Valent Functions Assoicated with Generalized Linear - Downloads ...

Hindawi Publishing Corporation Journal of Mathematics Volume 2013, Article ID 692045, 8 pages http://dx.doi.org/10.1155/2013/692045

Research Article Differential Subordination for Certian Subclasses of 𝑝-Valent Functions Assoicated with Generalized Linear Operator R. M. El-Ashwah,1 M. K. Aouf,2 and S. M. El-Deeb1 1 2

Department of Mathematics, Faculty of Science at Damietta, University of Mansoura, New Damietta 34517, Egypt Department of Mathematics, Faculty of Science, University of Mansoura, Mansoura 33516, Egypt

Correspondence should be addressed to S. M. El-Deeb; [email protected] Received 14 October 2012; Accepted 20 January 2013 Academic Editor: Liwei Zhang Copyright © 2013 R. M. El-Ashwah et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 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.

Now, we introduce the linear multiplier operator L𝑚,𝛿 𝑐,𝑝,𝜆 : A(𝑝) → A(𝑝) by

1. Introduction Let A(𝑝) denote the class of functions of the form 𝑝

∞

𝑘+𝑝

𝑓 (𝑧) = 𝑧 + ∑ 𝑎𝑘+𝑝 𝑧

(𝑝 ∈ N = {1, 2, . . .}) ,

(1)

𝑘=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) = 𝑧𝑝 + ∑ 𝑏𝑘+𝑝 𝑧𝑘+𝑝

(𝑝 ∈ N) ,

the Hadamard product (or convolution) of 𝑓 and 𝑔 is defined by ∞

𝑘=1

= (𝑔 ∗ 𝑓) (𝑧)

∞

= 𝑧𝑝 + ∑ (1 + 𝑘=1

𝛿 𝑐+𝑝 𝜆𝑘 𝑚 ) ( ) 𝑎𝑘+𝑝 𝑧𝑘+𝑝 𝑝 𝑐+𝑘+𝑝

(4)

(𝛿 ≥ 0; 𝑐 > −𝑝; 𝜆 ≥ 0; 𝑚 ∈ N0 ) . It is easily verified from (4) that

(2)

𝑘=1

(𝑓 ∗ 𝑔) (𝑧) = 𝑧𝑝 + ∑ 𝑎𝑘+𝑝 𝑏𝑘+𝑝 𝑧𝑘+𝑝

L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧)

(3)

(𝑧 ∈ 𝑈; 𝑝 ∈ N) .

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]).

󸀠

𝑚,𝛿−1 𝑚,𝛿 𝑧(L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧)) = (𝑐 + 𝑝) L𝑐,𝑝,𝜆 𝑓 (𝑧) − 𝑐L𝑐,𝑝,𝜆 𝑓 (𝑧) ,

(5) 󸀠

𝑚+1,𝛿 𝑚,𝛿 𝜆𝑧(L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧)) = 𝑝L𝑐,𝑝,𝜆 𝑓 (𝑧) − 𝑝 (1 − 𝜆) L𝑐,𝑝,𝜆 𝑓 (𝑧) . (6)

By specializing the parameters 𝑐, 𝜆, 𝛿, 𝑝, and 𝑚, we obtain the following operators: 𝛿 (i) L1,𝛿 𝑐,𝑝,0 𝑓(𝑧) = 𝐾𝑐,𝑝 𝑓(𝑧) (see [3]); 𝛿 (ii) L0,𝛿 𝑎−1,1,𝜆 𝑓(𝑧) = 𝐿 𝑎 𝑓(𝑧) (see [4, 5]); 𝛿 (iii) L1,𝛿 1,𝑝,0 𝑓(𝑧) = I𝑝 𝑓(𝑧) (see [6, 7]); 𝑚 (iv) L𝑚,0 𝑐,𝑝,𝜆 𝑓(𝑧) = 𝐷𝜆,𝑝 𝑓(𝑧) (see [8]);

2

Journal of Mathematics 𝛿 (v) L1,𝛿 𝑐,1,0 𝑓(𝑧) = P𝑐 𝑓(𝑧) (see [9, 10]);

Remark 2. The function 𝑓 ∈ A(𝑝) is in the class S𝑚,𝛿 𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵) if and only if

(vi) L1,1 𝑐,1,0 𝑓(𝑧) = L𝑐 𝑓(𝑧) (see [11]);

󵄨󵄨 󵄨󵄨 󸀠 󵄨 󵄨󵄨 𝑧(L𝑚,𝛿 󵄨󵄨 1 𝑐,𝑝,𝜆 𝑓 (𝑧)) 1 − 𝐴𝐵 󵄨󵄨󵄨󵄨 𝐴 − 𝐵 󵄨󵄨 − 𝛼) − , ( 󵄨< 󵄨󵄨 𝑝 − 𝛼 1 − 𝐵2 󵄨󵄨󵄨󵄨 1 − 𝐵2 (9) L𝑚,𝛿 𝑓 (𝑧) 󵄨󵄨 𝑐,𝑝,𝜆 󵄨󵄨 󵄨󵄨

𝛿 (vii) L1,𝛿 1,1,0 𝑓(𝑧) = I 𝑓(𝑧) (see [12]); 𝑚 (viii) L𝑚,0 𝑐,1,𝜆 𝑓(𝑧) = D𝜆 𝑓(𝑧) (see [13]);

(ix)

L𝑚,0 𝑐,1,1 𝑓(𝑧)

𝑧 ∈ 𝑈, for 𝐵 ≠ − 1,

𝑚

= D 𝑓(𝑧) (see [14]).

󸀠

By using the multiplier operator L𝑚,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 S𝑚,𝛿 𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵) if it satisfies the following subordination condition: 1 ( 𝑝−𝛼

𝑧(L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧)) L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧)

󸀠

− 𝛼) ≺

1 + 𝐴𝑧 . 1 + 𝐵𝑧

(7)

𝑧(L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧)) 1−𝐴 1 − 𝛼) > ( , Re 𝑚,𝛿 𝑝−𝛼 2 L𝑐,𝑝,𝜆 𝑓 (𝑧) 𝑧 ∈ 𝑈, for 𝐵 = −1.

Definition 3. The function 𝑓 ∈ A(𝑝) is in the class S𝑚,𝛿 𝑐,𝑝,𝜆 (𝛼; 𝜌) if it satisfies the following the inequality: 󸀠

𝑧(L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧)) 1 Re ( − 𝛼) > 𝜌, 𝑝−𝛼 L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧)

𝑚,𝛿 S𝑚,𝛿 𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵) ⊂ S𝑐,𝑝,𝜆 (𝛼;

∗ (i) S1,0 1,𝑝,0 (𝛼; 𝐴, 𝐵) = S𝑝 (𝛼; 𝐴, 𝐵) was studied by Cho et al. [15]; 𝛿 (ii) S1,𝛿 1,𝑝,0 (𝛼; 𝐴, 𝐵) = S𝑝 (𝛼; 𝐴, 𝐵) was studied by Aouf et al. [16];

𝑚,𝛿 S𝑚,𝛿 𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵) = S𝑐,𝑝,𝜆 (𝛼;

For complex numbers 𝑎, 𝑏, and 𝑐, the Gaussian hypergeometric function is defined by

∞

𝑎 ⋅ 𝑏 𝑧 𝑎 (𝑎 + 1) ⋅ 𝑏 (𝑏 + 1) 𝑧2 + + ⋅⋅⋅ 𝑑 1! 𝑑 (𝑑 + 1) 2!

(𝑎)𝑘 (𝑏)𝑘 𝑧𝑘 , (𝑑)𝑘 𝑘! 𝑘=0

=∑

(11)

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 Re 𝑤 > (1 − 𝐴)/2, hence we obtain the following.

1−𝐴 ), 1−𝐵

1−𝐴 ) ⇐⇒ 𝐵 = −1. 1−𝐵

(12)

We note that when 𝑐 = 1 and 𝑚 = 0, the class S1,𝛿 1,𝑝,0 (𝛼; 𝜌) = S𝛿𝑝 (𝛼; 𝜌) was studied by Aouf et al. [16]. Let us consider the first-order differential subordination

𝑚 (iii) S𝑚,0 𝑐,1,𝜆 (0; 𝐴, 𝐵) = S𝜆 (𝐴, 𝐵) was studied by Patel [17].

(𝑎, 𝑏, 𝑑; 𝑧) = 1 +

𝑧 ∈ 𝑈,

where 𝜌 < 1; from (9) and (10) it follows, respectively, that

We note that

2 𝐹1

(10)

𝐻 (𝜑 (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 𝑞̃ is called the best dominant, if 𝑞̃(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 S𝑚,𝛿 𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵) and

S𝑚,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 with Re [𝛽ℎ (𝑧) + 𝛾] > 0,

𝑧 ∈ 𝑈.

(14)

If 𝑝 is analytic in 𝑈, with 𝑝(0) = ℎ(0), then 𝑝 (𝑧) +

𝑧𝑝󸀠 (𝑧) ≺ ℎ (𝑧) 󳨐⇒ 𝑝 (𝑧) ≺ ℎ (𝑧) . 𝛽𝑝 (𝑧) + 𝛾

(15)

Journal of Mathematics

3

Lemma 5 (see [20]). Let 𝛽 > 0, 𝛽 + 𝛾 > 0, and consider the integral operator 𝐽𝛽,𝛾 defined by 𝐽𝛽,𝛾 (𝑓) (𝑧) = [

𝑧

𝛽+𝛾 ∫ 𝑓𝛽 (𝑡) 𝑡𝛾−1 𝑑𝑡] 𝑧𝛾 0

1/𝛽

,

(16)

where the powers are the principal ones. If 𝜎 ∈ [−𝛾/𝛽, 1) then the order of starlikeness of the class 𝐽𝛽,𝛾 (𝑆∗ (𝜎)), that is, the largest number 𝛿(𝜎; 𝛽, 𝛾) such that 𝐽𝛽,𝛾 (𝑆∗ (𝜎)) ⊂ 𝑆∗ (𝛿), is given by the number 𝛿(𝜎; 𝛽, 𝛾) = inf {Re q(𝑧) : 𝑧 ∈ 𝑈}, where 𝑞 (𝑧) =

𝛾 1 − , 𝛽𝑄 (𝑧) 𝛽

1

1−𝑧 ) 𝑄 (𝑧) = ∫ ( 1 − 𝑡𝑧 0

2𝛽(1−𝜎)

(17) 𝑑𝑡.

𝑧𝑔󸀠 (𝑧) > 𝛿 (𝜎; 𝛽, 𝛾) 𝑔 (𝑧)

(𝑧 ∈ 𝑈) ,

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 𝑧𝜙󸀠 (𝑧) 1 + 𝐴𝑧 ≺ 1+ 𝛾𝜙 (𝑧) 1 + 𝐵𝑧

(20)

𝜙 (𝑧) ≺ (1 + 𝐵𝑧)𝛾(𝐴−𝐵)/𝐵 ,

(21)

then

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

(22)

then 𝜙 (𝑧) ≺ 𝑒𝛾𝐴𝑧 ,

(23)

and this is the best dominant.

(2) Moreover, if we suppose in addition that 𝐴≤1+(

𝑐−𝑝 1 1−𝐵 + ), ) (𝛼 + 𝑝−𝛼 2 2

(26)

𝑚,𝛿 S𝑚,𝛿−1 𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵) ⊂ S𝑐,𝑝,𝜆 (𝛼; 𝜌 (𝑝, 𝛼, 𝑐, 𝐴, 𝐵)) ,

(27)

𝜌 (𝑝, 𝛼, 𝑐, 𝐴, 𝐵) =

1 𝑝−𝛼 𝑝+𝑐 𝐹 (1, 2 (𝑝 − 𝛼) − 𝐵) / (1 − 𝐵) , 𝑝 + 𝑐 + 1; 1/2) (𝐴 2 1 − (𝛼 + 𝑐) ] , (28)

is the best possible. Proof. Let 𝑓 ∈ S𝑚,𝛿−1 𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵), and put 𝑔 (𝑧) = 𝑧(

L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧) 𝑧𝑝

1/(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 󸀠

𝑧(L𝑚,𝛿 𝑧𝑔󸀠 (𝑧) 𝑐,𝑝,𝜆 𝑓 (𝑧)) 1 𝜙 (𝑧) = = ( − 𝛼) 𝑔 (𝑧) 𝑝−𝛼 L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧)

(30)

(𝑧 ∈ 𝑈) , then, using (5) in (30), we obtain (𝑐 + 𝑝)

L𝑚,𝛿−1 𝑐,𝑝,𝜆 𝑓 (𝑧) L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧)

(31)

= (𝑝 − 𝛼) 𝜙 (𝑧) + (𝛼 + 𝑐) . By differentiating both sides of (31) logarithmically with respect to 𝑧 and multiplying by 𝑧, we have

2. Inclusion Relationships Unless otherwise mentioned, we assume throughout this paper that −1 ≤ 𝐵 < 𝐴 ≤ 1,

𝑝 > 𝛼,

𝛿 ≥ 0,

𝑐 > −𝑝,

𝑝 ∈ N,

𝑚∈Z

𝜆 ≥ 0,

𝑚,𝛿 then S𝑚,𝛿−1 𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵) ⊂ S𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵).

×[

𝛽+𝛾 1 − 𝛾] . [ 𝛽 2 𝐹1 (1, 2𝛽 (1 − 𝜎) , 𝛽 + 𝛾 + 1; 1/2) (19)

1+

̇ (1) Supposing that L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓(𝑧) ≠ 0 for all 𝑧 ∈ 𝑈 = 𝑈 \ {0},

(18)

where 𝛿 (𝜎; 𝛽, 𝛾) =

(25)

where the bound

Moreover, if 𝜎 ∈ [𝜎0 , 1), where 𝜎0 = max{(𝛽 − 𝛾 − 1)/2𝛽; −𝛾/𝛽} and 𝑔 = 𝐽𝛽,𝛾 (𝑓)(𝑧) with 𝑓 ∈ 𝑆∗ (𝜎), then Re

(𝑝 − 𝛼) (1 − 𝐴) + (𝛼 + 𝑐) (1 − 𝐵) ≥ 0.

then

𝛽+𝛾−1

𝑡

Theorem 7. Let

and the power is the principal one.

(24)

󸀠

𝑧(L𝑚,𝛿−1 𝑐,𝑝,𝜆 𝑓 (𝑧)) 1 ( − 𝛼) 𝑝−𝛼 L𝑚,𝛿−1 𝑐,𝑝,𝜆 𝑓 (𝑧) = 𝜙 (𝑧) +

󸀠

𝑧𝜙 (𝑧) . (𝑝 − 𝛼) 𝜙 (𝑧) + (𝛼 + 𝑐)

(32)

4


Combining (32) together with 𝑓 ∈ S𝑚,𝛿−1 𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵), we obtain that the function 𝜙 satisfies the Briot-Bouquet differential subordination as follows:

(33)

1 + 𝐴𝑧 ≺ ≡ ℎ (𝑧) . 1 + 𝐵𝑧

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 𝑧 ∈ 𝑈,

(34)

whenever (25) holds, we have 𝜙(𝑧) ≺ ℎ(𝑧); that is, 𝑓 ∈ S𝑚,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 S𝑚,𝛿−1 𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵) ⊂ S𝑚,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.

⊂

(𝛼; 𝜌 (𝑝, 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 L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓(𝑧) ≠ 0 for all 𝑧 ∈ 𝑈, then 𝑚,𝛿 S𝑚,𝛿−1 𝑐,𝑝,𝜆 (𝛼; 𝑎) ⊂ S𝑐,𝑝,𝜆 (𝛼; 𝑎) .

2𝛼 + 𝑐 − 𝑝 + 1 𝛼+𝑐 ;− }, 2 (𝑝 − 𝛼) (𝑝 − 𝛼)

(37)

− (𝛼 + 𝑐) ] , is the best possible. Theorem 10. Let 𝑓 ∈ S𝑚,𝛿 𝑐,𝑝,𝜆 (𝛼; 𝜌), where 𝜌 < 1 and 𝜌 > 𝛼, then 𝑓 ∈ S𝑚,𝛿−1 𝑐,𝑝,𝜆 (𝛼; 𝜌) for |𝑧| < 𝑅(𝑝, 𝛼, 𝑐, 𝜌), where

𝑅 (𝑝, 𝛼, 𝑐, 𝜌) = min {𝑟 > 0 : 𝜃 (𝑝, 𝛼, 𝑐, 𝜌, 𝑟) = 0} , 𝜃 (𝑝, 𝛼, 𝑐, 𝜌, 𝑟) = 2𝑟 ((1 − 𝑟) (𝑝 − 𝛼)

(41)

󵄨 × 󵄨󵄨󵄨(1 − 𝜌) (1 − 𝑟) 󵄨 −1 󵄨 󵄨 − 󵄨󵄨󵄨𝜌 + (𝛼 + 𝑐) / (𝑝 − 𝛼)󵄨󵄨󵄨 (1 + 𝑟)󵄨󵄨󵄨) .

𝑧(L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧)) 1 𝑢 (𝑧) = − 𝛼) (𝑧 ∈ 𝑈) , ( 𝑝−𝛼 L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧) (42)

󸀠

𝑧(L𝑚,𝛿−1 𝑐,𝑝,𝜆 𝑓 (𝑧)) 1 ( − 𝛼) 𝑝−𝛼 L𝑚,𝛿−1 𝑐,𝑝,𝜆 𝑓 (𝑧) = 𝑢 (𝑧) +

(43)

𝑧𝑢󸀠 (𝑧) . (𝑝 − 𝛼) 𝑢 (𝑧) + (𝛼 + 𝑐)

If we denote 𝑘(𝑧) = (𝑢(𝑧) − 𝜌)/(1 − 𝜌), then 𝑘(0) = 1 and Re 𝑘(𝑧) > 0, 𝑧 ∈ 𝑈 and substituting in (43) we obtain 𝑧(L𝑚,𝛿−1 𝑐,𝑝,𝜆 𝑓 (𝑧)) 1 ( − 𝛼) − 𝜌 𝑝−𝛼 L𝑚,𝛿−1 𝑐,𝑝,𝜆 𝑓 (𝑧) = (1 − 𝜌)

(38)

then

× [𝑘 (𝑧) +

𝑚,𝛿 S𝑚,𝛿−1 𝑐,𝑝,𝜆 (𝛼; 𝑎) ⊂ S𝑐,𝑝,𝜆 (𝛼; 𝜌 (𝑝, 𝛼, 𝑐, 𝑎)) ,

(40)

󸀠

(2) If we suppose in addition that 𝑎 ≥ max {−

𝑝+𝑐 𝐹 (1, 2 (𝑝 − 𝛼) − 𝑎) , 𝑝 + 𝑐 + 1; 1/2) (1 2 1

is analytic in 𝑈 with 𝑢(0) = 1 and Re 𝑢(𝑧) > 𝜌. Using (5) in (42) and taking the logarithmic differentiation in the resulting equation, we obtain

(2) If we suppose in addition that (26) holds, then

S𝑚,𝛿 𝑐,𝑝,𝜆

×[

󸀠

1−𝐴 ). 1−𝐵 (35)

𝑚,𝛿 S𝑚,𝛿−1 𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵) ⊂ S𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵)

1 𝑝−𝛼

Proof. Since 𝑓 ∈ S𝑚,𝛿 𝑐,𝑝,𝜆 (𝛼; 𝜌), the function 𝑢 given by

̇ (1) Supposing that L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓(𝑧) ≠ 0 for all 𝑧 ∈ 𝑈, then 𝑚,𝛿 𝑚,𝛿 S𝑚,𝛿−1 𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵) ⊂ S𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵) ⊂ S𝑐,𝑝,𝜆 (𝛼;

𝜌 (𝑝, 𝛼, 𝑐, 𝑎) =

𝑧𝜙󸀠 (𝑧) 𝜙 (𝑧) + (𝑝 − 𝛼) 𝜙 (𝑧) + (𝛼 + 𝑐)

Re {(𝑝 − 𝛼) ℎ (𝑧) + (𝛼 + 𝑐)} > 0,

where the bound

(39)

𝑧𝑘󸀠 (𝑧) ], (𝑝 − 𝛼) (1 − 𝜌) 𝑘 (𝑧) + 𝜌 (𝑝 − 𝛼) + (𝛼 + 𝑐) (44)


5 Theorem 11. Let 𝛼 + 𝜇 > 0 and

hence 󸀠

(𝛼 + 𝜇) (1 − 𝐵) + (𝑝 − 𝛼) (1 − 𝐴) ≥ 0.

𝑧(L𝑚,𝛿−1 𝑐,𝑝,𝜆 𝑓 (𝑧)) 1 Re − 𝛼) − 𝜌 ( 𝑚,𝛿−1 𝑝−𝛼 L𝑐,𝑝,𝜆 𝑓 (𝑧)

(i) Supposing that 𝐹𝜇,𝑝 (𝑓(𝑧)) ≠ 0 for all 𝑧 ∈ 𝑈̇ = 𝑈 \ {0}, then

≥ (1 − 𝜌) 󵄨 󵄨 × [Re 𝑘 (𝑧) − (󵄨󵄨󵄨󵄨𝑧𝑘󸀠 (𝑧)󵄨󵄨󵄨󵄨) ((𝑝 − 𝛼) 󵄨󵄨 −1 󵄨 󵄨 × 󵄨󵄨󵄨(1 − 𝜌) 𝑘 (𝑧) − 󵄨󵄨󵄨𝜌 + ((𝛼 + 𝑐) / (𝑝 − 𝛼))󵄨󵄨󵄨󵄨󵄨󵄨) ] . (45)

𝑚,𝛿 𝐹𝜇,𝑝 (S𝑚,𝛿 𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵)) ⊂ S𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵) .

𝐴≤1+(

Re 𝑘 (𝑧) ≥

1−𝑟 , 1+𝑟

(46)

|𝑧| = 𝑟 < 1,

𝜇 + 2𝛼 − 𝑝 + 1 1−𝐵 ) min { ; 𝜇 + 𝛼} , 𝑝−𝛼 2

then 𝑚,𝛿 𝐹𝜇,𝑝 (S𝑚,𝛿 𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵)) ⊂ S𝑐,𝑝,𝜆 (𝛼; 𝑟 (𝑝, 𝛼, 𝜇, 𝐴, 𝐵)) ,

(47)

≥ (1 − 𝜌) [1 − 𝜃 (𝑝, 𝛼, 𝑐, 𝜌, 𝑟)] Re 𝑘 (𝑧) , |𝑧| = 𝑟.

=

1 𝑝−𝛼

×[

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 ∈ S𝑚,𝛿−1 𝑐,𝑝,𝜆 (𝛼; 𝜌) for |𝑧| < 𝑅(𝑝, 𝛼, 𝑐, 𝜌). For a function 𝑓 ∈ A(𝑝), let the integral operator 𝐹𝜇,𝑝 : A(𝑝) → A(𝑝) defined by Saitoh [23] and Saitoh et al. [24]

𝑝+𝜇 2 𝐹1 (1, 2 (𝑝 − 𝛼) (𝐴 − 𝐵) / (1 − 𝐵) , 𝜇 + 𝑝 + 1; 1/2) − (𝜇 + 𝛼) ] , (54)

is the best possible. Proof. Let 𝑓 ∈ S𝑚,𝛿 𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵), and suppose that 𝐹𝜇,𝑝 (𝑓(𝑧)) ≠ 0 for all 𝑧 ∈ 𝑈.̇ Let

𝜇 + 𝑝 𝑧 𝜇−1 ∫ 𝑡 𝑓 (𝑡) 𝑑𝑡 𝑧𝜇 0 ∞

𝜇+𝑝 𝑘 𝑎𝑧 𝑘+𝜇 𝑘 𝑘=𝑝+1

= 𝑧𝑝 + ∑

𝑔 (𝑧) = 𝑧(

∞

𝜇+𝑝 𝑘 = (𝑧𝑝 + ∑ 𝑧 ) ∗ 𝑓 (𝑧) 𝑘+𝜇 𝑘=𝑝+1

(𝑧 ∈ 𝑈; 𝜇 > −𝑝) .

(55)

󸀠

𝑧(L𝑚,𝛿 𝑧𝑔󸀠 (𝑧) 𝑐,𝑝,𝜆 𝐹𝜇,𝑝 (𝑓 (𝑧))) 1 = 𝜙 (𝑧) = − 𝛼) ( 𝑚,𝛿 𝑔 (𝑧) 𝑝−𝛼 L𝑐,𝑝,𝜆 𝐹𝜇,𝑝 (𝑓 (𝑧))

󸀠

(𝑧 ∈ 𝑈) . −

𝜇L𝑚,𝛿 𝑐,𝑝,𝜆 𝐹𝜇,𝑝

𝑚,𝛿 L𝑚,𝛿 𝑐,𝑝,𝜆 𝐹𝜇,𝑝 (𝑓 (𝑧)) = 𝐹𝜇,𝑝 (L𝑐,𝑝,𝜆 𝑓 (𝑧)) ,

(𝑓 (𝑧)) ,

𝑓 ∈ 𝐴 (𝑝) .

(56)

(49) Now, by using (49) in (56), we obtain (𝑝 + 𝜇)

We now prove the next result.

)

then 𝑔 is analytic in 𝑈, with 𝑔(0) = 0 and 𝑔󸀠 (0) = 1. Taking the logarithmic differentiation in (55), we have

From (4) and (48), we have

= (𝜇 +

𝑧𝑝

1/(𝑝−𝛼)

(𝑧 ∈ 𝑈) ,

= 𝑧 2 𝐹1 (1, 𝜇 + 𝑝; 𝜇 + 𝑝 + 1; 𝑧)

𝑝) L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧)

L𝑚,𝛿 𝑐,𝑝,𝜆 𝐹𝜇,𝑝 (𝑓 (𝑧))

(48)

𝑝

𝑧(L𝑚,𝛿 𝑐,𝑝,𝜆 𝐹𝜇,𝑝 (𝑓 (𝑧)))

(53)

𝑟 (𝑝, 𝛼, 𝜇, 𝐴, 𝐵)

󸀠

𝑧(L𝑚,𝛿−1 𝑐,𝑝,𝜆 𝑓 (𝑧)) 1 Re − 𝛼) − 𝜌 ( 𝑝−𝛼 L𝑚,𝛿−1 𝑐,𝑝,𝜆 𝑓 (𝑧)

∗ 𝑓 (𝑧)

(52)

where the bound

together with the inequality (45), we get

𝐹𝜇,𝑝 (𝑓 (𝑧)) =

(51)

(ii) Moreover, if we suppose in addition that

By using the well-known results [22] 2𝑟 󵄨󵄨 󸀠 󵄨󵄨 󵄨󵄨𝑧𝑘 (𝑧)󵄨󵄨 ≤ 󵄨 󵄨 1 − 𝑟2 Re 𝑘 (𝑧) ,

(50)

L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧) L𝑚,𝛿 𝑐,𝑝,𝜆 𝐹𝜇,𝑝 (𝑓 (𝑧))

= (𝑝 − 𝛼) 𝜙 (𝑧) + (𝜇 + 𝛼) . (57)

6


By differentiating in both sides of (57) logarithmical with respect to 𝑧 and multiplying by 𝑧, we have 󸀠

𝑧(L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧)) 1 − 𝛼) ( 𝑝−𝛼 L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧) = 𝜙 (𝑧) +

(58)

󸀠

𝑧𝜙 (𝑧) . (𝑝 − 𝛼) 𝜙 (𝑧) + (𝜇 + 𝛼)

Since 𝑓 ∈ S𝑚,𝛿 𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵), from (58), we obtain that the function 𝜙 satisfies the Briot-Bouquet differential subordination 𝑧𝜙󸀠 (𝑧) 1 + 𝐴𝑧 ≺ ≡ ℎ (𝑧) . 𝜙 (𝑧) + (𝑝 − 𝛼) 𝜙 (𝑧) + (𝜇 + 𝛼) 1 + 𝐵𝑧

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 L𝑚,𝛿 𝑐,𝑝,𝜆 𝐹𝜇,𝑝 (𝑓(𝑧)) ≠ 0 for all 𝑧 ∈ 𝑈, then (𝛼; 𝑎)) ⊂

S𝑚,𝛿 𝑐,𝑝,𝜆

(𝛼; 𝑎) .

𝜇 + 2𝛼 − 𝑝 + 1 𝜇 + 𝛼 ;− }, 𝑝−𝛼 2 (𝑝 − 𝛼)


(


(61)

{(1 + 𝐵𝑧) 𝑞1 (𝑧) = { 𝑒𝜈(𝑐+𝑝)𝐴𝑧 {

) ≺ 𝑞1 (𝑧) ,

(66)

𝜈(𝑐+𝑝)(𝐴−𝐵)/𝐵

if 𝐵 ≠ 0, if 𝐵 = 0,

(67)

is the best dominant. Proof. Let us put 𝜙 (𝑧) = (


𝜈

)

(𝑧 ∈ 𝑈) ,

L𝑚,𝛿−1 𝑧𝜙󸀠 (𝑧) 𝑐,𝑝,𝜆 𝑓 (𝑧) = 1+ 𝜈 (𝑐 + 𝑝) 𝜙 (𝑧) L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧)

(68)

(69)

1 + 𝐴𝑧 . ≺ 1 + 𝐵𝑧

(62)

Now the assertions of Theorem 13 follows by using Lemma 6 for the special case 𝛾 = 𝜈(𝑐 + 𝑝).

𝑟 (𝑝, 𝛼, 𝜇, 𝑎)

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

1 𝑝−𝛼 ×[

𝜈

then 𝜙 is analytic in 𝑈, with 𝜙(0) = 1 and 𝜙(𝑧) ≠ 0 for all 𝑧 ∈ 𝑈.̇ By differentiating both sides of (68) logarithmical with respect to 𝑧 and using (5), we have

where the bound

=

(65)

where

then 𝑚,𝛿 𝐹𝜇,𝑝 (S𝑚,𝛿 𝑐,𝑝,𝜆 (𝛼; 𝑎)) ⊂ S𝑐,𝑝,𝜆 (𝛼; 𝑟 (𝑝, 𝛼, 𝜇, 𝑎)) ,

1 + 𝐴𝑧 1 + 𝐵𝑧

≺

implies

(60)

(2) If we suppose in addition that 𝑎 ≥ max {−

̇ If 𝑓 ∈ A(𝑝) with L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓(𝑧) ≠ 0 for all 𝑧 ∈ 𝑈, then

(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, 𝐹𝜇,𝑝 ∈ S𝑚,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 𝑚,𝛿 the inclusion 𝐹𝜇,𝑝 (S𝑚,𝛿 𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵)) ⊂ S𝑐,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.

𝐹𝜇,𝑝 (S𝑚,𝛿 𝑐,𝑝,𝜆

Theorem 13. Let 𝜈 ∈ C∗ , and let 𝐴, 𝐵 ∈ C with 𝐴 ≠ 𝐵 and |𝐵| ≤ 1. Suppose that 󵄨󵄨 𝜈 (𝑐 + 𝑝) (𝐴 − 𝐵) 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ≤ 1 or − 1 󵄨󵄨 󵄨󵄨 𝐵 󵄨 󵄨 󵄨󵄨 𝜈 (𝑐 + 𝑝) (𝐴 − 𝐵) 󵄨󵄨 󵄨󵄨 󵄨󵄨 (64) 󵄨󵄨 󵄨󵄨 ≤ 1, if 𝐵 ≠ 0, + 1 𝐵 󵄨󵄨󵄨 󵄨󵄨󵄨 𝜋 , if 𝐵 = 0. |𝜈𝐴| ≤ 𝑐+𝑝

Corollary 14. Assume that 𝜈 ∈ C∗ satisfies either 󵄨󵄨󵄨2𝜈 (𝑐 + 𝑝) (1 − 𝜌) − 1󵄨󵄨󵄨 ≤ 1 or 󵄨 󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨2𝜈 (𝑐 + 𝑝) (1 − 𝜌) + 1󵄨󵄨 ≤ 1, if 𝐵 ≠ 0,

𝜇+𝑝 2 𝐹1 (1, 2 (𝑝 − 𝛼) (1 − 𝑎) , 𝜇 + 𝑝 + 1; 1/2)

̇ If 𝑓 ∈ A(𝑝) with L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓(𝑧) ≠ 0 for all 𝑧 ∈ 𝑈, then

− (𝜇 + 𝛼) ] , (63) is the best possible.

(70)

Re


>𝜌

(71)


7 Corollary 16. If 𝑓 ∈ S𝑚,𝛿 𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵), then for |𝑧| = 𝑟 < 1, the next inequalities hold

implies (


𝜈

) ≺ 𝑞2 (𝑧) = (1 − 𝑧)−2𝜈(𝑐+𝑝)(1−𝜌) ,

(72)

and 𝑞2 is the best dominant.

𝑟𝑝 (1 − 𝐵𝑟)(𝑝−𝛼)(𝐴−𝐵)/𝐵 , for 𝐵 ≠ 0, 󵄨 󵄨󵄨 𝑚,𝛿 󵄨󵄨L𝑐,𝑝,𝜆 𝑓 (𝑧)󵄨󵄨󵄨 ≥ { 𝑝 󵄨 󵄨 𝑟 exp [− (𝑝 − 𝛼) 𝐴𝑟] , for 𝐵 = 0,

3. Properties Involving the Multiplier Operator L𝑚,𝛿 𝑐,𝑝,𝜆 Theorem 15. If 𝑓 ∈ S𝑚,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 1 + 𝐵𝑧𝑠 (𝑝−𝛼)(𝐴−𝐵)/𝐵 { , for 𝐵 ≠ 0 ) {( ≺ { 1 + 𝐵𝑧𝑡 { {exp [(𝑝 − 𝛼) 𝐴𝑧 (𝑠 − 𝑡)] , for 𝐵 = 0. (73)

𝑝

𝑡 L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧𝑠) 𝑠𝑝 L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧𝑡) Proof. If 𝑓 ∈

S𝑚,𝛿 𝑐,𝑝,𝜆 (𝛼; 𝐴, 𝐵), 󸀠

𝑧(L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧))

≺

L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧)

from (7) it follows that

𝑝 + [𝑝𝐵 + (𝑝 − 𝛼) (𝐴 − 𝐵)] 𝑧 ≡ 𝑘 (𝑧) . 1 + 𝐵𝑧 (74)

Moreover, the function 𝑘 defined by (74) and the function ℎ given by 𝑧 𝑠 𝑡 (75) − ) 𝑑𝑢 ℎ (𝑧) ≡ ℎ (𝑧; 𝑠, 𝑡) = ∫ ( 1 − 𝑠𝑢 1 − 𝑡𝑢 0 are convex in 𝑈. By combining a general subordination theorem [25, Theorem 4] with (74), we get 󸀠

(

𝑧(L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧)) L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧)

− 𝑝) ∗ ℎ (𝑧) (76)

(𝑝 − 𝛼) (𝐴 − 𝐵) 𝑧 ∗ ℎ (𝑧) . 1 + 𝐵𝑧 For every analytic function 𝜙 in 𝑈 with 𝜙(0) = 0, we have 𝑠𝑧 𝜙 (𝑢) (77) 𝜙 (𝑧) ∗ ℎ (𝑧) = ∫ 𝑑𝑢, 𝑢 𝑡𝑧 and thus, from (76) and (77), we deduce ≺

𝑠𝑧

(L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑢))

𝑡𝑧

L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑢)

∫ (

󸀠

−

𝑝 ) 𝑑𝑢 𝑢

≺ (𝑝 − 𝛼) (𝐴 − 𝐵) ∫

𝑠𝑧

𝑡𝑧

(78)

𝑑𝑢 . 1 + 𝐵𝑢

(L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑢))

𝑡𝑧

L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑢)

exp (∫ (

󸀠

𝑝 ) 𝑑𝑢) 𝑢

−

≺ exp ((𝑝 − 𝛼) (𝐴 − 𝐵) ∫

𝑠𝑧

𝑡𝑧

𝑑𝑢 ), 1 + 𝐵𝑢

and by simplification, we get the assertion of Theorem 15.

(79)

(81)

󵄨 󵄨󵄨 L𝑚,𝛿 𝑓 (𝑧) 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨arg 𝑐,𝑝,𝜆 󵄨󵄨 󵄨󵄨 𝑧𝑝 󵄨󵄨 󵄨󵄨 󵄨 󵄨 (𝑝 − 𝛼) (𝐴 − 𝐵) −1 { sin (|𝐵| 𝑟) , for 𝐵 ≠ 0, { |𝐵| ≤{ { for 𝐵 = 0. {(𝑝 − 𝛼) 𝐴𝑟,

(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 L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓 (𝑧) 𝑧𝑝

(𝑝−𝛼)(𝐴−𝐵)/𝐵 , {(1 + 𝐵𝑤(𝑧)) ={ exp [(𝑝 − 𝛼) 𝐴𝑤 (𝑧)] , {

for 𝐵 ≠ 0, for 𝐵 = 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 󵄨󵄨 𝑚,𝛿 󵄨 󵄨󵄨 L𝑐,𝑝,𝜆 𝑓 (𝑧) 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 = exp [ (𝑝 − 𝛼) (𝐴 − 𝐵) log |1 + 𝐵𝑤 (𝑧)|] 󵄨󵄨 𝑝 󵄨󵄨󵄨 𝑧 𝐵 󵄨󵄨 󵄨󵄨 󵄨 = |1 + 𝐵𝑤 (𝑧)|(𝑝−𝛼)(𝐴−𝐵)/𝐵 ≤ (1 + 𝐵𝑟)(𝑝−𝛼)(𝐴−𝐵)/𝐵 . (84) (ii) If 𝐵 < 0, we can easily obtain 󵄨󵄨 𝑚,𝛿 󵄨 󵄨󵄨 L𝑐,𝑝,𝜆 𝑓 (𝑧) 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 = |1 + 𝐵𝑤 (𝑧)|(𝑝−𝛼)(𝐴−𝐵)/−𝐵 𝑝 󵄨󵄨󵄨 󵄨󵄨󵄨 𝑧 󵄨󵄨 󵄨󵄨 ≤ [(1 + 𝐵𝑟)−1 ]

This last subordination implies 𝑠𝑧

𝑟𝑝 (1 + 𝐵𝑟)(𝑝−𝛼)(𝐴−𝐵)/𝐵 , for 𝐵 ≠ 0, 󵄨 󵄨󵄨 𝑚,𝛿 󵄨󵄨L𝑐,𝑝,𝜆 𝑓 (𝑧)󵄨󵄨󵄨 ≤ { 𝑝 (80) 󵄨 󵄨 𝑟 exp [(𝑝 − 𝛼) 𝐴𝑟] , for 𝐵 = 0,

(𝑝−𝛼)(𝐴−𝐵)/−𝐵

(85)

= (1 + 𝐵𝑟)(𝑝−𝛼)(𝐴−𝐵)/𝐵 . 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 󵄨 󵄨󵄨 L𝑚,𝛿 𝑓 (𝑧) 󵄨󵄨󵄨 (𝑝 − 𝛼) (𝐴 − 𝐵) 󵄨 󵄨󵄨 󵄨󵄨 = 󵄨󵄨arg 𝑐,𝑝,𝜆 󵄨󵄨arg (1 + 𝐵𝑤 (𝑧))󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 𝑧𝑝 |𝐵| 󵄨󵄨 󵄨󵄨 󵄨 󵄨 (86) (𝑝 − 𝛼) (𝐴 − 𝐵) −1 ≤ sin (|𝐵| 𝑟) , |𝐵| and for 𝐵 = 0, (82) is a direct consequence of (83).

8


It is easy to see that all of the estimates in Corollary 16 are sharp, being attained by the function 𝑓0 defined by L𝑚,𝛿 𝑐,𝑝,𝜆 𝑓0 (𝑧) = {

𝑧𝑝 (1 + 𝐵𝑧)(𝑝−𝛼)(𝐴−𝐵)/𝐵 , 𝑧𝑝 exp [(𝑝 − 𝛼) 𝐴𝑧] ,

for 𝐵 ≠ 0, for 𝐵 = 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.

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