Differential Subordination for Certian Subclasses of 𝑝-Valent Functions Assoicated with Generalized Linear Operator

Abstract

By making use of the differential subordination analytic functions, we investigate inclusion relationships among certain classes of analytic and 𝑝-valent functions defined by generalized linear operator.

1. Introduction

Let 𝒜(𝑝) denote the class of functions of the form 𝑓(𝑧)=𝑧^𝑝+^∞_𝑘=1𝑎_𝑘+𝑝𝑧^𝑘+𝑝𝑝∈ℕ={1,2,…}, (1) which are analytic and 𝑝-valent in the open unit disc 𝑈={𝑧∈ℂ∶|𝑧|<1}.

For functions 𝑓∈𝒜(𝑝), given by (1) and 𝑔 given by 𝑔(𝑧)=𝑧^𝑝+^∞_𝑘=1𝑏_𝑘+𝑝𝑧^𝑘+𝑝𝑝∈ℕ, (2) the Hadamard product (or convolution) of 𝑓 and 𝑔 is defined by 𝑓∗𝑔(𝑧)=𝑧^𝑝+^∞_𝑘=1𝑎_𝑘+𝑝𝑏_𝑘+𝑝𝑧^𝑘+𝑝=𝑔∗𝑓(𝑧)𝑧∈𝑈;𝑝∈ℕ. (3)

For 𝑓,𝑔∈𝒜(𝑝), we say that the function 𝑓 is subordinate to 𝑔 denoted by 𝑓(𝑧)≺𝑔(𝑧), if there exists a Schwarz function 𝑤, that is, 𝑤∈𝒜 with 𝑤(0)=0 and |𝑤(𝑧)|<1,𝑧∈𝑈, such that 𝑓(𝑧)=𝑔(𝑤(𝑧)) for all 𝑧∈𝑈. It is well known that, if the function 𝑔 is univalent in 𝑈, then 𝑓(𝑧)≺𝑔(𝑧) is equivalent to 𝑓(0)=𝑔(0) and 𝑓(𝑈)⊂𝑔(𝑈) (see [1, 2]).

Now, we introduce the linear multiplier operator ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}∶𝒜(𝑝)→𝒜(𝑝) by ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)=𝑧^𝑝+^∞_𝑘=11+𝜆𝑘𝑝^𝑚𝑐+𝑝𝑐+𝑘+𝑝^𝛿𝑎_𝑘+𝑝𝑧^𝑘+𝑝𝛿≥0;𝑐>−𝑝;𝜆≥0;𝑚∈ℕ₀. (4) It is easily verified from (4) that 𝑧ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)^′=𝑐+𝑝ℒ^{𝑚,𝛿−1}_{𝑐,𝑝,𝜆}𝑓(𝑧)−𝑐ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧),𝜆𝑧ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)^′=𝑝ℒ^𝑚+1,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)−𝑝(1−𝜆)ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧). (5)(6) By specializing the parameters 𝑐, 𝜆, 𝛿, 𝑝, and 𝑚, we obtain the following operators:(i)ℒ^1,𝛿_𝑐,𝑝,0𝑓(𝑧)=𝐾^𝛿_𝑐,𝑝𝑓(𝑧) (see [3]);(ii)ℒ^0,𝛿_{𝑎−1,1,𝜆}𝑓(𝑧)=𝐿^𝛿_𝑎𝑓(𝑧) (see [4, 5]); (iii)ℒ^1,𝛿_1,𝑝,0𝑓(𝑧)=ℐ^𝛿_𝑝𝑓(𝑧) (see [6, 7]); (iv)ℒ^𝑚,0_{𝑐,𝑝,𝜆}𝑓(𝑧)=𝐷^𝑚_𝜆,𝑝𝑓(𝑧) (see [8]); (v)ℒ^1,𝛿_𝑐,1,0𝑓(𝑧)=𝒫^𝛿_𝑐𝑓(𝑧) (see [9, 10]);(vi)ℒ^1,1_𝑐,1,0𝑓(𝑧)=ℒ_𝑐𝑓(𝑧) (see [11]); (vii)ℒ^1,𝛿_1,1,0𝑓(𝑧)=ℐ^𝛿𝑓(𝑧) (see [12]); (viii)ℒ^𝑚,0_𝑐,1,𝜆𝑓(𝑧)=𝒟^𝑚_𝜆𝑓(𝑧) (see [13]); (ix)ℒ^𝑚,0_𝑐,1,1𝑓(𝑧)=𝒟^𝑚𝑓(𝑧) (see [14]).

By using the multiplier operator ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧), we define the following classes of functions.

Definition 1. For fixed parameters 𝐴 and 𝐵, with −1≤𝐵<𝐴≤1 and 𝑝>𝛼, we say that the function 𝑓∈𝒜(𝑝) is in the class 𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵) if it satisfies the following subordination condition: 1𝑧ℒ𝑝−𝛼^{𝑚,𝛿𝑐,𝑝,𝜆}𝑓(𝑧)^′ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)−𝛼≺1+𝐴𝑧.1+𝐵𝑧 (7) We note that(i)𝒮^1,0_1,𝑝,0(𝛼;𝐴,𝐵)=𝒮^∗_𝑝(𝛼;𝐴,𝐵) was studied by Cho et al. [15]; (ii)𝒮^1,𝛿_1,𝑝,0(𝛼;𝐴,𝐵)=𝒮^𝛿_𝑝(𝛼;𝐴,𝐵) was studied by Aouf et al. [16]; (iii)𝒮^𝑚,0_𝑐,1,𝜆(0;𝐴,𝐵)=𝒮^𝑚_𝜆(𝐴,𝐵) was studied by Patel [17].
For complex numbers 𝑎, 𝑏, and 𝑐, the Gaussian hypergeometric function is defined by ₂𝐹₁(𝑎,𝑏,𝑑;𝑧)=1+𝑎⋅𝑏𝑑𝑧+𝑎()()1!𝑎+1⋅𝑏𝑏+1𝑑()𝑧𝑑+1²=2!+⋯_∞𝑘=0(𝑎)_𝑘(𝑏)_𝑘(𝑑)_𝑘𝑧^𝑘{},𝑘!,𝑎,𝑏∈ℂ,𝑑∈ℂ⧵0,−1,−2,… (8) where (𝜆)_𝑘=𝜆(𝜆+1)⋯(𝜆+𝑘−1) and (𝜆)₀=1. The series (8) converges absolutely for 𝑧∈𝑈, hence it represents an analytic function in 𝑈 (see [18, Chapter 14]).
If 𝐵≠−1, from the fact that ℎ(𝑧)=ℎ(𝑧),𝑧∈𝑈, we deduce that the image ℎ(𝑈) is symmetric with respect to the real axis, and that ℎ maps the unit disc 𝑈 onto the disc |𝑤−(1−𝐴𝐵)/(1−𝐵²)|<(𝐴−𝐵)/(1−𝐵²). If 𝐵=−1, the function ℎ maps the unit disc 𝑈 onto the half plan Re𝑤>(1−𝐴)/2, hence we obtain the following.

Remark 2. The function 𝑓∈𝒜(𝑝) is in the class 𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵) if and only if 1𝑧ℒ𝑝−𝛼^{𝑚,𝛿𝑐,𝑝,𝜆}𝑓(𝑧)^′ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)−𝛼−1−𝐴𝐵1−𝐵²<𝐴−𝐵1−𝐵²,𝑧∈𝑈,for𝐵≠−1,1Re𝑧ℒ𝑝−𝛼^{𝑚,𝛿𝑐,𝑝,𝜆}𝑓(𝑧)^′ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)−𝛼>1−𝐴2,𝑧∈𝑈,for𝐵=−1. (9)(10)

Definition 3. The function 𝑓∈𝒜(𝑝)is in the class 𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝜌) if it satisfies the following the inequality: 1Re𝑧ℒ𝑝−𝛼^{𝑚,𝛿𝑐,𝑝,𝜆}𝑓(𝑧)^′ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)−𝛼>𝜌,𝑧∈𝑈, (11) where 𝜌<1; from (9) and (10) it follows, respectively, that 𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵)⊂𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}𝛼;1−𝐴,𝒮1−𝐵_{𝑚,𝛿𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵)=𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}𝛼;1−𝐴1−𝐵⟺𝐵=−1. (12) We note that when 𝑐=1 and 𝑚=0, the class 𝒮^1,𝛿_1,𝑝,0(𝛼;𝜌)=𝒮^𝛿_𝑝(𝛼;𝜌) was studied by Aouf et al. [16].
Let us consider the first-order differential subordination 𝐻𝜑(𝑧),𝑧𝜑^′(𝑧)≺ℎ(𝑧). (13)
A univalent function 𝑞 is called its dominant, if 𝜑(𝑧)≺𝑞(𝑧) for all analytic functions 𝜑 that satisfy this differential subordination. A dominant 𝑞 is called the best dominant, if 𝑞(𝑧)≺𝑞(𝑧) for all dominants 𝑞. For the general theory of the first-order differential subordination and its applications, we refer the reader to [1, 2].
The object of the present paper is to obtain several inclusion relationships and other interesting properties of functions belonging to the subclasses 𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵) and 𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝜌) by using the theory of differential subordination.
To establish our main results, we will require the following lemmas.

Lemma 4 (see [19]). Let 𝛽,𝛾∈ℂ, and let ℎ be a convex function withRe𝛽ℎ(𝑧)+𝛾>0,𝑧∈𝑈. (14)

If 𝑝 is analytic in 𝑈, with 𝑝(0)=ℎ(0), then 𝑝(𝑧)+𝑧𝑝^′(𝑧)(𝑧)(𝑧)(𝑧)(𝑧).𝛽𝑝+𝛾≺ℎ⇒𝑝≺ℎ (15)

Lemma 5 (see [20]). Let 𝛽>0,𝛽+𝛾>0, and consider the integral operator 𝐽_𝛽,𝛾 defined by 𝐽_𝛽,𝛾𝑓(𝑧)=𝛽+𝛾𝑧^𝛾𝑧₀𝑓^𝛽(𝑡)𝑡^𝛾−1𝑑𝑡^1/𝛽, (16) where the powers are the principal ones.
If 𝜎∈[−𝛾/𝛽,1) then the order of starlikeness of the class 𝐽_𝛽,𝛾(𝑆^∗(𝜎)), that is, the largest number 𝛿(𝜎;𝛽,𝛾) such that 𝐽_𝛽,𝛾(𝑆^∗(𝜎))⊂𝑆^∗(𝛿), is given by the number 𝛿(𝜎;𝛽,𝛾)=inf{Req(𝑧)∶𝑧∈𝑈}, where 𝑞(𝑧)=1(𝑧)−𝛾𝛽𝑄𝛽,𝑄(𝑧)=¹₀1−𝑧1−𝑡𝑧^{2𝛽(1−𝜎)}𝑡^{𝛽+𝛾−1}𝑑𝑡. (17)
Moreover, if 𝜎∈[𝜎₀,1), where 𝜎₀=max{(𝛽−𝛾−1)/2𝛽;−𝛾/𝛽} and 𝑔=𝐽_𝛽,𝛾(𝑓)(𝑧) with 𝑓∈𝑆^∗(𝜎), then Re𝑧𝑔^′(𝑧)𝑔(𝑧)>𝛿𝜎;𝛽,𝛾(𝑧∈𝑈), (18) where 𝛿=1𝜎;𝛽,𝛾𝛽𝛽+𝛾₂𝐹₁().1,2𝛽1−𝜎,𝛽+𝛾+1;1/2−𝛾 (19)

Lemma 6 (see [21]). Let 𝜙  be analytic in 𝑈 with 𝜙(0)=1 and 𝜙(𝑧)≠0 for 0<|𝑧|<1, and let 𝐴,𝐵∈ℂ with 𝐴≠𝐵, |𝐵|≤1.

(i) Let 𝐵≠0 and 𝛾∈ℂ^∗=ℂ⧵{0} satisfy either |(𝛾(𝐴−𝐵)/𝐵)−1|≤1 or |(𝛾(𝐴−𝐵)/𝐵)+1|≤1. If 𝜙 satisfies 1+𝑧𝜙^′(𝑧)(𝑧)≺𝛾𝜙1+𝐴𝑧1+𝐵𝑧 (20) then 𝜙(𝑧)≺(1+𝐵𝑧)^{𝛾(𝐴−𝐵)/𝐵}, (21) and this is the best dominant.

(ii) Let 𝛾∈ℂ^∗ be such that |𝛾𝐴|<𝜋, and if 𝜙 satisfies 1+𝑧𝜙^′(𝑧)𝛾𝜙(𝑧)≺1+𝐴𝑧 (22) then 𝜙(𝑧)≺𝑒^𝛾𝐴𝑧, (23) and this is the best dominant.

2. Inclusion Relationships

Unless otherwise mentioned, we assume throughout this paper that −1≤𝐵<𝐴≤1,𝑝>𝛼,𝛿≥0,𝑐>−𝑝,𝜆≥0,𝑝∈ℕ,𝑚∈ℤ (24) and the power is the principal one.

Theorem 7. Let 𝑝−𝛼(1−𝐴)+(𝛼+𝑐)(1−𝐵)≥0. (25)(1)Supposing that ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)≠0 for all 𝑧∈𝑈̇=𝑈⧵{0}, then 𝒮^{𝑚,𝛿−1}_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵)⊂𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵).(2)Moreover, if we suppose in addition that 𝐴≤1+1−𝐵𝑝−𝛼𝛼+𝑐−𝑝2+12, (26)then 𝒮^{𝑚,𝛿−1}_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵)⊂𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}𝛼;𝜌𝑝,𝛼,𝑐,𝐴,𝐵, (27) where the bound 𝜌=1𝑝,𝛼,𝑐,𝐴,𝐵×𝑝−𝛼𝑝+𝑐₂𝐹₁()/()−(,1,2𝑝−𝛼𝐴−𝐵1−𝐵,𝑝+𝑐+1;1/2𝛼+𝑐) (28) is the best possible.

Proof. Let 𝑓∈𝒮^{𝑚,𝛿−1}_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵), and put 𝑔(𝑧)ℒ=𝑧^{𝑚,𝛿𝑐,𝑝,𝜆}𝑓(𝑧)𝑧^{𝑝1/(𝑝−𝛼)}(𝑧∈𝑈), (29) the function 𝑔 is analytic in 𝑈, with 𝑔(0)=0 and 𝑔^′(0)=1. Differentiating (29) logarithmically with respect to 𝑧, we have 𝜙(𝑧)=𝑧𝑔^′(𝑧)𝑔(𝑧)=1𝑧ℒ𝑝−𝛼^{𝑚,𝛿𝑐,𝑝,𝜆}𝑓(𝑧)^′ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)(),−𝛼𝑧∈𝑈 (30) then, using (5) in (30), we obtain ℒ𝑐+𝑝^{𝑚,𝛿−1𝑐,𝑝,𝜆}𝑓(𝑧)ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)=𝑝−𝛼𝜙(𝑧)+(𝛼+𝑐). (31) By differentiating both sides of (31) logarithmically with respect to 𝑧 and multiplying by 𝑧, we have 1𝑧ℒ𝑝−𝛼^{𝑚,𝛿−1𝑐,𝑝,𝜆}𝑓(𝑧)^′ℒ^{𝑚,𝛿−1}_{𝑐,𝑝,𝜆}𝑓(𝑧)(𝑧)+−𝛼=𝜙𝑧𝜙^′(𝑧)𝜙(𝑧)+().𝑝−𝛼𝛼+𝑐 (32) Combining (32) together with 𝑓∈𝒮^{𝑚,𝛿−1}_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵), we obtain that the function 𝜙 satisfies the Briot-Bouquet differential subordination as follows: 𝜙(𝑧)+𝑧𝜙^′(𝑧)𝜙(𝑧)+()≺𝑝−𝛼𝛼+𝑐1+𝐴𝑧(𝑧).1+𝐵𝑧≡ℎ (33)
Now we will use Lemma 4 for the special case 𝛽=𝑝−𝛼 and 𝛾=𝛼+𝑐. Since ℎ is a convex function in 𝑈, a simple computation shows that Re𝑝−𝛼ℎ(𝑧)+(𝛼+𝑐)>0,𝑧∈𝑈, (34) whenever (25) holds, we have 𝜙(𝑧)≺ℎ(𝑧); that is, 𝑓∈𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵). If in addition, we suppose that the inequality (26) holds, then all the assumptions of Lemma 5 are verified for the above values of 𝛽, 𝛾, and 𝜎=(1−𝐴)/(1−𝐵). Then it follows the inclusion 𝒮^{𝑚,𝛿−1}_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵)⊂𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝜌(𝑝,𝛼,𝑐,𝐴,𝐵)), where the bound 𝜌(𝑝,𝛼,𝑐,𝐴,𝐵) given by (28) is the best possible.
From Theorem 7, according to the definitions (7) and (11), we deduce the next inclusions.

Corollary 8. Let 𝑝>𝛼, such that (25) holds.(1)Supposing that ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)≠0 for all 𝑧∈𝑈̇, then 𝒮^{𝑚,𝛿−1}_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵)⊂𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵)⊂𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}𝛼;1−𝐴.1−𝐵 (35)(2)If we suppose in addition that (26) holds, then 𝒮^{𝑚,𝛿−1}_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵)⊂𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵)⊂𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}𝛼;𝜌𝑝,𝛼,𝑐,𝐴,𝐵, (36)where 𝜌(𝑝,𝛼,𝑐,𝐴,𝐵) is given by (28). As a consequence of the last inclusion, one has 𝜌(𝑝,𝛼,𝑐,𝐴,𝐵)≥(1−𝐴)/(1−𝐵).
For the special case 𝐵=−1, Theorem 7 reduces to the following.

Corollary 9. Let 𝑎≥−(𝑐+𝛼)/(𝑝−𝛼).(1)Supposing that ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)≠0 for all 𝑧∈𝑈̇, then𝒮^{𝑚,𝛿−1}_{𝑐,𝑝,𝜆}(𝛼;𝑎)⊂𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝑎). (37)(2)If we suppose in addition that 𝑎≥max−2𝛼+𝑐−𝑝+12𝑝−𝛼;−𝛼+𝑐,𝑝−𝛼 (38)then 𝒮^{𝑚,𝛿−1}_{𝑐,𝑝,𝜆}(𝛼;𝑎)⊂𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}𝛼;𝜌𝑝,𝛼,𝑐,𝑎, (39) where the bound 𝜌=1𝑝,𝛼,𝑐,𝑎×𝑝−𝛼𝑝+𝑐₂𝐹₁()−(,1,2𝑝−𝛼1−𝑎,𝑝+𝑐+1;1/2𝛼+𝑐) (40) is the best possible.

Theorem 10. Let 𝑓∈𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝜌), where 𝜌<1 and 𝜌>𝛼, then 𝑓∈𝒮^{𝑚,𝛿−1}_{𝑐,𝑝,𝜆}(𝛼;𝜌) for |𝑧|<𝑅(𝑝,𝛼,𝑐,𝜌), where 𝑅,𝜃()×()−()/𝑝,𝛼,𝑐,𝜌=min𝑟>0∶𝜃𝑝,𝛼,𝑐,𝜌,𝑟=0𝑝,𝛼,𝑐,𝜌,𝑟=2𝑟1−𝑟𝑝−𝛼1−𝜌1−𝑟𝜌+𝛼+𝑐𝑝−𝛼(1+𝑟)⁻¹. (41)

Proof. Since 𝑓∈𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝜌), the function 𝑢 given by 𝑢(𝑧)=1𝑧ℒ𝑝−𝛼^{𝑚,𝛿𝑐,𝑝,𝜆}𝑓(𝑧)^′ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)−𝛼(𝑧∈𝑈), (42) is analytic in 𝑈 with 𝑢(0)=1 and Re𝑢(𝑧)>𝜌. Using (5) in (42) and taking the logarithmic differentiation in the resulting equation, we obtain 1𝑧ℒ𝑝−𝛼^{𝑚,𝛿−1𝑐,𝑝,𝜆}𝑓(𝑧)^′ℒ^{𝑚,𝛿−1}_{𝑐,𝑝,𝜆}𝑓(𝑧)(𝑧)+−𝛼=𝑢𝑧𝑢^′(𝑧)𝑢(𝑧)+().𝑝−𝛼𝛼+𝑐 (43) If we denote 𝑘(𝑧)=(𝑢(𝑧)−𝜌)/(1−𝜌), then 𝑘(0)=1 and Re𝑘(𝑧)>0,𝑧∈𝑈 and substituting in (43) we obtain 1𝑧ℒ𝑝−𝛼^{𝑚,𝛿−1𝑐,𝑝,𝜆}𝑓(𝑧)^′ℒ^{𝑚,𝛿−1}_{𝑐,𝑝,𝜆}𝑓(𝑧)=×𝑘(𝑧)+−𝛼−𝜌1−𝜌𝑧𝑘^′(𝑧)𝑘(𝑧)+(),𝑝−𝛼1−𝜌+𝜌𝑝−𝛼𝛼+𝑐 (44) hence 1Re𝑧ℒ𝑝−𝛼^{𝑚,𝛿−1𝑐,𝑝,𝜆}𝑓(𝑧)^′ℒ^{𝑚,𝛿−1}_{𝑐,𝑝,𝜆}𝑓(𝑧)≥×(𝑧)−−𝛼−𝜌1−𝜌Re𝑘𝑧𝑘^′(𝑧×𝑘(𝑧)−()/)𝑝−𝛼1−𝜌𝜌+𝛼+𝑐𝑝−𝛼⁻¹. (45) By using the well-known results [22] 𝑧𝑘^′(𝑧)≤2𝑟1−𝑟²Re𝑘(𝑧),Re𝑘(𝑧)≥1−𝑟,|𝑧|1+𝑟=𝑟<1, (46) together with the inequality (45), we get 1Re𝑧ℒ𝑝−𝛼^{𝑚,𝛿−1𝑐,𝑝,𝜆}𝑓(𝑧)^′ℒ^{𝑚,𝛿−1}_{𝑐,𝑝,𝜆}𝑓(𝑧)≥(𝑧),|𝑧|−𝛼−𝜌1−𝜌1−𝜃𝑝,𝛼,𝑐,𝜌,𝑟Re𝑘=𝑟. (47) Since the right hand side term of the inequality (47) is nonnegative whenever |𝑧|≤𝑅(𝑝,𝛼,𝑐,𝜌) is given by (41), using the fact that the real part of an analytic function is harmonic, we deduce that 𝑓∈𝒮^{𝑚,𝛿−1}_{𝑐,𝑝,𝜆}(𝛼;𝜌) for |𝑧|<𝑅(𝑝,𝛼,𝑐,𝜌).
For a function 𝑓∈𝒜(𝑝), let the integral operator 𝐹_𝜇,𝑝∶𝒜(𝑝)→𝒜(𝑝) defined by Saitoh [23] and Saitoh et al. [24] 𝐹_𝜇,𝑝𝑓(𝑧)=𝜇+𝑝𝑧^𝜇𝑧₀𝑡^𝜇−1𝑓(𝑡)𝑑𝑡=𝑧^𝑝+^∞_𝑘=𝑝+1𝜇+𝑝𝑎𝑘+𝜇^𝑘𝑧^𝑘=𝑧^𝑝+^∞_𝑘=𝑝+1𝜇+𝑝𝑧𝑘+𝜇^𝑘∗𝑓(𝑧)=𝑧^𝑝₂𝐹₁(𝑧).1,𝜇+𝑝;𝜇+𝑝+1;𝑧∗𝑓𝑧∈𝑈;𝜇>−𝑝 (48) From (4) and (48), we have 𝑧ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝐹_𝜇,𝑝𝑓(𝑧)^′=𝜇+𝑝ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)−𝜇ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝐹_𝜇,𝑝𝑓(𝑧),ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝐹_𝜇,𝑝𝑓(𝑧)=𝐹_𝜇,𝑝ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧),𝑓∈𝐴𝑝. (49)

We now prove the next result.

Theorem 11. Let 𝛼+𝜇>0 and 𝛼+𝜇(1−𝐵)+𝑝−𝛼(1−𝐴)≥0. (50)(i)Supposing that 𝐹_𝜇,𝑝(𝑓(𝑧))≠0 for all 𝑧∈𝑈̇=𝑈⧵{0}, then 𝐹_𝜇,𝑝𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵)⊂𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵). (51)(ii)Moreover, if we suppose in addition that 𝐴≤1+1−𝐵𝑝−𝛼min𝜇+2𝛼−𝑝+12;𝜇+𝛼, (52)then 𝐹_𝜇,𝑝𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵)⊂𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}𝛼;𝑟𝑝,𝛼,𝜇,𝐴,𝐵, (53) where the bound 𝑟=1𝑝,𝛼,𝜇,𝐴,𝐵×𝑝−𝛼𝑝+𝜇₂𝐹₁()/()−,1,2𝑝−𝛼𝐴−𝐵1−𝐵,𝜇+𝑝+1;1/2𝜇+𝛼 (54) is the best possible.

Proof. Let 𝑓∈𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵), and suppose that 𝐹_𝜇,𝑝(𝑓(𝑧))≠0 for all 𝑧∈𝑈̇. Let 𝑔(𝑧)ℒ=𝑧^{𝑚,𝛿𝑐,𝑝,𝜆}𝐹_𝜇,𝑝𝑓(𝑧)𝑧^{𝑝1/(𝑝−𝛼)}(𝑧∈𝑈), (55) then 𝑔 is analytic in 𝑈, with 𝑔(0)=0 and 𝑔^′(0)=1. Taking the logarithmic differentiation in (55), we have 𝜙(𝑧)=𝑧𝑔^′(𝑧)𝑔(𝑧)=1𝑧ℒ𝑝−𝛼^{𝑚,𝛿𝑐,𝑝,𝜆}𝐹_𝜇,𝑝𝑓(𝑧)^′ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝐹_𝜇,𝑝𝑓(𝑧)().−𝛼𝑧∈𝑈 (56) Now, by using (49) in (56), we obtain ℒ𝑝+𝜇^{𝑚,𝛿𝑐,𝑝,𝜆}𝑓(𝑧)ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝐹_𝜇,𝑝𝑓(𝑧=𝜙(𝑧)+.)𝑝−𝛼𝜇+𝛼 (57) By differentiating in both sides of (57) logarithmical with respect to 𝑧 and multiplying by 𝑧, we have 1𝑧ℒ𝑝−𝛼^{𝑚,𝛿𝑐,𝑝,𝜆}𝑓(𝑧)^′ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)(𝑧)+−𝛼=𝜙𝑧𝜙^′(𝑧)𝜙(𝑧)+.𝑝−𝛼𝜇+𝛼 (58) Since 𝑓∈𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵), from (58), we obtain that the function 𝜙 satisfies the Briot-Bouquet differential subordination 𝜙(𝑧)+𝑧𝜙^′(𝑧)𝜙(𝑧)+≺𝑝−𝛼𝜇+𝛼1+𝐴𝑧(𝑧).1+𝐵𝑧≡ℎ (59)
Now we will use Lemma 4 for the special case 𝛽=𝑝−𝛼 and 𝛾=𝜇+𝛼; we have 𝜙(𝑧)≺ℎ(𝑧), that is, 𝐹_𝜇,𝑝∈𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵). If we suppose in addition that the inequality (52) holds, then all the assumptions of the Lemma 5 are satisfied for 𝛽,  𝛾, and 𝜎=(1−𝐴)/(1−𝐵), hence it follows the inclusion 𝐹_𝜇,𝑝(𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵))⊂𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝑟(𝑝,𝛼,𝜇,𝐴,𝐵)), and the bound 𝑟(𝑝,𝛼,𝜇,𝐴,𝐵) given by (54) is the best possible.

Taking 𝐵=−1 in Theorem 11, we obtain the next corollary.

Corollary 12. Let 𝛼+𝜇>0 and 𝑎≥−(𝜇+𝛼)/(𝑝−𝛼).(1)Supposing that ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝐹_𝜇,𝑝(𝑓(𝑧))≠0 for all 𝑧∈𝑈̇, then𝐹_𝜇,𝑝𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝑎)⊂𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝑎). (60)(2)If we suppose in addition that 𝑎≥max−𝜇+2𝛼−𝑝+12𝑝−𝛼;−𝜇+𝛼,𝑝−𝛼 (61)then 𝐹_𝜇,𝑝𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝑎)⊂𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}𝛼;𝑟𝑝,𝛼,𝜇,𝑎, (62) where the bound 𝑟=1𝑝,𝛼,𝜇,𝑎×𝑝−𝛼𝜇+𝑝₂𝐹₁()−,1,2𝑝−𝛼1−𝑎,𝜇+𝑝+1;1/2𝜇+𝛼 (63) is the best possible.

Theorem 13. Let 𝜈∈ℂ^∗, and let 𝐴,𝐵∈ℂ with 𝐴≠𝐵 and |𝐵|≤1. Suppose that 𝜈𝑐+𝑝(𝐴−𝐵)𝐵𝜈()−1≤1𝑜𝑟𝑐+𝑝𝐴−𝐵𝐵||≤𝜋+1≤1,𝑖𝑓𝐵≠0,𝜈𝐴𝑐+𝑝,𝑖𝑓𝐵=0. (64) If 𝑓∈𝒜(𝑝) with ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)≠0 for all 𝑧∈𝑈̇, then ℒ^{𝑚,𝛿−1}_{𝑐,𝑝,𝜆}𝑓(𝑧)ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)≺1+𝐴𝑧1+𝐵𝑧 (65) implies ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)𝑧^𝑝𝜈≺𝑞₁(𝑧), (66) where 𝑞₁(𝑧)=(1+𝐵𝑧)^{𝜈(𝑐+𝑝)(𝐴−𝐵)/𝐵}𝑒𝑖𝑓𝐵≠0,_{𝜈(𝑐+𝑝)𝐴𝑧}𝑖𝑓𝐵=0, (67) is the best dominant.

Proof. Let us put 𝜙(𝑧)=ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)𝑧^𝑝𝜈(𝑧∈𝑈), (68) then 𝜙 is analytic in 𝑈, with 𝜙(0)=1 and 𝜙(𝑧)≠0for all 𝑧∈𝑈̇. By differentiating both sides of (68) logarithmical with respect to 𝑧 and using (5), we have 1+𝑧𝜙^′(𝑧)𝜈𝜙(𝑧)=ℒ𝑐+𝑝^{𝑚,𝛿−1𝑐,𝑝,𝜆}𝑓(𝑧)ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)≺1+𝐴𝑧.1+𝐵𝑧 (69) Now the assertions of Theorem 13 follows by using Lemma 6 for the special case 𝛾=𝜈(𝑐+𝑝).

Putting 𝐵=−1 and 𝐴=1−2𝜌,0≤𝜌<1, in Theorem 13, we obtain the following corollary.

Corollary 14. Assume that 𝜈∈ℂ^∗ satisfies either 2𝜈𝑐+𝑝1−𝜌−1≤1𝑜𝑟2𝜈𝑐+𝑝1−𝜌+1≤1,𝑖𝑓𝐵≠0, (70) If 𝑓∈𝒜(𝑝) with ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)≠0 for all 𝑧∈𝑈̇, then ℒRe^{𝑚,𝛿−1𝑐,𝑝,𝜆}𝑓(𝑧)ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)>𝜌 (71) implies ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)𝑧^𝑝𝜈≺𝑞₂(𝑧)=(1−𝑧)^{−2𝜈(𝑐+𝑝)(1−𝜌)}, (72) and 𝑞₂ is the best dominant.

3. Properties Involving the Multiplier Operator ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}

Theorem 15. If 𝑓∈𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵), then for all 𝑠, 𝑡∈ℂ with |𝑠|≤1,  |𝑡|≤1 and 𝑠≠𝑡, the next subordination holds 𝑡^𝑝ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧𝑠)𝑠^𝑝ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓()≺𝑧𝑡1+𝐵𝑧𝑠1+𝐵𝑧𝑡^{(𝑝−𝛼)(𝐴−𝐵)/𝐵}(,𝑓𝑜𝑟𝐵≠0exp𝑝−𝛼𝐴𝑧𝑠−𝑡),𝑓𝑜𝑟𝐵=0. (73)

Proof. If 𝑓∈𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵), from (7) it follows that 𝑧ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)^′ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)≺𝑝+𝑝𝐵+𝑝−𝛼(𝐴−𝐵)𝑧(𝑧).1+𝐵𝑧≡𝑘 (74) Moreover, the function 𝑘 defined by (74) and the function ℎ given by ℎ(𝑧)≡ℎ(𝑧;𝑠,𝑡)=^𝑧₀𝑠−𝑡1−𝑠𝑢1−𝑡𝑢𝑑𝑢 (75) are convex in 𝑈. By combining a general subordination theorem [25, Theorem 4] with (74), we get 𝑧ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)^′ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)(𝑧)≺()𝑧−𝑝∗ℎ𝑝−𝛼𝐴−𝐵(𝑧).1+𝐵𝑧∗ℎ (76) For every analytic function 𝜙 in 𝑈 with 𝜙(0)=0, we have 𝜙(𝑧)∗ℎ(𝑧)=^𝑠𝑧_𝑡𝑧𝜙(𝑢)𝑢𝑑𝑢, (77) and thus, from (76) and (77), we deduce ^𝑠𝑧_𝑡𝑧ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑢)^′ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑢)−𝑝𝑢≺()𝑑𝑢𝑝−𝛼𝐴−𝐵^𝑠𝑧_𝑡𝑧𝑑𝑢.1+𝐵𝑢 (78) This last subordination implies exp^𝑠𝑧_𝑡𝑧ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑢)^′ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑢)−𝑝𝑢)𝑑𝑢≺exp𝑝−𝛼(𝐴−𝐵^𝑠𝑧_𝑡𝑧𝑑𝑢,1+𝐵𝑢 (79) and by simplification, we get the assertion of Theorem 15.

Corollary 16. If 𝑓∈𝒮^𝑚,𝛿_{𝑐,𝑝,𝜆}(𝛼;𝐴,𝐵), then for |𝑧|=𝑟<1, the next inequalities hold ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧≤𝑟)^𝑝(1+𝐵𝑟)^{(𝑝−𝛼)(𝐴−𝐵)/𝐵}𝑟,𝑓𝑜𝑟𝐵≠0,_𝑝exp𝑝−𝛼𝐴𝑟,𝑓𝑜𝑟𝐵=0,ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧≥𝑟)^𝑝(1−𝐵𝑟)^{(𝑝−𝛼)(𝐴−𝐵)/𝐵}𝑟,𝑓𝑜𝑟𝐵≠0,_𝑝exp−𝑝−𝛼𝐴𝑟,𝑓𝑜𝑟𝐵=0,ℒarg^{𝑚,𝛿𝑐,𝑝,𝜆}𝑓(𝑧)𝑧^𝑝≤𝑝−𝛼(𝐴−𝐵)|𝐵|sin⁻¹(|𝐵|𝑟),𝑓𝑜𝑟𝐵≠0,𝑝−𝛼𝐴𝑟,𝑓𝑜𝑟𝐵=0. (80)(81)(82) All of the estimates asserted here are sharp.

Proof. Taking 𝑠=1 and 𝑡=0 in (73) and using the definition of subordination, we obtain ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)𝑧^𝑝=(1+𝐵𝑤(𝑧))^{(𝑝−𝛼)(𝐴−𝐵)/𝐵}(𝑧,𝑓𝑜𝑟𝐵≠0,exp𝑝−𝛼𝐴𝑤),𝑓𝑜𝑟𝐵=0, (83) where 𝑤 is analytic function in 𝑈 with 𝑤(0)=0 and |𝑤(𝑧)|≤1 for 𝑧∈𝑈. According to the well-known Schwarz's theorem, we have |𝑤(𝑧)|≤|𝑧| for all 𝑧∈𝑈.(i)If 𝐵>0, then we find from (83) that ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)𝑧^𝑝()=exp𝑝−𝛼𝐴−𝐵𝐵|(𝑧=|(𝑧log1+𝐵𝑤)|1+𝐵𝑤)|^{(𝑝−𝛼)(𝐴−𝐵)/𝐵}≤(1+𝐵𝑟)^{(𝑝−𝛼)(𝐴−𝐵)/𝐵}. (84)(ii)If 𝐵<0, we can easily obtain ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓(𝑧)𝑧^𝑝=|1+𝐵𝑤(𝑧)|^{(𝑝−𝛼)(𝐴−𝐵)/−𝐵}≤(1+𝐵𝑟)^{−1(𝑝−𝛼)(𝐴−𝐵)/−𝐵}=(1+𝐵𝑟)^{(𝑝−𝛼)(𝐴−𝐵)/𝐵}. (85)This proves the inequality (80) for 𝐵≠0. Similarly, we can prove the other inequalities in (80) and (81). Now, for |𝑧|=𝑟 and 𝐵≠0, we observe from (83) that ℒarg^{𝑚,𝛿𝑐,𝑝,𝜆}𝑓(𝑧)𝑧^𝑝=𝑝−𝛼(𝐴−𝐵)|𝐵|((𝑧≤()arg1+𝐵𝑤))𝑝−𝛼𝐴−𝐵|𝐵|sin⁻¹(|𝐵|𝑟), (86) and for 𝐵=0, (82) is a direct consequence of (83).
It is easy to see that all of the estimates in Corollary 16 are sharp, being attained by the function 𝑓₀ defined by ℒ^𝑚,𝛿_{𝑐,𝑝,𝜆}𝑓₀(𝑧)=𝑧^𝑝(1+𝐵𝑧)^{(𝑝−𝛼)(𝐴−𝐵)/𝐵}𝑧,𝑓𝑜𝑟𝐵≠0,_𝑝exp𝑝−𝛼𝐴𝑧,𝑓𝑜𝑟𝐵=0. (87)

Remark 17. (i) Putting 𝑐=1 and 𝑚=0 in our results, we obtain the results obtained by Aouf et al. [16].
(ii) By specializing the parameters 𝑐,𝜆,𝛿,𝑝, and 𝑚, we obtain various results for different operators defined in Section 1.

Acknowledgment

The authors thank the referees for their valuable suggestions which led to improvement of this paper.