# On Starlike and Convex Functions with Respect tok-Symmetric Points

Authors: Afaf A. Ali Abubaker; Maslina Darus
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

We introduce new subclassesSkσ,s(λ,δ,ϕ) and Kkσ,s(λ,δ,ϕ)
of analytic functions with respect to k-symmetric points defined by differential
operator. Some interesting properties for these classes are obtained.


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## Body

## 1. Introduction

LetA denote the class of functions of the form (1.1)f(z)=z+∑n=2∞anzn,
which are analytic in the unit disk U={z∈ℂ:|z|<1}.Also let℘ be the class of analytic functions p with p(0)=1, which are convex and univalent in U and satisfy the following inequality: (1.2)R{p(z)}>0,z∈U.
A function f∈A is said to be starlike with respect to symmetrical points in U if it satisfies (1.3)R{zf′(z)f(z)-f(-z)}>0,z∈U.
This class was introduced and studied by Sakaguchi in 1959 [1]. Some related classes are studied by Shanmugam et al. [2].In 1979, Chand and Singh [3] defined the class of starlike functions with respect to k-symmetric points of order α (0≤α<1). Related classes are also studied by Das and Singh [4].Recall that the functionF is subordinate to G if there exists a function ω, analytic in U, with ω(0)=0 and |ω(z)|<1, such that F(z)=G(w(z)), z∈U. We denote this subordination by F(z)≺G(z). If G(z) is univalent in U, then the subordination is equivalent to F(0)=G(0) and F(U)⊂G(U).A functionf∈A is in the class Sk(ϕ) satisfying (1.4)zf′(z)fk(z)≺ϕ(z),z∈U,
where ϕ∈℘, k is a fixed positive integer, and fk(z) is given by the following:(1.5)fk(z)=1k∑ν=0k-1ε-νf(ενz)=z+∑ι=2∞ak(ι-1)+1zk(ι-1)+1,(ε=exp⁡(2πik),z∈U).
The classes Sk(ϕ) of starlike functions with respect to k-symmetric points and Kk(ϕ) of convex functions with respect to k-symmetric points were considered recently by Wang et al. [5]. Moreover, the special case (1.6)ϕ(z)=1+βz1-αβz,0≤α≤1,0<β≤1
imposes the class Sk(α,β), which was studied by Gao and Zhou [6], and the class S1(ϕ)=S*(ϕ) was studied by Ma and Minda [7].Let two functions given byf(z)=z+∑n=2∞anzn and g(z)=z+∑n=2∞bnzn be analytic in U. Then the Hadamard product (or convolution) f*g of the two functions f, g is defined by (1.7)f(z)*g(z)=z+∑n=2∞anbnzn,
and for several function f1(z),…,fm(z)∈A, (1.8)f1(z)*⋯*fm(z)=z+∑n=2∞(a1n⋯amn)zn,z∈U.
The theory of differential operators plays important roles in geometric function theory. Perhaps, the earliest study appeared in the year 1900, and since then, many mathematicians have worked extensively in this direction. For recent work see, for example, [8–12].We now define differential operator as follows:(1.9)Dλ,δσ,sf(z)=z+∑n=2∞ns(C(δ,n)[1+λ(n-1)])σanzn,
where λ≥0, C(δ,n)=(δ+1)n-1/(n-1)!, for δ,σ,s∈N0={0,1,2,…}, and (x)n is the Pochhammer symbol defined by (1.10)(x)n=Γ(x+n)Γ(x)={1,n=0,x(x+1)⋯(x+n-1),n={1,2,3,…}.
Here Dλ,δσ,sf(z) can also be written in terms of convolution as (1.11)ψ(z)=[λz(1-z)2-λz1-z+z1-z]*z(1-z)δ+1,z∈U,Dλ,δσ,sf(z)=ψ(z)*⋯*ψ(z)︸σ-times*∑n=1∞nszn*f(z)=Dδ*⋯*Dδ︸σ-times*Dλσ,sf(z),
where Dδ=z+∑n=2∞C(δ,n)zn and Dλσ,s=z+∑n=2∞ns[1+λ(n-1)]σzn.Note thatDλ,δ0,1f(z)=D1,01,0f(z)=zf′(z) and Dλ,δ0,0f(z)=f(z). When σ=0, we get the Sǔlǔgean differential operator [9], when λ=s=0, σ=1 we obtain the Ruscheweyh operator [8], when s=0, σ=1, we obtain the Al-Shaqsi and Darus [11], and when δ=s=0, we obtain the Al-Oboudi differential operator [10].In this paper, we introduce new subclasses of analytic functions with respect tok-symmetric points defined by differential operator. Some interesting properties of Skσ,s(λ,δ,ϕ) and Kkσ,s(λ,δ,ϕ) are obtained.Applying the operatorDλ,δσ,sf(z)(1.12)Dλ,δσ,sfk(z)=1k∑ν=0k-1ε-νDλ,δσ,sf(ενz),εk=1,
where k is a fixed positive integer, we now define classes of analytic functions containing the differential operator.Definition 1.1.
LetSkσ,s(λ,δ,ϕ) denote the class of functions in A satisfying the condition
(1.13)z(Dλ,δσ,sf(z))′Dλ,δσ,sfk(z)≺ϕ(z),
where ϕ∈℘.Definition 1.2.
LetKkσ,s(λ,δ,ϕ) denote the class of functions in A satisfying the condition
(1.14)(z(Dλ,δσ,sf(z))′)′(Dλ,δσ,sfk(z))′≺ϕ(z),
where ϕ∈℘.In order to prove our results, we need the following lemmas.Lemma 1.3 (see [13]).
Letc>-1, and let Ic:A→A be the integral operator defined by F=Ic(f), where
(1.15)F(z)=c+1zc∫0ztc-1f(t)dt.
Let ϕ be a convex function, with ϕ(0)=1 and ℜ{ϕ(z)+c}>0 in U. If f∈A and zf′(z)/f(z)≺ϕ(z), then zF′(z)/F(z)≺q(z)≺ϕ(z), where q is univalent and satisfies the differential equation
(1.16)q(z)+zq′(z)q(z)+c=ϕ(z).Lemma 1.4 (see [14]).
Letκ, υ be complex numbers. Let ϕ be convex univalent in U with ϕ(0)=1 and ℜ[κϕ+υ]>0, z∈U, and let q(z)∈A with q(0)=1 and q(z)≺ϕ(z). If p(z)=1+p1z+p2z2+⋯∈℘ with p(0)=1, then
(1.17)p(z)+zp′(z)κq(z)+υ≺ϕ(z)⟹p(z)≺ϕ(z).Lemma 1.5 (see [15]).
Letf and g, respectively, be in the classes convex function and starlike function. Then, for every function H∈A, one has
(1.18)(f(z)*g(z)H(z))f(z)*g(z)∈co¯(H(U)),z∈U,
where co¯(H(U)) denotes the closed convex hull of H(U).


## 2. Main Results

Theorem 2.1.
Letf∈Skσ,s(λ,δ,ϕ). Then fk defined by (1.5) is in S1σ,s(λ,δ,ϕ)=Sσ,s(λ,δ,ϕ).Proof.
Letf∈Skσ,s(λ,δ,ϕ), then by Definition 1.1 we have
(2.1)z(Dλ,δσ,sf(z))′Dλ,δσ,sfk(z)≺ϕ(z).
Substituting z by ενz, where εk=1 (ν=0,1,…,k-1) in (2.1), respectively, we have
(2.2)ενz(Dλ,δσ,sf(ενz))′Dλ,δσ,sfk(ενz)≺ϕ(z).
According to the definition of fk and εk=1, we know fk(ενz)=ενfk(z) for any ν=0,1,…,k-1, and summing up, we can get
(2.3)1k∑ν=0k-1ε-ν[z(Dλ,δσ,sf(ενz))′Dλ,δσ,sfk(z)]=z[(1/k)∑ν=0k-1ε-νDλ,δσ,sf(ενz)]′Dλ,δσ,sfk(z)=z(Dλ,δσ,sfk(z))′Dλ,δσ,sfk(z).
Hence there exist ζν in U such that
(2.4)z(Dλ,δσ,sfk(z))′Dλ,δσ,sfk(z)=1k∑ν=0k-1ϕ(ζν)=ϕ(ζ0),
for ζ0∈U since ϕ(U) is convex. Thus fk∈Sσ,s(λ,δ,ϕ).Theorem 2.2.
Letf∈A and ϕ∈℘. Then
(2.5)f∈Kkσ,s(λ,δ,ϕ)⟺zf′∈Skσ,s(λ,δ,ϕ).Proof.
Let(2.6)g(z)=z+∑n=2∞ns(C(δ,n)[1+λ(n-1)])σzn,
and the operator Dλ,δσ,sf can be written as Dλ,δσ,sf=g*f.
Then from the definition of the differential operatorDλ,δσ,s, we can verify
(2.7)(z(Dλ,δσ,sf(z))′)′(Dλ,δσ,sfk(z))′=(z(g*f)′)′(z)(g*f)k′(z)=z(g*zf′)′(z)(g*zf′)k(z)=z(Dλ,δσ,szf′(z))′Dλ,δσ,szfk′(z).
Thus f∈Kkσ,s(λ,δ,ϕ) if and only if zf′∈Skσ,s(λ,δ,ϕ).By using Theorems2.2 and 2.1, we get the following.Corollary 2.3.
Letf∈Kkσ,s(λ,δ,ϕ). Then fk defined by (1.5) is in K1σ,s(λ,δ,ϕ)=Kσ,s(λ,δ,ϕ).Proof.
Letf∈Kkσ,s(λ,δ,ϕ). Then Theorem 2.2 shows that zf′∈Skσ,s(λ,δ,ϕ). We deduce from Theorem 2.1 that (zf′)k∈Sσ,s(λ,δ,ϕ). From (zf′)k=zfk′, Theorem 2.2 now shows that fk∈K1σ,s(λ,δ,ϕ)=Kσ,s(λ,δ,ϕ).Theorem 2.4.
Letϕ∈℘, λ>0 with ℜ[ϕ(z)+(1/λ)-1]>0. If f∈Skσ,s(λ,δ,ϕ), then
(2.8)z(Dλ,δσ-1,s(Dδfk(z)))′Dλ,δσ-1,s(Dδfk(z))≺q(z)≺ϕ(z),
where Dλ,δσ-1,s(Dδfk(z))=Dλ,δσ-1,s*Dδfk(z) and q is the univalent solution of the differential equation
(2.9)q(z)+zq′(z)q(z)+(1/λ)-1=ϕ(z).Proof.
Letf∈Skσ,s(λ,δ,ϕ). Then in view of Theorem 2.1, fk∈Sσ,s(λ,δ,ϕ), that is,
(2.10)z(Dλ,δσ,sfk(z))′Dλ,δσ,sfk(z)≺ϕ(z).
From the definition of Dλ,δσ,s, we see that
(2.11)Dλ,δs,σfk(z)=(1-λ)(Dλ,δs,σ-1*Dδfk(z))+λz(Dλ,δs,σ-1*Dδfk(z))′
which implies that
(2.12)Dλ,δs,σ-1*Dδfk(z)=1λz(1/λ)-1∫0zt(1/λ)-2Dλ,δs,σfk(t)dt.
Using (2.10) and (2.12), we see that Lemma 1.3 can be applied to get (2.8), where c=(1/λ)-1>-1 and ℜ{ϕ}>0 with ℜ[ϕ(z)+(1/λ)-1]>0 and q satisfies (2.9). We thus complete the proof of Theorem 2.4.Theorem 2.5.
Letϕ∈℘ and s∈N0. Then
(2.13)Skσ,s+1(λ,δ,ϕ)⊂Skσ,s(λ,δ,ϕ).Proof.
Letf∈Skσ,s+1(λ,δ,ϕ). Then
(2.14)z(Dλ,δσ,s+1f(z))′Dλ,δσ,s+1fk(z)≺ϕ(z).
Set
(2.15)p(z)=z(Dλ,δσ,sf(z))′Dλ,δσ,sfk(z),
where p is analytic function with p(0)=1. By using the equation
(2.16)z(Dλ,δσ,sf(z))′=Dλ,δσ,s+1f(z),
we get
(2.17)p(z)=Dλ,δσ,s+1f(z)Dλ,δσ,sfk(z)
and then differentiating, we get
(2.18)z(Dλ,δσ,s+1f(z))′=zDλ,δσ,sfk(z)p′(z)+z(Dλ,δσ,sfk(z))′p(z),
Hence
(2.19)z(Dλ,δσ,s+1f(z))′Dλ,δσ,s+1fk(z)=Dλ,δσ,sfk(z)Dλ,δσ,s+1fk(z)zp′(z)+z(Dλ,δσ,sfk(z))′Dλ,δσ,s+1fk(z)p(z).
Applying (2.16) for the function fk we obtain
(2.20)z(Dλ,δσ,s+1f(z))′Dλ,δσ,s+1fk(z)=Dλ,δσ,sfk(z)Dλ,δσ,s+1fk(z)zp′(z)+p(z).
Using (2.20) with q(z)=(Dλ,δσ,s+1fk(z))/(Dλ,δσ,sfk(z)), we obtain
(2.21)z(Dλ,δσ,s+1f(z))′Dλ,δσ,s+1fk(z)=zp′(z)q(z)+p(z).
Since f∈Skσ,s+1(λ,δ,ϕ), then by using (2.14) in (2.21) we get the following.
(2.22)zp′(z)q(z)+p(z)≺ϕ(z).
We can see that q(z)≺ϕ(z), hence applying Lemma 1.4 we obtain the required result.By using Theorems2.2 and 2.5, we get the following.Corollary 2.6.
Letϕ∈℘ and s∈N0. Then
(2.23)Kkσ,s+1(λ,δ,ϕ)⊂Kkσ,s(λ,δ,ϕ).Now we prove that the classSkσ,s(λ,δ,ϕ), ϕ∈℘, is closed under convolution with convex functions.Theorem 2.7.
Letf∈Skσ,s(λ,δ,ϕ), ϕ∈℘, and φ is a convex function with real coefficients in U. Then f*φ∈Skσ,s(λ,δ,ϕ).Proof.
Letf∈Skσ,s(λ,δ,ϕ), then Theorem 2.1 asserts that Dλ,δσ,sfk(z)∈S*(ϕ), where ℜ{ϕ}>0. Applying Lemma 1.5 and the convolution properties we get
(2.24)z(Dλ,δσ,s(f*φ)(z))′Dλ,δσ,s(fk*φ)(z)=z(Dλ,δσ,sf(z)*φ(z))′φ(z)*Dλ,δσ,sfk(z)=φ(z)*(z(Dλ,δσ,sf(z))′/fDλ,δσ,sfk(z))Dλ,δσ,sfk(z)φ(z)*Dλ,δσ,sfk(z)∈co⁡¯(z(Dλ,δσ,sf)′Dλ,δσ,sfk(U))⊆ϕ(U).Corollary 2.8.
Letf∈Kkσ,s(λ,δ,ϕ), ϕ∈℘, and φ is a convex function with real coefficients in U. Then f*φ∈Kkσ,s(λ,δ,ϕ).Proof.
Letf∈Kkσ,s(λ,δ,ϕ), ϕ∈℘. Then Theorem 2.2 shows that zf′∈Skσ,s(λ,δ,ϕ). The result of Theorem 2.7 yields (zf′)*φ=z(f*φ)′∈Skσ,s(λ,δ,ϕ), and thus f*φ∈Kkσ,s(λ,δ,ϕ).Some other works related to other differential operators with respect to symmetric points for different types of problems can be seen in ([16–21]).



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