𝑛-Bazilevic Functions

Abstract

The aim of this paper is to define and study a class of Bazilevic functions using the generalized Salagean operator. Some properties of this class are investigated: inclusion relation, some convolution properties, coefficient bounds, and other interesting results.

1. Introduction

Let 𝐻 be the set of analytic functions in the open unit disc 𝐸={𝑧∶|𝑧|<1}. Let 𝐴 be the set of functions 𝑓∈𝐻, with 𝑓(𝑧)=𝑧+∑∞𝑘=2𝑎𝑘𝑧𝑘, and let 𝐴0 be the set of functions 𝑓∈𝐻, with 𝑓(0)=1. Let 𝑆 be the class of functions 𝑓∈𝐴, which are univalent in 𝐸. Denote by 𝑆𝑇(𝛾)(𝐶𝑉(𝛾))(𝐾(𝛾)), 𝛾<1, the class of starlike (convex)(close-to-convex) functions of order 𝛾. Note that when 0≤𝛾<1, then 𝑆𝑇(𝛾)(𝐶𝑉(𝛾))(𝐾(𝛾))⊂𝑆, let 𝑆𝑇(0)(𝐶𝑉(0))(𝐾(0))≡𝑆𝑇(𝐶𝑉)(𝐾). A function 𝐹∈𝐴0, where 𝐹≠0 belongs to the Kaplan class 𝐾(𝛼,𝛽), 𝛼≥0, 𝛽≥0, [1] if−𝛼𝜋+12(𝛼−𝛽)(𝜃2−𝜃1)≤arg𝐹(𝑟𝑒𝑖𝜃2)−arg𝐹(𝑟𝑒𝑖𝜃1)≤12(𝛼−𝛽)(𝜃2−𝜃1)+𝛽𝜋,(1.1) for 𝜃1<𝜃2<𝜃1+2𝜋 and 0<𝑟<1.

The Dual of 𝜈⊂𝐴0 is defined as𝜈⋆={𝑔∈𝐴0∶𝑓∗𝑔≠0inΔ,𝑓∈𝜈},(1.2) where ∗ denotes Hadamard product (convolution).

A set 𝜈⊂𝐴0 is called a test set for 𝑢⊂𝐴 (denoted by 𝜈⇝𝑢) if 𝜈⊂𝑢⊂𝜈⋆⋆. Note that if 𝜈⇝𝑢, then 𝜈⋆⊂𝑢⋆.

Denote by 𝑃 the class of functions 𝑝∈𝐴0, such that Re𝑝>0, in 𝐸, and let 𝑃𝛼={𝑓𝛼∈𝐴0,𝑓∈𝑃}. Note that for 0≤𝛼≤𝛽,𝐾(𝛼,𝛽)=𝑃𝛼⋅𝐾(0,𝛽−𝛼),(1.3) and that 𝑓∈𝑆𝑇(𝛼), 𝛼<1, if, and only if, 𝑓/𝑧∈𝐾(0,2−2𝛼).

For 𝛼≥0 and 𝛽≥0, define the class 𝑇(𝛼,𝛽) as𝑇(𝛼,𝛽)={(1+𝑥𝑧)[𝛼](1+𝑦𝑧)𝛼−[𝛼](1+𝑢𝑧)𝛽∶|𝑥|=|𝑦|=|𝑢|=1}.(1.4) Note that 𝑇(𝛼,𝛽)⇝𝐾(𝛼,𝛽), 𝛼≥1, 𝛽≥1.

A function 𝑓∈𝐴0, is called prestarlike of order 𝛼, 𝛼≤1, (denoted by 𝑅(𝛼)) if and only if 𝑓/𝑧∈𝑇(0,3−2𝛼)⋆, or 𝑓∈𝑅(𝛼) if and only if𝑓∗𝑧(1−𝑧)2−2𝛼∈𝑆𝑇(𝛼)𝛼<1,Re𝑓𝑧>12,𝑧∈𝐸𝛼=1.(1.5)

Let 𝐵(𝛼,𝛽), 𝛼>0, 𝛽∈ℝ, denote the class of Bazilevic functions in 𝐸, introduced by Bazilevic [2], 𝑓∈𝐵(𝛼,𝛽), 𝛼>0, 𝛽∈ℝ, if and only if there exists a 𝑔∈𝑆𝑇(1−𝛼), such that for 𝑧∈𝐸𝑧𝑓′𝑔(𝑓𝑧)𝛼+𝑖𝛽−1∈𝑃,(1.6) where (𝑓/𝑧)𝛼+𝑖𝛽=1 at 𝑧=0. We denote 𝐵(𝛼,0) by 𝐵(𝛼). Bazilevic shows that 𝐵(𝛼,𝛽)⊂𝑆, for 𝛼>0, 𝛽∈ℝ. Note that𝐶𝑉⊂𝑆𝑇⊂𝐾⊂𝐵(𝛼,𝛽).(1.7)

For further information, see [3–7].

The generalized Salagean operator 𝐷𝑛𝜆𝑓∶𝐴→𝐴, 𝑛∈ℕ0=ℕ∪{0}, 𝜆≥0, is defined [8] as𝐷𝑛𝜆𝑓(𝑧)=ℎ𝑛,𝜆(𝑧)∗𝑓(𝑧),(1.8) whereℎ𝑛,𝜆(𝑧)=ℎ𝜆(𝑧)∗⋯∗ℎ𝜆(𝑧)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟(𝑛-times),ℎ𝜆(𝑧)=𝑧−(1−𝜆)𝑧2(1−𝑧)2=𝑧+∞∑𝑘=2[1+𝜆(𝑘−1)]𝑧𝑘.(1.9)

The Operator 𝐷𝑛𝜆𝑓 satisfies the following identity:𝐷𝑛+1𝜆𝑓(𝑧)=(1−𝜆)𝐷𝑛𝜆𝑓(𝑧)+𝜆𝑧(𝐷𝑛𝜆𝑓(𝑧))′.(1.10) Not that for 𝜆=1, 𝐷𝑛1𝑓(𝑧)≡𝐷𝑛𝑓(𝑧), Salagean differential operator [9].

Let𝑘𝑛,𝜆(𝑧)=ℎ(−1)𝑛,𝜆(𝑧)=𝑧+∞∑𝑘=2𝑧𝑘[1+𝜆(𝑘−1)]𝑛,𝜆>0,(1.11) we mean by 𝑓(−1), the solution of 𝑓∗𝑓(−1)=𝑧/(1−𝑧). Hence𝑘𝑛,𝜆(𝑧)=𝑘𝜆(𝑧)∗⋯∗𝑘𝜆(𝑧)⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟(𝑛-times),(1.12) where 𝑘𝜆(𝑧)=ℎ(−1)𝜆(𝑧). It is known [10] that 𝑘𝜆/𝑧∈𝑇(1,1+1/𝜆)∗, hence 𝑘𝜆∈𝑅(1−1/𝜆), and that𝑘𝜆(𝑧)∗𝑧(1−𝑧)(1/𝜆)+1=𝑧(1−𝑧)1/𝜆.(1.13)

The class 𝑆𝑇𝑛(𝛾), 𝛾≤1 is defined as 𝑓∈𝑆𝑇𝑛(𝛾) if and only if 𝐷𝑛𝜆𝑓∈𝑆𝑇(𝛾). For 𝜆=1, we get Salagean-type 𝑛-starlike functions [9].

The operator 𝐷𝑛𝜆𝑓 is now called “Al-Oboudi Operator” and has been extensively studied latly, [5, 11, 12].

In this paper we define and study a class of Bazilevic functions using the operator 𝐷𝑛𝜆𝑓 and study some of its basic properties, inclusion relation, convolution properties coefficient bounds, and other interesting results.

2. Definition and Preliminaries

In this section, the class of 𝑛-Bazilevic functions 𝐵𝑛𝜆, 𝜆>0, where 𝐵0𝜆≡𝐵(1/𝜆,0), is defined and some preliminary lemmas are given.

2.1. Definition

Let 𝑓∈𝐴. Then 𝑓∈𝐵𝑛𝜆, 𝑛∈𝑁0, 𝜆>0, if and only if there exists a 𝑔∈𝑆𝑇𝑛(1−1/𝜆), such that𝐷𝑛+1𝜆𝑧(𝑓/𝑧)1/𝜆𝐷𝑛𝜆𝑔(𝑧)∈𝑃,𝑧∈𝐸,(2.1) where the power (𝑓/𝑧)1/𝜆is chosen as a principal one.

Denote by 𝐵𝑛1𝜆 the class of functions 𝑓∈𝐵𝑛𝜆, where 𝑔≡𝑧.

Using (1.10), we see that 𝐷𝑛+1𝜆𝑧(𝑓/𝑧)1/𝜆=𝑓′(𝑧)𝐷𝑛𝜆𝑧(𝑓/𝑧)1/𝜆−1, from which the following special cases are clear.

2.1.1. Special Cases

(1)For 𝑛=0, 𝐵01/𝛼≡𝐵(𝛼), 𝛼>0, Bazilevic [2].(2)For 𝜆=1, 𝐵𝑛1≡𝐾𝑛(0), Salagean type close to convex functions, Blezu [13].(3)For 𝑛=0, 𝜆=1, 𝐵01≡𝐾, Kaplan [14].(4)For 𝜆=1, 𝐵𝑛11≡𝐵𝑛+1(1), Abdul Halim [15] and 𝐵𝑛11≡𝑇1𝑛+1(0), Opoola [16].

2.2. Lemmas

The following lemmas are needed to prove our results.

Lemma 2.1 (see [10]). Let 𝛼,𝛽≥1 and 𝑓∈𝑇(𝛼,𝛽)∗, 𝑔∈𝐾(𝛼−1,𝛽−1). Then for 𝐹∈𝐴(𝜑∗𝑔𝐹)(𝜑∗𝑔)(𝐸)⊂𝐶𝑂(𝐹(𝐸)).(2.2)

Lemma 2.2. If 𝐷𝑛+1𝜆𝑓∈𝑆𝑇(1−1/𝜆), 𝜆>0, then 𝐷𝑛𝜆𝑓∈𝑆𝑇(1−1/𝜆).

Proof. Since 𝐷𝑛𝜆𝑓=𝑘𝜆∗𝐷𝑛+1𝜆𝑓, we will show that (𝑘𝜆∗𝐷𝑛+1𝜆𝑓)∈𝑆𝑇(1−1/𝜆). Now 𝐷𝑛+1𝜆𝑓∈𝑆𝑇(1−1/𝜆)⊂𝑆𝑇(1/2−1/2𝜆), implies (𝑧(1−𝑧)1/𝜆+1)(−1)∗𝐷𝑛+1𝜆𝑓(𝑧)∈𝑅(12−12𝜆)⊂𝑅(1−12𝜆).(2.3) Hence (𝑧(1−𝑧)1/𝜆)(𝑧(1−𝑧)1/𝜆+1)(−1)∗𝐷𝑛+1𝜆𝑓(𝑧)∈𝑆𝑇(1−12𝜆)⊂𝑆𝑇(1−1𝜆).(2.4) From (1.13), we get the required result.

Lemma 2.3 (see [1]). Let 𝛼,𝛽≥1 and 𝑓∈𝑇(𝛼,𝛽)∗, 𝑔∈𝐾(𝛼,𝛽). Then 𝑓∗𝑔∈𝐾(𝛼,𝛽).

For 𝑋⊂𝐴, let 𝑟𝑐(𝑋) denote the largest positive number so that every 𝑓∈𝑋 is convex in |𝑧|<𝑟𝑐(𝑋). The following result is due to Al-Amiri [17].

Lemma 2.4. One has 𝑟𝑐(ℎ𝜆)=𝑟𝑐=11+|𝑐|+√1+|𝑐|+|𝑐|2,𝑐=2𝜆−1,0<𝜆≤1.(2.5)

Lemma 2.5 (see [18]). Let 𝑓,𝑔∈𝐻, with 𝑓(0)=𝑔(0)=0 and 𝑓′(0)𝑔′(0)≠0. Let 𝜑∈𝜈⋆ in |𝑧|<𝑟<1, where 𝜈={1+𝑥𝑧1+𝑦𝑧𝑔(𝑧)∶|𝑥|=|𝑦|=1}.(2.6) Then for each 𝐹∈𝐻, (𝜑∗𝐹𝑔)(𝜑∗𝑔)(|𝑧|<𝑟)⊂𝐶𝑂(𝐹(𝐸)),(2.7) where 𝐶𝑂 stands for closed convex hull.

Remark 2.6. In [1], it was shown that condition (2.6) is satisfied for all 𝑧 in 𝐸 whenever 𝜑 is in 𝐶𝑉 and 𝑔 is in 𝑆𝑇.

Lemma 2.7 (see [10]). Let 𝛼,𝛽,𝛾,𝛿,𝜇,𝜈∈ℝ be such that 0≤𝛾≤𝛼−1,0≤𝛿≤𝛽−1,0≤𝜇≤𝛼−𝛾,0≤𝛾≤𝛽−𝛿,(2.8) and let 𝑓∈𝑇(𝛼,𝛽)∗, 𝑔∈𝐾(𝛾,𝛿), 𝐹∈𝐾(𝜇,𝜈). Then (𝑓∗𝑔𝐹)(𝑓∗𝑔)∈𝑃max{𝜇,𝜈}.(2.9)

From (1.12) and (1.13), we immediately have;

Lemma 2.8. One has 𝑘𝑛+1,𝜆=𝑧(1−𝑧)(1/𝜆)−𝑛∗(𝑧(1−𝑧)(1/𝜆)+1)(−1),(2.10)

Lemma 2.9 (see [10]). Let 𝑓∈𝐾(𝛼,𝛽), 𝛼,𝛽≥1. Then 𝑓≪(1+𝑧)𝛼(1−𝑧)𝛽,(2.11) where ≪stands for coefficient majorization.

3. Main Results

Theorem 3.1. One has 𝐵𝑛+1𝜆⊂𝐵𝑛𝜆,𝜆>0.(3.1)

Proof. Let 𝑓∈𝐵𝑛+1𝜆. Then there exists 𝑔∈𝑆𝑇𝑛+1(1−1/𝜆), such that 𝐷𝑛+2𝜆𝑧(𝑓𝑧)1/𝜆=𝐷𝑛+1𝜆𝑔(𝑧)⋅𝑝(𝑧),𝑝∈𝐾(1,1).(3.2) Hence (𝑓𝑧)1/𝜆=𝑘𝑛+1,𝜆𝑧∗𝑘𝜆𝑧∗𝐷𝑛+1𝜆𝑔𝑧⋅𝑝,𝑝∈𝐾(1,1).(3.3) Since 𝐷𝑛+1𝜆𝑔/𝑧∈𝐾(0,2/𝜆), and 𝑘𝜆/𝑧∈𝑇(1,1+1/𝜆)∗, application of Lemma 2.3 gives 𝑘𝜆𝑧∗𝐷𝑛+1𝜆𝑔𝑧⋅𝑝=(𝑘𝜆𝑧∗𝐷𝑛+1𝜆𝑔𝑧)𝑝0,𝑝0∈𝐾(1,1)(3.4) hence (𝑓𝑧)1/𝜆=𝑘𝑛+1,𝜆𝑧∗𝐷𝑛𝜆𝑔𝑧𝑝0,𝑝0∈𝐾(1,1)(3.5) Using Lemma 2.2 we deduce that 𝑓∈𝐵𝑛𝜆.
As a consequence of (3.1) we immediately have the following.Corollary 3.2. One has 𝐵𝑛𝜆⊂𝑆.(3.6)

Corollary 3.3. If 𝑓∈𝐵𝑛𝜆, 𝑛∈𝑁0, 𝜆>0, then, for 𝑧∈𝐸(𝑓𝑧)1/𝜆∈𝐾(1,1+2𝜆).(3.7)

Proof. Since 𝑓∈𝐵𝑛𝜆, there exists a 𝑔∈𝑆𝑇𝑛(1−1/𝜆) or 𝐷𝑛𝜆𝑔/𝑧∈𝐾(0,2/𝜆) such that (𝑓𝑧)1/𝜆=𝑘𝑛+1,𝜆𝑧∗𝐷𝑛𝜆𝑔𝑧⋅𝑝,𝑝∈𝐾(1,1),(3.8)=k𝑛+1,𝜆𝑧∗𝐹,𝐹∈𝐾(1,1+2𝜆),(3.9) using (1.3). From (3.6), we conclude that (𝑓𝑧)1/𝜆=𝑘𝑛+1,𝜆𝑧∗𝐹≠0,0<|𝑧|<1,(3.10) which implies that 𝑘𝑛+1,𝜆𝑧∈𝐾(1,1+2𝜆)∗≡𝑇(1,1+2𝜆)∗,(3.11) Applying Lemma 2.3 to (3.9), we get the result.

Theorem 3.4. Let 𝑓∈𝐵𝑛𝜆. Then (𝐷𝑛+1𝜆𝑧(𝑓/𝑧)1/𝜆𝑧)𝜆∈𝐾(𝜆,𝝀+𝟐).(3.12)

Proof. From (2.1), we see that 𝐷𝑛+1𝜆𝑧(𝑓𝑧)1/𝜆=𝐷𝑛𝜆𝑔(𝑧)⋅𝑝(𝑧),𝑝∈𝐾(1,1).(3.13) Since (𝐷𝑛𝜆𝑔/𝑧)𝜆∈𝐾(0,2), and 𝑝(𝑧)𝜆∈𝐾(𝜆,𝜆), then (𝐷𝑛+1𝜆𝑧(𝑓/𝑧)1/𝜆𝑧)𝜆=(𝐷𝑛𝜆𝑔𝑧)𝜆𝑝(𝑧)𝜆∈𝐾(𝜆,𝝀+𝟐),(3.14) which is the required result.

In the following we prove the converse of Theorem 3.1, for 0<𝜆≤1.

Theorem 3.5. Let 𝑓∈𝐵𝑛𝜆, 0<𝜆≤1. Then 𝑓∈𝐵𝑛+1𝜆 in |𝑧|<𝑟𝑐, where 𝑟𝑐 is given by (2.5)

Proof. 𝑓∈𝐵𝑛𝜆 implies (2.1), where 𝐷𝑛𝜆𝑔∈𝑆𝑇(1−1/𝜆)⊂𝑆𝑇.
Now 𝐷𝑛+2𝜆𝑧(𝑓/𝑧)1/𝜆𝐷𝑛+1𝜆𝑔=ℎ𝜆∗𝐷𝑛+1𝜆𝑧(𝑓/𝑧)1/𝜆ℎ𝜆∗𝐷𝑛𝜆𝑔.(3.15) Using Lemma 2.4, we see that ℎ𝜆∈𝐶𝑉 in |𝑧|<𝑟𝑐, for 0<𝜆≤1, where 𝑟𝑐 is given by (2.5).
From Remark 2.6, we conclude ℎ𝜆∗{1+𝑥𝑧1+𝑦𝑧𝐷𝑛𝜆𝑔∶|𝑥|=|𝑦|=1}≠0.(3.16)
Applying Lemma 2.5, we deduce (ℎ𝜆∗(𝐷𝑛+1𝜆𝑧(𝑓/𝑧)1/𝜆/𝐷𝑛𝜆𝑔)𝐷𝑛𝜆𝑔ℎ𝜆∗𝐷𝑛𝜆𝑔)(|𝑧|<𝑟𝑐)⊂𝐶𝑂(𝐷𝑛+1𝜆𝑧(𝑓/𝑧)1/𝜆𝐷𝑛𝜆𝑔)(𝐸),(3.17) hence 𝐷𝑛+2𝜆𝑧(𝑓/𝑧)1/𝜆/𝐷𝑛+1𝜆𝑔∈𝑃 in |𝑧|<𝑟𝑐, as required.

Corollary 3.3 can be improved for 0<𝜆≤1, as follows.

Theorem 3.6. Let 𝑓∈𝐵𝑛𝜆, 0<𝜆≤1. Then 𝑓𝑧∈𝐾(1,2).(3.18)

Proof. We will use Ruscheweyh’s.method of proof [10]. 𝑓∈𝐵𝑛𝜆 implies (3.8), where 𝐷𝑛+1𝜆𝑔/𝑧∈𝐾(0,2/𝜆), 𝑘𝑛+1,𝜆/𝑧∈𝑇(1,1+2/𝜆)∗.
Let 𝐷𝑛𝜆𝑔/𝑧=𝑙⋅𝑚, where 𝑙=(𝐷𝑛𝜆𝑔/𝑧)(1+𝜆)/2 and 𝑚=(𝐷𝑛𝜆𝑔/𝑧)(1−𝜆)/2.
Then 𝑙∈𝐾(0,1/𝜆+1)⋅𝑚∈𝐾(0,1/𝜆−1) and 𝑚⋅𝑝=𝐹∈𝐾(1,1/𝜆). This implies (𝑓𝑧)1/𝜆=𝑘𝑛+1,𝜆𝑧∗𝑙𝐹.(3.19)
Using Lemma 2.7, we get (𝑘𝑛+1,𝜆/𝑧)∗𝑙𝐹(𝑘𝑛+1,𝜆/𝑧)∗𝑙=𝐹0∈𝐾(1𝜆,1𝜆),0<𝜆≤1.(3.20)
Hence (𝑓𝑧)1/𝜆=(𝑘𝑛+1,𝜆𝑧∗𝑙)𝐹0,𝐹0∈𝐾(1𝜆,1𝜆).(3.21) To prove that (𝑓/𝑧)1/𝜆∈𝐾(1/𝜆,2/𝜆), we have to show that ((𝑘𝑛+1,𝜆/𝑧)∗𝑙)∈𝐾(0,1/𝜆), or equivalently 𝑘𝑛+1,𝜆∗𝑧𝑙∈𝑆𝑇(1−1/2𝜆).
Since 𝑧𝑙∈𝑆𝑇((𝜆−1)/2𝜆), then from (1.5) (𝑧(1−𝑧)1/𝜆+1)(−1)∗𝑧𝑙∈𝑅(𝜆−12𝜆)⊂𝑅((𝑛+2)𝜆−12𝜆),𝑧(1−𝑧)1/𝜆−𝑛∗(𝑧(1−𝑧)1/𝜆+1)(−1)∗𝑧𝑙∈𝑆𝑇((𝑛+2)𝜆−12𝜆)⊂𝑆𝑇(1−12𝜆).(3.22) From Lemma 2.8, (1.13), and (3.22), we see that ((𝑘𝑛+1,𝜆/𝑧)∗𝑙)∈𝐾(0,1/𝜆). Using (3.21), we obtain (𝑓/𝑧)1/𝜆∈𝐾(1/𝜆,2/𝜆). From (1.1) we get the required result.

Remark 3.7. For 𝑛=0, 𝜆=1/𝛼 Theorem 3.6 and other stronger results depending on 𝛼, are proved by Sheil-Small [7].

For the coefficient bounds of 𝑓∈𝐵𝑛𝜆, Theorem 3.6 is not strong enough to settle this problem for 0<𝜆<1, In 1962, Zamorski [19] proved the Bieberbach conjecture for 𝑓∈𝐵(𝛼), when 𝛼=1,1/2,1/3,…, in the following we prove this result for 𝑓∈𝐵𝑛𝜆, using the extreme points of Kaplan class 𝐾(𝛼,𝛽).

Theorem 3.8. For 𝑓∈𝐵𝑛𝜆, 𝜆=𝑚∈ℕ, 𝑓≪𝑧(1−𝑧)2−𝑚𝑛.(3.23)

Proof. From (3.9) and Lemma 2.9, we get (𝑓𝑧)1/𝜆≪𝑘𝑛+1,𝜆𝑧∗1+𝑧(1−𝑧)1+2/𝜆,=1(1−𝑧)2/𝜆−𝑛,(3.24) using (2.10). Raising both sides of (3.24) to the 𝑚th power, where 𝜆=𝑚∈ℕ, we get the required result.

Remark 3.9. For 𝑛=0, we get the result of Zamorski [19], and the result of Sheil-Small [7], from which we get the idea of proof.