Abstract

We introduce new classes, Σ∗,𝜇,𝑘𝑛,𝜆(𝛼,𝛽,𝜌) and Σ𝜇,𝑘𝑛,𝜆(𝛼,𝛽,𝜌), of meromorphic functions defined by means of the Hadamard product of Cho-Kwon-Srivastava operator, and we define here a similar transformation by means of an operator given by Ghanim and Darus, in 𝑈={𝑧∶0<|𝑧|<1}, and investigate a number of inclusion relationships of these classes. We also derive some interesting properties of these classes.

1. Introduction

Let Σ denote the class of meromorphic functions 𝑓(𝑧) normalized by1𝑓(𝑧)=𝑧+∞𝑛=1𝑎𝑛𝑧𝑛,(1.1) which are analytic in the punctured unit disk 𝑈={𝑧∶0<|𝑧|<1}. For 0≤𝛽, we denote by 𝑆∗(𝛽) and 𝑘(𝛽), the subclasses of Σ consisting of all meromorphic functions, which are, respectively, starlike of order 𝛽 and convex of order 𝛽 in 𝑈.

For functions 𝑓𝑗(𝑧)(𝑗=1;2) defined by𝑓𝑗(1𝑧)=𝑧+∞𝑛=1𝑎𝑛,𝑗𝑧𝑛,(1.2) we denote the Hadamard product (or convolution) of 𝑓1(𝑧) and 𝑓2(𝑧) by𝑓1∗𝑓2=1𝑧+∞𝑛=1𝑎𝑛,1𝑎𝑛,2𝑧𝑛.(1.3)

Let 𝑃𝑛(𝜌) be the class of functions 𝑝(𝑧) analytic in 𝑈 satisfying the properties 𝑝(0)=1 and02𝜋||||ℜ𝑝(𝑧)−𝜌||||1−𝜌𝑑𝜃≤𝑛𝜋,(1.4) where 𝑧=𝑟𝑒𝑖𝜃,𝑛≥2, and 0≤𝜌<1. This class has been introduced in [1]. We note that 𝑃𝑛(0)=𝑃𝑛 and 𝑃2(𝜌)=𝑃(𝜌), see ([2, 3]), the class of analytic functions with positive real part greater than 𝜌 and 𝑃2(0)=𝑃, the class of functions with positive real part. From (1.6), we can write 𝑝∈𝑃𝑛(𝜌) as𝑛𝑝(𝑧)=4+12𝑝11(𝑧)+2−𝑛4𝑝2(𝑧),(1.5) where 𝑝𝑖(𝑧)∈𝑃(𝜌),𝑖=1,2 and 𝑧∈𝑈.

Let us define the function 𝜙(𝛼,𝛽;𝑧) by1𝜙(𝛼,𝛽;𝑧)=𝑧+∞𝑛=0||||(𝛼)𝑛+1(𝛽)𝑛+1||||𝑧𝑛,(1.6) for 𝛽≠0,−1,−2,…, and 𝛼∈ℂ/{0}, where (𝜆)𝑛=𝜆(𝜆+1)𝑛+1 is the Pochhammer symbol. We note that1𝜙(𝛼,𝛽;𝑧)=𝑧2𝐹1(1,𝛼,𝛽;𝑧),(1.7) where2𝐹1(𝑏,𝛼,𝛽;𝑧)=∞𝑛=0(𝑏)𝑛(𝛼)𝑛(𝛽)𝑛𝑧𝑛𝑛!(1.8) is the well-known Gaussian hypergeometric function.

Let us put𝑞𝜆,𝜇(1𝑧)=𝑧+∞𝑛=1𝜆𝑛+1+𝜆𝜇𝑧𝑛,(𝜆>0,𝜇≥0).(1.9) Corresponding to the functions 𝜙(𝛼,𝛽;𝑧) and 𝑞𝜆,𝜇(𝑧), and using the Hadamard product for 𝑓(𝑧)∈Σ, we define a new linear operator 𝐿(𝛼,𝛽,𝜆,𝜇) on Σ by𝐿(𝛼,𝛽,𝜆,𝜇)𝑓(𝑧)=𝑓(𝑧)∗𝜙(𝛼,𝛽;𝑧)∗𝑞𝜆,𝜇=1(𝑧)𝑧+∞𝑛=1||||(𝛼)𝑛+1(𝛽)𝑛+1||||𝜆𝑛+1+𝜆𝜇𝑎𝑛𝑧𝑛.(1.10) The meromorphic functions with the generalized hypergeometric functions were considered recently by Dziok and Srivastava [4, 5], Liu [6], Liu and Srivastava [7–9], and Cho and Kim [10].

For a function 𝑓∈𝐿(𝛼,𝛽,𝜆,𝜇)𝑓(𝑧), we define𝐼𝜇,0𝛼,𝛽,𝜆=𝐿(𝛼,𝛽,𝜆,𝜇)𝑓(𝑧)(1.11) and for 𝑘=1,2,3,…,𝐼𝜇,𝑘𝛼,𝛽,𝜆𝐼𝑓(𝑧)=𝑧𝑘−1𝐿(𝛼,𝛽,𝜆,𝜇)𝑓(𝑧)+2𝑧=1𝑧+∞𝑛=1𝑛𝑘||||(𝛼)𝑛+1(𝛽)𝑛+1||||𝜆𝑛+1+𝜆𝜇𝑎𝑛𝑧𝑛.(1.12) Note that if 𝑛=𝛽,𝑘=0, the operator 𝐼𝜇,0𝛼,𝑛,𝜆 has been introduced by Cho et al. [11] for 𝜇∈ℕ0=ℕ∪0. It was known that the definition of the operator 𝐼𝜇,0𝛼,𝑛,𝜆 was motivated essentially by the Choi-Saigo-Srivastava operator [12] for analytic functions, which includes a simpler integral operator studied earlier by Noor [13] and others (cf. [14–16]). Note also the operator 𝐼0,𝑘𝛼,𝛽 has been recently introduced and studied by Ghanim and Darus, [17], [18] and [19], respectively. To our best knowledge, the recent work regarding operator 𝐼𝜇,0𝛼,𝑛,𝜆 was charmingly studied by Piejko and Sokół [20]. Moreover, the operator 𝐼𝜇,𝑘𝛼,𝛽,𝜆 was defined and studied by Ghanim and Darus [21]. In the same direction, we will study the operator 𝐼𝜇,𝑘𝛼,𝛽,𝜆 given in (1.12).

Now, it follows from (1.10) and (1.12) that𝑧𝐼𝜇,𝑘𝛼,𝛽,𝜆𝑓(𝑧)=𝛼𝐼𝜇,𝑘𝛼+1,𝛽,𝜆𝑓(𝑧)−(𝛼+1)𝐼𝜇,𝑘𝛼,𝛽,𝜆𝑓(𝑧).(1.13)

In the present paper, we will use of the operator 𝐼𝜇,𝑘𝛼,𝛽,𝜆𝑓(𝑧) and introduce some new classes of meromorphic functions.

Definition 1.1. Let 𝑓∈Σ. Then 𝑓∈Σ∗,𝜇,𝑘𝑛,𝜆(𝛼,𝛽,𝜌) if and only if −(1−𝛼)𝑧2𝐼𝜇,𝑘𝛼,𝛽,𝜆𝑓(𝑧)−𝛼𝑧2𝐼𝜇,𝑘𝛼,𝛽,𝜆𝑓(𝑧)∈𝑃𝑛(𝜌),(1.14) where 𝛼>0,𝑛≥2,0≤𝜌<1, and 𝑧∈𝑈.

Definition 1.2. Let 𝑓∈Σ. Then 𝑓∈Σ𝜇,𝑘𝑛,𝜆(𝛼,𝛽,𝜌) if and only if 𝐼(1−𝛼)𝑧𝜇,𝑘𝛼,𝛽,𝜆𝑓(𝑧)𝐼−𝛼𝑧𝜇,𝑘𝛼,𝛽,𝜆𝑓(𝑧)∈𝑃𝑛(𝜌),(1.15) where 𝛼>0,𝑛≥2,0≤𝜌<1, and 𝑧∈𝑈.

2. Preliminary Results

Lemma 2.1 (see [22]). If 𝑝(𝑧) is analytic in 𝑈 with 𝑝(0)=1, and if 𝛼 is a complex number satisfying ℜ(𝛼)≥0, then ℜ𝑝(𝑧)+𝛼𝑧𝑝(𝑧)>𝛽(0≤𝛽<1)(2.1) implies ℜ𝑝(𝑧)>𝛽+(1−𝛽)(2𝛾−1),(2.2) where 𝛾 is given by 𝛾=𝛾(𝛼)=101+𝑡ℜ𝛼−1𝑑𝑡,(2.3) which is an increasing function of ℜ(𝛼) and 1/2≤𝛾<1. The estimate is sharp in the sense that the bound cannot be improved.

Lemma 2.2 (see [23]). If 𝑝(𝑧) is analytic in 𝑈, 𝑝(0)=1, and ℜ𝑝(𝑧)>1/2,𝑧∈𝑈, then for any function 𝐹 analytic in 𝑈, the function 𝑝∗𝐹 takes values in the convex hull of the image of 𝑈 under 𝐹.

Lemma 2.3 (see [24]). Let 𝑝(𝑧)=1+𝑏1𝑧+𝑏2𝑧2+⋯∈𝑃(𝜌). Then ℜ𝑝(𝑧)≥2𝜌−1+2(1−𝜌)1+|𝑧|.(2.4)

3. Main Results

Theorem 3.1. Let 𝑓∈Σ∗,𝜇,𝑘𝑛,𝜆(𝛼,𝛽,𝜌). Then −𝑧2𝐼𝜇,𝑘𝛼,𝛽,𝜆𝑓(𝑧)∈𝑃𝑛𝜌1,(3.1) where 𝜌1 is given by 𝜌1=𝜌+(1−𝜌)(2𝛾−1),(3.2)𝛾=101+𝑡ℜ𝛼−1𝑑𝑡.(3.3)

Proof. Let −𝑧2𝐼𝜇,𝑘𝛼,𝛽,𝜆𝑓(𝑧)𝑛=𝑝(𝑧)=4+12𝑝11(𝑧)+2−𝑛4𝑝2(𝑧).(3.4) Then 𝑝(𝑧) is analytic in 𝑈 with 𝑝(0)=1. Applying the identity (1.13) in (3.4) and differentiating the resulting equation with respect to 𝑧, we have −(1−𝛼)𝑧2𝐼𝜇,𝑘𝛼,𝛽,𝜆𝑓(𝑧)−𝛼𝑧2𝐼𝜇,𝑘𝛼,𝛽,𝜆𝑓(𝑧)=𝑝(𝑧)+𝛼𝑧𝑝(𝑧).(3.5) Since 𝑓∈Σ∗,𝜇,𝑘𝑛,𝜆(𝛼,𝛽,𝜌), so {𝑝(𝑧)+𝛼𝑧𝑝(𝑧)}∈𝑃𝑛(𝜌) for 𝑧∈𝑈. This implies that ℜ𝑝𝑖(𝑧)+𝛼𝑧𝑝𝑖(𝑧)>𝜌,𝑖=1,2.(3.6) Using Lemma 2.1, we see that ℜ{𝑝𝑖(𝑧)}>𝜌1, where 𝜌1 is given by (3.2).
Consequently, 𝑝∈𝑃𝑛(𝜌1) for 𝑧∈𝑈, and the proof is complete.

Theorem 3.2. Let 𝑓∈Σ∗,𝜇,𝑘𝑛,𝜆(0,𝛽,𝜌) for 𝑧∈𝑈. Then 𝑓∈Σ∗,𝜇,𝑘𝑛,𝜆(𝛼,𝛽,𝜌) for |𝑧|<ℜ(𝛼), where 1ℜ(𝛼)=√𝛼+1+𝛼2.(3.7)

Proof. Set −𝑧2𝐼𝜇,𝑘𝛼,𝛽,𝜆𝑓(𝑧)=(1−𝜌)ℎ(𝑧)+𝜌,ℎ∈𝑃𝑛.(3.8) Now proceeding as in Theorem 3.1, we have −(1−𝛼)𝑧2𝐼𝜇,𝑘𝛼,𝛽,𝜆𝑓(𝑧)−𝛼𝑧2𝐼𝜇,𝑘𝛼,𝛽,𝜆𝑓(𝑧)−𝜌=(1−𝜌)ℎ(𝑧)+𝛼𝑧ℎ𝑛(𝑧)=(1−𝜌)4+12ℎ1(𝑧)+𝛼𝑧ℎ1+1(𝑧)2−𝑛4ℎ2(𝑧)+𝛼𝑧ℎ2,(𝑧)(3.9) where we have used (1.6) and ℎ1,ℎ2∈𝑃,𝑧∈𝑈. Using the following well-known estimate [25] ||𝑧ℎ𝑖||≤(𝑧)2𝑟1−𝑟2ℜℎ𝑖(𝑧)(|𝑧|=𝑟<1,𝑖=1,2),(3.10) we have ℜℎ𝑖(𝑧)+𝛼𝑧ℎ𝑖ℎ(𝑧)≥ℜ𝑖||(𝑧)+𝛼𝑧ℎ𝑖||(𝑧)≥ℜℎ𝑖(𝑧)1−2𝛼𝑟1−𝑟2.(3.11) The right-hand side of this inequality is positive if 𝑟<𝑅(𝛼), where ℜ(𝛼) is given by (3.7). Consequently, it follows from (3.9) that 𝑓∈Σ∗,𝜇,𝑘𝑛,𝜆(𝛼,𝛽,𝜌) for |𝑧|<𝑅(𝛼). Sharpness of this result follows by taking ℎ𝑖(𝑧)=(1+𝑧)/(1−𝑧) in (3.9), 𝑖=1,2.

Theorem 3.3. Let 𝑓∈Σ∗,𝜇,𝑘𝑛,𝜆(0,𝛽,𝜌), and let 𝐹𝛿𝛿(𝑓(𝑧))=𝑧𝛿+1𝑧0𝑡𝛿𝑓(𝑡)𝑑𝑡(𝛿>0,𝑧∈𝑈).(3.12) Then −𝑧2𝐼𝜇,𝑘𝛼,𝛽,𝜆𝐹(𝑓(𝑧))∈𝑃𝑛𝜌2,(3.13) where 𝜌2 is given by 𝜌2=𝜌+(1−𝜌)2𝛾1𝛾−1,(3.14)1=101+𝑡ℜ(1/𝛿)−1𝑑𝑡.(3.15)

Proof. Setting −𝑧2𝐼𝜇,𝑘𝛼,𝛽,𝜆𝐹(𝑓(𝑧))𝑛=𝑝(𝑧)=4+12𝑝11(𝑧)+2−𝑛4𝑝2(𝑧),(3.16) then 𝑝(𝑧) is analytic in 𝑈 with 𝑝(0)=1. Using the following operator identity: 𝑧𝐼𝜇,𝑘𝛼,𝛽,𝜆𝐹(𝑓(𝑧))𝐼=𝛿𝜇,𝑘𝛼,𝛽,𝜆𝐼𝐹(𝑓(𝑧))−(𝛿+1)𝜇,𝑘𝛼,𝛽,𝜆𝐹(𝑓(𝑧)),(3.17) in (3.16), and differentiating the resulting equation with respect to 𝑧, we find that −𝑧2𝐼𝜇,𝑘𝛼,𝛽,𝜆𝑓(𝑧)=1𝑝(𝑧)+𝛿𝑧𝑝(𝑧)∈𝑃𝑛(𝜌).(3.18) Using Lemma 2.1, we see that −𝑧2(𝐼𝜇,𝑘𝛼,𝛽,𝜆𝐹(𝑓(𝑧)))∈𝑃𝑛(𝜌2) for 𝑧∈𝑈, where 𝜌2 is given by (3.14).
Hence, the proof is complete.

Theorem 3.4. Let 𝜙(𝑧)∈Σ satisfy the following inequality: 1ℜ(𝑧𝜙(𝑧))>2,𝑧∈𝑈.(3.19) If 𝑓∈Σ𝜇,𝑘𝑛,𝜆(𝛼,𝛽,𝜌), then 𝜙∗𝑓∈Σ𝜇,𝑘𝑛,𝜆(𝛼,𝛽,𝜌).
Proof. Let 𝐺=𝜙∗𝑓. Then 𝐼(1−𝛼)𝑧𝜇,𝑘𝛼,𝛽,𝜆𝐼𝐺(𝑧)+𝛼𝑧𝜇,𝑘𝛼,𝛽,𝜆𝐼𝐺(𝑧)=(1−𝛼)𝑧𝜇,𝑘𝛼,𝛽,𝜆𝐼(𝜙∗𝑓)(𝑧)+𝛼𝑧𝜇,𝑘𝛼,𝛽,𝜆(𝜙∗𝑓)(𝑧)=𝑧𝜙(𝑧)∗ℎ(𝑧),ℎ∈𝑃𝑛(=𝑛𝜌)4+12(1−𝜌)𝑧𝜙(𝑧)∗ℎ1+1(𝑧)+𝜌2−𝑛4(1−𝜌)𝑧𝜙(𝑧)∗ℎ2(𝑧)+𝜌,ℎ1,ℎ2∈𝑃.(3.20) Since ℜ(𝑧𝜙(𝑧))>1/2 and by using Lemma 2.2, we can conclude that 𝐺=𝜙∗𝑓∈Σ𝜇,𝑘𝑛,𝜆(𝛼,𝛽,𝜌).

Theorem 3.5. Let 𝜙(𝑧)∈Σ satisfy the inequality (3.19) and 𝑓∈Σ∗,𝜇,𝑘𝑛,𝜆(0,𝛽,𝜌). Then 𝜙∗𝑓∈Σ∗,𝜇,𝑘𝑛,𝜆(0,𝛽,𝜌).

Proof. We have −𝑧2𝐼𝜇,𝑘𝛼,𝛽,𝜆(𝜙∗𝑓)(𝑧)=−𝑧2𝐼𝜇,𝑘𝛼,𝛽,𝜆𝑓(𝑧)∗𝑧𝜙(𝑧),𝑧∈𝑈.(3.21) Now the remaining part of Theorem 3.5 follows by employing the techniques that we used in proving Theorem 3.4 above.

Theorem 3.6. Let 𝑓∈Σ𝜇,𝑘𝑛,𝜆(𝛼,𝛽,𝜌3) and 𝑔∈Σ𝜇,𝑘𝑛,𝜆(𝛼,𝛽,𝜌4), and let 𝐹=𝑓∗𝑔. Then 𝐹∈Σ𝑛(𝛼,𝛽,𝑘,𝜌5), where 𝜌5=1−41−𝜌31−𝜌411−𝛼10𝑢(1/(1−𝛼))−11+𝑢𝑑𝑢.(3.22) This result is sharp.

Proof. Since 𝑓∈Σ𝜇,𝑘𝑛,𝜆(𝛼,𝛽,𝜌3) and 𝑔∈Σ𝜇,𝑘𝑛,𝜆(𝛼,𝛽,𝜌4), it follows that 𝐼𝑆(𝑧)=(1−𝛼)𝑧𝜇,𝑘𝛼,𝛽,𝜆𝐼𝑓(𝑧)+𝛼𝑧𝜇,𝑘𝛼,𝛽,𝜆𝑓(𝑧)∈𝑃𝑛𝜌3,𝑆∗𝐼(𝑧)=(1−𝛼)𝑧𝜇,𝑘𝛼,𝛽,𝜆𝐼𝑔(𝑧)+𝛼𝑧𝜇,𝑘𝛼,𝛽,𝜆𝑔(𝑧)∈𝑃𝑛𝜌4,(3.23) and so using identity (1.13) in the above equation, we have 𝐼𝜇,𝑘𝛼,𝛽,𝜆1𝑓(𝑧)=𝛼𝑧−1−1/𝛼𝑧0𝑡1/𝛼−1𝐼𝑆(𝑡)𝑑𝑡,𝜇,𝑘𝛼,𝛽,𝜆1𝑔(𝑧)=𝛼𝑧−1−1/𝛼𝑧0𝑡1/𝛼−1𝑆∗(𝑡)𝑑𝑡.(3.24) Using (3.24), we have 𝐼𝜇,𝑘𝛼,𝛽,𝜆1𝑓(𝑧)=𝛼𝑧−1−1/𝛼𝑧0𝑡1/𝛼−1𝑄(𝑡)𝑑𝑡,(3.25) where 𝑛𝑄(𝑧)=4+12𝑞11(𝑧)+2−𝑛4𝑞21(𝑧)=𝛼𝑧−1−1/𝛼𝑧0𝑡1/𝛼−1𝑆∗𝑆∗(𝑡)𝑑𝑡.(3.26) Now 𝑛𝑆(𝑧)=4+12𝑠11(𝑧)+2−𝑛4𝑠2𝑆(𝑧),(3.27)∗𝑛(𝑧)=4+12𝑠∗11(𝑧)+2−𝑛4𝑠∗2(𝑧),(3.28) where 𝑠𝑖∈𝑃(𝜌3) and 𝑠∗𝑖∈𝑃(𝜌4),𝑖=1,2. Since 𝑃∗𝑖𝑠(𝑧)=∗𝑖(𝑧)−𝜌421−𝜌4+121∈𝑃2,𝑖=1,2,(3.29) we obtain that (𝑠𝑖∗𝑝∗𝑖)(𝑧)∈𝑃(𝜌3), by using the Herglots formula. Thus 𝑠𝑖∗𝑠∗𝑖𝜌(𝑧)∈𝑃5(3.30) with 𝜌5=1−21−𝜌31−𝜌4.(3.31)
Using (3.25), (3.26), (3.28), (3.31), and Lemma 2.3, we have ℜ𝑞𝑖1(𝑧)=𝛼10𝑢1/𝛼−1ℜ𝑠𝑖∗𝑠∗𝑖1(𝑢𝑧)𝑑𝑢≥𝛼10𝑢1/𝛼−12𝜌52−1+1−𝜌5≥11+𝑢|𝑧|𝑑𝑢𝛼10𝑢1/𝛼−12𝜌52−1+1−𝜌51+𝑢𝑑𝑢=1−41−𝜌31−𝜌411−𝛼10𝑢(1/(1−𝛼))−1.1+𝑢𝑑𝑢(3.32) From this, we conclude that 𝐹∈Σ𝜇,𝑘𝑛,𝜆(𝛼,𝛽,𝜌5), where 𝜌5 is given by (3.22). We discuss the sharpness as follows: we take 𝑛𝑆(𝑧)=4+121+1−𝜌3𝑧+11−𝑧2−𝑛41−1−𝜌3𝑧,𝑆1+𝑧∗𝑛(𝑧)=4+121+1−𝜌4𝑧+11−𝑧2−𝑛41−1−𝜌4𝑧.1+𝑧(3.33) Since 1+1−𝜌3𝑧∗1−𝑧1+1−𝜌4𝑧1−𝑧=1−41−𝜌31−𝜌4+41−𝜌31−𝜌41−𝑧,(3.34) it follows from (3.26) that 𝑞𝑖1(𝑧)=𝛼10𝑢1/𝛼−11−41−𝜌31−𝜌4+41−𝜌31−𝜌41−𝑧𝑑𝑢⟶1−41−𝜌31−𝜌411−𝛼10𝑢(1/(1−𝛼))−11+𝑢𝑑𝑢(3.35) as 𝑧→−1. This completes the proof.

Note 1. Some other works related to Cho-Kwon-Srivastava operator can also be referred to [26–29].

Acknowledgments

The work presented here was fully supported by UKM-ST-06-FRGS0244-2010. The authors also would like to thank the referees for some suggestions to improve the content of the paper.