1)) is a constant function, namely,

(F (N ) , G(N ) ) = (2, 0). Now we try to find N in terms of ε, and we try to find the value of
Ψ(F (0) , G(0) ).
(1)

Condition (1.6.9) implies that y3

(0)

= y3 (1 − ε(p + ps0 − 2)) − 2εg(s0 ) + O(ε2 ). We

denote δ0 = p + ps0 − 2 > 0. Therefore, after the N -th iteration we obtain
(N )

y3

(N )

The requirement y3

= (1 − εδ0 )N

(0)

y3 +

2g(s0 )
δ0

= g(−1) implies that

75

−

2g(s0 )
+ O(ε).
δ0

(0)

(1 − εδ0

)−N

=

2g(s0 )
δ0
2g(s )
g(−1) + δ 0
0

y3 +

+ O(ε).

This implies that lim supε→0 ε · N = lim supε→0 ε · N (ε) < ∞. Therefore, we get
(0)

eεδ0 N

=

2g(s0 )
δ0
2g(s0 )
g(−1) + δ
0

y3 +

+ O(ε).

(1.6.10)

Also note that

Ψ(F (0) , G(0) ) = Ψ(F, G) = εΨ(Fε , Gε ) + (1 − ε)Ψ(F1−ε , G1−ε ) =

1 + εs0 p
˜
Ψ(F˜ , G)
1−ε

εf (s0 ) + (1 − ε)Ψ(F1−ε , G1−ε ) = εf (s0 ) + (1 − ε)
= εf (s0 ) + (1 − ε)(1 − ε)

1 + εs0 p
˜ 1−δ )
δf (s0 ) + (1 − δ)Ψ(F˜1−δ , G
1−ε

= 2εf (s0 ) + (1 + εδ0 )Ψ(F (1) , G(1) ) + O(ε2 ).

Therefore, after the N -th iteration (and using the fact that Ψ(F (N ) , G(N ) ) = f (−1)) we
obtain

Ψ(F (0) , G(0) ) = (1 + εδ0 )N
eεδ0 N

f (−1) +

2f (s0 )
δ0

−

f (−1) +

2f (s0 )
δ0

−

2f (s0 )
+ O(ε) =
δ0

2f (s0 )
+ O(ε).
δ0

The last equality follows from the fact that lim supε→0 ε · N (ε) < ∞.

76

(1.6.11)

Therefore (1.6.10) and (1.6.11) imply that


(0)
y3



2g(s )
+ δ 0
(0)
(0)
0
 f (−1) + 2f (s0 )
Ψ(F , G ) = 
2g(s )
δ0
g(−1) + δ 0
0


2f (s0 )
f (−1) + δ
(0)
0
 + O(ε).
f (−1) + (y3 − g(−1)) 
2g(s0 )
g(−1) + δ
0

−

2f (s0 )
+ O(ε) =
δ0

Now we recall (1.6.6). So if we show that
f (−1) +
g(−1) +

2f (s0 )
δ0
2g(s0 )
δ0

=

f (−1) − f (s0 )
g (−1) − g (s0 )

(1.6.12)

(0)

then (1.6.12) will imply that Ψ(F (0) , G(0) ) = B(−1, y3 ) + O(ε). So choosing ε sufficiently
small we can obtain the extremizer (F (0) , G(0) ) for the point (−1, y3 ). Therefore, we need
only to prove equality (1.6.12). It will be convenient to make the following notations: set
f− = f (−1), f− = f (−1), f = f (s0 ), f = f (s0 ), g− = g(−1), g− = g (−1), g = g(s0 ) and
g = g(s0 ). Then the equality (1.6.12) turns into the following one
f g − f g − f− g + f g
δ0
= −
.
2
g f − − f g−
This simplifies into the following one

s0 − yp =

f g − f g − f− g + f g
2 f g− − f g − f− g + f g
·
= −
p
g f− − f g−
−g f− + f g−

which is true by (1.4.24).

77

(1.6.13)

Chapter 2
Hessian of Bellman functions and
uniqueness of Brascamp–Lieb
inequality
2.0.3 Brascamp–Lieb inequality
The classical Young’s inequality for convolutions on the real line asserts that for any f ∈
Lp (R) and g ∈ Lq (R) where p, q ≥ 1, we have an inequality

f ∗g r ≤ f p g q

(2.0.1)

1 1
1
+ =1+ .
p q
r

(2.0.2)

if and only if

In what follows f ∗ g denotes convolution i.e. f ∗ g(y) = R f (x)g(y − x)dx. The necessity
of (2.0.2) follows immediately: by stretching the functions f and g as fλ (x) = λ1/p f (λx),
1 + 1 −1− 1
q
r

gλ (x) = λ1/q g(λx) corresponding norms do not change. Since fλ ∗gλ r = λ p

f ∗g r

we obtain (2.0.2). Beckner (see [22]) found the sharp constant C = C(p, q, r) of the inequality
78

f ∗ g r ≤ C f p g q . At the same time (see [23]) Brascamp and Lieb derived more general
inequality, namely, let a1 , .., an be the vectors of Rk where 1 ≤ k ≤ n, let uj ∈ L1 (R) be
nonnegative functions, where 1 ≤ pj ≤ ∞ and
n

1/pj

Rk j=1

uj

n
1
j=1 pj

= k, then we have a sharp inequality
n

( aj , x )dx ≤ D(p1 , . . . , pn )

1/pj

uj (x)dx

,

(2.0.3)

R

j=1

where ·, · denotes scalar product in Euclidian space,
n

D(p1 , . . . , pn ) =

sup

b1 ,..,bn >0 Rk j=1

1/pj

gj

( aj , x )dx

(2.0.4)

2
1/2
and gj (x) = bj e−πx bj . In other words, supremum in (2.0.3) is achieved by centered,

normalized (i.e.

gj 1 = 1) gaussian functions. Usually inequality (2.0.3) is written as

follows:
n

n
Rk j=1

wj ( aj , x )dx ≤ D(p1 , . . . , pn )

wj pj .
j=1

for uj ∈ Lpj (R). Surely the above inequality becomes the same as (2.0.3) after introducing
1/pj

the functions wj (x) = uj

(y).

It is clear that Brascamp–Lieb inequality (2.0.3) implies sharp Young’s inequality for
convolutions. Indeed, just take n = 3, k = 2, a1 = (1, 0), a2 = (1, −1), a3 = (0, 1) and use
duality argument.
The next natural question which arose was the following one: what conditions should
the vectors aj and the numbers pj satisfy in order for the constant D(p1 , . . . , , pn ) to be
finite. It turns out that the answer has simple geometrical interpretation which was first

79

found by Barthe (see [24]): we consider all different k-tuples of vectors (aj1 , .., ajk ) such that
they create basis in Rk . All we need from these tuples are the numbers j1 , . . . , jk . Each
k-tuple defines a unique vector v ∈ Rn with entries 0 and 1 so that ji -th component is 1
(i = 1, . . . , k) and the rests are zeros. Finally we take convex hull of the vectors v and
1
1
p1 , . . . , pn

denote it by K. The constant D(p1 , . . . , pn ) is finite if and only if

∈ K. In

other words, in order to make the set K large we want the vectors a1 , . . . , an to be more
linearly independent. Later the proof of the Brascamp–Lieb inequality (2.0.3) was simplified
(see [25]) by heat flow method. The idea of the method is quite similar to Bellman function
technique which we are going to discuss in the current article. The same idea was used in [26]
in order to derive general rank Brascamp–Lieb inequality (see also [27]): let Bj : Rk → Rkj
be a surjective linear maps, uj : Rkj → R+ , kj , k ∈ N, and pj ≥ 1 are such that

n
j=1

kj
pj

=k

then we have a sharp inequality

1/p
1/p
u1 1 (B1 x) · · · un n (Bn x)dx
k
R

1/pn

1/p1

≤C

Rk1

···

u1

Rk n

un

where

C=

sup

1/p1

A1 ,...,An >0 Rk

G1

1/pn

(B1 x) · · · Gn

(Bn x)dx

(2.0.5)

and Gj (y) = e−π Aj y,y (det Aj )1/2 . Supremum in (2.0.5) is taken over all positive definite
kj × kj matrices Aj . One of the main result obtained in [26] describes finiteness of the
number C, namely, C is finite if and only if
n

dim(V ) ≤

j=1

dim (Bj V )
pj

for all subspaces V ⊂ Rk .

80

After this result the original inequality (2.0.3) got a name rank 1 Brascamp–Lieb inequality.
If k = 1 the inequality (2.0.3) becomes usual multilinear H¨older’s inequality
n
R j=1

1/p
uj j (x)dx

n

≤

1/pj

⇐⇒

uj
R

j=1

j

1
= 1.
pj

(2.0.6)

From the Bellman function point of view the multilinear H¨older’s inequality holds because
the following function

1/p1

B(x1 , . . . , xn ) = x1

n
1
j=1 pj

is concave in the domain xj ≥ 0 for

1/pn

· · · xn

(2.0.7)

≤ 1 (we assume that pj > 0).

This Bellman function point of view asks us to look for the description of functions B
such that
n
Rk

R
1/p1

Function B(x1 , . . . , xn ) = x1

1/pn

· · · xn

,

n
1
j=1 pj

.

ui (x)dx

B(u1 ( a1 , x ), . . . , un ( an , x )dx is estimated in terms of

(2.0.8)

i=1

= 1, is an example of such a function for

k = 1. But for k = 1 one can easily get the full description of “Bellman functions” that give
inequality (2.0.9) below.
The equality

n
1
j=1 pj

= 1 was needed because the function B(x1 , . . . , xn ) has to be

homogeneous of degree 1 i.e., B(λx) = λB(x). This allows us to write integral over the real
line. Indeed, if the nonnegative functions uj are integrable then Jensen’s inequality implies
1
B(u1 , . . . , un )dx ≤ B
|I| I

1
1
u1 dx, . . . ,
un dx
|I| I
|I| I

81

where I is any subinterval of the real line. Since the function B is 1-homogeneous we can
rewrite the above inequality as follows

I

B(u1 , . . . , un )dx ≤ B

I

u1 dx, . . . ,

un dx .
I

Take I = [−R, R] and send R to infinity. B is continuos, so that

B
I

u1 (x)dx, . . . ,

un (x)dx
I

→B

u1 dx, . . . ,
R

un dx

.

R

Continuity of B and monotone convergence theorem implies that

R

B(u1 (x), . . . , un (x))dx ≤ B

u1 dx, . . . ,
R

un dx

(2.0.9)

R

It is worth to formulate the following lemma. Set Rn
+ = {(x1 , . . . , xn ) : xj ≥ 0}.
Lemma 15. Let uj be nonnegative integrable functions j = 1, . . . , n on the real line. If B is
1-homogeneous concave function on Rn
+ , then (2.0.9) holds. Equality is achieved in (2.0.9)
if (u1 , . . . , un ) are all proportional.
Proof. As we just saw, the proof follows from showing that I B(u1 , . . . , un ) → R B(u1 , . . . , un ).
We are going to find now a summable amjorant. Take any point x0 from the interior of Rn
+.
Consider any subgradient v = (v1 , . . . , vn ) at point x0 i.e. B(x) ≤ v, x − x0 + B(x0 ). Take
x = λx0 and use the homogeneity of B. Thus we obtain (λ − 1)B(x0 ) ≤ (λ − 1) v, x0 for
any λ ≥ 0. This means that B(x0 ) = v, x0 and, therefore, B(x) ≤ v, x . On the other
hand let ej = (0, . . . , 1, . . . , 0) be a basis vectors (j-th component entry is 1 and the rests
are zero). Consider any point x = (x1 , . . . . , xn ) ∈ Rn
+ . Concavity and homogeneity of B
82

implies that B(x) ≥

n
j=1 xj B(ej ).

|B(x)| ≤ max


 n


j=1

So we obtain the majorant
n

xj |B(ej )|,

j=1



xj |vj |


for any x ∈ Rn
+.

Plugging uj for xj we see that the use of Lebesgue’s dominated convergence theorem is
justified.
The above Lemma says that homogeneity and concavity of the function implies the
inequality (2.0.9). The converse is also true.
Now the following question becomes quite natural:
Question. Assume a1 , . . . , an ∈ Rk , B is continuous function defined on Rn
+ and uj , j =
1, . . . , n are nonnegative integrable functions. What is the sharp estimate of the expression

Rk

B(u1 ( a1 , x ), . . . , un ( an , x ))dx

(2.0.10)

in terms of R uj ?
In other words, along with Young’s functions

1/p1

B(x1 , . . . , xn ) = x1

1/pn

· · · · · xn

,

1
= k,
pj

what can be other Brascamp–Lieb Bellman functions that would give us sharp estimates of
(2.0.10)?
We give partial answer on this question. It turns out that if one requires function B is
homogeneous of degree k and in addition it satisfies some mild assumptions (smoothness and
exponential integrality given below), then we can find the sharp estimate of the expression
83

(2.0.10) in terms of R uj , if B satisfies an interesting concavity condition. In a trivial case
k = 1 our theorem gives us of course inequality (2.0.9).
In the trivial case k = 1 we already saw that the interesting concavity condition mentioned above is precisely the usual concavity of B. In another trivial case k = n, the
interesting concavity condition mentioned above becomes “separate concavity” of B in each
of its variables.
For 1 < k < n our concavity condition is, in fact, some compromise between these two
concavities.
As we will see k = n − 1 and k = n this concavity condition (plus k-homogeneity and
mild regularity) imply that Brascamp–Lieb Bellman functions B can be only the standard
1/p1

ones: B(x1 , . . . , xn ) = x1

1/pn

· · · · · xn

,

1
pj

= k.

Before we start formulating our results, we will explain that there are many 1-homogeneous
concave functions B on Rn
+.
Lemma 16. Function B is continuous, concave and homogeneous of degree 1 on Rn
+ if and
˜
˜ (λy)
only if there exists continuous, concave function B(y)
on Rn−1
such that limλ→∞ λ1 B
+
˜ x2 , . . . , xn
exists, it is continuous with respect to y and B(x1 , . . . , xn ) = x1 B
x1
x1
Proof. Indeed, if B is continuous, concave and homogeneous of degree 1 then B(x1 , . . . , xn ) =
x1 B 1, xx21 , . . . , xxn1

˜ = B(1, y1 , . . . , yn−1 ) is continuous and concave in
and the function B

n−1
1 ˜
Rn−1
+ . Moreover, for each y ∈ R+ , limλ→∞ λ B(λy) exists and it is continuos with respect

to y.
˜ satisfies the conditions of the Lemma. Consider
Assume B

˜
B(x1 , . . . , xn ) = x1 B
84

x2
xn
,...,
x1
x1

.

It is clear that B is continuous on Rn
+ and it is homogeneous of degree 1. We will show that
B is concave in the interior of Rn
+ and hence by continuity it will be concave on the closure
¯ = (x2 , . . . , xn ), y
¯ = (y2 , . . . , yn )
as well. Let x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) ∈ Rn . Let x
and α + β = 1 for α, β ∈ [0, 1]. Then we have
B(αx + βy) = (αx1 + βy1 )B 1, α
αB 1,

¯
x
αx1 + βy1

+ βB 1,

¯
¯
y
x
+β
αx1 + βy1
αx1 + βy1

¯
y
αx1 + βy1

≥ (αx1 + βy1 )×

= αB(x) + βB(y).

2.0.4 Bellman function in Brascamp–Lieb inequality
2
n
In what follows we assume that B ∈ C(Rn
+ )∩C (int(R+ )). In order for the quantity (2.0.10)

to be finite it is necessary to assume that 1 ≤ k ≤ n. Fix some vectors aj = (aj1 , . . . , ajn ) ∈
Rk and k × k symmetric matrix C such that Caj , aj > 0 for j = 1, . . . , n. Let A be a k × n
matrix constructed by columns aj i.e. A = (a1 , . . . , an ). We denote A∗ transpose matrix of
A.
Let uj : R → R+ be such that 0 < R uj < ∞. Let uj (y, t) solves the heat equation
∂uj
∂t

∂2u

− Caj , aj ∂y2j = 0 with the initial value uj (y, 0) = uj (y). Let Hess B(y) denotes

Hessian matrix of the function B at point y.
For two square matrices of the same size P = {pij } and Q = {qij } , let P • Q = {pij qij }
be Hadamard product. Denote by symbol

u(x, t) = (u1 ( a1 , x , t), . . . , un ( an , x , t))

85

and denote
u (x, t) = u1 ( a1 , x , t), . . . , un ( an , x , t) ,
where uj ( aj , x , t) =

∂uj (y,t)
∂y
y= aj ,x

.

Lemma 17. For any 0 < t < ∞ and any x ∈ Rk we have


∂ −
∂t

k



∂2

cij

∂xi ∂xj

i,j=1

 B(u(x, t)) = − (A∗ CA) • Hess B(u(x, t))u (x, t), u (x, t) .
(2.0.11)

Proof. First we show that the functions u ( a , x , t) satisfy the following heat equation


∂ −
∂t

k

cij
i,j=1

Indeed, let uj ( aj , x , t) =

k

cij
i,j=1
k
i,j=1

∂2
∂xi ∂xj


 u ( a , x , t) = 0,

∂ 2 uj (y,t)
∂y 2

for any

= 1, . . . , n.

. Then
y= aj ,x

∂2
u ( a , x , t) =
∂xi ∂xj

k

cij
i,j=1

∂
a j u ( a , x , t) =
∂xi

∂
cij a j a i u ( a , x , t) = Ca , a u ( a , x , t) = u ( a , x , t).
∂t

86

Let u = u(x, t) and u = u ( a , x , t). Then


∂ −
∂t

k

∂2



n

k

n

∂
∂B ∂u
∂B ∂u
−
cij
=
∂xi ∂xj
∂u ∂t
∂xi
∂u ∂xj
i,j=1
i,j=1
=1
=1




n
k
k
n
n
2
2
2
∂
∂B 
∂ B ∂um ∂u
∂B ∂ u 
cij
u −
cij 
+
=
∂u
∂xi ∂xj
∂um ∂u ∂xi ∂xj
∂u ∂xi ∂xj
i,j=1
i,j=1
=1
,m=1
=1
k

−
−

cij

n

cij
i,j=1 ,m=1
n

Cam , a
,m=1

 B(u) =

∂ 2 B ∂um ∂u
=−
∂um ∂u ∂xi ∂xj

k

n

cij
i,j=1 ,m=1

∂ 2B
a a u u =
∂um ∂u mi j m

∂ 2B
u u = − (A∗ CA) • Hess B(u)u , u .
∂um ∂u m

Remark 13. Let us denote by ∆C :=
∂
− ∆C
∂t

k
∂2
i,j=1 cij ∂xi ∂xj .

We used above that

u ( a , x , t) = 0 .

This is exactly the equality that implies

∂
− ∆C
∂t

B(u ( a , x , t)) = − (A∗ CA) • Hess B(u(x, t))u (x, t), u (x, t) .

In other words, we look at the natural “energy” of the problem

(2.0.12)

B(u ( a , x , t)) dx at

time t, and differentiate it in t. Replacing d/dt by d/dt − ∆C does not change the result
because when we integrate the above equality over x varibales, we should expect the term
∆C B(u ( a , x , t)) dx to disappear (and this is exactly what happens below). But the
definite sign in the right hand side of (2.0.11) guarantees us now the monotonicity property
of the energy.
So composing of the special heat flow e−t∆C and special function B seems like a good idea
87

exactly because of the monotonicity formula, which we are going to obtain shortly below.
Further we make several assumptions on the function B. The assumption L3 is exactly
the concavity we were talking about above.
2
n
L1. B ∈ C(Rn
+ ) ∩ C (int(R+ )).

L2. B(λy) = λk B(y) for all λ ≥ 0 and y ∈ Rn
+.
L3. There exists k × k symmetric matrix C such that (A∗ CA) • Hess B(y) ≤ 0 for y ∈
int(Rn
+ ), and Caj , aj > 0 for all j = 1, . . . , n.
L4. B ≥ 0 and B is not identically 0.
L5.

Rk

2
2
B(e− a1 ,x , . . . , e− an ,x )dx < ∞.

(2.0.13)

We make several observations: properties L3 and L4 imply that the function B is separately concave (i.e. concave with respect to each variable) and increasing with respect to
each variable, moreover, B > 0 in int(Rn
+ ). The above properties imply that

Rk

2
2
B(b1 e−δ1 a1 ,x , . . . , bn e−δ2 an ,x )dx < ∞

(2.0.14)

for any positive numbers bj , δj > 0.
Consider the following class of functions E(R): u ∈ E(R) if and only if there exist
2

constants b, δ > 0 such that |u(y)| ≤ be−δy . It is clear that if u ∈ E(R) then u(y, t) ∈
E(R) for any t ≥ 0 where u(y, t) denotes heat extension of u(y) i.e. u(y, t) = u(y, 0) and
88

∂
∂t u(y, t)

2

∂ u(y, t) with some σ > 0. Note that E(R) contains the functions with
= σ ∂y
2

compact support. Also note that if nonnegative functions uj belong to the class E(R) then
the following function

B(t) =

Rk

B(u1 ( a1 , x , t), . . . , un ( an , x , t))dx.

(2.0.15)

is finite for any t ≥ 0.
Lemma 18. Let uj be nonnegative functions from E(R). Then for any t ∈ (0, ∞) we have

k

lim

r→∞ V
r i,j=1

cij

∂2
B(u(x, t))dx = 0 where Vr = {x ∈ Rk : x ≤ r}.
∂xi ∂xj

2
2
Proof. Let F (x) = B(e− a1 ,x , . . . , e− an ,x ). Let x = rσ where σ ∈ Sk−1
1 . Since B is

increasing with respect to each components, for each σ the function F (rσ) is decreasing
with respect to r. Therefore the function F˜ (r) = Sk−1 F (rσ)dσ1 is decreasing. Here σr
1
or radius r. Since 0∞ F˜ (r)rk−1 dr < ∞ we
denotes surface measure of the sphere Sk−1
r
obtain

Rk F˜ (2R) ≤ R

min

R≤r≤2R

F˜ (r)rk−1 ≤

2R

F˜ (r)rk−1 dr.

R

This implies that limr→∞ rk F˜ (r) = 0.
By Stokes’ formula we have
∂
∂2
B(u(x, t)) =
B(u(x, t))ni dσr
Vr ∂xi ∂xj
∂Vr ∂xj
89

where ni is the i-th component of the unit normal vector to the boundary of the ball Vr .
Homogeneity of B implies that

n
∂
j=1 ∂yj B(y)yj

∂ B(y) ≥ 0 we obtain
= kB(y). Since ∂y
j

∂ B(y)y ≤ kB(y). Also we note that for each t > 0 there exists a constant L
estimate ∂y
j
j
∂ u (y, t) ≤ Lyu (y, t) for all y ∈ R.
depending on the parameters t, uj such that ∂y
j
j

So we obtain that
n

∂
∂B ∂u ( a , x , t)
B(u(x, t))ni dσr ≤
dσr ≤
∂u
∂x
j
∂Vr ∂xj
∂V
r
=1
c
∂Vr

rB(u(x, t))dσr ≤ C1 rk F˜ (C2 r)

where constants C1 , C2 do not depend on r. Since B is homogeneous and it is increasing with
respect to each components the last inequality follows from the the observation B(u(x, t)) ≤
C3 F (C2 x) where C3 , C2 do not depend on x. So the lemma follows.

Remark 14. Lemma 18 holds even if we take supremum with respect to t over any compact
subset of (0, ∞).
Corollary 6. The function B(t) is increasing for t > 0, and it is constant if and only if
(A∗ CA) • Hess B(u(x, t))u (x, t) = 0 for all x ∈ Rn and any t > 0.
Proof. First we integrate (2.0.11) with respect to t (over any closed interval [t1 , t2 ] ⊂ (0, ∞))
and then we integrate other the balls Vr . Thus the corollary is immediate consequence of
Lemmas 17, 18 and Remark 14.
Thus we obtain an inequality B(t1 ) ≤ B(t2 ) for 0 < t1 ≤ t2 < ∞ and we want to pass
to the limits.

90

Lemma 19. Let B satisfies assumptions L1 − L5 and let uj ∈ E(R) be nonnegative (not
identically zero) functions. Then the following equalities hold

lim B(t) =

t→0

Rk

B(u1 ( a1 , x ), . . . , un ( an , x ))dx,

2

(2.0.16)
a ,x 2

a ,x

lim B(t) =

t→∞

Rk


B

− 1
e Ca1 ,a1

u1 dx, . . . ,

π Ca1 , a1



− n
e Can ,an

π Can , an

R

R


un dx dx.

(2.0.17)

Proof. Take any nonnegative (not identically zero) functions uj ∈ E(R). Then there exist
2

positive numbers βj , δj such that uj (y) ≤ βj e−δj y for all j = 1, . . . , n. Note that
uj (y, t) =

(y−x)2

1
(4πt Caj , aj )1/2 R

uj (x)e

− 4t Ca ,a
j j

dx ≤

y2 δ

j
− 1+4tδ Ca
j
j ,aj

βj

e

1 + 4tδj Caj , aj
So the first limit (2.0.16) follows immediately from Lebesgue’s dominated convergence theorem. For the second limit (2.0.17) we use homogeneity of the function B. So by changing
√
variable x = y t we obtain

Rk

B . . . ,

1

uj (x)e
(4πt Caj , aj )1/2 R


Rk


B
. . . ,

( a ,x −x)2
− 4t jCa ,a
j j

e

aj ,y 2
− 4 Ca
j ,aj

(4π Caj , aj )1/2 R

uj (x)e

91

√
2 t aj ,y x−x2
4t Caj ,aj


dx, . . . dx =


dx, . . .
 dy.

It is clear that that for each fixed y integrand tends to


B
. . . ,

e

aj ,y 2
− 4 Ca
j ,aj



(4π Caj , aj )1/2 R


uj (x)dx, . . .
.

2

Since uj (x) ≤ bj e−δj x we obtain

uj (x)e

√
2 t aj ,y x
4t Caj ,aj

δ

δ

j 2
j 2
≤ bj e− 2 x emaxx≥0 [− 2 x +αj (t)x] ,

a ,y
. Hence
where αj (t) := √ j
2 t Caj ,aj

uj (x)e

√
2 t aj ,y x
4t Caj ,aj

≤

α (t)2
δj 2 1 j
2 δj
−
x
bj e 2 e

=

aj ,y 2
δj 2
2
bj e− 2 x e 8tδj Caj ,aj ,

Now we can apply Lebesgue’s dominated convergence theorem twice.
The last display estimate gives us a summable majorant for the integration in x. On the
other hand,
2

e

aj ,y
− 4 Ca
j ,aj

e

aj ,y 2
8tδj Caj ,aj 2

2

≤e

aj ,y
− 8 Ca
j ,aj

for all t ≥ tC . Thus we get the uniform in t estimate for the jth argument of function B:
2

aj ,y
− 8 Ca
j ,aj

γj e

(4π Caj , aj )1/2
δj 2

,

where γj := R bj e− 2 x dx. Therefore we have the summable majorant (that it is summable

92

follows from L3)
2





aj ,y
− 8 Ca
j ,aj
e



γj
,
B
.
.
.
,
,
.
.
.


(4π Caj , aj )1/2
and the lemma is proved.
Corollary 6 and Lemma 19 imply the following theorem.
Theorem 2.0.1. Let B satisfies assumptions L1 − L5 and let uj ∈ E(R) be nonnegative
(not identically zero) functions. Then we have

Rk

B(u1 ( a1 , x ), . . . , un ( an , x ))dx ≤

2

a ,x 2

a ,x

Rk


B

− 1
e Ca1 ,a1

π Ca1 , a1

(2.0.18)

u1 , . . . ,
R



− n
e Can ,an

π Can , an

R


un  dx.

Equality holds if and only if

(A∗ CA) • Hess B(u(x, t))u (x, t) = 0

for all x ∈ Rn

and any

t > 0.

(2.0.19)

Remark 15. So any function satisfying our strange concavity condition L3, homogeneity
condition L2 and some mild conditions L1, L4, L5 gives a certain Brascamp–Lieb inequality.
Our next goal will be to show that in interesting cases the finiteness of (2.0.13) implies that
there is basically only one such B.
In the Bellman function technique theorems of the above type are known as a first part
of the Bellman function method which is usually simple. Any function B that satisfies
properties L1 − L5 will be called Bellman function of Brascamp–Lieb type.
The difficult technical part is how to find such Bellman functions. It is worth mentioning
93

that the property L3 in principle requires solving partial differential inequalities. We are
going to give partial answer on this question in the following section.

2.1 How to find the Bellman function
Definition 12. Let y = (y1 , y2 , . . . , yn ) ∈ int(Rn
+ ) and let D(y) be a diagonal square matrix
y

such that on the diagonal it has the terms Ca j,a , j = 1, . . . , n.
j j
Theorem 2.1.1. If the function B satisfies assumptions L1 − L5 then we have
AD(y)[A∗ CA • Hess B(y)] = 0

for all

y ∈ int(Rn
+ ).

(2.1.1)

Remark 16. Equality (2.1.1) is a second order partial differential equation on B. However,
assumptions L1 − L5 are either of quantitative nature, or in the form of partial differential
inequalities. So it is quite surprising that based only on assumptions L1 − L5 one can expect
equality (2.1.1).
The proof of the above equality is interesting in itself.
Proof. We saw in the previous section that assumptions L1−L5 imply the inequality (2.0.18).
One can easily observe that the following functions
2

uj (y) = bj

e

− Cay ,a
j j

π Caj , aj

,

bj > 0.

−2 D(u(x, t))A∗ x, Theorem 2.0.1
give equality in the inequality (2.0.18). Since u (x, t) = 4t+1

94

implies that

A∗ CA • Hess B(u(x, t))D(u(x, t))A∗ x = 0
Choose any x ∈ Rk , any y ∈ int(Rn
+ ) and any t > 0. We can find b1 , . . . , bn > 0 such that
a ,x 2

uj ( aj , x , t) = bj

e

− Ca ,aj (4t+1)
j j

π Caj , aj (4t + 1)

= yj ,

j = 1, . . . , n.

Hence we obtain

[A∗ CA • Hess B(y)]D(y)A∗ x = 0,

∀x ∈ Rk ,

∀y ∈ int(Rn
+ ).

So equality (2.1.1) follows.
Theorem 2.1.1 implies the following corollary.
Corollary 7. For any y ∈ int(Rn
+ ) we have
rank(A∗ CA • Hess B(y)) ≤ n − k.

(2.1.2)

The above corollary immediately follows from the fact that rank(AD(y)) = rank(A) = k
and, for example, from the Sylvester’s rank inequality.
Thus for each fixed n we have a range of admissible dimensions 1 ≤ k ≤ n. For the
boundary cases k = 1 and k = n, we find the Bellman function with the properties L1 − L5.
For the intermediate cases 1 < k < n we partially find the function B.

95

2.1.1 Case k = 1. Jointly concave and homogeneous function
We want to see that in this case L1 − L5 gives us precisely convex and 1-homogeneous
functions. In the case k = 1 we have A = (a1 , . . . , an ) ∈ Rn . Since the condition Caj , aj >
0 must hold, the 1 × 1 matrix C must be a positive number and aj = 0 for all j = 1, . . . , n.
The fact that B is homogeneous of degree 1 and B is increasing with respect to each variable
immediately imply L5. The only property we left to ensure is L3. For v = (v1 , . . . , vn ) ∈ Rn
let d(v) denotes n × n diagonal matrix with entries vj on the diagonal.
A∗ CA • Hess B(y) = C · A∗ A • Hess B(y) = C · d(A)Hess B(y)d(A) .
So the inequality A∗ CA • Hess B(y) ≤ 0 is equivalent to the inequality Hess B(y) ≤ 0,
because C is just a number. Thus we obtain the following lemma.
Lemma 20. If the function B satisfies assumptions L1 − L5 then aj = 0 for all j, C is
2
n
any positive number and B ∈ C(Rn
+ ) ∩ C (int(R+ )) is a concave homogeneous function of
2
n
degree 1. Conversely, if aj = 0 for all j and B ∈ C(Rn
+ ) ∩ C (int(R+ )) is a nonnegative,

not identically zero, concave, homogeneous function of degree 1 then B satisfies assumptions
L1 − L5.
The above lemma gives complete characterization of the Bellman function in the case
k = 1, and the inequality (2.0.18) is the same as inequality (2.0.9) (see Lemma 15).

2.1.2 Case k = n. B(y) = Const · y1 · · · yn
We show that in the case k = n the assumptions L1 − L5 are satisfied if and only if
B(y) = M y1 · · · yn where M is a positive number. We present 3 different proofs (according
96

to their chronological order), each of them uses different assumptions on B in necessity part.
Sufficiency follows immediately. Indeed, if B = M y1 · · · yn then all the assumptions L1 − L5
are satisfied except that one has to check existence of the symmetric matrix C such that
A∗ CA • Hess B ≤ 0 and Caj , aj > 0. But it is enough to take C = (AA∗ )−1 . Now we go
to proving necessity.

2.1.2.1 First proof
n
As we already mentioned the assumptions L1 − L5 imply that B ∈ C 2 (int(Rn
+ )) ∩ C(R+ ) is

nonnegative, separately concave, and it is homogeneous of degree n. We need to show that
such B then must have the form B(y) = M y1 · · · yn . To show this, we consider a function
G such that G(ln z1 , . . . , ln zn ) =

B(z1 ,...,zn )
z1 ···zn

for zj > 0. Homogeneity of order 0 of B implies

∂G + ∂ 2 G ≤ 0
that div G = 0, and concavity of B with respect to each variable implies that ∂y
2
j

∂yj

for j = 1, . . . , n. After summation of the last inequalities we obtain that G is superharmonic
function on Rn . But then it is easy to check that if ∆G ≤ 0, then ∆g ≥ 0, where g := e−G .
We get a bounded subharmonic function g, 0 ≤ g ≤ 1, in the whole space. It is well known
that then g must be constant. This implies implies that G is a constant.

2.1.2.2 Second proof
The second proof immediately follows from the following lemma which does not use any
assumptions regarding smoothness of B.
n
Lemma 21. If a function B defined on Rn
+ is nonnegative on the boundary of R+ , and it is

separately concave and homogeneous of degree n then B(y1 , . . . , yn ) = M y1 · · · yn for some
real number M .

97

Proof. The idea is almost as follows: we are going to construct superharmonic function in
the bounded domain such that it is nonnegative on the boundary and it achieves zero value
at an interior point of the domain. This implies that the constructed function is identically
zero.
Consider a function G(y1 , . . . , yn ) = B(y) − B(1, . . . , 1)y1 · · · yn . Take any cube Q =
[0, R]n where R > 1. The function G is separately concave and it is zero on the diagonal
of the cube Q i.e. G(y, . . . , y) = 0 for y ∈ [0, R]. G is nonnegative on the whole boundary
of the cube Q. Indeed, G is zero at the point (R, . . . , R) and it is nonnegative at point
(R, . . . , R, 0), so separate concavity implies that G is nonnegative on the set (R, . . . , R, t)
where t ∈ [0, R]. Similar reasoning implies that G is nonnegative on the whole boundary of
the cube Q.
Suppose now G is not zero at some interior point of the cube Q, say at point W . Take any
interior point A0 of the cube Q such that G(A0 ) = 0. Take a sequence of points A1 , ..., An
belonging to the interior of Q such that the segments Aj Aj+1 (the segment in Rn with the
endpoints Aj , Aj+1 ) are collinear to one of the vector ek = (0, . . . , 1, . . . , 0) (on the k-th
position we have 1 and the rest of the components are zero) for all j = 0, .., n − 1, and the
same is true for the segment An W . Then clearly G is zero on the segment A0 A1 . Indeed, It
is zero at point A0 . Take a line joining the points A0 , A1 . This line intersects the boundary
of the cube Q, and G is concave on the line. Since G is nonnegative at the points of the
intersection and it is zero at point A0 we obtain that G is zero on the part of the line which
lies in the cube Q. In particular, it is zero at A1 . By induction we obtain that G is zero at
the points A2 , .., An , W . So the lemma follows.

98

2.1.2.3 Third proof
In this proof let us assume that B is infinitely differentiable in

(Rn
+ ). The assumptions

L1 − L5 imply that B must be a separately concave. Therefore for the assumption L3 we
2
can choose C = (AA∗ )−1 . Then (2.1.2) implies that ∂ B2 = 0 for all j = 1, . . . , n. We claim

∂yj

2
that if B satisfies the system of differential equations ∂ B2 = 0 for all j = 1, . . . , n then it

∂yj

has a form
n

c0 +



n

(2.1.3)

yij 

ci1 ...ik


k=1



k
j=1

ip =iq , i1 ,...,ik =1

where the second summation is taken over the pairwise different indexes. Indeed, proof is by
2
induction over the dimension n. If n = 1 the claim is trivial. Since ∂ B2 = 0 we have B(y) =

y1 B1 (y2 , . . . , yn ) + B2 (y2 , . . . , yn ). The condition

∂ 2 B1
∂yj2

=

∂y1
3
∂
B =
∂y1 ∂ 2 yj

0 for j = 2, . . . , n

implies that B1 satisfies hypothesis of the claim. On the other hand, B2 = B(0, y2 , . . . , yn ),
and so B2 has less variables, but satisfies the same system of differential equations.
Homogeneity of B implies that B(y) = cy1 · · · yn .
Remark 17. The second proof is a modification of the proof shown to us by Bernd Kirchheim,
we express our gratitude to him.

2.1.3 Case k = n − 1. Young’s function.
Theorem 2.1.2. If B satisfies assumptions L1 − L5 and Byi yj = 0 in int(Rn
+ ) for all
i, j = 1, . . . , n then B(y) = M y1α1 · · · ynαn for some M > 0 and 0 < αj < 1, j = 1, . . . , n
In the end of the section we present F. Nazarov’s examples which show that if we remove
the condition Bxi xj = 0 in the Theorem 2.1.2 then the conclusion of the theorem does not
99

hold. It is also worth mentioning that in the classical case when n = 3 and k = 2 we obtain
that under the assumptions L1 − L5 and Byi yj = 0 there are only Young’s inequalities for
convolution.
Proof. Equality (2.1.1) is the same as
n

yj ajs By yj
j=1

Ca , aj
= 0,
Caj , aj

∀ = 1, . . . , n,

∀s = 1, . . . , k.

(2.1.4)

We introduce a vector function P (x) = (p1 (x), . . . , pn (x)), where x ∈ Rn , such that
P (ln y1 , . . . , ln yn ) = ∇B(y).
Then equality (2.1.4), the fact that Byi yj = Byj yi and homogeneity of B combined imply
the following

∇p , w s = 0,

∀ = 1, . . . , n,
j

e−xj pixj = e−xi pxi ,
div p = (k − 1)p ,

∀s = 1, . . . , k;

i, j = 1, . . . , n;

(2.1.5)
(2.1.6)

= 1, . . . , n.

(2.1.7)

where

ws=

a1s

Ca , a1
Ca , an
, . . . , ans
Ca1 , a1
Can , an

,

∀ = 1, . . . , n,

∀s = 1, . . . , k.

(2.1.8)

Now we show that the assumptions Byi yj = 0 imply that Cai , aj = 0 for all i, j = 1, . . . , n.
Indeed, suppose that Cai0 , aj0 = 0 for some i0 and j0 . Assumption L3 implies that

100

i0 = j0 . Since C is symmetric we get that Caj0 , ai0 = 0. Corollary 7 says now that
rank(A∗ CA•Hess B(y)) ≤ 1. So the determinant of any 2×2 submatrix of A∗ CA•Hess B =
Cai , aj Byi yj

ij

(2-minor) is zero. Consider 2 × 2 submatrix of A∗ CA • Hess B with the

following entries: (i0 , i0 ), (i0 , j0 ), (j0 , i0 ) and (j0 , j0 ). Since its determinant is zero and we
assumed Cai0 , aj0 = 0, we get that Cai0 , ai0 Caj0 , aj0 = 0. This contradicts to our
assumption L3.
Thus we obtain that for each fixed

the vectors w s , s = 1, . . . , n, span k = n −

1 dimensional subspace W . Therefore, equality (2.1.5) implies that ∇p (x) = λ (x)v
where λ (x) is a nonvanishing scalar valued function in int(Rn
+ ), v ⊥ W and none of the
components of v is zero.
The equality (2.1.7) implies that we can choose v so that v , 1 = k − 1 (here 1 =
(1, . . . , 1) ∈ Rn ) and so that λ (x) = p (x) for all

= 1, . . . , n.

Hence the equation ∇p (x) = p (x)v easily implies that p (x) = e v ,x p (0) for all .
The equalities (2.1.6) imply that

v = (q1 , . . . , q −1 , q − 1, q +1 , . . . , qn ),

∀ = 1, . . . .n.

where q = (q1 , . . . , qn ) ∈ Rn . It also follows that P (0) = kq for some number k = 0. Thus
q

q

we get that By = kq y11 · · · ynn /y and this proves Theorem 2.1.2.

101

2.1.3.1 Example of necessity of the assumption Byi yj = 0 in Theorem 2.1.2
Let n = 3, k = 2 and B(x1 , x2 , x3 ) = ϕ(x1 , x2 )x3 where ϕ ∈ C 2 (int R2+ ) ∩ C(R2+ ) is an
arbitrary concave function and homogeneous of degree 1. Let

√
 0 0 1/ 2 
A=
,
1 1
0







 2 0 
C=
.
0 1

Then




 1 1 0 




A∗ CA =  1 1 0  .




0 0 1
Since ϕ satisfies homogeneous Monge–Amp`ere equation we have A∗ CA•Hess B ≤ 0. Clearly
all the assumptions L1 − L5 are satisfied.

2.1.3.2 Theorem 2.1.2 does not hold in the case 1 < k < n − 1
It turns out that even if Byi yj = 0 and 1 < k < n − 1 then it is not necessarily true that a
function B which satisfies assumptions L1 − L5 has a form B = M y1α1 · · · ynαn . This means
that Theorem 2.1.2 cannot be improved. We give an example in a general case.
Assume that 1 < k < n − 1 and n > 3 (case n = 3 was already discussed above).
Take arbitrary nonnegative ϕ ∈ C 2 (int(R2+ )) ∩ C(R2+ )) so that ϕ is a concave function and
homogeneous of degree one. We choose ϕ so that it has nonzero second derivatives. We

102

consider the following function

α

n−2
B(y) = y1α1 · · · yn−2
· ϕ(yn−1 , yn ),

y ∈ Rn
+.

(2.1.9)

Let an−1 = an = (0, . . . , 0, 1) ∈ Rk and let a1 = (˜
a1 , 0), . . . , an−2 = (˜
an−2 , 0) ∈ Rk . We
choose vectors a
˜1 , . . . , a
˜n−2 ∈ Rk−1 in the following way. First of all they span Rk−1 . Intersection of the interior of the convex hull K (described in the Introduction and constructed
by the vectors a
˜1 , . . . , a
˜n−2 ) with the hyperplane {y1 + y2 + · · · + yn−2 = k − 1} is nonempty.
We choose a point (α1 , . . . , αn−2 ) from this intersection.
Then there exists (k − 1) × (k − 1) symmetric, positive semidefinite matrix C˜ such that
˜ ≤ 0 where A˜ = (˜
A˜∗ C˜ A˜ • Hess B
a1 , . . . , a
˜n−2 ) and
˜ 1 , . . . , yn−2 ) = y α1 · · · y αn−2 .
B(y
1
n−2
˜ aj , a
Moreover, we have C˜
˜j > 0 for j = 1, . . . , n − 2. The existence of such a matrix C˜
follows from the solution of the Euler–Lagrange equation for the right side of (2.0.18) (see
[25], Theorem 5.2), see also Subsection 2.1.4 below. It is clear that the function B satisfies
all properties L1 − L5 except one has to check the property L3. We choose C as follows



˜
 C 0 
C=
.
T
0
1
Function B from (2.1.9) satisfies L3 (and of course it can easily be made to satisfy all
other properties L1 − L5), but it is not a Young function.

103

2.1.4 Case of Young’s function
1/p1

In this subsection we consider classical case when B(y) = y1

1/pn

· · · yn

where 1 ≤ pj < ∞.

Assumptions 1 ≤ pj follows from the assumption L3 (which implies in particular that the
function B is separately concave) and the assumption pj < ∞ was made because otherwise
we have a function of less variables m < n. Note that we also must require that

1
pj

= k.

This function satisfies all assumptions of L1 − L5 except of L3. We try to understand for
which matrix A and numbers pj there is a matrix C mentioned in the assumption L3. The
answer on this question was obtained in [25] by using Euler–Lagrange equation.
We will obtain equation on the matrix C.
Note that Hess B = B · pi pj1yi yj

δij
pi yi2

−B·

where δij = 1 if i = j, and otherwise it

is zero. Therefore equality (2.1.1) becomes

A

yi
Cai , ai

A∗ CA •

1
pi pj y i y j

B·

−B·

δij
pi yi2

=0

After simplification we obtain

1
pi Cai , ai

A

Notice that the rank of A
matrix A p Ca1 ,a
i
i i

1
pi Cai ,ai

√

A∗ C = Ik×k

(2.1.10)

is k because the rank of A is k. Then k × k

A∗ is invertible by Binet–Cauchy formula. Then we can find C from

(2.1.10) by the following obvious formula

C=

A diag

1
pi Cai , ai

104

A∗

−1

(2.1.11)

if we can solve the following system of non-linear equations fefining Caj , aj , j = 1, . . . , n:

Caj , aj =

1
pi Cai , ai

A diag

A∗

−1

aj , aj .

(2.1.12)

aj , aj , j = 1, . . . , n,

(2.1.13)

Using the notations
s2j :=

1
, j = 1, . . . , n,
pj Caj , aj

we readily transfer (2.1.12) to

1
= s2j
pj

−1

A diag s2j A∗

which is precisely equation (3.12) of [25]. In [24], [25] it is proved that for { p1j }n
j=1 in the
interior of the convex set K from [24], [25] this system (2.1.13) has a solution. In particular,
C as in (2.1.11) does exist.
Notice also, that the Young’s functions found by Brascamp–Lieb [23] and corresponding
to the interior of the convex set K from [24], [25], do satisfy all properties L1 − L5. Only L3
is interesting because we need to show that there exists a certain matrix C. We just found a
certain C in (2.1.11) (when the system (2.1.12) has a solution). This matrix C will satisfy L3
1/p1

when B is the Young’s function B(y) = y1

1/pn

· · · yn

where 1 < pj < ∞,

In fact, A∗ CA • HessB(y) ≤ 0 for such a B is equivalent to

diag

1
y j pj

A∗ CA diag

1
y j pj

≤ diag

This is immediately equivalent to

A∗ CA ≤ diag{1/s2j }.
105

Caj , aj pj
yj2 p2j

.

n
j=1 1/pj

= k.

But denoting S = diag{sj } we make this inequlity (AS)∗ C(AS) ≤ In×n , which holds
because (AS)∗ C(AS) is an orthogonal projection onto the span of the columns of S(A∗ ).
So, we repeat, that the Young’s functions found by Brascamp–Lieb [23] and corresponding
to the interior of the convex set K from [24], [25], do satisfy all properties L1 − L5. But it is
more interesting that, as we have shown above, in certain situations all functions satisfying
L1 − L5 must be of the form of a Young function found by Brascamp and Lieb.

106

Chapter 3
Harmonic analysis, PDE and
differential geometry
3.0.5 Short review of some harmonic analysis problems
3.0.5.1 John–Nirenberg inequality
For a measurable function f , and measurable set K we set

f

def
K

=

1
f.
|K| K

We say that the measurable function f belongs to BM O(Rn ) if

def

f BM O = sup |f − f
n
Q⊂R

2 1/2

Q|

Q

< ∞,

where the supremum is take over all hypercubes Q. Theorem about equivalence of BMO
norms states that
Theorem 3.0.3. For any p ∈ (0, ∞) there exist c1 , c2 > 0 such that:
c1 f BM O ≤ sup |f − f
n
Q⊂R

107

p 1/p

Q|

Q

≤ c2 f BM O .

The celebrated John–Nirenberg inequality describes growth of the distributions of the
function from BM O:
Theorem 3.0.4. There exist constants c1 , c2 > 0 such that for any f ∈ BM O we have
1
|Q|

x : |f (x) − f

Q| > λ

−c2 f λ
BM O .

≤ c1 e

John–Nirenberg inequality in integral form states that:
Theorem 3.0.5. There exists ε0 > 0 and a positive function C(ε), 0 < ε < ε0 such that

eϕ

I

≤ C(ε)e ϕ

I

ϕ ∈ BM O(I) :

for all

ϕ BM O ≤ ε.

Interesting question is to find the best possible C(ε) and ε0 .
Closely related classes to BMO are Ap classes, reverse H¨older classes and Gehring classes.

3.0.5.2 Uniform convexity
def

Let I be an interval of the real line. For an integrable function f over I, we set f p =
1/p

|f |p I . We recall the definition of uniform convexity of a normed space (X, · ) (see [29]).
Definition 13. (Clarkson ’36 ) X is uniformly convex if ∀ε > 0, ∃δ > 0 s.t. if x = y =
1 and x − y ≥ ε, then

x+y
2

≤ 1 − δ.

Modulus of convexity of the normed space X is defined as follows:

δX (ε) = inf

1−

f +g
2

:

f = 1, g = 1, f − g ≥ ε .

Remark 18. (X, · ) space is uniformly convex iff δX (ε) > 0.
108

O. Hanner (see [28]) gave an elegant proof of finding the constant δLp (ε) in Lp ([0, 1])
space for p ∈ (1, ∞) in 1955. He proved two necessary inequalities (further called Hanner’s
inequalities) in order to obtain constant δLp (ε). Namely,

p

p

f + g p + f − g p ≥ ( f p + g p )p + | f p − g p |p ,

p ∈ [1, 2],

(3.0.1)

and the inequality (3.0.1) is reversed if p ≥ 2. Hanner mentions in his note [28] that his
proof is a reconstruction of some Beurling’s ideas given at a seminar in Upsala in 1945.
In [30] non-commutative case of Hanner’s inequalities was investigated. Namely, Hanner’s
inequality holds for p ∈ [1, 3/4] ∪ [4, ∞), and the case p ∈ (3/4, 4) (where p = 2) was left
open.

3.0.5.3 Brunn–Minkowski and isoperimetric inequalities
Let A and B be nonempty compact subsets of Rn .
Theorem 3.0.6. The following sharp Brunn–Minkowski inequality holds

|A + B|1/n ≥ |A|1/n + |B|1/n ,
where n ≥ 1 and |A| denotes Lebesgue measure of the set A.
The Brunn-Minkowski inequality is a consequence of its multiplicative version:
Theorem 3.0.7. Let λ ∈ (0, 1). Then for any compact measurable sets U, V ⊂ Rn we have
|λU + (1 − λ)V | ≥ |U |λ |V |1−λ .

109

(3.0.2)

Indeed, if one sets U λ = A and (1 − λ)V = B then inequality (3.0.2) takes the form
|A|λ |B|1−λ
|A + B| ≥ λ
.
λ (1 − λ)1−λ

(3.0.3)

By maximizing the right hand side of (3.0.3) over λ ∈ (0, 1) we obtain the Brunn–Minkowski
inequality.
Brunn–Minkowski inequality implies the classical isoperimetric inequality:
Theorem 3.0.8. Among all simple closed surfaces with given surface area, the sphere encloses a region of maximal volume. In other words

1

1

|∂A| ≥ n|A|1− n |B(0, 1)| n .
Where |∂A| means surface area of the boundary of the body A. |A| denotes volume of the
body and B(0, 1) denotes the ball of radius 1 at center 0.
Indeed, let us sketch the proof: Since |A + B(0, ε)| = |A| + ε|∂A| + O(ε2 ), we have
1
1
(|A|1/n + |B(0, ε)|1/n )n − |A|
|A + B(0, ε)| − |A|
≥ lim
= n|A|1− n |B(0, 1)| n .
|∂A| = lim
ε
ε
ε→0
ε→0

3.0.5.4 Sobolev inequality
It is known that the classical isoperimetric inequality is equivalent to its functional version,
to Sobolev inequality on Rn with optimal constant

Rn

|∇f | ≥

1
n|B(0, 1)| n

110

Rn

n
|f | n−1

1− n1

.

(3.0.4)

Indeed, testing (3.0.4) over characteristic functions f (x) = ✶A (x) we obtain implication in
one direction. Opposite direction follows from Coarea formula: assume f ≥ 0 is sufficiently
nice compactly supported function. Then by coarea formula we have

Rn

∞

|∇f |dx =

0

∞

1

|{x : f (x) = t}|dt ≥ n|B(0, 1)| n

0

1

|{x : f (x) ≥ t}|1− n dt.

It is left to show that
∞
0

|{x : f (x) ≥

n−1
t}| n dt

n
n−1

≥

∞
1
n
|{x : f (x) ≥ t}|t n−1 dt
n−1 0

This follows from the following observation
∞

F

ϕ

∞

=

0

0

d
F
dt

∞

t

ϕ dt =
0

0

∞

t

F
0

ϕ ϕdt ≥

F (tϕ(t))ϕ(t)dt,

0

n

where ϕ is decreasing and F is increasing (F (t) = t n−1 , ϕ(t) = |{x : f (x) ≥ t}|

n−1
n ).

So

the claim follows.

3.0.5.5 Prekopa–Leindler inequality
Multiplicative Brunn–Minkowski inequality follows from its functional version, so called
Prekopa–Leindler inequality.
Theorem 3.0.9. Let h, f, g be positive measurable functions and λ ∈ (0, 1). If
h(λx + (1 − λ)y) ≥ f (x)λ g(y)1−λ

111

(3.0.5)

Then
λ
Rn

h≥

Rn

1−λ

f

Rn

g

.

If one takes h(x) = ✶λU +(1−λ)V (x), f (x) = ✶U (x) and g(x) = ✶V (x) then clearly the
assumption (3.0.5) is satisfied and one obtains multiplicative version of Brunn–Minkowski inequality. It is not quite clear what will be the functional analog of original Brunn–Minkowski
inequality.
Straightforward generalization of Prekopa–Leindler inequality takes the following form:
Theorem 3.0.10. Let fj : Rn → R+ be integrable functions, and let

m
j=1 λj

= 1, 0 <

λj < 1. If




m

h
j=1

λ j xj  ≥

m

f (xj )λj ,
j=1

then
m
Rn

h≥

λj

j=1

Rn

fj

.

The above inequality can be treated as reverse to H¨older’s inequality:

Rn

sup


m


j=1

f (xj

)λj

:

xj λj = z





m

dz ≥

j=1

m

λj
Rn

fj

≥

n
j=1 R

fj (xj )λj .

where integral in the left hand side is understood as upper Lebesgue integral.
One of the other applications of Prekopa–Leindler inequality in probability is that:

112

Corollary 8. If F (x, y) : Rn × Rm → R+ is log-concave distribution i.e.,
F (λu + (1 − λ)v) ≥ F (u)1−λ F (v)λ

for all

u, v ∈ Rn+m ,

then H(x) = Rm F (x, y)dy is log-concave distribution.
The corollary immediately follows from application of Prekopa–Leindler inequality to the
functions F (x, λy1 + (1 − λ)y2 ), F (x, y1 ) and F (x, y2 ).

3.0.5.6 Borell–Brascamp–Lieb inequality
We also mention Borell–Brascamp–Lieb inequality since it generalizes Prekopa–Leindler inequality
Theorem 3.0.11. Let h, f, g be nonnegative functions, 0 < λ < 1 and − n1 ≤ p ≤ ∞.
Suppose

h(λx + (1 − λ)y) ≥ Mp (f (x), g(y), λ),

where

def

def

Mp (a, b, λ) = (λap + (1 − λ)bp )1/p ,

M0 = (a, b, λ) = aλ b1−λ .

Then

Rn

h≥M

p
np+1

Rn

113

f,

Rn

g, λ .

3.0.5.7 Ehrhard’s inequality
The condition of Preko–Leindler type appears in Ehrhards inequality:
−|x|2 /2
x
Theorem 3.0.12. Let dγ(x) = e n/2 dx be the Gaussian measure. And let Φ(x) = −∞
dγ.

(2π)

Then for any measurable compact sets A, B ⊂ Rn and any numbers λ, µ ≥ 0, such that
λ + µ ≥ 1 and |λ − µ| ≤ 1 we have
Φ−1 (|λA + µB|γ ) ≥ λΦ−1 (|A|γ ) + µΦ−1 (|B|γ ),
where |A|γ denotes Gaussian measure of A i.e., |A|γ = A dγ.
The inequality initially was stated for convex sets A and B. Later it was improved in the
sense that only one of them has to be convex and it was conjectured that the inequality is
true in general for any measurable sets, and the conjecture was proved recently. Ehrhrad’s
inequality is consequence of its functional version:
Theorem 3.0.13. Let h, f, g : Rn → [0, 1] be functions such that
Φ−1 (h(λx + µy)) ≥ λΦ−1 (f (x)) + µΦ−1 (g(y)),

for all x, y ∈ Rn ,

where λ, µ ≥ 0, λ + µ ≥ 1 and |λ − µ| ≤ 1 then
Φ−1

Rn

hdγ

≥ λΦ−1

Rn

114

f dγ

+ µΦ−1

Rn

gdγ .

3.0.5.8 Borell’s Gaussian noise “stability”
2

−|x| /2
x
Let γn = e n/2 be a standard Gaussian measure on Rn and let Φ = −∞
dγ1 . Borell’s
(2π)

Gaussian noise “stability” states that
Theorem 3.0.14. If A, B are measurable subsets of Rn . Then if X = (X1 , . . . , Xn ), Y =
(Y1 , . . . , Yn ) are independent Gaussian standard random variables, and p ∈ (0, 1) then
P(X ∈ A, pX +

1 − p2 Y ∈ B) ≤ P(X1 ≤ Φ−1 (γn (A)), pX1 +

1 − p2 Y1 ≤ Φ−1 (γn (B))).

The functional version of the above inequality can be stated as follows:
Theorem 3.0.15. Let p ∈ (0, 1), f, g : Rn → (0, 1) and let
B(u, v) = P(X1 ≤ Φ−1 (u), pX1 +

1 − p2 Y2 ≤ Φ−1 (v)).

Then

R2n

B f (x), g(px +

1 − p2 y) dγdγ ≤ B

Rn

f dγ,

Rn

gdγ .

3.0.5.9 Hypercontractivity
Let

Pt f (x) =

Rn

f (e−t x +

be Ornstein–Uhlenbeck semigroup where t ≥ 0.
Uhlenbeck semigroup means that
115

1 − e−2t y)dγ(y)
The hypercontractivity for Ornstein–

q−1
≥ e−2t . Then
Theorem 3.0.16. Let p, q > 1 be such that p−1

Pt f Lp (dγ) ≤ f Lq (dγ) .

3.0.5.10 Logarithmic Sobolev inequalities
2

−|x| /2
Let dγ(x) = e k/2 dx. Logarithmic Sobolev inequality can be stated as follows

(2π)

Theorem 3.0.17. For any positive function f on Rk we have

Rk

f 2 ln f 2 dγ −

Rk

f 2 dγ ln

Rk

f 2 dγ

≤2

Rk

|∇f |2 dγ.

(3.0.6)

3.0.5.11 Beckner–Sobolev inequality
W. Beckner proved tho following Sobolev inequality for Gaussian measure:
Theorem 3.0.18. Let 1 ≤ p ≤ 2. Then
f 2L2 (dγ) − f 2Lp (dγ) ≤ (2 − p) ∇f 2L2 (dγ) .

3.0.5.12 L´evy–Gromov’s isoperimetric inequlity
Let M be a compact connected Riemannian manifold of dimension n ≥ 2, and of Ricci
curvature bounded below by R > 0. Let µ be normalized Riemannian measure on M . Let
σ(r) be normalized volume of a geodesic ball of radius r ≥ 0 on the n-sphere with curvature
R > 0. L´evy–Gromov’s isoperimetric inequality states that:

116

Theorem 3.0.19. For every set A in M with smooth boundary ∂A we have

σ (σ −1 (µ(A))) ≤ µs (∂A),
where µs (∂A) stands for the surface measure of the boundary ∂A.

3.0.5.13 Bobkov’s inequality
In particular since the spherical measures converge to Gaussian distributions one expects to
obtain L´evy–Gromov’s isoperimetric inequlity in infinitely dimensional setting
x
Theorem 3.0.20. Let Φ(x) −∞
dγ and ϕ(x) = Φ (x). Then for every Borel set A ∈ Rn

with smooth boundary,

ϕ(Φ−1 (γ(A))) ≤ γs (∂A).

(3.0.7)

The celebrated functional version of the above inequality was obtained and proved by
Bobkov.
Theorem 3.0.21. Let U (x) = ϕ(Φ−1 (x)). Then for any differentiable f : Rn → (0, 1)
U

Rn

f dγ

≤

Rn

U 2 (f ) + |∇f |2 dγ.

(3.0.8)

Testing (3.0.8) on the characteristic functions f = ✶A and noticing that U (0) = U (1) = 0
gives the desired result (3.0.7).
It is worth mentioning Brascamp–Lieb inequality. We refer the reader to the Chapter 2.

117

3.0.6 Relation to PDEs
3.0.6.1 Prekopa–Leindler, Ehrhard’s inequality and its underlying PDE
One can see that in general inequalities of these types (Prekopa–Leindler, Ehrhard) can
be formulated as follows. Let B(x1 , . . . , xm ) be a smooth real valued function defined on
Ω ⊂ Rm . And let Aj : Rn → Rnj be matrices j = 1, . . . , m. Let (u1 , . . . , um ) : Rn1 × . . . ×
Rnm → Ω be smooth functions.
Question. Under what conditions on B it is true that whenever

B(u1 (A1 x), . . . , um (Am x)) ≥ 0,

for all

x ∈ Rn

(3.0.9)

we have that

B

Rn 1

u1 dγn1 , . . . ,

Rn m

unm dγnm

≥ 0.

(3.0.10)

2

−|x| /2
where dγk (x) = e k/2 dx denotes k dimensional Gaussian measure.
(2π)

Note that for the function u : Rk → R its heat extension can be written as follows
Pt u(x) = u(x, t) =

Rk

u(x +

√

2t y)dγk (y) t ≥ 0.

Notice that inequality (3.0.9) must remain true under the shifts and dilation of the variable

118

x. In particular this implies that (3.0.10) can be written as follows

def

C(x, t) = B(u1 (A1 x, t), . . . , um (Am x, t)) =
B

Rn1

u1 (A1 x +

√

2t y1 )dγn1 (y), . . . ,

Rnm

unm (Anm x +

√

2t ym )dγnm (y)

≥ 0.

Remark 19. The same is true for Ornstein–Uhlenbeck extensions.
Now the question is under what conditions on C(x, t) the inequality C(x, 0) ≥ 0 implies
that C(x, t) ≥ 0 for all t > 0. Further we always assume that
lim inf

|x|→∞

≥ 0 for all T > 0.

inf C(x, t)

T ≥t≥0

There is a simple way to check this condition: maximum principle for elliptic operator. If the
action of an elliptic operator is nonpositive then the infimum is attained on the boundary.
In other words, if there exists positive semi-definite matrix {aij (x, t)}n
i,j=1 such that


n

aij (x, t)


i,j=1

∂2
∂xi ∂xj



n

+

bj (x, t)
j=1

∂
∂
−  C(x, t) ≤ 0
∂xj
∂t

(3.0.11)

for some vector b(x, t) = (b1 (x, t), . . . , bn (x, t)) then infimum is attained on the boundary
t = 0. Indeed, the condition (3.0.11) implies that “whenever we are on the hill we go up”.
∂ C(x, t) ≥ 0.
In other words if whenever Hessx C(x, t) ≥ 0 and ∇x C(x, t) = 0 then ∂t

Now the last conclusion gives the desired result. Indeed (see also [32]), take any ε > 0
and set Cε (x, t) = C(x, t) + εt. If Cε (x, t) < 0 at some point then since Hessx Cε (x, t) =
Hessx C(x, t) ≥ 0 and ∇x Cε (x, t) = ∇x C(x, t) = 0 at that point then we must have
∂
∂t Cε (x, t)

∂ C(x, t) = −ε < 0, but this contradicts to the fact
= 0. This means that ∂t

119

that whenever we are o the hill we must go up.
Definition 14. We say that the function C(x, t) satisfies ellipticity if (3.0.11) holds for
n
some nonnegative matrix {aij (x, t)}n
i,j=1 and the vector {bj (x, t)}j=1 . We say that the func-

tion C(x, t) satisfies hill property if whenever Hessx C(x, t) ≥ 0 and ∇x C(x, t) = 0 then
∂
∂t C(x, t)

≥ 0.

Remark 20. As we already mentioned ellipticity implies the hill property, however the converse is not true. Ehrhard’s inequality is an example when the desired elliptic operator does
not exists, however the hill property holds.
In the following particular case we will describe all functions B for which C(x, t) satisfies
hill property and thus we will answer on our question. We consider the following case

B(u1 (a1 x + a2 y), u2 (x), u3 (y)) ≥ 0,

(3.0.12)

where a1 , a2 are real numbers and x, y ∈ Rn . If B is sufficiently nice then one can rewrite
the pointwise inequality (3.0.12) as follows

u1 (a1 x + a2 y) ≥ H(u2 (x), u3 (y))
fore some function H.
Theorem 3.0.22. Let Ω be a rectangular domain in R2 . Let a1 , a2 ∈ R be such that
|a1 | + |a2 | ≥ 1 and 1 ≥ ||a1 | − |a2 ||. Let the function H(x, y) : Ω → R be such that
H1 , H2 = 0 and

H1 H2 H12 (1 − a21 − a22 ) + a21 H22 H11 + a22 H12 H22 ≥ 0.
120

(3.0.13)

If the real-valued functions u1 , u2 , u3 are such that (u2 , u3 ) : Rn → Ω and
u1 (a1 x + a2 y) ≥ H(u2 (x), u3 (y)) for all x, y ∈ Rn ,

then

Rn

u1 dγ ≥ H

Rn

u2 dγ,

Rn

u3 dγ .

Proof. It is enough to prove the theorem for the case n = 1. Indeed, for arbitrary n proof
goes by induction. Consider the functions

h(z) = u1 (z, a1 x2 + a2 y2 , . . . , a1 xn + a2 yn )
f (x1 ) = u2 (x1 , x2 , . . . xn )
g(y1 ) = u3 (y1 , . . . , yn )

Then the theorem implies that

R

h(z)dγ(z) ≥ H

u2 (x1 )dγ(x1 ),

u3 (y1 )dγ(y1 )
R

R

After that we apply the theorem to the new functions

˜
h(z)
=

u1 (x, z, a1 x3 + a2 y3 , . . . , a1 xn + a2 yn )dγ(x)
R

f˜(x2 ) =

u2 (x, x2 , x3 , . . . xn )dγ(x)
R

g˜(y2 ) =

u3 (x, y2 . . . , yn )dγ(x)
R

121

and by iterating this process we see that it is sufficient to prove the theorem for the case
n = 1.
We will check the hill property. We would like to show that the following quantity is
nonnegative

∂
(u (a x + a2 y, t) − H(u2 (x, t), u3 (y, t))) = u1 − H1 u2 − H2 u3 ,
∂t 1 1

(3.0.14)

under the assumptions that




u1 a21

− H11 (u2

)2

− H1 u2

u1 a1 a2 − H12 u2 u3

u1 a1 a2 − H12 u2 u3
u1 a22

− H22 (u3

)2

− H2 u3



 ≥ 0,

(3.0.15)

and

a1 u1 − H1 u2 = 0 and a2 u1 − H2 u3 = 0.

(3.0.16)

Expressing u2 , u3 from (3.0.16) and substituting into (3.0.15) gives





u1 a21

−

H11 2
a (u )2
H12 1 1

− H1 u2

u1 a1 a2 − HH12
a a (u )2
1 H2 1 2 1

u1 a1 a2 −

H12
2
H1 H2 a1 a2 (u1 )

u1 a22 − H222 a22 (u1 )2 − H2 u3
H2



 ≥ 0.


Further we assume that H1 , H2 = 0. We will treat derivatives u1 , u1 , u2 , u3 as independent
variables. We multiply the above matrix by (u2 )−2 (and (3.0.14) as well) and we introduce
u

H u

H u

the new variables (u 1)2 = x, (u1 )22 = ya21 and (u2 )23 = za22 (further we assume that a1 , a2 = 0).
1
1
1
Then positive definiteness of the above matrix is the same as positive definiteness of the
following matrix (after conjugating by 2×2 diagonal matrix with the elements on the diagonal
122

a1 , a2 )




H11
H12
 x − H 2 − y x − H1 H2 
 ≥ 0.

1


H22
x
−
−
z
x − HH12
2
H
1

(3.0.17)

H2

2

Desired expression (3.0.14) takes the following form

x − ya21 − za22 = (x, y, z) · (1, −a21 , −a22 ).

(3.0.18)

We denote H112 = p, H222 = q, and HH12
= r. Then Condition (3.0.17) is equivalent to the
1 H2
H1
H2
following two inequalities

(x, y, z) · (2, −1, −1) ≥ p + q;
yz − xy − xz + pz + qy − x(p + q − 2r) + pq − r2 ≥ 0.
We shift coordinates x → x + r, y → y + r − p, z → z + r − q and we obtain that the
above conditions take the following form:

(x, y, z) · (2, −1, −1) ≥ 0;

(3.0.19)

yz − xy − xz ≥ 0;

(3.0.20)

and the desired inequality becomes

(x, y, z) · (1, −a21 , −a22 ) + r − a21 (r − p) − a22 (r − q) ≥ 0.

123

(3.0.21)

Testing (3.0.21) for x = 0, y = 0, z = 0 we obtain the necessary condition that

r − a21 (r − p) − a22 (r − q) ≥ 0.
By stretching the point (x, y, z) → λ(x, y, z) (where λ > 0) we see that conditions (3.0.19,
3.0.20) remain true and we can get rid off the term r − a21 (r − p) − a22 (r − q). So it remains to
understand when the quantity (x, y, z) · (1, −a21 , −a22 ) is nonnegative under the assumptions
(3.0.19, 3.0.20). Since the cone (3.0.19, 3.0.20) must lie below the hyperplane (x, y, z) ·
(1, −a21 , −a22 ) = 0 we obtain that we must have
|a1 | + |a2 | ≥ 1;
1 ≥ ||a1 | − |a2 ||.

Corollary 9. If a function H satisfies conditions of Theorem 3.0.22 and in addition it is
1-homogeneous then the conclusion of the theorem holds for Lebesgue measure, i.e.,

Rn

u1 dx ≥ H

Rn

u2 dx,

Rn

u3 dx

Corollary 10. If H is convex then conclusion of Theorem 3.0.22 holds.
Corollary 11. If H(x, y) = xa1 y a2 then

H1 H2 H12 (1 − a21 − a22 ) + a21 H22 H11 + a22 H12 H22 = x3a1 −2 y 3a2 −2 a21 a22 (1 − a1 − a2 )
so the hill property holds if a1 + a2 ≥ 1 and ||a1 | − |a2 || ≤ 1.
124

x
Corollary 12. Set Φ = −∞
ϕ(s)ds, where ϕinC 1 is positive and integrable function at

negative infinity, and let Φ−1 (x) be inverse function of Φ. Take

H(x, y) = Φ(a1 Φ−1 (x) + a2 Φ−1 (y)) where (x, y) ∈ (0, 1)2 .
Denote Λ = a1 Φ−1 (x) + a2 Φ−1 (y). Then direct computation shows that

H1 =

a1 ϕ(Λ)
;
ϕ(Φ−1 (x))

a2 ϕ(Λ)
;
ϕ(Φ−1 (y))
a21 ϕ (Λ)
a1 ϕ(Λ)ϕ (Φ−1 (x))
−
;
H11 =
ϕ(Φ−1 (x))2
ϕ(Φ−1 (x))3
H2 =

H12 =

a1 a2 ϕ (Λ)
;
−1
ϕ(Φ (x))ϕ(Φ−1 (y))

H22 =

a22 ϕ (Λ)
a2 ϕ(Λ)ϕ (Φ−1 (y))
−
.
ϕ(Φ−1 (y))2
ϕ(Φ−1 (y))3

So we obtain

H1 H2 H12 (1 − a21 − a22 ) + a21 H22 H11 + a22 H12 H22 =

a21 a22 ϕ(Λ)3
×
ϕ(Φ−1 (x))2 ϕ(Φ−1 (y))2

ϕ (a1 Φ−1 (x)) + a2 Φ−1 (y))
ϕ (Φ−1 (x))
ϕ (Φ−1 (y))
− a1
− a2
ϕ(a1 Φ−1 (x)) + a2 Φ−1 (y))
ϕ(Φ−1 (x))
ϕ(Φ−1 (y))

.

Denote Φ−1 (x) = u, Φ−1 (y) = v. Then we see that if the logarithmic derivative of the density
ϕ satisfies concavity condition

f (a1 u + a2 v) ≥ a1 f (u) + a2 f (v),
ϕ (x)

where f (x) = ϕ(x) then the hill property holds. In particular this implies Ehrhard’s inequality
125

2
if ϕ = e−|x| /2 and it implies Prekopa–Leindler inequality if ϕ = ex .

3.0.6.2 log-Sobolev inequality, Beckner–Sobolev inequality, Bobkov’s inequality and its underliyng PDE
The inequalities mentioned in the title are related to some PDEs which we are going to
´
describe shortly. Firstly we will need some preliminaries from Bakry–Emery
“Gamma calculus”. Complete description of the required material the reader can find in [48].
Mostly we will be working with Ornstein–Uhlenbeck semigroup Pt on the Euclidean
space whose generator is L = ∆ − x · ∇, however the most part of the results (thanks to the
Bochner–Lichnerowicz formula) can be pushed forward to the diffusion semigroups Pt on the
some weighted Riemannian manifolds (M, g) with uniformly bounded below Ricci curvature.
We remind that the differential operator L is a diffusion operator if for every C ∞ function
Ψ on Rk and every finite family F = (f1 , . . . , fk ) from some suitable algebra A we have
LΨ(F ) =
j

∂Ψ
Lf +
∂fj j

i,j

∂ 2Ψ
Γ(fi , fj ),
∂fi ∂fj

where the so-called carr´e du champ operator Γ is defined as follows

2Γ(f, g) = L(f g) − Lf · g − f · Lg.
For example, if

L=

aij (x)
i,j

∂2
+
∂xi ∂xj

126

bj (x)
j

∂
,
∂xj

where aij (x) is symmetric matrix then

Γ(f, g) =

aij (x)
i,j

∂f ∂g
,
∂xi ∂xj

for f, g : Rn → R. One can define iterated carr´e du champ operator
2Γ2 (f, g) = LΓ(f, g) − Γ(f, Lg) − Γ(g, Lf ).
Sometimes we will write just Γ(f, f ) = Γ(f ) and Γ2 (f, f ) = Γ2 (f ). It is important to notice
that for the Ornstein–Uhlenbeck generator L = ∆ − x · ∇ we have
Γ(ui , uj ) = ∇ui ∇uj ;
Γ2 (uj ) = |∇uj |2 + |Hess uj |2HS ;
Γ(ui , Γuj ) = 2 Hess uj ∇uj , ∇ui ;
Γ(Γui , Γuj ) = 4(Hess ui ∇ui )T Hess uj ∇uj .
where |Hess uj |HS denotes Hilbert–Schmidt norm of the matrix Hess uj . For the C 2 function
∇y M
we denote 2n × 2n diagonal
y
2n×2n
My
entries yi i and the rest is zero. For the vector

M : R2n → R, say M (x1 , . . . , xn , y1 , . . . , yn ) by
matrix so that in the first n × n block it has

v we write v ≤ 0 if each component of v is nonpositive.
Theorem 3.0.23. Let Ω be a convex subset of Rn . Let (u1 , . . . , un ) : Rk → Ω be sufficiently
nice. Let M : Ω × Rn
+ → R, say M (x1 , . . . , xn , y1 , . . . , yn ) where yj ≥ 0, satisfy
Hess M +

∇y M
≤ 0,
y
2n×2n
127

and

∇y M ≤ 0,

(3.0.22)

then

Rk

M (u1 , . . . , un , |∇u1 |, . . . , |∇un |)dγ ≤ M

Rk

u1 dγ, . . . ,

Rk

un dγ, 0, . . . , 0 . (3.0.23)

If the inequalities (3.0.22) are reversed then we have reversed inequality in (3.0.23). In the
case n = 1 we do not need the condition ∇y M ≤ 0 (or ∇y M ≥ 0).
Proof. Let uj : Rk → R for all j = 1, . . . , n. Take
B(x1 , . . . , xn , y12 , . . . , yn2 ) = M (x1 , . . . , xn , y1 , . . . , yn ).

Then we are going to show that Pt B − B(Pt ) ≤ 0 for all t ≥ 0 where
Pt B = Pt B(u1 , . . . , un , Γu1 , . . . , Γun )

and B(Pt ) = B(Pt u1 , . . . , Pt un , ΓPt u1 , . . . , ΓPt un ) then by sending t → ∞ and noticing that
ΓPt u = e−2t Pt Γu we obtain the desired result thanks to the ergodicity of the semigroup Pt
i.e., Pt u →

udγ.

Remark 21. The condition Pt B − B(Pt ) ≤ 0 can be verified by showing that the function
C(x, t) = Pt B − B(Pt ) satisfies ellipticity (or even hill property see Definition 14) i.e.,
∂ )(B(P ) − P B) ≤ 0 for some elliptic operator L. We will check its ellipticity by
(L − ∂t
t
t

choosing L to be Ornstein–Uhlenbeck generator. Note that
t
∂
d
Ps B(Pt−s )ds =
Ps L −
∂t
0
0 ds
t

Pt B − B(Pt ) =

B(Pt−s ).

Therefore the reader can notice that since Ps is positive operator, the next computations are
128

the same as checking ellipticity for the function B(Pt ) − Pt B where L is Orsntein–Uhlenbeck
generator.

Pt B(u1 , . . . , un , Γu1 , . . . , Γun ) − B(Pt u1 , . . . , Pt un , ΓPt u1 , . . . , ΓPt un ) =
t

d
Ps B(Pt−s u1 , . . . , Pt−s un , ΓPt−s u1 , . . . , ΓPt−s un ) =
ds
0 

n
n
t
∂B
∂B
LPt−s uj −
2Γ(LPt−s uj , Pt−s uj ) =
Ps LB −
∂uj
∂vj
0
j=1
j=1

n
n
n
t
∂B
∂B
∂ 2B
Ps 
LPt−s uj +
LΓPt−s uj +
Γ(Pt−s ui , Pt−s uj )+
∂uj
∂vj
∂ui uj
0

=

j=1

n
i,j=1

j=1

∂ 2B
Γ(ΓPt−s ui , ΓPt−s uj ) + 2
∂vi vj

n

n

i,j=1

n

∂ 2B
Γ(Pt−s ui , ΓPt−s uj )
∂ui ∂vj
i,j=1


∂B
∂B
−
LPt−s uj −
2Γ(LPt−s uj , Pt−s uj ) =
∂uj
∂vj
j=1
j=1

n
n
t
∂ 2B
∂B 2

Ps
Γ (Pt−s uj ) +
Γ(Pt−s ui , Pt−s uj )+
∂vj
∂ui uj
0
j=1

n
i,j=1

∂ 2B
∂vi vj

i,j=1

n

Γ(ΓPt−s ui , ΓPt−s uj ) + 2
i,j=1

129

∂ 2B
∂ui ∂vj


Γ(Pt−s ui , ΓPt−s uj ) .

Now notice that
n
j=1
n

∂B 2
Γ (Pt−s uj ) +
∂vj

i,j=1
n

n
i,j=1

∂ 2B
Γ(Pt−s ui , Pt−s uj )+
∂ui uj

∂ 2B
Γ(ΓPt−s ui , ΓPt−s uj ) + 2
∂vi vj

n
i,j=1

∂ 2B
Γ(Pt−s ui , ΓPt−s uj ) =
∂ui ∂vj

∂B
(2|∇Pt−s uj |2 + 2|Hess Pt−s uj |2HS ) +
∂vj

j=1
n

4
i,j=1
n

4
i,j=1

n
i,j=1

∂ 2B
∇Pt−s ui ∇Pt−s uj +
∂ui ∂uj

∂ 2B
(Hess Pt−s ui ∇Pt−s ui )T Hess Pt−s uj ∇Pt−s uj +
∂vi ∂vj
∂ 2B
Hess Pt−s uj ∇Pt−s uj , ∇Pt−s ui .
∂ui ∂vj

We will check that the last expression is nonpositive. We need the following technical lemma
Lemma 22. For the given x the image of Ax as A runs over all matrices such that A HS =
r is the the set B(0, x r).
Proof. If x = 0 there is nothing to proof. Assume x = 0. It is clear that {Ax : A HS =
r} ⊂ B(0, x r) because Ax ≤ A x ≤ A HS x = r x . Let y ∈ B(0, x r). Take
T

the matrix A+ = yx
. Clearly A+ x = y and
x 2
1
y
A+ HS = (T r[A+ (A+ )T ])1/2 =
(T r[yxT xy T ])1/2 =
≤ r.
2
x
x
T
˜ = 0. Take the matrix f (λ) = A+ +
Consider the matrix A˜ = I − xx
. Then Ax
x 2

˜ Clearly f (λ)x = y for all λ ∈ R and f (λ) HS = (T r[A+ (A+ )T ] + 2λT r[A+ A˜∗ ] +
λA.
λ2 T r[A˜A˜∗ ])1/2 . And f (0) HS ≤ r, and limλ→∞ f (λ) HS → ∞. Hence there exists λ∗
such that f (λ∗ ) HS = r.

130

The above lemma implies that we should study the sign of the expression
n

∂B
(2|∇Pt−s uj |2 + 2|Hess Pt−s uj |2HS ) +
∂vj

j=1
n

4
i,j=1

∂ 2B
z , z |∇Pt−s ui ||∇Pt−s uj | + 4
∂vi ∂vj i j

n
i,j=1
n
i,j=1

∂ 2B
∇Pt−s ui ∇Pt−s uj +
∂ui ∂uj
∂ 2B
|∇Pt−s uj |. zj , ∇ui
∂ui ∂vj

for any vectors zi so that they satisfy the condition zi ≤ |Hess ui |HS . Going back to the
function B(x1 , . . . , xn , y12 , . . . , yn2 ) = M (x1 , . . . , xn , y1 , . . . , yn ) we obtain

n
j=1
n

∂M
|∇Pt−s uj | +
∂yj

i,j=1
n

n

n
∂M |Hess Pt−s uj |2HS − zj 2
∂ 2M
∇Pt−s ui ∇Pt−s uj +
+
∂yj
|∇Pt−s uj |
∂xi ∂xj
j=1
i,j=1

∂ 2M
z ,z + 2
∂yi ∂yj i j

∂M
|∇Pt−s uj | +
∂yj

j=1
n

2
i,j=1

∂ 2M
∂xi ∂yj

n

∂ 2M
z , ∇Pt−s ui ≤
∂xi ∂yj j

i,j=1
n
∂ 2M

i,j=1

∂xi ∂xj

n

∇Pt−s ui ∇Pt−s uj +

i,j=1

∂ 2M
z ,z +
∂yi ∂yj i j

zj , ∇Pt−s ui .

Let w1 = ∇Pt−s u1 , . . . , wn = ∇Pt−s un , wn+1 = z1 , . . . , w2n = zn be columns. And let W
be the corresponding matrix constructed by these columns. Let wj be the columns of the
transpose matrix W . Then the above expression can be written as
T



wj 


j



Hess M +



∇y M

wj  .
y
2n×2n
j

The last expression is nonpositive and since the operator Pt is positive semidefinite we obtain
the desired result.

131

Corollary 13. Take

M (x, y) = x ln x −

1 y2
.
2 x

Then clearly M satisfies conditions of Theorem 3.0.23, i.e., Ω = R+ ,


1
y M2

 M11 +

M12


M12 
 ≤ 0,
M22

and

M2 ≤ 0.

Therefore

Rk

u ln u −

1 |∇u|2
2 u

dγ =

Rk

M (f, |∇u|)dγ ≤ M

Rk

udγ, 0

=

Rk

udγ ln

Rk

udγ .

This is log-Sobolev inequality (3.0.6) after introducing the function u(x) = f (x)2 .
Corollary 14. Take

2

M (x, y) = x p −

2 − p p2 −2 2
x
y ,
p2

and Ω = R+ . Clearly M satisfies condition of Theorem 3.0.23. This function gives Beckner–
Sobolev inequality for f ≥ 0.
Corollary 15. Take

M (x, y) =

U 2 (x) + y 2 ,

x
where Ω = (0, 1), U (x) = ϕ(Φ−1 (x)) , Φ(x) = −∞
dγ and ϕ(x) = Φ (x). Clearly M satisfies

132

opposite inequalities of Theorem 3.0.23. Therefore it gives Bobkov’s inequality

Rk

M (f, |∇f |)dγ ≥ M

Rk

f dγ, 0 .

Thus we see that some special functions which satisfy underling PDE with prescribed
obstacle conditions solve the problems.

3.0.6.3 Brascamp–Lieb, hypercontractivity, Borell’s Gaussian noise “stability” and its underlying PDE
Let L =

k
∂2
i,j=1 cij ∂xi ∂xj

be an elliptic operator (i.e., C = {cij }ki,j=1 ≥ 0). Let aj =

(aj1 , . . . , ajk ) ∈ Rk for j = 1, . . . , n so that the matrix A = (a1 , . . . , an ) has full rank
(n ≥ k). Let Pt be a semigroup with generator L and let
Pt f (y) =

Rk

f (x)pC
t (y, x)dx,

for all t ≥ 0.

−|x−y|2 /4t . Also note that if u : R → R, and g (x) = u (a · x)
√1
Note that pid
j
j j
t (y, x) = 4πt e

then

Pt gj (x) =
R

u(aj · y + x 2t Caj , aj )dγ(x).

Theorem 3.0.24. If AT CA • Hess B ≤ 0 then

Rk

B(u1 (a1 · x), . . . , un (an · x))pC
t (y, x)dx ≤

B
R

u1 (a1 · y + x 2t Ca1 , a1 )dγ, . . . ,

133

R

un (an · y + x 2t Can , an )dγ .

If AT CA • Hess B ≥ 0 then the inequality is reversed.
Proof. First we refer the reader to Remark 21.
t
d
Ps B(Pt−s uj ) =
Ps
0
0 ds
i,j,p,q
t

Pt B(uj ) − B(Pt uj ) =
t

∂ 2B
cpq aip ajq Pt−s ui Pt−s uj
∂ui ∂uj

Ps
0

i,j,p,q

t

=

Ps
0

∂ 2B
∂Pt−s ui ∂Pt−s uj
cpq
∂ui ∂uj
∂xp
∂xq

=

A∗ CA • Hess BPt−s u , Pt−s u

.

Corollary 16. If we choose C = id, t = 1/2 and y = 0 then we obtain that if A∗ A•Hess B ≤
0 then

Rk

B(u1 (a1 · x), . . . , un (an · x))dγ ≤ B

R

u1 (x|a1 |)dγ, . . . ,

R

un (x|an |)dγ .

Corollary 17. The similar reasoning shows that if Aj are kj ×k matrices i.e., Aj : Rk → Rkj
then

Rk

B

B(u1 (A1 x), . . . , un (An x))pC
t (y, x)dx ≤
A1 CA∗1

Rk1

u1 (x)pt

(A1 y, x)dx, . . . ,

An CA∗n

Rk n

un (x)pt

(An y, x)dx .

If

ACA∗ • Hess B ≤ 0
where A∗ = (A∗1 , . . . , A∗n ) and Hadamard product is understood as

(3.0.24)

2

B
Aj CA∗i ∂x∂ ∂x
i

j

k
i,j=1

.

Before we continue to the applications we have to mention some elementary facts from
134

the linear algebra. If A is n × n positive semidefinite matrix then by spectral decomposition
we have

A=
j

λj vj ⊗ vj ,

vj ∈ Rn ,

λj ≥ 0.

Proposition 4. If by using the n × n matrix A one constructs k copies of the matrix A, i.e.,
block matrix kn × kn then it is positive semidefinite if and only if A is positive semidefinite.
Proof. In one direction claim is obvious, in another direction claim follows from the spectral
decomposition.
Proposition 5. Let M ≥ 0 be kn × kn matrix constructed by k copies of the matrix A ≥ 0.
Let U ≥ 0 be k × k matrix. Then U ⊗ A ≥ 0.
Proof. The claim follows by linearity. It is enough to consider only the case A = vv T and
U = uuT and M is k copy of vv T . In this case

M ⊗ U = (vu1 , vu2 , . . . , vuk ) ⊗ (vu1 , vu2 , . . . , vuk ) ≥ 0.
So the claim follows.
Corollary 18. In Borell’s Gaussian noise “stability”, take A1 = (In , 0n ), A2 = (pIn , (1 −
p2 )1/2 In ), C = In . Then

A1 A∗1 = In ;
A1 A∗2 = A2 A∗1 = pIn ;
A2 A∗2 = In .
135

Therefore the condition ACA∗ • Hess B ≤ 0 becomes




 B11 pB12 
 ⊗ In ≤ 0.

pB12 B22
So in order to prove the functional version of Borell’s Gaussian noise “stability” it is enough
to check that the symmetric function

B(u, v) = P(X1 ≤ Φ−1 (u), pX1 +

1 − p2 Y2 ≤ Φ−1 (v)),

satisfies the condition




 B11 pB12 

 ≤ 0.
pB12 B22
Note that

136

B(u, v) = P X1 ≤ Φ−1 (u), Y2 ≤
Φ−1 (u)
−∞

B1 =

Φ−1
√ (v)−ps
1−p2

1 − p2

ϕ(t)ϕ(s)dtds =

−∞

−1 (u)
Φ−1 (v)−pΦ
√
2
1−p

−∞

B12 = ϕ

Φ−1 (v) − pΦ−1 (u)

1 − p2

−∞

B22 = ϕ

=
Φ−1
√ (u)−ps
1−p2

ϕ(t)ϕ(s)dtds;

−∞

ϕ(t)dt;

B11 = ϕ

1 − p2

−1 (v)
Φ−1 (u)−pΦ
√
1−p2

Φ−1 (v)
−∞

Φ−1 (v) − pΦ−1 (u)

B2 =

Φ−1 (v) − pX

−p

(1 − p2 )1/2 ϕ(Φ−1 (u))
1
(1 − p2 )1/2 ϕ(Φ−1 (v))

;
;

ϕ(t)dt;

Φ−1 (u) − pΦ−1 (v)
1 − p2

−p

(1 − p2 )1/2 ϕ(Φ−1 (v))

;

It is clear that B11 , B22 ≤ 0 and
2 = 0.
B11 B22 − p2 B12

Thus we obtain Borell’s Gaussian noise “stability”.
Corollary 19. With the same tensor trick we see that we have inequality

R2n

B(ϕ(x), ψ(e−t x +

1 − e−2t y))dγ ≤ B

137

Rn

ϕdγ,

Rn

ψdγ .

If the function B satisfies the following condition




B11

e−t B12

e−t B12

B22



 ≤ 0.

Taking B(u, v) = u1/a v 1/b we see that the above condition holds if and only if 1 ≤ a, 1 ≤ b
and

(a − 1)(b − 1) − e−2t ≥ 0.
This means that if we denote ϕ = f a , ψ = g b then we have inequality

R

f · Pt gdγ =
n

R2n

f (x)g(e−t x +

1 − e−2t y)dγ ≤

Rn

f a dγ

1/a
Rn

g b dγ

1/b

.

Taking supremum over f ∈ La (dγ) we obtain hypercontractivity for Ornstein–Uhlenbeck
semigroup Pt :

Pt g Lq (dγ) ≤ g Lp (dγ) ,
a is dual exponent to a, p = b under the condition (a − 1)(b − 1) − e−2t ≥ 0
where q = a−1

which can be rewritten in terms of p, q ≥ 1 as follows
p−1
− e−2t ≥ 0.
q−1

138

3.0.6.4 Extremal problems on BM O, Ap classes, Gehring classes and its
underlying PDE
PDE in this subsection corresponds to homogeneous Monge–Amp`ere equation (surfaces with
zero Gaussian curvature). These extremal problems correspond to finding minimal concave
function over an obstacle. We will start with the simplest example which correspond to the
convex domains.
Let Ω ⊂ Rn , m : Ω → Rk and H : Ω → R. Let Ω(I) denotes class of vector-valued
functions ϕ : I → Ω, and let conv(Ω) be the convex hull of the set Ω. We define the Bellman
function as follows

B(x) = sup { H(ϕ)
ϕ∈Ω(I)

I

:

m(ϕ)

I

= x}.

Theorem 3.0.25. The following properties hold:
1. B is defined on the convex set conv[m(Ω)];
2. B(m(y)) ≥ H(y) for all y ∈ Ω;
3. B is concave function;
4. B is minimal among those who satisfy properties 1,2 and 3.
Proof. Fist we show the property 1. Let Dom B denotes the domain where B is defined. Since
m(ϕ) ∈ conv[m(Ω)] we have m(ϕ)

I

∈ conv[m(Ω)]. Therefore Dom B ⊆ conv[m(Ω)]. Now

we show the opposite inclusion. Carath´eodory’s theorem implies that for any x ∈ conv[m(Ω)]
we have x =

n+1
j=1 aj xj ,

where aj ≥ 0,

n+1
j=1 aj

= 1 and xj ∈ m(Ω). Let the points yj be

such that m(yj ) = xj . We choose ϕ so that |{t ∈ I : ϕ(t) = yj }| = aj |I|. Then ϕ ∈ Ω(I).
139

Hence,
n+1

m(ϕ)

I

n+1

1
1
1
m(ϕ(t))dt =
m(ϕ(t))dt =
x |I|aj = x
=
|I| I
|I| {t:ϕ(t)=yj }
|I| j
j=1
j=1

Now we show the property 2. Let ϕ0 (t) = y, t ∈ I. Then m(ϕ0 )
B(m(y)) =

sup

H(ϕ)

ϕ∈Ω(I): m(ϕ) I =m(y)

I

≥ H(ϕ0 )

I

I

= m(y). Thus

= H(y).

Now we show the property 3. It is enough to show that B(θx + (1 − θ)y) ≥ θB(x) + (1 −
θ)B(y) for all x, y ∈ conv[m(Ω)] and θ ∈ [0, 1]. There exist functions ϕ, ψ ∈ Ω(I) such that
m(ϕ)

I

= x, m(ψ)

I

= y and

H(ϕ)

I

≥ B(x) − ε,

H(ψ)

I

> B(y) − ε.

We split interval I by two disjoint subintervals I1 and I2 so that |I1 | = θ|I|. Let Lj : Ij → I
be a linear bijections. We consider the concatenation as follows

η(t) =

Clearly η(t) ∈ Ω(I), m(η)

I

B(θx + (1 − θ)y) ≥ H(η)





ϕ(L1 (t)),

t ∈ I1 ,




ϕ(L2 (t)),

t ∈ I2 .

= θx + (1 − θ)y and

I

= θ H(ϕ)

I

+ (1 − θ) H(ψ)

I

> θB(x) + (1 − θ)B(y) − ε.

Now we show the property 4. Let G satisfies properties 1,2 and 3. Then Jensen’s

140

inequality implies that for any ϕ ∈ Ω(I) we have
H(ϕ)

I

≤ G(m(ϕ))

I

≤ G( m(ϕ) I ) = G(x)

Corollary 20. The definition of the modulus of uniform convexity tells us to consider the
following function

Bθ (x1 , x2 , x3 ) = sup{ |θf + (1 − θ)g|p I , (|f |p , |g|p , |f − g|p )
f,g

I

= (x1 , x2 , x3 )}. (3.0.25)

Theorem 3.0.25 implies that B is a minimal concave function over the obstacle. It is clear
that

δLp (ε) = 1 −

sup

2p ≥x

3

≥εp

(B1/2 (1, 1, x3 ))1/p .

There are different modulus of uniform convexity. One of them corresponds to the complex case. Let (X, ·, ) be a normed space. For 0 < p < ∞ we set
hX
p (ε) = inf 1 − x :

2π
1
x + eiθ y p dθ ≤ 1, y = ε .
2π 0

Corollary 21. Theorem 3.0.25 implies that

B(x, y) = sup
f,g

|f |p

I

:

2π
1
|f + eiθ g|p
= x, |g|p
2π 0
I

I

=y ,

corresponds to the minimal concave function over the obstacle. By knowing B one finds the

141

p
value hL
p (ε) i.e.,

1/p
p
hL
p (ε)

=1−

B(x, εp )

sup

.

0≤x≤1

Corollary 22. It is worth mentioning that H¨older’s inequality corresponds to the function

1/p1

B(x1 , . . . , xn ) = x1

1/pn

· · · xn

,

which coincides with its obstacle. Minkowski inequality corresponds to the function

B(x, y) = (x1/p + y 1/p )p ,

which again coincides with its obstacle.
It turns out that if one imposes extra condition — “boundedness of the mean oscillation”
on the class of functions Ω(I), one again obtains minimal concave function over an obstacle
but in a different domain (not necessarily convex domain as it was before). Let us illustrate
example on the BM O class.
Let Ωε be a parabolic strip

Ωε = {(x, y) ∈ R2 : x2 ≤ y ≤ x2 + ε2 },
and let let γ(t) = (t, t2 ). Then

BM Oε (J) = {ϕ : γ(ϕ)

142

I

∈ Ωε ∀I ⊂ J }.

(3.0.26)

In other words for general ϕ the points γ(ϕ(x)), x ∈ J belong to the parabola Γ = {(x, y) :
y = x2 } (a convex curve) and the integral average (or convex hull of these points) γ(ϕ)

I

always will belong to the convex full of the parabola i.e., conv(Γ) = {(x, y) : y ≥ x2 }. Therefore extremal problems on this class of functions would correspond to the minimal concave
functions on the convex hull of the parabola (which is convex set). However, BM Oε (J) class
can be seen as those functions ϕ ∈ L1 so that the convex hull of the points γ(ϕ) belong to
the parabolic strip Ωε and therefore extremal problems on this class of functions correspond
to the minimal concave functions on the parabolic strip.
Theorem 3.0.26. . Let f ∈ C 3 (R) be sufficiently nice. Then
Bε (x, y) = sup{ f (ϕ)
ϕ

I

: ϕ ∈ BM Oε , γ(ϕ)

I

= (x, y)}

is a minimal concave function defined in the parabolic strip Ωε with the obstacle condition
B(γ(t)) ≥ f (t). If the sign of the torsion of the space curve (t, t2 , f (t)) changes finitely many
times then there we have the finite algorithm which finds the function B.
For the proof of the theorem we refer the reader to [9, 10, 11, 12]. What is more we
also describe evolution of the function Bε as ε changes. The important part of the theorem
is that we find the function B, and the fact that it is minimal concave function is just the
corollary of this result.
In other words one can think about the problem as follows. The function B(t, t2 ) is given
and it is equal to f (t). This is what we call obstacle condition, or boundary condition. Now
the question is to find its minimal concave extension in the parabolic strip Ωε , and the result
we will denote by the same function B(x, y). This means that we are looking for the function

143

Bε (x, y) = sup{ B(γ(ϕ))
ϕ

I

: ϕ ∈ BM Oε , γ(ϕ)

I

= (x, y)},

where B(γ(t)) = f (t). Before we continuo let us mention some applications.
Corollary 23. For every p ∈ (0, ∞) the p-(quasi)norm on the BM O(I) is equivalent to the
2-norm:

cp ϕ BM O(I) ≤ sup |ϕ − ϕ J |p 1/p
≤ Cp ϕ BM O(I) .
J

(3.0.27)

J⊂I

In term of the Bellman function this is the same as

if

if

B(t, t2 ) = |t|p ,

then

B(t, t2 ) = −|t|p ,

then

( sup Bε (0, ))1/p ≤ Cp ε,
≤ε2

( inf −Bε (0, ))1/p ≥ cp ε,
≤ε2

∀ε > 0.

∀ε > 0.

which is true by Theorem 3.0.26.
What is also important in these corollaries is that these inequalities (3.0.27) are equivalent
to the statements about estimates for the minimal concave extensions Bε (x, y). This means
that if one tries to prove these inequalities (3.0.27), one implicitly tries to obtain estimates
for the functions B.
Corollary 24. There exist some constants c1 , c2 > 0 such that for all ϕ ∈ BM O(I) we have

144

John–Nirenberg inequality

1
|I|

t ∈ I : |ϕ(t) − ϕ I | > λ

≤ c1 exp −

c2 λ
ϕBM O(I)

.

In terms of the Bellman function it is the same as to show that

if

B(t, t2 ) = ✶(−∞,−λ)∪(λ,∞) ,

sup Bε (0, ) ≤ c1 e−c2 λ/ε ,

then

≤ε2

∀ε, λ > 0,

which is true by Theorem 3.0.26.
Corollary 25. There exist ε0 > 0 and a positive function C(ε), 0 < ε < ε0 , such that

eϕ

I

≤ C(ε)e ϕ I ,

for all ϕ ∈ BM Oε (I).

This is the same as to show that

if

B(t, t2 ) = et ,

then

Bε (x, y) ≤ C(ε)ex ,

0 < ε < ε0 ,

(x, y) ∈ Ωε ,

which is true by Theorem 3.0.26.
One can find similar characterization for the Reverse H¨older classes, Gehring classes, Ap
a classes etc. We formulate our abstract theorem and then we mention that these examples
become corollaries of our theorem. It is important to notice that BM Oε (I) defines parabolic
strip Ωε which uniquely reconstructs the class BM Oε (I). On the other hand notice that the
parabolic strip Ωε is difference of two convex domains, of the convex hull of the parabola
{y = x2 } and the convex hull of another parabola {y = x2 + ε2 }, and second convex set
conv({y = x2 + ε2 }) belongs to first convex set conv({y = x2 }). This idea suggest to
145

introduce the following class of functions. Let γ1 , γ2 ∈ C 3 be a strictly convex curves in R2
such that conv(γ2 ) ⊆ conv(γ1 ). Set
def

A(γ1 , γ2 , I) = {ϕ : ∀J ⊂ I γ1 (ϕ)

J

∈ conv(γ1 ) \ conv(γ2 )}.

In some sense A(γ1 , γ2 , I) represents class of functions with “small mean oscillation”. Under
some mild assumptions on γ1 , γ2 , f we have the following theorem
Theorem 3.0.27. If the space curve (γ1 (t), f (t)) has nowhere vanishing curvature and its
torsion changes sign finitely many times then we present an algorithm which finds expression
for the function B where

def

B(x, y) =

sup
ϕ∈A(γ1 ,γ2 ,I)

{ f (ϕ)

I

: γ1 (ϕ)

I

= (x, y)}.

B(x, y) is minimal concave function defined in conv(γ1 ) \ conv(γ2 ) such that B(γ1 ) = f .
For the proof we refer the reader to [9, 10, 11, 12]. It is worth mentioning that we
can describe dynamics of the function B (which we call evolution) if the domain conv(γ2 )
increases by inclusion. Note that

A((t, t2 ), (t, t2 + ε2 ), I) = BM Oε (I).

Let p1 > p2 and Q ≥ 1. Then
A((t1/p2 , t1/p1 ), (Qt1/p2 , t1/p1 ), I) = Ap1 ,p2 (Q),

146

where

def
1 ϕp2 −1/p2 ≤ Q}.
Ap1 ,p2 (Q) = {ϕ : [ϕ]Ap ,p = sup ϕp1 1/p
J
J
1 2
J⊂I

If p ∈ (1, ∞) then A1,−

1
p−1

= Ap where Ap stands for the classical Muckenhoupt class. When

p2 = 1 and p1 > 1, the class Ap1 ,p2 coincides with the so-called Gehring class (sometimes
the Gehring class is called reverse-H¨older class).
Finally we finish this subsection by mentioning underlying PDE in these extremal problems. The concave function B has prescribed boundary condition f (t) and it satisfies homogeneous Monge–Amp`ere equation i.e.,

B(γ1 (t)) = f (t);
Hess B ≤ 0 in conv(γ1 ) \ conv(γ2 );

(3.0.28)

det(Hess B) = 0 in conv(γ1 ) \ conv(γ2 ).

3.0.7 Relation to differential geometry
If one tries to obtain sharp inequalities in the theorems 3.0.22, 3.0.23,3.0.24 and 3.0.27 then
besides of solving the corresponding partial differential inequalities (3.0.13), (3.0.22), (3.0.24)

147

and (3.0.28) one also has to solve corresponding partial differential equations

H1 H2 H12 (1 − a21 − a22 ) + a21 H22 H11 + a22 H12 H22 = 0.
Hess M +

∇y M
≤ 0 and
y
2n×2n

ACA∗ • Hess B ≤ 0 and
Hess B ≤ 0 and

det Hess M +

(3.0.29)
∇y M
= 0.
y
2n×2n

det (ACA∗ • Hess B) = 0.

det(Hess B) = 0.

(3.0.30)
(3.0.31)
(3.0.32)

And find the minimal (or maximal if the inequalities are reversed) solution if possible with
prescribed obstacle conditions.

3.0.7.1 Developable surfaces, concavity and the torsion of the space curve
Solution of Theorem 3.0.27 includes finding the following function B: given f (t) and the
strictly convex curves γ1 , γ2 on the plane R2 such that conv(γ2 ) ⊂ conv(γ1 ), find B in the
domain conv(γ1 ) \ conv(γ2 ) such that
B(γ1 (t)) = f (t);
Hess B ≤ 0 in conv(γ1 ) \ conv(γ2 );
det(Hess B) = 0.

(3.0.33)

And among all possible solutions choose the minimal one. This means that in general we do
not have uniqueness results for B. This is so because the domain of B is not convex.
These surfaces (x, y, B(x, y)) are known as developable surfaces because they have almost everywhere zero Gaussian curvature. It turns out that torsion of the boundary data
(γ1 (t), f (t)) appears as the main object in studying concavity of the surfaces with zero Gaus148

sian curvature (see Chapter 1, see also [9, 10, 11, 12]). It is worth mentioning that some flaw
of 4 vertex theorem and bitangent line (see [56, 58, 58]) appeared in the work [13] in order
to find the function B in uniform convexity.
Some abstract works have been done regarding existence of such solutions in the works
of Caffarelli, Nirenberg and Spruck (see [2]). The main difference between these results is
that we explicitly construct solutions for each given boundary data (which as a corollary
gives answer about existence and uniqueness of the solution), however we do it only in two
dimensional case whereas in the work of [2] the proof of existence of smooth solutions (for
the smooth boundary data) in any dimension is given. Theory in [9] was developed also for
piecewise continuous boundary data f (t), and in particular it explicitly shows what happens
with the continuity of the global concave solutions of homogeneous Monge–Amp`ere equation
once we remove smoothness of the boundary data. We also developed dynamic for the
solutions once we change smoothly domain of the function B. This is what we call evolution
of the Bellman function over its domain.
There are still some open problems left:
Problem 1. Find the function B in some adequate way if the torsion of the space curve
(γ1 (t), f (t)) changes sign infinitely many times.
Problem 2. Find the similar characterizations in the high dimensions n ≥ 3.

149

3.0.7.2 Modified Monge–Amp`ere equation
Finding sharp inequalities in Theorem 3.0.24 includes solving the following differential equations

A∗ CA • Hess B ≤ 0 and

det(A∗ CA • Hess B) = 0,

where B ∈ C 2 is given in some parallelepiped, for example, say Rk+ , A = (a1 , . . . , an ) is k × n
matrix with full rank.
If B satisfies assumptions L1-L5 (see Chapter 2) then we gave complete characterization
of such functions B in the case k = 1, k = n and in the case k = n − 1 if in addition Bij = 0.
General questions still remains open. Let A be k × n matrix.
Problem 3. Describe all possible solutions of the partial differential inequalities

A∗ A • Hess B ≤ 0 and

det(A∗ A • Hess B) = 0.

Let us consider the following particular. Let B ∈ C 2 be given in some rectangular
domain. Let n = k = 2 and take A = (a1 , a2 ) where a1 , a2 = 0. Then we must have


A∗ A • Hess B = 



|a1 |2 B11

a1 · a2 B12 
 ≤ 0 and
2
a1 · a2 B12 |a2 | B22

det(A∗ A • Hess B) = 0.

If a1 · a2 = 0 then B has to be separate concave functions such that B11 B22 = 0 and these
are the all possible solutions. Therefore we assume that a1 · a2 = 0. Then we see that B

150

must be separate concave function and moreover
|a1 |2 |a2 |2
2 = 0.
B11 B22 − B12
|a1 · a2 |2
So in the case n = k = 2 the problem reduces to the following one
Problem 4. Let |c| ∈ [1, ∞) and let B ∈ C 2 be given on some rectangular domain in R2 .
Characterize all possible separately concave functions B such that

2 = 0.
c2 B11 B22 − B12

(3.0.34)

The trivial case |c| = 1 corresponds to developable surface and the characterization of
these surfaces are mostly known. For general |c| > 1 we can give local characterization.
Namely, we will show that the above equation can be reduced to the following one

∂f
= f¯
∂ z¯

for some appropriate f (see below).
For separately concave B(x, y) set Bxx = −p2 , Byy = −q 2 . Then equation (3.0.34)
implies that Bxy = cpq. We also have

− 2ppy = cqpx + cpqx ,

(3.0.35)

− 2qqx = cqpy + cpqy .

(3.0.36)

Further we assume that p, q = 0. Assume that the locally the map p, q : (x, y) → R2 is

151

invertible, and let (x, y) be its inverse map. Then






−1

 p x p y   xp xq 

=

qx qy
yp yq



=



1
 yq −xq 
·
.
det(Jacob(x, y))
−yp xp

Therefore equations (3.0.35) and (3.0.36) take the following form

2pxq = cqyq − cpyp ,
2qyp = −cqxq + cpxp .
This can be written as follows

2(px)q = c(qy)q − c(py)p ,
2(qy)p = −c(qx)q + c(px)p .
We set U˜ (p, q) = px(p, q) and V˜ (p, q) = qy(p, q). Then we obtain
V˜ p
q

2U˜q = cV˜q − c
2V˜p = −c

U˜ q
p

,
p

+ cU˜p .
q

After the logarithmic substitution U (p, q) = M (ln p, ln q) and V (p, q) = N (ln p, ln q) then we

152

obtain the linear equation

2M2 = c(N2 − N − N1 ),
2N1 = c(−M − M2 + M1 ).
By setting k = 2/c ∈ (−2, 2), this can be rewritten as follows
















 N   −1 1   N1   0 −k   M1 

=

+

.
M
−k 0
N2
1 −1
M2

We need the following technical lemma.
Lemma 23. If the vector function N (x, y) = (N, M ) : Ω ⊂ R2 → R2 satisfies the following
first order system of linear differential equations

N = P N1 + QN2

for some invertible 2 × 2 matrices P, Q where


 −2t
QP −1 = 
−1

153

δ2
0





for some t ∈ (−δ, δ), δ > 0 then after making change of variables N (x) = B U (Ax) where

√
2
2
δ −t 
 t
B=

1
0





and

− √ 2t 2
δ −t

1
 −1
AT = P −1 
2
0 − √ 21

δ −t2

and we obtain

∂f
= f¯,
∂ z¯

where f = U + iV .
Proof. Set P = (P1 , P2 ), Q = (Q1 , Q2 ) where Pi , Qj are columns.




 N 

 = N1 P1 + N2 P2 + M1 Q1 + M2 Q2 .
M
˜ (α1 x + α2 y, β1 x + β2 y) then
Now let N (x, y) = N

˜1 + β1 N
˜2 ;
N1 = α1 N
˜1 + β2 N
˜2 ;
N2 = α2 N
˜ 1 + β1 M
˜ 2;
M1 = α1 M
˜ 1 + β2 M
˜ 2.
M2 = α2 M

154



,

So we obtain



˜
 N 
˜1 + (P1 β1 + P2 β2 )N
˜2 +

 =(P1 α1 + P2 α2 )N
˜
M
˜ 1 + (Q1 β1 + Q2 β2 )M
˜ 2.
(Q1 α1 + Q2 α2 )M

Finally we set

˜ = a1 U + b 1 V
N
˜ = a2 U + b2 V.
M

and




 a1 b 1 
B=
.
a2 b 2
Thus we obtain




 U 

=
V
B −1 [a1 (P1 α1 + P2 α2 ) + a2 (Q1 α1 + Q2 α2 )]U1 +
B −1 [a1 (P1 β1 + P2 β2 ) + a2 (Q1 β1 + Q2 β2 )]U2 +
B −1 [b1 (P1 α1 + P2 α2 ) + b2 (Q1 α1 + Q2 α2 )]V1 +
B −1 [b1 (P1 β1 + P2 β2 ) + b2 (Q1 β1 + Q2 β2 )]V2 .

155

And we would like to see that
















 U  1  1 0   U1  1  0 −1   V1 

= 

+ 

.
2
2
V
0 −1
U2
−1 0
V2

This can hold if and only if




1  1 0  −1 1 + −1
(P α, Qα) = B 
 B = BI B ;
2
2
0 −1


1  0 −1  −1 1 − −1
(P β, Qβ) = B 
 B = BI B ,
2
2
−1 0
where






 α1 
α=
;
α2



 β1 
β=
.
β2

Let e1 = (1, 0), e2 = (0, 1), and let
1
BI + B −1 e1 , BI − B −1 e1 ,
2
1
B2 =
BI + B −1 e2 , BI − B −1 e2 .
2

B1 =

156

Then the above conditions hold iff

P (α, β) = B1 ;
Q(α, β) = B2 .

Therefore if B is chosen in such a way that QP −1 = B2 B1−1 then by setting (α, β) = P −1 B1
we obtain the desired result. But note that if a and b are corresponding rows of the matrix
B then


B2 B1−1 = 

a·b
−2 |b|
2

|a|2
|b|2

−1

0







,

so the claim follows.
In our case of P, Q we have


− 2c



1 
 −k 1  
QP −1 = 
=
,
−1 0
−1 0
therefore we can apply the lemma and we see that taking t = 1/c ∈ (−1, 1) and δ = 1 we
have

 t
B=
1

√

1 − t2
0









A=

157

0
√1
4t 1−t2

− 12
2t2 −1

− 14 √

t 1−t2



.

This means that if we set






 N (x, y)   t

=
M (x, y)
1

√

1 − t2
0


 U

V

− y2 , √x 2
4t 1−t
− y2 , √x 2
4t 1−t

−

2
y 2t
√ −1
4 t 1−t2

2
√ −1
− y4 2t
2



,

t 1−t

where by setting z = x + iy for the function f (z, z¯) = U (x, y) + iV (x, y) we have

∂f
= f¯.
∂ z¯
It is known that all C 1 solutions of the above equation are real analytic and they can be
represented in terms of power series

f (z) =

∞

ck J (k) (z z¯)z k + c¯k J (k+1) (z z¯)¯
z k+1 ,

k=0

where J(r) is modified Bessel I-functions whose series representation is

f (r) =

∞
j=0

158

rj
.
(j!)2

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