ISSN: 1821-1291, URL: http://www.bmathaa.org Volume 2 Issue 4(2010), Pages 1-6.
SOME SUFFICIENT CONDITIONS FOR STARLIKENESS USING SUBORDINATION CRITERIA
(DEDICATED IN OCCASION OF THE 70-YEARS OF PROFESSOR HARI M. SRIVASTAVA)
DEEPAK BANSAL, R. K. RAINA
Abstract. In this paper we derive certain sufficient conditions for the star- likeness of functions analytic in the open unit disk. Our main result generalizes some previously established results. We also present some useful consequences of the main result and point out relevances with some of the known results.
1. Introduction Let𝒜denote the class of functions of the form
𝑓(𝑧) =𝑧+
∞
∑
𝑘=2
𝑎𝑘𝑧𝑘, (1.1)
which are analytic and univalent in the open unit disk𝕌={𝑧:𝑧∈ℂand∣𝑧∣<1}.
A function 𝑓(𝑧) in 𝒜 is said to be in class 𝒮∗(𝛼) of starlike functions of order 𝛼(0≦𝛼 <1) in𝕌, if it satisfies the following inequality:
ℜ
(𝑧𝑓′(𝑧) 𝑓(𝑧)
)
> 𝛼(𝑧∈𝕌; 0≦𝛼 <1). (1.2) A function𝑓(𝑧)∈ 𝒜is said to be strongly starlike of order𝛼in 𝕌, if it satisfies
arg𝑧𝑓′(𝑧) 𝑓(𝑧)
<𝛼𝜋
2 (𝑧∈𝕌; 0< 𝛼≦1), (1.3) for some𝛼(0< 𝛼≦1).We denote by𝒮˜∗(𝛼) (0< 𝛼≦1) the subclass of𝒜consist- ing of all functions which are strongly starlike of order𝛼in 𝕌. Also, we denote by 𝒮∗(0) =𝒮˜∗(1) =𝒮∗.
If the functions 𝑓 and 𝑔 are analytic in𝕌, then we say that 𝑓 is subordinate to 𝑔 in 𝕌, and write 𝑓 ≺ 𝑔, if there exists a function 𝑤(𝑧) analytic in 𝕌 such that
∣𝑤(𝑧)∣<1, 𝑧∈𝕌,and 𝑤(0) = 0 with𝑓(𝑧) =𝑔(𝑤(𝑧)) in𝕌. If𝑔 is univalent in𝕌,
2000Mathematics Subject Classification. 30C45.
Key words and phrases. Analytic function, univalent function, differential subordination, best dominant, starlike function.
c
⃝2010 Universiteti i Prishtin¨es, Prishtin¨e, Kosov¨e.
Submitted June 15, 2010. Published August 24, 2010.
1
then𝑓 ≺𝑔 is equivalent to𝑓(0) =𝑔(0) and 𝑓(𝕌)⊂𝑔(𝕌).
For−1≦𝑏 < 𝑎≦1, the function 𝑓(𝑧) in 𝒜is said to be in the class𝒮∗[𝑎, 𝑏], if it satisfies
𝑧𝑓′(𝑧)
𝑓(𝑧) ≺ 1 +𝑎𝑧
1 +𝑏𝑧 (𝑧∈𝕌). (1.4)
The class𝒮∗[𝑎, 𝑏] (−1≦𝑏 < 𝑎≦1) can be identified with a known class of starlike functions by giving particular values to the parameters𝑎and𝑏. Indeed, we have
𝒮∗[1−2𝛼,−1]≡ 𝒮∗(𝛼) (0≦𝛼 <1).
Throughout this paper, the power functions, wherever occurring, are interpreted to be in terms of their principal values. In order to prove our main result (Theorem 2.1 below), we need the following well known result due to Miller and Mocanu [2]
see also [3, p. 132].
Lemma 1.1. Let the function 𝑞(𝑧) be univalent in𝕌, and let the functions𝜃 and 𝜙be analytic in a domain 𝔻containing𝑞(𝕌), with𝜙(𝑤)∕= 0when 𝑤∈𝑞(𝕌). Set
𝑄(𝑧) =𝑧𝑞′(𝑧)𝜙(𝑞(𝑧)) 𝑎𝑛𝑑 ℎ(𝑧) =𝜃(𝑞(𝑧)) +𝑄(𝑧) and suppose that
(i)𝑄(𝑧)is univalent and starlike in 𝕌 (ii) ℜ(𝑧ℎ′(𝑧)
𝑄(𝑧)
)
=ℜ(𝜃′(𝑞(𝑧))
𝜙(𝑞(𝑧)) +𝑧𝑄𝑄(𝑧)′(𝑧))
>0 (𝑧∈𝕌).
If 𝑝(𝑧)is analytic in𝕌, with 𝑝(0) =𝑞(0), 𝑝(𝕌)⊂𝔻 and
𝜃(𝑝(𝑧)) +𝑧𝑝′(𝑧)𝜙(𝑝(𝑧))≺𝜃(𝑞(𝑧)) +𝑧𝑞′(𝑧)𝜙(𝑞(𝑧)) =ℎ(𝑧), (1.5) then𝑝(𝑧)≺𝑞(𝑧) (𝑧∈𝕌), and𝑞(𝑧)is the best dominant.
2. MAIN RESULT
Theorem 2.1. Let 𝑓(𝑧)∈ 𝒜 satisfy 𝑓(𝑧)∕= 0 (𝑧∈𝕌). Also, let the function 𝑞(𝑧) be univalent in 𝕌, with 𝑞(0) = 1 and𝑞(𝑧)∕= 0,such that
ℜ (
1 + (𝛽−1)𝑧𝑞′(𝑧)
𝑞(𝑧) +𝑧𝑞′′(𝑧) 𝑞′(𝑧)
)
>0 (𝑧∈𝕌) (2.1) and
ℜ (
[1 + (𝜆−1)𝛽] + (𝛽+ 1)𝑞(𝑧) + (𝛽−1)𝑧𝑞′(𝑧)
𝑞(𝑧) +𝑧𝑞′′(𝑧) 𝑞′(𝑧)
)
>0 (𝑧∈𝕌) (2.2) for𝜆 >0and∣𝛽∣≦1.
If
(𝑧𝑓′(𝑧) 𝑓(𝑧)
)𝛽(
𝜆+𝑧𝑓′′(𝑧) 𝑓′(𝑧)
)
≺ℎ(𝑧) (𝑧∈𝕌), (2.3)
where
ℎ(𝑧) = [𝑞(𝑧)]𝛽+1+ (𝜆−1)[𝑞(𝑧)]𝛽+𝑧𝑞′(𝑧)[𝑞(𝑧)]𝛽−1, (2.4) then
𝑧𝑓′(𝑧)
𝑓(𝑧) ≺𝑞(𝑧) (𝑧∈𝕌), and𝑞(𝑧)is the best dominant of (2.3).
Proof. We first choose 𝑝(𝑧) =𝑧𝑓′(𝑧)
𝑓(𝑧) , 𝜃(𝑤) =𝑤𝛽(𝜆−1 +𝑤) and𝜙(𝑤) =𝑤𝛽−1, (2.5) then 𝜃(𝑤) and 𝜙(𝑤) are analytic inside the domain𝔻∗ =ℂ∖ {0}, which contains 𝑞(𝕌), 𝑞(0) = 1, and𝜙(𝑤)∕= 0 when𝑤∈𝑞(𝕌).
Now, if we define the functions𝑄(𝑧) andℎ(𝑧) by
𝑄(𝑧) =𝑧𝑞′(𝑧)𝜙(𝑞(𝑧)) =𝑧𝑞′(𝑧)[𝑞(𝑧)]𝛽−1 and
ℎ(𝑧) =𝜃(𝑞(𝑧)) +𝑄(𝑧) = [𝑞(𝑧)]𝛽+1+ (𝜆−1)[𝑞(𝑧)]𝛽+𝑧𝑞′(𝑧)[𝑞(𝑧)]𝛽−1, then it follows from (2.1) and (2.2) that𝑄(𝑧) is starlike in𝕌and
ℜ
(𝑧ℎ′(𝑧) 𝑄(𝑧)
)
>0 (𝑧∈𝕌).
We also note that the function 𝑝(𝑧) is analytic in 𝕌, with 𝑝(0) =𝑞(0) = 1. Since 0∈/ 𝑝(𝕌), therefore,𝑝(𝕌)⊂𝔻∗. and hence, the hypothesis of Lemma 1 are satisfied.
Applying Lemma 1.1, we find that (𝑧𝑓′(𝑧)
𝑓(𝑧)
)𝛽(
𝜆+𝑧𝑓𝑓′′′(𝑧)(𝑧)
)
= [𝑝(𝑧)]𝛽+1+ (𝜆−1)[𝑝(𝑧)]𝛽+𝑧𝑝′(𝑧)[𝑝(𝑧)]𝛽−1
=𝜃(𝑝(𝑧)) +𝑧𝑝′(𝑧)𝜙(𝑝(𝑧))≺ℎ(𝑧)
= [𝑞(𝑧)]𝛽+1+ (𝜆−1)[𝑞(𝑧)]𝛽+𝑧𝑞′(𝑧)[𝑞(𝑧)]𝛽−1
=𝜃(𝑞(𝑧)) +𝑧𝑞′(𝑧)𝜙(𝑞(𝑧)) (𝑧∈𝕌),
which implies that 𝑧𝑓𝑓(𝑧)′(𝑧)≺𝑞(𝑧) (𝑧∈𝕌) and𝑞(𝑧) is the best dominant of (2.3). □ 3. SOME CONSEQUENCES OF THEOREM 2.1
In this section we present some useful corollaries of Theorem 2.1 and also men- tion their relevances with some of the known results.
Let us set 𝜆= 1
𝛾(𝛾 >0) and𝑞(𝑧) =
(1 +𝑎𝑧 1 +𝑏𝑧
)𝛼
(−1≦𝑏 < 𝑎≦1, 0< 𝛼≦1).
It is easily verified that the above function𝑞(𝑧) being analytic and convex in𝕌is such that𝑞(𝕌) is symmetric with respect to the real axis and satisfies the inequality (see [5, p. 16]):
0≤ (1−𝑎
1−𝑏 )𝛼
<ℜ(𝑞(𝑧)) (𝑧∈𝕌;−1≤𝑎 < 𝑏≤1; 0< 𝛼≤1). Thus, (2.1) is equivalent to
ℜ [
1 + (𝛼𝛽−1) 𝑎𝑧
1 +𝑎𝑧−(𝛼𝛽+ 1) 𝑏𝑧 1 +𝑏𝑧
]
=ℜ [
−1 + (1−𝛼𝛽) 1
1 +𝑎𝑧+ (1 +𝛼𝛽) 1 1 +𝑏𝑧
]
>−1 + (1−𝛼𝛽) 1
1 +∣𝑎∣ + (1 +𝛼𝛽) 1 1 +∣𝑏∣ ≧0,
provided that
(1−𝛼𝛽) 1
1 +∣𝑎∣+ (1 +𝛼𝛽) 1 1 +∣𝑏∣ ≧1.
Similarly, (2.2) is satisfied if 𝛽
(1−𝛾 𝛾
)
+ (1 +𝛽) (1−𝑎
1−𝑏 )𝛼
+ (1−𝛼𝛽) 1
1 +∣𝑎∣+ (1 +𝛼𝛽) 1
1 +∣𝑏∣−1≧0.
With the above specialization of𝑞(𝑧) and the corresponding related conditions, Theorem 2.1 gives the following result:
Corollary 3.1. Let 𝑓(𝑧)∈ 𝒜satisfy 𝑓(𝑧)∕= 0 (𝑧∈𝕌). Also, let 𝛾 >0,
∣𝛽∣≦1; −1≦𝑏 < 𝑎≦1 and0< 𝛼≦1 be so chosen that (1−𝛼𝛽) 1
1 +∣𝑎∣+ (1 +𝛼𝛽) 1
1 +∣𝑏∣−1≧0 (3.1)
and 𝛽
(1−𝛾 𝛾
)
+ (1 +𝛽) (1−𝑎
1−𝑏 )𝛼
+ (1−𝛼𝛽) 1
1 +∣𝑎∣+ (1 +𝛼𝛽) 1
1 +∣𝑏∣−1≧0. (3.2) If
(𝑧𝑓′(𝑧) 𝑓(𝑧)
)𝛽(
1 +𝛾𝑧𝑓′′(𝑧) 𝑓′(𝑧)
)
≺ℎ(𝑧) (𝑧∈𝕌), (3.3)
where ℎ(𝑧) =
(1 +𝑎𝑧 1 +𝑏𝑧
)𝛼𝛽−1[
(1−𝛾)1 +𝑎𝑧 1 +𝑏𝑧 +𝛾
(1 +𝑎𝑧 1 +𝑏𝑧
)𝛼+1
+𝛼𝛾𝑧(𝑎−𝑏) (1 +𝑏𝑧)2
]
, (3.4)
then 𝑧𝑓𝑓(𝑧)′(𝑧) ≺(
1+𝑎𝑧 1+𝑏𝑧
)𝛼
(𝑧∈𝕌).
Remark. It may be observed that if we set𝑏=−1 and𝛼= 1, then Corollary 3.1 corresponds to the known result of Aghalary and Jahangiri [1, p. 2054, Theorem 2.2]. On the other hand, if we set 𝑏 = 0 and 𝛼 = 1, then Corollary 3.1 readily gives another known result [1, p. 2056, Theorem 2.5]. Also, by setting𝛽= 1, 𝛾=
1
𝜆 (𝜆 > 0) and 𝛼 = 1 in Corollary 3.1, and simplyfying the conditions (3.1) and (3.2), we get the result which was earlier obtained by Xu and Yang [6, p. 581, Theorem 1].
A much simpler consequence of Corollary 3.1 occurs when we assign in it the values𝛽 = 1, 𝛾 = 𝜆1 (𝜆 >0) 𝑎= 1 and𝑏=−1, so that the conditions (3.1) and (3.2) are both satisfied provided that𝜆≧1. Also, we note that (see [6, p. 589])
𝑧𝑓′(𝑧) 𝑓(𝑧) ≺
(1 +𝑧 1−𝑧
)𝛼
⇒𝑎𝑟𝑔
(𝑧𝑓′(𝑧) 𝑓(𝑧)
)
=𝛼 𝑎𝑟𝑔
(1 +𝑤(𝑧) 1−𝑤(𝑧) )
(where𝑤(0) = 0 and∣𝑤(𝑧)∣≦∣𝑧∣<1 (𝑧∈𝕌))
⇒
𝑎𝑟𝑔
(𝑧𝑓′(𝑧) 𝑓(𝑧)
)
=𝛼
𝑎𝑟𝑔
(1 +𝑤(𝑧) 1−𝑤(𝑧)
)
→ 𝛼𝜋
2 𝑎𝑠 𝑤→𝑖.
Corollary 3.1 then eventually gives the following result:
Corollary 3.2. Let 𝑓(𝑧)∈ 𝒜satisfy 𝑓(𝑧)∕= 0 (𝑧∈𝕌)and 𝑧2𝑓′′(𝑧)
𝑓(𝑧) +𝜆𝑧𝑓′(𝑧)
𝑓(𝑧) ≺ℎ(𝑧) (𝑧∈𝕌;𝜆≥1), (3.5) where
ℎ(𝑧) = (𝜆−1) (1 +𝑧
1−𝑧 )𝛼
+ (1 +𝑧
1−𝑧 )2𝛼
+2𝛼𝑧(1 +𝑧)𝛼−1
(1−𝑧)𝛼+1 , (3.6) then 𝑧𝑓𝑓(𝑧)′(𝑧) ≺(
1+𝑧 1−𝑧
)𝛼
(𝑧∈𝕌) ⇒𝑓(𝑧)∈𝒮˜∗(𝛼).
If we put𝛽= 0 and𝜆= 1 in Theorem 2.1, then we get the following interesting corollary:
Corollary 3.3. Let 𝑓(𝑧)∈ 𝒜satisfy 𝑓(𝑧)∕= 0 (𝑧∈𝕌). Also let that the function 𝑞(𝑧)be univalent in 𝕌, with𝑞(0) = 1 and𝑞(𝑧)∕= 0 (𝑧 ∈𝕌), satisfying each of the following inequlities:
ℜ (
1−𝑧𝑞′(𝑧)
𝑞(𝑧) +𝑧𝑞′′(𝑧) 𝑞′(𝑧)
)
>0 (3.7)
and
ℜ (
1 +𝑞(𝑧)−𝑧𝑞′(𝑧)
𝑞(𝑧) +𝑧𝑞′′(𝑧) 𝑞′(𝑧)
)
>0. (3.8)
If
1 + 𝑧𝑓′′(𝑧)
𝑓′(𝑧) ≺𝑞(𝑧) +𝑧𝑞′(𝑧)
𝑞(𝑧) (𝑧∈𝕌), (3.9)
then 𝑧𝑓𝑓(𝑧)′(𝑧) ≺𝑞(𝑧) (𝑧∈𝕌).
Remark. Corollary 3.3 can also be derived from a known result due to Srivastava and Attiya [4, p. 1154, Theorem 1 (when𝑝=𝜆= 1)].
By putiing𝛽=−1, 𝜆= 1 and𝑞(𝑧) = 1+𝑎𝑧1+𝑏𝑧 in Theorem 2.1, we get the following result.
Corollary 3.4. Let 𝑓(𝑧)∈ 𝒜satisfy 𝑓(𝑧)∕= 0 (𝑧∈𝕌). If−1≦𝑏 < 𝑎≦1 and 1 +𝑧𝑓′′(𝑧)/𝑓′(𝑧)
𝑧𝑓′(𝑧)/𝑓(𝑧) ≺1 + 𝑧(𝑎−𝑏)
(1 +𝑎𝑧)2 (𝑧∈𝕌), (3.10) then
𝑧𝑓′(𝑧)
𝑓(𝑧) ≺ 1 +𝑎𝑧
1 +𝑏𝑧 (𝑧∈𝕌). (3.11)
Remark. For𝑏=−1, Corollary 3.4 corresponds to the result [1, p. 2055, Corollary 2.3].
References
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[3] S. S. Miller and P. T. Mocanu,Differential Subordinations: Theory and Applications, Series in Pure and Applied Mathematics, No. 225, Marcel Dekker, New York, (2000).
[4] H.M. Srivastava and A.A. Attiya,Some applicatios of differential subordination, App. Math.
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Deepak Bansal
Department of Mathematics, College of Engineering and Tecnology, Bikaner, Ra- jasthan, India
E-mail address:deepakbansal [email protected]
R. K. Raina
M.P. University of Agriculture and Technology, Udaipur, India.
Present Address: 10/11 Ganpati Vihar, Opposite Sector 5, Udaipur 313002, Rajasthan, India
E-mail address:rkraina [email protected]