# Notes on Certain Multivalent Analytic Functions Associated with a Linear Operator

Authors: Jin-Lin Liu
Journal: Journal of Complex Analysis
 (2013)

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## Abstract

The purpose of the present paper is to investigate some properties of multivalent analytic functions.


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## Body

## 1. Introduction

LetAp(n) denote the class of the functions of the form
(1)f(z)=zp+∑k=n∞ap+kzp+k(n,p∈N={1,2,3,…}),
which are analytic in the open unit disk U={z:z∈Cand|z|<1}. Also let the Hadamard product (or convolution) of two functions
(2)fj(z)=zp+∑k=1∞ak,jzp+k(j=1,2)
be given by
(3)(f1*f2)(z)=zp+∑k=1∞ak,1ak,2zp+k.In [1], Liu introduced the following generalized Srivastava-Attiya operator ℑs,b:
(4)ℑs,b:Ap(1)→Ap(1)(5)ℑs,bf(z)=Gp,s,b(z)*f(z)(z∈U;b∈C∖Z0-,Z0-={0,-1,-2,…};s∈C),
where
(6)Gp,s,b(z)=(1+b)s[Φp(z,s,b)-b-s],Φp(z,s,b)=1bs+zp(1+b)s+zp+1(2+b)s+⋯.
It is not difficult to see from (5) and (6) that
(7)ℑs,bf(z)=zp+∑k=1∞‍(1+bk+1+b)sap+kzp+k.
When p=1, the operator ℑs,b is the well-known Srivastava-Attiya operator ℒs,b [2]. The generalized Srivastava-Attiya operator has been studied by several authors (see [1–5]).In this investigation, we focus on certain inequalities consisting of the following differential operator:(8)Qp,s,b(δ,μ;f)(z)=μz(ℑs,bf(z))′ℑs,bf(z)+δ(1+z(ℑs,bf(z))′′(ℑs,bf(z))′).
Our results generalize the recent results obtained by Irmak et al. [6].In order to prove our main results, we need the following lemmas.Lemma 1 (see [7]).
LetΩ⊂C and suppose that the function ψ:C2×U→C satisfies ψ(Meiθ,Keiθ;z)∉Ω for all K≥Mn, θ∈R, and z∈U. If h(z)=a+hnzn+⋯ is analytic in U and ψ(h(z),zh′(z);z)∈Ω for all z∈U, then |h(z)|<M(z∈U).Lemma 2 (see [7]).
LetΩ⊂C and suppose that the function ψ:C2×U→C satisfies ψ(ix,y;z)∉Ω for all x∈R, y≤-n(1+x2)/2, and z∈U. If h(z)=a+hnzn+⋯ is analytic in U and ψ(h(z),zh′(z);z)∈Ω for all z∈U, then Re{h(z)}>0(z∈U).


## 2. Main Results

We now state and then prove each of our main results given by Theorems3 and 4 below.Theorem 3.
Letf(z)∈Ap(n) with (ℑs,bf(z))′ℑs,bf(z)/z2p-1≠0 for all z∈U, and also let δ∈R and μ∈R. If
(9)Re{Qp,s,b(δ,μ;f)(z)}<p(μ+δ)+nMM+1(z∈U),
where M≥1, then
(10)|[(ℑs,bf(z))′pzp-1]δ[ℑs,bf(z)zp]μ-1|<M(z∈U).Proof.
Put(11)h(z)=[(ℑs,bf(z))′pzp-1]δ[ℑs,bf(z)zp]μ-1.
Then the function h(z) is analytic in U with h(0)=0. A simple computation shows that
(12)Qp,s,b(δ,μ;f)(z)=p(μ+δ)+zh′(z)h(z)+1.
Now letting(13)ψ(r,s;z)=p(μ+δ)+sr+1,Ω={w∈C:Re(w)<p(μ+δ)+nMM+1},
we obtain that ψ(h(z),zh′(z);z)=Qp,s,b(δ,μ;f)(z)∈Ω for all z∈U. Further, for any θ∈R, K≥nM, and z∈U, since M≥1, we also have
(14)Re{ψ(Meiθ,Keiθ;z)}=p(μ+δ)+KRe(1M+e-iθ)≥p(μ+δ)+nMM+1,
which shows that ψ(Meiθ,Keiθ;z)∉Ω. Therefore, according to Lemma 1, we obtain |h(z)|<M(z∈U). This completes the proof of Theorem 3.Theorem 4.
Letf(z)∈Ap(n) with (ℑs,bf(z))′ℑs,bf(z)/z2p-1≠0 for z∈U, and also let μ,δ∈R and γ∈[0,1). If
(15)Re{Qp,s,b(δ,μ;f)(z)}>β(p,δ,μ;γ)(z∈U),
where
(16)β(p,δ,μ;γ)={p(μ+δ)-nγ2(1-γ),γ∈[0,12],p(μ+δ)-n(1-γ)2γ,γ∈[12,1),
then
(17)Re{[(ℑs,bf(z))′pzp-1]δ[ℑs,bf(z)zp]μ}>γ(z∈U).Proof.
Suppose that(18)h(z)=11-γ([(ℑs,bf(z))′pzp-1]δ[ℑs,bf(z)zp]μ-γ).
Then h(z)=1+hnzn+⋯ is analytic in U. It is easily seen from (18) that
(19)Qp,s,b(δ,μ;f)(z)=p(μ+δ)+(1-γ)zh′(z)(1-γ)h(z)+γ.
Further, since
(20)ψ(r,s;z)=p(μ+δ)+(1-γ)s(1-γ)r+γ,Ω={w∈C:Re(w)>β(p,δ,μ;γ)},
it leads to ψ(h(z),zh′(z);z)=Qp,s,b(δ,μ;f)(z)∈Ω for z∈U. Also, for any x∈R, y≤-n(1+x2)/2 and z∈U, we have
(21)Re{ψ(ix,y;z)}=p(μ+δ)+γ(1-γ)y(1-γ)2x2+γ2≤p(μ+δ)-nγ(1-γ)2x2+1(1-γ)2x2+γ2≡q(x)≤β(p,δ,μ;γ)={limx→+∞q(x),ifγ∈[0,12],q(0),ifγ∈[12,1),
that is, ψ(ix,y;z)∉Ω. Finally, by Lemma 2, we obtain that Reh(z)>0(z∈U). The proof of Theorem 4 is completed.Remark 5.
Forp=1 and s=0, Theorems 3 and 4 reduce to the results obtained by Irmak et al. [6].



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