Three Extremal Problems for Hyperbolically Convex Functions

Roger W. Barnard, Kent Pearce, G. Brock Williams

Texas Tech University

[Computational Methods and Function Theory 4

(2004) pp 97-109]

Notation & Definitions

{ : | | 1}D z z

Notation & Definitions

{ : | | 1}D z z

2

2 | |( ) | |

1 | |

dzz dz

z

hyperbolic metric

Notation & Definitions

Hyberbolic Geodesics

{ : | | 1}D z z

Notation & Definitions

Hyberbolic Geodesics

Hyberbolically Convex Set

{ : | | 1}D z z

Notation & Definitions

Hyberbolic Geodesics

Hyberbolically Convex Set

Hyberbolically Convex Function

{ : | | 1}D z z

Notation & Definitions

Hyberbolic Geodesics

Hyberbolically Convex Set

Hyberbolically Convex Function

Hyberbolic Polygono Proper Sides

{ : | | 1}D z z

Classes

{ ( ) : ( ) is hyp. convex,

(0) 0, (0) 0}

H f D f D

f f

A

Classes

{ ( ) : ( ) is hyp. convex,

(0) 0, (0) 0}

H f D f D

f f

A

{ : ( ) is hyp. polygon}polyH f H f D

Classes

{ ( ) : ( ) is hyp. convex,

(0) 0, (0) 0}

H f D f D

f f

A

{ : ( ) is hyp. polygon}polyH f H f D

{ : ( ) has at most

proper sides}

n polyH f H f D

n

Classes

{ ( ) : ( ) is hyp. convex,

(0) 0, (0) 0}

H f D f D

f f

A

{ : ( ) is hyp. polygon}polyH f H f D

{ : ( ) has at most

proper sides}

n polyH f H f D

n

2 32 3{ : ( ) }H f H f z z a z a z

Examples

2 2

2( )

(1 ) (1 ) 4

zk z

z z z

k

Problems

1.

Fix 0 1 and let . For \{0}, findf H z D ( )

min Ref H

f zz

Problems

1.

2.

Find

2 32 3Fix 0 1 and let ( ) .f H with f z z a z a z

Fix 0 1 and let . For \{0}, findf H z D ( )

min Ref H

f zz

3max Ref H

a

Problems

1.

2.

Find

3.

2 32 3Fix 0 1 and let ( ) .f H with f z z a z a z

Fix 0 1 and let . For \{0}, findf H z D ( )

min Ref H

f zz

3max Ref H

a

2 32 3Let ( ) . Findf H with f z z a z a z

3max Ref H

a

Theorem 1

0 1 \{0}. ( ) ( ) / . ,Let and let z D Let L f f z z Then ( )the extremal value maximum or minimum for L over

H is obtained from a hyperbolically convex

function f which maps D onto a hyperbolic polygon

. ,with exactly one proper side Specifically

( )( )max Re , | |f H

k rf zr z

z r

and( )( )

min Re , | |f H

k rf zr z

z r

Theorem 2

Remark Minda & Ma observed that cannot be extremal for

30 1. , ( ) ReLet Then the maximal value for L f a

over H is obtained by a hyperbolically convex

2 32 3( )function f z z a z a z which maps D onto

.a hyperbolic polygon with at most two proper sides

k1

2

Theorem 3

3( ) ReThe maximal value for L f a over H is

obtained by a hyperbolically convex function2 3

2 3( )f z z a z a z which maps D onto a

.hyperbolic polygon with at most two proper sides

Julia Variation

Let be a region bounded by piece-wise analytic curve. Let be non-negative piece-wise continuous function

on . For let ( ) denote the outward normalw n w to at . For small let { ( ) ( ) : }w w w n w w and let be the region bounded by .

Julia Variation (cont.)

onto

Let be a 1 1 conformal map, : , with (0) 0.f f D f Suppose has a continuous extension to . Letf D f

ontobe a 1 1 conformal map, : , with (0) 0. Then,f D f

( ) 1( ) ( ) ( )

2 1D

zf z zf z f z d o

z

where

( ( ))

| ( ) |

fd d

f

for ie

Julia Variation (cont.)

Furthermore, the change in the mapping radius between

and is given byf f

(0)( , ) ( )

2 D

fmr f f d o

Variations for (Var. #1)

polyH

Suppose , not constant. If ( ) is a propernf H f f

side of ( ), then for small there exists af D variation which " " either in or outnf H pushes to a nearby geodesic . Furthermore, agreesf with the Julia variational formula up to ( ) terms.o

Variations for (Var. #2)

polyH

Suppose , not constant. If ( ) is a propernf H f f *side of ( ) which meets a side . Then, forf D

1small , there exists a variation which adds anf H side to ( ) by pushing one end of in to af D nearby side . Furthermore, agrees with the Juliaf variational formula up to ( ) terms.o

Proof (Theorem 1)

Step 1. Reduction to at most two sides.

Step 2. Reduction to one side.

Proof (Theorem 1)

Let . Suppose is extremal in for somen n nH H H f H

j3 and ( ) has (at least) 3 proper sides, say ( ),jn f D f

1, 2, 3. For each side apply the variation #1 withjj

control . Let be the varied function. Then,j j f

3

1

( ) ( ) 1( ) ( )

2 1j

j

j

f z f z zf z d o

z z z

Step 1. Reduction to at most two sides.

Proof (Theorem 1)

3

1

( , ) ( )2

j

j

j

mr f f d o

3

1

1( ) ( ) Re ( ) ( )

2 1j

j

j

zL f L f f z d o

z

Hence,

and

Proof (Theorem 1)

From the Calculus of Variations:

0

( , )If 0, then for small.nmr f f

f H

0

( )If 0, then the value of ( ) can be

L fL f

made smaller than the value of ( ).L f

Proof (Theorem 1)

3

10

( , )

2j

j

j

mr f fd

3

10

( ) 1Re ( )

2 1j

j

j

L f zf z d

z

and

1Let ( ) ( ) . Then, using the Mean Value Theorem

1

zQ f z

z

3

10

( )Re ( )

2j

jj

j

L fQ d

We have

Proof (Theorem 1)

1 2 3Then, we will push in and out (not vary ) so that

( ) is smaller than ( ). Specifically, chooseL f L f

1 2 30 ( =0) so that

Since is bilinear in , not all three of the points ( ) canjQ Q

1 2have the same real part. Wolog, Re ( ) Re ( ). Q Q

1 2

1 2

0

( , )0.

2 2

mr f fd d

Proof (Theorem 1)

then ( ) can have at most two proper sides.f D

Consequently, if is extremal in , 3,nf H n

Then,

1 2

1 2

1 21 2

0

1 21

( )Re ( ) Re ( )

2 2

Re ( ) 02 2

L fQ d Q d

Q d d

Proof (Theorem 1)

Step 2. Reduction to one side.

Suppose is extremal in for some 3. By the above argumentnf H n

( ) can have at most 2 proper sides. Suppose ( ) has exactly f D f D

2 proper sides, say ( ), 1, 2. As above apply variation #1j jf j

jto each side with control and let be the variedj j f

0

( )function. If in the formulation for we were to have

L f

1 2Re ( ) Re ( ), then would not be extremal.Q Q f

Proof (Theorem 1)

1 2 0Thus, we must have Re ( ) Re ( ) . Further, we must Q Q x

have that maps the pre-image arcs to arcs which overlapjQ

0the line { : Re }. l z z x

Proof (Theorem 1)

* 11We consider the vertex whose image under lies to the rightz Q f

*1of . Apply variation #2 to near to add another side to ( )l z f D

1- making sure the side is short enough so that its image under Q f

lies to the right of .l

1Q f

Proof (Theorem 1)

* 11We consider the vertex whose image under lies to the rightz Q f

*1of . Apply variation #2 to near to add another side to ( )l z f D

1- making sure the side is short enough so that its image under Q f

lies to the right of .l

1Q f

Proof (Theorem 1)

* 11We consider the vertex whose image under lies to the rightz Q f

*1of . Apply variation #2 to near to add another side to ( )l z f D

1- making sure the side is short enough so that its image under Q f

lies to the right of .l

1Q f

2At the same time push out to preserve the mapping radius.

Proof (Theorem 1)

* 11We consider the vertex whose image under lies to the rightz Q f

*1of . Apply variation #2 to near to add another side to ( )l z f D

1- making sure the side is short enough so that its image under Q f

lies to the right of .l

1Q f

2At the same time push out to preserve the mapping radius.

Proof (Theorem 1)

The varied function and by the above variationalnf H argument has a smaller value for than .L f

Hence, if is extremal for in , 3, then ( ) cannotnf L H n f D have two proper sides, , ( ) must have exactly one proper side. i.e. f D

2Since for all 3, if is extremal for in n nH H n f L H 2(and since by the above ), then must be extremal f H f

2for in as well.L H

Proof (Theorem 1)

Finally, since , the extremal value for over isn

nH H L H

achieved by a function for which ( ) has exactly one proper side.f f D

We note the range of ( ) / is symmetric about the realk z z

axis. Also, for fixed , 0 1, Re ( ( ) / ) is ai ir r k re re

monotonically decreasing function of , 0 .

Proofs (Theorem 2 & 3)

52

3 210

( )Re (3 4 2 )

2j

j

j

L fa a d

Step 1. Reduction to at most four sides.

Step 2. Reduction to at most two sides.