# n-Bazilevic Functions

Authors: F. M. Al-Oboudi
Journal: Abstract and Applied Analysis
 (2012)

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## 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.


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## Body

## 1. Introduction

LetH 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=2∞akzk, 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(1.1)-απ+12(α-β)(θ2-θ1)≤argF(reiθ2)-argF(reiθ1)≤12(α-β)(θ2-θ1)+βπ,
for θ1<θ2<θ1+2π and 0<r<1.The Dual ofν⊂A0 is defined as(1.2)ν⋆={g∈A0:f*g≠0inΔ,f∈ν},
where * denotes Hadamard product (convolution).A setν⊂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 byP 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,(1.3)K(α,β)=Pα⋅K(0,β-α),
and that f∈ST(α), α<1, if, and only if, f/z∈K(0,2-2α).Forα≥0 and β≥0, define the class T(α,β) as(1.4)T(α,β)={(1+xz)[α](1+yz)α-[α](1+uz)β:|x|=|y|=|u|=1}.
Note that T(α,β)⇝K(α,β), α≥1, β≥1.A functionf∈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 if(1.5)f*z(1-z)2-2α∈ST(α)α<1,Refz>12,z∈Eα=1.LetB(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∈E(1.6)zf′g(fz)α+iβ-1∈P,
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 that(1.7)CV⊂ST⊂K⊂B(α,β).For further information, see [3–7].The generalized Salagean operatorDλnf:A→A, n∈ℕ0=ℕ∪{0}, λ≥0, is defined [8] as(1.8)Dλnf(z)=hn,λ(z)*f(z),
where(1.9)hn,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.The OperatorDλnf satisfies the following identity:(1.10)Dλn+1f(z)=(1-λ)Dλnf(z)+λz(Dλnf(z))′.
Not that for λ=1, D1nf(z)≡Dnf(z), Salagean differential operator [9].Let(1.11)kn,λ(z)=hn,λ(-1)(z)=z+∑k=2∞zk[1+λ(k-1)]n,λ>0,
we mean by f(-1), the solution of f*f(-1)=z/(1-z). Hence(1.12)kn,λ(z)=kλ(z)*⋯*kλ(z)︸(n-times),
where kλ(z)=hλ(-1)(z). It is known [10] that kλ/z∈T(1,1+1/λ)*, hence kλ∈R(1-1/λ), and that(1.13)kλ(z)*z(1-z)(1/λ)+1=z(1-z)1/λ.The classSTn(γ), γ≤1 is defined as f∈STn(γ) if and only if Dλnf∈ST(γ). For λ=1, we get Salagean-type n-starlike functions [9].The operatorDλnf 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 operatorDλnf 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 ofn-Bazilevic functions Bλn, λ>0, where Bλ0≡B(1/λ,0), is defined and some preliminary lemmas are given.


### 2.1. Definition

Letf∈A. Then f∈Bλn, n∈N0, λ>0, if and only if there exists a g∈STn(1-1/λ), such that(2.1)Dλn+1z(f/z)1/λDλng(z)∈P,z∈E,
where the power (f/z)1/λis chosen as a principal one.Denote byB1λn the class of functions f∈Bλn, where g≡z.Using (1.10), we see that Dλn+1z(f/z)1/λ=f′(z)Dλnz(f/z)1/λ-1, from which the following special cases are clear.


#### 2.1.1. Special Cases

(1)
Forn=0, B1/α0≡B(α), α>0, Bazilevic [2].(2)
Forλ=1, B1n≡Kn(0), Salagean type close to convex functions, Blezu [13].(3)
Forn=0, λ=1, B10≡K, Kaplan [14].(4)
Forλ=1, B11n≡Bn+1(1), Abdul Halim [15] and B11n≡Tn+11(0), Opoola [16].


### 2.2. Lemmas

The following lemmas are needed to prove our results.Lemma 2.1 (see [10]).
Letα,β≥1 and f∈T(α,β)*, g∈K(α-1,β-1). Then for F∈A(2.2)(φ*gF)(φ*g)(E)⊂CO¯(F(E)).Lemma 2.2.
IfDλn+1f∈ST(1-1/λ), λ>0, then Dλnf∈ST(1-1/λ).Proof.
SinceDλnf=kλ*Dλn+1f, we will show that (kλ*Dλn+1f)∈ST(1-1/λ). Now Dλn+1f∈ST(1-1/λ)⊂ST(1/2-1/2λ), implies
(2.3)(z(1-z)1/λ+1)(-1)*Dλn+1f(z)∈R(12-12λ)⊂R(1-12λ).
Hence
(2.4)(z(1-z)1/λ)(z(1-z)1/λ+1)(-1)*Dλn+1f(z)∈ST(1-12λ)⊂ST(1-1λ).
From (1.13), we get the required result.Lemma 2.3 (see [1]).
Letα,β≥1 and f∈T(α,β)*, g∈K(α,β). Then f*g∈K(α,β).ForX⊂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(2.5)rc(hλ)=rc=11+|c|+1+|c|+|c|2,c=2λ-1,0<λ≤1.Lemma 2.5 (see [18]).
Letf,g∈H, with f(0)=g(0)=0 and f′(0)g′(0)≠0. Let φ∈ν⋆ in |z|<r<1, where
(2.6)ν={1+xz1+yzg(z):|x|=|y|=1}.
Then for each F∈H,
(2.7)(φ*Fg)(φ*g)(|z|<r)⊂CO¯(F(E)),
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]).
Letα,β,γ,δ,μ,ν∈ℝ be such that
(2.8)0≤γ≤α-1,0≤δ≤β-1,0≤μ≤α-γ,0≤γ≤β-δ,
and let f∈T(α,β)*, g∈K(γ,δ), F∈K(μ,ν). Then
(2.9)(f*gF)(f*g)∈Pmax⁡{μ,ν}.From (1.12) and (1.13), we immediately have;Lemma 2.8.
One has(2.10)kn+1,λ=z(1-z)(1/λ)-n*(z(1-z)(1/λ)+1)(-1),Lemma 2.9 (see [10]).
Letf∈K(α,β),  α,β≥1. Then
(2.11)f≪(1+z)α(1-z)β,
where ≪stands for coefficient majorization.


## 2.1. Definition

Letf∈A. Then f∈Bλn, n∈N0, λ>0, if and only if there exists a g∈STn(1-1/λ), such that(2.1)Dλn+1z(f/z)1/λDλng(z)∈P,z∈E,
where the power (f/z)1/λis chosen as a principal one.Denote byB1λn the class of functions f∈Bλn, where g≡z.Using (1.10), we see that Dλn+1z(f/z)1/λ=f′(z)Dλnz(f/z)1/λ-1, from which the following special cases are clear.


### 2.1.1. Special Cases

(1)
Forn=0, B1/α0≡B(α), α>0, Bazilevic [2].(2)
Forλ=1, B1n≡Kn(0), Salagean type close to convex functions, Blezu [13].(3)
Forn=0, λ=1, B10≡K, Kaplan [14].(4)
Forλ=1, B11n≡Bn+1(1), Abdul Halim [15] and B11n≡Tn+11(0), Opoola [16].


## 2.1.1. Special Cases

(1)
Forn=0, B1/α0≡B(α), α>0, Bazilevic [2].(2)
Forλ=1, B1n≡Kn(0), Salagean type close to convex functions, Blezu [13].(3)
Forn=0, λ=1, B10≡K, Kaplan [14].(4)
Forλ=1, B11n≡Bn+1(1), Abdul Halim [15] and B11n≡Tn+11(0), Opoola [16].


## 2.2. Lemmas

The following lemmas are needed to prove our results.Lemma 2.1 (see [10]).
Letα,β≥1 and f∈T(α,β)*, g∈K(α-1,β-1). Then for F∈A(2.2)(φ*gF)(φ*g)(E)⊂CO¯(F(E)).Lemma 2.2.
IfDλn+1f∈ST(1-1/λ), λ>0, then Dλnf∈ST(1-1/λ).Proof.
SinceDλnf=kλ*Dλn+1f, we will show that (kλ*Dλn+1f)∈ST(1-1/λ). Now Dλn+1f∈ST(1-1/λ)⊂ST(1/2-1/2λ), implies
(2.3)(z(1-z)1/λ+1)(-1)*Dλn+1f(z)∈R(12-12λ)⊂R(1-12λ).
Hence
(2.4)(z(1-z)1/λ)(z(1-z)1/λ+1)(-1)*Dλn+1f(z)∈ST(1-12λ)⊂ST(1-1λ).
From (1.13), we get the required result.Lemma 2.3 (see [1]).
Letα,β≥1 and f∈T(α,β)*, g∈K(α,β). Then f*g∈K(α,β).ForX⊂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(2.5)rc(hλ)=rc=11+|c|+1+|c|+|c|2,c=2λ-1,0<λ≤1.Lemma 2.5 (see [18]).
Letf,g∈H, with f(0)=g(0)=0 and f′(0)g′(0)≠0. Let φ∈ν⋆ in |z|<r<1, where
(2.6)ν={1+xz1+yzg(z):|x|=|y|=1}.
Then for each F∈H,
(2.7)(φ*Fg)(φ*g)(|z|<r)⊂CO¯(F(E)),
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]).
Letα,β,γ,δ,μ,ν∈ℝ be such that
(2.8)0≤γ≤α-1,0≤δ≤β-1,0≤μ≤α-γ,0≤γ≤β-δ,
and let f∈T(α,β)*, g∈K(γ,δ), F∈K(μ,ν). Then
(2.9)(f*gF)(f*g)∈Pmax⁡{μ,ν}.From (1.12) and (1.13), we immediately have;Lemma 2.8.
One has(2.10)kn+1,λ=z(1-z)(1/λ)-n*(z(1-z)(1/λ)+1)(-1),Lemma 2.9 (see [10]).
Letf∈K(α,β),  α,β≥1. Then
(2.11)f≪(1+z)α(1-z)β,
where ≪stands for coefficient majorization.


## 3. Main Results

Theorem 3.1.
One has(3.1)Bλn+1⊂Bλn,λ>0.Proof.
Letf∈Bλn+1. Then there exists g∈STn+1(1-1/λ), such that
(3.2)Dλn+2z(fz)1/λ=Dλn+1g(z)⋅p(z),p∈K(1,1).
Hence
(3.3)(fz)1/λ=kn+1,λz*kλz*Dλn+1gz⋅p,p∈K(1,1).
Since Dλn+1g/z∈K(0,2/λ), and kλ/z∈T(1,1+1/λ)*, application of Lemma 2.3 gives
(3.4)kλz*Dλn+1gz⋅p=(kλz*Dλn+1gz)p0,p0∈K(1,1)
hence
(3.5)(fz)1/λ=kn+1,λz*Dλngzp0,p0∈K(1,1)
Using Lemma 2.2 we deduce that f∈Bλn.
As a consequence of (3.1) we immediately have the following.Corollary 3.2.
One has(3.6)Bλn⊂S.Corollary 3.3.
Iff∈Bλn, n∈N0, λ>0, then, for z∈E(3.7)(fz)1/λ∈K(1,1+2λ).Proof.
Sincef∈Bλn, there exists a g∈STn(1-1/λ) or Dλng/z∈K(0,2/λ) such that
(3.8)(fz)1/λ=kn+1,λz*Dλngz⋅p,p∈K(1,1),(3.9)=kn+1,λz*F,F∈K(1,1+2λ),
using (1.3). From (3.6), we conclude that
(3.10)(fz)1/λ=kn+1,λz*F≠0,0<|z|<1,
which implies that
(3.11)kn+1,λz∈K(1,1+2λ)*≡T(1,1+2λ)*,
Applying Lemma 2.3 to (3.9), we get the result.Theorem 3.4.
Letf∈Bλn. Then
(3.12)(Dλn+1z(f/z)1/λz)λ∈K(λ,λ+2).Proof.
From (2.1), we see that
(3.13)Dλn+1z(fz)1/λ=Dλng(z)⋅p(z),p∈K(1,1).
Since (Dλng/z)λ∈K(0,2), and p(z)λ∈K(λ,λ), then
(3.14)(Dλn+1z(f/z)1/λz)λ=(Dλngz)λp(z)λ∈K(λ,λ+2),
which is the required result.In the following we prove the converse of Theorem3.1, for 0<λ≤1.Theorem 3.5.
Letf∈Bλn, 0<λ≤1. Then f∈Bλn+1 in |z|<rc, where rc is given by (2.5)Proof.
f∈Bλn implies (2.1), where Dλng∈ST(1-1/λ)⊂ST.
Now(3.15)Dλn+2z(f/z)1/λDλn+1g=hλ*Dλn+1z(f/z)1/λhλ*Dλng.
Using Lemma 2.4, we see that hλ∈CV in |z|<rc, for 0<λ≤1, where rc is given by (2.5).
From Remark2.6, we conclude
(3.16)hλ*{1+xz1+yzDλng:|x|=|y|=1}≠0.
Applying Lemma2.5, we deduce
(3.17)(hλ*(Dλn+1z(f/z)1/λ/Dλng)Dλnghλ*Dλng)(|z|<rc)⊂CO¯(Dλn+1z(f/z)1/λDλng)(E),
hence Dλn+2z(f/z)1/λ/Dλn+1g∈P in |z|<rc, as required.Corollary3.3 can be improved for 0<λ≤1, as follows.Theorem 3.6.
Letf∈Bλn, 0<λ≤1. Then
(3.18)fz∈K(1,2).Proof.
We will use Ruscheweyh’s.method of proof [10]. f∈Bλn implies (3.8), where Dλn+1g/z∈K(0,2/λ), kn+1,λ/z∈T(1,1+2/λ)*.
LetDλng/z=l·m, where l=(Dλng/z)(1+λ)/2 and m=(Dλng/z)(1-λ)/2.
Thenl∈K(0,1/λ+1)·m∈K(0,1/λ-1) and m·p=F∈K(1,1/λ). This implies
(3.19)(fz)1/λ=kn+1,λz*lF.
Using Lemma2.7, we get
(3.20)(kn+1,λ/z)*lF(kn+1,λ/z)*l=F0∈K(1λ,1λ),0<λ≤1.
Hence(3.21)(fz)1/λ=(kn+1,λz*l)F0,F0∈K(1λ,1λ).
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).
Sincezl∈ST((λ-1)/2λ), then from (1.5)
(3.22)(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).
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.
Forn=0, λ=1/α Theorem 3.6 and other stronger results depending on α, are proved by Sheil-Small [7].For the coefficient bounds off∈Bλn, 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∈Bλn, using the extreme points of Kaplan class K(α,β).Theorem 3.8.
Forf∈Bλn, λ=m∈ℕ,
(3.23)f≪z(1-z)2-mn.Proof.
From (3.9) and Lemma 2.9, we get
(3.24)(fz)1/λ≪kn+1,λz*1+z(1-z)1+2/λ,=1(1-z)2/λ-n,
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.
Forn=0, we get the result of Zamorski [19], and the result of Sheil-Small [7], from which we get the idea of proof.



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