ON A FAMILY OF INTEGRAL OPERATORS ON THE BALL

By

Wenchuan Tian

A DISSERTATION

Submitted to

Michigan State University

in partial fulfillment of the requirements

for the degree of

Mathematics – Doctor of Philosophy

2021

ABSTRACT

ON A FAMILY OF INTEGRAL OPERATORS ON THE BALL

By

Wenchuan Tian

In this dissertation, we transform the equation in the upper half space first studied by

Caffarelli and Silvestre to an equation in the Euclidean unit ball Bn. We identify the Poisson

kernel for the equation in the unit ball. Using the Poisson kernel, we define the extension

operator. We prove an extension inequality in the limit case and identify the extremal

functions using the method of moving spheres. In addition we offer an interpretation of the

limit case inequality as a conformally invariant generalization of Carleman’s inequality.

Copyright by
WENCHUAN TIAN
2021

To my parents.

iv

ACKNOWLEDGEMENTS

I would like to thank my advisor Prof. Xiaodong Wang for many things. Above all, I’d

like to thank him for guiding me through the research project that became my dissertation.

He pointed me to this interesting problem and showed me how to handle some technical

calculations at the beginning. Then he introduced me to the powerful moving sphere method

as a potential way to solve the problem. Looking back I am amazed at my own progress

along the way, and for that I am greatly indebted to my advisor.

I would like to thank Prof. Christina Sormani from the City University of New York.

I first met Prof. Sormani at the Fields Institute during a summer school in 2017. At that

time Prof. Sormani invited all the students in the summer school to join groups to work on

research problems. I was scared to join but felt obliged to participate. Luckily the research

project turn out to be successful, and I also became friends with my collaborators Changliang

Wang and Jiewon Park.

I would like to thank all of my committee members Prof. Ben Schmidt, Prof. Jun

Kitagawa and Prof. Willie Wong, who were very supportive of me during graduate school.

I learned the geometry knowledge from Prof. Schmidt. I learned the PDE knowledge that

is most relevant to my dissertation from Prof. Kitagawa and Prof. Wong. I also would like

to thank Prof. Baisheng Yan, Prof. Dapeng Zhan, Prof. Tom Parker, Prof. Jeff Schenker

and Prof. Thomas Walpuski for excellent graduate courses. Their patience and rigor are

constant inspiration for both my research and teaching.

I had a great time teaching at Michigan State University, for that I need to thank Tsveta

Sendova, Andy Krause and Ryan McCombs. Tsveta and Andy helped me build up my

teaching skills when I first arrived at Michigan State University. Later on, I had several

delightful collaborations with Ryan as his teaching assistant.

I would like to thank Samuel Lin for many interesting discussions about the details

of this dissertation. I learned a lot from my friends at Michigan State University namely

v

Leonardo Abbrescia, Rodrigo Matos, Christos Grigoriadis, Ioannis Zachos, Abhishek Mallick,

Gorapada Bera, Arman Tavakoli, Keshav Sutrave, Zhe Zhang and Wenzhao Chen. I’d like

to thank all of them for interesting discussions and fun time together.

Lastly, I would like to thank my parents. I changed my major several times, from art and

design to physics to math. My parents were concerned about this, but they chose to support

me and let me make my own mistakes. I had a great time during all these experiences.

vi

TABLE OF CONTENTS

CHAPTER 1

INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1 Sharp Integral Inequalities for Harmonic Functions
. . . . . . . . . . . . . .
1.2 An Extension Problem Related to the Fractional Laplacian . . . . . . . . . .
1.3 Sharp Integral Inequalities with α . . . . . . . . . . . . . . . . . . . . . . . .
1.4 Fractional Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4.1 Conformally Compact Manifold . . . . . . . . . . . . . . . . . . . . .
1.4.2 Connection with Scattering Theory . . . . . . . . . . . . . . . . . . .
Sobolev Trace Inequality in 4-dimensional Unit Ball . . . . . . . . . .
1.4.3
1.4.4
Sobolev Trace Inequality in Even-dimensional Unit Ball . . . . . . . .
1.5
Introduction to the Method of Moving Spheres . . . . . . . . . . . . . . . . .
1.6 Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7 Conformally Invariant Generalization of Carleman’s Inequality . . . . . . . .

CHAPTER 2 FROM THE UPPER HALF SPACE TO THE UNIT BALL . . . . .
2.1 Chapter Outline and Notation . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 The Equation in the Unit Ball . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Poisson Kernel in the Unit Ball
. . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Poisson Kernel under Conformal Transformation . . . . . . . . . . . . . . . .
2.5 Compactness
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6 Extremal Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

CHAPTER 3 LIMIT CASE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1 Chapter Outline . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Proof of Theorem 1.13: Limit Case Inequality . . . . . . . . . . . . . . . . .

3.3 The Function !In . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.1 Conformal Transformation of !In . . . . . . . . . . . . . . . . . . . . .
Simplify the Function !In . . . . . . . . . . . . . . . . . . . . . . . . .

3.3.2
3.3.3
Induction Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.4
Supplementary Calculation . . . . . . . . . . . . . . . . . . . . . . . .
3.3.5 Hyperbolic Harmonic Through Induction . . . . . . . . . . . . . . . .
3.3.6 Boundary Value Through Induction . . . . . . . . . . . . . . . . . . .
3.4 Conformal Trasformation of the Inequality . . . . . . . . . . . . . . . . . . .
3.5 Uniqueness Through the Method of Moving Spheres . . . . . . . . . . . . . .
3.5.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5.2
Inversion with Respect to Spheres . . . . . . . . . . . . . . . . . . . .
3.5.3
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Start the Sphere
3.5.4 The Case ¯λ0 = ∞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5.5 The Case ¯λ0 < ∞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5.6 Proof of Theorem 1.14 . . . . . . . . . . . . . . . . . . . . . . . . . .

1
1
3
5
6
6
7
9
10
11
13
15

18
18
20
23
27
29
37

41
41
41
43
44
46
48
51
52
56
59
60
60
61
65
69
70
75

APPENDICES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

78

vii

79
APPENDIX A CONFORMAL TRANFORMATIONS . . . . . . . . . . . . . .
82
APPENDIX B HARMONIC FUNCTIONS IN HYPERBOLIC SPACE . . . . .
APPENDIX C REGULARITY . . . . . . . . . . . . . . . . . . . . . . . . . .
90
APPENDIX D CONFORMALLY COMPACT MANIFOLD . . . . . . . . . . . 102

BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

viii

CHAPTER 1

INTRODUCTION

1.1 Sharp Integral Inequalities for Harmonic Functions

The Laplace equation with Dirichlet boundary in the unit ball is a well known equation.

We can solve it with the help of the Poisson kernel. We use Bn to denote the unit ball in Rn

and Sn−1 to denote the unit sphere in Rn. For !f : Sn−1 → R regular enough, the equation

has the unique solution

"##$##%

∆!u = 0, in Bn
!u = !f , on Sn−1
|Sn−1|&Sn−1

1

!u(x) =

1 − |x|2
|x − ξ|n!f (ξ)dξ.

We can consider the extension operation as an operator that maps functions on the unit

sphere to functions in the unit ball. For any !f define
|Sn−1|&Sn−1

1

then it is easy to see that

1 − |x|2
!P!f (x) =
|x − ξ|n!f (ξ)dξ,
’’’!P!f’’’L∞(Bn) ≤’’’!f’’’L∞(Sn−1)

.

This is the well known maximum principle.

In general we want to ask whether we can prove that the extension operator !P is a

bounded operator from Lp(Sn−1) to Lq(Bn) for some p and q. Hang, Wang and Yan con-

sidered this problem in both the upper half space and the unit ball.

In the upper half space, Hang, Wang and Yan [17] prove that the extension operator

np

n−1(Rn
P : Lp(Rn−1) → L
follows, for any f : Rn−1 → R
P f (y) =

+) is bounded for 1 < p < ∞. Here the operator P is defined as
nωn&Rn−1

f (w)dw.

yn

2

(|y′ − w|2 + y2

n)n/2

1

Here y = (y′, yn) denote points in the upper half space such that y′ ∈ Rn−1 and yn > 0,
w ∈ Rn−1, dw denotes the volume form on Rn−1 with the Euclidean metric. When p =
2(n−1)
n−2 , there is conformal symmetry in the system, they even identify the sharp constant
and classify all the extremal functions. In particular they proved the following

Theorem 1.1. [17, Theorem 1.1] Assume n ≥ 3; then for any f ∈ L

2(n−1)
n−2 (Rn−1)

’P f’L

2n
n−2 (Rn

2(n−1) ω− n−2
2n(n−1)

n

+) ≤ n− n−2

’f’

2(n−1)
n−2 (Rn−1)

L

.

(1.1.1)

Moreover, equality holds if and only if

f (ξ) =

c

(λ2 + |ξ − ξ0|2)(n−2)/2

for some constant c, positive constant λ, and ξ0 ∈ Rn−1. Here ωn is the Euclidean volume
of the unit ball in Rn.

Hang, Wang and Yan [17] proved the existence of the maximizer using two different

approaches. In the first approach, they proved it with the concentration compactness prin-

ciple. In the second approach they proved the existence of the maximizer with the method

of symmetrization as in [21].

Hang, Wang and Yan [17] also considered the variational problem

cn = sup{’P f’ 2n

n−2 (Rn
+)

: f ∈ L

2(n−2)
n−2 (Rn−1),’f’

2(n−1)
n−2 (Rn−1)

L

= 1},

derived the Euler-Lagrange equation for nonnegative critical function f

f (ξ)

n

n−2 =&Rn

+

P (x, ξ)(P f )(x)

n+2
n−2 dx,

(1.1.2)

and proved the classification result for positive critical functions as follows

Theorem 1.2. [17, Theorem 1.2] Assume n ≥ 3, f ∈ L
identically zero and satisfies (1.1.2), then for some λ > 0 and ξ0 ∈ Rn−1,

2(n−1)
n−2 (Rn−1) is nonnegative, not

f (ξ) = c(n)*

λ

λ2 + |ξ − ξ0|2+ n−2

2

,

where c(n) is determined by the scaling.

2

In the unit ball, Hang, Wang and Yan also considered the same extension problem. In

[18] they proved the following:

Theorem 1.3. [18, Theorem 3.1] Assume n ≥ 3, then for every !f ∈ L

n−2 (Bn) ≤ n− n−2

2n

2(n−1) ω− n−2
2n(n−1)

n

2(n−1)
n−2 (Sn−1)

’!P!f’L
Here for any x ∈ Bn

2(n−1)
n−2 (Sn−1),

.

(1.1.3)

(1.1.4)

L

’!f’
1 − |x|2
|x − ξ|n!f (ξ)dξ

1

|Sn−1|&Sn−1

!P!f (x) =

is the harmonic extension of !f , ωn is the volume of the unit ball in Rn with the Euclidean
metric and |Sn−1| is the volume of the unit sphere in Rn with the standard metric. Equality
holds if and only if

!f (ξ) = c(1 + ξ · ζ)− n−2

2 ,

for some c ∈ R and ζ ∈ Bn. Note that here ξ · ζ denotes the inner product between ξ and ζ
with respect to the Euclidean metric.

Inequality (1.1.3) can be considered as a direct consequence of inequality (1.1.1), but in

[18] Hang, Wang and Yan proved inequality using the Kazdan-Warner type argument.

1.2 An Extension Problem Related to the Fractional Laplacian

In [3] Caffarelli and Silvestre considered an interesting generalization of Laplace equation

in the upper half space.

For a given parameter α such that 2−n ≤ α < 1, in Rn

consider the following equation

+ =,y =(y′, yn) ∈ Rn−1 × R : yn > 0-,

div (yα

n∇u) = 0, for y ∈ Rn
+,
u(y′, 0) = f (y′), for y′ ∈ Rn−1

"##$##%

3

(1.2.1)

For the case −1 < α < 1, Caffarelli and Silvestre [3] developed the Poisson formula for this
equation

where

(Pαf )(y′, yn) = cn,α&Rn−1
n,α =000Sn−2000& ∞

c−1

0

rn−2dr

(1 + r2) n−α

2

y1−α
n

.|y′ − w|2 + y2

n/ n−α

2

f (w) dw,

(1.2.2)

(1.2.3)

=

Γ( 1−α

2 )Γ( n−1
2 )
2Γ( n−α
2 )

000Sn−2000 .

Caffarelli and Silvestre [3] also found an interesting relation between the Poisson formula
(1.2.2) and the fractional Laplacian (−∆)(1−α)/2 on Rn−1.

For a function f : Rn−1 → R regular enoght, the fractional Laplacian of f can be defined

using the formula

(−∆)(1−α)/2f (w) = C(n, α)&Rn−1

f (w) − f (v)
|w − v|n−α dv,

where −1 < α < 1, and C(n, α) is some normalization constant. Note that in this article
we used the notation C(n, α) to denote any constant that only depends on n and α, and the

constant changes through out the article.

In [3], Caffarelli and Silvestre proved that for the extension u = Pαf given in (1.2.2) we

have

C(n, α)(−∆)(1−α)/2f (y′) =

=

n uyn(y′, yn)

lim

yn→0+ −yα
lim
1 − α
yn→0+

1

u(y′, yn) − u(y′, 0)

y1−α
n

.

In Proposition B.4 and Remark 13 in the appendix, we offer an alternative explanation

of why the function

is a fundamental solution to the equation

Γ(y) = C(n, α)

1

|y|n−2+α

div(yα

n∇u) = 0

4

as observed by Caffarelli and Silvestre in [3]. Our explanation does not involve fractional di-

mension and suggests that the hyperbolic space plays a special role in the family of equations

(1.2.1).

1.3 Sharp Integral Inequalities with α

Chen [8] considered the extension problem related to the equation (1.2.1). He defined the
2(n−1)
n−2+α (Sn−1) he defined the extension

1

n+α−2

2)

.!Pα!f/ ◦ φ(y′, yn) =0000*y′, yn +

extension operator !Pα in the unit ball. For any !f ∈ L
in the unit ball to be !Pα!f such that
2+0000
+ and w ∈ Rn−1. Here φ : Rn
+ → Bn is the projection map
φ(y′, yn) = .y′, yn + 1
2/
000.y′, yn + 1
2/000

Pα1 !f ◦ φ(w, 1

2)|n+α−22 ,

where (y′, yn) ∈ Rn

2 − (0, 1),

|(w, 1

(1.3.1)

(1.3.2)

2 )

(1.2.2).

where 0 is assumed to be the origin in Rn−1. The extension Pα* !f◦φ(w, 1
2 )|n+α−2+ is as in
Note that Chen [8] did not give an explicit formula for the operator !Pα. We will use
a slightly different projection map to define a similar extension operator !Pα and give the
2(n−1)
Theorem 1.4. [8, Theorem 1] For any ˜f ∈ L
n−2+α (Sn−1), n ≥ 2, 2 − n < α < 1, we have
the sharp inequality

explicit formula in (2.3.1). Chen [8] used the definition (1.3.1) to prove the following:

|(w, 1

’!Pα ˜f’L

2n

n−2+α (Bn) ≤ Sn,α’ ˜f’

2(n−1)
n−2+α (Sn−1)

L

,

where the sharp constant Sn,α depends only on n and α. The optimizers are unique up to a

conformal transform and include the constant function f = 1.

The proof of this theorem uses similar ideas as in [17].

Chen [8] also considered the limit case α → 2 − n, and proved the following.

5

Theorem 1.5. [8, Theorem 2] For any F such that eF ∈ Ln−1(Sn−1), n > 2, we have

’e!In+!PαF’Ln(Bn) ≤ Sn’eF’Ln−1(Sn−1).

conformal transform any constant is an optimizer.

where Sn is a sharp constant that only depends on n, and !In is a radial function. Up to a
In [8] Chen gave a way to calculate !In but did not give the general formula for it. He
General formula for !In in the even dimension can be found in [25, formula (3.35)]. We
will also prove an induction relation for !In in all dimensions in Lemma 3.2.

also did not prove the uniqueness of optimizers in the limit case.

1.4 Fractional Laplacian

In this section I will introduce the fractional Laplacian in conformally compact manifolds

following [7] and define the adapted metic following [1] and [5]. The adapted metric in

the limit case is as in the work of Fefferman and Graham [11]. I will follow [1] and refer

to it as the Fefferman-Graham metric.

I will also introduce the Sobolev trace inequality

of higher orders proved by Ache and Chang [1] and Yang [25]. The higher order Sobolev

trace inequality suggests that there is a natural relation between the hyperbolic harmonic

extension of functions and the Fefferman-Graham metric in the unit ball.

The fractional Laplacian in conformally compact manifolds is closely related to the work

of Caffarelli and Silvestre as we will see in Lemma 1.6. The Fefferman-Graham metric in the

unit ball shows up naturally in the proof of Theorem 1.13 which is one of the main theorem

of this dissertation.

1.4.1 Conformally Compact Manifold

The notion of conformally compact manifold generalizes the relation between the Euclidean

unit ball and the ball model of hyperbolic space. Given any compact n−dimensional manifold
with interior M and boundary ∂M = Σn−1. A function ρ : M → R is called a defining
M

n

6

function of Σ in M if

ρ > 0 in M, ρ = 0 on Σ, dρ ∕= 0 on Σ.

Suppose g+ is a metric on M , we say that g+ is a conformally compact metric on M with

conformal infinity (Σ, [h]) if there exists a defining function ρ such that the manifold (M , g)
is compact with g = ρ2g+, and g|Σ ∈ [h]. Here g|Σ denotes the restriction of g to the
submanifold Σ, and [h] denotes the conformal class containing the metric h.

As an example we can look at the hyperbolic space. Take M = Bn with boundary
(1−|x|2)2 dx2, where dx2 is the standard metric in the Euclidean

Σ = Sn−1. We can take g+ =
space and |x| denotes the standard Euclidean norm. We can choose the ρ = 1−|x|2
defining function, then we have g = dx2 and h is the standard metric on the unit sphere

as the

4

2

Sn−1.

1.4.2 Connection with Scattering Theory

In a given conformally compact manifold, Case and Chang [5] defined the adapted metric

using scattering theory.

It is well know that (Mazzeo and Melrose [22], Graham and Zworski [15]) given f ∈
/∈ N ∪ {0}, and s(n − 1 − s) is not in the

C∞(Σ) and s ∈ C such that Re(s) > n−1
2 , s − n−1
pure point spectrum of ∆g+, the eigenvalue problem

2

−∆g+u − s(n − 1 − s)u = 0 in M

(1.4.1)

has a unique solution of the form

u = F ρn−1−s + Hρs for F, H ∈ C∞(M ) and F|Σ = f.

Consider the case f = 1, and denote the corresponding solution to (1.4.1) as vs. When
vs > 0 in M (this is the case when M is the unit ball and g+ is the hyperbolic metric),
s g+ is called the

n−1−s as a defining function, and the metric gs = y2

we can use ys = (vs)

1

7

adapted metric. In the limiting case when n is even and s → n − 1 we can define

τ = −

vs

d

ds0000s=n−1

then the metric g∗ = e2τ g+ is as in the work of Fefferman and Graham [11], we will refer to

it as the Fefferman-Graham metric.

In [7] Chang and Gonz´alez found a relation between the scattering theory and the ex-

tension problem related to the fractional Laplacian. They proved the following

Lemma 1.6. [7, Lemma 4.1] Let (M n, g+) be any conformally compact Einstein manifold
with boundary Σn−1. For any γ ∈ (0, n−1
α = 1 − 2γ. For any defining function ρ of Σ, the equation

2 ) such that γ is not an integer, denote s = n−1

2 +γ,

is equivalent to

where

−∆g+u − s(n − 1 − s)u = 0 in (M, g+)

−div(ρα∇U ) + E(ρ)U = 0 in (M, g),

g = ρ2g+, U = ρs−n+1u

(1.4.2)

(1.4.3)

and the derivatives in (1.4.3) are taken with respect to the metric g. The lower order term

is given by

E(ρ) = −∆g(ρ

α
2 )ρ

α

2 +*γ2 −

E(ρ) = −∆g+(ρ

n−2+α

2

)ρ−n−2+α

2

1

4+ ρ−2+α +
−* (n − 1)2

4

n − 2
4(n − 1)

Rgρa,

− γ2+ ρ−2+α.

here Rg is the scalar curvature of g. Or equivalently writing everything in g+, we have

Chang and Gonz´alez also proved a similar results for hyperbolic upper half space in

theorem 3.1 of their paper [7].

As we will see in Remark 2, equation (2.2.1) obtained by me is a special case of equation

(1.4.3).

8

1.4.3 Sobolev Trace Inequality in 4-dimensional Unit Ball

Ache and Chang found an explicit formula for the Fefferman-Graham metric in 4-dimensional

hyperbolic unit ball (proposition 2.2 in [1]). Their calculation make use of the connection

(Lemma 1.6) discovered by Chang and Gonz´alez [7]. Ache and Chang also proved a very

interesting Sobolev trace inequality of order 4:

Theorem 1.7. [1, Theorem B ] Given f ∈ C∞(S3), suppose u is a smooth extension of f
to the Euclidean unit ball (B4, dx2). If u satisfies the Neumann boundary condition

= 0,

(1.4.4)

here ν is the outer unit normal vector with respect to the Euclidean metric, then we have the

∂
∂ν

u0000S3
16π2&B4

3

inequality

log1 1

2π2&S3

e3(f−f )dξ2 ≤

(∆u)2dx +

(1.4.5)

3

8π2&S3 |!∇f|2dξ.

Here, dξ is the volume form on the standard sphere, f = 3S3 f dξ is the average of f , ∆
is the Laplacian in the Euclidean unit ball and !∇ is the gradient on the standard sphere.
Moreover, equality holds if and only if u is a biharmonic extension of a function of the form
fζ (ξ) = − log |1 − 〈ζ, ξ〉| + c, where c ∈ R is a constant, ζ ∈ B4 is a fixed point, ξ ∈ S3, and
u satisfies the Neumann boundary condition (1.4.4).

Their proof suggests that there is a natural relation between the hyperbolic harmonic

extension of functions and the Fefferman-Graham metric in the unit ball. They start with
a smooth function f ∈ C∞(S3) and consider the biharmonic extension to the unit ball
satisfying the boundary condition (1.4.4). Then they do integration by part using the bihar-

monic extension functions and the Fefferman-Graham metric. Eventually they establish the

inequality (1.4.5) using Beckner’s generalization of Moser-Trudinger inequality [2].

Based on the calculation in the appendix B, we know that: in B4 biharmonic extension

of a given function satisfying Neumann boundary condition (1.4.4) is exactly the hyperbolic

harmonic extension of the same function.

9

1.4.4 Sobolev Trace Inequality in Even-dimensional Unit Ball

Yang [25] generalized Ache and Chang’s result. He found an explicit formula for the

Fefferman-Graham metric in all even-dimensional unit ball and used the integration by parts

argument in even-dimensional unit ball to prove the following:

Theorem 1.8. [25, Theorem 1.7] Let n ≥ 4 be an even integer. Given f ∈ C∞(Sn−1),
suppose u is a smooth extension of f to the unit ball Bn which also satisfies the Neumann

boundary condition

∆ku|Sn−1 = (−1)k

and

∂
∂ν

∆ku0000Sn−1

2) Γ. n−1
2 − k/
Γ( n
Γ. n−1
2 / Γ( n
2 − k)
2 − k/
2) Γ. n−1
Γ( n
2 / Γ( n
Γ. n−1
2 − k)

Pn−1
Pn−1−2k

f, for 0 ≤ k ≤ [

n − 2

4

],

(1.4.6)

= (−1)k+1

Pn−1
Pn−1−2k

f, for 0 ≤ k ≤ [

n − 4

4

].

(1.4.7)

Here ∆ denotes the Laplacian in the Euclidean unit ball, ν is the outer unit normal vector on
the boundary Sn−1 with respect to the Euclidean metric in Bn. For any α ∈ R, [α] denotes
the largest integer that is less than or equal to α. P2γ are operators on Sn−1 defined by
Beckner [2], such that for any 0 < γ < n−1

2 we have

P2γ =

, B =4−!∆ +* n − 2
2 +2
here !∆ is the Laplace-Beltrami operator on the standard sphere Sn−1.

Γ(B + 1/2 + γ)
Γ(B + 1/2 − γ)

Then we have the inequality

,

e(n−1)(f−f )dξ+

log* 1
|Sn−1|&Sn−1
2nπn/2Γ(n/2)*&Bn |∇n/2u|2dx +&Sn−1

n − 1

≤

f dξ+ .

f Tn−2

2

(1.4.8)

Here

∇m ="##$##%

∆

m
2 , when m is even,
m−1
2 , when m is odd.

∇∆

10

Tm is an operator of order 2m on Sn−1 defined as follows: when m is an odd integer

Tm =

n − 2

2

Γ(m + 1)Γ(1/2)

P2m+1

Γ(m + 1/2)

P1

+

(m−1)/25k=1

(m − 2k)

Γ(m + 1)2Γ(k + 1/2)Γ(m − k + 1/2)
Γ(m + 1/2)2Γ(k + 1)Γ(m − k + 1)

P2

2m+1

P2m+1−2kP2k+1

,

and when m is an even integer

Tm =

n − 2

2

Γ(m + 1)Γ(1/2)

P2m+1

Γ(m + 1/2)

P1

+

+

n − 2 − m

2

m/2−15k=1

(m − 2k)

67

2

m+1

P2
2m+1
P2

Γ(m + 1)Γ. m+1
2 /
Γ(m + 1/2)Γ(m/2 + 1)89
Γ(m + 1)2Γ(k + 1/2)Γ(m − k + 1/2)
Γ(m + 1/2)2Γ(k + 1)Γ(m − k + 1)
2 )&Sn−1

Moreover, equality holds if and only if

u(x) = π− n−1

2

Γ(n − 1)
2n−1Γ( n−1

where c ∈ R and x0 ∈ Bn.

P2

2m+1

P2m+1−2kP2k+1

.

(1 − |x|2)n−1
|x − ξ|2n−2 (− ln|1 − 〈x0, ξ〉| + c)dξ,

(1.4.9)

Note that if we use 00Sn−100 to denote the volume of the unit sphere with the standard

metric, then we have

π− n−1

2

Γ(n − 1)
2n−1Γ( n−1
2 )

−1

.

=000Sn−1000

We obtain the same explicit formula for Fefferman-Graham metric in all even dimensional

unit ball using a different method in Theorem 3.1. Moreover, we obtain the same boundary

conditions (1.4.6) and (1.4.7) using a different method in Proposition B.2 in the appendix

Yang’s proof further suggests that there is a natural relation between the hyperbolic

harmonic extension of functions and the Fefferman-Graham metric in the unit ball.

In

particular, (1.4.9) is the hyperbolic harmonic extension of the function − ln|1 − 〈x0, ξ〉| + c.
1.5

Introduction to the Method of Moving Spheres

The method of moving spheres is a powerful tool to prove uniqueness of solutions to

equations that have conformal symmetry. The method relies on maximum principle and

11

the conformal symmetry of the equation. The method of moving spheres can be considered

as a powerful generalization of the method of moving planes, but this dissertation will not

talk more about the method of moving planes. For more information about the method of

moving planes we refer the readers to the articles [13] [9] [16].

The method of moving spheres is closely related to conformal geometry. In [10] Escobar

proved an important result in conformal geometry:

Theorem 1.9. [10, Theorem 3.1] In Rn

+ = {(x′, xn) : x′ ∈ Rn−1, xn ≥ 0} with the Euclidean

metric. Let u be a positive solution to the problem

∆u = 0 on Rn
+,

∂u
∂xn

+ (n − 2)un/(n−2) = 0 on Rn−1,

(1.5.1)

u(x) = O(|x|2−n) near ∞.
u(x′, xn) =*

|x′ − x0|2 + (- + xn)2+(n−2)/2

-

,

"######$######%

Then

where - > 0 and x0 ∈ Rn−1.

This uniqueness result is very important in conformal geometry, because it entails the

fact that the only metric in Rn

+ that is conformal to the Euclidean metric and has zero scalar

curvature and constant mean curvature is

g’ =*

|x′ − x0|2 + (- + xn)2+2

-

where dx2 denotes the Euclidean metric.

dx2,

In [20] Li and Zhu used the method of moving spheres to prove the uniqueness of solutions

to the equation (1.5.1). They proved the following:

Theorem 1.10. [20] For any integer n ≥ 3, let u ∈ C2(Rn
solution of

+) ∩ C1(R

"##$##%

−∆u = 0 in Rn
+,

∂u
∂xn

= cun/(n−2) on Rn−1,

12

n
+) be any nonnegative

where c < 0. Then either u = 0 or

u(x′, xn) =*
for some t0 > 0 and x0 ∈ Rn−1.

|x′ − x0|2 + (xn + t0)2+(n−2)/2
−(n − 2)t0c−1

Note that in Theorem 1.10 Li and Zhu dropped the asymptotic assumption u(x) =
O(|x|2−n). But more importantly, the method of moving sphere is very general; it can be
applied to a wide range of equations with conformal symmetry. It can be applied to differ-

ential equations as well as integral equations. For example in [19] Li proved the following:

Theorem 1.11. [19, Theorem 1.1] For any integer n ≥ 3, 0 < α < n let u ∈ L∞loc(Rn) be a
positive solution to the equation

then

u(x) =&Rn
u(x) =*
for some a, d > 0 and some x0 ∈ Rn.

u(y)

n+α
n−α

|x − y|n−α dy, for all x ∈ Rn,
d + |x − x0|2+(n−α)/2

a

,

In this dissertation I will also use the method of moving spheres to prove Theorem 1.14.

My proof combines both approaches for differential equations as in [20] and approaches for

integral equations as in [19].

1.6 Main Result

In this article, we revisit the extension problem studied in [8] by a different approach.

We firstly derive an explicit formula for !Pα in (2.3.1) and then carry out the analysis on Bn.

In Theorem 2.12, we prove that the following inequality has constant function as opti-

mizers.

Theorem 1.12. Assume n ≥ 3 and α ∈ (2 − n, 1). For every f ∈ L

2(n−1)
n−2+α (Sn−1), we have

’’’!Pαf’’’L

2n

n−2+α (Bn) ≤ Sn,α ’f’

L

.

2(n−1)
n−2+α (Sn−1)

13

Where Sn,α is a constant that only depends on n and α. Up to conformal transformation

any constant is an optimizer..

Our proof of the existence of optimizer relies on subcritical analysis as in [18], while our

proof of uniqueness is the same as [8].

In the limit case α → 2 − n. We prove

.

(1.6.1)

Theorem 1.13. For dimension n ≥ 2, and any function !F ∈ L∞(Sn−1) we have
’’’e!In+!P2−n!F’’’Ln(Bn) ≤ Sn’’’e!F’’’Ln−1(Sn−1)
dα 00α=2−n. When n is even we have
Where !In(x) = 2 d!Pα1
Γ. n−2
2 / Γ (n − k − 1)
n/2−15k=1
!In(x) =
2 − k)
Γ(n − 2)Γ( n
The sharp constant Sn = """"e!In""""Ln(Bn)
###Sn−1###

(1 − |x|2)k.

1
2k ·

1
n−1

Our proof of the limit case inequality is very similar to [8]. When n is even, in addition

induction. The induction formula is in proved in Lemma 3.2. When n is odd, we can

to proving the inequality, we also found an explicit formula for the function !In through
calculate !In by change of variable. We don’t have an explicit formula for !In when n is odd,
In particular, we used the induction relation to prove that the function !In + ln 1−2xn+|x|2

but the induction relation in Lemma 3.2 is true for both when n is even and when n is odd.

is hyperbolic harmonic in Lemma 3.6.

2

We also considered the variational problem

and derived the Euler Lagrange equation

Sn = sup:’’’e!In+!P2−n!f’’’ : !f ∈ L∞(Sn−1),’’’e!f’’’Ln−1(Sn−1)
en!In+n!P2−n!f!p2−n(x, ξ)dx.

e(n−1)!f (ξ) =&Bn

= 1; ,

We prove the following uniqueness result

14

Theorem 1.14. For any integer n ≥ 2, if !f ∈ L∞(Sn−1) satisfies the equation

e(n−1)!f (ξ) =&Bn
!f (ξ) = ln

en!In+n!P2−n!f!p2−n(x, ξ)dx,
1 − |ζ|2
|ξ − ζ|2 + Cn,

then for all ξ ∈ Sn−1

where ζ ∈ Bn and Cn = − 1
the standard sphere.

n−1 ln00Sn−100 is a constant. Here 00Sn−100 denotes the volume of

Note that the choice of Cn is to make sure that ’’’e!f’’’Ln−1(Sn−1)

theorem using the moving sphere method in Section 3.5.

= 1. We prove this

1.7 Conformally Invariant Generalization of Carleman’s Inequal-

ity

In this section we offer an interpretation of the limit case inequality as a conformally

invariant generalization of Carleman’s inequality.

In [4] Carleman proved the following:

Theorem 1.15. ([4]) For any u ∈ C∞(B2

the Euclidean metric then we have

) such that u is harmonic in B2 with respec to

&B2

e2udx ≤

1

4π*&S1

eudθ+ .

(1.7.1)

Where equality holds u(x) = c or u(x) = −2 ln|x − x0| + c where c ∈ R is any constant and
x0 ∈ R2\B2

.

Note that the inequality (1.7.1) is conformally invariant and that it also holds for sub-

harmonic functions. We will refer to (1.7.1) as Carleman’s inequality through out the dis-

sertation.

Hang, Wang and Yan [18] proved a higher dimensional generalization of Carleman’s

inequality for harmonic functions. As an application of inequality (1.1.3), they proved the

following result:

15

Corollary 1.16. [18, Corollary 3.1] Assume n ≥ 3, then for !f ∈ L∞(Sn−1),
Here !P!f is the harmonic extension of !f as defined in (1.1.4), ωn is the volume of the unit
ball in Rn with the Euclidean metric. Moreover, equality holds if and only if !f is constant.

n ’’’e!f’’’L1(Sn−1)

Note that the inequality in the corollary also works for subharmonic functions but it is

’’’e!P!f’’’L

n

n−1 (Bn) ≤ n−1ω− 1

n

.

not invariant under conformal transformation.

In [8] Chen proposed another way to generalize the Carleman’s inequality in dimension

4, he proved the following:

Corollary 1.17. [8, Corollary 1] For any u : B4 → R satisfying ∆2u ≤ 0 and − ∂u

∂ν ≤ 1,

*&B4

4

e4udx+ 1

≤ S*&S3

3

e3udξ+ 1

.

Note that here ∆ is the Laplacian in B4 with the Euclidean metric, ν is the outer unit normal

vector with respect to the Euclidean metric. Here S is the sharp constant, and is assumed by

the solution to the equation

∆2u = 0, in B4

u = 0, on S3
− ∂u
∂ν = 1, on S3.

(1.7.2)

"######$######%

Note that Chen did not prove uniqueness of extremal function for this generalization, and

as he pointed out at the end of [8] that this generalization works well because the Green’s

function of equation (1.7.2) is positive. As a result will be difficult for us to find similar

generalizations in higher dimensions.

As an application of inequality (1.6.1), we propose another way to generalize the Carle-

man’s inequality:

Corollary 1.18. Assume n ≥ 3, then for any !f ∈ L∞(Sn−1)
’’’e!In+!P2−n!f’’’Ln(Bn) ≤ Sn’’’e!f’’’Ln−1(Sn−1)

16

.

The sharp constant Sn = """"e!In""""Ln(Bn)
###Sn−1###

1
n−1

− ln|1 − ζ · ξ| + C. Where C ∈ R is a constant and ζ ∈ Bn.

. Moreover, equality holds if and only if !f (ξ) =

Note that this inequality is invariant under conformal transformation and that it also

holds for hyperbolic subharmonic functions.

The proof of this corollary is simply a combination of Theorem 1.13, Theorem 1.14

and regularity results Proposition C.6 from the appendix. When n is even, we can think

of ’e!In+!P2−n!f’Ln(Bn) as the Ln norm of e!P2−n!f measured using the Fefferman-Graham

metric [1][11]

where dx2 is the standard Euclidean metric on the unit ball.

g = e2!Indx2

17

CHAPTER 2

FROM THE UPPER HALF SPACE TO THE UNIT BALL

2.1 Chapter Outline and Notation

In this chapter we transform the equation (1.2.1) from the upper half space to the unit

ball. We also identify the Poisson kernel of the corresponding equation in the unit ball and

study how the Poisson kernel transforms under conformal changes. At the end of this chapter

we formulate a conjecture inspired by the Martin boundary theory.

Through out this dissertation, we let

+ = {(y′, yn) ∈ Rn such that y′ ∈ Rn−1, yn > 0},
Rn

and

Bn = {x ∈ Rn such that |x| < 1},

here |x| denotes the norm of x with respect to the Euclidean metric. We also use the notation

Sn−1 = {x ∈ Rn such that |x| = 1}

to denote the unit sphere in Sn−1.

Through out this dissertation we use notations like (y′, yn) and (x′, xn) to denote points

in Rn where y′, x′ ∈ Rn−1 and yn, xn ∈ R.
Definition 2.1. For r ∈ (0, 1], define Sn−1
r}.

r = {x ∈ Bn : |x| = r} and Bn

r = {x ∈ Bn : |x| <

Let Ψ : Rn

+ → Bn be the projection map defined by

2y′

1 + 2yn + |y|22
1 + 2yn + |y|2 , −1 + |y|2
(1 + yn)2 + |y′|22

2 (1 + yn)

2y′

(1 + yn)2 + |y′|2 , 1 −

(2.1.1)

(x′, xn) = Ψ(y′, yn) =1

=1

18

with the inverse Ψ−1 : Bn → Rn

+

(y′, yn) = Ψ−1(x′, xn) =1

2x′

(1 − xn)2 + |x′|2 ,

1 − |x|2

(1 − xn)2 + |x′|22 .

It is useful to record

Which means that if we define [Ψ(y)]n to be the n−th component of Ψ(y) (note that by
definition [Ψ(y)]n = −1+|y|2

1 − 2xn + |x|2

2

=

2

1 + 2yn + |y|2

1+2yn+|y|2 ), then we have
1 − 2[Ψ(y)]n + |Ψ(y)|2

2

2

=

1 + 2yn + |y|2 .

(2.1.2)

The restriction of Ψ on yn = 0 is the stereographic projection Rn−1 → Sn−1. From the

calculation in proposition 2.2, in particular (2.2.2), we see that

Ψ∗dx2 =

4dy2

.1 + 2yn + |y|2/2 .

It means that Ψ : Rn
+ → Bn is a conformal transformation. Here the conformal factor is
very important for our calculation, through out this article we will use |Ψ′(y)| to denote the
conformal factor, in particular for any y ∈ Rn

+ we have

|Ψ′(y)| =

and for any w ∈ Rn−1 = ∂Rn

+ we have

2

1 + 2yn + |y|2 ,

|Ψ′(w)| =

2

1 + |w|2 .

(2.1.3)

(2.1.4)

We will discuss more about Ψ and other conformal transformation in the appendix.

For a function !f on Bn or Sn−1 we define
f(y′, yn) = !f ◦ Ψ(y′, yn)1

2

1 + 2yn + |y|22 n−2+α

2

It is easy to check that this map is an isometry from L

and from L

2n

n−2+α (Bn) to L

2n

n−2+α(Rn

+). The inverse map is

.

(2.1.5)

2(n−1)

n−2+α(Sn−1) to L
1 − 2xn + |x|22 n−2+α

2

2

.

2(n−1)

n−2+α(Rn−1)

(2.1.6)

!f(x′, xn) = f ◦ Ψ−1(x′, xn)1

19

2.2 The Equation in the Unit Ball

Now we are ready to transform the equation (1.2.1) from the upper half space to the unit
ball. For any 2− n ≤ α < 1 define the operator L in Bn such that for any x = (x′, xn) ∈ Bn

and for any!u ∈ C2(Bn)

2

α (2 − n − α)

2(n+2−α)/2
L!u = 1 1 − 2xn + |x|2
·<=div>1 1 − |x|2
∇!u? +
2 2α
Note that in Bn we have L!u = 0 if and only if
∇!u? +
div>1 1 − |x|2
2 2α
Proposition 2.2. (How the operator tranforms) For any 2−n ≤ α < 1 and any u ∈ C2(Rn
+)
define !u using (2.1.6) then we have

1 1 − |x|2
2 2α−1

2 2α−1

1 1 − |x|2

α (2 − n − α)

!u = 0.

!u@A .

(2.2.1)

div(yα

n∇u) = 0, in Rn

+

2

2

if and only

L!u = 0, in Bn

Proof. In the following, for any y = (y′, yn) ∈ Rn, let ρ = .1 + 2yn + |y|2/ /2. Then we
have ∇ρ =(y′, 1 + yn). Let a, b, c, d = 1, 2, ..., n be indices. Suppose x = Ψ(y), then by

direct calculation we have, when c ∕= n,

∂xc
∂ya

=

and when c = n

∂xn
∂ya

"#####$#####%
= "#$#%

−

4yayc

(1+2yn+|y|2)2 ,
4y2
c

2

1+2yn+|y|2 −

(1+2yn+|y|2)2 ,

− 4yc(yn+1)
(1+2yn+|y|2)2 ,

a ∕= c and a ∕= n,

c = a ∕= n,

a = n,

2ya

1+2yn+|y|2 + 2ya(1−|y|2)
a ∕= n,
(1+2yn+|y|2)2 ,
1+2yn+|y|2 + 2(1−|y|2)(1+yn)
(1+2yn+|y|2)2 , a = n.

2yn

20

From it we have

n5a=1

∂xc
∂ya

∂xd
∂ya

= "#$#%

ρ−2, c = d,
c ∕= d.

0,

(2.2.2)

n5a=1

∂xc
∂ya

∂ρ
∂ya

= "#$#%
= "#$#%
n5a=1
= ρ−(n+α)/2Bρ
∂!u

∂xc
∂y2
a

∂xc

−ρ−1yc,
c ∕= n,
ρ−1 (1 + yn) c = n.

,

− (n − 2) ρ−2yc,
c ∕= n,
(n − 2) ρ−2 (1 + yn) c = n.
∂yaC ,

+*1 −

∂xc
∂ya

n + α

∂ρ

2 +!u

We calculate

∂u
∂ya

21

div(yα

n∇u) =

∂

∂2xc
∂y2
a
n + α

n + α

∂ρ

n + α

2 +!uC
2 +!u
+*1 −
2 +!u
∂ynC
yj+ + n*1 −
2 +!u
∂ynC
yj+ − α*1 −
∂!u

(1 + yn) −

∂xj

∂ρ

n + α

∂xj

∂ρ

∂yaC

n + α

2 +!uC

n + α

2 +!uC
yj+

∂xc

2

n

∂xc

∂xc

n + α

n + α

∂xn

n + α

= yα

+ ρ

∂xc

∂u

n

= yα

∂ya

∂xc

∂xc
∂ya

∂xc∂xd

∂xc
∂yn

∂xd
∂ya

+αyα−1

∂2xc
∂y2
a

(1 + yn) −

−
+αyα−1

∂!u
+ n*1 −
∂!u
+*1 −

∂ya+
∂ya*yα
n ρ−(n+α)/2Bρ
∂2!u
∂xc
∂ya
2 + ∂ρ
+*2 −
∂!u
∂xc
∂ya
∂yaBρ
n ρ−(n+α)/2−1 ∂ρ
yα
ρ−(n+α)/2Bρ
∂!u
n ρ−(n+α)/2Bρ−1∆!u + ρ
∂!u
+ (2 − n − α) ρ−1* ∂!u
∂!u
2 + yα
− (n + α)*1 −
n ρ−(n+α)/2!u
+*1 −
ρ−(n+α)/2Bρ
∂!u
·Bρ−1∆!u − αρ−1* ∂!u
∂!u
ρ−(n+α)/2Bρ−1 (1 + yn)* ∂!u
2 +!u (1 + yn)C
+*1 −
∂!u
n ρ−(n+α)/2ρ−1∆!u + αyα−1
yj+ −
·Bρ−1* ∂!u
∂!u
ρ−(n+α)/2> 1 − |x|2
∆!u − αxa
2 2α−11 1 − 2xn + |x|2
= 1 1 − |x|2
·> 1 − |x|2

−
= yα

(1 + yn) −

(1 + yn) −

n ρ−(n+α)/2

= yα−1

+αyα−1

ρ−(n+α)/2

∂xc
∂yn

= yα

n

n

n

∂xj

∂xj

∂xn

n + α

∂xn

∂xn

∂xn

2

∆!u − αxa

∂!u

∂xa

2

2

∂xc

n

+*1 −

n + α

2 +!uC
!u?

α (2 − n − α)

2

+

∂xa

∂xn

∂!u
∂!u
2(n+2−α)/2
!u?

2

α (2 − n − α)

+

Remark 1. For any integer n ≥ 2 and any α ∈ (2 − n, 0), we can apply theorem 1.1 in [24]

22

to show that solution to the equation

Lu = 0, in Bn

u = f, in Sn−1

"##$##%

is unique in C2(Bn) ∩ C0(Bn).

Remark 2. Note that the equation (2.2.1) is a special case of equation (4.2) in [7] if we

take g as the Euclidean metric in the unit ball, g+ as the hyperbolic metric in the unit ball,
and ρ = 1−|x|2

as the defining function.

2

2.3 Poisson Kernel in the Unit Ball

Caffarelli and Silvestre [3] found a Poisson kernel that solves the Dirichlet problem (1.2.1).

For any −1 < α < 1, y ∈ Rn

+ and any ξ ∈ Rn−1

y1−α
n
|y − w|
For any f : Rn−1 → R regular enough, we can define

pα(y, w) = cn,α

.

n−α
2

Pαf =&Rn−1

pα(y, ξ)f (w)dw,

such that Pαf solves the Dirichlet problem (1.2.1).

We want to find the corresponding Poisson kernel in the unit ball. Define

and for any function !φ : Sn−1 → R, define

then we have the following proposition:

(1 − |x|2)1−α
|x − ξ|n−α ,

!pα(x, ξ) = 2α−1cn,α
!Pα!φ(x) =&Sn−1!pα(x, ξ)!φ(ξ)dξ.

(2.3.1)

23

Proposition 2.3. For any integer n ≥ 2, any α ∈ (2 − n, 1), any f ∈ L

2(n−1)
n−2+α (Rn−1),

define !f as in (2.1.6), then we have

DPαf = !Pα!f .

in the unit ball as defined in (2.3.1).

Here DPαf is the transformation of Pαf as defined in (2.1.6), and !Pα!f is the extension of !f
Proof. The proof is by direct calculation. Note that for any w ∈ Rn−1 using the fact that
(2.1.5) and (2.1.6) are inverse to each other, we have

f (w) = !f ◦ Ψ(w)*

As a result, we have

2

1 + |w|2+ n−2+α

2

.

2x′

2

2

dw

(Pαf ) ◦ Ψ−1(x′, xn)
* 1−|x|2
1−2xn+|x|2+1−α
= cn,α&Rn−1
10000
1−2xn+|x|2+22 n−α
(1−xn)2+|x′|2 − w0000
+* 1−|x|2
1 + |w|22 n−2+α
·!f ◦ Ψ (w)1 2
* 1−|x|2
1−2xn+|x|2+1−α
= cn,α&Sn−1
10000
1−ξn0000
(1−xn)2+|x′|2 − ξ′
·!f (ξ) (1 − ξn)
= 2(α−n)/2cn,α.1 − 2xn + |x|2/ n−2+α

1−2xn+|x|2+22 n−α
+* 1−|x|2
&Sn−1.1 − |x|2/1−α

|x − ξ|n−α

α−n
2 dξ

2x′

2

2

2

2

Divide both sides by* 1−2xn+|x|2

2

2

+ n−2+α

then we are done.

24

!f (ξ) dξ.

then the same calculation as in the previous proposition show that

(P2−nf ) ◦ Ψ−1 = !P2−n.f ◦ Ψ−1/ = !P2−n!f .

Remark 4. We note that

!f = f ◦ Ψ−1,

&Sn−1!pα (x, ξ) dξ

Remark 3. For any integer n ≥ 2 and α = 2 − n we can prove similar result. For any
f ∈ L∞(Rn−1), define

is not constant in x except when α = 0 or 2 − n.

Proposition 2.4. For any f ∈ C(Sn−1)

u (x) :="##$##%

3Sn−1!pα (x, ξ) f (ξ) dξ, x ∈ Bn

f (x), x ∈ Sn−1

defines a continuous function on Bn which is smooth in Bn and satisfies Lu = 0.

rn−13Sn−1

r

that for r ∈ [0, 1]

Proof. The integral 3Sn−1!pα(x, ξ)dξ = 1
!pα(x, ξ)dx is a function that only de-
pends on |x|. Define h(|x|) =3Sn−1!pα(x, ξ)dξ, then by Lemma 2.7, and remark 6 we know
2 /
2 ) Γ. n−1
2 / ≤ h(r) ≤* 2
Γ(n − 1)Γ. 1−α

1 + r+n−2+α Γ( n−α
* 2

1 + r+n−2+α

≤ 2n−2+α.

By dominated convergence theorem as in Remark 7, we know that for r ∈ [0, 1], h(r) is

continuous and that

h(r) = 1.

lim
r→1

By the continuity of f on Sn−1, we can choose δ > 0 small, such that when |ξ1 − ξ2| ≤ δ,
we have |f (ξ1) − f (ξ2)| ≤ -. By the continuity of h(r) on the interval [0, 1] and the fact

25

that it is strictly positive on [0, 1] we can choose δ > 0 smaller if needed, such that when

|x − x0| ≤ δ we have000

1

h(|x|) − 1

M := ’f’L∞(Sn−1)

h(|x0|)000 < -. Define
Γ(n − 1)Γ. 1−α
2 /
2 ) Γ. n−1
2 / .
Γ( n−α
2 ) Γ. n−1
2 /
Γ( n−α
2 / ≤ h(r),
Γ(n − 1)Γ. 1−α

0 <

Note that we have

for all r ∈ [0, 1].

Suppose x0 ∈ Sn−1, and x ∈ Bn such that |x − x0| < δ/2 consider

|u(x) − u(x0)| = 0000&Sn−1!pα (x, ξ) f (ξ) dξ −&Sn−1!pα (x, ξ)
≤ &|ξ−x0|≤δ!pα(x, ξ)*|f (ξ) − f (x0)| +0000f (x0) −
h(|x|)0000 dξ
+&|ξ−x0|>δ!pα(x, ξ)0000f (ξ) −

C(n, α)M (1 − |x|2)1−α

f (x0)

u(x0)
h(|x|)

f (x0)

dξ0000
h(|x|)0000+ dξ

≤ C(n, α)- +

δn−α

As a result u(x) is continuous at x0.

The part that Lu = 0 follows from dominated convergence theorem and direct calcula-

tion.

Based on the Martin theory for harmonic functions, we make the following conjecture:

Conjecture 1. (The representation theorem) Let !u : Bn → R be a positive solution of

Then there exists a Borel measure ν on Sn−1 s.t.

L!u = 0.

!f (x) =&Sn−1!pα (x, ξ) dν (ξ) .

26

2.4 Poisson Kernel under Conformal Transformation

We also want to know how the Poisson kernel transforms under conform transformation.

We prove the following:

Proposition 2.5. For any integer n ≥ 2, any α ∈ [2− n, 1), any y ∈ Rn
we have

+ and any w ∈ Rn−1

!pα(Ψ(y), Ψ(w)) = pα(y, w)|Ψ′(y)|(2−n−α)/2|Ψ′(w)|(α−n)/2.

Here Ψ is the conformal transformation defined in (2.1.1), |Ψ′(y)| and |Ψ′(w)| are the con-
formal factors in (2.1.3) and (2.1.4) respectively.

(2.4.1)

Proof. For any x ∈ Bn and any ξ ∈ Sn−1, by definition we have
(1 − |x|2)1−α
|x − ξ|n−α .

From this we have for any y ∈ Rn

+ and any w ∈ Rn−1

!pα(x, ξ) = 2α−1cn,α
!pα(Ψ(y), Ψ(w)) = 2α−1cn,α

(1 − |Ψ(y)|2)1−α
|Ψ(y) − Ψ(w)|n−α .

Through direct calculation we have

1 − |Ψ(y)|2 =

4yn

1 + 2yn + |y|2 = 2yn|Ψ′(y)|

and

|Ψ(y) − Ψ(w)|2 = 0000

=

4

8

2

2

2

4

1 + |w|2

1 + |w|2 (w,−1)0000
1 + 2yn + |y|2 (y′,−yn − 1) −
1 + 2yn + |y|2 +
(1 + |w|2)(1 + 2yn + |y|2)(〈w, y′〉 + 1 + yn)
−
(1 + |w|2)(1 + 2yn + |y|2) ·
.1 + |w|2 + 1 + 2yn + |y|2 − 2〈w, y′〉 − 2 − 2yn/
(1 + |w|2)(1 + 2yn + |y|2)|y − w|2
= |y − w|2|Ψ′(w)||Ψ′(y)|

=

=

4

4

27

Note that here we use the notation (y′,−yn − 1) and (w,−1) to denote points in Rn and
use the notation 〈w, y′〉 to denote the Euclidean inner product in Rn−1. Combine these
calculations together, then we can get (2.4.1).

We also want to consider how the Poisson kernel transform under the isometry group of

(Bn, gh). Here gh =

4

(1−|x|2)

dx2 denotes the hyperbolic metric in the unit ball. We use the

notation SO(n, 1) to denote the isometry group of the unit ball with the hyperbolic metric.
For any Φ ∈ SO(n, 1) any x ∈ Bn and any ξ ∈ Sn−1 we use |Φ′(x)| and |Ψ′(ξ)| to denote
the conformal factors in Bn and Sn−1 respectively. We prove the following:

Proposition 2.6. For any integer n ≥ 2 any α ∈ [2 − n, 1), any x ∈ Bn, any ξ ∈ Sn−1 and
any Φ ∈ SO(n, 1) we have

.

(2.4.2)

!pα (Φ (x) , Φ (ξ)) =!pα (x, ξ)00Φ′ (x)00(2−n−α)/200Φ′ (ξ)00(α−n)/2

4

Proof. For any Φ ∈ SO (n, 1), since it is an isometry of Bn with the hyperoblic metric
$1−|x|2%2 dx2, it is a conformal transformation with respect to the Euclidean metric.
gh =
We have for any x ∈ Bn
4

4

4

Φ∗*

(1 − |x|2)2 dx2+ =

(1 − |Φ(x)|2)2 Φ∗(dx2) =

(1 − |x|2)2 dx2,

and

Φ∗dx2 = |Φ′(x)|2dx2.

Here |Φ′(x)| is the conformal factor, it is a notation similar to |Ψ′(y)|. From this we conclude
that for any x ∈ Bn

.

(2.4.3)

For any x, z ∈ Bn, define d(x, z) as the distance between the two points measured by

00Φ′ (x)00 =

1 − |Φ (x)|2
1 − |x|2

the hyperbolic metric. Then we have

cosh d (x, z) = 1 + 2

|x − z|2

.1 − |x|2/.1 − |z|2/ .

28

Since Φ is an isometry with respect to the hyperbolic metric, we have Thus

|x − z|2

.1 − |x|2/.1 − |z|2/ =

Letting z → ξ ∈ Sn−1 yields

|Φ (x) − Φ (z)|2

.1 − |Φ (x)|2/.1 − |Φ (z)|2/

|Φ (x) − Φ (ξ)|2 =00Φ′ (ξ)0000Φ′ (x)00|x − ξ|2 .

Plug (2.4.3) and (2.4.4) into !pα(Φ(x), Φ(ξ)) we have
(1 − |Φ(x)|2)1−α
|Φ(x) − Φ(ξ)|n−α

!pα (Φ (x) , Φ (ξ)) = 2α−1cn,α

= 2α−1cn,α

(1 − |x|2)1−α|Φ′(x)|1−α

|x − ξ|n−α|Φ′(x)|(n−α)/2|Φ′(ξ)|(n−α)/2

= !pα(x, ξ)|Φ′(x)|(2−n−α)/2|Φ′(ξ)|(α−n)/2.

For any Φ ∈ SO (n, 1) and any !f ∈ L

2(n−1)
n−2+α (Sn−1) we define

2(n−1)
n−2+α (Sn−1) and that

!fΦ (ξ) = !f ◦ Φ (ξ)00Φ′ (ξ)00(n−2+α)/2
Then it is easy to see that !fΦ ∈ L
’!fΦ’
!Pα.!fΦ/ (x) =.!Pα!f/ ◦ Φ (x)00Φ′ (x)00(n−2+α)/2

Moreover, the extension of !fΦ as defined in (2.3.1) transforms in the following way:

2(n−1)
n−2+α (Sn−1)

= ’!f’

2(n−1)
n−2+α (Sn−1)

L

L

.

.

.

2.5 Compactness

(2.4.4)

(2.4.5)

(2.4.6)

The goal of this section is to prove that the extension operator !Pα : Lp(Sn−1) → Lq(Bn)

is compact for certain choices of p and q. This is done in Corollary 2.10. Before we can prove

Corollary 2.10, we need to prove an estimate in Proposition 2.9. The proof of Proposition

2.9 depends on the following important technical lemma:

29

Lemma 2.7. For any n ≥ 2, α ∈ [2 − n, 1) and r ∈ (0, 1)

&Sn−1

r

!pα(x, ξ)dx ≤ 1,

is as in definition 2.1.

The notation Sn−1

r

Proof.

Using u-substitution, take u = tan(φ/2), we get

r

0

!pα(x, ξ)dx

&Sn−1
= 2α−1cn,α|Sn−2|rn−1(1 − r2)1−α& π
& π
(r2 + 1 − 2r cos(φ))(n−α)/2
1 + u2+n−2
0 * 2u
=& ∞
(1 − r)n−α& ∞

sinn−2(φ)

(1 + u2)(n−2+α)/2

2n−1

dφ

=

0

0

1

sinn−2(φ)

(r2 + 1 − 2r cos(φ))(n−α)/2

dφ.

2du
1 + u2

(2.5.1)

un−2

.r2 + 1 − 2r. 1−u2
1+u2//(n−α)/2
*. 1+r
1−r/2

du

u2 + 1+(n−α)/2

Using u-substitution again, take v = 1+r

0

1−r u, we have
dφ

sinn−2(φ)

& π
2n−1(1 − r)α−1

(1 + r)n−1

(r2 + 1 − 2r cos(φ))(n−α)/2
*1 +. 1−r
1+r/2

& ∞
& ∞
(v2 + 1)(n−α)/2
2 )Γ( n−1
2n−2(1 − r)α−1Γ( 1−α
2 )
(1 + r)n−1Γ( n−α
2 )

2n−1(1 − r)α−1

(1 + r)n−1

vn−2dv

0

0

vn−2

=

≤

=

dv

(v2 + 1)(n−α)/2

v2+(n−2+α)/2

Overall, we have

&Sn−1

r

!pα(x, ξ)dx ≤

=

2n−3+αrn−1cn,α|Sn−2|Γ( 1−α
2 )Γ( n−1
2 )
(1 + r)n−2+αΓ( n−α
2 )

2n−2+αrn−1
(1 + r)n−2+α

≤ 1.

30

Here we used (1.2.3) and the fact that the function

r for 2 − n ≤ α < 1 and 0 < r < 1.

rn−1

(1+r)n−2+α is an increasing function of

Remark 5. From (2.5.1) we see that when α = 2 − n

0

sinn−2(φ)

& π
(r2 + 1 − 2r cos(φ))n−1 dφ
0 * 2u
1 + u2+n−2
=& ∞
(1 − r)2n−2& ∞

2n−1

=

0

1

1+u2//n−1

.r2 + 1 − 2r. 1−u2
u2 + 1+n−1 .
*. 1+r
1−r/2

un−2du

2du
1 + u2

Now if we take v = 1+r

& π

sinn−2(φ)

1−r u then we have
(r2 + 1 − 2r cos(φ))n−1 dφ =

(1 − r2)n−1& ∞
2n−2.Γ. n−1
2 //2
(1 − r2)n−1Γ(n − 1)
where in the last step we used (1.2.3). As a result, for any x ∈ Bn

2n−1

=

0

0

,

vn−2dv

(1 + v2)n−1

&Sn−1!p2−n(x, ξ) = 1.

Remark 6. From the calculation in Lemma 2.7 we can also get a lower bound for the

integration. Note that

0

0

dφ

sinn−2(φ)

& π
2n−1(1 − r)α−1

(1 + r)n−1

(r2 + 1 − 2r cos(φ))(n−α)/2
*1 +. 1−r
1+r/2

& ∞
& ∞
2n−1(1 − r)α−1
2n−2(1 − r)α−1(Γ( n−1
(1 + r)n−1Γ(n − 1)

(v2 + 1)n−1
2 ))2

(1 + r)n−1

vn−2dv

0

.

vn−2

v2+(n−2+α)/2

=

≥

=

dv

(v2 + 1)(n−α)/2

As a result, we have

31

&Sn−1

r

!pα(x, ξ)dx ≥

2n−2+αrn−1Γ( n−α
2 )Γ( n−1
2 )
(1 + r)n−2+αΓ(n − 1)Γ( 1−α
2 )

Remark 7. In the calculation of Lemma 2.7, we have

*1 +. 1−r
1+r/2

vn−2

v2+(n−2+α)/2

vn−2

(v2 + 1)(n−α)/2

,

≤

(v2 + 1)(n−α)/2

for all r ∈ [0, 1] and all v ≥ 0. As a result, by dominated convergence theorem, for any
r0 ∈ [0, 1] we have

vn−2dv

0

lim

r→r0& ∞
= & ∞

0

vn−2dv

*1 +. 1−r
v2+(n−2+α)/2
1+r/2
v2+(n−2+α)/2
*1 +. 1−r0
1+r0/2
r→1& ∞
= & ∞

*1 +. 1−r
1+r/2

v2+(n−2+α)/2

(v2 + 1)(n−α)/2

lim

0

vn−2dv

.

vn−2dv

0

(v2 + 1)(n−α)/2

,

(v2 + 1)(n−α)/2

(v2 + 1)(n−α)/2

In particular, we have

Before we can prove the estimate in Proposition 2.9 we need to define the weak norm:

Definition 2.8. Define the weak norm Lp

W (Bn) , such that

|u|L

p
W (Bn)

= sup
t>0

t||u| > t|

1
p .

Here ||u| > t| is the measure of the set {|u| > t}.

We are now ready to prove the following estimates, the proof uses the same method as

in [18].

32

and

000!Pαf000L
’’’!Pαf’’’L

W (Bn) ≤ C(n, α)’f’L1(Sn−1),
n
n−1

np

n−1 (Bn) ≤ C(n, α, p)’f’Lp(Sn−1)

Proposition 2.9. For any n ≥ 2 and any 2− n ≤ α < 1 the extension operator !Pα satisfies

Proof. Note that we only need to prove the weak estimate. The strong estimate follows from
Marcinkiewicz interpolation theorem and the fact that for any x ∈ Bn

|!Pαf (x)| ≤ |f|L∞(Sn−1)&Sn−1!pα(x, ξ)dξ ≤ C(n, α)|f|L∞(Sn−1),

here the last step follows from Lemma 2.7. The constant C(n, α) here only depends on n

and α. Note that it is different from the notation cn,α, and that C(n, α) changes through out
the dissertation. To prove the weak type estimate. Assume that f ≥ 0 and |f|L1(Sn−1) = 1.
Note that

(2.5.2)

(2.5.3)

(1 − |x|2)1−α
|x − ξ|n−α
(1 − |x|2)1−α
(1 − |x|)n−α

!pα(x, ξ) = 2α−1cn,α
≤ C(n, α)
(1 − |x|)n−1 .

C(n, α)

≤

As a result, we have

From (2.5.3) we conclude that

0 ≤ !Pαf ≤

C(n, α)

(1 − |x|)n−1 .

|!Pαf > λ| = |{x ∈ Bn : 1 − |x| < C(n, α)λ− 1

n−1 ,!Pαf > λ}|

33

If 1 ≤ C(n, α)λ− 1

n−1 , then we have

|!Pαf > λ| ≤

≤

1

λ&Bn !Pαf dxdξ
λ&Sn−1

1

f (ξ)&Bn!pα(x, ξ)dxdξ

C(n, α)

λ

≤
≤ C(n, α)λ− n
≤ C(n, α)λ− n
n−1 .

n−1 λ

1
n−1

Note that here we used Lemma 2.7 and the fact that λ

constant C(n, α) changes along the argument.

1

n−1 ≤ C(n, α). Note also that the

If C(n, α)λ− 1

n−1 , then we have

n−1 < 1 then we can define r0 = 1 − C(n, α)λ− 1
|!Pαf > λ| ≤

1

1

λ&Bn\Bn
λ&Sn−1

r0 !Pαf dxdξ
f (ξ)&Bn\Bn

≤

r0!pα(x, ξ)dxdξ

C(n, α)(1 − r0)
≤
≤ C(n, α)λ− n
n−1 .

λ

This finishes the proof.

With the help of Proposition 2.9 we can prove the following:

Corollary 2.10. For any n ≥ 2, 2 − n ≤ α < 1, 1 ≤ p < ∞, and 1 ≤ q < np
!Pα : Lp(Sn−1) → Lq(Bn) is compact.
Proof. First assume 1 < p < ∞. Suppose we have a sequence of function fi ∈ Lp(Sn−1)
such that |fi|Lp(Sn−1) ≤ 1. Then from (2.5.3) we have for all i and all x ∈ Bn

n−1 the operator

|!Pαfi(x)| ≤

C(n, α)

(1 − |x|)n−1 .

34

By Schauder estimate, there exists u ∈ C2(Bn) such that !Pαfi → u in C2

we have: for r ∈ (0, 1)

loc(Bn). As a result

|!Pαfi − !Pαfj|Lq(Bn) ≤ |!Pαfi − !Pαfj|Lq(Bn
≤ |!Pαfi − !Pαfj|Lq(Bn
+|!Pαfi − !Pαfj|L
≤ |!Pαfi − !Pαfj|Lq(Bn

r )

np

where we used Holder inequality and Proposition 2.9. Hence

r ) + |!Pαfi − !Pαfj|Lq(Bn\Bn

r )

1

q− n−1

np

r )|Bn\Bn
r|

n−1 (Bn\Bn
r ) + C(n, α, p)|Bn\Bn
r|

1

q− n−1
np ,

lim sup
i,j→∞

|!Pαfi − !Pαfj|Lq(Bn) ≤ C(n, α, p)|Bn\Bn

r|

1

q− n−1
np .

Lq(Bn) is compact.

Letting r → 1, we see that !Pαfi is a Cauchy sequence in Lq(Bn), hence !Pα : Lp(Sn−1) →
We can see that !Pα : L

n−2+α (Bn) is not compact in the following

example, which is similar to the example given in [6, Chapter 1].

2(n−1)
n−2+α (Sn−1) → L

2n

Remark 8. We consider a sequence of conformal transformation Φa : Bn

→ Bn

defined by

Φa(x) =

a|x − a|2 + (1 − |a|2)(a − x)

|a|2|a∗ − x|2

.

Here a ∈ Bn such that a = (0, ..., 0, 1 − -) for some - ∈ (0, 1), and a∗ = a
we see that for any x ∈ Bn

|a|2 . From (2.4.3)

|Φ′a(x)| =

1 − |Φa(x)|2
1 − |x|2 =

1 − |a|2
|a|2|a∗ − x|2 ,

take limit x → ξ for some ξ ∈ Sn−1 we get
1 − |a|2
|a|2|a∗ − ξ|2 =

|Φ′a(ξ)| =

-(2 + -)

(1 − -)2|a∗ − ξ|2 .

If ξ ∕= (0, ..., 0, 1), then it is easy to see that lim’→0 |Φ′a(ξ)| = 0. If ξ = (0, ..., 0, 1) = a
|a|

,

then we have

0000Φ′a* a
|a|+0000 =

35

-(2 + -)

-2

,

hence lim’→0000Φ′a. a
Now consider the function !f : Sn−1 → R such that !f = 1. Define !fΦa as in (2.4.5), then

|a|/000 = ∞.

it is easy to see that

,

L

L

’!fΦa’

2(n−1)
n−2+α (Sn−1)

2(n−1)
n−2+α (Sn−1)
2(n−1)
n−2+α (Sn−1). For any given x ∈ Bn,

= ’!f’
and that !fΦa weakly converges to the zero function in L
think of !pα(x, ξ) as a function of ξ, using the L∞ bound (2.5.2) we can show that
’→0!Pα!fΦa(x) = 0.
Now we can show that !Pα!fΦa weakly converges to the zero function in L
n+2−α (Bn) and any r ∈ (0, 1), we have

any function in the dual space h ∈ L

n−2+α (Bn). For

lim

2n

2n

&Bn !Pα!fΦa(x)h(x)dx
=&Bn\Bn
≤’’’!Pα!fΦa’’’L

r !Pα!fΦa(x)h(x)dx +&Bn
n−2+α (Bn) ’h’L

2n

r !Pα!fΦa(x)h(x)dx
+&Bn

n+2−α (Bn\Bn
r )

2n

r !Pα!fΦa(x)h(x)dx,

where the second step follows from H¨older’s inequality. Note that from (2.4.6) we can see

that

n−2+α (Bn)
By dominated convergence theorem we have

’’’!Pα!fΦa’’’L

2n

=’’’!Pα!f’’’L

.

2n

n−2+α (Bn)

Combine the L∞ bound (2.5.3) with dominated convergence theorem we see that for any
r ∈ (0, 1)

Now for any δ > 0 small, we can choose r ∈ (0, 1) such that

lim

r→1’h’L

2n

n+2−α (Bn\Bn
r )

= 0.

’→0&Bn

lim

r !Pα!fΦa(x)h(x)dx = 0.
’h’L

n+2−α (Bn\Bn
r )

< δ.

2n

36

For this given r, we can choose - > 0 small such that

Combine these results, we see that

&Bn
r !Pα!fΦa(x)h(x)dx < δ.
’→0&Bn !Pα!fΦa(x)h(x)dx = 0

lim

2n

n+2−α (Bn).

for any h ∈ L
But since

we conclude that !Pα : L

2.6 Extremal Function

2n

’’’!Pα!fΦa’’’L

n−2+α (Bn)
2(n−1)
n−2+α (Sn−1) → L

2n

=’’’!Pα!f’’’L

n−2+α (Bn) ∕= 0,
n−2+α (Bn) is not compact.

2n

Using Corollary 2.10, we can identify the extremal function in the same way as in [18].

In order to do so we need the help of the Kazdan-Warner type condition in the following

lemma

Lemma 2.11. Suppose α ∈ (2 − n, 1), and K, f ∈ C1(Sn−1) such that for any ξ ∈ Sn−1

Let X be a conformal vector field in Bn

K(ξ)f (ξ)

n−2+α dx.

n−α

n−2+α =&Bn!pα(x, ξ).!Pαf (x)/ n+2−α
&Sn−1

2(n−1)
n−2+α dξ = 0.

, then we have

XK · f

Proof. Consider the functional

I(K, f ) =

|!Pαf|L
13Sn−1 K · f

37

2n

n−2+α (Bn)
2(n−1)

n−2+α dξ2 n−2+α

2(n−1)

with Euler-Lagrange equation

K(ξ)f (ξ)

n−α

n−2+α =&Bn!pα(x, ξ).!Pαf (x)/ n+2−α

n−2+α dx.

Consider Φt as the 1-parameter family of conformal group generated by X. Define

fΦt = f ◦ Φt(ξ)|Φ′t(ξ)|

n−2+α

2

.

Since f is a critical function for the functional I(K, f ), we have

Where according to the calculation of conformal invariance in the beginning, we have

I(K, fΦt) = 0.

d

dt0000t=0

I(K, fΦt) =

=

=

2n

n−2+α (Bn)
2(n−1)

n−2+α |Φ′t(ξ)|n−1dξ2 n−2+α

2(n−1)

2n

2(n−1)

n−2+α (Bn)
2(n−1)
n−2+α
Φt

dξ2 n−2+α
|!Pαf|L

|!PαfΦt|L
13Sn−1 K · f
13Sn−1 K ◦ Φ−t ◦ Φt(ξ) · (f ◦ Φt(ξ))
|!Pαf|L
n−2+α dξ2 n−2+α
13Sn−1 K ◦ Φ−t(ξ) · f

n−2+α (Bn)
2(n−1)

2(n−1)

2n

= I(K ◦ Φ−t, f ).

As a result, we have

I(K, fΦt) =

d

dt0000t=0

From this we can conclude that

d

dt0000t=0

I(K ◦ Φ−t, f ) = 0.

&Sn−1

2(n−1)
n−2+α dξ = 0.

XK · f

38

Now we can find the extremal function and the sharp constant using subcritical approx-

imation as in [18].

Theorem 2.12. Assume n ≥ 3 and α ∈ (2 − n, 1). For every f ∈ L

2(n−1)
n−2+α (Sn−1), we have

’’’!Pαf’’’L

2n

n−2+α (Bn) ≤ Sn,α ’f’

L

.

2(n−1)
n−2+α (Sn−1)

Where Sn,α is a constant that only depends on n and α. Up to conformal transformation

any constant is an optimizer.

Proof. For p > 2(n−1)

n−2+α , by corollary 2.10, the operator

is compact. Consider the variational problem

!Pα : Lp(Sn−1) → L

2n

n−2+α (Bn)

Sn,α = sup"$%’’’!Pαf’’’L

.

= 1EFG

2n

n−2+α (Bn)

: f ∈ Lp(Sn−1) such that ’f’

2(n−1)
n−2+α (Sn−1)

L

We show that the supremum is achieved as follows:

Consider a maximizing sequence fi ∈ Lp(Sn−1), with fi ≥ 0,

’fi’Lp(Sn−1) = 1

and

By uniform boundedness of Lp norm, we know that there exists a subsequence fi weakly

lim

i→∞’fi’Lp(Sn−1) = Sn,α.

n

converges to some function fp ∈ Lp(Sn−1). By compactness of !Pα we also know that there
exists a subsequence fi such that !Pαfi converges to v in L
Lp convergence of fi to fp, we also have !Pαfi converges to !Pαfp pointwise. As a result we
have !Pαfp = v and the supremum Sn,α is achieved at fp.
Replacing fp by f∗p if necessary, we may assume that fp is radial symmetric and decreas-
ing. Meaning that for ξ = (ξ1, ..., ξn) ∈ Rn such that |ξ| = 1, the function f (ξ) only depends
on ξn and that ∂f
∂ξn

n−2+α (Bn) norm. By the weak

(ξ) ≤ 0.

39

After rescaling, we may assume fp satisfy the Euler-Lagrange equation

n+2−α
n−2+α dx = fp(ξ)p−1 = fp(ξ)

n−α
n−2+α fp(ξ)

p−

2(n−1)
n−2+α

Apply Proposition C.5 from the appendix, we know that fp ∈ C1(Sn−1). By Lemma 2.11,
we have

&Bn!pα(x, ξ)(!Pαfp)(x)
&Sn−1〈∇fp(ξ)
& π

0

Consider the function gp(r) = fp(0, ..., 0, sin r, cos r) for r ∈ [0, π]. The equality becomes

p−

2(n−1)
n−2+α ,∇ξn〉fp(ξ)

2(n−1)
n−2 dξ = 0.

g′p(r)gp(r)p−1 sinn−2(r)dr = 0.

Note that g′p = −∂nf sin(r) ≥ 0. Hence we know that fp is actually a constant.

40

CHAPTER 3

LIMIT CASE

3.1 Chapter Outline

In this chapter we want to take limit α → 2 − n and study the limit case inequality. We

take the limit in section 3.2. In the process of taking the limit, a very special function !In
shows up. The property of the function!In is crucial in the study of the limit case inequality.
We prove several important properties of the function!In in section 3.3. We prove uniqueness

of the limit case inequality using the moving sphere method in section 3.5, but before that

we need to transform the limit case inequality to the upper half space in section 3.4.

Through out this chapter, we still use notations !f and f to denote functions on Sn−1 and

Rn−1 respectively, but the relation between them is different from the relation discussed in

Chapter 2. We will specify their relation in (3.4.2) below.

3.2 Proof of Theorem 1.13: Limit Case Inequality

We consider the limit case α → 2 − n in the same way as [8], our statement and proof

are slightly different.

For any !F ∈ L∞(Sn−1), define !f = 1 + n−2+α
α ∈ (2 − n, 1). We prove the following theorem for !F .
Theorem 3.1. For dimension n ≥ 2, and any function !F ∈ L∞(Sn−1) we have

!F . We have !f ∈ L

2

2(n−1)
n−2+α (Sn−1) for all

(3.2.1)

’e!In+!P2−n!F’Ln(Bn) ≤ Sn’e!F’Ln−1(Sn−1).
dα 00α=2−n. When n is even we have
Where !In(x) = 2 d!Pα1
2 / Γ (n − k − 1)
Γ. n−2
!In(x) =
2 − k)
Γ(n − 2)Γ( n

n/2−15k=1

1
2k ·

(1 − |x|2)k.

41

The sharp constant Sn = (e!In(Ln(Bn)
Proof. For any !F ∈ L∞(Sn−1), define !f = 1 + n−2+α

2.12, we have

|Sn−1|

1
n−1

2

!F . Define - = n− 2 + α, from theorem

’!Pα(1 + -!F )’L

which is equivalent to

2(n−1)
n−2+α (Sn−1)

L

,

2n

n−2+α (Bn) ≤ Sn,α’1 + -!F’
’89
!Pα1 2 n
-!Pα!F

≤ (Sn,α)

1

n

1
n

(!Pα1)

’ 11 +

67&Bn
’ + 1
As in [8], we need to find a lower bound for !Pα1 and an upper bound for (!Pα1)
We handle the lower bound for !Pα1 firstly. From remark 6, we know that

’*&Sn−1

(1 + -!F )

n−1

n−1

.

n
’ .

2 ) Γ. n−1
2 /
Γ( n−α
Γ(n − 1)Γ. 1−α
2 / .

!Pα1 ≥

Since

2 %
Γ$ n−α
2 %Γ$ n−1
Γ(n−1)Γ$ 1−α
2 % is a continuous function of α for all α ∈ [2 − n, 1), and

for all α such that 2 − n ≤ α < α0 < 1. Here m will be the lower bound for the function

for α < 1. As a result for some 0 < α0 < 1, there exists m > 0 such that

2 ) Γ. n−1
2 /
Γ( n−α
Γ(n − 1)Γ. 1−α
2 / > 0
2 /
2 ) Γ. n−1
Γ( n−α
Γ(n − 1)Γ. 1−α
2 / ≥ m > 0
!Pα1(x). Note that it does not depend on α or x.
Now we consider the upper bound for (!Pα1)
!pα(x, ξ)dx

≤* 2

!Pα1 = 3Sn−1

rn−1

r

42

n
’ . From Lemma 2.7, we have

1 + r+n−2+α

,

dα |α=2−n, then with the help of the lower bound
n
’ , we can apply dominated convergence theorem

n

to get

(!Pα1)

1 + r+n

for all α ∈ [2 − n, 1). As a result we have an upper bound
≤ 2n,

’ ≤* 2
for all α ∈ [2 − n, 1). If we define !In = 2 d!Pα1
for !Pα1(x) and the upper bound for (!Pα1)
’ 11 +
!Pα1 2 n
=&Bn
-!Pα!F
’ =&Sn−1
(1 + -!F )

(!Pα1)
’→0&Sn−1

For the right hand side of the inequality we can get

’→0&Bn

n−1

lim

lim

n

’

In order to find the limit lim’→0(Sn,α)

en!In+n!P2−n!F .

e(n−1)!F .

1
n , first note that since constant is an optimizer in

theorem 2.12. As a result, if we take !F = 0, then we can have
= ’e!In’Ln(Bn)
|Sn−1|

’→0.3Bn(!Pα1)

’/ 1

1
’ = lim

|Sn−1|

lim
’→0

1
n−1

1
n−1

(Sn,α)

n

n

.

When n is an even integer, using mathematical induction and (3.3.10) it is easy to prove

that

n/2−15k=1

1
2k ·

2 / Γ (n − k − 1)
Γ. n−2
Γ(n − 2)Γ( n
2 − k)

(1 − |x|2)k.

This finishes the proof.

!In(x) =
3.3 The Function !In
The function !In naturally appears in the process of taking limit; its properties are very
important for subsequent analysis. Yang [25] found an explicit formula for the function !In
when n is an even integer. In the previous section we use a different method to find !In when
When n is odd we do not have an explicit formula for !In, but the same induction relation

n is even. Our method relies on the induction relation (3.3.10) to be proved in this section.

still apply.

43

Subsection 3.3.3 we prove a very useful induction relation, which helps us to determine the

In Subsection 3.3.1 we consider how !In transforms under conformal transformation. In
explicit formula for !In in even dimensional unit ball. The induction relation is the basis for

the proofs in Subsection 3.3.5 and Subsection 3.3.6

3.3.1 Conformal Transformation of !In
In this subsection we want to take a closer look at how the function !In changes under
conformal transformation Φ : Bn → Bn, as well as how the it changes under the projection
map Ψ : Rn

+ → Bn. These are given in (3.3.2) and (3.3.3) respectively.

To begin with, we have

Taking derivative with respect to α at α = 2 − n, we get

(1 − |x|2)1−α
|x − ξ|n−α dξ
(2α−1cn,α)+&Sn−1

!Pα1 = 2α−1cn,α&Sn−1
dα0000α=2−n
= * d
−21−ncn,2−n&Sn−1
+21−ncn,2−n&Sn−1
dα0000α=2−n
(2α−1cn,α)&Sn−1

dα|α=2−n(2α−1cn,α)

d

d

(1 − |x|2)n−1
|x − ξ|2n−2 dξ
(1 − |x|2)n−1 ln(1 − |x|2)
(1 − |x|2)n−1 ln|x − ξ|

|x − ξ|2n−2

dξ

dξ

|x − ξ|2n−2

(1 − |x|2)n−1
|x − ξ|2n−2 dξ

dα 0000α=2−n
d!Pα1

Note that

=

21−ncn,2−n
ψ0(n − 1)

2

= ln(2) −

+

ψ0( n−1
2 )

2

,

where ψ0(x) = d

dx ln(Γ(x)) is the polygamma function. Hence the result simplifies to

dα 0000α=2−n
d!Pα1

= 21−ncn,2−n&Sn−1
− ln.1 − |x|2/ + ln(2) −

(1 − |x|2)n−1 ln|x − ξ|

dξ

|x − ξ|2n−2
ψ0(n − 1)
+

2

(3.3.1)

ψ0( n−1
2 )

2

44

Under conformal transformation Φ, we can see that

&Sn−1!pα(Φ(x), ξ) ln|Φ(x) − ξ|dξ
= &Sn−1
(!pα(x, ξ) ln|x − ξ|)|Φ′(x)|
+&Sn−1
2 )|Φ′(x)|
(!pα(x, ξ) ln|Φ′(ξ)|
+&Sn−1
(!pα(x, ξ) ln|Φ′(x)|
2 )|Φ′(x)|

1

1

2−n−α

2

n−2+α

2

dξ

|Φ′(ξ)|

2−n−α

2

2−n−α

2

|Φ′(ξ)|
|Φ′(ξ)|

n−2+α

2

dξ

n−2+α

2

dξ.

When we take limit α → 2 − n, we get

1
2

ln|Φ′(x)|.

&Sn−1!p2−n(Φ(x), ξ) ln|Φ(x) − ξ|dξ = &Sn−1!p2−n(x, ξ) ln|x − ξ|dξ

+

1

2!P2−n(ln|Φ′(ξ)|) +
dα 00α=2−n
From which we can get the conformal change for d!Pα1
dα 0000α=2−n
d!Pα1
d- 0000α=2−n
d!Pα1

dα 0000α=2−n ◦ Φ(x) +
d!Pα1

Recall that for any x ∈ Bn, we define

!In(x) = 2

ln|Φ′(x)| =

then we have

(x) +

(x),

1
2

1

2!P2−n(ln|Φ′(ξ)|).

!In ◦ Φ(x) + ln|Φ′(x)| =!In(x) + !P2−n(ln|Φ′(ξ)|).

From this we see that!In is a radial function, since when Φ is a rotation we have ln|Φ′(x)| = 0
for any x ∈ Bn and ln|Φ′(ξ)| = 0 for any ξ ∈ Sn−1. As a result, we sometimes think of!In(x)
as !In(r) for r = |x|.
We also want to consider how !In changes under the transformation Ψ : Rn
+ → Bn. For
any y ∈ Rn

+ define

(3.3.2)

In(y) = 2

dPα1

dα 0000α=2−n
= −2&Rn−1

p2−n(y, u) ln

yn

|y − u|

du − ψ0(n − 1) + ψ0* n − 1
2 + .

45

Note that since

Pα1 = cn,α&Rn−1
= cn,α&Rn−1

through change of variable, we get

Pα1 = cn,α&Rn−1

1
(|u|2 + 1)

n−α
2

In(y) = 2

dPα1

for all y ∈ Rn

+. Using (2.4.1) we can show that

du
yn−1
n

,

rn−2dr
n−α
2

(r2 + 1)

= 1.

du

1

n−α
2

n−α
2

y1−α
n
(|y′ − u|2 + y2
n)
(| y′
yn − u
yn|2 + 1)
du = cn,α|Sn−2|& ∞
dα 0000α=2−n

= 0

0

Note that in the last step we used (1.2.3). From this it is easy to see that

!In ◦ Ψ(y) + ln|Ψ′(y)| = In(y) + P2−n ln|Ψ′(w)| = P2−n ln|Ψ′(w)|.

Note that here w ∈ Rn−1, while |Ψ′(y)| and |Ψ′(w)| are as in (2.1.3) and (2.1.4) respectively.

(3.3.3)

3.3.2 Simplify the Function !In
In this subsection we want to further simplify the function !In. We write the integration in

the polar coordinate in the Euclidean ball Bn, then (3.3.1) becomes

dα 0000α=2−n
d!Pα1

=

(1 − r2)n−1& π
ψ0(n − 1)

21−nΓ(n − 1)
Γ. n−1
2 /2
− ln.1 − r2/ + ln(2) −

2

0

sinn−2 φ ln(1 − 2r cos φ + r2)dφ

(1 − 2r cos φ + r2)n−1
ψ0( n−1
2 )

+

.

2

Note that here we also used the explicit formula for the constant cn,2−n from (1.2.3).

Polygamma functions have two special properties that are useful to us. The first property

is

ψ0(n) = −γ +

n−15k=1

1
k

46

, where n ∈ N+.

(3.3.4)

Here γ is the Euler Mascheroni constant. The second property is

ψ0(2z) =

1

2*ψ0(z) + ψ0(z +

1
2

)+ + ln(2), where z ∈ C∗.

For any n ∈ N+ such that n ≥ 2, plug z = n−1

2

into (3.3.5) we get

2/ − ln(4).
Combine (3.3.6) and (3.3.4) we get that when n ∈ N+ is an even integer then

2 + − ψ0(n − 1) = ψ0(n − 1) − ψ0. n

ψ0* n − 1

ψ0* n − 1

2 + = − ln(4) − γ +

n−25k=n/2

1
k

+

n−25k=1

1
k

(3.3.5)

(3.3.6)

(3.3.7)

Now combine (3.3.4), (3.3.6) and (3.3.7) we see that for any n ≥ 2 such that n is an even
integer:

ψ0* n − 1

2 + − ψ0(n − 1) = − log(4) +

n−25k=n/2

1
k

.

Using (3.3.4) we can see that for any n ≥ 2 such that n is an odd integer:

ψ0* n − 1

2 + − ψ0(n − 1) = −

n−25k=(n−1)/2

1
k

.

As a result, when n ≥ 2 is an even integer, equation (3.3.1) simplifies to

dα 0000α=2−n
d!Pα1

=

dα 0000α=2−n
d!Pα1

=

When n ≥ 2 is an odd integer, equation (3.3.1) simplifies to

1
2

0

0

1
k

.

(1 − 2r cos φ + r2)n−1

2 /2 & π
21−nΓ(n − 1)
(1 − r2)n−1 sinn−2 φ ln(1 − 2r cos φ + r2)
Γ. n−1
n−25k=n/2
− ln.1 − r2/ +
2 /2 & π
21−nΓ(n − 1)
Γ. n−1
− ln.1 − r2/ + ln 2 −

(1 − r2)n−1 sinn−2 φ ln(1 − 2r cos φ + r2)

(1 − 2r cos φ + r2)n−1

n−25k=(n−1)/2

1
k

1
2

.

dφ

dφ

(3.3.8)

(3.3.9)

47

3.3.3

Induction Relation

If we take derivative of !In with respect to r, then we have the following induction relation.
Lemma 3.2. For n ∈ N+ such that n > 3, !In satisfies the induction relation

(3.3.10)

Proof. The main calculation here is to use integration by parts to evaluate the integral

1 − r4
4r(n − 3)

d
dr

!In =
& π

0

(!In−2) +!In−2 +
sinn−2 φ ln(1 − 2r cos φ + r2)

(1 − 2r cos φ + r2)n−1

1 − r2
2(n − 3)

.

dφ.

Take

then we have

Take

then we have

v = sinn−3 φ

dv = (n − 3) sinn−4 φ cos φdφ.

dw =

sin φ ln(1 − 2r cos φ + r2)
(1 − 2r cos φ + r2)n−1 dφ,

w = −

1

2r*

ln(1 − 2r cos φ + r2)

(n − 2)(1 − 2r cos φ + r2)n−2 +

(n − 2)2(1 − 2r cos φ + r2)n−2+ .

1

As a result, we have

0

=

(1 − 2r cos φ + r2)n−2

(1 − 2r cos φ + r2)n−1

& sinn−4 φ ln(1 − 2r cos φ + r2)dφ
(1 − 2r cos φ + r2)n−3

& π
sinn−2 φ ln(1 − 2r cos φ + r2)dφ
(n − 3)(1 + r2)
(n − 2)4r2
(n − 2)4r2& sinn−4 φ ln(1 − 2r cos φ + r2)dφ
(n − 3)
(n − 2)24r2 &
(n − 3)(1 + r2)
(n − 2)24r2&

(1 − 2r cos φ + r2)n−2
sinn−4 φdφ
(1 − 2r cos φ + r2)n−3 .

sinn−4 φdφ

n − 3

−

+

−

48

Note that here we used

& vdw = vw −& wdv = −& wdv,

and that sin 0 = sin π = 0. We evaluate each one of the integrals separately in the next

subsection. Using the results from next subsection, namely by (3.3.14), (3.3.15) and (3.3.16),

we have

0

(1 − 2r cos φ + r2)n−1

& π
sinn−2 φ ln(1 − 2r cos φ + r2)dφ
(1 + r2)
(n − 2)4r2
(n − 3)(1 + r2)

(1 − r2)
1

=

+

d

r

dr*& sinn−4 φ ln(1 − 2r cos φ + r2)dφ
1 − r2& sinn−4 φ ln(1 − 2r cos φ + r2)dφ

(1 − 2r cos φ + r2)n−3

(1 − 2r cos φ + r2)n−3

(n − 2)4r2

+

+

−

+

−

(n − 3)(1 − r2)n−1

2r2

2 )2

2n−4Γ( n−3
Γ(n − 3)

(n − 2)4r2
(n − 3)

(n − 3)(1 + r2)
(n − 2)4r2& sinn−4 φ ln(1 − 2r cos φ + r2)dφ
(n − 3)(1 + r2)
(n − 2)24r2
n − 3

2n−4Γ( n−3
Γ(n − 3)
2 )2

(1 − 2r cos φ + r2)n−3

(1 − r2)n−1
1
(1 − r2)n−3 .

2n−4Γ( n−3
Γ(n − 3)

1 + r2

2 )2

(n − 2)24r2
After reordering, we have

sinn−2 φ ln(1 − 2r cos φ + r2)dφ

(1 − 2r cos φ + r2)n−1
(1 + r2)

d

0

& π
(n − 2)4r(1 − r2)

=

dr*& sinn−4 φ ln(1 − 2r cos φ + r2)dφ

(1 − 2r cos φ + r2)n−3

+

(n − 3)

2(n − 2)(1 − r2)& sinn−4 φ ln(1 − 2r cos φ + r2)dφ

(1 − 2r cos φ + r2)n−3
1

2n−4Γ( n−3

1 + r2

1

2 )2

Γ(n − 3) * n − 3

n − 2

+

2 +

(1 − r2)n−1

+

+

If we define

n − 2

an =

21−nΓ(n − 1)
Γ. n−1
2 /2

(1 − r2)n−1& sinn−2 φ ln(1 − 2r cos φ + r2)dφ

(1 − 2r cos φ + r2)n−1

,

49

then we have the following induction relation

an =

1 − r4
4r(n − 3)

d
dr

(an−2) + an−2 +

1

2(n − 2)

+

1 + r2
4(n − 3)

When n is even, from (3.3.8) we see that

d- |’=0 = 2an − 2 ln.1 − r2/ +
d!Pα1

n−25k= n

2

1
k

.

!In = 2

Taking derivative with respect to r we see that

Combine (3.3.11), (3.3.12) and (3.3.13) we get that

d
dr

(!In−2) = 2

d
dr

(an−2) +

4r

1 − r2 .

(3.3.11)

(3.3.12)

(3.3.13)

(2an−2) + 2an−2 +

1

n − 2

+

1 + r2
2(n − 3)

1
k

d
dr

1 − r4
4r(n − 3)

!In =
n−25k=n/2
− 2 ln.1 − r2/ +
4r(n − 3)* d
1 − r4
(!In−2) −
+ 2an−2 − 2 ln.1 − r2/ +

dr

=

4r

1 − r2+
n−45k=(n−2)/2

1
k

+

1

+
n − 3
1 − r4
4r(n − 3)

1 + r2
2(n − 3)
d
dr

1 − r2
2(n − 3)
This is the end of the calculation for the case when n is even.

(!In−2) +!In−2 +

=

.

In the case when n is odd, from (3.3.9) we have

d- |’=0 = 2an − 2 ln.1 − r2/ + ln 4 −
d!Pα1

n−25k=(n−1)/2

1
k

.

!In = 2

Going through similar calculations as in the case when n is even, we can see that for the

case n is odd we have exactly the same induction relation.

Using the induction relation (3.3.10) it is easy to find an explicit formula for !In when n

is even.

50

3.3.4 Supplementary Calculation

In this subsection we continue several calculations from previous subsection. We use k to

denote any positive integer.

Lemma 3.3. For any k ∈ N+ such that k ≥ 2

& π

sink−1 φdφ

=

2k

1

|Sk−1|ck+1,1−k

(1 − r2)k

.

(3.3.14)

(1 − 2r cos φ + r2)k
Proof. This follows directly from Remark 5.

0

Lemma 3.4. For any k ∈ N+ such that k ≥ 2

& π

0

sink−1 φdφ

(1 − 2r cos φ + r2)k+1

=

2k−1(Γ(k/2))2

1 + r2

Γ(k)

(1 − r2)k+2

(3.3.15)

Proof. By taking derivative with respect to r, we get

r

d

sink−1 φdφ

dr1& π
(1 − 2r cos φ + r2)k+1 −

(1 − 2r cos φ + r2)k2
1 − r2& π

sink−1 φdφ

1

0

0

k(1 − r2)
=& π

0

sink−1 φdφ

(1 − 2r cos φ + r2)k

.

Combine this with (3.3.14) then we are done.

Lemma 3.5. For any k ∈ N+ such that k ≥ 2

sink−1 φ ln(1 − 2r cos φ + r2)dφ

(1 − 2r cos φ + r2)k+1

d

dr1& π

0

sink−1 φ ln(1 − 2r cos φ + r2)dφ

(1 − 2r cos φ + r2)k
sink−1 φ ln(1 − 2r cos φ + r2)dφ

(1 − 2r cos φ + r2)k

2

0

& π
k(1 − r2)

=

r

1

1 − r2& π

0

+

+

2k−1(Γ(k/2))2

2r2

.

Γ(k)

k(1 − r2)k+2

51

with respect to r, we can get

Proof. Take derivative of3 π

0

sink−1 φ ln(1−2r cos φ+r2)dφ

(1−2r cos φ+r2)k

sink−1 φ ln(1 − 2r cos φ + r2)dφ

(1 − 2r cos φ + r2)k+1

0

& π
k(1 − r2)

=

r

+

−

+

0

1

1

1 − r2& π
k(1 − r2)& π
k& π

1

0

d

dr1& π

sink−1 φ ln(1 − 2r cos φ + r2)dφ

0

(1 − 2r cos φ + r2)k
sink−1 φ ln(1 − 2r cos φ + r2)dφ

2

(1 − 2r cos φ + r2)k
sink−1 φdφ

(1 − 2r cos φ + r2)k

0
sink−1 φdφ

.

(1 − 2r cos φ + r2)k+1

By (3.3.14) and (3.3.15), we have

1

k&

=

sink−1 φdφ

(1 − 2r cos φ + r2)k+1 −
2k−1(Γ(k/2))2
k(1 − r2)k+2

Γ(k)

2r2

1

k(1 − r2)&

sink−1 φdφ

(1 − 2r cos φ + r2)k

Combine these two equations we can get (3.3.16).

3.3.5 Hyperbolic Harmonic Through Induction

Using the induction relation (3.3.10), we can prove that!In◦Ψ(y)+ln|Ψ′(y)| is harmonic with

respect to the standard hyperbolic metric. We prove it in the unit ball model of hyperbolic

space.

Lemma 3.6. For n ≥ 2, in Bn we have

Here ∆H is the Laplacian in hyperbolic space. For any function u ∈ C∞(Bn) we have

∆H*!In + ln
+2
∆Hu =* 1 − |x|2

2

1 − 2xn + |x|2

2

+ = 0.

∆u + (n − 2)

1 − |x|2

2

〈x,∇u〉.

Where ∆ is the Laplacian in Euclidean space and 〈x,∇u〉 is the inner product in Euclidean
space.

52

Proof. Through direct calculation we have

1 − 2xn + |x|2

2

∇ ln

=

2x − 2en
1 − 2xn + |x|2 ,

here en means the unit vector in the direction xn.

As a result, we have

∆ ln

1 − 2xn + |x|2

2

=

2n − 4

1 − 2xn + |x|2 ,

and

∆H ln

1 − 2xn + |x|2

2

=

=

Next we want to show that

= * 1 − |x|2

2

+2
1 − |x|2

2n − 4

2

2x − 2en

1 − 2xn + |x|2I

1 − 2xn + |x|2
Hx,
2(1 − 2xn + |x|2)

+(n − 2)
(1 − |x|2)2(n − 2) + (n − 2)(1 − |x|2)(2|x|2 − 2xn)
(n − 2)(1 − |x|2)

2

∆H!In = −

(n − 2)(1 − |x|2)

2

.

(3.3.16)

Since !In is a radial function, we can verify this in polar coordinates in the Euclidean unit

ball. Where we have

∂2

∆H!In =* 1 − r2
2 +2

r!In +* 1 − r2
2 +2
For n = 2 it is easy to see that !I2 = 0, and for n = 3, we can integrate by part to get
!I3 = ln(4) + (1−r)2 ln(1−r)−(1+r)2 ln(1+r)

. So it is easy to verify by direct calculation that

(3.3.16) is true for n = 2 and n = 3.

(n − 2)r(1 − r2)

∂r!In +

∂r!In

n − 1

r

2

2r

53

∂2

When n > 3, suppose (3.3.16) is true for !In−2, from which we have
∂r!In−2

r!In−2 +* 1 − r2
2 +2
∆H!In−2 = * 1 − r2
2 +2
(n − 4)r(1 − r2)
∂r!In−2
= * 1 − r2
2 +2
1 − r2
r!In−2 +
4r
(n − 4)(1 − r2)

n − 3

∂2

+

2

r

.

= −

2

((n − 3) + (n − 5)r2)∂r!In−2

After rearranging we get

Consider (3.3.10), take derivative with respect to r, we get

∂2

r!In−2 = −
1 − r4
4r(n − 3)

2(n − 4)
1 − r2 −
r!In−2 +*1 −

∂2

∂r!In =

2(n − 4)r

n − 3

r

1 + 3r4

∂r!In−2.
1 − r2 ∂r!In−2 −
4r2(n − 3)+ ∂r!In−2 −
2r2(n − 3)+ ∂2
r!In−2

1 + 3r4

1

r

n − 3

(3.3.17)

,

∂2

r!In =

Using (3.3.17), we have

∂3

r!In−2 = −

.

n − 3

1 − r4
∂3
4r(n − 3)
1 − 3r4
+
2r3(n − 3)

r!In−2 +*1 −
∂r!In−2 −
(1 − r2)2 +* n − 3
−* n − 3

1 − r2 + ∂2

2(n − 4)r

4(n − 4)r

r2 −

+

r

and

and

+ ∂r!In−2

2(n − 4)(1 + r2)
(1 − r2)2
r!In−2,

= −

∂3

1 − r4
r!In−2
4r(n − 3)
+* 1 − r4
(n − 4)(1 + r2)
4r3 −
(n − 3)(1 − r2)
−* 1 − r4
(n − 4)(1 + r2)

4r2 +

2(n − 3)

2r(n − 3)(1 − r2)+ ∂r!In−2
(n − 4)(1 + r2)2
+ ∂2
r!In−2

54

As a result, we have

∂2

4r2(n − 3)

r!In = *−(n + 1)r4 + 2(n − 2)r2 − (n − 1)

+ ∂2
r!In−2
4r3(n − 3)(1 − r2)

+* (n − 1) − (3n − 9)r2 − (5n − 13)r4 − (n − 11)r6
(n − 3) + (n − 5)r2
(n − 3)(1 − r2)

−

+ ∂r!In−2

Hence

∂2

r!In
4r2(n − 3)

* 1 − r2
2 +2
+* 1 − r2
2 +2
= *−(n + 1)r4 + 2(n − 2)r2 − (n − 1)
+* (n − 1) − (3n − 9)r2 − (5n − 13)r4 − (n − 11)r6
8r3(n − 3)
* 1 − r2
2 +
(n − 3) + (n − 5)r2

2(n − 3)

−

∂2

r!In−2
+ 1 − r2

2

∂r!In−2

=

n − 1

* 1 − r2
2 +2
∂r!In
r
4r2(n − 3) * 1 − r2
2 +2
(n − 1)(1 − r4)
+* (n − 1)(1 − r2)2
(n − 1)(1 − r2)2(1 + 3r4)
(n − 1)(1 − r2)2

r!In−2
16r3(n − 3)

−

∂2

4r

−

4(n − 3)

+ ∂r!In−2

=

∂r!In
* 1 − r2
2 +2

∂2

r!In−2
4(n − 3)r2+ (n − 2)r(1 − r2)

2

2

(n − 2)r(1 − r2)
(n − 2)(1 + r2)
+*1 −
1 + 3r4
(n − 2)r2(1 − r2)

2(n − 3)

−

2(n − 3)

∂r!In−2

55

Adding them up, we get

∆H!In =

∂2

n − 3

* 1 − r2
2 +2

(n − 2) − r2
r!In−2
+−(n − 5)r4 + (n2 − 8n + 13)r2 + (n − 2)(n − 3)
2r(n − 3)
(1 − r2)((n − 2) + (n − 4)r2)

−

2(n − 3)

* 1 − r2
2 + ∂r!In−2

Note that we have

(n − 2) − r2

(n − 3)

(n − 3) + (n − 5)r2

2r

·

= −(n − 5)r4 + (n2 − 8n + 13)r2 + (n − 2)(n − 3)

,

2r(n − 3)

as a result, we have

=

∆H!In
(n − 2) − r2
n − 3
(1 − r2)((n − 2) + (n − 4)r2)

1* 1 − r2
2 +2

∂2

r!In−2 +

−

2(n − 3)

(n − 3) + (n − 5)r2

2r

* 1 − r2

2

∂r!In−2+2

Use induction assumption in the form

* 1 − r2
2 +2

Then we get

∂2

r!In−2 +

Which finishes the proof.

1 − r2
4r

((n − 3) + (n − 5)r2)∂r!In−2 = −
∆H!In = −

(n − 2)(1 − r2)

2

.

(n − 4)(1 − r2)

2

,

3.3.6 Boundary Value Through Induction

Using the induction relation (3.3.10), we can find boundary value for !In. We have the

following lemma.

56

Lemma 3.7. For n > 2 we have

and

lim

r→1!In = 0,
∂r!In = −1,

lim
r→1

lim

r→1!In − 0

r − 1

= −1.

Proof. For n = 2, we have !In = 0. For n = 3 we have

(1 − r)2 ln(1 − r) − (1 + r)2 ln(1 + r)

.

2r

!I3 = ln 4 +

It is easy to see that

and

In general we use induction to prove that

lim

r→1!I3 − 0

r − 1

= −1.

lim

r→1!I3 = 0,
∂r!I3 = −1,

lim
r→1

lim

r→1!In = 0
∂r!In = C.

lim
r→1

lim
r→1

lim

r→1!In−2 = 0
∂r!In−2 = C.
∂r!In−2 + lim

57

and

and

Here C = 0 for n = 2 and C = −1 for n ≥ 3. Suppose the induction assumption holds for
n − 2, then we have

Now consider !In. By (3.3.10), we have

1 − r4
4r(n − 3)

lim
r→1

lim

r→1!In = lim

r→1

r→1!In−2 + lim

r→1

1 − r2
2(n − 3)

.

Using the induction assumption for !In−2, it is easy to see that

Start with (3.3.10), take derivative with respect to r, we get

∂r!In =
∂r!In =

Use (3.3.17) to substitute ∂2
4r(n − 3)*−
1 − r4
+*1 −

1 + 3r4

4r2(n − 3)+ ∂r!In−2 −
1 − r2 ∂r!In−2 −

2(n − 4)r

r

n − 3

r

.

n − 3

r

n − 3

.

∂r!In−2+

Use the induction assumption and take limit we have

∂2

lim

r→1!In = 0.
r!In−2 +*1 −
1 − r4
4r(n − 3)
r!In−2, then we get
2(n − 4)
1 − r2 −
4r2(n − 3)+ ∂r!In−2 −
∂r!In = −1.
r→1!In(r) − 0

1 + 3r4

r − 1

lim
r→1

lim

= C,

Now we use induction to show that

again, we have C = 0 for n = 2 and C = −1 for n ≥ 3. Suppose it is true for n − 2 then we
have the induction assumption

Use (3.3.10), then we have

lim

r→1!In−2(r) − 0

r − 1

= C.

!In − 0

r − 1

= −

(1 + r)(1 + r2)

4r(n − 3)

∂r!In−2 + !In−2 − 0
r − 1 −

1 + r
2(n − 3)

.

Now take limit and use the induction assumption, then we have

lim

r→1!In(r) − 0

r − 1

= −

C
n − 3

1

.

+ C −

When n ≥ 5, we have limr→1 ∂r!In−2 = −1 and limr→1 !In−2−0

have

n − 3
r−1 = −1, and as a result we

lim

r→1!In(r) − 0

r − 1

= −1.

58

When n = 4, we have limr→1 ∂r!In−2 = 0 and limr→1 !In−2−0

1

r−1 = 0, and

lim

r→1!In(r) − 0

r − 1

= −

n − 3

= −1.

This concludes the proof.

3.4 Conformal Trasformation of the Inequality

We will prove Theorem 1.14 in the next section.

In this section we consider how the

inequality (3.2.1) and its corresponding Euler Lagrange equation transforms under conformal

transformation Ψ : Rn

+ → Bn.

Note that in (3.2.1) we have the inequality

’e!In+!P2−n!f’Ln(Bn) ≤ Sn’e!f’Ln−1(Sn−1),

with Euler-Lagrange euquation

e(n−1)!f (ξ) =&Bn

Consider change of variable

&Sn−1

If we define

e(n−1)!f dξ =&Rn−1

en!In+n!P2−n!f!p2−n(x, ξ)dx
e(n−1)(!f◦Ψ+ln|Ψ′(w)|)dw.

(3.4.1)

(3.4.2)

then we have ’’’e!f’’’Ln−1(Sn−1)

change of variable, we have

f (w) = !f ◦ Ψ(w) + ln|Ψ′(w)|,
= ’’’ef’’’Ln−1(Rn−1)

. On the other hand, using the same

&Bn

en!In+n!P2−n!f dx = &Rn
= &Rn

+

+

en!In◦Ψ+n(!P2−n!f )◦Ψ+n ln|Ψ′(y)|dy
enP2−n(!f◦Ψ+ln|Ψ′(w)|)dy.

59

Note that in the last step we used (2.4.1) and (3.3.3). Which means we have

’e!In+!P2−n!f’Ln(Bn) = ’eP2−nf’Ln(Rn
+).

As a result, for any !f ∈ L∞(Bn), if we define a corresponding f as in (3.4.2), then we

have the corresponding inequality

’eP2−nf’Ln(Rn

+) ≤ Sn’ef’Ln−1(Rn−1).

(3.4.3)

If we further suppose !f satisfies the Euler-Lagrange equation (3.4.1), then f satisfies the

Euler-Lagrange equation

e(n−1)f (w) =&Rn

+

enP2−nf p2−n(y, w)dy.

(3.4.4)

3.5 Uniqueness Through the Method of Moving Spheres

In this section, we prove the uniqueness of solutions to equation (3.4.1). We define several

notation related to the inversion with respect to a sphere in Subsection 3.5.1.

We prove Theorem 1.14 from Subsection 3.5.2 to Subsection 3.5.5. We will start with a

smooth solution !f ∈ C∞(Sn−1) to (3.4.1), then define the corresponding f as in (3.4.2). We

will prove that the function f is unique up to conformal transformation using the method

of moving sphere.

Note that this is different from directly proving uniqueness of smooth solutions to (3.4.4).
Since by starting with smooth solutions of (3.4.1) we gain the asymptotic behavior ln|Φ′(u)|
as in (3.4.2). This asymptotic behavior helps us to start the sphere in subsequent parts.

3.5.1 Notation

For any w ∈ Rn−1 and λ ∈ R, define v = (w, 0) ∈ Rn. For all y ∈ Rn such that y ∕= v define
the inversion with respect to a sphere centered at v with radius λ as

φλ,v(y) = v +

λ2
|y − v|2 (y − v).

60

(3.5.1)

Note that φλ,v(y) maps the upper half space into the itself, and φλ,v(y) : Rn

+ → Rn
conformal transformation. We use |φ′λ,v(y)| to denote the conformal factor such that

+ is a

φ∗λ,vdy2 = |φ′λ,v(y)|2dy2

Note that |φ′λ,v(y)|n is the Jacobian in Rn
see that

+. Through direct calculation using (3.5.1), we can

|φ′λ,v(y)|n =

λ2n
|y − v|2n ,

and

|φ′λ,v(y)| =

λ2

|y − v|2

Note that since Ψ ◦ φλ,v ◦ Ψ−1 is a conformal transformation that maps Bn to itself, we
have Ψ ◦ φλ,v ◦ Ψ−1 ∈ SO(n, 1). For any f : Rn−1 → R, we can define fλ,v as the conformal
transformation of f under φλ,v such that

(3.5.2)

fλ,v = f ◦ φλ,v + ln|φ′λ,v|.

(3.5.3)

In addition we define

Bλ,v = {y ∈ Rn : |y − v| < λ},
λ,v denotes the closure of B+

λ,v in Rn. We also use Rn

+ to denote

λ,v = Bλ,v ∩Rn

with B+
the closure of Rn

+ in Rn.

+ and B+

3.5.2

Inversion with Respect to Spheres

We now start the proof of Theorem 1.14.

Suppose !f ∈ C∞(Sn−1) is a solution to (3.4.1), for w ∈ ∂Rn
f (w) = !f ◦ Ψ(w) + ln|Ψ′(w)|.

+ = Rn−1 define

61

Through the discussion in Section 3.4 we know that f is a solution to (3.4.4). Under trans-

formation φλ,v0

as defined in (3.5.1) we can define fλ,v0

as in (3.5.3)

fλ,v0

(w) = f ◦ φλ,v0

(w) + ln|φ′λ,v0

(w)|

= !f ◦ Ψ ◦ φλ,v0
= !f ◦ Ψ ◦ φλ,v0

(w) + ln|Ψ′(φλ,v0

(w))| + ln|φ′λ,v0
2λ2

(w)|

(w) + ln

(1 + |v0|2)|w − v0|2 + λ4 + 2λ2〈v0, w − v0〉

.

(w)| is as in (3.5.2). The notation

Note that here |Ψ′(φλ,v0
(w))| is as in (2.1.4) and |φ′λ,v0
〈v0, u − v0〉 denotes the Euclidean inner product in Rn.

In the special case v0 = 0, we have

fλ,0(v) = !f ◦ Ψ ◦ φλ,0(v) + ln

2λ2
λ4 + |v|2 .

(3.5.4)

Remark 9. Note that for any !f ∈ C∞(Sn−1) and any λ > 0 the function fλ,0 is smooth in

Rn−1. We can see this using the definition of φλ,v0

and Ψ, for any w ∈ Rn−1

Ψ ◦ φλ,0(w) = Ψ* λ2
|w|2 w+
=6J7
2λ2
|w|2 w
1 + λ4
|w|2
=* 2λ2w
λ4 + |w|2 ,

8K9
, −1 + λ4
|w|2
1 + λ4
|w|2
λ4 + |w|2+ .
λ4 − |w|2

If we define Ψ ◦ φλ,0(0) = (0, ..., 0, 1) ∈ Rn then it is easy to see that the map

Ψ ◦ φλ,0 : Rn−1 → Sn−1

is smooth. The other parts of the function fλ,0 is also smooth.

Through similar calculation we can also conclude that for any !f ∈ C∞(Sn−1), any λ > 0
and any v0 ∈ Rn−1, the function fλ,v0
On the other hand, for any !f ∈ C∞(Sn−1) define f as in (3.4.2) then we have

is smooth in Rn−1.

P2−n(f ) = P2−n(!f ◦ Ψ) + P2−n(ln|Ψ′(w)|).

62

Using (3.3.3) and Remark 3, we see that

P2−n(f ) = P2−n(!f ◦ Ψ) +!In ◦ Ψ(y) + ln|Ψ′(y)|
= (!P2−n!f ) ◦ Ψ +!In ◦ Ψ(y) + ln|Ψ′(y)|.

, using (A.0.2) and (A.0.4), we have for any y ∈ Rn

+

For fλ,v0

P2−nfλ,v0

(y) = P2−n(f ) ◦ φλ,v0

(y)|

= P2−n(!f ◦ Ψ) ◦ φλ,v0
+ ln|Ψ′(φλ,v0
= P2−n(!f ◦ Ψ) ◦ φλ,v0

(y) + ln|φ′λ,v0
(y) +!In ◦ Ψ ◦ φλ,v0
(y) +!In ◦ Ψ ◦ φλ,v0

(y))| + ln|φ′λ,v0

(y)|

2λ2

+ ln

(y)

(y)

(1 + |v0|2)|y − v0|2 + λ4 + 2λ2〈v0, y − v0〉 + λ2yn

(3.5.5)

.

Using result from Remark 3 we have

P2−nfλ,v0

(y) =.!P2−n!f/ ◦ Ψ ◦ φλ,v0

+ ln

(y) +!In ◦ Ψ ◦ φλ,v0

2λ2

(y)

(1 + |v0|2)|y − v0|2 + λ4 + 2λ2〈v0, y − v0〉 + λ2yn

(3.5.6)

.

Remark 10. Note that when !f ∈ C(Sn−1), both functions P2−nf and P2−nfλ,v0

+. We can see this from (3.5.5) and (3.5.6) using similar calculations as in

tinuous in Rn

are con-

Remark 9.

Again, note that . When v0 = 0, we have

We can show the following result, which will be used in subsequent parts.

2λ2

|y|2 + λ4 + λ2yn

(3.5.7)

P2−nfλ,0(y) =.!P2−n!f/ ◦ Ψ ◦ φλ,0(y) +!In ◦ Ψ ◦ φλ,0(y) + ln
Lemma 3.8. Suppose !f ∈ C(Sn−1). Define f , fλ,v0

Define the coresponding extensions P2−nf and P2−nfλ,v0
P2−nfλ,v0
f and fλ,v0

are harmonic in Rn

respectively.

as in (3.4.2) and (3.5.3) repectively.

as in (3.5.5), then P2−nf and
+ with the standard Hyperbolic metric and with boundary values

63

Proof. We consider P2−nf firstly. Using (3.5.5) and (2.4.1) we can see that

+ = Rn−1.

+ with the Hyperbolic

harmonic. Since Ψ : Rn

P2−n(f ) = P2−n(!f ◦ Ψ) +!In ◦ Ψ(y) + ln|Ψ′(y)|
= (!P2−n!f ) ◦ Ψ +!In ◦ Ψ(y) + ln|Ψ′(y)|.
From Proposition 2.4 we can see that (!P2−n!f ) ◦ Ψ is harmonic in Rn
metric and that !P2−n!f ◦ Ψ(u) = !f ◦ Ψ(u) for all u ∈ ∂Rn
From Lemma 3.6 we see that in Bn the fucntion !In(x) + ln 1−2xn+|x|2
have !In ◦ Ψ(y) + ln|Ψ′(y)| is also harmonic in Rn
From Lemma 3.7 we see that!In is continuous up to the boundary and that!In◦ Ψ(w) = 0
+ = Rn−1. As a result, we have !In ◦ Ψ(w) + ln|Ψ′(w)| = ln|Ψ′(w)| for all
for all w ∈ ∂Rn
+ = Rn−1. This finishes the proof for P2−nf .
w ∈ ∂Rn
. By Remark 9, fλ,v0

is hyperbolic
+ → Bn is an isometry between two models of hyperbolic spaces, we
+ with the Hyperbolic metric. Note that

here we used the relation (2.1.2).

is continuous on ∂Rn

Now we consider P2−nfλ,v0

+. Using the

2

definition of fλ,v0

, (A.0.2) and (A.0.4), we have

P2−nfλ,v0

= P2−nf ◦ φλ,v0

+ ln|φ′λ,v0|.

+ with the hyperbolic metric. On the other hand, using (3.5.2), by direct calculation as

+ with hyperbolic metric, we have P2−nf ◦ φλ,v0

is harmonic

is an isometry of Rn

Since φλ,v0
in Rn
in (B.3.1) we can see that ln|φ′λ,v0| is harmonic in Rn

+ with the hyperbolic metric.

From (3.5.6) and similar calculations as in Remark 9 we see that P2−nfλ,v0
+. As a result, P2−nfλ,v0

is hyperbolic harmonic in Rn

+ with boundary value fλ,v0

n

.

is continuous

in R

Note that in this proof we only need the asymptotic behavior (3.5.4) and (3.5.7) and

Proposition 2.4. We also need the regularity !f ∈ C(Sn−1) in order to use Proposition 2.4,
but we do not need to assume !f to be a solution of (3.4.1).

64

3.5.3 Start the Sphere

Now we start the moving sphere argument. The first step is called start the sphere; and

we prove it in the following two propositions. The proof of Proposition 3.9 relies on the

asymptotic behavior (3.5.4) while the proof of Proposition 3.10 relies on the asymptotic

behavior (3.5.7) and the maximum principle.

Proposition 3.9. For any !f ∈ C∞(Sn−1) any λ > 0 and any v0 ∈ ∂Rn

as in (3.4.2) and (3.5.3) respectively. Then there exists λ0 > 0 depending only on

+ = Rn−1, define f

and fλ,v0
v0, such that for all 0 < λ < λ0

(n−1)fλ,v0 (w) < e(n−1)f (w),

e

for all |w − v0| > λ.

Proof. Our proof uses similar ideas as in Lemma 3.1 of [19] and Lemma 2.1 of [20]. Our

proof is even simpler since we have the asymptotic behavior (3.5.4).

Consider in polar coordinate of Rn−1, w = (r, θ) ∈ Rn−1 where r ≥ 0 and θ ∈ Sn−2.
Since f ∈ C∞(Rn−1) (actually we only need f to be C1) and e(n−1)f > 0, there exists
r0 > 0 such that

d

dr.rn−1e(n−1)f (r, θ)/ > 0

for all 0 < r < r0 and all θ ∈ Sn−2.

As a result, when 0 < λ < |w| < r0 we have

* λ2
|w|+n−1

e(n−1)f* λ2w

|w|2+ < |w|n−1e(n−1)f (w),

and hence using (3.5.3) we have

(n−1)fλ,0(w) < e(n−1)f (w)

e

for all 0 < λ < |w| < r0. (Note that here we used the fact that when 0 < λ < |w|, we have

00 λ2w
|w|002 = λ2

|w|

< |w|.)

65

For |u| > r0, from (3.4.2) and (3.5.4) we have

and

e

Define m = infξ∈Sn−1 e!f (ξ) and M = supξ∈Sn−1 e!f (ξ). Note that we have 0 < m ≤ M .

Choose λ0 small enough such that 0 < λ0 < r0 and

2

e(n−1)f (w) =*
(n−1)fλ,0(w) =* 2λ2

1 + |w|2+n−1
λ4 + |w|2+n−1

e(n−1)!f◦Ψ(w)
e(n−1)!f◦Ψ* λ2w
|w|2+ .

and

m − λ2

0M > m/2,

r2
0m/2 > λ2

0M.

Then we have for all |w| > r0 and 0 < λ < λ0

(m − λ2M )|w|2 > (m − λ2
0/2

> mr2

0M )r2
0

> λ2

0M

> λ2M
> λ2(M − λ2m).

As a result, we have for all 0 < λ < λ0 < r0 < |w|,

2m

1 + |w|2 >

2λ2M
|w|2 + λ4 ,

66

and hence

e(n−1)f (w) = *

e(n−1)!f◦Ψ(w)

2

1 + |w|2+n−1
1 + |w|2+n−1
|w|2 + λ4+n−1
|w|2 + λ4+n−1

≥ * 2m
> * 2λ2M
≥ * 2λ2

= e

(n−1)fλ,0(w).

e(n−1)!f◦Ψ* λ2w
|w|2+

With the chosen λ0, combining the two cases together, we get for all 0 < λ < λ0 and all

|w| > λ, we have

e(n−1)f (w) > e

(n−1)fλ,0(w).

Remark 11. Note that

(n−1)fλ,v0 (w) < e(n−1)f (w)

e

for all |w − v0| > λ is equivalent to

(n−1)fλ,v0 (w) > e(n−1)f (w)

e

for all |w− v0| < λ. Since for any w such that |w− v0| < λ, we can define wλ,v0
Then we have

= φλ,v0

(w).

φλ,v0

(wλ,v0

) = φλ,v0 ◦ φλ,v0

(w) = w,

(3.5.8)

and

|wλ,v0 − v0| = |φλ,v0

(w) − v0| =

λ2

|w − v0|

> λ.

67

As a result, using Theorem 3.9, we have for |w − v0| < λ

e

(n−1)fλ,v0 (w) = |φ′λ,v0
> |φ′λ,v0
= |φ′λ,v0
= e(n−1)f (w).

)

(w)|n−1e(n−1)f (wλ,v0
(w)|n−1e
(w)|n−1|φ′λ,v0

(wλ,v0

(n−1)fλ,v0 (wλ,v0

)

)|n−1e(n−1)f (w)

Note that in the last step we used

|φ′λ,v0

(w)|n−1|φ′λ,v0

(wλ,v0

)|n−1 = 1,

(3.5.9)

which follows from (3.5.8) and chain rule.

With the help of the previous remark, we can use maximum principle to prove similar

results for P2−nf .

respectively.

Proposition 3.10. For any !f ∈ C∞(Sn−1), define f and fλ,v0
For any v0 ∈ ∂Rn

+ = Rn−1 there exists λ0 > 0 such that for all 0 < λ < λ0 we have

as in (3.4.2) and (3.5.3)

enP2−nf (y) < e

nP2−nfλ,v0 (y)

for all |y − v0| < λ.

Proof. From Lemma 3.8 we see that P2−nf and P2−nfλ,v0
boundary values f and fλ,v0
four functions f , fλ,v0
Choose the same λ0 as in Theorem 3.9. Using the previous remark we have

respctivly. By the discussion at the beginning of this section, all
+ = Rn−1.

are continuous at the point v0 ∈ ∂Rn

, P2−nf and P2−nfλ,v0

are Hyperbolic harmonic with

e(n−1)f (w) < e

(n−1)fλ,v0 (w),

for any 0 < λ < λ0 and for all w ∈ Rn−1 such that |w − v0| < λ. Which is equivalent to

f (w) < fλ,v0

(w)

68

for all w ∈ Rn−1 such that |w − v0| < λ. On the other hand, using the definition of fλ,v0
(A.0.2) and (A.0.4), we can see

,

P2−nf (y) = P2−nfλ,v0

(y)

for all y ∈ Rn

+ such that |y − v0| = λ. By maximum principle we can see

P2−nf (y) < P2−nfλ,v0

(y)

for all |y − v0| < λ.

Note that in this subsection we only used asymptotic behavior (3.5.4), (3.5.7) and the

maximum principle. We did not use the fact that !f is a solution to the integral equation

(3.4.1).

3.5.4 The Case ¯λ0 = ∞
In the previous subsection we showed that λ0 > 0 exists. In this subsection we will show

that it can not go to infinity.

Define

¯λ0 = sup{λ > 0 such that e(n−1)f (w) < e

(n−1)fλ,v0 (w) for all |w − v0| < λ.}

(3.5.10)

In the following lemma we show that ¯λ0 can not equal to ∞.
Lemma 3.11. For any f : Rn−1 → R, such that ef ∈ Ln−1(Rn−1), and for all v0 ∈ ∂Rn
Rn−1, we have ¯λ0 < ∞.
Proof. We can prove this by contradiction following [12]. Suppose for some v0 ∈ ∂Rn
Rn−1, we have ¯λ0 = ∞, then we can find a sequence λi → ∞ such that

+ =

+ =

e(n−1)f (w) < e

(n−1)fλi,v0 (w)

69

for all |w− v0| < λi and for all i. For a given i, by the previous inequality, (3.5.9) and change
of variable we have

&B+

λi,v0

e(n−1)f (w)dw < &B+
= &Rn

λi,v0

(n−1)fλi,v0 (w)dw

e

e(n−1)f (w)dw.

+\B+

λi,v0

As a result, we have

0 <

1

2&Rn−1

e(n−1)f (w)dw <&|w−v0|>λi

e(n−1)f (w)dw.

But the right hand side of the inequality goes to zero as λi → ∞ by dominated convergence
theorem.

Note that in the this proof we only need ef ∈ Ln−1(Rn−1) in order to use dominated
convergence theorem. We don’t need !f to be a solution to the integral equation (3.4.1).
3.5.5 The Case ¯λ0 < ∞
Note that since ¯λ0 > 0, this is the last case we need to consider. In this subsection we will
show that at the critical value ¯λ0, we have

But firstly we need to show three lemmas. The first lemma is about the kernel function

f¯λ0,v0

= f.

p2−n(y, u).
Lemma 3.12. For any v0 ∈ ∂Rn
for any y ∈ B+
and any w ∈ ∂Rn
λ,v0

+ = Rn−1 and any λ > 0, define φλ,v0
+ = Rn−1 such that |w − v0| < λ, we have

as in (3.5.1). Then

p2−n(y, w) − p2−n(φλ,v0

(y), w) > 0

(3.5.11)

Proof. By (A.0.1), we have

p2−n(φλ,v0

(y), w) = p2−n(y, φλ,v0

(w))|φ′λ,v0

(w)|n−1,

70

so we only need to prove

1

|y − w|2n−2 −

|φ′λ,v0
|y − φλ,v0

(w)|n−1
(w)|2n−2 > 0,

for any y ∈ B+
λ,v0
(3.5.1), we have

and any w ∈ Rn−1 such that |w − v0| < λ. By direct calculation using

So the proof follows from

|φ′λ,v0

(w)| =

λ2

|w − v0|2 .

|y − φλ,v0

(w)|2|w − v0|2

λ2

− |y − w|2 =

(λ2 − |w − v0|2)(λ2 − |y − v0|2)

λ2

,

which is positive when both |w − v0| < λ and |y − v0| < λ.

If we define

K(v0, λ; y, w) = p2−n(y, w) − p2−n(φλ,v0

(y), w)

as in [19] (right before Lemma 3.1 in [19]), then we can show the following result about the

derivative of K with respect to w (as in the proof of Lemma 3.2 in [19]).

Lemma 3.13.

〈∇wK(v0, λ; y, w), w−v0〉00|w−v0|=λ = −

2(n − 1)cn,2−nyn−1

n

|w − y|2n

(|w−v0|2−|y−v0|2), (3.5.12)

for all y ∈ Rn
+.
Proof. Here 〈∇wK(v0, λ; y, w), w−v0〉 denote the inner product in Rn−1 (or Rn) with respect
to the Euclidean metric. And ∇w denotes the gradient in Rn−1 with the Euclidean metric.
The subscript w emphasizes the fact that the derivative is taken with respect to w. The

proof follows from direct calculation. As in the previous lemma, use A.0.1 we can have

K(v0, λ; y, w)

= cn,2−nyn−1

n 6JJJ7

1

|w − y|2n−2 −*

λ

|w − v0|+2n−2

71

1

0000y − v0 − λ2(w−v0)
|w−v0|2 0000

2n−2

.

8KKK9

As a result, we can calculate ∇uK to be

∇wK

= −

n

cn,2−nyn−1
2(n − 1)(w − y′)
(n − 1)λ2n−2|w − v0|2n−2(4|y − v0|2|w − v0|2 − 4λ2〈y − v0, w − v0〉)

|(y − v0)|w − v0|2 − λ2(w − v0)|2n−2 (w − v0)
|(y − v0)|w − v0|2 − λ2(w − v0)|2n

2(n − 1)λ2n−2|w − v0|2n−4

|w − y|2n

−

+

(w − v0)

2(n − 1)λ2n|w − v0|2n−2

−

Here 〈y − v0, w − v0〉 denotes inner product in Rn. Then for the inner product we have

|(y − v0)|w − v0|2 − λ2(w − v0)|2n (|w − v0|2(y − v0) − λ2(w − v0)).
〈∇wK, w − v0〉00|w−v0|=λ = −2(n − 1)cn,2−nyn−1

(|w − v0|2 − |y − v0|2).

|w − y|

n

In the next lemma, we use change of variable to rewrite equation (3.4.4) into a new form.

Lemma 3.14. For any v0 ∈ ∂Rn
For any !f ∈ C∞(Rn−1) that is a solution to (3.4.1). Define f and fλ,v0

+ and any λ > 0, define the inversion φλ,v0

(3.5.3) respectively, then we have

as in (3.5.1).

as in (3.4.2) and

e

(n−1)fλ,v0 − e(n−1)f
λ,v0*e
=&B+

nP2−nfλ,v0 − enP2−nf+ (p2−n(y, w) − p2−n(φλ,v0

(y), w))dy

(3.5.13)

Proof. Since !f is a solution to 3.4.1, through change of variable we can see that f is a solution

to (3.4.4). Using equation (3.4.4) we have

where the second step follows from change of variable and (A.0.4). Similarly

e(n−1)f = &B+
= &B+
(n−1)fλ,v0 = &B+
= &B+

λ,v0

+\B+
λ,v0
nP2−nfλ,v0 p2−n(φλ,v0

enP2−nf p2−n(y, w)dy +&Rn
λ,v0.enP2−nf p2−n(y, w) + e
nP2−nfλ,v0 p2−n(y, w)dy +&Rn
nP2−nfλ,v0 p2−n(y, w) + enP2−nf p2−n(φλ,v0

+\B+
λ,v0

e

e

λ,v0

λ,v0.e

e

enP2−nf p2−n(y, w)dy
(y), w)/ dy,

(y), w)/ dy.

nP2−nfλ,v0 p2−n(y, w)dy

72

Subtracting the first inequality from the second one, we get the desired result.

We now prove f = fλ0,v0

by contradiction.

Proposition 3.15. Suppose f ∈ C1(Rn−1) is a solution to (3.4.4). For any v0 ∈ ∂Rn
Rn−1, with λ0 defined as in (3.5.10), f and fλ0,v0
tively, we have

defined as in (3.4.2) and (3.5.3) respec-

+ =

for all w ∈ Rn−1.

f (w) = fλ0,v0

(w),

Proof. We prove this by contradiction. Our proof is similar to a combination of Lemma 2.4
in [20] and Lemma 3.2 in [19]. Suppose there exists v ∈ ∂Rn

+ = Rn−1 such that

f (v) < fλ0,v0

(v),

then by maximum principle (note that by Lemma 3.8 we can use maximum principle here),

we have

P2−nf (y) < P2−nfλ0,v0

(y),

for all y ∈ B+

λ0,v0

. As a result, using (3.4.4) and Lemma 3.12 we see that

f (w) < fλ0,v0

(w)

for all w ∈ ∂Rn
+. Using compactness of B+

∂Rn

+ = Rn−1 such that |w − v0| < λ0. By (3.5.4), fλ0,v0

, there exists γ > 0 such that

λ0/2,v0

(3.5.14)

is continuous on

P2−nfλ0,v0

(y) − P2−nf (y) ≥ γ > 0

for all y ∈ B+

λ0/2,v0

.

Consider a sequence {λi}∞i=1 such that for each i we have

λi > λi+1 > λ0,

73

and that

as i → ∞. We can also require that

λi → λ0

P2−nfλi,v0

(y) − P2−nf (y) ≥

γ
2

> 0,

(3.5.15)

since by (3.5.7) we know that P2−nfλ,v0
λi,v0

For each i by compactness of B+

(y) is continuous function of λ.

, there exists wi ∈ B+

λi,v0

such that

P2−nfλi,v0

(wi) − P2−nf (wi) = inf
B+
λ0,v0

P2−nfλi,v0 − P2−nf.

Since λ0 is the critical value, we must have

P2−nfλi,v0

(wi) − P2−nf (wi) < 0.

By maximum principle, we also know that wi ∈ B+

λi,v0 ∩ ∂Rn

+, and that

fλi,v0

(wi) − f (wi) = P2−nfλi,v0

(wi) − P2−nf (wi) < 0.

In addition, by 3.5.15 we have

λ0
2 ≤ |wi − v0| < λi. We have strict inequality because

fλi,v0

(w) = f (w) for all |w − v0| = λi.
Since wi is an interior minimum for fλi,v0 − f in B+

λi,v0 ∩ ∂Rn

+, we have

∇(fλi,v0 − f )(wi) = 0.

Here we use ∇ to denote the gradient in Rn−1 with the Euclidean metric. It is the same
notation as in Lemma 3.13.
By compactness of B+

, we can choose a subsequence (still denote it {wi}∞i=1 and
+. By (3.5.4) we can see that both
(w) are continuous for λ > 0 and |w| ∕= 0. As a result, we can take limit

λ0,v0 ∩ ∂Rn

{λi}∞i ) such that wi → w0 for some w0 ∈ B+
fλ,v0
i → ∞ to get

(w) and ∇fλ,v0

λ1,v0

fλ0,v0

(w0) − f (w0) ≤ 0

(3.5.16)

74

and

∇(fλ0,v0 − f )(w0) = 0.

(3.5.17)

Because of (3.5.16) and (3.5.14) we have |w0 − v0| = λ0. But by Lemma 3.14 and Lemma
3.13, we can see that

(n−1)f

= 〈∇(e

λ0,v0 − e(n−1)f )(w0), w0 − v0〉
nP2−nf

(n − 1)e(n−1)f (w0)〈∇(fλ0,v0 − f )(w0), w0 − v0〉
= &B+
= −2(n − 1)cn,2−n&B+

λ0,v0 − enP2−nf (y)+〈∇wK(v0, λ0; y, w0), w0 − v0〉dy
λ0,v0*e
|w0 − y|2n (λ

λ0,v0 − enP2−nf (y)+ yn−1

λ0,v0*e

nP2−nf

n

2

0 − |y − v0|2)dy

< 0.

Which is a contradiction to (3.5.17).

3.5.6 Proof of Theorem 1.14

For the proof of Theorem 1.14 we need the following Lemma proved by Li and Zhu [20]

Lemma 3.16. [20, Lemma 2.5] For any integer n ≥ 3, suppose f ∈ C1(Rn−1) satisfying:
for any b ∈ Rn−1, there exists λb ∈ R such that

f (w) =

λn−2
b

|w − b|n−2 f1 λ2

|w − b|2 (w − b) + b2 , for all w ∈ Rn−1\{b}.

b

Then for some a ≥ 0, d > 0, w0 ∈ Rn−1,

a

f (w) =*
f (w) = −*

|w − w0|2 + d+(n−2)/2
|w − w0|2 + d+(n−2)/2

a

, for all w ∈ Rn−1,

, for all w ∈ Rn−1.

or

We restate Theorem 1.14 here for the convenience of the reader:

75

Theorem 3.17. For any integer n ≥ 2, if !f ∈ L∞(Sn−1) satisfies the equation

e(n−1)!f (ξ) =&Bn

then for all ξ ∈ Sn−1

en!In+n!P2−n!f!p2−n(x, ξ)dx,
1 − |ζ|2
|ξ − ζ|2 + Cn,

!f (ξ) = ln

where ζ ∈ Bn and Cn = − 1
the standard sphere.

n−1 ln00Sn−100 is a constant. Here 00Sn−100 denotes the volume of
Proof. By Proposition C.6 in the appendix we know that !f ∈ C1(Sn−1). For w ∈ Rn−1
define f (w) as in (3.4.2) then we have f ∈ C1(Rn−1). By discussion at the end of Section
3.4 we know that f satisfies the Euler-Lagrange equation (3.4.4). Then by Proposition 3.15
we know that for any v0 ∈ ∂Rn

+ = Rn−1, there exists λ0 > 0 depending on v0, such that

f (w) = fλ0,v0

(w),

for all w ∈ Rn−1. Note that here

(w) = f ◦ φλ0,v0

(w) + ln|φ′λ0,v0

(w)|

= f1 λ

|w − v0|2 (w − v0) + v02 + ln

2
0

2
0

λ

|w − v0|2

fλ0,v0

is as in (3.5.3).

Define

ϕ(w) = e

n−2
2 f (w),

then it is easy to check that ϕ satisfies the assumption of Lemma 3.16. As a result we have

a

ϕ(w) =*

|w − w0|2 + d+(n−2)/2
for some a > 0, d > 0 and w0 ∈ Rn−1. From this we know that
f (w) = − ln.|w − w0|2 + d/ + ln a.

76

Define

ζ = Ψ(w0,√d),

+. Then we have

1 − |ζ|2
|ξ − ζ|2 + Cn, for all ξ ∈ Sn−1.
n−1 ln00Sn−100 is determined by the restriction
’’’e!f’’’Ln−1(Sn−1)

= 1.

where (w0,√d) is a point in Rn
!f (ξ) = ln

Note that here the constant Cn = − 1

This finishes the proof.

77

APPENDICES

78

APPENDIX A

CONFORMAL TRANFORMATIONS

Note that we use three different types of conformal transformation in this part

Ψ : Rn

+ → Bn,

defined in section 2. Φ ∈ SO(n, 1), and the corresponding φ : Rn

+ → Rn

+ defined by

φ = Ψ−1 ◦ Φ ◦ Ψ.

Using the definition of Ψ, through direct calculation, we have

|Ψ′(y)| =

1 − |Ψ(y)|2

2yn

.

Lemma A.1. For any Φ ∈ SO(n, 1), define

φ = Ψ−1 ◦ Φ ◦ Ψ,

then for any y ∈ Rn

+ and any u ∈ Rn−1 we have

p2−n(φ(y), φ(w)) = p2−n(y, u)|φ′(w)|1−n,

as a result, for any f : Rn

+ → R we have

Proof. Since we have

using chain rule, we have

(P2−nf ) ◦ φ = P2−n(f ◦ φ)

Ψ ◦ Ψ−1 = Id,

Ψ′(Ψ−1(x)) · (Ψ−1)′(x) = Id

(A.0.1)

(A.0.2)

for all x ∈ Bn. Note that we think of the left hand side of the equation as matrix multipli-
cation. As a result we have

79

|Ψ′(Ψ−1(x))| = |(Ψ−1)′(x)|−1

(A.0.3)

Using both (2.4.1) and (2.4.2) we have

p2−n(φ(y), φ(w)) = p2−n(Ψ−1 ◦ Φ ◦ Ψ(y), Ψ−1 ◦ Φ ◦ Ψ(w))

= !p2−n(Φ ◦ Ψ(y), Φ ◦ Ψ(w))|Ψ′(Ψ−1 ◦ Φ ◦ Ψ(w))|n−1
= !p2−n(Ψ(y), Ψ(w))|Φ′(Ψ(w))|1−n|Ψ′(Ψ−1 ◦ Φ ◦ Ψ(w))|n−1
= p2−n(y, u)|Ψ′(w)|1−n|Φ′(Ψ(w))|1−n|Ψ′(Ψ−1 ◦ Φ ◦ Ψ(w))|n−1
= p2−n(y, w)|φ′(w)|1−n.

Note that in the last step we used chain rule and (A.0.3) by plug in x = Φ ◦ Ψ(w).

Using (A.0.1) we have

(P2−nf ) ◦ φ(y) = cn,2−n&Rn−1
= cn,2−n&Rn−1
= cn,2−n&Rn−1

= P2−n(f ◦ φ)(y).

p2−n(φ(y), w)f (w)dw
p2−n(φ(y), φ(w))f (φ(w))|φ′(w)|n−1dw
p2−n(y, w)f (φ(w))dw

Lemma A.2.

ln|φ′(y)| = P2−n ln|φ′(w)|

(A.0.4)

Proof. Since Φ = Ψ ◦ φ ◦ Ψ−1, by (3.3.2), we have

!In ◦ Φ(x) + ln|Φ′(x)| =!In(x) + !P2−n(ln|Φ′(ξ)|).

If we define y = Ψ−1(x), then using (3.3.3) we have

!In(x) = !In ◦ Ψ(y)

= P2−n(ln|Ψ′(u)|) − ln|Ψ′(y)|,

80

and

!In ◦ Φ(x) = !In ◦ Φ ◦ Ψ(y)
= !In ◦ Ψ(φ(y))
= (P2−n ln|Ψ′(w)|) ◦ φ(y) − ln|Ψ′(φ(y))|
= P2−n ln|Ψ′(φ(w))|(y) − ln|Ψ′(φ(y))|,

where the last step follows from (A.0.2).

On the other hand, using chain rule, we have

ln|Φ′(x)| = ln|Ψ′(φ ◦ Ψ−1(x))| + ln|φ′(Ψ−1(x))| + ln|(Ψ−1)′(x)|

= ln|Ψ′(φ(y))| + ln|φ′(y)| + ln|(Ψ−1)′(Ψ(y))|.

If we define w = Ψ−1(ξ), then using chain rule, we have

ln|Φ′(ξ)| = ln|Ψ′(φ ◦ Ψ−1(ξ))| + ln|φ′(Ψ−1(ξ))| + ln|(Ψ−1)′(ξ)|,

and by (2.4.1) we have

(!P2−n ln|Φ′(ξ)|) ◦ Ψ = P2−n ln|Ψ′(φ(w))| + P2−n ln|φ′(w)| + P2−n ln|(Ψ−1)′(Ψ(w))|.

Putting everything together and using (A.0.3) we have

ln|φ′(y)| = P2−n ln|φ′(w)|.

81

APPENDIX B

HARMONIC FUNCTIONS IN HYPERBOLIC SPACE

B.1

Induction Formula

In this section we prove a simple induction formula for hyperbolic harmonic functions.

This induction formula will help us relate harmonic functions in hyperbolic space and poly-

harmonic functions in Euclidean space in subsequent sections.

For an integer n ≥ 2, consider the unit ball model of hyperbolic space (Bn, gh), where
(1−|x|2)2 dx2. Define ρ = 1−|x|2

, then we have

2

gh =

4

and

∇ρ = −x,

∆ρ = −n,

Note that here ∇ and ∆ are the gradient and Laplacian with respect to the Euclidean metric.

We prove the following induction formula

Lemma B.1. For k ∈ N+ we have

ρ∆ku = −(k − 1)(n − 2k)∆k−1u − (n − 2k)〈x,∇∆k−1u〉.

(B.1.1)

Note that here the inner product is with respect to the Euclidean metric.

Proof. When k = 1 the equation (B.1.1) holds because u is harmonic with respect to the

hyperbolic metric.

Now suppose (B.1.1) holds for k. Apply ∆ on both sides, we get

and

∆.ρ∆ku/ = −n∆ku + ρ∆k+1u − 2〈x,∇∆ku〉,
∆.−(k − 1)(n − 2k)∆k−1u − (n − 2k)〈x,∇∆k−1u〉/
= −(k − 1)(n − 2k)∆ku − (n − 2k)(2∆ku + 〈x,∇∆ku〉).

82

Rearrange terms then we have

ρ∆k+1u = −(k)(n − 2(k + 1))∆ku − (n − 2(k + 1))〈x,∇∆ku〉.

This finishes the proof.

B.2 Hyperbolic Harmonic Functions in the Unit Ball

In this section we only consider even dimensional unit ball. From (B.1.1) we see that for

any u ∈ C∞(Bn) if

then

∆Hu = 0,

∆n/2u = 0.

Here ∆ denotes the Laplacian with respect to the Euclidean metric, whereas ∆H denotes

the Laplacian with respect to the hyperbolic metric.

In [25] Yang found the boundary condition satisfied by hyperbolic harmonic extension of

smooth functions. Suppose n ≥ 2 is even and f ∈ C∞(Sn−1), if we define

then u is hyperbolic harmonic, and satisfies the boundary conditions

(1 − |x|2)n−1
|x − ξ|2n−2 f dξ

u(x) = cn&Sn−1
2 − k/
2) Γ. n−1
Γ( n
2 / Γ( n
Γ. n−1
2 − k)
2 − k/
2) Γ. n−1
Γ( n
Γ. n−1
2 / Γ( n
2 − k)

∆ku0000Sn−1
∆ku0000Sn−1

and

∂
∂ν

= (−1)k

= (−1)k+1

Pn−1
Pn−1−2k

f ; for 0 ≤ k ≤ [

n − 2

4

];

(B.2.1)

Pn−1
Pn−1−2k

f ; for 0 ≤ k ≤ [

n − 4

4

].

(B.2.2)

Here ν is the outer normal unit vector. The notation [α] denotes the largest integer that

is less than or equal to α. For example [1/2] = 0 and [1] = 1. P2γ denotes a family of
operators on Sn−1 such that

P2γ =

Γ(B + 1
Γ(B + 1

2 + γ)
2 − γ)

, B =L−!∆ +

83

(n − 2)2

4

,

here !∆ is the Laplace-Beltrami opertor on the standard sphere Sn−1.

Note that using the property of gamma function, namely Γ(x+1)

Γ(x) = x, we can get

=

Pn−1
Pn−1−2k

Plug this into (B.2.1), we have

Γ(B + n/2)

Γ(B − n/2 + 1 + k)

= Πk

= Πk

Γ(B − n/2 + 1)
i=1(B − n/2 + i)

Γ(B + n/2 − k)
i=1(B + n/2 − i)Πk
i=1(−!∆ + (i − 1)(n − i − 1)).

∆ku0000Sn−1

= (−1)k Πk

i=1(n − 2i)
Πk
i=1(n − 2i − 1)

Πk

i=1[−!∆ + (i − 1)(n − i − 1)]u.

(B.2.3)

B.2.1 Recover the Boundary Conditions

In this subsection we will show that we can recover the boundary conditions (B.2.1) and

(B.2.2) using only the induction formula (B.1.1) and results from [23].

Proposition B.2. For any even integer n ≥ 2 and any f ∈ C∞(Sn−1), if we define

u(x) = cn&Sn−1

(1 − |x|2)n−1
|x − ξ|2n−2 f dξ

n

) and satisfies the boundary conditions (B.2.1) and (B.2.2). Here cn =

then u ∈ C∞(B

2Γ(n−1)

###Sn−2###Γ$ n−1

1 % is the normalization constant.

Proof. In proposition 5.1 of [23], in particular (5.14) of the note, Michael Taylor showed that

when the unit ball is even dimensional, the hyperbolic extension maps smooth function to

functions that are smooth up to the boundary. As a result we have u ∈ C∞(Bn

).

Now we showed that u satisfies the boundary conditions (B.2.1) and (B.2.2). It will be

easier for us to work in the polar coordinate in the Euclidean unit ball. In this coordinate

(B.1.1) becomes

1 − r2

2

∆ku = −(k − 1)(n − 2k)∆k−1u − (n − 2k)r

∂
∂r

∆k−1u.

(B.2.4)

84

We want to take limit r → 1 to obtain the boundary conditions. Since u ∈ C∞(Bn
k ∈ N+ both ∆ku and r ∂

∂r ∆ku are continuous in Bn

.

), for any

This leads to two simple results, firstly when k = 1, by taking limit r → 1 in (B.2.4) we

have

and for any k ∈ N+ by taking limit r → 1 in (B.2.4) we have

= 0,

∂
∂r

u0000Sn−1
= −k∆ku0000Sn−1
∆ku0000Sn−1

.

∂
∂r

1 − r2

2

∆u = −(n − 2)r

∂
∂r

u,

So now in order to finish the calculation, we only need to calculate ∆ku0000Sn−1

Consider the case when k = 1, from (B.2.4) we have

for any k ∈ N+.

divide both sides by (1 − r2)/2, and take limit r → 1 (using the boundary condition that

= 0), we have

∂

∂r u0000Sn−1

= (n − 2)

∂

∂r*r

∂
∂r

∆u0000Sn−1

In Euclidean polar coordinate we have

u+0000Sn−1
+ !∆u

∂u
∂r

r2 ,

∆u =

∂2u
∂r2 +

n − 1

r

= (n − 2)

∂2u
∂r2 .

(B.2.5)

here !∆ is the Laplace-Beltrami operator in the standard sphere Sn−1. Taking limit r → 1

we have

where the last step follows from (B.2.5). From which we have

,

∂2u

∂r20000Sn−1

∂2u

∂r20000Sn−1

= (n − 2)

+!∆f = ∆u0000Sn−1
!∆f = (n − 3)
∆u0000Sn−1

=

85

∂2u

∂r20000Sn−1
n − 3!∆f.

n − 2

and

This is exactly (B.2.3) for k = 1.
Calculation for 1 < k ≤ [ n−2

4 ] can be done using induction with (B.2.3), we omit the

calculation here. This finishes the proof.

B.2.2 The Other Direction

On the contrary we can prove the following

Proposition B.3. For any even integer n ≥ 2 and any u ∈ C(Bn
)∩ C∞(Bn) with boundary
value f ∈ C∞(Sn−1) such that u satisfies the boundary conditions (B.2.1) and (B.2.2), we
have

Recall that ∆H is the Laplacian in the hyperbolic unit ball.

∆Hu = 0, in Bn.

Proof. We consider the hyperbolic harmonic extension of f , define!u : Bn → R by

!u(x) = cn&Sn−1

(1 − |x|2)n−1
|x − ξ|2n−2 f dξ.

From the previous subsection we know that !u satisfies the same boundary conditions as u.

Since the equation and boundary conditions are all linear, we only need to show that if

There are two cases, the first case is when n/2 is even. By doing integration by parts

repeatedly, using the boundary condition ∂

= 0, we get3Bn(∆n/4v)2 = 0, which

v∆n/2v = 0.

∂ν ∆kv0000Sn−1

86

= 0; for 0 ≤ k ≤ [ n−2
4 ];

= 0; for 0 ≤ k ≤ [ n−4
4 ],

"#######$#######%

∆n/2v = 0,

∂

∆kv0000Sn−1
∂ν ∆kv0000Sn−1
&Bn

then v = 0.

This can be done by considering the integration

means ∆n/4−1v is harmonic with boundary condition ∆n/4−1v0000Sn−1
Now use all the boundary conditions ∆kv0000Sn−1

that v = 0.

= 0, so ∆n/4−1v = 0.

= 0 we eventually arrive at the conclusion

The second case is when n/2 is odd. This is similar, the only difference is in the last

step, we only get

&Bn |∇∆(n−2)/4v|2 = 0,
which mean ∆(n−2)/4v is constant, but since ∆(n−2)/4v0000Sn−1

0 in Bn. This finishes the proof.

B.3 Calculation in the Upper Half Space

= 0, we know that ∆(n−2)/4v =

We also include some calculation in the upper half space model of hyperbolic space.

Consider the space *Rn

n+. Here dy2 denotes the Euclidean metric; yn is the last
+, dy2
y2

component;

+ = {(y′, yn) ∈ Rn such that y′ ∈ Rn−1, yn > 0}.
Rn

We show that ln|y|2 is hyperbolic harmonic in the upper half space. Note that here |y|2
denotes the norm of y with respect to the Euclidean norm.

This is a straight forward calculation. We use ∆H to denote the hyperbolic Laplacian in

87

the upper half space.

∂2
i ln|y|2 + (2 − n)yn∂n ln|y|2
2(2 − n)y2

4y2
i

n

= y2
n

n
y−n
n

∂i ln|y|2)

i=1 ∂i(y2−n
∆H ln|y|2 = Mn
n5i=1
|y|42 +
n5i=11 2
|y|2 −
|y|4 + +
n* 2n
4|y|2
|y|2 −
2(2 − n)y2
(2n − 4)y2
|y|2

= y2
n

|y|2

= y2

n

+

=

|y|2
2(2 − n)y2

n

|y|2

n

(B.3.1)

= 0.

Remark 12. The calculation in (B.3.1) offers an alternative explanation of the origin of

It is easy to expect that (B.3.1) is invariant under translation in Rn−1. In the sense that

the function !In as discussed in Section 3.3.
for any v ∈ Rn−1 we have

∆H ln|y − v|2 = 0.

It is also easy to expect that (B.3.1) is not invariant under translation in yn direction, in the

sense that for any λ < 0

∆H ln.(yn − λ)2 + |y′|2/ ∕= 0.
Note that here y = (y′, yn) where y′ ∈ Rn−1 and yn > 0.

But instead we have when λ = −1

∆H.!In ◦ Ψ(y) − ln.(yn + 1)2 + |y′|2// = 0.

Note that this is the consequence of Lemma 3.6, where we also use the relation (2.1.2).

The calculation (B.3.1) also leads to an interesting observation

Proposition B.4. For any y ∈ Rn

+ define u(y) = ln|y| then

div.yα

n∇e(2−n−α)u/ = 0

88

in Rn

+. Where the divergence and ∇ are all defined with respect to the Euclidean metric.

Proof. This is a straight forward calculation. From (B.3.1) we have

∆u =

(n − 2)∂nu

yn

,

where ∆u denotes the Laplacian with respect to the Euclidean metric. As a result, we have

n∇e(2−n−α)u/
div.yα
= (2 − n − α)yα
= (2 − n − α)yα

n e(2−n−α)u*∆u + α
n e(2−n−α)u* (n − 2)∂nu

∂nu
yn

yn

+ (2 − n − α)|∇u|2+

+ α

∂nu
yn

+ (2 − n − α)|∇u|2+ .

We conclude the proof by pointing out the fact that

∂nu
yn

= |∇u|2.

Remark 13. Note that Proposition B.4 offers an alternative explanation of why the function

Γ(y) = C(n, α)

1

|y|n−2+α

is a fundamental solution to the equation

div(yα

n∇u) = 0

as observed by Caffarelli and Silvestre in [3].

We think of it as the consequence of the fact that ln|y| is hyperbolic harmonic in the

upper half space and of the coincidence that ∂n ln|y|

yn

= |∇ ln|y||2.

89

APPENDIX C

REGULARITY

For any integer n ≥ 3 any α ∈ (2 − n, 1) and any p > 2(n−1)
n−2+α we want to prove regularity of
the function !f ∈ Lp(Sn−1) such that !f is a solution to the integral equation for any ξ ∈ Sn−1

Here !pα(x, ξ) is the Poisson kernel .For any x ∈ Bn and any ξ ∈ Sn−1 we have

&Bn!pα(x, ξ).!Pα!f/ (x)
!pα(x, ξ) = 2α−1cn,α

n+2−α

n−2+α = !f (ξ)p−1.
(1 − |x|2)1−α
|x − ξ|n−α .

The normalizing constant cn,α is as in (1.2.3), we restate it here:

c−1

n,α =000Sn−2000& ∞

0

rn−2dr

(1 + r2) n−α

2

In upper half space the kernel function is

=

Γ( 1−α

2 )Γ( n−1
2 )
2Γ( n−α
2 )

000Sn−2000 .

pα(y, w) = cn,α

y1−α
n

|y − w|n−α

with

Pαf (y′, yn) = cn,α&Rn−1

y1−α
n

(|y′ − w|2 + y2

n) n−α

2

f (w)dw

The corresponding integral equation in the upper half space is

.!f ◦ Ψ(w)/p−1

or equivalently

=&Rn

+

pα(y, w)|Ψ′(w)|

α−n
2

((Pαf )(y))

n+2−α
n−2+α dy,

(f (w))p−1 |Ψ′(w)|

n−α
2 −

(p−1)(n−2+α)

2

=&Rn

+

pα(y, w) ((Pαf )(y))

n+2−α
n−2+α dy,

(C.0.1)

90

C.1 From Lp to L∞

n−2+α < p < ∞

Proposition C.1. For any integer n ≥ 3, any α ∈ (2 − n, 1) and any 2(n−1)
suppose !f ∈ Lp(Sn−1) is a solution to the Euler-Lagrange equation
n−2+α dx = !f (ξ)p−1,

&Bn!pα(x, ξ)..!Pα!f/ (x)/ n+2−α
then we have !f ∈ L∞(Sn−1)
Proof. Note that the operator !Pα : Lp(Sn−1) → L
n−2+α < p < ∞. For!u ∈ Lq(Bn) where 1 ≤ q < n, define the operator Tα

2(n−1)

2n

n−2+α (Bn) is bounded and compact when

Using a duality argument as in [18] we can prove

!Tα!u =&Bn!pα(x, ξ)!u(x)dx.
’!Tα!u’
≤ ’!u’Lq(Bn)

(n−1)q
n−q (Sn−1)

L

(C.1.1)

Suppose !f ∈ Lp(Sn−1) is a solution to the Euler-Lagrange equation
n−2+α dx = !f (ξ)p−1.

&Bn!pα(x, ξ)..!Pα!f/ (x)/ n+2−α

From Proposition 2.9 we see that

As a result if we define γ = n−2+α

Using (C.1.1) we have

and hence

np
n−1 (Bn).

!Pα!f ∈ L
n+2−α , then we have
(!Pα!f )1/γ ∈ L
!f p−1 ∈ L
!f ∈ L

n−1−γp

npγ
n−1 (Bn).

γp(n−1)
n−1−γp (Sn−1),

γp(p−1)(n−1)

(Sn−1)

If we keep going we can have !f ∈ L∞(Sn−1).

91

C.2 Derivative of !pα with respect to x

Next, we want to prove regularity for f in using the same idea as in the book by Gilbart-

Trudinger [14, Chapter 4]. (The way they handle the Newtonian Potential) So firstly, we

take derivative of!pα with respect to x and with respect to ξ. Through direct calculation we

get

∂yipα(y, w) ="##$##%

−(n − α)
−(n − α)

∂wipα(y, w) = (n − α)

y2−α
n

(yi−wi)

y1−α
n
|y−w|n−α+2 , i ∕= n
|y−w|n−α+2 + (1 − α)
y1−α
n
|y − w|n−α+2 , i = 1, 2, ..., n − 1

(yi − wi)

|y−w|n−α , i = n

y−α
n

C.3 C β Regularity for F

Suppose we have the integral equation

y1−α
n

F (w) = cn,α&Rn
+ and y′, w ∈ Rn−1. We assume U (y) ∈ L∞(Rn

|y − w|n−α U (y)dy,

+

+) and U (y) > 0

where y = (y′, yn) ∈ Rn

for all y ∈ Rn
+.

For any R > 0 we can write

F (w) = cn,α& R
+ cn,α& ∞

0 &Rn−1
R &Rn−1

y1−α
|y − w|n−α U (y)dy′dyn
n
y1−α
|y − w|n−α U (y)dy′dyn.
n

Define

FR(w) = cn,α& R

0 &Rn−1

y1−α
|y − w|n−α U (y)dy′dyn.
n

(C.3.1)

It is easy to see that F − FR ∈ C∞(Rn−1), since there is no local singularity, and the

singularity at ∞ is still summable after taking derivatives.

92

Lemma C.2. For any 2 − n ≤ α < 1. For any R > 0 and for U ∈ L∞(Rn
define FR as in (C.3.1). Then for any β such that 0 < β < 1, we have FR ∈ Cβ
Proof. For any v, w ∈ Rn−1, consider
FR(w) − FR(v) = cn,α& R

0 &Rn−1* y1−α

|y − v|n−α+ U (y)dy′dyn.

y1−α
n

n

+), and U > 0,

loc(Rn−1).

|y − w|n−α −
2 . Define

Define r = |v − w|. Suppose we have 0 < r < R

A = {y ∈ Rn

+ : such that |y − v| < 2r and |y − w| < 2r}.

Then we can write

FR(w) − FR(v) = cn,α& R
+ cn,α& R

0 &A* y1−α
|y − w|n−α −
0 &Rn−1\A* y1−α

n

n

|y − w|n−α −

y1−α
n

|y − v|n−α+ U (y)dy′dyn

y1−α
n

|y − v|n−α+ U (y)dy′dyn

:= I + II

For I we have

|y − v|n−α+ U (y)dy′dyn0000

U (y)dy′dyn

n

y1−α
n

0000cn,α& R
0 &A* y1−α
|y − w|n−α −
≤ C(n, α)& R
0 &A
y1−α
|y − w|β
n
|y − w|n−α+β
+C(n, α)& R
0 &A
y1−α
|y − v|β
n
|y − v|n−α+β
+)*& R
0 &A
≤ 2β|w − v|βC(n, α)’U’L∞(Rn
dy′dyn+,
+& R
0 &A

|y − v|n−α+β

y1−α
n

U (y)dy′dyn

y1−α
n

|y − w|n−α+β

dy′dyn

in the last inequality we used the fact that |y − w| < 2r = 2|v − w|, |y − w| < 2|v − w| and

93

then we have

2

n−α+β
rn−2dr

with dz′ = dy′
yn−1
n
dz′
(|z′|2 + 1)
n & ∞
Γ. 1−(α−β)
/ Γ( n−1
/
2Γ. n−(α−β)

(r2 + 1)

dyn
yβ

0

2

2

n−(α−β)

2

2 )

.

= |Sn−2|& R
= |Sn−2|

0

R1−β
1 − β ·

U > 0 in Rn

+. By change of variable, choose z′ = y′−w
yn
& R
0 &A

dy′dyn ≤ & R

|y − w|n−α+β

dyn
yβ

y1−α
n

n &Rn−1

0

Note that when 2 − n ≤ α < 1 and 0 < β < 1, we have 1−(α−β)

2

> 0 and n−(α−β)

2

> 0. As

a result, we have

Similarly for v we have

As a result, we have

& R
0 &A
& R
0 &A

y1−α
n

|y − w|n−α+β

y1−α
n

|y − v|n−α+β

dy′dyn ≤ C(n, α, β, R).

dy′dyn ≤ C(n, α, β, R).

I ≤ C(n, α, β, R)’U’L∞(Rn

+)|w − v|β.

Now we consider II. Notice that

|Dwpα(y, w)| ≤ (n − α)cn,α

y1−α
n

|y − w|n−α+1 .

Using mean value theorem we have: for some w0 lying on the line segment between v and w

|pα(y, w) − pα(y, v)| ≤ |Dwpα(y, w0)||w − v|.

As a result, we have

|II| ≤ C(n, α)’U’L∞(Rn

= C(n, α)’U’L∞(Rn

+)& R
0 &Rn−1\A |Dwpα(y, w0)||w − v|dy′dyn
+)& R
0 &Rn−1\A

|y − w0|n−α+1|w − v|1−β|w − v|βdy′dyn.

y1−α
n

In Rn

+\A, we have

|y − w0| ≥ |w − v| = r.

94

We can prove this by contradiction. Suppose we have

then by triangle inequality, we have

|y − w0| < r,

|y − w| ≤ |w − w0| + |y − w0| ≤ |w − v| + |y − w0| ≤ 2r.

Similarly for v, we have

|y − v| ≤ 2r.
As a result, we have y ∈ A, which is a contradiction.

Now we have

|II| ≤ C(n, α)’U’L∞(Rn

y1−α
n

+)& R
0 &Rn−1\A
+)|w − v|β& R

|y − w0|n−α+1|y − w0|1−β|w − v|βdy′dyn
0 &Rn−1\A

|y − w0|n−α+β

dy′dyn.

y1−α
n

= C(n, α)’U’L∞(Rn

By previous calculation we have

& R
0 &Rn−1\A

y1−α
n

|y − w0|n−α+β

As a result, we have

dy′dyn ≤ & R

0 &Rn−1

y1−α
n

|y − w0|n−α+β

≤ C(n, α, β, R)

dy′dyn.

|II| ≤ C(n, α, β, R)’U’L∞(Rn

+)|w − v|β.

All together, for any 0 < β < 1 we have

|FR(w) − FR(v)| ≤ C(n, α, β, R)’U’L∞(Rn

+)|w − v|β,

which means FR ∈ Cβ

loc(Rn−1).

95

C.4 C β Regularity for U

Lemma C.3. For some 2 − n ≤ α < 1, 0 < β < 1 such that α + β < 1 and for some
f ∈ Cβ

loc(Rn−1) ∩ L∞(Rn−1), define

u(y) = Pαf (y) = cn,α&Rn−1

y1−α
n

|y − w|n−α f (w)dw.

Then for any y′ ∈ Rn−1 we have

u(y′, yn) = f (y′),

lim
yn→0

and u ∈ Cβ

loc(Rn
+).

Proof. Through change of variable we see that

|u(y)| = cn,α000000

From which we can get

&Rn−1

f (ynw + y′)
n−α
(|w|2 + 1)
2

≤ ’f’L∞(Rn−1)cn,α&Rn−1

dw
(|w|2 + 1)

.

n−α
2

dw000000

+) ≤ ’f’L∞(Rn−1).
In addition, by dominated convergence theorem, we get

’u’L∞(Rn

u(y′, yn) = cn,α&Rn−1
Since it is easy to see that U ∈ C∞(Rn

lim
yn→0

lim
yn→0

f (ynw + y′)
n−α
(|w|2 + 1)
2

dw = f (y′).

+), we only need to show that U is H¨older continuous

up to the boundary.
For any y ∈ Rn

+ and any v ∈ Rn−1, define D = {w ∈ Rn−1 such that |(ynw+y′)−v| < 1}.
. Also, note that for any w ∈ D

with radius 1
yn

Note that D is a ball in Rn−1 centered at v−y′
yn
we have

|ynw + y′| ≤ 1 + |v|

Choose R > 0 large enough such that

1 + |v| < R

96

then for all w ∈ D we have

|ynw + y′| ≤ 1 + |v| < R.

Since f ∈ Cβ

loc(Rn−1), we have

sup

|w1|<R, |w2|<R

|f (w1) − f (w2)|

|w1 − w2|β

= κ < ∞.

Consider

000000

000000

000000

dw

dw

n−α
2

(|w|2 + 1)

n−α
2

(|w|2 + 1)

n−α
2

(|w|2 + 1)
dw

f (ynw + y′) − f (v)

dw

f (ynw + y′) − f (v)

f (ynw + y′) − f (v)

|u(y) − u(v)| = cn,α&Rn−1000000
≤ cn,α&D000000
+cn,α&Rn−1\D000000
≤ cn,ακ&D
|ynw + (y′ − v)|β
n−α
(|w|2 + 1)
2
+2cn,α’f’L∞(Rn−1)&Rn−1\D
≤ C(n, α, β, κ,’f’L∞)&Rn−1
≤ C(n, α, β, κ,’f’L∞)&Rn−167 |yn|β|w|β
n−α
2
≤ C(n, α, β, κ,’f’L∞)|y − v|β&Rn−1
|w|β
(|w|2 + 1)
+C(n, α, β, κ,’f’L∞)|y − v|β&Rn−1
1
(|w|2 + 1)
Γ. 1−(α+β)
/ Γ. n+β−1
2Γ( n−α
2 )
2 / Γ. n−1
Γ. 1−α
2 /
2Γ( n−α
2 )

&Rn−1
|w|β
(|w|2 + 1)
&Rn−1

|ynw + (y′ − v)|β
n−α
(|w|2 + 1)
2
|ynw|β + |y′ − v|β
n−α
(|w|2 + 1)
2

|w|β
(|w|2 + 1)

dw = |Sn−2|

dw = |Sn−2|

n−α
2

(|w|2 + 1)

dw

dw

2

2

n−α
2

97

Note that in the third inequality, we used subadditivity of concave function. Also, note that

and

+

|y′ − v|β
(|w|2 + 1)

2 89 dw

n−α

dw

n−α
2

dw.

n−α
2

/

,

.

Both integrals are finite when 2− n ≤ α < 1, 0 < β < 1 and α + β < 1. As a result, we have

|u(y) − u(v)| ≤ C(n, α, β, κ,’f’L∞)|y − v|β

for all v ∈ Rn−1 such that |v| + 1 < R and for all y ∈ Rn
the function u is only locally H¨older continuous. We have u ∈ Cβ

loc(Rn
+).

+. Note that since κ depends on R,

C.5 C 1 Regularity for F

R,v = BR,v ∩ Rn

For any v ∈ Rn−1 = ∂Rn and for any R > 0 define BR,v = {y ∈ Rn : |y − v| < R},
R,v = {w ∈ Rn−1 :
R,0 and Bn−1
R,0

+. We also define the n − 1 dimensional ball as Bn−1

and B+
|w − v| < R}. We also use notation BR, B+
respectively.

to denote BR,0, B+

R and Bn−1

R

We prove C1

loc(Rn−1) regularity using the same argument as in [14, Lemma 4.2].
+) ∩ Cβ

Lemma C.4. For any 2 − n ≤ α < 1, 0 < β < 1 and for any U ∈ L∞(Rn
U > 0 such that

loc(Rn
+),

F (w) = cn,α&Rn

+

y1−α
n

|y − w|n−α U (y)dy

is well defined. We have F ∈ C1
Proof. Note that for any R > 0 we can write

loc(Rn−1).

where

F (w) = cn,α&B+
cn,α&Rn

R

y1−α
n

|y − w|n−α U (y)dy + cn,α&Rn

+\B+

R

y1−α
n

|y − w|n−α U (y)dy,

y1−α
|y − w|n−α U (y)dy ∈ C∞loc(Bn−1
n

R ),

+\BR,w

so we only need to consider

cn,α&B+

R

y1−α
n

|y − w|n−α U (y)dy.

Extend U (y) to Rn by defining

U (y′, yn) ="##$##%

U (y′, yn), for yn ≥ 0,
U (y′,−yn), for yn < 0.

98

for all y′ ∈ Rn−1. Then it is easy to see that U ∈ L∞(Rn) ∩ Cβ
Rn in the y variable by defining

loc(Rn). Extend pα(y, w) to

pα(y, w) = cn,α |yn|1−α
|y − w|n−α .

As a result we have

for all w ∈ Bn−1

argument as in Lemma 4.2 of [14] to prove that

1

R

&B+

pα(y, w)U (y)dy =

2&BR
R . Now we only need to consider3BR
pα(y, w)U (y)dy2 = &BR
Di1&BR

pα(y, w)U (y)dy,

pα(y, w)U (y)dy. We can use the same

Dipα(y, w)(U (y) − U (w)) dy

pα(y, w)νi(y)dSy,

−U (w)&∂BR

for i = 1, 2, ..., n− 1. Here derivative is taken with respect to w, ∂BR is the boundary of BR
in Rn, dSy is the standard measure on ∂BR.

C.6 Application to the Non-limit case

Now we are ready to prove regularity results which was used in Theorem 2.12.

Proposition C.5. For any integer n ≥ 3, for any 2−n < α < 1 and any p > 2(n−1)
we have !f ∈ L∞(Sn−1), !f ≥ 0 and that !f is a solution to the Euler-Lagrange equation

n−2+α suppose

&Bn!pα(x, ξ)..!Pα!f/ (x)/ n+2−α

n−2+α dx = !f (ξ)p−1,

then !f ∈ C1(Sn−1).
Proof. If for any w ∈ Rn−1 we define

f (w) = !f ◦ Ψ (w)1 2

1 + |w|22 n−2+α

2

99

as in 2.1.5, then by Proposition 2.5 and change of variable we can see that f satisfies the the

integral equation

(f (w))p−1 |Ψ′(w)|

n−α
2 −

(p−1)(n−2+α)

2

=&Rn

+

pα(y, w) ((Pαf )(y))

n+2−α
n−2+α dy.

As an immediate result, we have f (w) > 0 for all w ∈ Rn−1, since otherwise we have f = 0.
Where ’Pαf’L∞(Rn

n+2−α
n−2+α ∈ L∞(Rn
+).
Using Lemma C.2, choose some β such that 0 < β < 1, β < p − 1 and β

+) ≤ ’f’L∞(Rn−1), and hence ((Pαf )(y))

p−1 + α < 1, we

get that

(f (w))p−1 |Ψ′(w)|

n−α
2 −

(p−1)(n−2+α)

2

∈ Cβ

loc(Rn−1).

Since |Ψ′(w)| = 2

1+|w|2 is smooth as a function of w and that it is always positive, we have

and as a reuslt

Now apply Lemma C.3, we get

f p−1 ∈ Cβ

loc(Rn−1),

β
p−1
loc

f ∈ C

(Rn−1).

β
p−1
loc

(Rn

+).

Pαf ∈ C

Finally apply Lemma C.4 to get

(f (w))p−1 |Ψ′(w)|

n−α
2 −

(p−1)(n−2+α)

2

∈ C1

loc(Rn),

and hence

(f (w))p−1 ∈ C1
Lastly, since f (w) > 0 for all w ∈ Rn−1, we have

loc(Rn).

f (w) ∈ C1
Transform f back to the unit ball we see that

loc(Rn−1).

!f ∈ C1(Sn−1).

100

C.7 Application to the Limit Case α = 2 − n
Proposition C.6. For any integer n ≥ 3, suppose we have !f ∈ L∞(Sn−1) and that !f is a

solution to the Euler-Lagrange equation

(C.7.1)

e(n−1)!f (ξ) =&Bn

en!In+n!P2−n!f!p2−n(x, ξ)dx

then !f ∈ C1(Sn−1).
Proof. Suppose !f ∈ L∞(Sn−1) satisfies the integral equation (C.7.1). Define

f (w) = !f ◦ Ψ(w) + ln|Ψ′(w)|,

then as discussed in Section 3.4 we see that f (w) satisfy the following integral equation:

e(n−1)f (w) =&Rn

+

enP2−nf p2−n(y, w)dy.

(C.7.2)

+), we can apply Lemma C.2 to

get

Since enP2−nf (y) = |Ψ′(y)|nen!In+n!P2−n!f ◦ Ψ(y) ∈ L∞(Rn
loc(Rn−1).

e(n−1)f (w) ∈ Cβ
Since e(n−1)f (w) > 0 for all w ∈ Rn−1, we have
f (w) ∈ Cβ

loc(Rn−1)

Now from Lemma C.3 we know that P2−nf ∈ Cβ
(C.7.2) we eventually get f ∈ C1
get !f ∈ C1(Sn−1).

+). Using Lemma C.4 and equation
loc(Rn−1). Eventually, transform back to the unit ball we

loc(Rn

101

APPENDIX D

CONFORMALLY COMPACT MANIFOLD

A natural way to generalize observations in the hyperbolic space is to consider the confor-

mally compact manifolds. In this appendix we consider the solvability of Laplace equation

with Dirichlet boundary condition. Firstly we define some notations that will be used through

out this appendix.

Definition D.1. By (M, g) is conformally compact to (M , g), we mean: M is a compact

manifold with nonempty boundary, and M is the interior of M . We assume g to be smooth.
With a fixed defining function r : M → R such that r is smooth and positive in M , where
r = 0 on ∂M and |dr|g = 1 on ∂M . Such that g = r−2g. Note that here we use ∇ and
∆ = trg ∇2 to denote operators on (M, g) and use ∇ and ∆ = trg ∇2
to denote operators

on (M , g). Moreover, Sobolev spaces on M and M are defined using measure from g and g

respectively.

Note that this definition is closely related to Subsection 1.4.1. In conformally compact

manifold there exists a special neighborhood called the collar neighborhood. We discuss the

collar neighborhood in the following remark.

Remark 14. Note that by compactness of ∂M and the fact that |dr|g = 1, there exists a
collar neighborhood Ω = [0, -) × ∂M of ∂M , such that the metric g = r−2(dr2 + h) in the
collar neighborhood. Here - > 0 is a real number, and h is a metric on ∂M that depends on

r, we use the h0 to denote the metric on {0} × ∂M .

In order to prove the solvability of Laplace equation in conformally compact manifolds

with Dirichlet boundary condition, we firstly prove that we can extend a smooth function

on the boundary to be a nice function on the collar neighborhood in the following sense:

102

Proposition D.2. Suppose (M, g) is a connected n-dimensional manifold conformally com-
pact to (M , g) as in Definition D.1. For C∞ function f : ∂M → R, extend f to f0 : Ω → R
by f0(t, θ) = f (θ). Where (t, θ) ∈ [0, -)× ∂M . Then there exists functions ai : [0, -)× ∂M →
R for i = 2, ..., n, which only depends on θ ∈ ∂M , such that
airi2 = O(rn−1)

∆1f0 +

n−25i=2

Proof. In the collar neighborhood as mentioned in Remark 14, for any C2 function u : Ω → R,
we have

∆u = rn∂r(r2−n∂ru) + r2 ∂r det h
2 det h

∂ru + r2∆hu.

(D.0.1)

In particular, when u = rα, we have

∆rα = α(α − n + 1)rα + α

∂r det h
2 det h

rα+1 = α(α − n + 1)rα + O(rα+1)

(D.0.2)

If n < 4 then we are done.

If n ≥ 4, then we can take

Hence we have ∆f0 = r2∆hf .
a2 = 1

f0 such that

2(n−3) ∆h0

∆(f0 + a2r2) =

We can keep going until

∆h0

f0 + r4∆h∆h0

f0+ = O(r3).

1

det h

2(n − 2)*2r3 ∂r det h
∆1f0 +

airi2 = O(rn−1),

n−25i=2

(D.0.3)

where ai : [0, -) × ∂M → R does not depend on r, i = 2, ..., n. Note that this is the best we
can do, since by equation (D.0.2), we have for any an−1 : [0, -) × ∂M → R that does not
depend on r,

So in general we can not cancel the O(rn−1) term in the same way.

∆an−1rn−1 = O(rn).

(D.0.4)

Now we are ready to prove the solvability of Laplace equation in conformally compact

manifolds with Dirichlet boundary condition.

103

Proposition D.3. For any C∞ function f : ∂M → R, there exists a function ψ : M → R
such that ψ|∂M = f and that ∆ψ = 0 in M .

Proof. Extend f to f0 as in proposition D.2. By proposition D.2, there exists functions
ai : [0, -) × ∂M → R for i = 2, ..., n, which only depends on θ ∈ ∂M , such that

∆1f0 +

n−25i=2

airi2 = O(rn−1)

Choose a cut off function χ : ∂M → R such that 0 ≤ χ ≤ 1 and χ = 1 on [0, -/2] × ∂M and
that χ is supported in [0, -) × ∂M . Define

M’ = {x ∈ M| r(x) > -}.

Note that M ’ = {x ∈ M | r(x) ≥ -}. Define

U = χ1f0 +

airi2 ,

n−25i=2

then we can extend U to be a function on M such that U = 0 on M’ and that

∆U = O(rn−1).

Choose 0 < -0 < -/2 small enough such that

sup
∂M ’0

|U| ≤ 2 sup

∂M |f|

For any 0 < -′ < -0, in M ’′, by standard theorem on bounded domain, there exists a function
ϕ’′ such that

Since ∆(ϕ’′ + U ) = 0 in M’′, by maximum principle we have

"##$##%

∆ϕ’′ = −∆U, in M’′
ϕ’′ = 0, on ∂M ’′.

sup
M
’′

|ϕ’′ + U| ≤ sup
’′

∂M

|U| ≤ 2 sup

∂M |f|

104

As a result we have

|ϕ’′| ≤ 2 sup

∂M |f| + sup

M |U|

in M’′ for all -′ ∈ (0, -0). Define C = 2 sup∂M |f| + supM |U|. Choose α ∈ (0, n − 1), by
equation (D.0.2) we have

∆rα = α(α − n + 1)rα + O(rα+1).
Moreover, since ∆U = O(rn−1) there exists -1 ∈ (0, -0) such that

∆*C

rα
-α

1 ± U+ ≤ 0

for all (t, θ) ∈ [0, -1) × ∂M . Choose -′ ∈ (0, -1), define A’′,’1
then we have

= {x ∈ M| -′ < r(x) < -1},

. Which mean C rα
1 ± ϕ’′ is a super harmonic function in A’′,’1
’α

. By maximum

in A’′,’1
principle, we have

∆*C

rα
-α

1 ± ϕ’′+ ≤ 0

rα
1 ± ϕ’′ ≥ 0.
-α

C

A

inf
’′,’1

Note that when r = -′ we have C rα
’α
C ± ϕ’′ ≥ 0. As a result we have

1 ±ϕ’′ = C ’′α

’α
1

> 0, and when r = -1 we have C rα
1 ±ϕ’′ =
’α

|ϕ’′| ≤ C

rα
-α
1

for all x ∈ A’′,’1

and for all -′ ∈ (0, -1). Hence we have

|ϕ’′ + U| ≤ C* rα

-α
1

+ 1+ ≤ 2C

for all x ∈ A’′,’1
principle we have for any x ∈ M’′

and for all -′ ∈ (0, -1). Since ϕ’′ + U is harmonic in M’′, by maximum

By standard elliptic estimate we have

|ϕ’′ + U| ≤ 2C.

’ϕ’′ + U’C2,α(M

’′ ) ≤ C1.

105

By Arzela-Ascoli, and a diagonalization argument, there exists a sequence -i → 0 such that
ϕ’i + U converges to some function ψ : M → R in C2 norm. As a result, we have ψ is
harmonic in M and that ψ = f on ∂M .

106

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