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 π-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 Β© 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 π-valent functions defined by generalized linear operator.
Now, we introduce the linear multiplier operator Lπ,πΏ π,π,π : A(π) β A(π) by
1. Introduction Let A(π) denote the class of functions of the form π
β
π+π
π (π§) = π§ + β ππ+π π§
(π β N = {1, 2, . . .}) ,
(1)
π=1
which are analytic and π-valent in the open unit disc π = {π§ β C : |π§| < 1}. For functions π β A(π), given by (1) and π given by β
π (π§) = π§π + β ππ+π π§π+π
(π β N) ,
the Hadamard product (or convolution) of π and π is defined by β
π=1
= (π β π) (π§)
β
= π§π + β (1 + π=1
πΏ π+π ππ π ) ( ) ππ+π π§π+π π π+π+π
(4)
(πΏ β₯ 0; π > βπ; π β₯ 0; π β N0 ) . It is easily verified from (4) that
(2)
π=1
(π β π) (π§) = π§π + β ππ+π ππ+π π§π+π
Lπ,πΏ π,π,π π (π§)
(3)
(π§ β π; π β N) .
For π, π β A(π), we say that the function π is subordinate to π denoted by π(π§) βΊ π(π§), if there exists a Schwarz function π€, that is, π€ β A with π€(0) = 0 and |π€(π§)| < 1, π§ β π, such that π(π§) = π(π€(π§)) for all π§ β π. It is well known that, if the function π is univalent in π, then π(π§) βΊ π(π§) is equivalent to π(0) = π(0) and π(π) β π(π) (see [1, 2]).
σΈ
π,πΏβ1 π,πΏ π§(Lπ,πΏ π,π,π π (π§)) = (π + π) Lπ,π,π π (π§) β πLπ,π,π π (π§) ,
(5) σΈ
π+1,πΏ π,πΏ ππ§(Lπ,πΏ π,π,π π (π§)) = πLπ,π,π π (π§) β π (1 β π) Lπ,π,π π (π§) . (6)
By specializing the parameters π, π, πΏ, π, and π, we obtain the following operators: πΏ (i) L1,πΏ π,π,0 π(π§) = πΎπ,π π(π§) (see [3]); πΏ (ii) L0,πΏ πβ1,1,π π(π§) = πΏ π π(π§) (see [4, 5]); πΏ (iii) L1,πΏ 1,π,0 π(π§) = Iπ π(π§) (see [6, 7]); π (iv) Lπ,0 π,π,π π(π§) = π·π,π π(π§) (see [8]);
2
Journal of Mathematics πΏ (v) L1,πΏ π,1,0 π(π§) = Pπ π(π§) (see [9, 10]);
Remark 2. The function π β A(π) is in the class Sπ,πΏ π,π,π (πΌ; π΄, π΅) if and only if
(vi) L1,1 π,1,0 π(π§) = Lπ π(π§) (see [11]);
σ΅¨σ΅¨ σ΅¨σ΅¨ σΈ σ΅¨ σ΅¨σ΅¨ π§(Lπ,πΏ σ΅¨σ΅¨ 1 π,π,π π (π§)) 1 β π΄π΅ σ΅¨σ΅¨σ΅¨σ΅¨ π΄ β π΅ σ΅¨σ΅¨ β πΌ) β , ( σ΅¨< σ΅¨σ΅¨ π β πΌ 1 β π΅2 σ΅¨σ΅¨σ΅¨σ΅¨ 1 β π΅2 (9) Lπ,πΏ π (π§) σ΅¨σ΅¨ π,π,π σ΅¨σ΅¨ σ΅¨σ΅¨
πΏ (vii) L1,πΏ 1,1,0 π(π§) = I π(π§) (see [12]); π (viii) Lπ,0 π,1,π π(π§) = Dπ π(π§) (see [13]);
(ix)
Lπ,0 π,1,1 π(π§)
π§ β π, for π΅ =ΜΈ β 1,
π
= D π(π§) (see [14]).
σΈ
By using the multiplier operator Lπ,πΏ π,π,π π(π§), we define the following classes of functions. Definition 1. For fixed parameters π΄ and π΅, with β1 β€ π΅ < π΄ β€ 1 and π > πΌ, we say that the function π β A(π) is in the class Sπ,πΏ π,π,π (πΌ; π΄, π΅) if it satisfies the following subordination condition: 1 ( πβπΌ
π§(Lπ,πΏ π,π,π π (π§)) Lπ,πΏ π,π,π π (π§)
σΈ
β πΌ) βΊ
1 + π΄π§ . 1 + π΅π§
(7)
π§(Lπ,πΏ π,π,π π (π§)) 1βπ΄ 1 β πΌ) > ( , Re π,πΏ πβπΌ 2 Lπ,π,π π (π§) π§ β π, for π΅ = β1.
Definition 3. The function π β A(π) is in the class Sπ,πΏ π,π,π (πΌ; π) if it satisfies the following the inequality: σΈ
π§(Lπ,πΏ π,π,π π (π§)) 1 Re ( β πΌ) > π, πβπΌ Lπ,πΏ π,π,π π (π§)
π,πΏ Sπ,πΏ π,π,π (πΌ; π΄, π΅) β Sπ,π,π (πΌ;
β (i) S1,0 1,π,0 (πΌ; π΄, π΅) = Sπ (πΌ; π΄, π΅) was studied by Cho et al. [15]; πΏ (ii) S1,πΏ 1,π,0 (πΌ; π΄, π΅) = Sπ (πΌ; π΄, π΅) was studied by Aouf et al. [16];
π,πΏ Sπ,πΏ π,π,π (πΌ; π΄, π΅) = Sπ,π,π (πΌ;
For complex numbers π, π, and π, the Gaussian hypergeometric function is defined by
β
π β
π π§ π (π + 1) β
π (π + 1) π§2 + + β
β
β
π 1! π (π + 1) 2!
(π)π (π)π π§π , (π)π π! π=0
=β
(11)
π, π β C,
π β C \ {0, β1, β2, . . .} , (8) where (π)π = π(π+1) β
β
β
(π+πβ1) and (π)0 = 1. The series (8) converges absolutely for π§ β π, hence it represents an analytic function in π (see [18, Chapter 14]). If π΅ =ΜΈ β 1, from the fact that β(π§) = β(π§), π§ β π, we deduce that the image β(π) is symmetric with respect to the real axis, and that β maps the unit disc π onto the disc |π€ β (1 β π΄π΅)/(1 β π΅2 )| < (π΄ β π΅)/(1 β π΅2 ). If π΅ = β1, the function β maps the unit disc π onto the half plan Re π€ > (1 β π΄)/2, hence we obtain the following.
1βπ΄ ), 1βπ΅
1βπ΄ ) ββ π΅ = β1. 1βπ΅
(12)
We note that when π = 1 and π = 0, the class S1,πΏ 1,π,0 (πΌ; π) = SπΏπ (πΌ; π) was studied by Aouf et al. [16]. Let us consider the first-order differential subordination
π (iii) Sπ,0 π,1,π (0; π΄, π΅) = Sπ (π΄, π΅) was studied by Patel [17].
(π, π, π; π§) = 1 +
π§ β π,
where π < 1; from (9) and (10) it follows, respectively, that
We note that
2 πΉ1
(10)
π» (π (π§) , π§πσΈ (π§)) βΊ β (π§) .
(13)
A univalent function π is called its dominant, if π(π§) βΊ π(π§) for all analytic functions π that satisfy this differential subordination. A dominant πΜ is called the best dominant, if πΜ(π§) βΊ π(π§) for all dominants π. 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 Sπ,πΏ π,π,π (πΌ; π΄, π΅) and
Sπ,πΏ π,π,π (πΌ; π) by using the theory of differential subordination. To establish our main results, we will require the following lemmas. Lemma 4 (see [19]). Let π½, πΎ β C, and let β be a convex function with Re [π½β (π§) + πΎ] > 0,
π§ β π.
(14)
If π is analytic in π, with π(0) = β(0), then π (π§) +
π§πσΈ (π§) βΊ β (π§) σ³¨β π (π§) βΊ β (π§) . π½π (π§) + πΎ
(15)
Journal of Mathematics
3
Lemma 5 (see [20]). Let π½ > 0, π½ + πΎ > 0, and consider the integral operator π½π½,πΎ defined by π½π½,πΎ (π) (π§) = [
π§
π½+πΎ β« ππ½ (π‘) π‘πΎβ1 ππ‘] π§πΎ 0
1/π½
,
(16)
where the powers are the principal ones. If π β [βπΎ/π½, 1) then the order of starlikeness of the class π½π½,πΎ (πβ (π)), that is, the largest number πΏ(π; π½, πΎ) such that π½π½,πΎ (πβ (π)) β πβ (πΏ), is given by the number πΏ(π; π½, πΎ) = inf {Re q(π§) : π§ β π}, where π (π§) =
πΎ 1 β , π½π (π§) π½
1
1βπ§ ) π (π§) = β« ( 1 β π‘π§ 0
2π½(1βπ)
(17) ππ‘.
π§πσΈ (π§) > πΏ (π; π½, πΎ) π (π§)
(π§ β π) ,
Lemma 6 (see [21]). Let π be analytic in π with π(0) = 1 and π(π§) =ΜΈ 0 for 0 < |π§| < 1, and let π΄, π΅ β C with π΄ =ΜΈ π΅, |π΅| β€ 1. β
(i) Let π΅ =ΜΈ 0 and πΎ β C = C \ {0} satisfy either |(πΎ(π΄ β π΅)/π΅) β 1| β€ 1 or |(πΎ(π΄ β π΅)/π΅) + 1| β€ 1. If π satisfies π§πσΈ (π§) 1 + π΄π§ βΊ 1+ πΎπ (π§) 1 + π΅π§
(20)
π (π§) βΊ (1 + π΅π§)πΎ(π΄βπ΅)/π΅ ,
(21)
then
and this is the best dominant. (ii) Let πΎ β Cβ be such that |πΎπ΄| < π, and if π satisfies π§πσΈ (π§) βΊ 1 + π΄π§ πΎπ (π§)
(22)
then π (π§) βΊ ππΎπ΄π§ ,
(23)
and this is the best dominant.
(2) Moreover, if we suppose in addition that π΄β€1+(
πβπ 1 1βπ΅ + ), ) (πΌ + πβπΌ 2 2
(26)
π,πΏ Sπ,πΏβ1 π,π,π (πΌ; π΄, π΅) β Sπ,π,π (πΌ; π (π, πΌ, π, π΄, π΅)) ,
(27)
π (π, πΌ, π, π΄, π΅) =
1 πβπΌ π+π πΉ (1, 2 (π β πΌ) β π΅) / (1 β π΅) , π + π + 1; 1/2) (π΄ 2 1 β (πΌ + π) ] , (28)
is the best possible. Proof. Let π β Sπ,πΏβ1 π,π,π (πΌ; π΄, π΅), and put π (π§) = π§(
Lπ,πΏ π,π,π π (π§) π§π
1/(πβπΌ)
)
(π§ β π) ,
(29)
the function π is analytic in π, with π(0) = 0 and πσΈ (0) = 1. Differentiating (29) logarithmically with respect to π§, we have σΈ
π§(Lπ,πΏ π§πσΈ (π§) π,π,π π (π§)) 1 π (π§) = = ( β πΌ) π (π§) πβπΌ Lπ,πΏ π,π,π π (π§)
(30)
(π§ β π) , then, using (5) in (30), we obtain (π + π)
Lπ,πΏβ1 π,π,π π (π§) Lπ,πΏ π,π,π π (π§)
(31)
= (π β πΌ) π (π§) + (πΌ + π) . By differentiating both sides of (31) logarithmically with respect to π§ and multiplying by π§, we have
2. Inclusion Relationships Unless otherwise mentioned, we assume throughout this paper that β1 β€ π΅ < π΄ β€ 1,
π > πΌ,
πΏ β₯ 0,
π > βπ,
π β N,
πβZ
π β₯ 0,
π,πΏ then Sπ,πΏβ1 π,π,π (πΌ; π΄, π΅) β Sπ,π,π (πΌ; π΄, π΅).
Γ[
π½+πΎ 1 β πΎ] . [ π½ 2 πΉ1 (1, 2π½ (1 β π) , π½ + πΎ + 1; 1/2) (19)
1+
Μ (1) Supposing that Lπ,πΏ π,π,π π(π§) =ΜΈ 0 for all π§ β π = π \ {0},
(18)
where πΏ (π; π½, πΎ) =
(25)
where the bound
Moreover, if π β [π0 , 1), where π0 = max{(π½ β πΎ β 1)/2π½; βπΎ/π½} and π = π½π½,πΎ (π)(π§) with π β πβ (π), then Re
(π β πΌ) (1 β π΄) + (πΌ + π) (1 β π΅) β₯ 0.
then
π½+πΎβ1
π‘
Theorem 7. Let
and the power is the principal one.
(24)
σΈ
π§(Lπ,πΏβ1 π,π,π π (π§)) 1 ( β πΌ) πβπΌ Lπ,πΏβ1 π,π,π π (π§) = π (π§) +
σΈ
π§π (π§) . (π β πΌ) π (π§) + (πΌ + π)
(32)
4
Journal of Mathematics
Combining (32) together with π β Sπ,πΏβ1 π,π,π (πΌ; π΄, π΅), we obtain that the function π satisfies the Briot-Bouquet differential subordination as follows:
(33)
1 + π΄π§ βΊ β‘ β (π§) . 1 + π΅π§
Now we will use Lemma 4 for the special case π½ = π β πΌ and πΎ = πΌ + π. Since β is a convex function in π, a simple computation shows that π§ β π,
(34)
whenever (25) holds, we have π(π§) βΊ β(π§); that is, π β Sπ,πΏ π,π,π (πΌ; π΄, π΅). If in addition, we suppose that the inequality (26) holds, then all the assumptions of Lemma 5 are verified for the above values of π½, πΎ, and π = (1 β π΄)/ (1 β π΅). Then it follows the inclusion Sπ,πΏβ1 π,π,π (πΌ; π΄, π΅) β Sπ,πΏ π,π,π (πΌ; π(π, πΌ, π, π΄, π΅)), where the bound π(π, πΌ, π, π΄, π΅) 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 π > πΌ, such that (25) holds.
β
(πΌ; π (π, πΌ, π, π΄, π΅)) ,
(36)
where π(π, πΌ, π, π΄, π΅) is given by (28). As a consequence of the last inclusion, one has π(π, πΌ, π, π΄, π΅) β₯ (1 β π΄)/(1 β π΅). For the special case π΅ = β1, Theorem 7 reduces to the following. Corollary 9. Let π β₯ β(π + πΌ)/(π β πΌ). Μ (1) Supposing that Lπ,πΏ π,π,π π(π§) =ΜΈ 0 for all π§ β π, then π,πΏ Sπ,πΏβ1 π,π,π (πΌ; π) β Sπ,π,π (πΌ; π) .
2πΌ + π β π + 1 πΌ+π ;β }, 2 (π β πΌ) (π β πΌ)
(37)
β (πΌ + π) ] , is the best possible. Theorem 10. Let π β Sπ,πΏ π,π,π (πΌ; π), where π < 1 and π > πΌ, then π β Sπ,πΏβ1 π,π,π (πΌ; π) for |π§| < π
(π, πΌ, π, π), where
π
(π, πΌ, π, π) = min {π > 0 : π (π, πΌ, π, π, π) = 0} , π (π, πΌ, π, π, π) = 2π ((1 β π) (π β πΌ)
(41)
σ΅¨ Γ σ΅¨σ΅¨σ΅¨(1 β π) (1 β π) σ΅¨ β1 σ΅¨ σ΅¨ β σ΅¨σ΅¨σ΅¨π + (πΌ + π) / (π β πΌ)σ΅¨σ΅¨σ΅¨ (1 + π)σ΅¨σ΅¨σ΅¨) .
π§(Lπ,πΏ π,π,π π (π§)) 1 π’ (π§) = β πΌ) (π§ β π) , ( πβπΌ Lπ,πΏ π,π,π π (π§) (42)
σΈ
π§(Lπ,πΏβ1 π,π,π π (π§)) 1 ( β πΌ) πβπΌ Lπ,πΏβ1 π,π,π π (π§) = π’ (π§) +
(43)
π§π’σΈ (π§) . (π β πΌ) π’ (π§) + (πΌ + π)
If we denote π(π§) = (π’(π§) β π)/(1 β π), then π(0) = 1 and Re π(π§) > 0, π§ β π and substituting in (43) we obtain π§(Lπ,πΏβ1 π,π,π π (π§)) 1 ( β πΌ) β π πβπΌ Lπ,πΏβ1 π,π,π π (π§) = (1 β π)
(38)
then
Γ [π (π§) +
π,πΏ Sπ,πΏβ1 π,π,π (πΌ; π) β Sπ,π,π (πΌ; π (π, πΌ, π, π)) ,
(40)
σΈ
(2) If we suppose in addition that π β₯ max {β
π+π πΉ (1, 2 (π β πΌ) β π) , π + π + 1; 1/2) (1 2 1
is analytic in π with π’(0) = 1 and Re π’(π§) > π. 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
Sπ,πΏ π,π,π
Γ[
σΈ
1βπ΄ ). 1βπ΅ (35)
π,πΏ Sπ,πΏβ1 π,π,π (πΌ; π΄, π΅) β Sπ,π,π (πΌ; π΄, π΅)
1 πβπΌ
Proof. Since π β Sπ,πΏ π,π,π (πΌ; π), the function π’ given by
Μ (1) Supposing that Lπ,πΏ π,π,π π(π§) =ΜΈ 0 for all π§ β π, then π,πΏ π,πΏ Sπ,πΏβ1 π,π,π (πΌ; π΄, π΅) β Sπ,π,π (πΌ; π΄, π΅) β Sπ,π,π (πΌ;
π (π, πΌ, π, π) =
π§πσΈ (π§) π (π§) + (π β πΌ) π (π§) + (πΌ + π)
Re {(π β πΌ) β (π§) + (πΌ + π)} > 0,
where the bound
(39)
π§πσΈ (π§) ], (π β πΌ) (1 β π) π (π§) + π (π β πΌ) + (πΌ + π) (44)
Journal of Mathematics
5 Theorem 11. Let πΌ + π > 0 and
hence σΈ
(πΌ + π) (1 β π΅) + (π β πΌ) (1 β π΄) β₯ 0.
π§(Lπ,πΏβ1 π,π,π π (π§)) 1 Re β πΌ) β π ( π,πΏβ1 πβπΌ Lπ,π,π π (π§)
(i) Supposing that πΉπ,π (π(π§)) =ΜΈ 0 for all π§ β πΜ = π \ {0}, then
β₯ (1 β π) σ΅¨ σ΅¨ Γ [Re π (π§) β (σ΅¨σ΅¨σ΅¨σ΅¨π§πσΈ (π§)σ΅¨σ΅¨σ΅¨σ΅¨) ((π β πΌ) σ΅¨σ΅¨ β1 σ΅¨ σ΅¨ Γ σ΅¨σ΅¨σ΅¨(1 β π) π (π§) β σ΅¨σ΅¨σ΅¨π + ((πΌ + π) / (π β πΌ))σ΅¨σ΅¨σ΅¨σ΅¨σ΅¨σ΅¨) ] . (45)
π,πΏ πΉπ,π (Sπ,πΏ π,π,π (πΌ; π΄, π΅)) β Sπ,π,π (πΌ; π΄, π΅) .
π΄β€1+(
Re π (π§) β₯
1βπ , 1+π
(46)
|π§| = π < 1,
π + 2πΌ β π + 1 1βπ΅ ) min { ; π + πΌ} , πβπΌ 2
then π,πΏ πΉπ,π (Sπ,πΏ π,π,π (πΌ; π΄, π΅)) β Sπ,π,π (πΌ; π (π, πΌ, π, π΄, π΅)) ,
(47)
β₯ (1 β π) [1 β π (π, πΌ, π, π, π)] Re π (π§) , |π§| = π.
=
1 πβπΌ
Γ[
Since the right hand side term of the inequality (47) is nonnegative whenever |π§| β€ π
(π, πΌ, π, π) is given by (41), using the fact that the real part of an analytic function is harmonic, we deduce that π β Sπ,πΏβ1 π,π,π (πΌ; π) for |π§| < π
(π, πΌ, π, π). For a function π β A(π), let the integral operator πΉπ,π : A(π) β A(π) defined by Saitoh [23] and Saitoh et al. [24]
π+π 2 πΉ1 (1, 2 (π β πΌ) (π΄ β π΅) / (1 β π΅) , π + π + 1; 1/2) β (π + πΌ) ] , (54)
is the best possible. Proof. Let π β Sπ,πΏ π,π,π (πΌ; π΄, π΅), and suppose that πΉπ,π (π(π§)) =ΜΈ 0 for all π§ β π.Μ Let
π + π π§ πβ1 β« π‘ π (π‘) ππ‘ π§π 0 β
π+π π ππ§ π+π π π=π+1
= π§π + β
π (π§) = π§(
β
π+π π = (π§π + β π§ ) β π (π§) π+π π=π+1
(π§ β π; π > βπ) .
(55)
σΈ
π§(Lπ,πΏ π§πσΈ (π§) π,π,π πΉπ,π (π (π§))) 1 = π (π§) = β πΌ) ( π,πΏ π (π§) πβπΌ Lπ,π,π πΉπ,π (π (π§))
σΈ
(π§ β π) . β
πLπ,πΏ π,π,π πΉπ,π
π,πΏ Lπ,πΏ π,π,π πΉπ,π (π (π§)) = πΉπ,π (Lπ,π,π π (π§)) ,
(π (π§)) ,
π β π΄ (π) .
(56)
(49) Now, by using (49) in (56), we obtain (π + π)
We now prove the next result.
)
then π is analytic in π, with π(0) = 0 and πσΈ (0) = 1. Taking the logarithmic differentiation in (55), we have
From (4) and (48), we have
= (π +
π§π
1/(πβπΌ)
(π§ β π) ,
= π§ 2 πΉ1 (1, π + π; π + π + 1; π§)
π) Lπ,πΏ π,π,π π (π§)
Lπ,πΏ π,π,π πΉπ,π (π (π§))
(48)
π
π§(Lπ,πΏ π,π,π πΉπ,π (π (π§)))
(53)
π (π, πΌ, π, π΄, π΅)
σΈ
π§(Lπ,πΏβ1 π,π,π π (π§)) 1 Re β πΌ) β π ( πβπΌ Lπ,πΏβ1 π,π,π π (π§)
β π (π§)
(52)
where the bound
together with the inequality (45), we get
πΉπ,π (π (π§)) =
(51)
(ii) Moreover, if we suppose in addition that
By using the well-known results [22] 2π σ΅¨σ΅¨ σΈ σ΅¨σ΅¨ σ΅¨σ΅¨π§π (π§)σ΅¨σ΅¨ β€ σ΅¨ σ΅¨ 1 β π2 Re π (π§) ,
(50)
Lπ,πΏ π,π,π π (π§) Lπ,πΏ π,π,π πΉπ,π (π (π§))
= (π β πΌ) π (π§) + (π + πΌ) . (57)
6
Journal of Mathematics
By differentiating in both sides of (57) logarithmical with respect to π§ and multiplying by π§, we have σΈ
π§(Lπ,πΏ π,π,π π (π§)) 1 β πΌ) ( πβπΌ Lπ,πΏ π,π,π π (π§) = π (π§) +
(58)
σΈ
π§π (π§) . (π β πΌ) π (π§) + (π + πΌ)
Since π β Sπ,πΏ π,π,π (πΌ; π΄, π΅), from (58), we obtain that the function π satisfies the Briot-Bouquet differential subordination π§πσΈ (π§) 1 + π΄π§ βΊ β‘ β (π§) . π (π§) + (π β πΌ) π (π§) + (π + πΌ) 1 + π΅π§
Taking π΅ = β1 in Theorem 11, we obtain the next corollary. Corollary 12. Let πΌ + π > 0 and π β₯ β(π + πΌ)/(π β πΌ). Μ (1) Supposing that Lπ,πΏ π,π,π πΉπ,π (π(π§)) =ΜΈ 0 for all π§ β π, then (πΌ; π)) β
Sπ,πΏ π,π,π
(πΌ; π) .
π + 2πΌ β π + 1 π + πΌ ;β }, πβπΌ 2 (π β πΌ)
Lπ,πΏβ1 π,π,π π (π§) Lπ,πΏ π,π,π π (π§)
(
Lπ,πΏ π,π,π π (π§) π§π
(61)
{(1 + π΅π§) π1 (π§) = { ππ(π+π)π΄π§ {
) βΊ π1 (π§) ,
(66)
π(π+π)(π΄βπ΅)/π΅
if π΅ =ΜΈ 0, if π΅ = 0,
(67)
is the best dominant. Proof. Let us put π (π§) = (
Lπ,πΏ π,π,π π (π§) π§π
π
)
(π§ β π) ,
Lπ,πΏβ1 π§πσΈ (π§) π,π,π π (π§) = 1+ π (π + π) π (π§) Lπ,πΏ π,π,π π (π§)
(68)
(69)
1 + π΄π§ . βΊ 1 + π΅π§
(62)
Now the assertions of Theorem 13 follows by using Lemma 6 for the special case πΎ = π(π + π).
π (π, πΌ, π, π)
Putting π΅ = β1 and π΄ = 1β2π, 0 β€ π < 1, in Theorem 13, we obtain the following corollary.
1 πβπΌ Γ[
π
then π is analytic in π, with π(0) = 1 and π(π§) =ΜΈ 0 for all π§ β π.Μ By differentiating both sides of (68) logarithmical with respect to π§ and using (5), we have
where the bound
=
(65)
where
then π,πΏ πΉπ,π (Sπ,πΏ π,π,π (πΌ; π)) β Sπ,π,π (πΌ; π (π, πΌ, π, π)) ,
1 + π΄π§ 1 + π΅π§
βΊ
implies
(60)
(2) If we suppose in addition that π β₯ max {β
Μ If π β A(π) with Lπ,πΏ π,π,π π(π§) =ΜΈ 0 for all π§ β π, then
(59)
Now we will use Lemma 4 for the special case π½ = π β πΌ and πΎ = π + πΌ; we have π(π§) βΊ β(π§), that is, πΉπ,π β Sπ,πΏ π,π,π (πΌ; π΄, π΅). If we suppose in addition that the inequality (52) holds, then all the assumptions of the Lemma 5 are satisfied for π½, πΎ, and π = (1 β π΄)/(1 β π΅), hence it follows π,πΏ the inclusion πΉπ,π (Sπ,πΏ π,π,π (πΌ; π΄, π΅)) β Sπ,π,π (πΌ; π(π, πΌ, π, π΄, π΅)), and the bound π(π, πΌ, π, π΄, π΅) given by (54) is the best possible.
πΉπ,π (Sπ,πΏ π,π,π
Theorem 13. Let π β Cβ , and let π΄, π΅ β C with π΄ =ΜΈ π΅ and |π΅| β€ 1. Suppose that σ΅¨σ΅¨ π (π + π) (π΄ β π΅) σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ β€ 1 or β 1 σ΅¨σ΅¨ σ΅¨σ΅¨ π΅ σ΅¨ σ΅¨ σ΅¨σ΅¨ π (π + π) (π΄ β π΅) σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ (64) σ΅¨σ΅¨ σ΅¨σ΅¨ β€ 1, if π΅ =ΜΈ 0, + 1 π΅ σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨ π , if π΅ = 0. |ππ΄| β€ π+π
Corollary 14. Assume that π β Cβ satisfies either σ΅¨σ΅¨σ΅¨2π (π + π) (1 β π) β 1σ΅¨σ΅¨σ΅¨ β€ 1 or σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨2π (π + π) (1 β π) + 1σ΅¨σ΅¨ β€ 1, if π΅ =ΜΈ 0,
π+π 2 πΉ1 (1, 2 (π β πΌ) (1 β π) , π + π + 1; 1/2)
Μ If π β A(π) with Lπ,πΏ π,π,π π(π§) =ΜΈ 0 for all π§ β π, then
β (π + πΌ) ] , (63) is the best possible.
(70)
Re
Lπ,πΏβ1 π,π,π π (π§) Lπ,πΏ π,π,π π (π§)
>π
(71)
Journal of Mathematics
7 Corollary 16. If π β Sπ,πΏ π,π,π (πΌ; π΄, π΅), then for |π§| = π < 1, the next inequalities hold
implies (
Lπ,πΏ π,π,π π (π§) π§π
π
) βΊ π2 (π§) = (1 β π§)β2π(π+π)(1βπ) ,
(72)
and π2 is the best dominant.
ππ (1 β π΅π)(πβπΌ)(π΄βπ΅)/π΅ , for π΅ =ΜΈ 0, σ΅¨ σ΅¨σ΅¨ π,πΏ σ΅¨σ΅¨Lπ,π,π π (π§)σ΅¨σ΅¨σ΅¨ β₯ { π σ΅¨ σ΅¨ π exp [β (π β πΌ) π΄π] , for π΅ = 0,
3. Properties Involving the Multiplier Operator Lπ,πΏ π,π,π Theorem 15. If π β Sπ,πΏ π,π,π (πΌ; π΄, π΅), then for all π , π‘ β C with |π | β€ 1, |π‘| β€ 1 and π =ΜΈ π‘, the next subordination holds 1 + π΅π§π (πβπΌ)(π΄βπ΅)/π΅ { , for π΅ =ΜΈ 0 ) {( βΊ { 1 + π΅π§π‘ { {exp [(π β πΌ) π΄π§ (π β π‘)] , for π΅ = 0. (73)
π
π‘ Lπ,πΏ π,π,π π (π§π ) π π Lπ,πΏ π,π,π π (π§π‘) Proof. If π β
Sπ,πΏ π,π,π (πΌ; π΄, π΅), σΈ
π§(Lπ,πΏ π,π,π π (π§))
βΊ
Lπ,πΏ π,π,π π (π§)
from (7) it follows that
π + [ππ΅ + (π β πΌ) (π΄ β π΅)] π§ β‘ π (π§) . 1 + π΅π§ (74)
Moreover, the function π defined by (74) and the function β given by π§ π π‘ (75) β ) ππ’ β (π§) β‘ β (π§; π , π‘) = β« ( 1 β π π’ 1 β π‘π’ 0 are convex in π. By combining a general subordination theorem [25, Theorem 4] with (74), we get σΈ
(
π§(Lπ,πΏ π,π,π π (π§)) Lπ,πΏ π,π,π π (π§)
β π) β β (π§) (76)
(π β πΌ) (π΄ β π΅) π§ β β (π§) . 1 + π΅π§ For every analytic function π in π with π(0) = 0, we have π π§ π (π’) (77) π (π§) β β (π§) = β« ππ’, π’ π‘π§ and thus, from (76) and (77), we deduce βΊ
π π§
(Lπ,πΏ π,π,π π (π’))
π‘π§
Lπ,πΏ π,π,π π (π’)
β« (
σΈ
β
π ) ππ’ π’
βΊ (π β πΌ) (π΄ β π΅) β«
π π§
π‘π§
(78)
ππ’ . 1 + π΅π’
(Lπ,πΏ π,π,π π (π’))
π‘π§
Lπ,πΏ π,π,π π (π’)
exp (β« (
σΈ
π ) ππ’) π’
β
βΊ exp ((π β πΌ) (π΄ β π΅) β«
π π§
π‘π§
ππ’ ), 1 + π΅π’
and by simplification, we get the assertion of Theorem 15.
(79)
(81)
σ΅¨ σ΅¨σ΅¨ Lπ,πΏ π (π§) σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨arg π,π,π σ΅¨σ΅¨ σ΅¨σ΅¨ π§π σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨ (π β πΌ) (π΄ β π΅) β1 { sin (|π΅| π) , for π΅ =ΜΈ 0, { |π΅| β€{ { for π΅ = 0. {(π β πΌ) π΄π,
(82)
All of the estimates asserted here are sharp. Proof. Taking π = 1 and π‘ = 0 in (73) and using the definition of subordination, we obtain Lπ,πΏ π,π,π π (π§) π§π
(πβπΌ)(π΄βπ΅)/π΅ , {(1 + π΅π€(π§)) ={ exp [(π β πΌ) π΄π€ (π§)] , {
for π΅ =ΜΈ 0, for π΅ = 0, (83)
where π€ is analytic function in π with π€(0) = 0 and |π€(π§)| β€ 1 for π§ β π. According to the well-known Schwarzβs theorem, we have |π€(π§)| β€ |π§| for all π§ β π. (i) If π΅ > 0, then we find from (83) that σ΅¨σ΅¨ π,πΏ σ΅¨ σ΅¨σ΅¨ Lπ,π,π π (π§) σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ = exp [ (π β πΌ) (π΄ β π΅) log |1 + π΅π€ (π§)|] σ΅¨σ΅¨ π σ΅¨σ΅¨σ΅¨ π§ π΅ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ = |1 + π΅π€ (π§)|(πβπΌ)(π΄βπ΅)/π΅ β€ (1 + π΅π)(πβπΌ)(π΄βπ΅)/π΅ . (84) (ii) If π΅ < 0, we can easily obtain σ΅¨σ΅¨ π,πΏ σ΅¨ σ΅¨σ΅¨ Lπ,π,π π (π§) σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ = |1 + π΅π€ (π§)|(πβπΌ)(π΄βπ΅)/βπ΅ π σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨ π§ σ΅¨σ΅¨ σ΅¨σ΅¨ β€ [(1 + π΅π)β1 ]
This last subordination implies π π§
ππ (1 + π΅π)(πβπΌ)(π΄βπ΅)/π΅ , for π΅ =ΜΈ 0, σ΅¨ σ΅¨σ΅¨ π,πΏ σ΅¨σ΅¨Lπ,π,π π (π§)σ΅¨σ΅¨σ΅¨ β€ { π (80) σ΅¨ σ΅¨ π exp [(π β πΌ) π΄π] , for π΅ = 0,
(πβπΌ)(π΄βπ΅)/βπ΅
(85)
= (1 + π΅π)(πβπΌ)(π΄βπ΅)/π΅ . This proves the inequality (80) for π΅ =ΜΈ 0. Similarly, we can prove the other inequalities in (80) and (81). Now, for |π§| = π and π΅ =ΜΈ 0, we observe from (83) that σ΅¨ σ΅¨σ΅¨ Lπ,πΏ π (π§) σ΅¨σ΅¨σ΅¨ (π β πΌ) (π΄ β π΅) σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ = σ΅¨σ΅¨arg π,π,π σ΅¨σ΅¨arg (1 + π΅π€ (π§))σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨ π§π |π΅| σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨ (86) (π β πΌ) (π΄ β π΅) β1 β€ sin (|π΅| π) , |π΅| and for π΅ = 0, (82) is a direct consequence of (83).
8
Journal of Mathematics
It is easy to see that all of the estimates in Corollary 16 are sharp, being attained by the function π0 defined by Lπ,πΏ π,π,π π0 (π§) = {
π§π (1 + π΅π§)(πβπΌ)(π΄βπ΅)/π΅ , π§π exp [(π β πΌ) π΄π§] ,
for π΅ =ΜΈ 0, for π΅ = 0.
(87)
Remark 17. (i) Putting π = 1 and π = 0 in our results, we obtain the results obtained by Aouf et al. [16]. (ii) By specializing the parameters π, π, πΏ, π, and π, 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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