the verification of an inequality roger w. barnard, kent pearce, g. brock williams texas tech...
TRANSCRIPT
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
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
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
Extremal Problems for
Euclidean Convexity Nehari (1976):
Spherical Convexity Mejía, Pommerenke (2000):
( ) convex || || 2f Df D S
( ) convex || || 2f Df D S
|| ||f DS
Extremal Problems for
Euclidean Convexity Nehari (1976):
Spherical Convexity Mejía, Pommerenke (2000):
Hyperbolic Convexity Mejía, Pommerenke Conjecture (2000):
( ) convex || || 2f Df D S
( ) convex || || 2f Df D S
( ) 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
Verification of M/P Conjecture
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
Verification of M/P Conjecture
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
Strategy #1 – Case 1. (cont)
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
Strategy #1 – Case 1. (cont)
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
Case 2. Case (negative real axis)
Case 3. Case originally resolved.
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.
Make a change of variable Rewrite p1 as
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
22cosh 1 2sinh ( )ss
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)
We have reduced our problem to showing that
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
Strategy #2 – Case 1-a. (cont)
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
Strategy #2 – Case 1-a. (cont)
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
Strategy #2 – Case 1-a. (cont)
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
Strategy #2 – Case 1-a. (cont)
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
Strategy #2 – Case 1-a. (cont)
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
Strategy #2 – Case 1-a. (cont)
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
Case 2. Case (negative real axis)
Case 3. Case originally resolved.
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
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 #2 – Case 1-b. (cont)
We have reduced our problem to showing that
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
Strategy #2 – Case 1-b. (cont)
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
Strategy #2 – Case 1-b. (cont)
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
Strategy #2 – Case 1-b. (cont)
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
Strategy #2 – Case 1-b. (cont)
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
Strategy #2 – Case 1-b. (cont)
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
Strategy #2 – Case 1-b. (cont)
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
Strategy #2 – Case 1-b. (cont)
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
Strategy #2 – Case 1-b. (cont)
6. Partition the parameter space R* intosubregions where the quadratic q* hasspecified 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 non-negative
Strategy #2 – Case 1-b. (cont)
7. Find bounding curves for D1 2andl l
Strategy #2 – Case 1-b. (cont)
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
Case 2. Case (negative real axis)
Case 3. Case originally resolved.
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
Strategy #2 – Case 2. (cont)
Let for The
numerator p1 is a reflexive 4th-degree polynomial in r.
Make a change of variable Rewrite p1 as
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
22cosh 1 2sinh ( )ss
22 23 2( ,sinh( ), ) ( ,1 2sinh ( ), )s sp x p x
2sinh( ) .s
2sinh( )st
4 ( , , ) .p t x
1 2
Strategy #2 – Case 2. (cont)
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
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)
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
Case 2. Case (negative real axis)
Case 3. Case originally resolved.
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