ON THE MTW CONDITIONS OF

MONGE-AMP`ERE TYPE EQUATIONS

By

Seonghyeon Jeong

A DISSERTATION

Submitted to

Michigan State University

in partial fulfillment of the requirements

for the degree of

Mathematics – Doctor of Philosophy

2021

ABSTRACT

ON THE MTW CONDITIONS OF

MONGE-AMP`ERE TYPE EQUATIONS

By

Seonghyeon Jeong

The MTW condition was introduced in [9] to study the regularity theory of the optimal

transportation problem, and the MTW condition was used by many researchers to study

other regularity properties of the optimal transportation problem. For example, the MTW

condition was used by G. Loeper, A. Figalli, Y-H. Kim, R. McCann and other researchers

to show H¨older regularity of the potential function, and by A. Figalli and De Phillippis to

show Sobolev regularity of the potential function.

I present two of my results about the MTW condition in this dissertation. The first result

concerns about the synthetic expressions of the MTW condition. The cost function of the

optimal transportation problem need a high regularity assumption (C 4) to define the MTW

condition. There are some expressions of MTW condition, however, which only need much

weaker regularity assumption to define, but equivalent to the MTW condition when the

cost function has enough regularity. We call these conditions synthetic MTW conditions.

Although the synthetic MTW conditions are equivalent to the MTW condition under some

assumption, it was not shown that if the synthetic MTW conditions are equivalent under

weak regularity assumption which is not enough to define the MTW condition. I present

a proof of the equivalence of the synthetic MTW conditions under C 2,1 assumption on the

cost function in chapter 3.

The other result is about the H¨older regularity of solutions to generated Jacobian equations.

In generated Jacobian equations, we study more general structure than the optimal trans-

portation problem. Some examples of generated Jacobian equations which is more compli-

cated than the optimal transportation problem can be found in geometric optics problems.

The H¨older regularity result was proved by G. Loeper in [8] in the optimal transporta-

tion problem case and this can be generalized to generated Jacobian equations. Since the

structure of generated Jacobian equations has more non-linearlity than the structure of the

optimal transportation problem, however, there are some difficulties to apply Loeper’s idea

to generated Jacobian equations. We discuss about the difficulties and suggest a way to go

around the problems in chapter 4. Then I generalize Loeper’s idea to more general generated

Jacobian equations and show that we can have a similar local H¨older regularity result.

TABLE OF CONTENTS

Chapter 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Chapter 2 Optimal transportation problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.1 Optimal transportation problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2.2 Monge-Amp`ere type equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Chapter 3 The synthetic MTW conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.1 The synthetic MTW conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.2 Equivalence of the Synthetic MTW conditions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .16

Chapter 4 Local H¨older regularity of generated Jacobian equations . . . . . . . . 33

4.1 Generated Jacobian equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
4.2 Structure of generated Jacobian equation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .34
4.3 Quantitative Loeper’s condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
4.4 G-convex functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .50
4.5 Proof of the local Holder regularity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .60

BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

iv

Chapter 1

Introduction

The Monge-Amp`ere equation is a Partial Differential Equation (PDE) of the form

det(cid:0)D2φ(x)(cid:1) = f (x).

(MA)

It is known that the Monge-Amp`ere equation is elliptic over the family of convex functions,

and it is fully non-linear. Moreover, the Monge-Amp`ere equation is degenerate elliptic,

so that the methods which are developed for uniformly elliptic equations, for example the

Evans-Krylov theorem, do not work for the Monge-Amp`ere equation.

It is well-known that the Monge-Amp`ere equation is closely related to optimal trans-

portation problems and geometric optics problems. In fact, the potential functions of the

solutions to optimal transportation problems satisfy PDEs of the form

det(cid:0)D2φ(x) − A(x, Dφ(x))(cid:1) = ψ(x, Dφ(x))

(c-MA)

for some ψ, where A(x, p) = −D2

xxc(x, expc

x(p)) is a matrix valued function defined using

the cost function from the optimal transportation problem. Also, the solutions from the

geometric optics problems have potentials that satisfy PDEs of the form

det(cid:0)D2φ(x) − A(x, Dφ(x), φ(x))(cid:1) = ψ(x, Dφ(x), φ(x))

for some ψ where A(x, p, u) = D2

xxG(cid:0)x, expG

x,u(p), Zx(p, u)(cid:1) is a matrix valued function

(GJE)

defined using the generating function from the geometric optics problems.

The difficulties that arise from fully non-linearity and degenerate ellipticity along with

application to optimal transportation problems and geometric optics problems made research

1

about Monge-Amp`ere type equations very attractive and active.

To study regularity theory of Monge-Amp`ere type equations, it is needed to define notions

of convexity for each Monge-Amp`ere type equation, which are called c-convexity and G-

convexity. Like the Monge-Amp`ere equation case, (c-MA) and (GJE) are elliptic over c-

convex functions and G-convex functions respectively. Moreover, some additional structural

conditions are needed. The MTW condition is one of these structural conditions and the

MTW condition is a very important condition for studies about H¨older regularity theory of

Monge-Amp`ere type equations. The MTW condition is a condition about sign of a (2,2)-

tensor, which is called the MTW tensor, that contains 4th order derivative of c or G. The

MTW condition is first discovered in [9] and used to prove the regularity result in the paper.

Meaning of the MTW condition was not clear when it was first discovered, but what the

MTW condition means geometrically was found later, for instance in [6] and [8]. What is

more, it is proved that the MTW condition is a necessary and sufficient condition for H¨older

regularity of solutions to (c-MA) in [8].

In this thesis, I present my works regarding the MTW condition. In the next chapter,

connections of optimal transportation problems and (c-MA) will be introduced with struc-

tural conditions on (c-MA).

In chapter 3, we discuss synthetic expressions of the MTW

condition, and prove that these synthetic expressions are equivalent even when the cost fuc-

tion c does not have enough regularity to define the MTW tensor. In the last chapter, the

result of Loeper in [8] about the H¨older regularity of solutions to (c-MA) with certain density

conditions on the source measure µ will be generalized to a similar result about the H¨older

regularity of solutions to (GJE).

2

Chapter 2

Optimal transportation problem

2.1 Optimal transportation problem

In this section, we introduce the optimal transportation problem and c-convexity. Let X
and Y be two compact sets with non-empty interior in Rn, and let µ ∈ P(X) and ν ∈ P(Y )
where P(X) is the set of Borel probability measures on X. We first define the push-forward

of a measure.

Definition 2.1.1. Let T : X → Y be a measurable function. We define the push-forward

measure T(cid:93)µ by

for any measurable set A ⊂ Y .

T(cid:93)µ[A] = µ(cid:2)T −1(A)(cid:3)

Let c : X × Y → R be a continuous function, which we will call the cost function. The

optimal transportation problem which was introduced by G. Monge in 1781 asks to find a

function T which minimizes the total transportation cost caused by distributing mass µ to

ν.

Problem 1 (Monge problem). Find a measurable function T : X → Y which minimizes the

following quantity:

(cid:90)

c(x, S(x))dµ,

(2.1)

among the family of functions S(µ, ν) = {S : X → Y |S(cid:93)µ = ν}.

X

The Monge problem can be easily applied to real situations such as delivering some

products from factories to customers. However, there was not a lot of progress until the

1940s due to high non-linearity of the problem. It was L. Kantorovich who made a break

3

through in the optimal transportation problem in 1942. He introduced a relaxed version of

Monge problem, which we call the Kantorovich problem.

Problem 2 (Kantorovich problem). Find a measure π which minimizes the following quan-

tity:

(cid:90)

X×Y

c(x, y)dγ

(2.2)

among the family of measures

Γ(µ, ν) =(cid:8)γ ∈ P(X × Y )|ProjX (cid:93)[γ] = µ and ProjY (cid:93)[γ] = ν(cid:9) ,

where ProjX : X × Y → X and ProjY : X × Y → Y are the projections onto X and Y
respectively.

Kantorovich used measures γ on X × Y instead of functions S : X → Y , and let the total

cost (2.2) depend on γ linearly. What is more, the Kantorovich problem always admits a

solution while the Monge problem does not admit a solution in some cases (for example, µ =

δ0 and ν = 1

2δ−1 + 1

2δ1). We call solutions to the Kantorovich problem and Monge problem,

Kantorovich solutions and Monge solutions respectively. Since Kantorovich solutions always

exist, we can try to find information about Monge solutions from Kantorovich solutions.
In fact, if a Kantorovich solution is of the form π = (Id × T )(cid:93)µ, then T will be a Monge

solution. Therefore, to deduce the existence of a Monge solution from a Kantorovich solution,

we should observe the support of the Kantorovich solution. To achieve this, we introduce

another result of Kantorovich called Kantorovich duality.

Problem 3 (Dual problem). Find a pair of functions (φ, ψ) such that φ ∈ L1(dµ), ψ ∈

L1(dν), and maximizes the following

−

(cid:90)

(cid:90)

φ(cid:48)dµ −

X

Y

ψ(cid:48)dν,

4

(2.3)

among the family of pairs of functions

(φ(cid:48), ψ(cid:48)) ∈ L1(dµ) × L1(dν)

φ(cid:48)(x) + ψ(cid:48)(y) ≥ −c(x, y),
dµ ⊗ dν a.e. (x, y) ∈ X × Y

 .

(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)

Φc(µ, ν) =

Theorem 2.1.2 (Kantorovich duality). The miminum total Kantorovich cost (2.2) equals

the maximum of (2.3).

(cid:90)

(cid:90)

(cid:18)

−

(cid:90)

(cid:19)

.

ψ(cid:48)dν

φ(cid:48)dµ −

X

Y

(2.4)

c(x, y)dγ =

sup

(φ(cid:48),ψ(cid:48))∈Φc(µ,ν)

inf

γ∈Γ(µ,ν)

X×Y

By Kantorovich duality, we can expect to obtain some information about the Kantorovich

solution by solving the dual problem. In the dual problem, we consider pairs of L1 functions

φ and ψ. However, we can reduce the family of pairs of functions Φc(µ, ν) to a smaller set.

Note that

φ(x) + ψ(y) ≥ −c(x, y)
⇒φ(x) ≥ −c(x, y) − ψ(y)
⇒φ(x) ≥ sup
y∈Y

{−c(x, y) − ψ(y)} = ψc(x).

Therefore, we have

−

(cid:90)

X

(cid:90)

Y

φ(x)dµ −

(cid:90)

X

(cid:90)

Y

ψ(y)dν.

ψc(x)dµ −

ψ(y)dν ≤ −

Hence, we only need to consider pairs of the form (ψc, ψ). Similarly, we define φc(y) =
supx∈X{−c(x, y) − φ(x)} for φ : X → R, then a similar argument shows that we only need
to consider the pairs (φcc, φc).

5

Definition 2.1.3. A function φ : X → R is called c-convex if

φ(x) = sup
y∈Y

{−c(x, y) − ψ(y)}

(2.5)

for some ψ : Y → R ∪ {∞} such that ψ (cid:54)≡ ∞ . A function ψ : Y → R is called c∗-convex if

ψ(y) = sup
x∈X

{−c(x, y) − φ(x)}

(2.6)

for some φ : X → R ∪ {∞} such that φ (cid:54)≡ ∞.

Hence, we only need to consider pairs of c and c∗-convex functions for the dual problem.

In fact, existence of a solution to the dual problem of the form (φ, φc) can be proved by
considering maximizing sequence of pairs of c and c∗-convex functions. See, for instance, [12]

Chapter 1.

As the name and the definition of c(c∗)-convex function suggest, there are many properties

analogous to properties of convex functions. For example, the analogy of the subdifferential

is the c-subdifferential:
Definition 2.1.4. Let φ : X → R be a c-convex function, and let x0 ∈ X. Then there exists
some y0 ∈ Y such that

φ(x) ≥ −c(x, y0) + c(x0, y0) + φ(x0).

(2.7)

We say y0 belongs to the c-subdifferential of φ at x0, and we denote

∂cφ(x0) = {y0 ∈ Y |φ(x) ≥ −c(x, y0) + c(x0, y0) + φ(x0),∀x ∈ X}.

If A ⊂ X, we denote ∂cφ(A) =
Proposition 2.1.5. Let φ : X → R be a c-convex function. Let x ∈ X, then

∂cφ(x).

(cid:91)

x∈A

y ∈ ∂cφ(x) ⇔ φ(x) + φc(y) = −c(x, y).

(2.8)

6

Equation (2.8) and the Kantorovich duality provides very important information about

the support of the Kantorovich solution. Let π be a Kantorovich solution and (φ, φc) be a
pair of c and c∗-convex functions that solves the dual problem. Then the marginal condition

on the Kantorovich solution π implies that we have

(cid:90)

X×Y

φ(x) + φc(y) + c(x, y)dπ = 0.

This shows that the Kantorovich solution π is concentrated in the set {φ(x) + φc(y) =
−c(x, y)}. Noting (2.8), we obtain

spt(π) ⊂ {(x, y)|y ∈ ∂cφ(x)}.

Therefore, as we discussed earlier, if ∂cφ is single valued, then the Monge solution is given

by T (x) = ∂cφ(x). Moreover, if c is C 1, then (2.7) shows

Dφ(x0) = −Dxc(x0, y0).

Hence if y (cid:55)→ −Dxc(x, y) is injective, then we obtain

∂cφ(x0) = y0 = [−Dxc(x0,·)]−1(Dφ(x0))

(2.9)

which implies single valuedness of the c-subdifferential ∂cφ at each differentiable points of φ.

If we assume more differentiability on the cost function c, then (2.7) implies semi-convexity

of φ so that φ is differentiable almost everywhere. Then the Monge solution can be defined

dx a.e. in X.

7

2.2 Monge-Amp`ere type equations

In this section, we explain the connection between the optimal transportation problem and

the c-Monge-Amp`ere equation, and we present the structural conditions for the c-Monge-

Amp`ere equation.

To see the relation between optimal transportation problem and the c-Monge-Amp`ere

equation, let us derive (c-MA) formally from the optimal transportation problem. Let dµ =
f (x)dx and dν = g(y)dy, and let T : X → Y be a Monge solution. From the push-forward

condition, we obtain

(cid:90)

(cid:90)

u(T (x))f (x)dx =

u(y)g(y)dy,

for any continuous function u ∈ C(Y ). On the other hand, we use the change of variable

X

Y

formula with y = T (x) to obtain

(cid:90)

u(y)g(y)dy =

Y

Therefore, we obtain

(cid:90)

X

u(T (x))g(T (x)) det(DT (x))dx.

det(DT (x)) =

f (x)

g(T (x))

.

(2.10)

Noting that T is given by a c-subdifferential of a c-convex function φ and equation (2.9), we

obtain the expression

DT (x) =(cid:2)−D2

xyc(x, T (x))(cid:3)−1(cid:0)D2φ(x) + D2

xxc (x, T (x))(cid:1) .

Denoting A(x, p) = −D2

xxc (x, [Dxc(x,·)]−1(p)), we obtain

det(cid:0)D2φ(x) − A(x, Dφ(x))(cid:1) =

det(cid:0)−D2

xyc(x, T (x))(cid:1) .

f (x)

g(T (x))

(2.11)

From the above formal derivation of the c-Monge-Amp`ere equation (2.11), we observe

8

that the cost function c should be at least C 2, and we can deduce that we need the following

conditions on the cost function.

y (cid:55)→ −Dxc(x, y) is injective ∀x ∈ X,

det(cid:0)−D2

xyc(x, y)(cid:1) (cid:54)= 0.

(Twist)

(Non-deg)

Note that we can define a condition which is symmetric to the condition (Twist).

x (cid:55)→ −Dyc(x, y) is injective ∀y ∈ Y.

(Twist*)

(Twist) and (Twist*) condition imply inverse functions of −Dxc(x,·) and −Dyc(·, y). We

call these inverse functions c-exponential maps.

Definition 2.2.1. Define Y ∗

x ⊂ Rn by

x = −Dxc(x, Y ).
Y ∗

Then −Dxc(x,·) : Y → Y ∗
x → Y by the inverse function of −Dxc(x,·):
Y ∗

x is bijective by (Twist). We define the c-exponential map expc

x :

−Dxc(x, expc

x(p)) = p.

(2.12)

We call expc

x the c-exponential map focused at x. Similarly, define X∗

y ⊂ Rn by

y = −Dyc(X, y).
X∗

Then −Dyc(·, y) : X → X∗

y is bijective. We define the c∗-exponential map expc∗

y : X∗

y → X

9

by the inverse function of −Dyc(·, y):

−Dyc(expc∗

y (q), y) = q.

(2.13)

We call expc∗

y the c∗-exponential map focused at y.

Definition 2.2.2. Let x ∈ X and y0, y1 ∈ Y , and let pi = −Dxc(x, yi). The c-segment
{yθ|θ ∈ [0, 1]} focused at x that connects y0 and y1 is the image of the segment [p0, p1] under
the c-exponential map expc
x:

{yθ|θ ∈ [0, 1]} = expc

x([p0, p1]).

We say that yθ is a c-segment if there is no confusion.

Remark 2.2.3. (Non-deg) implies that the c-exponential maps are differentiable and

x(p) =(cid:2)−D2

x(p))(cid:3)−1 .

Dp expc

xyc(x, expc

Moreover, compactness of X and Y with (Non-deg) implies that we have a constant λ such

that

≤ |D2

xyc| ≤ λ,

1
λ

|y1 − y0| ≤ | − Dxc(x, y1) + Dxc(x, y0)| ≤ λ|y1 − y0|,
1
λ
|x1 − x0| ≤ | − Dyc(x1, y) + Dyc(x0, y)| ≤ λ|x1 − x0|.
1
λ

The conditions (Twist), (Twist*), and (Non-deg) are enough to obtain a Monge solution

that is defined almost everywhere, and observe the relation with the Monge-Amp`ere equation.

To study regularity theory, however, we need one more condition called the MTW condition.

This condition first appeared in [9] in (A3s) form and is named after the authors of the paper.

10

The (A3w) form is appeared in [11]. To define the MTW condition, we need to assume more

regularity on the cost function c.

c ∈ C 4(X × Y ).

(Regular)

Next, we define the MTW tensor, which is a (2,2) tensor that contains 4th derivatives of the
cost function. Let A(x, p) = −D2

x(p)). Then the MTW tensor is

xxc(x, expc

M T W = D2

ppA(x, p).

The MTW condition is a sign condition on the MTW tensor in some directions.

M T W [η, η, ξ, ξ] ≥ 0, ∀η ⊥ ξ.

(2.14)

(A3w)

If the cost function c satisfies (A3w) with strict inequality, we say that c satisfies (A3s). In

this case, from the compactness of X and Y and the tensorial nature, we obtain a constant

α > 0 such that

M T W [η, η, ξ, ξ] > α|η|2|ξ|2, ∀η ⊥ ξ.

(A3s)

To study the c-Monge-Amp`ere equation, we should define a weak solution for the equation

like other PDEs. For the c-Monge-Amp`ere equation, we can define several different weak

solutions. The first weak solution is defined using the mass balance condition (push-forward

condition) of the optimal transportation problem.

Definition 2.2.4. A function φ : X → R is called a Brenier solution of (2.11) if φ satisfies

∂cφ(cid:93)µ = ν.

(2.15)

Another weak solution can be defined using equation (2.10). If we integrate equation

11

(2.10) on a measurable set A ⊂ X, then we obtain

(cid:90)

A

(cid:90)

det(DT (x))dx =

f (x)

dx.

g(T (x))

A

T = ∂cφ, we obtain

|∂cφ(A)| =

f (x)

dx.

g(T (x))

A

We use this equation to define another weak solution.

(cid:90)

(cid:90)

Using the change of variable formula with y = T (x) on the left hand side and noting that

Definition 2.2.5. A c-convex function φ : X → R is called an Alexandrov solution of (2.11)

if φ satisfies

|∂cφ(A)| =

f (x)

dx.

g(T (x))

A

Note that, in contrast to the Brenier solution, an Alexandrov solution can be defined

with more general formulas on the right hand side of (2.10) or (2.11). In addition, a Brenier

solution does not have to be an Alexandrov solution. To observe this, suppose f and g are
bounded away from 0 and ∞ on each support, and let φ be a Brenier solution. Then we

have T(cid:93)f (x)dx = g(y)dy, and

(cid:90)

(cid:90)

f (x)dx =

A

T (A)

g(y)dy,

so that we have |A| ∼ |T (A) ∩ Y |. If φ was an Alexandrov solution, however, Definition
2.2.5 shows that we should have |A| ∼ |T (A)|. An explicit counter example is explained in
[12] and [1]. A Brenier solution becomes an Alexandrov solution when ∂cφ(X) ⊂ Y . This,

in fact, can be deduced if we add some geometric conditions on X and Y .

Definition 2.2.6. Let x ∈ X, y ∈ Y and A ⊂ X, B ⊂ Y . B is called c-convex with respect

to x if the set

x = −Dxc(x, B)
B∗

12

is convex. We say that B is c-convex with respect to A if B is c-convex with respect to x for
any x ∈ A. Similarly, A is called c∗-convex with respect to y if the set

y = −Dyc(A, y)
A∗

is convex, and we say that A is c∗-convex with respect to B if A is C∗-convex with respect
to y for any y ∈ B.

We add the following conditions on X and Y .

Y is c-convex with respect to X.

(DomConv)

X is c-convex with respect to Y .

(DomConv*)

It is proved, for example in [8], that if a cost function c satisfies (Twist), (Twist*), (Non-deg),
(A3w) and (DomConv), then a c-subdifferential ∂cφ(x) at a point x ∈ X of a c-convex
function φ is c-convex with respect to x. Then we obtain that ∂cφ ⊂ Y , and the Brenier

solution becomes an Alexandrov solution.

13

Chapter 3

The synthetic MTW conditions

3.1 The synthetic MTW conditions

In [8], Loeper suggested a condition that is equivalent to the MTW condition when the cost

function is C 4. The condition is the following

Definition 3.1.1 (Loeper’s condition). Let x0, x1 ∈ X and define a function F (p) =
−c(x1, expc

x0(p)). Then the cost function c is said to satisfy Loeper’s condi-

x0(p)) + c(x0, expc

tion if

F (tp1 + (1 − t)p0) ≤ max{F (p0), F (p1)}

for any p0, p1 ∈ Y ∗

x0 and for any x0, x1 ∈ X.

Technically, we only need C 1 cost function with twisted condition to form Loeper’s con-

dition. Therefore, Loeper’s condition can be viewed as a synthetic expression of the MTW

condition. Moreover, Loeper’s condition implies that the c-subdifferential of a c-convex func-

tion at a point x0 is c-convex with respect to x0. As we can see from Definition 3.1.1, Loeper’s

condition means quasi-convexity of the function F . We introduce notations for level sets and

sublevel sets of the function F :

Lp0 = {p ∈ Y ∗

x0|F (p) = F (p0)}, SLp0 = {p ∈ Y ∗

x0|F (p) ≤ F (p0)}.

(3.1)

Then SLp0 is a convex set, and Lp0 is a C 1 manifold.

It is proved in [8] that Loeper’s

condition is equivalent to the MTW condition when the cost function is C 4.

In [2], Kitagawa and Guillen suggested another condition that is equivalent to the MTW

condition when the cost function is C 4.

14

Definition 3.1.2 (Quantitative quasi-convexity (QQconv)). Let x0, x1 ∈ X and define
F (p) = −c(x1, expc
if there exists a constant M ≥ 1 such that

x0(p)). Then the cost function c is said to satisfy QQconv

x0(p)) + c(x0, expc

F (tp1 + (1 − t)p0) − F (p0) ≤ M t(F (p1) − F (p0))+

(3.2)

for any p0, p1 ∈ Y ∗

x0 and for any x0, x1 ∈ X.

Like Loeper’s condition, QQconv makes sense when the cost function is only C 1 with

twisted condition. Therefore, QQconv is another synthetic expression of the MTW condition.

In fact, QQconv implies Loeper’s condition.

Lemma 3.1.3. Suppose the cost function c satisfies QQconv, Then c also satisfies Loeper’s

condition.

Proof. If the cost function c satisfies QQconv, then we have (3.2). If F (p1) ≤ F (p0), then

we have

F (tp1 + (1 − t)p0) ≤ F (p0) = max{F (p1), F (p0)}.

(3.3)

If F (p1) ≥ F (p0), then we switch the role of p1 and p0 in (3.3), and we obtain the same

inequality.

Although both Loeper’s condition and QQconv are equivalent to MTW condition when

the cost function is C 4, it is not clear if the two synthetic MTW conditions are equivalent

under weaker regularity assumptions on the cost function c. The main theorem of this chapter

shows that the two synthetic MTW conditions are equivalent under weaker assumption.

(cid:12)(cid:12)D2

xyc(x1, y1)(cid:12)(cid:12) ≤ Λ|(x1, y1) − (x0, y0)|

xyc(x0, y0) − D2

(Lip hessian)

(Lip hessian) condition with non-degeneracy implies Lipschitzness of the inverse matrix of

15

the mixed hessian(cid:12)(cid:12)(cid:12)(cid:2)D2

xyc(x0, y0)(cid:3)−1 −(cid:2)D2

xyc(x1, y1)(cid:3)−1(cid:12)(cid:12)(cid:12) ≤ Λ(cid:48)|(x0, y0) − (x1, y1)|

(3.4)

Enlarging Λ or Λ(cid:48) if necessary, we can assume Λ = Λ(cid:48).

Now we state the main theorem of this chapter.

Theorem 3.1.4 (Main theorem of Chapter 3). Let c : X × Y → R be a C 2 cost function

that satisfies (Twist), (Twist*), (Non-deg), and (Lip hessian). Suppose c satisfies Loeper’s

condition, then c also satisfies QQconv.

We give the proof of the main theorem in the next section.

3.2 Equivalence of the Synthetic MTW conditions

We start with showing that (Lip hessian) condition implies Lipschitzness of the gradient of
the function F (p) = −c(x1, expc

x0(p)) + c(x0, expc

x0(p)).

Lemma 3.2.1. For any x0, x1 ∈ X and p0, p1 ∈ Y ∗

x0, we have

|∇F (p0) − ∇F (p1)| ≤ C|x0 − x1||p0 − p1|

(3.5)

for some constant C that depends on λ and Λ

Proof. Note that F is C 1 with

∇F (p) = [−D2

yxc(x0, y)]−1(−Dyc(x1, y) + Dyc(x0, y)),

(3.6)

where y = expc

x0(p). Therefore,

∇F (p1) − ∇F (p0)

16

= [−D2
− [−D2

xyc(x0, y1)T ]−1 (−Dyc(x1, y1) + Dyc(x0, y1))
xyc(x0, y0)T ]−1 (−Dyc(x1, y0) + Dyc(x0, y0))

(3.7)

where yi = expc

x0(pi). Let L1 and L2 be the second and third line in (3.7) and let

L1(cid:48) = L1 − [−D2
L2(cid:48) = L2 + [−D2

xyc(x0, y0)T ]−1 (−Dyc(x1, y1) + Dyc(x0, y1)) ,
xyc(x0, y0)T ]−1 (−Dyc(x1, y1) + Dyc(x0, y1)) ,

so that ∇F (p1) − ∇F (p0) = L1(cid:48) + L2(cid:48). (Lip hessian) implies

|L1(cid:48)| =(cid:12)(cid:12)[−D2

xyc(x0, y1)T ]−1 + [D2

xyc(x0, y0)T ]−1(cid:12)(cid:12)|Dyc(x1, y1) − Dyc(x0, y1)|

≤ Λ|y1 − y0| × λ|x1 − x0|
≤ λ2Λ|x0 − x1||p0 − p1|.

(3.8)

To get an estimate for L2(cid:48), we use the fundamental theorem of calculus.

|L2(cid:48)|

xyc(x0, y0)T ]−1(cid:12)(cid:12)

=(cid:12)(cid:12)[−D2
(cid:12)(cid:12)(cid:12)(cid:12)(cid:90) 1
=(cid:12)(cid:12)[−D2
(cid:12)(cid:12)(cid:12)(cid:12)(cid:90) 1

×

0

0

× |−Dyc(x1, y1) + Dyc(x0, y1) + Dyc(x1, y0) − Dyc(x0, y0)|
= [−D2
×

xyc(x0, y0)T ]−1
[−D2

xyc(xs, y1)T ](q1 − q0)ds − (q1 − q0)

(cid:12)(cid:12)(cid:12)(cid:12)

xyc(xs, y0)T ]−1[−D2

xyc(x0, y0)T ]−1(cid:12)(cid:12)
xyc(xs, y0)T ]−1(cid:0)[−D2

[−D2

xyc(xs, y1)T ] − [−D2

(cid:12)(cid:12)(cid:12)(cid:12)
xyc(xs, y0)T ](cid:1) ds(q1 − q0)

where qi = −Dyc(xi, y0) and xs is the c∗-segment focused at y0. Then (Non-deg) with

17

(Lip hessian) implies

|L2(cid:48)| ≤ λ3Λ|x0 − x1||p0 − p1|.

(3.9)

Combining the two estimate (3.8) and (3.9) with ∇F (p1) − ∇F (p0) = L1(cid:48) + L2(cid:48), we obtain
the Lipschitzness of ∇F

|∇F (p1) − ∇F (p0)| ≤ C|x1 − x0||p1 − p0|

(3.10)

where C = λ2Λ + λ3Λ.

Note that (3.6) with (Non-deg) condition implies the following

|∇F (p)| ∼ |x1 − x0|.

In particular, we have a constant C1 such that

|∇F (p)| ≥ C1|x1 − x0|.

(3.11)

(3.12)

By Lemma 3.1.3, we only need to consider the case c satisfies Loeper’s condition, and show
that c satisfies QQconv. However, we do not have to consider arbitrary points p0, p1 ∈ Y ∗
x0.

We use the notation

pt = (1 − t)p0 + tp1.

Lemma 3.2.2. Suppose the cost function c satisfies Loeper’s condition. If c satisfies (3.2)
for any pair of points p0, p1 ∈ Y ∗

x0 such that p1 ∈ B+

r (p0) where

r (p0) = {p||p − p0| < r,(cid:104)p − p0,∇F (p0)(cid:105) ≥ 0},
B+

(3.13)

then c satisfies QQconv.

Proof. We divide the proof into three steps.

18

Step 1 ) Claim: We only need to consider the case F (p1) > F (p0).
If F (p1) ≤ F (p0), then we obtain (F (p1)− F (p0))+ = 0. However, by Loeper’s condition, we

have

F (pt) − F (v0) ≤ max{F (p1), F (p0)} − F (p0)

= F (p0) − F (p0) = 0 = M t(F (p1) − F (p0))+.

Hence (3.2) always holds when F (p1) ≤ F (p0), and we only need to check the case F (p1) >

F (p0).
Step 2 ) Claim: If there exist r > 0 such that (3.2) holds whenever |p1 − p0| < r, then c

satisfies QQconv.
Suppose (3.2) holds whenever |p1 − p0| < r. We choose M(cid:48) > 1
x0), and suppose
we have p0 and p1 which does not satisfy (3.2) with M(cid:48) instead of M . Note that by step
1, we can assume F (p1) > F (p0). Then by quasi-convexity of F , we have F (p1) ≥ F (pt).

r diam(Y ∗

Therefore

M(cid:48)t(F (p1) − F (p0)) < F (pt) − F (p0) ≤ F (p1) − F (p0).

(3.14)

This implies 0 < t < 1

M(cid:48) . We choose t(cid:48) ∈ (t, 1

M(cid:48) ] such that

1

t(cid:48) (F (pt(cid:48)) − F (p0)) = M(cid:48)(F (p1) − F (p0)).

(3.15)

Note that such t(cid:48) exists by the intermediate value theorem. Let q1 = pt(cid:48) and q0 = p0. Then

we have

|q1 − q0| = t(cid:48)|p1 − p0| ≤ 1

M(cid:48) diam(Y ∗

x0) < r.

(3.16)

Therefore, by assumption, we obtain

F (qs) − F (q0) ≤ M s(F (q1) − F (q0))

19

= M s(F (pt(cid:48)) − F (p0)) = M M(cid:48)st(cid:48)(F (p1) − F (p0)).

where s = t

t(cid:48) so that qs = pt. Hence, we have

F (pt) − F (p0) ≤ M M(cid:48)t(F (p1) − F (p0)).

(3.17)

r (p0).

r (p0)) ∩ Y ∗

Since M M(cid:48) does not depend on p0, p1, it shows that c satisfies QQconv.
Step 3 ) Claim: We only need to consider the case p1 ∈ B+
Suppose we have (3.2) for p1 ∈ (B+
p1 ∈ Br(p0)\ B+
r (p0).
If F (p1) ≤ F (p0), there is nothing to show by Step 1, therefore we assume F (p1) > F (p0).
x0 → R is C 1, (cid:104)p1−p0,∇F (p0)(cid:105) < 0 implies that F (pt) < F (p0) when
Since the function F : Y ∗
t is small enough. In addition, convexity of the sublevel sets of the function F implies that
there exists t(cid:48) such that F (pt(cid:48)) = F (p0), with (cid:104)p1− pt(cid:48),∇F (pt(cid:48))(cid:105) > 0. Therefore p1 ∈ B+

r (p0) = {p||p− p0| < r,(cid:104)p− p0,∇F (p0)(cid:105) < 0}. Assume p1 ∈ Br(p0)\ B+

x0. By Step 2, we only need to show (3.2) when

r (pt(cid:48)),

and we obtain

F (ps) − F (p0) = F (ps) − F (pt(cid:48))

≤ M

s − t(cid:48)
1 − t(cid:48) (F (p1) − F (pt(cid:48))) ≤ M s(F (p1) − F (p0)).

for any s ∈ [t(cid:48), 1]. If s < t(cid:48), then F (pt(cid:48)) ≤ F (p0) so that (3.2) holds.

Before we start the next proof, we introduce a notation.

(cid:26)
p(cid:12)(cid:12)(cid:104)p − p0,∇F (p0)(cid:105) ≥ 1

k

Ck,p0 =

(cid:27)

|p − p0||∇F (p0)|

.

(3.18)

Lemma 3.2.3. For any k ∈ N, there exists rk > 0 such that if p0 ∈ Y ∗
Brk(p0) ∩ Y ∗

x0, then

x0 and p1 ∈ Ck,p0 ∩

F (pt) − F (p0) ≤ 5t(F (p1) − F (p0)).

(3.19)

20

Proof. We choose rk = C1
any p ∈ Brk(p0) ∩ Y ∗
x0,

2kC where C1 is from (3.12) and C is from Lemma 3.2.1. Then, for

|∇F (p) − ∇F (p0)| ≤ C|x1 − x0||p − p0| ≤ C1
2k

|x1 − x0| ≤ 1
2k

|∇F (p0)|.

(3.20)

Let p1 ∈ (Ck,p0 ∩ Brk(p0) ∩ Y ∗

x0) \ C k

2 ,p0

, and let v = p1−p0

|p1−p0|. Then (3.20) gives

(cid:104)∇F (p1), v(cid:105) = (cid:104)∇F (p1) − ∇F (p0), v(cid:105) + (cid:104)∇F (p0), v(cid:105)

≥ − 1
2k

|∇F (p0)| +

1
k

|∇F (p0)| =

|∇F (p0)|.

1
2k

(3.21)

Note that p0 + sv ∈ (Ck,p0 ∩ Brk(p0) ∩ Y ∗
so that we can apply (3.21) with p0 + sv instead of p1. Therefore, we have

for 0 ≤ s ≤ |p1 − p0| by convexity of Y ∗
x0,

2 ,p0

F (p1) − F (p0) =

(cid:104)∇F (p0 + sv), v(cid:105)ds

0

≥ 1
2k

|∇F (p0)||p1 − p0|.

(3.22)

Moreover, we also have

(cid:104)∇F (p0 + sv), v(cid:105) = (cid:104)∇F (p0 + sv) − ∇F (p0), v(cid:105) + (cid:104)∇F (p0), v(cid:105)
|∇F (p0)|

|∇F (p0)| =

|∇F (p0)| +

≤ 1
2k

2
k

5
2k

x0) \ C k
(cid:90) |p1−p0|

for 0 ≤ s ≤ |p1 − p0|, where we have used that p1 /∈ C k

and (3.20). Therefore,

F (p0 + t(p1 − p0)) − F (p0) =

(cid:104)∇F (p0 + sv), v(cid:105)ds

2 ,p0

(cid:90) t|p1−p0|

0

≤ t|p1 − p0| × 5
2k

|∇F (p0)|.

(3.23)

21

We combine (3.22) and (3.23).

F (pt) − F (p0) ≤ 5t
2k

|p1 − p0||∇F (p0)| ≤ 5t(F (p1) − F (p0)).

Now note that rk increases as k decreases. Moreover,

Ck,p0 ∩ Brk(p0) =

∩ Brk(p0)

2i p0

(cid:104)(cid:16)C k

∞(cid:91)

i=0

(cid:17) \ C k

2i+1 ,p0

(cid:105)

.

Therefore, for any p1 ∈ Ck,p0 ∩ Brk(p0), we can repeat the proof with k replaced by k
some i.

2i for

Note that rk varies as we choose k. We will choose k later in this chapter.

Remark 3.2.4. Let ρ0 = |∇F (p0)|. Then (3.20) implies that

∀p ∈ Brk(p0) ∩ Y ∗

x0, ∇F (p) ∈ B ρ0

2k

(∇F (p0)).

(3.24)

Let ∇F (p) = ∇F (p0) + v, and consider ξ such that ξ + p0 ∈ Ck(cid:48),p0 with |ξ| = 1 where k(cid:48) ∈ N.
Then

(cid:104)ξ,∇F (p)(cid:105) = (cid:104)ξ,∇F (p0)(cid:105) + (cid:104)ξ, v(cid:105) ≥ 1

k(cid:48)|∇F (p0)| − 1

2k

|∇F (p0)|.

(3.25)

Therefore, once we fix k and k(cid:48) < 2k, we obtain (cid:104)ξ,∇F (p)(cid:105) ∼ |∇F (p0)| for any p ∈ Brk(p0)∩
x0 and ξ such that ξ + p0 ∈ Ck(cid:48),p0.
Y ∗
Remark 3.2.5. Quasi-convexity of the function F implies that if p ∈ B+
F (p) ≥ F (p0).

r (p0) ∩ Y ∗

x0, then

We introduce another notation for a cone:

(cid:26)

p(cid:12)(cid:12)(cid:104)p − p1,∇F (p0)(cid:105) ≤ − 1

k

(cid:27)

|p − p1||∇F (p0)|

.

(3.26)

Ck,p0(p1) =

22

Lemma 3.2.6. Let 4 ≤ k(cid:48) ≤ k, and suppose Brk(p0) ⊂ Y ∗
have

x0. Then, for any p1 ∈ B+

rk

(p0), we

F (pt) − F (p0) ≤ Mk(cid:48)t(F (p1) − F (p0))

where Mk(cid:48) is some constant that depends on k(cid:48)(which will be decided later).
Proof. Note that by Lemma 3.2.3, we only need to check when p1 /∈ Ck,p0. Fix p1 ∈ B+
Ck,p0. Consider the cone Ck(cid:48),p0(p1) defined in (3.26). We divide the proof into three steps.
Step 1 ) Ck(cid:48),p0(p1) ∩ Lp0 ∩ Brk(p0) (cid:54)= ∅.
Note that we have chosen k(cid:48) ≥ 4. Then

rk

(p0)\

(cid:104)(p0 − rk
2ρ0

∇F (p0)) − p1,∇F (p0)(cid:105) = −1
2

rk|∇F (p0)| − (cid:104)p1 − p0,∇F (p0)(cid:105)

≤ −1
2
≤ −1
4

rk|∇F (p0)|

(cid:12)(cid:12)(cid:12)(cid:12)p0 − rk

2ρ0

∇F (p0) − p1

(cid:12)(cid:12)(cid:12)(cid:12)|∇F (p0)|

(3.27)

where ρ0 = |∇F (p0)|. Note that we have used p1 ∈ B+
p0 − rk

(p0) in the first inequality and
∇F (p0), p1 ∈ Brk(p0) in the second inequality. (3.27) implies that the point p0 −
∇F (p0) is in the cone Ck(cid:48),p0(p1). In addition, (3.24) shows that (cid:104)∇F (p),∇F (p0)(cid:105) > 0 for

2ρ0

rk

rk
2ρ0

any p ∈ Brk(p0). Therefore,

(cid:18)

F (p0) − F

p0 − rk
2ρ0

(cid:19)

∇F (p0)

(cid:90) 0

− rk
2ρ0

=

≥ 0,

(cid:104)∇F (p0 + t∇F (p0)),∇F (p0)(cid:105) dt

2ρ0

so that the point p0 − rk
∇F (p0) is in the sublevel set SLp0. Therefore, by the intermediate
∇F (p0)] such that F (q1) = F (p0)
value theorem, there is a point q1 in the segment [p1, p0− rk
i.e. q1 ∈ Lp0. By convexity of Ck(cid:48),p0(p1) ∩ Brk(p0), q1 is also in Ck(cid:48),p0(p1) ∩ Brk(p0). This
concludes Step 1.

2ρ0

Step 2 )Utilizing convexity of SLp0.

23

Let ξ = (p1−q1)

|p1−q1| and consider pt − sξ. If we set s = t|q1 − p1|, then we have

pt − sξ = tq1 + (1 − t)p0 ∈ SLp0,∀t ∈ [0, 1].

Therefore, by intermediate value theorem, for each t ∈ [0, 1], we obtain st ∈ [0, t|q1−p1|] such
that pt − stξ ∈ Lp0. Now, up to an isometry, we can set ξ = −en, p0 = 0, and p1 = ae1 + ben
for some a, b ∈ R, a > 0. Then we can view the set {pt−stξ|t ∈ [0, 1]} as a graph of a function
g on [0, 1]. Since SLp0 is a convex set, g is a convex function. Note that st = g(at) − bt so
that st is also a convex function of t on [0, 1]. Convexity of st with s0 = 0 implies

|pt − qt| = st ≤ ts1 = t|p1 − q1|

(3.28)

where qt = pt − stξ ∈ Lp0.
Step 3 ) Estimate on the segment [qt, pt].
Note that ξ + p0 ∈ Ck(cid:48),p0 and pt − sξ ∈ Brk(p0) for s ∈ [0, st]. By Remark 3.2.4 and the
fundamental theorem of calculus, we obtain

F (pt) − F (p0) = F (pt) − F (qt)

(cid:90) st

=

0

from (3.24), and

(cid:104)∇F (qt + sξ), ξ(cid:105)ds ≤ st

2k + 1

2k

|∇F (p0)|

(3.29)

F (p1) − F (p0) = F (p1) − F (q1)

(cid:90) s1

=

0

(cid:104)∇F (q1 + sξ), ξ(cid:105)ds ≥ s1

(cid:19)

(cid:18) 1
k(cid:48) − 1

2k

|∇F (p0)|

(3.30)

from (3.25). We combine (3.29), (3.30) with (3.28) to obtain

F (pt) − F (p0) ≤ 2st|∇F (p0)|

24

≤ 2ts1|∇F (p0)| ≤ 4k(cid:48)F (p1) − F (p0).

Note that we have used
Mk(cid:48) = 4k(cid:48).

2k + 1

2k

≤ 2 and

1

k(cid:48) − 1

2k

≥ 1

2k(cid:48) . Hence, we obtain the lemma with

Lemma 3.2.6 shows that we can obtain (3.2) with a uniform constant M when we only

consider the points that stay away from the boundary. When the point p0 is close to the

boundary, however, the proof of Lemma 3.2.6 may not work. The problematic part in the
|∇F (p0)| may not be in
proof of Lemma 3.2.6 is the Step 1, because the point p0 − rk
2ρ0
Y ∗
x0. Therefore it is not clear that the point q1 and the direction ξ exist. To go around this

problem, we need to introduce another argument when p0 is close to the boundary.

The idea in Lemma 3.2.6 is to find a direction ξ so that we can view the level set Lp0 as

a convex function over the segment [p0, p1] with ξ as a vertical direction. When we can not

find such a direction ξ, we try to look at the opposite direction, and view the level set Lp1

as a function over the segment [p0, p1]. In this case, the function will be a concave function.

We use this idea in the next lemma

Lemma 3.2.7. Let p0 ∈ Y ∗
(p0) \ Ck,p0. Let k(cid:48) < k and fix ξ such that
ξ+p0 ∈ Ck(cid:48),p0 and |ξ| = 1. Suppose for any t ∈ [0, 1],∃qt ∈ Lp1∩Brk(p0) such that qt = pt+stξ
for some st ∈ R. Then

x0 and let p1 ∈ B+

rk

F (pt) − F (p0) ≤ Mk,k(cid:48)t(F (p1) − F (p0))

for some constant Mk,k(cid:48).

Proof. Up to an isometry, let ξ = −en, p0 = 0 and p1 = ae1 + ben for some a, b ∈ R, a > 0.
Then the set {qt|t ∈ [0, 1]} can be viewed as a graph of a C 1 convex function g on [0, 1]
and st = bt − g(t). Therefore, st is a concave function of t.
In addition, s1 = b − g(1),

and ae1 + g(1)en = q1 = p1 = ae1 + ben so that s1 = 0. Moreover, qt = ate1 + g(t)en and

25

F (qt) = F (p1) imply that

F (qt) = (cid:104)∇F (qt), ae1 + g(cid:48)(t)en(cid:105) = 0 ⇒ g(cid:48)(t) =

d
dt

a(cid:104)∇F (qt), e1(cid:105)
(cid:104)∇F (qt), ξ(cid:105) .

(3.31)

Note that from our choice of ξ, for any q ∈ Brk(p0),

(cid:104)∇F (q), ξ(cid:105) = (cid:104)∇F (q) − ∇F (p0), ξ(cid:105) + (cid:104)∇F (p0), ξ(cid:105)

≥ − 1
2k

|∇F (p0)| +

1

k(cid:48)|∇F (p0)|,

(3.32)

where we have used (3.24) and (3.25). Therefore we combine (3.32) and (3.24) to obtain an
upper bound for |g(cid:48)|

|g(cid:48)(t)| ≤ k(cid:48)(2k + 1)rk

2k − k(cid:48) ≤ (2k + 1)rk.

(3.33)

(3.34)

Next, we use concavity of st with s1 = 0 to obtain

|qt − pt| = st ≥ (1 − t)s0 = (1 − t)|q0 − p0|.

Now we observe that

F (pt) − F (p0) = (F (p1) − F (p0)) − (F (p1) − F (pt))
= (F (q0) − F (p0)) − (F (qt) − F (pt))

(cid:90) 1

(cid:104)∇F (p0 + (s0ξ)s), s0ξ(cid:105)ds −

(cid:104)∇F (pt + (stξ)s), stξ(cid:105)ds

0

(cid:104)∇F (p0 + (s0ξ)s) − ∇F (pt + (stξ)s), ξ(cid:105)s0ds

0

(cid:90) 1
(cid:90) 1
(cid:90) 1

0

=

=

+

(cid:104)∇F (pt + (stξ)s), ξ(cid:105)(s0 − st)ds

0

=: I1 + I2.

26

(3.24) and (3.34) imply that

I2 =

(cid:104)∇F (pt + (stξ)s), ξ(cid:105)ds × (s0 − st) ≤ 2|∇F (p0)| × ts0.

2kC and(cid:12)(cid:12) d

dtst

(3.35)

(cid:12)(cid:12) = |b−g(cid:48)(t)| ≤ 2(k+1)rk.

To estimate I1, we use Lemma 3.2.1. Recall that rk = C1

(cid:90) 1

0

I1 =

≤

≤

0

(cid:90) 1
(cid:90) 1
(cid:90) 1
(cid:90) 1

0

0

(cid:104)∇F (p0 + (s0ξ)s) − ∇F (pt + (stξ)s), ξ(cid:105)s0ds

C|x1 − x0||p0 − pt + (s0 − st)sξ|s0ds

(cid:18)

(cid:12)(cid:12)(cid:12)(cid:12)s0 − st

(cid:12)(cid:12)(cid:12)(cid:12)(cid:19)

C|x1 − x0|t

|p0 − p1| +

t
C|x1 − x0|t × 2(k + 1)rks0ds

≤
≤ 2|∇F (p0)| × ts0.

0

sξ

s0ds

(3.36)

Finally, we combine (3.36), (3.35), (3.34), and (3.32) to obtain

F (pt) − F (p0) = I1 + I2 ≤ 4|∇F (p0)| × ts0

(cid:90) s0

(cid:104)∇F (p0 + sξ), ξ(cid:105)ds

0

2kk(cid:48)
≤ 4t
2k − k(cid:48)
8kk(cid:48)
2k − k(cid:48) t(F (q0) − F (p0))
=
= Mk,k(cid:48)t(F (p1) − F (p0)).

Finally, what is left is to show that one of the cases in Lemma 3.2.3, Lemma 3.2.6, and
Lemma 3.2.7 always holds when |p1− p0| ≤ rk. To achieve this goal, we should discuss about
the boundary of Y ∗

p0. We first show local Lipschitzness of convex functions.

Lemma 3.2.8. Let g : Bl(0) → R be a bounded convex function. Then for x, y ∈ B l

2

(0), we

27

have

|g(x) − g(y)| ≤ 4(cid:107)g(cid:107)L∞(Bl)

l

|x − y|.

Proof. Let x ∈ B l

2

(0) and p be a subdifferential of g at x. Let v =

x +

l
2

v ∈ Bl(0), and

(3.37)

p
|p| be a unit vector, then

l
2
l
2

g(x +
⇒g(x +
⇒4(cid:107)g(cid:107)L∞

l

(cid:104)p, v(cid:105)
|p|

v) ≥ g(x) +
l
2
v) − g(x) ≥ l
2
≥ |p|.

Now, for any y ∈ B l

2

(0), we have

g(y) ≥ g(x) + (cid:104)p, y − x(cid:105) ≥ g(x) − |p||x − y|
⇒g(x) − g(y) ≤ |p||x − y| ≤ 4(cid:107)g(cid:107)L∞
|x − y|.

l

Note that we can change the role of x and y, and that finishes the proof.

Lemma 3.2.8 shows that the boundary ∂Y ∗

x0 is Lipschitz. However, the Lipchitz constant

can vary with x0. In the next lemma, we show that we can chose the Lipschitz constants
uniform over x0 ∈ X, and therefore, at each point on the boundary ∂Y ∗

x0, we can obtain an

interior cone which has uniform opening.

Lemma 3.2.9. There exist ρ > 0 and 0 < σ < 1 that satisfy the following : For any x0 ∈ X
and q ∈ ∂Y ∗

x0, there exists a unit vector v such that for any p0 ∈ Bρ(q) ∩ Y ∗

x0, we have

{p ∈ Bρ(q)|(cid:104)p − p0, v(cid:105) ≥ σ|p − p0|} ⊂ Y ∗
x0.

(3.38)

28

Proof. Fix y ∈ Int(Y ) and Bl(y) ⊂ Y . Then from the bi-Lipschitzness of Dxc, we obtain

(py) ⊂ −Dxc(x0, Bl(y)) ⊂ Y ∗

x0

B l
λ

(3.39)

where py = −Dxc(x0, y). Denote Hq = (py − q)⊥ + py, a hyperplane that is perpendicular
to py − q and containing py. Let Bn−1
p = {p + (py − q)s|s ≤ 0}.

(py) ∩ Hq and L−

(py) = B l

λ

Define

Dq,l =

l
λ ,q

(cid:91)

p∈Bn−1

l
λ

,q

p , Yq,l = Dq,l ∩ ∂Y ∗
L−
x0.

(py)

(3.40)

p is a ray with starting point p in the interior of Y ∗

Since L−
convexity of Y ∗
viewed as a graph of a convex function g on Bn−1
Lipschitz constant of g on Bn−1
l
2λ ,q

x0 is a singleton by
x0. Therefore, letting py = 0 and (py − q)//en up to an isometry, Yq,l can be
l,q (0) ⊂ Rn−1. Lemma 3.2.8 shows that the

(0) is bounded by

p ∩ ∂Y ∗

x0, L−

2(cid:107)g(cid:107)L∞
l/(2λ)

≤ 4λdiam(Y ∗
x0)

l

≤ 4λ2diam(Y )

l

= L.

This shows that for any p0 ∈ Dq,l/2 ∩ Y ∗
L√
|p − p0|} does not intersect with Yq, l
L2 + 1

2

(cid:26)

(cid:27)

x0, the upper Lipschitz cone {p|(cid:104)p − p0, en(cid:105) >

, the graph of g on Bn−1
1
2λ ,q

(0) :

p|(cid:104)p − p0, en(cid:105) >

L√
L2 + 1

|p − p0|

∩ Yq, l

2

= ∅.

Noting the definition of Yq, l

2

, we get the proof with ρ = l

2λ, σ = L√

L2+1

, and v = en.

Now we show that under appropriate choice of k and k(cid:48), one of the cases in Lemma 3.2.3,

Lemma 3.2.6, and Lemma 3.2.7 must hold.

Lemma 3.2.10. Let p0 ∈ Y ∗
and 1

√
1 − σ2 where ρ and σ are from Lemma 3.2.9. Suppose B rk

x0 and take k, k(cid:48) big enough so that 2 ≤ k(cid:48) < 4

(p0) ∩ ∂Y ∗

7k, 2rk < ρ,
x0 (cid:54)= ∅. If

k(cid:48) <

4

29

p1 ∈ B+

rk
4

(p0) \ Ck,p0, then

for some constant Mk,k(cid:48).

F (pt) − F (p0) ≤ Mk,k(cid:48)t(F (p1) − F (p0))

k(cid:48)|p|} for some unit vectors v

Proof. We divide the proof into three steps.
Step 1 ) Let K(v) = {p|(cid:104)p, v(cid:105) ≥ σ|p|} and K(w) = {p|(cid:104)p, w(cid:105) ≥ 1
and w such that (cid:104)v, w(cid:105) ≥ 0. In the first step, we show that K(v) ∩ K(w) (cid:54)= ∅.
WLOG, we can assume that v = en and w = ae1 + ben for some a ≥ 0. If b > 1
(cid:104)w, en(cid:105) = b ≥ 1

k(cid:48) , then
k(cid:48) and we obtain that en ∈ K(v)∩ K(w). Otherwise, denote w⊥ = −be1 + aen.
(cid:42)

k(cid:48) w + (1 − 1
k(cid:48)2 )
(cid:19)1/2

2 w⊥ ∈ K(w). Moreover,

Then we can check that

(cid:43)

(cid:18)

1

1

1
k(cid:48) w +

1 − 1
k(cid:48)2

w⊥, en

(cid:18)
(cid:18)

=

=

1
k(cid:48) b +

1
k(cid:48) b +

1 − 1
k(cid:48)2
1 − 1
k(cid:48)2

(cid:19)1/2
(cid:19)1/2

a

(1 − b2)1/2.

(cid:19) 1

2

(cid:18)

1 − 1
k(cid:48)2

The last formula is a concave function of b on [0, 1
k(cid:48) ], hence it attains minimum value at the
√
1 − σ2, we obtain that

boundary b = 0 or b = 1

> σ,

(cid:18)

(cid:19) 1
k(cid:48) . Since k(cid:48) ≥ 2 and 1

2

k(cid:48) <

4

1
k(cid:48) w +

1 − 1
k(cid:48)2
(p0) ∩ ∂Y ∗

w⊥ ∈ K(v).
x0 (cid:54)= ∅ and p1 ∈ B+

and hence
Now suppose q ∈ B rk
(p0) \ Ck,p0. Then there is a unit vector v
which satisfies (3.38). Note K(v) + p0 = {p|(cid:104)p − p0, v(cid:105) > σ|p − p0|}. We consider the cases
(cid:104)∇F (p0), v(cid:105) ≥ 0 and (cid:104)∇F (p0), v(cid:105) ≤ 0.
Step 2 ) Suppose (cid:104)∇F (p0), v(cid:105) ≥ 0. Then, by step 1, we have Ck(cid:48),p0 ∩ (K(v) + p0) (cid:54)= ∅.
Let ξ be a unit vector such that p0 + ξ ∈ Ck(cid:48),p0 ∩ (K(v) + p0). Then by Lemma 3.2.9,
pt + 1
2rk],

x0 ∩ Bρ(q), ∀t ∈ [0, 1]. Moreover, noting that |pt + sξ − p0| ≤ rk,∀s ∈ [0, 1

2rkξ ∈ Y ∗

rk
4

F (pt +

rkξ) − F (pt) =

1
2

(cid:104)∇F (pt + sξ), ξ(cid:105)ds

(cid:90) 1

2 rk

0

30

(cid:104)∇F (p0), ξ(cid:105)ds

0

2 rk

(cid:90) 1
(cid:90) 1
(cid:18) 1
2k(cid:48) − 1

2 rk

0

2k

=

+

≥

(cid:19)

|∇F (p0)|rk

(cid:104)∇F (pt + sξ) − ∇F (p0), ξ(cid:105)ds

4 and (p1 − pt) + p0 =

where we have used (3.24). In addition, p1 /∈ Ck,p0 with |p1 − p0| ≤ rk
(1 − t)(p1 − p0) + p0 ∈ Ck,p0 implies
(cid:90) 1
(cid:90) 1
(cid:90) 1
(cid:18) 1

(cid:104)∇F (pt + s(p1 − pt)), p1 − pt(cid:105)ds

(cid:104)∇F (p0), p1 − pt(cid:105)ds

F (p1) − F (pt) =

(cid:19)

+

=

0

0

0

|∇F (p0)||p1 − pt|

(cid:104)∇F (pt + s(p1 − pt)) − ∇F (p0), p1 − pt(cid:105)ds

≤

+

1
2k

k
|∇F (p0)|rk
≤ 3
8k

7k, we obtain F (pt + 1

where we have used (3.24). Noting that k(cid:48) < 4
2rkξ) > F (p1) and this
implies that there exists wt ∈ Lp1 such that wt = pt + stξ for some st. Therefore, we can
apply Lemma 3.2.7 to obtain the desired inequality.
Step 3 ) Suppose (cid:104)∇F (p0), v(cid:105) ≤ 0. Then by Step 1, Ck(cid:48),p0(p1) ∩ (K(v) + p1) (cid:54)= ∅. Let ξ be
a unit vector such that p1 − ξ ∈ Ck(cid:48),p0(p1) ∩ (K(v) + p1). Then Lemma 3.2.9 and convexity
of Y ∗
In addition,
p1 − ξ ∈ Ck(cid:48),p0(p1) implies p0 + ξ ∈ Ck(cid:48),p0. Therefore, using (3.24) again,

x0 ∩ Bρ(q) shows that pt − sξ ∈ Y ∗
x0 ∩ Bρ(q), ∀t ∈ [0, 1], ∀s ∈ [0, 1
(cid:90) 0
(cid:90) 0

F (pt) − F (pt − 1
2

(cid:104)∇F (pt + sξ), ξ(cid:105)ds

rkξ) =

2rk].

− 1

2 rk

(cid:104)∇F (p0), ξ(cid:105)ds

=

− 1

2 rk

31

(cid:90) 0
(cid:18) 1
2k(cid:48) − 1

− 1

2 rk

2k

(cid:19)

+

≥

(cid:104)∇F (pt + sξ) − ∇F (p0), ξ(cid:105)ds

|∇F (p0)|rk.

(cid:104)∇F (p0 + s(pt − p0)), pt − p0(cid:105)

(cid:104)∇F (p0), pt − p0(cid:105)ds

(cid:104)∇F (p0 + s(pt − p0)) − ∇F (p0), pt − p0)ds

(cid:19)

|∇F (p0)||pt − p0|

Moreover, p1 /∈ Ck,p0 implies that pt /∈ Ck,p0. Therefore

F (pt) − F (p0) =

=

+

0

(cid:90) 1
(cid:90) 1
(cid:90) 1
(cid:18) 1

0

0

≤

+

1
2k

k
|∇F (p0)|rk.
≤ 3
8k

2rkξ) < F (p0) which implies that ∃wt ∈ Lp0 such that
Like in Step 2, we obtain F (pt − 1
wt = pt − stξ for some st. Therefore, we can apply Step 2 and Step 3 of the proof of Lemma

3.2.6 to obtain the desired inequality.

Finally, we obtain the proof for the main theorem of this chapter.

Proof of the main theorem of chapter 3. By Lemma 3.2.2, We only need to consider the case
p1 ∈ B+

4 . If p1 ∈ Ck,p0, then we obtain (3.2) from Lemma

r (p0) for some r > 0. Let r = rk

3.2.3. Otherwise, we can apply Lemma 3.2.10 and we obtain (3.2).

32

Chapter 4

Local H¨older regularity of generated Jacobian equations

4.1 Generated Jacobian equations

Generated Jacobian equations are Monge-Amp`ere type equations of the form

det(D2φ(x) − A(x, Dφ(x), φ(x))) = ψ(x, Dφ(x), φ(x))

(GJE)

where A(x, p, u) = D2
xxG(x, T (x, p, u), Z(x, p, u)) is a matrix valued function. The matrix
valued function A has an extra dependency on u compared to the matrix valued function

from the c-Monge-Amp`ere equation. In fact, the c-Monge-Amp`ere equation is a special case

of generated Jacobian equations. It is easy to see that we can obtain the c-Monge-Amp`ere
equation by setting G(x, y, z) = −c(x, y) − z.

Generated Jacobian equations have an application in some geometric optic problem. For

example, a generated Jacobian equation was derived in [4] for the near field refractor case

and in [5] for the reflector shape design. Like equation (2.10) of c-Monge-Amp`ere equation,

the generated Jacobian equations are derived from the following equations which is called

Prescribed Jacobian Equation (PJE):

det (Dx(T (x, Dφ(x), φ(x)))) = ψ(cid:48)(x, Dφ(x), φ(x)).

(PJE)

We can derive generated Jacobian equations from (PJE) if there are functions G and Z that

satisfy

 DxG(x, T (x, p, u), Z(x, p, u)) = p

G(x, T (x, p, u), Z(x, p, u)) = u

.

33

Like the optimal transportation problem, the second boundary conditions on (PJE) can be

defined using two probability measures µ and ν:

T (·, Dφ(·), φ(·))(cid:93)µ = ν.

In case of generated Jacobian equations, the above condition can be written in terms of the

G-subdifferential. Therefore, we define the weak solutions of generated Jacobian equations

using the G-subdifferentials.

The main theorem in this chapter is local H¨older regularity of solutions to (GJE). This result

is proved by Loeper in [8] for the c-Monge-Amp`ere equation case. we generalize the result

in [8] to generated Jacobian equation case. Obtaining the general proof for the local H¨older

regularity is not trivial because of the extra non-linearity that comes from the dependency
of the matrix valued function A on the scalar variable u. We discuss this in the next section.

4.2 Structure of generated Jacobian equation

We add some conditions on the generating function G.

G ∈ C 4(X × Y × R),

DzG < 0.

(Regular)

(G-mono)

The (Regular) condition is imposed on the set X × Y × R for simplicity. However, there are
some examples of the generating functions which are not defined on whole X × Y × R, for

example, see [7]. Since the argument in this chapter is local, the result of this chapter can
be applied to the cases when the generating function is not defined on whole X × Y × R.
From (G-mono), we see that there exists a function H : X × Y × R → R such that

G(x, y, H(x, y, u)) = u.

(4.1)

34

The implicit function theorem implies that H ∈ C 4 and (G-mono) implies

DuH < 0.

(H-mono)

We also need some conditions on the generating function G which corresponds to (Twist)

and (Non-deg) conditions in the optimal transportation problem. However, in contrast to

the optimal transportation problem, the structural conditions do not necessarily hold on the
whole domain X × Y × R, but the structural conditions hold on a subset g of X × Y × R.
Therefore, we assume that there exists a set g ⊂ X × Y × R such that

g is relatively open with respect to X × Y × R

(DomOpen)

and we assume the following:

(y, z) (cid:55)→ (DxG(x, y, z), G(x, y, z)) is injective on gx.

(cid:18)

x (cid:55)→ −DyG
DzG
xyG − D2
D2

det

(x, y, z) is injective on gy,z.
(cid:54)= 0 on g.

xzG ⊗ DyG
DzG

(cid:19)

(G-twist)
(G∗-twist)

(G-nondeg)

where gx = {(y, z)|(x, y, z) ∈ g} and gy,z = {x|(x, y, z) ∈ g}. We denote

E = D2

xyG − D2

xzG ⊗ DyG
DzG

.

The conditions (G-twist), (G∗-twist), and (G-nondeg) can be written in terms of H
instead of G. In fact, we can see that (G-twist) and (G∗-twist) are symmetric conditions

like in optimal transportation case by writing the conditions in terms of H. Let gx,y =
{z|(x, y, z) ∈ g} ⊂ R and define h by

hx,y = G(x, y, gx,y),

(4.2)

35

h = {(x, y, u)|u ∈ hx,y}.

(4.3)

We denote hx,u = {y|(x, y, u) ∈ h} and hy = {(x, u)|(x, y, u) ∈ h}. Note that (DomOpen)

implies

h is relatively open with respect to X × Y × R.

(DomOpen*)

(G-twist), (G∗-twist), and (G-nondeg) become (H-twist), (H∗-twist), and (H-nondeg) re-

spectively when we rewrite the conditions in terms of H.

(x, y, u) is injective on hx,u,

y (cid:55)→ − DxH
DuH

(cid:18)

(x, u) (cid:55)→ (DyH(x, y, u), H(x, y, u)) is injective on hy,

det

yxH − D2
D2

yuH ⊗ DxH
DuH

(cid:54)= 0 on h.

(cid:19)

(H∗-twist)

(H-twist)

(H-nondeg)

The conditions (G-twist) and (G∗-twist) allow us to define the inverse maps of the func-

tions in (G-twist) and (G∗-twist).

Definition 4.2.1. We define the maps expG

x,u and Zx by

 DxG(x, expG

G(x, expG

x,u(p), Zx(p, u)) = p

x,u(p), Zx(p, u)) = u

.

We call expG

x,u the G-exponential map with focus (x, u). We define another map expG∗

y,z by

−DyG
DvG

(expG∗

y,z(q), y, z) = q.

We call expG∗

y,z the G∗-exponential map with focus (y, z).

36

Remark 4.2.2. The G-exponential map can be also defined in the following way

− DxH
DuH

(x, expG

x,u(p), u) = p.

Note that by the implicit function theorem, the functions expG

x,u, Zx and expG∗

y,v are C 3 on

the domain of each function. In fact, computing the derivative of G-exponential map expG
x,u

shows that

x,u(p) = E−1(cid:0)x, expG

x,u(p), Zx(p, u)(cid:1)

DpexpG

where E is the matrix defined above.

Now we impose one more condition on the generating function G which corresponds

to (A3s) condition of optimal transportation problem. We first define the Tru tensor of

the generating function G. The Tru tensor generalize the MTW tensor of the optimal

transportation problem. The Tru Tensor of the generating function G was introduced by

Trudinger in [10]. The Tru tensor is a (2,2)-tensor of the form

Tru(x, p, u) = D2

ppA(x, p, u)

where A(x, p, u) = D2

xxG(x, expG

x,u(p), Zx(p, u)) is a matrix valued function. We impose a

sign condition on this Tru tensor, which we call (G3s).

M T W [ξ, ξ, η, η] > 0 for any ξ ⊥ η.

(G3s)

In addition to the conditions that we have imposed on the generating function G, We

also need to impose some conditions about convexity of the domains X and Y .

Definition 4.2.3. We define the sets g∗

y,z and h∗

x,u by

y,z = −DyG
g∗
DzG

(gy,z, y, z),

(4.4)

37

x,u = − DxH
h∗
DuH

(x, hx,u, u).

(4.5)

X is said to be G-convex if g∗
if h∗

x,u is convex for any (x, u) ∈ X × R.

y,z is convex for any (y, z) ∈ Y ×R and Y is said to be G∗-convex

We assume that the sets X and Y satisfy G-convex and G∗-convex respectively.

X is G-convex,
Y is G∗-convex.

(hDomConv)

(vDomConv)

Definition 4.2.4. For x ∈ X and u ∈ R, let y0, y1 ∈ hx,u. Let pi = − DxH
DuH

(x, yi, u). The

G-segment that connects y0 and y1 with focus (x, u) is the image of [p0, p1] under the map

expG

x,u:

{expG

x,u((1 − θ)p0 + θp0)|θ ∈ [0, 1]}.

For y ∈ Y and z ∈ R, let x0, x1 ∈ gy,z and let qi = −DyG
DzG
connects x0 and x1 with focus (y, z) is the image of [q0, q1] under the map expG∗
y,z:

(xi, y, z). The G∗-segment that

{expG∗

y,z((1 − θ)q0 + θq0)|θ ∈ [0, 1]}.

Definition 4.2.5. A function φ : X → R is called G-convex if, for any x0 ∈ X, there exist
y0 ∈ Y and w0 ∈ R such that

φ(x0) = G(x0, y0, z0),
φ(x) ≥ G(x, y0, z0).

Definition 4.2.6 (G-subdifferential). Let φ : X → R be a G-convex function. The G-

38

subdifferential of φ at x0 ∈ X is defined by

y0 ∈ Y

∂Gφ(x0) =

(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) φ(x) ≥ G(x, y0, H(x0, y0, φ(x0))),

(x0, y0, φ(x0)) ∈ h

 .

Proposition 4.2.7. If a G-affine function G(·, y0, z0) supports a G-convex function φ at x0

locally,

φ(x0) = G(x0, y0, z0),

φ(x) ≥ G(x, y0, z0) on some neighborhood of x0,

and if (x0, y0, z0) ∈ g, then y0 ∈ ∂Gφ(x0).

This is proved in [3], but under extra conditions which are called (unif) and (nice).

There exist a, b ∈ R such that [a, b] ⊂ hx,y,

The solution φ is bounded by a and b, : a < φ < b.

(unif)

(nice)

In [3], these condition are used to check that the G-exponential maps they used in the proof
are well-defined. In addition, compactness of the set X × Y × [a, b] ensures that the norms of

derivatives of the generating function are bounded. However, we can weaken the conditions

(unif) and (nice) by replacing the constants a and b with some continuous functions a(x, y)

and b(x, y).

There exist continuous functions a, b : X × Y → R

such that [a(x, y), b(x, y)] ⊂ hx,y,

(unifw)

The solution φ satisfies a(x, y) < φ(x) < b(x, y)

39

for any y ∈ ∂Gφ(x).

(nicew)

With these conditions (unifw) and (nicew), the G-exponential maps used in [3] are still

well-defined, and we get compact sets

Φ = {(x, y, u)|u ∈ [a(x, y), b(x, y)]} (cid:98) h,

Ψ = {(x, y, z)|z ∈ H(x, y, [a(x, y), b(x, y)])} (cid:98) g.

On these compact sets Φ and Ψ, we can bound the norms of derivatives of the generating

function G and the function H. Hence Proposition 4.2.7 is still true under the conditions
(unifw) and (nicew). In addition, since X×Y ×[min a, max b] is compact, we have a constant

β > 0 such that

DzG < −β

(4.6)

on X × Y × [min a, max b].
Remark 4.2.8. Let S ⊂ h be a compact set. Then (G-nondeg) implies that we have a constant

Ce that depends on S such that

≤ (cid:107)E(cid:107) ≤ Ce

1
Ce

on S where (cid:107)E(cid:107) is the operator norm of E. This implies that the G-exponential map expG

x,u

is Ce-Lipschitz :

1
Ce

|p1 − p0| ≤ |expG

x,u(p1) − expG

x,u(p0)| ≤ Ce|p1 − p0|

(4.7)

when (x, expG

x,u(pθ), u) ∈ S for any θ ∈ [0, 1] where pθ = (1 − θ)p0 + θp1. Also, compactness

of S with (G3s) implies that we have a constant α > 0 that depends on S such that

Tru[ξ, ξ, η, η] > α|ξ|2|η|2, ∀ξ ⊥ η

(4.8)

40

on S.

Proposition 4.2.9. The subdifferential of φ at a point x is a closed subset of Y , and is

compactly contained in hx,φ(x) :

∂Gφ(x) (cid:98) hx,φ(x).

Proof. We first show that the G-subdifferential ∂Gφ(x) is closed. Suppose y ∈ ∂Gφ(x), then
∃yi ∈ ∂Gφ(x) such that limi→∞ yi = y. (unifw) and (nicew) implies

φ(x) ∈

(a(x, yi), b(x, yi))

∞(cid:92)
⊂(cid:104)

i=1

⊂ [sup a(x, yi), inf b(x, yi)]

lim
i→∞ a(x, yi), lim

i→∞ b(x, yi)

(cid:105)

= [a(x, y), b(x, y)] ⊂ hx,y.

Therefore, (x, y, φ(x)) ∈ h. In addition, from Definition 4.2.6,

G(x(cid:48), yi, H(x, yi, φ(x))) ≤ φ(x(cid:48)),∀x(cid:48) ∈ X.

Taking i → ∞, we obtain

G(x(cid:48), y, H(x, y, φ(x))) ≤ φ(x(cid:48)),∀x(cid:48) ∈ X.

Hence, y ∈ ∂Gφ(x) and the G-subdifferential at x is closed, and therefore compact. Noting

that the set hx,φ(x) is open, we obtain the desired result.

Proposition 4.2.10. Let φ be a G-convex function with (nicew). Let x ∈ X, then for 
 > 0,
there exists δ > 0 such that if |x(cid:48) − x| ≤ δ, then

∂Gφ(x(cid:48)) ⊂ N
(∂Gφ(x)).

41

Proof. Suppose the proposition is not true. Then there exist sequences xi ∈ X and yi ∈
∂Gφ(xi) such that xi → x as i → ∞, but yi /∈ N
 (∂Gφ(x)) for any i. Since Y is compact,
we can assume that yi → y for some y ∈ Y . Then y /∈ N
 (∂Gφ(x)) and (nicew) implies
(x, y, φ(x)) ∈ hx,φ(x). Since yi ∈ ∂Gφ(xi), we have

φ(x(cid:48)) ≥ G(x(cid:48), yi, H(xi, yi, φ(xi))),∀x(cid:48) ∈ X.

Taking i → ∞, we obtain

φ(x(cid:48)) ≥ G(x(cid:48), y, H(x,y, φ(x))),∀x(cid:48) ∈ X.

Hence y ∈ ∂Gφ(x), which contradicts to y /∈ N
 (∂Gφ(x)).

Now we define the weak solutions to generated Jacobian equation. Like in the optimal

transportation problem, we define two weak solutions.

Definition 4.2.11. Let φ : X → R be a G-convex function. Then

1. φ is called a weak Alexandrov solution to (GJE) if

µ(A) = ν(∂Gφ(A)),∀A ⊂ X.

2. φ is called a weak Brenier solution to (GJE) if

ν(B) = µ(∂Gφ−1(B)),∀B ⊂ Y.

Now we state the main theorem of this chapter.

Theorem 4.2.12 (Main theorem of Chapter 4). Suppose X and Y are compact domains in
Rn and let let µ and ν be probability measures on X and Y respectively. Let G : X × Y ×
R → R be the generating function satisfying (Regular), (G-mono), (G-twist), (G∗-twist),

42

(G-nondeg), (G3s), and (unifw). Assume also that X and Y satisfy (hDomConv) and
(vDomConv) and the target measure ν is bounded away from 0 and ∞ with respect to the

Lebesgue measure on Y . Let φ be a weak Alexandrov solution to equation (GJE) that satisfies

(nicew). Then we have the following:

1.

2.

If there exist p ∈ (n,∞] and Cµ such that µ(Br(x)) ≤ Cµrn(1− 1
for all r ≥ 0, x ∈ X, then φ ∈ C 1,σ
loc (X).
If there exist f : R+ → R+ such that lim
r→0
f (r)rn(1− 1
Here, σ = ρ

n ) for all r ≥ 0, x ∈ X, then φ ∈ C 1

f (r) = 0 and µ(Br(x)) ≤

4n−2+ρ where ρ = 1 − n
p .

loc(X).

p )

4.3 Quantitative Loeper’s condition

Before we start this section, we decide some notations. xm is a point in X, u is a real number,
and y0, y1 ∈ hxm,u. We denote the G-segment that connects y0 and y1 with focus (xm, u)
by yθ and we denote zθ = H(xm, yθ, u). Let pθ = DxG(xm, yθ, zθ). Then note that we have
pθ = (1 − θ)p0 + θp1.

In this section, we will assume that we have a compact set S (cid:98) h such that (xm, yθ, u) ∈ S.

Then by Remark 4.2.8, we obtain constants Ce and α which depend on S in the remark.

Lemma 4.3.1. For some constant C1 that depend on the C 3 norm of G, C 1 norm of H,

xxG(xm, yθ, zθ) − D2

xxG(xm, yθ(cid:48), zθ(cid:48))(cid:1) [ξ, ξ](cid:12)(cid:12) ≤ C1|θ − θ(cid:48)||p1 − p0||ξ|2

(4.9)

and Ce, we have

(cid:12)(cid:12)(cid:0)D2

Proof.

(cid:107)D2
≤ (cid:107)D3
≤ ((cid:107)D3

xxG(xm, yθ, H(xm, yθ, u)) − D2

xxG(xm, yθ(cid:48), H(xm, yθ(cid:48), u))(cid:107)

xxyG(cid:107)|yθ − yθ(cid:48)| + (cid:107)D3
xxyG(cid:107) + (cid:107)D3

xxzG(cid:107)(cid:107)DyH(cid:107)|yθ − yθ(cid:48)|

xxzG(cid:107)(cid:107)DyH(cid:107))Ce|θ − θ(cid:48)||p1 − p0|.

43

We set C1 = ((cid:107)D3

xxyG(cid:107) + (cid:107)D3

xxzG(cid:107)(cid:107)DyH(cid:107))Ce.

Lemma 4.3.2. Let ξp = Projp1−p0(ξ), where Projp is the orthogonal projection on to p.

Then for some constants ∆1 and ∆2 which depend on α, and the C 4 norm of G, we have

xxG(xm, yθ, zθ)[ξ, ξ]

D2

≤(cid:0)(1 − θ)D2

xxG(xm, y0, z0) + θD2

xxG(xm, y1, z1)(cid:1) [ξ, ξ]

+ θ(1 − θ)|p1 − p0|2(−∆1|ξ|2 + ∆2|ξp|2).

Proof. Define Fξ : [0, 1] → R by

Fξ(θ) = D2

xxG(xm, yθ, zθ)[ξ, ξ].

Let ξ(cid:48) = ξ − ξp so that ξ(cid:48) ⊥ ξp. Then (4.8) implies

F(cid:48)(cid:48)
ξ(cid:48)(θ) ≥ α|p1 − p0|2|ξ(cid:48)|2,

which shows that Fξ(cid:48) is uniformly convex. Therefore, we obtain

Fξ(cid:48)(θ) ≤ θFξ(cid:48)(1) + (1 − θ)Fξ(cid:48)(0) − 1
2

α|p1 − p0|2|ξ(cid:48)|2θ(1 − θ).

(4.10)

Let Gξ = Fξ − Fξ(cid:48). Then

ξ (θ) = F(cid:48)(cid:48)
G(cid:48)(cid:48)
= D2

ξ(cid:48)(θ)

ξ (θ) − F(cid:48)(cid:48)
ppA[ξ, ξ, p1 − p0, p1 − p0] − D2
ppA[ξ(cid:48), ξp, p1 − p0, p1 − p0] + D2

= 2D2

ppA[ξ(cid:48), ξ(cid:48), p1 − p0, p1 − p0]

ppA[ξp, ξp, p1 − p0, p1 − p0]

44

where D2

ppA is evaluated at (xm, yθ, zθ). We bound |ξ(cid:48)| by |ξ| to obtain

Gξ(θ) ≤ θGξ(1) + (1 − θ)Gξ(0) +

|D2

ppA||p1 − p0|2|ξ||ξp|θ(1 − θ).

3
2

(4.11)

We combine (4.10) and (4.11) to obtain

D2
xxG(xm, yθ, zθ)[ξ, ξ]
= Gξ + Fξ(cid:48)
≤ θGξ(1) + (1 − θ)G(0)

We use the weighted Young’s inequality in the last line of above equation.

(cid:18)3

(cid:19)

|D2

ppA| + α

|ξ||ξp| ≤ α
4

|ξ|2 + α−1

|D2

ppA| + α

(cid:19)2 |ξp|2.

2

Then we obtain

3
2

|D2

ppA||p1 − p0|2|ξ||ξp|θ(1 − θ)

+
+ θFξ(cid:48)(1) + (1 − θ)Fξ(cid:48)(0) − 1
2

α|p1 − p0|2|ξ(cid:48)|2θ(1 − θ)

xxG(xm, y1, z1)[ξ, ξ] + (1 − θ)D2
|D2
xxG(xm, y1, z1)[ξ, ξ] + (1 − θ)D2
xxG(xm, y0, z0)[ξ, ξ]
|ξ||ξp|
ppA| + α
|D2

(cid:19)
xxG(xm, y0, z0)[ξ, ξ]
ppA||ξ||ξp|
(cid:19)

= θD2
+ θ(1 − θ)|p1 − p0|2
≤ θD2
+ θ(1 − θ)|p1 − p0|2

(cid:18)
(cid:18)

(cid:18)3

|ξ(cid:48)|2 +

|ξ|2 +

−α
2

(cid:19)

3
2

.

−α
2

2

(cid:18)3

2

D2
xxG(xm, yθ, zθ)[ξ, ξ]
≤ θD2

xxG(xm, y1, z1)[ξ, ξ] + (1 − θ)D2
|ξ|2 + α−1

+ θ(1 − θ)|p1 − p0|2

(cid:32)

(cid:18)3

−α
4

xxG(xm, y0, z0)[ξ, ξ]

|D2

ppA| + α

2

(cid:19)2 |ξp|2

(cid:33)

.

45

Therefore, we obtain the inequality with constants ∆1 = α

4 and ∆2 = α−1(cid:0) 3

2|D2

ppA| + α(cid:1)2.

The next lemma is the quantitative version of Loeper’s condition. We will use (G3s)

condition through Lemma 4.3.3 later.

Lemma 4.3.3. Define φ(x) : X → R by

φ(x) = max{G(x, y0, z0), G(x, y1, z1)}

Then we have the quantitative Loeper’s condition :

φ(x) ≥ G(x, yθ, zθ) + δ0θ(1 − θ)|y1 − y0|2|x − xm|2 − γ|x − xm|3

(4.12)

for any 
 ∈ (0, 1

2) and θ ∈ [
, 1 − 
] and |x − xm| ≤ C
 for some constants δ0, γ, C.

Proof. Note that the Taylor expansion theorem yields

G(x, yi, zi) =u + (cid:104)DxG(xm, yi, zi), (x − xm)(cid:105)

xxG(x, yi, zi)[x − xm, x − xm] + o(|x − xm|2).
D2

+

1
2

Therefore,

φ(x) ≥ θG(x, y0, z0) + (1 − θ)G(x, y1, z1)

= u + (cid:104)θp1 + (1 − θ)p0, x − xm(cid:105)

xxG(x, y0, z0) + (1 − θ)D2

(cid:0)θD2

1
2

+
+ o(|x − xm|2).

xxG(x, y1, z1)(cid:1) [x − xm, x − xm]

46

We apply Lemma 4.3.2 to obtain

φ(x) ≥u + (cid:104)θp1 + (1 − θ)p0, x − xm(cid:105) +

θ(1 − θ)|p1 − p0|2(cid:0)−∆1|x − xm|2 + ∆2|(x − xm)p|2(cid:1)

xxG(xm, yθ, zθ)[x − xm, x − xm]
D2

1
2

− 1
2
+ o(|x − xm|2).

(4.13)

Since (4.13) is true for any θ ∈ [0, 1], we can replace θ with θ(cid:48) in (4.13). Let us call the
inequality that we obtain from (4.13) by replacing θ with θ(cid:48) (4.13’). We now add and

subtract the right hand side of the inequality (4.13) to the right hand side of (4.13’), and

rearrange some terms to obtain

xxG(xm, yθ, zθ)[x − xm, x − xm]
D2

θ(1 − θ)∆2|p1 − p0|2|(x − xm)p|2

xxG(xm, yθ, zθ)(cid:1) [x − xm, x − xm]

(4.14)

φ(x) ≥u + (cid:104)θp1 + (1 − θ)p0, x − xm(cid:105) +

1
2
∆1θ(1 − θ)|p1 − p0|2|x − xm|2

+
+ (θ(cid:48) − θ)(cid:104)p1 − p0, x − xm(cid:105) − 1
2

1
2

xxG(xm, yθ(cid:48), zθ(cid:48)) − D2

(cid:0)D2

+

+

+

1
2
1
2
1
2

∆1 ((θ(cid:48)(1 − θ(cid:48)) − θ(1 − θ))|p1 − p0|2|x − xm|2
∆2 ((θ(1 − θ) − θ(cid:48)(1 − θ(cid:48)))|p1 − p0|2|(x − xm)p|2 + o(|x − xm|2).

Let Li be the i-th line of the right hand side of (4.14). Note that by definition of (x − xm)p,
we have |p1 − p0||(x − xm)p| = |(cid:104)p1 − p0, x − xm(cid:105)|. Therefore, we can rewrite the third line

(cid:18)

θ(cid:48) − θ − 1
2

L3 as

L3 =

We choose

(cid:19)

θ(1 − θ)∆2(cid:104)p1 − p0, x − xm(cid:105)

(cid:104)p1 − p0, x − xm(cid:105).

θ(cid:48) = θ +

1
2

θ(1 − θ)∆2(cid:104)p1 − p0, x − xm(cid:105)

(4.15)

so that we have L3 = 0. To ensure θ(cid:48) ∈ [0, 1], we first assume that θ is away from 0 and 1,

47

i.e. we assume θ ∈ [
, 1 − 
] for 
 > 0. Then we make the second term in (4.15) small by

assuming

|x − xm| ≤

4


∆2|p1 − p0| ≤




θ(1 − θ)∆2|p1 − p0|.

Then we obtain that θ(cid:48) ∈ [0, 1] and L3 = 0. We apply Lemma 4.3.1 and (4.15) on the forth

line L4 of (4.14) to obtain

xxG(xm, yθ(cid:48), zθ(cid:48)) − D2

xxG(xm, yθ, zθ)(cid:1) [x − xm, x − xm]

(cid:0)D2

L4 =

1
2

≥ −C1|θ − θ(cid:48)||p1 − p0||x − xm|2
≥ −C1
2
≥ −C1
8

θ(1 − θ)∆2|p1 − p0|2|x − xm|3
∆2|p1 − p0|2|x − xm|3.

For the fifth line L5 and sixth line L6, note that (4.15) implies

θ(cid:48)(1 − θ(cid:48)) − θ(1 − θ) = (θ − θ(cid:48))(θ + θ(cid:48) − 1)

= −1
2

θ(1 − θ)∆2(cid:104)p1 − p0, x − xm(cid:105)(θ + θ(cid:48) − 1),

so that we can obtain

|L5| =(cid:12)(cid:12)∆1 (θ(cid:48)(1 − θ(cid:48)) − θ(1 − θ))|p1 − p0|2|x − xm|2(cid:12)(cid:12)

θ(1 − θ)(θ + θ(cid:48) − 1)∆1∆2|p1 − p0|3|x − xm|3
∆1∆2|p1 − p0|3|x − xm|3,

≤ 1
2
≤ 1
8

|L6| =(cid:12)(cid:12)∆2 (θ(1 − θ) − θ(cid:48)(1 − θ(cid:48)))|p1 − p0|2|(x − xm)p|2(cid:12)(cid:12)

θ(1 − θ)(θ + θ(cid:48) − 1)(∆2)2|p1 − p0|3|x − xm|3
(∆2)2|p1 − p0|3|x − xm|3.

≤ 1
2
≤ 1
8

48

(4.16)

(4.17)

(4.18)

We combine (4.16), (4.17), and (4.18) to bound (4.14) from below

φ(x) ≥u + (cid:104)θp1 + (1 − θ)p0, x − xm(cid:105) +
1
2
+ ∆1θ(1 − θ)|p1 − p0|2|x − xm|2
− C2(|p1 − p0|2 + |p1 − p0|3)|x − xm|3 + o(|x − xm|2)

xxG(xm, yθ, zθ)[x − xm, x − xm]
D2

(4.19)

where C2 depends on C1, ∆1, and ∆2. We apply Taylor’s theorem on the first line of (4.19)
to obtain G(x, yθ, zθ) with o(|x − xm|2) term. Note that the little o term o(|x − xm|2) is at
least O(|x− xm|3) because the generating function G is C 4. Therefore, we can put |x− xm|3
term in place of o(|x − xm|2), and we obtain

φ(x) ≥G(x, yθ, zθ) + ∆1θ(1 − θ)|p1 − p0|2|x − xm|2

(cid:0)1 + |p1 − p0|2 + |p1 − p0|3(cid:1)|x − xm|3

− C2

possibly taking larger value for C2. Finally, we bound |p1 − p0| by Cediam(Y ) from above
and 1
Ce

|y1 − y0| from below to obtain

φ(x) ≥ G(x, yθ, zθ) + δ0θ(1 − θ)|y1 − y0|2|x − xm|2 − γ|x − xm|3

where δ0 = ∆1
C2
e

and γ = C2 (1 + C 2

e diam(Y )2 + C 3

e diam(Y )3).

Remark 4.3.4. Lemma 4.3.3 implies G-convexity of G-subdifferentials of a G-convex function
φ with respect to (x, φ(x)). Suppose y0, y1 ∈ ∂Gφ(xm), then G(x, yi, zi) supports φ at xm

where zi = H(xm, yi, φ(xm)). Let yθ be the G-segment connecting y0 and y1 with respect to
(xm, φ(xm)). Fix θ and 
 such that θ ∈ [
, 1 − 
]. Then Lemma 4.3.3 shows that we have

(4.3.3), which implies that G(x, yθ, zθ) is a G-affine function that supports φ locally. Then
Proposition 4.2.7 shows that yθ ∈ ∂Gφ(xm).

49

4.4 G-convex functions

Proposition 4.4.1. Let φ be a G-convex function that satisfies (nicew). Then φ is semi

convex:

φ(xt) ≤ (1 − t)φ(x0) + tφ(x1) +

t(1 − t)(cid:107)D2

xxG(cid:107)|x0 − x1|2

1
2

(4.20)

where xt = (1 − t)x0 + tx1.

Proof. Since φ is G-convex, we have y ∈ Y and z ∈ R such that (xt, y, z) ∈ Ψ and

φ(xt) = G(xt, y, z),

φ(x) ≥ G(x, y, z),∀x ∈ X.

Moreover, we have

G(x, y, z) ≥ φ(xt) + (cid:104)pt, x − xt(cid:105) − 1
2

(cid:107)D2

xxG(cid:107)|x − xt|2

where pt = DxG(xt, y, z). Evaluate this at x = x0 and x = x1 and add them with weight
(1 − t) and t respectively.

(1 − t)φ(x0) + tφ(x1) ≥ (1 − t)G(x0, y, z) + tG(x1, y, z)

≥ φ(xt) + (cid:104)pt, (1 − t)(x0 − xt) + t(x1 − xt)(cid:105)

xxG(cid:107)(cid:0)(1 − t)|x0 − xt|2 + t|x1 − xt|2(cid:1) .

(cid:107)D2

− 1
2

(4.21)

Note that by the choice of xt, we have (1 − t)(x − xt) + t(x1 − xt) = 0 and

|x0 − xt| = t|x0 − x1| and |x1 − xt| = (1 − t)|x0 − x1|.

50

Then (4.21) becomes

(1 − t)φ(x0) + tφ(x1) ≥ φ(xt) − 1
2

t(1 − t)(cid:107)D2

xxG(cid:107)|x0 − x1|2.

Note that this inequality shows that φ(x) + 1

2(cid:107)D2

xxG(cid:107)|x|2 is convex.

When we use norms of some derivatives of G and H, we need to check that the points we
are using are in some compact subset of X × Y × R so that we can have a finite value for the

norms. If the point (x, y, u) is on the graph of the G-subdifferential of a (nicew) G-convex

function, then the point will be in the set Φ, and therefore, we can use the norms on the

set Φ (or Ψ). Later in this section, however, we will need to use some points which might

not be in the set Φ (or Ψ). In Lemma 4.4.2, we show that for each point, we can choose a
compact subset S ⊂ h which contains the points that we use later. With Lemma 4.4.2, we

will be able to obtain finite valued norms.

Lemma 4.4.2. Let φ be a G-convex function with (nicew) and let xc be an interior point of
X. Then there exists δ(xc) > 0 and S (cid:98) h such that if x0, x1 ∈ Bδ(xc)(xc), then

(xt, yθ, G(xt, y0, H(x0, y0, φ(x0)))), (xt, yθ, φ(xt)) ∈ S

(4.22)

for any xt = (1 − t)x0 + tx1, t ∈ [0, 1] and yθ, the G-segment connecting y0 and y1 with focus
(xt, φ(xt)) where y0 ∈ ∂Gφ(x0) and y1 ∈ ∂Gφ(x1).

Proof. Note that by (nicew), we have that (xc, yc, φ(xc)) is in the interior of h for any
yc ∈ ∂Gφ(xc). Therefore, we have r1, r2, r3 > 0 such that

S := Br1(xc) × (Nr2 (∂Gφ(xc)) ∩ Y ) × (φ(xc) − r3, φ(xc) + r3) (cid:98) h,

(4.23)

that is, S is compact and S is contained in the interior of h. Therefore, we obtain the

51

constant Ce from Remark 4.2.8. We define

Gφ(xc) =(cid:0)expG

∂∗

xc,φ(xc)

(cid:1)−1

(∂Gφ(xc)).

Note that ∂∗

Gφ(xc) is convex by Remark 4.3.4. From Remark 4.2.8, we obtain

N r2

Gφ(xc)) ∩ h∗
(∂∗

Ce

Noting that DxG(x,·, H(x,·, u)) =(cid:0)expG

x,u

xc,φ(xc) ⊂(cid:0)expG

(cid:1)−1
(cid:1)−1 and the map

xc,φ(xc)

(Nr2 (∂Gφ(xc)) ∩ Y ).

(x, y, u) (cid:55)→ DxG(x, y, H(x, y, u))

is uniformly continuous on S, there exist δx, δu > 0 such that if |x−xc| < δx and |u−φ(xc)| <

δu, then

|DxG(x, y, H(x, y, u)) − DxG(xc, y, H(xc, y, φ(xc)))| <

r2
4Ce

for any y ∈ Nr2 (∂Gφ(xc)) ∩ Y . Hence, for any y ∈ Nr2 (∂Gφ(xc)) ∩ Y such that

(cid:1)−1

(cid:0)expG
(cid:0)expG

xc,φ(xc)

(y) ∈ N r2

4Ce

(∂∗

Gφ(xc)) ,

(cid:1)−1

(y) ∈ N r2

2Ce

(∂∗

Gφ(xc))

(4.24)

x,u

we have

shows that

if |x − xc| < δx and |u − φ(xc)| < δu. Note that N r2

(∂∗

Gφ(xc)) is convex. Remark 4.2.8

(cid:0)expG

(cid:1)−1(cid:16)N r2

4C2
e

2Ce

(cid:17) ⊂ N r2

4Ce

xc,φ(xc)

(∂Gφ(xc))

(∂∗

Gφ(xc)) .

By Proposition 4.2.10, there exists δ1 such that if |x − xc| < δ1, then

∂Gφ(x) ⊂ N r2

4C2
e

(∂Gφ(xc)) .

52

Moreover, by continuity of G, H and φ, and (nicew) which implies that the range of φ is
compact, we have δ2 such that if |x − xc| < δ2 and |x0 − xc| < δ2, then for any y0 ∈ ∂Gφ(x0),

|G(x, y0, H(x0, y0, φ(x0))) − φ(xc)| < min{δu, r3},

|φ(x) − φ(xc)| < min{δu, r3}.

We take δ(xc) small enough so that δ(xc) ≤ min{δx, δ1, δ2, r1}. Suppose x0, x1 ∈ Bδ(xc)(xc).
Then |xt − xc| < δ(xc) and therefore

|G(xt, y0, H(x0, y0, φ(x0))) − φ(xc)| < min{δu, r3},

|φ(xt) − φ(xc)| < min{δu, r3}.

where yi ∈ ∂Gφ(xi). Hence we obtain that (xt, yi, u) ∈ S ⊂ h for i = 0, 1 where u is either
φ(xt) or G(xt, y0, H(x0, y0, φ(x0))). Moreover, ∂Gφ(xi) ⊂ N r2

(∂Gφ(xc)) for i = 0, 1, (4.24)

(cid:0)expG

(cid:1)−1

xt,φ(xt)

(yi) ∈ N r2

2Ce

(∂∗

Gφ(xc)) .

4C2
e

(cid:1)−1

implies

Since N r2

2Ce

(∂∗

Gφ(xc)) is convex, the segment connecting the two points

(cid:0)expG

xt,φ(xt)

(cid:1)−1

(y0) and (cid:0)expG

xt,φ(xt)

(y1)

(∂∗

Gφ(xc)). Therefore, the G-segment yθ connecting y0, y1 with focus (xt, φ(xt))

lies in N r2
is in r2

2Ce

2 neighborhood of ∂Gφ(xc):

yθ ∈ N r2

2

(∂Gφ(xc)) .

Therefore, (4.23) implies (4.22).

Remark 4.4.3. The constant δ(xc) depends on the modulus of continuity of φ and ∂Gφ at xc.

53

Remark 4.4.4. In the proof of Lemma 4.4.2, we have used that X is a domain in Rn so that
we can use the identification T ∗X = X × Rn. In the general manifold setting, where we do

not have this trivialization, we should choose δ(xc) small enough so that we have the local
trivialization T ∗Bδ(xc)(xc) = Bδ(xc)(xc) × Rn.

Lemma 4.4.5. Let φ be a G-convex function and let xc be an interior point. Choose x0 and
x1 such that |xi − xc| < δ(x0). Let G(x, y0, z0) and G(x, y1, z1) be G-affine functions that
support φ at x0 and x1 respectively. Then there exists xt ∈ [x0, x1] such that

G(xt, y0, z0) = G(xt, y1, z1) =: u.

We assume |y1 − y0| ≥ |x0 − x1|. Then we have

φ(xt) − u ≤ C3|x1 − x0||y1 − y0|

(4.25)

where C3 depends on the C 2 norm of G.

Proof. First of all, we show the existence of the point xt. By the definition of supporting

function, we have

G(x0, y0, z0) − G(x0, y1, z1) = φ(x0) − G(x0, y1, z1) ≥ 0,
G(x1, y0, z0) − G(x1, y1, z1) = G(x1, y0, z0) − φ(x1) ≤ 0.

Therefore xt exists by the intermediate value theorem. If t was either 1 or 0, then the left

hand side of (4.25) is 0. Otherwise, by our choice of xt and u, we have

u = G(xt, y0, z0) = G(xt, y0, H(x0, y0, φ(x0))).

Then by Lemma 4.4.2, we know (xt, yi, u) ∈ S, i = 0, 1. We use Taylor expansion on

54

G(x, yi, zi) at x = xt to obtain

φ(xi) − u ≤ (cid:104)DxG(xt, yi, zi), xi − xt(cid:105) +

(cid:107)D2

xxG(cid:107)|xi − xt|2.

1
2

(4.26)

Also, from Proposition 4.4.1,

φ(xt) − u ≤ (1 − t)(φ(x0) − u) + t(φ(x1) − u) +
≤ (1 − t)(φ(x0) − u) + t(φ(x1) − u) +

1
2
1
8

xxG(cid:107)|x1 − x0|2

t(1 − t)(cid:107)D2
(cid:107)D2

xxG(cid:107)|x1 − x0|2.

(4.27)

If (1 − t)(cid:104)DxG(xt, y0, z0), x0 − xt(cid:105) + t(φ(x1) − u) ≤ 0, then from (4.26) and (4.27), we obtain

(cid:18)

φ(xt) − u ≤ (1 − t)

(cid:19)

(cid:107)D2

xxG(cid:107)|x0 − xt|2

1
2

(cid:104)DxG(xt, y0, z0), x0 − xt(cid:105) +
xxG(cid:107)|x1 − x0|2
(cid:107)D2

(cid:107)D2

1
8

+ t(φ(x1) − u) +
(1 − t)(cid:107)D2
(cid:107)D2

xxG(cid:107)|x0 − xt|2 +
xxG(cid:107)|y1 − y0||x1 − x0|.

1
8

≤ 1
2
≤ 5
8

xxG(cid:107)|x1 − x0|2

Otherwise, since t(1 − t) ≤ 1, we have

0 ≤ (1 − t)(cid:104)DxG(xt, y0, z0), x0 − xt(cid:105) + t(φ(x1) − u)
(φ(x1) − u),
≤ 1
t

(cid:104)DxG(xt, y0, z0), x0 − xt(cid:105) +

1
1 − t

so that

φ(xt) − u ≤ (1 − t)(cid:104)DxG(xt, y0, z0), x0 − xt(cid:105) + t(φ(x1) − u)

(cid:19)

(cid:18)1

8

+

1
2

+

≤ 1
t

(1 − t)

(cid:107)D2

xxG(cid:107)|x1 − x0|2

(cid:104)DxG(xt, y0, z0), x0 − xt(cid:105) +

1
1 − t

(φ(x1) − u)

55

(cid:18)1

8

+

+

1
2

(cid:19)

(1 − t)

(cid:107)D2

xxG(cid:107)|x1 − x0|2.

Here, we use t =

|xt − x0|
|x1 − x0|, 1 − t =

|x1 − xt|
|x1 − x0|, and Taylor expansion to obtain

φ(xt) − u ≤(cid:104)DxG(xt, y0, z0), x0 − x1(cid:105) + (cid:104)DxG(xt, y1, z1), x1 − x0(cid:105)
xxG(cid:107)|x1 − x0|2.

xxG(cid:107)|x1 − xt||x1 − x0| +

(cid:107)D2

(cid:107)D2

+

5
8

1
2

Now, we use the fundamental theorem of calculus to obtain

φ(xt) − u ≤

(cid:104)DxG(xt, yθ, H(xt, yθ, u)), x1 − x0(cid:105)dθ
xxG(cid:107)|x1 − x0|2

(cid:90) 1
(cid:90) 1

0

0

=

+

d
dθ
(cid:107)D2
9
8
(cid:104)E(xt, yθ, H(xt, yθ, u))
xxG(cid:107)|x1 − x0|2
(cid:107)D2
9
+
8
(cid:19)
(cid:18)
e|y1 − y0||x0 − xt| +
≤ C 3
≤

(cid:107)D2

xxG(cid:107)

e +

C 3

9
8

9
8

yθ, x1 − x0(cid:105)

d
dθ

(cid:107)D2

xxG(cid:107)|x1, x0|2

|y1 − y0||x1 − x0|

where yθ is the G-segment connecting y0 and y1 with focus (xt, u).

Lemma 4.4.6. Let xt be as in Lemma 4.4.5. There exist l, r that depend on |x1 − x0| and
|y1 − y0| and κ such that if Nr ([x0, x1]) ⊂ X and

|y1 − y0| ≥ max{|x1 − x0|, κ|x1 − x0|1/5}

(4.28)

then , choosing x1 close to x0 if necessary, we have

Nl

yθ,|θ ∈

∩ Y ⊂ ∂Gφ(Br(xt))

(cid:18)(cid:26)

(cid:21)(cid:27)(cid:19)

(cid:20)1

,

3
4

4

56

where yθ is the G-segment connecting y0 and y1 with focus (xt, u) as in the proof of Lemma

4.4.5.

Proof. Let u be as in the proof of Lemma 4.4.5:

G(xt, y0, z0) = G(xt, y1, z1) = u.

Note that by Lemma 4.4.2, (xt, yθ, u) ∈ S (cid:98) h. By G-convexity of φ and Lemma 4.3.3, we

have

φ(x) ≥ G(x, yθ, zθ) +

δ0|y1 − y0|2|x − xt|2 − γ|x − xt|3

3
16

(4.29)

for θ ∈(cid:2) 1

4, 3

4

(cid:3), |x − xt| ≤ 1

4C (notice that we can choose in 4.3.3) where zθ = H(xt, yθ, u).

Next, we look at the G-affine function G(x, y, H(xt, y, φ(x))) on the boundary of the ball

Br(xt) to compare the value with φ. Let z(y, u) = H(xt, y, u).

G(x, y, z(y, φ(xt))) =G(x, y, z(y, φ(xt))) − G(x, yθ, z(yθ, φ(xt)))
+ G(x, yθ, z(yθ, φ(xt))) − G(x, yθ, z(yθ, u))

(4.30)

+ G(x, yθ, z(yθ, u)).

For the first line, noting that φ(xt) = G(xt, y, z(y, φ(xt))), we have

G(x, y, z(y, φ(xt))) − φ(xt)

(cid:90) 1
(cid:90) 1

0

=

=

(G(xt + s(x − xt), y, z(y, φ(xt)))) ds

d
ds
(cid:104)DxG(xt + s(x − xt), y, z(y, φ(xt))), x − xt(cid:105)ds

(4.31)

0

and similar equation holds for G(x, yθ, z(yθ, φ(xt))) − φ(xt). Therefore, we have

G(x, y, z(y, φ(xt))) − G(x, yθ, z(yθ, φ(xt)))

57

0

(cid:90) 1
(cid:90) 1
(cid:90) 1
(cid:90) 1

0

0

=

=

=

=

−DxG(xt + s(x − xt), yθ, z(yθ, φ(xt)))

 , x − xt
(cid:43)

 xt + s(x − xt), yθ + s(cid:48)(y − yθ),
 , x − xt

(cid:42) DxG(xt + s(x − xt), y, z(y, φ(xt)))
(cid:90) 1
(cid:90) 1
(cid:90) 1

(DxyG + DxzG ⊗ DyH) [y − yθ, x − xt]ds(cid:48)ds

(cid:42)DxG

z(yθ + s(cid:48)(y − yθ), φ(xt))

(cid:19)

(cid:18)

d
ds(cid:48)

0

0

ds

DxyG + DxzG ⊗ DyG
−DzG

0

0

[y − yθ, x − xt]ds(cid:48)ds

(cid:43)

ds(cid:48)ds

(4.32)

≤ C4|x − xt||y − yθ|,

where C4 depends on the C 2 norm of G and β. Note that the functions in the last integral

are evaluated at different points so that we can not simply bound the functions by Ce. For

the second line of (4.30), we use (4.25).

(G(x, yθ, z(yθ, u + s(φ(xt) − u))) ds

G(x, yθ, z(yθ, φ(xt))) − G(x, yθ, z(yθ, u))

0

=

=

(cid:90) 1
(cid:90) 1
d
ds
DzGDuH(φ(xt) − u)ds
5|φ(xt) − u|

≤ C(cid:48)
≤ C5|x1 − x0||y1 − y0|,

0

(4.33)

where C5 depends on the C 1 norm of G, β, and C3. We apply (4.32) and (4.33) to (4.30) to

obtain

G(x, y, z(y, φ(xt))) ≤G(x, yθ, z(yθ, u)) + C4|x − xt||y − yθ|

+ C5|x1 − x0||y1 − y0|.

(4.34)

58

We compare (4.29) and (4.34). If we have

C4|x − xt||y − yθ| + C5|x1 − x0||y1 − y0|
δ0|y1 − y0|2|x − xt|2 − γ|x − xt|3,
≤ 3
16

then we obtain G(x, y, z(y, φ(xt))) ≤ φ(x). Note that (4.35) is satisfied if we have

C5|x1 − x0||y1 − y0| ≤ 1
δ0|y1 − y0|2|x − xt|2,
16
δ0|y1 − y0|2|x − xt|2,
C4|x − xt||y − yθ| ≤ 1
16
δ0|y1 − y0|2|x − xt|2.

γ|x − xt|3 ≤ 1
16

(4.35)

(4.36)

(4.37)

(4.38)

Therefore, we choose

r2 =

16C5

δ0

|x1 − x0|
|y1 − y0| ,

l =

δ0

16C4

r|y1 − y0|2,

κ =

(cid:18)163γ2C5

(cid:19) 1

5

δ3
0

.

(4.39)

Note that the choice of r gives (4.36), then the choice of l gives (4.37). The choice of r and

κ gives (4.38). Therefore we obtain

(cid:0)(cid:8)yθ|θ ∈ [ 1

for y ∈ Nl

G(x, y, z(y, φ(xt))) ≤ φ(x)

4](cid:9)(cid:1) and x ∈ ∂Br(xt). Note that κ does not depend on x0 and x1.

4, 3

From the condition (4.28), we have

r2 ≤ 16C5
κδ0

|x1 − x0| 4

5 , l ≤

√

δ0C5
4C4

|x1 − x0| 1

2 diam(Y )

3

2 .

Hence, if we choose x1 so that |x1 − x0| ≤

4C 2

4 r2
2

diam(Y )3δ0C5

, we have

l ≤ r2
2

59

(4.40)

where r2 is from the proof of Lemma 4.4.2. Since G(xt, y, z(y, φ(xt))) = φ(xt), we get a local
maximum of G(·, y, z(y, φ(xt))) − φ(·) at some point xy ∈ Br(xt) with non-negative value.

Then from the proof of Lemma 4.4.2, we obtain

Nl ({yθ|θ ∈ [0, 1]}) ∩ Y ⊂ hxy,φ(xt).

If G(xy, y, z(y, φ(xt))) = φ(xy), then G(·, y, z(y, φ(xt))) is a local support of φ at xy and
Proposition 4.2.7 implies that y ∈ ∂Gφ(xy) ⊂ ∂Gφ(Br(xt)). Otherwise, we have the strict
inequality G(xy, y, z(y, φ(xt))) > φ(xy). In addition, we know φ(·) ≥ G(·, y, z(y, u)). We

define a function uy by

uy(h) = max
x∈Br(xt)

{G(x, y, z(y, h)) − φ(x)}

= max
x∈Br(xt)

{G(x, y, H(xt, y, h)) − φ(x)}.

Then uy(φ(xt)) > 0 and uy(u) ≤ 0. Moreover, since (cid:107)DvG(cid:107) and (cid:107)DuH(cid:107) are finite on Ψ and
Φ, {G(x, y, z(y,·)) − φ(x)|x ∈ Br(xt)} is a family of equicontinuous functions. Therefore,
uy is a continuous function and there exists hy ∈ [u, φ(xt)] at which we have uy(hy) = 0.
In other words, G(·, y, z(y, hy)) supports φ at some point x(cid:48) in Br(xt). From (4.40) and the
proof of Lemma 4.4.2, we have (x(cid:48), y, hy) ∈ h. Hence we obtain y ∈ ∂Gφ(Br(xt)).

4.5 Proof of the local Holder regularity

The idea of the proof of the main theorem is to use Lemma 4.4.6 to compare the volume of
Nl
Nl

(cid:3)(cid:9)(cid:1) ∩ Y and Br(xt). Therefore, we should estimate the volume of the set
(cid:3)(cid:9)(cid:1) ∩ Y .

(cid:0)(cid:8)yθ,|θ ∈(cid:2) 1
(cid:0)(cid:8)yθ,|θ ∈(cid:2) 1

4

4, 3
4, 3

4

Remark 4.5.1. Lemma 3.2.8 and the proof of Lemma 3.2.9 shows that for a compact convex
set A ⊂ Rn, there exist rA > 0 and LA such that for any p ∈ ∂A there exists a unit vector vp
such that ∂A∩BrA(p) can be written as a graph of a convex function with Lipschitz constant

60

LA in a coordinate system that has vp as a vertical axis. Moreover, we can choose rA smaller
so that for any r(cid:48) ≤ rA and p(cid:48) ∈ A∩ BrA(p), Br(cid:48)(p(cid:48))∩ A contains a conical sector of a certain
size.

(cid:40)
q ∈ Br(cid:48)(p(cid:48))(cid:12)(cid:12)(cid:104)q − p(cid:48), vp(cid:105) ≥

(cid:41)

.

LA(cid:112)L2

A + 1

Br(cid:48)(p(cid:48)) ∩ A ⊃

This implies that if r(cid:48) ≤ rA then we have

Vol(Br(cid:48)(p) ∩ A) ≥ CAVol(Br(cid:48)(p))

(4.41)

for any p ∈ A.

Lemma 4.5.2. Let A be a compact convex set and let γ : [0, 1] → A be a bi-Lipschitz curve,

that is

L|s − t| ≤ |γ(s) − γ(t)| ≤ L|s − t|

for some constants L and L. Then there exist KA and lA > 0 that depend on A, L, and L
such that for any l ≤ lA, we have

Vol(Nl (γ) ∩ A) ≥ KALln−1.

(4.42)

Before we start the proof, note that the curve γ does not have to be differentiable.

Therefore, we give the next definition.

Definition 4.5.3. Let γ : [0, 1] → Rn be a continuous curve. We define its length by

(cid:40) n(cid:88)

|γ(ti) − γ(ti−1)|(cid:12)(cid:12)a = t0 ≤ t1 ≤ ··· ≤ tn = b

(cid:41)

.

Length(γ) = sup

i=1

It is well known that this definition preserves many properties of arclength of C 1 curves.

Note that the length is finite if the curve is Lipschitz.

Proof. We assume l <

rA
4

where rA is from Remark 4.5.1. Let m ∈ N be the smallest number

61

such that 2L ≤ rAm. Then by taking rA smaller if necessary, we have m ≤ 4L
rA
we have Length(γ) ≤ L. We define

. Note that

(cid:18)

γi(t) = γ

(1 − t)

i

2m

+ t

i + 1
2m

(cid:19)

.

L
2m

L
2m

Then γi is bi-Lipschitz with Lipschitz constants
of m, we have Length(γi) ≤ rA
4

. In addition, by our choice
. Suppose Nl (γi) ∩ ∂A (cid:54)= ∅. Then we can write ∂A as a

and

graph of a Lipschitz convex function fA with Lipschitz constant LA in BrA(p) for some point
p ∈ Nl (γi) ∩ ∂A. Moreover, for any p(cid:48) ∈ Nl (γi), there are some t, s ∈ [0, 1] such that

|p(cid:48) − p| ≤ |p(cid:48) − γi(t)| + |γi(t) − γi(s)| + |γi(s) − p|

≤ rA
4

+

rA
4

+

rA
4

=

3
4

rA.

Therefore Nl (γi) ∩ A lies in the epigraph of fA in BrA(p). Then Remark 4.5.1 shows that
at each t, there exists a conical sector Secγi(t) with vertex γi(t) in Bl(γi(t)) ∩ A which is a
translation of the conical sector Sec0:

(cid:40)
p(cid:48)(cid:12)(cid:12)(cid:104)p(cid:48), vp(cid:105) ≥

(cid:41)

.

LA(cid:112)L2

A + 1

Sec0 = Bl(0) ∩

Note that the inscribed ball in this conical sector has radius L(cid:48)

Al where L(cid:48)

A is a constant that
Al(q) for some q ∈ Sec0. Therefore, for each

depends on LA so that the inscribed ball is BL(cid:48)
0 ≤ i ≤ 2m − 1, we get qi such that

NL(cid:48)

Al (γi + qi) ⊂ Nl (γ) ∩ A.

Now, if we have l < L

4m, then for any p ∈ Nl (γi) and p(cid:48) ∈ Nl (γj) where |i− j| ≥ 2, we obtain

62

that for some s, t ∈ [0, 1],

|p − p(cid:48)| ≥ |γi(t) − γj(s)| − (|p − γi(t)| + |p(cid:48) − γj(s)|)

≥ L
2m

− 2l > 0.

Therefore, NL(cid:48)
by (L(cid:48)

Al)n−1|γi(0) − γi(1)| ≥ (L(cid:48)

A)n−1L
2m ln−1 so that

Al (γi) ∩ NL(cid:48)

Al (γj) = ∅. Note that each NL(cid:48)

Al (γi) has a volume bounded below

Vol(Nl (γ) ∩ A) ≥ Vol

NL(cid:48)

Al (γi + qi)

(cid:32)2m−1(cid:91)
(cid:32)m−1(cid:91)

i=0

≥ Vol
≥ (L(cid:48)

i=0

A)n−1L
2m

(cid:33)
(cid:33)

NL(cid:48)

Al (γ2i + q2i)

ln−1 × m =

(L(cid:48)

A)n−1Lln−1.

1
2

Therefore we get the lemma with lA = min{ rA

4m} and KA = 1

2(L(cid:48)

4 , L

A)n−1.

Lemma 4.5.4. Let yθ be as in Lemma 4.4.6 and let A = h∗

xc,φ(xc). If l ≤ lA, then we have

(cid:18)

Vol

Nl

(cid:18)(cid:26)

(cid:12)(cid:12)θ ∈

yθ

(cid:20)1

,

3
4

4

(cid:21)(cid:27)(cid:19)

(cid:19)

∩ Y

≥ CV ln−1|y0 − y1|

(4.43)

where CV depends on xc, h, and Ce.

Proof. Note that θ (cid:55)→ yθ is a bi-Lipschitz curve with

|y1 − y0||θ − θ(cid:48)| ≤ |yθ − yθ(cid:48)| ≤ C 2

e|y1 − y0||θ − θ(cid:48)|.

1
C 2
e

Therefore, the reparametrized curve θ (cid:55)→ y(1−θ) 1
2
C 2
e

|y1 − y0| and 2C 2

e|y1 − y0|. Then the curve
(cid:16)

4 +θ 3

4

θ (cid:55)→ − DxH
DuH

xc, y(1−θ) 1

4 +θ 3

4

(cid:17)

, φ(xc)

is bi-Lipschitz with Lipschitz constants

63

is bi-Lipschitz in h∗

xc,φ(xc) with Lipschitz constants

L =

2
C 3
e

|y1 − y0| and L = 2C 3

e|y1 − y0|.

Moreover, the function − DxH
DuH

(xc,·, φ(xc)) = expG

xc,φ(xc)

−1

(·) is bi-Lipschitz with Lipschitz

constants 1
Ce

and Ce, so that we have

(cid:18)

N l

Ce

expG

xc,φ(xc)

⊂ expG

xc,φ(xc)

−1

(cid:18)(cid:26)
(cid:20)1
(cid:12)(cid:12)θ ∈
(cid:18)(cid:26)
(cid:12)(cid:12)θ ∈

Nl

yθ

yθ

4

−1

(cid:18)

(cid:21)(cid:27)(cid:19)(cid:19)
(cid:21)(cid:27)(cid:19)

,

3
4

,

3
4

(cid:20)1

4

∩ h∗

xc,φ(xc)

(cid:19)

∩ Y

.

Note that by (vDomConv), h∗
that depends on h∗

xc,φ(xc) such that

xc,φ(xc) is convex. From Lemma 4.5.2, we obtain a constant Kxc

(cid:18)

(cid:18)

Vol

N l

Ce

expG

xc,φ(xc)

(cid:18)(cid:26)

−1

(cid:20)1
(cid:21)(cid:27)(cid:19)(cid:19)
yθ,(cid:12)(cid:12)θ ∈
(cid:19)n−1
(cid:18) l

3
4

4

,

.

≥ Kxc

2
C 3
e

|y1 − y0|

Ce

(cid:19)

∩ h∗

xc,φ(xc)

We use bi-Lipschitzness once more to obtain

(cid:18)

Nl

(cid:18)(cid:26)

(cid:12)(cid:12)θ ∈

yθ

(cid:20)1

,

3
4

4

(cid:21)(cid:27)(cid:19)

(cid:19)

∩ Y

≥ CV ln−1|y1 − y0|,

Vol

with CV =

2Kxc
C 2n+2

e

.

Remark 4.5.5. The constant CV in Lemma 4.5.4 depends on the set hx0,φ(x0), in particular,

on rA and Lipschitz constant LA of the boundary ∂hxc,φ(xc). Therefore, if we assume that
the constants rA and LA are uniform over {h∗
x,u}(x,u)∈X×[min a,max b], the constant CV does not

depend on xc and φ(xc).

Remark 4.5.6. In the proof of the main theorem of this chapter, we use Lemma 4.4.2, Lemma

64

4.4.6, and Lemma 4.5.4. Therefore, we should choose xc in X and choose x0 and x1 close
enough to xc so that |xi − xc| satisfies the assumptions for the lemmas. In particular, we
assume |xi − xc| < δ(xc) to use Lemma 4.4.2, |xi − xc| <
4.4.6, and |xi − xc| <
C5diam(Y )3 so that l from (4.39) is smaller than lA in Lemma
x0,φ(x0). On the other hand, we also need to assume that |y1 − y0| ≥
4.5.4 with A = h∗
max{|x1 − x0|, κ|x1 − x0|1/5} to use Lemma 4.4.6. Note that if we have points (x0, y0) and
(x1, y1) that do not satisfy this assumption, then we already obtain an inequality for 1

diam(Y )3δ0C5

to use Lemma

2C 2

4 r2
2

8C 2

4 l2
A

5-H¨older

regularity at these points.

Proof of the main theorem of chapter 4. Let xc be an interior point of X and choose x0 and
x1 close to xc as we have discussed in Remark 4.5.6. Let yi ∈ ∂Gφ(xi).
Case 1 ) We deal with the first case of the Theorem. We separate the case p = ∞ and p < ∞.
If p = ∞, then we have

µ(Br(xt)) ≤ CVol(Br(xt)) ≤ C(cid:48)rn

for some C and C(cid:48). Moreover, since φ is an Alexandrov solution, Lemma 4.4.6 and Lemma

4.5.4 imply

µ(Br(xt)) = ν(∂Gφ(Br(xt))) ≥ ν

(cid:18)

Nl

(cid:18)(cid:26)

(cid:12)(cid:12)θ ∈

yθ

(cid:20)1

4

,

3
4

(cid:21)(cid:27)(cid:19)

(cid:19)

∩ Y

≥ νCV ln−1|y1 − y0|,

(4.44)

where ν > 0 is a lower bound of ν with respect to the Lebesgue measure dy, that is ν ≥ νdy.
Therefore, we obtain C(cid:48)rn ≥ νCV ln−1|y1 − y0|. We plug the values of r and l from (4.39)

and rearrange to obtain

|y1 − y0| ≤ C|x1 − x0|

1

4n−1

for some constant C. Note that this implies single valuedness and H¨older continuity of ∂Gφ.

65

Next, we see the case p < ∞. In this case, we define

Fµ(V ) = sup{µ(B)|B ⊂ X a ball of volume V }.

(4.45)

Then we have Fµ(Vol(Br(xt))) ≥ µ(Br(xt)) = ν(∂Gφ(Br(xt))). Thus (4.44) implies

(cid:18)

(cid:19)

Fµ

C

|x1 − x0| n
|y1 − y0| n

2

2

≥ C(cid:48)|x1 − x0| n−1

2 |y1 − y0| 3n−1

2

(4.46)

for some constants C and C(cid:48). From the condition we have imposed on µ, we have F (V ) ≤
C(cid:48)(cid:48)V 1− 1

p for some C(cid:48)(cid:48). This inequality and (4.46) shows that

|y1 − y0|2n−1+ 1

2(1− n

p ) ≤ C|x1 − x0| 1

2(1− n

p ).

Therefore, since p > n, we get

|y1 − y0| ≤ C|x1 − x0|

ρ

4n−2+ρ

where ρ = 1 − n

p . Therefore, we have that for any interior point xc of X, there exists

some constants rxc and Cxc that depends on xc, φ(xc), continuity of φ at xc such that if
|xi − xc| < rxc,i = 0, 1, we have

|y1 − y0| ≤ Cxc|x1 − x0|

ρ

4n−2+ρ .

(4.47)

Note that this inequality shows the single valuedness of G-subdifferential ∂Gφ. Therefore,

for any xc in the interior of X, there exists a ball around xc on which the function ∂Gφ is

H¨older continuous, hence ∂Gφ is locally H¨older continuous. To obtain the H¨older regularity

of the potential φ, we note that ∂Gφ(x) = expG
Case 2 ) Now we prove the second part of the theorem. Suppose we have f : R+ → R+ such

x,φ(x)(Dxφ(x)), and Remark 4.2.8.

66

that lim
r→0

f (r) = 0 and for any x ∈ X and r ≥ 0 we have µ(Br(x)) ≤ f (r)rn(1− 1

n). Note that

we can choose f strictly increasing. Then by (4.46), we have

(cid:32)(cid:18) 1

(cid:19) 1

n

ωn

(cid:33)

1
n

V

×

(cid:18) 1

(cid:19)1− 1

n

ωn

Fµ(V ) ≤ f

V 1− 1

n

(4.48)

where ωn is the volume of the unit ball in Rn. Define f by

Then (4.48) becomes

We combine (4.49) with (4.46) to obtain

f (V )2n−1 =

ωn

(cid:18) 1
Fµ(V ) ≤(cid:16)
(cid:18)

C(cid:48)|x1 − x0|
|y1 − y0|

f

(cid:19) 1

n

(cid:33)

1
2

V

.

f

(cid:32)(cid:18) 1
(cid:17)(cid:17)2n−1

ωn

(cid:19)1− 1

n

(cid:16)

2
n

V

f

V 1− 1
n .

(cid:19)

≥ C(cid:48)(cid:48)|y1 − y0|

(4.49)

(4.50)

for some constants C(cid:48), C(cid:48)(cid:48) > 0. Note that we can assume that

|x1 − x0|
|y1 − y0| → 0 as |x1− x0| → 0

because otherwise, we obtain a Lipschitz estimate. Then (4.50) implies that ∂Gφ is a single

valued map. Let g be the modulus of continuity of G-subdifferential map ∂Gφ at x0. Note
5}, then we get g(u) → 0 as u → 0. In the other case, the pairs
that if g(u) ≤ max{u, κu 1

(x0, y0) and (x1, y1) satisfies the assumption of Lemma 4.4.6 and we can apply (4.50) to

obtain

f

(cid:18)

(cid:19)

C(cid:48) u
g(u)

≥ C(cid:48)(cid:48)g(u).

Since f is strictly increasing, so is f , so that f is invertible. Therefore, we obtain

u ≥ f

−1

(C(cid:48)(cid:48)g(u))

g(u)
C(cid:48)

.

Let ω be the inverse of z (cid:55)→ f

−1

(C(cid:48)(cid:48)z) z

C(cid:48) . Note that ω is strictly increasing. Therefore,

67

composing ω with the above inequality shows that

g(u) ≤ ω(u).

is strictly increasing and has limit 0 as z → 0, ω(u)
Since the function z (cid:55)→ f
also has limit 0 as u → 0. Therefore the above inequality implies that g(u) → 0 as u → 0.

(C(cid:48)(cid:48)z) z
C(cid:48)

−1

Hence the modulus of continuity of ∂Gφ has limit 0 as the variable tends to 0 so that ∂Gφ is

continuous at x0.

68

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69

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