Majorization-subordination theorems for
locally univalent functions. IV
A Verification of Campbell’s Conjecture
Roger W. Barnard, Kent Pearce
Texas Tech University
Presentation: May 2008
Notation

D  {z : | z |  1}
Notation
D  {z : | z |  1}
 A (D)

Notation
D  {z : | z |  1}
 A (D)

 Schwarz Function

  A (D)
 : D  D, |  ( z ) |  | z | on D
Notation
D  {z : | z |  1}
 A (D)

 Schwarz Function
  A (D)
 : D  D, |  ( z ) |  | z | on D
 Majorization: f
F on | z |  r


| f ( z ) |  | F ( z ) | on | z |  r
Notation
D  {z : | z |  1}
 A (D)

 Schwarz Function
  A (D)
 : D  D, |  ( z ) |  | z | on D
 Majorization: f
F on | z |  r


| f ( z ) |  | F ( z ) | on | z |  r
 Subordination: f

F
f  F  for some Schwarz 
Notation
 S : Univalent Functions

K : Convex Univalent Functions:
Notation
 S : Univalent Functions

K : Convex Univalent Functions:
 U: Linearly Invariant Functions of order 
K==U1 , S  U2
Notation
 S : Univalent Functions

K : Convex Univalent Functions:
 U: Linearly Invariant Functions of order 
K==U1 , S  U2
2
 Footnote: S, K and U are normalized by f ( z )  z  a2 z 
Majorization-Subordination
 Classical Problems (Biernacki, Goluzin, Tao Shah,
Lewandowski, MacGregor)
Let F  S
Majorization-Subordination
 Classical Problems (Biernacki, Goluzin, Tao Shah,
Lewandowski, MacGregor)
Let F  S

A. If f
F on D find r so that f 
M (1967) : r  2  3
F  on | z |  r
Majorization-Subordination
 Classical Problems (Biernacki, Goluzin, Tao Shah,
Lewandowski, MacGregor)
Let F  S

A. If f
F on D find r so that f 
F  on | z |  r
M (1967) : r  2  3

B. If f
F on D find r so that f 
TS (1958) : r  3  8
F  on | z |  r
Majorization-Subordination
 Campbell (1971, 1973, 1974)
Let F  U
Majorization-Subordination
 Campbell (1971, 1973, 1974)
Let F  U

A. If f
F on D, then f 
F  on | z |  n( )
(  1)   1
where n( ) 
for   1
1

(  1)  1
1
Majorization-Subordination
 Campbell (1971, 1973, 1974)
Let F  U

A. If f
F on D, then f 
F  on | z |  n( )
(  1)   1
where n( ) 
for   1
1

(  1)  1
1

B. If f
F on D, then f 
F  on | z |  m( )
where m( )    1   2  2 for   1.65
Campbell’s Conjecture
 Let F  U
If f
F on D, then f 
F  on | z |  m( )
where m( )    1   2  2 for 1    1.65
Campbell’s Conjecture
 Let F  U
If f
F on D, then f 
F  on | z |  m( )
where m( )    1   2  2 for 1    1.65
 Footnote: Barnard, Kellogg (1984) verified Campbell’s for K= =U1
Summary of Campbell’s Proof
 Let F  U and suppose that f
for some Schwarz 
F so that f  F

Summary of Campbell’s Proof
 Let F  U and suppose that f
F so that f  F
for some Schwarz 
 Suppose that f has been rotated so that a  f (0)
satisfies 0  a  1

Summary of Campbell’s Proof
 Let F  U and suppose that f
F so that f  F
for some Schwarz 
 Suppose that f has been rotated so that a  f (0)
satisfies 0  a  1
a   ( z)
 Note we can write  ( z )  z
where
1  a ( z )
 is a Schwarz function

Summary of Campbell’s Proof
 Let F  U and suppose that f
F so that f  F
for some Schwarz 
 Suppose that f has been rotated so that a  f (0)
satisfies 0  a  1
a   ( z)
 Note we can write  ( z )  z
where
1  a ( z )
 is a Schwarz function
ac
i
 Let c   ( z )  re . We can write  ( z )  z
1  ac

Summary of Campbell’s Proof
 Let F  U and suppose that f
F so that f  F

for some Schwarz 
 Suppose that f has been rotated so that a  f (0)
satisfies 0  a  1
a   ( z)
 Note we can write  ( z )  z
where
1  a ( z )
 is a Schwarz function
ac
i
 Let c   ( z )  re . We can write  ( z )  z
1  ac
 For x  | z |  m( ) we have 0  r  x  m( )
Summary of Proof (Campbell)
 Fundamental Inequality [Pommerenke (1964)]

f ( z )
1  x  |1   ( z )z |  |  ( z)  z | 

 |  ( z ) |
2 
F ( z ) 1 |  ( z ) |  |1   ( z )z |  |  ( z )  z | 
2
(*)
Summary of Proof (Campbell)
 Fundamental Inequality [Pommerenke (1964)]

f ( z )
1  x  |1   ( z )z |  |  ( z)  z | 

 |  ( z ) |
2 
F ( z ) 1 |  ( z ) |  |1   ( z )z |  |  ( z )  z | 
2
 Two lemmas for estimating |  ( z ) |
(*)
“Small” a
 Campbell used “Lemma 2” to obtain

f ( z )
ba  1  b  a 



F ( z )
b  a  b 1 
1  x2
where b 
  1
2x
 k (a,  , b)  k (a,  ,   1)
“Small” a
 Campbell used “Lemma 2” to obtain

f ( z )
ba  1  b  a 



F ( z )
b  a  b 1 
 k (a,  , b)  k (a,  ,   1)
1  x2
where b 
  1
2x
 He showed there is a set R on which k is increasing in a
A1  {(a,  )  R : k (a,  ,   1)  1}
 Let C1  {(a,  )  R : k (a,  ,   1)  1}
 Let
“Small” a
“Small” a
“Large” a
 Campbell used “Lemma 1” to obtain
 1
f ( z )
 1 G 
 

F ( z )
 1 G 

2 C 
(1

x
) 2   G H  L

B 
where G,C,B are functions of c, x and a
(**)
“Large” a
 Campbell used “Lemma 1” to obtain
 1
f ( z )
 1 G 
 

F ( z )
 1 G 

2 C 
(1

x
) 2   G H  L

B 
(**)
where G,C,B are functions of c, x and a
 He showed there is a set S on which L maximizes
at c=r
 He showed that L(r,x,a) increases on S in a and
that L (r , x,1)  1
“Large” a
L
 Let A2  {(a,  )  S :
(r , x, a )  0}
a
L
(r , x, a)  0}
 Let C2  {(a,  )  S :
a
“Large” a
“Large” a
Combined Rectangles
Problematic Region
 Parameter space below   1.65
Verification of Conjecture
 Campbell’s estimates valid in A1 union A2
Verification of Conjecture
 Find L1 in A1 and L2 in A2
Verification of Conjecture
 Reduced to verifying Campbell’s conjecture on T
Step 1
 Consider the inequality
 1
f ( z )
 1 G 
 

F ( z )
 1 G 

2 C 
(1

x
) 2   G H  L

B 
(**)
 Show for ( a,  )  T that
x(1  a) |1  c |
G(c, x, a) 
|1  ac  x 2 (a  c) |
6 
maximizes at G (m( ), m( ), l1 ( )) 
6  9
Step 2
 Consider the inequality
 1
f ( z )
 1 G 
 

F ( z )
 1 G 

2 C 
(1

x
) 2   G H  L

B 
(**)
 1
 1 y 
6 
 Show at y 
that g ( y )  

6  9
 1 y 
is bounded above by
l ( y )  1  2.1(  1)(1 
 1
4
)y
Step 3
 Consider the inequality
 1
f ( z )
 1 G 
 

F ( z )
 1 G 

2 C 
(1

x
) 2   G H  L

B 
(**)
 Show for ( a,  )  T that
H (c, x, a)  (1  x 2 )
| a  2c  ac 2 | (1  x 2 )  ( x 2  r 2 )(1  a 2 )
|1  ac  x 2 (a  x) |  x(1  a) |1  c |
is bounded above by
4
13
13
2
h3 ( )  1  (  1)  (  1)  (  1)3
5
10
10
2
Step 4
 Consider the inequality
 1
f ( z )
 1 G 
 

F ( z )
 1 G 
 Let

2 C 
(1

x
) 2   G H  L

B 
 6  
g3 ( )  l 

6

9



and
4
13
13
h3 ( )  1  (  1)  (  1) 2  (  1)3
5
10
10
 Show that
g3 ( )h3 ( )  1
(**)