The Verification of an Inequality
Roger W. Barnard, Kent Pearce, G. Brock Williams
Texas Tech University
Leah Cole
Wayland Baptist University
Presentation: Chennai, India
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
Examples
k ( z ) 
2 z
(1  z )  (1  z )  4 z
2
k
2
Examples

f ( z )  tan  

where  
z
 (1  2
2
1
4  2
cos 2   )
0

2 , 0  
2
K (cos )
f

d 

Schwarz Norm || S f ||D
For f  A( D) let

Sf  

2

f   1  f  
  

f   2 f  
and
|| S f ||D  sup{D2 ( z) | S f ( z) |: z  D}
where
1
D ( z) 
1 | z |2
Extremal Problems for || S f ||D
 Euclidean Convexity

Nehari (1976):
f ( D) convex  || S f ||D  2
Extremal Problems for || S f ||D
 Euclidean Convexity

Nehari (1976):
f ( D) convex  || S f ||D  2
 Spherical Convexity

Mejía, Pommerenke (2000):
f ( D) convex  || S f ||D  2
Extremal Problems for || S f ||D
 Euclidean Convexity

Nehari (1976):
f ( D) convex  || S f ||D  2
 Spherical Convexity

Mejía, Pommerenke (2000):
f ( D) convex  || S f ||D  2
 Hyperbolic Convexity

Mejía, Pommerenke Conjecture (2000):
f ( D) convex  || S f ||D  2.3836
Verification of M/P Conjecture
 “The Sharp Bound for the Deformation of a Disc
under a Hyperbolically Convex Map,”
Proceedings of London Mathematical Society
(accepted), R.W. Barnard, L. Cole, K.Pearce,
G.B. Williams.
http://www.math.ttu.edu/~pearce/preprint.shtml
Verification of M/P Conjecture
 Invariance of hyperbolic convexity under disk
automorphisms
 Invariance of || S f ||D under disk automorphisms
 For f ( z )   ( z  a2 z 2  a3 z 3 
S f (0)  6(a3  a22 )
),
Verification of M/P Conjecture
 Classes H and Hn
 Julia Variation and Extensions
 Two Variations for the class Hn
 Representation for S f (0)
 Reduction to H2
Computation in H2
 Functions whose ranges are convex domains
bounded by one proper side (k )
 Functions whose ranges are convex domians
bounded by two proper sides which intersect
inside D
 Functions whose ranges are odd symmetric
convex domains whose proper sides do not
intersect ( f )
Leah’s Verification
 For each fixed  that (1  r )2 S f ( ) (r ) is
maximized at r = 0, 0 ≤ r < 1
 The curve S f ( ) (0)  2(c   2 ) is unimodal, i.e.,
there exists a unique  *  0.2182 so that
S f ( ) (0) increases for 0     * and S f  (0)
decreases for  *     2 . At  ,
( )
*
Sf
 ( * )
 2.3836
Graph of S f
 ( )
*
(0)  2(c   )
2
Innocuous Paragraph
 “Recall that D2 ( z) | S f
is invariant under precomposition with disc automorphisms. Thus by
pre-composing with an appropriate rotation, we
can ensure that the sup in the definition of the
Schwarz norm occurs on the real axis.”

( z) |
Graph of c  
2
0
|| S f ||D
(1 | z |)2 | S f ( z ) |  (1 | z |) 2 2(c   2 )
1  2dz  z 2
1  2cz  z 
where

2
2
3

2

c

c
2 , d
c  cos 2 ,  
K (cos  )
2(c   2 )
and
z  rei  D 
2 2
|| S f ||D
θ = 0.1π /2
|| S f ||D
θ = 0.3π /2
|| S f ||D
θ = 0.5π /2
|| S f ||D
θ = 0.7π /2
|| S f ||D
θ = 0.9π /2
Locate Local Maximi
For fixed  let h (r,  )  | (1  r )2 S f (rei ) |2

Solve
 h
 r  0

 h  0
 
For sin   0 there exists unique
solution (r ,  ) which satisfies

c  c2  3  d
cos  

2

 2
2
r

(1

d
(
c

c
 3)r  1  0

Let r0  min r . Claim r0  2 5

Strategy #1
 Case 1. Show (1  r )2 | S f (rei ) |  2

for
r  2 , 0  
5
 Case 2. Case    (negative real axis)
 Case 3. Case   0 originally resolved.
Strategy #1 – Case 1.
Let 4  | (1  r ) 2 S f (rei ) |2  p1 ( , r , x)

q1 ( , r , x)
where x  cos  . The
numerator p1 is a reflexive 8th-degree polynomial in r.
Make a change of variable r  e s . Rewrite p1 as
e4 s p1 ( , e  s , x)  p2 ( , cosh s, x)
where p2 is 4th-degree in cosh s . Substitute cosh s  1  2sinh 2 ( s 2)
to obtain
p3 ( ,sinh( s 2 ), x)  p2 ( ,1  2sinh 2 ( s 2), x)
which is an even 8th-degree polynomial in sinh( s 2 ) .
Substituting t  sinh( s 2 ) we obtain a 4th-degree polynomial
p4 ( , t , x) .
Strategy #1 – Case 1. (cont)
We have reduced our problem to showing that
225
p4 (t )  0 for 0  t  sinh (log 2 ) 
1000
2
5
Write
p4 (t )  c4t 4  c3t 3  c2t 2  c1t  c0
It suffices to show that p4 is totally monotonic, i.e.,
that each coefficient c j  0 , j  0 4
Strategy #1 – Case 1. (cont)
It can be shown that c3, c1, c0 are non-negative.
However,
c4  16(c   2  1)(c   2  1)
which implies that for 0    0 that c4 is negative.
Strategy #1 – Case 1. (cont)
In fact, the inequality p4 (t )  0 is false;
or equivalently,
the original inequality (1  r )2 | S f (rei ) |  2
is not valid for r  2 5 , 0    
Problems with Strategy #1
 The supposed local maxima do not actually exist.
 For fixed  near 0, the values of (1  r )2 | S f (rei ) |
stay near S f (0)  2(c   2 ) for large values of r ,
i.e., the values of (1  r )2 | S f (rei ) | are not
bounded by 2 for r  2 5 .
Problems with Strategy #1
(1  r )2 | S f (rei ) |
  0.01

2
cos   0.9999
Strategy #2
 Case 1-a. 0    0

Show (1  r )2 | S f (rei ) |  S f (0) for
0 
 Case 1-b. 0     2

Show (1  r )2 | S f (rei ) |  2 for r  2 5 , 0    
 Case 2. Case    (negative real axis)
 Case 3. Case   0 originally resolved.
Strategy #2 – Case 1-a.
Let | S f (0) |2  | (1  r ) 2 S f (rei ) |2  p1 ( , r , x) where


q1 ( , r , x)
x  cos  .
The numerator p1 is a reflexive 6th-degree polynomial in r.
Make a change of variable r  e s . Rewrite p1 as
e3s p1 ( , e  s , x)  p2 ( , cosh s, x)
where p2 is 3rd-degree in cosh s . Substitute cosh s  1  2sinh 2 ( s 2)
to obtain
p3 ( ,sinh( s 2 ), x)  p2 ( ,1  2sinh 2 ( s 2), x)
which is an even 6th-degree polynomial in sinh( s 2 ) .
Substituting t  sinh( s 2 ) we obtain a 3rd-degree polynomial
p4 ( , t , x) .
Strategy #2 – Case 1-a. (cont)
We have reduced our problem to showing that
p4 (t )  c3t 3  c2t 2  c1t  c0  0
for t > 0 under the assumption that 0    0
It suffices to show that p4 is totally monotonic, i.e.,
that each coefficient c j  0 , j  0 3
Strategy #2 – Case 1-a. (cont)
c3  16[(d  2c) x  1]
c2  4[(1  4c 2 ) x 2  (12c  2d ) x  4  2c 2  d 2 ]
c1  8[(1  cx)( x  c) 2 ]
c0  ( x  c)
4
c3 is linear in x. Hence,
c3  min c3
x 1
 16(2c  1  d ), c3
x 1
 16(2c  1  d )
Strategy #2 – Case 1-a. (cont)
It is easily checked that
3c  2c(c   )  2(c   )  3
2c  1  d 
2(c   2 )
2
2
3c  2c  1
0.1


2
2
2(c   )
2(c   )
2
2
Strategy #2 – Case 1-a. (cont)
2c  1  d  1  c  d  c  0
write
3  2 2c  c 2
3 1  c2
d c 
c 
0
2
2
2(c   )
2 c 
Strategy #2 – Case 1-a. (cont)
c2 is quadratic in x. It suffices to show that the vertex
of c2 is non-negative.
2(4c 4  9c 2  2d 2 c 2  6dc  d 2  2)
c2 vertex  
1  4c 2
(1  c) 2 (1  c) 2 (14c 2  40 2 c  9  8 4 )

2(1  4c 2 )(c   2 ) 2
Strategy #2 – Case 1-a. (cont)
The factor in the numerator satisifes
14c 2  40 2 c  9  8 4 
8(c   2 ) 2  8  6c 2  24 2 c  1 
6c 2  24 2 c  1 
6c(c   )  1  18 c 
2
6c  1  1.1
2
Strategy #2 – Case 1-a. (cont)
Finally, clearly
c1  8[(1  cx)( x  c)2 ]
c0  ( x  c)4
are non-negative
Strategy #2
 Case 1-a. 0    0

Show (1  r )2 | S f (rei ) |  S f (0) for
0 
 Case 1-b. 0     2

Show (1  r )2 | S f (rei ) |  2 for r  2 5 , 0    
 Case 2. Case    (negative real axis)
 Case 3. Case   0 originally resolved.
Strategy #2 – Case 1-b.
Let 4  | (1  r ) 2 S f (rei ) |2  p1 ( , r , x)

q1 ( , r , x)
where x  cos  . The
numerator p1 is a reflexive 8th-degree polynomial in r.
Make a change of variable r  e s . Rewrite p1 as
e4 s p1 ( , e  s , x)  p2 ( , cosh s, x)
where p2 is 4th-degree in cosh s . Substitute cosh s  1  2sinh 2 ( s 2)
to obtain
p3 ( ,sinh( s 2 ), x)  p2 ( ,1  2sinh 2 ( s 2), x)
which is an even 8th-degree polynomial in sinh( s 2 ) .
Substituting t  sinh( s 2 ) we obtain a 4th-degree polynomial
p4 ( , t , x) .
Strategy #2 – Case 1-b. (cont)
We have reduced our problem to showing that
p4 (t )  c4t 4  c3t 3  c2t 2  c1t  c0  0
225
under the assumption
for 0  t  sinh 2 (log 5 2 )  1000
that 0    
2
It suffices to show that p4 is totally monotonic, i.e.,
that each coefficient c j  0 , j  0 4
Strategy #2 – Case 1-b. (cont)
It can be shown that the coefficients c4, c3, c1, c0 are
non-negative.
Given,
c4  16(c   2  1)(c   2  1)
and that
0     2
, it follows that c4 is positive.
Coefficients c3, c1, c0
c3  ( 2c 2  3 2  c3  2 4c  c) x  2 4  4  4 2c  2c 2
Since c3 is linear in x, it suffices to show that
c3 x1  (1  c)(c2   2c  3c  2 4  3 2  4)  (1  c)q p  0
c3 x1  (1  c)(c2   2c  3c  2 4  3 2  4)  (1  c)qm  0
Rewriting qp we have
q p  c2  3(c   2 )  2  (1   2 (c   2 ))  (1   4 )  0
Coefficients c3, c1, c0 (cont.)
2
Making a change of variable c  2 y  1 we have
qm  4 y 4  10 y 2  8  2 2 y 2  2 2  2 4

where   K (2y) . Since all of the coefficients of α are negative,
then we can obtain a lower bound for qm by replacing α with
an upper bound
1
4
  p8  1  y 2 
1 4
7 6 19 8
y 
y 
y
16
192
768
Hence, qm  qm*  qm   p . qm* is a 32nd degree polynomial
in y with rational coefficients. A Sturm sequence
argument shows that qm* has no roots (i.e., it is positive).
8
Coefficients c3, c1, c0 (cont.)
The coefficients c1 and c0 factor
c1  (1  xc)( x  c) 2  0
c0  ( x  c) 4  0
Strategy #2 – Case 1-b. (cont)
However, c2 is not non-negative.
Since c4, c3, c1, c0 are non-negative, to show that
p4 (t )  c4t 4  c3t 3  c2t 2  c1t  c0  0
for 0 < t < ¼ it would suffice to show that
pa (t )  c2t  c1
or
pb (t )  c2t 2  c1t  c0
was non-negative – neither of which is true.
Strategy #2 – Case 1-b. (cont)
We note that it can be shown that
c2  (12c 2  8  4 4  8 2 c) x 2
 (4 2 c 2  36c  8 4c  12 2  4c 3 ) x
 c 4 7  14c 2  4 4 c 2  12 2 c  4 2 c3
is non-negative for -0.8 < x < 1 and 0     2
Strategy #2 – Case 1-b. (cont)
We will show that
q  c3 ( , x)t  c2 ( , x)t  c1 ( , x)  0
2
is non-negative for -1 < x < -0.8 and 0     2
and 0 < t < ¼ from which will follow that
p4 (t )  c4t 4  c3t 3  c2t 2  c1t  c0  0
Strategy #2 – Case 1-b. (cont)
 1. Expand q in powers of α
q  d 4 4  d 2 2  d 0
 2. Show d4 and d2 are non-positive
 3. Replace c  2 y 2  1 and use the upper bound
2
y
  1  4 where y  cos  to obtain a
lower bound q* for q
q  q( , x, t )  q *( y, x, t )  q *
which has no α dependency
Strategy #2 – Case 1-b. (cont)
 4. Expand q* in powers of t
q*  q *(t )  e2 ( y, x)t 2  e1 ( y, x)t  e0 ( y, x)
where
( y, x)  R  {( y, x) : 0  y  cos 0 and  1  x  0.8}
Note: e0(y,x) ≥ 0 on R.
Recall 0 < t < ¼
Strategy #2 – Case 1-b. (cont)
 5. Make a change of variable (scaling)
q*  q *(t )  e2 ( y, w)t  e1 ( y, w)t  e0 ( y, w)
2
where
( y, w)  R*  {( y, w) : 0  y  cos 0 and 0  w  1}
Strategy #2 – Case 1-b. (cont)
 6. Partition the parameter space R* into
subregions where the quadratic q* has
specified properties
Strategy #2 – Case 1-b. (cont)
 Subregion A
 e2(y,w)

<0
Hence, it suffices to verify that q*(0) > 0 and
q*(0.25) > 0
Strategy #2 – Case 1-b. (cont)
 Subregion B
 e2(y,w)

> 0 and e1(y,w) > 0
Hence, it suffices to verify q*(0) > 0
Strategy #2 – Case 1-b. (cont)
 Subregion C
 e2(y,w)
> 0 and e1(y,w) < 0 and the location of the
vertex of q* lies to the right of t = 0.25

Hence, it suffices to verify that q*(0.25) > 0
Strategy #2 – Case 1-b. (cont)
 Subregion D
 e2(y,w)
> 0 and e1(y,w) < 0 and the location of the
vertex of q* lies between t = 0 and t = 0.25

Required to verify that the vertex of q* is nonnegative
Strategy #2 – Case 1-b. (cont)
 7. Find bounding curves l1 and l2 for D
Strategy #2 – Case 1-b. (cont)
 8. Parameterize y between l1 and l2
by
y  l1 (w)  z (l2 (w)  l1 (w)) , 0  z  1

Note: q* = q*(z,w,t) is polynomial in z, w, t
with rational coefficients, 0 < z < 1, 0 < w < 1,
0 < t < 0.25, which is quadratic in t
 9. Show that the vertex of q* is non-negative,
i.e., show that the discriminant of q* is negative.
Strategy #2
 Case 1-a. 0    0

Show (1  r )2 | S f (rei ) |  S f (0) for
0 
 Case 1-b. 0     2

Show (1  r )2 | S f (rei ) |  2 for r  2 5 , 0    
 Case 2. Case    (negative real axis)
 Case 3. Case   0 originally resolved.
Strategy #2 – Case 2.

2
2
2(
c


)(1

2
dx

x
)
2
2
(1  x) S fa ( x)  (1  x)
(1  2cx  x 2 )2
 Show there exists 1  0.598 which is the unique
solution of d = 2c + 1 such that for   1
(1  x)2 S f ( x) is strictly decreasing, i.e., for
2
  1 we have (1  x) S f ( x) takes its
maximum value at x = 0.
Note: 1  0
Strategy #2 – Case 2. (cont)
Let
2  (1  x) 2 S f ( x) 
p1 ( , r , x)
q1 ( , r , x)
for 1     2
The
numerator p1 is a reflexive 4th-degree polynomial in r.
Make a change of variable r  e s . Rewrite p1 as
e2 s p1 ( , e  s , x)  p2 ( , cosh s, x)
where p2 is 2nd-degree in cosh s . Substitute cosh s  1  2sinh 2 ( s 2)
to obtain
p3 ( ,sinh( s 2 ), x)  p2 ( ,1  2sinh 2 ( s 2), x)
which is an even 4th-degree polynomial in sinh( s 2 ) .
Substituting t  sinh( s 2 ) we obtain a 2nd-degree polynomial
p4 ( , t , x) .
Strategy #2 – Case 2. (cont)
 Show that the vertex of p4 is non-negative
p4 vertex
(c   2 )[(c   2 )d 2  (4  2 2  2c)d  4c 2   2  5c]

2(1  (c   2 ))
(c   2 )

q1
2
2(1  (c   ))
 Rewrite
(1  c) 2 [4 4  (4c  12) 2  c 2  18c  15]
q1 
4(c   2 )
(1  c) 2

q2
2
4(c   )
 Show q2  0
Strategy #2 – Case 2. (cont)
Since all of the coefficients of α in q2 are negative,
then we can obtain a lower bound for q2 by replacing α with
an upper bound (also writing c = 2y2-1)
1
4
  p8  1  y 2 
1 4
7 6 19 8
y 
y 
y
16
192
768
Hence, q2  q2*  q2   p . q2* is a 32nd degree polynomial
in y with rational coefficients. A Sturm sequence
argument shows that q2* has no roots (i.e., it is positive).
8
Strategy #2
 Case 1-a. 0    0

Show (1  r )2 | S f (rei ) |  S f (0) for
0 
 Case 1-b. 0     2

Show (1  r )2 | S f (rei ) |  2 for r  2 5 , 0    
 Case 2. Case    (negative real axis)
 Case 3. Case   0 originally resolved.
Innocuous Paragraph
 “Recall that D2 ( z) | S f
is invariant under precomposition with disc automorphisms. Thus by
pre-composing with an appropriate rotation, we
can ensure that the sup in the definition of the
Schwarz norm occurs on the real axis.”

( z) |
New Innocuous Paragraph
 Using an extensive calculus argument which
considers several cases (various interval ranges for
|z|, arg z, and α) and uses properties of
polynomials and K, one can show that this
problem can be reduced to computing
sup (1  x) 2 | S f ( x) |
0 x 1