Subordination Properties of Multivalent Functions Defined by Generalized Multiplier Transformation

Abstract

The main object of the present paper is to investigate several interesting subordination properties and a sharp inclusion relationship for certain subclass of multivalent analytic functions, which are defined here by the generalized multiplier transformation. Relevant connections of the results which are presented in this paper with various known results are also considered.

1. Introduction and Definitions

Let 𝒜 denote the class of analytic functions in the open unit disk 𝕌∶={𝑧∶𝑧∈ℂ,|𝑧|<1}. (1) If 𝑓 and 𝑔 are in 𝒜, we say that the function 𝑓 is said to be subordinate to 𝑔 or (equivalently) 𝑔 is said to be superordinate to 𝑓, written symbolically as 𝑓≺𝑔in𝕌or𝑓(𝑧)≺𝑔(𝑧)(𝑧∈𝕌), (2) if there exists a Schwarz function  𝑤 analytic in 𝕌, with 𝑤(0)=0 and |𝑤(𝑧)|<1, for all 𝑧∈𝕌, such that 𝑓(𝑧)=𝑔(𝑤(𝑧))(𝑧∈𝕌). (3) In particular, if the function 𝑔 is univalent in 𝕌, then we have the following equivalence (cf. [1, 2]): 𝑓≺𝑔⟺𝑓(0)=𝑔(0),𝑓(𝕌)⊂𝑔(𝕌). (4)

Let 𝒜(𝑝)(𝑝∈ℕ={1,2,…}) be the subclass of 𝒜 consisting of functions 𝑓 defined by 𝑓(𝑧)=𝑧^𝑝+^∞_𝑘=𝑝+1𝑎_𝑘𝑧^𝑘(𝑧∈𝕌), (5) which are analytic and 𝑝-valent in the open unit disk 𝕌. We note that 𝒜(1)≡𝒜.

For a function 𝑓∈𝒜(𝑝) given by (5) and 𝑔∈𝒜(𝑝) defined by 𝑔(𝑧)=𝑧^𝑝+^∞_𝑘=𝑝+1𝑏_𝑘𝑧^𝑘, (6) the Hadamard product (or convolution) of 𝑓 and 𝑔 is given by 𝑓∗𝑔(𝑧)∶=𝑧^𝑝+^∞_𝑘=𝑝+1𝑎_𝑘𝑏_𝑘𝑧^𝑘=∶𝑔∗𝑓(𝑧). (7)

Recently, Cătaş [3] defined the generalized multiplier transformation  𝐼^𝑚_𝑝(𝜆,𝑙) on 𝒜(𝑝) by the following infinite series: 𝐼^𝑚_𝑝(𝜆,𝑙)𝑓(𝑧)=𝑧^𝑝+^∞_𝑘=𝑝+1𝑝+𝜆𝑘−𝑝+𝑙𝑝+𝑙^𝑚𝑎_𝑘𝑧^𝑘𝜆>0,𝑙≥0,𝑝∈ℕ,𝑚∈ℕ₀=ℕ∪{0},𝑧∈𝕌. (8) We note that 𝐼⁰_𝑝(1,0)𝑓(𝑧)=𝑓(𝑧),𝐼¹_𝑝(1,0)𝑓(𝑧)=𝑧𝑓^′(𝑧)𝑝. (9) It is easily verified from (8) that 𝜆𝑧𝐼^𝑚_𝑝(𝜆,𝑙)𝑓^′(𝑧)=𝑝+𝑙𝐼_𝑝^𝑚+1(𝜆,𝑙)𝑓(𝑧)−𝑝(1−𝜆)+𝑙𝐼^𝑚_𝑝(𝜆,𝑙)𝑓(𝑧). (10)

The generalized multiplier transformation 𝐼^𝑚_𝑝(𝜆,𝑙) reduces several familiar operators by specializing the parameters 𝑚,𝜆,𝑙, and 𝑝.(1)For the choice of 𝜆=1, the operator defined by (8) reduces the operator 𝐼_𝑝(𝑚,𝑙), studied by Srivastava et al. [4] and Sivaprasad Kumar et al. [5].(2)By taking 𝜆=𝑝=1, the generalized multiplier transformation 𝐼^𝑚_𝑝(𝜆,𝑙) yields the operator 𝐼^𝑚_𝑙, which was investigated by Cho et al. [6, 7].(3)For 𝜆=1 and 𝑙=0, the operator 𝐼^𝑚_𝑝(𝜆,𝑙) reduces the differential operator 𝐷^𝑚_𝑝 studied by Kamali and Orhan [8] and Orhan and Kiziltunç [9] and also, for 𝜆=𝑝=1 and 𝑙=0, it yields the differential operator 𝐷^𝑚 introduced by Sălăgean [10].(4)As a special case of this operator 𝐼^𝑚_𝑝(𝜆,𝑙) for 𝑙=0 and 𝑝=1, it reduces the generalized Sălăgean operator 𝐷^𝑚_𝜆 studied by Al-Oboudi [11] and also earlier, for 𝜆=𝑙=𝑝=1, it gives the operator 𝐼^𝑚 investigated by Uralegaddi and Somanatha [12].

Now, we introduce a new subclass of functions in 𝒜(𝑝), by making use of the generalized multiplier transformation 𝐼^𝑚_𝑝(𝜆,𝑙) as follows.

Definition 1. Let 𝑚∈ℕ₀, 𝐴, 𝐵, 𝛼, 𝜆, 𝑙 be arbitrary fixed real numbers such that −1≤𝐵<𝐴≤1, 0≤𝛼<𝑝, 𝜆>0, 𝑝∈ℕ, and 𝑙≥0. A function 𝑓∈𝒜(𝑝) is said to be in the class ℛ^𝑚_𝑝(𝜆,𝑙,𝛼;𝐴,𝐵), if it satisfies the following subordination condition: 1𝑧𝐼𝑝−𝛼^𝑚𝑝(𝜆,𝑙)𝑓^′(𝑧)𝐼^𝑚_𝑝()𝑓(𝑧)≺𝜆,𝑙−𝛼1+𝐴𝑧,().1+𝐵𝑧𝑧∈𝕌 (11)

In particular, for 𝐴=1 and 𝐵=−1, we write ℛ^𝑚_𝑝(𝜆,𝑙,𝛼;1,−1)=ℛ^𝑚_𝑝(𝜆,𝑙,𝛼), where ℛ^𝑚_𝑝()=𝑝𝑧𝐼𝜆,𝑙,𝛼𝑓∈𝒜∶Re^𝑚𝑝(𝜆,𝑙)𝑓^′(𝑧)𝐼^𝑚_𝑝()𝑓(𝑧).𝜆,𝑙>𝛼,𝑧∈𝕌 (12)

Motivated by the recent work of Bulboacă et al. [13] and Patel and Mishra [14], we investigate the subordination properties of the generalized multiplier transformation 𝐼^𝑚_𝑝(𝜆,𝑙) defined by (8) and obtain a sharp inclusion relationship for the multivalent analytic function class ℛ^𝑚_𝑝(𝜆,𝑙,𝛼;𝐴,𝐵). We also derive a number of sufficient conditions for functions belonging to the subclass ℛ^𝑚_𝑝(𝜆,𝑙,𝛼) which satisfy certain subordination properties. Relevant connections of the results presented in this paper with earlier sequels are also pointed out.

2. Preliminaries

To prove our results, we will need the following lemmas.

Lemma 2 (see [1, 2]). Let a function ℎ be analytic and convex (univalent) in 𝕌, with ℎ(0)=1. Suppose also that the function 𝜑 given by 𝜑(𝑧)=1+𝑏₁𝑧+𝑏₂𝑧²+⋯ (13) is analytic in 𝕌. If 𝜑(𝑧)+𝑧𝜑^′(𝑧)𝑐≺ℎ(𝑧)(Re𝑐≥0,𝑐≠0), (14) then 𝜑(𝑧)(𝑧)=𝑐≺𝜓𝑧^𝑐𝑧₀𝑡^𝑐−1ℎ(𝑡)𝑑𝑡≺ℎ(𝑧), (15) where 𝜓 is the best dominant of (14).

We denote by 𝑃(𝛾) the class of functions 𝜑 given by (13) which are analytic in 𝕌 and satisfy the following inequality: Re𝜑(𝑧)>𝛾,0≤𝛾<1,𝑧∈𝕌. (16)

Lemma 3 (see [15]). Let the function 𝜑 given by (13) be in the class 𝑃(𝛾). Then (𝑧)2Re𝜑≥2𝛾−1+1−𝛾|𝑧|.1+0≤𝛾<1,𝑧∈𝕌 (17)

Lemma 4 (see [16]). For 0≤𝛾₁<𝛾₂<1, 𝑃𝛾₁∗𝑃𝛾₂⊂𝑃𝛾₃,𝑤ℎ𝑒𝑟𝑒𝛾₃=1−21−𝛾₁1−𝛾₂. (18) The result is the best possible.

For any complex numbers 𝑎, 𝑏, and 𝑐(𝑐∉ℤ⁻₀∶={0,−1,−2,…}), the Gaussian hypergeometric function is defined by ₂𝐹₁(𝑎,𝑏;𝑐;𝑧)=1+𝑎𝑏𝑐𝑧+𝑎()𝑏()1!𝑎+1𝑏+1𝑐()𝑧𝑐+12²!+⋯. (19)

Lemma 5 (see [17]). For any complex numbers 𝑎, 𝑏, 𝑐 (𝑐∉ℤ⁻₀), one has ¹₀𝑡^𝑏−1(1−𝑡)^{𝑐−𝑏−1}(1−𝑧𝑡)^−𝑎=Γ(𝑏)Γ()𝑑𝑡𝑐−𝑏Γ(𝑐)²𝐹₁(𝑎,𝑏;𝑐;𝑧),Re𝑐>Re𝑏>0,₂𝐹₁(𝑎,𝑏;𝑐;𝑧)=(1−𝑧)₂^−𝑎𝐹₁𝑧𝑎,𝑐−𝑏;𝑐;,(),𝑧−1𝑧∉1,∞₂𝐹₁(𝑎,𝑏;𝑐;𝑧)=₂𝐹₁(),()𝑏,𝑎;𝑐;𝑧𝑏+1₂𝐹₁(1,𝑏;𝑏+1;𝑧)=(𝑏+1)+𝑏𝑧₂𝐹₁(1,𝑏+1;𝑏+2;𝑧). (20)

Lemma 6 (see [18]). If −1≤𝐵<𝐴≤1, 𝛽>0, and the complex number 𝛾 is constrained by Re(𝛾)≥−𝛽(1−𝐴)/(1−𝐵), then the following differential equation: 𝑞(𝑧)+𝑧𝑞^′(𝑧)(𝑧)=𝛽𝑞+𝛾1+𝐴𝑧()1+𝐵𝑧𝑧∈𝕌 (21) has a univalent solution in 𝕌 given by 𝑞(𝑧)=𝑧^𝛽+𝛾(1+𝐵𝑧)^{𝛽(𝐴−𝐵)/𝐵}𝛽∫^𝑧₀𝑡^{𝛽+𝛾−1}(1+𝐵𝑡)^{𝛽(𝐴−𝐵)/𝐵}−𝛾𝑑𝑡𝛽𝑧,𝑖𝑓𝐵≠0,_𝛽+𝛾exp𝛽𝐴𝑧𝛽∫^𝑧₀𝑡^{𝛽+𝛾−1}−𝛾exp𝛽𝐴𝑡𝑑𝑡𝛽,𝑖𝑓𝐵=0. (22) If the function 𝜑(𝑧) given by (13) is analytic in 𝕌 and satisfies the following subordination: 𝜑(𝑧)+𝑧𝜑^′(𝑧)(𝑧)≺𝛽𝜑+𝛾1+𝐴𝑧(),1+𝐵𝑧𝑧∈𝕌 (23) then 𝜑(𝑧)≺𝑞(𝑧)≺1+𝐴𝑧(),1+𝐵𝑧𝑧∈𝕌 (24) and 𝑞(𝑧) is the best dominant of (23).

Lemma 7 (see [19]). Let 𝑣 be a positive measure on the interval [0,1]. Let ℎ(𝑧,𝑡) be a complex-valued function defined on 𝕌×[0,1] such that ℎ(⋅,𝑡) is analytic in 𝕌 for each 𝑡∈[0,1] and ℎ(𝑧,⋅) is 𝑣-integrable on [0,1] for each 𝑧∈𝕌. In addition, suppose that Re(ℎ(𝑧,𝑡))>0, ℎ(−𝑟,𝑡) is real, and 1Reℎ()≥1𝑧,𝑡ℎ()𝑧|[.−𝑟,𝑡(|≤𝑟<1;𝑡∈0,1]) (25) If the function 𝐻(𝑧) is defined by 𝐻(𝑧)=¹₀ℎ(𝑧,𝑡)𝑑𝑣(𝑡), (26) then 1Re𝐻(𝑧)≥1𝐻()𝑧|).−𝑟(|≤𝑟<1 (27)

Lemma 8 (see [20]). Let 𝜆≠0 be a real number, 𝑎/𝜆>0, and 0≤𝛽<1. Let 𝑔(𝑧)=1+𝑐_𝑛𝑧^𝑛+𝑐_𝑛+1𝑧^𝑛+1+⋯ be analytic in 𝕌 and 𝑔(𝑧)≺1+𝑎𝑀𝑧(),𝑛𝜆+𝑎𝑛∈ℕ (28) where |𝜆)𝑀=1−𝛽|(1+𝑛𝜆/𝑎1−𝜆+𝜆𝛽+1+(1+𝑛𝜆/𝑎)². (29) If 𝑃(𝑧)=1+𝑑_𝑛𝑧^𝑛+𝑑_𝑛+1𝑧^𝑛+1+⋯ is analytic in 𝕌 and satisfies the subordination relation 𝑔(𝑧)1−𝜆+𝜆1−𝛽𝑃(𝑧)+𝛽≺1+𝑀𝑧, (30) then Re𝑃(𝑧)>0 for 𝑧∈𝕌.

Lemma 9 (see [2]). Suppose that the function Ψ∶ℂ²×𝕌→ℂ satisfies the following condition: ReΨ𝑖𝑥,𝑦;𝑧≤𝜀, (31) for all 𝑥∈ℝ and 𝑦≤−(1/2)(1+𝑥²) and for all 𝑧∈𝕌. If the function 𝜑 of the form (13) is analytic in 𝕌 and ReΨ𝜑(𝑧),𝑧𝜑^′(𝑧);𝑧>𝜀, (32) then Re𝜑(𝑧)>0(𝑧∈𝕌). (33)

3. Subordination Properties of 𝐼^𝑚_𝑝(𝜆,𝑙)

Unless otherwise mentioned, we assume throughout this paper that 𝜆>0,𝑙≥0,𝑚∈ℕ₀,0≤𝛼<𝑝,𝑝∈ℕ,−1≤𝐵<𝐴≤1. (34)

Theorem 10. Let 𝛽>0 and −1≤𝐵_𝑗<𝐴_𝑗≤1, 𝑗=1,2. If the functions 𝑓_𝑗∈𝒜(𝑝) satisfy the following subordination condition: 𝐼1−𝛽^𝑚𝑝(𝜆,𝑙)𝑓_𝑗(𝑧)𝑧^𝑝𝐼+𝛽^𝑝𝑚+1(𝜆,𝑙)𝑓_𝑗(𝑧)𝑧^𝑝≺1+𝐴_𝑗𝑧1+𝐵_𝑗𝑧,𝑗=1,2, (35) then 𝐼1−𝛽^𝑚𝑝(𝜆,𝑙)𝐹(𝑧)𝑧^𝑝𝐼+𝛽^𝑝𝑚+1(𝜆,𝑙)𝐹(𝑧)𝑧^𝑝≺1+(1−2𝛿)𝑧,1−𝑧 (36) where 𝐹=𝐼^𝑚_𝑝(𝜆,𝑙)(𝑓₁∗𝑓₂) and 4𝐴𝛿=1−¹−𝐵₁𝐴₂−𝐵₂1−𝐵₁1−𝐵₂×11−2²𝐹₁1,1;𝑝+𝑙1𝜆𝛽+1;2. (37) The result is the best possible when 𝐵₁=𝐵₂=−1.

Proof. Let the functions 𝑓_𝑗∈𝒜(𝑝), 𝑗=1,2, satisfy the subordination condition (35). Then, by setting 𝜑_𝑗(𝑧)=𝐼1−𝛽^𝑚𝑝(𝜆,𝑙)𝑓_𝑗(𝑧)𝑧^𝑝𝐼+𝛽^𝑝𝑚+1(𝜆,𝑙)𝑓_𝑗(𝑧)𝑧^𝑝≺1+𝐴_𝑗𝑧1+𝐵_𝑗𝑧,𝑗=1,2, (38) we have 𝜑_𝑗∈𝑃𝛾_𝑗,𝛾_𝑗=1−𝐴_𝑗1−𝐵_𝑗,𝑗=1,2. (39) By making use of (10) and (38), we obtain 𝐼^𝑚_𝑝(𝜆,𝑙)𝑓_𝑗(𝑧)=𝑝+𝑙𝑧𝜆𝛽^{𝑝−(𝑝+𝑙)/𝜆𝛽𝑧}₀𝑡^{(𝑝+𝑙)/𝜆𝛽−1}𝜑_𝑗(𝑡)𝑑𝑡,𝑗=1,2. (40) Now, if we let 𝐹=𝐼^𝑚_𝑝(𝜆,𝑙)(𝑓₁∗𝑓₂), then by using (40) and the fact that 𝐼^𝑚_𝑝(𝜆,𝑙)𝐹(𝑧)=𝐼^𝑚_𝑝(𝜆,𝑙)𝐼^𝑚_𝑝(𝜆,𝑙)𝑓₁∗𝑓₂(𝑧)=𝐼^𝑚_𝑝(𝜆,𝑙)𝑓₁(𝑧)∗𝐼^𝑚_𝑝(𝜆,𝑙)𝑓₂(𝑧), (41) a simple computation shows that 𝐼^𝑚_𝑝(𝜆,𝑙)𝐹(𝑧)=𝑝+𝑙𝑧𝜆𝛽^{𝑝−(𝑝+𝑙)/𝜆𝛽𝑧}₀𝑡^{(𝑝+𝑙)/𝜆𝛽−1}𝜑₀(𝑡)𝑑𝑡, (42) where 𝜑₀(𝑧)=𝐼1−𝛽^𝑚𝑝(𝜆,𝑙)𝐹(𝑧)𝑧^𝑝𝐼+𝛽^𝑝𝑚+1(𝜆,𝑙)𝐹(𝑧)𝑧^𝑝=𝑝+𝑙𝑧𝜆𝛽^{−(𝑝+𝑙)/𝜆𝛽𝑧}₀𝑡^{(𝑝+𝑙)/𝜆𝛽−1}𝜑₁∗𝜑₂(𝑡)𝑑𝑡. (43) Since 𝜑_𝑗∈𝑃(𝛾_𝑗), 𝑗=1,2, it follows from Lemma 4 that 𝜑₁∗𝜑₂∈𝑃𝛾₃,where𝛾₃=1−21−𝛾₁1−𝛾₂, (44) and the bound 𝛾₃ is the best possible. Hence, by using Lemma 3 in (43), we deduce that Re𝜑₀(𝑧)=𝑝+𝑙𝜆𝛽¹₀𝑢^{(𝑝+𝑙)/𝜆𝛽−1}Re𝜑₁∗𝜑₂()≥𝑢𝑧𝑑𝑢𝑝+𝑙𝜆𝛽¹₀𝑢^{(𝑝+𝑙)/𝜆𝛽−1}2𝛾₃2−1+1−𝛾₃|𝑧|>1+𝑢𝑑𝑢𝑝+𝑙𝜆𝛽¹₀𝑢^{(𝑝+𝑙)/𝜆𝛽−1}2𝛾₃2−1+1−𝛾₃4𝐴1+𝑢𝑑𝑢=1−¹−𝐵₁𝐴₂−𝐵₂1−𝐵₁1−𝐵₂×1−𝑝+𝑙𝜆𝛽¹₀𝑢^{(𝑝+𝑙)/𝜆𝛽−1}(),1+𝑢𝑑𝑢=𝛿𝑧∈𝕌 (45) where 𝛿 is given by (37).
When 𝐵₁=𝐵₂=−1, we consider the functions 𝑓_𝑗∈𝒜(𝑝)  (𝑗=1,2) which satisfy the hypothesis (35) and are given by 𝐼^𝑚_𝑝(𝜆,𝑙)𝑓_𝑗(𝑧)=𝑝+𝑙𝑧𝜆𝛽^{𝑝−(𝑝+𝑙)/𝜆𝛽𝑧}₀𝑡^{(𝑝+𝑙)/𝜆𝛽−1}1+𝐴_𝑗𝑡1−𝑡𝑑𝑡,𝑗=1,2. (46) Since 1+𝐴₁𝑧∗1−𝑧1+𝐴₂𝑧1−𝑧=1−1+𝐴₁1+𝐴₂+1+𝐴₁1+𝐴₂,1−𝑧 (47) it follows from (43) that 𝜑₀(𝑧)=𝑝+𝑙×𝜆𝛽₁₀𝑢^{(𝑝+𝑙)/𝜆𝛽−1}1−1+𝐴₁1+𝐴₂+1+𝐴₁1+𝐴₂1−𝑢𝑧𝑑𝑢=1−1+𝐴₁1+𝐴₂+1+𝐴₁1+𝐴₂(1−𝑧)²𝐹₁1,1;𝑝+𝑙𝑧𝜆𝛽+1;.𝑧−1 (48) Therefore 𝜑₀(𝑧)⟶1−1+𝐴₁1+𝐴₂+121+𝐴₁1+𝐴₂×₂𝐹₁1,1;𝑝+𝑙1𝜆𝛽+1;2as𝑧⟶−1, (49) which evidently completes our proof of Theorem 10.

By setting 𝛽=1, 𝐵_𝑗=−1, and 𝐴_𝑗=1−2𝛿_𝑗, 𝑗=1,2, in Theorem 10, we have the following corollary.

Corollary 11. If the functions 𝑓_𝑗∈𝒜(𝑝) satisfy the following subordination condition: 𝐼_𝑝^𝑚+1(𝜆,𝑙)𝑓_𝑗(𝑧)𝑧^𝑝≺1+1−2𝛿_𝑗𝑧1−𝑧,𝑗=1,2, (50) then 𝐼Re^𝑝𝑚+1(𝜆,𝑙)𝐹(𝑧)𝑧^𝑝>1−21−𝛿₁1−𝛿₂×2−₂𝐹₁1,1;𝑝+𝑙𝜆1+1;2(𝑧∈𝕌), (51) where 𝐹=𝐼^𝑚_𝑝(𝜆,𝑙)(𝑓₁∗𝑓₂).

In Theorem 12, we have determined the sufficient condition for the functions 𝐼^𝑚_𝑝(𝜆,𝑙)𝑓(𝑧)/𝑧^𝑝 to be a member of the class 𝑃(𝜂).

Theorem 12. If 𝑓∈𝒜(𝑝) satisfy the following subordination condition: 𝐼1−𝛽^𝑚𝑝(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝𝐼+𝛽^𝑝𝑚+1(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝≺1+𝐴𝑧,1+𝐵𝑧 (52) then 𝐼Re^𝑚𝑝(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝>𝜂(𝑧∈𝕌), (53) where 𝐴𝜂=𝐵+𝐴1−𝐵(1−𝐵)⁻¹×₂𝐹₁1,1;𝑝+𝑙𝐵𝜆𝛽+1;𝐵−1,𝑖𝑓𝐵≠0,1−𝑝+𝑙𝑝+𝑙+𝜆𝛽𝐴,𝑖𝑓𝐵=0. (54) The result is the best possible.

Proof. Let 𝑔(𝑧)=𝐼^𝑚_𝑝(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝for𝑓∈𝒜𝑝. (55) Then, the function 𝑔 is of the form (13). Differentiating (55) with respect to 𝑧 and using the identity (10), we obtain 𝐼_𝑝^𝑚+1(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝(𝑧)+𝜆=𝑔𝑝+𝑙𝑧𝑔^′(𝑧). (56) By using (52), (55), and (56), we get 𝑔(𝑧)+𝜆𝛽𝑝+𝑙𝑧𝑔^′(𝑧)≺1+𝐴𝑧.1+𝐵𝑧 (57) Now, by applying Lemma 2, we have 𝑔(𝑧)≺𝑄(𝑧)=𝑝+𝑙𝑧𝜆𝛽^{−(𝑝+𝑙)/𝜆𝛽𝑧}₀𝑡^{(𝑝+𝑙)/𝜆𝛽−1}1+𝐴𝑡1+𝐵𝑡𝑑𝑡. (58) By using Lemma 5, we get 𝑄(𝑧)=𝐴𝐵+𝐴1−𝐵(1+𝐵𝑧)⁻¹×₂𝐹₁1,1;𝑝+𝑙𝜆𝛽+1;𝐵𝑧𝐵𝑧+1,if𝐵≠0,1+𝑝+𝑙𝑝+𝑙+𝜆𝛽𝐴𝑧,if𝐵=0. (59) Now, we will show that inf{Re𝑄(𝑧)∶|𝑧|<1}=𝑄(−1). (60) We have Re1+𝐴𝑧≥1+𝐵𝑧1−𝐴𝑟|𝑧|1−𝐵𝑟=𝑟<1, (61) and setting ℎ(𝑠,𝑧)=1+𝐴𝑧𝑠(),(𝑠)=1+𝐵𝑧𝑠0≤𝑠≤1𝑑𝜇𝑝+𝑙𝑠𝜆𝛽^{(𝑝+𝑙)/𝜆𝛽}𝑑𝑠, (62) which is a positive measure on the closed interval [0,1], we get 𝑄(𝑧)=¹₀ℎ(𝑠,𝑧)𝑑𝜇(𝑠), (63) so that Re𝑄(𝑧)≥¹₀1−𝐴𝑠𝑟(𝑠)(),|𝑧|1−𝐵𝑠𝑟𝑑𝜇=𝑄−𝑟=𝑟<1. (64) As 𝑟→1⁻ in (64), we obtain the assertion (60). Now, by using (59) and (60), we get 𝐼Re^𝑚𝑝(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝>𝜂, (65) where 𝜂 is given by (54).
To show that the estimate (54) is the best possible, we consider the function 𝑓∈𝒜(𝑝) defined by 𝐼^𝑚_𝑝(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝=𝑝+𝑙𝜆𝛽¹₀𝑢^{(𝑝+𝑙)/𝜆𝛽−1}1+𝐴𝑢𝑧1+𝐵𝑢𝑧𝑑𝑢. (66) For the above function, we find that 𝐼1−𝛽^𝑚𝑝(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝𝐼+𝛽^𝑝𝑚+1(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝=1+𝐴𝑧,𝐼1+𝐵𝑧_𝑚𝑝(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝⟶𝑝+𝑙𝜆𝛽¹₀𝑢^{(𝑝+𝑙)/𝜆𝛽−1}1−𝐴𝑢=𝐴1−𝐵𝑢𝑑𝑢𝐵+𝐴1−𝐵(1−𝐵)⁻¹×₂𝐹₁1,1;𝑝+𝑙𝐵𝜆𝛽+1;𝐵−1,if𝐵≠0,1−𝑝+𝑙𝑝+𝑙+𝜆𝛽𝐴,if𝐵=0 (67) as 𝑧→−1, and the proof of the Theorem 12 is completed.

In its special case when 𝐴=1−2𝛾, 𝐵=−1 and 𝛽=1, Theorem 12 yields the following corollary.

Corollary 13. If 𝑓∈𝒜(𝑝) satisfy the following condition: 𝐼_𝑝^𝑚+1(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝≺1+1−2𝛾𝑧(),1−𝑧𝑧∈𝕌 (68) then 𝐼Re^𝑚𝑝(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝×>𝛾+1−𝛾₂𝐹₁1,1;𝑝+𝑙𝜆1+1;2−1(𝑧∈𝕌). (69) The result is the best possible.

For a function 𝑓∈𝒜(𝑝), the integral operator 𝐹_𝜇,𝑝𝑓(𝑧)=𝜇+𝑝𝑧^𝜇𝑧₀𝑡^𝜇−1𝑓(𝑡)𝑑𝑡=𝑧^𝑝₂𝐹₁(𝑧).1,𝜇+𝑝;𝜇+𝑝+1;𝑧∗𝑓𝜇>−𝑝,𝑧∈𝕌 (70)

Also, it is easily verified from (70) that 𝑧𝐼^𝑚_𝑝(𝜆,𝑙)𝐹_𝜇,𝑝𝑓^′(𝑧)=𝜇+𝑝𝐼^𝑚_𝑝(𝜆,𝑙)𝑓(𝑧)−𝜇𝐼^𝑚_𝑝(𝜆,𝑙)𝐹_𝜇,𝑝𝑓(𝑧). (71)

In the next Theorem 14, by using the integral operator defined in (70), we established the sufficient condition for the functions 𝐼^𝑚_𝑝(𝜆,𝑙)𝐹_𝜇,𝑝𝑓(𝑧)/𝑧^𝑝 belongs to 𝑃(𝜌₀).

Theorem 14. If 𝑓∈𝒜(𝑝) and 𝐹_𝜇,𝑝𝑓 is given by (70), satifies the subordination condition: 𝐼1−𝛽^𝑚𝑝(𝜆,𝑙)𝐹_𝜇,𝑝𝑓(𝑧)𝑧^𝑝𝐼+𝛽^𝑚𝑝(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝≺1+𝐴𝑧,1+𝐵𝑧 (72) then 𝐼Re^𝑚𝑝(𝜆,𝑙)𝐹_𝜇,𝑝𝑓(𝑧)𝑧^𝑝>𝜌₀(𝑧∈𝕌), (73) where 𝜌₀=𝐴𝐵+𝐴1−𝐵(1−𝐵)⁻¹×₂𝐹₁1,1;𝜇+𝑝𝛽𝐵+1;𝐵−1,𝑖𝑓𝐵≠0,1−𝜇+𝑝𝑝+𝑙+𝛽𝐴,𝑖𝑓𝐵=0. (74) The result is the best possible.

Proof. Let ℎ(𝑧)=𝐼^𝑚_𝑝(𝜆,𝑙)𝐹_𝜇,𝑝𝑓(𝑧)𝑧^𝑝. (75) Then by using the hypothesis (72) together with (71) and (75), we obtain ℎ(𝑧)+𝛽𝜇+𝑝𝑧ℎ^′(𝑧)=𝐼1−𝛽^𝑚𝑝(𝜆,𝑙)𝐹_𝜇,𝑝𝑓(𝑧)𝑧^𝑝𝐼+𝛽^𝑚𝑝(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝≺1+𝐴𝑧.1+𝐵𝑧 (76) The remaining part of the proof of Theorem 14 is similar to that of Theorem 12 and hence we omit the details.

4. Inclusion Relationship for the Class ℛ^𝑚_𝑝(𝜆,𝑙,𝛼;𝐴,𝐵)

Theorem 15. If 𝑓(𝑧)∈ℛ_𝑝^𝑚+1(𝜆,𝑙,𝛼;𝐴,𝐵) and 𝜆𝑝−𝛼(1−𝐴)+𝛼𝜆+𝑝(1−𝜆)+𝑙(1−𝐵)≥0, (77) then 1𝑧𝐼𝑝−𝛼^𝑚𝑝(𝜆,𝑙)𝑓^′(𝑧)𝐼^𝑚_𝑝()𝑓(𝑧)(𝑧)≺𝜆,𝑙−𝛼≺𝑞1+𝐴𝑧,(),1+𝐵𝑧𝑧∈𝕌 (78) where 𝑞(𝑧)=1𝜆1𝑝−𝛼𝑄(𝑧)−(),𝑄(𝑧)=∫𝛼𝜆+𝑝1−𝜆+𝑙₁₀𝑡^{𝑝+𝑙−1}1+𝐵𝑡𝑧1+𝐵𝑧^{𝜆(𝑝−𝛼)(𝐴−𝐵)/𝐵}∫𝑑𝑡,𝑖𝑓𝐵≠0,₁₀𝑡^{𝑝+𝑙−1}exp𝜆𝑝−𝛼𝐴𝑧(𝑡−1)𝑑𝑡,𝑖𝑓𝐵=0 (79) and 𝑞(𝑧) is the best dominant of (78). If, in addition to (77), ()𝐴>−𝐵𝛼𝜆+𝑝1−𝜆+𝑙+1𝜆𝑝−𝛼𝑤𝑖𝑡ℎ−1≤𝐵<0, (80) then ℛ_𝑝^𝑚+1(𝜆,𝑙,𝛼;𝐴,𝐵)⊂ℛ^𝑚_𝑝𝜆,𝑙,𝛼;1−2𝜌,−1, (81) where 1𝜌=𝜆××𝑝−𝛼𝑝+𝑙₂𝐹₁𝜆()1,𝑝−𝛼𝐵−𝐴𝐵𝐵;𝑝+𝑙+1;𝐵−1⁻¹−𝛼𝜆+𝑝(1−𝜆)+𝑙. (82) The bound on 𝜌 is the best possible.

Proof. Let 𝑓(𝑧)∈ℛ_𝑝^𝑚+1(𝜆,𝑙,𝛼;𝐴,𝐵). Define the function 𝑔 by 𝑔(𝑧)𝐼=𝑧^𝑚𝑝(𝜆,𝑙)𝑓(𝑧)𝑧^{𝑝1/(𝑝−𝛼)} (83) and 𝑟₁=sup{𝑟∶𝑔(𝑧)≠0,0<|𝑧|<𝑟<1}. Then 𝑔(𝑧) is single-valued and analytic function in |𝑧|<𝑟₁. By logarithmic differentiation in (83), it follows that the function 𝜙(𝑧) given by 𝜙(𝑧)=𝑧𝑔^′(𝑧)𝑔(𝑧)=1𝑧𝐼𝑝−𝛼^𝑚𝑝(𝜆,𝑙)𝑓^′(𝑧)𝐼^𝑚_𝑝()𝑓(𝑧)≺𝜆,𝑙−𝛼1+𝐴𝑧1+𝐵𝑧 (84) is analytic in |𝑧|<𝑟₁ and 𝜙(0)=1. Using the identity (10) in (84) and logarithmic differentiation of the resulting equation yields the following: 1𝑧𝐼𝑝−𝛼^𝑝𝑚+1(𝜆,𝑙)𝑓^′(𝑧)𝐼_𝑝^𝑚+1()𝑓(𝑧)(𝑧)+𝜆,𝑙−𝛼=𝜙𝜆𝑧𝜙^′(𝑧)𝜆𝜙(𝑧)()≺𝑝−𝛼+𝛼𝜆+𝑝1−𝜆+𝑙1+𝐴𝑧|𝑧|1+𝐵𝑧<𝑟₁. (85) Hence, by using Lemma 6 with 𝛽=𝜆(𝑝−𝛼) and 𝛾=𝛼𝜆+𝑝(1−𝜆)+𝑙, we find that 𝜙(𝑧)≺1𝜆1𝑝−𝛼𝑄(𝑧)−()(𝑧)≺𝛼𝜆+𝑝1−𝜆+𝑙=𝑞1+𝐴𝑧|𝑧|1+𝐵𝑧<𝑟₁, (86) where 𝑞(𝑧) is the best dominant of (86) and 𝑄(𝑧) is given by (79). Since Re1+𝐴𝑧(),1+𝐵𝑧>0−1≤𝐵<𝐴≤1;𝑧∈𝕌 (87) by (86), we have Re(𝜙(𝑧))>0(|𝑧|<𝑟₁). Now, (84) shows that 𝑔(𝑧) is starlike (univalent) in |𝑧|<𝑟₁. Thus, it is not possible that 𝑔(𝑧) vanishes on |𝑧|=𝑟₁ if 𝑟₁<1. So, we conclude that 𝑟₁=1, and, therefore, 𝜙(𝑧) is analytic in 𝕌. Hence, (86) implies that 𝜙(𝑧)≺𝑞(𝑧)≺1+𝐴𝑧().1+𝐵𝑧𝑧∈𝕌 (88) This proves the assertion (78) of Theorem 15.
In order to establish (81), we have to find the greatest lower bound of 𝜌(0<𝜌<1) such that 𝜙(𝑧)≺1+1−2𝜌𝑧().1−𝑧𝑧∈𝕌 (89) By (86), we have to show that 𝜌=inf_𝑧∈𝕌Re𝑞(𝑧)=𝑞(−1). (90) To prove (90), we need to show that inf_𝑧∈𝕌1Re𝑄(𝑧)=1𝑄().−1 (91) From (79), we see that, for 𝐵≠0, 𝑄(𝑧)=(1+𝐵𝑧)^𝑎1₀𝑡^𝑏−1(1−𝑡)^{𝑐−𝑏−1}(1+𝐵𝑧𝑡)^−𝑎(),𝑑𝑡𝑧∈𝕌 (92) where 𝜆()𝑎=𝑝−𝛼𝐵−𝐴𝐵,𝑏=𝑝+𝑙,𝑐=𝑝+𝑙+1. (93) Since 𝑐>𝑏>0, by using Lemma 5, we get following: 𝑄(𝑧)=(1+𝐵𝑧)^𝑎Γ(𝑏)Γ(𝑐−𝑏)Γ(𝑐)²𝐹₁()=Γ(𝑏)𝑎,𝑏;𝑐;−𝐵𝑧Γ(𝑐)²𝐹₁𝑎,𝑐−𝑏;𝑐;𝐵𝑧=Γ(𝑏)𝐵𝑧+1Γ(𝑐)²𝐹₁1,𝑎;𝑐;𝐵𝑧.𝐵𝑧+1 (94) Since ()𝐴>−𝐵𝛼𝜆+𝑝1−𝜆+𝑙+1𝜆𝑝−𝛼with−1≤𝐵<0 (95) implies that 𝑐>𝑎>0, by using Lemma 5, we find from (94) that 𝑄(𝑧)=¹₀ℎ(𝑧,𝑡)𝑑𝑣(𝑡), (96) where ℎ(𝑧,𝑡)=1+𝐵𝑧()(),(𝑡)=Γ(𝑏)1+1−𝑡𝐵𝑧0≤𝑡≤1𝑑𝑣Γ(𝑎)Γ()𝑡𝑐−𝑎^𝑎−1(1−𝑡)^{𝑐−𝑎−1}𝑑𝑡, (97) which is positive measure on [0,1]. For −1≤𝐵<0, it may be noted that Re(ℎ(𝑧,𝑡))>0 and ℎ(−𝑟,𝑡) is real for 0≤|𝑧|≤𝑟<1 and 𝑡∈[0,1]. Hence, by using Lemma 7, we have 1Re𝑄(𝑧)≥1𝑄()𝑧|),−𝑟(|≤𝑟<1inf_𝑧∈𝕌1Re𝑄(𝑧)=inf_−1<𝑟<11𝑄()=1−𝑟∫¹₀ℎ()(𝑡)=1−1,𝑡𝑑𝑣𝑄().−1 (98) We note that 𝑄(−1)≠0. Thus, by using (90) and (94), we have 1𝜌=𝜆××𝑝−𝛼𝑝+𝑙₂𝐹₁𝜆()1,𝑝−𝛼𝐵−𝐴𝐵𝐵;𝑝+𝑙+1;𝐵−1⁻¹−𝛼𝜆+𝑝(1−𝜆)+𝑙, (99) when 𝐴>−𝐵[𝛼𝜆+𝑝(1−𝜆)+𝑙+1]/𝜆(𝑝−𝛼). Further by taking 𝐴→(−𝐵[𝛼𝜆+𝑝(1−𝜆)+𝑙+1]/𝜆(𝑝−𝛼))⁺  for the case 𝐴=−𝐵[𝛼𝜆+𝑝(1−𝜆)+𝑙+1]/𝜆(𝑝−𝛼) and using (78), we get (81). The result is the best possible as the function 𝑞(𝑧) is the best dominant of (78). This completes the proof of Theorem 15.

In the following section, we obtain the sufficient condition for the function 𝑓 to be a member of the class ℛ^𝑚_𝑝(𝜆,𝑙,𝛼).

5. Sufficient Conditions for the Class ℛ^𝑚_𝑝(𝜆,𝑙,𝛼)

Theorem 16. If 𝑓∈𝒜(𝑝) satisfy the following subordination condition: 𝐼1−𝛽^𝑚𝑝(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝𝐼+𝛽^𝑝𝑚+1(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝≺1+𝑀₁𝑧, (100) where 𝑀₁=𝜆𝛽𝑝−𝛼1+𝜆𝛽/𝑝+𝑙𝑝+𝑙−𝜆𝛽𝑝−𝛼+𝑝+𝑙²+𝑝+𝑙+𝜆𝛽², (101) then 𝑓∈ℛ^𝑚_𝑝(𝜆,𝑙,𝛼).

Proof. Let 𝑔(𝑧)=𝐼^𝑚_𝑝(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝. (102) Then, the function 𝑔 is of the form (13) and is analytic in 𝕌. From Theorem 12 with 𝐴=𝑀₁ and 𝐵=0, we have 𝑔(𝑧)≺1+𝑝+𝑙𝑀𝑝+𝑙+𝜆𝛽¹𝑧, (103) which is equivalent to 𝑔(𝑧)−1<𝑝+𝑙𝑀𝑝+𝑙+𝜆𝛽¹=𝑁<1(𝑧∈𝕌). (104) If we set 𝑃(𝑧)=1𝑧𝐼𝑝−𝛼^𝑚𝑝(𝜆,𝑙)𝑓^′(𝑧)𝐼^𝑚_𝑝()𝑓(𝑧).𝜆,𝑙−𝛼0≤𝛼<𝑝 (105) Then, by using the identity (10) followed by (102), we obtain 𝐼_𝑝^𝑚+1(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝=𝜆1−𝑝−𝛼+𝜆𝑝+𝑙𝑝−𝛼𝑃(𝑧)𝑔(𝑧).𝑝+𝑙 (106) In view of (106), the hypothesis (100) can be written as follows: 1−𝜆𝛽𝑝−𝛼𝑔(𝑧)+𝑝+𝑙𝜆𝛽𝑝−𝛼𝑃(𝑧)𝑔(𝑧)𝑝+𝑙−1<𝑀₁(𝑧∈𝕌). (107) We need to show that (107) yields Re𝑃(𝑧)>0(𝑧∈𝕌). (108) Suppose that this is false. Since 𝑃(0)=1, there exists a point 𝑧₀∈𝕌 such that 𝑃(𝑧₀)=𝑖𝑥 for some 𝑥∈ℝ. Therefore, in order to show that (108), it is sufficient to obtain the contradiction from the inequality 𝐸=1−𝜆𝛽𝑝−𝛼𝑔𝑧𝑝+𝑙⁰+𝜆𝛽𝑝−𝛼𝑃𝑧𝑝+𝑙⁰𝑔𝑧₀−1≥𝑀₁. (109) If we let 𝑔(𝑧₀)=𝑢+𝑖𝑣, then, by using (104) and the triangle inequality, we obtain that 𝐸²=1−𝜆𝛽𝑝−𝛼𝑔𝑧𝑝+𝑙⁰+𝜆𝛽𝑝−𝛼𝑃𝑧𝑝+𝑙⁰𝑔𝑧₀−1²=𝑢²+𝑣²𝜆𝛽𝑥𝑝−𝛼𝑝+𝑙²+2𝑣𝜆𝛽𝑥𝑝−𝛼+𝑝+𝑙1−𝜆𝛽𝑝−𝛼𝑔𝑧𝑝+𝑙⁰−1²≥𝑢²+𝑣²𝜆𝛽𝑥𝑝−𝛼𝑝+𝑙²+2𝑣𝜆𝛽𝑥𝑝−𝛼+𝑝+𝑙𝜆𝛽(𝑝−𝛼)−𝑝+𝑙1−𝜆𝛽(𝑝−𝛼)𝑁𝑝+𝑙². (110) If we let Ψ(𝑥)=𝐸²−𝑀²₁≥𝑢²+𝑣²𝜆𝛽𝑥𝑝−𝛼𝑝+𝑙²+2𝑣𝜆𝛽𝑥𝑝−𝛼+𝑝+𝑙𝜆𝛽𝑝−𝛼−𝑝+𝑙1−𝜆𝛽𝑝−𝛼𝑁𝑝+𝑙²−𝑁²𝑝+𝑙+𝜆𝛽𝑝+𝑙², (111) then (109) holds true if Ψ(𝑥)≥0, for any 𝑥∈𝕌. Since (𝑢²+𝑣²)(𝜆𝛽(𝑝−𝛼)/(𝑝+𝑙))²>0, the inequality Ψ(𝑥)≥0 holds true if the discriminant Δ≤0; that is, Δ=4𝜆𝛽𝑝−𝛼𝑝+𝑙²×𝑣²−𝑢²+𝑣²×𝜆𝛽𝑝−𝛼−𝑝+𝑙1−𝜆𝛽𝑝−𝛼𝑁𝑝+𝑙²−𝑁²𝑝+𝑙+𝜆𝛽𝑝+𝑙²≤0, (112) which is equivalent to 𝑣²1−𝜆𝛽𝑝−𝛼−𝑝+𝑙1−𝜆𝛽𝑝−𝛼𝑁𝑝+𝑙²+𝑁²𝑝+𝑙+𝜆𝛽𝑝+𝑙²≤𝑢²𝜆𝛽𝑝−𝛼−𝑝+𝑙1−𝜆𝛽𝑝−𝛼𝑁𝑝+𝑙²−𝑁²𝑝+𝑙+𝜆𝛽𝑝+𝑙². (113) After a simple computation, by using (104), we obtain the inequality 𝑣²𝑢²≤𝜌²1−𝜌²≤𝑁²1−𝑁²≤𝜆𝛽𝑝−𝛼−𝑝+𝑙1−𝜆𝛽𝑝−𝛼𝑁𝑝+𝑙²−𝑁²𝑝+𝑙+𝜆𝛽𝑝+𝑙²×1−𝜆𝛽𝑝−𝛼−𝑝+𝑙1−𝜆𝛽𝑝−𝛼𝑁𝑝+𝑙²+𝑁²𝑝+𝑙+𝜆𝛽𝑝+𝑙²⁻¹, (114) which yields Δ≤0. Therefore 𝐸≥𝑀₁, which contradicts (107). It follows that Re𝑃(𝑧)>0, and 𝑓∈ℛ^𝑚_𝑝(𝜆,𝑙,𝛼).

Theorem 17. Let 𝛽>0 and if 𝑓∈𝒜(𝑝) such that 𝐼^𝑚_𝑝(𝜆,𝑙)𝑓(𝑧)/𝑧^𝑝≠0, 𝑧∈𝕌, satisfies the following differential subordination: 𝐼1−𝛽^𝑚𝑝(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝𝜎𝐼+𝛽^𝑚𝑝(𝜆,𝑙)𝑓^′(𝑧)𝑝𝑧^𝑝−1𝐼^𝑚_𝑝(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝𝜎−1≺1+𝑀₂𝑧, (115) where the powers are understood as the principle value, and 𝑀₂=𝑝−𝛼𝛽1+𝛽/𝜎𝑝𝑝−𝑝−𝛼𝛽+𝑝²+𝑝+𝛽/𝜎²𝛽,𝑖𝑓𝜎≠0,𝑝−𝛼𝑝,𝑖𝑓𝜎=0, (116) then 𝑓∈ℛ^𝑚_𝑝(𝜆,𝑙,𝛼).

Proof . If 𝜎=0, then the condition (115) is equivalent to 𝑧𝐼^𝑚_𝑝(𝜆,𝑙)𝑓^′(𝑧)𝐼^𝑚_𝑝()𝑓(𝑧)().𝜆,𝑙−𝑝<𝑝−𝛼𝑧∈𝕌 (117) The above equation (117) implies that 𝑓∈ℛ^𝑚_𝑝(𝜆,𝑙,𝛼).
If we consider 𝜎>0 and suppose that 𝑔(𝑧)=𝐼^𝑚_𝑝(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝𝜎(𝑧∈𝕌). (118) Choosing the principal value in (118), we note that 𝑔 is of the form (13) and is analytic in 𝕌. Differentiating (118) with respect to 𝑧, we obtain 𝑔(𝑧)+𝛽𝜎𝑝𝑧𝑔^′(𝑧)=𝐼1−𝛽^𝑚𝑝(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝𝜎𝐼+𝛽^𝑚𝑝(𝜆,𝑙)𝑓^′(𝑧)𝑝𝑧^𝑝−1𝐼^𝑚_𝑝(𝜆,𝑙)𝑓(𝑧)𝑧^𝑝𝜎−1, (119) which, in view of Lemma 2 with 𝑐=𝜎𝑝/𝛽, yields 𝑔(𝑧)≺1+𝜎𝑝𝑀𝜎𝑝+𝛽²𝑧. (120) Also, with the aid of (118), (115) can be written as follows: 𝑔(𝑧𝛼)1−𝛽+𝛽1−𝑝𝑃(𝑧)+𝛼𝑝≺1+𝑀₂𝑧, (121) where 𝑃 is given by (105). Therefore, by Lemma 8, we find that Re𝑃(𝑧)>0(𝑧∈𝕌); (122) that is 𝑧𝐼Re^𝑚𝑝(𝜆,𝑙)𝑓^′(𝑧)𝐼^𝑚_𝑝()𝑓(𝑧),𝜆,𝑙>𝛼0≤𝛼<𝑝,𝑧∈𝕌 (123) which completes the proof of Theorem 17

By taking 𝑚=0 in Theorem 17, we get the following corollary due to Patel et al. [21].

Corollary 18. Let 𝛽>0, 0≤𝛼<𝑝, 𝑝∈ℕ, and 𝜎≥0. If 𝑓∈𝒜(𝑝) such that 𝑓(𝑧)/𝑧^𝑝≠0  (𝑧∈𝕌), and satisfies the following subordination condition: 𝑓(𝑧)1−𝛽𝑧^𝑝𝜎𝑓+𝛽^′(𝑧)𝑝𝑧^𝑝−1𝑓(𝑧)𝑧^𝑝𝜎−1≺1+𝑀₂𝑧(𝑧∈𝕌), (124) where the powers are understood as the principal value and 𝑀₂ is given by (116), then 𝑓∈𝑆^∗_𝑝(𝛼).

Remark 19. Taking 𝑚=0 and 𝑝=1 in Theorem 17, we get the result of Liu [20, Theorem  2.2, with 𝑛=1].

Remark 20. Putting 𝑚=0 and 𝛽=𝑝=1 in Theorem 17, we obtain the result of Liu [20, Corollary  2.1, with 𝑛=1].

Remark 21. Putting 𝑚=0 and 𝑝=𝛽=𝜎=1, in Theorem 17, we get the result of Mocanu and Oros [22, Corollary  2.2, with 𝑛=1].

Remark 22. Taking 𝑚=𝛼=0, and 𝑝=𝛽=𝜎=1 in Theorem 17, we obtain the result of Mocanu [23, with 𝑛=1].

Taking 𝑚=0 and 𝛽=1/(𝑝−𝛼), 0≤𝛼<𝑝 in Theorem 17, we obtain the following corollary.

Corollary 23. Let 𝜎>0, 0≤𝛼<𝑝, and 𝑝∈ℕ. If 𝑓∈𝒜(𝑝) such that 𝑓(𝑧)/𝑧^𝑝≠0(𝑧∈𝕌) satisfies the inequality: 𝑓(𝑧)𝑝−𝛼−1𝑧^𝑝𝜎+𝑓^′(𝑧)𝑝𝑧^𝑝−1𝑓(𝑧)𝑧^𝑝𝜎−1<+𝛼−𝑝𝑝−𝛼/𝑝𝑝𝜎𝑝−𝛼+1𝜎𝑝−𝛼𝑝−1+𝑝𝜎𝑝−𝛼²+𝑝𝜎𝑝−𝛼+1²(𝑧∈𝕌), (125) where the powers are the principal value ones, then 𝑓∈𝑆^∗_𝑝(𝛼).

Remark 24. Taking 𝑝=1 in Corollary 23, we obtain the result of Liu [20, Corollary  2.2, with 𝑛=1].

Remark 25. Putting 𝑝=𝜎=1 in Corollary 23, we get the result of Mocanu and Oros [22, Corollary  2.4, with 𝑛=1].

In Theorem 26, we obtain the necessary condition for functions belonging to class 𝑃(𝛾_𝑗).

Theorem 26. Let 𝑓_𝑗∈𝒜(𝑝) (𝑗=1,2). If the functions 𝐼_𝑝^𝑚+1(𝜆,𝑙)𝑓_𝑗(𝑧)/𝑧^𝑝∈𝑃(𝛾_𝑗)  (0≤𝛾_𝑗<1), then the function 𝑔=𝐼^𝑚_𝑝(𝜆,𝑙), (𝑓₁∗𝑓₂) satisfies the following inequality: 𝐼Re^𝑝𝑚+1(𝜆,𝑙)𝑔(𝑧)𝐼^𝑚_𝑝()𝑔(𝑧)(),𝜆,𝑙>0𝑧∈𝕌 (126) provided that 21−𝛾₁1−𝛾₂<𝜆+2𝑝+𝑙𝜆₂𝐹₁1,1;𝑝+𝑙/𝜆+1;1/2−1².+2𝑝+𝑙 (127)

Proof. Let 𝑔=𝐼^𝑚_𝑝(𝜆,𝑙)(𝑓₁∗𝑓₂). By using Lemma 4 and (10), we obtain 𝐼Re^𝑝𝑚+1(𝜆,𝑙)𝑓₁(𝑧)𝑧^𝑝∗𝐼_𝑝^𝑚+1(𝜆,𝑙)𝑓₂(𝑧)𝑧^𝑝𝐼=Re^𝑝𝑚+1(𝜆,𝑙)𝑔(𝑧)𝑧^𝑝+𝜆𝑧𝐼𝑝+𝑙^𝑝𝑚+1(𝜆,𝑙)𝑔(𝑧)𝑧^𝑝′>1−21−𝛾₁1−𝛾₂. (128) Then, by using Lemma 2 with 𝑐=(𝑝+𝑙)/𝜆, 𝐴=−1+4(1−𝛾₁)(1−𝛾₂), and 𝐵=−1, we have 𝐼Re^𝑝𝑚+1(𝜆,𝑙)𝑔(𝑧)𝑧^𝑝>1+21−𝛾₁1−𝛾₂×₂𝐹₁1,1;𝑝+𝑙𝜆1+1;2−1. (129) Again, from (129) and Corollary 13, we obtain 𝐼Re^𝑚𝑝(𝜆,𝑙)𝑔(𝑧)𝑧^𝑝>1−21−𝛾₁1−𝛾₂×₂𝐹₁1,1;𝑝+𝑙𝜆1+1;2−1². (130) Now, if we let 𝜑(𝑧)=𝐼_𝑝^𝑚+1(𝜆,𝑙)𝑔(𝑧)/𝐼^𝑚_𝑝(𝜆,𝑙)𝑔(𝑧)  (𝑧∈𝕌) then 𝜑 is of the form (13) and is analytic in 𝕌. A simple computation shows that 𝐼_𝑝^𝑚+1(𝜆,𝑙)𝑔(𝑧)𝑧^𝑝+𝜆𝑧𝐼𝑝+𝑙^𝑝𝑚+1(𝜆,𝑙)𝑔(𝑧)𝑧^𝑝′=𝐼^𝑚_𝑝(𝜆,𝑙)𝑔(𝑧)𝑧^𝑝𝜑²(𝑧)+𝑧𝜆𝜑^′(𝑧)𝜑(𝑧)𝑝+𝑙=Ψ,𝑧𝜑^′(𝑧);𝑧, (131) where Ψ(𝑢,𝑣;𝑧)=(𝐼^𝑚_𝑝(𝜆,𝑙)𝑔(𝑧)/𝑧^𝑝)(𝑢²+𝜆𝑣/(𝑝+𝑙)).
Thus, by using (128), (131) can be written as follows: ReΨ𝜑(𝑧),𝑧𝜑^′(𝑧);𝑧>1−21−𝛾₁1−𝛾₂. (132) For all real 𝑥 and 𝑦≤−(1/2)(1+𝑥²), we have Ψ=Re𝑖𝑥,𝑦;𝑧𝜆𝑦𝑝+𝑙−𝑥²𝐼Re^𝑚𝑝(𝜆,𝑙)𝑔(𝑧)𝑧^𝑝𝜆≤−22𝑝+𝑙1+1+𝑝+𝑙𝜆𝑥²𝐼×Re^𝑚𝑝(𝜆,𝑙)𝑔(𝑧)𝑧^𝑝𝜆≤−2𝐼𝑝+𝑙Re^𝑚𝑝(𝜆,𝑙)𝑔(𝑧)𝑧^𝑝<1−21−𝛾₁1−𝛾₂(𝑧∈𝕌), (133) where we used (127) and (130). Thus, by Lemma 9, we get Re𝜑(𝑧)>0(𝑧∈𝕌), (134) which completes the proof of Theorem 26.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The authors thank the referees for their insightful suggestions.