(?) Extremal Analytic Functions in the Unit Circle

Extremal Analytic Functions in the Unit Circle

J O H N L. L E W I S

l~oyal Institute of Technology, Stockholm, S w e d e n

1. Introduction

L e t n be s u b h a r m o n i c in { ] z ] < l } a n d p u t

A(r)---infM=ru(z ), B ( r ) =

maxl~I= ,

u(z)

w h e n 0 < r < 1. L e t ~ be a f i x e d n u m b e r in (0, 1) a n d suppose t h a t

A(r) <

cos

azXB(r),

0 < r < 1 (1.1)

u < c < + ~ . (1.2)

H e r e c is a positive c o n s t a n t . U n d e r t h e s e a s s u m p t i o n s IIellsten, K j e l l b e r g a n d

~ o r s t a d [4] p r o v e d

THEOREM A.

There exists a subharmonic function

U(z~

= - - t a n

2~

z

f t~,-~ _ t~-~, I~,e

1 - - t 2 0

dt,

!argz[ < a ,

i n {Lzl < 1}

for which

(1.1)

holds with equality and such that for 0 < r < 1

(?)

B(r) < U(r) < ~

t a n ~ .

W e n o t e t h a t t h e a u t h o r [5] a n d E s s 6 n [1] h a v e p r o v e d r e l a t e d t h e o r e m s . K e r e we consider a similar p r o b l e m for a n a l y t i c f u n c t i o n s . More specifically, let a be a step f u n c t i o n on [0, 1], i.e.; a piecewise c o n s t a n t f u n c t i o n w i t h a f i n i t e n u m b e r o f jumps, w h i c h satisfies

(*) a is u p p e r s e m i c o n t i n u o u s ,

(**) - - 1 < a ( r ) < 1 w h e n 0 < r < 1 .

174 Jo~N L. L]~WlS

L e t f b e a n a l y t i c in { [z[ < 1} a n d p u t re(r) = mini,l= ~ If(z)1, M ( r ) = maxl=[= ~ [f(z) t w h e n 0 < r < 1. L e t a a n d r o be f i x e d n u m b e r s s a t i s f y i n g 0 < a < 1 a n d 0 < r 0 < 1. F i n a l l y , we suppose t h a t

m ( r ) M ( r ) @ < a ~+~(0, 0 < r < r o , (1.3)

Ifl _< 1. (1.4)

T h e n we shall p r o v e

TI~EOI~E~ 1. There exists an analytic f u n c t i o n F -= F(., a, ~, ro) i n {]z[ < 1}

satisfying (1.3) and (1.4) with the following properties:

(vii) (viii)

(i) I f f satisfies (1.3) and (1.4), then M ( r , f ) < F(r), 0 < r < 1,

(ii) F is the unique analytic f u n c t i o n in {[z[ < 1} f o r which (1.3), (1.4) and (i) are true,

(iii) all the zeros of F are negative and simple,

(iv) between two zeros of F in [ - - ro, 0] there exists at least one p o i n t - - r where i F ( - - r)IF(r) ~ = a l§

(v) F has at m o s t one zero -- t f o r which r o <_t < 1,

(vi) i f F has zeros - - s, - - t, such that O < s < r o < t < 1, then IF(-- r) lF(r) "(~) = a ~+4r) f o r some r C (s, ro],

F(0) = a,

.F is a f i n i t e or infinite Blaschke product depending on whether r o < 1 or r o = l .

N o w let u be s u b h a r m o n c in { [z[ < 1} a n d s a t i s f y (1.1) a n d (1.2). I n a d d i t i o n a s s u m e t h a t

u = log tgi, w h e r e g is a n a l y t i c in {[z I < 1}. (1.5) P u t a = e -c, f - ~ - a g , a n d a = - - e o s a 2 . A p p l y i n g T h e o r e m 1 w i t h F = F(., a, ~, 1) we o b t a i n M ( r , f ) < F(r) w h e n 0 < r < 1 a n d t h e r e u p o n t h a t

B(r) < c @ l o g F ( r ) , 0 < r < 1 . (1.6) Since all steps are reversible, it follows t h a t U = log IF[ ~- c is t h e u n i q u e m e m b e r o f t h e class o f s u b h a r m o n i c f u n c t i o n s u s a t i s f y i n g (1.1), (1.2) a n d (1.5) for w h i c h

(1.6) is t r u e .

T o o b t a i n a m o r e g e n e r a l t h e o r e m let 2(r), 0 < r < 1, be a n u p p e r semi- c o n t i n u o u s step f u n c t i o n o n [0, 1] a n d suppose t h a t 0 < 2 ( r ) < 1. L e t u be s u b h a r m o n i e in { [zj < 1} a n d s a t i s f y (1.2), (1.5) a n d t h e c o n d i t i o n

A ( r ) < cos ~ ( r ) B ( r ) , (1.7)

w h e n 0 ~ r < 1. Again, we a p p l y T h e o r e m 1 w i t h a -~ e -c, f = ag a n d g(r) = - - cos z~(r) for 0 < r < i. Using this t h e o r e m we see t h a t (1.6) is t r u e w h e n

E X T R E M A L A N A L Y T I C F U N C T I O N S I N T H E U N I T C I I { O L E 175 F = F(-, a, a, 1). I t n o w follows t h a t in t h e class o f s u b h a r m o n i c f u n c t i o n s u s a t i s f y i n g (1.2), (1.5) a n d (1.7), t h e r e exists a m e m b e r w i t h largest m a x i m u m m o d u l u s .

H e r e we r e m a r k t h a t t h e c o r r e s p o n d i n g p r o b l e m for t h e class o f s u b h a r m o n i e f u n c t i o n s s a t i s f y i n g o n l y (1.2) a n d (1.7), has n o t b e e n solved. T h a t is, we do n o t k n o w w h e t h e r t h e r e exists a m e m b e r o f this class w i t h l a r g e s t m a x i m u m m o d u l u s . W e n o t e t h a t l~teins (see [2, w 7] a n d [3, T h e o r e m 3.2]) p r o v e d T h e o r e m 1 w h e n

~ 0. H e also g a v e t w o m e t h o d s for d e t e r m i n i n g F w h e n a = 0 a n d r o = 1.

I n w 9 we shall discuss t h e p r o b l e m o f d e t e r m i n i n g F w h e n a is c o n s t a n t on [0, r0].

N o w let n be a positive i n t e g e r a n d suppose t h a t F~ is t h e set o f all a n a l y t i c f u n c t i o n s b~ in { Izl < 1} w h i c h c a n be w r i t t e n in t h e f o r m

T h e n we p r o v e

- - (~, real; lc~k] < 1, 1 < k < n ) . (1.8)

T a E O R E ~ 2. L e t 0 < r o <_ 1. L e t

a n d p u t a 1+~ = # ( n ) .

f e

I ~. a n d s u p o < r < r ~ ~ = a 1+~, have f = e~~

(~ a n d ro be f i x e d n u m b e r s s a t i s f y i n g - - 1 < a < 1 a n d

#(n) -~ i n f sup re(r, b , ) M ( r , b~) ~

b n e F n r e(O, ro]

T h e n F = F ( . , a, o, ro) is a m e m b e r o f l~n. M o r e o v e r , i f then f o r some O, 0 < 0 < 2~, we

H e r e we r e m a r k t h a t T h e o r e m 2 is similar to T h e o r e m 7.1 o f H e i n s [3] w h e n

= 0. F i n a l l y , t h e a u t h o r w o u l d like t o t h a n k P r o f e s s o r H e i n s for suggesting t h e s e p r o b l e m s t o h i m a n d P r o f e s s o r M a r t s Ess@n for several helpful suggestions in pre- s e n t i n g this p a p e r .

2. Preliminary reductions

L e t f # 0 be as in (1.3) a n d (1.4). T h e n it is well k n o w n (see N e v a n l i n n a [6]) t h a t f c a n be w r i t t e n in t h e f o r m ,

2 z

f ( z ) = e x p - - 2~ e ~ ~

0

H e r e ~ is a n o n d e c r e a s i n g f u n c t i o n o n [0, 2~], 0 < ~ < 2g, a n d e i t h e r B is a B l a s c h k e p r o d u c t or B ~- 1.

L e t

176 JOHN L. LEWIS

2~

f*(z)

= e x p + 1 2 z

0

I f B ~ 1, t h e n B * = 1. O t h e r w i s e if (as) d e n o t e s t h e zeros o f B, t h e n B * is t h e associated B l a s c h k e p r o d u c t w i t h zeros a t { - - [a~l}.

W e h a v e t h e following inequalities b e t w e e n f a n d f * :

I f * ( - - r)[ < [f*(z)i

<f*(r),

lz] : r (0 < r < 1), (2.3)

If*(-- r)I ~_ m(r,f) ~_ M ( r , f ) < f*(r),

0 < r < 1 , (2.4)

If*(-- r)lf*(r) ~(r) ~ m ( r , f ) i ( r , f )

~(r ), 0 < r < 1 , (2.5)

If*(-- r)ff*(r) ~(r) ~ If(-- z)[[f(z)[ ~

Izl = r (0 < r < 1 ) . (2.6) I-Iere a is as in T h e o r e m 1. (2.3)--(2.6) are easily v e r i f i e d (see for e x a m p l e }Iellsten, K j e l l b e r g a n d N o r s t a d [4]).

F r o m (2.3) a n d (2.5) we see t h a t f * also satisfies (1.3) a n d (1.4). I n v i e w o f (2.4) it follows t h a t it suffices t o p r o v e T h e o r e m I for f * .

3. Proof of Theorem 1

T h e p r o o f of T h e o r e m 1 is long. I n this section we c o n s t r u c t F for r 0 < 1. I n w 4 we show F is a f i n i t e B l a s c h k e p r o d u c t w i t h n e g a t i v e zeros. I n w 5 we p r o v e L e m m a 2. Using this l e m m a we show in w 6 t h a t F has p r o p e r t i e s (iii)--(viii) w h e n r 0 < 1. I n w 7 we p r o v e o u r m a i n l e m m a . W e t h e n d e d u c e p r o p e r t i e s (i) a n d (ii) o f F for r 0 < 1 f r o m this l e m m a . I n w 8 we p r o v e T h e o r e m 1 for r 0 = 1.

F i r s t a s s u m e t h a t 0 < r o < 1. I n this case we let E d e n o t e t h e class of a n a l y t i c f u n c t i o n s in {lz[ < 1} s a t i s f y i n g (1.3) a n d (1.4). T h e n f = a is a m e m b e r o f E , a n d h e n c e E has a n o n z e r o m e m b e r . U s i n g this f a c t a n d a n o r m a l f a m i l y a r g u m e n t it follows for f i x e d ~, 0 < ~ < 1, t h a t t h e r e exists F E E for which

0 < F(~) = sup

M ( ~ , f ) .

(3.1)

f e z

Moreover, if F * is t h e f u n c t i o n associated w i t h F as in (2.2), t h e n F * = F since o t h e r w i s e F * E E a n d 2'(~) < F*(~). This i n e q u a l i t y follows f r o m t h e f a c t t h a t s t r i c t i n e q u a l i t y holds in (2.4) unless f =

e~~ *

for s o m e real 0. H e n c e f r o m (2.3) we h a v e

m ( r , F ) =

I F ( - - r ) [ a n d

M ( r , F ) = F ( r )

w h e n 0 < r < 1.

W e assert t h a t

_F(0) > 0 . (3.2)

T o v e r i f y this a s s e r t i o n we s h a l l w a n t t h e following l e m m a .

L~MMA 1. Let 0 ~_u < 1 and put

E X T R E I ~ I A L A N A L Y T I C F U N C T I O N S I N T H E U N I T C I R C L E 1 7 7

z + u

O(z, u) - - l + u z ' ]zl < l . T h e n i f 0 ~ o~ 1 < or < 1 we have

10(-- r, ~)lO(r, o&(o < lO(-- r, o~l)]O(r, o~1) +) w h e n o~ < r < 1 while i f 0 < r < a 1, the reverse i n e q u a l i t y holds.

P r o o f : T h e l e m m a is a d i r e c t c o n s e q u e n c e o f t h e f o r m u l a

~-d log {1o(- r, u)lO(r, u)+)} =

O

( 1 - - #)[r(1 - - a(r))u 2 + (1 + a(r))(1 + r~)u + (1 - - a(r))r]

(u s - - rZ)(1 - - u~#) W e o m i t t h e details.

T h e following r e m a r k will b e u s e d b o t h in t h e p r o o f o f (3.2) a n d in t h e p r o o f o f L e m m a 3.

r + e

r - + - is a n i n c r e a s i n g f u n c t i o n on [0, 1] . (3.2a) 1 + r e

W e n o w p r o v e (3.2). I f

(3.2)

is false, t h e n f r o m (2.2) w e see t h a t F ( 0 ) = 0.

W e write, F ( z ) - ~ z h ( z ) , w h e r e h is a n a l y t i c in {]z I < 1} a n d ]h] < 1 . L e t 0 < c~ < 1 a n d p u t H -~ 0(., a)h. H e r e 0 is as in L e m m a 1. U s i n g t h e l e m m a w i t h a 1 - - 0 , we f i n d t h a t

IH(- r)I//(r)+) _< IF(-- r)iF(r) ~

w h e n ~ < r < 1. I t follows f o r ~ n e a r 0 t h a t H E E . A l s o b y ( 3 . 2 a ) w e s e e t h a t _F(Q) < H(9). W e h a v e r e a c h e d a c o n t r a d i c t i o n . H e n c e (3.2) is t r u e .

W e also a s s e r t t h a t

IF(- r)IF(r)+)

s u p - - 1 (3.3)

O<r<ro a l + ~ - - .

I n d e e d , s u p p o s e t h a t

I F ( - - r ) I F ( r ) "(0

s u p al+~(r ) ~ e 1 < 1 .

0 < r _ < r o

I n t h i s ease l e t n > 2 be a p o s i t i v e i n t e g e r a n d p u t f , -~ F o 0(', l / n ) . H e r e 0 is as in L e m m a 1. T h e s e q u e n c e (f,)~o c o n v e r g e s u n i f o r m l y to F on [ - - to, r0]

a n d since F ( 0 ) > 0, f , ( r ) / J F ( r ) - - ~ l u n i f o r m l y f o r 0 < r < r 0. L e t s > 0 b e a small p o s i t i v e n u m b e r . Choose n o large e n o u g h s u c h t h a t f o r n > n o a n d 0 < r < r o,

1 7 8 J O H ~ L . L E W I S

L ( r )

1 - - e _ < F ~ _ < 1 - t - 8 ,

IF(-- r)] -- s __ Ifn(-- r) l _< IF(-- r)] § s . T h e n i f 8 is small e n o u g h we have

rf,,(r)l~ o(,..

If,(--

r)lfn(r)4") <

( I F ( - - r)l § s)[F--~J

F(r) '

< ( I F ( - - r)l § ~)(1 --

e),lF(r)~

_< Cl(1 -- e) -131+"(') -t- e(1 - - 8)-1F(0) -1 < a l+"(r)

w h e n n > n o a n d 0 < r < r o. H e n c e for n > n o , f n E E . Since F(@) <fn(O), we h a v e reached a contradiction. W e conclude f r o m this c o n t r a d i c t i o n t h a t (3.3) is true.

F r o m (3.3) we see there exists a sequence (rn)~, 0 _< rn _< r 0, such t h a t r n - ~ s (0 < s < r e ) and

I F ( - r.)IF(r.)"('0

a l + ~ ( r n ) - - ~ 1 .

Since IE(--

ro) l _< F(r.)

a n d a is a step f u n c t i o n it follows t h a t F(s) > a. l ] s i n g t h i s i n e q u a l i t y a n d t h e f a c t t h a t a is u p p e r semieontinuous, we o b t a i n

I F ( - - s)IF(s) 48) IF( -- rn)If(rn) "('0

a~+~(, ) >__ lira a~+O(,~) = 1 .

rn--~s

Since 0 < s < r 0 a n d F satisfies (1.3) we conclude t h a t

I F ( - s ) I / ~ ( s ) ~ = d +~ . ( 3 . 4 )

W e h a v e s h o w n t h a t _~ satisfies (3.4) for a t least one p o i n t in [0, re]. N e x t we s h o w t h e r e are a t m o s t a f i n i t e n u m b e r o f d i s t i n c t points in [0, r0] for which (3.4) is true. I n d e e d if (3.4) were t r u e for a n i n f i n i t e n u m b e r of points in [0, re], t h e n b y t h e I d e n t i t y T h e o r e m for a n a l y t i c f u n c t i o n s we would h a v e for some b E (-- 1, 1),

.~(-- Z).~(Z) b = -~-a l+b, z ~{IZI <

1 } - (-- 1,0].

Here F ~ denotes t h e a n a l y t i c b-th power of F in { Iz I < 1} -- (-- 1, 0] for which F(89 b > 0. Since b y (2.2), lim=_~io I F ( - - z)llF(z)l ~ = 1, o < 101 < ~, we w o u l d t h e n h a v e a contradiction. W e let {p~}~, d e n o t e t h e f i n i t e set of points in [-- ro, r0]

for which (3.4) is t r u e w h e n either s = p l or s ~ - - - p ~ .

4. A property of F

N e x t we shall show that F is a finite Blaschke product. Since F ---- F * and Y(0) ~ 0, it then follows that Y has negative real zeros.

~X~E~AL xxx~Y~c ru~cTm~s x~ T~E c ~ T c~csE 1~9 W e consider t w o possibilities. F i r s t i f ~ ~ {Pi}~, p u t # = F(~). Using t h e same a r g u m e n t as I t e i n s [2, p. 353--354], we f i n d t h a t either there exists a u n i q u e a n a l y t i c f u n c t i o n r in {]z I < 1} such t h a t

(i) 1r < 1

(fi) ~ ( p ~ ) = F ( p ~ ) , 1 < i < n , ~b(~) = F ( ~ ) ,

or t h e r e exists ~u*, # < g* < I, a n d a n a n a l y t i c f u n c t i o n ~o in {Iz I < 1} such t h a t

(i') l~ol < 1

(ii') ~0(1o~) = F ( ~ ) , 1 < i < n, ~o(e) = Ix*.

I n t h e f i r s t ease it follows as in I t e i n s [2, p. 353] t h a t F = ~b a n d 2' is a f i n i t e Blaschke p r o d u c t .

I n t h e second case following I t e i n s [2, (2.14)] we i n t r o d u c e t h e f u n c t i o n ~(., z) d e f i n e d for 0 < ~ < 1 b y

# + # *

~ ( z , ~) = vF(z) -4- (1 -- T)fiq~o(Z), Izt < 1, where f l - - 2#*

We shall show for ~1 n e a r 1 t h a t

I ~ ( - r, ~)ll

T(r, ~)I

~ < a ~+~ (4.1)

w h e n e v e r r E [ 0 , ro] a n d z E [ ~ l , 1). To do this we let r I E { p l ) ~ a n d r 1 > 0 . 2qext for small ~ > 0 we let

I ( r ~ ) = [r~-- ~, r~ + ~] w h e n r ~ # 0, r o,

~-[r~, r ~ + ~] w h e n r ~ = 0 , --- [rx -- ~,

rl]

w h e n rx = r o . P u t

~o(-- ~) ~o(-- ~)

+ e(r) = 1 + e(r), r ~ I(r~) (4.2)

F ( - r) = F ( - r~) a n d

~o(r) r

F ( r ) - - F(r~) + ~(r) = 1 + v(r), r e I(r~) . (4.3) T h e n i f r ~ I(rx) a n d l(rx) ~ [0, to], we h a v e since _~ e E ,

a-0+o(o)lk~(-- r, ~)1 l~(r, ~)1 ~ _<

F ( - - r ) 1 + r - F ( r ) (4.4)

= 1 + (fl - - 1)(1 + a(r))(1 -- ~) + fl[e(r) + a(r)~(v)](1 - - ,) + 0((1 - - ,)~)

1 8 0 J O H N L . L E W I S

as 7 - + 1. T h e t e r m 0[(1 - - 3) 8] is i n d e p e n d e n t o f r w h e n 7 is n e a r 1. Since fl ~ 1 a n d e(r), ~(r), are u n i f o r m l y small w h e n ~ ~ 0 is small, it follows f o r T 0 n e a r l t h a t (4.1) is t r u e w h e n

r 6 1 ( r l )

a n d 7 o ~ 3 ~ 1.

L e t A : U I ( r l ) , r 1 ~ 0 , r 1E{p~}~. N o w

sup I F ( - r)[F(r)~ -(~+~ < 1,

r e [0, ro]--A

since o t h e r w i s e t h e r e w o u l d e x i s t s in t h e closure o f [0, r e ] - A for w h i c h IF( - s)[-~(s) "(s) ----a 1+0(0. Using this f a c t a n d t h e f a c t t h a t F ( 0 ) > 0 we f i n d (4.1) is t r u e w h e n r E [ 0 , ro] a n d 3 1 ~ 7 ~ 1.

L e t ~ * ( . , 7) be t h e f u n c t i o n a s s o c i a t e d w i t h ~ ( . , 7) as in (2.2). T h e n f r o m (2.3), (2.6) a n d (4.1) we see t h a t ~ * ( . , 3 ) E E f o r 7 1 _ ~ 3 < 1. Since

/ z + # *

F(~) = / t < 7/t -~ (1 - - 3) 2 - - T(~, 3) ~ W*(e , 3)

we h a v e r e a c h e d a c o n t r a d i c t i o n . H e n c e i f ~ ~ {p~}~, t h e n F : ~ a n d F is a f i n i t e B l a s e h k e p r o d u c t .

N o w suppose t h a t e E{p~}~. W e assume, as we m a y , t h a t ~ : l o , . L e t

~'(Q) = #. T h e n as before e i t h e r t h e r e exists a u n i q u e a n a l y t i c f u n c t i o n r in {Izl < 1} satisfying [r < 1 a n d

r = ~ ( p , ) (1 < i < n - 1), r = ~ ,

or t h e r e are a n i n f i n i t e n u m b e r o f such f u n c t i o n s . I f r is u n i q u e , t h e n as p r e v i o u s l y , 2' : ~b a n d F is a f i n i t e B l a s e h k e p r o d u c t . Otherwise we d e f i n e ~b o, /~*, a n d c o r r e s p o n d i n g t o {pl}~ -1 as p r e v i o u s l y . W e also d e f i n e e(r) a n d ~(r) c o r r e s p o n d i n g t o r 1 = ~ as in (4.2) a n d (4.3)

P r o c e e d i n g as in t h e f i r s t case we o b t a i n f o r r n e a r ~ a n d 0 < r ~ r 0, (see (4.4))

a - ( ~ + ( r ) ) l ~ ( - - r,

T)I[T(r,

3)l ~(') < 1 § [fl -[- o ' ( r ) ~ x - 1 - - (1 + if(r))](1 ^{- -} 7) +

(4.5)

-t- fi(e(r) q- a(r)~(r))(1 -- 3) q-

0[(1 - - 3) 8]

w h e r e x = / t / # * . Since fi = 1(# -t- #*)/#*, we h a v e

fl(1 ~-

a(r)x -1)

: 1(1 ~- x)(1 +

a(r)/x) : g(x).

L e t a o : max0_<,_<1 [a(r)l. F r o m t h e d e f i n i t i o n o f a we see t h a t a o ~ 1. F r o m H e i n s a r g u m e n t [2, p. 353], it is clear t h a t # * m a y b e chosen such t h a t ~ a 0 x ~ 1. Since

g ' ( x ) : 8 9 a(r)x-2),

we h a v e

g(1) - -

g(x) >

89 - - aox-~)(1 - - x) > 0 . (4.6) H e n c e g(1) - -

g(x)

is b o u n d e d below b y a p o s i t i v e c o n s t a n t w h i c h does n o t d e p e n d on r. Since g(1) : 1 ~-

a(r),

it follows f r o m (4.6) a n d (4.5) t h a t for r n e a r ~,

E X T I ~ E i g A L A N A L Y T I C F U N C T I O N S I N T H E U N I T CIP.CLE 181 T 1 n e a r 1 a n d r E [0, r0] t h a t (4.1) is t r u e . Using this fact ~nd a r g u i n g as in t h e previous case we o b t a i n a contradiction. Hence in b o t h cases r ---- F a n d F is a finite Blaschke product.

5. A L e m m a

W e w a n t to prove t h a t F satisfies (i)-- (viii) of T h e o r e m 1. T h e following lemm~

will p l a y a f u n d a m e n t a l role in t h i s proof. This l e m m a m a y b e c o m p a r e d w i t h l e m m a 7.1 in I t e i n s [2].

L E ~ M A 2. Let 0 ~ u ~ v ~ 1 and p u t

z § z + v

r u, v) = 1 - ~ u z 1-Jr v z ~ ]zl < 1 .

Let 0 ~ o h ~ ~ < fl < fil ~ 1 a n d s u p p o s e f o r f i x e d Q, 0 ~ Q < 1 that r ~) ---- r al, fll). T h e n i f either 0 ~ r ~ o h or fll ~ r < 1 we have

while i f a < r < fl the opposite inequality holds.

Proof. Consider t h e set

F={(u,v):~l<u<~, t ~ < v < ~ l , and r 1 6 2

which defines a f u n c t i o n v---> u(v) a n d a f u n c t i o n u----> v(u). T h e n we shall show for f i x e d r t h a t

d

d-v []r r, u(v), v)Ir u(v), v) ~(')] < 0 (5.1)

w h e n either 0 < r ~ a l d dv

or f l i ~ r ~ 1 a n d

[1r

r, u(v),

v)lr u(v),

vy(0] > 0

(5.2)

w h e n a < r ~ ft. To prove (5.1) a n d (5.2) we observe t h a t d u (Q + u)(1 + uo)

dv (e + v)(1 + re)

a n d hence t h a t

d

(1 - - s~)-l(e -[- v)(1 ~- re) dvv log [r u, v)] = N ( s , v) - - N ( s , u) ,

(5.3)

182 zo~r L. LEWIS

w h e r e At(s, x) -= (O -1- x)(1 +

xO)/(s +

x)(1 -}-

sx).

I t follows for f i x e d r w h e r e O < r < 1 a n d r ~ { u , v } t h a t

d

(1 - -

r2)-l(O +

v)(1 -t- v0) dvv log{ [r

r, u, v)]r u,

v) "(r)} = to(r, v) -- to(r, u ) , (5.4) w h e r e

to(r, x) = yY(-- r, x) -]- (r(r)N(r, x).

W e shall p r o v e t h a t

(a) W h e n 0 < r < or log IF(r, ")1 is a decreasing function a n d ~o(r, .) is a positive function on (o% 1),

(b) W h e n ~ < r < f l , ~ ( r , u ) < 0 < ~ o ( r , v ) ,

(c) W h e n fll ~< r < 1, log ]to(r, .)[ is an increasing function a n d to(r,.) is a n e g a t i v e f u n c t i o n on [0,

ill)"

Since c h < u < ~ < fi < v < ill, we h a v e t h e following consequences of (5.4), (a), (b), a n d (c), respectively.

(a') to(r, .) is a decreasing f u n c t i o n on (c% 1) a n d (5.1) is t r u e for 0 < r < c%

(b') (5.2) is t r u e w h e n c~ < r < fi,

(c') ~v(r, .) is a decreasing f u n c t i o n on [0, ill) a n d (5.1) is t r u e w h e n fll < r < 1.

To complete t h e proofs o f (5.1) a n d (5.2), it only remains t o p r o v e (a), (b), a n d (c). To p r o v e (a) we write

to(r, x) = tol(x)W,(r, xho3(r, x) , w h e r e tel(x) = (q + x)(1 +

xQ),

to2(r, x) = {(x 2 - - r~)(1 --

x2r2)} -1

tea(r, x) = r(1 - -

a(r))x 2 -]=

(1 -t- a(r))(1 +

r2)x -t-

(1 - -

a(r))r.

W e claim t h a t

d 1

- -

dx

log l~ol(x)l < x ' - - 0 < x < l

0 - - 2

0-x loglto2(r,x)] < - - x ' 0 < r < x < l ,

0 1

- - 0x log %(r, x) < - - x ' 0 < x < l .

The proofs of these s t a t e m e n t s are straight f o r w a r d a n d are omitted. Using these inequalities w e see t h a t (a) is true.

T h e p r o o f o f (b) is i m m e d i a t e from t h e definition o f to. The p r o o f of (e) is also simple a n d w e o m i t t h e details. This proves (5.1) a n d (5.2).

Using (5.1) a n d (5.2) we conclude t h a t

E X T R E M A I ~ A N A L Y T I C F U N C T I O N S I N T H E U N I T C I R C L E 183

1r r, 0% 81)]r o~, ill) ~(') = lim ]~b(-- r, u, v)l~(r, u, v) ~(~) <

< lim [4(-- r, u, v)[r u, v) ~ = [4(-- r, ~, fl)lr ~, 8) ~(~)

v.-~-fl

w h e n e i t h e r 0 < r < a 1 or fll ~ r < 1, a n d t h a t t h e r e v e r s e i n e q u a l i t y holds w h e n a < r < 8- This c o m p l e t e s t h e p r o o f o f L e m m a 2.

6. Descriptive properties of F

T o c o n t i n u e t h e p r o o f o f T h e o r e m 1, we recall t h a t we are a s s u m i n g r 0 < 1.

T h e n in w 4 we s h o w e d t h a t F is a f i n i t e B l a s c h k e p r o d u c t w i t h n e g a t i v e zeros.

I n this section we p r o v e t h a t F satisfies (iii)--(vii) o f T h e o r e m 1. T h e o r e m 1 is t h e r e b y e s t a b l i s h e d f o r r 0 < 1 save f o r (i) a n d (ii) w h i c h we t r e a t in w 7. W e b e g i n b y p r o v i n g (iii). L e t 0 < ~ < 1 a n d s u p p o s e t h a t - - ~ is a m u l t i p l e zero o f F .

W e w r i t e

.F(z) ~- \ ~ ] g(z), lzl < 1 ,

w h e r e g is a n a l y t i c in {[zl < 1} a n d tgl -< 1. L e t r u, v), 0 < u < v < 1, be as in L e m m a 2 a n d p u t c~ = ft. Choose al, ill, such t h a t 0 < ~1 < a -~ fl < fix < 1 a n d r S x ) = r 8). Since F e E , it follows b y L e m m a 2 t h a t G = r ~1, fll)g satisfies

IG(- r)IG(r) ~') < I F ( - - r)fF(r) ~ < a l+~

w h e n r E [ 0 , a l ] O [ 8 1 , 1 ) , 0 < r < r o , a n d F ( - - r ) ~ 0 . U s i n g this i n e q u a l i t y we f i n d t h a t if al a n d 8~ are n e a r ~, t h e n G E E a n d

[G(-- r)[G(r) "(r) < a 1+~

for 0 < r < r 0. H o w e v e r , since F(Q) -~ G(~) a n d G C E we m u s t h a v e (see (3.4)), IG(-- s)]G(s) ~(~) = a l+"(s) for some s E [0, r0]. A g a i n we h a v e r e a c h e d a c o n t r a - diction. ~ e n c e (iii) o f T h e o r e m 1 is t r u e .

T h e p r o o f o f (iv) is similar to t h e p r o o f o f (iii). L e t - - r 0 < - - 8 < - - a < 0 be t w o zeros o f F a n d suppose t h a t

[_F(- r)lF(r)~(r) < a~+~ ~ < r < 8 . (6.1) T h e n F = r w h e r e f is a n a l y t i c in {Izi < 1} a n d If[ ~ 1. Choose

~1,81, such t h a t 0 < ~ l < a < 8 < f i l < 1 a n d r = r L e t I = ~(', al, fll)f. ~Tsing ( 6 . 1 ) a n d L e m m a 2 we f i n d f o r a I n e a r ~ a n d fll n e a r 8 t h a t I E E a n d

1I(-- r)II(r) ~(*) < a ~+"(~), 0 < r < r o .

184 ^{J O H N} L . L E ~ v l S

H o w e v e r , s i n c e I ( p ) - - - - F ( ~ ) w e m u s t h a v e b y (3.4), [ I ( - - 8 ) ] 1 ( 8 ) ~ = a I+@

f o r s o m e s E [0, r0]. A g a i n w e h a v e r e a c h e d a c o n t r a d i c t i o n . H e n c e (iv) is t r u e . T h e p r o o f s o f (v) a n d (vi) a r e e x a c t l y t h e s a m e as t h e p r o o f o f (iv). W e o m i t t h e d e t a i l s .

T o p r o v e (viii), l e t t 1 ---- m i n {r : F ( - r) = 0}. F r o m (iii) w e see t h a t t 1 > 0.

W e a s s e r t t h a t

I F ( - - r)IF(r) @ ~ F ( 0 ) 1+~ (6.2)

w h e n 0 < r < t 1. I n d e e d , f o r f i x e d a(r) t h e f u n c t i o n g(z) = log

IF(-- z)[IF(z)[~

]z 1 < tl, is h a r m o n i c i n {[z I < tl}. ~r f r o m (2.6) w e see t h a t mini=i=, g(z) = g ( - - 8) f o r 0 < 8 <: t 1. S i n c e mini=I= ~ g(z) is a n o n i n c r e a s i n g f u n c t i o n o f 8, it f o l l o w s t h a t (6.2) is t r u e .

N e x t w e n o t e t h a t sup0 <,_<,, { [ F ( - - r ) ] F ( r ) ~(') 9 a -(1+~ = 1. I n d e e d , o t h e r w i s e t h e f u n c t i o n G(z) = F(z)O(z, t)/O(z, tl) is i n E f o r t n e a r tl, t > t 1, a n d G(O ) >

F ( ~ ) , as f o l l o w s e a s i l y f r o m L e m m a 1. T h i s f a c t a n d (6.2) i m p l y F ( 0 ) -= a. H e n c e (vii) i s t r u e .

7. The final proof for r o < 1

T o p r o v e (i) a n d (if) o f T h e o r e m 1 w e shall w a n t t h e f o l l o w i n g l e m m a .

L~.MMA 3. Let n and m be two positive integers. Let 0 ~ tl ~ . . . ~ tn ~ 1 and 0 ~ _ s l ~ . . . ~ s m < 1. P u t f(z) = \ ~ z ] ' and g(z) = , , \1 + 8 , z / "

1

Let t o = O . F o r given e E ( 0 , 1 ) suppose that f(~) < g ( e ) . T h e n either f - - - - g or there exists a positive integer j , 1 < j ~_ n, such that

I f ( - - r)lf(r) @ < It(-- r)lg(r) ~('), r E [tj_ 1, tj) . Proof. W e f i r s t a s s u m e t h a t n--- 1.

I f m = 1, t h e n f r o m (3.2a) a n d t h e f a c t t h a t f(Q) < g(~), w e d e d u c e t 1 _< 81.

I f t 1 = 8 1 , t h e n c l e a r l y f - - - - g . I f t l < s 1 , w e a p p l y L e m m a l w i t h a l = t l , a n d

= 81. U s i n g t h i s l e m m a , w e o b t a i n

I f ( - - r)jf(r) ~(') < I t ( - - r) lg(r) ~(') (7.1) w h e n r E [ 0 , tl]. H e n c e L e m m a 3 is t r u e f o r n = 1 a n d m = 1.

L e t ]c b e a p o s i t i v e i n t e g e r a n d s u p p o s e t h a t L e m m a 3 is t r u e w h e n e v e r n = 1 a n d l ~ m ~ ] c . T h e n i f m = k ~ - 1 w e see f r o m (3.2a) t h a t t l ~ s l _ ~ s 2 ~ 1.

L e t r u, v), 0 < u < v < 1, b e as i n L e m m a 2. W e also c o n s i d e r t h e f u n c t i o n Fo d e f i n e d b y

T' o ---- {(u, v) : t 1 < u < v < 1 a n d ~(e, u, v) = 9)(~, 81, 82) } .

EXTREIKAL ANALYTIC FUNOTIONS IN THE UNIT CIRCLE 185 Since dv/du < 0 (see (5.3)), t h e r e e x i s t x0, Y0, such t h a t (xo, Yo) E F o a n d

x o < u < v < y o w h e n ( u , v ) E / ' o . (7.2) W e claim tha~ Yo = h Clearly e i t h e r x o = t 1 or Yo = 1, since o t h e r w i s e (7.2) w o u l d be c o n t r a d i c t e d . I f x o = t: t h e n since f(~) < g(9) we m u s t h a v e Yo = 1.

H e n c e Yo = 1.

W e consider t h e f u n c t i o n

\ 1 + X 0 ~ ] r 81, 82) ' [Z[ < 1 9

U s i n g L e m m a 2 w i t h a : = x 0 , ~ = 8 : , f l = s u , a n d f l : = 1, we o b t a i n

Ih(-- r)th(r) ~(r) < Ig(-- r)lg(r)"(r), 0 < r < t:. (7.3) M o r e o v e r h has k zeros a n d

h(e)= g(e)-

H e n c e e i t h e r h - - ~ f or

I f ( - - r)[f(r) "(r) < Ih(-- r)lh(r) "(0, 0 _< r < t : .

I n e i t h e r case we c o n c l u d e f r o m (7.3) t h a t L e m m a 3 is t r u e w h e n n = 1 a n d m - - k @ h H e n c e b y i n d u c t i o n L e m m a 3 is t r u e f o r n = 1.

L e t q be a p o s i t i v e i n t e g e r a n d suppose t h a t

L e m m a 3 is t r u e for n < q . (@)

T h e n i f n = q @ 1, we f i r s t a s s u m e t h a t m = h I n this e a s e i f t : ~ s : , t h e n f r o m L e m m a 1 w i t h a 1 = t 1 a n d a = Sl, we see t h a t (7.1) is v a l i d f o r 0 _< r < t r I f t: > s 1, t h e n f r o m L e m m a 1 w i t h ~ = t: a n d al = s:, we see t h a t (7.1) is v a l i d f o r t l < r < t ~ . H e n c e L e m m a 3 is t r u e f o r m = 1 a n d n~---q@4- h

L e t k be a p o s i t i v e i n t e g e r a n d suppose in a d d i t i o n t o (@) t h a t

L e m m a 3 is t r u e for n = q @ 1 a n d m _ < k . ( @ + ) T h e n i f n = q @ 1 a n d r e = k @ 1, we assume, as we m a y , t h a t f a n d g h a v e no c o m m o n zeros. I n d e e d , i f s, 0 < s < 1, is such t h a t f ( - - s) = g ( - - s) = 0, t h e n

1 @ sz 1 @ sz

f : ( z ) - - z @ s f(z) a n d g : ( z ) - - z @ s g(z), ]z I < 1 h a v e q a n d k zeros r e s p e c t i v e l y in {Izl < 1}. U s i n g t h i s f a c t a n d (@) we o b t a i n t h a t L e m m a 3 is t r u e for f:, g:, a n d t h e r e u p o n for f, g.

W e p r o c e e d u n d e r t h e a b o v e a s s u m p t i o n s . W e consider t h e s i t u a t i o n ,

(a) F o r some p o s i t i v e i n t e g e r i (1 < i < n - - 1) t h e r e exists a p o s i t i v e i n t e g e r ' p (1 < p _ < m - - 1) such t h a t s e , % + : are in (t~,t~+:).

I f s i t u a t i o n (a) occurs, we d e f i n e t h e f u n c t i o n /'1 b y / " l = { ( u , v ) : t , < u < v < t , + l a n d r 1 6 2

T h e n since dv/du < 0 we see t h e r e e x i s t xl, y:, such t h a t ( % y:) E F: a n d x l < u < v < y l w h e n ( u , v ) E F 1 . (7.4)

186 jor~N L. LEWIS

Clearly e i t h e r xl = ti o r Yl -= ti+l, since o t h e r w i s e (7.4) w o u l d be c o n t r a d i c t e d . I f x l = t ~ , a n d Y l < t ~ + i , we p u t

(l + t'z 1

A(z) = \ z + t, / / ( z ) , Izl < 1 ,

(z + y~)g(z)

gz(z) ----

(1 q-

yaz)4(z, at, sp+x) '

Izl < 1.

T h e n f2(e)--<g2(e) a n d

f2, gz,

h a v e q a n d k zeros r e s p e c t i v e l y in {Izl < x}.

Using t h e s e f a c t s a n d (q-) we f i n d t h a t e i t h e r f~ = g~ in w h i c h case n = m = 2 or f o r some positive i n t e g e r j (1

< j <_ n, j =/= i -}-

1)

If2(-- r)lfu(r) ~ < [!lu(-- r){g2(r)

~('),

tj_ 1 "~

r "<: tj . (7.5) I f f~ = g~, t h e n f = ~(., tl, t2) a n d g = ~(., sl, sz). A p p l y i n g L e m m a 2 w i t h

~1 = tx, or = 81, fl = au, a n d /71 =

tz,

we o b t a i n t h a t (7.1) is v a l i d f o r 0 < r < t 1.

I f (7.5) is t r u e , we a g a i n a p p l y L e m m a 2 w i t h a l = ti, c~ ~-- sp, fl = se+ 1 a n d fll = Y~.

T h e n b y this l e m m a a n d (7.5) we h a v e for

ti_l <_ r < tj,

r--t, ( r + t, l~

If(-- r)[f(r) ~(r) ^{= 1 -}

rti \ ~ - ~ /

if2(-- r)IA(r) "(') <

Ir r, ti, y~)lq~(r, t,, yl) o(r)

< 1~(--

r, sp, Sp+l)I~(r, s~, sp+~) ~ " Ig(-- r)Ig(r) ~(~) < lg(-- r)Ig(r) ~

H e n c e in e i t h e r case L e m m a 3 is v a l i d f o r Xl ----

ti.

T h e p r o o f f o r y~ --- t~+l is similar. W e o m i t t h e details. W e c o n c l u d e t h a t i f s i t u a t i o n (a) occurs, t h e n L e m m a 3 is v a l i d f o r n = q q- I a n d m = k q- 1.

W e n o w s u p p o s e t h a t s i t u a t i o n (a) does n o t occur. W e f i r s t a s s u m e t h a t m = k q- I < n ---- q q- 1. I n this case we claim t h e r e exists a p o s i t i v e i n t e g e r i

(1 < i < n ) s u c h t h a t

sj~[t,_~,t,]

w h e n 1 < j < m . (7.6)

I f m <

n,

t h e n clearly (7.6) is t r u e . I f m = n, t h e n (7.6) is t r u e , since o t h e r w i s e i t follows t h a t s i < tj, 1 ~ j < n, a n d t h e r e u p o n f r o m (3.2a) t h a t g(~) < f(~).

L e t lo be t h e m i n i m u m o f t h e set o f p o s i t i v e i n t e g e r s i f o r w h i c h (7.6) is t r u e . W e shall p r o v e t h a t (7.1) is v a l i d for tp_, < r < te. I f 29 = 1, t h e n t / < sj f o r 1 < j ___< m. U s i n g this f a c t a n d a p p l y i n g L e m m a 1 w i t h al ---- tj a n d ~ =

si,

l < j < m , we o b t a i n for 0 < r < t 1

m l

if(-- r)lf(r) ~(') <-- ]-[ [0(-- r, ti)[O(r,

tj) ~(') <

1 (7.7)

7rt

< ~ 10(-- r, 8j)10(r , sj) ~ -~ ]g(-- r)[g(r) ~

1

E X T R E M A L A I ~ A L Y T I C FUI~CTIO:N-S I N T H E U N I T C I R C L E 187 I f ~ = m + 1, t h e n s i < t i for 1 ~ j _~ m. U s i n g this f a c t a n d a p p l y i n g L e m m a 1 w i t h c r i a n d c r 1 ~ j _ ~ m , we f i n d t h a t ( 7 . 7 ) h o l d s for

tm < r _< tm+1.

I f 1 < p ~ m , t h e n f r o m t h e d e f i n i t i o n o f io we see t h a t s i < t i w h e n 1 ~ j ~_ p - - 1 a n d tj < s 1 w h e n ~ ~_ j ~ m. W e a p p l y L e m m a 1 w i t h cr = s i,

= t i, w h e n 1 ~ j ~ p - - 1 , a n d w i t h a l = t i , ~ " - 8 i , w h e n p ~ j _ ~ m . W e o b t a i n t h a t (7.7) is t r u e f o r tp_l _~ r ~ t e. H e n c e , i f s i t u a t i o n (a) does n o t o c c u r a n d i f m = k + l ~ n ~ - - q ~ - l , t h e n L e m m a 3 is t r u e .

~ i n a l l y s u p p o s e t h a t s i t u a t i o n (a) does n o t o c c u r a n d n ---- q + 1 <: m --~ k + 1.

I n this case we a s s e r t t h a t t, ~ sin_ I ~ sm ~ 1. I n d e e d , o t h e r w i s e it follows t h a t s i ~ ti, 1 _~ j ~ n, a n d t h e r e u p o n t h a t g(q) < f(q).

W e let F~ be t h e f u n c t i o n d e f i n e d b y

_r', = {(u, v } : t , < u < v < 1 a n d r s~_~, s~) = r u, v)}.

Since d v / d u < 0 we see t h e r e e x i s t x2, Y2, such t h a t (x2, Y2) E / " 2 a n d

x s < u < v < y 2 w h e n ( u , v ) E / ~ . (7.8) Clearly e i t h e r x 2 = t~ or Y2 = 1, since o t h e r w i s e (7.8) w o u l d be c o n t r a d i c t e d . I f Y 2 = 1, we p u t

z + x~ g(z)

gs(z) - - 1 + x~z ~(z, s,~_~, 8m) ' Izl < 1 .

T h e n ga(~)----g(~) a n d g3 has k zeros in {[z] ~ 1}. U s i n g ( - ~ - - ~ ) i t follows t h a t e i t h e r f ~ g3 o r f o r some p o s i t i v e i n t e g e r j (1 _ ~ j ~ n)

l f ( - - r)If(r) ~(~) < lg3(-- r)Iga(r) ~(0, ty, ~ _~ r < t i .

A p p l y i n g L e m m a 2 w i t h al ---- x:, a = sin_l, fl -~ s~ a n d fll = 1, we f i n d in e i t h e r case t h a t L e m m a 3 is t r u e for y2 = 1.

I f x 2 ~ t ~ a n d y ~ l , we let

( ~ + y~] ~(~)

ya(z) ~-- [1 + yez] r s~_~, s~) ' ]z] < I .

T h e n fa, ga, h a v e q a n d k zeros r e s p e c t i v e l y in {tzl ~ 1 } a n d f a ( ~ ) ~ g a ( ~ ) . Using t h e s e facts a n d (-t-) we see t h a t e i t h e r fa = ga or for some p o s i t i v e i n t e g e r j ,

~ < j < n - ~ ,

[fa(-- r)l"(0fa(r) ~ < [ga(-- r)["(r)ga(r)~('), t~_l _~ r < t~.

A p p l y i n g L e m m a 2 w i t h ~1 = t., ~ = s~_~, fl ---- s . , a n d fll ~- Y2, we f i n d in e i t h e r case t h a t L e m m a 3 is v a l i d for x: --- t~. I-Ience i f s i t u a t i o n (a) does n o t occur, t h e n

188 JotiN L. LEVy'IS

L e m m a 3 is t r u e f o r n = q -]- 1 a n d m = / c -t- 1. Since we h a v e a l r e a d y c o n s i d e r e d t h e p o s s i b i l i t y t h a t s i t u a t i o n (a) does occur, we conclude t h a t L e m m a 3 h o l d s f o r n - ~ q - t - 1 a n d m = k + l .

N o w b y i n d u c t i o n on m we see t h a t L e m m a 3 is v a l i d f o r n---q-t- I a n d m a p o s i t i v e i n t e g e r . T h e n b y i n d u c t i o n o n n it follows t h a t L e m m a 3 is t r u e w h e n - e v e r n a n d m a r e p o s i t i v e i n t e g e r s . T h i s c o m p l e t e s t h e p r o o f of L e m m a 3.

T o c o n t i n u e t h e p r o o f o f T h e o r e m 1, let E b e as in w 3 a n d s u p p o s e t h a t F E E s a t i s f i e s (3.1). T h e n in w 6 we p r o v e d t h a t F h a s p r o p e r t i e s (iii) - - (viii) of T h e o r e m

1 f o r r o < 1. W e n o w use L e m m a 3 t o s h o w t h a t F satisfies (i) a n d (ii).

L e t ~ b e a f i x e d n u m b e r a n d 0 < ~ ~ 1. L e t G b e a n a n a l y t i c f u n c t i o n in {lzl < 1} for w h i c h G E E a n d

G(~) = sup M(T, f ) .

f e n

W e see t h a t G satisfies (iii)--(viii) o f T h e o r e m 1 f o r r o < 1. A p p l y i n g L e m m a 3 w i t h f = F , g = G, a n d @ = 3, we o b t a i n t h a t e i t h e r F = G or f o r s o m e p o s i t i v e i n t e g e r j ,

IF(-- r) lf(r) ~ < IG(-- r) lG(r) "(e, tj_l < r < t~. (7.9) T h e l a t t e r s i t u a t i o n c a n n o t occur. I n d e e d , i f (7.9) w e r e t r u e , t h e n f r o m p r o p e r t i e s ( i v ) - - ( v i i ) o f F a n d t h e f a c t t h a t G E E , we w o u l d h a v e f o r s o m e r E [0, ro] fl

[tj--1, ~]),

a 1+~(~) = I F ( - - r)]F(r) ~(~) <~ ]G(-- r)IG(r) "(~) ~ a 1+~(~) . H e n c e , F ---~ G. F r o m t h i s e q u a l i t y we c o n c l u d e t h a t F satisfies (i).

P r o p e r t y (ii) o f F (unicity) is n o w a n i m m e d i a t e c o n s e q u e n c e o f (i) a n d t h e I d e n t i t y T h e o r e m f o r a n a l y t i c f u n c t i o n s .

T h i s c o m p l e t e s t h e p r o o f o f T h e o r e m 1 f o r r 0 < 1.

8. Proof of Theorem 1 for r o ~ 1

W e w i s h t o p r o v e T h e o r e m 1 w h e n r o = 1. T o do so it will b e n e c e s s a r y t o i n d i c a t e t h e d e p e n d e n c e o f F on r o w h e n 0 < r o < 1. T h e r e f o r e we shall o f t e n w r i t e F ( - , r o ) f o r F .

L e t 0 < r 0 < r o * < 1. T h e n since

I F ( - r, r*)[F(r, r*) ~ < a l+~

w h e n 0 ~ r ~ r o, we see f r o m T h e o r e m l t h a t F(r,r~) ~ F ( r , ro) f o r 0 ~ r ~ 1.

H e n c e , F(r, 1) ~ lim,o~l F(r, to) e x i s t s f o r e a c h r E (0, 1). U s i n g t h i s f a c t a n d a n o r m a l f a m i l y a r g u m e n t , w e see t h a t 2'(z, 1) ---- limro~l F(z, to) e x i s t s a n d is a n a l y t i c f o r [z] < 1. ~r F(., 1) satisfies (1.3) a n d (1.4) w h e n r o = 1. Also, F(-, 1)

E X T R E I ~ I A L A l f f A L Y T I C ~ ' U 1 V C T I O : N S I N T H E U ~ I T C I R C L E 189 satisfies (i) of T h e o r e m 1 w h e n r 0 ~ 1. I n d e e d , if f satisfies (1.3) a n d (1.4) for r 0 = 1, t h e n

M ( r , f ) ~ F ( r , ro)

w h e n 0 < r < 1 a n d 0 < r 0 < 1. ~ e n c e

M(r,f) ~ F(r,

1). P r o p e r t y (ii) o f E(., 1) is n o w a n i m m e d i a t e c o n s e q u e n c e o f (i) a n d t h e I d e n t i t y T h e o r e m for a n a l y t i c f u n c t i o n s .

T o p r o v e (viii) we f i r s t n o t e f r o m (2.2) tha$

F(z, 1) = e x p ~

B*(z),

Izl < 1. (8.1)

z-i-

H e r e 0 ~ a < o~ a n d e i t h e r / 3 " = 1 or 23* is a B l a s c h k e p r o d u c t . I f c r f h e n f r o m (8.1) we see t h a t limr+ 1 I F ( - - r, 1)IF(r, 1) @ = 0 . Using this fact, we f i n d t h a t

IF(--

r, 1)IF(r, 1)+)

sup - - 1

O < r < l C $ 1 + q ( r ) - - ~"

since o t h e r w i s e for small ~ ~ 0 t h e f u n c t i o n

f(z) ~--

F((z ~ e)/(I -~ ez), 1), Izl ~ 1 , w o u l d s a t i s f y (1.3) a n d (1.4) for r 0 ~-- 1, a n d

F(r,

1) ~

f(r),

0 < r < 1, (see

3.3)).

I t also follows as in w 3 t h a t t h e r e are a t m o s t a f i n i t e n u m b e r ( ~ 0) o f p o i n t s r in [0, 1) w h e r e I F ( - - r, 1)IF(r, 1) "(0 --~

a l+~

U s i n g t h e a r g u m e n t o f H e i n s as in w 4 we n o w o b t a i n t h a t F is a f i n i t e B l a s c h k e p r o d u c t . Since ~ ~ 0, we h a v e r e a c h e d a c o n t r a d i c t i o n . H e n c e ~ ~ 0.

T o c o m p l e t e t h e p r o o f o f (vii) for r o ~ 1, we o b s e r v e f r o m (8.1) t h a t F(., 1) has a n i n f i n i t e n u m b e r o f zeros, since o t h e r w i s e we w o u l d h a v e lim~+ 1 I F ( - -

r)[F(r) ~(~)

1, in c o n t r a d i c t i o n t o (1.3). H e n c e , F ( . , 1) has p r o p e r t y (viii).

T h e proofs o f (iii), (iv), a n d (vii) are e x a c t l y t h e same as in t h e ease r 0 ~ 1.

W e o m i t t h e details. H e r e (v) a n d (vi) are t r i v i a l l y t r u e . This completes t h e p r o o f o f T h e o r e m 1.

9. Remark

W e r e m a r k t h a t H e i n s [3, w167 4 - - 6 ] g a v e t w o m e t h o d s for d e t e r m i n i n g F w h e n r o ~ 1 a n d

a(r)

= 0, 0 < r ~ 1. H e r e we consider t h e p r o b l e m o f d e t e r m i n i n g F w h e n

a(r),

0 < r < l , is c o n s t a n t on [0,1]. I n this case we p u t

a(r)-~a,

0 < r < l .

F i r s t suppose t h a t r 0 < 1. L e t (--t~)~, w h e r e 0 ~ t i < t ~ + l , d e n o t e t h e zeros o f F a n d p u t t 0 = 0 . W e assert t h a t

(*) I f

f E E

a n d f satisfies (iii)--(viii) o f T h e o r e m 1 for r 0 < 1, t h e n f_----E.

T h i s assertion is v e r i f i e d b y using L e m m a 3 as in t h e p r o o f o f (7.9). W e also assert that.

(**) I f 1 < i < n , t h e n log { }F(-- r) ]F(r) ~} is a c o n c a v e f u n c t i o n o f l o g r o n

(t,_l, t,).

T h i s assertion is v e r i f i e d b y a r g u i n g as in t h e p r o o f of (6.2).

190 J-O]XN L. LEW:tS H e n c e (**) is true.

Using (**) a n d (iv)-- (vii) o f T h e o r e m 1 we deduce t h a t

I f 1 < i < n and t~ 6 [0, r0], t h e n I F ( - - r ) l F ( r ) ~ = 6b 1+•

("~)

for e x a c t l y one p o i n t in [t~_l, t~].

I f t, 6 (r 0, 1), t h e n t._, 6 [0, r0) a n d either IF( - r ) l F ( r ) ~ = a x+~ ( + + ) for e x a c t l y one p o i n t r 6 [t,_l, t,] or [ F ( - - re)IF(re) ~ --- a 1+~.

W e observe f r o m (vii) t h a t r = 0 is t h e u n i q u e p o i n t in [0, tl] for which ( + ) is true. F r o m ( + ) a n d ( + + ) we see t h a t F is t h e solution of t h e equations (A) F ( 0 ) = a,

(B) I f n > 2 , if 1 < i < n - - 1, a n d if t~+16(0, t0], t h e n for some r6[t~,t~+l]

we h a v e IF(-- r ) i F ( r ) ~ = a 1+~ a n d d / d r ] F ( - - r ) ] F ( r ) ~ = O.

(C) I f n>__2 a n d if t ~ 6 ( r 0,1), t h e n either [ F ( - - r ) l F ( r ) ~ = a 1+~ a n d d / d r I F ( - - r ) I F ( r ) ~ = 0 for some r satisfying t,_ 1 < r < r o, or IF(--r0)IF(re) ~ = a ~ a n d for some r 6 (r0, t,), I F ( - - r ) I F ( r ) ~ : a 1+~.

F r o m (A), (B), (C), ( + ) a n d ( + + ) we see t h a t F is t h e solution o f 2n -- 1 equations in 2 n - 1 u n k n o w n s .

Conversely, suppose f is a finite Blaschke p r o d u c t w i t h simple n e g a t i v e zeros (-- tl) ? a n d t h a t f is a solution o f (A), (B), a n d (C) w i t h f = F . T h e n f r o m (**) w i t h f - - - - F we see t h a t f 6 E a n d f satisfies (iii)--(viii) of T h e o r e m 1. H e n c e b y (*), f = F . W e conclude t h a t i f we can f i n d a solution f o f (A), (B), a n d (C), t h e n we h a v e f o u n d F .

I n practice a n explicit d e t e r m i n a t i o n o f F m a y be quite difficult even in v e r y simple cases. Consider for e x a m p l e t h e associated p r o b l e m of d e t e r m i n i n g for f i x e d r a, a n d k t h e largest possible value of r 0 for w h i c h F(., a, r r0) has k zeros.

I f k = 1, t h e n f r o m (A) we see t h a t F(z) = (z + a)/(1 + az). Moreover, r 0 = r0((r, 1) is t h e u n i q u e p o i n t in (0, 1) for which

( ro - - a ~ ( ro + a l "

I f a --- 0, we see t h a t r0(0, 1) = 2a/(1 -}- uP). Also, i f a is replaced b y -- l a n d 1 in t h e above equation, a n d i f t h e resulting e q u a t i o n s are soIved, t h e n we o b t a i n

~v/2a2/(1 ~- a 4) < r0(a, 1) < 1 .

Of course a n explicit d e t e r m i n a t i o n of re(a, 1) for - - 1 < a < 1 is difficult.

If k = 2 , t h e n

F(z) = ( z -~ t

1~(. + t. ~

O < t i < t . < x

%

\ 1 + t ~ z ] \ l + t 2 z ] ' can be f o u n d b y solving t h e equations

E X T ] ~ E Y l i L AIqALIrTIC F U N C T I O N S I N T H E U N I T C I R C L E 1 9 1

t 1 9 t 2 = a (9.1)

d

- - _F(-- r ) F ( r ) ~ = a 1+~, d r 2 ' ( - - r ) F ( r ) ~ -~ O , (9.2) f o r some r 6 (t 1, t2), ro(a, 2), 0 < r o < 1, is t h e n f o u n d b y solving t h e e q u a t i o n , 2 , ( - - ro)2,(ro) ~ -~ a l+~. F o r a ---- 0 this solution is g i v e n b y

+ ~/-~2 a 2,(z)

^{- -}

H e r e # = 2a/(1 + a2). M o r e o v e r , ro(0, 2) - - 2 W/~/(1 + #).

Again a n explicit d e t e r m i n a t i o n o f 2, a n d ro(a , 2 ) is difficult w h e n - - 1 < : a < 1. H o w e v e r it can b e s h o w n t h a t

]/2 v *l(1 + < r0( , 2) < 1.

I-Iere /** : 2a~/1 + a a.

F i n a l l y we r e m a r k t h a t H e i n s [3, (4.9)] d e t e r m i n e d 2, a n d r o e x p l i c i t l y w h e n a = 0 a n d k - ~ 2 ". H e r e n is a p o s i t i v e integer.

Suppose n o w t h a t a a n d a are g i v e n a n d r o ---- 1. L e t ( - - t~)~, 0 < ti < t~+~ < 1, d e n o t e t h e zeros o f 2,(-, a, a, 1). T h e n as in t h e case r o < 1, we see t h a t 2,(0) = a a n d f o r some r 6 [tt, t~+1],

[2,(-- r)]2,(r) ~ = a l+~ - - 2 , ( - - r ) F ( r ) " d ---- 0 . (9.3)

'

d r

Conversely, let f be a n i n f i n i t e B l a s c h k e p r o d u c t w i t h simple n e g a t i v e zeros ( - - s~)T, 0 < s~ < s~+l < 1. W e r e p l a c e 2' b y

f

in

(9.3). If f(0)

= a a n d

(9.3)

is t r u e for some r E [s t, st+l], I __< i < oo, t h e n f r o m (**) we see t h a t f satisfies (1.3) a n d (1.4). W e a s s e r t t h a t f - - - - 2 , . I n d e e d , i f f # 2,, t h e n f o r some Q, 0 < ~ < 1, we w o u l d h a v e f(Q) < 2,(~). L e t n a n d /r b e positive i n t e g e r s a n d suppose t h a t n > k. P u t

( z + t , ~

1

Since 2,n(z) --> 2,(z) a n d fk(z) -->f(z),

~ ( z + s t I

a n d fk(z) ---- U \ 1 + stz ] "

]z] < l , we see f o r k s u f f i c i e n t l y large t h a t fk(~) < 2,,(~) w h e n e v e r n > k. A p p l y i n g L e m m a 3 we o b t a i n for s o m e p o s i t i v e

i n t e g e r j t h a t

Ifk(-- r)]f~(r) ~ <_ I F , ( - - r ) I F , ( r ) ~, sj_ 1 _< r < s i . H e r e i f j = 1, t h e n s o - - 0 .

1 9 2 ~ O H N L . L E W I S

L e t t i n g n - + oo t h r o u g h a p r o p e r l y chosen s e q u e n c e we o b t a i n for s o m e p o s i t i v e i n t e g e r i t h a t

[ A ( - - r) lA(r) ~ < IF( - r) lF(r)L 8i_: ~ r < s l . H o w e v e r for s o m e r 6 [8i__1, 8i) we h a v e

a:+~ = If(-- r ) l f ( r F < I A ( - r) lA(rF < I F ( - r ) [ F ( r F < a :+~ 9 W e h a v e r e a c h e d a c o n t r a d i c t i o n . H e n c e F = f.

W e conclude t h a t F is u n i q u e l y d e t e r m i n e d b y t h e e q u a t i o n s (9.3) a n d t h e c o n d i t i o n t h a t F ( 0 ) = a.

lO. Proof of Theorem 2

L e t o, r 0, a n d /'n b e as in t h e s t a t e m e n t o f T h e o r e m 2. W e recall t h a t /'= is t h e set o f all a n a l y t i c f u n c t i o n s bn in {Izl < 1} w h i c h c a n b e r e p r e s e n t e d in t h e f o r m (1.8).

L e t

At(n) = i n f s u p re(r, b,)M(r, b,) ~ a n d p u t a :+" = # ( n ) .

bnEl'rt 0 < r <-- ro

W e r e p l a c e n b y k in (1.8). T h e n t h e cla~s o f f u n c t i o n s w h i c h c a n b e w r i t t e n in t h e f o r m (1.8) w h e r e 0 < k < n is c o m p a c t . H e n c e b y a n o r m a l f a m i l y a r g u m e n t t h e r e exists a m e m b e r G o f t h i s class w i t h k zeros (0 < k < n) f o r w h i c h

s u p re(r, G)M(r, G) ~ = At(n) = a :+" .

O < r < _ r o

N o w k = n, since o t h e r w i s e f(z) = z~-~G(z) is in _F a n d s u p m ( r , f ) M ( r , f)~ < At(n) .

O < : r ~ < r o

L e t G* be t h e f u n c t i o n a s s o c i a t e d w i t h G as in (2.2). T h e n f r o m (2.3) a n d (2.5) we see t h a t G ' E / ' = a n d

s u p ]G*(-- r)lG*(r) ~ = a :+~ = At(n). (10.1)

O < r < r o

W e c l a i m t h a t G* satisfies tiii) - - (viii) o f T h e o r e m 1. W e f i r s t s h o w t h a t G* satisfies (vii).

I f G*(0) < a, let

< a : < 1 a n d p u t

H e r e

= m i n { r > 0 : G * ( , r) = 0}.

O(z, ~l)G*(z)

g(z) - - O(z,cr , [z[ < 1 . 0 is as in L e m m a 1. F r o m L e m m a 1 we see t h a t

m a x [g(-- r)[g(r)" <

L e t ~1 b e s u c h t h a t

c6---r< ro

EXTI~E~CIAL AI~rALYTIC FUNCTIONS I N T]~E UNIT CIRCLE 193

Also f r o m (**) o f w 9 we d e d u c e lg(-- r)tg(r) ~ <_ g(0) 1+~ w h e n 0 ~ r ~ a. Since G*(0) ~ a, it follows f o r a 1 n e a r ~ t h a t

m a x t g ( - r) lg(r) ~ < a 1+~ .

O ~ r ~ _ r o

Since g E / ' n we h a v e r e a c h e d a c o n t r a d i c t i o n . H e n c e , G*(0) = a.

Also, G* h a s no zeros in ( - - 1, - - to]. O t h e r w i s e u s i n g L e m m a 1, w e c o u l d o b t a i n a c o n t r a d i c t i o n . H e n c e , (v) a n d (vi) a r e t r i v i a l l y t r u e . M o r e o v e r , L e m m a 1 a n d (**) o f w 9 i m p l y t h a t

[ r So)IG*(ro) ~ = a I+~ 9 ( l O . 2 ) F i n a l l y , we s h o w t h a t G* h a s p r o p e r t y (iv). L e t - - a, - - fl, 0 ~ c~ ~ fl ~ 1, b e t w o zeros o f G*. Choose ~l, fil, 0, s u c h t h a t 0 ~ 0 ~ 1 a n d 0 ~ a l ~ a _ ~ t~ < fli < 1. F u r

r 0r 81)

h ( z ) - - r ~ , t~) G * ( z ) , lzl < 1 .

H e r e r a n d a a r e as i n L e m m a 2. F r o m L e m m a 2 w e see t h a t ]h(-- r)Ih(r)" < [G*(-- r)lG*(r) ~

w h e n r e [ 0 , a l ] U [ f l l , 1) a n d G * ( - - r ) % 0 . T h e n if m a x [G*(-- r)lG*(r) ~ < a 1+~ , it follows f o r ~1, ill, n e a r ~, fi, t h a t

m a x l h ( - - r ) I h ( r ) ~ < a ~§ .

O ~ r < _ r o

Since h C F , , we h a v e a c o n t r a d i c t i o n . H e n c e , G* h a s p r o p e r t y (iv) o f T h e o r e m 1.

t ) r o p e r t y (iii) o f G* is a n o b v i o u s c o n s e q u e n c e o f p r o p e r t i e s (iv) a n d (vii). W e conclude t h a t G* satisfies (iii)--(vii) o f T h e o r e m 1.

L e t F ---- F ( . , a, ~, r0) be as in T h e o r e m 1. T h e n f r o m t h e d i s c u s s i o n in w 9 we see t h a t F - - ~ G * . I t r e m a i n s t o s h o w t h a t f o r s o m e 0, 0 _ ~ 0 ~ 2 ~ , we h a v e G* = e~~ T o p r o v e t h i s we let a ( r ) - ~ 1 in (2.5). W e o b t a i n t h a t

[G*(-- r)lG*(r ) ~ m(r, G)M(r, G), 0 ~ r < 1 . (10.3) T h e n i f G r e~~ * f o r s o m e 0, 0 ~ 0 ~ 2z~, we see f r o m (2.3) t h a t M ( r , G) ~ G*(r) f o r 0 ~ r ~ 1. U s i n g t h i s i n e q u a l i t y , (10.2), a n d (10.3), we o b t a i n

a ~+" --~ [G*(-- ro)[G*(ro) ~ ~-

[G*(-- ro)]G*(ro) 9 G*(ro) ~-1 ~ m(ro, G)M(ro, G) 9 M ( r o, G) "-1 ~_ a t+~ . W e h a v e r e a c h e d a co~ltradiction. H e n c e , for s o m e r e a l 0 we h a v e G ~ e~~

T h i s concludes t h e p r o o f o f T h e o r e m 2.

194 JOHI~ L. LEWIS

References

1. Ess#.~, M., Theorems of (A, @)-type on the growth of subharmonio functions, to appear.

2. tIEI~s, M., On a problem of Walsh concerning the H a d a m a r d three circles theorem, Trans.

Amer. Math. Soe. 55 (1944), 349--372.

3. --~)-- The problem of Milloux for functions analytic throughout the interior of the u n i t circle, Amer. J . Math. 67 (1945), 212--234.

4. HELLSTEI% U., KJELLBEI~G, B., & NOI~STAD, F., Subharmonie functions in a circle. Ark.

Mat. 8 (1969) 185--193.

5. LEwis, J., Some theorems on the cos ~ inequality, Trans. Amer. Math. Soc: 167 (1972) 171--189.

6. lXlEVAI~LI~I~A, R., Eindeutige analytische FunIctionen, Berlha, Springer-Verlag, 1953.

Received June 24, 1971 J o h n Lewis

D e p a r t m e n t of Mathematics University of K e n t u c k y Lexington K Y 40506 USA