# Properties of Certain Subclasses of Meromorphicallyp-Valent Functions Associated with Certain Integral Operator

Authors: Fatma El-Emam
Journal: Journal of Mathematics
 (2020)

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## Abstract

The object of the this paper is to derive some interesting properties of certain subclasses of meromorphicallyp-valent functions which are defined by using an integral operator.


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## Body

## 1. Introduction

Let∑p,m denote the class of functions of the form(1)fz=1zp+∑k=m∞akzk,m>−p,p∈ℕ=1,2,3,…,which are analytic and p-valent in the punctured unit disc U∗=U/0, where U=z:z∈ℂ,z<1. For convenience, we write ∑p,1−p=∑p.For functionsfz∈∑p,m given by (1) and gz∈∑p,m defined by(2)gz=1zp+∑k=m∞bkzk,m>−p,p∈ℕ,the Hadamard product (or convolution) of fz and gz is given by(3)f∗gz=1zp+∑k=m∞akbkzk=g∗fz.Letf and g be analytic in U. The function f is said to be subordinate to g, or g is superordinate to f, written f≺gz∈U, if there exists a Schwarz function wz in U with w0=0 and wz<1z∈U, such that fz=gwz,z∈U. If gz is univalent in U, then the equivalence (cf., e.g., [1, 2])(4)fz≺gz⇔f0=g0,fU⊂gU.For0≤μ,α≤1,m>−p,p∈ℕ, and f∈∑p,m, Saleh et al. [3] introduced the p-valent Rafid operator Sμ,pα:∑p,m⟶∑p,m as follows:(5)Sμ,pαfz=11−μα+1Γα+1∫0∞tα+pe−t/1−μfztdt=1zp+∑k=m∞1−μk+pα+1k+pakzk,where νk is the Pochhammer symbol defined, in terms of the Gamma function Γ, by(6)νk=Γν+kΓν=1,ifk=0,ν∈ℂ/0,νν+1,⋯,ν+k−1,ifk∈ℕ,ν∈ℂ.Note thatSμ,1αfz=Sμαfz (see [4]).It follows from (5) that(7)zSμ,pαfz′=α+1Sμ,pα+1fz−p+α+1Sμ,pαfz,0≤μ,α≤1.By using the integral operatorSμ,pαfz, we define a subclasses of ∑p,m as follows.Definition 1.
For fixed parametersAandB, we say that a function fz∈∑p,m is in the class ∑p,mSα,μ,λ,A,B if it satisfies the following condition:(8)−zp+1p1−λSμ,pαfz′+λSμ,pα+1fz′≺1+Az1+Bz,z∈U,p∈ℕ,0≤μ,α≤1,λ≥0and−1≤B<A≤1.
There are many papers about some subclasses of meromorphic functions associated with several families of linear operators (see, for example, [5–11]). In this paper, we obtain some properties of the class ∑p,mSα,μ,λ,A,B.


## 2. Preliminary Lemmas

To establish our main results, in this paper, we shall need the following lemmas.Lemma 1 (see [12] and [2]).
Suppose that the functionhz is analytic and convex (univalent) in U with h0=1 and ϕz given by(9)ϕz=1+cp+mzp+m+cp+m+1zp+m+1+⋯.
If(10)ϕz+zϕ′zγ≺hz,ℜγ≥0,γ≠0,z∈U,then(11)ϕz≺ψz=γp+mz−γp+m∫0ztγp+m−1htdt≺hz,z∈U,and ψz is the best dominant of (10).LetPγ be the class of analytic in U of the form(12)φz=1+b1z+b2z2+⋯,which satisfies the following inequality:(13)ℜφz>γ,0≤γ<1,z∈U.Lemma 2 (see [13]).
Let the functionφz, given by (12), be in the class Pγ. Then,(14)ℜφz≥2γ−1+21−γ1+z,0≤γ<1,z∈U.Lemma 3 (see [14]).
Ifφj∈Pγj,0≤γj<1,j=1,2, then(15)φ1∗φ2∈Pγ3,γ3=1−21−γ11−γ2.The result is the best possible.Leta,b, and c be any real or complex numbers with c∉Z0−=0,−1,−2,…, and consider the function given by(16)F12a,b,c,z=∑k=0∞akbkckzkk!.This function, called the Gauss hypergeometric function, is analytic and converges absolutely forz∈U (see [15]).Lemma 4 (see [15]).
Leta,b, and c any real or complex numbers with c∉Z0−. Then,(17)∫01tb−11−tc−b−11−zt−adt=ΓbΓc−bΓcF12a,b,c,z,ℜc>ℜb>0,(18)F12a,b,c,z=1−z−aF12a,c−b,c,zz−1,(19)F12a,b,a+b+12,12=πΓa+b+1/2Γa+1/2Γb+1/2,(20)F121,1,2,zz+1=z+1zln1+z,z≠0.


## 3. Main Results

Unless otherwise mentioned, we shall assume throughout the sequel that(21)m>−p,p∈ℕ,0≤μ,α≤1,λ>0,z∈Uand−1≤B<A≤1.Theorem 1.
Iff∈∑p,mSα,μ,λ,A,B, then(22)−zp+1Sμ,pαfz′p≺q1z≺1+Az1+Bz,where the function q1z given by(23)q1z=AB+1−AB1+Bz−1F121,1,α+1λp+m+1,BzBz+1,B≠0,1+α+1Aλp+m+α+1z,B=0,is the best dominant of (22). Furthermore,(24)ℜ−zp+1Sμ,pαfz′p>ρ,where(25)ρ=AB+1−AB1−B−1F121,1,α+1λp+m+1,BB−1,B≠0,1−α+1λp+m+α+1A,B=0.
The result is the best possible.Proof.
Set(26)ϕz=−zp+1Sμ,pαfz′p.
Then, the functionϕz is of form (9) and is analytic in U. Differentiating (26) and with the aid of identity (7), we obtain(27)ϕz+λzϕ′zα+1=−zp+1p1−λSμ,pαfz′+λSμ,pα+1fz′≺1+Az1+Bz.
Now, by using Lemma1 for γ=α+1/λ, we deduce that(28)ϕz≺q1z≺1+Az1+Bz,where q1z is the best dominant of (22) given by(29)q1z=α+1λp+mz−α+1/λp+m∫0ztα+1/λp+m−11+At1+Btdt=AB+1−AB1+Bz−1F121,1,α+1λp+m+1,BzBz+1,B≠0,1+α+1λp+m+α+1Az,B=0,by change of variables followed by the use of identities (17) and (18) (with a=1, b=α+1/λp+m, and c=b+1). This proves assertion (22) of Theorem 1. Next, in order to prove assertion (24) of Theorem 1, it suffices to show that(30)infz<1ℜq1z=q1−1.
Indeed, forz≤r<1,(31)ℜ1+Az1+Bz≥1−Ar1−Br.
Setting(32)Gs,z=1+Asz1+Bsz,dμs=α+1λp+msα+1λp+m−1ds,0≤s≤1,which is a positive measure on [0, 1], we obtain(33)q1z=∫01Gs,zdμsso that(34)ℜq1z≥∫011−Asr1−Bsrdμs=q1−r,z≤r<1.
Lettingr⟶1− in the above inequality, we obtain assertion (30). The result in (24) is best possible as the function q1z is the best dominant of (22).
Puttingλ=σα+1/1−σp+10<σ<1/p+1 in Theorem 1, we obtain the following corollary.Corollary 1.
Iffz∈∑p,m satisfies(35)−zp+1Sμ,pαfz′+σzSμ,pαfz″p1−σp+1≺1+Az1+Bz,then(36)−zp+1Sμ,pαfz′p≺q2z≺1+Az1+Bz,where the function q2z given by(37)q2z=AB+1−AB1+Bz−1F121,1,1−σ1−mσp+m,BzBz+1,B≠0,1+1−σp+11−σ1−mAz,B=0,is the best dominant of (36). Furthermore,(38)ℜ−zp+1Sμ,pαfz′p>ρ,where(39)ρ=AB+1−AB1−B−1F121,1,1−σ10mσp+m,BB−1,B≠0,1−1−σp+11−σ1−mA,B=0.
The result is the best possible.Remark 1.
Form=0 and p=1, the result (asserted by Corollary 1) was also obtained by Patel and Sahoo [16] and Lashin [17].
Applying Theorem1 with A=1−2δ/p,0≤δ<p,B=−1, m=2−p, and λ=α+1α>−1 and making use of (19), we obtain the following corollary.Corollary 2.
Iffz∈∑p,2−p satisfies the following inequality(40)ℜ−zp+1p+2Sμ,pαfz′+zSμ,pαfz″>δ,0≤δ<p,then(41)ℜ−zp+1Sμ,pαfz′>δ+p−δπ2−1.
The result is the best possible.Remark 2.
The result (asserted by Corollary2) was also obtained by Srivastava and Patel [18].
Takingδ=−pπ−2/4−π in Corollary 2, we have the following corollary.Corollary 3.
Iffz∈∑p,2−p satisfies the following inequality(42)ℜ−zp+1p+2Sμ,pαfz′+zSμ,pαfz″>−pπ−24−π,by Replacing Sμ,pαfz by fz, then(43)ℜ−zp+1Sμ,pαfz′>0.
The result is the best possible.Remark 3.
The result (asserted by Corollary3) was also obtained by Pap [19].
Applying Theorem1 with A=1−2δ/p,0≤δ<p,B=−1, m=1−p, and λ=α+1 and making use of (20), we obtain the following corollary.Corollary 4.
Iffz∈∑p satisfies the following inequality(44)ℜ−zp+1p+2Sμ,pαfz′+zSμ,pαfz″>δ,0≤δ<p,then(45)ℜ−zp+1Sμ,pαfz′>p+2p−δln2−1.The result is the best possible.Replacingϕz by zpSμ,pαfz in (26) and applying the same method and technique as the proof of Theorem 1, we can prove the following result.Theorem 2.
Iff∈∑p,m satisfies(46)zp1−λSμ,pαfz+λSμ,pα+1fz≺1+Az1+Bz,then(47)zpSμ,pαfz≺q1z≺1+Az1+Bz,(48)ℜzpSμ,pαfz>ρ,where q1 and ρ are given as in Theorem 1. The result is the best possible.Theorem 3.
Let−1≤Bj<Aj≤1j=1,2. If each of the functions fjz∈∑p satisfies the following subordination condition(49)zp1−λSμ,pαfjz+λSμ,pα+1fjz≺1+Ajz1+Bjz,j=1,2,then(50)zp1−λSμ,pαHz+λSμ,pα+1Hz≺1+1−2γ/pz1−z,where(51)Hz=Sμ,pαf1∗f2z,(52)γ=1−4A1−B1A2−B21−B11−B21−12F121,1,α+1λ+1,12.
The result is the best possible whenB1=B2=−1.
Proof. If we let(53)φjz=zp1−λSμ,pαfjz+λSμ,pα+1fjz,j=1,2,then, by the hypothesis of Theorem 3, we have(54)φjz∈Pγj,γj=1−Aj1−Bj,j=1,2.
Using identity (7), (53) can be written as(55)Sμ,pαfjz=α+1λz−α+1λ−p∫0ztα+1λ−1φjtdt,j=1,2.
From (51) and (55), we obtain(56)Sμ,pαHz=α+1λz−α+1λ−p∫0ztα+1λ−1φ0tdt,where(57)φ0z=zp1−λSμ,pαHz+λSμ,pα+1Hz=α+1λz−α+1λ∫0ztα+1λ−1φ1∗φ2tdt.Sinceφ1z∈Pγ1 and φ2z∈Pγ2, it follows from Lemma 3 that(58)φ1∗φ2z∈Pγ3,γ3=1−21−γ11−γ2.According to Lemma2, we have(59)ℜφ1∗φ2z≥2γ3−1+21−γ31+z.Now, by using (59) in (57) and then appealing to Lemma 4, we obtain(60)ℜφoz=α+1λ∫01uα+1/λ−1ℜφ1∗φ2uzdu≥α+1λ∫01uα+1/λ−12γ3−1+21−γ31+uzdu>α+1λ∫01uα+1/λ−12γ3−1+21−γ31+udu=1−4A1−B1A2−B21−B11−B21−α+1λ∫01uα+1/λ−11+u−1du=1−4A1−B1A2−B21−B11−B21−12F121,1,α+1λ+1,12=γ,which completes the proof of assertion (49).WhenB1=B2=−1, we consider the functions fjz∈∑p,mj=1,2 defined by(61)Sμ,pαfjz=α+1λz−α+1λ−p∫0ztα+1λ−11+Ajt1−tdt,j=1,2.Now, by using Lemma4 and (57), we have(62)φoz=α+1λ∫01uα+1/λ−11−1+A11+A2+1+A11+A21−uzdu=1−1+A11+A2+1+A11+A21−z−1F121,1,α+1λ+1,zz−1⟶1−1+A11+A2+121+A11+A2F121,1,α+1λ+1,12,as z⟶−1, which ends the proof of Theorem 3.PuttingAj=1−2ηj/p,0≤ηj<p,Bj=−1j=1,2, and λ/α+1=τ, in Theorem 3, we get the following result.Corollary 5.
Iff∈∑p satisfies(63)ℜzp1+pτSμ,pαfjz+τzSμ,pαfjz′>ηj,j=1,2,then(64)ℜzp1+pτSμ,pαf1z∗Sμ,pαf2z+τzSμ,pαf1z∗Sμ,pαf2z′>γ,where(65)γ=1−41−η1p1−η2p1−12F121,1,1τ+1,12.Remark 4.
Forp=1, the result (asserted by Corollary 5) was also obtained by Yang [20].



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