reflects, in fact, a special case of the Julia-Wolff Theorem. It is known as. Jack's Lemma C261: ...... 1 z l a = ~ l . Note that in this range. (2.38). %CS+Ra. 2. 1.
CONVOLUTIONS IN GEOMETRIC FUNCTION THEORY
Notes du cours de Monsieur Stephan Ruscheweyh a la vingtitme session du SCminaire de mathkmatiques supCrieures ISCminaire scientifique OTAN (AS1 81 /62), tenue au DCpartement de mathkmatiques et de statistique de 1'Universit6 de MontrCal du 3 au 21 aoDt 1981. Cette session avait pour titre general ( 1.
for a certain
c R
A
Thus c o n s i d e r
A,
C
The
C V**,
*
1, Then
lzl
i s analytic in Since
G * fx)(l)
f)(x) =
( " R * f ) ( z ) # 0.
This proves ( 1 . 3 ) .
Our assumption on
i n particular i n a point
f ) (xo) # 0.
we g e t
lzl < R .
A
and
By compactness we conclude t h a t t h e same i s
t r u e i n a c e r t a i n neighborhood o f 1c x
1 A([/**)
t h e n shows t h a t
i s a compact s e t of a n a l y t i c f u n c t i o n s i n (1.1) g i v e s
1 z1
To prove t h e i n v e r s e i n -
Now assume
V
x
0
with and f o r
5 1
f C V
i s comoletq we have f o r
# 0.
Thus
2
The c h o i c e
-
E V*
and f o r a r b i -
z = l/x
co (V) # G ( v * * ) .
0
then gives
Then by a sepa-
r a t i o n theorem i n l o c a l l y convex s e p a r a t e d t o p o l o g i c a l v e c t o r spaces (compare C30, s e c t i o n 20, 7 . ( 1 ) ; s e c t i o n 16, 3 , ( 1 ) 1 ) t h e r e e x i s t s ments of
V**
from
V.
X E A
which s e p a r a t e s e l e -
This i s i m p o s s i b l e by ( 1 . 3 ) .
Although (1.4) i n d i c a t e s a c e r t a i n connection of d u a l i t y and c o n v e x i t y i t t u r n s out t h a t t h e second d u a l
V**
i s i n g e n e r a l more c l o s e l y r e l a t e d t o a s e t
V
than
=(V).
The f o l l o w i n g c o r o l l a r y i s f a l s e with
Let
COROLLARY 1.1:
f h2(V).
0
PROOF: h = h f
0
C
1
V
Then doh any
-
r e p l a c e d by
Fob
hl,h2 C h
co(V).
Moume
.them e u l x an f o C V ouch .that
f C V**
h2 C h
(hl(f)/h2(f)) with
be compacX and complete.
V
A2(f) # 0
F i r s t note t h a t
V**
by ( 1 . 3 ) .
we have
For t h e f u n c t i o n a l
0 C h(V**).
Again u s i n g (1.3) we f i n d
A(fo) = 0.
The f o l l o w i n g two theorems d e a l with t h e r e l a t i o n between second d u a l s and convex h u l l s .
I n t h i s d i s c u s s i o n we u s e some elementary c o n v e x i t y t h e o r y as
d i s c u s s e d i n C70, Appendix A]. a given s e t .
By
E(-)
Note t h a t f o r compact s e t s
we d e n o t e t h e s e t of extreme p o i n t s of
V c A.
we have (Krein-Milman Theorem)
= cow)
,
(1 6 )
co
(~(coV))
€ ( c o ( ~ ) )c V
(1.7) THEOREM 1 . 2 C741:
L&
.
g(zl,z2,...,zn)
be andy.tic i n .the p l q d i 6 c
un
and
Then doti any
PROOF:
V
f C V**
.thtl..ehe exina2 a ptiobabdLty m m m e
i s compact and complete.
it s u f f i c e s t o show t h a t
u on
( 3 ~ ) " ouch .that
I n view of ( 1 . 4 ) , (1.5) and Choquetls Theorem
Consider
g(xlz,
. . . ,xnz)
t V
with
Without l o s s of g e n e r a l i t y assume
Ixk/ < 1 f o r a c e r t a i n k = 1.
l a t h a t t h e r e i s a p r o b a b i l i t y measure
For
.,Zn
z2 ,
The c h o i c e
Since not i n
v
z
j
C U
J
€(=(I)))
1)
Let
( 1 € (f
. ., n ,
on
aU
such t h a t
then gives
(excluding t h e t r i v i a l c a s e when
li c A.
Let
f C V*.
be t h e union of
*
g) (U)
g
does not depend on t h e
(1.7) now g i v e s ( 1 . 8 ) .
f o r any
be compuc.t and complete.
We wish t o show t h a t t h e r e i s a
Ig F U:
(1.11) A
I t i s a consequence of H e r g l o t z ' formu-
i s n o t c o n c e n t r a t e d i n one p o i n t we conclude t h a t t h e f u n c t i o n (1.10) i s
THEOREM 1 . 3 :
Let
. . ,n}.
f i x e d we o b t a i n by convolution with (1.9) t h e r e l a t i o n
= x.z, j = 2,.
f i r s t variable).
PROOF:
v
k F {I,.
(f
*
Re eiY(f
*
g)(U), g C U,
g C If).
Then
y = y(f) F R
If = V * * ,
such t h a t
g ) ( z ) > 0, z C U. and n o t e t h a t
A
Assume (1.11) t o be f a l s e .
i s a domain Then t h e r e i s a
straight line origin.
X~'Yo
(f
Let
E U.
p
through t h e o r i g i n which i n t e r s e c t s
* g) (xo) ,
(f
2)
V**
€
U
From (1.11) we deduce
by Theorem 1.1, f o r
*
h ) (yo)
Then t h e r e e x i s t s
But (1.12) i s i m ~ o s s i b l es i n c e
Re eiY(f
*
(tg
+
i s convex.
g E V**.
If
(1 - t)h))(z) > 0
A
on both s i d e s of t h e
p,
be such p o i n t s on
(0,l)
g,h € U ,
such t h a t
i s complete and Re eiY (f
*
f C V*.
g) (2) > 0, z € U,
g,h € V**, t € C O , l l , in
where
U,
for
g € V
and,
we t h e r e f o r e have
and t h i s shows
tg
+
(1
-
t ) h f V**:
From (1.4) we o b t a i n
and t h u s t h e a s s e r t i o n .
Note t h a t C o r o l l a r y 1.1 a p p l i e s t o t h e s i t u a t i o n d e s c r i b e d i n Theorem 1.3.
I t i m p l i e s t h a t i n complete and compact convex s e t s i n
t h e form
X,/X2,
X 1, A 2
€
A,
Ao,
f u n c t i o n a l s of
a r e extremized by convex l i n e a r combinations of a t
most two extreme 7 o i n t s and t h e i r r o t a t i o n s .
A similar result for analytic
S t i e l t j e s i n t e g r a l s i n [51] h a s found many a p p l i c a t i o n s ( f o r i n s t a n c e C871, C791).
1.2.
Test s e t s The most d i f f i c u l t p a r t i n t h e a p p l i c a t i o n o f t h e d u a l i t y p r i n c i ~ l ei s
t h e d e t e r m i n a t i o n of t h e second d u a l f o r a given interesting set be a p p l i e d , i . e . ,
U c A.
t o f i n d a (small) s e t
U c V**.
V c A. V c U,
o r , i n t u r n , f o r a given such t h a t Theorem 1.1 can
This l a t t e r problem i s somewhat e a s i e r t o h a n d l e and
l e a d s t o t h e i n t r o d u c t i o n of t e s t s e t s . DEFINITION:
Let
If c Ao.
Then
T c A.
i s called a X a X
for
U
(written
T -> U)
if
The f o l l o w i n g s i m p l e o b s e r v a t i o n w i l l be u s e f u l :
a set
U c A.
is a
d u a l s e t i f and o n l y i f
In particular,
for arbitrary
V c Ao.
THEOREM 1 . 4 :
The i(ol?.louLng
k e e a t i o m hold
c T => 2
(1.16)
T
(1.17)
T1 -> T ---, T* = T* 2 1 2 ' T1 c T2 c T3
(1.18)
and
T* 1
( U Tk)* =
(1.20)
3
Tk, Uk
T
T
1
-> T2 -> T3
1 -> Tg
T1 = T2
o f (1.15) g i v e s ( 1 . 1 7 ) .
c T** 2
Mi
,
9
and (1.16) i m p l i e s
From ( 1 . 1 6 ) we o b t a i n
Ao:
,
(1.16) f o l l o w s from t h e d e f i n i t i o n o f d u a l s e t s .
(1.17) we have
C
T* 2 '
T1 -> T => 3
T1 -> T2 -> T => 3
(1.19)
PROOF:
1
604
T* 1
T; 3
3
T* 2
From t h e assumption i n T* 2 3
3
T;,
T***. 1
An a p p l i c a t i o n
w h i l e (1.17) g i v e s
T* = T* 1 3'
Thus
T2 c T;*
= T** 1
T; = T I
T i * = T** = T;* 2
and
T3 c T** = T**. 3 2
and
From t h e a s s u m p t i o n we have T3 c T i *
( m k ) *
3
(
serve t h a t
U Ti*
3
(
U Ti)*
u T;)** (
3
U Tk)**
and t h u s
T I * c T**** = TT*. 1
T3 c T i * . Since
( 1 . 2 0 ) i s immediate from t h e d e f i n i t i o n o f d u a l i t y ,
fl
=
U T;,
= (
To Drove ( 1 . 1 9 ) we h a v e t o show
T2 c T;"
t h e r e s u l t follows.
and t h i s i m p l i e s
and (1.18) f o l l o w s from
Ti*
n Tk.
3
An a p p l i c a t i o n of (1.16) g i v e s
which i s ( 1 . 2 1 ) .
n Ti)*
3
F o r t h e p r o o f o f (1.22) we ob-
U T i * by ( 1 . 2 0 ) , ( 1 . 2 1 ) .
( 1 . 2 2 ) f o l l o w s from
U u,. For
U , V c A.
THEORE!!
(1.24)
1.5 :
Tk
let
Let
U
V
Tk, Ifk,
-> U k , k = 1 , 2
be t h e d i r e c t product
V
C
AO, Tk
=>
T
cumpXete.
Then
T2 -> U1 ' U 2
1
.
I n t h e proof o f t h i s theorem and on o t h e r o c c a s i o n s we u s e t h e n o t a t i o n
* z ' * x' i f t h e c o n v o l u t i o n i s t o be performed w . r .
etc.
t o the variable
z , x,
etc.
Note t h a t
convolution involving v a r i o u s v a r i a b l e s is a s s o c i a t i v e :
f ( x ) *x ( F ( z , x ) * z g ( z 1 ) = ( f ( x 1 *x F ( z , x ) ) * z g ( z )
PROOF o f Theorem 1 . 5 :
To p r o v e ( 1 . 2 3 ) we may assume t h a t
j u s t one e l e m e n t , s a y
g.
f 6 T1, h C (TI
such t h a t f o r any
V)*
. V
contains
The g e n e r a l c a s e t h e n f o l l o w s by a p p l y i n g ( 1 . 2 2 ) .
1x1
5 1
(completeness!)
Let
This shows t h a t t h e f u n c t i o n
(z
fixed) i s i n
T;.
+
C,
f C U1 c T i *
For a r b i t r a r y
Thus (1.25) h o l d s w i t h t h e new
x,z C U .
with Hurwitz' theorem g i v e s f g € (T1
V)**
(h
and f i n a l l y
c a t i o n o f (1.23) :
1.3.
U
FZ:
T1
*
T2 -> T,
FZ(x) *x f ( x ) # 0,
t o o , and t h e l i m i t
g ) ) ( z ) # 0, z 6 U.
(f
U1
f,
we o b t a i n
V c (TI
V)**.
U 2 -> U1
U2.
(1.24)
x
+
1 together
This i m p l i e s
i s an i t e r a t e d a p p l i -
S p e c i a l c a s e s (1) In t h e next two s e c t i o n s we s h a l l d e t e r m i n e a f a i r l y b i g c l a s s o f s e t s i n
A.
t o which t h e above c o n c e p t s a n p l y . THEOREM 1 . 6 :
LeA
V = ((1
+
A s i m p l e but c r u c i a l r e s u l t i s :
xz)/(l
+
yz)
I
1x1 = l y l = 1 ) .
V** = H.
H denotes t h e c l a s s of functions
for a certain PROOF:
y C
R.
We w r i t e -l t-x z -
ltyz
f C A.
f C A.
is in
V*
i f and only i f
(1
-
x 1 X t-. Y l+yz Y
-)-
such t h a t
T~QM
f
for
z E U , 1x1 =
For
y
lyl
* -lt-xz
ltyz
= 1,
f i x e d and v a r y i n g
t
X
-2 Y
0
t h e r i g h t hand s i d e of (1.27) r e p r e s e n t s t h e
Thus
f (U)
Re f (2) > $, z € U.
f (0) = 1:
X
( 1 - --)f(-yz)
or
x,
i.
Re w =
straigth line
-
cannot i n t e r s e c t t h a t l i n e and because of f 6 V*.
This c o n d i t i o n i s a l s o s u f f i c i e n t f o r
Any such (and no o t h e r ) f u n c t i o n has a H e r g l o t z r e p r e s e n t a t i o n
where
i s a p r o b a b i l i t y measure on
such t h a t t h e range implies
0
1
(f
*
(f
g)(U)
*
g ) (U)
satisfies
0 C (fo
*
1-1 0
g) (U)
g € tf
Now i f
we have
i s c o n t a i n e d i n t h e i n t e r i o r of
and t h u s :
f i n d a two-point measure
aU.
g C V**.
If
g C
A.
is not i n
such t h a t t h e c o r r e s p o n d i n g f u n c t i o n
and t h i s shows
g
{
V**.
g(U).
U,
This
one can
f o € V*
We omit t h e d e t a i l s .
The f o l l o w i n g r e s u l t which g e n e r a l i z e s Theorem 1 . 6 w i l l be r e f i n e d i n t h e next s e c t i o n .
T h e r e f o r e we s t a t e it a s a lemma.
Note t h a t f o r
a , @> 0
we have
f f a - u B = t f a+B ,
For
a
7
0
we u s e t h e n o t a t i o n
and t h a t 1+xz 1t y z
Fah
LEMMA 1 . 1 :
a
2
c H, x,y c U
.
we have
1
Cal PROOF: 1.6,
TI V1 and V1 k=l An i t e r a t e d a p p l i c a t i o n of (1.23) g i v e s We have
Va = Va-Cal
complete w i t h (la
->
va-Cal
V1 -> H
ffw
If
by Theorem F C Haw1,
we have
and t h u s
(1.18) i m p l i e s
Va -> V1
f o l l o w s now from (1.19) THEOREM 1 . 7 :
p-1
Fa& al ,. . .
n
(1tx.z) 3 j =1
PROOF:
For
f C V**, x C U,
x C U
let
V1
Ha-1
-
Ha.
.
V={n Then do& any
and (1.23) g i v e s
ak
c c ee/t CAo
we have
1
xj
cUa
j = 1 ,..., n ] .
The r e s u l t
be automorphisms of
x. C
U,
Note t h a t
C
For
J
w € U,
U.
These f u n c t i o n s a r e c o r r e l a t e d by
put
a(wx.) I
=
such t h a t with (1.31)
with
i s independent of
-
z.
Thus f o r a r b i t r a r y
a
(ltXZ) *z 1-b(z)w
Since t h i s i s t r u e f o r a r b i t r a r y
a
n
t
n
( i t X . W )j w j=1 I
z,w C U
g € V*
we o b t a i n from
.
we deduce t h a t t h e f u n c t i o n
FZ(w) C AO
with
F (w) = z
is in
V*.
We n o t e t h a t
Now l e t
f # 0
f € V**
in
(ltxqa g ( z ) *z 1-b(z)w g(z) *z (1+xda
such t h a t
U by t h e d u a l i t y p r i n c i p l e , i n p a r t i c u l a r
f (Y) # 0 .
Thus we may a p p l y Hurwitzl theorem (1.33)
g(z)
*
(1
-t
1)
t o deduce
~ z ) ~ f ( b ( z )# ) 0, z € U ,
t
# 0
because t h e f u n c t i o n (1.33) i s g t V*
(w
in
z = 0.
(1.33) h o l d s f o r a r b i t r a r y
and t h i s i m p l i e s
This r e s u l t h a s a number of u s e f u l a p p l i c a t i o n s .
A v e r y important
s p e c i a l case is contained i n t h e next c o r o l l a r y . COROLLARY 1 . 2 :
k = 1,
...,m,
PROOF:
Since
Let
a
Let
(1
be
an in inheaxem
Bk t C,
Fax ce.ktcLin
1.7.
anawne
f € U.
Using (1.31) we can w r i t e t h e f u n c t i o n (1.34) a s
i s an automorphism o f
U
t h e proof i s complete.
An i m ~ r e s s i v edemonstration of t h e power of C o r o l l a r y 1 . 2 i s t h e proof of o u r n e x t theorem which, i n f a c t , i s e q u i v a l e n t t o SzegB1s theorem ( 0 . 1 ) . p, deg p 5 n ,
denote t h e s e t of polynomials P 6 pn¶ p
nonvanishing i n THEOREM 1 . 8 :
p
and
is in
pn
i f and only i f
U.
Fan n
€
N let
Vn = ( ( 1
t
xzln
I
x t
v).
Let
Then
Pn
PROOF:
1)
Using t h e f u n c t i o n a l s V** c Pn U0. n
p r i n c i p l e we s e e :
h k ( f ) = f ( k ) (0) C A , k > n , The f u n c t i o n a l
V** n
used t o show t h a t t h e f u n c t i o n s i n remains t o show:
Pn n
A(f) = f (zo) , z
cannot v a n i s h i n
V* ;
U:
0
C U,
c
can be
nnAo.
It
Ao.
The n e x t s t e p i s t o prove
2) in
Vn ->
and t h e d u a l i t y
V;
3
V;.
In f a c t , i f
f € V;,
we have
Thus
lfr(0)151
U:
(1.35)implies
z k 1 , k = l
.
n,
n
1 fr(0)=;
and
1
zk.
k=1 and (1
trivial. therefore
Assume i t h o l d s f o r (1
+ V
y z ) C Vi*, y C
n A
(1.37)
Vn-l
f = 1
+
x f l ( 0 ) z # 0, z C U
Now we proceed by mathematical i n d u c t i o n .
(1.36) Since
*
xz)
,
f € V;.
which g i v e s 3)
+
->
-> { ( l
n
vl
+
"
n - 1.
n = 1 t h e claim i s
For
V;* c V** n "
From 2) we know t h a t
U. C o r o l l a r y 1.2 a p p l i e d t o
xz)"-l(l
+
yz)
I
x,y C U } =
v1
V =
Vn-l
and
gives Vn ' U = V**, 1
.
we can apply (1.23) t o o b t a i n A
-> Vl
Pn-ln
A
A ~ =) pn
n
.
The r e s u l t f o l l o w s from (1.19).
C 741.
The i d e a t o t h i s proof a s w e l l a s C o r o l l a r y 1 . 2 a r e due t o Sheil-Small n k Note t h a t a polynomial akz C A. i s in i f and o n l y i f k= 0
Vi
Thus Theorem 1 . 8 shows t h a t f o r every This i s SzegE's Theorem 0 . 1 .
q C
Pn
fl A.
we have
*
p
q
# 0, z
C U.
Of c o u r s e , Theorem 1 . 7 c a r r i e s more i n f o r m a t i o n
since the duality principle applies t o t h i s situation.
1.4.
S p e c i a l c a s e s (2) For
a,f3
r 0 let
T(a,B) = (
( l t x z ) Cal ( l t y z ) a - ~ a l ( l + u z )B
I
x,y,uCm
The aim of t h i s s e c t i o n i s t o d e t e r m i n e f a i r l y l a r g e s e t s T(a,B)
are test sets.
.
K(a,G)
f o r which
The f i n a l r e s u l t (Theorem 1 . 9 ) i s due t o Sheil-Small [ 7 4 ]
and s l i g h t l y weaker f o r m u l a t i o n s a r e i n C581.
In both p r e v i o u s approaches a geo-
a",
m e t r i c p r o p e r t y of f u n c t i o n s , " s t a r l i k e of o r d e r
was a c r u c i a l i n g r e d i e n t .
The proof p r e s e n t e d i n t h i s s e c t i o n makes no u s e of t h a t r e s u l t .
We s t a r t with a p r e l i m i n a r y o b s e r v a t i o n . LEMMA 1 . 2 :
fok
6
z
1
we have
T(1,B
-
1)*
3
T(l,B)*.
I n t h e proof we need a method which r e c e n t l y found many a p p l i c a t i o n s and r e f l e c t s , i n f a c t , a s p e c i a l c a s e of t h e J u l i a - W o l f f Theorem. J a c k ' s Lemma C261:
I t i s known a s
PROOF of Lemma 1 . 2 : of t h e functions
1)
f 6 T(l,y)*,
F i r s t we g i v e an a l t e r n a t i v e c h a r a c t e r i z a t i o n A s l i g h t m o d i f i c a t i o n of t h e d e f i n i t i o n i s
which i s e q u i v a l e n t t o t h e s t a t e m e n t t h a t t h e l e f t hand s i d e of (1.38) has r e a l part
>
2
in
U.
Let
Then t h e i d e n t i t y
leads t o t h e r e l a t i o n
A combination of (1.38) and (1.40) shows:
Note t h a t t h i s h o l d s f o r
2)
Now l e t
T(1,1)** = ff e x i s t s an
3
T(1,O)
62
f C T(l,y)*
i f and o n l y i f
y 1 0.
1.
If
and t h u s
f € T(l,B)*\T(l,P
P
= 1
we conclude from Theorem 1 . 6 t h a t
T ( 1 , 1 ) * c T ( l Y 0 ) * . For
- I)*.
(3 > 1 assume t h e r e
I f we w r i t e
1
"'6-1
w(z)
[~E B- 1T = 1-w( z ) , B #
t h e n i f f o l l o w s from o u r a s s u m p t i o n s t h a t that there exists
z0 C U
such t h a t
w
1 w(zo) 1
i s meromorphic i n = 1,
x = z w ' ( z )/w(z ) r 1.
From Lemma 1 . 3 we g e t
0
0
2 ,
I w(z) 1
5 1
U , w(0) = 0,
for
1z(
5
and
IzoI.
Taking t h e l o g a r i t h m i c d e r i v a t i v e
0
o f ( 1 . 4 3 ) and u s i n g ( 1 . 4 1 ) we o b t a i n a f t e r some m a n i p u l a t i o n zf' (z)
1 B -
6-1
Since
it is c l e a r t h a t
f C T(l,B)*
T h i s shows t h a t
fS("
PROOF:
y
f
B
# 0
in
U,
1 -
in particular,
f B ( z o ) # 0.
w(zo) # 1 and t h u s
(1.45) c o n t r a d i c t s (1.42) with For
( 1 - 1-zw' ( z ) / w ( z ) B- 1 1-w(z)
, w(z)
7
0
y = B.
The p r o o f i s c o m p l e t e .
we d e f i n e
The p r o o f c o n s i s t s o f a l a r g e number of t e s t s e t o p e r a t i o n s a s d e s c r i b e d
i n Theorems 1 . 4 , 1 . 5 .
We s t a r t w i t h t h e c a s e
T(1,B - I ) * * c T ( l , B ) * * ,
a
= 1.
and t h u s , by C o r o l l a r y 1 . 2 ,
From Lemma 1 . 3 we h a v e
(compare Theorem 1 . 6 ) .
An i n d u c t i v e argument g i v e s contains
0
-
1
( I t x z ) / q € ff
Since
T(1,B) -> D(0,CBI)
( 1 1 )
+
x z ) ( 1 t v z ) Cal-1 €
The l a t t e r set c o n t a i n s
This s e t contains
D(a
a = 1.
COROLLARY 1 . 3 :
FCal n
D(1,O)
-
1,O)
Let
q C Dl
T(1,P
-
we o b t a i n
[Dl).
The l a t t e r s e t
and t h u s
This is t h e d e s i r e d r e s u l t f o r ((1
for
a > 1.
Now l e t
From Theorem 1 . 7 w e g e t
AO)
T(a
-
T(l,B),
15 a 5
B.
1,B),
and a n i n d u c t i v e argument g i v e s
and t h u s
Then
This i s an obvious consequence o f Lemmas 1.1, 1 . 4
since (in the f i r s t case):
The second c a s e i s similar. S i n c e second d u a l s a r e c l o s e d we may improve C o r o l l a r y 1 . 3 by t a k i n g t h e
ffa
c l o s u r e s of t h e r i g h t hand s i d e s o f (1.47).
is already closed.
Let
Then we have Re
:
-Re[
q (2)
m
1 k= 1
yk GIxkz k
-
2
I
2 '
Z
z
u
.
I t i s w e l l known t h a t t h e f u n c t i o n s
a r e dense i n t h e s e t of f u n c t i o n s Thus
D(0,y)
and
with
i s dense i n t h e ( c l o s e d ) s e t
Re (zg' ( z ) / g ( z ) ) > (1.48)
f f A
f (0) = 0
K(0,y)
and
of f u n c t i o n s
Y7 . Now l e t K(a,B) =
K(O,B
-
a),
Re f 2
O5a.E
B ,
g E
A.
5
in with
U.
Thus we g e t THEOREM 1.9:
K(a,B) the fact that
r 1 we have
a 1 1,
fok
a r e called the Kapkn
clanae,h of t y p e
This i s due t o
(q,B).
i s t h e c l a s s o f d e r i v a t i v e s of t h e s o - c a l l e d c l o s e - t o -
K(1,3)
I n h i s work,
convex f u n c t i o n s , f i r s t i n t r o d u c e d by Kaplan C291 ( s e e Chapter 2 ) . Kaplan used an i n t r i n s i c d e f i n i t i o n of t h i s c l a s s , namely
i9
This e x t e n d s t o
) 2 -IT
+
8
1
2
. : a1
- 82
+
is in
K(1,3)
2n, 0 < r < 1,
.
K(a,B) :
THEOREM 1.10:
f C A.
i9 (1.51)
i8
2, - a r e f ( r e
arg f (re
1 9 ~< 9
and f o r
U
i f and o n l y i f it i s nonvanishing i n
f € A.
arg f(re
2,
-
and nonvanisklng i n
i9 arg f ( r e
) 2 -an
U
- f (a -
i.A i n K(a,B), a,B
B)(O1
-
02)
.
For a p r o o f , u s i n g Kaplan's o r i g i n a l i d e a , s e e Sheil-Small C741. I s it perhaps t r u e t h a t
seems t o b e a weakness i n Theorem 1 . 9 .
2 0,
T(a,B)
There can be
r e p l a c e d by t h e s e t s
The answer i s n o t known b u t a h i n t i n t h i s d i r e c t i o n i s c o n t a i n e d i n t h e f o l l o w i n g theorem. THEOREM 1.11: p o b a U y
L a a
meanwe p on
2 1,
6
2 1,
auch
Mat
and
f € K(a,B).
Then Xhehe
a
PROOF:
I t f o l l o w s from Theorem 1 . 9 and t h e d u a l i t y p r i n c i p l e (compare Theorem
1 . 2 ) t h a t e v e r y extreme p o i n t
f (z) = (
where we may assume
f €
/((a,@)
h a s t h e form
l+xz)["l (l+yz)"-[al
0 c y = a
n
- [a]
W
and
Then (1.56) hala2 doh PROOF: (f
*
Let
*
g)
f , g € V*, h € W .
f
*
Vn
that
V** = n
f
*
U
(f € V*,g
+
X Z ) (~ x 6
(g
*
*
h) =
h f VX*),
g € V*** = V*. From Theorem 1 . 8 we have f o r
EXAMPLE:
relation
I f we u s e (1.57) t w i c e , we o b t a i n
S i n c e t h i s f u n c t i o n i s nonvanishing i n
h € W.
we conclude
U = V*.
Cn
!l Ao.
Vn = { (1
U) t h e
Thus it f o l l o w s a l r e a d y from t h e d e f i n i t i o n of d u a l i t y
h a s p r o p e r t y (1.57) and we conclude t h a t
Vi
i s c l o s e d under convo-
Note t h a t t h i s i s e x a c t l y SzegS's Theorem 0.1.
lution.
T(a,B)*, a,B r 1.
Next we s t u d y t h e s e t s
To show t h a t t h e s e s e t s a r e
i n v a r i a n t under c o n v o l u t i o n we need some p r e l i m i n a r y r e s u l t s which w i l l be u s e f u l also in other situations. THEOREM 1.12:
Then doh
Fotr
let
g f AO
evmy f C V* and F
€
A we have co(F(U))
PROOF:
V -> H for
We have {g)
H € H.
f
*
g # 0
and t h u s For
y € U
in
f € (H
U.
S i n c e from Theorems 1.5, 1 . 6 we o b t a i n
{g)) *,
f i x e d and
a
1 H = (1-yz
i s in
.
€
we conclude t h a t
R
i+
(f
*
(Hg))/ ( f
*
g) # 0
t h e function ia)/(i
+
ia)
H and i n s e r t i n g t h i s i n t o t h e above i n e q u a l i t y we g e t
Re F Z ( y ) >
1
for
H e r g l o t z l formula i m p l i e s t h e e x i s t e n c e of a measure
and t h u s f o r
For
aU
such t h a t
F € A
(1.58) i s t h e l i m i t i n g c a s e
PROOF:
1-Iz on
x,y E
y
+
1.
we have
l t X Zg ltyz
Theorem 1.9 g i v e s
T(a,B)* = K(a,B)*
€ K(a,B)
.
and t h u s Theorem 1 . 1 2 a p p l i e s .
The n e x t two theorems a r e g e n e r a l i z a t i o n s and r e f i n e m e n t s of C o r o l l a r y
S p e c i a l c a s e s o f Theorem 1 . 1 3 a r e i n Sheil-Small C741 and i n C581. PROOF:
F i r s t assume
p 2 V.
If
1-1 5 1 t h e a s s e r t i o n i s a s p e c i a l c a s e o f
Corollary 1.5.
Now let
u
7
There are functions R E K(y and
Q = R
For k = O,l,
1 and without loss of generality assume v
- v,O), S E K(v,v)
with
f = R
2
1.
S. Let m = [p]
such that
...,m
- 1 we have
and thus by Corollary 1.5
Multiplication of all these functions yields
v1 Nowlet
n = C v 3 and
v1 n = H cff.
P=snV ] ? , $ ' ( K C
For k = O
,...,n
we
have
and by Corollary 1.5
...,n
Multiplication of these functions for k = O,l,
- 1 gives
v- 1 Finally, (1.61) for k = n
shows gRSv
K(a
- 1,B -1) and since sl/'
6
H
we
conclude from C o r o l l a r y 1.5 t h a t
A m u l t i p l i c a t i o n of (1.60)'
(1.62) and (1.63) g i v e s t h e r e s u l t .
The c a s e
1-1 < v
can b e proved by e x a c t l y t h e same method.
FOR
THEOREM 1.14: Then
f
PROOF:
*
g E K(a - 1'0)
Re F2 5 ; ( 1
Now l e t
-
a)
z g l / g = g1
-
-
is in
with
F1' F2
in
1 1
K(0,B
F E A.
A function
exist functions
a,B
RQA: f
g E K(a
E T(a,B)*,
-
1,O)
K(0,B
-
1)
g2
where
-
1).
i f and o n l y i f t h e r e
F1(0) = F2(0) = 0, Re F1 2 !(1
such t h a t
l,O)*K(O,B
1).
K(a
U,
-
g l , g2
-
B),
s a t i s f y t h e c o n d i t i o n s mentioned above.
The i d e n t i t y
g i v e s t h e r e s u l t once we have shown t h a t U.
Re hl 2 i ( 1
But t h i s f o l l o w s from t h e assumptions f o r
g € K(a- 1,B
-
PROOF:
f 3 ) , Re h 2 2 i ( 1
-
a)
in
and C o r o l l a r y 1 . 5 s i n c e
1' 82
1).
THEOREM 1.15:
Le,t
i)
h E T(a,B)* -->
ii)
h E K(a,B) =>
iii)
g
-
h E T(a,B)** =>
a,B 1 1 and
*
f f
f C T(a,B)*.
Then
h E T(a,B)*,
*
h € K(a,B),
f
*
h E T(a,B)**.
Without l o s s of g e n e r a l i t y we assume
15
5
a.
According t o Lemma 1 . 5
r e l a t i o n i) f o l l o w s from i i ) . g f K(a - f3,O), F f K(f3,P)
Let
with
h € K(a,f3) h = gF.
be such t h a t t h e r e a r e f u n c t i o n s
From Theorem 1.13 we o b t a i n
f * h = f * (gF)= ( f * g ) Fo E K(f3,B).
with and
i i ) f o l l o w s from Theorem 1.14 s i n c e
f E T ( a , @ ) *c T ( a
T(a,B)*. Since
-
6 +l,l)*.
From i ) we o b t a i n
fo
0
is arbitrary i n
To prove i i i ) l e t
*
0
f € T(a,B)*
T(a,B)*
fo
and t h u s
we conclude
f
*
-
g f K(a
6,O)
K(0,O)
be a second f u n c t i o n i n fo
*
f
*
h # 0
in
U.
h 6 T(a,B)**.
Theorem 1.15 has f i r s t been proved i n C581 and by Sheil-Small [74]. t h a t t h e example given above s t a t e s t h a t Theorem 1.15 h o l d s f o r a s well.
For
T(O,n),
however, it f a i l s .
Note
T(n,O), n C N ,
The e x a c t range o f t h e parameters
a,P
f o r which Theorem 1.15 i s v a l i d i s unknown. To conlude t h i s s e c t i o n we prove t h a t under c e r t a i n c i r c u m s t a n c e s , convol u t i o n i n v a r i a n c e of a s e t t r a n s f e r s t o l a r g e r s e t s . THEOREM 1.16:
h
*
f
*
q f Ao.
I t w i l l be s u f f i c i e n t t o prove t h a t
f , g € V**, h € V*.
Let g # 0
in
U.
For
z
XI f A
fixed l e t
Now choose
such t h a t
X2(q) = (h
t h e d u a l i t y p r i n c i p l e shows t h a t a r e l e f t with t h e proof o f
h
*
*
go
*
X 2 ( f ) = X 2 ( f 0) fo
*
go # 0
f i n i t i o n o f d u a l i t y and t h e assumption
fo
*
in
q) (z)
.
h
*
*
q)(z),
*
f
go # 0
in
U.
Another a p p l i c a t i o n of
for a certain U.
f
such t h a t
go C V
and t h e r e s u l t f o l l o w s i f we can show t h a t
X2 E h
*
Al(q) = (h
with
From t h e d u a l i t y p r i n c i p l e we o b t a i n a f u n c t i o n
hl (g) = h1 (go)
V -il closed
26
&ue do& V**.
unda conuolu;tiovln Xhe dame PROOF:
be complete and compacX.
L e t V c A.
f
0
E V
and we
But t h i s f o l l o w s from t h e de-
go E V .
A s i m i l a r s t a t e m e n t d e a l s w i t h convex s e t s i n
Ao.
Although t h i s r e s u l t
i s n o t d i r e c t l y r e l a t e d t o d u a l i t y we p r e f e r t o mention it a t t h i s s t a g e .
Let
THEOREM 1.17:
~uncaXon h C A.
V
6uch ;that
AO with W =
C
604
in
Let
V,
Vc
cornpa&.
&nwnefimeLh a
a l l f , g 6 V we have
Then (1.64) h o l d 604 & f , g C PROOF:
GV
W.
d e n o t e t h e s e t of f i n i t e convex l i n e a r combinations of f u n c t i o n s
such t h a t
= W.
C
holds f o r a r b i t r a r y
Since
f ,g € Vc.
W
i s convex we f i r s t conclude t h a t (1.64)
Since
W
i s compact, (1.64) h o l d s f o r
f ,g C
7
a s well.
1.6.
Additional information 1)
We wish t o mention two more s t r u c t u r a l p r o p e r t i e s o f d u a l i t y .
have seen (Theorem 1 . 8 ) t h a t a s i n g l e f u n c t i o n (namely s e t f o r a large set. such p r o p e r t i e s .
(1
+
We
can b e a t e s t
z)*)
I t would be very i n t e r e s t i n g t o determine a l l f u n c t i o n s with
A negative r e s u l t i n t h i s d i r e c t i o n i s contained i n t h e next
theorem.
Let
THEOREM 1.18:
V = {f),
Here we d e n o t e by
Clearly, i f only i f i )
f =
1;
akr
k
f(-')
C A.
ak # 0, k z 0 ,
whehe
f € A.
and
f
ex,&%.
Then
A.
i f and
t h e s o l u t i o n of
t h e e q u a t i o n (1.65) can b e s o l v e d i n
and i i )
lak/l'k
+ 1,
k +
m.
PROOF:
Under t h e assumptions we have
h C VX* such t h a t
Now l e t U.
The f u n c t i o n s
(h
*
f (-I))
*
g # 0
g C Ao,
for arbitrary
g # 0
in
g € T ( l , $ ) , $ 2 1, have t h i s p r o p e r t y and t h u s
I n Chapter 2 (Theorem 2.3) we s h a l l prove t h a t t h e l a t t e r s e t c o n s i s t s of t h e functions
1
-
x ) , x C
.
Thus we have
L e t T1 ,T2
THEOREM 1.19 :
c A.
h = f
*
(1
-
xz)-l,
the result.
be complete and compaot.
Fon y
C
R let
Then
PROOF:
Let
g = ygl
+
-
y)g2, g j C Tt*. 7 duality principle the existence of f . C T J j we have
In p a r t i c u l a r , f o r is i n
A
and t h u s
(1
h C V*
Y
(h
*
and
z C U
g) ( z ) = (h
*
If
h C A,
we conclude from t h e
such t h a t f o r
f = yf
fixed, the functional
f ) ( z ) # 0.
This implies
+
(l-y)f2 c V Y
X(q) = (h g C Vj*
*
q) ( z )
which i s
the result. 2)
The f o l l o w i n g c o r o l l a r y t o t h e d u a l i t y p r i n c i p l e h a s a number of s u r -
p r i s i n g a p p l i c a t i o n s s i n c e it p e r m i t s t o t r a n s f e r c e r t a i n e x t r e m a l problems f o r second d u a l s t o d i f f e r e n t extremal problems f o r n o t r e l a t e d t e s t s e t s .
L e t Tj
THEOREM 1.20:
C
Ao, j = 1 , 2 ,
be compact and comflete,
g c Ao.
T h w we have
i.6 and o n l y i.6
PROOF:
F i r s t we prove t h e theorem with (1.67) r e p l a c e d by
(1.68)
g*hCT;
I n f a c t , assume
h C Ti*
for arbitrary (1.68).
f C Tf *.
for a l l
Then (1.66) shows t h a t
*
g
and t h u s
.
h C T;* g
f C T;** = T;.
*
*
h
f = (g
*
f)
To prove (1.67) =>
f C T2.
From t h e d u a l i t y p r i n c i p l e it i s c l e a r t h a t
h C T;*
i f t h e same i s t r u e f o r a l l
h € T1.
h # 0
T h e r e f o r e (1.66) i m p l i e s
The o t h e r d i r e c t i o n f o l l o w s by i n t e r c h a n g i n g t h e s u b s c r i p t s
Obviously, (1.68) i m p l i e s (1.67).
*
1,2.
(1.68) choose an a r b i t r a r y g
*
f
*
h # 0
for a l l
The proof i s complete.
Some a p p l i c a t i o n s w i l l be given i n Chapter 2 , s e c t i o n 8; compare [SO].
3)
We r e t u r n t o Theorem 1.11 and g i v e an a l t e r n a t e proof which, however,
works o n l y f o r
a r 2.
In f a c t , consider
TO(a,B) = {
Writing
Tl (y, 0) = { ( 1
This i m p l i e s f o r
t
xzlY
a,@r 1
I
(ltxz) (ltyz)a-l (ltuz)B
x
C
we have
1
XYYYU C
According t o Theorem 1.1 t h e extreme p o i n t s of sets,
To(ci,B)
and
T(a,B).
For
a
2 2,
-
c o K(a,B)
a r e c o n t a i n e d i n both
however, t h e i n t e r s e c t i o n c o n s i s l s of
the functions
S i n c e f u n c t i o n s with
x € U
or
y € U
cannot b e extreme p o i n t s , t h e c o n c l u s i o n
follows. Comparison of (1.69) with (1.50) l e a d s t o t h e f o l l o w i n g problem: I s it t r u e t h a t i f t h e compact and complete s e t s f o r t h e same s e t
U,
the intersection
The answer i s unknown.
T1 0 T2
are test sets
is also a t e s t s e t f o r
U?
I f it i s a f f i r m a t i v e , we would have a proof f o r
t h e problem mentioned a f t e r Theorem 1 . 1 0 , a t l e a s t f o r (1.50).
TI, T2
a r 2,
u s i n g (1.69) and
Chapter 2 APPLICATIONS TO GEOMETRIC FUNCTION THEORY
2.1.
I n t r o d u c t o r y remarks I n t h i s c h a p t e r we s h a l l a p p l y t h e d u a l i t y t h e o r y t o c o n c r e t e s i t u a t i o n s
i n geometric f u n c t i o n t h e o r y , i n p a r t i c u l a r t o ( c l a s s e s o f ) u n i v a l e n t f u n c t i o n s . Most of t h e f u n c t i o n s
f € A
of i n t e r e s t i n t h i s c o n t e x t a r e normalized by t h e
conditions
and t h e c o l l e c t i o n of t h e s e f u n c t i o n s i s denoted by with f C A1
A.
A1.
Since d u a l i t y is dealing
a d i r e c t a p p l i c a t i o n of t h e p r e v i o u s r e s u l t s i s n o t p o s s i b l e . i f and o n l y i f
f / z 6 A.
and f o r
f , g C A1
However,
we have
i f and o n l y i f
and s o t h e r e i s an obvious t r a n s f o r m a t i o n of d u a l i t y t o A function
f C Al
is called nfat.tibe
A1.
06 oadm a
5 1
i f and only if
The s e t of t h e s e f u n c t i o n s is denoted by usual notation
Si).
S
a
( f o r obvious r e a s o n s we avoid t h e
In p a r t i c u l a r ,
and t h e s e f u n c t i o n s p l a y an important r o l e i n extremal problems f o r w e l l known t h a t
S
c S
a
A1.
univalent functions i n
A function f E
exists
A1
0 5 a 5 1,
i f and o n l y i f
S
.
It i s
i s t h e s e t of a l l
I t i s c l e a r from (2.1) t h a t
i s s a i d t o be i n t h e c l a s s
g C Say p 6 R ,
where
S a
a
1
i f and only i f t h e r e
such t h a t Re e
icp zf ( z )
g(z>
>O,ZCU,
which i s e q u i v a l e n t t o
Co
The f u n c t i o n s i n subclass of
S
a r e c a l l e d close-to-convex and t h e y form an important
so>
( l a r g e r than
Another even l a r g e r s u b s e t of B(a,B), a > 0, (3 C
R.
such t h a t f o r a c e r t a i n
where
Here cp C
( f ( z ) /z) a+iB-1 = 1 a t
f C B(a,B)
S
i s formed by t h e
EuzXevi: a u n c t i o ~ n
i f and o n l y if t h e r e e x i s t s
R
z = 0.
An e q u i v a l e n t c o n d i t i o n i s
g C Sl-a
Another f r e q u e n t l y s t u d i e d e x t e n s i o n of t h e close-to-convex f u n c t i o n s a r e t h e done-.to-convex ~unc.tioltd 04 ondeh only i f t h e r e e x i s t
g E So,
E R,
8.
f C A1
i s such a f u n c t i o n i f and
such t h a t
which i s e q u i v a l e n t t o
A function
k 2 2, of
f(U)
f € A1
i s s a i d t o be
06 boundany hoXalAon at mont
IT,
i f i n a l i m i t i n g s e n s e t h e v a r i a t i o n of t h e t a n g e n t a n g l e a t t h e boundary i s a t most
k i ~ , s e e C70,p.231.
These f u n c t i o n s a r e c h a r a c t e r i z e d by
the representation
and t h u s
We s e e t h a t t h e n o t i o n of Kaplan c l a s s e s describing various geometrical s i t u a t i o n s . K(a,B)
unifies a l l these definitions
Since t h e d u a l i t y theory a p p l i e s t o
we can expect t o o b t a i n some v a l u a b l e i n f o r m a t i o n r e g a r d i n g t h e above-
mentioned f u n c t i o n s .
Some o t h e r s e t s of f u n c t i o n s , d i r e c t l y r e l a t e d t o
w i l l be d i s c u s s e d a s w e l l .
Rike
K(a,B)
06 andm
a 5 1
In p a r t i c u l a r , a function
i f and o n l y i f
f E A1
K(a,@),
i s c a l l e d pted&xh-
Although it i s n o t immediately c l e a r from t h e d e f i n i t i o n what t h e p a r t i c u l a r i n t e r e s t of t h e s e c l a s s e s may be, o u r r e s u l t s w i l l show t h a t t h e y p l a y a c e n t r a l r o l e i n some s i t u a t i o n s .
F i n a l l y we wish t o mention t h e c l a s s
A s we s h a l l see, M l a r g e s u b s e t of
2.2.
M of f u n c t i o n s
f C A1
such t h a t
c o n t a i n s o n l y u n i v a l e n t f u n c t i o n s and seems t o be a f a i r l y
S.
Prestarlike functions Let
(2.11).
Ua
b e t h e c l a s s of p r e s t a r l i k e f u n c t i o n s o f o r d e r
A simple c a l c u l a t i o n u s i n g ( l . 3 9 ) ,
(1.42) shows t h a t
a
a s defined i n
f 6 Ra,
a
5 1,
and o n l y i f
Note t h a t t h e " f a c t o r " name " p r e s t a r l i k e " .
z/(l
-
z ) 2-2a
i s itself in
I f we i n t r o d u c e t h e o p e r a t o r
we deduce from (1.38) t h e e q u i v a l e n t c o n d i t i o n f o r
Sa
.
(2.13) j u s t i f i e s t h e
DY: A1
f € Ra:
-+
A1
with
if
Since
..
("1, n = 0 ~ 1 , .
~"+'f = 2 n!( z n - ' f )
(2.16)
t h e r e l a t i o n (2.15) t a k e s a p a r t i c u l a r l y simple form i f
, 2
-
2a C N.
The s p e c i a l
c a s e s . a = 0 1, ~g i v e
f E R
(2.18)
R1
Thus we have members map
2
u
=
0
zf " Re(-+ f'
S 1 and Ro
= K
2
where
0'
1) 7 0 , z C U .
KO
S
i s t h e s u b c l a s s of
whose
o n t o convex domains.
The f o l l o w i n g theorem i s b a s i c f o r t h e t h e o r y of p r e s t a r l i k e f u n c t i o n s . THEOREM 2 . 1 : ii) PROOF:
i)
Fa& a c 6
Let 5 1
a 5 1
we have
and
f , g € Ra.
Then
f
*
g C
Ra.
Ra c R6.
i ) i s a r e f o r m u l a t i o n of Theorem 1 . 1 5 , i ) u s i n g d e f i n i t i o n ( 2 . 1 1 ) .
prove i i ) , n o t e t h a t
K(1,3 - 2a)
3
K(1,3 - 26)
To
and t h u s by Theorem 1 . 9 and
(1 171,
P a r t i ) of Theorem 2 . 1 h a s t h r e e c a s e s of p a r t i c u l a r i n t e r e s t (2.19)
f , g c A1,
f Re - > z
1,
Reg> z
1=>
Re
f > g z
(in
U)
,
(a=1,;,0):
Although (2.19) can e a s i l y be o b t a i n e d from t h e H e r g l o t z i n t e g r a l r e p r e s e n t a t i o n f o r such f u n c t i o n s , (2.20) and (2.21) a r e much s t r o n g e r . c o n j e c t u r e of ~ 6 l y aand Schoenberg C421 i s v a l i d .
(2.21) s t a t e s t h a t t h e
Theorem 2.1, i i ) i m p l i e s t h a t
an o l d r e s u l t due t o S t r o h h s c k e r C811. For
a = 0, q ,
Theoren 2.1, i ) was f i r s t proved i n C621.
p l e t e proof of Theorem 2 . 1 h a s been given by S u f f r i d g e 1831.
The first com-
He proved a deep and
much s t r o n g e r theorem on t h e composition of polynomials with c e r t a i n r e s t r i c t i o n s on t h e i r z e r o s ( s e e Chapter 4, Theorems 4.14-4.171, relations
-
e q u i v a l e n t t o Theorem 2 . 1
-
and showed t h a t t h e f o l l o w i n g
a r e a l i m i t i n g c a s e of h i s r e s u l t :
Let
Then i f
(2.22)
we have
Furthermore, i f
Compare ( 2 . 2 2 ) ,
a < 6
5 1,
then
(2.23) with SzegZ1s theorem ( 0 . 1 ) !
2 . 1 i s due t o Lewis C 331.
Yet a n o t h e r proof of Theorem
S i n c e we have a H e r g l o t z formula f o r
i s a consequence of Theorem 2 . 1 ,
COROLLARY 2 . 1 : mmute
on
au
Let f
t h e following c o r o l l a r y
f C R1,
ii).
Ra,
C
a 5 1.
Then ;them ex&&
a pobabiecty
nuch ;that
Using t h e c h a r a c t e r i z a t i o n (2.13) we o b t a i n
COROLLARY 2 . 2 : on
m e a w e 1-1
au
Let f
C Sa,
a 5 1.
Then ;thehe e x & a
a pmbabUy
nuch Rhat z
f (z) =
dlJ(T)
au
z
f (z) Q
(l-z)2-2a
'
This r e p r e s e n t a t i o n h a s f i r s t been proved (by a d i f f e r e n t method) by Brickman, Hallenbeck, MacGregor and Wilken C121. i n (2.27) d e c r e a s e f o r
of measures
l~
Theorem 2 . 1 ,
i i ) and ( 2 . 2 5 ) .
The f u n c t i o n s i n
%,
a z
!,
I t i s remarkable, however, t h a t t h e s e t s
a
decreasing.
can be r e p r e s e n t e d with t h e a i d of a
c e r t a i n c o n v o l u t i o n i n t e g r a l and f u n c t i o n s i n for
a ,
1 we have
T h i s i s a consequence of
R , = S T . To t h i s end we n o t e t h a t Z
Z
Since
(1
have f o r
-
z ) - ~= 2Fl(ay 1.1, z)
,
where
i s t h e hypergeometric f u n c t i o n , we
2F1
a > 1
For t h e n o t a t i o n
f
s e e (1.65).
A combination of t h e s e formulas and (2.13)
gives THEOREM 2.2:
exin& a g
Let a
.
Then we have
f C
and anLg
id
them
duch ;thCLt
€ 2
A s a r e s u l t of Theorem 2 . 1 , i i ) we s e e t h a t t h e c l a s s e s decreasing
!ah
Ra
a.
become narrower f o r
The n e x t r e s u l t shows t h a t t h e i r i n t e r s e c t i o n i s v e r y small C491.
THEOREM 2.3: z / ( l - xz), x C
Ra
The i n t m e c t i o n od
Rap a 5 1,
eonhha2 oh t h e Quncfionn
U.
PROOF:
I t w i l l be enough t o s t u d y t h e i n t e r s e c t i o n of
n CN.
Let
W e wish t o show t h a t f o r
k 5 n
the estimate
Ra
with
a
= (1 - n ) / 2 ,
h o l d s with
~
[/I::[
~ =3
(
~
)
)
ntk- 1 k-1
'
Let
be t h e f u n c t i o n a p p e a r i n g i n t h e r e p r e s e n t a t i o n of l a t i o n shows t h a t
r (g/z)ntl
where
q
a
2
= b2.
The
f
k-th c o e f f i c i e n t
by ( 2 . 3 1 ) .
dk
A simple c a l c u -
of t h e expansion of
h a s t h e form
i s a polynomial w i t h non-negative c o e f f i c i e n t s i n i t s v a r i a b l e s .
We
obtain
Idk since
q(1,
..., 1 )
n t l ,,k-1 2
- (k-l]
1
'9(1,....1)
i s the
t
=
]
(
ntk- 1 k-l -
k - t h c o e f f i c i e n t of
1
(z] - z ) .
But from
(2.31) we deduce
which i s (2.32).
Now assume t h a t f i x e d and or
n
k.
Since
f E R
1-n -
f o r every
n E N.
2 l i m y ( n , k ) = 1 we o b t a i n n-
Then (2.32) h o l d s f o r l a k - ak-'1 2
= 0
for this
k f
which i s a f u n c t i o n of t h e form mentioned i n t h e a s s e r t i o n . (2.33) with
l a 2 / 5 1 belongs t o
Ra'
a 5 1
That any f u n c t i o n
i s an immediate consequence of t h e
d e f i n i t i o n of p r e s t a r l i k e f u n c t i o n s . CO
REMARK:
For
1
f (z) =
akz
k
F Ra
one can u s e t h e s t a n d a r d t e c h n i q u e s
k= 1 t o obtain t h e s h a m inequality
which, i n f a c t , d e s c r i b e s t h e e x a c t c o e f f i c i e n t body Another consequence of Theorem 2 . 1 , This i s no l o n g e r t r u e f o r THEOREM 2.3:
1
(a2,a3)
i i ) i s that
R
a
Ran
in c S
a
for
5
;.
.= a 5 1.
%cS
hoh!A eds{ and o d y ir( a '. f
.
This r e s u l t h a s f i r s t been p u b l i s h e d by Silverman and S i l v i a C771. g i v e a proof i n t h e c o n t e x t of d u a l i t y . g € T(1,3
-
2a)**
I n f a c t , t h e d u a l i t y p r i n c i p l e shows t h a t
implies
Thus
g(z) = (1
-
T(1,3
-
such t h a t
2a)**)
We
z)-'
f T ( 1 , 3 - 2a)**, f 1( e ) = ( 1 -
1
2 s
a 5 1,
212 *
(f/z)
and we f i n d
f
has a zero i n
c
pa
U.
(f/z This
e f
i s n o t even l o c a l l y u n i v a l e n t .
The f o l l o w i n g c o n v o l u t i o n r e s u l t f o r p r e s t a r l i k e and s t a r l i k e f u n c t i o n s i s a r e f o r m u l a t i o n o f a s p e c i a l c a s e of C o r o l l a r y 1 . 5 .
THEOREM 2.4:
Fon a 5
1
let
f € Pa, g € Sa, F € A .
Then
(2.35) i s fundamental f o r p r e s t a r l i k e f u n c t i o n s and c o ~ i t a i n sTheorem 2 . 1 .
Another
( e q u i v a l e n t ) s t a t e m e n t i s even more u s e f u l f o r c e r t a i n a p p l i c a t i o n s , i n p a r t i c u l a r because it c l e a r l y p o i n t s out t h e advantage compared with c o n v e x i t y t h e o r y . THEOREM 2.5 : pa
= z(1 - z )
PROOF:
Let
2a- 2
.
Fob
and
g F Sa, F C A,
Then
pa
be t h e s o l u t i o n of
From our assumptions we have f o l l o w s from (2.35)
a < 1 let f C Ra,
g
*
p
a
(-1) pa
*
pa
Ra and
= (
1 - z).
*
pa 6 Sa.
f
Thenwehave
The r e s u l t
. f 5 z
Note t h a t (2.36) with t h e c h o i c e emphasizes t h e r o l e of
pa
i s equivalent t o (2.27).
.Ya.
a s an extremal element i n t h e c l a s s
(2.36)
Extremal
problems of v a r i o u s f i e l d s a r e covered by Theorem 2.5, and a number of a p p l i c a t i o n s w i l l be given w i t h i n t h e n e x t s e c t i o n s .
A t p r e s e n t we c o n f i n e o u r s e l v e s
t o two examples. THEOREM 2.6:
PROOF:
LeX
x , y C U, 0 5 t
I n Theorem 2.5 we choose
a =
i,
r 1, and
g F
R1.2
z F ( z ) = (1-xz) (1-yz) -
The1
and
56
f(z) = z(l
-
(xt t y ( l
-
t))z)-'.
Note t h a t
The l a t t e r s e t i s i n t h e h a l f p l a n e
Re w 2
Theorem 2.6 i s due t o V . Singh.
f C R,. Z
Then we o b t a i n
and t h i s g i v e s (2.37) t = 0
The s p e c i a l c a s e
. i s i n [62].
1 z l a = ~ l Note .
The n e x t r e s u l t d e a l s with t h e c o n v o l u t i o n of f u n c t i o n s i n t h a t i n t h i s range
% C S + 2R a .
(2.38)
Let
THEOREM 2.7:
nMkeneno z
06
f
*
g
-LA
1
2 5
a
5
8
< 1, f F SB, g C Sa.
at lean*
ohden
06
Then Xhe ahdeh
nXan&kenan
ud
on * o6
06
=
F (2-2a,2-28,1,z). 2 1
PROOF:
We a p p l y (2.36) t w i c e t o o b t a i n
V . Singh (unpublished) used t h e continued f r a c t i o n s expansion of o b t a i n t h e f o l l o w i n g lower bound f o r t h e o r d e r of s t a r l i k e n e s s of
pa
* pp,
2F1
to
We omit t h e d e t a i l s .
Note t h a t t h e same method p e r m i t s one t o s t u d y t h e equiva-
l e n t problem f o r c o n v o l u t i o n s of f i n i t e l y many f u n c t i o n s i n
S
a,
z1 5 a
r
1.
The f o l l o w i n g a p p l i c a t i o n of Theorem 2.4 h a s a f l a v o r s i m i l a r t o t h a t of Theorem 2.7.
such t h a t
C(f)
l i e s i n the halfplane
THEOREM 2.8:
Let
(2.39) Theorem 2 . 4 y i e l d s
D B g F Sa.
since
of
-
I
=
f € Ra
let
1
2.
Then
.
n
~ ( f ) ~ ( g )
(6 = 2 - 2a)
a = 0
1 c 1 we conclude t h a t
KO
Re w 1
I n t e r c h a n g i n g t h e r o l e s of
I f , f o r example,
Iw
* g)
For
5 1.
f , g € Ra, a C 1. c(f
PROOF:
a
I t i s v a l i d , however, f o r
C(f
C(f) , C(g)
and
*
c o n s i s t i n g of a l l f u n c t i o n s
f,g
g)
gives t h e r e s u l t . a r e contained i n t h e d i s c
l i e s i n t h e same d i s c :
f F A1
with
the subclass
T
i s c l o s e d under c o n v o l u t i o n . f C A1
(a =
The same h o l d s
4)
f o r t h e c l a s s of f u n c t i o n s
with
(2.41)
I
zf' (z) 1 - 1 / < 2 , Z F U . f (z)
I t i s c l e a r how Theorem 2.8 can b e a p p l i e d i n o t h e r s i m i l a r s i t u a t i o n s . I t i s w e l l known t h a t if have
g ( z ) = fi f ( 6 ) C S1.
i s an odd s t a r l i k e f u n c t i o n o f o r d e r
f
0 we
This can e a s i l y be extended t o t h e f o l l o w i n g
2
statement:
if
we have
This f a c t can be used t o o b t a i n t h e f o l l o w i n g i n t e r e s t i n g s t r u c t u r a l formula f o r
R
a'
a =1
THEOREM 2.9:
PROOF:
The f u n c t i o n
Theorem 2 . 1 ,
-
n/2.
LeZ
n € N
s(z) = z/(l
and
-
i ) and (2.13) we s e e t h a t
The c o n c l u s i o n f o l l o w s from ( 2 . 4 3 ) .
is in
zn) s
*
f
S 1- ;n
and by an a p p l i c a t i o n of
i s a f u n c t i o n of t h e form ( 2 . 4 2 ) .
We conclude t h i s s e c t i o n with t h e d e t e r m i n a t i o n o f a number of s p e c i a l t a r l i k e and p r e s t a r l i k e f u n c t i o n s .
An important s u b c l a s s of
R1
i s described
i n t h e f o l l o w i n g c l a s s i c a l r e s u l t o f F e j 6 r C211: THEOREM 2.10:
Assume
a l = 1,
and Xhat
ak 2 0
doh k t 2 ,
such .that
{ak} -i6 a convex decheanhg sequence, i.e.,
Then
Let
X >
Clearly C
Ri-A.
-;,
-1 5 x 5 1,
G(* , x ) € Sl-A
and
and
G(-,1) =
such t h a t
This l a t t e r statement i s equivalent t o
-1
P:"'")
(x)
k=O P ( X , u ( l ) z k where
G(z,l)
Pk(a'B)
k
cRf-A
3
a r e t h e J a c o b i polynomials
Lewis C341 h a s given a f a r - r e a c h i n g e x t e n s i o n o f (2.44): THEOREM 2.11:
FOX
5
, 1
x 5 1
We have
*
G(z,x)
H i s proof u s e s a c a r e f u l s t u d y of t h e f u n c t i o n (2.45) w r i t t e n i n terms of t h e I t i s t o o complicated t o b e reproduced h e r e .
hypergeometric f u n c t i o n .
x =
-1
of (2.45) i s of p a r t i c u l a r i n t e r e s t s i n c e it shows ( a 1
IB])
The c a s e that
The s t u d y of hypergeometric f u n c t i o n s i s v e r y n a t u r a l i n t h i s c o n t e x t , compare The f o l l o w i n g e x t e n s i o n of (2.46) i s u s e f u l .
Theorem 2.7.
PROOF:
.a,B,y € R, a t z 2
$1.
+
Then
We s h a l l prove t h e e q u i v a l e n t s t a t e m e n t
We f o l l o w t h e argument i n C341.
Since
i s t r i v i a l , we may assume
+
and
IP
1
LeZ
THEOREM 2.12:
F (z) =
F (a , b , c , z) 2 1
t h e meromorphic f u n c t i o n
.
a = 1
a
1
t
+ a +
7
0.
Let
We have t o show t h a t w (z) , w (0) = 0,
r 0 and t h e c a s e
(3
b = 1
+
1
+ a +
(3 t i y , c = 1 t
Re (zF1 ( z ) / F ( z ) )
7
-a/2
f3 = 0
a + iy
or that
d e f i n e d by
zF1 ( z ) = a w(z> F (z) 1-w(z) i s bounded by
1 in
U.
I f t h i s i s f a l s e we f i n d
Iw(z)l 5 ~ w ( z o )=~ 1, l z l 5 I z O I . F(z)
z0 E U
with
According t o Lemma 1.3, x = z 0w l ( z 0 )/w(z 0 ) 2 1.
s a t i s f i e s t h e d i f f e r e n t i a l equation
(2. 50)
z(1
-
z)F"
+
Cc - ( a
+
b
+
l)z]F1
Taking t h e d e r i v a t i v e of (2.49) and e l i m i n a t i n g t o the relation
F"
-
abF = 0
.
l e a d s a f t e r some manipulation
(2.51) with
I n s e r t i n g t h e v a l u e s of A(zo) # 0
a,b ,c
I A ( Z ~1 )= 1 B(z0) 1 .
we f i n d
a,&
under t h e r e s t r i c t i o n s f o r
Furthermore,
Thus (2.51) g i v e s
1 zOl
= 1,
a con-
tradiction.
a
-1.
and maps
onto t h e convex dorrain
U
We d e n o t e t h e c l a s s o f t h e s e f u n c t i o n s by implies 0 < t
6
ft
convex u n i v a l e n t i n
f(U).
K(a).
Since
w = ra(f)
U, i t i s c l e a r t h a t f o r
g C K
we have f o r
-
K(a)
1, u s i n g Theorem 2 . 1 ,
convex u n i v a l e n t i n
-
U.
Thus
r,(f
*
g ) C K(a)
by t h e above d i s c u s s i o n .
We
have t h u s proved t h e f o l l o w i n g e x t e n s i o n of t h e P6lya-Schoenberg c o n j e c t u r e :
THEOREM 5 . 6 :
I n p u c l L e a n , id K(a)
Fok
~ , ( f * g ) C K(a).
w = -ra(f) C K(a), g C K , we have
w l = ?,(f),
w2 = ~ , ( g ) C K(a), ,then
w = w1@w2
= ~
~ * g) ( cf
. I n t h e f o l l o w i n g theorem we determine a s p e c i f i c c l a s s of f u n c t i o n s i n
K(a).
This r e s u l t i s w e l l known f o r
a
w
THEOREM 5 . 7 :
Let
f (z) =
0.
a k zk C A l
be 6uch .th&
k= 1
Then ~ ~ ( C f K(a). ) PROOF:
Without l o s s of g e n e r a l i t y we may assume
f(z)
z.
A look i n t o t h e
second o r d e r d i f f e r e n t i a l e q u a t i o n s s a t i s f i e d by t h e f u n c t i o n s k C N,
dk
/dl
Y
shows t h a t
Hence, f o r
0
