Hindawi Publishing Corporation Journal of Mathematics Volume 2014, Article ID 749251, 4 pages http://dx.doi.org/10.1155/2014/749251
Research Article Inclusion Properties of Certain Subclasses of 𝑝-Valent Functions Associated with the Integral Operator Tamer M. Seoudy Department of Mathematics, Faculty of Science, Fayoum University, Fayoum 63514, Egypt Correspondence should be addressed to Tamer M. Seoudy;
[email protected] Received 2 May 2014; Accepted 18 May 2014; Published 29 May 2014 Academic Editor: Ming-Sheng Liu Copyright © 2014 Tamer M. Seoudy. 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. The purpose of the present paper is to introduce two subclasses of 𝑝-valent functions by using the integral operator and to investigate various properties for these subclasses.
1. Introduction Let A(𝑝) denote the class of functions of the following form: 𝑝
∞
𝑝+𝑗
𝑓 (𝑧) = 𝑧 + ∑𝑎𝑝+𝑗 𝑧 𝑗=1
,
(𝑝 ∈ N = {1, 2, 3, ...}) ,
(1)
which are analytic and 𝑝-valent in the open unit disc U = {𝑧 ∈ C : |𝑧| < 1}. Let P𝑘 (𝑝, 𝛾) be the class of functions 𝑔 analytic in U satisfying 𝑔(0) = 𝑝 and 2𝜋 R {𝑔 (𝑧)} − 𝛾 𝑑𝜃 ≤ 𝑘𝜋, ∫ 𝑝 − 𝛾 0
(𝑧 = 𝑟𝑒𝑖𝜃 ; 𝑘 ≥ 2; 0 ≤ 𝛾 < 𝑝) . (2)
The class P𝑘 (𝑝, 𝛾) was introduced by Aouf [1] and we note the following: (i) the class P𝑘 (1, 𝛾) = P𝑘 (𝛾) was introduced by Padmanabhan and Parvatham [2]; (ii) the class P𝑘 (1, 0) = P𝑘 was introduced by Pinchuk [3];
From (1), we have 𝑔 ∈ P𝑘 (𝑝, 𝛾) if and only if there exists 𝑔1 , 𝑔2 ∈ P(𝑝, 𝛾) such that 𝑘 1 𝑘 1 𝑔 (𝑧) = ( + ) 𝑔1 (𝑧) − ( − ) 𝑔2 (𝑧) , 4 2 4 2
It is known that [4] the class P𝑘 (𝛾) is a convex set. Motivated essentially by Jung et al. [5], Liu and Owa [6] 𝛼 : A(𝑝) → A(𝑝) (𝛼 ≥ introduced the integral operator 𝑄𝛽,𝑝 0; 𝛽 > −𝑝; 𝑝 ∈ N) as follows: 𝛼 𝑄𝛽,𝑝 𝑓 (𝑧)
𝑝+𝛼 + 𝛽−1 𝛼 𝑧 𝑡 𝛼−1 𝛽−1 { { {( 𝑝+𝛽−1 ) 𝑧𝛽 ∫ (1− 𝑧 ) 𝑡 𝑓 (𝑡) 𝑑𝑡 0 ={ {𝑓 (𝑧) { {
(v) P2 (1, 0) = P is the class of functions with positive real part.
(𝛼 > 0) , (𝛼 = 0) . (4)
For 𝑓 ∈ A(𝑝) given by (1) and then from (4), we deduce that 𝛼 𝑄𝛽,𝑝 𝑓 (𝑧) = 𝑧𝑝 +
Γ (𝛼+𝛽+𝑝) ∞ Γ (𝛽+𝑝+𝑗) 𝑎 𝑧𝑝+𝑗 ∑ Γ (𝛽+𝑝) 𝑗=1 Γ (𝛼+𝛽+𝑝+𝑗) 𝑝+𝑗 (𝛼 ≥ 0; 𝛽 > −𝑝) . (5)
(iii) P2 (𝑝, 𝛾) = P(𝑝, 𝛾) is the class of functions with positive real part greater than 𝛾 (0 ≤ 𝛾 < 𝑝); (iv) P2 (1, 𝛾) = P(𝛾) is the class of functions with positive real part greater than 𝛾 (0 ≤ 𝛾 < 1);
(𝑧 ∈ U) . (3)
It is easily verified from (5) that (see [6])
𝛼+1 𝛼 𝛼 𝑧(𝑄𝛽,𝑝 𝑓(𝑧)) = (𝛼 + 𝛽 + 𝑝) 𝑄𝛽,𝑝 𝑓 (𝑧) − (𝛼 + 𝛽) 𝑄𝛽,𝑝 𝑓 (𝑧) . (6)
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We note that (i) the one-parameter family of integral operator 𝛼 = 𝑄𝛽𝛼 was defined by Jung et al. [5] and studied by Aouf 𝑄𝛽,1 [7] and Gao et al. [8]. (ii) Consider 1 𝑓 (𝑧) = 𝐹𝑐,𝑝 (𝑓) (𝑧) = 𝑄𝑐,𝑝
𝑐 + 𝑝 𝑐−1 ∫ 𝑡 𝑓 (𝑧) 𝑑𝑡, 𝑧𝑐
(7)
𝑧𝑓 (𝑧) ∈ P𝑘 (𝑝, 𝛾) , 𝑧 ∈ U} , 𝑓 (𝑧)
𝑟0 =
𝜇 + 1 , √𝐴 + (𝐴2 − 𝜇2 − 1)1/2
(𝑧𝑓 (𝑧)) { } C𝑘 (𝑝, 𝛾) = {𝑓 ∈ A (𝑝) : ∈ P𝑘 (𝑝, 𝛾) , 𝑧 ∈ U} . 𝑓 (𝑧) { } (8) we introduce Next, by using the integral operator the following classes of analytic functions for 0 ≤ 𝛾 < 𝑝 and 𝑘 ≥ 2: 𝛼 𝑓 (𝑧) ∈ S𝑘 (𝑝, 𝛾)} , S𝑘 (𝑝, 𝛼; 𝛾) = {𝑓 ∈ A (𝑝) : 𝑄𝛽,𝑝
C𝑘 (𝑝, 𝛼; 𝛾) = {𝑓 ∈ A (𝑝) :
(12)
and this radius is the best possible. Lemma 3 (see [12]). Let 𝜓 be convex and let 𝑔 be starlike in U. Then, for 𝐹 analytic in U with 𝐹(0) = 1, ((𝜓 ∗ 𝐹𝑔)/(𝜓 ∗ 𝑔)) is contained in the convex hull of 𝐹(U). In this paper, we obtain several inclusion properties of the classes S𝑘 (𝑝, 𝛼; 𝛾) and C𝑘 (𝑝, 𝛼; 𝛾) associated with the 𝛼 . operator 𝑄𝛽,𝑝
2. Main Results Unless otherwise mentioned, we assume throughout this paper that 𝑘 ≥ 2, 𝛼 ≥ 0, 𝛽 > 0, 0 ≤ 𝛾 < 𝑝, and 𝑝 ∈ N. Theorem 4. One has
𝛼 , 𝑄𝛽,𝑝
𝛼 𝑄𝛽,𝑝 𝑓 (𝑧)
2 𝐴 = 2(𝑠 + 1)2 + 𝜇 − 1,
(𝑐 > −𝑝) ,
where the operator 𝐹𝑐,𝑝 is the generalized Bernardi-LiberaLivingston integral operator (see [9]). We have the following known subclasses S𝑘 (𝑝, 𝛾) and C𝑘 (𝑝, 𝛾) of the class A(𝑝) for 0 ≤ 𝛾, 𝜂 < 𝑝, and 𝑘 ≥ 2 which are defined by S𝑘 (𝑝, 𝛾) = {𝑓 ∈ A (𝑝) :
where 𝑟0 is given by
∈ C𝑘 (𝑝, 𝛾)} .
S𝑘 (𝑝, 𝛼 + 1; 𝛾) ⊂ S𝑘 (𝑝, 𝛼; 𝛾) . Proof. We begin by setting
(9)
𝛼+1 𝑧(𝑄𝛽,𝑝 𝑓(𝑧)) 𝛼+1 𝑄𝛽,𝑝 𝑓 (𝑧)
= (𝑝 − 𝛾) ℎ (𝑧) + 𝛾 𝑘 1 = ( + ) {(𝑝 − 𝛾) ℎ1 (𝑧) + 𝛾} 4 2
We also note that 𝑓 ∈ C𝑘 (𝑝, 𝛼; 𝛾) ⇐⇒
𝑧𝑓 ∈ S𝑘 (𝑝, 𝛼; 𝛾) . 𝑝
(13)
(14)
𝑘 1 − ( − ) {(𝑝 − 𝛾) ℎ2 (𝑧) + 𝛾} , 4 2
(10)
In particular, we set S𝑘 (1, 𝛼; 𝛾) = S𝑘 (𝛼; 𝛾) and C𝑘 (1, 𝛼; 𝛾) = C𝑘 (𝛼; 𝛾). The following lemma will be required in our investigation.
where ℎ𝑖 is analytic in U with ℎ𝑖 (0) = 1, 𝑖 = 1, 2. Using the identity (6) in (14) and differentiating the resulting equation with respect to 𝑧, we obtain
Lemma 1 (see [10]). Let 𝑢 = 𝑢1 + 𝑖𝑢2 and V = V1 + 𝑖V2 and let Ψ(𝑢, V) be a complex-valued function satisfying the following conditions:
𝛼 𝑓(𝑧)) 𝑧(𝑄𝛽,𝑝
𝛼 𝑄𝛽,𝑝 𝑓 (𝑧)
= {𝛾 + (𝑝 − 𝛾) ℎ (𝑧)
(i) Ψ(𝑢, V) is continuous in a domain 𝐷 ∈ C2 ;
+
(ii) (0, 1) ∈ 𝐷 and Ψ(1, 0) > 0; (iii) R{Ψ(𝑖𝑢2 , V1 )} > 0 whenever (𝑖𝑢2 , V1 ) ∈ 𝐷 and V1 ≤ −(1/2)(1 + 𝑢22 ). If ℎ(𝑧) = 1 + 𝑐1 𝑧 + 𝑐2 𝑧2 + ⋅ ⋅ ⋅ is analytic in U such that (ℎ(𝑧), 𝑧ℎ (𝑧)) ∈ 𝐷 and R{Ψ(ℎ(𝑧), 𝑧ℎ (𝑧))} > 0 for 𝑧 ∈ U, then R{Ψ(ℎ(𝑧), 𝑧ℎ (𝑧))} > 0 in U. Lemma 2 (see [11]). Let 𝑝(𝑧) be analytic in U with 𝑝(0) = 𝑎 and R{𝑝(𝑧)} > 0, 𝑧 ∈ U. Then, for 𝑠 > 0 and 𝜇 ∈ C \ {−1}, R {𝑝 (𝑧) +
𝑠𝑧𝑝 (𝑧) } > 0, 𝑝 (𝑧) + 𝜇
(|𝑧| < 𝑟0 ) ,
(11)
(𝑝 − 𝛾) 𝑧ℎ (𝑧) } ∈ P𝑘 (𝑝, 𝛾) . (𝑝 − 𝛾) ℎ (𝑧) + 𝛾 + 𝛼 + 𝛽 (15)
This implies that ℎ𝑖 (𝑧) +
𝑧ℎ𝑖 (𝑧) ∈ P, (𝑝 − 𝛾) ℎ𝑖 (𝑧) + 𝛾 + 𝛼 + 𝛽
(𝑧 ∈ U; 𝑖 = 1, 2) . (16)
We form the functional Ψ(𝑢, V) by choosing 𝑢 = ℎ𝑖 (𝑧) and V = 𝑧ℎ𝑖 (𝑧): Ψ (𝑢, V) = 𝑢 +
V . (𝑝 − 𝛾) 𝑢 + 𝛾 + 𝛼 + 𝛽
(17)
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3
Clearly, the first two conditions of Lemma 1 are satisfied. Now, we verify condition (iii) as follows:
{ℎ𝑖 (𝑧) +
V1 } R {Ψ (𝑖𝑢2 , V1 )} = R { (𝑝 − 𝛾) 𝑖𝑢2 + 𝛾 + 𝛼 + 𝛽 ≤−
(𝛾 + 𝛼 + 𝛽) (1 +
2
2 [(𝑝 − 𝛾) 𝑢22 + (𝛾 + 𝛼 + 𝛽) ]
Theorem 5. One has C𝑘 (𝑝, 𝛼 + 1; 𝛾) ⊂ C𝑘 (𝑝, 𝛼; 𝛾) .
(19)
𝑓 ∈ C𝑘 (𝑝, 𝛼 + 1; 𝛾) 𝑧𝑓 𝑧𝑓 ∈ S𝑘 (𝑝, 𝛼 + 1; 𝛾) ⇒ ∈ S𝑘 (𝑝, 𝛼; 𝛾) (20) 𝑝 𝑝
⇐⇒ 𝑓 ∈ C𝑘 (𝑝, 𝛼; 𝛾) ,
Ψ (𝑢, V) = 𝑢 +
Theorem 6. If 𝑓 ∈ S𝑘 (𝑝, 𝛼; 𝛾), then 𝐹𝑐,𝑝 (𝑓) ∈ S𝑘 (𝑝, 𝛼; 𝛾) (𝑐 ≥ 0), where the generalized Libera integral operator 𝐹𝑐,𝑝 is defined by (7).
Theorem 7. If 𝑓 ∈ C𝑘 (𝑝, 𝛼; 𝛾), then 𝐹𝑐,𝑝 (𝑓) ∈ C𝑘 (𝑝, 𝛼; 𝛾) (𝑐 ≥ 0), where 𝐹𝑐,𝑝 is defined by (7).
𝑓 ∈ C𝑘 (𝑝, 𝛼; 𝛾) ⇐⇒
⇐⇒
−
𝛼 𝑐𝑄𝛽,𝑝 𝐹𝑐,𝑝
(𝑓) (𝑧) . (22)
= (𝑝 − 𝛾) ℎ (𝑧) + 𝛾 + 𝑐.
(23)
Taking the logarithmic differentiation on both sides of (23) with respect to 𝑧 and multiplying by 𝑧, we have 1 𝑝−𝛾
𝛼 𝑓 (𝑧)) 𝑧(𝑄𝛽,𝑝 ( 𝛼 𝑄𝛽,𝑝 𝑓 (𝑧)
− 𝛾)
𝑧ℎ (𝑧) ∈ P𝑘 . = ℎ (𝑧) + (𝑝 − 𝛾) ℎ (𝑧) + 𝛾 + 𝑐
𝑝
∈ S𝑘 (𝑝, 𝛼; 𝛾)
which proves Theorem 7. Theorem 8. If 𝑓 ∈ C𝑘 (𝑝, 𝛼 + 1; 𝛾), for 𝑧 ∈ U, then 𝑓 ∈ C𝑘 (𝑝, 𝛼; 𝛾) for |𝑧| < 𝑟0 =
𝜇 + 1 , √𝐴 + (𝐴2 − 𝜇2 − 1)1/2
(28)
where 𝐴 = 2(𝑠 + 1)2 +|𝜇|2 −1, with 𝜇 = ((𝛾+𝛼+𝛽)/(𝑝−𝛾)) ≠ −1 and 𝑠 = (1/(𝑝 − 𝛾)). This radius is the best possible.
Then, by using (21) and (22), we obtain 𝛼 𝑄𝛽,𝑝 𝐹𝑐,𝑝 (𝑓) (𝑧)
𝑧(𝐹𝑐,𝑝 (𝑓))
⇐⇒ 𝐹𝑐,𝑝 (𝑓) ∈ C𝑘 (𝑝, 𝛼; 𝛾) ,
where ℎ is analytic in U with ℎ(0) = 1. From (21), we have 𝛼 𝑝) 𝑄𝛽,𝑝 𝑓 (𝑧)
(27)
𝑘 1 − ( − ) {(𝑝 − 𝛾) ℎ2 (𝑧) + 𝛾} , 4 2 (21)
(𝑐 + 𝑝)
𝑧𝑓 ) ∈ S𝑘 (𝑝, 𝛼; 𝛾) 𝑝
(by Theorem 5)
𝑘 1 = ( + ) {(𝑝 − 𝛾) ℎ1 (𝑧) + 𝛾} 4 2
𝛼 𝑄𝛽,𝑝 𝑓 (𝑧)
𝑧𝑓 ∈ S𝑘 (𝑝, 𝛼; 𝛾) 𝑝
⇒ 𝐹𝑐,𝑝 (
= (𝑝 − 𝛾) ℎ (𝑧) + 𝛾
(𝑓) (𝑧)) = (𝑐 +
(26)
Then clearly Ψ(𝑢, V) satisfies all the properties of Lemma 1. Hence, ℎ𝑖 ∈ P (𝑖 = 1, 2) and consequently ℎ ∈ P𝑘 for 𝑧 ∈ U, which implies that 𝐹𝑐,𝑝 (𝑓) ∈ S𝑘 (𝑝, 𝛼; 𝛾).
Proof. Let 𝑓 ∈ S𝑘 (𝑝, 𝛼; 𝛾) and set
V . (𝑝 − 𝛾) 𝑢 + 𝛾 + 𝑐
Proof. By applying Theorem 6, it follows that
which evidently proves Theorem 5.
𝛼 𝐹𝑐,𝑝 𝑧(𝑄𝛽,𝑝
(25)
Next, we derive an inclusion property for the subclass C𝑘 (𝛼; 𝛾) involving 𝐹𝑐,𝑝 (𝑓), which is given by the following theorem.
Proof. Applying (10) and Theorem 4, we observe that
𝛼 𝐹𝑐,𝑝 (𝑓) (𝑧)) 𝑧(𝑄𝛽,𝑝 𝛼 𝑄𝛽,𝑝 𝐹𝑐,𝑝 (𝑓) (𝑧)
(𝑧 ∈ U; 𝑖 = 1, 2) .
We form the functional Ψ(𝑢, V) by choosing 𝑢 = ℎ𝑖 (𝑧) and V = 𝑧ℎ𝑖 (𝑧):
< 0.
Therefore applying Lemma 1, ℎ𝑖 ∈ P (𝑖 = 1, 2) and consequently ℎ ∈ P𝑘 for 𝑧 ∈ U. This completes the proof of Theorem 4.
⇐⇒
𝑧ℎ𝑖 (𝑧) } ∈ P, (𝑝 − 𝛾) ℎ𝑖 (𝑧) + 𝛾 + 𝑐
(18)
𝑢22 )
2
This implies that
(24)
Proof. Let 𝑓 ∈ C𝑘 (𝑝, 𝛼 + 1; 𝛾) for 𝑧 ∈ U and let
𝛼+1 𝑧(𝑄𝛽,𝑝 𝑓(𝑧)) 𝛼+1 𝑄𝛽,𝑝 𝑓 (𝑧)
= (𝑝 − 𝛾) ℎ (𝑧) + 𝛾 𝑘 1 = ( + ) {(𝑝 − 𝛾) ℎ1 (𝑧) + 𝛾} 4 2 𝑘 1 − ( − ) {(𝑝 − 𝛾) ℎ2 (𝑧) + 𝛾} , 4 2
(29)
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where ℎ𝑖 is analytic in U with ℎ𝑖 (0) = 1 and R{ℎ𝑖 (𝑧)} > 0 for 𝑖 = 1, 2. Using the identity (6) in (29) and differentiating the resulting equation with respect to 𝑧, we obtain
𝛼
} 1 { 𝑧(𝑄𝛽,𝑝 𝑓 (𝑧)) − 𝛾} { 𝛼 𝑝−𝛾 𝑄𝛽,𝑝 𝑓 (𝑧) { }
(1/ (𝑝 − 𝛾)) 𝑧ℎ1 (𝑧) 𝑘 1 } = ( + ) {ℎ1 (𝑧) + 4 2 ℎ1 (𝑧) + ((𝛾 + 𝛼 + 𝛽) / (𝑝 − 𝛾)) 𝑘 1 − ( − ) {ℎ2 (𝑧) 4 2 (1/ (𝑝 − 𝛾)) 𝑧ℎ2 (𝑧) }, ℎ2 (𝑧) + ((𝛾 + 𝛼 + 𝛽) / (𝑝 − 𝛾)) (30)
where R{ℎ𝑖 (𝑧)} > 0 for 𝑖 = 1, 2. Applying Lemma 2 with 𝑠 = ((𝛾 + 𝛼 + 𝛽)/(𝑝 − 𝛾)) and 𝜇 = ((𝛾 + 𝛼 + 𝛽)/(𝑝 − 𝛾)) ≠ − 1, we get R {ℎ𝑖 (𝑧) +
(1/ (𝑝 − 𝛾)) 𝑧ℎ𝑖 (𝑧) }>0 ℎ𝑖 (𝑧) + ((𝛾 + 𝛼 + 𝛽) / (𝑝 − 𝛾))
(31)
for |𝑧| < 𝑟0 , where 𝑟0 is given by (28). This completes the proof of Theorem 8. Theorem 9. Let 𝜙 be a convex function and 𝑓 ∈ S2 (𝛼; 𝛾). Then 𝐺 ∈ S2 (𝛼; 𝛾), where 𝐺 = 𝜙 ∗ 𝑓. Proof. Let = 𝜙 ∗ 𝑓. Then 𝛼 𝛼 𝛼 𝑄𝛽,𝑝 𝐺 (𝑧) = 𝑄𝛽,𝑝 (𝜙 ∗ 𝑓) (𝑧) = 𝜙 (𝑧) ∗ 𝑄𝛽,𝑝 𝑓 (𝑧) .
(32)
𝛼 Also, 𝑓 ∈ S2 (𝛼; 𝛾). Therefore, 𝑄𝛽,𝑝 𝑓 ∈ S2 (𝛾). By logarithmic differentiation of (32) and after some simplification, we obtain
𝛼 𝐺 (𝑧)) 𝑧(𝑄𝛽,𝑝 𝛼 𝑝𝑄𝛽,𝑝 𝐺 (𝑧)
=
The author declares that there is no conflict of interests regarding the publication of this paper.
References
(1/ (𝑝 − 𝛾)) 𝑧ℎ (𝑧) = ℎ (𝑧) + ℎ (𝑧) + ((𝛾 + 𝛼 + 𝛽) / (𝑝 − 𝛾))
+
Conflict of Interests
𝛼 𝜙 (𝑧) ∗ 𝐹 (𝑧) 𝑄𝛽,𝑝 𝑓 (𝑧) 𝛼 𝜙 (𝑧) ∗ 𝑄𝛽,𝑝 𝑓 (𝑧)
,
(33)
𝛼 𝛼 where 𝐹 = 𝑧(𝑄𝛽,𝑝 𝑓(𝑧)) /𝑝𝑄𝛽,𝑝 𝑓(𝑧) is analytic in U and
𝛼 𝐺(𝑧)) / 𝐹(0) = 1. From Lemma 3, we can see that 𝑧(𝑄𝛽,𝑝 𝛼 𝑝𝑄𝛽,𝑝 𝐺(𝑧) is contained in the convex hull of 𝐹(U). Since
𝛼 𝛼 𝐺(𝑧)) /𝑝𝑄𝛽,𝑝 𝐺(𝑧) is analytic in U and 𝑧(𝑄𝛽,𝑝
𝛼 𝑤 (𝑧)) 𝑧(𝑄𝛽,𝑝 } { ∈ P (𝛾)} , 𝐹 (U) = Ω = {𝑤 : 𝛼 𝑝𝑄𝛽,𝑝 𝑤 (𝑧) } {
(34)
𝛼 𝛼 𝐺(𝑧)) /𝑝𝑄𝛽,𝑝 𝐺(𝑧) lies in Ω; this implies that 𝐺 = then 𝑧(𝑄𝛽,𝑝 𝜙 ∗ 𝑓 ∈ S2 (𝛼; 𝛾).
Remark 10. Putting 𝑝 = 1 in the above results, we obtain corresponding results for the operator 𝑄𝛽𝛼 .
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