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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. Notation & Definitions. Notation & Definitions. - PowerPoint PPT Presentation
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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 dzz
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 Df f
A
Classes
{ ( ) : ( ) is hyp. convex,(0) 0, (0) 0}
H f D f Df f
A
{ : ( ) is hyp. polygon}polyH f H f D
Classes
{ ( ) : ( ) is hyp. convex,(0) 0, (0) 0}
H f D f Df f
A
{ : ( ) is hyp. polygon}polyH f H f D
{ : ( ) has at mostproper sides}
n polyH f H f Dn
Classes
{ ( ) : ( ) is hyp. convex,(0) 0, (0) 0}
H f D f Df f
A
{ : ( ) is hyp. polygon}polyH f H f D
{ : ( ) has at mostproper sides}
n polyH f H f Dn
2 32 3{ : ( ) }H f H f z z a z a z
Examples
2 2
2( )(1 ) (1 ) 4
zk zz z z
k
Problems 1.
Fix 0 1 and let . For \{0}, findf H z D ( )min Re
f 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 Re
f 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 Re
f 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 z r zz r
and( )( )min Re , | |
f H
k rf z r zz 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 oz
where
( ( ))| ( ) |
fd df
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 1
j
j
j
f z f z zf z d oz 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 1
j
j
j
zL f L f f z d oz
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 beL f L f
made smaller than the value of ( ).L f
Proof (Theorem 1)
3
10
( , )2
j
j
j
mr f f d
3
10
( ) 1Re ( )2 1
j
j
j
L f zf z dz
and
1Let ( ) ( ) . Then, using the Mean Value Theorem1
zQ f zz
3
10
( ) Re ( )2
j
jj
j
L f Q 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 f d 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 f Q 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 haveL 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 )2
j
j
j
L f a a d
Step 1. Reduction to at most four sides.
Step 2. Reduction to at most two sides.
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