Differential Subordinations for Certain Class of Symmetric Functions
Abbas Kareem Wanas Department of Mathematics College of Computer Science and Mathematics University of Al-Qadisiya Diwaniya – Iraq E-Mail:
[email protected] ,
[email protected]
Abstract The object of this paper is to introduce and study a new class 𝐿𝑚 (𝜂; ) of symmetric analytic functions in the open unit disk. Also we obtain some results for this class.
2000 Mathematics Subject Classification: 30C45, 30C50. Keywords: Analytic function, Symmetric function, Hadamard product, Convex univalent, Subordination. .
1. Introduction and Preliminaries Let 𝐴𝑚 denote the class of functions 𝑓 of the form: ∞
𝑎𝑛 𝑧 𝑛
𝑓 𝑧 =𝑧+
𝑚 ∈ ℕ = {1,2,3, … } ,
1.1
𝑛=𝑚 +1
which are analytic in the open unit disk 𝑈 = 𝑧 ∈ ℂ: 𝑧 < 1 . For two functions ƒ and 𝑔 analytic in 𝑈, we say that the function 𝑓 is subordinate to 𝑔, written ƒ ≺ 𝑔 or ƒ 𝑧 ≺ 𝑔 𝑧 (𝑧 ∈ 𝑈), if there exists a Schwarz function 𝑤(𝑧) analytic in 𝑈 with 𝑤(0) = 0 and 𝑤 𝑧 < 1 (𝑧 ∈ 𝑈) such that ƒ(𝑧) = 𝑔(𝑤(𝑧)). In particular, if the function 𝑔 is univalent in 𝑈, then ƒ ≺ 𝑔 if and only if ƒ(0) = 𝑔(0) and ƒ(𝑈) ⊂ 𝑔(𝑈). The Hadamard product (or convolution) (𝑓1 ∗ 𝑓2 )(𝑧) of two functions
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Abbas Kareem Wanas et.al, Journal of Global Research in Mathematical Archives, 2(3), March 2014, 143-149
∞
𝑎𝑛,𝑗 𝑧 𝑛 ∈ 𝐴𝑚 𝑗 = 1,2
𝑓𝑗 𝑧 = 𝑧 + 𝑛=𝑚 +1
is given by
∞
𝑎𝑛,1 𝑎𝑛,2 𝑧 𝑛 .
𝑓1 ∗ 𝑓2 𝑧 = 𝑧 + 𝑛=𝑚 +1
A function 𝑓 ∈ 𝐴𝑚 is said to be starlike of order 𝛼 in 𝑈 if and only if 𝑧𝑓 ′ 𝑧 >𝛼 0 ≤ 𝛼 < 1 ;𝑧 ∈ 𝑈 . 𝑓 𝑧 Denote the class of all starlike functions of order 𝛼 in 𝑈 by 𝑆 ∗ (𝛼). 𝑅𝑒
A function 𝑓 ∈ 𝐴𝑚 is said to be prestarlike of order 𝛼 in 𝑈 if 𝑧 ∗ 𝑓 𝑧 ∈ 𝑆∗ 𝛼 2(1−𝛼) (1 − 𝑧)
𝛼 0, 𝛿 > 0 and 𝑓 ∈ 𝐿𝑚 (𝜂; 𝛿 + 1 − 𝛿). If 𝛿 ≤ 𝛿0 , where 𝛿0 =
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1 1 1− 2 𝜂
1 0
1 −1 𝜂 𝑢
1+𝑢
−1
𝑑𝑢
,
(2.5)
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Abbas Kareem Wanas et.al, Journal of Global Research in Mathematical Archives, 2(3), March 2014, 143-149
1
then 𝑓 ∈ 𝐿𝑚 (0; ). The bound 𝛿0 is the sharp when 𝑧 = 1−𝑧 .
Proof. Suppose that 𝑔 𝑧 =
𝑓 𝑧 − 𝑓 −𝑧 . 2𝑧
2.6
Let 𝑓 ∈ 𝐿𝑚 (𝜂; 𝛿 + 1 − 𝛿) with 𝜂 > 0 and 𝛿 > 0. Then we have ′
𝑔 𝑧 + 𝜂𝑧𝑔 𝑧 = 1 − 𝜂
𝑓 𝑧 − 𝑓 −𝑧 2𝑧
𝑓 ′ 𝑧 − 𝑓 ′ −𝑧 +𝜂 2
≺ 𝛿(𝑧) + 1 − 𝛿. An application of Lemma 1.2, we obtain 𝑔 𝑧 ≺
𝛿 −1 𝑧 𝜂 𝜂
𝑧 1 −1 𝑡𝜂
𝑡 𝑑𝑡 + 1 − 𝛿 = ∗ 𝜙 𝑧 ,
(2.7)
0
where 𝛿 −1 𝜙 𝑧 = 𝑧 𝜂 𝜂
𝑧 0
1
−1
𝑡𝜂 𝑑𝑡 + 1 − 𝛿 . 1−𝑡
(2.8)
If 0 < 𝛿 ≤ 𝛿0 , where 𝛿0 > 1 is given by (2.5), then it follows from (2.8) that 𝑅𝑒 𝜙 𝑧
𝛿 = 𝜂
1
1 −1 𝑢𝜂 𝑅𝑒
0
1 𝛿 𝑑𝑢 + 1 − 𝛿 > 1 − 𝑢𝑧 𝜂
1 0
1
−1
𝑢𝜂 𝑑𝑢 + 1 − 𝛿 1+𝑢
1 ≥ . 2 Now, by using the Herglotz representation for 𝜙 z , from (2.6) and (2.7), we get 𝑓 𝑧 − 𝑓 −𝑧 ≺ ∗𝜙 𝑧 ≺ 𝑧 . 2𝑧 Since is convex univalent in 𝑈, then 𝑓 ∈ 𝐿𝑚 (0; ). 1
For 𝑧 = 1−𝑧 and 𝑓 ∈ 𝐴𝑚 defined by 𝑓 𝑧 − 𝑓 −𝑧 𝛿 −1 = 𝑧 𝜂 𝜂 2𝑧
𝑧 0
1
−1
𝑡𝜂 𝑑𝑡 + 1 − 𝛿 , 1−𝑡
we have 1−𝜂
𝑓 𝑧 − 𝑓 −𝑧 2𝑧
𝑓 ′ 𝑧 − 𝑓 ′ −𝑧 +𝜂 2
= 𝛿(𝑧) + 1 − 𝛿.
Thus, 𝑓 ∈ 𝐿𝑚 𝜂; 𝛿 + 1 − 𝛿 .
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Abbas Kareem Wanas et.al, Journal of Global Research in Mathematical Archives, 2(3), March 2014, 143-149
Also, for 𝛿 > 𝛿0 , we have 𝑓 𝑧 − 𝑓 −𝑧 𝑅𝑒 2𝑧
1
𝛿 ⟶ 𝜂
0
1
−1
1 𝑢𝜂 𝑑𝑢 + 1 − 𝛿 < 2 1+𝑢
𝑧→1 ,
which implies that 𝑓 ∉ 𝐿𝑚 0; . Therefore, the bound 𝛿0 cannot be increased 1 when 𝑧 = 1−𝑧 and this completes the proof of the theorem.
Theorem 2.3. Let 𝑓 ∈ 𝐿𝑚 𝜂; , 𝑔 ∈ 𝐴𝑚 and 𝑔 𝑧 𝑧
𝑅𝑒
>
1 . 2
(2.9)
Then 𝑓 ∗ 𝑔 ∈ 𝐿𝑚 𝜂; .
Proof. Let 𝑓 ∈ 𝐿𝑚 𝜂; and 𝑔 ∈ 𝐴𝑚 . Then, we have 1−𝜂 = 1−𝜂
𝑓 ∗ 𝑔 𝑧 − 𝑓 ∗ 𝑔 −𝑧 2𝑧 𝑔 𝑧 𝑧
∗
+𝜂
𝑓 𝑧 − 𝑓 −𝑧 2𝑧 𝑔 𝑧 𝑧
=
+𝜂
𝑓∗𝑔
′
𝑔 𝑧 𝑧
𝑧 − 𝑓∗𝑔 2 ∗
′
−𝑧
𝑓 ′ 𝑧 − 𝑓 ′ −𝑧 2
∗𝜓 𝑧 ,
2.10
where 𝜓 𝑧 = 1−𝜂
𝑓 ′ 𝑧 − 𝑓 ′ −𝑧 +𝜂 2
𝑓 𝑧 − 𝑓 −𝑧 2𝑧
From (2.9), note that the function
𝑔 𝑧 𝑧
2.11
has the Herglotz representation
𝑑𝜇(𝑥) 𝑥 =1 1 − 𝑥𝑧
𝑔 𝑧 = 𝑧
≺ 𝑧 .
𝑧∈𝑈 ,
2.12
where 𝜇(𝑥) is a probability measure defined on the unit circle x = 1 and 𝑑𝜇 𝑥 = 1. 𝑥 =1
Since is convex univalent in 𝑈, it follows from (2.10) to (2.12) that 1−𝜂
𝑓 ∗ 𝑔 𝑧 − 𝑓 ∗ 𝑔 −𝑧 2𝑧
+𝜂
𝑓∗𝑔
′
𝑧 − 𝑓∗𝑔 2
′
−𝑧
𝜓 𝑥𝑧 𝑑𝜇 𝑥 ≺ 𝑧 .
= 𝑥 =1
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Abbas Kareem Wanas et.al, Journal of Global Research in Mathematical Archives, 2(3), March 2014, 143-149
Therefore, 𝑓 ∗ 𝑔 ∈ 𝐿𝑚 𝜂; .
Theorem 2.4. Let 𝑓 ∈ 𝐿𝑚 𝜂; and let 𝑔 ∈ 𝐴𝑚 be prestarlike of order 𝛼 𝛼 < 1 . Then
𝑓 ∗ 𝑔 ∈ 𝐿𝑚 𝜂; .
Proof. Let 𝑓 ∈ 𝐿𝑚 𝜂; and 𝑔 ∈ 𝐴𝑚 . Then, we have 1−𝜂
𝑓 𝑧 − 𝑓 −𝑧 2𝑧
+𝜂
𝑓 ′ 𝑧 − 𝑓 ′ −𝑧 2
≺ 𝑧 .
(2.13)
Hence 𝑓 ∗ 𝑔 𝑧 − 𝑓 ∗ 𝑔 −𝑧 2𝑧
1−𝜂 = 1−𝜂
𝑔 𝑧 𝑧
∗
𝑓 𝑧 − 𝑓 −𝑧 2𝑧 =
+𝜂 +𝜂
𝑔 𝑧 ∗ (𝑧𝜓 𝑧 ) 𝑔 𝑧 ∗𝑧
𝑓∗𝑔 𝑔 𝑧 𝑧
′
𝑧 − 𝑓∗𝑔 2 ∗
′
−𝑧
𝑓 ′ 𝑧 − 𝑓 ′ −𝑧 2
𝑧∈𝑈 ,
(1.14)
where 𝜓 𝑧 is defined as in (2.11). Since is convex univalent in 𝑈, 𝜓 𝑧 ≺ 𝑧 , 𝑔 𝑧 ∈ ℜ(𝛼) and 𝑧 ∈ 𝑆 ∗ 𝛼 , 𝛼 < 1 , it follows from (2.14) and Lemma 1.3, we obtain the result.
Theorem 2.5. Let 𝑓 ∈ 𝐿𝑚 𝜂; be defined as in (1.1). Then 𝑐+1 𝑘 𝑧 = 𝑐 𝑧
𝑧
𝑡 𝑐−1 𝑓 𝑡 𝑑𝑡, (𝑅𝑒 𝑐 > −1)
0
is also in the class 𝐿𝑚 𝜂; .
Proof. Let 𝑓 ∈ 𝐿𝑚 𝜂; be defined as in (1.1). Then, we have 1−𝜂
𝑓 𝑧 − 𝑓 −𝑧 2𝑧
+𝜂
𝑓 ′ 𝑧 − 𝑓 ′ −𝑧 2
≺ 𝑧 .
2.15
Note that 𝑐+1 𝑘 𝑧 = 𝑐 𝑧
∞
𝑧
𝑡
𝑐−1
𝑓 𝑡 𝑑𝑡 = 𝑧 +
0
𝑛=𝑚 +1
𝑐+1 𝑎𝑛 𝑧 𝑛 . 𝑐+𝑛
2.16
We find from (2.16) that 𝑘 ∈ 𝐴𝑚 and 𝑓 𝑧 =
𝑐𝑘 𝑧 + 𝑧𝑘 ′ 𝑧 . 𝑐+1
(2.17)
Define the function 𝑝 by
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Abbas Kareem Wanas et.al, Journal of Global Research in Mathematical Archives, 2(3), March 2014, 143-149
𝑘 𝑧 − 𝑘 −𝑧 2𝑧
𝑝 𝑧 = 1−𝜂
𝑘 ′ 𝑧 − 𝑘 ′ −𝑧 +𝜂 2
.
(2.18)
By using (2.17) and (2.18), we get 𝑝 𝑧 +
1 𝑐 1 𝑧𝑝′ 𝑧 = 𝑝 𝑧 + 𝑧𝑝′ 𝑧 + 𝑝 𝑧 𝑐+1 𝑐+1 𝑐+1
= 1−𝜂
+𝜂
𝑐𝑘 𝑧 + 𝑧𝑘 ′ 𝑧
𝑐𝑘 𝑧 + 𝑧𝑘 ′ 𝑧
− 𝑐𝑘 −𝑧 + 𝑧𝑘 ′ −𝑧 2𝑧(𝑐 + 1)
′
− 𝑐𝑘 −𝑧 + 𝑧𝑘 ′ −𝑧 2(𝑐 + 1)
𝑓 𝑧 − 𝑓 −𝑧 2𝑧
= 1−𝜂
+𝜂
′
𝑓 ′ 𝑧 − 𝑓 ′ −𝑧 2
.
2.19
From (2.15) and (2.19), we arrive at 𝑝 𝑧 +
1 𝑧𝑝′ 𝑧 ≺ 𝑧 , 𝑐+1
𝑅𝑒 𝑐 > −1 .
An application of Lemma 1.2, we obtain 𝑝 𝑧 ≺ 𝑧 . By (2.18), we get 1−𝜂
𝑘 𝑧 − 𝑘 −𝑧 2𝑧
+𝜂
𝑘 ′ 𝑧 − 𝑘 ′ −𝑧 2
≺ 𝑧 .
Therefore, 𝑘 ∈ 𝐿𝑚 𝜂; .
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