Punishing Factors for Angles

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Computational Methods and Function TheoryVolume 3 (2003), No. 1, 127–141

Punishing Factors for Angles

Farit G. Avkhadiev and Karl-Joachim Wirths

(Communicated by Walter K. Hayman)

Dedicated to the memory of D. Gaier

Abstract. Let Ω and Π be two simply connected domains in the complexplane C which are not equal to the whole plane C. We consider functionsf : Ω → Π analytic in Ω and we get estimates for |f (n)(z)|, z ∈ Ω, which aresharp in the following sense. Let λΩ(z) and λΠ(w) denote the reciprocal ofthe conformal radius of Ω in z and of Π in w, respectively. Inequalities of thetype

|f (n)(z)|n!

≤Mn(z,Ω,Π)(λΩ(z))n

λΠ(f(z)), z ∈ Ω,

are considered where Mn(z,Ω,Π) does not depend on f and represents thesmallest value possible at this place. We especially consider cases where Ωor Π is an angular domain Hα with opening angle απ, 1 ≤ α ≤ 2. Wedetermine Mn(z,∆,Hα) where ∆ denotes the unit disk.

Furthermore, we prove identities and inequalities for

Cn(Ω,Π) := supMn(z,Ω,Π) | z ∈ Ωfor several cases where Hα plays the role of Ω or Π.

Keywords. Angular domains, derivatives of arbitrary order, conformal ra-dius.

2000 MSC. Primary 30C80; Secondary 30C55.

1. Introduction and statement of results

Let Ω and Π be two simply connected domains in the complex plane and let ∆denote the unit disk. If z0 ∈ Ω and

ΦΩ,z0 : ∆→ Ω

Received February 14, 2003, in revised form August 7, 2003.This work was supported by a grant of the Deutsche Forschungsgemeinschaft for F. G.Avkhadiev.

ISSN 1617-9447/$ 2.50 c© 2003 Heldermann Verlag

128 F. G. Avkhadiev and K.-J. Wirths CMFT

is the conformal mapping of ∆ onto Ω, normalized by taking ΦΩ,z0(0) = z0 andΦ′Ω,z0(0) > 0, then Φ′Ω,z0(0) is called the conformal radius of Ω in z0. We preferto work with its reciprocal

λΩ(z0) :=1

Φ′Ω,z0(0)

which is called, especially in articles on differential geometry and complex analy-sis, the density of the Poincare metric of Ω at z0. For the set A(Ω,Π) of functionsf : Ω → Π analytic in Ω we consider the quantities Ln(f, z,Ω,Π), Mn(z,Ω,Π)and Cn(Ω,Π) defined by

1

n!

∣∣f (n)(z)∣∣ = Ln(f, z,Ω,Π)

(λΩ(z))n

λΠ(f(z)), n ∈ N, f ∈ A(Ω,Π), z ∈ Ω,

Mn(z,Ω,Π) := supLn(f, z,Ω,Π) | f ∈ A(Ω,Π),Cn(Ω,Π) := supMn(z,Ω,Π) | z ∈ Ω.

After a colloquium talk of the second author on the results of this paper Ch.Pommerenke ([14]) suggested looking at the above definitions in the followingway. The quotient (λΩ(z))n/λΠ(f(z)) reflects the influence of the positions of thepoints z and f(z) in Ω and Π on the nth derivative f (n)(z), whereas the quantitiesCn(Ω,Π) are factors punishing bad behavior of Ω or Π at the boundary. Thismotivated the title of the present paper.

In this paper we are concerned with the above quantities where one or both ofthe domains in question belong to the class of angular domains Hα with openingangle απ, 1 ≤ α ≤ 2, which means that there exists a linear transformationT (z) = Az +B such that

Hα = T(z∣∣∣ | arg z| < απ

2

).

For the convenience of the reader we give a short outline of the method whichwill be used here (compare [3], [4] and [20]). For f ∈ A(Ω,Π) and z0 ∈ Ω weconsider the functions

s(ζ) := (ΦΩ,z0(ζ)− z0)λΩ(z0), ζ ∈ ∆,

t(ζ) :=(ΦΠ,f(z0)(ζ)− f(z0)

)λΠ(f(z0)), ζ ∈ ∆.

Both of them belong to the class S of functions univalent in ∆ and normalizedin the origin as usual. The fact that f(Ω) is a subset of Π may be expressed interms of the function

u(ζ) := (f(ΦΩ,z0(ζ))− f(z0))λΠ(f(z0)), ζ ∈ ∆.

It means that u(ζ) is subordinate to t(ζ). This will be denoted by the abbreviationu ≺ t.

3 (2003), No. 1 Punishing Factors for Angles 129

Using the Taylor expansion

u(ζ) =∞∑k=1

akλΠ(f(z0))ζk

and the function ΨΩ,z0 inverse to ΦΩ,z0 we get

f(z) = f(z0) +∞∑k=1

ak (ΨΩ,z0(z))k ,

and thereforef (n)(z0)

n!=

n∑k=1

ak1

n!

((ΨΩ,z0(z))k

)(n)∣∣∣∣z=z0

.

If we denote by s−1(w) the function inverse to s(ζ), define(s−1(w)

)k=∞∑n=k

An,k(z0)wn

and use

s−1 (λΩ(z0)(z − z0)) = ΨΩ,z0(z),

we see that1

n!

((ΨΩ,z0(z))k

)(n)∣∣∣∣z=z0

= An,k(z0) (λΩ(z0))n .

Hence, we get the following formula

(1)f (n)(z0)

n!=

n∑k=1

ak An,k(z0) (λΩ(z0))n

which will be central in what follows.

There are many theorems in geometric function theory which allow us to getupper bounds on the |ak| by subordination theorems and upper bounds for the|An,k(z0)| by theorems on the inverse coefficients of univalent functions. Thisconstitutes a basic principle which is behind many inequalities in the geometrictheory of analytic functions. We mention as the oldest of these theorems theSchwarz-Pick Lemma, which says that

M1(z,∆,∆) = C1(∆,∆) = 1

and in turn

M1(z,Ω,Π) = C1(Ω,Π) = 1

for all simply connected domains Ω and Π. St. Ruscheweyh proved in [16] that

Mn(z,∆, H1) = (1 + |z|)n−1

and in [17] that

Mn(z,∆,∆) = (1 + |z|)n−1.

130 F. G. Avkhadiev and K.-J. Wirths CMFT

It may be deduced from the results of Yamashita in [20] that

Cn(H2, H1) =

(2n− 1

n

)and that for simply connected Ω the inequality

Cn(Ω, H1) ≤ Cn(H2, H1)

is valid. In [3] the authors of the present article showed that

Mn(z,∆,Π) = (1 + |z|)n−1

for any convex domain Π. Using results from [2], [6] and [7] we shall prove thefollowing theorem.

Theorem 1. Let 1 ≤ α ≤ 2, n ∈ N and z0 ∈ ∆. Then

Mn(z0,∆, Hα) =n−1∑k=0

(1 + |z0|)k(n− 1

k

)(α

n− k

)2n−k−1

α,

where (α

ν

)=

1

ν!

ν−1∏µ=0

(α− µ).

When α = 2, more can be said.

Theorem 2. Let Π be a simply connected domain and z0 ∈ ∆. Then

Mn(z0,∆,Π) ≤Mn(z0,∆, H2) = (n+ |z0|)(1 + |z0|)n−2.

E. Landau remarked in [12] that this inequality for univalent functions is a con-sequence of the validity of the Bieberbach Conjecture.

If we let |z0| → 1 in Theorems 1 and 2 and use results from [3], then we imme-diately obtain the following results on the quantities Cn(Ω,Π).

Theorem 3.

a) Let Π be a convex domain and n ∈ N. Then (see [3])

Cn(∆,Π) = 2n−1.

b) Let 1 ≤ α ≤ 2 and n ∈ N. Then

Cn(∆, Hα) =2n−1

α

(n+ α− 1

n

).

c) Let Π be a simply connected domain and n ∈ N. Then

Cn(∆,Π) ≤ (n+ 1)2n−2.

Using theorems on inverse coefficients of conformal mappings we generalize The-orem 3 and Yamashita’s results in [20] as follows.

3 (2003), No. 1 Punishing Factors for Angles 131

Theorem 4.

a) Let 1 ≤ α, β ≤ 2 and n ∈ N. Then

Cn(Hβ, Hα) =(2β)n

2α

(αβ

+ n− 1

n

).

b) Let 1 ≤ β ≤ 2, n ∈ N and Π be a convex domain. Then

Cn(Hβ,Π) = Cn(Hβ, H1).

c) Let 1 ≤ β ≤ 2, n ∈ N and Π ( C be a simply connected domain. Then

Cn(Hβ,Π) ≤ Cn(Hβ, H2).

d) Let 1 ≤ α ≤ 2, n ∈ N and Ω ( C be a simply connected domain. Then

Cn(Ω, Hα) ≤ Cn(H2, Hα).

In Theorem 5 below we are concerned with analytic functions f : Ω → Hα overdomains Ω in C not equal to C and simply connected with respect to C. It turnsout that sharp estimates for |f (n)(z0)| at the point z0 ∈ Ω \ ∞ depend onthe Gaussian curvature A of the Poincare metric and on the hyperbolic distanceD(z0,∞) (see [1], [10] and [11]). We define this distance by

D(z0,∞) := inf

∫Γ

λΩ(z) |dz|,

where the infimum is taken over all piecewise smooth curves Γ ⊂ Ω joining z0

to ∞. For technical reasons we choose the Poincare metric with the constantcurvature A = −4. This means that for the unit disk ∆ the density λ∆(ζ), asdefined earlier, is (1− |ζ|2)−1. Therefore, for ζ1, ζ2 ∈ ∆, the hyperbolic distanceis given in the form

D(ζ1, ζ2) =1

2ln

1 +∣∣∣ ζ1−ζ2

1−ζ1ζ2

∣∣∣1−

∣∣∣ ζ1−ζ21−ζ1ζ2

∣∣∣ .Later we also use the quantity p ∈ (0, 1] such that D(0, p) in ∆ equals D(z0,∞)in Ω. If we take the Gaussian curvature as above, then

p := p(z0) = tanhD(z0,∞),

where, as usual, tanhx = (ex − e−x)/(ex + e−x). We put p = 1, if ∞ 6∈ Ω. Ourresult concerning sharp estimates for |f (n)(z0)| with z0 ∈ Ω \ ∞, for the casesin question, is given by the following theorem.

132 F. G. Avkhadiev and K.-J. Wirths CMFT

Theorem 5. Let Ω be as above, z0 ∈ Ω \ ∞, 1 ≤ α ≤ 2 and n ≥ 2. Then forany analytic function f : Ω→ Hα, the inequality

1

n!|f (n)(z0)| ≤ (λΩ(z0))n

λHα(f(z0))

n∑k=1

4k

2α

(n− 1

n− k

)(α2

k

)(p(z0) +

1

p(z0)+ 2

)n−kis valid. Further, for each n ≥ 2, each α ∈ [1, 2] and each p ∈ (0, 1], thereexist Ω, Hα, z0 ∈ Ω \ ∞ and f as above such that equality is attained in theabove inequality.

This theorem closes two gaps in the results of [4]. On the one hand a similarresult was proved there for convex Π without giving an explicit bound. It is easyto see that the bound given by the above theorem, in the case α = 1, is therelevant upper bound. On the other hand, in [4], such an inequality was provedfor domains Π which are simply connected (with respect to C and with respectto C). Thus Theorem 5 forms a bridge between the results for Π convex and Πsimply connected.

2. Proofs

For the proof of Theorem 1 we need a number of facts about the Taylor co-efficients of some special functions. If the functions f and g are analytic in aneighborhood of the origin, then we write

f(z) =∞∑k=0

akzk g(z) =

∞∑k=0

bkzk

if and only if for all k ∈ N ∪ 0 the inequalities

|ak| ≤ |bk|

are valid. We use this abbreviation in the following theorem to state some crucialfacts we shall need later.

Theorem 6. Let φ be a function analytic in ∆. If α ≥ 1 and

φ(z) ≺ 1 + cz

1− zfor some constant c ∈ ∆, then there exists a probability measure µ : [0, 2π] → R

such that

φ(z)α =

∫ 2π

0

(1 + ceitz

1− eitz

)αdµ(t), z ∈ ∆,

and such that

φ(z)α (

1 + z

1− z

)α.

3 (2003), No. 1 Punishing Factors for Angles 133

In particular, for α ≥ 1 and |c| ≤ 1, we have(1 + cz

1− z

)α(

1 + z

1− z

)α.

Short proofs of these facts can be found in [19]. The original proofs are givenin [2], [6] and [7].

Using Theorem 6 and the cited proofs in [6] and [19] we get the following result.

Theorem 7. If c ∈ ∆ \ −1 and α ≥ 1, then

1

c+ 1

((1 + c z

1− z

)α− 1

) 1

2

((1 + z

1− z

)α− 1

).

Proof of Theorem 7. For c ∈ ∆\−1 let Tc(z) = (1+cz)/(1−z). Since Tc(∆)is a half-plane whose boundary cuts the real axis at a point of the interval [0, 1)and 1 ∈ Tc(∆), there exists, for any c ∈ ∆ \ −1 and α ≥ 1, a point c1 ∈ ∆such that

φ1(z) := Tc(z)1−1/αT1(z)1/α ≺ 1 + c1z

1− z.

According to Theorem 6 this implies

φ1(z)α T1(z)α.

Using this and the non-negativity of the Taylor coefficients of the functions T1(z)and 1/(1− z2) we get(

1 + cz

1− z

)α−11

(1− z)2= φ1(z)α

1

1− z2(

1 + z

1− z

)α1

1− z2.

By integration we obtain the assertion of Theorem 7.

We now prove a subordination theorem which we need for the application of themethod explained in the introduction.

Theorem 8. For any angular domain Hα with 1 ≤ α ≤ 2, and any function ganalytic in ∆ with g(∆) ⊂ Hα, the assertion

(g(ζ)− g(0))λHα(g(0)) 1

2α

((1 + ζ

1− ζ

)α− 1

), ζ ∈ ∆,

is valid.

Proof of Theorem 8. Let

Gα(ζ) :=(ΦHα,g(0)(ζ)− g(0)

)λHα(g(0)), ζ ∈ ∆.

The function Gα belongs to the class S and maps ∆ univalently onto an angulardomain Hα. This yields the existence of a complex number c ∈ ∆ \ −1 suchthat

Gα(ζ) =1

(c+ 1)α

((1 + c ζ

1− ζ

)α− 1

), ζ ∈ ∆.

134 F. G. Avkhadiev and K.-J. Wirths CMFT

Therefore, it follows from our earlier assumption that

(g(ζ)− g(0))λHα(g(0)) ≺ Gα(ζ), ζ ∈ ∆,

and in turn that

φ(ζ) := (1 + (c+ 1)α(g(ζ)− g(0))λHα(g(0)))1/α ≺ 1 + cζ

1− ζ, ζ ∈ ∆.

According to Theorem 6, this implies that

(g(ζ)− g(0))λHα(g(0)) =1

(c+ 1)α

∫ 2π

0

((1 + cζ

1− ζ

)α− 1

)dµ(t), ζ ∈ ∆.

This, together with Theorem 7, yields the result of Theorem 8.

We are now ready to prove Theorem 1.

Proof of Theorem 1. We first use a fact proved by O. Szasz in [18], namelythat for Ω = ∆ in (1) we may insert the identities

An,k(z0) =

(n− 1

n− k

)z0n−k.

Next, in Theorem 8, we set

g(ζ) = f

(ζ + z0

1 + z0ζ

),

and define the Taylor coefficients dk(α) by

1

2α

((1 + ζ

1− ζ

)α− 1

)=∞∑k=1

dk(α)ζk.

If we use the fact that the coefficients dk(α) are non-negative in our cases (com-pare [2]), we see that in (1) we may insert the inequalities

|ak|λHα(f(z0)) ≤ dk(α)

to get

(2) Mn(z0,∆, Hα) ≤n∑k=1

(n− 1

n− k

)|z0|n−kdk(α).

It is easy to see that the upper bound is attained in (2) for z0 = r0eiθ, r0 ≥ 0, if

we choose f to be given by the identity(f

(ζ + z0

1 + z0ζ

)− f(z0)

)λHα(f(z0)) =

eiθ

2α

((1 + e−iθζ

1− e−iθζ

)α− 1

).

3 (2003), No. 1 Punishing Factors for Angles 135

It remains to show that the upper bound for Mn(z0,∆, Hα) in (2) equals theupper bound given in Theorem 1. To show this we remark that the left side ofthe inequality (2) is the nth Taylor coefficient of the function

(1 + |z0|z)n−1

2α

(1 + z

1− z

)α.

Applying the Binomial Theorem to the expression

(1 + |z0|z)n−1 = ((1 + |z0|)z + 1− z)n−1

shows that the said nth Taylor coefficient equals the nth Taylor coefficient of theproduct

1

2α(1 + z)α

n∑k=1

(n− 1

n− k

)zk−1(1 + |z0|)k−1(1− z)n−k−α.

Hence, we have to sum the (n− k + 1)th Taylor coefficients of the functions

1

2α(1 + z)α

(n− 1

n− k

)(1 + |z0|)k−1(1− z)n−k−α, k = 1, . . . , n.

These coefficients may be written in the form(n− 1

n− k

) n−k+1∑q=0

(α

q

)(−1)n−k+1−q

2α

(n− k − α

n− k + 1− q

)(1 + |z0|)k−1

= (1 + |z0|)k−1

(n− 1

n− k

)(α

n− k + 1

)2n−k

α, k = 1, . . . , n.

This proves the result of Theorem 1 if we replace k by (k − 1) and sum up fromk = 0 to k = (n− 1).

Proof of Theorem 2. Here, we use the fact that de Branges’ proof of theBieberbach Conjecture (see [8]) implies a proof of the generalized Bieberbachor Rogosinski Conjecture (compare [9]). This means that the Taylor coefficientsof a function subordinate to a univalent function are dominated by the Taylor co-efficients of the Koebe function. Therefore, we see that in (1), for any domain Π,simply connected with respect to C but not equal to C, the inequalities

|ak|λΠ(f(z0)) ≤ k

are valid. Since1

4

((1 + z

1− z

)2

− 1

)=∞∑k=1

kzk,

we may use the last inequality, formula (1) and the result of Theorem 1 in thecase α = 2 to prove the assertion of Theorem 2.

Since the proof of Theorem 3 needs no further explanation, we shall now provethe following result in preparation for proving Theorem 4.

136 F. G. Avkhadiev and K.-J. Wirths CMFT

Theorem 9. For 1 ≤ α ≤ 2, let

hα(z) :=1

2α

(1−

(1− z1 + z

)α), z ∈ ∆.

Further let h ∈ S and h(∆) be an angular domain Hα. We denote by h−1 thefunction inverse to h and by h−1

α the function inverse to hα. If we let h−1(w)and h−1

α (w) represent the Taylor expansions of h−1 and h−1α in a neighborhood of

the origin, thenh−1(w) h−1

α (w).

Proof of Theorem 9. Since there exists a complex number c ∈ ∆ \ −1 suchthat any function h of the above type may be written in the form

h(z) =1

(c+ 1)α

(1−

(1− z

1 + c z

)α), z ∈ ∆,

we get by a straightforward computation that

h−1(w) =1− (1− (c+ 1)αw)1/α

1 + c(1− (c+ 1)αw)1/α.

Now, we use the expansion

(1− (c+ 1)αw)1/α =∞∑k=0

(−α)k(

1α

k

)(c+ 1)kwk

and the fact that for k ≥ 1 and α ≥ 1

Dk(α) := − (−α)k

(c+ 1)k

(1α

k

)≥ 0

to get

h−1(w) =

∑∞k=1 Dk(α)(c+ 1)k−1wk

1− c∑∞

k=1 Dk(α)(c+ 1)k−1wk

∑∞

k=1 Dk(α)2k−1wk

1−∑∞

k=1 Dk(α)2k−1wk= h−1

α (w).

This completes the proof of Theorem 9.

Proof of Theorem 4a. First, note that the nth Taylor coefficient of (h−1β (w))k

is non-negative, since the Taylor coefficients of h−1β (w) are all non-negative. This

fact and Theorem 9 imply that we may now use as an upper bound for thequantity |An,k(z0)|, 1 ≤ k ≤ n, in formula (1) this nth Taylor coefficient of(h−1

β (w))k. For |ak| we use the same inequality as in the proof of Theorem 1. Toprove that the resulting inequality is sharp and to compute the explicit formulagiven in the assertion we use the conformal map fα,β defined by

(3) fα,β(z) := gα(h−1β (z)

)

3 (2003), No. 1 Punishing Factors for Angles 137

where

gα(ζ) :=1

2α

((1 + ζ

1− ζ

)α− 1

), ζ ∈ ∆.

One the one hand, fα,β maps the special angular domain

Hβ =

z

∣∣∣∣∣∣∣∣∣arg

(z − 1

2β

)∣∣∣∣ < β π

2

conformally onto the special angular domain

Hα =

z

∣∣∣∣∣∣∣∣∣arg

(z +

1

2α

)∣∣∣∣ < απ

2

such that fα,β(0) = 0. Now, we compute f

(n)α,β(0) starting with formula (3) in the

same way as we obtained (1). Using the fact that

λHβ(0) = λHα(0) = 1,

we observe that the resulting sum is just the sum that we derived as an upperbound in the beginning of the proof of Theorem 4a.

On the other hand, a direct computation of fα,β shows that

fα,β(z) =1

2α

((1− 2β z)−α/β − 1

).

Hence,

f(n)α,β(0)

n!=

(2β)n

2α

(αβ

+ n− 1

n

).

This proves the assertion of Theorem 4a.

Proof of Theorem 4b. The inequality

Cn(Hβ,Π) ≤ Cn(Hβ, H1)

for a convex domain Π follows in our case from Rogosinski’s Theorem on convexsubordination (compare [15], [10] and [9]) according to which we may use

|ak|λΠ(z0) ≤ 1

in (1) in a way analogous to that of the proof of Theorem 4a.

The proof ofCn(Hβ,Π) ≥ Cn(Hβ, H1)

for a convex domain follows the proof of

Mn(z,∆,Π) = (1 + |z|)n−1

for convex Π in [3]. Therefore, we only describe the crucial steps here. Withoutloss of generality, we may assume that

∆1 = w | |w − 1| < 1 ⊂ Π ⊂ H1 = w | Rew > 0.

138 F. G. Avkhadiev and K.-J. Wirths CMFT

For the special angular domain Hβ chosen as in the proof of Theorem 4a weconsider the analytic functions

fk(z) =2

1 + k(1− 2βz)1/β, k ∈ N,

which map Hβ conformally onto ∆1 such that fk(0) = 2/(k + 1) =: wk. Wecompute

λHβ(0) = 1, λ∆1(wk) =(k + 1)2

4k.

Since the sequences (kfk)k∈N are uniformly convergent in a neighborhood of theorigin we see that

limk→∞

kf(n)k (0) = 2

dn

dzn(1− 2β z)−1/β

∣∣∣∣z=0

= 4(n!)Cn(Hβ, H1).

The last equation follows from the proof of Theorem 4a. Using the ComparisonPrinciple for densities of the Poincare metric (see [1]) as in the proof of thecorresponding relation in [3] it is easily seen that

limk→∞

λΠ(wk)

λ∆1(wk)= 1.

Putting together the above relations we get the equation

limk→∞

Ln(fk, 0, Hβ,Π) = limk→∞

(f (n)(0)λ∆1(wk)

n!(λHβ(0)

)n λΠ(wk)

λ∆1(wk)

)= Cn(Hβ, H1),

which immediately implies the assertion.

Proof of Theorem 4c. In this proof we only have to use, once again, the va-lidity of the Rogosinski or the generalized Bieberbach Conjecture together withTheorem 4a analogously to the proofs of Theorems 2 and 3c.

Proof of Theorem 4d. Here, in the application of (1), we use the fact thataccording to Lowner’s Theorem (see [13]) the Taylor coefficients of the inversesof univalent functions and their kth powers, k ∈ N, are dominated by the related(positive) coefficients of the inverse of the Koebe function z/(1 + z)2 and its kthpowers. This results in the assertion. For a similar proof in the case that Ωand Π are simply connected, compare this proof with that of Theorem 2 in [3].

3 (2003), No. 1 Punishing Factors for Angles 139

Proof of Theorem 5. We again start with the fundamental equality (1). Let

p := p(z0), a := p+1

p, κp(ζ) :=

ζ

(1 + pζ)(1 + ζp), ζ ∈ ∆,

and let Kp be the function inverse to κp, having an expansion

Kp(z) := z +∞∑ν=2

Bνzν ,

valid in some neighborhood of the origin. Considering the class Sp of functionsmeromorphic and univalent in ∆ which have a pole at a point b, with |b| = p,and an expansion

m(ζ) = ζ +∞∑ν=2

Aνζν , |ζ| < p,

Baernstein and Schober showed in [5] that

|Aν | ≤ Bν , ν ≥ 2.

This implies that the nth Taylor coefficients of the kth powers of m ∈ Sp aredominated by the nth Taylor coefficients Bn,k of the kth powers of the func-tion Kp. By a procedure analogous to that which led us to formula (1), we get,for the meromorphic functions f in question, the inequality

(4)|f (n)(z0)|

n!≤ (λΩ(z0))n

λHα(f(z0))

n∑k=1

Bn,kdk(α).

To show that this inequality is sharp and that the right hand side has the formgiven in the assertion we proceed in principle as in the proof of Theorem 4a(compare also [4]). We consider in the present case the function

fα,p(z) := gα(Kp(z))

which maps the domain

Ω0 = C \[

1

a+ 2,

1

a− 2

]conformally onto the special angular domain Hα chosen as in the proof of The-orem 4a. We see that in this case fα,p(0) = 0 and λHα(0) = λΩ0(0) = 1 and

we deduce the sharpness of (4) considering f(n)α,p (0) as above. To get the explicit

form of the upper bound we consider

fα,p(z) =1

2α

((1− (a− 2)z

1− (a+ 2)z

)α/2− 1

)=:

∞∑n=1

Cn(α, p)zn.

140 F. G. Avkhadiev and K.-J. Wirths CMFT

This implies (1 +

∞∑k=1

4(a+ 2)k−1zk

)α/2

= 1 + 2α∞∑n=1

Cn(α, p)zn.

With the abbreviation Z := z(a+ 2) this is equivalent to the equation

1 +∞∑k=1

(α2

k

)(4

a+ 2

)k (Z

1− Z

)k= 1 + 2α

∞∑n=1

Cn(α, p)

(a+ 2)nZn.

If we then expand Zk/(1 − Z)k in powers of Z and compare the coefficients onboth sides of this equation, we establish the assertion of Theorem 5.

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3 (2003), No. 1 Punishing Factors for Angles 141

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Farit G. Avkhadiev E-mail: nina@dionis.kfti.kcn.ruAddress: Chebotarev Research Institute, Kazan State University, 420008 Kazan, Russia.

Karl-Joachim Wirths E-mail: kjwirths@tu-bs.deAddress: Institut fur Analysis, Technische Universitat Braunschweig, 38106 Braunschweig,Germany.