# Some Properties of Certain Multivalent Analytic Functions Involving the Cătas Operator

Authors: E. A. Elrifai; H. E. Darwish; A. R. Ahmed
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

---

## Abstract

We introduce a certain subclass of multivalent analytic functions by making use of the principle of subordination between these functions and Cătas operator. Such results as subordination and superordination properties, convolution properties, inclusion relationships, distortion theorems, inequality properties, and sufficient conditions for multivalent starlikeness are provide. The results presented here would provide extensions of those given in earlier works. Several other new results are also obtained.


---

## Body

## 1. Introduction

LetAp(n) denote the class of functions of the following form:(1.1)f(z)=zp+∑k=n∞ap+kzp+k(p,n∈N:={1,2,3,…}),
which are analytic in the open unit disk U:={z:z∈ℂand|z|<1}.For simplicity, we write(1.2)Ap(1):=Ap,A1(1)=A.A functionf(z)∈Ap(n) is said to be in the class Sp,n*(γ) of p-valent starlike functions of order γ in U if it satisfies the following inequality: (1.3)Re{zf′(z)f(z)}>γ(0≤γ<p;z∈U).
Let H[a,n] be the class of analytic functions of the following form: (1.4)F(z)=a+anzn+an+1zn+1+⋯(z∈U).
Let f,g∈Ap(n), where f(z) is given by (1.1) and g(z) is defined by (1.5)g(z)=zp+∑k=n∞bp+kzp+k.
Then the Hadanard product (or convolution) f*g of the functions f(z) and g(z) is defined by (1.6)(f*g)(z):=zp+∑k=n∞ap+kbp+kzp+k=:(g*f)(z).
We consider the following multiplier transformations.Definition 1.1 (see [1]).
Letf(z)∈Ap(n). For p,n∈N,δ,λ≥0,ℓ≥0, define the multiplier transformations Ip(δ,λ,ℓ) on Ap(n) by the following infinite series:
(1.7)Ip(δ,λ,l)f(z):=zp+∑k=n∞[p+λk+lp+l]δak+pzk+p.
It is easily verified from (1.7), that
(1.8)(p+l)Ip(δ+1,λ,l)f(z)=[p(1-λ)+l]Ip(δ,λ,l)f(z)+λz(Ip(δ,λ,l)f(z))′.
It should be remarked that the class of multiplier transformsIp(δ,λ,ℓ) is a generalization of several other linear operators considered, in earlier investigations (see [2–12]).
Iff(z) is given by (1.1), then we have
(1.9)Ip(δ,λ,l)f(z)=(f*φp,λ,lδ)(z),
where
(1.10)φp,λ,lδ(z)=zp+∑k=n∞[p+λk+lp+l]δzp+k.
In particular, we set
(1.11)I1(δ,λ,l)f(z):=I(δ,λ,l)f(z).
For two functionsf(z) and g(z), analytic in U, we say that the function f(z) is subordinate to g(z) in U, and write f(z)≺g(z)(z∈U) if there exists a Schwarz function w(z), which is analytic in U with(1.12)w(0)=0,|w(z)|<1(z∈U)
such that
(1.13)f(z)=g(w(z))(z∈U).
Indeed, it is known that
(1.14)f(z)≺g(z),(z∈U)⟹f(0)=g(0),f(U)⊂g(U).
Furthermore, if the function g is univalent in U, then we have the following equivalence:
(1.15)f(z)≺g(z),(z∈U)⟺f(0)=g(0),f(U)⊂g(U).
By making use of the linear operator Ip(δ,λ,ℓ) and the above-mentioned principle of subordination between analytic functions, we introduce and investigate the following subclass of the class Ap(n) of p-valent analytic functions.Definition 1.2.
A functionf(z)∈Ap(n) is said to be in the class ßpα,β(δ,λ,ℓ,n;A,B) if it satisfies the following subordination condition:
(1.16)(1-α)(Ip(δ,λ,l)f(z)zp)β+α(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β≺1+Az1+Bz(z∈U),
where (and throughout this paper unless otherwise mentioned) the parameters α,β,p,n,λ,δ,ℓ,A, and B are constrained as follows:
(1.17)α∈C,Re(β)>0,λ,l≥0,λ,l∈R,δ≥0,-1≤B≤1,A≠B∈R,p,n∈N.For simplicity, we write(1.18)ß11,β(1,1,1;1;1,-1)=ß(β).
Clearly, the class ß(β) is a subclass of the familiar class of Bazilevič functions of type β.If we setδ=0,λ=ℓ=p=1 in the class ßpα,β(δ,λ,ℓ,n;A,B), which was studied by Liu [13]. In particular, Zhu [14] determined the sufficient conditions such that ßn(α,β,A,0)⊂Sp,n*(ρ).Cătas [1, 5, 15], Cho and Srivastava [6], Cho and Kim [7], and Kumar et al. [10] obtained many interesting results associated with the multiplier operator.In the present paper, we aim at proving such results as subordination and superordination properties, convolution properties, inclusion relationships, distortion theorems, inequality properties, and sufficient conditions for multivalent starlikeness of the classßpα,β(δ,λ,ℓ,n;A,B). The results presented here would provide extensions of those given in earlier works. Several other new results are also obtained.


## 2. Preliminary Results

In order to establish our main results, we need the following definition and lemmas.Definition 2.1 (see [16]).
Denote byQ the set of all functions f(z) that are analytic and injective on U¯-E(f), where
(2.1)E(f)={ε∈∂U:lim⁡z→ɛf(z)=∞},
and such that f′(ε)≠0 for ε∈∂U-E(f).Lemma 2.2 (see [17]).
Let the functionh be analytic and univalent (convex) in U with h(0)=1. Suppose also that the function k given by
(2.2)k(z)=1+cnzn+cn+1zn+1+⋯
is analytic in U. If
(2.3)k(z)+zk′(z)ζ≺h(z)(Re(ζ)>0;ζ≠0;z∈U),
then
(2.4)k(z)≺χ(z)=ζnz-ζ/n∫0zt(ζ/n)-1h(t)dt≺h(z)(z∈U),
and χ(z) is the best dominant of (2.3).Lemma 2.3 (see [18]).
Letq(z) be a convex univalent function in U and let σ,η∈ℂ with
(2.5)Re(1+zq′′(z)q′(z))>max{0,-Re(ση)}.
If the function p is analytic in U and
(2.6)σp(z)+ηzp′(z)≺σq(z)+ηzq′(z),
then p(z)≺q(z) and q(z) is the best dominant.Lemma 2.4 (see [16]).
Letq be convex univalent in U and k∈C. Further assume that Re(k¯)>0 if
(2.7)p(z)∈H[q(0),1]∩Q,
and p(z)+kzp′(z) is univalent in U, then
(2.8)q(z)+kzq′(z)≺p(z)+kzp′(z)
implies that q(z)≺p(z) and q(z) is the best subdominant.Lemma 2.5 (Jach’s Lemma [19]).
Letw(z) be a noncostant analytic function in U with w(0)=0. If |w| attains its maximum value on the circle |z|=r<1 at z0, then
(2.9)z0w′(z0)=kw(z0),
where k≥1 is a real number.Lemma 2.6 (see [20]).
LetF be analytic and convex in U. If f(z),g(z)∈A and f(z),g(z)≺F(z); then
(2.10)λf(z)+(1-λ)g(z)≺F(z)(0≤λ≤1).Lemma 2.7 (see [21, 22]).
Letk,v∈ℂ. Suppose also that m is convex and univalent in U with
(2.11)m(0)=1,Re(km(z)+v)>0(z∈U).
If u is analytic in U with u(0)=1, then the following subordination:
(2.12)u(z)+zu′(z)ku(z)+v≺m(z),(z∈U)
implies that
(2.13)u(z)≺m(z)(z∈U).Lemma 2.8 (see [23]).
Letf(z)=1+∑k=1∞akzk analytic in U and g(z)=1+∑k=1∞bkzk be analytic and convex in U. If f(z)≺g(z), then |ak|≤|b1|,(k∈N).Lemma 2.9 (see [24]).
Letδ≠0,δ∈R,v/δ>0,0≤ρ<1,p∈H[1,n], and p(z)≺1+kz(k:=vM/(nδ+v)), where
(2.14)M:=Mn(δ,ν,ρ)=(1-ρ)|δ|(1+nδ/v)|1-δ+ρδ|+1+(nδ/v)2.
If q(z)∈H[1,n] satisfies the following subordination condition:
(2.15)p(z)[1-δ+δ((1-ρ)q(z)+ρ)]≺1+Mz,
then
(2.16)Re(q(z))>0(z∈U).


## 3. Main Results

We begin by presenting our first subordination property given by Theorem3.1 below.Theorem 3.1.
Letf(z)∈ßpα,β(δ,λ,ℓ;n;A,B) with Re(α)>0. Then
(3.1)(Ip(δ,λ,l)f(z)zp)β≺(p+l)βλnα∫011+Azu1+Bzuu(p+l)β/λnα-1du≺1+Az1+Bz(z∈U).Proof.
Define the functionP(z) by
(3.2)P(z):=(Ip(δ,λ,l)f(z)zp)β(z∈U).
Then P(z) is analytic in U with P(0)=1. By taking the derivatives in the both sides in equality (3.2) and using (1.8), we get
(3.3)(1-α)(Ip(δ,λ,l)f(z)zp)β+α(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β=P(z)+λαzP′(z)β(p+l)≺1+Az1+Bz(z∈U).
An application of Lemma 2.2 to (3.3) yields
(3.4)(Ip(δ,λ,l)f(z)zp)β≺(p+l)βλnαz-(p+l)β/λnα∫0zt(p+l)β/λnα-11+At1+Btdt=ζn∫01uζ/n-11+Azu1+Bzudu≺1+Az1+Bz(z∈U),
where
(3.5)ζ=(p+l)βλα.The proof of Theorem 3.1 is thus completed.Theorem 3.2.
Letq(z) be univalent in U, 0≠α∈ℂ. Suppose also that q(z) satisfies
(3.6)Re(1+zq′′(z)q′(z))>max{0,-Re((p+l)βλα)}.
If f(z)∈Ap(n) satisfying the following subordination:
(3.7)(1-α)(Ip(δ,λ,l)f(z)zp)β+α(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β≺q(z)+αλzq′(z)(p+l)β,
then
(3.8)(Ip(δ,λ,l)f(z)zp)β≺q(z),
and q(z) is the best dominant.Proof.
Let the functionP(z) be defined by (3.2). We know that (3.3) holds true. Combining  (3.3) and (3.7), we find that
(3.9)P(z)+λαzP′(z)(p+l)β≺q(z)+λαzq′(z)(p+l)β.By Lemma2.3 and (3.9), we easily get the assertion of Theorem 3.2.Takingq(z)=(1+Az)/(1+Bz) in Theorem 3.2, we get the following result.Corollary 3.3.
Letα∈ℂ and -1≤B<A≤1. Suppose also that (1+Az)/(1+Bz) satisfies the condition (3.6). If f(z)∈Ap(n) satisfies the following subordination:
(3.10)(1-α)(Ip(δ,λ,l)f(z)zp)β+α(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β≺1+Az1+Bz+λα(A-B)z(p+l)β(1+Bz)2,
then
(3.11)(Ip(δ,λ,l)f(z)zp)β≺1+Az1+Bz,
and (1+Az)/(1+Bz) is the best dominant.
Iff(z) is subordinate to F(z), then F(z) is superordinate to f(z). We now derive the following superordination result for the class ßpα,β(δ,λ,ℓ;n;A,B).Theorem 3.4.
Letq(z) be convex univalent in U,α∈ℂ, with Re(α)>0. Also let
(3.12)(Ip(δ,λ,l)f(z)zp)β∈H[q(0),1]∩Q,(1-α)(Ip(δ,λ,l)f(z)zp)β+α(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β
be univalent in U. If
(3.13)q(z)+λαzq′(z)(p+l)β≺(1-α)(Ip(δ,λ,l)f(z)zp)β+α(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β,
then
(3.14)q(z)≺(Ip(δ,λ,l)f(z)zp)β,
and q(z) is the best subdominant.Proof.
Let the functionP(z) be defined by (3.2). Then
(3.15)q(z)+λαzq′(z)(p+l)β≺(1-α)(Ip(δ,λ,l)f(z)zp)β+α(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β=P(z)+λαzP′(z)(p+l)β.
An application of Lemma 2.4 yields the assertion of Theorem 3.4.Takingq(z)=(1+Az)/(1+Bz) in Theorem 3.4, we get the following corollary.Corollary 3.5.
Letq(z) be convex univalent in U and -1≤B<A≤1,α∈ℂ with Re(α)>0. Also let
(3.16)(Ip(δ,λ,l)f(z)zp)β∈H[q(0),1]∩Q,(1-α)(Ip(δ,λ,l)f(z)zp)β+α(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β
be univalent in U. If
(3.17)1+Az1+Bz+λα(A-B)z(p+l)β(1+Bz)2≺(1-α)(Ip(δ,λ,l)f(z)zp)β+α(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β,
then
(3.18)1+Az1+Bz≺(Ip(δ,λ,l)f(z)zp)β,
and (1+Az)/(1+Bz) is best subdominant.Combining the above results of subordination and superordination. We easily get the following “Sandwich-type result”.Corollary 3.6.
Letq1(z) be convex univalent and let q2(z) be univalent in U, α∈ℂ, Re(α)>0. Let q2(z) satisfies (3.6). If
(3.19)0≠(Ip(δ,λ,l)f(z)zp)β∈H[q1(0),1]∩Q,(1-α)(Ip(δ,λ,l)f(z)zp)β+α(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β
is univalent in U, also
(3.20)q1(z)+λαzq1′(z)(p+l)β≺(1-α)(Ip(δ,λ,l)f(z)zp)β+α(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β≺q2(z)+λαzq2′(z)(p+l)β,
then
(3.21)q1(z)≺(Ip(δ,λ,l)f(z)zp)β≺q2(z),
and q1(z),q2(z) are, respectively, the best subordinate, and dominant.Theorem 3.7.
Letf(z)∈Ap(n),ξ∈ℂ∖{0}, and 0≤γ<1. Also let the function φ be defined by
(3.22)φ(z)=z((Ip(δ+1,λ,l)f(z)/Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)/zp)β-1)′(Ip(δ+1,λ,l)f(z)/Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)/zp)β-1(z∈U).
If φ satisfies one of the following conditions:
(3.23)Re(φ(z)){<1|ξ|2Re(ξ)(Re(ξ)>0),≠0(Re(ξ)=0),>1|ξ|2Re(ξ)(Re(ξ)<0),
or
(3.24)I(φ(z)){>-1|ξ|2I(ξ)(I(ξ)>0),≠0(I(ξ)=0),<-1|ξ|2I(ξ)(I(ξ)<0),
then
(3.25)|((Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β-1)ξ|<1-γ.Proof.
We define the functionϕ(z) by
(3.26)((Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β-1)ξ=(1-γ)ϕ(z)(0≤γ<1;ξ∈C∖{0};z∈U).
It is easy to see that the function ϕ is analytic in U with ϕ(0)=0.
Differentiating both sides of (3.26) with respect to z logarithmically, we get(3.27)zϕ′(z)ϕ(z)=ξz((Ip(δ+1,λ,l)f(z)/Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)/zp)β-1)′(Ip(δ+1,λ,l)f(z)/Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)/zp)β-1(z∈U;ξ∈C∖{0}).
We now consider the function φ defined by
(3.28)φ:=ξ¯|ξ|2zϕ′(z)ϕ(z)(z∈U,ξ∈C∖{0}).
Assume that there exists a point z0∈U such that
(3.29)max⁡|z|≤|z0||ϕ(z)|=|ϕ(z0)|=1,
by Lemma 2.5, we know that
(3.30)z0ϕ′(z)=kϕ(z0)(k≥1).
If follows from (3.28) and (3.30) that
(3.31)Re(φ(z0))=Re(ξ¯|ξ|2zϕ′(z0)ϕ(z0))=k|ξ|2Re(ξ¯)=k|ξ|2Re(ξ){≥1|ξ|2Re(ξ)(Re(ξ)>0),=0(Re(ξ)=0)≤1|ξ|2Re(ξ)(Re(ξ)<0),(3.32)I(φ(z0))=I(ξ¯|ξ|2zϕ′(z0)ϕ(z0))=k|ξ|2I(ξ¯)=-k|ξ|2I(ξ){≤-1|ξ|2I(ξ)(I(ξ)>0),=0(I(ξ)=0),≥-1|ξ|2I(ξ)(I(ξ)<0).
But the inequalities in (3.31) and (3.32) contradict, respectively, the inequalities in (3.23) and (3.24). Therefore, we can conclude that
(3.33)|ϕ(z)|<1(z∈U),
which implies that
(3.34)|((Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β-1)ξ|=(1-γ)|ϕ(z)|<1-γ.
We thus complete the proof of Theorem 3.7.From Theorem3.7, we easily get the following result for the class
ß(β) of Bazilevič functions of type β.Corollary 3.8.
Letf(z)∈A,δ=0,p=λ=ℓ=ξ=1, and γ=0. Also let the function φ be defined by (3.22). If φ satisfies one of the following conditions:
(3.35)Re(φ(z))<1orI(φ(z))≠0,
then f(z)∈ß(β).Theorem 3.9.
LetRe(α)>0,β>0, and f(z)∈ßp0,β(δ,λ,ℓ;n;1-2ρ,-1)(0≤ρ<1). Then f(z)∈ßpα,β(δ,λ,ℓ;n,1-2ρ,-1) for |z|<R(α,β,ℓ,λ,p), where
(3.36)R(α,β,l,λ,p)=-λα+λ2α2+(p+l)2β2(l+ρ)β.
The bound R(α,β,ℓ,λ,p) is the best possible.Proof.
Suppose that(3.37)(Ip(δ,λ,l)f(z)zp)β=ρ+(1-ρ)h(z)(z∈U;0≤ρ<1),
where h is analytic and has a positive real part in U. By taking the derivatives in the both sides in equality (3.37) and using (1.9), we get
(3.38)Re((1-α)(Ip(δ,λ,l)f(z)zp)β+α(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β)-ρ=(1-ρ)Re(h(z)+λαzh′(z)(p+l)β)≥(1-ρ)Re(h(z)-λα|zh′(z)|(p+l)β).
By making use of the following well-known estimate (see [25]):
(3.39)|zh′(z)|≤2r1-r2Re(h(z))(|z|=r<1)
in (3.38), we obtain that
(3.40)Re[(1-α)(Ip(δ,λ,l)f(z)zp)β+α(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β-ρ]=(1-ρ)(1-2λαr(p+l)β(1-r2))Re(h(z))>0
for r<R(α,β,ℓ,λ,p), where R(α,β,ℓ,λ,p) is given by (3.36).
To show that the boundR(α,β,ℓ,λ,p) is the best possible, we consider the function f(z)∈Ap(n) defined by(3.41)(Ip(δ,λ,l)f(z)zp)β=ρ+(1-ρ)1+z1-z(z∈U).
By noting that
(3.42)Re((1-α)(Ip(δ,λ,l)f(z)zp)β+α(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β)-ρ=(1-ρ)Re(1+z1-z+2λα(p+l)βz(1-z)2)=0
for z=R(α,β,ℓ,λ,p), we conclude that the bound is the best possible. Theorem 3.9 is thus proved.Theorem 3.10.
Letf(z)∈ßpα,β(δ,λ,ℓ;n;A,B) with Re(α)>0. Then
(3.43)f(z)=(zp(1+Aw(z)1+Bw(z))1/β)*(zp+∑k=n∞(p+λk+lp+l)-δzk+p),
where w(z) is analytic in U with w(0)=0 and |w(z)|<1(z∈U).Proof.
Suppose thatf(z)∈ßpα,β(δ,λ,ℓ;n;A,B). It follows from (3.1) that
(3.44)(Ip(δ,λ,l)f(z)zp)β=1+Aw(z)1+Bw(z),
where w(z) is analytic in U with w(0)=0 and |w(z)|<1(z∈U).
By virtue of (3.44), we easily find that(3.45)Ip(δ,λ,l)f(z)=zp(1+Aw(z)1+Bw(z))1/β.
Combining (1.10), (1.16), and (3.45), we have
(3.46)(zp+∑k=n∞(p+λk+lp+l)δzk+p)*f(z)=zp(1+Aw(z)1+Bw(z))1/β.The assertion (3.43) of Theorem 3.10 can now easily be derived from (3.46).Theorem 3.11.
Letf(z)∈ßpα,β(δ,λ,ℓ;n;A,B) with Re(α)>0. Then
(3.47)1z[(1+Biθ)1/β(zp+∑k=n∞(p+λk+lp+l)-δzk+p)*f(z)-zp(1+Aeiθ)1/β]≠0(z∈U;0<θ<2π).Proof.
Suppose thatf(z)∈ßpα,β(δ,λ,ℓ;n;A,B) with Re(α)>0. We know that (3.1) holds true, which implies that
(3.48)(Ip(δ,λ,l)f(z)zp)β≠1+Aeiθ1+Beiθ(z∈U;0<θ<2π).
It is easy to see that the condition (3.48) can be written as follows:
(3.49)1z[Ip(δ,λ,l)f(z)(1+Beiθ)1/β-zp(1+Aiθ)1/β]≠0(z∈U;0<θ<2π).
Combining (1.9), (1.10), and (3.49), we easily get the convolution property (3.47) asserted by Theorem 3.11.Theorem 3.12.
Letα2≥α1≥0 and -1≤B1≤B2<A2<A1≤1. Then
(3.50)ßpα2,β(δ,λ,l;n;A2,B2)⊂ßpα1,β(δ,λ,l;n;A1,B1).Proof.
Suppose thatf(z)∈βpα2,β(δ,λ,ℓ;n;A2,B2). We know that
(3.51)(1-α2)(Ip(δ,λ,l)f(z)zp)β+α2(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β≺1+A2z1+B2z.
Since -1≤B1≤B2<A2≤A1≤1, we easily find that
(3.52)(1-α2)(Ip(δ,λ,l)f(z)zp)β+α2(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β≺1+A2z1+B2z≺1+A1z1+B1z
that is, f(z)∈ßpα2,β(δ,λ,ℓ;n;A1,B1). Thus the assertion of Theorem 3.12 holds for α2=α1≥0.
Ifα2>α1≥0, by Theorem 3.1 and (3.52), we know that f(z)∈ßp0,β(δ,λ,ℓ;n;A1,B1), that is,(3.53)(Ip(δ,λ,l)f(z)zp)β≺1+A1z1+B1z.
At the same time, we have
(3.54)(1-α1)(Ip(δ,λ,l)f(z)zp)β+α1(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β=(1-α1α2)(Ip(δ,λ,l)f(z)zp)β+α1α2[(1-α2)(Ip(δ,λ,l)f(z)zp)β+α2(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β].Moreover, since 0≤(α1/α2)<1 and h1(z):=(1+A1z)/(1+B1z), is analytic and convex in U. Combining (3.52)–(3.54) and Lemma 2.6, we find that
(3.55)(1-α1)(Ip(δ,λ,l)f(z)zp)β+α1(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β≺1+A1z1+B1z,
that is, f(z)∈ßpα1,β(δ,λ,ℓ;n;A1,B1), which implies that the assertion (3.50) of Theorem 3.12 holds.
Letρ denote the class of functions of the following form:(3.56)t(z)=1+∑k=n∞tkzk(n∈N),
which are analytic and convex in U and satisfy the following condition:
(3.57)Re(t(z))>0(z∈U).
By making use of the principle of subordination between analytic functions, we introduce the subclasses Sp,n*(μ,ϕ) and Cp,n(μ,ϕ) of the class Ap(n):
(3.58)Sp,n*(μ,ϕ):={f∈Ap(n):1p-μ(zf′(z)f(z)-μ)≺ϕ(z)(ϕ∈ρ;0≤μ<p;z∈U)},Cp,n(μ;ϕ)={f(z)∈Ap(n):1p-μ(1+zf′′(z)f′(z)-μ)≺ϕ(z)(ϕ∈ρ;0≤μ<p;z∈U)}.
Next, by using the operator defined by (1.7), we define the following two subclasses Sp,n(δ,λ,ℓ;μ;ϕ) and Cp,n(δ,λ,ℓ;μ;ϕ) of the class Ap(n):
(3.59)Sp,n(δ,λ,l;μ;ϕ):={f(z)∈Ap(n):Ip(δ,λ,l)f(z)∈Sp,n*(μ;ϕ)},Cp,n(δ,λ,l;μ;ϕ):={f(z)∈Ap(n):Ip(δ,λ,l)f(z)∈Cp,n(μ;ϕ)}.
Clearly, we know that
(3.60)f∈Cp,n(δ,λ,l;μ;ϕ)⟺zf′(z)p∈Sp,n(δ,λ,l;μ;ϕ).
We now derive some inclusion relationships for the classesSp,n(δ,λ,ℓ;μ;ϕ) and Cp,n(δ,λ,ℓ;μ;ϕ), by similarly applying the method of proof of Proposition 1 obtained by Cho et al. [26] and Wang et al. [27].Theorem 3.13.
Let0≤μ<p,λ>-p, and ϕ∈ρ with
(3.61)Re(ϕ(z))>max{0,-δ+μp-μ,-λ+μ-pp-μ}(z∈U).
Then
(3.62)Sp,n(δ+1,λ,l;μ;ϕ)⊂Sp,n(δ,λ,l;μ;ϕ)⊂Sp,n(δ,λ+1,l;μ;ϕ).Theorem 3.14.
Let0≤μ<p,λ>-p and ϕ∈ρ with (3.61) holds. Then
(3.63)Cp,n(δ+1,λ,l;n;μ;ϕ)⊂Cp,n(δ,λ,l;μ;ϕ)⊂Cp,n(δ,λ+1,l;μ;ϕ).Proof.
By virtue of (3.60) and Theorem 3.13, we observe that
(3.64)f(z)∈Cp,n(δ+1,λ,l;μ;ϕ)⟺Ip(δ+1,λ,l)f(z)∈Cp,n(μ;ϕ)⟺z(Ip(δ+1,λ,l)f(z))′p∈Sp,n*(μ;ϕ)⟺Ip(δ+1,λ,l)(zf′(z)p)∈Sp,n*(μ;ϕ)⟹zf′(z)p∈Sp,n(δ+1,λ,l;μ;ϕ)⟺zf′(z)p∈Sp,n(δ,λ,l;μ;ϕ)⟺Ip(δ,λ,l)(zf′(z)p)∈Sp,n*(μ,ϕ)⟺z(Ip(δ,λ,l)f(z))′p∈Sp,n*(μ;ϕ)⟺Ip(δ,λ,l)f(z)∈Cp,n(μ;ϕ)⟺f(z)∈Cp,n(δ,λ,l;μ;ϕ),f(z)∈Cp,n(δ,λ,l;μ;ϕ)⟺zf′(z)p∈Sp,n(δ,λ,l;μ;ϕ)⟹zf′(z)p∈Sp,n(δ,λ+1,l;μ;ϕ)⟺z(Ip(δ,λ+1,l)f(z))′p∈Sp,n*(μ;ϕ)⟺Ip(δ,λ+1,l)f(z)∈Cp,n(μ;ϕ)⟺f(z)∈Cp,n(δ,λ+1,l;μ;ϕ).
From (3.64), we conclude that the assertion of Theorem 3.14 holds true.Takingϕ(z)=(1+Az)/(1+Bz) in Theorems 3.13 and 3.14, we get the following results.Corollary 3.15.
Let0≤μ<p,λ>-p, and -1≤B<A≤1. Then
(3.65)Sp,n(δ+1,λ,l;μ;1+Az1+Bz)⊂Sp,n(δ,λ,l;μ,1+Az1+Bz)⊂Sp,n(δ,λ+1,l;μ;1+Az1+Bz),Cp,n(δ+1,λ,l;μ;1+Az1+Bz)⊂Cp,n(δ,λ,l;μ,1+Az1+Bz)⊂Cp,n(δ,λ+1,l;μ;1+Az1+Bz).Theorem 3.16.
Letf(z)∈ßpα,β(δ,λ,ℓ;n;A,B) with Re(α)>0 and -1≤B<A≤1. Then
(3.66)(p+l)βλnα∫011-Au1-Buu(p+l)β/λnα-1du<Re(Ip(δ,λ,l)f(z)zp)β<(p+l)βλnα∫011+Au1+Buu(p+l)β/λnα-1du.
The extremal function of (3.66) is defined by
(3.67)Ip(δ,λ,l)Fα,β,A,B(z)=zp((p+l)βλαn∫011+Azu1+Bzuu(p+l)β/λnα-1du)1/β.Proof.
Letf(z)∈ßpα,β(δ,λ,ℓ;n;A,B) with Re(α)>0. From Theorem 3.1, we know that (3.1) holds, which implies that
(3.68)Re(Ip(δ,λ,l)f(z)zp)β<sup⁡z∈URe((p+l)βλαn∫011+Azu1+Bzuu(p+l)β/λnα-1du)≤(p+l)βλnα∫01sup⁡z∈URe(1+Azu1+Bzu)u(p+l)β/λnα-1du<(p+l)βλnα∫011+Au1+Buu(p+l)β/λnα-1du,Re(Ip(δ,λ,l)f(z)zp)β>inf⁡z∈URe((p+l)βλnα∫011+Azu1+Bzuu(p+l)β/λnα-1du)≥(p+l)βλnα∫01inf⁡z∈URe1+Azu1+Bzuu(p+l)β/λnα-1du>(p+l)βλnα∫011-Au1-Buu(p+l)β/λnα-1du.
Combining both equations of (3.68), we get (3.66). By noting that the function Ip(δ,λ,ℓ)Fα,β,A,B(z) defined by (3.67) belongs to the class ßpα,β(δ,λ,ℓ;n;A,B), we obtain that the equality (3.66) is sharp. The proof of Theorem 3.16 is evidently completed.By similarly applying the method of proof of Theorem3.16, we easily get the following result.Corollary 3.17.
Letf(z)∈ßpα,β(δ,λ,ℓ;n;A,B) with Re(α)>0 and -1≤A<B≤1. Then
(3.69)(p+l)βλnα∫011+Au1+Buu(p+l)β/λnα-1du<Re(Ip(δ,λ,l)f(z)zp)β<(p+l)βλnα∫011-Au1-Buu(p+l)β/λnα-1du.
The extremal function of (3.69) is defined by (3.67).In view of Theorem3.16 and Corollary 3.17, we easily derive the following distortion theorems for the class ßpα,β(δ,λ,ℓ;n;A,B).Corollary 3.18.
Letf(z)∈ßpα,β(δ,λ,ℓ;n;A,B) with Re(α)>0 and-1≤B<A≤1. Then for |z|=r<1, we have
(3.70)rp((p+l)βλnα∫011-Aur1-Buru(p+l)β/λnα-1du)1/β<|Ip(δ,λ,l)f(z)|<rp((p+l)βλnα∫011+Aur1+Buru(p+l)β/λnα-1du)1/β.
The extremal function of (3.70) is defined by (3.67).Corollary 3.19.
Letf(z)∈ßpα,β(δ,λ,ℓ;n;A,B) with Re(α)>0 and-1≤A<B≤1. Then for |z|=r<1, we have
(3.71)rp((p+l)βλnα∫011+Aur1+Buru(p+l)β/λnα-1du)1/β<|Ip(δ,λ,l)f(z)|<rp((p+l)βλnα∫011-Aur1-Buru(p+l)β/λnα-1du)1/β.
The extremal function of (3.71) is defined by (3.67).By noting that(3.72)(Re(ν))1/2≤Re(ν1/2)≤|ν|1/2(ν∈C;Re(ν)≥0).
From Theorem 3.16 and Corollary 3.17, we easily get the following results.Corollary 3.20.
Letf(z)∈ßpα,β(δ,λ,ℓ;n;A,B) with Re(α)>0 and -1≤B<A≤1. Then
(3.73)((p+l)βλnα∫011-Au1-Buu(p+l)β/λnα-1du)1/2<Re(Ip(δ,λ,l)f(z)zp)β/2<((p+l)βλnα∫011-Au1-Buu(p+l)β/λnα-1du)1/2.Corollary 3.21.
Letf(z)∈ßpα,β(δ,λ,ℓ;n;A,B) with Re(α)>0 and -1≤A<B≤1. Then
(3.74)((p+l)βλnα∫011+Au1+Buu(p+l)β/λnα-1du)1/2<Re(Ip(δ,λ,l)f(z)zp)β/2<((p+l)βλnα∫011-Au1-Buu(p+l)β/λnα-1du)1/2.Theorem 3.22.
Let(3.75)f(z)=zp+∑k=n∞ap+kzp+k∈ßpα,β(δ,λ,l;n;A,B).
Then(3.76)|ap+n|≤(l+p)δ+1(p+λn+l)δ|A-B(l+p)β+λnα|.
The inequality (3.76) is sharp, with extremal function defined by (3.67).Proof.
Combining  (1.16) and (3.75), we obtain
(3.77)(1-α)(Ip(δ,λ,l)f(z)zp)β+α(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β=1+(1+λnα(p+l)β)(p+λn+lp+l)δβap+nzn⋯≺1+Az1+Bz.
An application of Lemma 2.8 to (3.77) yields
(3.78)|(1+λnα(p+l)β)(p+λn+lp+l)δβap+n|≤|A-B|.
Thus, from (3.78), we easily arrive at (3.76) asserted by Theorem 3.22.Theorem 3.23.
Let0≠α∈ℝ,β∈ℝ,β/α>0,λ>-p and 0≤ρ<1. If f(z)∈ßpα,β(δ,λ,ℓ;n;A,0) with
(3.79)A=An(α,β,λ,ρ,p)=(1-ρ)|α|(1+λnα/(p+l)β)|1-α+ρα|+1+(1+λnα/(p+l)β)2,
then
(3.80)Ip(δ,λ,l)f(z)∈Sp,n*(ρp-(1-ρ)λ).Proof.
Suppose thatf(z)∈ßpα,β(δ,λ,ℓ;n;A,0). By definition, we have
(3.81)(1-α)(Ip(δ,λ,l)f(z)zp)β+α(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))(Ip(δ,λ,l)f(z)zp)β≺1+Az(z∈U).
Let the function P(z) be defined by (3.2). We then find from (3.1) and (3.81) that
(3.82)P(z)≺(p+l)βλnαz-(p+l)β/λnα∫0z(1+At)t(p+l)β/λnα-1dt=1+(l+p)βAλnα+(p+l)βz.
We now suppose that
(3.83)Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z)=(1-ρ)q(z)+ρ(0≤ρ<1;z∈U).
Then q(z)∈H[1,n]. It follows from (3.81) and (3.83) that
(3.84)P(z){(1-α)+α[(1-ρ)q(z)+ρ]}≺1+Az(z∈U).
An application of Lemma 2.9 to (3.84) yields
(3.85)Re(q(z))>0(z∈U).
Combining (3.83) and (3.85), we find that
(3.86)Re(Ip(δ+1,λ,l)f(z)Ip(δ,λ,l)f(z))=(1-ρ)Re(q(z))+ρ>ρ(z∈U).
The assertion of Theorem 3.23 can now easily be derived from (1.8) and (3.86).Theorem 3.24.
Letf(z)∈ßρα,β(δ,λ,ℓ;n;A,0) with β>0,A>0,Re(α)>0, and |α|(n+Re(p+ℓ)β/λα)>A((p+ℓ)β/λ). Then
(3.87)|z(Ip(δ,λ,l)f(z))′Ip(δ,λ,l)f(z)-p|<A(p+l)β[λ|α|(n+Re((p+l)β/λα))+(p+l)β]λ|α|[λ|α|(n+Re((p+l)β/λα))-A(p+l)β].Proof.
Let the functionP(z) be defined by (3.2). It follows from (3.3) that
(3.88)p(z)+λαzp′(z)(p+l)β=1+Aw(z),
where
(3.89)w(z)=∑k=n∞wkzk(n∈N)
is analytic in U with |w(z)|<1(z∈U). From (3.88), we easily get
(3.90)p(z)=1+A(p+l)βλα∫01t(p+l)β/λα-1w(tz)dt=1+A(p+l)βλα∑k=n∞1k+(p+l)β/λαwkzk.
It follows from (3.90) that
(3.91)(zp(z))′=1+A(p+l)βλα∑k=n∞k+1k+(p+l)β/λαwkzk=1+A(p+l)βλα∑k=n∞1k+(p+l)β/λαwkzk+A(l+p)βλα(w(z)-(p+l)βλα∫01t(l+p)/λα-1w(tz)dt).
We next find from (3.90) and (3.91) that
(3.92)zp′(z)=A(p+l)βλα(w(z)-(p+l)βλα∫01t(l+p)β/λα-1w(tz)dt)=A(p+l)βλα∑k=n∞kk+(p+l)β/λαwkzk.
Now from (3.88) we get(3.93)|zp′(z)p(z)|<[(n+Re((p+l)β/λα))+(p+l)β/λ|α|]|p(z)|,
and from (3.90) we get
(3.94)|p(z)|=|1+A(p+l)βλα∑k=n∞1k+(p+l)β/λαwkzk|>[λ|α|(n+Re((p+l)β/λα))-A(p+l)β]-λ|α|(n+Re((p+l)β/λα)),
from (3.93) and (3.94), we get
(3.95)|zp′(z)p(z)|<A(p+l)β[λ|α|(n+Re((p+l)β/λα))+(p+l)β]λ|α|[λ|α|(n+Re((p+l)β/λα))-A(p+l)β].
Thus, from (3.2) and (3.95), we easily arrive at the assertion of Theorem 3.24.



---

*Source: 752341-2011-07-12.xml*