# Differential Subordinations for Certain Meromorphically Multivalent Functions Defined by Dziok-Srivastava Operator

Authors: Ying Yang; Yu-Qin Tao; Jin-Lin Liu
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

By making use of the Dziok-Srivastava operator, we introduce a new class of meromorphically multivalent functions. Some inclusion properties of functions belonging to this class are derived.


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## Body

## 1. Introduction

LetΣ(p) denote the class of functions of the form (1.1)f(z)=z-p+∑n=1∞anzn-p(p∈N:={1,2,3,…}),					which are analytic in the punctured open unit disk 𝕌0={z:0<|z|<1} with a pole at z=0. Also let the Hadamard product (or convolution) of the following functions: (1.2)fj(z)=z-p+∑n=1∞an,jzn-p(j=1,2)
be given by (1.3)(f1*f2)(z):=z-p+∑n=1∞an,1an,2zn-p=(f2*f1)(z).Given two functionsf(z) and g(z), which are analytic in 𝕌=𝕌0∪{0}, we say that the function g(z) is subordinate to f(z) and write g≺f or (more precisely) g(z)≺f(z)(z∈𝕌), if there exists a Schwarz function w(z), analytic in 𝕌 with w(0)=0 and |w(z)|<1(z∈𝕌) such that g(z)=f(w(z))(z∈𝕌). In particular, if f(z) is univalent in 𝕌, we have the following equivalence: (1.4)g(z)≺f(z)(z∈U)⟺g(0)=f(0),g(U)⊂f(U).LetA be the class of functions of the form (1.5)f(z)=z+∑n=2∞bnzn,
which are analytic in 𝕌. A function f(z)∈A is said to be in the class S*(α) if (1.6)Re{zf′(z)f(z)}>α(z∈U)
for some α(α<1). When 0≤α<1, S*(α) is the class of starlike functions of order α in 𝕌. A function f(z)∈A is said to be prestarlike of order α in 𝕌 if (1.7)z(1-z)2(1-α)*f(z)∈S*(α)(α<1),
where the symbol * means the familiar Hadamard product (or convolution) of two analytic functions in 𝕌. We denote this class by R(α) (see [1]). Clearly a function f(z)∈A is in the class R(0) if and only if f(z) is convex univalent in 𝕌 and R(1/2)=S*(1/2).For complex parameters(1.8)α1,…,αqandβ1,…,βs(βj∉Z0-:={0,-1,-2,…};j=1,2,…,s),
we define the generalized hypergeometric function qFs(α1,…,αq;β1,…,βs;z) by (1.9)qFs(α1,…,αq;β1,…,βs;z):=∑n=0∞(α1)n⋯(αq)n(β1)n⋯(βs)n⋅znn!(q≤s+1;q,s∈N0:=N∪{0};z∈U),
where (x)n is the Pochhammer symbol defined, in terms of the Gamma function Γ(z),by (1.10)(x)n:=Γ(x+n)Γ(x)={1(n=0;x∈C∖{0})x(x+1)⋯(x+n-1)(n∈N;x∈C).
Corresponding to a function hp(α1,…,αq;β1,…,βs;z) defined by (1.11)hp(α1,…,αq;β1,…,βs;z)=z-pqFs(α1,…,αq;β1,…,βs;z),
we now consider a linear operator (1.12)Hp(α1,…,αq;β1,…,βs):Σ(p)⟶Σ(p),
defined by means of the Hadamard product (or convolution) as follows: (1.13)Hp(α1,…,αq;β1,…,βs)f(z):=hp(α1,…,αq;β1,…,βs;z)*f(z).
For convenience, we write (1.14)Hp,q,s(α1):=Hp(α1,…,αq;β1,…,βs).
Thus, after some calculations, we have (1.15)z(Hp,q,s(α1)f(z))′=α1Hp,q,s(α1+1)f(z)-(α1+p)Hp,q,s(α1)f(z).The operatorHp,q,s(α1) is popularly known as the generalized Dziok-Srivastava operator. Many interesting subclasses of multivalent functions, associated with the operator Hp,q,s(α1) and its various special cases, were investigated recently by (e.g.) Dziok and Srivastava [2–4], Liu [5], Liu and Srivastava [6, 7], Patel et al. [8], Wang et al. [9], and others.LetP be the class of functions h(z) with h(0)=1, which are analytic and convex univalent in 𝕌.Definition 1.1.
A functionf(z)∈Σ(p) is said to be in the class Tp,q,s(α1,λ;h) if it satisfies the subordination condition
(1.16)(λ-1)pzp+1(Hp,q,s(α1)f(z))′+λp(p+1)zp+2(Hp,q,s(α1)f(z))′′≺h(z),
where λ is a complex number and h(z)∈P.The main object of this paper is to present a systematic investigation of the classTp,q,s(α1,λ;h) defined above by means of the generalized Dziok-Srivastava operator Hp,q,s(α1).For our purpose, we shall need the following lemmas to derive our main results for the classTp,q,s(α1,λ;h).Lemma 1.2 (see [10]).
Letg(z) be analytic in 𝕌 and h(z) be analytic and convex univalent in 𝕌 with h(0)=g(0). If
(1.17)g(z)+1μzg′(z)≺h(z),
where Reμ>0, then
(1.18)g(z)≺h̃(z)=μz-μ∫vztμ-1h(t)dt≺h(z)
and h̃(z) is the best dominant of (1.17).Lemma 1.3 (see [1]).
Letα<1, f(z)∈S*(α) and g(z)∈R(α). Then, for any analytic function F(z) in 𝕌,
(1.19)g*(fF)g*f(U)⊂co⁡¯(F(U)),
where co¯(F(𝕌)) denotes the closed convex hull of F(𝕌).


## 2. Properties of the ClassTp,q,s(α1,λ;h)

Theorem 2.1.
Letλ1<λ2≤0. Then Tp,q,s(α1,λ1;h)⊂Tp,q,s(α1,λ2;h).Proof.
Letλ1<λ2≤0 and suppose that
(2.1)g(z)=-zp+1(Hp,q,s(α1)f(z))′p
for f(z)∈Tp,q,s(α1,λ1;h). Then the function g(z) is analytic in 𝕌 with g(0)=1. Differentiating both sides of (2.1) with respect to z and using (1.16), we have
(2.2)(λ1-1)pzp+1(Hp,q,s(α1)f(z))′+λ1p(p+1)zp+2(Hp,q,s(α1)f(z))′′=g(z)-λ1p+1zg′(z)≺h(z).
Hence an application of Lemma 1.2 yields
(2.3)g(z)≺h(z).
Noting that 0<λ2/λ1<1 and that h(z) is convex univalent in 𝕌, it follows from (2.1) to (2.3) that
(2.4)(λ2-1)pzp+1(Hp,q,s(α1)f(z))′+λ2p(p+1)zp+2(Hp,q,s(α1)f(z))′′=λ2λ1((λ1-1)pzp+1(Hp,q,s(α1)f(z))′+λ1p(p+1)zp+2(Hp,q,s(α1)f(z))′′)+(1-λ2λ1)g(z)≺h(z).
Thus f(z)∈Tp,q,s(α1,λ2;h) and the proof of Theorem 2.1 is completed.Theorem 2.2.
Let0<b1<b2. Then Tp,q,s(b2,λ;h)⊂Tp,q,s(b1,λ;h).Proof.
Define a functiong(z) by
(2.5)g(z)=z+∑n=1∞(b1)n(b2)nzn+1(z∈U;0<b1<b2).
Then
(2.6)zp+1hp(b1,α2,…,αs,1;b2,α2,…,αs;z)=g(z)∈A,
where
(2.7)hp(b1,α2,…,αs,1;b2,α2,…,αs;z)
is defined as in (1.11), and
(2.8)z(1-z)b2*g(z)=z(1-z)b1.
By (2.8), we see that
(2.9)z(1-z)b2*g(z)∈S*(1-b12)⊂S*(1-b22)(0<b1<b2),
which implies that
(2.10)g(z)∈R(1-b22).
Letf(z)∈Tp,q,s(b2,λ;h). It is easy to verify that
(2.11)zp+1(Hp,q,s(b1)f(z))′=(zphp(b1,α2,…,αs,1;b2,α2,…,αs;z))*(zp+1(Hp,q,s(b2)f(z))′)(2.12)zp+2(Hp,q,s(b1)f(z))′′=(zphp(b1,α2,…,αs,1;b2,α2,…,αs;z))*(zp+2(Hp,q,s(b2)f(z))′′).
From (2.11), (2.12), and (2.6), we deduce that
(2.13)(λ-1)pzp+1(Hp,q,s(b1)f(z))′+λp(p+1)zp+2(Hp,q,s(b1)f(z))′′=g(z)z*w(z)=g(z)*(zw(z))g(z)*z,
where
(2.14)w(z):=(λ-1)pzp+1(Hp,q,s(b2)f(z))′+λp(p+1)zp+2(Hp,q,s(b2)f(z))′′≺h(z).
Since the function z belongs to the function class S*(1-b2/2) and h(z) is convex univalent in 𝕌, it follows from (2.12), (2.13), (2.14), and Lemma 1.3 that
(2.15)(λ-1)pzp+1(Hp,q,s(b1)f(z))′+λp(p+1)zp+2(Hp,q,s(b1)f(z))′′≺h(z).
Thus f(z)∈Tp,q,s(b1,λ;h) and the proof of Theorem 2.2 is completed.Theorem 2.3.
Letf(z)∈Tp,q,s(α1,λ;h),g(z)∈Σ(p) and
(2.16)Re{zpg(z)}>12(z∈U).
Then
(2.17)(f*g)(z)∈Tp,q,s(α1,λ;h).Proof.
Forf(z)∈Tp,q,s(α1,λ;h) and g(z)∈Σ(p), we have
(2.18)(λ-1)pzp+1(Hp,q,s(α1)(f*g)(z))′+λp(p+1)zp+2(Hp,q,s(α1)(f*g)(z))′′=(λ-1)p(zpg(z))*(zp+1(Hp,q,s(α1)f(z))′)+λp(p+1)(zpg(z))*(zp+2(Hp,q,s(α1)f(z))′′)=(zpg(z))*ψ(z),
where
(2.19)ψ(z)=(λ-1)pzp+1(Hp,q,s(α1)f(z))′+λp(p+1)zp+2(Hp,q,s(α1)f(z))′′.
In view of (2.16), the function zpg(z) has the Herglotz representation
(2.20)zpg(z)=∫|x|=1dμ(x)1-xz(z∈U),
where μ(x) is a probability measure defined on the unit circle |x|=1 and
(2.21)∫|x|=1dμ(x)=1.
Since h(z) is convex univalent in 𝕌, it follows from (2.18) to (2.20) that
(2.22)(λ-1)pzp+1(Hp,q,s(α1)(f*g)(z))′+λp(p+1)zp+2(Hp,q,s(α1)(f*g)(z))′′=∫|x|=1ψ(xz)dμ(x)≺h(z).
This shows that (f*g)(z)∈Tp,q,s(α1,λ;h) and the theorem is proved.Theorem 2.4.
Letf(z)∈Tp,q,s(α1,λ;h),g(z)∈Σ(p) and
(2.23)zp+1g(z)∈R(α)(α<1).
Then
(2.24)(f*g)(z)∈Tp,q,s(α1,λ;h).Proof.
Forf(z)∈Tp,q,s(α1,λ;h) and g(z)∈Σ(p), from (2.18) we have
(2.25)(λ-1)pzp+1(Hp,q,s(f*g)(z))′+λp(p+1)zp+2(Hp,q,s(f*g)(z))′′=(zp+1g(z))*(zψ(z))(zp+1g(z))*z(z∈U),
where ψ(z) is defined as in (2.19).
Sinceh(z) is convex univalent in 𝕌,
(2.26)ψ(z)≺h(z),zp+1g(z)∈R(α),z∈S*(α)(α<1),
it follows from (2.25) and Lemma 1.3 the desired result.Theorem 2.5.
Letλ<0,β>0 and f(z)∈Tp,q,s(λ;βh+1-β). If β≤β0, where
(2.27)β0=12(1+p+1λ∫01u-((p+1)/λ)-11+udu)-1,
then f(z)∈Tp,q,s(0;h). The bound β0 is sharp when h(z)=1/(1-z).Proof.
Let us define(2.28)g(z)=-zp+1(Hp,q,s(α1)f(z))′p
for f(z)∈Tp,q,s(λ;βh+1-β) with λ<0 and β>0. Then we have
(2.29)g(z)-λp+1zg′(z)=(λ-1)pzp+1(Hp,q,s(α1)f(z))′+λp(p+1)zp+2(Hp,q,s(α1)f(z))′′≺βh(z)+1-β.
Hence an application of Lemma 1.2 yields
(2.30)g(z)≺-β(p+1)λz(p+1)/λ∫0zt-((p+1)/λ)-1h(t)dt+1-β=(h*ψ)(z),
where
(2.31)ψ(z)=-β(p+1)λz(p+1)/λ∫0zt-((p+1)/λ)-11-tdt+1-β.
If0<β≤β0, where β0(>1) is given by (2.27), then it follows from (2.31) that
(2.32)Reψ(z)=-β(p+1)λ∫01u-((p+1)/λ)-1Re(11-uz)du+1-β>-β(p+1)λ∫01u-((p+1)/λ)-11+udu+1-β≥12(z∈U;λ<0).
Now, by using the Herglotz representation for ψ(z), from (2.28) and (2.30), we arrive at
(2.33)-zp+1(Hp,q,s(α1)f(z))′p≺(h*ψ)(z)≺h(z)
because h(z) is convex univalent in 𝕌. This shows that f(z)∈Tp,q,s(0;h).
Forh(z)=1/(1-z) and f(z)∈Σ(p) defined by
(2.34)-zp+1(Hp,q,s(α1)f(z))′p=-β(p+1)λz(p+1)/λ∫0zt-((p+1)/λ)-11-tdt+1-β,
it is easy to verify that
(2.35)(λ-1)pzp+1(Hp,q,s(α1)f(z))′+λp(p+1)zp+2(Hp,q,s(α1)f(z))′′=βh(z)+1-β.
Thus f(z)∈Tp,q,s(λ;βh+1-β). Also, for β>β0, we have
(2.36)Re{-zp+1(Hp,q,s(α1)f(z))′p}⟶-β(p+1)λ∫01u-((p+1)/λ)-11+udu+1-β<12(z⟶-1),
which implies that f(z)∉Tp,q,s(0;h). Hence the bound β0 cannot be increased when h(z)=1/(1-z).



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