the verification of an inequality roger w. barnard, kent pearce, g. brock williams texas tech...

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

{ : | | 1}D z z


{ : | | 1}D z z

2

2 | |( ) | |

1 | |

dzz dz

z

hyperbolic metric


Hyberbolic Geodesics

{ : | | 1}D z z



Hyberbolically Convex Set

{ : | | 1}D z z




Hyberbolically Convex Function

{ : | | 1}D z z




Hyberbolically Convex Function

Hyberbolic Polygono Proper Sides

{ : | | 1}D z z

Examples

2 2

2( )

(1 ) (1 ) 4

zk z

z z z

k

Examples

12 4 2

0

( ) tan (1 2 cos2 )

2where , 0 2(cos )

z

f z d

K

f

Schwarz Norm

For let

and

where

( )f A D

21

2f

f fS

f f

2|| || sup{ ( ) | ( ) |: }f D D fS z S z z D

2

1( )

1 | |D zz

|| ||f DS

Extremal Problems for

Euclidean Convexity Nehari (1976):

( ) convex || || 2f Df D S

|| ||f DS



Spherical Convexity Mejía, Pommerenke (2000):



|| ||f DS



Spherical Convexity Mejía, Pommerenke (2000):

Hyperbolic Convexity Mejía, Pommerenke Conjecture (2000):



( ) convex || || 2.3836f Df D S

|| ||f DS

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


Invariance of hyperbolic convexity under disk automorphisms

Invariance of under disk automorphisms

For

|| ||f DS

23 2(0) 6( )fS a a

2 32 3( ) ( ) ,f z z a z a z


Classes H and Hn

Julia Variation and Extensions

Two Variations for the class Hn

Representation for

Reduction to H2

(0)fS

Computation in H2

Functions whose ranges are convex domains bounded by one proper side

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

( )k

( )f

Leah’s Verification

For each fixed that is maximized at r = 0, 0 ≤ r < 1

The curve is unimodal, i.e., there exists a unique so that

increases for and

decreases for At ( )

(0)fS

( )

2(1 ) ( )fr S r

( )

2(0) 2( )fS c

* 0.2182

*0 * .2

* ,

*( )

2.3836fS

( )(0)fS

Graph of

*

( )

2(0) 2( )fS c

Innocuous Paragraph

“Recall that is invariant under pre-composition 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.”

2 ( ) | ( ) |D fz S z

Graph of

0

2c

where

and

|| ||f DS

2

2 2 222

1 2(1 | |) | ( ) | (1 | |) 2( )

1 2f

dz zz S z z c

cz z

2 2

2

3 22cos 2 , ,(cos ) 2( )

c cc d

K c

iz re D

θ = 0.1π /2

|| ||f DS

θ = 0.3π /2

|| ||f DS

θ = 0.5π /2

|| ||f DS

θ = 0.7π /2

|| ||f DS

θ = 0.9π /2

|| ||f DS

Locate Local Maximi

For fixed let

Solve For there exists unique solution which satisfies

Let Claim0 min .r r

2 2( , ) | (1 ) ( ) |ifh r r S re

0

0

h

rh

sin 0

2

2 2

3cos

2

(1 ( 3) 1 0

c c d

r d c c r

250r

( , )r

Strategy #1

Case 1. Show

for

Case 2. Case (negative real axis)

Case 3. Case originally resolved.

2 , 05r

0

2(1 ) | ( ) | 2ifr S re

Strategy #1 – Case 1.

Let where The

numerator p1 is a reflexive 8th-degree polynomial in r.

Make a change of variable Rewrite p1 as

where p2 is 4th-degree in cosh s . Substituteto obtain

which is an even 8th-degree polynomial in Substituting we obtain a 4th-degree polynomial

2 2 1

1

( , , )4 | (1 ) ( ) |

( , , )i

f

p r xr S re

q r x

cos .x

.sr e4

1 2( , , ) ( ,cosh , )s se p e x p s x

22cosh 1 2sinh ( )ss

22 23 2( ,sinh( ), ) ( ,1 2sinh ( ), )s sp x p x

2sinh( ) .s

2sinh( )st

4 ( , , ) .p t x

Strategy #1 – Case 1. (cont)

We have reduced our problem to showing that

Write

It suffices to show that p4 is totally monotonic, i.e.,

that each coefficient

4 3 24 4 3 2 1 0( )p t c t c t c t c t c

2 524

225( ) 0 for 0 sinh (log )

1000p t t

0 , 0 4jc j


It can be shown that c3, c1, c0 are non-negative.

However,

which implies that for that c4 is negative.

2 24 16( 1)( 1)c c c

00


In fact, the inequality is false;

or equivalently,

the original inequality

is not valid for

4 ( ) 0p t

2(1 ) | ( ) | 2ifr S re

2 , 05r

Problems with Strategy #1

The supposed local maxima do not actually exist.

For fixed near 0, the values of

stay near for large values of r , i.e., the values of are not bounded by 2 for

2(1 ) | ( ) |ifr S re

2(0) 2( )fS c

25 .r

2(1 ) | ( ) |ifr S re

Problems with Strategy #1

0.012

cos 0.9999

2(1 ) | ( ) |ifr S re

Strategy #2

Case 1-a. Show for

Case 1-b. Show for



2 , 05r

0

2(1 ) | ( ) | 2ifr S re

00

0 2

2(1 ) | ( ) | (0)if fr S re S

0

Strategy #2 – Case 1-a.

Let where

The numerator p1 is a reflexive 6th-degree polynomial in r.


where p2 is 3rd-degree in cosh s . Substituteto obtain

which is an even 6th-degree polynomial in Substituting we obtain a 3rd-degree polynomial

2 2 2 1

1

( , , )| (0) | | (1 ) ( ) |

( , , )i

f f

p r xS r S re

q r x

cos .x

.sr e3

1 2( , , ) ( ,cosh , )s se p e x p s x


22 23 2( ,sinh( ), ) ( ,1 2sinh ( ), )s sp x p x

2sinh( ) .s

2sinh( )st

4 ( , , ) .p t x

Strategy #2 – Case 1-a. (cont)


for t > 0 under the assumption that

It suffices to show that p4 is totally monotonic, i.e.,

that each coefficient 0 , 0 3jc j

00

3 24 3 2 1 0( ) 0p t c t c t c t c


c3 is linear in x. Hence,

3

2 2 2 22

21

40

16[( 2 ) 1]

4[(1 4 ) ( 12 2 ) 4 2 ]

8[(1 )( ) ]

( )

c d c x

c c x c d x c d

c cx x c

c x c

3 3 31 1min 16(2 1 ), 16( 2 1 )

x xc c c d c c d


It is easily checked that2 2 2

2

2

2 2

3 2 ( ) 2( ) 32 1

2( )

3 2 1 0.1

2( ) 2( )

c c c cc d

c

c c

c c


write

2 2 2

2 2

3 2 3 10

2( ) 2

c c cd c c

c c

2 1 1 0c d c d c


c2 is quadratic in x. It suffices to show that the vertex

of c2 is non-negative. 4 2 2 2 2

2 2

2 2 2 2 4

2 2 2

2( 4 9 2 6 2)

1 4

(1 ) (1 ) (14 40 9 8 )

2(1 4 )( )

vertex

c c d c dc dc

c

c c c c

c c


The factor in the numerator satisifes

2 2 4

2 2 2 2

2 2

2 2

14 40 9 8

8( ) 8 6 24 1

6 24 1

6 ( ) 1 18

6 1 1.1

c c

c c c

c c

c c c

c


Finally, clearly

are non-negative

21

40

8[(1 )( ) ]

( )

c cx x c

c x c

Strategy #2

Case 1-a. Show for

Case 1-b. Show for



2 , 05r

0

2(1 ) | ( ) | 2ifr S re

00

0 2

2(1 ) | ( ) | (0)if fr S re S

0

Strategy #2 – Case 1-b.

Let where The



where p2 is 4th-degree in cosh s . Substituteto obtain

which is an even 8th-degree polynomial in Substituting we obtain a 4th-degree polynomial

2 2 1

1

( , , )4 | (1 ) ( ) |

( , , )i

f

p r xr S re

q r x

cos .x

.sr e4

1 2( , , ) ( ,cosh , )s se p e x p s x


22 23 2( ,sinh( ), ) ( ,1 2sinh ( ), )s sp x p x

2sinh( ) .s

2sinh( )st

4 ( , , ) .p t x

Strategy #2 – Case 1-b. (cont)


under the assumption that

It suffices to show that p4 is totally monotonic, i.e.,that each coefficient

2 22552 1000for 0 sinh (log )t

0 , 0 4jc j

4 3 24 4 3 2 1 0( ) 0p t c t c t c t c t c

0 2


It can be shown that the coefficients c4, c3, c1, c0 are

non-negative.

Given,

and that , it follows that c4 is positive.

2 24 16( 1)( 1)c c c

0 2

Coefficients c3, c1, c0

Since c3 is linear in x, it suffices to show that

Rewriting qp we have

2 2 2 3 4 4 2 23 ( 3 2 ) 2 4 4 2c c c c c x c c

2 2 4 23 1

(1 )( 3 2 3 4) (1 ) 0mxc c c c c c q

2 2 4 23 1

(1 )( 3 2 3 4) (1 ) 0pxc c c c c c q

2 2 2 2 43( ) 2 (1 ( )) (1 ) 0pq c c c

Coefficients c3, c1, c0 (cont.)

Making a change of variable we have

where 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

Hence, is a 32nd degree polynomial

in y with rational coefficients. A Sturm sequence

argument shows that has no roots (i.e., it is positive).

22 1c y

4 2 2 2 2 44 10 8 2 2 2mq y y y

2 .( )K y

2 4 6 88

1 1 7 191

4 16 192 768p y y y y

8

* *.m m m mpq q q q

*mq

Coefficients c3, c1, c0 (cont.)

The coefficients c1 and c0 factor

21 (1 )( ) 0c xc x c

40 ( ) 0c x c


However, c2 is not non-negative.

Since c4, c3, c1, c0 are non-negative, to show that

for 0 < t < ¼ it would suffice to show that

or

was non-negative – neither of which is true.

22 1 0( )bp t c t c t c

4 3 24 4 3 2 1 0( ) 0p t c t c t c t c t c

2 1( )ap t c t c


We note that it can be shown that

is non-negative for -0.8 < x < 1 and

2 4 2 22

2 2 4 2 3

4 2 4 2 2 2 3

(12 8 4 8 )

(4 36 8 12 4 )

7 14 4 12 4

c c c x

c c c c x

c c c c c

0 2


We will show that

is non-negative for -1 < x < -0.8 and

and 0 < t < ¼ from which will follow that

0 2

23 2 1( , ) ( , ) ( , ) 0q c x t c x t c x

4 3 24 4 3 2 1 0( ) 0p t c t c t c t c t c


1. Expand q in powers of α

2. Show d4 and d2 are non-positive 3. Replace and use the upper

bound where to obtain a lower bound q* for q

which has no α dependency

4 24 2 0q d d d

2

1 4y cosy

22 1c y

( , , ) *( , , ) *q q x t q y x t q


4. Expand q* in powers of t

where

Note: e0(y,x) ≥ 0 on R.

Recall 0 < t < ¼

22 1 0* *( ) ( , ) ( , ) ( , )q q t e y x t e y x t e y x

0( , ) {( , ) : 0 cos and 1 0.8}y x R y x y x


5. Make a change of variable (scaling)

where

22 1 0* *( ) ( , ) ( , ) ( , )q q t e y w t e y w t e y w

0( , ) * {( , ) : 0 cos and 0 1}y w R y w y w


6. Partition the parameter space R* intosubregions where the quadratic q* hasspecified properties


Subregion A

e2(y,w) < 0

Hence, it suffices to verify that q*(0) > 0 and q*(0.25) > 0


Subregion B

e2(y,w) > 0 and e1(y,w) > 0

Hence, it suffices to verify q*(0) > 0


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


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 non-negative


7. Find bounding curves for D1 2andl l


8. Parameterize y between by

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.

1 2andl l

1 2 1( ) ( ( ) ( )) , 0 1y l w z l w l w z

Strategy #2

Case 1-a. Show for

Case 1-b. Show for



2 , 05r

0

2(1 ) | ( ) | 2ifr S re

00

0 2

2(1 ) | ( ) | (0)if fr S re S

0

Strategy #2 – Case 2.

Show there exists which is the unique solution of d = 2c + 1 such that for is strictly decreasing, i.e., for we have takes its maximum value at x = 0.

Note:

1

2 22 2

2 2

2( )(1 2 )(1 ) ( ) (1 )

(1 2 )af

c dx xx S x x

cx x

1 0.598

1

2(1 ) ( )fx S x

2(1 ) ( )fx S x

1 0


Let for The



where p2 is 2nd-degree in cosh s . Substituteto obtain

which is an even 4th-degree polynomial in Substituting we obtain a 2nd-degree polynomial

2 1

1

( , , )2 (1 ) ( )

( , , )f

p r xx S x

q r x

.sr e2

1 2( , , ) ( ,cosh , )s se p e x p s x


22 23 2( ,sinh( ), ) ( ,1 2sinh ( ), )s sp x p x

2sinh( ) .s

2sinh( )st

4 ( , , ) .p t x

1 2


Show that the vertex of p4 is non-negative

Rewrite

Show

2 2 2 2 2 2

4 2

2

12

( )[ ( ) (4 2 2 ) 4 5 ]

2(1 ( ))

( )

2(1 ( ))

vertex

c c d c d c cp

c

cq

c

2 4 2 2

1 2

2

22

(1 ) [ 4 (4 12) 18 15]

4( )

(1 )

4( )

c c c cq

c

cq

c

2 0q


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)

Hence, is a 32nd degree polynomial

in y with rational coefficients. A Sturm sequence

argument shows that has no roots (i.e., it is positive).

2 4 6 88

1 1 7 191

4 16 192 768p y y y y

8

* *2 2 2 2.

pq q q q

*2q

Strategy #2

Case 1-a. Show for

Case 1-b. Show for



2 , 05r

0

2(1 ) | ( ) | 2ifr S re

00

0 2

2(1 ) | ( ) | (0)if fr S re S

0

Innocuous Paragraph

“Recall that is invariant under pre-composition 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.”

2 ( ) | ( ) |D fz S 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

2

0 1sup (1 ) | ( ) |f

xx S x

the verification of an inequality roger w. barnard, kent pearce, g. brock williams texas tech...

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