the verification of an inequality roger w. barnard, kent pearce, g. brock williams texas tech...
TRANSCRIPT
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The Verification of an Inequality
Roger W. Barnard, Kent Pearce, G. Brock Williams
Texas Tech University
Leah Cole
Wayland Baptist University
Presentation: Chennai, India
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Notation & Definitions
{ : | | 1}D z z
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Notation & Definitions
{ : | | 1}D z z
2
2 | |( ) | |
1 | |
dzz dz
z
hyperbolic metric
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Notation & Definitions
Hyberbolic Geodesics
{ : | | 1}D z z
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Notation & Definitions
Hyberbolic Geodesics
Hyberbolically Convex Set
{ : | | 1}D z z
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Notation & Definitions
Hyberbolic Geodesics
Hyberbolically Convex Set
Hyberbolically Convex Function
{ : | | 1}D z z
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Notation & Definitions
Hyberbolic Geodesics
Hyberbolically Convex Set
Hyberbolically Convex Function
Hyberbolic Polygono Proper Sides
{ : | | 1}D z z
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Examples
2 2
2( )
(1 ) (1 ) 4
zk z
z z z
k
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Examples
12 4 2
0
( ) tan (1 2 cos2 )
2where , 0 2(cos )
z
f z d
K
f
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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
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Extremal Problems for
Euclidean Convexity Nehari (1976):
( ) convex || || 2f Df D S
|| ||f DS
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Extremal Problems for
Euclidean Convexity Nehari (1976):
Spherical Convexity Mejía, Pommerenke (2000):
( ) convex || || 2f Df D S
( ) convex || || 2f Df D S
|| ||f DS
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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
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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
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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
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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
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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
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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
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Graph of
*
( )
2(0) 2( )fS c
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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
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Graph of
0
2c
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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
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θ = 0.1π /2
|| ||f DS
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θ = 0.3π /2
|| ||f DS
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θ = 0.5π /2
|| ||f DS
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θ = 0.7π /2
|| ||f DS
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θ = 0.9π /2
|| ||f DS
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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
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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
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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
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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
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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
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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
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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
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Problems with Strategy #1
0.012
cos 0.9999
2(1 ) | ( ) |ifr S re
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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
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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
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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
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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
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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
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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
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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
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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
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Strategy #2 – Case 1-a. (cont)
Finally, clearly
are non-negative
21
40
8[(1 )( ) ]
( )
c cx x c
c x c
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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
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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
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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
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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
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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
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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
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Coefficients c3, c1, c0 (cont.)
The coefficients c1 and c0 factor
21 (1 )( ) 0c xc x c
40 ( ) 0c x c
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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
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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
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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
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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
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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
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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
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Strategy #2 – Case 1-b. (cont)
6. Partition the parameter space R* intosubregions where the quadratic q* hasspecified properties
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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
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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
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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
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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
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Strategy #2 – Case 1-b. (cont)
7. Find bounding curves for D1 2andl l
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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
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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
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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
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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
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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
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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
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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
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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
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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