THE ALEXANDER TRANSFORMATION OF A SUBCLASS OF SPIRALLIKE FUNCTIONS OF TYPE β

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Alexander Transformation Qinghua Xu and Sanya Lu vol. 10, iss. 1, art. 17, 2009

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THE ALEXANDER TRANSFORMATION OF A SUBCLASS OF SPIRALLIKE FUNCTIONS

OF TYPE β

QINGHUA XU SANYA LU

School of Mathematics and Information Science Department of Science

Jiangxi Normal University Nanchang Institute of Technology

Jiangxi, 330022, China Jiangxi, 330099, China

EMail:xuqhster@gmail.com EMail:yasanlu@163.com

Received: 13 August, 2008

Accepted: 27 December, 2008 Communicated by: gkohr@math.ubbcluj.ro 2000 AMS Sub. Class.: 30C45.

Key words: Univalent functions, Starlike functions of orderα, spirallike functions of typeβ, Integral transformations.

Abstract: In this paper, a subclass of spirallike function of typeβdenoted bySˆ^βαis introduced in the unit disc of the complex plane. We show that the Alexander transformation of class ofSˆ_α^β is univalent whencosβ≤ _2(1−α)¹ , which generalizes the related results of some authors.

Acknowledgements: This research has been supported by the Jiangxi Provincial Natural Science Foun- dation of China (Grant No. 2007GZS0177) and Specialized Research Fund for the Doctoral Program of JiangXi Normal University.

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1. Introduction

LetAdenote the class of analytic functionsfon the unit diskD={z ∈C:|z|<1}

normalized by f(0) = 0and f⁰(0) = 1, S denote the subclass ofA consisting of univalent functions, andS^∗denote starlike functions onD. Obviously,S^∗ ⊂S ⊂A holds.

In [1], Robertson introduced starlike functions of orderαonD.

Definition 1.1. Letα∈[0,1),f ∈Sand

<e

zf⁰(z) f(z)

> α, z ∈D.

We say thatf is a starlike function of order α. LetS^∗(α)denote the whole starlike functions of orderαonD.

Spaˇcek [2] extended the class ofS^∗,and obtained the class of spirallike functions of type β. In the same article, the author gave an analytical characterization of spirallikeness of typeβonD.

Theorem 1.2. Let f ∈ S andβ ∈ (−^π₂,^π₂). Then f(z) is a spirallike function of typeβ onDif and only if

<e

e^iβzf⁰(z) f(z)

>0, z ∈D.

We denote the whole spirallike functions of typeβ onDbySˆ_β.

From Definition 1.1and Theorem 1.2, it is easy to see that starlike functions of orderαand spirallike functions of typeβhave some relationships on geometry. Spi- rallike functions of typeβ mapDinto the right half complex plane by the mapping

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e^{iβ zf}_f(z)⁰^(z), while starlike functions of orderαmapDinto the right half complex plane whose real part is greater thanαby the mapping ^zf_f(z)⁰^(z). Sincelim

z→0e^{iβ zf}_f(z)⁰^(z) =e^iβ, we can deduce that if we restrict the image of the mappinge^{iβ zf}_f_(z)⁰^(z) in the right complex plane whose real part is greater than a certain constant, then the constant must be smaller thancosβ. According to this, we introduce the functions classSˆ_α^β onD.

Definition 1.3. Letα∈[0,1),β ∈(−^π₂,^π₂),f ∈S, thenf ∈Sˆ_α^β if and only if

<e

e^iβzf⁰(z) f(z)

> αcosβ, z ∈D.

Obviously, whenβ = 0,f ∈S^∗(α); whileα= 0, f ∈Sˆ_β. Example 1.1. Letf(z) = ^z

(1−z)

2(1−α) 1+itanβ

, z ∈D. The branch of the power function is chosen such that

[(1−z)]

2(1−α) 1+itanβ

z=0 = 1.

It is easily proved thatf ∈Sˆ_α^β. We omit the proof.

For our applications, we setSˆ=S

βSˆ_α^β.

In this paper, we first establish the relationships amongSˆ_α^β and some important subclasses ofS, then investigate the Alexander transformation ofSˆ_α^β preserving uni- valence. Furthermore, some other properties of the class ofSˆ_α^β are obtained. These results generalize the related works of some authors.

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2. Integral Transformations and Lemmas

Integral Transformation 1. The integral transformation J[f](z) =

Z z

0

f(ζ) ζ dζ

is called the Alexander Transformation and it was introduced by Alexander in [4].

Alexander was the first to observe and prove that the Integral transformationJmaps the classS^∗of starlike functions onto the classK of convex functions in a one-to-one fashion.

In 1960, Biernacki conjectured that J(S) ⊂ S , but Krzyz and Lewandowski disproved it in 1963 by giving the examplef(z) = z(1−iz)ⁱ⁻¹, which is a spirallike function of type ^π₄ but is transformed into a non-univalent function byJ [4]. In 1969, Robertson studied the Alexander Integral Transformation of spirallike functions of typeβ. The author showed thatJ( ˆS_β)⊂ Sholds whenβ satisfies a certain condition, that is cosβ ≤ x₀ (a constant). Robertson also noticed that x₀ cannot be replaced by any number greater than ¹₂ and asked about the best value for this [3]. In 2007, Y.C. Kim and T. Sugawa proved thatJ( ˆS_β)⊂ Sholds precisely when cosβ ≤ ¹₂ orβ = 0[4].

Integral Transformation 2. Let γ ∈ C, f(z) ∈ A be locally univalent, and the Integral transformationI_γ[5] be defined by

I_γ[f](z) = Z z

0

[f⁰(ζ)]^γdζ =z Z 1

0

[f⁰(tz)]^γdt.

Based on the definition ofI_γ, we may easily show thatI_γ◦I_γ⁰ =I_γγ⁰.

LetA(F) = {γ ∈ C :I_γ(F) ⊂ S}, F ⊂ Abe locally univalent. According to the definition of theA(F),J( ˆS_α^β)⊂Sis equivalent to1∈A(J( ˆS_α^β)).

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For the proof of the theorems in this paper, we need the following lemma, which establishes the relationships amongSˆ_α^β and some important subclasses ofS.

Lemma 2.1. Forα ∈ [0,1), β ∈ (−^π₂,^π₂), c= e^−iβcosβ, the following assertions hold:

(i) ([6,7])f ∈S^∗(α)if and only if f(z)

z = u(z)

z ^1−α

, z ∈D,

where u(z) ∈ S^∗. The branch of the power function is chosen such that hu(z)

z

i1−α z=0

= 1.

(ii) f ∈Sˆ_α^β if and only if

f(z) z =

g(z) z

c

, z ∈D,

where g(z) ∈ S^∗(α). The branch of the power function is chosen such that hg(z)

z

ic z=0

= 1.

(iii) f ∈Sˆ_α^β if and only if

f(z) z =

s(z) z

^(1−α)c

, z ∈D,

where s(z) ∈ S^∗. The branch of the power function is chosen such that hs(z)

z

i(1−α)c z=0

= 1.

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Now we give the proof of (ii) and (iii).

Proof. (ii). First, assume that f(z) ∈ Sˆ_α^β. Settingg(z) = zh

f(z) z

i_cosβ^eiβ

, through simple calculations we may obtain the equality

zg⁰(z)

g(z) = (1 +itanβ)zf⁰(z)

f(z) −itanβ.

Therefore the following inequality holds,

<e

zg⁰(z) g(z)

= 1

cosβ<e

e^iβzf⁰(z) f(z)

> αcosβ cosβ =α, namelyg(z)∈S^∗(α).

Conversely, suppose g(z) ∈ S^∗(α), then according to the above calculation, we have the inequality

1 cosβ<e

e^iβzf⁰(z) f(z)

=<e

zg⁰(z) g(z)

> α.

This implies

<e

e^iβzf⁰(z) f(z)

> αcosβ,

i.e.,f(z)∈Sˆ_α^β.

(iii). It is easy to see from (ii) thatf∈Sˆ_α^β if and only ifg∈S^∗(α)such that ^f^(z)_z = h_g(z)

z

ic

, herec=e^−iβcosβ. Noting thatg(z)∈S^∗(α)if and only if s(z)∈S^∗ such that ^g(z)_z =h

s(z) z

i1−α

which holds in (i), we may obtain an important relationship

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between the class of Sˆ_α^β and the class of S^∗ : f ∈ Sˆ_α^β if and only if there exists s(z)∈ S^∗ such that ^f^(z)_z =h

s(z) z

i(1−α)c

. Here,c= e^−iβcosβ and the branch of the power function is chosen such that

hs(z) z

i(1−α)c z=0

= 1.

Lemma 2.1 expresses the relations of the Sˆ_α^β and S^∗ classes, which play a key role in this paper.

Lemma 2.2 ([5], [8]). A(K) =

|γ| ≤ ¹₂ ∪₁

2,³₂ . Lemma 2.3. Forα∈[0,1),β ∈ −^π₂,^π₂

,J( ˆS_α^β) = I_(1−α)e^−iβ^cos^β(K).

Proof. Let f ∈ J( ˆS_α^β), then there exists g(z) ∈ Sˆ_α^β such that f(z) = Rz 0

g(ζ) ζ dζ.

According to (iii) of Lemma2.1there iss(z)∈S^∗such that g(z) =z

s(z) z

(1−α)e^−iβcosβ

, therefore

f(z) = Z z

0

s(ζ) ζ

(1−α)e^−iβcosβ

dζ.

By the relationship of theS^∗ class and theK class, there existsu(z) ∈K such that s(z) =zu⁰(z), thus

f(z) = Z z

0

[u⁰(ζ)]^(1−α)e^−iβ^cos^βdζ,

i.e.,f(z)∈I_(1−α)e^−iβ_cos_β(K). As a result,J( ˆS_α^β)⊂I_(1−α)e^−iβ_cos_β(K)holds.

Conversely, whenf(z)∈ I_(1−α)e^−iβ_cos_β(K), we can trace back the above procedure to getf ∈J( ˆS_α^β), soI_(1−α)e^−iβ_cos_β(K)⊂J( ˆS_α^β).

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From the above proof, we obtain the assertion.

Remark 1. If, in the hypothesis of Lemma2.3, we setα= 0, we arrive at Lemma 4 of [4].

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3. The Main Results and Their Proofs

In this section, we let[z, w]denote the closed line segment with endpoints z andw forz, w ∈C.

Now we give the main results and their proofs.

Theorem 3.1. Forα∈[0,1),β ∈(−^π₂,^π₂), A(J( ˆS_α^β)) =

|γ| ≤ 1

2(1−α) cosβ

[

e^iβ

2(1−α) cosβ, 3e^iβ 2(1−α) cosβ

.

Proof. By Lemma2.3, we have

I_γ(J( ˆS_α^β)) =I_γ(I_(1−α)e^−iβ_cos_β(K)) =I_γ(1−α)e^−iβ_cos_β(K).

Therefore,γ ∈A(J( ˆS_α^β))if and only ifγ(1−α)e^−iβcosβ ∈A(K), and by Lemma 2.2we may easily get the result.

Remark 2. In this theorem, if we setα= 0, we obtain Theorem 3 of [4].

Theorem 3.2. Forα∈[0,1),β ∈(−^π₂,^π₂), the inclusion relationJ( ˆS_α^β)⊂S holds precisely if eithercosβ ≤ _2(1−α)¹ orα=β = 0.

Proof. Asα =β = 0, the result holds evidently by Integral transformation1; while forα = 0 andβ 6= 0, the result is Theorem 1 of [4] and was proved by Y.C. Kim and T. Sugawa [4].

Ifα6= 0andβ = 0, thenf(z)∈S^∗(α). By Lemma2.1(i), there existsu(z)∈S^∗ such thatu(z) = z

f(z) z

_1−α¹

. The branch of the power function is chosen such that f(z)

z

_1−α¹ z=0

= 1. From Integral transformation 1, we can easily see that there

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existsg(z)∈J( ˆS_α^β)such that g(z) =

Z z

0

f(ζ) ζ

_1−α¹ dζ.

For

<e

1 + zg⁰⁰(z) g⁰(z)

=<e 1

1−α

zf⁰(z) f(z)

and<eh

zf⁰(z) f(z)

i

> α, we can deduce that<eh

1 + ^zg_g0⁰⁰(z)^(z)

i

>0. This impliesg(z) ∈ K andJ(S^∗(α))⊂S.

Now letα 6= 0andβ 6= 0. SinceJ( ˆS_α^β) ⊂S is equivalent to1∈ A(J( ˆS_α^β))and 1 ∈/ h

e^iβ

2(1−α) cosβ,_{2(1−α) cos}^3e^iβ _βi

, by Theorem3.1, we deduce that 1 ≤ _{2(1−α) cos}¹ _β , i.e., cosβ ≤ _2(1−α)¹ .

Summarizing the above procedure, forα∈[0,1),β ∈(−^π₂,^π₂),J( ˆS_α^β)⊂Sholds whencosβ ≤ _2(1−α)¹ orα=β = 0.This completes the proof.

Remark 3. This theorem is an extension of Theorem 1 of [4]. Indeed, if we set α= 0, we will obtain the result of [4].

Theorem 3.3. Forα∈[0,1),β ∈(−^π₂,^π₂), A(J( ˆS)) =

|γ| ≤ 1

2(1−α) cosβ

. Proof. In view of Sˆ = S

βSˆ_α^β and A(F) = {γ ∈ C : I_γ(F) ⊂ S}, we deduce A(J( ˆS)) = T

β(J( ˆS_α^β)). With the aid of Theorem 3.1, a simple observation gives A(J( ˆS)) =n

|γ| ≤ _{2(1−α) cos}¹ _βo

. Thus the proof is now complete.

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Remark 4. Forα =β = 0, Theorem3.3implies the Theorem 2 of [4].

At the end of this paper, we mention the norm estimate of pre-Schwarzian derivatives. The hyperbolic norm of the pre-Schwarzian derivativeT_f = f⁰⁰/f⁰ off ∈ A is defined to be

kfk= sup

|z|<1

(1− |z|²)|T_f(z)|.

It is known thatf is bounded ifkfk < 2and the bound depends only on the value ofkfk([9]). Since

kI_γ[f]k= sup

|z|<1

(1− |z|²)

Rz

0[f⁰(ζ)]^γdζ00

Rz

0[f⁰(ζ)]^γ0

= sup

|z|<1

(1− |z|²)

([f⁰(z)]^γ)⁰ f⁰(z)]^γ

= sup

|z|<1

(1− |z|²)

γf⁰⁰(z) f⁰(z)

=|γ|kfk.

We obtain the following assertion.

Proposition 3.4. For each α ∈ [0,1), β ∈ (−^π₂,^π₂), the sharp inequality kfk ≤ 4(1−α) cosβholds for f ∈ J( ˆS_α^β). Moreover, ifcosβ < _2(1−α)¹ , then a function in J( ˆS_α^β)is bounded by a constant depending onαandβ.

Proof. For each f ∈ J( ˆS_α^β), by Lemma 2.3, there is a function k ∈ K such that f = I_γ(k), where γ = (1−α)e^−iβcosβ. Noting thatkkk ≤ 4[10], we obtain the following inequality

kfk=|γ|kkk ≤4|γ|= 4(1−α) cosβ.

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Since the inequalitykkk ≤ 4is sharp, the above inequality is also sharp. Ifcosβ <

1

2(1−α), the above inequality implieskfk ≤4(1−α) cosβ < 2, sof is bounded by a constant depending onαandβ.

Remark 5. If, in the statement of Proposition 3.4, we set α = 0, we arrive at the result of [4].

In the above proposition, the bound ¹₂ cannot be replaced by any number greater than √ ¹

2(1−α). Indeed, by the Alexander transformation, if the function g(z) =z(1−z)^{−2(1−α)e}^−iβ^cos^β ∈Sˆ_α^β,

then the function

f(z) = (1−z)^{1−2(1−α)e}^−iβ^cos^β−1

2(1−α)e^−iβcosβ−1 ∈J( ˆS_α^β), and we may verify that the latter is unbounded whencosβ > √ ¹

2(1−α).

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References

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[2] L. SPA ˇCEK, Contribution à la théorie des fonctions univalentes, Casopis Pˇest Math., 62 (1932), 12–19, (in Russian).

[3] M.S. ROBERTSON, Univalent functions f(z) for which zf⁰(z) is spirallike, Michigan Math. J., 16 (1969), 97–101.

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[9] Y.C. KIM AND T. SUGAWA, Growth and coefficient estimates for uniformly locally univalent functions on the unit disk, Rocky Mountain J. Math., 32 (2002), 179–200.

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