# Generalizedk-Uniformly Close-to-Convex Functions Associated with Conic Regions

Authors: Khalida Inayat Noor
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We define and study some subclasses of analytic functions by using a certain multiplier transformation. These functions map the open unit disc onto the domains formed by parabolic and hyperbolic regions and extend the concept of uniformly close-to-convexity. Some interesting properties of these classes, which include inclusion results, coefficient problems, and invariance under certain integral operators, are discussed. The results are shown to be the best possible.


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## Body

## 1. Introduction

LetA denote the class of analytic functions f defined in the unit disc E={z:|z|<1} and satisfying the condition f(0)=0,f′(0)=1. Let S,S*(γ),C(γ) and K(γ) be the subclasses of A consisting of functions which are univalent, starlike of order γ, convex of order γ, and close-to-convex of order γ, respectively, 0≤γ<1. Let S*(0)=S*,C(0)=C and K(0)=K.For analytic functionsf(z)=∑n=0∞anzn and g(z)=∑n=0∞bnzn, by f*g we denote the convolution (Hadamard product) of f and g, defined by(1.1)(f*g)(z)=∑n=0∞anbnzn.We say that a functionf∈A is subordinate to a function F∈A and write f(z)≺F(z) if and only if there exists an analytic function w(z),w(0)=0,|w(z)|<1 for z∈E such that f(z)=F(w(z)),z∈E.IfF is univalent in E, then(1.2)f(z)≺F(z)⟺f(0)=F(0),f(E)⊂F(E).Fork∈[0,1], define the domain Ωk as follows, see [1]:(1.3)Ωk={u+iv:u>k(u-1)2+v2}.
For fixed k,Ωk represents the conic region bounded, successively, by the imaginary axis (k=0), the right branch of hyperbola (0<k<1), a parabola (k=1).Related withΩk, the domain Ωk,γ is defined in [2] as follows:(1.4)Ωk,γ=(1-γ)Ωk+γ,(0≤γ<1).The functions which play the role of extremal functions for the conic regionsΩk,γ are denoted by pk,γ(z) with pk,γ(0)=1, and pk,γ′(0)>0 are univalent, map E onto Ωk,γ, and are given as(1.5)pk,γ(z)={1+(1-2γ)z(1-z),k=0,1+2(1-γ)π2(log⁡1+z1-z)2,k=1,1+2(1-γ)1-k2sinh⁡2⁡[(2πarccos⁡⁡k)arctanhz],(0<k<1).It has been shown [3, 4] that pk,γ(z) is continuous as regards to k and has real coefficients for all k∈[0,1].LetP(pk,γ) be the class of functions p(z) which are analytic in E with p(0)=1 such that p(z)≺pk,γ(z) for z∈E. It can easily be seen that P(pk,γ)⊂P, where P is the class of Caratheodory functions of positive real part.The classPm(pk,γ) is defined in [5] as follows.Letp(z) be analytic in E with p(0)=1. Then p∈Pm(pk,γ) if and only if, for m≥2,0≤γ<1,k∈[0,1],z∈E,(1.6)p(z)=(m4+12)p1(z)-(m4-12)p2(z),p1,p2∈P(pk,γ).Fork=0,γ=0, the class Pm(p0,0) coincides with the class Pm introduced by Pinchuk in [6]. Also P2=P.The generalized Harwitz-Lerch Zeta function [7] ϕ(z,λ,μ) is given as(1.7)ϕ(z,λ,μ)=∑n=0∞zn(μ+n)λ,(λ∈C,μ∈C∖Z-={-1,-2,…}).Using (1.7), the following family of linear operators, see [7–9], is defined in terms of the Hadamard product as(1.8)Jλ,μf(z)=Hλ,μ(z)*f(z),
where f∈A,(1.9)Hλ,μ(z)=(1+μ)λ[ϕ(z,λ,μ)-μ-λ],(z∈E),andϕ(z,λ,μ) is given by (1.7).From (1.7) and (1.8), we can write(1.10)Jλ,μf(z)=z+∑n=2∞(1+μn+μ)λanzn.For the different permissible values of parametersλ and μ, the operator Jλ,μ has been studied in [3, 4, 7, 10–12].We observe some special cases of the operator (1.10) as given below(i)J0,μf(z)=f(z),(ii)J1,0f(z)=z+∑n=2∞(an/n)zn=∫0z(f(t)/t)dt,(iii)J1,μf(z)=z+∑n=2∞((1+μ)/(n+μ))anzn=((1+μ)/zμ)∫0ztμ-1f(t)dt,(μ>-1).We remark thatJ1,1f(z) is the well-known Libera operator and J1,μf(z) is the generalized Bernardi operator, see [13, 14]. Also Jλ,1f(z)=Lλf(z) represents the operator closely related to the multiplier transformation studied by Flett [3].We define the operatorIλ,μ:A→A as(1.11)Iλ,μf(z)*Jλ,μf(z)=z(1-z),(λreal,μ>-1),see [15]. This gives us(1.12)Iλ,μf(z)=z+∑n=2∞(n+μ1+μ)λanzn,(λreal,μ>-1).From (1.12), the following identity can easily be verified(1.13)z(Iλ,μf(z))′=(μ+1)Iλ+1,μf(z)-μIλ,μf(z).Remark 1.1.
(i)
Fork∈(0,1), we note that the domain Ωk given by (1.3) represents the following hyperbolic region:(1.14)(u+k21-k2)2-k21-k2v2>(k1-k2)2,u>kk+1.
The extremal functionpk,γ(z), for 0<k<1, can be written as(1.15)pk,γ(z),=(1-γ)pk(z)+γ,where pk(z), in a simplified form, is given below
(1.16)pk(z)=1+12sin2σ{(1+z1-z)2σ/π+(1-z1+z)2σ/π-2},=1+12sin2σ∑n=1∞(∑j=02n(-1)j(2σ/πj)(-2σ/π2n-j))zn,=1+8(σπsin⁡σ)2z+⋯,(z∈E,σ=arccos⁡k),
and the branch of z is chosen such that Im⁡z≥0.
It is easy to see that, forh∈P(pk,γ),Reh(z)>(k+γ)/(k+1),k∈(0,1). That is(1.17)P(pk,γ)⊂P(k+γ1+k),and the order (k+γ)/(1+k) is sharp with the extremal function p(z)=(1-γ)pk(z)+γ, where pk(z) is given by (1.16).(ii)
Fork=1, the extremal function(1.18)p1(z)=1+2π2(log⁡1+z1-z)2,=1+8π2∑n=1∞(1n∑j=0n-112j+1)zn,=1+8π2(z+23z2+2345z3+44105z4+⋯),(z∈E),
maps E conformally onto the parabolic region Ω1={u+iv:u>(v2+1)/2}.
It can easily be verified thatRep1(z)>1/2 and, in this case, the order 1/2 is sharp.We now define the following.Definition 1.2.
Letf∈A and let the operator Iλ,μf be defined by (1.12). Then f∈k-∪Rmγ(λ,μ) for m≥2,k∈[0,1] and γ∈[0,1) if and only if
(1.19){z(Iλ,μf(z))′Iλ,μg(z)}∈Pm(pk,γ),z∈E.
We note the following.
(i)
Form=2,k=0, and λ=0, the class k-∪Rmγ(λ,μ) reduces to S*(γ), and λ=0 gives us the        class k-∪ST of uniformly starlike functions, see [2, 16].(ii)
0-∪Rm0(0,μ)=Rm is the class of functions of bounded radius rotation, see [13, 14].(iii)
We denotek-∪Rmγ(0,μ) as k-∪Rmγ, see [5].(iv)
Letm=2. Then f∈k-∪R2γ(λ,μ) implies that(1.20)Re{z(Iλ,μf(z))′Iλ,μf(z)}>k|z(Iλ,μf(z))′Iλ,μf(z)-1|+γ,
and we note that, for 0≤k2<k1, k1-∪R2γ(λ,μ)⊂k2-∪R2γ(λ,μ).Definition 1.3.
Letf∈A. Then f∈k-∪Tmγ(λ,μ) if and only if there exists g∈k-∪R2γ(λ,μ) such that(1.21){z(Iλ,μf(z))′Iλ,μg(z)}∈Pm(pk,γ)inE.Special Cases.
(i)
0-∪T2γ(0,μ)=K(γ).(ii)
Fork=γ=λ=0, we obtain the class Tm introduced and discussed in [17].(iii)
When we takem=2 and λ=0, then k-∪T2γ(0,μ)=k-∪Kγ, the class of uniformly close-to-convex functions, see [2].


## 2. Preliminary Results

We need the following results in our investigation.Lemma 2.1 (see [18]).
Letq(z) be convex in E and j:E→ℂ with Re[j(z)]>0,z∈E. If p(z), analytic in E with p(0)=1, satisfies
(2.1)(p(z)+j(z)zp′(z))≺q(z),
then
(2.2)p(z)≺q(z).In the following, one gives an easy extension of a result proved in [1].Lemma 2.2 (see [5]).
Letk≥0 and let β,δ be any complex numbers with β≠0 and Re((βk/(k+1))+δ)>γ. If h(z) is analytic in E,h(0)=1 and satisfies
(2.3)(h(z)+zh′(z)βh(z)+δ)≺pk,γ(z),
and qk,γ(z) is an analytic solution of
(2.4)qk,γ(z)+zqk,γ′(z)βqk,γ(z)+δ=pk,γ(z),then qk,γ(z) is univalent,
(2.5)h(z)≺qk,γ(z)≺pk,γ(z),
and qk,γ(z) is the best dominant of (2.3).Lemma 2.3 (see [19]).
Iff∈ℂ,g∈S*, then for each h analytic in E with h(0)=1,
(2.6)(f*hg)(E)(f*g)(E)⊂Co¯h(E),where Co¯h(E) denotes the convex hull of h(E).Lemma 2.4 (see [18]).
Letu=u1+iu2,v=v1+iv2 and let ψ(u,v) be a complex-valued function satisfying the conditions:(i)
ψ(u,v) is continuous in a domain D⊂ℂ2,(ii)
(1,0)∈D and Reψ(1,0)>0,(iii)
Reψ(iu2,v1)≤0, whenever (iu2,v1)∈D and v1≤-(1/2)(1+u22).If h(z)=1+c1z+c2z2+⋯ is a function analytic in E such that (h(z),zh′(z))∈D and Reψ(h(z),zh′(z))>0 for z∈E, then Reh(z)>0 in E.Lemma 2.5 (see [20]).
Leth∈Pm(ρ),0≤ρ<1. Then, with =reiθ,z∈E, one has(i)
(1/2π)∫02π|h(reiθ)|2dθ≤(1-[m2(1-ρ)2-1]r2)/(1-r2),(ii)
(1/2π)∫02π|h′(reiθ)|dθ≤m(1-ρ)/(1-r2).Lemma 2.6 (see [5]).
Letf∈k-∪Rmγ. Then there exist s1,s2∈k-∪R2γ such that(2.7)f(z)=(s1(z))(m+2)/4(s2(z))(m-2)/4,k≥0,m≥2,z∈E.Lemma 2.7.
Letp∈Pm(pk,γ) and p(z)=1+∑n=1∞cnzn. Then
(2.8)|cn|≤m2|δk,γ|,n≥1,
where
(2.9)δk,γ={8(1-γ)(cos⁡-1⁡k)2π2(1-k2),0≤k<1,8(1-γ)π2,k=1.Proof.
Letp(z)=(m/4+1/2)p1(z)-(m/4-1/2)p2(z). Then,
(2.10)pi(z)≺pk,γ(z)=1+δk,γz+⋯,i=1,2.
Now the proof follows immediately by using the well-known Rogosinski’s result, see [21].


## 3. Main Results

We shall assume throughout, unless stated otherwise, thatk∈[0,1],m≥0,0≤γ<1,λ∈ℂ,μ>-1 and z∈E.Theorem 3.1.
Letf∈k-∪Rmγ(λ,μ). Let, for α,β>0,
(3.1)F(z)=[(1+β)z-β∫0ztβ-1′fα(t)dt]1/α.
Then, F∈k-∪Rmγ(λ,μ) in E.Proof.
Set(3.2)z(Iλ,μF(z))′Iλ,μF(z)=H(z)=(m4+12)H1(z)-(m4-12)H2(z).
We note H(z) is analytic in E with H(0)=1.
From (3.1), we have(3.3){zβ(Iλ,μF(z))α}′=zβ-1(Iλ,μf(z))α.
That is(3.4)(Iλ,μF(z))α[β+αH(z)]=(Iλ,μf(z))α.
Logarithmic differentiation of (3.4) and simple computations give us(3.5)H(z)+zH′(z)αH(z)+β=z(Iλ,μf(z))′Iλ,μf(z)∈Pm(pk,γ).
Define(3.6)ϕa,b(z)=11+bz(1-z)a+1+bb+1z(1-z)a+2,(a>0,b≥0),
then, with a=1/α,b=β/α, we have
(3.7)H(z)*(ϕa,b(z)z)={H(z)+zH′(z)αH(z)+β}.
From (3.2), (3.5), and (3.7), it follows that(3.8){Hi(z)+zHi′(z)αHi(z)+β}∈P(pk,γ),z∈E,i=1,2.
On applying Lemma2.2, we obtain(3.9)Hi(z)≺qk,γ(z)≺pk,γ(z)inE,where qk,γ(z) is the best dominant and is given as
(3.10)qk,γ(z)=[α∫01(tβ+α-1exp⁡∫ztzpk,γ(u)-1udu)αdt]-1-βα.
Consequently it follows, from (3.2), that H∈Pm(pk,γ) and F∈k-∪Tmγ(λ,μ) in E.Fork=0,γ=0, we have the following special case.Corollary 3.2.
Letf∈0-∪Rm0(λ,μ)=Rm(λ,μ) and let F(z) be defined by (3.1). Then, F∈Rmγ1(λ,μ), where(3.11)γ1=2{(1+2β)+(1+2β)2+8α}.Proof.
We write(3.12)z(Iλ,μF(z))′Iλ,μF(z)=(1-γ1)H(z)+γ1=(m4+12){(1-γ1)H1(z)+γ1}-(m4-12){(1-γ1)H2(z)+γ1},
and proceeding as in Theorem 3.1, we obtain
(3.13)z(Iλ,μf(z))′Iλ,μf(z)=(m4+12)[(1-γ1){H1(z)+αzH1′(z)αH1(z)+(αγ1+β)/(1-γ1)}+γ1]-(m4-12)[(1-γ1){H2(z)+αzH2′(z)αH2(z)+(αγ1+β)/(1-γ1)}+γ1].
We construct the functionalψ(u,v) by taking u=Hi(z),  v=zHi′(z), as(3.14)ψ(u,v)=u+vαu+(αγ1+β)/(1-γ1)+γ11-γ1.
The first two conditions of Lemma2.4 can easily be verified. For condition (iii), we proceed as follows:(3.15)Reψ(iu2,v1)=γ11-γ1+Re(1-γ1)v1αγ1+β+iα(1-γ1)u2,=11-γ1[γ1+(1-γ1)(αγ1+β)v1(αγ1+β)2+α2(1-γ1)2u22],≤11-γ1[γ1-(1-γ1)(αγ1+β)(1+u22)2{(αγ1+β)2+α2(1-γ1)2u22}],=11-γ1[A+Bu222C],(v1≤-(1+u22)2),
where A=2γ1(αγ1+β)2-(1-γ1)(αγ1+β),B=2γ1α2(1-γ1)2-(1-γ1)(αγ1+β),C={(αγ1+β)2+α2(1-γ1)2u22}>0.
The right-hand side of (3.15) is less than equal to zero when A≤0 and B≤0. From A≤0, we obtain γ1 as given by (3.11), and B≤0 ensures that γ1∈[0,1).
This shows that all the conditions of Lemma2.4 are satisfied and therefore ReHi(z)>0. This implies H∈Pm and consequently Iλ,μF∈Rmγ1. That is F∈Rmγ1(λ,μ) as required.By takingα=1,β=0,λ=0, and m=2, we obtain a well-known result that every convex function is starlike of order 1/2. Also, for β=1,λ=0,α=1, and m=2, we obtain from (3.1) the Libera operator and in this case we obtain a known result with γ1=2/(3+17) for starlike functions, see [18].Assigning permissible values to different parameters, we obtain several new and known results from Theorem3.1 and Corollary 3.2.Theorem 3.3.
Letf∈k-∪Tmγ(λ,μ) and let F(z) be defined by (3.1). Then F∈k-∪Tmγ(λ,μ).Proof.
We can write (3.1) as
(3.16)Iλ,μF(z)=[(1+β)z-β∫0ztβ-1′(Iλ,μf(t))αdt]1/α,f∈k-∪Tmγ(λ,μ),=[(Iλ,μf(z)z)α*hα,β(z)z]1/α,where
(3.17)hα,β(z)=∑n=1∞znn+α+β,
is convex in E.
Letf∈k-∪Tmγ(λ,μ). Then there exists some g∈k-∪R2γ(λ,μ) such that(3.18){z(Iλ,μf(z))′Iλ,μg(z)}∈Pm(pk,γ).(3.19)G(z)=[(β+1)z-β∫0ztβ-1gα(t)dt]1/α.
From Theorem3.1, it follows that G∈k-∪R2γ(λ,μ). We can write (3.19) as(3.20)Iλ,μG(z)=z[Iλ,μg(z)z*hα,β(z)z]1/α,
where hα,β(z), given by (3.17), is convex in E.
Sinceg∈k-∪R2γ(λ,μ), so Iλ,μg∈S*((k+γ)/(k+1))⊂S*. It can easily be shown that z(Iλ,μg/z)α and z(Iλ,μG/z)α are in the class S*.
Now, from (3.1), we have(3.21)z(Iλ,μF(z))′(Iλ,μF(z))α-1(Iλ,μG(z))α=hα,β(z)*z(Iλ,μg(z)/z)α(z(Iλ,μf(z))′(Iλ,μf(z))α-1/(Iλ,μg(z))α)hα,β(z)*z(Iλ,μg(z)/z)α=(m4+12)[hα,β(z)*z(Iλ,μg(z)/z)αh1(z)]hα,β(z)*z(Iλ,μg(z)/z)α-(m4-12)[hα,β(z)*z(Iλ,μg(z)/z)αh2(z)]hα,β(z)*z(Iλ,μg(z)/z)α.
We use Lemma2.3 with hi≺pk,γ,i=1,2, to have(3.22){hα,β(z)*z(Iλ,μg(z)/z)αhi(z)hα,β(z)*z(Iλ,μg(z)/z)α}≺pk,γ(z)inE.
Thus from (3.19), (3.21), and (3.22) we obtain the required result that F∈k-∪Tmγ(λ,μ). This completes the proof.As a special case we note that, forλ=0=k, the subclass Tmγ⊂Tm is invariant under the integral operator defined by (3.1).Theorem 3.4.
One has(3.23)k-∪Rmγ(λ+1,μ)⊂k-∪Rmγ(λ,μ).Proof.
Letf∈k-∪Rmγ(λ+1,μ) and let
(3.24)z(Iλ,μf(z))′Iλ,μf(z)=H(z),
where H(z) is analytic in E and is defined by (3.2).
Then, from (1.13), we have(3.25)z(Iλ+1,μf(z))′Iλ+1,μf(z)={H(z)+zH′(z)H(z)+μ}∈Pm(pk,γ).
Applying similar technique used before, we have from (3.2) and (3.7) for i=1,2(3.26){Hi(z)+zHi′(z)Hi(z)+μ}≺pk,γ.
Thus, using Lemma2.2, it follows that Hi≺pk,γ,i=1,2 and z∈E, consequently H∈Pm(pk,γ) in E and this completes the proof.As special cases, we have the following.(i)
Letm=2,λ≥0. Then, from Theorem 3.4, it easily follows that(3.27)k-∪R2γ(λ,μ)⊂k-∪R2γ(0,μ)⊂S*(k+γ1+k)⊂S*.(ii)
Letk=0 and λ≥0. Then f∈0-∪Rmγ(λ,μ) implies f∈Rmγ⊂Rm, that is, f(z) is a function of bounded radius rotation in E.Theorem 3.5.
One has(3.28)k-∪Tmγ(λ+1,μ)⊂k-∪Tmγ(λ,μ),μ,λ≥0.Proof.
Letf∈k-∪Tmγ(λ+1,μ). Then, for z∈E,
(3.29)z(Iλ+1,μf(z))′Iλ+1,μg(z)∈Pm(pk,γ),for some g∈k-∪R2γ(λ+1,μ).
We define an analytic functionh(z) in E such that(3.30)z(Iλ,μf(z))′Iλ,μg(z)=h(z)=(m4+12)h1(z)-(m4-12)h2(z),
where h(0)=1. We shall show that h∈Pm(pk,γ) in E.
Sinceg∈k-∪R2γ(λ+1,μ) and k-∪R2γ(λ+1,μ)⊂k-∪R2γ(λ,μ), we have(3.31)z(Iλ,μg(z))′Iλ,μg(z)=h0(z),h0∈P(pk,γ),z∈E.
Now, on using (1.13), we have(3.32)z(Iλ+1,μf(z))′Iλ+1,μg(z)=(1/(μ+1))z[Iλ,μ(zf′(z))′]+(μ/(μ+1))[Iλ,μzf′(z)](1/(μ+1))z(Iλ,μg(z))′+(μ/(μ+1))Iλ,μg(z),=[(z[Iλ,μ(zf′(z))′]/Iλ,μg(z))+μh(z)]h0(z)+μ.
Differentiation of (3.30) gives us(3.33)z(z(Iλ,μf(z))′)′Iλ,μg(z)=zh′(z)+(h(z))(h0(z)),
and using (3.33) in (3.32), we obtain
(3.34)z(Iλ+1,μf(z))′Iλ+1,μg(z)=h(z)+zh′(z)h0(z)+μ=(m4+12)[h1(z)+zh1′(z)h0(z)+μ]-(m4-12)[h2(z)+zh2′(z)h0(z)+μ].
Sincef∈k-∪Tmγ(λ+1,μ), we have with 1/H0(z)={h0(z)+μ}∈P,(3.35){hi(z)+H0(z)[zhi′(z)]}≺pk,γ(z)inE,
and thus, applying Lemma 2.1, we have hi(z)≺pk,γ(z) in E. This shows h∈Pm(pk,γ) in E and consequently f∈k-∪Tmγ(λ,μ).As a special case, we note thatf∈k-∪T2γ(λ,μ) is a close-to-convex function for z∈E.Theorem 3.6.
Letf∈k-∪R2γ(λ,μ) and let ϕ(z) be convex in E. Then (f*ϕ)∈k-∪R2γ(λ,μ) for z∈E.Proof.
We have(3.36)z(Iλ,μ(ϕ*f))′Iλ,μ(ϕ*f)=ϕ*z[Iλ,μf]′ϕ*Iλ,μf,=ϕ*[z(Iλ,μf)′/Iλ,μf]Iλ,μfϕ*Iλ,μf.
Nowk-∪R2γ(λ,μ)⊂S*((k+γ)/(1+k))⊂S*, and ϕ is a convex in E, we use Lemma 2.3 to (3.36) and conclude that (ϕ*f)∈k-∪R2γ(λ,μ) for z∈E. This completes the proof.Remark 3.7.
Following the similar technique, we can easily extend Theorem3.6 to the class k-∪Tmγ(λ,μ), that is, k-∪Tmγ(λ,μ), is invariant under convolution with convex function.


### 3.1. Applications of Theorem3.6

The classesk-∪R2γ(λ,μ) and k-∪Tmγ(λ,μ) are preserved under the following integral operators:(1)f1(z)=∫0z(f(t)/t)dt=(ϕ1*f)(z), where ϕ1(z)=-log⁡(1-z),(2)f2(z)=(2/z)∫0zf(t)dt=(ϕ2*f)(z), where ϕ2(z)=-2[z-log⁡(1-z)]/z,(3)f3(z)=∫0z((f(t)-f(tx))/(t-tx))dt=(ϕ3*f)(z),|x|≤1,x≠1, where ϕ3(z)=(1/(1-x))log⁡((1-xz)/(1-z)),|x|≤1,x≠1,(4)f4(z)=((1+c)/zc)∫0ztc-1f(t)dt=(ϕ4*f)(z),Rec>0, where ϕ4(z)=∑n=1∞((1+c)/(n+c))zn,Rec>0,The proof is immediate sinceϕi(z) is convex in E for i=1,2,3,4.With essentially the same method together with Lemma2.7, we can easily prove the following sharp coefficient results.Theorem 3.8.
Letf∈k-∪Rmγ(λ,m) and let it be given by
(3.37)f(z)=z+∑n=2∞anzn.
Then
(3.38)|an|≤m2[(n-1)!][(1+μn+μ)λ(|δk,γ|)n-1],(n≥2),
where (ρ)n is Pochhamer symbol defined, in terms of Gamma function Γ, by
(3.39)(ρ)n=Γ(n+ρ)Γ(ρ)={1,n=0,ρ(ρ+1)(ρ+2)⋯(ρ+n-1),n∈N,
and δk,γ is as given by (2.9).
As special case, one notes that
(i)
λ=0,m=2, then one has(3.40)|an|≤(|δk,γ|)n-1(n-1)!,
see [2].(ii)
Letλ=0,n=2. Then,(3.41)|an|≤m2|δk,γ|.
This coefficient bound is well known form=2, see [2].Using Theorem3.8 with m=2, the following result can easily be proved.Theorem 3.9.
Letf:f(z)=z+∑n=2∞anzn∈k-∪Tmγ(λ,μ). Then, for n≥2(3.42)|an|≤(1+μn+μ)λ[(|δk,γ|)n-1n!+m|δk,γ|2n∑j=1n-1(|δk,γ|)j-1(j-1)!],where δk,γ is as given by (2.9).By assigning different permissible values to the parameters, we obtain several known results, see [2, 22].We now prove the following.Theorem 3.10.
Letf:f(z)=z+∑n=2∞anzn∈k-∪Rmγ(λ,μ) with (m+2)(1-γ)/2(1+k)>1. Then, for n≥1,
(3.43)||an+1|-|an||≤c1(m,γ,k,λ,μ)n[{((m/2)+1)((1-γ)/(1+k))}-(2+λ)],
where c1(m,γ,k,λ,μ) is a constant.Proof.
LetF(z)=Iλ,μf(z)=z+∑n=2∞Anzn,An=((n+μ)/(1+μ))λan. Since f∈k-∪Rmγ(λ,μ), we can write
(3.44)zF′(z)=F(z)h(z),h∈Pm(pk,γ).
That is(3.45)z(zF′(z))′=F(z)[h2(z)+zh′(z)].
Now,F∈k-∪Rmγ⊂Rm((k+γ)/(1+k)), and it follows from a result proved in [5] that there exist s1,s2∈S*((k+γ)/(k+1)) such that(3.46)F(z)=(s1(z))(m+2)/4(s2(z))(m-2)/4,m≥2,=((g1(z))(1-(k+γ)/(k+1)))(m+2)/4((g2(z))(1-(k+γ)/(k+1)))(m-2)/4,g1,g2∈S*,=(g1(z))((1-γ)/(k+1))((m+2)/4)(g2(z))((1-γ)/(k+1))((m-2)/4).Thus, for ξ∈E and n≥1,
(3.47)|(n+1)2ξAn+1-n2An|≤12πrn+2∫02π|z-ξ||z(zF′(z))′|dθ=12πrn+2∫02π|z-ξ||F(z)||h2(z)+zh′(z)|dθ=12πrn+2∫02π|z-ξ||(g1(z))(m/4+1/2)(g2(z))(m/4-1/2)|((1-γ)/(k+1))|h2(z)+zh′(z)|dθ,
where g1,g2∈S* and h∈Pm(pk,γ) in E.
Let0<r<1. Then, by a result [23], there exists a ξ with |ξ|=r such that, for z,|z|=r(3.48)|z-ξ||g1(z)|≤2r21-r2.
We now use (3.48), distortion theorems for starlike functions g1,g2 for ((m/2)+1)((1-γ)/(1+k))>1, and Lemma 2.5 with ρ=(k+ρ)/(k+1),r=(1-1/n),n→∞ and obtain from (3.47),
(3.49)|(n+1)2ξ(n+μ+11+μ)λan+1-n2(n+μ1+μ)λan|≤C(m,γ,k)n{((m/2)+1)((1-γ)/(1+k))}.
From (3.48), we easily obtain the required result given by (3.43), (n→∞).
This completes the proof.Using the similar technique, we can easily prove the following.Theorem 3.11.
Letf:f(z)=z+∑n=2∞anzn∈k-∪Tmγ(λ,μ). Then
(3.50)an=O(1)n{2((1-γ)/(1+k))-λ-1},(n⟶∞),
where O(1) is a constant depending on γ,k,m,μ, and λ only. The exponent {2((1-γ)/(1+k))-λ-1} is best possible.


## 3.1. Applications of Theorem3.6

The classesk-∪R2γ(λ,μ) and k-∪Tmγ(λ,μ) are preserved under the following integral operators:(1)f1(z)=∫0z(f(t)/t)dt=(ϕ1*f)(z), where ϕ1(z)=-log⁡(1-z),(2)f2(z)=(2/z)∫0zf(t)dt=(ϕ2*f)(z), where ϕ2(z)=-2[z-log⁡(1-z)]/z,(3)f3(z)=∫0z((f(t)-f(tx))/(t-tx))dt=(ϕ3*f)(z),|x|≤1,x≠1, where ϕ3(z)=(1/(1-x))log⁡((1-xz)/(1-z)),|x|≤1,x≠1,(4)f4(z)=((1+c)/zc)∫0ztc-1f(t)dt=(ϕ4*f)(z),Rec>0, where ϕ4(z)=∑n=1∞((1+c)/(n+c))zn,Rec>0,The proof is immediate sinceϕi(z) is convex in E for i=1,2,3,4.With essentially the same method together with Lemma2.7, we can easily prove the following sharp coefficient results.Theorem 3.8.
Letf∈k-∪Rmγ(λ,m) and let it be given by
(3.37)f(z)=z+∑n=2∞anzn.
Then
(3.38)|an|≤m2[(n-1)!][(1+μn+μ)λ(|δk,γ|)n-1],(n≥2),
where (ρ)n is Pochhamer symbol defined, in terms of Gamma function Γ, by
(3.39)(ρ)n=Γ(n+ρ)Γ(ρ)={1,n=0,ρ(ρ+1)(ρ+2)⋯(ρ+n-1),n∈N,
and δk,γ is as given by (2.9).
As special case, one notes that
(i)
λ=0,m=2, then one has(3.40)|an|≤(|δk,γ|)n-1(n-1)!,
see [2].(ii)
Letλ=0,n=2. Then,(3.41)|an|≤m2|δk,γ|.
This coefficient bound is well known form=2, see [2].Using Theorem3.8 with m=2, the following result can easily be proved.Theorem 3.9.
Letf:f(z)=z+∑n=2∞anzn∈k-∪Tmγ(λ,μ). Then, for n≥2(3.42)|an|≤(1+μn+μ)λ[(|δk,γ|)n-1n!+m|δk,γ|2n∑j=1n-1(|δk,γ|)j-1(j-1)!],where δk,γ is as given by (2.9).By assigning different permissible values to the parameters, we obtain several known results, see [2, 22].We now prove the following.Theorem 3.10.
Letf:f(z)=z+∑n=2∞anzn∈k-∪Rmγ(λ,μ) with (m+2)(1-γ)/2(1+k)>1. Then, for n≥1,
(3.43)||an+1|-|an||≤c1(m,γ,k,λ,μ)n[{((m/2)+1)((1-γ)/(1+k))}-(2+λ)],
where c1(m,γ,k,λ,μ) is a constant.Proof.
LetF(z)=Iλ,μf(z)=z+∑n=2∞Anzn,An=((n+μ)/(1+μ))λan. Since f∈k-∪Rmγ(λ,μ), we can write
(3.44)zF′(z)=F(z)h(z),h∈Pm(pk,γ).
That is(3.45)z(zF′(z))′=F(z)[h2(z)+zh′(z)].
Now,F∈k-∪Rmγ⊂Rm((k+γ)/(1+k)), and it follows from a result proved in [5] that there exist s1,s2∈S*((k+γ)/(k+1)) such that(3.46)F(z)=(s1(z))(m+2)/4(s2(z))(m-2)/4,m≥2,=((g1(z))(1-(k+γ)/(k+1)))(m+2)/4((g2(z))(1-(k+γ)/(k+1)))(m-2)/4,g1,g2∈S*,=(g1(z))((1-γ)/(k+1))((m+2)/4)(g2(z))((1-γ)/(k+1))((m-2)/4).Thus, for ξ∈E and n≥1,
(3.47)|(n+1)2ξAn+1-n2An|≤12πrn+2∫02π|z-ξ||z(zF′(z))′|dθ=12πrn+2∫02π|z-ξ||F(z)||h2(z)+zh′(z)|dθ=12πrn+2∫02π|z-ξ||(g1(z))(m/4+1/2)(g2(z))(m/4-1/2)|((1-γ)/(k+1))|h2(z)+zh′(z)|dθ,
where g1,g2∈S* and h∈Pm(pk,γ) in E.
Let0<r<1. Then, by a result [23], there exists a ξ with |ξ|=r such that, for z,|z|=r(3.48)|z-ξ||g1(z)|≤2r21-r2.
We now use (3.48), distortion theorems for starlike functions g1,g2 for ((m/2)+1)((1-γ)/(1+k))>1, and Lemma 2.5 with ρ=(k+ρ)/(k+1),r=(1-1/n),n→∞ and obtain from (3.47),
(3.49)|(n+1)2ξ(n+μ+11+μ)λan+1-n2(n+μ1+μ)λan|≤C(m,γ,k)n{((m/2)+1)((1-γ)/(1+k))}.
From (3.48), we easily obtain the required result given by (3.43), (n→∞).
This completes the proof.Using the similar technique, we can easily prove the following.Theorem 3.11.
Letf:f(z)=z+∑n=2∞anzn∈k-∪Tmγ(λ,μ). Then
(3.50)an=O(1)n{2((1-γ)/(1+k))-λ-1},(n⟶∞),
where O(1) is a constant depending on γ,k,m,μ, and λ only. The exponent {2((1-γ)/(1+k))-λ-1} is best possible.



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*Source: 274985-2012-04-03.xml*