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Thesis fat The Degree of Ph D l~;:.‘-Eg.;.-_:.¢..}€
, MECHEGAN STATE UNWERSETY  ~ ; _
KENNETH P. GOLDBERG ‘ -- -
1973

 

 

 

gnaw-ll

K m
f3 LIBRARY :
Michigan State
University

«a.-

     

This is to certify that the

..,:.r ”‘3’ '- ‘ thesis entitled
. f . ' g _ J T"

. ~ “V435

"Qua51symmetr1c Functlons

and Quasiconformal Mappings"

presented by

Kenneth Philip Goldberg

has been accepted towards fulfillment
of the requirements for

Ph u D . deg-6e in Mathematics

 

   

Major professor

Date June 13, 1973 t-

 

 

ABSTRACT
QUASISYMMETRIC FUNCTIONS AND PLANE QUASICONFORMAL MAPPINGS

BY

Kenneth P. Goldberg

In 1928 H. Grotzsch gave a definition of quasiconformal
mappings which sought to generalize the concept of conformal
mappings. However, these mappings satisfy neither the
reflection principle nor the normal family property. both
of which are satisfied by conformal mappings. In 1954
L. V. Ahlfors gave a new definition for quasiconformal mappings
which in fact extends the class of mappings that are quasi-
conformal in the sense of Grotzsch. These new'mappings do
satisfy the above two properties. in addition to having many
other properties of conformal mappings.

The dilatation Dgz 1 of a differentiable topological
mapping f: (x.y)-—€>(u.v) of one plane domain onto another
is determined by

U 2+U 2+V 2+V 2
_Z;, V

Ivay - vax

D+D-l=

 

Geometrically. D represents the ratio between major and

minor axes of the infinitesimal ellipse Obtained by mapping

Kenneth P. Goldberg

an infinitesimal circle of center (x.y). A mapping is said
to be Quasiconformal in the sense of Ahlfors if D is bounded.
The least upper bound of D is called the maximal dilatation.
Beurling and Ahlfors showed (Acta Math. 1956) that
there exists a quasiconformal mapping of the upper half plane
onto itself with boundary correspondence x-——>¢(x) if and

only if

g; op (x+t) - cP (XL

(1) h 3 com -<O(x-t) S h

for some constant h, lgh<m. and for all real x and t.
A function m which satisfies (1) is said to be guasisymmetric
and the least upper bound of h is called the guasisymmetric
dilatation of T and is denoted by P(¢).

In Chapter I we show that (1) is really a generalized
convexity-concavity condition, and that the assumptions in
the definition of quasisymmetry can be significantly weakened
'without altering the class of such functions. We then use
this fact to obtain sharp bounds for the dilatation of the
sums. products. inverses and compositions of quasisymmetric
functions on (0.»).

In Chapter II we introduce. by means of a differentiability
condition,a subclass of the quasisymmetric functions which
we call ratio-bounded. We study closure properties of this

class under sums. products. compositions and the taking of

Kenneth P. Goldberg

inverses and find sharp bounds for the quasisymmetric dilatation
of such functions.

As already indicated, quasisymmetric functions of
(“”0”) onto itself can be extended to quasiconformal mappings
of the upper half plane. but until now only one such extension
had been given explicitly, namely that of Beurling and Ahlfors
in 1956, and hardly anything was known about extremal exten-
sions (that is, quasiconformal extensions with minimal maximal
dilatation). In Chapters III and IV we use the results of
Chapter II about ratio-bounded functions to obtain explicit
extremal extensions for each quasisymmetric function in a
certain class.

In Chapter V we continue our study of extremal quasi-
conformal extensions and generalize a result of Reich and

Strebel (Comment. Math. Helv., 1970).

QUASISYMMETRIC FUNCTIONS AND PLANE QUASICONFORMAL MAPPINGS

BY
, , T,"
jl‘i 1‘

Kenneth Py‘Goldberg

A THESIS

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY
Department of Mathematics

1973

7.553.
w

-.

H»:

‘5. ,r‘
’ A

. "m
3.; “
‘31

TO
NEAL AND MYRM.
JOAN AND SID:
AND ESPECIALLY TO MY MOTHER
DOLORES

ii

ACKNOWLEDGEMENTS

I would like to thank my major
Professor Glen Anderson for his help
and encouragement in the writing of this

dissertation.

iii

CHAPTER

TABLE OF CONTENTS

I INTRODUCTION . . . . . .

CHAPTER II QUASISYMMETRIC FUNCTIONS

CHAPTER

Section
Section

Section

Section

Section

Section

Section

1:

2:

3:

A new definition . . . . .

Upper and lower bounds . .

The power functions

u(x) = xa.

The sum of OS
The product of
The inverse of a

The composition

III RATIO BOUNDED FUNCTIONS

Section
Section
Section
Section
Section
Section
Section
Section

Section

Section 10 :

l:

2:

3:

Introduction . .

Notation . . . .

Alternative defi

The sum of RB

The

The

The

class T

The L.M

The

Convex and concave functions . . . . .

a > O . . . . .

functions
05
OS

of OS

nitions .

functions

of RB functions

functions

function

product of RB functions

relationship between RB

functions

inverse of an RB function

composition of RB functions

and

OS

14

18

20

23

25

28

33

33

34

35

39

4O

41

42

43

47

59

CHAPTER IV RADIAL EXTENSIONS . . . . . . . . . . . . . . . 62
Section 1: Notation . . . . . . . . . . . . . . . . 62
Section 2: The radial extension . . . . . . . . . . 63
Section 3: The odd radial extension . . . . . . . . 73
Section 4: Extremal radial extensions . . . . . . . 76

Section 5: An application of the extremal
radial extension . . . . . . . . . . . . 93

Section 6: General radial mappings . . . . . . . . 94

CHAPTER V ADDITIONAL EXTREMAL EXTENSIONS: A GENERALIZATION
OF REICH AND STREBEL . . . . . . . . . . . . . 105

Figure 1.1

Figure 5.1

Figure 5.2

LIST OF FIGURES

vi

CHAPTER I

INTRODUCTION

A well-known theorem of Riemann [2. p.172] states that
any simply-connected domain 0 whose boundary consists of
more than one point can be mapped conformally onto the unit
disk. Thus any two such domains 0. 0' can be mapped
conformally onto each other.

In 1928 H. Grdtzsch [8] posed the following problem:
given a square D and a rectangle R which is not a square,
can D be mapped conformally onto R so that the vertices
correspond?

As was subsequently shown by Grotzsch [8]. no such
conformal mapping exists. He then asked for the most nearly
conformal map of D onto R with vertices corresponding. In
order to answer this new question one needs a method of
measuring approximate (or quasi) conformality. It was in
attempting to supply such a measure that Grotzsch laid the
foundations for the modern theory of plane quasiconformal
mappings [8].

Let Q be a domain. A quadrilateral in O is a Jordan

domain Q. 5 C'Q. together with a pair of disjoint, closed

arcs on the boundary of Q (called the b-arcs). If we map
Q conformally onto a rectangle with side lengths a and b,

‘with the b-arcs going onto the sides of length b (see Figure 1.1),

then the modulus of Q is defined uniquely as

 

 

 

 

 

 

E.
(1.1) mod Q — b'
2
3 w W
z“ 4 3
Q b
b-arc
b-arc
Z] 22 WIW “Ta
a
Figure 1.1

The modern definition of quasiconformality. as first

given by L. V. Ahlfors in 1954 [4] is:

Definition 1.1. Let w = f(z) be a sense-preserving
homeomorphism from a domain 0 onto a domain 0'. Then f
is said to be quasiconformal on Q. or 99, if there is some

K, 1gg K < a. satisfying

g1 mod Q'
(1'2) KS mon SK

for all quadrilaterals Q in O with f(Q) = Q'. We define

the maximal, or 99, dilatation K(f) g; f ggg Q to be the

infimum of all numbers K satisfying (1.2). ;§_ K(f) = KO

we say that f ig Kofiggasiconformal (or KO-QC) ggn Q.

The above definition. as well as several equivalent ones,
is given in [6].
As is easily shown [4, p.8]. f is conformal if and only

if K(f) = 1. Thus K(f) can be used as a measure of approximate

conformality.

We will also find the following definition quite useful

later in this paper.

Definition 1.2. Let f = u + iv be a sense-preserving
homeomorphism of the domain 0 onto the domain 0'. and let
(xo.yo) be a point of D at which both u and v have
continuous partial derivatives. Let

_1_ . __ _1_ .
fz — 2(fx ify), f2 — 2(fX + ify).

Then

f-(x IY )
x (xo'yo) = FZ'EQ—QT
z O'YO

is called the complex dilatation of f at the point (xo.yo)

and
1+‘X(x009)‘

D(X o ) =
0 YO 1-|x(xo.yoll

is called the pgint_§i1§§atign_of f at (xo,yo). Moreover,

D(xo.yo) satisfies
2 2 2 2
tgéxo.yo)+uy(xo.y0)+vxtxo.yo)+vy(xo.yo)

)vx(xo.yo)

 

(1.3) D(x ,y )+ = _
o o D(xo.yo) ‘ux(xo.yo)vy(xooyo) uy(XooYo

Remark: It is clear, from Definition B on page 24 of [3],
that if f is a QC map of 0 onto 0' then
K(f) = ess sup D(z).
z E 0

A few of the more important properties of QC maps.

whose proofs may be found in [3], are:

(i) f is conformal if and only if K(f) = 1.
(ii) f is QC if and only if f.1 is QC, and
-1
K(f ) = K(f).
(iii) The QC dilatation K(f) is invariant under
composition with conformal mappings. (I.e., whenever f is

QC, 9 is conformal, and fog (gof) is well-defined, then

K(f°g) = K(f) (K(9°f) = K(f))).

By (iii) above and the theorem of Riemann mentioned at
the beginning of this chapter. it can be assumed without loss
of generality that both 0 and 0' are the upper half plane

H = {z = x+iyly>0}.

A problem that aroused considerable interest during the
early research in QC mappings was to determine necessary and
sufficient conditions on a homeomorphism u of (-o,m) onto
itself which would allow u to be extended to a QC map
of H onto itself.

This problem was completely solved by Ahlfors and Beurling

in 1956 [5]. They proved that u can be extended to a QC

map of H onto itself if and only if u satisfies a condition
now referred to as quasisymmetry. The definition of quasisymmetry

is as follows.

Definition 1.3. Let u be a continuous, strictly
increasing function defined on an interval (a.b) with
-agg a < bgg n. Then u is said to be guasisygmetric on (a,b),

or 98 on (a.b). if there is some p, lgg p < a, satisfying

_1_ u (x+t) - u(x)

(1’4) p S u(x)-u(x-t) S p

for all x and t with a < x-t < x < x+t < b. The 95

dilatation p(u) f u 0 (a,b) is defined as the infimum

‘ of all numbers p satisfying (1.4). g_§ p(u) = po we say

that u gig po-QS 29g (a.b).

In Chapter 2 we give an alternative definition of 08

and prove it equivalent to Definition 1.3. Using the new def-

inition the class of 98 functions is then shown to be closed

gppggr the operations of addition! multiplication, composition
ppgythe takingiof invepggp. The remainder of Chapter 2

deals with the prOblem of finding explicit, sharp bounds

for the OS dilatation of sums. products. compositions

and inverses in terms of the OS dilatation of the original
functions.

One of the difficulties frequently encountered when
working with OS functions is that of determining whether or
not a particular function u is QS and. if it is. finding
p(u). With this problem in mind we begin Chapter 3 by making

the following definition.

Definition 1.4. Let u be a strictly increasing
self-homeomorphism of [0.o). Then u is said to be ratio
bounded on [0.o), or 3p, if there are numbers L, M.

O ( nggM < a, such that u satisfies

(1.5) nggm a.e. on (0...).

u(x)

The lower (uppgr) ratio bound L(u) (M(u)) _g, u _p_ [0,.)
is defined as the supremum (infimum) of all numbers L (M)

satisfying (1.5).

In the first part of Chapter 3 it is shown that the class

pf RB functions is closed under the opgrations of addition,
mgltipligatipn, comppsition and the taking of inverses. Sharp

bounds are found for the ratio bounds of these sums. products.
compositions and inverses.

It is then shown that if a function u is RB on [0,m)
gppgp u must also be 05 on (0.») and sharp bounds are found
for p(u) on (0.») in terms of L(u) and M(u).

At the end of Chapter 3 we prove that if_ u is either

;

ggpygxggr concave on [0,a) gppgp u is QS on (O,m) if and
only if it is RB on [0,m).

In [3] Ahlfors and Beurling give an explicit extension
for a OS self-homeomorphism u of (-a,o) to a QC self-
homeomorphism of H.

We begin Chapter 4 by defining a new extension for u
p__ H. This extension is called the radial extension and is
shown to be 99 if and only if u is RB on [0.a) and the
function v given by v(x) = -u(-x) is RB on [0,.). Sharp
bounds are found for the QC dilatation of the radial extension
in terms of L(u). M(u). L(v) and M(v).

The remainder of Chapter 4 is concerned with the
following generalization of the prdblem of Gr6tzsch: ggiygp

a homeomorphism u between the boundaries of two domains,

find the extension to the interiors of these domains which

 

is most nearly conformal. Such an extension is said to be

extremal for the given boundary homeomorphism u.

Conditions are given on u for whiCh the gadial:exten-
sion is extremal. Questions of uniqueness and non-uniqueness
are also investigated.

In Chapter 5 we continue our study of extremal

mappings and generalize a result of Reich and Strebel [14].

CHAPTER I I

QUASISYMMETRIC FUNCTIONS
1. A new definition.

As was pointed out in the introduction, a homeomorphism
u of (-a.m) onto itself can be extended to a QC map of
H onto itself if and only if u is QS according to
Definition 1.3. The following theorem gives an alternative
formulation of quasisymmetry which is equivalent to
Definition 1.3 but better suited to the estimates we want to

make in this chapter.

Theorem 2.1. Let u be a non-constant fupction
defined on an interval (a,b) with -agg a < bgg n. Then u

is 98 on (a.b) if and only if

(i) u is linear
9; (ii) there is some x. 1/2 < l < 1. such that

X1+X2
(2.1) Au(x1) + (1-l)u(x2) gu( 2 )g (l-x)u(x1) + lu(x2)

 

fgr all x1.x2 with a < x1 < x2 < b.

10

Definition 2.1. If u is a nonlinear QS function on
(a,b) we define the midpoint dilatation 1(u) of u to
be the infimum of all numbers I for which (2.1) holds.
The relation (2.1) is called the midpoint condition. The

relationship between 1(u) and p(u) is

 

_ JUL _ .2111.
Proof of Theorem 2.1.
(i) Let u be 08 on (a,b). By Definition 1.2
;L_ u(x+t) u(x x)
(2'3) Po g 11(X) - u(x-t) 3 p0

for all x,t satisfying a < x-t < x < x+t < b, where

p0 = p(u). Multiplying (2.3) by the positive expression

u(x) - u(x-t) and solving for u(x) gives

—9—u (x-t) +

“90 —u(X+t) S u(x) 3—“1+10pfl(x't) + —9—u(x+t),

1+1pO 1+pO

This double inequality becomes (2.1) if we set x1 = x-t,

x2 = x+t. x(u) = po/(1+po).

(ii) Let u satisfy (2.1) with 10 = 1(u). It must

be shown that

11

a) u is continuous on (a.b).
b) u is strictly increasing on (a,b).

c) u satisfies (1.4) for some p, lgg p < m.

Proof of a: In [3, p.66] it is proved that (2.1)

implies the continuity of u.

Proof of b: Let x1. x2 be given with a < xl < x2 < b.

Then by (2.1) it is clear that the inequality

Aou(x1) + (l-l0)u(x2) g Aou(x2) + (l-lo)u(xl)

implies (210-1)u(x1) < (ZlO-l)u(x2) or equivalently, since
1/2 < A0 < l, u(x1)gg u(x2). Hence u is non-decreasing.
To complete the proof of (b) assume there are points

x x with a < x1 < x2 < b and u(xl) = u(xz) = M. Since u

1' 2
is non-decreasing this would imply u(x) = M identically on

[x1.x2]. By assumption u is not constant. Hence there is

some x3 6 (a.b) for which u(x3) # M. Without loss of

generality it can be assumed that x2 < x3. Then by the
monotonicity of u, u(xz) = M < u(x3).

Let S be defined as

(2.4) S = {xlx2 g xgg x3 with u(x) > u(x2) = M].

Clearly x3 6 S so that S is not empty. In addition x2 is

12

a lower bound for S by (2.4). Thus S must have a

greatest lower bound (g.l.b.) i with ngg i. If u(x) > M
then by continuity there is some e > 0 such that u(x) ; M

on (i-e.§). But this contradicts the assumption that i

is the g.1.b. of S. Hence u(i) = M and so u(x) = M
identically on [xl.§]. Pick 3 > 0 so small that u(i+e) > M
and u(i-e) = M. This is possible because i is the g.1.b.

of S. Then by (2.1), using the points i-e. i. §+e. we

obtain
Xou(x-e) + (l-Ao)u(x+e)gg u(x) g (l-lo)u(x-e) + Aou(x+e).
The left inequality implies that
u(xlgz lou(X-e) + (l-lo)u(x+e) > 10M + (l-lolM = M-

But this is obviously a contradiction of u(x) = M. That is,

there cannot be any x1 < x with u(xl) = u(xz). Hence u

2

must be strictly increasing on (a,b).

Proof of c: The proof is immediate since all the steps

in the proof of (i) are reversible.

Remark: Looking at the statement of Theorem 2.1 it is
reasonable to ask if condition (i) can be omitted by simply
changing condition (ii) to allow' 1/2 g 10 < 1. It is obvious

that if u is 08 and linear then (2.1) does hold with IO = 1/2.

13

The converse. however. is not true. It is possible to have a
non-constant function which satisfies (2.1) with AC = 1/2
but is not 08 on (a.b). To show this we will exhibit an
example .

In [7. p.150] a function f is constructed. using a
Hamel basis x1.x2.....xa,.... a E Q, with the property
f(x+y) = f(x) + f(y) for all x.y. Taking x = y we obtain
f(2x) = 2f(x). which leads to f(x) = f(2x)/2 or f(x/2) = f(x)/2.

Thus for any x,y.

«as = fee = ME) + res = + Ew-

But this is just (2.1) with A0 = 1/2. As indicated in
[7, p.150]. f can be defined arbitrarily on the Hamel basis.
Since a Hamel basis has an infinite number of elements. we can

choose three elements x x x from this particular basis.

1' 2' 3
Assume without loss of generality that in the usual ordering

of the reals we have x1 < x2 < x3. and define f(xl) = 0,

f(xz) = l. f(x3) = 0 and f = O for all the other elements
in this basis. Let u = f. Then u satisfies (2.1) with

A0 = 1/2 because f does. The function u is not constant
since u(xl) ¥ u(xz). Yet u cannot be QS because it is not
even monotonic since x1 < x2 < x3 with u(xl) < u(x2) and

u(x2) > u(x3). Hence condition (i) cannot be omitted in the

statement of Theorem 2.1.

l4

2. Upper and lower bounds.

In order to Obtain upper and lower bounds for a QS
function u on the interior of an interval when the function
values are known at the endpoints. we define the following
two functions P(l) and p(l) on [0.1]. If

fi+i+ +9
2 22 2

A:

a]:
4.

(ai = 0 or 1)

is the binary expansion of A E [0.1]. then

(205)

90.) = P (I) = X [6 +x 9+). A e +...+>. x ...). 9+...].
A0 0 1 e1 2 e1 e2 3 e1 e2 en-1 n

p(1)=p (M =l[e+l_ 6+X_ A_ 9+...+)._ ...)._ e+...],
x0 1119121911923 191 len_ln

where 10. 1/2 g x0 < l, is the midpoint dilatation of u

(Cf. Definition 2.1) and l = 1-10. When there is no chance

1

of confusion we will use P. p in place of PA . Pl .
0 0

respectively.

Theorem 2.2. Let u be a Q§ function on (a.b) and
le x1. x2 6 (a.b) be given. Then for any 1 6 [0.1],

u[(l->.)x1 + 1x2] g [1-P(l)]u(x1) + P(l)u(x2).

(2.6)
u[(l-l)x1 + 1x2] 2_[1-p(l)]u(xl) + p(l)u(x2).

lS

gpggf: In a paper by R. Salem [15] P and p are
shown to be continuous. strictly increasing functions mapping
[0.1] onto itself. If v is QS with v(0) = 0, v(l) = 1,
then by (2.1) and the method of construction of P and p
in [15] we must have p(i) g v0.) g pm for all A e [0.1]
with a finite binary expansion. But the set of these numbers

is dense in [0.1]. Hence. by the continuity of v.P and

13. pm g v0.) g P(x) for all I e [0.1].

If we now take any 08 function u and set

u[(l-l)x1+lx;l-u(xl)

 

V(X) = u(x2)-u(x1)

then v is QS with v(0) = 0. v(l) = 1. Hence

u[(1-1)xl+lx2]-u(xl)

P0.) g v(l) = u;fi(x2)-u(’xT) g P(l).

Solving for u[(l-l)xl+lx2] gives the desired bounds.
Theorem 2.3. P(t) + p(l-t) = l identically on [0,1].

3%: Define f(t) = P(t) + p(l-t) on [0.1]. Since
P and p are both continuous [15], so is f. It will be
shown by induction on n that f(t) = 1 when t is of the
form t = m/Zn, n = 0.1.2.... and m = 0.1,....2n. Since
the set of all such t is dense in [0,1] the continuity

of f will then imply f(x) = 1 identically on [0,1].

16

(i) Let n = 0. Then m can be either 0 or 1

and f(0) = P(O) + p(l-O) = 0 + l = l, f(1) = P(l) + p(l-l) = 1.
(ii) Let n = 1. Then m can be either 0.1, or 2.

But the cases m = 0 and m = 2 are covered by (i), while

m = 1 gives f(l/2) = P(1/2) + p(l-(l/2)) = 10 + l = 1.

1
(iii) Assume f(t) = 1 for all t: of the form
N N +1 N +1
t=m/20and1et t0=m/2°, ogmgzo. If m=O
NO+1
or m = 2 then t = 0 or 1, respectively. These cases

have been treated in (i).

N
Let m be an even number. Then m = 2k, lgg k < 2 0-1,
N +1 N +1 N
and so t = m/2 O = 2k/2 O = k/2 0. By the induction
N0

hypothesis this would imply f(t) = f(k/Z ) = 1.
Let m be an odd number. Then m-l and m+1 are

both even. Hence m-l = 2k m+1 = 2k for some k , k
N

1' 2 1 2
with ogkl<k2g2°. Then

NO+1
(2.7) f(t) = f(m/Z )
_ m-l m+1
—f((N+l N+l/2)
2 ° 2 °
-1 m+1 m+1
= p((—"‘———+ -————)/2) + p(1-<——- + —->/2)
N0+1 N0+1N0+1 Nb+l

2 2 2 2

17

NO NO NO NO
P([(kl/2 ) + (kz/z )1/2) + p([(1-k1/2 )+(1-k2/2 )1/2)

No No
[119(k1/2 ) + 109(k2/2 )]

NO NO
+ [Alp(1-kl/2 ) + Xop(l-k2/2 )]

NO NO
xlf(k1/2 ) + Aof(k2/2 ).

Hence, by the induction hypothesis, (2.7) reduces to

11 + 10 = 1. The induction proof is completed.

Corollapy 2.3. p'1(1/2) + p'l(1/2) = 1.

Proof: Since P(O) = 0. P(l) = 1 and P is continuous.
there must be some t1 6 (0.1) with P(tl) = 1/2 or. equi-
valently. t = P-1(l/2). Similarly p(O) = 0. p(1) = 1

1

and p continuous imply the existence of a t2 6 (0,1) with

t2 = p-1(1/2). By Theorem 2.3. 1/2 = P(tl) = 1-p(1-tl).

Hence p(l-tl) = 1/2 = P(t2)° Since p is strictly increasing,

p(l-tl) = p(tz) implies 1-tl = t2 or tl + t2 = 1.

Substituting t = P.1(1/2), t = p-1(1/2) gives

1
p71(1/2) + p'1(1/2) = 1.

2

18

3. The powe; fupctions u(x) = x0, a > 0.

Corollary 2.3 will be found to be very useful later

in this chapter. Also quite useful is

Lemma 2.1. Let u(x) = xa, a > 0, on [0,w). Then

u is OS on (o,e) with

 

1
(2.8) p(u) = max 2“ - 1, .
1 }

Proof: It is clear that

1 2a-1 if o.2 1
max {2a-1. } = 1. .
2a-1 _ET—' if 0 < o < 1
2 -1

 

The cases qu 1 and O < o < 1 will be treated separately.

(i) Assume a.2 1 and let x,t be given arbitrarily

with x > O. 0 < t < x. Then

 

 

 

x+t o _ 1
(2 9) u(x+t) -ULX)=(X+§£'£=(X) = 1+Sa-l
u(x) - u(x-t) x01 _ (x-t)a 1 _ (my 1 .. (1-s)°‘
X
1+3,“ - 1
=—-—§——— with O<s=§<lo
l - l-s o
8

Clearly. ((1+S)o - l)/s is just the slope of the secant line

connecting the points (1. lo) and (1+s. (1+S)o) on the

l9

graph of the convex function y = xa. Similarly (l - (l-s)a)/s

is the slope of the secant connecting (l-s, (1_s)a) and

(10 10). By [10, p.3], ((l+s)a - l)/s is non-decreasing

 

 

 

 

and (l - (1-s)q)/s non-increasing as 8 goes from 0 to 1.
Hence
(l+s)q - 1
(1+s)0 - 1 _ s
1 - (1-3)” 1 - (1-3)”
s

is non-decreasing for s 6 (0,1), and (2.9) together with

 

 

 

lim _(l+s)0' - 1 = 1 lim (1+SLQ -:_1_ = 2a _ 1
340+ l - (l-s)a 541— 1 - (1-s)O
gives
____. u(x+t) - u(x)
(2.10) a ggl gg u(x) _ u(x-t)‘g 2O - 1.

2 -1

Since the upper bound in (2.10) is actually approached as

O.

t approaches x, 2 - 1 must be the best possible QS bound.

That is, p(u) = 2“ - 1.

(ii) Assume O < a < 1. Then the function y =
is concave instead of convex, as in (i). For concave functions,
however, the double inequality in [10, p.3] is just reversed.

The rest of the proof is the same as in (i) and we find that

20

u(x-tt) - u (x)
u(x) - u(x-t)

     

2"-1 g
2 ”-1

Here the lower bound is approached as t approaches x so

that 1/(20_1) must be the best possible 08 bound. That is,

p(u) = 1/(20-1).

CLOSURE PROPERTIES

4. The sum of 98 functions.

 

. u be 98 functions on

u .0.
2' n

Theorem 2.4. Let ul,

(a,b) ‘with 95 dilatations pl. p2. ..., pn and midpoint

dilatations 1(1), 1(2). ..., 1(n). respectively. Then thg

 

n
function v defined as v(x) = Z) ui(x) is also 98 on (a,b)
=1

P.

and

PM 5. max {PP}
» lgign

This result is sharp:

Proof: The proof is by induction on n.

(1) 1(2)}.

(1) Assume n = 2 and let c = max [A

Then for x , x 6 (a,b) (2.1) gives

 

 

1 2
X +x x +x x +x
1.2. _ 1 2 1 2
V(2)—u1(2)+u2(2)
_<. [ (1-1‘“)u1(x1)+x‘1’u1(x2) 1+1 (1-1‘2’)u2(x1)+1‘2’u2(x2) 1

21

gg [(1-c)ul(xl)+cu1(x2)]+[(l-c)u2(x1)+cu2(x2)]

= (l-c)v(xl)+cv(x2).

 

 

 

Similarly.
X +x X +X X +X
v(l 2) =u1(l 2) +u2(l 2)
2 2 2
1 l
2 [1‘ ’ul(xl)+(1-1‘ ))u1(x2)1+[1‘2’u2(x1)+(1-1‘2’)u2(x2)J

g2 cv(x1)+(l-c)v(x2).

Hence

x +x
cv(xl)+(l-c)v(x )gW'L-z)_g(l-c)v(x )+cv(x ).
2 2 1 2

By Theorem 2.1 v is 08 on (a,b) while Definition 2.1

(1), 1(2)}. Since

shows that 1(ng c = max {A
p(u) = l(u)/(l-l(u)) is an increasing function of 1(u).

this is equivalent to p(v) g max [(1, 92}.

(ii) Assume the theorem true for n = N and let

N 0
0 ° A
-i§aui. Then v = v + uNb+1. But uN,0+1 is assumed QS
and 9 is 08 by the induction hypothesis. with p(e) s max [pi].
. A 1.<.i.<.No
Hence. by part (i). v = v + u is also 08 on
Nb+l

(a,b) ‘with

P(V) gmax {9(0), PM +1} 3 max {pi}.
0 lgigNO+1

22

Equality holds when u1 = 112 = ... = un.

An obvious analog of Theorem 2.4 would be to show that

p(v) 2 min [pi].
lgign

This statement. however, is false even in the simple case

n = 2. as the following Theorem shows.

Thegrem 2.5. There exist functions u and u

1 2'

S on (-a, a), such that p(ul + u2) < min (pl.p2).

 

Proof: For an arbitrary 10, 1/2 < X0 < l. and

_ __ . A A
A1 — 1 10, define u1 and u2 on [0,1] as
21 x if Ofixgl/Z
Gl(x) ={: 0
(2-210)x+(210-1) if 1/2<xg1
211x if qugl/Z
u2(X) ={

(2-2xl)x+(211-1) if 1/2<xgl

G
l

A . A A A

112 on [0.1] ‘Wlth p = 10/11. and that v(x) = u1(x)-+u2(x) = 2x

for all x 6 [0.1]. Next let us define functions u1 and

It is easy to see that (1.4) is satisfied for both and

u on (-o, a) by

u.(x) = 81(x) for qug1. ui(x+2) = ui(x)+2. i = 1.2.

23

Then by Theorem 5 of [9. p.239]. ul and u2 are both OS on

(-a,¢). A180

 

1
Q . .

if i = l

.3. .1. A '
ui(4)"“i(2)_ 1
.1. 1. ‘ x

“i‘2’ “1(4) -—-1- if 1: 2.
"0

Therefore p1 = p(ul)g2 10/11,P2==p(u2) 2.10/11. But by the
construction of v. v(x) = u1(x) + u2(x) = 2x identically
on (-c.a). so that p(v) = 1. Hence p(v) = 1 < lo/kl <

min {pl'p2].

5. The product of QS functions.

Theorem 2.6. Let ul. uz. .... un be Q§ functions

0 for each i. Then the fupgtion

pp, (a,b) with ui(a)

v defined as v(x) = 3 ui(x) is also 93 on (a,b) gppg
i=1
n
P(V) g [.11 (1+pi)] - 1.

i=1

This result is sharp:

Proof: The proof is by induction on n.

 

‘1’)(1-1‘2’).

(i) Assume n = 2 and let c = 1-(1-1

x 6 (a,b) we have by (2.1)

Then for any x1, 2

24

l:

x +x x H:
_1_—2. _1_2.
ul( 2 )u2( 2 )

s f (1-1‘“) 111 (x1)+1‘1’u1(x2) 1r (1-1‘2’)u2(x1)+1‘2’u2(x2) 1

= (1-1‘“) (1-1‘2) (1)) 1(2)u (x )u (x )

)ul(x1) u2(x1)+(l-l l 1 2 2

dun-1(2)) m1 (x2) u2 (x1) +1‘1)1(2)u1 (x2) u2 (x2)

(1) ( ) (1) (2)

(l) )12v(x2)+l (l-A )v(x2)

g (1-1 1(2)

)(1- )V(x1)+(1-A

+).(1)).(2)v (x2)

(1)) (1&2) (1) (2)

(1-1 )v(x1)+(1-(1-1 )(1-1

))V(x2)
= (1-c)v(xl)+cv(x2).

Similarly

X +x X +x x 4-3

2_cv(x1)+(1-c)v(x2).
Hence

x +x
cv(x1)+(l-c)v(x2)$y(-l——z)g(l-c)v(xl)+cv(x2).
2
By Theorem 2.1 v is 08 on (a,b). while 1(v) g c. Changing

to the OS dilatation. this becomes p(v)gg (p1+l)(P2+l) - l.

25

(ii) Assume the theorem true for n = N and let

N ’0
A O A .
v = OH ul. Then v = V°uN +1. But uN +1 is assumed QS and
i=1 0 O NO
0 is QS by the induction hypothesis. with p(o) g [H (pi+1)]-1.
i=1
Hence. by part (i). v = G-uN +1 is also 05 on (a,b) with
0 NO-l-l
_ A _ _ A ,_
p(v)g(p(v)+1)(pN +1+1) 1 — p(v)(pN +1+1)+pN +1-S [.n (pi+1)J 1.
O O 0 i=1
a.
i

Equality holds when ui(x) = x . i = l,2,....n. on

[0.m) for any choice of a1,a ..an all greater than or

2'00

equal to 1. By Lemma 2.1 this choice of the ui gives

oi Edi Zai
2 )-1=2 -l=p(x )=p(v).

l

[

i

":19
":15

(P.+l)]-l = (
1 1 i

6. The inverse of a 98 function.

 

Theorem 2.7. Let u be a 95 function from (a,b)
onto (c.d) ‘with midpgint dilatation 10 = 1(u). Then u-1
is a 98 function of (c.d) onto (a,b) whose 95 dilatation

satisfies p;1(1/2)
P(u-l) 3'1??—
9 (1/2)

10

where p and P dengte the Salem functions in (2.5). This

result is sharp:

26

Proof: Let y1,y3 E (c.d) with yl < y3 and let
-1 .

X = p’1(1/2) = 1-p'1(1/2). Then
u((i-X)xl+1x3] g (1-P(T))u(x1)+P(X)u(x3>
= (u(x1)+U(x3))/2 = Y2
by Theorem 2.2. Since u is monotone increasing this gives
x2 2_(l-):)xl + Xx3.

Similarly,

u(ix1+(1-X)x3) = u((l-(l—anl + (l-X)x3)

2.<1-p(1-X))u(xl) + p(l-X)u<x3)

(u(xl) + u(x3))/2 = Y2-

This gives, by the monotonicity of u. ngg Xxl + (l-I)x3.

Hence
- -1 - -l -l - -l - -l
(l-A) u (Y1) +(1-(1-A) )u (y3\_<.u (y2)g(l-(1-A) )u (yl) +(1-A) u (y3)

and by Theorem 2.1 u.1 must be QS on (c.d) with
1(u-1) g l-I = p-1(1/2). By Definition 2.1 this becomes. in

terms of the Q8 dilatation.

-1
P(u-l) S LML .

P 1(1/2)

27

Equality holds when u(x) = xl/n on [0,a) for any
. 1/n
integer n 2_1. By Lemma 2.1. p(u) = 1/(2 -l). Hence.
. . . l/n . .
by Definition 2.1. 1(u) = 1/2 . From (2.5) it is clear
that
n n 1/n n
p(1-(1/2 )) = 1-(A(u)) = 1-(1/2 ) = 1-1/2 = 1/2.
-1 _
or p (1/2) = l-l/Zn. For this choice of u, u is given
by u-l(y) = yn and. as above. it is easy to see that
p(u-l) = Zn-l. Hence
p(u-l) _ 2n__1 _ 1-1g2n _ 2'1(14 ) _ 2'1(142)

For computations the following bound for

though no longer sharp.

bound in Theorem 2.7.

Corollary 2.7.

P(u-l).S

Proof: Let A0
P(l/Zn) = A: for each

there exists some integer

k+l
)

(2.11) P(l/2

1 +1.3 1/2 < A:

1/2n 1-p'l(1/2) P'1(1/2)

1

P(u- ).

is probably easier to use than the

Under the conditions of Theorem 2.7.

log 2
2,21<>g(1+1/p(11)) _ 1.

K(u) = P .
10
n 2_0. and P(l/ZO) =

and P Since

P(l) = 1 > 1/2.

k such that

k —

k
0 P(l/2 ).

28

Since P is strictly increasing (2.11) is equivalent to

1/2k+1 g P-1(1/2) < (l/2k), which reduces easily to
(2.12) 1 - 1/2k < 1-P'1(1/2) = p'1(1/2) g 1-1/2k+1.

Solving (2.11) for k gives

 

 

10g 2 = log 2g k
k 3 log (1/10) log (1+1/p (u)) < “‘1'
and thus log 2
ZR‘S 2 log(l+1/p(u)) .

Hence, by Theorem 2.7 and (2.12) we find the desired bound
lo

_ _____;a_2______
-1 2 1(142) _g _—__/_2:+_ = = S m+1/mm.
p (1/2) 1/2k+l

7. The composition of 98 functions.

Theorem 2.8. Let ul, u2. .... un be Q§ functions

such that the domain of each ui+1 is contained in the range
of thegpgeceding ui. Then the composed function u defined
gag u(x) = u oun_lo...ou1(x) is also 95 and

pn(1/2)
P ‘u’ 5 I-EIn (1/2)

 

= ... . ' ' t
where Pn PMun)oPMu )0 °PA(u1) This result is sharp

n-l

29

Proof: The proof is by induction on n.

(i) Assume n = 2 and let x .x 6 (a,b) be given.

1 2

‘where (a,b) is the domain of u . Let c = PA

1 (1(u1)).

(“2)

Then by Theorem 2.1 and Theorem 2.2

X1+X2 xl+x2

“ “2‘“1‘ 2

 

 

V
I

))
g_u2((1-A(u1))ul(x1)+A(u1)u1(x2))

(u )(A(u1))u(x2)

gg (l-Px(u2)(A(ul)))u(xl)+Px 2

(l-c)u(x1)+cu(x2).

Similarly
x1+x xl+x
tit—jg'z) - u2(u1(-3fjh)

2.(l-Px(u2)(1-X(ul)))u(x1)+PA(u2)(1-X(u1))11(x2)

= P )(A(ul))u(x1)+(1-P

A(u2 u )(A(u1)))u(x2)

A( 2

= cu(x1)+(l-c)u(x2).

Hence

x+x
cu(x1)+(l-c)u(x2)$u(—;5-z)g(l-c)u(xl)+cu(x2).

By Theorem 2.1 u is 08 on (a,b) and A(u)gc = PA (A(u ))
(u2) l

Pl(u2)(Pl(u1)1/2)) = P2(l/2). Changing to the QS dilatation,

this becomes

3O

92(1/2)

p(u)-S 1-P2(1/2)

(ii) Assume the theorem true for n = NO and let

G u u u Then u u G But u is
= o o. o o O O = . o
N -1 +1 1
Nb 0 1 Nb Nb+

assumed QS and G is 08 by the induction hypothesis with

A(G)g P (1/2). Hence. by part (i). u = u .G is also
Nb NO+1

QS and

A
(1/2)) = Pl(u (1(u))
Nb+

Mu) Mu oG)gP (P

Nb+l) A

No+l A (u (3)

(1/2).

p
+1
N0

By Definition 2.1 this is equivalent to

 

PNo+l(l/2)
p(u).s _
1 PN +1(1/2)
O
”i
Equality holds when ui(x) = x . i = 1,2,....n. on

[0.«) for any choice of a1.a2.....ah with all gig 1.

 

F0.
Obviously u(x) = x l on [0,o). so that Lemma 2.1 gives
a.
P(u.) i
1(u.) =""""""1-"""==“Z“—l : 1 - l for each i.
i 1+p(u.) a. o.
i 2 i 2 i

77a.
It is now trivial to show by induction that Pn(1/2) = 1-1/2 1.

Hence
No

”“i 1-1

1 P (1/2)
p(u) = 2 -l = _n

Fa. = 1-p (1/2) '
1/2 1 n

 

31

Remark: It would be interesting. in Theorem 2.8. to
see how p(u) depends on the individual p(ui). For simplicity
we will restrict the investigation to the case n = 2.

If n = 2 then by Theorem 2.8

92(1/2) P1(u2)(*(“1’)
(2°13) P(u) = P(uzoul) g I:P;7I7§T =

 

l-P1(u2)(x(ul))

since l/2g1(ul)<l there is some integer k 2_1 with

1 - 1/2k g A(u1) g 1 - 1/2k+l.

Solving for k and using Definition 2.1 gives

109(1+P(u1))
k 3 log 2 g k+l.

Thus

k+1 k+1
PMuz) (Muln g PMuz) (1-1/2 ) = l-(l-Muzl)

and using this in (2.13) gives
log(2+2p(ul))

 

(2.14) P(uzoull S (1+P(u2))k+1-1 S (1+P(u2)) log 2 - 1..

Now suppose the function 111 is fixed and let

log (2+2p(u1))/log 2. Then 2'g a < a and (2.14) shows

a

that

a
(2.15) p(u2°u1) g (l+p(u2)) - 1.

32

A simple expansion of the right-hand side in (2.15) shows
that

(2.16) p(uzou1)gg p(u2)a + 0[p(u2)a-1] as p(uz) approaches a.

Suppose now that 112 is fixed instead of ul. Then

using (2.14) and the identity A1°g B = 3109 A

,‘with
A = (1+p(u2)). B = (2+Zp(ul)). we obtain

log (2+2p(ul))

 

log 2
(2.17) p(u2°u1)-$ (l+p(u2)) - l

109(1+P(u2))

 

 

= (2+2p(u1)) 1°9 2 - 1.

Let B = log(l+p(u2))/log 2. Then lgg B < a and (2.17) gives
. B-
(2.18) P(u2 u1)gg (2+2p(u1)) 1.

A simple expansion of the right-hand side in (2.18) shows

that
P(u on ).g (2p(u ))B + O[p(u )B-l] as p(u ) approaches w.
2 l l 1 1

The inequalities (2.15) and (2.18) show that if
u1 (u2) is fixed. then p(uzoul) is bounded by a power
function of] p(uz) (p(u1)) as p(uz) (p(u1)) tends to
infinity, the power depending on the finite constant

P(u1)(P(u2)).

CHAPTER III

RATIO BOUNDED FUNCTIONS

1. Introduction.

Although the definition of quasisymmetry is quite
simple. in practice it is often very difficult to determine
whether or not a particular function u is 08, and to obtain
bounds on its 08 dilatation p(u). Since all 08 functions
are by definition strictly increasing. however. and a monotone
function must be differentiable almost everywhere. it would
certainly be helpful to have a simple differentiability
condition for quasisymmetry.

'With this in mind we begin Chapter 3 by defining a new
class of functions called ratio bounded or RB functions.
This class is defined by a relatively simple differentiability
condition. The major portion of this chapter is spent in
an investigation of closure and other properties of this
class, much as we spent Chapter 2 studying the properties of
the class of 08 functions.

The first application made of this class comes at the

end of Chapter 3 when we show that the RB functions actually

33

34

form a subclass of thepQS fpnctigpg and then use the
differentiability condition to obtain sharp bounds for the
OS dilatation p(u) when u is RE.

A second. and more important. application of RB
functions is their use in the problem of extending a 08
function on (-a.a) to a QC mapping of the upper half

plane H. In particular we are interested in finding such

an extension that is gpasiconformally extremal. I.e.. Qne

that is as close to being conformal as ppssible. This

 

second application will be treated in the next chapter.

however.

2. Notation.

A function u will be said to be normalized if it
satisfies

(3.1) ‘ u(0) = o. 110) = 1. u(a) = ...

whenever the points 0, 1. m are in the domain of u.
Throughout the remainder of this paper it will be assumed
that all 08 functions are normalized according to (3.1).
Furthermore. if u is a 08 function of (-wow) onto itself
with 08 dilatation p(u) as given in Definition 1.3,then
3(u) will mean the OS dilatation of the restriction of u

to (0,»). Obviously 8(u) g p(u) always holds.

35

3. Alternative definitions.

For ease of reference we repeat here the definition

of ratio boundedness (Cf. Definition 1.4).

Definition 3.1. Let u be a strictly increasing
self-homeomorphism of [0,m). Then u is said to be

ratio bounded 0n [0.«). or RB. if there are numbers L, M.

O < ng M < m, such that

xu' x
u(x)

(3.2) Lg gm a.e. on (0.69).

The lower (upper) ratio bound L(u) (M(u)) g_§ u gg_
[0,a) is defined as the supremum (infimum) of all numbepg

L (M) satisfying (3.2).

 

Our first goal is to give two equivalent. and very

useful. alternative definitions for ratio boundedness.

Theorem 3.1. A function u pf, [0,») onto itself

is RB with ratio bounds Lb' MO if and only if

x .
(3.3) u(x) = exp(J %E§i§%’€f) for all x > O. u(O) = 0,
l

 

where X is some real valued measurable function on (0,o)

with
(3.4) sup |x(s)| < 1. ess sup lx(S)l = 372% < 1.

0<s<¢ O<8<¢

36

for some Q 2_l. MoreoverI if u satisfies (3.3). then

Q = max [M0, l/LO}.

Proof:
(i) Assume that u is of the form (3.3). Then
x measurable implies that (1+x(s))/(l-x(s)) is also measurable
since (3.4) shows that the denominator in the integral in
x
(3.3) is never zero. Hence lilléL.Q§' and thus also u in
11-x(s) s
(3.3) must be continuous on (0,m). It is also clear. taking
a limit in (3.3) as x approaches 0. that u is continuous

at 0 as well. By the second inequality in (3.4),

0 < l/Q g (1+x(s))/(l-x(s)) a.e. on (0.a). so that

x

21+3(s).g§
1’X(3) s
"1
x1 < x2 =9 u(x2) - u(xl) = e > 0 =; u(x2) > u(xl).

Thus u is strictly increasing. It is also trivial to show
from the form of u (3.3) that u is normalized according

to (3.1). Finally,

a. e. on (00 “)0

. _J-JML}.
u (x) - 1'X(x) x u(x)

whence

o<_1_qu x _1+X(81

u(x) - l-x(x)'S'Q < a a.e. on (030).

37

By definition 3.1 u is RB with 1/QgLOgMOg Q, and either

L0 = 1/0 or MO

in the second part of (3.4). Hence Q = max[MO.1/LO}.

= 0: otherwise there could not be equality

(ii) Assume u is RB with ratio bounds L0 and

MO. Then since u is monotonic. u' exists a.e. on (0,o)

and is measurable. Let Q = max[MO,l/LO] 2_l and define X by

su' s —u s

. wherever u' exists
su (s)+u(s)

(3-5) x(8) =
O elsewhere

Then X is measurable on (0,») and

sup |x(s)l = ess sup |x(s)l = (Q-l)/(Q+1) < 1.
0<s<s 0<s<s

From (3.5). fl: w}; a.e. on (0,:9). Integrating

and exponentiating gives

“1:13). is
1l-x(s) s
u(x) = e .

and by continuity u(0) = 0.

Theorem 3.2. Let u be a function mapping [0,»)
onto itself and normalized according to (3.1). Then u
is RB on [0,.). with rapio pgundg L0 and MO. if and

only if

38

L0 Mo
19. 11119.). .11
(3'6) )a) 'S u(a)‘S )a)

for all a. b E (0,») with a < b.

Proof:

(i) Assume u is RB with ratio bounds LO' MO. By

Theorem 3.1

le-t-X‘s) (lg
11-x(s) s
u(x) = e for all x > 0, u(O) = O,

‘which implies that

M'MSMO a.e. on (0,0)-

LO'S l-x(s) - u(s)
Hence
b L b 13M
L() I '—Q' ds I i: (S) _d_s_ I "Q ds MO
p_ _ a £5 a X s 8 __11fln e a S _ jp
(a) - e 'g e - u(a)S - 1a)
(ii) Assume
L0 “0
(3.7) (‘3) g fi‘fifi' g ('5') for all O<a<b<co.
L

0

Then L > 0 implies u(b)/u(a) 2_(b/a) > 1 for any a < b.

O
'which means u is strictly increasing. Hence u must also
'be differentiable a.e. on (0.»). If x e (0.w) is a point of

differentiability for u. then for any h > 0 (3.7) gives

39

 

 

 

LO LO M0 M0
g_ _ u(x+h) x+h _ [h
1+x) - Kt?!) g u(x) 3 K x ) — (1+x) a.e.,
or L M
h o h 0
(l+-) -1 (l+—) -1
x u(x+h)-u(x) x
(3.8) Jh u(x) g h g h u(x) a.e.
Letting h approach 0. (3.8) reduces easily to
L0 MO
3"" u(x) g u' (x) g '3"— u(x) a.e. on (0, no) ,

‘which is equivalent to

xu'(x)

Log u(x) 3M0 a.e. on (0.»).

Therefore u is RB.

CLOSURE PROPERTIES

4. The sum of RB functions.

Theorem 3.3. Let ul, u2. ....un be RB functions
with ratio bounds L(ui) = Li' M(ui) = Mi, i = 1. 2.....n. Then

n
the function v =(l/n) Z) ui is also RB, and
i=1

L(v) 2 min {L.}. M(v) 3 max {Mi}.
lgign 1 Igign

This result is sharp!

40

Proof: Let L = min [Li]. M0 = max [M.]. Then
1313!! 1 £1 gn 1
(3-9) Loui(X) g xui(x) ggMoui(x) a.e. on (0.m)

for each i. By the subadditivity of measure. however, the

set of points at which any of the ui does not satisfy (3.9)

has measure 0. Hence

n n n
‘33 Loui XV. x xx? ui (x) i3 M011i
Loz—rT—S v(x) =-l'-1——S—I-l———=MO a.e. on (0.ao).
ZDu. Z)u.(x) Z3u.
1 1 1 1 1 1

The proofs of continuity. monotonicity and normality follow

trivially from the definition of v.

(I

Equality holds when ui(x) = x on [0,a). i = 1.2.....n,

for some a 2,1.

5. The product of RB functions.

Thegpem 3.4. Let u1.u2.....un be RB functions.

. n
Then thegggnction v = H 111 is also RB, and
n n
L(v) 22L” M(v) $2M..
1 1 1 1

This result is sharp!

41

n n
Proof: Let L = Z L.. M = 2311,. By the subadditivity
“———' O 1 i O 1 i

of measure.

Liui(X)'S xui(x) g Miui(X) a.e. on (O,a)
for each 1 implies

n
x Z)(u!(x) H uj(x))

 

I ' '- (X)
xv'(x) _ i=1 1 1¢i fi___ n xu
Lo.g v(x) —- n — .Ei u (x) g MO a.e. on (0.x).
H 11.(x) l-
i=1 3

The proofs of continuity. monotonicity and normality follow
trivially from the definition of v.
i

Equality holds when ui(x) = u on [0.o). i=1.2.....n,

for any choice of the ai all greater than 0.

6. The inverse gf an 33 function.

Theorem 3.5. f u is an RB function then so is u .

 

L(u-l) = l/M(u), M(u-l) = l/L(u).

Proof: Let v = u . It is obvious that the continuity.
monotonicity and normality of u imply the same properties
for v. Since v is monotonic it is differentiable a.e. Let

yo be a point at which v is differentiable. Then also u

42

:must be differentiable at x0, where x0 = v(yo). Also

v'(yo) = l/u'(xo) implies that

1

. ‘u(x l'fi“"'
1 S.y v (YO) _ 0 u (x0)._ u(xo) 1 a e on (0 w)
.M(u) v(yo) ’ x0 " xou'(xo)‘g L(u) ° ° '

 

 

 

7. The composition of RB fupctions.

Theorem 3.6._ Let ul.u2.....un be RB functions.

Then the compgsition function v = unoun_lo...ou1 is also

RB, and
n n

L(v) 2 n L.. M(v) g n Mi.
i=1 1 i=1

This result is sharp:

2599;: The proof is by induction on n.

(i) Assume n = 2. Then v = uzoul. The proofs of
continuity. monotonicity and normality follow trivially from
the definition of v. Hence v is differentiable a.e. on
(0.»). Let xo be a point of differentiability for v and
let yo = u1(xo). Then v'(xo) = ui(yo)ui(xo). Since

v(xo) = u2(yo) and y0 = u1(xb) we thus have

L L .g XOV'(xg) = xgué(yg)ui(xo) y = Ygué(yg) . xoui(xo)
1 2 v(xo) u2(yo) u1(xo) u2(yo) u1(xo)

S M1142 a.e. on (O. a) o

43

(ii) Assume the theorem true for n = NO and let

A .
u ov. But u 15
+1 +1

N0 N0

0 = ‘JNoouNo-1Oo o o oulo Then V =
RB by assumption and 0 is RB by the induction hypothesis

wi th

A NC) A NO
L(v); H L.. M(v) g;n M..
i i
l 1
Hence. by part (i). v = uN +lo9 is also RB and
O
A A No+l A
L(v) = L(uN +1av) Z'LN +1°L(v)g2 II Li' M(v) = M(uN +1ov)
O O l O
A NO+1
3 MN +1°MM 3 n Mi'
0 1
“i
Equality holds when ui(x) = x on [0,o). i = l,2,...,n.

for any choice of the “i all greater than 0.

8. The class TL M of RB functions.
0

Definition 3.2. For given L. M E (0.a) with L i M,
let TL M denote the class of RB functions u with L(u) 2 L.
I

and M(u) 3M.

______Lema 3-1- La 11 W tom) and
su se {ui} converges to u pgintwise on [0.«) and
uniformly on compact subsets. where 111 6 TL M for each i.

0

Then u e T also.

L,M

44

Proof: Since ui(0) = 0 and ui(l) = l for each i.
we obviously have u(O) = 0. u(l) = 1 as well. by the pointwise
convergence of the ui. Also. ui(x) > 0 for each i and

any x > 0, so that u(x) 2_0. Suppose there exist x .x e (0.«)

l
‘with u(xl) > u(xz). Let d = u(xl) - u(xz) > 0. Since

2

[uil converges uniformly to u on the compact set [XI/2. 2x2].

there exists an Nb such that

)uNo(x1)-u(xlll < 6/4,] uNb(x2)-u(x2)|< d/4.

Then

uNo(x1)g2 u(x1)-d/4 = u(x2)+3d/4 > u(x2)+d/4 > uno(x2).

Therefore (x ) > (x ), which is impossible since u
“No 1 uuo 2 NO

is strictly increasing. Therefore u is non-decreasing.

The proof will be completed by showing that for any

 

x1.x2 6 (0.o) ‘with x1 < x2.
x L u(x ) x M
_Z. 2 .4;
(3°10) )xl)'s u(xl)‘S 1x1) ,

Given any integer No, let 6 = u(x1)/2(2Nb+l). By the uniform
convergence of [ui} to u on [XI/2' 2x2], there exists

an n such that

0 < un(x1)-e g u(xl) g un(xl)+e. un(x2)-e g u(xz) g un(x2)+e.

45

Thus

Mx) u(X)+e u(x) 11(X)

__2__S_B__2___=_B__2_ __n___2___
u(x ) un(x1)-e un(x1) 1 g

 

1 u (x )
___£___.
3 :2 M 1 un(x1) g :2 M l+u(x1)-e:
(x1) 1 g le) 1 a

un (x1) u(X1)-e

 

= K22)!“ u(xul ()x- 1:2)6 k:l)M[1+—_]

Since No can be chosen arbitrarily large. this means that
u(x2)/u(x1) g (xz/xl)M. Similarly, we find that
u(x2)/u(x1) 2.(x2/x1)L. Thus (3.10) holds. By Theorem 3.2

u must be RB with L(u) 2 L, M(u) g_M. Hence u E T

L.M'
Theorem 3.7. TL M is compact under uniform convergence
I

 

on compact subsets of (0.m).

Proof: We begin by showing that T is equicontinuous

L,M

on compact subsets of (O.¢).
Let S be a compact subset of (O.o). Then there

exist positive numbers a,b 'with o<a<l<b<o such that

M)l/2_

agxg b for all x E 8. Given any e > 0. let = a((1+e/2bM 1).

Let u 6 T and x .x 6 S ‘with x < x Then by Theorem 3.2.

L,M 1 2 l 2

u(b) = u(b)/u(l) g,b W(u)'s . Hence

46

Ixz-x1| < 6 implies that xz/xl < l + (6/x1)‘g l + (6/a),

‘which implies further that

u(x )
— -- - = _—-'2_-
|u(x2) u(x1)| — u(xz) u(xl) u(xl) Lu(x1) lJ

gnu»): :ZM-l ngL1+9-M-1
kal) J k a) J

=1, is... ._._.e<€.

"LZbMJ 2
Thus T is equicontinuous on compact subsets of (O.a).

L.M
Now let x E (O.w). It must be shown that

S = {u(x)lu E T } is bounded. If x = 1 then S = {l} by

X LoM

the normality condition. and {l} is Obviously bounded.

Suppose O < x < 1. Then by Theorem 3.2. using a = x and

M L].

b = l, we Obtain x" g_u(x) g xL. so Sx c:[x ,x Hence

Sx is bounded. Finally, if x > 1 then by Theorem 3.2
‘with a = l and b = x ‘we find xL.g’u(x)‘g xM, so
Sx C’fo, xM] is bounded.

Hence T satisfies al’ the conditions of Ascoli's

L.M

of elements from T has

Theorem, so any sequence (ui} L.M

a subsequence which converges to a continuous function u
pointwise on (0,») and uniformly on compact subsets of

(0.»). By Lemma 3.1 u e T

L,M

. Therefore T is compact.
, L.M

47

Corollary 3.7. For any a > 0, let ua(x) = xa

gr; [0.en). Then for any LI“! 6 (0,...) with LgM,

 

uL E TL,M and uM E TL,M° Also, for any_ u E TL.M we have
uM(x) g u(x) g uL(x) for all x 6 [0.1],
uL(x) S u(x) g uM(x) for all x E [l,m).

Proof: The proof is immediate from Theorem 3.2. If
x E [0.1] take a = x, b = 1. If x E [l,m) take a = l,

b = x.

FIRST APPLICATION

9. The relationship between RB and Q5.

 

It will be shown in this section that any function u
which is RB on [0,m) must also be 05 on (O.c). Sharp
bounds on p(u) 'will be found in terms of L(u) and M(u).

The following lemmas will be found quite useful.

Lemma 3.2. Let T > 0. Then T > 2T - 1 .;§

O<T<1, T=2T-1 i_ T Lang T<2T-l i_§ T>l.

T

Proof: Let f(T) 2 - l - T for T 2_O. Clearly

f(0) = o and £(1) 0. Furthermore, f'(T) = 2T log 2 - 1

has only one zero T with O<To<1. and f"(T) = 2T(log 2)2 > 0

0

everywhere. Therefore the critical point To must be a

48

minimum and so f is decreasing on (0.1) and increasing

on (l,o). This completes the proof.

Lemma 3.3. f T > 0 and O<sgl then

min{T.

T
ZT-l} 314351—23 max (T, 2T-l}.

Proof: Let f(s)=((1+s)T—1)/s for «331. Then

f'(s) _ Ts(l+s)T-1 + l - (1+5)T
_ a O
s

u 2 1+3 T - 2Ts l-o-s)T-1 + T(T-1)§2(l+slT-2 - 2
f (s) = 3 .
8

If there is some 51 6 (0,1) with f'(sl) = 0 then

T T-l
(1+sl) - l — Tsl(1+sl) .

 

Therefore
'r('r-1)(1+:=.1)T'2
f"(81) = 8 o
1
so
< 0 if O<T<1
(3.11) f"(s) is = 0 if T = 1

> 0 if T > 1 .

Case 1: If O<T<l then by (3.11) 31 is a maximum

point of f and

 

 

(1+sl)T-1 Tsl(1+sl)T-1
(3.12) f(sl) = s _ S
1 1
= T(l+sl)T--1 = T l-
l+s )
( 1
g T

Since any maximum or minimum of f on [0.1] must occur

either at one of the endpoints or at an interior point 31

with f' (51) O. (3.12) shows that for any 5 E (0.1],

T

f(s) g max {f(O). f(l), T} = max (T, 2 - l},

f(s) 2 min {f(O), f(l)} = min (T. 2T - 1}.

Case 2: If T = 1 then
f(s)= =1=T=2-1.

Therefore, once again,

T

min {T.2T - 1}.g f(s) g max (T. 2 - 1].

Case 3: If T > 1 then by (3.11) 51 is a minimum

point of f and

 

T T-l
(Sl - 81 - sl - s1 2 °

Hence

50

f(s) 2 min [f(O). f(l), T} = min (T. 2T - l},
f(s) g.max [f(O). f(l)} = max[T,2T-1}.

Lemma 3.4. f T > O and O<sgl then

T .
min {l,T] gw—gmax {l,T}.

Proof: Let

 

 

 

T
9(8) = l - £5135) .
Then
(3.13) g. (s) = 18.11.2le +2 LL—s)T - 1
s
and
(3.14) g"(s) ’2 1-5 T- s l- T-:;T(T-l)§2(1-s)T-2+2 .

S

If there is some 5

1 E (0.1) with g'(sl) = 0 then by (3.13)

T T-l
l-(l-sl) — Tsl(l-sl) .

Therefore, by (3.14).

2 T-2
'T(T'1)Sl(1-sl)

9" (s1) = 7———-— .
s1

 

H¢n°e > o if O<T<l

0 if T = 1
< 0 if T > 1.

(3.15) 9"(51) is

51

Case 1: If O<T<l then by (3.15) 51 is a minimum
point of g on (0.1) and
T
l-(l-sl)

-l
g<s1)=-—s—-—-=T(1-s1)T = T _Tz'r-

1 (l-sl)1

Thus

9(8) 2min {9(0). 9(1). T} = min [LT].
9(5) Smax [9(0). 9(1)} = max {LT}.
w: If T = 1 then 9(3) = 1 = T. so again
min [1. T}‘g g(s) ngax [1, T}.

Case 3: If T > 1 then by (3.15) s1 is a maximum

point for g and

Hence
9(8) gmax (9(0). 9(1). T} = max {LT}.
9(8) 2min (9(0). 9(1)} = min {1. T}.

Theorem 3.8. Let u. be, RB 9g, [0,o) with ratio

bounds L = L(u), M = M(u). Then u must also be 98 on

52

(i) 2L-1 g 80:) g 2M-1 if 1991,

 

 

 

.. 1 A 1 .
(11) T's p(u) g L if Lgmgl,
2 -l 2 -l
A M M M 2 M-L M-l 1-L
... _ _. _.__—. m, _
(m) p(u) smax { L . 2 1. L. L5H) < 1) (1 L) 1
2 -1
if L<1<M.
These results are sharp!
Proof: The proof will be in two parts. In (a) we
will treat cases (i) and (ii). In (b) we will treat case
(iii).
(a) To find lower bounds for $(u), we notice that by
the definition of 8(u), if it exists.
u(x+t)_l
l u(x+t)-u(x) _ u(x) A
$(u) S'u(x)-u(x-t) - 1_u(x-t] .g_p(u) for all x>0. O<t<x.
u(x)

Letting t-—>x, and remembering that u is continuous on

[0.m) with u(O) = O, we find that

l u(2x) A
6“) S u(x) - 1 S P(LI).

 

Thus

8m); M_12KZ£)L_1=2L_L

u(x) x

1 u(2x) _ 2;,M _ _ M _
Tongue) lskx) 1-2 1.

53

Hence 8(u) 2 max{2L-1, 1/(2M_1) }. The lower bounds in cases (i)
and (ii) follow simply from this inequality.
To find upper bounds for 3(u) let x 6 (O.m)o

O<t<x. and define s as O < s = t/x < 1. Then by Theorem 3.2

u(x+t) x+t L

 

 

 

Ei’fi't) ‘11(X) = u(x) )_ x ) _1' 1+3 L-l
u(x)-u(x-t)1u(x-t) 2 1_ _x_-_t_ M=1-(l-s)M '
u(X) ) x )

(3. 16)

u(x+t)_ x+tfl
2(x+t) -u(x) _ u(x) “(Z—SS 21 (1+ 32M
u(x)-u(x-t) 1_ u(x- t) 1_ _Kx J)f 1_ (1_S)L
x

u(x)

But by Lemma 3.3 and Lemma 3.4. if O<s<l then

M

1+s -1 M
.L1_§lM_-1._ ___L_ W11 _
LS min{1 L] - a.
1-(1-8)L1-(1-8)L '
s
(3.17)
(l+s)L- l

1.1-321:1... _____§__2minlL,2L-l| = B
1-(1-3)"1.(1-82M max(1.M}

Thus. if c = max {ml/B). then by (3.16) and (3.17)

1 (+t)-()
:Sux uxSc'

u(x)-u(x-t)

so 3(u) g c. Now by Leanna 3.2. 1% implies a = ZM-l,

B = L/M 2 l/M. Hence c = max [ZM-l.M} = 2M-1. This is just

 

Fain. 'y. g..-

 

54

the bound we are looking for in case (i). Similarly in case
(ii), ngugl implies by Lemma 3.2 that a = M/L g l/L.
a = 2L-1. Hence c = max {1/L. 1/(2L-1) } = 1/(2L-1). Again, this

is the bound we want.

(b) For any x 6 (O.m). O<t<x. let s be defined as

O < s = t/x < lo

M L

f(s) = 1+5 'i . g(s) = 1+5 ”i . for s e (0.1).
1-(1-s) 1-(1-3)
and a==sup f(s). B = inf g(s). Then as we showed in (a).
0<s<1 O<s<1

_.(.__1:__.L.L
539(81'Sux+t ux

u(x)-u(x-t)

.S f(s) S a.

If, as before, we set c = max [a.l/B}. then we have 8(u) g c
in case (iii) as well. Our prdblem is to estimate a and
B in this case.
It is clear that both f and g are continuous on

(0,1) and that by l'Hepital's rule

1im+f(s) =%. lim f(s) = 2M-1, 1339(3) = f;- 1im_g(s) = 2L-1.

340 s41- s40 s41
Hence, if we define f and g at 0 and 1 by their limit
values. then f and 9 will be continuous on [0.1]. Since

[0.1] is compact f and 9 must actually attain their

maximum and minimum values on [0.1]. That is.

55

a = max f(s). B = min g(S).

ogsgl ogsgl

Since we already know the values of f and g at the end-
points 0 and 1 we need only investigate possible interior
maxima for f and interior minima for g to obtain bounds

on a and B.

 

 

 

Let so, O<so<l. be any critical point for f. Then
L M-l M L-l
(1-(1-s ) )M(1+s ) -((l+s ) -1)L(l-s )
O 0 O O
O = f' (so) = J L 2
(1-(1-30) )
implies
L M-l M L-l
(l-(l-so) )M(l+so) - ((1+so) -1)L(l-so) .
which in turn implies
M M-l
f(s ) = meg) -1 = mush) = 94mg )M'lo (l-s )l’L
o 1 L o o ’

1-(l-so)L L(l-so)L-

If we define h(so) =(M/flX1+so)M-l(1-so)l-L for O<so<1.

then we find that f(so) g sup h(s ). But now

 

O
O<so<1
h' (s )
0 M-1 l-L 1
h(so) l+so 1 s0 l-s02 0

If M+L-ng then h is a decreasing function of s so

00
that h(so) gM/L on o<so<1. But if M+L-2>O then h'(so) = o

for so= (M+L-2)/(M-L) E (0.1] since L31 implies M+L-2§M-L.

56
And in this case h increases from S0 = O to S0 = (M+L-2)/(M-L)

and then decreases until s0 = 1. So if M+L-2>O then the

supremum of h on (0.1) is

M+L~2 _ M-l l-L
h(——M_L) - ;(—;-M L)” 1‘04 1) (1 L) .
Thus. in any case, if 50 is any critical point for f on
(0.1) then

mo) 3 mam-”If. ”-(-%-)“'L(M-1)M’1(1-L)1'1‘).

This means. finally. that

_2__

M-L M-l l-L}
M-L

) (M-l) (l-L)

a = max£f<0). f(l). f(son s mam-E. 2M-1. %

(

Now suppose s O<so<l, is a critical point for g.

00
Then in a manner similar to our treatment of f, we must have

_ L-l _ l-M
g(so) - (L/M) (1+so) (1 so) .

. . L-l l-M
Setting g(so) — (L/M)(1+so) (l-so) for O<so<l and

differentiating j logarithmically as we did with h. we
find that M+L-229 implies j increasing so that j(so)2j(0) = L/M

for s0 6 (0.1). And. as before. if .M+L-2<O then j

decreases from so = 0 to 30 = (M+L-2)/(L-M). then increases

until so = 1. Hence we must have

“L --;-;2) = (- Wkl- (1-L) L"104-1) 1’".

j(so ) 23(_— M-L

57

Thus. in any case, if s is any critical point of g with

}.

 

O
O<SO<1 then
._L_;._2L-M_L-1_1—M
g(so) 2m1n{M. M(M-L) (1 L) (M 1) }
and so, finally,
. ._L_L£2L-M_-1_l-M
a _ mn(g(0). g(1).g(so)) 2mm (W2 1.M(—M_L) (1 L)” (M 1)
And. at long last.
A(u) c - max( ‘1'} = max[2M-1 l 24" fl("'2—')M_L(l\i-l)M-l(l-L)luL}
P .S - 0'5 ' 2L_1' L' L M-L °

Equality will hold in cases (i), (ii) and (iii) by

the following examples 3.1, 3.2 and 3.3, respectively.

Example 3.1. Let u(x) = xa on [0.m) with a.2 1.

Then u is differentiable on (O.m) and

1 0-1
X11 X = m— = a 2 1 for all X 6 (00”)-
u(x) xa

Therefore u is RB with L = M = a. Hence the lower and
upper bounds in case (i) of Theorem 3.8 are the same and so
both inequalities must actually be equalities. Thus

8(u) = Za-l as was previously shown in Lemma 2.1.

Example 3.2. Let u(x) = xa on [0.m). with

O<qgl. As in the previous example we find that u is RB

on (O.m) with L = M = a. But since a.g 1 ‘we must use

58

case (ii) of Theorem 3.8. Once again both the lower and upper

bounds are the same. so both inequalities must be equalities

a
and. 3(u)==l/(2 -1).again as was shown in Lemma 2.1.

Example 3.3. Define u on [0,w) as

1/2

x if qugl
u(x) ={
x3/2 if 1<x.

Then
,1/2 if O<x<l

xu' x
u(x)

3/2 if 1<x.

Hence u is RB with L = 1/2 < l. M = 3/2 > 1 and so we

are in case (iii). Looking at the upper bound in case (iii)

of Theorem 3.8 we see that

M 23/2-1 = 2¢2 -1 < 2(3/2)-1 = 3-1 = 2.

N
I
..a
ll

l/(ZL-l) = 1/(21/2-1) < l/(1.4-1) = l/.4 = 5/2.

M/L (3/2)/(1/2) = 3.

L 1/2 /2 =

(m/L) (2m—L))”"(u-1)M'1(1-L)1' (3) (2) (1/2) (1/2)1 3.

Hence the upper bound says 8(u) g 3. But if we take x = 1

in the definition of 08 then we find

u(l+t)-u(l) _ 1+t 3/2-1
u(l)-u(l-t) “ 1/ ‘

1-(1-t)

59

u(l+t) -u(lL _

. A
Thus 11m¥+u(l)-u(l-t) - 3 so that p(u) 2_3. Hence we must

1?—>O

have 3(u) = 3. which shows that equality holds in case (iii).

10. Convex anchoncave functions.

If we look back at Theorem 2.1 we can think of (2.1)
as a generalized convexity-concavity condition. For if
10 = 1/2 then the left inequality becomes the condition for
the concavity of u and the right inequality becomes the
condition for the convexity of u. Then it is natural to
ask:, If u is concave or convex. what else is needed to

show that u is QS? The answer is contained in the next

two theorems.

Theogem 3.9. Let u §§,a continuous. monotone function

23_ [0.w). and normalized according to (3.1). Furthermore.

su se u is convexggn (O.w). Then u is 98 if and

 

gnly if it is RB. Moreover, if u i§_ 05. then

 

(3 . 18) 1gL(u) . 2L-1g8 (u) gzM-l.

This result is sharp:

Proof:

 

(1) Assume u RB and let xo be any point of
differentiability for u. (Since u is monotonic it must

be differentiable a.e.). since u is assumed to be convex

we can use statement (1.4) of [10, p.3] if we simply reverse

60
both of the inequality symbols. Hence

u(xo)-u(xO-h) u(xO)-u(0) _ u(xo)
h 2' xO-O -' xO

 

 

Letting h-€>O+. we find that

 

 

 

u(x )-u(x -h) u(x )
lim+ O h9 =u'(x0)>_ x .
1v->O O
or
x u'(x )
O O
u(xo) 2'1.

Therefore L(u) 2.1. Case (i) of Theorem 3.8 now shows that

u is 08. and gives the desired bounds.

(ii) Now assume that u is 05 on (O.m) with OS
dilatation 6(u). Then again by (1.4) of [10, p.3] with
both inequalities reversed,

u(xQ)-u(x -h) u(2xo)-u(xo)

 

 

 

 

O
h 'S xO
_ u(2xo)-u(xgl . u(xo)-u(0)
u(xo)-u(0) xO
u(x )
g 301) x0 .
o

Letting h-€>O+, we find that u'(xo) g_8(u) u(xo)/xo or

x u'(x )
O Q A
u(xo) S P (u) .

61

And. as we showed in part (i) of this proof,

x u'(x )
O O
__57;;T—'2'1’

Therefore u is RB. Again, since L(u) 2.1 we may apply
case (i) of Theorem 3.8. and the bounds follow. Equality
holds in (3.18) if we take u(x) = x for Q§x<m. In this
case we have L(u) = M(u) = 1 and all the inequalities

in Theorem 3.9 become equalities.

Theorem 3.10. Let u be a continuous. monotone func-
tion on [0.m), normalized according to (3.1). Furthermore,
su se u i§ concave on (0.0). Then u is 98 if and

only if it is RB. MoreovegLiif u is 98, then

 

1 A
M(u) S 1.““'.S P(u).S
2M-1 2L-1

This result is sharp!

Proof: The proof is the same as for Theorem 3.9
except that here. since u is assumed to be concave. we use
(1.4) of [10. p.3] exactly as it is given. Equality again

holds when u(x) = x. Q§x<m.

CHAPTER IV

RADIAL EXTENSIONS

1. Notation.

For a given 08 function u mapping (-m.m) onto
itself let

(i) ul(x) =u(x) on [0.00).

(ii) u2(x) = -u(-x) on [0.m).

(iii) U(x)

log(u1(ex)) for -m<x<m.

(iv) V(x) log(u2(ex)) for -~<x<e.

(V) Ni!)

(V (x) -U(x) ) /1r.
Furthermore we say that

(vi) u is RB on [0.6) if u1 if RB on [0.w).

and its ratio bounds are L1 = L(ul). M1 = M(ul).
(vii) u is RB on (-m.O] if u2 is RB on

[0.m). and its ratio bounds are L2 = L(uz). M2 = M(uz).

Lemma 4.1. f u is RB on (-m.0] and on [0.a)

then

0<LISUKSM1<o for almost all x. -m<x<o,

O<L2§VKSM2<co for almost all x. -~<x<m.

62

63

Proof: The proof follows immediately from the definition

of RB and the fact (cf. (iii) and (iv)) that

x x x x
e u'(e ) e u'(e )
1 2
U. (X) = I V. (X) =
x x
u1(e ) u2(e )

whenever the right-hand sides exist.
If we let R be the subset of [0.m) where either

ul or 112 fails to exist. then since u1 and u2 are
monotone functions we know that L(R) = 0. where L is
one-dimensional Lebesgue measure. Finally. if we let R'
be the set of points in (-m.m) where either U or V

is not differentiable. we must have also L(R') = 0 since

U and V are strictly increasing.
2. The radial extension.

Definition 4.1. Let u be a 08 function on

(-m.m). Then the function fu defined as

(4.1) fu (z) = fu(reie) = [ul(r;7T-YVTT.u2(r)Y/Tf]eie

for r > 0. O<9<F. is called the radial extension of u.

Lemma 4.g. The function fu is an ex ension of u

to a homeomorphism of the upper half plane H = {z = x+iy|y > 0}

onto itself.

64

Proof: It is Obvious from the definition of fu that
arg fu(z) = arg z for Im z > 0; hence fu maps each ray
with argument between 0 and F onto itself. But
lfu(reie) l = ul(r)(7T--y)/Trou2(fly/Tr increases monotonically
from 0 to m as r increases from 0 to m. Hence fu
is a one-to-one map of H onto itself. Since arg fu and
Iful are both continuous functions of arg z and lzl.
fu must be a bi-continuous mapping. Finally. taking 6 = 0.
e = F shows that fu = u on (-m.m). so that fu is an

extension of u to H.

Theorem 4.1. Let u be a 08 map of (-a.o) onto
itself. Then the radial extension fu is C if and only
if u is RB on both (-m.0] and [0.m). Furthermore.

if_ f i§_ QC. then

 

 

u
2
_ _ + -4
Q - K(f“) - 2
where
6 = ess sup 6(x)’
-a<x<m
and
2 2
A _ , l+h (x) , 1+h (x)
Q(x) - max[U (x) + U'(x) . V (x) + V'(x) }

for all x with -c<x<w.

Proof: since. by Lemma 4.2. fu is a homeomorphism.

the proof will depend primarily on a study of the point

65

dilatation (cf. Definition 1.2). In order to simplify the

proof we begin by forming the composed mapping 9 = f ofuof

2 1'
'where f1(z) = ez and f2(z) = log 2. Letting
Q = {x+iy|0<y<w} and 0' = {S+iT|0<T<n}. g is clearly a

homeomorphism of 0 onto 0' given by

(4.2) g:x+iy —-—> [I]? U(x) + %v (x)] + iy
for ~m<x<m. 0<y<v. Therefore

(4.3) sow) = "—fuuc) + 330:). T(x.y) = y.

and by the invariance of the QC dilatation under composition
with conformal maps we can study the simpler function 9
instead of fu.

In order to prove that g. and hence also fu' is QC.
we will use the analytic definition as given in [3. p.24].
According to this definition we must show that

(i) g is ACL (absolutely continuous on lines)
in 0.

(ii) D's K a.e. in 0. where D is the point

dilatation of g and K.2 l is a finite constant.

(i) To show that g is absolutely continuous on
every horizontal line in a. fix yo. 0<y0<v. and let
x1 + iyo. x2 + iyO be any two points on this horizontal line

with x1 < x Then

2.

66

(4.4) Ig(x2+iyo) - g(x1+iyo)| = Eflung) - U(x1)] + fIV(x2) - V(Xi’]°

But by Theorem 3.2.

x2 x1 L1
(4.5) L1(x2-xl) = log (e /e ) gU(x2) - U(xl)

X X

= log u e 2) - log ul(e 1)

1(

X X

= log (e 2)/ul(e 1)

u1

x x M
2 l l _ _
‘g log (e /e ) — Ml(x2 x1).

Similarly.

(4.6) L2(x2-x1) log (e 2/e 1) 2 g V(x2) - v(xl)
.g log (e 2/e 1) 2 = M2(x2-xl).
Using (4.5) and (4.6) in (4.4) gives
|g(x2+iyo) - g(xl+iyo)| = E§¥{U(x2) - U(x1)] + §IV(x2) - V(x1)]
-SLE#XM1(22"21) + %M2(Xz‘xi)
g max[Ml.M2}o(x2-x1).

Thus

|9(x2+iyo) - g(x1+iyo)|
|(x2+iyo) - (xl+iyo)|

 

g max{M1.M2} < a.

Therefore 9 is Lipschitz continuous. hence absolutely

continuous. on each horizontal line in Q.

67

To show that g is absolutely continuous on every

vertical line in 0. fix x0. -o<xo<m. and let x0+iyl.

xo+iy2 be any two points on this vertical line with
o<ylgy2<w. Then
Y

Y
a ' _ _2- _ -_1 - . -
|9(xo+ly2) - g(XO+ly1) l - I” [v(xo) U(XO) 1+7, [U(Xo) v(xo) ]+1(Y2 yl) l

Yz'yi
.s I F [V(xO)-U(xo)n + Iyz-yll

 

Y "Y
= 2” 1[V(x0) - U(xo) + w]

 

.s (yz-y1)[1 + l9fi-91.

This last statement is true because. from the quasisymmetry

of u.
x
u1(e 0)
U(xo) - v(xo) = log'-—-;;—gg log p(u).
u2(ex )
u2(e 0)
v(xo) - U(xo) = log -*——;;- g log p(u).
u1(e )
Therefore,

|g(xo+1yg)- 9(x0+1y1)l S 1 + 10 E < a
|(x0+1y ) - (xo+iy1)| F
Thus 9 is Lipschitz continuous. and hence absolutely
continuous. on each vertical line in 0.

(ii) Finally we show that D.g K a.e. in Q for some

finite constant K 2_l. As was shown in the proof of Lemma 4.1.

68

the set R' of points in (~w.w) at which either U' or
\7' does not exist has one-dimensional Lebesgue measure 0.
frherefore the set of points obtained by the cartesian product
inx(O.W) must have two-dimensional Lebesgue measure 0.
'Thus. for investigating the QC dilatation of 9 we need only
look at the points in Q of the form xO+iyO where U'(xO)
and V'(xo) exist and O<y0<v.

Let xO be any fixed point for which U' and V'
exist. and let y satisfy 0<y<v. Then. recalling that

h(xo) = (V(xO)-U(x0))/W. we have

 

 

Sx(x0'y0) = U (xo)+yh (x0). Sy(x0'Y0) = h(xo). Tx = 0. TY = 1.
Hence

1 8:+S:+T:+T: 1+h2(xo)
(4'7) D + 3 = s T -s T = (U (X0)+yh (2(0)) + U' (xo)+yh7(x0)

XYYX

‘We wish to consider D + l/D as a function of y. 0<y<w.
Clearly (4.7) is a continuous function of y on [0.w]. and

the values at the end-points are

 

2
(D +l)(0) - U'( )+ “h (X0)
D — xO U'(xo) '
l l+h2(xo) l+h2(x )
(D + 3) m = U. 0‘0””. (x0) + U' (x0)+yh' (x0) = V. (2‘0” V' (x0) '

To check on possible interior extrema. we observe that

69

1+h2(x0)
U'(x0)+yh'(xo)

 

.41. _1_- . _
dy(12+D)- h (x0)(1 ).

2
d2 1 2 1+h (xo)

--§(n +'B) = 2(h'(x )) ( ) > o.
(U'(x0)+yh'(xo))

dy 0

'Therefore. any interior extremum will be a minimum. not a

Inaximum. So

 

l+h2(x ) l+h2(x )
sup (D(xO .y)+D-'(—"---'-X1 )) = mafo' (X0)+—IT'_(—x—)2-' V' “(OH-17%}
o<y<v 0y 0 o
= 6(x0)-
‘we know. from the analytic definition of quasiconformality
[3. p.24]. that
K(g) + K(1)=6=ess sup 6(x).
g -m<x<m
and solving this for K(g). or equivalently for K(fu). we
get the desired result.
Finally. if u is not RB on both (-m.0] and
[0.m) it is because one of the following is true: L1 = 0. M1 = m,
L2 = 0. or M2 = a. Since all the cases are similar we will
treat only M1 = w. Since
I
M1 = ess sup x:(x) = ess sup U'(x).
O<x<w , -w<x<m
‘we see immediately that if M = a then for each A > 0 there

1

exists a set ‘W c (-w.m) with positive one-dimensional measure

70

on which U'(x‘ > A. But then the set Wx(0,n/2) has
positive two-dimensional measure and for any (x0,y0) in
Wx(0,v/2) we have
2
1+h (x0)
U (x0)+yh (x0)

+ ____1_
D(xoo YO)

 

D(xo.y0) (U'(x0)+yh'(x0)) +

2
1+h
(x0)

 

_ m . +x .
— ( U (X ) V (x )) + _
7’ ° 7’ ° (EfUWxOHffoOM
1r
"’2'
2,(-;-11 (xo)+0) + o

U'(xO)/2 > A/2.

Therefore. for any A > 0 there exists a set of positive
measure in Q on which D + l/D > A/Z. Thus D + l/D. and
hence also D, cannot be bounded a.e. in 0. This means

that g, and hence also fu' is not QC.

Corollary 4.1. Under the hypgtheses of Theorem 4.1

...a
(D

L1,M1 and L2.M2 be the ratio bounds of u .gg [0,m)

and (-m,0], respectiyelv. Also let p0 = po(u) be the

smallest constant greater than or eggal to l for which

 

 

 

(t)
u(t}-u§02_ u1
(4.8) —.g _ ‘g p0 for all t > 0.
p0 u(O) U(- 1:)= u21(t) ““““
Then
109 P
K(fu) + I“: ) g max[L + l(1+(-——-Q)2). M + -(14(1-—-p£)2 )}

u

‘where L = min[L1.L2}, M = max{M1.M2}.

71

Proof: Consider the function x + a/x, where a is

a positive constant. Then

6 a. a d2 a. 2a
.E;(x+;) = l-‘33 -—31x+;) = -§-> o for x > O.
x dx x

So if this function is defined on an interval, any maximum

must occur at an endpoint. Now apply this reasoning to the

 

expression
l+h2(xo)
U! (x0) + ”_U' (x0) .
van—v(x) 1 3.25:). 333.32
since ngUKng and h(x) = =‘— log .g .
n F v
u (e )
l

we have 2

1+h (x ) log p log p
(4.9) U (x0)+ U'(xo) .g max[L1+Ll(l+( 1r ) ), M1+M1(l+( "_ ) )}.
Similarly 2

1+h (x ) log p log p

. ____Q._ _1_ ___9. 2 A _J 2

(4.10) V (xo)+ V'(xo) ‘g,max{L2+L2(l+( 1r ) ), M2+M2(l+( 1r ) )}.

By Theorem 4.1 we are looking for the maximum of the four
right-hand terms in (4.9) and (4.10). But these are all
of the form x + a/x with a = 1 + (iii—{22¢ and, as we have
already shown. any maximum must occur at an endpoint. Since

the values L1.M L2,M2 are contained in the interval

10
[L.M], where L = min{L1.L2} and M = max{M1.M2}, all four
terms can be replaced by the maximum of the two terms using

L and M. The desired bound follows easily.

72

Theorem 4.2. Let u be S on (-m.m) and RB

on both [0.w) and (~m.0]. Let p1 = p(ul) and p2 = p(uz).

 

_1_:

(i) u1 and u2 are convex. then

1+<<log p0)/v)2

 

log p 2
)

K(fu)+l/K(fu) g max{2+("'—n'_'—Q , max{pl.p }+ ).

 

 

2 maXIPlapz}
(ii) u1 and u2 are concave, then
log 9 109 P
0 2 1 Q 2
K(fu)+1/K(fu) _{max{2+("'—""-7r ) . max{p1'pz}+(max{91.92}) (1+( 1r ) )}

(iii) u1 convex and u2 concave. then

2
1+((log 9 )/1r) 1 109 P
K(fu)+1/K(fu) s max£p1+ 91 9-——. E'; + p2<1+(-——9)2) 1.

 

1T

(iv) u1 concave and u2 ggpvex. then

 

2
1+((1°9 P )/F) log p 2
K(fu)+1/K(fu) g max(pz+ P2 0 . J; + (mm—1:9) )1.

Proof: The proof for each of the four cases listed
above depends on Theorem 4.1 and Theorems 3.9 and 3.10.
Since all the cases are similar, we will prove only (iii).

By Theorem 3.9, since u1 is convex on (O.m). we
L M

have llg L1 and max{M1.2 1-1} g p(ul) S 2 1-1. Hence
' LIgU{gM1 implies lgUflnggpl. Similarly by Theorem 3.10.
‘since u is concave on (O,m). we have “2-3 l and

2
M L
max(1/L2.1/(2 2-1)]g p(uz) g 1/(2 2-1). Hence. l/ngngvmgl.

73

Therefore
1/p2 3 U' g p1 and 1/p2 g v' g p1,

and we already know that h‘g (log pO)/w,
As in the proof of Corollary 4.1, we can get an upper bound for

both

+ I + I
1U? and V' + 1V?

 

 

UI

by taking the maximum of x + a/x with x = l/p2 and
a = ((log pO)/F)2, or x = p1 and a = ((log pO)/v)2. Hence,

by Theorem 4.1,

 

2
109 P 1+((log P )/W)
K(fu)+l/K(fu) _<. max{p2+P2( 1T ) . Pl + p1 }.

A special case of the radial extension and of the
results we have just obtained occurs when the 08 function
u is odd, i.e.. when u(-x) = -u(x) for all x 2_O. In

this case = 1 so that log p0 = O, whence L = L

1 2

Thus all the calculations and numerical results

po
and M1 = M2.
become much simpler. In this case we will call fu the odd

radial extension. We now investigate this special case.

3. The odd radial extension.

Theorem 4.3. Let u be an odd Q§ mapping of

(-O.~) onto itself. Then the odd radial extension

74

. . . ar 2
fu(z) = fu(rele) = eleu(r) = e1 «g u(|zI) for z 6 H,
is 99 in H if and only if u is RB on [0.m). Moreover,

 

 

i fu is QC. then

K(fu) = Q = max{M,l/L}.

Proof: Since u is odd we have u1 = uz, which
implies that L1 = L2 = L, M1 = M2 = M, U = V and h = 0.

Hence, by Theorem 4.1,

 

600 = U' (x) + U'(x)

whenever U'(x) exists and so

1
max[M,l/L}

 

max{:M,l/L} +

6 =ess sup 6(x)
-co<x<m

Thus

Q = K(fu) max[M,l/L}

and the result then follows easily from Theorem 4.1.

Corollary 4.3. f u is RB on [0,”) then u

 

must also be 98 on (O.m).

Remark: In Theorem 3.8 we proved this corollary and
in addition found bounds on p(u) in terms of L(u) and
M(u). we (now give a much simpler proof of the corollary

making use of Theorem 4.3.

75

groof of Corollagy_4.3: Define a function v on

(-ml a) by

u(x) if x 2.0

v (x) =

-u(-x) if x < 0.
Then v is clearly odd on (-m,m) and RB on [0,m) with
ratio bounds L(v) = L(u) and M(v) = M(u). By Theorem 4.3
fV is a QC extension of v to H. But by Theorem 1 of
Ahlfors and Beurling [5, p.126]. the existence of the QC

extension fV implies that v is 08 on (-a,w). Since

v = u on (O.w) u must be QS on (O,m).

Theorem 4.4. Let u be odd on (-m.m) and RB

‘93 [0.m). f u is either convex or concave on (0.x) then
A
K(fu).S P(u).S P(u)o

where 8(u) is the 95 dilatation of the restriction of u

132 (00 m) .

Egggf: .As we saw in the proof of Theorem 4.2, if
u is odd on (-m,w) and either convex or concave on (O.m)
then maxIM.1/L] g Q(u). Furthermore it is clear that we
always have 3(u).g p(u). Therefore, from Theorem 4.3,

K(fu) = max{M,l/L} and the result follows trivially.

From [12] we get the following very useful theorem

dealing with odd radial extensions.

76

Theorem 4.5. Let f be a Qequasiconformal mapping

of the upper half plane H onto itself with f(0) = O and

 

f(1) = 1. Then the following statements are equivalent:
(i) f is an odd radial mapping. i.e.. there is some
odd QS function u defined on (-m,m) for which f = fu'
(ii) The complex dilatation X ‘gf f in. H satisfies

= 621 arg z

X(Z) x(|z|) a.e. in H.

(iii) f is given by the formula

2|
mg+ iargz) for 2 EH.

|
f(2) = exka' l-x(r) :-

with boundary values f(0) = O and f(m) = m, where X, the

complex dilatation, is measurable with

Q—l
Q+1

sup IX(r)| < 10 ess SUP ‘X(r)|.S
0<r<m O<r<w

4. Extremal radial extensions.

Having found necessary and sufficient conditions for
a radial extension to be QC in the upper half plane. as
well as a means for determining the QC dilatation when the
extension is QC. we now seek conditions under which the
radial extension is extremal. That is, we wish to know when
the radial extension is as close to being conformal as possible.

For this purpose the following definition is helpful.

Definition 4.2. Let u be a QS function on (-«,w).

Then we define

-n --.: -.1

n. A. :3)

 

 

77
K = inf K(f )
u u

where the infimum is taken over all QC extensions fu of

u to the upper half plane. If f* is any extension of

u satisfying K(f*) Ku' then f* is called an extremal

extension of u.

On the problem of finding extremal extensions. we
need a lower bound on Ku for an arbitrary u. The following

theorem will give such a bound.

Theorem 4.6. Let u be a 95 function definedton

1%— S minffiTn 101011: ' Wu 101011: ' m 101011: ' m 1010“: }'
u xem g x4-m g x40+ 9 x40" 9
K 2_max{lim log u(x). lim log u(x). lim log u(x), lim log u(x)},
u --' log x -——' log x "'71 log x '——: log x
xao xa-m x40 x40

where lim denotes the limit superigt and lim denotes

the limit inferior.

Both bounds are sharp:

Proof: For any b < 0 we can consider the upper half
plane H, with the points b. 0. x. m as vertices. as a
quadrilateral (cf. page 1). Thus, if f* is any extension

of u to H whatsoever, it must map the quadrilateral H

78

onto the quadrilateral H' consisting of the upper half
plane with the vertices u(b), u(O) = O. u(x) and u(m) = m.

Thus {3, p.217 we must have

l/K(f*) g mod H'/mod H‘g K(f*).

Now by [5, p.130] we see that

mod H = P(x/-b) and mod H' = P(u(x)/—u(b)),

where P is a function such that P(O) = O, P(l) = 1,
P(l/l) = l/P(X). P(m) = m and P(X) = l + e(1)log A for
1.2 1. where 9(1) increases from 9(1) = .2284... to
6(a) = l/v = .3183... . Now, as we let x approach m.

we see that

-7- mod H' _ _T' P(u(x)/—u(b))
11m mod H _ 11m P(x/-b)
X4m x4e

‘T— lfe(u(x)[ru(b)[log(ufng-ugb)2

 

= 11’“ l+e(x/-b)log(x/-b)
X400
_ iii 10 u X -u b
x4e log(x/-b)
= i:; log u(x) .
log x
x4e
Similarly,
11,“ M = lim £5.22). .
"— mod H —" log x

X4” x-Ocn

79

 

 

 

 

Thus
fl)- *
K(f*) ngm SKCE):
(4.11)
1 log u(x) *
K(f*) S l_i__m log x S K(f )'
x-m
On the other hand. since P(l/l) = l/P(x), we have as
x approaches 0+
mod H -—.— P(X/-b)
lim ————;' = 11m
x 0+ 11106 H x40+ P(u(X)/ u(b))
1—.- PL—u(b)[u(x))
= 1m P(—b/x)
x40+
:11?“ 1+ -ub ux lo -ub ux
X40+ 1+ e (-b/x) log (-b/X)
= ET. __9.__LL1‘;O ‘1 x .
x40+ g x

Similarly,

lim modH ___ 1im log u(x) .
"—4- mod H _1_ log x

 

 

x40 x-OO
Thus
1 Sm lo ung(f*)
I
K(f*) ”.0... log x
(4.12)
l . log u(x) *
K(f*) 3,23%. log x S K(f )°

80

Of course we could just as easily have used b > O and

x < O. In this case we would have ended up with x 4 - m
and x 4 0’ instead of x 4 a and x a 0+ in (4.11) and
(4.12), respectively. This completestfluaproof. As we shall
soon see, equality holds when u(x) = (sign x)|x|a, a > O.
Equality holds in the lower bound for o<qg1; in the upper

bound for a 2.1.

Corollary 4.6. Under the same hypotheses as for

Theorem 4.6,

—l g min( lim X , xlim M. lim x . lim —‘(-J-}.
K + u(x) 40- u(x) (x)
u x40 x+m xq-o

I I
K >_ max{ lim M, 11m '——(—L . lim xu x
u x + ) - u(x )

u x x
40 x*O x4» x4-

whenever any of these limits exists. Both tgungs are sharp:

2399:: Since all the cases are similar we will treat
only x 4 0. Assume
lim xu'(x)/u(x) = A
xaa
exists, where A can be finite or infinite. This implies
that xu'(x)/u(x) is actually defined on some interval
(a,m). But we know that x and u(x) exist on (a.w).
Thus u' must also exist on (a.o). That is. u is differentiable

on (a.«). Thus we can apply l'HSpital's rule to the ratio

81

log u(x)/log x to obtain

0 I ' O .

11m log u(x) = 11m u (x)(u(x) = 11m xu (x) .
log x 1/x u(X)

x4» x+m x4e

Equality again holds for u(x) = (sign x)|xla. a > 0.

Theorem 4.7. Let u Qt OS on (-m,m) and RB

on both (-o.O] and [0.m). Let

I I
B=limxux, c=1im"“",
x40+ u(x) x-oO- u(x)

I I
D=1imxux, E=limxux.
x4” u(x) xe-o u(x)

and put

1 1 1 1
A = max{B. 3' c. 3' D. 3. E, E}.

gt, A 2_Q.'where Q is defined as in Theorem 4,1, then the
radial extension fu is extremal for u, and hence

K = K(fu) = Q.

Proof: By Corollary 4.6 we must have Ku 2_A. Thus
K(fu) 2_A. If K(fu) = Q gLA, however. then by the definition
of Ku we must have K(fu) = Ku' Hence the radial extension

fu is extremal.

Theorem 4.8. Let u be Q§ on (-o,m). odd, and

RB on [0.6) with ratio bounds L and M. Let Q = max{M,l/L}.

If any one of the four conditions

82

(1) X131 *3}; = 0.
(iii) ii: x31xf = 0.
(iv) lim X3}; .= 215-,

xam
holds. then the tadial extension fu is extremal.

Proof: The proof follows immediately from Theorem 4.7

since any one of the four COnditions leads to A.2 Q.

ggrollaty 4.8. Let u be 98 on (-m,w) and odd.

‘Ihgg_ u satisfies the hypgtheses 9f Ihegtem 4.8 if and gnLy

'f u is of the form

l-X(s) s + i arg x). x a" O. u(O)

IXI
(4.13) u(x) = expkf Ell—8')— fig 0:
1

where X is real-valued and measurable on (O.m).
sup |X(s)l < l. ess sup |X(s)l = gi% < 1.
O<s<~ 0<8<co

and either

. u -l .. . Q-l
(1) 11m IX(s)| = Q‘— or (ii) lim |x(s)l =
8‘0... 0+1 84a: 0+

Proof: We know from Theorem 3.1 that u is RB on

[0.o) if and only if it has the form (3.3). But (3.3)

83

and (4.13) are identical on [0,m). The term i arg x
in (4.13) just extends the function u symmetrically about

the origin. Since

1+ 5 = su' s
1-X(S) 11(8)

for such an RB function we see that (i) of Corollary 4.8
corresponds to either (i) or (ii) of Theorem 4.8, while
(ii) of Corollary 4.8 corresponds to either (iii) or (iv)
of Theorem 4.8.

n c.
Example 4.1: Let u(x) = (sign x) ZEaix 1 where
i=1

ai 2,0. ci > 0. the a1 and ci not necessarily integers.
n

and 2231 = 1. We obviously may assume for simplicity that
i=1

c1 < c2 < ...< Cn' Then it is easy to see that u is odd,

u(O) = O. u(l) = l, and u is continuous and monotonic.

Now for any x > O.

 

n C
Z}a.c x 1
xu' x _ l
u(x) n ci '
ZDaix
1
Thus. since
n ci :1 ci :1 c,
c Za.x g Zc.a.x _gc Ea.x 1 ,
1 i i i n i
l l 1
we easily see that
xu' x
L — c1.g u(x) ‘g cn — M.

84.
Thus u is RB on [0,m). Also

lim ' X - c lim xu' x - c
_ O - l
x-m u (x) n x + u (x) 1

so that one of the conditions of Theorem 4.8 must be
satisfied -- condition (iii) if cn 2_l/c1. condition (ii)

if cn‘g l/cl. Hence the radial extension is extremal.

Theorem 4.9. Let u satisfy the hypgtheses of

 

 

Theorem 4.7 so that the radiaifextension is extremal, and

let Q(x), 6 and Q be defined as in Theorem 4.1. If there

 

 

is some 6 > O and somefiinterval (c.d) c (-w.°) on which
Q(x) g 6 - 9 then the radial extension fu is not unique

extremal.

Proof: As in the proof of Theorem 4.1, if we use
the auxiliary conformal mappings e2 and log 2, then we

need only consider functions of 0 onto O' where
Q = {x+iy‘O<Y<F] and Q' = {S+iT|O<T<W} .
The boundary conditions then become
(4.14) x + Oi -—€> U(x) + Oi, x + vi -—e> V(x) + vi,

for -m<x<w and O<y<W. Furthermore the radial extension

fu then corresponds to the function fu:0 -—€>Q' given by

85

fu:x + iy —-> (Ki-Z U(x) + f V(x)) + iy.
Let W(x) = (U(x) + V(x))/2 + U(x) where 0 20

is a function to be defined later in the proof, and let

g:Q -€>Q' be defined as

[13:21 U(x) +11} W(x)] + iy if O<y<1r/2
g(x+iy) =

[12:31 W(X) +2371". V(x)] + iy if 1r/23y<1r.

Obviously 9 satisfies (4.14). Furthermore it is clear that
the proof of Theorem 4.1 can be used here if the cases
o<ygw/2 and v/23y<v are treated separately. Hence g is
QC.

In order to calculate K(g) we will use the fact that

K(g) = ess sup D(z). where D is the point dilatation of
z 6 0

g on Q [3. p.24]. Moreover, since the cases O<ygr/2 and
v/zgy<w are similar, we will treat only O<ygn/2. Let

xo 6 (c.d) be a point at which U and V are differentiable,
and consider D(xo.y) + l/D(xo.y) for O<y<W/2. As was

shown in the proof of Theorem.4.l, D + l/D must attain its

maximum at one of the endpoints y = O. y = W/Z. But

1 l+(h(x;)+20(xo))2
D(xo.0) + _—_D(xo.o) = U' (x0) + U' (x0)

86

l+h(x )2 4h(xo)o(xo)+40(x0)2

 

 

 

_ . 0
_ U (x0) + U' (x0) + U' (x0)
2
A 4h(x0)o(xo)+4o(xo)
,S Q - e + U'(xo)

and

l U' (x0) +V' (x0)

_ 1
D(XOOW/Z) - 2

D(xo.F/2) + +1r0'(x0)J
l+(h(x >+2o(x ))2
o o

(U'(xo)+V'(xo))+wc'(xO) '

 

+

Hence, if we choose 0 to be positive but sufficiently small
on (c.d), and with a sufficiently small derivative a.e. on

(c.d). then

___l___. I _r_JL__. .____1____
D(xo.y) + D(xo'y) g max[D(xO'O)+D(xo.O)' D(xO'F/2)+D(xo.1r/2)}S6

for all y with O<¥Sfl/2. A similar result holds for
v/Zgy<v. Choose 0 in this way on (c.d) and let 0 = O
everywhere else. Then g(x,y) = fu(x.y) for all (x,y)

with x £ (c.d). Hence

A .
D+-gQ a.e. in Q.
and so

K<g>.s o = K(Qu) = xu.

Hence 9 is also extremal. But 0 > O on (c.d) implies

that g # fu on (c.d)x(O,v). That is, @u is not unique

extremal.

 

 

III" I)! I

 

87

The statement of this theorem is much simpler in the

case where u is an odd function. In fact the result becomes:

Theorem 4.10. Let u atigfiy the hypotheses of

Theorem 4.8. so thgt the odd radial extension fu is extremal
and Q = max{M,l/L}. If there is some 5 > O and some

interval (a,b) c (O.w) on which

(4.15) _1_. S £13.99.

0%: u(x) 59“”

then the odd radial extension is not unique exttemal.

Proof: Suppose Q = 1. Then M = L = l as well

 

and thus Q = 1 = xu'(x)/u(x) a.e. on (O.w). That is,
(4.15) could not hold on the interval (a,b). Hence we can
assume that Q > 1. Pick e in (4.15) small enough that

Q - e > 1. Since the expression x + l/x is increasing for

x > 1, l<Q-e<Q implies

.1. _ _1_._’\
(Q + Q) (Q a + Q-e) — e > 0.

Since 11 is odd we must have h = 0 (Cf. (v), page 69.) and

hence 6(x) = U'(x) + l/U'(x). Furthermore. by (4.15).

l xu' x

Q-e S u(x) = U. (x) S Q-e'

so by the monotonicity of x + l/x we must also have

. 1 _ _1_
U (x) +U'(x)‘SQ€+Q-e°

 

88

Finally. by definition of 6 and Q. 6 Q + l/Q. Putting

all this together we find

 

.S Q-e +'—l‘ < Q + - 2 = 6 - G

A . t
Q(X) = U (X) + Q_€ Q

1
W (X)

for all x 6 (a,b). The result follows immediately from

Theorem 4.9.

Corollary 4.10. The condition for non-unigteness in

Theorem 4.10 is satisfied if there is any pgint xo 6 (O.o)

at which u' is continuous and

xou'(xo)

.1
‘1929 # Q or Q .

Proof: Since u' is continuous at xb. and x0 > 0.

also xu'(x)/u(x) is continuous at x0. If xu'(x)/u(x) is

either > Q or < l/Q for x = x0, then the same must be

true in some small interval containing x0. But this is
impossible. since Q = max{M.l/L}, where L and M are

the ratio bounds of u. Thus we must have

i. x u'(x )
'<'Jl———£L'<TQ.
Q u (x0)

on the entire interval [xo-t. xo+t]. By continuity the

function xu'(x)/u(x) must attain its maximum and minimum

89

on this compact set. Hence there is some 6 > 0 for which

_stso_e

u(x)

on (xo-t/Z. x0+t/2) c [xo-t, xo+t]. We can now appeal to

Theorem 4.10.

Example 4.2: Consider again the function
11 c1
u(x) = (sign x)23a.x
i
1
introduced in Example 4.1 on page 83. If n > 1 then the

radial extension is not unique extremal for this u. The

function u is obviously odd. and if n > 1 then

n n n
clflai Zaic. ana.
L_ ____1__<_1.:21.Lll__l___1.<_l_1__c =M
- c1- n u(l) -' n n - n '
23a 236. 233
1 1 1 1 1 1
so that
;_ l u'(l)

where Q = maxfcn, l/cl]. Also u' is clearly continuous

at x = 1, since u(x) is a linear combination of powers

of x. Therefore the hypotheses of Corollary 4.10 are fulfilled
by u. and we conclude that the radial extension is not unique

extremal.

Theorem 4.11. Let u be a 98 function of [-l,1]

WWW L1. M1. L2. M2

90

such that
O < L1.g xu'(x)/u(x) 3 M1 < m a.e. on (0.1).

O < L2{g_xu'(x)/u(x)‘g_M < m a.e. on (-l.O).

2
I I
Let C = lim_ x303: and D = lim w. and put
3‘0 x40+
1 l
A = max{C. E’ D. B}.

A
_g_ A + l/A 2 ess sup Q(x). where Q(x) i§ dgfined ag in
-e<x<0
Theorem 4.1. then the radialpgxtension of u 'tg

 

G = {zIIzI < 1. Im z > O}. (i.e. the restriction of fu in

(4.1) to 0<r<1). is extremal for the boundaty values

u(z). for z = x 6 [-l.l]
f(z) =
z. for [2] = 1. Im z 2 0.

Proof: Let g be any other QC map of G onto

 

itself with boundary values f. Let fu denote the radial
extension of u to G. and let K(f“) and K(g) denote the
QC dilatations of fu and g. respectively. By extending
both fu and g by reflection in the semicircle I2] = 1.
Im z 2_O we obtain QC mappings 9n and 8. respectively.
of the upper half plane onto itself. with dilatations

K(fu) = K(f“). K(e) = K(g). But it is easy to see that if

fu satisfies the hypotheses of this theorem on [-l.l]. then

91

the reflection fu satisfies the hypotheses of Theorem 4.7
on (-a.m). Therefore K(gu) is minimal for the boundary
values of f“ on (-o.m). But 9 = fu on [-l.l] implies

S = f“ on (-o.m): so we must have K(g) = K(e) 2’K(?u) = K(fu)'

Hence fu is extremal as a map of G onto itself.

Corollary 4.11. Let u satisfy the hypotheses of
Theorem 4.11. so that the radial extension is extremal.

If there is some 6 > O and some intetval (c.d) C'(-°.0)

 

on which Q(x) < Q -6. where Q = ess sup Q(x). then the
O 0 -w<x<0

radial extension is not unigge extremal.
Proof: The proof is basically the same as the proof

of Theorem 4.9.

Theorem 4.12. Let u be an odd Q§ function of
[—l.l] onto itself and suppgse there are numbers L and

M such that

xu' x

O < L‘s u(x)

gm < co a.e. on (0.1).

Let Q0 = max[M.l/L}. If either of the two conditions

. . xu' x
(1) ;::1 u(x) - QO'
.. xu' x 1
(ii) lim = '—
xw+ u(x) Qo

holdg, then thg radial extension of u to a map of G. the

92

uppgr half of the unit diskI ontg itself, is extremal for the

boundagy values it assumes.

Proof: The proof follows immediately from Theorem 4.11.

Corollagy 4.12. Let u satigfy the hypotheses of
Theorem 4.12. so that the radial extension is extremal. If
there is some 6 > O. and some interval (a.b) C’(O.1) ‘95

which
fi.s'--1-L.g QC-

0- u(X)

0

then the ragial extension is not unique extremal.

Proof: This theorem follows immediately from Corollary 4.11.

Theorem 4.13. Let u be an odd Q§ function of
[-1.1] onto itself. Then u satisfies the hypgtheses of

Theorem 4.12 if and only if u is of the form

le
W, = ..pr M92

1-X(8) s + i arg x).

o # x e [-1.1]. u(O) = o.
where x is real-valued and measurable on (0.1).

Qo-l

QO+l

 

sup |X(s)l < 1. ese sup |X(8)l =
0<s<1 O<s<l

< 1.

Q -1
11113.. |X(8)| = .2...

s40 QO+1

93

Proof: Except for slight modifications. the proof is

 

the same as that for Corollary 4.8.

S. An application of the extremal radial extension.

It is well known [11. p.411] that in the class of all
QC mappings of the upper half plane onto itself satisfying
a finite number of boundary conditions w(zi) = wi. i = 1.....n.
the extremal map has constant point dilatation D(z) = K in
the upper half plane. It is also well known. and quite evident
from Theorem 4.7 and Theorem 4.8. that when the entire
boundary correspondence u is specified. then an extremal
mapping need no longer have constant point dilatation. As
a side result we will use Theorem 4.8 to show that the extremal
mapping need not have constant point dilatation even when a

countably infinite number of boundary conditions is specified.

Theorem 4.14. There exists a mapping w' defined on

the non-negative integers, and an extremal 99 extension
f* ._§ 'w to the uppgr half plane. such that f* has non-

constant ppint dilatation.

Proof: Let ‘w(n) = min2 for n = 0.1.2.... and let

u(x) = (sign x)(|x| + |x|2) for all x e (-~.¢). Clearly

, 2
ISM-&=Lfl.¥—S2 for xe‘o'a)'

u (x) " x+x2 l+x

and

1511;451:2'

lim u (x)

x4e:
Hence condition (iii) of Theorem 4.8 is satisfied and so
the radial extension fu of u is extremal with K(fu) = 2.
But fu(n) = u(n) = w(n) for n = 1.2..... so that fu is
also an extension of 'w. Moreover. if g is any QC extension

of w then

 

 

 

log g(n) _ log(n+n2) _ tpg n2 + log (1+1ZQI = 2 + log (1+1Zn)

log n -' log n log n log n
. lo n
shows that lim log n = 2. Hence. by Theorem 4.6. K(g) 2 2.
nae

That is. K(f“) g K(g) for all QC extensions 9 of w. Thus
letting f* = fu we see that f* is an extremal extension
of w and f* has non-constant point dilatation since

9

D(rei ) = (l+2r)/(l+r).

6. General radial mappings.

Locking back at Theorem 4.5. we see that parts (ii)
and (iii) give a characterization of a general odd radial
extension--part (ii) by the explicit formula for the complex
dilatation and part (iii) by the explicit form of the function
itself. Notice. however. that nowhere in Theorem 4.5 is the
function X assumed to be real-valued.

In most of our applications of the radial mapping
we have taken x to be real-valued so that the half-lines
(-o.0] and [0.w) would be mapped by the radial mapping

f onto themselves. This was necessary if we wanted a map

95

of either the upper half plane onto itself or the upper half
of the unit disk onto itself. Theorem 4.5 continues to hold.
however. as we see in [12]. even if X is complex-valued.

We can therefore still talk about the radial mapping in this
case. although it is easy to see from (iii) of Theorem 4.5
that if x is not real-valued in [0.m) then f 'will not
map [0.m) onto itself. The most important property of the
radial mapping continues to hold even if x is complex-

valued. however.‘ This is property (iii) of Theorem 4.5.

= e2 iarg z

X(z) X(|Z|),

so that

|x(z) I = x(IZI)-

Property (iii) shows that we can find the QC dilatation
K(f) of the radial mapping just by investigating the point
dilatation D(z) on the ray arg z = 0. and taking the
essential supremum of these values. This certainly saves
a lot of work and is what enables us to find conditions under
which the radial mapping is extremal for the boundary homeo-
morphism it induces.

Since we are no longer considering the radial map as
the extension of a given u on (-o.m) to the upper half
plane. we can no longer denote it by fu. Henceforth we will

consider a map to be radial in the general sense if it satisfies

96

parts (ii) or (iii) of Theorem 4.5. in some domain 0 with

(0.0) E 0. but not necessarily satisfies part (i). We will

denote such a general radial mapping by f*(z) or f*.

We are now ready to continue our study of extremal radial maps.
Up to now. as can be seen in part (ii) of Theorem 4.5.

we have been restricting our investigation of extremal radial

maps to those having complex dilatation X(z) = x(rele) of
the form
(4.16) X(rele) = e21eX(r).

where X is real-valued and measurable on its domain of
definition. We now generalize this somewhat by considering

instead complex dilatations of the form

ie _ Zie g(r)-l+ig(r)
(4.17) X(re ) — e a(r)+l+i8(r) . O<ngl. QSG<2Wo

with a and B real-valued and measurable on their domains
of definition. If B(r) = O identically then (4.17) reduces
to (4.16) as a special case with X = (a-l)/(a+l). The

following lemma will prove quite useful.

Lemma 4.3. Let f be a QQ mapping of the open

unit disk A onto itself with f(0) = 0. Then

K(f) 2_max[ili

10 fr ff- logr }
r40+ log(r/lJi r431 log (|f(r)|/16) °

97

Proof: It is well known [13] that for any function

f satisfying the hypotheses of Lemma 4.3.

(4.18) |f(zl)-f(zz) | g 16|zl-zzl

for all 21.22 6 A. where K = K(f) is the QC dilatation.
Taking 21 = r 6 (0.1] and 22 = O in (4.18) gives

|f(r)| g 16r1/K and solving for K we find that

K‘Z log r/log (|f(r)|/16). which implies
‘-F’ -1 . .
K >_ 1111 log r/log (|f(r) |/16). Now f also satisfies the
r40
hypotheses of Lemma 4.3. Hence we can use f-1 in (4.18)
with 21 = r 6 (0.1] and 22 = 0. Again solving for K
gives K 2_log |f(r)|/log (r/16). which implies
K 2 linl log [f(r) I/log (r/16). This proves the lemma.
r40

Theorem 4.15. Let

ie _ 2ie g(r)-l+i§(r)
x(re ) — e a(r)+1+i5(r) for O<rgl. qge<2v.

where u(r) and 8(r) are teal-valued and measurable gn

 

(0.1]. and satis the followin conditions:

(i) O<A$a(r)$B<c. |B(r)|gc<co for some A. B. c.

(ii) a is cgntinuous in somg right-hand neighborhood
(0: M M

(iii) I linl u(r) = a0 2A>O existg.
r40

98

f* be the ragial mapping of A onto itself that leaves

° IE

and 1 fixed and has x as its complex dilatation. If

(4.19) ess sup |X(r)l S I; :Io
o<r$1

then f* is extremal in the class of all QC maps of A'

 

onto f*(A') that agree with f* on the boundary. Here

A' denotes the open unit disk minus the closed interval

[0.1] so that EN = [2| [2]: 1} U {z = x+iy|0gx$l. y = 0}.

Proof: Let g be any QC map of 11' onto f*(A')
that agrees with f* on the boundary of A'. Then. by (iii)

of Theorem 4.5.

g(r) = f*(r) = exp”: fig) = exp(- I: (a(s>+iB(s))—)

for r 6 (0.1].

Hence

1 ds
I9(r) I = epr-[ra(s) '2').

But by (ii) and (iii) of Theorem 4.15 we can use l'HSpital's

rule to find

-:[ u(s) 9:

lgg |g(r)|

1m.)- log (r/16) 3 I113. log (r/16)
r40
= 11a}r Elf-1; = 111:; a(r)
r40 r40
= a

99

and

lim log r = li log,r
+ 109 (I9(r) I/16) “I 1
r40 r40 ‘gg
-[ u(s) S-log 16
r

 

 

_ . 11: __ . l
_ :31 u(r)/r 1' r133- a(r)
__ A.
_ a0 .
Hence. by Lemma 4.3 and (4.19).
x
1+I(a0-1)/(ao+l)l 1*egerEP I (1"
= _-_- 'I: .
mg) 2 1r11"“"I"‘o' 1/‘101 1-|(ao-l)/(ao+l) | >- l-ess sup |x(r)| “1 I
O<qgl

That is. f* is extremal.

Corollary 4.15. Let

9)

)((re1 =X(r)e2:"e for 0<rgl. 0ge<21r.

where

(i) x(r) is real-valued and measurable on (0.1].
(ii) sup |X(r)l < 1. ess sup IX(r)I = M < l.
O<rgl 0<rgl
(iii) x(r) is continuous in some right-hang neighborhood

(0.1) of the origin.

(iv) 1' xm exists.
r331 —"

Let f* be the radial map of A onto itself with 0 and

1 fixed and with x as its complex dilatation. If

100

th Ix(r)| = M then f* is extremal in the class of all
r40
99 maps of A' onto f*(A') that agree with f* on the

boundaty.

Proof: The proof follows immediately from Theorem 4.15

if we take B(r) = o and u(r) = (1+X(r))/(1-X(r)).
We shall need the following result of T. Sasaki [16].

Theorem of §asaki. Let f be a QC map of the open

unit disk onto itself with f(0) = O and f(1) = 1. Then

'1-' gtg f(r) .; _ 1
11:1 log r IS 2 [11(1) K(f) °

Thegrem 4.16. Let

ie _ 2ie Q(r)-l+ifl(r]
x(re ) — e a(r)+l+iB(r) for O<rgl. oge<2n.

where u(r) and 8(r) are real-valued and measurable on
(0.1] and satiety the following conditions:

(i) O<Aga(r)_§B<~. o<ngs(r)gc<o for some A. B. C. D.

(ii) B(r) is continuous in some right-hand neighborhood

(0.1) of the ori in.

(iii) lira. 8(r) = 60 2D>O eggists.
r-oo

Let f* bg the gagigl pgpptgg gt A ggtg itgglf that leaves

0 and 1 fixed and has X ag its complex dilatation. If

101

 

 

(4.20) ess sup . r - ;o(r) +1 + r .3 BO

O<rgl

then f* is extremal in the class of all QC maps of
13' t9 f*(A') that agree with f* on the boundary.

Proof: Let g be any QC map of A' onto f*(A')
that agrees with f* on the boundary of A'. Then. by

(iii) of Theorem 4.5.

r r r
.. .. ,_. film-1152s ._._. Mel. . £18.).
g(r) - f (r) expk'fllrxw) s) expu‘1 s ds+ ijl 3 ds)

for r 6 (0.1].

r
so that arg g(r) = I a{Elam But by (ii) and (iii) of
1
Theorem 4.16 we can use 1'H3pital's rule to evaluate
Ir s .
'QfirLds
lim 1

:rKfI log r

H I-11‘1-‘1111a I
r451 log r

M21411-

1i — li B(r)
ragg' Lfir rag)

Bo -

Hence. using the Theorem of Sasaki and (4.20). we find that

102

.r-l +..r2 W+s

 

(4.21) '*[K(g)- _] 2_B ess sup
K (19) o2 o (1, $1 2am
1+ess sup |x(r)| l-ess sup |x(r)|
_ ;,ogrg1 ogrgl
- 2 l-ess sup IX(r)I 1+ess sup |x(r)|
0<t$1 O<rgl
l l

 

= 311111”) " K(f*)] °

Since x - l/x is a non-decreasing function of x if
x > O. (4.21) implies that ‘K(g) 2 K(f*). Hence f* is

extremal.

Corollary 4.16. Let

' M).
19) - e219 1 1+1 for «:31. oge<21r.

x(re - a(r)+l+i8(r)

‘with g(r) = v/B(r)2+l. and such that
(i) B(r) is real-valued and measurable on (0.1].
(ii) B(r) is continugus in sgme right-hand neighborhood

(0.1) of the origin.
(iii) o<ngs(r)gc<e 39; r e (0.1].

Let f* be the radial map gf A onto itself leaving 0
and 1 fixed and having X as its complex dilatation. If

(4.22) r1111; B(r) 2 M =-- ess sup 8(r).
0<rgl

r

 

I III. Ill-[III In. III III II II

103

then f* is extremal in the class of all 99 maps of A'
ontg f*(A') thgt agtee with f* on the boungaty gf A'.
Moreover. if (4.22) holds. then K(f*) = M + M2+1.

Proof: The relation g(r) = V’B(r)2+l in the hypothesis

of Corollary 4.16 implies that

 

 

2 2 2 2
ess sup r -1 + r r +1 + r = ess sup 8(r) = M.

0<rgl 20(r) O<rgl
Hence (4.22) of Corollary 4.16 implies (4.20) of Theorem 4.16.
All the other hypotheses of Theorem 4.16 are obviously
satisfied. and it follows that f* is extremal. To find
K(f*) we notice that with the given relation between a and

B. we easily get

2 2
° 1111:..111 W
lx(reieH = I r 1+1 r r l + r

 

 

 

 

a(r)+l+iB(r) ’ v/1a(r)+1)2+a(r)2
= :Z (g(r) -1) 2ig(r) 2-l ,1 g(r)-1
v/Io(r)+l)2+o(r)2-1 a r +

Hence

nd_l___fii_l
D(rele) __.. gill—(3.4.1. = M = r +1+ r _._ 0.0“ + g(r)

l-|x(re‘16) l 1,112.11. g(r) +1-am

a(r)+1

104

and so

K(f*) = ess sup D(reie) = ess sup (a(r)+B(r)) = V’M2+l + M.
O<rgl 0<rgl

CHAPTER V

ADDITIONAL EXTREMAL EXTENSIONS?
A GENERALIZATION OF A RESULT OF REICH AND STREBEL

Although we now leave the study of radial maps. we
are far from finished with the study of extremal QC mappings
with given boundary homeomorphisms. We continue with the

following definition.

pefinitigp 5.1. A function U mapping [0.«) onto

itself will be called linegr tgdial. gr LB. if
U(x) = log u(ex) for all x6[0.¢)
where u is any function satisfying the conditions

(i) u RB on [0.c) with u(l) = 1.

(ii) O<Lgxu'(x)/u(x)gM<o a.e. on (l.o).

(iii) lim‘ffigLLEL— = Q = max [M.l/L].

x u (x)

emar : Since an explicit formula is given in (iii)

105

106

of Theorem 4.5 for a function u satisfying the above condi-

tions. we know that U is LR if and only if

x
_ x__mxie1e_e .
U(x) — log u(e ) - ' 1-x(s) s . x6[O. )

where X is a real-valued. measurable function on (0.9) with
-l
sup |X(s)l<l. ess sup IX(s)|g_g:I'.
O<S<a O<S<O
and
- _ 9:1
lim IX(s)I — 0+1
s-Oco
by condition (iii) of Definition 5.1.
By its definition U is a continuous. strictly
increasing map of [0.«) onto itself with
0<L§U'§M<w a.e. on (0.~)
and
lim U'(x) = Q = max [M.l/L}.
x40
Lemma 5.1. Let U be LR on [0.0). Then for each

e>0 there exists an N(e)>0 such that

(l-e)Qx$U(x) for all x>N(e).

107

Prggf: Consider

 

 

 

 

 

1
Q(x) = ' ds
Qx Qx
Since lim U'(x) = Q. there is some Q>O such that x29
x-Om
implies that U'(x)}Q-é. where Q = 06/2. Therefore
{2 x
[X] = IOU (s)ds + .121 U (s)ds
Qx QX 0x
x x A
IA U' (s)ds IA (Q-e)ds
2 X 2 X
QX Ox
’3 x-Q
= (l- a) (T) .
_A
Now since lim x X = 1. there exists an N(e) for which
x-Oco
Q x-Q Q 9
(l-Q)(—;—)2(l- 6) - 6': l-e for x>N(e).
Hence
uéitzl-e for x>N(e).

and the lemma is proved.

rem 1. Let h be a normalized RB function on
[0.») with ratio bounds L. M. where 1<LgM. Let G .23
the dgma in

G = {z = X+iy I -~<X<°. y>h(IXI)}.

108

W
F(z) = F(x+iy) = (sign x)U(Ix|) + iy
12. G. whare U is L3, If G' = F(G) then F i§ extremal
in tha glaaa at aLI QC mapa tram. G (t9 G' a a r wi
F on dar f G. Matagver.

K(F) = Q = max [M.l/L}.

(See Figure 5.1).

 

y " h(x)

 

 

 

 

Figure 5.1

2119.92: Let f:G->G' be any K-QC map of G to G'
that agrees with F on the boundary of G. Now choose e>0.
Then by Lemma 5.1 there exists an N(e)>0 such that
U(x)}(l-e)Qx whenever x>N(e). Let yo = h(N(e)). Then for

any n>yo. we have h-1(n)2h-l(yo) = N(e). so that

109

-1

_ _ h (n)

(5.1) 2(l-e)Qh 1m) _<. 2110: 1m) s L(n) =1 _1 Ifz+f§|dg .
h (n)

for any n > yo. where L(n) is the length of the f—image

of the segment

yn = {lem z = n. -h-1(n) S,Re 2.3 h-1(n)}.

(See figure 5.2)

m)

GI

 

 

 

 

 

 

 

 

 

Figure 5.2

Integrating (5.1) with respect to n from 0 to y for

any y > yO gives

_1 y b 1(7))
2(1-emjyh (man s] L(mdn g [’1‘ lfz+f§|d§-
YO YO O -1
-h (n)

110

Squaring and applying the Schwarz inequality gives

(2(1-e)oj:h' 1(T1)d1'1)2 g (I:L(mdn)2 g Ldegdn ”lg—I- dgdn .

1- N2

H)

H}
N In:

where J(z) = IfZI2 - Ilez and x(z) = are the Jacobian

and complex dilatation. respectively. of f. and
<3y = GnIZIIImz <y}.

Clearly
x x

U(x) = I U' (s)ds g! Qsds = Qx for any x >_ O.
O 0

Hence. if 6(y) 2_O is the maximal upper deviation of f(yy)

above the horizontal line Im z = y. i.e. 6(y) = sup [Im f(x)-y}.

26
Yy

then it is easy to see by considering the relevant areas

(cf. Figure 5.2) that

+6<y> _1 +6(y)_1
JIJdgdn g 2} U(h (mm?) g 20f h (mdn
Gy o o

and. using the same reasoning as on page 354 of [14].

Y -1
LII—ELL :dgdn g ZKJ h (mdn.
1-|x|20

where (K-l)/(K+l) = ess sup |X(z)l. Therefore

266
Y

y _ y _ +6(y)._
(Zn-emf h 1(n)dn)2 g 41(th 1(n)dnjr h 1(11) dn
0 o
0

or

111

 

y _1 +6(y)_1
[h (mdnfy h (mdn
(5.2) Q 3‘41"; 0 L -
(1-6) ([yh-1(n)dn)2
yo

If we can show that the term in brackets in (5.2)
approaches 1 as y tends to m. then we will have Q.g K/(l-e).
After this we can let 6 tend to O and achieve ng,K. from
which it will follow that F is extremal for its boundary

values. For convenience we write

_ +6(y)_ _
[ya 1mm]y h 1(n)dn [yh 1(n)dn

 

 

 

 

(5.3) 0 ° = 0

Y -1 2 y -1

(J h (man) Ln (man

yo 0
y _ y+6(y)_
[hlmdn I h1(n)dn
o y

+ Y o

[Yb-lmmn L h'l(n)dn
yo 0

In order to simplify calculations. we let h-1(n) = g(n) in
the rest of the proof. Now by Theorem 3.5. since h is an

RB function with

(5.4) 1<Lg32—2-SI1-gm,

it follows that g is also RB with

112

 

Clearly
y Yo
[9mm] I g(n)dn
lim —9——— = 1 + lim 0 = 1
y-IQ Y y-ooa Y
Lem)“ I g(n)dn
0 yo

and. using (5.4) and integration by parts.

+6 (y) y+6 (y) y+6 (y) Y+6 (y)
[y 90an S M] 719' men = M11900] - MI g(n)dn .
Y y Y

Thus
+6(y) y+6(y)
(M+l) [y g(n)dn g MTIgUD]

y y

or

+6 (y) M
[y g(r)) dn 3 gfl (y+6 (y))g(y+6 (y) ) -y9 (y) 1.
y

Similarly.

._L_ _
£23de 2 L” [yg(y) yogwo) ]-
0

Therefore. if we let C = M(L+l)/L(M+l). then
y+6 (y)

 

 

I g(n)dn ))

y (y+6(y q(v+6(Y))tYQi!l-
(5.5) o g S C yg (y) "Yog (yo)
i: g(fi)dn
o

yoghro) Y 9‘1”

1_ _—

y9(y)

But by Theorem 3.2
..1. .1
q(v+5(y)l_ y+6(y) L _ g(y) L
g(y) 31y I-11+y)°

 

113

Hence (5.5) gives

 

 

+6(y)
[Y g(n)dn L+l
y C 6(2) L _
(5'6) 0.3 .g 1-(yogurovygw))L1l ) 1] '
£f9(n)dn
o
1/L

Now for any n > 1. Theorem 3.2 shows that g(n) g n . l/L < 1.

Hence. the proof on the top of page 355 in [14] that

lim - 0 still goes through and so
lim (1-+§€f1)1+1/L-1 = 0.

Moreover. since

C

11“ 1—(yog(yO)/yg(y))

y-Oo:

=C'

we conclude from (5.6) that

+6(y)
[y g(n)an

lim ”y = 0.

yI” [yg(n)dn
Yo

Hence. by (5. 3) a

_ +6(y)_
[Yh 1(n)dn [y b 1(n)dn

lim 0 0

1"” (Eh-1(7)) c111)2
0

so that Q S K.

1.

Therefore any QC mapping f:G-—€>G' agreeing with

F on the boundary of G must have QC dilatation K(f) 2_Q.

We“ _0‘; huh-m

I F .51! nevi;
J

114

But for the original function

U(x) + iy if x 2.0
F(x+iy) =
-U(-x) + iy if x < 0.

it is clear from (4.7) in the proof of Theorem 4.1 that the

point dilatation of F is

l

= U' (IXI) 'I' U. (|XI)

D( +i ) + ——-l‘-—'
X y D(x+iy)

wherever U' exists (i.e.. almost everywhere). Thus. since
O<LgUKgM<w. D(x+iy) glmax[M.l/L] = Q. But in fact. by taking

Im 2 large enough. we can find points in G with arbitrarily

large real part x. and hence lim U'(x) = Q implies that
xaa

K(F) = ess sup D(z) = Q g K(f).
ZEG

That is. F is extremal.

Remark: If. in Theorem 5.1 we take h(n) = ”L. L > 1.

and U(x) = Qx. Q 2_l. then we get the extremality results of

Reich and Strebel [14] as a special case.

As before. the question of uniqueness in Theorem 5.1

is partially resolved by the following corollary.

Corollaty 5.1. Let U and h satisty the hypgtheses

of Theorem 5.1. so that the map F is extremal for the boundary

115
values itgaaapmes. If there is some 6 > 0 and apmeyantepyal
(a.b) c (0.m) on which
—l— g U' (x) S Q-eo
Q-e
tpap_ F is not unique extremal.
-1

Proof: Let yO be so large that h (yo) > b.

Define a map g:G-—€>G' as

 

 

 

 

f ZYO-y y-y
I ( y U(x)+—y—Q W(x))+iy if yogyg2yo
o o
Bye-y y-Zyo
. = . .f 3
g(x+1y) I< Yo W(x)+—;o—U(x))+1y 1 woe/s Yo
\F(x+iy) if y I [y0. 3Y0]:
where W(x) = U(x) + U(x) for x 6 [0.m) and some function

0 2.0 to be picked. It is clear that between y = y0 and
y = 2yo. g is a linear combination of U and W. and the
same is true between y = 2yO and y = 3y0. Hence we have
the same kind of mapping as in the proof of Theorem 4.9. The
proof here is almost identical to the proof of Theorem 4.9
and it is easy to see that if we choose 0 so that U(x) = 0
for x t [a.b]. U(x) > o for x e (a.b). and both a and
Io'I are sufficiently small on [a.b]. then D(x+iy) g Q

for all x+iy with agxgb and yogygBYO. Since 9 = F

everywhere else. this implies K(g) = Q = K(F). Hence 9 is

116

also extremal. It is also clear that g(z) = F(z) on the
boundary of G but that g is not the same as F everywhere
on G since 0 > 0 on (a.b). Thus F is not unique

extremal.

BIBLIOGRAPHY

l.

10.

11.

12.

BIBLIOGRAPHY

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L.V.

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. 0n quasiconformal mappings. J. Analyse

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pp. 1-58.

Ahlfors and A. Beurling. The boundary correspondence
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pp. 125-142.

Gehring. Definitions for a class of plane

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pp. 175-184.

Nagoya Math. J. 29 (1967).

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J.A.

Kelingos. Boundary correspondence under guasiconformal

mappings. Michigan Math. J. 13 (1966).

Akad. Wiss.

pp. 235-249.

M.A. Krasnosel'skii and Ya.B. Rutickii. Convex
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NeW'York. 1961.

S.L. Krushkal. Extremal gpasiconformal mappings. Siberian

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pp. 309—346.

117

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118

Mori. 0n am absolgte congtant in the thegty ot
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Salem. On some singular monotonic functions which

are strictly increasing. Trans. Amer. Math. Soc.
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