three extremal problems for hyperbolically convex functions roger w. barnard, kent pearce, g. brock...
TRANSCRIPT
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.