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.
1. Introduction
Let A 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 functions f(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(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.
If F is univalent in E, thenf(z)≺F(z)⟺f(0)=F(0),f(E)⊂F(E).
For k∈[0,1], define the domain Ωk as follows, see [1]: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 with Ωk, the domain Ωk,γ is defined in [2] as follows:Ωk,γ=(1-γ)Ωk+γ,(0≤γ<1).
The functions which play the role of extremal functions for the conic regions Ωk,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 aspk,γ(z)={1+(1-2γ)z(1-z),k=0,1+2(1-γ)π2(log1+z1-z)2,k=1,1+2(1-γ)1-k2sinh2[(2πarccosk)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].
Let P(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 class Pm(pk,γ) is defined in [5] as follows.
Let p(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,p(z)=(m4+12)p1(z)-(m4-12)p2(z),p1,p2∈P(pk,γ).
For k=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 asϕ(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 asJλ,μf(z)=Hλ,μ(z)*f(z),
where f∈A,Hλ,μ(z)=(1+μ)λ[ϕ(z,λ,μ)-μ-λ],(z∈E),
and ϕ(z,λ,μ) is given by (1.7).
From (1.7) and (1.8), we can writeJλ,μf(z)=z+∑n=2∞(1+μn+μ)λanzn.
For the different permissible values of parameters λ 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
We remark that J1,1f(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 operator Iλ,μ:A→A asIλ,μf(z)*Jλ,μf(z)=z(1-z),(λreal,μ>-1),see [15]. This gives usIλ,μf(z)=z+∑n=2∞(n+μ1+μ)λanzn,(λreal,μ>-1).
From (1.12), the following identity can easily be verifiedz(Iλ,μf(z))′=(μ+1)Iλ+1,μf(z)-μIλ,μf(z).
Remark 1.1.
For k∈(0,1), we note that the domain Ωk given by (1.3) represents the following hyperbolic region:
(u+k21-k2)2-k21-k2v2>(k1-k2)2,u>kk+1.
The extremal function pk,γ(z), for 0<k<1, can be written aspk,γ(z),=(1-γ)pk(z)+γ,where pk(z), in a simplified form, is given below
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(σπsinσ)2z+⋯,(z∈E,σ=arccosk),
and the branch of z is chosen such that Imz≥0.
It is easy to see that, for h∈P(pk,γ),Reh(z)>(k+γ)/(k+1),k∈(0,1). That isP(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).
For k=1, the extremal function
p1(z)=1+2π2(log1+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 that Rep1(z)>1/2 and, in this case, the order 1/2 is sharp.
We now define the following.
Definition 1.2.
Let f∈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
{z(Iλ,μf(z))′Iλ,μg(z)}∈Pm(pk,γ),z∈E.
We note the following.
For m=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].
0-∪Rm0(0,μ)=Rm is the class of functions of bounded radius rotation, see [13, 14].
We denote k-∪Rmγ(0,μ) as k-∪Rmγ, see [5].
Let m=2. Then f∈k-∪R2γ(λ,μ) implies that
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.
Let f∈A. Then f∈k-∪Tmγ(λ,μ) if and only if there exists g∈k-∪R2γ(λ,μ) such that
{z(Iλ,μf(z))′Iλ,μg(z)}∈Pm(pk,γ)inE.
Special Cases.
0-∪T2γ(0,μ)=K(γ).
For k=γ=λ=0, we obtain the class Tm introduced and discussed in [17].
When we take m=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]).
Let q(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
(p(z)+j(z)zp′(z))≺q(z),
then
p(z)≺q(z).
In the following, one gives an easy extension of a result proved in [1].
Lemma 2.2 (see [5]).
Let k≥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
(h(z)+zh′(z)βh(z)+δ)≺pk,γ(z),
and qk,γ(z) is an analytic solution of
qk,γ(z)+zqk,γ′(z)βqk,γ(z)+δ=pk,γ(z),then qk,γ(z) is univalent,
h(z)≺qk,γ(z)≺pk,γ(z),
and qk,γ(z) is the best dominant of (2.3).
Lemma 2.3 (see [19]).
If f∈ℂ,g∈S*, then for each h analytic in E with h(0)=1,
(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]).
Let u=u1+iu2,v=v1+iv2 and let ψ(u,v) be a complex-valued function satisfying the conditions:
ψ(u,v) is continuous in a domain D⊂ℂ2,
(1,0)∈D and Reψ(1,0)>0,
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]).
Let h∈Pm(ρ),0≤ρ<1. Then, with =reiθ,z∈E, one has
(1/2π)∫02π|h(reiθ)|2dθ≤(1-[m2(1-ρ)2-1]r2)/(1-r2),
(1/2π)∫02π|h′(reiθ)|dθ≤m(1-ρ)/(1-r2).
Lemma 2.6 (see [5]).
Let f∈k-∪Rmγ. Then there exist s1,s2∈k-∪R2γ such that
f(z)=(s1(z))(m+2)/4(s2(z))(m-2)/4,k≥0,m≥2,z∈E.
Lemma 2.7.
Let p∈Pm(pk,γ) and p(z)=1+∑n=1∞cnzn. Then
|cn|≤m2|δk,γ|,n≥1,
where
δk,γ={8(1-γ)(cos-1k)2π2(1-k2),0≤k<1,8(1-γ)π2,k=1.
Proof.
Let p(z)=(m/4+1/2)p1(z)-(m/4-1/2)p2(z). Then,
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, that k∈[0,1],m≥0,0≤γ<1,λ∈ℂ,μ>-1 and z∈E.
Theorem 3.1.
Let f∈k-∪Rmγ(λ,μ). Let, for α,β>0,
F(z)=[(1+β)z-β∫0ztβ-1′fα(t)dt]1/α.
Then, F∈k-∪Rmγ(λ,μ) in E.
Proof.
Set
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{zβ(Iλ,μF(z))α}′=zβ-1(Iλ,μf(z))α.
That is(Iλ,μF(z))α[β+αH(z)]=(Iλ,μf(z))α.
Logarithmic differentiation of (3.4) and simple computations give usH(z)+zH′(z)αH(z)+β=z(Iλ,μf(z))′Iλ,μf(z)∈Pm(pk,γ).
Defineϕa,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
H(z)*(ϕa,b(z)z)={H(z)+zH′(z)αH(z)+β}.
From (3.2), (3.5), and (3.7), it follows that{Hi(z)+zHi′(z)αHi(z)+β}∈P(pk,γ),z∈E,i=1,2.
On applying Lemma 2.2, we obtainHi(z)≺qk,γ(z)≺pk,γ(z)inE,where qk,γ(z) is the best dominant and is given as
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.
For k=0,γ=0, we have the following special case.
Corollary 3.2.
Let f∈0-∪Rm0(λ,μ)=Rm(λ,μ) and let F(z) be defined by (3.1). Then, F∈Rmγ1(λ,μ), where
γ1=2{(1+2β)+(1+2β)2+8α}.
Proof.
We write
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
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 functional ψ(u,v) by taking u=Hi(z), v=zHi′(z), asψ(u,v)=u+vαu+(αγ1+β)/(1-γ1)+γ11-γ1.
The first two conditions of Lemma 2.4 can easily be verified. For condition (iii), we proceed as follows: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 Lemma 2.4 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 taking α=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 Theorem 3.1 and Corollary 3.2.
Theorem 3.3.
Let f∈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
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
hα,β(z)=∑n=1∞znn+α+β,
is convex in E.
Let f∈k-∪Tmγ(λ,μ). Then there exists some g∈k-∪R2γ(λ,μ) such that{z(Iλ,μf(z))′Iλ,μg(z)}∈Pm(pk,γ).G(z)=[(β+1)z-β∫0ztβ-1gα(t)dt]1/α.
From Theorem 3.1, it follows that G∈k-∪R2γ(λ,μ). We can write (3.19) asIλ,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.
Since g∈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 havez(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 Lemma 2.3 with hi≺pk,γ,i=1,2, to have{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, for λ=0=k, the subclass Tmγ⊂Tm is invariant under the integral operator defined by (3.1).
Theorem 3.4.
One hask-∪Rmγ(λ+1,μ)⊂k-∪Rmγ(λ,μ).
Proof.
Let f∈k-∪Rmγ(λ+1,μ) and let
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 havez(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{Hi(z)+zHi′(z)Hi(z)+μ}≺pk,γ.
Thus, using Lemma 2.2, 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.
Let m=2,λ≥0. Then, from Theorem 3.4, it easily follows that
k-∪R2γ(λ,μ)⊂k-∪R2γ(0,μ)⊂S*(k+γ1+k)⊂S*.
Let k=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 hask-∪Tmγ(λ+1,μ)⊂k-∪Tmγ(λ,μ),μ,λ≥0.
Proof.
Let f∈k-∪Tmγ(λ+1,μ). Then, for z∈E,
z(Iλ+1,μf(z))′Iλ+1,μg(z)∈Pm(pk,γ),for some g∈k-∪R2γ(λ+1,μ).
We define an analytic function h(z) in E such thatz(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.
Since g∈k-∪R2γ(λ+1,μ) and k-∪R2γ(λ+1,μ)⊂k-∪R2γ(λ,μ), we havez(Iλ,μg(z))′Iλ,μg(z)=h0(z),h0∈P(pk,γ),z∈E.
Now, on using (1.13), we havez(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 usz(z(Iλ,μf(z))′)′Iλ,μg(z)=zh′(z)+(h(z))(h0(z)),
and using (3.33) in (3.32), we obtain
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].
Since f∈k-∪Tmγ(λ+1,μ), we have with 1/H0(z)={h0(z)+μ}∈P,{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 that f∈k-∪T2γ(λ,μ) is a close-to-convex function for z∈E.
Theorem 3.6.
Let f∈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
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.
Now k-∪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 Theorem 3.6 to the class k-∪Tmγ(λ,μ), that is, k-∪Tmγ(λ,μ), is invariant under convolution with convex function.
3.1. Applications of Theorem 3.6
The classes k-∪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 since ϕi(z) is convex in E for i=1,2,3,4.
With essentially the same method together with Lemma 2.7, we can easily prove the following sharp coefficient results.
Theorem 3.8.
Let f∈k-∪Rmγ(λ,m) and let it be given by
f(z)=z+∑n=2∞anzn.
Then
|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
(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
λ=0,m=2, then one has
|an|≤(|δk,γ|)n-1(n-1)!,
see [2].
Let λ=0,n=2. Then,
|an|≤m2|δk,γ|.
This coefficient bound is well known for m=2, see [2].
Using Theorem 3.8 with m=2, the following result can easily be proved.
Theorem 3.9.
Let f:f(z)=z+∑n=2∞anzn∈k-∪Tmγ(λ,μ). Then, for n≥2|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.
Let f:f(z)=z+∑n=2∞anzn∈k-∪Rmγ(λ,μ) with (m+2)(1-γ)/2(1+k)>1. Then, for n≥1,
||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.
Let F(z)=Iλ,μf(z)=z+∑n=2∞Anzn,An=((n+μ)/(1+μ))λan. Since f∈k-∪Rmγ(λ,μ), we can write
zF′(z)=F(z)h(z),h∈Pm(pk,γ).
That isz(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 thatF(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,
|(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.
Let 0<r<1. Then, by a result [23], there exists a ξ with |ξ|=r such that, for z,|z|=r|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),
|(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.
Let f:f(z)=z+∑n=2∞anzn∈k-∪Tmγ(λ,μ). Then
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.
Acknowledgment
The authors are grateful to Dr. S. M. Junaid Zaidi, Rector, COMSATS Institute of Information Technology, Islamabad, Pakistan, for providing excellent research facilities and environment.
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