The Verification of an Inequality
Roger W. Barnard, Kent Pearce, G. Brock Williams
Texas Tech University
Leah Cole
Wayland Baptist University
Presentation: Pondicherry, 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{(1 | z | ) | S f ( z) |: z D}
2 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 3 Jan 2006), R.W. Barnard, L. Cole,
K. Pearce, G.B. Williams.
http://www.math.ttu.edu/~pearce/preprint.shtml
Verification of M/P Conjecture
Preliminary Facts:
Invariance of hyperbolic convexity under disk
automorphisms
Verification of M/P Conjecture
Preliminary Facts:
Invariance of hyperbolic convexity under disk
automorphisms
Invariance of || S f ||D under disk automorphisms
For
we have
, Auto( D)
|| S
f
||D || S f ||D
|| S f ||D || S f || 1 ( D )
Classes H and Hn
H { f A( D) : f ( D) is hyp. convex,
f (0) 0, f (0) 0}
Classes H and Hn
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 and Hn
H { f A( D) : f ( D) is hyp. convex,
f (0) 0, f (0) 0}
poly
{ f H : f ( D) is hyp. polygon}
H
H n { f H poly : f ( D ) has at most
n proper sides}
Reduction to Hn
Lemma 1. To determine the extremal value of
|| S f ||D over H, it suffices to determine the value
over each Hn. Moreover, each Hn is pre-compact.
Reduction to Hn
Lemma 1. To determine the extremal value of
|| S f ||D over H, it suffices to determine the value
over each Hn. Moreover, each Hn is pre-compact.
A. Hn
{0}
is compact
Reduction to Hn
Lemma 1. To determine the extremal value of
|| S f ||D over H, it suffices to determine the value
over each Hn. Moreover, each Hn is pre-compact.
A. Hn {0} is compact
B. H n H
n
Reduction to Hn
Lemma 1. To determine the extremal value of
|| S f ||D over H, it suffices to determine the value
over each Hn. Moreover, each Hn is pre-compact.
A. Hn {0} is compact
B. H n H
n
C. Schwarz norm is lower semi-continuous
Examples
k ( z )
2 z
(1 z ) (1 z ) 4 z
2
k
2
Reduction to Re Sf (0)
Lemma 2. For each n > 2,
sup || S f ||D max | S f (0) | max Re S f (0)
f H n
A.
B.
C.
f H n
f H n
Schwarz Norm || S f ||D
For f A( D) let
Sf
2
f 1 f
f 2 f
and
|| S f ||D sup{(1 | z |2 )2 | S f ( z) |: z D}
Reduction to Re Sf (0)
Lemma 2. For each n > 2,
sup || S f ||D max | S f (0) | max Re S f (0)
f H n
f H n
A. (Nehari)
f H n
n
k
1 n 1 k2
Sg ( z)
2
2 k 1 ( z ak ) k 1 z ak
implies
lim 4(Im z )2 | S g ( z ) | 2 for w U
z w
B.
C.
Reduction to Re Sf (0)
Lemma 2. For each n > 2,
sup || S f ||D max | S f (0) | max Re S f (0)
f H n
f H n
A. (Nehari)
f H n
n
k
1 n 1 k2
Sg ( z)
2
2 k 1 ( z ak ) k 1 z ak
implies
lim 4(Im z )2 | S g ( z ) | 2 for w U
z w
B. There exist f H n for which || S f ||D 2
C.
Reduction to Re Sf (0)
Lemma 2. For each n > 2,
sup || S f ||D max | S f (0) | max Re S f (0)
f H n
f H n
A. (Nehari)
f H n
n
k
1 n 1 k2
Sg ( z)
2
2 k 1 ( z ak ) k 1 z ak
implies
lim 4(Im z )2 | S g ( z ) | 2 for w U
z w
B. There exist f H n for which || S f ||D 2
C. Invariance under disk automorphisms
Julia Variation
Let Ω be a region bounded by a piece-wise analytic curve
Γ and φ(w) piece-wise C1 on Γ .
Julia Variation (cont)
Let Ω be a region bounded by a piece-wise analytic curve
Γ and φ(w) piece-wise C1 on Γ .
Julia Variation (cont)
Let Ω be a region bounded by a piece-wise analytic curve
Γ and φ(w) piece-wise C1 on Γ . At each point w on Γ
(where Γ is smooth), let n(w) denote the unit outward
normal to Γ. For small ε let
{w * w (w)n( w) : w }
and let Ωε be the region bounded by Γε.
Julia Variation (cont)
Let Ω be a region bounded by a piece-wise analytic curve
Γ and φ(w) piece-wise C1 on Γ . At each point w on Γ
(where Γ is smooth), let n(w) denote the unit outward
normal to Γ. For small ε let
{w * w (w)n( w) : w }
and let Ωε be the region bounded by Γε.
Julia Variation (cont)
Theorem. Let f be a conformal map from D on Ω
with f (0) = 0 and suppose f has a continuous
extension to ∂D. Then, for sufficiently small ε the
map fε from D on Ωε with fε (0) = 0 is given by
zf ( z ) 1 z
f ( z ) f ( z )
d o( )
2 D 1 z
where
d
( f ( ))
d and ei
| f ( ) |
Two Variations for Hn
Variation #1
Two Variations for Hn
Variation #1
Two Variations for Hn
Variation #1
Two Variations for Hn
Variation #1
f ( z ) f ( z ) o( )
*
Barnarnd & Lewis, Subordination theorems for
some classes of starlike functions, Pac. J. Math 56
(1975) 333-366.
Two Variations for Hn
Variation #2
Two Variations for Hn
Variation #2
Schwarzian and Julia Variation
Lemma 3. If
f ( z ) ( z a2 z 2 a3 z 3
) then
S f (0) 6(a3 a22 )
Lemma 4. If f ( z ) ( z a2 z 2 a3 z 3 ) H n and Var. #1 or
Var. #2 is applied to a side Γj, then
S f (0) 6 a3
2
a2
2
(3a3 4a2 )d
j
2
(2a2 2 )d
j
2
o( )
Schwarzian and Julia Variation
In particular,
6
2
2
Re S f (0)
Re
3
a
4
a
2
d
3
2
2
0
j
Re K ( j ) d
j
where
6
2
2
K ( )
3
a
4
a
2
3
2
2
Reduction to H2
Step #1. Reduction to H4
Reduction to H2
Step #1. Reduction to H4
Step #2. (Step Down Lemma) Reduction to H2
Reduction to H2
Step #1. Reduction to H4
Step #2. (Step Down Lemma) Reduction to H2
Step #3. Compute maximum in H2
Reduction to H2 – Step #1
Suppose f H n is extremal and maps D to a
region bounded by more than four sides.
Reduction to H2 – Step #1
Suppose f H n is extremal and maps D to a
region bounded by more than four sides.
Then, pushing Γ5 out using Var. #1, we have
Re S f (0)
Re K ( 1 ) d 0
0
5
Reduction to H2 – Step #1
Consequently, the image of each side γj under K
must intersect imaginary axis
Reduction to H2 – Step #1
Consequently, the image of each side γj under K
must intersect imaginary axis
Reduction to H2 – Step #2
Suppose f H n is extremal and maps D to a
region bounded by exactly four sides.
Reduction to H2 – Step #2
Suppose f H n is extremal and maps D to a
region bounded by exactly four sides.
Reduction to H2 – Step #2
Suppose f H n is extremal and maps D to a
region bounded by exactly four sides.
Reduction to H2 – Step #2
Suppose f H n is extremal and maps D to a
region bounded by exactly four sides.
Then, pushing in the end of Γ3 , near f (z*), using
Var. #2, we have
Re S f (0)
(1) Re K ( * ) d 0
0
*
Reduction to H2 – Step #2
Suppose f H n is extremal and maps D to a
region bounded by exactly two sides.
Computation in H2
Functions whose ranges are convex domains
bounded by one proper side (k )
Functions whose ranges are convex domains
bounded by two proper sides which intersect
Functions whose ranges are odd symmetric
convex domains whose proper sides do not
intersect ( f )
Computation in H2
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
Computation in H2
Verified
(1 r )2 S f ( ) (r )
A. For each fixed that
maximized at r = 0
B. 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
decreases for * 2 . At * ,
( )
( )
Sf
( * )
2.3836
is
Graph of S f
( )
*
(0) 2(c )
2
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