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

D  {z :| z |  1}
Notation & Definitions

D  {z :| z |  1}
hyperbolic metric
2 | dz |
 ( z ) | dz |
2
1 | z |
Notation & Definitions

D  {z :| z |  1}
 Hyberbolic Geodesics
Notation & Definitions

D  {z :| z |  1}
 Hyberbolic Geodesics
 Hyberbolically Convex Set
Notation & Definitions

D  {z :| z |  1}
 Hyberbolic Geodesics
 Hyberbolically Convex Set
 Hyberbolically Convex Function
Notation & Definitions

D  {z :| z |  1}
 Hyberbolic Geodesics
 Hyberbolically Convex Set
 Hyberbolically Convex Function
 Hyberbolic Polygon
o Proper Sides
Classes

H  { f A ( D ) : f ( D) is hyp. convex,
f (0)  0, f (0)  0}
Classes


H  { f A ( D ) : f ( D) is hyp. convex,
f (0)  0, f (0)  0}
H
poly
 { f  H : f ( D) is hyp. polygon}
Classes



H  { f A ( D ) : f ( D) is hyp. convex,
f (0)  0, f (0)  0}
H
poly
 { f  H : f ( D) is hyp. polygon}
H n  { f  H poly : f ( D) has at most
n proper sides}
Classes




H  { f A ( D ) : f ( D) is hyp. convex,
f (0)  0, f (0)  0}
H
poly
 { f  H : f ( D) is hyp. polygon}
H n  { f  H poly : f ( D) has at most
n proper sides}
H  { f  H : f ( z )   z  a2 z 2  a3 z 3  }
Examples

k ( z ) 
2 z
(1  z )  (1  z )  4 z
2
k
2
Problems
 1. Fix 0    1 and let f  H . For z  D \{0}, find
min Re
f  H
f ( z)
z
Problems
 1. Fix 0    1 and let f  H . For z  D \{0}, find
min Re
f  H
f ( z)
z
 2. Fix 0    1 and let f  H with f ( z )   z  a2 z 2  a3 z 3 
Find
max Re a3
f  H
.
Problems
 1. Fix 0    1 and let f  H . For z  D \{0}, find
min Re
f  H
f ( z)
z
 2. Fix 0    1 and let f  H with f ( z )   z  a2 z 2  a3 z 3 
Find
max Re a3
f  H
2
3
 3. Let f  H with f ( z )   z  a2 z  a3 z 
max Re a3
f H
. Find
.
Theorem 1
Let 0    1 and let z  D \{0}. 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 ( z ) k (r )

,
z
r
min Re
f ( z ) k (r )

,
z
r
f  H
r | z |
and
f  H
r | z |
Theorem 2
Let 0    1. Then, the maximal value for L( f )  Re a3
over H is obtained by a hyperbolically convex
function f ( z )   z  a2 z 2  a3 z 3 
which maps D onto
a hyperbolic polygon with at most two proper sides.
Remark Minda & Ma observed that k cannot be extremal
for   1
2
Theorem 3
The maximal value for L( f )  Re a3 over H is
obtained by a hyperbolically convex function
f ( z )   z  a2 z 2  a3 z 3 
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 w  let n(w) denote the outward normal
to  at w. For  small let   {w   ( w) n( w) : w  }
and let  be the region bounded by  .
Julia Variation (cont.)
Let f be a 1  1 conformal map, f : D , with f (0)  0.
onto
Suppose f has a continuous extension to D. Let f
be a 1  1 conformal map, f : D  , with f (0)  0. Then,
onto
 zf ( z ) 1   z
f ( z )  f ( z ) 
d   o( )

2 D 1   z
where
for   ei
 ( f ( ))
d 
d
| f ( ) |
Julia Variation (cont.)
Furthermore, the change in the mapping radius between
f and f is given by
 f (0)
mr ( f , f ) 
d   o( )

2 D
Variations for H
poly
(Var. #1)
Suppose f  H n , f not constant. If   f ( ) is a proper
side of   f ( D), then for small  there exists a
variation f  H n which "pushes"  either in or out
to a nearby geodesic  . Furthermore, f agrees
with the Julia variational formula up to o( ) terms.
Variations for H
poly
(Var. #2)
Suppose f  H n , f not constant. If   f ( ) is a proper
side of   f ( D) which meets a side *. Then, for
small  , there exists a variation f  H n1 which adds a
side to f ( D) by pushing one end of  in to a
nearby side  . Furthermore, f agrees with the Julia
variational formula up to o( ) terms.
Proof (Theorem 1)
Step 1. Reduction to at most two sides.
Step 2. Reduction to one side.
Proof (Theorem 1)
Step 1. Reduction to at most two sides.
Let Hn  H  H n . Suppose f is extremal in Hn for some
n  3 and f ( D) has (at least) 3 proper sides, say  j  f ( j ),
j  1, 2, 3. For each side  j apply the variation #1 with
control  j   j . Let f be the varied function. Then,
3 
f ( z ) f ( z )
1  z
j


f ( z )
d   o( )

z
z
1  z
j  1 2  j
Proof (Theorem 1)
Hence,
 3 j

1  z
L( f )  L( f )   Re  
f ( z )
d   o( ) 

 j  1 2 

1 z
j


and
 j
mr ( f , f )   
d   o( )

j  1 2 
3
j
Proof (Theorem 1)
From the Calculus of Variations:

mr ( f , f )
If
 0, then f  Hn for  small.

 0

L( f )
If
 0, then the value of L( f ) can be
  0
made smaller than the value of L( f ).
Proof (Theorem 1)
We have
 3 j

L( f )
1  z
 Re  
f ( z )
d 

 j  1 2 

  0
1 z
j


and
3  
mr ( f , f )
  j  d

j  1 2  j
 0
Let Q( )  f ( z )
1  z
. Then, using the Mean Value Theorem
1  z
 3 j

L( f )
 
Re Q( j )  d  

  0  j  1 2
j


Proof (Theorem 1)
Since Q is bilinear in  , not all three of the points Q( j ) can
have the same real part. Wolog, Re Q( 1 )  Re Q( 2 ).
Then, we will push 1 in and  2 out (not vary 3 ) so that
L( f ) is smaller than L( f ). Specifically, choose
1  0  2 (3 =0) so that
mr ( f , f )


 1  d   2  d   0.

2 1
2  2
 0
Proof (Theorem 1)
Then,
L( f )
1
2
 Re Q( 1 )
d   Re Q( 2 )
d


  0
2 1
2  2


2
1
 Re Q( 1 ) 
d 
d   0


 2 

2  2


1
Consequently, if f is extremal in Hn , n  3,
then f ( D) can have at most two proper sides.
Proof (Theorem 1)
Step 2. Reduction to one side.
Suppose f is extremal in Hn for some n  3. By the above argument
f ( D) can have at most 2 proper sides. Suppose f ( D) has exactly
2 proper sides, say  j  f ( j ), j  1, 2. As above apply variation #1
to each side  j with control  j   j and let f be the varied
function. If in the formulation for
L( f )
we were to have
  0
Re Q( 1 )  Re Q( 2 ), then f would not be extremal.
Proof (Theorem 1)
Thus, we must have Re Q( 1 )  Re Q( 2 )  x0 . Further, we must
have that Q maps the pre-image arcs  j to arcs which overlap
the line l  {z : Re z  x0 }.
Proof (Theorem 1)
We consider the vertex z* 1 whose image under Q f 1 lies to the right
of l. Apply variation #2 to 1 near z* to add another side to f ( D)
- making sure the side is short enough so that its image under Q f 1
lies to the right of l.
Q f 1
Proof (Theorem 1)
We consider the vertex z* 1 whose image under Q f 1 lies to the right
of l. Apply variation #2 to 1 near z* to add another side to f ( D)
- making sure the side is short enough so that its image under Q f 1
lies to the right of l.
Q f 1
Proof (Theorem 1)
We consider the vertex z* 1 whose image under Q f 1 lies to the right
of l. Apply variation #2 to 1 near z* to add another side to f ( D)
- making sure the side is short enough so that its image under Q f 1
lies to the right of l.
Q f 1
At the same time push  2 out to preserve the mapping radius.
Proof (Theorem 1)
We consider the vertex z* 1 whose image under Q f 1 lies to the right
of l. Apply variation #2 to 1 near z* to add another side to f ( D)
- making sure the side is short enough so that its image under Q f 1
lies to the right of l.
Q f 1
At the same time push  2 out to preserve the mapping radius.
Proof (Theorem 1)
The varied function f  Hn and by the above variational
argument has a smaller value for L than f .
Hence, if f is extremal for L in Hn , n  3, then f ( D) cannot
have two proper sides, i.e., f ( D) must have exactly one proper side.
Since H2  Hn for all n  3, if f is extremal for L in Hn
(and since by the above f  H2 ), then f must be extremal
for L in H2 as well.
Proof (Theorem 1)
Finally, since H 
n
Hn , the extremal value for L over H is
achieved by a function f for which f (D) has exactly one proper side.
We note the range of k ( z ) / z is symmetric about the real
axis. Also, for fixed r , 0  r  1, Re (k (rei ) / rei ) is a
monotonically decreasing function of  , 0     .
Proofs (Theorem 2 & 3)
Step 1. Reduction to at most four sides.
 5 j

L( f )
2
 Re  
(3a3  4a2  2 ) d  

 j  1 2 

  0
j


Step 2. Reduction to at most two sides.