mean p-valent functions with mini-gaps

Mean p-valent functions with mini-gaps

Herrn HERBERT GROTZSCH zum 65. Geburtstag am 21. Mai 1967 gewidmet

By W. K. HAYMAN, London

(Eingegangen am 20. 2.1967)

1. Introduction and statement of results

Suppose that ca

f (4 = c a, 2" n=O

( 1 . 1 )

is mean p-valentl) in 1x1 < 1, and that a, = 0 for a sequence n = n,, 0 I - v < 00, where

(1.2) Under these circumstances we can show that the growth of the maximum

modulus of f ( z ) and its coefficients is considerably less than it would be in the general case. Our result is

Theorem 1. I f f @ ) , given by ( l . l ) , i s meanp-valent in l x 1 < 1 and a," = 0

n, < n,,, 5 n, + C.

for a sequence nv satisfying (1.2) then we have

(1.3) and hence we have for n 2 1

1

H ( r , f ) < ~ , ( p ) p ~ ( l + v?) c'+"(I - r ) - P , 0 < r < 1

1 2

If p <: -, we have

and

(1.7) lan/ = o ( n ') as n -+ 00.

1 --

1) In this paper mean p-valent denotes areally mean p-valent in the sense of [2].

314 Hayman, Mean p-valent functions with mini-gaps

Here ,u =- max lafi l , M ( r , f ) = max / f ( z ) l . P

OgnSp 1zJ = r

Also A ( p ) denotes a constant depending only on p ; A , ( p ) , A 2 ( p ) , . . are particular such constants.

This paper is a sequel t o [4] where I consider the stronger gap condition

a,, = 0, except when n = m V )

m,+, 2 m, + C. where

The conclusions of Theorem 1 are there obtained if C = 2. Clearly in this case if n = n, denotes the complementary sequence of coefficients for which a,, == 0 then

n,,, - n, 5 2 , if n, 2 m,. Thus (1.2) holds with C = 2, and so the hypotheses of our present Theorem 1 are considerably weaker than those of [4] with C = 2 , while the conclusions are the same. I n [4] it was also noted that (1.4) to (1.7) follow from (1.3) by standard methods ( [ 2 ] , Theorems 3.3 and 3.4 and [S]), so that it is sufficient to prove (1.3) and further that (1.3), (1.4), (1.6) and (1.7) are best possible even in the special case of even functions (n, = 2v + 1). It should be noted that in the special case n, = Cv + B the conclusion of Theorem 1 was obtained previously. ([2], Theorem 3.9, p. 65).

Also if p = I, it was shownin [3], that IlapL+l I - la,ll < A p l , so that the conclusion of Theorem 1 follows in this case and with C instead of

Just as in [4] we can also investigate the class of those functions for which M ( r , f ) attains the extremal growth indicated by (1.3). If

N ( r , f ) = o ( l - Y I P >

I and p > 2 , we can deduce

lafil = o(n” - ‘ )

and this is best possible. In the contrary case we deduce

Theorem 2. Xuppose that f ( x ) satisjies the hypotheses of Theorem 1 and further that

p = l;m ( I - r)” N ( r , f ) > 0. r -> I

(1.8)

(i) Then there exist two radii, arg x = O,, O , , such that

Hayman, Mean p-valent functions with mini-gaps 315

I

2 2 (ii) I f further p > then

i01 i i ( r , f ) - If(r 2"") I - i f ( r e ) 1 - P(I - r ) - ~ , as r + 3

and

(iii) We have 0, - 0, = 2..cp, ___ where p , q are integers andq 5 C. 4

in the case p = 1, is given by the univalent functions A typical example of functions satisfying the conditions of Theorem 2

Here 0,) = - O1 = 0, so that O1 - 0,) = 2 n;p/q. It would be interesting to decide whether the conclusions (ii) and (iii) of Theorem 2 also hold if

p 5 ~. The method of proof, which is based on an asymptotic formula of

EKE [I] for the coefficients off ( z ) certainly does not extend.

1

2

2. Proof of Theorem 1.

The proof of Theorem 1 is very similar to that of the corresponding result in [4]. We shall employ a similar sequence of Lemmas. Only a slight modification is required from time to time. We need first [6 , Theorem 1, Corollary 21.

Lemma 1. Xuppose that g ( 0 ) , dejned in [0,2 571 has a uniformly convergent FouRIm expansion

+m

g ( 0 ) = bn eino,

and that b% = 0 f o r a sequence n = n, , - 00 < v < + 0.3, which satisjes (1.2). Then if i s real

where Ai = 4 ~/2/(l /2 - 1). We now write

m no (2.1) P ( z ) = 2 an z" , g ( z ) = 2 a, z",

so that f (z ) = P ( z ) + g(z ) in (1.1). We note that if n=O n=n,+l

0 2 2 01 + 2 z [I - 1/(4 C)]


then we deduce from Lemma 1, that 2n 0,

/g'(te'@)/'dO 5 A l C \ Ig'(te'@)j dO 0 6,

We write

S(r , g ) = X ( r , (I, 2 sd, g ) ,

and deduce that for O2 2 O1 i- 2 n [I - l ' ( 4 C)], we have

(2.2) S ( r , g ) < A i C S ( r , 01, 02, 9 ) . We now quote [4, Lemma 21

Lemma 2. Sqcppose that f (2) i s regular in 1 x 1 < 1, that M. and K are positice constants and that

(2.3) where 0 < rz < rl < 1. Then there exists r , such that rl < r (2.4)

(2 .5)

where r' = - (1 + r ) .

We have next

Lemma 3. If f ( x ) satisjies the hypotheses of Theorem 1 , and i f P ( z ) is

N ( r 1 7 f) c: K ( l - J f ( r 2 , f) 2 K ( l - r 2 ) - u .

r 2 , and

M ( r , p ) 2 K(1 - r ) - a ,

S ( r ' , f ) 2 ~ cc2 N ( r , f ) z 2 ~ cc2 K ~ ( I - r ) - 2 a , 7c x 4 4

1

2

given b y (2 . i ) , then

#(I, P) <. A 5 ( p ) v p p i .

We choose r = exp ( - l,lvo) YO YO co

S(i . P ) = C n / a , , / 2 < e2 C n /anjzrL'L < e2 C nIag, l~rz" n = l n i l n= 1

= e2S(r , f ) < 7cp ezM(r , f ) ' < A(p)p:(l - r ) - 4 p

since f (z ) is mean p-valent. This proves Lemma 3. We deduce

Lemma 4. Suppose that f (2) satisfies the hypotheses of Theorem I and that (2.3) holds with r = p , and


Let r , r’ satisfy the conclusion,s of Lemma 2 . Then i f 0 2 2 01 + 3 j z [I -- 1/(4 G)]

there exists x = Q e i M , with 0 5 Q 5 r‘, O1 5 0 5 O2 such that 1

The proof is similar to that of Lemma 6 of [a], but we include it for completeness. We have by (2.5)

J Z

X(r ’ , f )>-p2K2(1 - r ) - 2 p , 4

while by Lemma 3

S(r’, P) < A; ( p ) vp ,ui. We have by MINKOWSXI’S inequality

i 1 1

x (r’ , 1) 2 5 x (r’, PI“ + s (r’, g)2 , so that

if S(r‘, f ) 2 4 S(r’, P),

i. e. certainly if (2.6) holds. We now apply (2.2) and deduce that

?Cp2K2 >- (1 - r)-? 16A1C

Then we deduce from Lemma 3 and (2.6) that 1 4 X(r’, O t , 02, P ) < S(1, P ) <- S(r’, Oi, 02, 9 ) .

Thus in this case we have, since f = P + g,

Let M = sup If(z) I in the sector E of values z = teio, for which

0 5 t 5 r’. Oi 2 0 5 02,

318 Hayman, Mean p-Talent functions with mini-gaps

Since f ( z ) is mean p-valeat ill E', the area of the image of this set by f ( z ) is at most x p M', so that

Thus n-e can find z in E such that 1

provided that (2.6) holds. This proves Lemma 4. We quote finally ( [ a ] , Theorem 2 . 6 ) .

Lemma 5. Suppose that f ( x ) i s wiecrn p-valent in the disjoint d i sks

/ z - z: 1 < r v , v = 1, 2,

1 2

and that f ( z ) =!= 0 in j z - z:/ < - Y ~ . Suppose also that

If(4) ~ 5 x, > I f (%) I 2 R, >

whew ?", -- 1 zv - 2 )

(7 = ~~~ - > 0 , R, > eR,. T'V

Then

We can now complete our proof. Since f ( z ) is mean p-valent in 1 z 1 < 1, f ( z ) has q 5 p zeros there.

Hence f ( z ) =f 0 in a t least one of the annuli 4 - t 4 - ( t + l )

10 c I O C ' 1 - -< 121 < 1 - ~ t = 0 to q .

TI7e choose such an annulus, set

4-' 20 C

@=I- ~ = 1 - r O , say,

and note that f ( z ) has 110 zeros in the annulus

1 2

1 --- 2 r o < jzi < 1 - r,,.

We choose zi = r' eiol , such that

I f (z1) I = n/r(r ' : f ) 2 N ( r , f ) 2 K(I - r1 -P


by (2.4). Then by Lenima 4 we can find z2 = r2 eio’, such that

n p 1 G2 - 0, I 2 z/(4 C ) , 0 < r 2 5 r’ and

1

i 0, We set z: = Q e the points do’, for v = I. S, and

, v = 1, 2 , and rv = yo. The circles I z - z: 1 < ro contain

Thus the circles / z - z,l < ro are disjoint. I T e now apply Lemma 6, with

R, = N ( c , , ~ ) 2 max I f ( $ , ) i . v = I, 1

1

Then we have either

(2.9) R, < e R , , or !f(z,) I > R, , so that p < j z, 1 < r’ , v = 1 2, and we can apply Lemma 5. This gives

i. e.

Hence Ri < A ( p ) RL [i + C‘-’(l - Y)-”].

This inequality is trivial if (2.9) holds and so is true in any case. Also by the SPENCER-CARTWRIGHT Theorem (see e. g. [ 2 ] , Theorem 2.5) wc have, since f ( z ) is mean p-valent

R1 < d ( p ) p p ( 1 - e ) -‘IJ < A ( p ) ,up C” . Thus we deduce that

320 Hayman, Nean p-valent functions with mini-gaps

In view of (2.5) this yields 1

i . e. I 2 K < A ( P ) ,up GI”+ >

K = A,@) (1 + v?) p p c*+

if the conditions of Lemmas 2 and 4 are satisfied with 1 - 1,C < r < 1, i. e. certainly if 1 - f/C < rl < 1 in Lemma 2 . Tq7e take

1

where A , ( p ) is a sufficiently large constant depending on p . Then in view of the SrENcER-CART~--I1IGHT Theorern we certainly have

Z ( r i , f) K ( l - T , ) - ~ , rl 6 I - 1 C . Also (2.6) holds. Thus if for home r , . such that rl ,< r L < 1 we have

M ( r 2 , f ) > K ( l - r 2 ) P

we are led to a contradiction. This completes the proof of (1.3) and so of Theorem 1.

3. Preliminary results

We need the following result of EKE [I].

Lemma 6. Xuppose that f ( x ) = a,, 2’’ is mean p-valent in 1x1 < 1, and that for some integer k and positive numbers 6, yi, there exists a sequence r?,’ + 1, and points z ~ , ~ ‘ , v = 1 to k , such that lzV,, 1 = r,,

1 5 p” < v I: k , m 2 1;

( I

l ~ ~ , ? ~ - z,,~J 2 6 ,

I1 - rnbIk i f ( x V , d 2 q, 2P

12 v 2 E , m 2 I. Then there exist I% rays arg x = O,, where Oo < O1 < . . . < 0, = Oo + 2 x, with the following properties

as r --f 1. io, ,

(ii) Purther i f r = r , (R) i s so chosen that 1 f ( r e ) I = R, then

Haynian, Mean p-Talent functions with mini-gaps 321

as R --j 00, where 0 < pi < 00.

(iii) If we write cr,(n) = ,-’PIk

then u*e have i( e - eV) - ? p / k f (reie) = [I + 0(1)] q ( n ) [I - re 1 -

as 1% + m, uniformly provided that n ( l - r ) is bounded above and below and I L 10 - @,I i s bounded above. Also

f ’ (reio) 2 P f(rei“) IC[I - re

while 1’ + 1: and 10 - 0, j = 0(1 - r ) .

-- 1 i(0-0,)

(iv) W e have as r --f 1

(v) If in addition 41, > k , then we have as r + 00

where .(n) = sup .,(n).

v = 0 to k - I

It is easy to see that with the hypotheses of Theorem 2 , f ( x ) satisfies the conditions of EKE’S Lemma 6 with k = 2 . The details of the argument are given in [4] Lemma 9, pp. 1 1 - 13. We therefore only give a rough sketch. We first deduce from (1.8) that

G (1 - qp+’ N(r,f) 2 1 , p . r + l

This leads as in the proof of Lemma 2 to

(1 - r)’” X ( r , f ) = p’ > 0 . r + l

Hence if r is so chosen that

1 (I - r)’”X(r,f) > Zp’,

and if(reiel) j = M ( r , f), it follows from (2.2) _i that


and hence we can find x i = r2 e iM2, with 1 O2 - O1 1 2 n/(4 C) such that 0 < r2 5 rl and

1

I n view of Theorem 1, it follows that (1 - r ) = 0 (1 - r?) , and so

Thus if r +- 1 through a suitable sequence the points x1 : reio’ 2 ’ ’ x = reio, satisfy the conditions of Lemma 6.

~ f ( z i ) i = O I f ( re io l ) l .

4. Proof of Theorem 2

We proceed to prove Theorem 2 . The existence of the radii arg z = Oo , O1 follows from Lemma 6. Suppose that R is a positive number. If R is sufficiently large and y , , ( R ) , r l ( R ) are defined as in Lemma 6 we have

1

2

By Theorem 1 we have for v = 1 . 2

R’( 1 - (1 - r,)?’ > 18, .

where K is a positive constant. Thus if ro is sufficiently near 1

Similarly

R ( l - r 0 ) P > - , K

1. e. 1

1 - r , This proves (i).

1 We now suppose that p > so that we can apply Lemma 6 (v). We

iiotc that in view of Theorem 2 (i) 1 a. (n) 1 and I dcl (n) I are bounded above and below by positive constants. Thus Lemma 6 (v) can be written in the form

(4.1)

2 ’

~ ( p ) n l - ~ a, = q ( n ) e-z7’’o + q ( n ) e-”’@1 + o(1) .

Hayman, Mean p - d e n t functions with mini-gaps 323

Suppose now that a, = a,%+, = 0 for an infinite sequence of values of n and a fixed q . It follows from Lemma 6 (iii) that

.o(n) - .o(n -1- 4 ) . Thus we deduce that

+ 4 ) XO (n) .o(n + 4 )

+ o(1) = ~ ~~ + o(1 ) -~ - - e - z n ( O , - @ , )

+ o(1) . - - i(R +a)(@,- @,I -

Thus e L g ( @ O - @ 1 ) = 1 + o(1) = 1. since y, Go, Gi are constants, i. e.

where p in an integer. Since the conclusion holds for some y C by hypo- thesis, we deduce Theorem 2 (iii). Finally it follows from (4.i), that if a,, = 0 , we have

I ~ ( n ) I = h ( n ) ! + o ( l ) ,

lf(rei@O) I - If(rei@l) 1 ,

i. e. ! a o ( % ) ! N ! ~ f (n ) l , or

with r = 1 - l/n. Given r , such that 1 - l /no < r < 1, u-e choose n,, so that

1 - l/n, 5 r < 1 - l/n,,+l.

I n view of (1.2) and Lemma 8 (iii) we deduce that then

f(rei@") - f [(I - l /nv) k"] , so that

j f(re"0)I - 1 f ( re z@l ) I . Now let ro (R) , r l (Iz) be defined as in Lemma 6 (ii). Then it follows from Theorem 2 (i) that (1 - r0))'( 1 - r , ) is bounded above and below hy positive constants. Thus we have by Lemma 6 (iii)

324 Hayman, Nean p-valent functions with mini-gaps

Hence we deduce from Lemma 6 (iv), that 1

M(r,f‘) - p; (1 - Y)-r’, 1 __

so that p = P I 4 by (1.8). This completes the proof of Theorem 2 .

Bibliography

[I] B. EKE, The asymptotic behaviour of areally mean valent functions, J. analyse Math. XX, 147-212 (1967).

121 W. K. H-YMAN, Multivalent functions, Cambridge 1958. [3] -, On successive coefficients of univalent functions. J. London Math. SOC. 38,228-243

(1963). [4] -, Mean p-valent functions with gaps. Coll. Math. XVI, 1-21 (1967). [5] -, A mini-gap theorem for FOURIER series. Proc. Cam. Phil. Soc. 64, 61-66 (1968). [6] C’. POMMERENKE, Uber die Mittelwerte und Koeffizienten multivalenter Funktionen.

Math. Ann. 146, 285-296 (1961/62).

mean p-valent functions with mini-gaps

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