Convex Holes Produce Lower Bounds for Coefficients

Convex holes produ e lower bounds for oeÆ ients F. G. AVKHADIEV

Chebotarev Resear h Institute, Kazan State University R-420008 Kazan, Russia K.-J. WIRTHS

Institut fur Analysis, Te hnis he Universitat D-38106 Brauns hweig, Germany Dedi ated to E. Kreyszig on the o

asion of his 80th birthday Let D denote the open unit disk and f : D ! C be meromorphi and inje tive in D. Espe ially, we onsider su h f whi h have an expansion

f (z ) = z +

1

X

n=2

an (f )z n

in a neighbourhood of the origin and map D onto a domain whose omplement with respe t to C is onvex. Let the set of these fun tions be denoted by Co. We x jf 1 (1)j for f 2 Co and determine the inner and outer radius of the ring domain whi h is the domain of variability of a2 (f ) for su h f . Further, it is shown that f 2 Co implies that f 0 (z ) (z ) = z + 2 00 f (z ) is holomorphi in D and maps D into itself. This impli ation in turn implies the inequalities jan (f )j  1 for f 2 Co and n = 2; 3; 4:

Meromorphi onvex fun tion, Taylor oeÆ ients AMS Classi ation: 30C45, 30C50

Keywords:

1

INTRODUCTION

Let f be a fun tion meromorphi and univalent in the unit disk D = fz j jz j < 1g. Suppose that f (z ) = z +

1

X

n=2

an (f )z n

(1)

in some neighbourhood of the origin. In parti ular, this means that either f is holomorphi in D or f has a pole and b := f (1) 2 D n f0g: (2) A onsequen e of well-known fa ts (see [1℄,[2℄, [3℄, [4℄, [9℄) are the inequalities jan(f )j  n; n 2 N n f1g; (3) 1

1

if f is holomorphi in D, and a

ording to Jenkins' and de Brange's estimate ( ompare [5℄) n jan(f )j  1 jbj k ; n 2 N n f1g; (4) jbjn

1 X

1

2

k=0

if f is meromorphi in D and b 2 D n f0g. Note that bounds of the kind (3) and (4) are known for several lasses of univalent fun tions. In many ases, the oeÆ ient region is a disk of enter 0, therefore the best lower bound for jan (f )j is 0 ( ompare again the above referen es). Let Co be the set of all fun tions f meromorphi and univalent in D whi h satisfy (1) in fz j jzj < jbjg (see(2)), resp. in D, and whi h in addition have the property that C n f (D) is a onvex set. Note that the known lasses CV (jbj); jbj 2 (0; 1); (see ([4℄) are proper sub lasses of Co. In ontrast to the above ited results we prove that jan(f )j  1 (5) for any f 2 Co and n = 2; 3; 4. In fa t, we onje ture that the estimate (5) may be true for all n  2. Note that a fun tion f 2 Co may be holomorphi in D. Geometri ally, for f 2 Co, there are two dierent ases. If the onvex set C n f (D) is bounded, then f has a single pole in D be ause of the univalen e of f . If the onvex set C n f (D) is unbounded, then 1 2 f (D). Therefore, f is holomorphi in D. A meromorphi onvex fun tion F su h that F ( ) =  +

1

X

k=0

bk  k ;

j j > 1;

satis es Goluzin's inequality jF 0( ) 1j  j j ; j j > 1; (see [3℄ and [8℄). Equality o

urs if and only if F ( ) =  + b + ei  ; 2 R; j j > 1: This fa t and Lemma 1 below easily imply the following assertion. If f 2 Co, then f (D) is a losed Jordan urve in C or a straight segment [z ; z ℄  C. Therefore, there exists an unique point b = b(f ) 2 D su h that f has a

ontinuous ompletition f~ on D n fbg and f~(z) ! 1 as z ! b; z 2 D n fbg. In the sequel, we will denote by f (1) this point b 2 D. For xed jf (1)j we nd the sharp lower and upper bounds for ja (f )j for f 2 Co. Some other properties of Co are also onsidered. 2

1

0

0

1

1

1

2

2

2

BOUNDS FOR COEFFICIENTS

First, we onsider the upper and lower bounds of ja (f )j for f 2 Co. We are interested mainly in lower bounds, here, and in lude the known upper bounds ( ompare [5℄, [6℄ and [7℄) valid for all meromorphi univalent fun tions with expansion (1) for the sake of ompleteness. As extremal fun tions we shall need the following members of Co with f (1) = r 2 (0; 1℄: z kr (z ) = (6) (1 zr )(1 r z) with r r kr (D) = C n ; (1 r) (1 + r) and r z z r (7) lr (z ) = (1 zr )(1 r z) with 2r r lr (D) = C n z j Re z = 1 + r ; jIm zj  1 r with the obvious interpretation in the ase r = 1. This means that k (z) = z (1 z ) is the Koebe fun tion and that l (z ) = z (1 z ) maps D onto the halfplane fw j Re w > 1=2g. THEOREM 1 Let f 2 Co and r := jf (1)j 2 (0; 1℄. Then the estimates 1 + r  ja (f )j  r + 1 (8) r(1 + r ) r 2

1





2

2

2 1+ 2

2

(

)

2

2

2

4

1

1

1

1

4

2

2

are valid. Equality in the rst estimate o

urs if and only if f (z ) = e i lr (ei z );

 2 R;

and equality in the se ond estimate o

urs if and only if f (z ) = e i kr (ei z ); Proof Consider rst the ase f fr (z ) = ei f (e i z ) we have

1

 2 R:

(1) = r ei 2 D. For the fun tion

(1) = r and a (fr ) = e i a (f ): Let the fun tions F and be de ned in
= f j j j > 1g by 1 + r ;  2
; F ( ) = f r  +r fr

2 Co;

fr

1

2





3

2

(9) (10)

and

00 ( ) ; 1 +  FF 0(()) = 11 + ( ) The fun tion F has an expansion F ( ) = a  +

1

X

k=0



2
:

a 6= 0; 

bk  k ;

(11)

2
;

(12)

and maps
onto fr (D). Sin e the set C n fr (D) is onvex, the fun tion (12) satis es the inequality 00 Re 1 +  FF 0(()) > 0;  2
; (13) (see [8℄). A

ording to (11) this implies (14) j ( )j  1 ;  2
: 



j j

2

By Lindelof's prin iple, (14) is equivalent to the inequality F 00 ( ) 2  2j j ;  2
:  0 (15) F ( ) j j 1 j j 1 If equality o

urs at one point  2
n f1g in (14), respe tively in (15), then j ( )j  1, i. e. there exists a real onstant su h that ( ) = ei  ;  2
: (16) From (10) we obtain (1 r z)(r z) fr00(z) 2 1 r z ; z 2 D: (17) F 00 ( ) =  0 F ( )  r z 1 r fr0 (z ) 1 r z r Thus, 1 F 00( 1=r) = 2ra (fr ) 2 ; (18) r F 0 ( 1=r) 1 r 1 r and the inequality (15) at the point  = 1=r be omes



4

2

4

0

2

2



=

2

1

2

2

2



r a2 (fr )

1



r4 1 + r2 

2

r2 1 + r2 :

(19)

The formulae (9) and (19) imply the desired estimates (8). If in (8) equality o

urs on either side, this means that equality o

urs in (19) and that Im (a (fr )) = 0 and equality o

urs in (15) for  = 1=r. This together with (11), (16) and (18) implies that ei = +1 and ( ) =  ; F ( ) = a( +  ) + b; a 6= 0; b 2 C 2

2

1

4

or

= 1 and ( ) =  ; F ( ) = a(  ) + b; a 6= 0; b 2 C: Inserting of  = (1 r z)(z r) into these two possibilities a

ording to (10) leads to the two extremal fun tions kr ( ompare (6)) and lr ( ompare (7)). This ompletes the onsideration of the ase f (1) 2 D. For the dis ussion of f (1) 2 D we need the following lemma. LEMMA 1 If f 2 Co and f is holomorphi in D then there exists a sequen e fn 2 Co; n 2 N, su h that fn (1) 2 D and fn ! f uniformly in D as n ! 1. Proof Observe that the ompa t sets Kn = (C n f (D)) \ fz j jz j  ng; n 2 N; are onvex. Let gn be the Riemann mapping fun tion of D onto C n Kn ; gn (0) = 0; gn0 (0) > 0. It is obvious that Lemma 1 is a onsequen e of the Caratheodory kernel theorem for the sequen e fn = gn=gn0 (0) (see [2℄, [3℄ and [9℄). The assertion f 2 Co and f (1) 2 D ) 1  ja (f )j  2 (20) of Theorem 1 is proved by applying Lemma 1 and (8) for any r 2 (0; 1): As to the equality in (20) the Bieberba h theorem asserts that ja (f )j = 2 if and only if f (z) = e i k (ei z) ( ompare (6)). The proof for the fa t that ja (f )j = 1 in (20) if and only if f (z) = e i l (ei z) ( ompare (7)) is postponed to the proof of Theorem 3, below. THEOREM 2 If f 2 Co then the fun tion  de ned by the formula 2 f 0(z) (z ) = z + 00 (21) f (z ) ei

2

1

1

1

1

1

1

2

2

1

2

1

and the holomorphi ompletition (b) = b if b = f 1(1) 2 D has the following properties: (i)  is holomorphi in D and j(z)j  1 for z 2 D. (ii) The point b = f 1(1) is the attra tive xed point of . More pre isely, (b) = b; 0 (b) = 00 (b) = 0 if jbj < 1 and 0 (b) 2 [0; 1=3℄ if jbj = 1. (iii)  has no xed point in D n fbg:

Proof Suppose that f 00 (z0 ) and 0 = (1 r z0)(z0 r) 1

= 0 for some z 2 D. Inserting b = into (17) we obtain 00 Re 1 +  FF 0(( )) = Re 1 1r +r 2r z < 0 5 0





0

0

0

2

0

2

r ei

in ontradi tion to the inequality (13). Thus, f 00(z) 6= 0 for z 2 D and therefore  is holomorphi in D. For the proof of j(z)j  1; z 2 D, we rst onsider the ase r 2 (0; 1). It is suÆ ient to onsider fr (z) = ei f (e iz) with 2 f 0 (z ) r (z ) = ei (e i z ) = z + 00r : fr (z ) Suppose that there exists a z 2 D su h that r (z ) = 1=r. This implies 2r fr00(z ) = 0 fr (z ) 1 r z and for  = (1 r z )(z r) 00 Re 1 +  FF 0(( )) = 1 in ontradi tion to (13). Therefore, r (z) 6= 1=r for z 2 D and the fun tion 0

0

0

0

0

0

0

1

0





0

0

0





r ( ) =

r r r 1+  +r   1 r r 1+ +rr

(22)

is holomorphi in
, and r (1) = 0. A straightforward omputation on erning the fun tion F de ned in (10) shows that 00 1 +  FF 0(()) = 11 +  rr (( )) ;  2
: (23) From (13), (22) and (23) we obtain that jr (z)j < 1 for z 2 D. Using Lemma 1 we get j(z)j  1 for z 2 D and any f 2 Co. We now prove the assertion (ii). It is known that a holomorphi fun tion  : D ! D may have at most one attra tive xed point z 2 D (see [10℄ and [11℄). If b 2 D, then (21) implies that (b) = b and 0(b) = 00 (b) = 0. Thus, b is the attra tive xed point of . Let now r = fr (1) = 1. Applying (19) for the oeÆ ient a (g) of the fun tion x fr z xz fr (x) ; z 2 D; x 2 (0; 1) g(z ) = fr0 (x)(1 x ) we get 3  1; a (g ) 2 2 that is fr00 (x) 3 = 1 x 2x + 3  1 : (1 x) x (24) 0 2 fr (x) 2 r (x) x 2 2 0

1

2



+ 1+



2

2



2







6

2



By Julia's Lemma (see [10℄ and [11℄), there exists the angular limit 0r (1) > 0, if r (z) 6 1: From (24) we obtain 2 5  1: 0 1 r (1) 2 2 Hen e, 0  0r (1)  1=3. The fun tions 1 1 + z 1 ; 2 [1; 2℄ f (z ) = 2 1 z with 1 2 [0; 13 ℄ 0 (1) = +1 are examples for any 0 (1) 2 [0; 1=3℄. To prove (iii) it is suÆ ient to see that f 0(z) 6= 0 for z 2 D n fbg implies (z ) 6= z for z 2 D n fbg. This on ludes the proof of Theorem 2. REMARK 1 If b 2 D nf0g and  is a fun tion satisfying the onditions (i), (ii) and (iii) of Theorem 2, then the fun tion f from (21) with an expansion of the form (1) belongs to Co. This fa t is a simple onsequen e of (10), (11), (22) and (23). THEOREM 3 If f 2 Co and an(f ) = f n (0)=n!, then jan(f )j  1 (25) for n = 2; 3; 4: Equality in (25) o

urs if and only if z f (z ) = e i l (ei z ) = 1 ei z ;  2 R ( ompare (7)): (26)





!



( )

1

As we have seen in Theorem 2, we an use the equation f 00 (z )(z ) = z f 00(z ) + 2f 0 (z ) (27) in some neighbourhood of the origin. The fun tion  in (27) satis es j(z)j  1 for z 2 D and (0) 6= 0: (28) Setting 1 (z ) =

k z k ; z 2 D; Proof

X

we get from (27):

k=0

2 a (f ) = 2 0

2

7

(29)

and

m

X

(k + 2)(k + 1) m

k=0

( ) = (m + 2)(m + 1)am

k ak+2 f

+1

As an abbreviation we use

An = n an (f );

;

m 2 N:

(30)

n 2 N n f1g:

Formula (30) immediately yields the re urren e relation (m+1)Am

+2

0 = (m+2 m 1 )Am+1

whereas (29) means that

A2 =

m X2 k=0

(k+1) m

k Ak+2 ;

m 2 N;

(31)

2:

(32) Now we x j j =: x 2 (0; 1℄ and ompute a lower bound for the quantity jAn(x)j; n  3, whi h denotes the modulus of the oeÆ ient of zn in the Taylor expansion of f 0 with xed jf 00(0)=2j = 2=x a

ording to (32). This equation imediately implies jA j  2 with equality if and only if x = 1. A

ording to the maximum modulus prin iple this means that (z)  e i ;  2 R. Integrating (21) in this ase leaves us with the desired extremal fun tions (26). Obviously, they have janj = 1; n 2 N nf1g: So, we have done now the missing step in the proof of Theorem 2. The ase m = 1 of (31) implies 3 : A = A = (3 ) 2

We use j j  1 x to get jA (x)j  2 + x ; (33)

0

0

1

2

3

1

1

2

2 0

0

2

1

2

3

respe tively

x2

jA (x)j 3  2(1 x x ) 2

3

2

whi h immediately yields the desired results for a . In the ase m = 2, we obtain from (31) jA (x)j  31x (j4 2 jjA (x)j j j jA (x)j): Using (32), (33) and j k j  1 x ; k = 1; 2; we ompute jA (x)j 4  2(1 x)(2 +3xx + 4x x ) 8 3

4

1

3

2

2

2

2

4

3

3

whi h implies the assertions of Theorem 3 for a . REMARK 2 The integration of (21) shows that the Taylor oeÆ ients of f; f 2 Co, may be written as polynomials in the Taylor oeÆ ients of 1=. Unfortunately, the method of proof of Theorem 3 using this fa t does not work any longer for n = 5. Nevertheless we hope that a re ned onsideration of the oeÆ ient regions for unimodular bounded fun tions may reveal the truth of the CONJECTURE If f 2 Co, then jf n (0)=n!j  1 for all n 2 N n f1g. 4

( )

A knowledgment

During the work on this arti le F. G. Avkhadiev was supported by a grant of the Deuts he Fors hungsgemeins haft.

Referen es

[1℄ L. de Brange, A proof of the Bieberba h onje ture, A ta Math. 154(1985),137{152. [2℄ P. L. Duren, Univalent fun tions, Springer-Verlag, New York, 1980. [3℄ G. M. Goluzin, Geometri theory of fun tions of a omplex variable, Translations of mathemati al monographs 26, AMS, Providen e, 1969. [4℄ A. W. Goodman, Univalent fun tions, Mariner Publishing Company, 1983. [5℄ J. A. Jenkins, On a onje ture of Goodman on erning meromorphi univalent fun tions, Mi higan Math. J. 9(1962),25{27. [6℄ Y. Komatu, Note on the theory of onformal representation by meromorphi fun tions I, Pro . Japan A ad. 21(1945), 269{277. [7℄ K. Ladegast, Beitrage zur Theorie der s hli hten Funktionen, Math. Z. 58(1953),115{159. [8℄ Ch. Pommerenke, U ber einige Klassen meromorpher s hli hter Funktionen, Math. Z. 78(1962),263{284. [9℄ Ch. Pommerenke, Univalent fun tions, Vandenhoe k & Rupre ht, Gottingen, 1975. [10℄ Ch. Pommerenke, Boundary Behaviour of Conformal Maps, SpringerVerlag, New York, 1992. 9

[11℄ J. H. Shapiro, Composition Operators and Classi al Fun tion Theory, Springer-Verlag, New York, 1993.

10