The hyperbolic metric and two-point distortion theorems for univalent functions

by

Alison Jo O’Leary

A creative component submitted to my graduate committee

in partial fulfillment of the requirements for the degree of

MASTER OF SCIENCE

Major: Applied Mathematics

Program of Study Committee:Elgin Johnston, Co-major ProfessorGary Lieberman, Co-major Professor

James Wilson

Iowa State University

Ames, Iowa

2006

Copyright c© Alison Jo O’Leary, 2006. All rights reserved.

ii

TABLE OF CONTENTS

CHAPTER 1. INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . 1

CHAPTER 2. PRELIMINARIES . . . . . . . . . . . . . . . . . . . . . . . . . 2

2.1 Basic Concepts and Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2.2 Subclasses of Univalent Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.3 Some Classical Results for Univalent Functions . . . . . . . . . . . . . . . . . . 4

2.4 Classical Growth and Distortion Theorems . . . . . . . . . . . . . . . . . . . . 7

2.5 Motivation for Using the Hyperbolic Metric . . . . . . . . . . . . . . . . . . . . 10

CHAPTER 3. THE HYPERBOLIC METRIC . . . . . . . . . . . . . . . . . . 11

3.1 The Upper-Half Plane Model of the Hyperbolic Plane . . . . . . . . . . . . . . 11

3.2 Möbius Transformations and the General Möbius Group . . . . . . . . . . . . . 12

3.3 Hyperbolic Length and Distance in H . . . . . . . . . . . . . . . . . . . . . . . 16

3.4 The Poincaré Disk Model, D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

CHAPTER 4. TWO-POINT DISTORTION THEOREMS . . . . . . . . . . 25

4.1 An Invariant Koebe Distortion Theorem . . . . . . . . . . . . . . . . . . . . . . 25

4.2 Lower Bounds for |f(a)− f(b)| . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.3 Upper Bounds for |f(a)− f(b)| . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4.4 Comparisons Between Hyperbolic and Euclidean Geometries on Simply Con-

nected Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

1

CHAPTER 1. INTRODUCTION

Univalent function theory has been around since the early 1900’s. In 1916, Bieberbach

gave an estimate for the second Taylor coefficient of a normalized univalent function. Some

important consequences of this coefficient estimate are the growth and distortion theorems

discussed in section 2.4. The classical growth theorem (Theorem 2.5) discussed in chapter 2

gives necessary but not sufficient conditions for univalence. In 1978, Blatter published a paper

stating a two-point distortion theorem that gives both necessary and sufficient conditions for

univalence. Since then, Blatter’s method has been used to extend his result to a one-parameter

family of theorems containing Blatter’s original theorem as a special case. These theorems are

stated in terms of hyperbolic distance and are invariant under composition with automorphisms

of the unit disk.

In chapter 2 we discuss general properties of univalent functions and give some classical

results, including the classical growth and distortion theorems. As the hyperbolic metric plays

a prominent role in the two-point distortion theorems, we give a construction of the hyperbolic

metric in chapter 3. Finally, in chapter 4 we discuss two-point distortion theorems, using the

methods employed by Blatter [2], Kim and Minda [6], and Ma and Minda [8, 9]. We conclude

chapter 4 by using the two-point distortion theorems to make some comparisons between

hyperbolic and Euclidean geometries on simply connected regions.

2

CHAPTER 2. PRELIMINARIES

2.1 Basic Concepts and Definitions

Definition 2.1. Let D ⊆ C be a domain and f : D → C an analytic function. We say that f

is univalent in D if f(z1) 6= f(z2) whenever z1 and z2 are points in D with z1 6= z2.

We will mainly be concerned with functions analytic and univalent on the unit disk

{z : |z| < 1}, which we denote throughout this paper by D. In particular, we will consider

normalized univalent functions. The schlicht class S is the class of functions f univalent on D

and normalized by the conditions f(0) = 0 and f ′(0) = 1. As a result of these normalizations,

if f ∈ S, then f has a Taylor expansion of the form

f(z) = z + a2z2 + a3z3 + a4z4 + · · · , |z| < 1.

Example 2.1. The most important example of a function in class S is the Koebe function,

given by

k(z) =z

(1− z)2=

∞∑n=1

nzn.

It is easy to verify that k(z) is univalent. In fact, if z1, z2 ∈ D and k(z1) = k(z2), then:

z1(1− z1)2

=z2

(1− z2)2⇒ z1(1− z2)2 = z2(1− z1)2

⇒ z1(1− 2z2 + z22) = z2(1− 2z1 + z21)

⇒ z1 − 2z1z2 + z1z22 = z2 − 2z1z2 + z2z21

⇒ z1 + z1z22 = z2 + z2z21

⇒ z1(1− z1z2) = z2(1− z1z2)

⇒ z1 = z2.

3

A function of the form

e−iβk(eiβz) =z

(1− eiβz)2=

∞∑n=1

nei(n−1)βzn, β ∈ R,

is called a rotation of the Koebe function. The Koebe function is important because it and its

rotations exhibit many extremal properties within the class S.

Univalence is not preserved under addition, but is preserved under other elementary oper-

ations such as rotation and dilation. In particular, if f ∈ S and F is defined by

F (z) =f

(z + t1 + tz

)− f(t)

(1− |t|2)f ′(t)

for some t ∈ D, then F ∈ S. To see this, notice that the function T (z) = z + t1 + tz

is a conformal

mapping of D onto D. Therefore, f ◦ T is univalent on D. Since F (0) = 0 and F ′(0) = 1, we

have F ∈ S. A function F defined this way is called a Koebe transform of f .

Another related class of functions that will be of interest to us is the class Σ. A function

g that is analytic (except for a simple pole at ∞) and univalent on ∆ = {z : |z| > 1} is in Σ if

g is of the form

g(z) = z + b0 + b1z−1 + b2z−2 + b3z−3 + · · · , |z| > 1.

2.2 Subclasses of Univalent Functions

Here we will define some important subclasses of S and Σ.

We first define what it means for a set to be starlike. If A ⊂ C, then we say that A is

starlike with respect to the point w0 ∈ A if the line segment joining w0 to any other point of A

is contained in A. A function is said to be starlike if it maps the unit disk conformally onto a

set that is starlike with respect to the origin. The subclass of S consisting of starlike functions

is denoted by S∗.

If A is starlike with respect to each point of A, then we say that A is convex. In other

words, a set is convex if the line segments joining any two points of the set is entirely contained

in the set. A function is said to be convex if it maps the unit disk conformally onto a convex

set. The subclass of S consisting of convex functions is denoted by C. Note that C ⊂ S∗ ⊂ S.

4

Closely related to the class of convex function is the class of close-to-convex functions. We

say that a function f is close-to-convex if there is a convex function g (not necessarily in C)

such that Re{

f ′(z)g′(z)

}> 0 for all z ∈ D. Note that the definition of a close-to-convex function

does not require univalence; however, as shown by Duren in [3], every close-to-convex function

is univalent.

The subclass of functions that are close-to-convex with the normalizations f(0) = 0 and

f ′(0) = 1 is denoted K. The subclass K0 of K is the set of all functions f such that

Re{

f ′(z)g′(z)

}> 0 for all z ∈ D and some g ∈ C.

We are also interested in some subclasses of Σ. Define Σ′ to be the subclass of functions

g ∈ Σ such that g(z) 6= 0 in ∆. The subclass Σ0 is defined to be the class of functions g ∈ Σ

such that b0 = 0. Finally, we define Σ̃ to be the subclass of functions g ∈ Σ such that C\g(∆)

has two-dimensional Lebesgue measure zero.

2.3 Some Classical Results for Univalent Functions

We now consider some classical results for univalent functions. These are standard results

that can be found in books such as Duren [3] and Pommerenke [10]. These theorems are useful

in proving the results of section 2.4.

Theorem 2.1 (Area Theorem). If g(z) = z+b0 +b1z−1 +b2z−2 + · · · ∈ Σ, then∞∑

n=1

n|bn|2 ≤ 1.

Equality holds if and only if g ∈ Σ̃.

Proof. Let E = C\g(∆), Cr be the image under g of the circle |z| = r > 1,

Er = C\{g(z) : |z| > r}, and Dr = g−1(Er). Note that the Cr is given by

w(t) = g(reit) = u(t) + iv(t), 0 ≤ t ≤ 2π.

5

By Green’s Theorem, the area of Er is given by

area(Er) =12

∫Cr

(udv − vdu)

=12i

∫Cr

(iudv − ivdu)

=12i

∫Cr

(iudv − ivdu) + 12i

∫Cr

(udu + vdv)

=12i

∫Cr

(udu + iudv − ivdu + vdv)

=12i

∫Cr

(u− iv)(du + idv)

=12i

∫Cr

w dw

=12i

∫|z|=r

g(z)g′(z) dz

Letting z = reit, 0 ≤ t ≤ 2π, we get

area(Er) =12

∫ 2π0

reitg(reit)g′(reit) dt

=12

∫ 2π0

reit

[reit +

∞∑n=0

bnr−ne−int

][1 +

∞∑n=1

−nbnr−n−1ei(−n−1)t]

dt

=12

∫ 2π0

[re−it +

∞∑n=0

bnr−neint

][reit −

∞∑n=1

nbnr−ne−int

]dt

=12

∫ 2π0

(r2 −

∞∑n=1

nbnr−n+1e−i(n+1)t +

∞∑n=0

bnr−n+1ei(n+1)t −

∞∑n=1

n|bn|2r−2n)

dt

= π

(r2 −

∞∑n=1

n|bn|2r−2n)

Letting r → 1+, we get m∗(E) = π

(1−

∞∑n=1

n|bn|2)

, where m∗ denotes two-dimensional

Lebesgue outer measure. Since m∗(E) ≥ 0, we must have∞∑

n=1

n|bn|2 ≤ 1.

For the equality case, note that∞∑

n=1

n|bn|2 = 1 if and only if m∗(E) = 0, i.e. g ∈ Σ̃.

6

Corollary 2.1. If g ∈ Σ, then |b1| ≤ 1. Equality holds if and only if g has the form

g(z) = z + b0 + b1/z where |b1| = 1.

Proof. We have |b1|2 ≤∞∑

n=1

n|bn|2 ≤ 1. Therefore, |b1|2 ≤ 1 and it follows that |b1| ≤ 1.

If |b1| = 1, then we have 1 ≥∞∑

n=1

n|bn|2 = |b1| +∞∑

n=2

n|bn|2 = 1 +∞∑

n=2

n|bn|2. Therefore,

bn = 0 for all n ≥ 2, and so g(z) = z + b0 + b1/z with |b1| = 1. Conversely, if g is of the given

form, then g ∈ Σ and |b1| = 1.

Theorem 2.2 (Bieberbach’s Theorem). If f ∈ S with f(z) = z + a2z2 + a3z3 + · · · , then

|a2| ≤ 2. Equality holds if and only if f is a rotation of the Koebe function.

Proof. Define g(z) = [f(1/z2)]−1/2 = z − a22

z−1 + · · · ∈ Σ.

The corollary to Theorem 2.1 implies∣∣∣a2

2

∣∣∣ ≤ 1, and so we must have |a2| ≤ 2.Equality holds if and only if g(z) = z − b1/z, with |b1| = 1. Let b1 = eiα for some α ∈ R.

Then f(z) = [g(z−1/2)]−2 = (z−1/2 − eiαz1/2)−2 = [z−1/2(1 − eiαz)]−2 = z(1 − eiαz)−2, which

is a rotation of the Koebe function.

Theorem 2.3 (Koebe 1/4-Theorem). If f ∈ S, then the range of f contains the disk

{w : |w| < 1/4}.

Proof. Suppose w /∈ {f(z) : |z| < 1}. Then g(z) = wf(z)w − f(z)

is analytic and univalent in D.

Note that

g(0) =wf(0)

w − f(0)= 0,

g′(z) =(w − f(z))wf ′(z)− wf(z)(−f ′(z))

(w − f(z))2=

w2f ′(z)(w − f(z))2

,

g′(0) =w2

w2= 1,

g′′(z) =(w − f(z))2w2f ′′(z)− w2f ′(z)2(w − f(z))(−f ′(z))

(w − f(z))4,

and

g′′(0) =w4f ′′(0) + 2w3

w4= f ′′(0) +

2w

= 2a2 +2w

.

Note in particular that g(0) = 0 and g′(0) = 1, so that g ∈ S.

7

Thus, g(z) =∞∑

n=0

g(n)(0)n!

zn = z +12

(2a2 +

2w

)z2 + · · · = z +

(a2 +

1w

)z2 + · · · . Because

g ∈ S, Theorem 2.2 implies that∣∣∣∣a2 + 1w

∣∣∣∣ ≤ 2, and, because f ∈ S, we also have |a2| ≤ 2.Thus, we have

∣∣∣∣ 1w∣∣∣∣ = ∣∣∣∣ 1w + a2 − a2

∣∣∣∣ ≤ ∣∣∣∣ 1w + a2∣∣∣∣+ |a2| ≤ 2 + 2 = 4. Therefore, |w| ≥ 14.

Hence, if |w| < 14, then w ∈ {f(z) : |z| < 1}, i.e. w is in the range of f .

Lemma 2.1. If f ∈ S, then∣∣∣∣zf ′′(z)f ′(z) − 2|z|21− |z|2

∣∣∣∣ ≤ 4|z|1− |z|2 for all z ∈ D.Proof. Let f ∈ S and fix t ∈ D. Let F be defined by

F (z) =f

(z + t1 + tz

)− f(t)

(1− |t|2)f ′(t)= z + A2(t)z2 + A3(t)z3 + · · · .

Note that F is a Koebe transform, so F ∈ S. Therefore, Theorem 2.2 implies that

|A2(t)| ≤ 2 for all t ∈ D. A calculation shows that F ′′(0) = (1 − |t|2)f ′′(t)f ′(t)

− 2t. Since

F ′′(0) = 2A2(t), we must have∣∣∣∣(1− |t|2)f ′′(t)f ′(t) − 2t

∣∣∣∣ ≤ 4. Multiplying both sides by |t|1− |t|2results in

∣∣∣∣ tf ′′(t)f ′(t) − 2|t|21− |t|2∣∣∣∣ ≤ 4|t|1− |t|2 , which is the desired result.

2.4 Classical Growth and Distortion Theorems

We are especially interested in the classical growth and distortion theorems. The two-point

distortion theorems that we will explore later are closely related to these.

Theorem 2.4. If f ∈ S and z ∈ D, then 1− |z|(1 + |z|)3

≤ |f ′(z)| ≤ 1 + |z|(1− |z|)3

. Equality holds if

and only if f is a suitable rotation of the Koebe function.

Proof. Let f ∈ S and z ∈ D. From Lemma 2.1, we have∣∣∣∣zf ′′(z)f ′(z) − 2|z|21− |z|2

∣∣∣∣ ≤ 4|z|1− |z|2 .Therefore:

− 4|z|1− |z|2

≤ Re{

zf ′′(z)f ′(z)

− 2|z|2

1− |z|2

}≤ 4|z|

1− |z|2

⇒ − 4|z|1− |z|2

≤ Re{

zf ′′(z)f ′(z)

}− 2|z|

2

1− |z|2≤ 4|z|

1− |z|2

⇒ 2|z|2 − 4|z|

1− |z|2≤ Re

{zf ′′(z)f ′(z)

}≤ 2|z|

2 + 4|z|1− |z|2

.

8

Let log f ′(z) denote a single-valued branch of the logarithm of f ′ with log f ′(0) = 0. Note that

we can choose such a branch of the logarithm since f ′(z) 6= 0 and f ′(0) = 1.

Now, Re{

zf ′′(z)f ′(z)

}= r

∂

∂rRe{log f ′(z)} (where z = reiθ). Therefore, we have

2r2 − 4r1− r2

≤ r ∂∂r

Re{log f ′(z)} ≤ 2r2 + 4r

1− r2,

which implies that2r − 41− r2

≤ ∂∂r

Re{log f ′(z)} ≤ 2r + 41− r2

.

Holding θ fixed and integrating with respect to r from 0 to |z| results in

log∣∣∣∣ 1− |z|(1 + |z|)3

∣∣∣∣ ≤ log |f ′(z)| ≤ log ∣∣∣∣ 1 + |z|(1− |z|)3∣∣∣∣ .

Exponentiating this expression gives the desired result.

Suitable rotations of the Koebe function provide cases of equality. Conversely, if equality

holds in either the upper bound or the lower bound for some z = Reiθ, then

Re{

eiθf ′′(0)f ′(0)

}= ±4. In this case, we must have |a2| = 2, which implies that f is a rotation

of the Koebe function.

Theorem 2.5. If f ∈ S and z ∈ D, then |z|(1 + |z|)2

≤ |f(z)| ≤ |z|(1− |z|)2

. Equality holds if

and only if f is a suitable rotation of the Koebe function.

Proof. Let f ∈ S and z = reiθ with 0 < r < 1.

To obtain the upper bound, let Γ denote the line segment from 0 to z and write

f(z) =∫

Γf ′(ζ) dζ =

∫ r0

f ′(ρeiθ)eiθ dρ.

Then Theorem 2.4 implies that:

|f(z)| ≤∫ r

0|f ′(ρeiθ)| dρ ≤

∫ r0

1 + ρ(1− ρ)3

dρ =r

(1− r)2.

This establishes the upper bound.

To obtain the lower bound, first notice thatr

(1 + r)2<

14

for 0 < r < 1. Therefore, if

|f(z)| ≥ 14, then the result holds trivially. If |f(z)| < 1

4, Theorem 2.3 implies that the range

9

of f contains the line segment from 0 to f(z). Let this line segment be denoted Λ and let

λ = f−1(Λ) be its preimage under f . Then f(z) =∫

Λdw =

∫λf ′(ζ) dζ. Because f ′(ζ)dζ has

constant argument along λ, Theorem 2.4 implies:

|f(z)| =∫

λ|f ′(ζ)||dζ| ≥

∫λ

1− |ζ|(1 + |ζ|)3

d|ζ| =∫ r

0

1− ρ(1 + ρ)3

dρ =r

(1 + r)2.

The statement about equality follows from the case of equality in Theorem 2.4.

Theorem 2.6. If f ∈ S and z ∈ D, then 1− |z|1 + |z|

≤∣∣∣∣zf ′(z)f(z)

∣∣∣∣ ≤ 1 + |z|1− |z| . Equality holds if andonly if f is a suitable rotation of the Koebe function.

Proof. Let f ∈ S be given and define F (z) =f

(z + ζ1 + ζz

)− f(ζ)

(1− |ζ|2)f ′(ζ)∈ S. By Theorem 2.5, we

have|ζ|

(1 + |ζ|)2≤ |F (−ζ)| ≤ |ζ|

(1− |ζ|)2.

But note that F (−ζ) = f(0)− f(−ζ)(1− |ζ|2)f ′(−ζ)

=−f(ζ)

(1− |ζ|2)f ′(−ζ).

Therefore:

|ζ|(1− |ζ|2)(1 + |ζ|)2

≤∣∣∣∣ f(ζ)f ′(ζ)

∣∣∣∣ ≤ |ζ|(1− |ζ|2)(1− |ζ|)2⇒ 1− |ζ|

2

(1 + |ζ|)2≤∣∣∣∣ f(ζ)ζf ′(ζ)

∣∣∣∣ ≤ 1− |ζ|2(1− |ζ|)2⇒ (1− |ζ|)(1 + |ζ|)

(1 + |ζ|)2≤∣∣∣∣ f(ζ)ζf ′(ζ)

∣∣∣∣ ≤ (1− |ζ|)(1 + |ζ|)(1− |ζ|)2⇒ 1− |ζ|

1 + |ζ|≤∣∣∣∣ f(ζ)ζf ′(ζ)

∣∣∣∣ ≤ 1 + |ζ|1− |ζ|⇒ 1− |ζ|

1 + |ζ|≤∣∣∣∣ζf ′(ζ)f(ζ)

∣∣∣∣ ≤ 1 + |ζ|1− |ζ| .Cases of equality are provided by suitable rotations of the Koebe function. We next show

that rotations of the Koebe function provide the only cases of equality. To this end, suppose

f ∈ S satisfies ∣∣∣∣ζf ′(ζ)f(ζ)∣∣∣∣ = 1− |ζ|1 + |ζ|

for some ζ ∈ D. Note then that∣∣∣∣ f(ζ)f ′(ζ)

∣∣∣∣ = |ζ|1 + |ζ|1− |ζ| .

10

Now, consider F ∈ S defined by

F (z) =f

(z + ζ1 + ζz

)− f(ζ)

(1− |ζ|2)f ′(ζ).

Then we have

|F (−ζ)| =∣∣∣∣ f(0)− f(ζ)(1− |ζ|2)f ′(ζ)

∣∣∣∣=∣∣∣∣ −f(ζ)(1− |ζ|2)f ′(ζ)

∣∣∣∣=

1(1− |ζ|2)

∣∣∣∣ f(ζ)f ′(ζ)∣∣∣∣

=1

(1− |ζ|)(1 + |ζ|)|ζ|1 + |ζ|

1− |ζ|

=|ζ|

(1− |ζ|)2.

By Theorem 2.5, F is a rotation of the Koebe function. It then can be shown that f must also

be a rotation of the Koebe function.

A similar argument shows that rotations of the Koebe function provide the only cases of

equality in the upper bound as well.

2.5 Motivation for Using the Hyperbolic Metric

The theorems we have just seen give conditions that are necessary, but not sufficient, for

univalence. In fact, there are many functions that satisfy the theorems but fail to be univalent.

Blatter [2] wondered if there were similar theorems that would give sufficient conditions for

univalence. Blatter’s research led to the proof of a two-point distortion theorem that gives

both necessary and sufficient conditions for univalence.

Blatter’s theorem and related results are given in terms of hyperbolic distance. In partic-

ular, Blatter related |f(a) − f(b)| to the hyperbolic distance between a and b (a, b ∈ D). As

we will see, Blatter’s result is invariant under suitable compositions with automorphisms of D

and C. The hyperbolic metric is useful in these results because it is invariant under conformal

automorphisms of D.

11

CHAPTER 3. THE HYPERBOLIC METRIC

In this chapter, we will follow the construction of the hyperbolic metric as given by Anderson

[1], with some results given by Krantz [7]. The goal is to construct a metric that preserves

distances under conformal automorphisms of the unit disk. To do so we will first construct

the hyperbolic metric on the upper-half plane model for the hyperbolic plane. We will then

transform this model to the unit disk.

3.1 The Upper-Half Plane Model of the Hyperbolic Plane

The underlying space in the upper-half plane model is H = {z ∈ C : Im(z) > 0}. To

construct this model of the hyperbolic plane, we will first define a hyperbolic line.

Definition 3.1. A hyperbolic line in H is:

1. the intersection of H with a Euclidean line in C that is orthogonal to R, or

2. the intersection of H with a Euclidean circle centered on R

Remark. If L is a Euclidean line in C, we can view L∪ {∞} as a circle in Ĉ = C∪ {∞} (the

extended complex plane). Therefore, every hyperbolic line in H is contained in a circle in Ĉ

that is orthogonal to the extended real line R̂ = R ∪ {∞}.

Proposition 3.1. Every circle in Ĉ can be expressed as the set of solutions in Ĉ to an equation

of the form αzz + βz + βz + γ = 0, where α, γ ∈ R and β ∈ C.

Proof. Since a circle in Ĉ is a Euclidean line in C or a Euclidean circle in C, we will consider

these two cases.

12

Case 1: First consider a Euclidean line in C given by ax + by + c = 0. Let z = x + iy. Since

x = Re(z) = 12(z + z) and y = Im(z) = −i2(z − z), we have:

ax + by + c =a

2(z + z)− i b

2(z − z) + c

=12(a− ib)z + 1

2(a + ib)z + c

Taking α = 0, β = 12(a − ib), and γ = c, we see that the line is given by the set of points

{z : αzz + βz + βz + γ = 0}.

Case 2: Next consider the Euclidean circle given by (x−h)2+(y−k)2 = r2. Letting z0 = h+ik,

we have:

r2 = (x− h)2 + (y − k)2

= |z − z0|2

= zz − z0z − z0z + |z0|2.

Hence, zz − z0z − z0z + |z0|2 − r2 = 0, and so the circle is given by the set of points

{z : αzz + βz + βz + γ = 0}, where α = 1, β = −z0, and γ = |z0|2 − r2.

Definition 3.2. We say that two hyperbolic lines in H are parallel if they are disjoint.

This construction of the hyperbolic plane does satisfy the axioms of hyperbolic geometry.

In particular, given a hyperbolic line ` in H and any point P in H not on `, there are infinitely

many hyperbolic lines through P parallel to `.

3.2 Möbius Transformations and the General Möbius Group

The metric that we construct on H needs to be invariant under certain transformations

taking H to itself. Since a hyperbolic line in H is contained in a circle in Ĉ, we will consider

transformations taking circles in Ĉ to circles in Ĉ.

We will begin by considering the set L of linear fractional transformations (also called

Möbius transformations). These are functions of the form T (z) =az + bcz + d

, where a, b, c, d ∈ C

and ad− bc 6= 0. If T ∈ L is of this form, then define:

13

1. T (∞) = limz→∞

T (z) =a

c

2. T(−d

c

)= lim

z→−d/cT (z) = ∞

With these conventions, every function T ∈ L is a continuous, bijective mapping of Ĉ to Ĉ.

Note that L is a group under composition.

Proposition 3.2. The group L is generated by elements of the form m(z) = αz+β (α, β ∈ C)

and the function J(z) = −1z.

Proof. Let T (z) =az + bcz + d

∈ L. We will consider two cases.

Case 1: If c = 0, then T (z) =a

dz +

b

d, which is an element of the form m(z) = αz + β with

α =a

dand β =

b

d.

Case 2: If c 6= 0, then we can write

T (z) =az + bcz + d

=(az + b)c(cz + d)c

=acz + bcc2z + cd

=acz + ad− (ad− bc)

c2z + cd

=a(cz + d)c(cz + d)

− ad− bcc2z + cd

=a

c− ad− bc

c2z + cd

= f(J(g(z)),

where g(z) = c2z + cd and f(z) = (ad− bc)z + ac.

In either case, we have written T as a composition of elements of the form m(z) = αz + β

and the function J(z) = −1z.

14

Proposition 3.3. Every element T ∈ L maps circles in Ĉ to circles in Ĉ.

Proof. By Proposition 3.2, it is enough to consider functions of the form m(z) = az + b

(a, b ∈ C) and the function J(z) = −1z.

Recall that a circle in Ĉ can be expressed as the set of solutions to an equation of the form

αzz + βz + βz + γ = 0, where α, γ ∈ R and β ∈ C (Proposition 3.1).

Let A = {z ∈ C : αzz + βz + βz + γ = 0} be a circle in Ĉ.

First we consider m(z) = az + b. Let w = az + b. Then z =1a(w − b). If z ∈ A, then we

have

0 = αzz + βz + βz + γ

=α

|a|2(w − b)(w − b) + β 1

a(w − b) + β 1

a(w − b) + γ

=α

|a|2(ww − bw − bw + |b|2) + β 1

a(w − b) + β 1

a(w − b) + γ

=α

|a|2ww +

(β

a− αb|a|2

)w +

(β

a− αb|a|2

)w +

(α|b|2

|a|2− 2Re

(βba

)+ γ)

=α

|a|2ww +

(β

a− αb|a|2

)w +

(β

a− αb|a|2

)w +

(α|b|2

|a|2− 2Re

(βba

)+ γ)

,

which gives the equation of a circle in Ĉ. Therefore, the image of A under m is a circle in Ĉ.

Next consider J(z) = −1z. Let w = J(z) = −1

z. Then z = − 1

wand so for any z ∈ A, we

have

0 = αzz + βz + βz + γ

= α1w

1w

+ β1w

+ β1w

+ γ.

Multiplying through by ww results in

γww + βw + βw + α = 0.

Since this is again an equation of a circle in Ĉ, we see that the image of A under J is a circle

in Ĉ.

15

We now extend L to a larger group, the general Möbius group M. The group M is

generated by L and C, where C : Ĉ → Ĉ is defined by C(z) = z for z ∈ C and C(∞) = ∞.

We have seen that circles in Ĉ are mapped by elements of L to circles in Ĉ. This is also true

of M.

Proposition 3.4. Every element of M maps circles in Ĉ to circles in Ĉ.

Proof. Consider a circle in Ĉ given by the equation αzz +βz +βz + γ = 0. Let w = C(z) = z.

Then z = w, and so we have αww + βw + βw + γ = 0. Note that this is also the equation of

a circle in Ĉ. Therefore, C maps circles in Ĉ to circles in Ĉ. Now, since every element of a

generating set for M takes circles in Ĉ to circles in Ĉ, the property must also hold for M.

Since our goal is to construct a metric that is invariant under conformal automorphisms,

we are interested in conformal mapping properties.

Proposition 3.5. Every element of M preserves magnitudes of angles.

Proof. It is well known that the elements of L are conformal mappings, and so preserve mag-

nitudes of angles. Thus, we need only consider the function C(z) = z. Since the geometric

interpretation of C is reflection across the real axis, it is easy to see that C preserves magnitudes

of angles.

Since we are concerned with elements of M that map H to itself, we define the subgroup

M(H) := {m ∈ M : m(H) = H}. It can be shown that a generating group for M(H) is given

by elements of the form m(z) = az + b (a > 0, b ∈ R) and the functions J(z) = −1z

and

B(z) = −z. We will also make use of the subgroup L(H) := {T ∈ L : T (H) = H}. Note that

L(H) ⊂ M(H) ⊂ M. Additionally, L(H) ⊂ L ⊂ M. We have the following results for these

subgroups.

Proposition 3.6. Every element of M(H) maps hyperbolic lines in H to hyperbolic lines in

H.

Proof. Apply Propositions 3.4 and 3.5 and the fact that every hyperbolic line in H is contained

in a circle in Ĉ orthogonal to R̂.

16

3.3 Hyperbolic Length and Distance in H

If f : [a, b] → C is a piecewise C1 path with f(t) = x(t) + iy(t), then we have

f ′(t) = x′(t) + iy′(t) and |f ′(t)| =√

(x′(t))2 + (y′(t))2. Therefore, the Euclidean length of f is

given by

length(f) =∫ b

a

√(x′(t))2 + (y′(t))2 dt =

∫ ba|f ′(t)| dt =

∫f|dz|.

We say that the standard Euclidean element of arc length in C is |dz| = |f ′(t)| dt. For any

continuous function ρ : C → R, the path integral of ρ along f is given by∫fρ(z)|dz| =

∫ ba

ρ(f(t))|f ′(t)| dt.

Viewing ρ(z)|dz| as a new element of arc length motivates the following definition.

Definition 3.3. Let f : [a, b] → C be a piecewise C1 path and ρ : C → R a continuous

function. The length of f with respect to the element of arc length ρ(z)|dz| is given by

lengthρ(f) =∫

fρ(z)|dz|.

To measure hyperbolic arc length, we must find an appropriate element of arc length. In

view of the fact that we want our metric to be invariant under composition with elements of

M(H), we will look for an element of arc length ρ(z)|dz| so that lengthρ(f) = lengthρ(γ ◦ f)

holds for all γ ∈M(H) and all piecewise C1 functions f : [a, b] → H.

We begin by investigating the conditions imposed on ρ by elements of L(H) ⊂ M(H). To

this end, let γ ∈ L(H) and let f : [a, b] → H be a piecewise C1 path. Then we have:

lengthρ(f) = lengthρ(γ ◦ f) ⇒∫ b

aρ(f(t))|f ′(t)| dt =

∫ ba

ρ(γ(f(t))

)|γ′(f(t))||f ′(t)| dt

⇒∫ b

a

(ρ(f(t))− ρ

(γ(f(t))

)|γ′(f(t))|

)|f ′(t)| dt = 0

Now, define µγ(z) = ρ(z)− ρ(γ(z))|γ′(z)|. Then the above condition becomes∫fµγ(z)|dz| =

∫ ba

µγ(f(t))|f ′(t)| dt = 0

for every piecewise C1 path f : [a, b] → H and every γ ∈ L(H).

17

Lemma 3.1. Let D ⊆ C be open and µ : D → R continuous. If∫

fµ(z)|dz| = 0 for every

piecewise C1 path f : [a, b] → D, then µ ≡ 0 on D.

Proof. Suppose by way of contradiction that µ ≡/ 0. Then there is some z0 ∈ D with µ(z0) 6= 0.

Without loss of generality, assume µ(z0) > 0. Let � =12|µ(z0)| > 0. By the continuity of µ

and openness of D, there is a δ > 0 so that µ(w) ∈ B�(µ(z0)) whenever w ∈ Bδ(z0) ⊂ D.

Since µ(w) ∈ B�(µ(z0)), we have: |µ(z0)| − |µ(w)| ≤ |µ(w) − µ(z0)| <12|µ(z0)|. Therefore,

|µ(w)| > |µ(z0)| −12|µ(z0)| =

12|µ(z0)| > 0. Hence, µ(w) > 0 for all w ∈ Bδ(z0).

Since∫

fµ(z)|dz| = 0 for all C1 paths f : [a, b] → D, it must hold for the path f : [0, 1] → D

defined by f(t) = z0 +12δt. Note that f ′(t) =

δ

2> 0. Also, for all t ∈ [0, 1], we have

f(t) ∈ Bδ(z0). Then by the above argument, we have µ(f(t)) > 0 for all t ∈ [0, 1].

In that case,∫

fµ(z)|dz| =

∫ 10

µ(f(t))|f ′(t)| dt > 0, which contradicts our hypothesis.

Therefore, we must have µ ≡ 0 on D.

Applying Lemma 3.1 to the analysis for the conditions imposed on ρ by elements of L(H),

we see that 0 = µγ(z) = ρ(z)− ρ(γ(z))|γ′(z)|, and this is true for all z ∈ H.

Now if γ, ϕ ∈ L(H), we have:

µγ◦ϕ(z) = ρ(z)− ρ((γ ◦ ϕ)(z)

)|(γ ◦ ϕ)′(z)|

= ρ(z)− ρ(γ(ϕ(z))

)|γ′(ϕ(z))||ϕ′(z)|

= ρ(z)− ρ(ϕ(z))|ϕ′(z)|+(ρ(ϕ(z))− ρ

(γ(ϕ(z))

)|γ′(ϕ(z))|

)|ϕ′(z)|

= µϕ(z) + µγ(ϕ(z))|ϕ′(z)|

Notice that if µγ ≡ 0 and µϕ ≡ 0, then µγ◦ϕ ≡ 0. Therefore, instead of considering γ ∈ L(H),

we may consider γ in a generating set for L(H). It can be shown that a generating set for L(H)

is given by elements of the form m(z) = az + b (a > 0, b ∈ R) and the function J(z) = −1z.

We will consider the conditions imposed on ρ by the functions of this generating set.

First consider γ(z) = z + b (a = 1). Then we have γ′(z) = 1 so that

µγ(z) = ρ(z)− ρ(γ(z))|γ′(z)| = ρ(z)− ρ(z + b).

18

Now, if µγ ≡ 0, then we must have ρ(z) = ρ(z+b) for all z ∈ C, b ∈ R. In particular, this implies

that ρ depends only on Im(z). Let y = Im(z) and consider a function r : (0,∞) → (0,∞)

defined by r(y) = ρ(iy). Note that since ρ depends only on its imaginary part, we have

ρ(z) = ρ(iIm(z)) = r(Im(z)) = r(y).

Now consider γ(z) = az (a > 0, b = 0). Then γ′(z) = a so that

µγ(z) = ρ(z)− ρ(γ(z))|γ′(z)| = ρ(z)− aρ(az) = r(y)− ar(ay).

The condition µγ ≡ 0 implies that r(y) = ar(ay) for all y > 0. Letting y = 1, we get

r(1) = ar(a); therefore, r(a) =r(1)a

for all a > 0. Thus, ρ(z) = r(Im(z)) =r(1)

Im(z)=

c

Im(z)for some constant c. The value of c is arbitrary, so we will take c = 12 for later convenience.

In our construction of ρ, we only used elements of the generating set for L(H) of the form

m(z) = az + b, where a > 0 and b ∈ R. Therefore, lengthρ(f) = lengthρ(m ◦ f) for any m

of this form. The question we have yet to answer is: Does this ρ produce a length function

that is invariant under composition with elements of M(H)? This question is answered by the

following proposition.

Proposition 3.7. If f : [a, b] → H is a piecewise C1 path, then

lengthρ(f) = lengthρ(γ ◦ f)

for any γ ∈M(H).

Proof. We have already seen that lengthρ(f) = lengthρ(m◦f) for any m = az+b (a > 0, b ∈ R).

19

Next consider lengthρ(J ◦ f), where J(z) = −1z. Note that J ′(z) =

1z2

, and so we have

µJ(z) = ρ(z)− ρ(J(z))|J ′(z)|

= ρ(z)− ρ(−1

z

)1|z|2

= ρ(z)− ρ(− z|z|2

)1|z|2

=1

2Im(z)− 1

2Im (−z/|z|2)1|z|2

=1

2Im(z)− 1

2Im(−z)/|z|2· 1|z|2

=1

2Im(z)− 1

2Im(−z)

=1

2Im(z)− 1

2Im(z)= 0

Therefore, lengthρ(f)− lengthρ(J ◦ f) =∫

fµJ(z)|dz| =

∫f0|dz| = 0.

Hence, lengthρ(f) = lengthρ(J ◦ f).

Finally, consider lengthρ(B ◦ f), where B(z) = −z. Write f(t) = x(t) + iy(t), where

x : [a, b] → R and y : [a, b] → R. Note then that B ◦ f(t) = −x(t) + iy(t),

|(B ◦ f)′(t)| =√

(x′(t))2 + (y′(t))2 = |f ′(t)|, and Im((B ◦ f)(t)) = y(t) = Im(f(t)).

Therefore:

lengthρ(B ◦ f) =∫ b

a

12Im(B ◦ f)(t)

|(B ◦ f)′(t)| dt =∫ b

a

12Im(f(t))

|f ′(t)| dt = lengthρ(f).

Since the elements considered above form a generating set for M(H), we must have

lengthρ(f) = lengthρ(γ ◦ f) for every element of M(H).

Motivated by these results, we make the following definitions.

Definition 3.4. The hyperbolic element of arc length λH|dz| on H is defined to be

λH|dz| =1

2Im(z)|dz|.

Definition 3.5. Let f : [a, b] → H be a piecewise C1 path. The hyperbolic length of f is

defined to be

lengthH(f) =∫

fλH(z)|dz| =

∫f

12Im(z)

|dz| =∫ b

a

12Im(f(t))

|f ′(t)| dt.

20

Definition 3.6. For x, y ∈ H, define the hyperbolic distance between x and y by

dH(x, y) = inf{lengthH(f) : f ∈ Γ[x, y]},

where Γ[x, y] = {f : [a, b] → H : f is piecewise C1, f(a) = x, and f(b) = y}.

Note that we have now constructed length and distance functions that are invariant under

composition with elements of M(H).

3.4 The Poincaré Disk Model, D

Recall that our goal is to construct the hyperbolic metric on the Poincaré disk model of

the hyperbolic plane. The underlying space for this model is the unit disk D = {z : |z| < 1}.

To transform the model H of the hyperbolic plane into the Poincaré disk model D, consider an

element m ∈M taking D to H. Define a hyperbolic line in D to be the image under m−1 of a

hyperbolic line in H. We will use the following facts:

1. Every hyperbolic line in H is contained in a circle in Ĉ orthogonal to R (by the remark

following Definition 3.1);

2. Every element of M takes circles in Ĉ to circles in Ĉ (by Proposition 3.4);

3. Every element of M preserves magnitudes of angles (by Proposition 3.5); and

4. R is mapped by m−1 to the unit circle, {z : |z| = 1} since the boundary of H must be

mapped to the boundary of D by m−1.

Combined, these facts imply that a hyperbolic line in D is the intersection of D with a circle

in Ĉ orthogonal to the unit circle.

If f : [a, b] → D is a piecewise C1 path in D and ξ is any element of L taking D to H, then

the composition ξ ◦ f : [a, b] → H is a piecewise C1 path in H. Now, if ξ and η are any two

elements of M taking D to H, then q = η ◦ξ−1 is an element of M(H). Since hyperbolic length

is invariant under composition with elements of M(H), we have

lengthH(ξ ◦ f) = lengthH(q ◦ ξ ◦ f) = lengthH(η ◦ f).

21

Therefore, we make the following definition.

Definition 3.7. Let ξ be any element of L taking D to H and f : [a, b] → D be a piecewise C1

path into D. The hyperbolic length of f in D is given by

lengthD(f) = lengthH(ξ ◦ f)

By the above work, we see that this definition is independent of the choice of ξ. Therefore,

we can find an integral expression for lengthD(f) using any ξ ∈ L taking D to H. In particular,

we will use the element ξ defined by ξ(z) =i√2z + 1√

2

− 1√2z − i√

2

. Note that Im(ξ(z)) =1− |z|2

| − z − i|2and

|ξ′(z)| = 2|z + i|2

. Hence,

12Im(ξ(z))

|ξ′(z)| = 11− |z|2

.

Therefore,

lengthD(f) = lengthH(ξ ◦ f) =∫

ξ◦f

12Im(z)

|dz| =∫ b

a

12Im(ξ(f(t)))

|ξ′(f(t))||f ′(t)| dt

=∫

f

12Im(ξ(z))

|ξ′(z)||dz| =∫

f

11− |z|2

|dz|.

Thus, the hyperbolic element of arc length on D is given by λD(z)|dz| =1

1− |z|2|dz|. Therefore,

we make the following definition.

Definition 3.8. The hyperbolic length of any piecewise C1 path f : [a, b] → D is given by

lengthD(f) =∫

fλD(z)|dz| =

∫f

11− |z|2

|dz|.

This length function is invariant under composition with conformal automorphisms of D.

Lemma 3.2. Every conformal automorphism of D is the composition of maps of the form

(i) gτ (z) = eiτz, τ ∈ R, and

(ii) φa(z) =z − a1− az

, a ∈ D.

Proof. We first show that any function of form (ii) is a conformal automorphism of D. To that

end, first consider {z : |z| = 1}. Now, if |z| = 1, we have

|φa(z)| =∣∣∣∣ z − a1− az

∣∣∣∣ = ∣∣∣∣1z∣∣∣∣ ∣∣∣∣ z − a1− az

∣∣∣∣ = ∣∣∣∣ z − az − a|z|2∣∣∣∣ = ∣∣∣∣z − az − a

∣∣∣∣ = 1.

22

Thus, φa maps {z : |z| = 1} to {z : |z| = 1}. Moreover, φa(a) = 0. Therefore, φa maps D to

D. Also, a calculation shows that (ϕa)−1 = ϕ−a. Thus, ϕa is one-to-one and onto. Hence, ϕa

is a conformal automorphism of D.

Now, suppose that h is a conformal automorphism of D. Let a = h(0) and consider the

function G = ϕa ◦ h. Note that G is a composition of conformal automorphisms of D, so G

itself must be a conformal automorphism of D. Also, G(0) = ϕa(h(0)) = ϕa(a) = 0. Then

the Schwarz Lemma implies that |G′(0)| ≤ 1. Applying the Schwarz Lemma to G−1 results

in∣∣∣∣ 1G′(0)

∣∣∣∣ = |(G−1)′(0)| ≤ 1. Hence, we may conclude that |G′(0)| = 1, which implies thatG is of the form G(z) = eiτz = gτ (z) (τ ∈ R). Since G = ϕa ◦ h = gτ , we must have

h = gτ ◦ (ϕa)−1 = gτ ◦ ϕ−a. Thus, h can be written as a composition of functions of the form

(i) and (ii).

Proposition 3.8. Let h : D → D be a conformal automorphism of the disk and f : [a, b] → D

a piecewise C1 path. Then lengthD(h ◦ f) = lengthD(f).

Proof. By Lemma 3.2, we need to consider only two cases.

Case 1: If h is a rotation, then h(z) = eiτz for some τ ∈ R. Therefore, |h′(z)| = 1, so we have

λD(h(z))|h′(z)| = λD(eiτz) =1

1− |eiτz|2=

11− |z|2

= λD(z).

Hence:

lengthD(h◦f) =∫

h◦fλD(z)|dz| =

∫ ba

λD(h◦f(t))|h′(f(t))||f ′(t)| dt =∫

fλD(h(z))|h′(z)||dz|

=∫

fλD(z)|dz| = lengthD(f)

Case 2: If h(z) =z − a1− az

for some a ∈ D, then |h′(z)| = 1− |a|2

|1− az|2. Therefore:

λD(h(z))|h′(z)| = λD(

z − a1− az

)· 1− |a|

2

|1− az|2=

1

1−∣∣∣∣ z − a1− az

∣∣∣∣2· 1− |a|

2

|1− az|2

=1− |a|2

|1− az|2 − |z − a|2=

1− |a|2

1− |z|2 − |a|2 + |a|2|z|2

=1− |a|2

(1− |z|2)(1− |a|2)=

11− |z|2

= λD(z).

Then, as above, we must have lengthD(h ◦ f) = lengthD(f).

23

Since any conformal automorphism of D is the composition of maps of the above forms, we

have lengthD(h ◦ f) = lengthD(f) for any conformal automorphism h.

We are finally ready to define the hyperbolic distance between two points in D.

Definition 3.9. For P,Q ∈ D, define the hyperbolic distance between P and Q by

dD(P,Q) = inf{lengthD(f) : f ∈ Γ[P,Q]},

where Γ[P,Q] = {f : [a, b] → D : f is piecewise C1, f(a) = P, and f(b) = Q}.

In fact, we can find an explicit formula for the hyperbolic distance between two points.

Proposition 3.9. If P,Q ∈ D, then the hyperbolic distance from P to Q is given by

dD(P,Q) =12

log

1 +∣∣∣∣ P −Q1− PQ

∣∣∣∣1−

∣∣∣∣ P −Q1− PQ∣∣∣∣ .

Proof. Case 1: First consider the case where P = 0 and Q ∈ R.

We claim that among all C1 curves of the form f(t) = t + iw(t), 0 ≤ t ≤ 1 − �, that satisfy

f(0) = 0 and f(1− �) = 1− �, the one of least length is µ(t) = t. To see that this is the case,

consider any such f . We have:

lengthD(f) =∫

fλD(z)|dz| =

∫f

11− |z|2

|dz| =∫ 1−�

0

11− |f(t)|2

|f ′(t)| dt

=∫ 1−�

0

11− t2 − [w(t)]2

|1+iw′(t)| dt =∫ 1−�

0

11− t2 − [w(t)]2

(1+[w′(t)]2)1/2 dt.

Note that1

1− t2 − [w(t)]2≥ 1

1− t2and (1 + [w′(t)]2)1/2 ≥ 1.

Thus,

lengthD(f) ≥∫ 1−�

0

11− t2

dt = lengthD(µ).

Note that with small modifications, this result can be extended to piecewise C1 curves.

Moreover, if a piecewise C1 curve connecting 0 to 1− � is not of the form

f(t) = t + iw(t), (∗)

it may cross itself. A shorter curve can be obtained by eliminating the loops. If the resulting

curve is not of the form (∗), it can be shown that it will be longer than a curve of the form

24

(∗). Hence, the shortest curve connecting 0 to 1− � is µ(t) = t.

Note that

lengthD(µ) =∫ 1−�

0

11− t2

dt =12

∫ 1−�0

(1

1 + t− 1

1− t

)dt

=12

[log∣∣∣∣1 + t1− t

∣∣∣∣]1−�0

=12

log∣∣∣∣2− ��

∣∣∣∣.Taking 1− � = Q, we get dD(P,Q) = dD(0, Q) =

12

log∣∣∣∣1 + Q1−Q

∣∣∣∣.Case 2: Now we consider the general case where P,Q ∈ D.

Define φ(z) =z − P1− Pz

. Then Proposition 3.8 implies dD(P,Q) = dD(φ(P ), φ(Q)) = dD(0, φ(Q)).

Since we can find τ ∈ R so that |φ(Q)| = eiτφ(Q), Proposition 3.8 also implies

dD(0, φ(Q)) = dD(0, |φ(Q)|). Finally, note that |φ(Q)| =∣∣∣∣ Q− P1− PQ

∣∣∣∣ = ∣∣∣∣ P −Q1− PQ∣∣∣∣. Therefore,

Case 1 implies dD(P,Q) =12

log

1 +∣∣∣∣ P −Q1− PQ

∣∣∣∣1−

∣∣∣∣ P −Q1− PQ∣∣∣∣.

25

CHAPTER 4. TWO-POINT DISTORTION THEOREMS

We will now discuss two-point distortion theorems for univalent functions. The theorems

on which we will focus were proved by Kim and Minda in [6] and Ma and Minda in [8]. These

theorems are extensions of the work done by Blatter in [2].

4.1 An Invariant Koebe Distortion Theorem

We will begin by looking at a special case of Kim and Minda’s distortion theorem. The

general theorem will be discussed in section 4.2. This result is given in [6] and is also proved

by Graham and Kohr in [4].

Theorem 4.1. Suppose f is univalent on D. Then for a, b ∈ D,

|f(a)− f(b)| ≥ sinh(2dD(a, b))2 exp(2dD(a, b))

max{(1− |a|2)|f ′(a)|, (1− |b|2)|f ′(b)|

}.

Conversely, if a nonconstant analytic function f satisfies this inequality, then f is univalent

on D.

Proof. If g is a normalized univalent function, then Theorem 2.5 implies that

|g(z)| ≥ |z|(1 + |z|)2

.

Now, since dD(y, z) =12

log

1 +∣∣∣ y−z1−yz ∣∣∣

1−∣∣∣ y−z1−yz ∣∣∣

for any y, z ∈ D, we have

dD(0, z) =12

log

1 +∣∣∣ −z1−0z

∣∣∣1−

∣∣∣ 0−z1−0z

∣∣∣ = 1

2log(

1 + |z|1− |z|

).

26

Therefore, exp(2dD(0, z)) =1 + |z|1− |z|

.

Also, because sinh(t) =exp(t) + exp(−t)

2, we have

sinh(2dD(0, z)) =12

(exp

(log(1 + |z|

1− |z|

))− exp

(− log

(1 + |z|1− |z|

)))

=12

(1 + |z|1− |z|

− 1− |z|1 + |z|

)=

12

((1 + |z|)2 − (1− |z|)2

(1− |z|)(1 + |z|)

)=

12

(1 + 2|z|+ |z|2 − 1 + 2|z| − |z|2

(1− |z|)(1 + |z|)

)=

2|z|(1− |z|)(1 + |z|)

Therefore,sinh(2dD(0, z))2 exp(2dD(0, z))

=2|z|

(1− |z|)(1 + |z|)· 1− |z|2(1 + |z|)

=|z|

(1 + |z|)2.

Thus, the lower bound in Theorem 2.5 can be restated as

|g(z)| ≥ sinh(2dD(0, z))2 exp(2dD(0, z))

for any normalized univalent function g.

Now let f be any function univalent in D. Note in particular that f need not be normalized.

Let a, b ∈ D and T (z) = z + a1 + az

. Then T is a conformal automorphism of D mapping 0 to a.

Consider the function g defined by

g(z) =f ◦ T (z)− f ◦ T (0)

(f ◦ T )′(0)=

f ◦ T (z)− f(a)(1− |a|2)f ′(a)

.

Let z0 =b− a1− ab

. Then T (z0) = b and so we have:

|g(z0)| =∣∣∣∣f ◦ T (z0)− f(a)(1− |a|2)f ′(a)

∣∣∣∣ = ∣∣∣∣ f(b)− f(a)(1− |a|2)f ′(a)∣∣∣∣

Also, since g is a normalized univalent function and dD is invariant under conformal automor-

phisms of D, Theorem 2.5 implies that

|g(z0)| ≥sinh(2dD(0, z0))2 exp(2dD(0, z0))

=sinh(2dD(T (0), T (z0)))2 exp(2dD(T (0), T (z0)))

=sinh(2dD(a, b))2 exp(2dD(a, b))

.

27

Hence: ∣∣∣∣ f(b)− f(a)(1− |a|2)f ′(a)∣∣∣∣ ≥ sinh(2dD(a, b))2 exp(2dD(a, b)) ,

which implies that

|f(b)− f(a)| ≥ sinh(2dD(a, b))2 exp(2dD(a, b))

(1− |a|2)|f ′(a)|.

By a similar argument, we also have

|f(b)− f(a)| ≥ sinh(2dD(a, b))2 exp(2dD(a, b))

(1− |b|2)|f ′(b)|.

Taking the maximum of these two lower bounds results in

|f(b)− f(a)| ≥ sinh(2dD(a, b))2 exp(2dD(a, b))

max{(1− |a|2)|f ′(a)|, (1− |b|2)|f ′(b)|

},

which is the desired inequality.

Next, suppose f is a nonconstant analytic function on D which satisfies the inequality.

If f is not univalent, then we can find some a, b ∈ D with a 6= b and f(a) = f(b). Since

f(a) = f(b), we havesinh(2dD(a, b))2 exp(2dD(a, b))

max{(1 − |a|2)|f ′(a)|, (1 − |b|2)|f ′(b)|

}= 0. Note that

sinh(2dD(a, b))2 exp(2dD(a, b))

6= 0 since a 6= b. Hence, (1 − |a|2)|f ′(a)| = (1 − |b|2)|f ′(b)| = 0. Because

a, b ∈ D, we have |a| < 1 and |b| < 1, so that (1 − |a|2) > 0 and (1 − |b|2) > 0. Hence,

f ′(a) = f ′(b) = 0, and so f is not univalent in any neighborhood of a or b. Since f is not

univalent in any neighborhood of a, we can find two sequences {cn}∞n=1 and {dn}∞n=1 of distinct

points in D so that limn→∞

cn = limn→∞

dn = a and f(cn) = f(dn) for all n ∈ N. Apply the

inequality again to conclude that f ′(cn) = 0 for all n ∈ N, which implies that f must be

constant. But this contradicts the hypothesis that f is nonconstant. Therefore, f must be

univalent on D.

4.2 Lower Bounds for |f(a)− f(b)|

The work that led to many of the results regarding two-point distortion theorems was done

by Blatter in [2]. Blatter’s distortion theorem is as follows.

28

Theorem 4.2. Suppose f is univalent in D and a, b ∈ D. Then

|f(a)− f(b)|2 ≥ sinh2(2dD(a, b))

8 cosh(4dD(a, b))

(((1− |a|2)|f ′(a)|

)2 + ((1− |b|2)|f ′(b)|)2).Conversely, if a nonconstant analytic function f satisfies this inequality, then f is univalent

on D.

Notice in particular that Blatter’s theorem gives necessary and sufficient conditions for

univalence. Moreover, the theorem requires no normalization on f . Since Blatter’s distortion

theorem is a special case of the theorem proved by Kim and Minda, we will not give the proof

here. The method used by Kim and Minda to prove their distortion theorem is an extension

of the method used by Blatter. Before we can prove Kim and Minda’s distortion theorem, we

need some more preliminary results.

The first result that we need, the Minimum Principle, is an extension of a result proved by

Blatter in [2]. The other results are from [6].

Lemma 4.1 (Minimum Principle). Suppose that a function u : [−L,L] → R satisfies the

following two conditions:

(i) |u′| ≤ q,

(ii) u′′ ≤ p(q2 − (u′)2),

where p and q are positive constants. If v satisfies

(i) |v′| ≤ q,

(ii) v′′ = p(q2 − (v′)2),

(iii) v(L) = u(L), and

(iv) v(−L) = u(−L),

then u(s) ≥ v(s) for all s ∈ [−L,L].

29

Proof. First note that condition (i) implies

|u(L)− u(−L)| =∣∣∣∣∫ L−L

u′(s) ds∣∣∣∣ ≤ ∫ L

−L|u′(s)| ds ≤

∫ L−L

q ds = 2qL,

and so −2qL ≤ u(L)− u(−L) ≤ 2qL. Moreover, u(L)− u(−L) = 2qL if and only if u′(s) ≡ q

and u(L)− u(−L) = −2qL if and only if u′(s) ≡ −q.

Since v′′ = p(q2 − (v′)2), v is of the form

v(s) =1p

log(cosh(pqs) + τ sinh(pqs)) + C,

where τ, C ∈ R.

We next show that |τ | ≤ 1. To that end, note that

v(s) =1p

log[cosh(pqs) + τ sinh(pqs)] + log C

=1p

log[12(exp(pqs) + exp(−pqs)) + τ

2(exp(pqs)− exp(−pqs))

]+ log C

=1p

log(exp(pqs) + exp(−pqs) + τ(exp(pqs)− exp(−pqs))

)− 1

plog 2 + log C

=1p

log(

exp(−pqs)(exp(2pqs) + 1 + τ exp(2pqs)− τ

))− 1

plog 2 + log C

=1p

log(exp(2pqs) + 1 + τ exp(2pqs)− τ

)− qs− 1

plog 2 + log C

Therefore,

v′(s) =1p

(2pq exp(2pqs)(τ + 1)

exp(2pqs)(τ + 1)− τ + 1

)− q

=2q(exp(2pqs)(τ + 1)− τ + 1

)− 2q(−τ + 1)

exp(2pqs)(τ + 1)− τ + 1− q

= q +2q(τ − 1)

exp(2pqs)(τ + 1)− τ + 1

By hypothesis, we have |v′(s)| ≤ q for all s. Hence,∣∣∣∣q + 2q(τ − 1)exp(2pqs)(τ + 1)− τ + 1

∣∣∣∣ ≤ q,and so we must have

2q(τ − 1)exp(2pqs)(τ + 1)− τ + 1

≤ 0. Now, 2q(τ − 1) ≤ 0 if and only if τ ≤ 1.

Moreover, we have

exp(2pqs)(τ + 1)− τ + 1 < 0 ⇒ τ(exp(2pqs)− 1) + exp(2pqs) + 1 < 0

⇒ τ < 1 + exp(2pqs)1− exp(2pqs)

=−1

tanh(pqs)for all s

⇒ τ < −1.

30

Therefore,2q(τ − 1)

exp(2pqs)(τ + 1)− τ + 1≤ 0 only for |τ | ≤ 1. Hence, the condition |v′| ≤ q implies

|τ | ≤ 1.

Now, if |τ | = 1, we obtain v′(s) = q or v′(s) = −q. If v′(s) = q, then we have

v(L) − v(−L) =∫ L−L

v′(s) dx =∫ L−L

q ds = 2qL. Then also (by the boundary conditions)

u(L) − u(−L) = 2qL, and so u′(s) = q (by condition (i)). Similarly, if v′(s) = −q, then

u′(s) = −q. Hence, for |τ | = 1, we have u′(s) = v′(s), u(L) = v(L), and u(−L) = v(−L).

Thus, u ≡ v and the desired inequality is trivial.

Therefore, we may assume that |τ | < 1. Because |τ | < 1, we have 2q(τ − 1) < 0 and

exp(2pqs)(τ + 1)− τ + 1 > 0. Hence, 2q(τ − 1)exp(2pqs)(τ + 1)− τ + 1

< 0 and |v′(s)| < q on [−L,L].

Applying Rolle’s Theorem to u−v, we see that there is an s0 ∈ (−L,L) with u′(s0) = v′(s0).

We will prove that u′(s) ≤ v′(s) for all s ∈ [s0, L]. Suppose this is not the case. Let

s1 = inf{s ≥ s0 : u′(s) > v′(s)}. Then we have u′(s1) = v′(s1) ∈ (−q, q). Now, there is a δ > 0

with |u′(s)| < q on [s1, s1 + δ). Therefore, for t ∈ (s1, s1 + δ), we have

u′′(t)p(q2 − (u′)2(t))

≤ 1 = v′′(t)

p(q2 − (v′)2(t)).

For s ∈ [s1, s1 + δ), integrating from s1 to s results in

arctanhu′

q

∣∣∣∣ss1

≤ arctanhv′

q

∣∣∣∣ss1

.

Since u′(s1) = v′(s1), we must have u′(s) ≤ v′(s) on [s1, s1 + δ). But this contradicts the

definition of s1, so u′(s) ≤ v′(s) for all s ∈ [s0, L].

Now, u′(s) − v′(s) ≤ 0 for all s ∈ [s0, L], so u − v is monotonically decreasing on this

interval. Since u(L)− v(L) = 0, we must then have u(s)− v(s) ≥ 0 on [s0, L].

A similar argument can be used to show that u(s) ≥ v(s) on [−L, s0].

Note in particular that the solution v of the differential equation in the lemma can be

expressed as

v(s) =1p

log[cosh(pqs) + τ sinh(pqs)] + log C,

where τ ∈ [−1, 1] and C ∈ R. In fact, it can be shown that C =(

exp(pu(L)) + exp(pu(−L))2 cosh(pqL)

)1/p.

31

Lemma 4.2. For p > 1, q > 0 and τ ∈ [−1, 1] let B(τ) =∫ L−L

(cosh(pqs) + τ sinh(pqs))1/p ds.

Then for τ ∈ (−1, 1), B(τ) > B(±1) = 2q

sinh(qL).

Proof. Since B(τ) =∫ L−L

(cosh(pqs) + τ sinh(pqs))1/p ds, we have

B′(τ) =1p

∫ L−L

sinh(pqs)(cosh(pqs) + τ sinh(pqs)

)(1−p)/pds

and

B′′(τ) =1− pp2

∫ L−L

sinh2(pqs)(cosh(pqs) + τ sinh(pqs)

)(1−2p)/pds.

Since p > 1, we have B′′(τ) < 0. Thus, B is strictly concave on [−1, 1] and the minimum value

of B is either B(1) or B(−1). Since B(1) = B(−1) = 2q

sinh(qL), the result follows.

Theorem 4.3. If g(z) = z + a2z2 + a3z3 + · · · ∈ S, then

∣∣∣∣a3 − (3− p3)

a22

∣∣∣∣+ p3 |a2|2 ≤

1 + 2 exp(

2p− 3p

), 0 < p <

32;

8p− 33

,32≤ p.

Proof. It suffices to find an upper bound for the functional

Lp(g) = Re{

a3 −(

3− p3

)a22

}+

p

3|a2|2

over the family of normalized univalent functions.

Notice that

Lp(g) = Re{a3} −(

3− p3

)Re{a22}+

p

3|a2|2

= Re{a3} −(

3− p3

)[(Re{a2})2 − (Im{a2})2] +

p

3[(Re{a2})2 + (Im{a2})2]

= Re{a3} −3− 2p

3(Re{a2})2 + (Im{a2})2

Since replacing g(z) by −g(−z) does not change the value of the functional, we may assume

that Re{a2} ≥ 0. Moreover, since 0 ≤ Re{a2} ≤ 2, by Theorem 2.2, there is a unique λ ∈ [0, 2]

with Re{a2} = λ(

1 + log2λ

).

In 1960, Jenkins (see [10], p.120) proved that

Re{a3} ≤ (Re{a2})2 − (Im{a2})2 − 2λRe{a2}+ λ2 log2λ

+32λ2 + 1.

32

Therefore:

Lp(g) ≤2p3

(Re{a2})2 − 2λRe{a2}+ λ2 log2λ

+32λ2 + 1

=2p3

λ2(

1 + log2λ

)2− 2λ2

(1 + log

2λ

)+ λ2 log

2λ

+32λ2 + 1

=2p3

λ2

(1 + 2 log

2λ

+(

log2λ

)2)− 2λ2 − 2λ2 log 2

λ+ λ2 log

2λ

+32λ2 + 1

= λ2(

2p3− 2 + 3

2

)+ λ2 log

2λ

(4p3− 2 + 1

)+ λ2

(log

2λ

)2(2p3

)+ 1

=(

4p− 36

)λ2 +

(4p− 3

3

)λ2 log

2λ

+2p3

λ2(

log2λ

)2+ 1

= H(λ)

Note that H(0) = 1 and H(2) =(

4p− 36

)(4) + 1 =

8p− 33

.

Also,

H ′(λ) = 2λ(

4p− 36

)+ 2λ

(4p− 3

3

)log

2λ

+(

4p− 33

)λ2(−2/λ2

2/λ

)+ 2λ

(2p3

)(log

2λ

)2+

2p3

λ2(

2 log2λ

)(−2/λ2

2/λ

)=(

4p− 33

)λ + 2λ

(4p− 3

3

)log

2λ−(

4p− 33

)λ + 2λ

(2p3

)(log

2λ

)2− 2λ

(2p3

)log

2λ

= 2λ(

2p− 33

)log

2λ

+ 2λ(

2p3

)(log

2λ

)2=

2λ3

log2λ

[2p− 3 + 2p log 2

λ

]

If p ≥ 32, then H ′(λ) has no zeros in (0, 2). Thus, H(λ) is strictly increasing with maximum

value H(2) =8p− 3

3.

If 0 < p <32, then H ′(λ) has a zero at λ0 = 2 exp

(2p− 3

2p

)∈ (0, 2). Note that H(λ)

is strictly increasing on (0, λ0) and strictly decreasing on (λ0, 2). Therefore, H attains its

maximum at H(λ0) = 1 + 2 exp(

2p− 3p

).

33

Corollary 4.1. If g(z) = z + a2z2 + a3z3 + · · · ∈ S, then∣∣∣∣a3 − 12a22

∣∣∣∣+ 56 |a2|2 ≤ 133 .Proof. The result follows from Theorem 4.3 with p =

32

and from Theorem 2.2. In fact, for

p =32, we have:∣∣∣∣a3 − 12a22

∣∣∣∣+ 56 |a2|2 =∣∣∣∣a3 − (3− p3

)a22

∣∣∣∣+ p3 |a2|2 + 13 |a2|2 ≤ 8p− 33 + 13(2)2 = 133 .

The distortion theorems given by Kim and Minda in [6] and Ma and Minda in [8] use the

following notation:

• D1f(z) = (1− |z|2)f ′(z),

• D2f(z) = (1− |z|2)2f ′′(z)− 2z(1− |z|2)f ′(z),

• D3f(z) = (1− |z|2)3f ′′′(z)− 6z(1− |z|2)2f ′′(z) + 6z2(1− |z|)2f ′(z),

• Rf (z) =f ′′′(z)f ′(z)

− 32

(f ′′(z)f ′(z)

)2, and

• Qf (z) =D2f(z)D1f(z)

= (1− |z|2)f′′(z)

f ′(z)− 2z.

We are finally ready to state and prove Kim and Minda’s distortion theorem.

Theorem 4.4. Suppose that f is univalent in D. There is a constant P ∈ (1, 3/2] such that

for any p ≥ P and all a, b ∈ D,

|f(a)− f(b)| ≥ sinh(2dD(a, b))2[2 cosh(2pdD(a, b))]1/p

(|D1f(a)|p + |D1f(b)|p

)1/p.

Conversely, if a nonconstant analytic function f satisfies this inequality for all a, b ∈ D, then

f is univalent on D.

Proof. The proof that a function satisfying the inequality is univalent is the same as the proof

of sufficiency in Theorem 4.1.

Now assume that f is univalent and let a, b ∈ D. Suppose p ≥ 1 is any number such that∣∣∣(1− |z|2)2Rf (z) + p2(Qf (z))2∣∣∣+ p + 12 |Qf (z)|2 − 2 ≤ 16p (4.1)

34

for every univalent function f defined on D and all z ∈ D.

Note that if inequality (4.1) holds for one value of p ≥ 1, then it holds for all larger values

of p. Let P be the minimum of all p ≥ 1 such that (4.1) holds for all univalent functions f

defined in D. It suffices to establish (4.1) for z = 0 and normalized univalent functions.

Now, Corollary 4.1 implies that the inequality∣∣∣∣a3 − (3− p3)

a22

∣∣∣∣+ (p + 13)|a2|2 −

13≤ 8p

3

is valid for p =32. Hence, we know that P ≤ 3

2. With p = 1, the inequality becomes∣∣∣∣a3 − 23a22

∣∣∣∣ ≤ 3− 23 |a2|2.It has been shown that this inequality does not hold for all normalized univalent functions.

Thus, P > 1. Hence, (4.1) holds for all p ≥ P , where 1 < P ≤ 32.

We first consider the case where [f(a), f(b)] is contained in f(D). In this case, there is a

Jordan arc γ in D with hyperbolic arc length 2L joining a and b such that f maps γ injectively

onto the segment [f(a), f(b)]. Let γ be parameterized by hyperbolic arc length as γ : z = z(s),

s ∈ [−L,L]. Then z′(s) = (1−|z(s)|2)eiθ(s), where θ(s) = arg z′(s). Let u(s) = log |D1f(z(s))|.

Then u′(s) = Re{Qf (z(s))eiθ(s)

}, (u′)2(s) =

12Re{(Qf (z(s)))2e2iθ(s)

}+

12

∣∣Qf (z(s))∣∣2, andu′′(s) = Re

{(1− |z(s)|2)2Rf (z(s))e2iθ(s)

}+

12

∣∣Qf (z(s))∣∣2 − 2.Therefore:

u′′(s) + p(u′)2(s) = Re{(1− |z(s)|2)2Rf (z(s))e2iθ(s)

}+

12

∣∣Qf (z(s))∣∣2 − 2+ p

(12Re{(Qf (z(s)))2e2iθ(s)

}+

12

∣∣Qf (z(s))∣∣2)= Re

{[(1− |z(s)|2)2Rf (z(s)) +

p

2(Qf (z(s)))2

]e2iθ(s)

}+

p + 12

∣∣Qf (z(s))∣∣2 − 2≤∣∣∣(1− |z(s)|2)2Rf (z(s)) + p2(Qf (z(s)))2∣∣∣+ p + 12 ∣∣Qf (z(s))∣∣2 − 2.

Since p satisfies inequality 4.1, we have u′′(s) + p(u′)2(s) ≤ 16p, so u′′ ≤ p(42 − (u′)2). Apply

Lemma 4.1 with q = 4, letting v be as in the statement of the lemma. Then u(s) ≥ v(s) for

all s ∈ [−L,L]. Moreover, v(s) = 1p

log[cosh(4ps) + τ sinh(4ps)] + log C. Note that

C =(

exp(pu(L)) + exp(pu(−L))2 cosh(4pL)

)1/p=(|D1f(a)|p + |D1f(b)|p

2 cosh(4pL)

)1/p,

35

since u(L) = |D1f(z(L))| = |D1f(b)| and u(−L) = |D1f(z(−L))| = |D1f(a)|.

Therefore:

|f(b)− f(a)| =∫ L−L|f ′(z(s))||dz(s)|

=∫ L−L

(1− |z(s)|2)|f ′(z(s))| |dz(s)|1− |z(s)|2

=∫ L−L|D1f(z(s))| ds

=∫ L−L

exp(u(s)) ds

≥∫ L−L

exp(v(s)) ds

≥ C2

sinh(4L) (by Lemma 4.2)

=sinh(4L)

2[2 cosh(4pL)]1/p(|D1f(a)|p + |D1f(b)|p)1/p .

Note that the function h(t) =sinh t

[2 cosh(pt)]1/pis increasing and 2dD(a, b) ≤ 4L. Hence

|f(b)− f(a)| ≥ sinh(2dD(a, b))2[2 cosh(2pdD(a, b))]1/p

(|D1f(a)|p + |D1f(b)|p)1/p .

This establishes the inequality when [f(a), f(b)] is contained in f(D).

Next we find a limiting form for this inequality. Set Ω = f(D). If α ∈ ∂Ω with [f(a), α) ⊂ Ω

and b ∈ D is such that f(b) ∈ [f(a), α), then we have

|f(a)− f(b)| ≥ 2 sinh(2dD(a, b))2[2 cosh(2pdD(a, b))]1/p

|D1f(a)|.

Letting f(b) → ∂Ω along the segment [f(a), α) results in b → ∂D and dD(a, b) →∞. Now, for

the function h(t) =sinh t

[2 cosh(pt)]1/p, we have lim

t→∞h(t) =

12. Hence

|f(a)− α| ≥ 14|D1f(a)|.

Now, if [f(a), f(b)] does not lie in Ω, then there are points α, β ∈ ∂Ω such that [f(a), α)

and (β, f(b)] are disjoint and lie in Ω with [f(a), α) ∪ (β, f(b)] ⊆ [f(a), f(b)]. Then, by the

36

limiting form, we have |f(a)− α| ≥ 14|D1f(a)| and |f(b)− β| ≥

14|D1f(b)|. Hence:

|f(a)− f(b)| ≥ |f(a)− α|+ |f(b)− β|

≥ 14(|D1f(a)|+ |D1f(b)|)

≥ 14(|D1f(a)|p + |D1f(b)|p)1/p

Since limt→∞

h(t) =12

and h is strictly increasing, we obtain

|f(a)− f(b)| > sinh(2dD(a, b))2[2 cosh(2pdD(a, b))]1/p

(|D1f(a)|p + |D1f(b)|p)1/p.

Hence, the inequality has been established in all cases.

Remarks.

1. Actually, the occurrence of P in Kim and Minda’s distortion theorem is a result of the

method used, and is not inherent in the problem. Jenkins [5] later proved that the

inequality holds for all p ≥ 1 and fails for 0 < p < 1.

2. Note that the limiting form of this inequality (p = ∞) is Theorem 4.1, which is the

invariant form of the lower bound given in Theorem 2.5.

4.3 Upper Bounds for |f(a)− f(b)|

A theorem giving a family of upper bounds for |f(a)− f(b)| is given by Ma and Minda in

[8]. The proof relies on an integral inequality which was proved by Ma and Minda in [9].

Lemma 4.3. Suppose w ∈ C2[a, b], k > 0, and w′′ ≤ k2w. If w(a) ≥ 0 and w(b) ≥ 0, then

w ≥ 0 on [a, b].

Proof. Let m = min{w(s) : s ∈ [a, b]}. If m ≥ 0, there is nothing to prove, so we will assume

m < 0. First, find s0 ∈ (a, b) with w(s0) = m. Since w(s0) is the minimum of w on [a, b],

we have w′(s0) = 0 and w′′(s0) ≥ 0. But also w′′(s0) ≤ k2w(s0) = k2m < 0. This is a

contradiction, and so m ≥ 0.

37

Corollary 4.2. Suppose u, v ∈ C2[a, b], k > 0, v′′ ≤ k2v, and u′′ ≤ k2u. If v(a) ≥ u(a) and

v(b) ≥ u(b), then v ≥ u on [a, b].

Proof. The result follows immediately by applying Lemma 4.3 to w = v − u.

Lemma 4.4. Suppose v ∈ C2[−L,L], v > 0, k > 0, |v′| ≤ kv and v′′ ≤ k2v. Then∫ L−L

ds

v(s)≤ 4 cosh(kL) sinh(kL)

k[v(L) + v(−L)].

Proof. Let u ∈ C2[−L,L] satisfy u′′ = k2u with the boundary conditions u(L) = v(L) and

u(−L) = v(−L). The general solution of u′′ = k2u is u(s) = A cosh(ks) + B sinh(ks), where

A,B ∈ R. Now, the boundary conditions imply A = v(L) + v(−L)2 cosh(kL)

and B =v(L)− v(−L)

2 sinh(kL).

Let τ =B

A=

v(L)− v(−L)v(L) + v(−L)

· cosh(kL)sinh(kL)

. Then u(s) = A[cosh(ks) + τ sinh(ks)]. We will next

show that τ ∈ [−1, 1]. To that end, notice that |v′| ≤ kv implies −k ≤ v′

v≤ k. Integrating

over [−L,L] and exponentiating results in

exp(−2kL) ≤ v(L)v(−L)

≤ exp(2kL).

Now, the function h(t) =t− 1t + 1

is strictly increasing for t > −1. Therefore:

exp(−2kL)− 1exp(−2kL) + 1

≤

v(L)v(−L)

− 1

v(L)v(−L)

+ 1≤ exp(2kL)− 1

exp(2kL) + 1

⇒ exp(−kL)− exp(kL)exp(−kL) + exp(kL)

≤ v(L)− v(−L)v(L) + v(−L)

≤ exp(kL)− exp(−kL)exp(kL) + exp(−kL)

⇒ − sinh(kL)cosh(kL)

≤ v(L)− v(−L)v(L) + v(−L)

≤ sinh(kL)cosh(kL)

⇒ −1 ≤ v(L)− v(−L)v(L) + v(−L)

· cosh(kL)sinh(kL)

≤ 1

⇒ −1 ≤ τ ≤ 1

Since A > 0 and τ ∈ [−1, 1], we have u > 0 and |u′| ≤ ku. Hence, Corollary 4.2 implies v ≥ u

on [−L,L]. Then we must have∫ L−L

ds

v(s)≤∫ L−L

ds

u(s)=

1A

∫ L−L

ds

cosh(ks) + τ sinh(ks).

38

Let I(τ) =∫ L−L

ds

cosh(ks) + τ sinh(ks). Then I ′′(τ) = 2

∫ L−L

sinh2(ks)[cosh(ks) + τ sinh(ks)]2

> 0.

Thus, I(τ) is strictly concave up on [−L,L], so I(τ) ≤ I(±1) = 2 sinh(kL)k

for all τ ∈ [−1, 1].

Hence, ∫ L−L

ds

v(s)≤ 2 sinh(kL)

Ak=

4 cosh(kL) sinh(kL)k[v(L) + v(−L)]

.

The result that we will need for Ma and Minda’s distortion theorem is the following corol-

lary.

Corollary 4.3. Suppose v ∈ C2[−L,L], v > 0, k > 0, |v′| ≤ kv and v′′ ≤ k2v. Then for any

p ≥ 1 ∫ L−L

ds

v(s)≤ 2[2 cosh(pkL)]

1/p sinh(kL)k[v(L)p + v(−L)p]1/p

.

Proof. First note that2[2 cosh(pkL)]1/p sinh(kL)

k[v(L)p + v(−L)p]1/p=

2 exp(kL) sinh(kL)kv(L)

1 + (exp(−2kL))p1 +

(v(−L)v(L)

)p

1/p

.

Now, in the proof of Lemma 4.4, we saw thatv(−L)v(L)

≥ exp(−2kL). Moreover, for

0 < s < t ≤ 1, the function h(p) =(

1 + sp

1 + tp

)1/pis strictly increasing. Thus, if

v(−L)v(L)

≤ 1, the

corollary follows immediately from Lemma 4.4. Ifv(−L)v(L)

> 1, we factor out v(−L) instead.

We will also make use of the inequalities (1 − |z|2)2|Rf (z)| ≤ 6 (which is due to Kraus

and Nehari; see [3], p.263) and |Qf (z)| ≤ 4 (as shown by Blatter [2]). These inequalities hold

whenever f is a univalent function.

Theorem 4.5. Suppose f is univalent in D. Then for a, b ∈ D and p ≥ 1

|f(a)− f(b)| ≤ [2 cosh(2pdD(a, b))]1/p sinh(2dD(a, b))

2[1/|D1f(a)|p + 1/|D1f(b)|p

]1/p .Proof. Let a, b ∈ D. Let γ : z = z(s), s ∈ [−L,L], be a Jordan arc joining a and b parameterized

by hyperbolic arc length such that 2L = dD(a, b). We first assume only that f is locally

univalent on D so that f ′(z) 6= 0 for z ∈ D. Let v(s) = |D1f(z(s))|−1.

39

Sinced

ds|D1f(z(s))| = |D1f(z(s))|Re

{eiθ(s)Qf (z(s))

}, we have

v′(s) = −v(s)Re{eiθ(s)Qf (z(s))

}.

Then

v′′(s) = −v′(s)Re{eiθ(s)Qf (z(s))

}− v(s) d

dsRe{eiθ(s)Qf (z(s))

}= v(s)

[(Re{eiθ(s)Qf (z(s))

})2− d

dsRe{eiθ(s)Qf (z(s))

}].

A calculation yields

d

ds

[eiθ(s)Qf (z(s))

]= e2iθ(s)(1− |z(s)|2)2Rf (z(s)) +

12[eiθ(s)Qf (z(s))

]2 − 2.Therefore:

v′′(s) = v(s)[(

Re{eiθ(s)Qf (z(s))

})2− 1

2Re{[eiθ(s)Qf (z(s))]2

}− Re

{e2iθ(s)(1− |z(s)|2)2Rf (z(s))

}+ 2]

= v(s)[12

∣∣Qf (z(s))∣∣2 − Re{e2iθ(s)(1− |z(s)|2)2Rf (z(s))}+ 2]≤ v(s)

[12

∣∣Qf (z(s))∣∣2 + (1− |z(s)|2)2∣∣Rf (z(s))∣∣+ 2]Now assume f is univalent on D. Then |Qf (z)| ≤ 4 and (1 − |z|2)2|Rf (z)| ≤ 6, so that

|v′(s)| ≤ 4v(s) and v′′(s) ≤ 16v(s).

Since f ◦ γ is a path connecting f(a) to f(b), we have

|f(a)− f(b)| ≤∫

f◦γ|dw|

=∫

γ|f ′(z)||dz|

=∫ L−L|f ′(z(s))|(1− |z(s)|2) ds

=∫ L−L|D1f(z(s))| ds

=∫ L−L

ds

v(s).

Applying Corollay 4.3 with k = 4 establishes the desired result.

40

Remark. The case p = ∞ gives

|f(a)− f(b)| ≤ 12

exp(2dD(a, b)) sinh(2dD(a, b))min{|D1f(a)|, |D1f(b)|

},

which is an invariant form of the upper bound in Theorem 2.5. To see this, apply the bound

in Theorem 4.5 to g ∈ S with a = 0 and b = z.

We get |g(z)| ≤ |z|(1− |z|)2

min{1, |D1g(z)|} ≤|z|

(1− |z|)2.

4.4 Comparisons Between Hyperbolic and Euclidean Geometries on

Simply Connected Regions

The two-point distortion theorems discussed in the previous two sections yield comparisons

between hyperbolic and Euclidean geometries on simply connected regions. Before we state

the theorems, we will introduce some more terminology and notation.

We say that a region Ω in the complex plane is hyperbolic if C\Ω contains at lease two

points. Saying that a region is hyperbolic is equivalent to saying that there is a holomorphic

universal covering projection f : D → Ω. This f is a conformal mapping if Ω is simply

connected.

If Ω is a hyperbolic region, then the density of the hyperbolic metric on Ω is obtained from

λΩ(f(z))|f ′(z)| = λD(z),

where f : D → Ω is any holomorphic universal covering projection of D onto Ω. The density is

independent of the choice of f . Notice that λΩ(f(z)) =λD(z)|f ′(z)|

=1

(1− |z|2)|f ′(z)|=

1|D1f(z)|

.

The hyperbolic distance function on Ω is given by

dΩ(P,Q) = inf{∫

γλΩ(w)|dw| : γ is a path in Ω joining P and Q

}.

If f is a conformal mapping, then dΩ(f(a), f(b)) = dD(a, b).

Theorem 4.6. Let Ω be a simply connected hyperbolic region in C. Then for any p ≥ 1 and

all A,B ∈ Ω,

|A−B| ≥ sinh(2dΩ(A,B))2[2 cosh(2pdΩ(A,B))]1/p

(1

λpΩ(A)+

1λpΩ(B)

)1/p.

41

Proof. Let A,B ∈ Ω be given and let f be a conformal mapping of D onto Ω. Find a, b ∈ D

with f(a) = A and f(b) = B. Then the remark following Theorem 4.4 implies

|A−B| = |f(a)− f(b)|

≥ sinh(2dD(a, b))2[2 cosh(2pdD(a, b))]1/p

(|D1f(a)|p + |D1f(b)|p

)1/p=

sinh(2dΩ(f(a), f(b)))2[2 cosh(2pdΩ(f(a), f(b)))]1/p

(1

λpΩ(f(a))+

1λpΩ(f(b))

)1/p=

sinh(2dΩ(A,B))2[2 cosh(2pdΩ(A,B))]1/p

(1

λpΩ(A)+

1λpΩ(B)

)1/p

Theorem 4.7. Let Ω be a simply connected hyperbolic region in C. Then for any p ≥ 1 and

all A,B ∈ Ω,

|A−B| ≤ [2 cosh(2pdΩ(A,B))]1/p sinh(2dΩ(A,B))

2[λpΩ(A) + λpΩ(B)]1/p

.

Proof. The proof is similar to that of Theorem 4.6. It follows from applying Theorem 4.5 to a

conformal mapping f : D → Ω.

42

BIBLIOGRAPHY

[1] Anderson, James, Hyperbolic geometry, 2nd ed, Springer, 2005.

[2] Blatter, Christian, Ein verzerrungssatz für schlichte funktionen, Commentarii Mathe-

matici Helvetici, 53 (1978), 651–659.

[3] Duren, Peter L., Univalent functions, Springer-Verlag, New York, 1983.

[4] Graham, Ian and Gabriela Kohr, Geometric function theory in one and higher dimensions,

Marcel Dekker, Inc, New York, 2003.

[5] Jenkins, James A., On weighted distortion in conformal mapping II, Bulletin of the London

Mathematical Society, 30 (1998), 151–158.

[6] Kim, Seong-A and David Minda, Two-point distortion theorems for univalent functions,

Pacific Journal of Mathematics, 163 (1994), 137–157.

[7] Krantz, Steven G., Complex analysis: the geometric viewpoint, 2nd ed, Mathematical

Association of America, 2004.

[8] Ma, William and David Minda, Two-point distortion for univalent functions, Journal of

Computational and Applied Mathematics, 105 (1999), 385–392.

[9] Ma, William and David Minda, Two-point distortion theorems for strongly close-to-convex

functions, Complex Variables Theory and Application, 33 (1997), 185–205.

[10] Pommerenke, Christian, Univalent functions, Vandenhoeck and Ruprecht, Göttingen,

1975.