Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2010, Article ID 175369, 11 pages
doi:10.1155/2010/175369
Research Article
Some Starlikeness Criterions for
Analytic Functions
Gejun Bao,1 Lifeng Guo,1 and Yi Ling2
1
2
Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China
Department of Mathematical Science, Delaware State University, Dover, DE 19901, USA
Correspondence should be addressed to Gejun Bao, baogj@hit.edu.cn and
Yi Ling, yiling@desu.edu
Received 26 October 2010; Accepted 16 December 2010
Academic Editor: Vijay Gupta
Copyright q 2010 Gejun Bao 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.
We determine the condition on α, μ, β, and λ for which |1−αz/fzμ αzf z/fzz/fzμ −
1| < λ implies fz ∈ S∗ β, where S∗ β is the class of starlike functions of order β. Some results of
Obradović and Owa are extended. We also obtain some new results on starlikeness criterions.
1. Introduction
Let n be a positive integer, and let Hn denote the class of function
fz z ∞
ak1 zk1
1.1
kn
that are analytic in the unit disk U {z : |z| < 1}. For 0 ≤ β < 1, let
S β ∗
zf z
> β, z ∈ U
f ∈ H1 : Re
fz
1.2
denote the class of starlike function of order β and S∗ 0 S∗ .
Let fz and Fz be analytic in U; then we say that the function fz is subordinate
to Fz in U, if there exists an analytic function wz in U such that |wz| ≤ |z|, and fz ≡
Fwz, denoted that f ≺ F or fz ≺ Fz. If Fz is univalent in U, then the subordination
is equivalent to f0 F0 and fU ⊂ FU 1.
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Journal of Inequalities and Applications
Let
Sλ f ∈ H1 : f z − 1 < λ, z ∈ U .
1.3
√
Singh 2 proved that Sλ ⊂ S∗ if 0 < λ ≤ 2/ 5. More recently, Fournier 3, 4 proved that
2
Sλ ⊂ S∗ ⇐⇒ 0 ≤ λ ≤ √ ,
5
⎧
1 − λ1 − λ/2
⎪
⎪
,
⎪
⎪
⎨
1 − λ2 /4
ρλ ⎪
⎪
1/2 1 − 5/4λ2
⎪
⎪
⎩
,
1 − λ2 /4
if 0 ≤ λ ≤
if
2
,
3
1.4
2
≤ λ ≤ 1,
3
is the order of starlikeness of Sλ . Now, we define
U λ, μ, n f ∈ Hn
μ
zf z
z
:
− 1 < λ, z ∈ U .
fz fz
1.5
Clearly, Uλ, −1, 1 Sλ . In 1998, Obradović 5 proved that
U λ, μ, 1 ⊂ S∗
1.6
if 0 < μ < 1 and 0 < λ ≤ 1 − μ/ 1 − μ2 μ2 . Recently, Obradović and Owa 6 proved that
U λ, μ, n ⊂ S∗
1.7
if 0 < μ < 1 and 0 < λ ≤ n − μ/ n − μ2 μ2 .
In this paper we find a condition on α, μ, β, and λ for which
μ
μ
zf z
z
z
1 − α
α
− 1 < λ
fz
fz fz
1.8
implies fz ∈ S∗ β and extend some results of Obradović and Owa 5, 6. Also, we obtain
some new results on starlikeness criterions.
Journal of Inequalities and Applications
3
2. Main Results
For our results we need the following lemma.
Lemma 2.1 see 6. Let pz 1 pn zn pn1 zn1 · · · be analytic in U, n ≥ 1, and satisfy the
condition
pz −
1 zp z ≺ 1 λz,
μ
0 < μ < 1, 0 < λ ≤ 1.
2.1
Then
pz ≺ 1 λ1 z,
2.2
where
λ1 λμ
.
n−μ
2.3
Theorem 2.2. Let 0 ≤ μ < 1, nα > μ, 0 ≤ β < 1, and
⎧ α nα − μ 1 − β
⎪
⎪
⎪
,
⎪
⎪
⎪
α n μ − μβ − 2μ
⎪
⎪
⎪
⎪
⎪
⎪
⎪ ⎪
⎪
⎪
2α 1 − β − 1
nα
−
μ
⎨
Mn α, β, μ ,
⎪
n2 α2 2 μ2 1 − β − nμ α
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
nα − μ 1 − β
⎪
⎪
⎪
,
⎪
⎪
⎪
⎩ n−μ 1−β
if α ≥ α2 ,
if α1 < α < α2 ,
2.4
if 0 < α ≤ α1 ,
where
α1 n−μ 1−β
,
n 1−β
2
− 8nμ 1 − β
n 3μ 1 − β n 3μ 1 − β
α2 .
2n 1 − β
2.5
If pz 1 pn zn pn1 zn1 · · · and qz 1 qn zn qn1 zn1 · · · are analytic in U, satisfy
μλ
z,
nα − μ
qz 1 − α αpz ≺ 1 λz,
qz ≺ 1 2.6
2.7
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Journal of Inequalities and Applications
where 0 < λ ≤ Mn α, β, μ, then
Re pz > β,
for z ∈ U.
2.8
Proof. If μ 0, it is easy to see the result is true. Now, assume μ > 0. Let
N
μλ
.
nα − μ
2.9
If there exists z0 ∈ U, such that Re pz0 β, then we will show that
qz0 1 − α αpz0 − 1 ≥ λ
2.10
for 0 < λ ≤ Mn α, β, μ. Note that |qz0 − 1| ≤ N for z ∈ U; it is sufficient to show that
αpz0 − 1 − N 1 − α αpz0 ≥ λ
2.11
for 0 < λ ≤ Mn α, β, μ. Let pz0 β iy, y ∈ R; then, the left-hand side of 2.11 is
2
2
α β − 1 y2 − N αβ 1 − α α2 y2
2
2
α β y 1 − 2β − N α2 β2 α2 y2 2α1 − αβ 1 − α2 .
2.12
Suppose that x β2 y2 and note that nα − μN μλ; then inequality 2.11 is equivalent to
αμ x 1 − 2β
N≤
nα − μ μ α2 x 2α1 − αβ 1 − α2
2.13
for all x ≥ β2 and 0 < λ ≤ Mn α, β, μ. Now, if we define
ϕx x 1 − 2β
,
nα − μ μ α2 x 2α1 − αβ 1 − α2
x ≥ β2 ,
2.14
x > β2 ,
2.15
then we have
nα − μ ψx μ 1 − 2α 1 − β
ϕ x 2 ,
2ψx x 1 − 2β nα − μ μψx
where
ψx α2 x 2α1 − αβ 1 − α2 .
2.16
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5
Since
ψx α2 x2 2α1 − αβ 1 − α2 > 1 − α 1 − β ,
for x > β2 ,
2.17
the denominator of ϕ x is positive. Further, let
T x nα − μ ψx μ 1 − 2α 1 − β ,
x ≥ β2 .
2.18
T x ≥ nα − μ 1 − α 1 − β μ 1 − 2α 1 − β .
2.19
1
≤ α,
1−β
2.20
T x ≥ nα2 1 − β − n 3μ 1 − β α 2μ
n 1 − β α − r1 α − r2 ,
2.21
2
− 8nμ 1 − β
n 3μ 1 − β −
n 3μ 1 − β
r1 ,
2n 1 − β
2
− 8nμ 1 − β
n 3μ 1 − β n 3μ 1 − β
r2 .
2n 1 − β
2.22
We have
If
we get
where
Note that
r1 <
1
< r2 .
1−β
2.23
We obtain
T x ≥ 0 for α ≥ α2 r2 .
2.24
1
1
,
≤α<
1−β
2 1−β
2.25
If
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Journal of Inequalities and Applications
we have
T x ≥ α n − μ 1 − β − n 1 − β α .
2.26
1
T x ≥ 0 for ≤ α ≤ α1 ,
2 1−β
2.27
n−μ 1−β
1
.
α1 <
1−β
n 1−β
2.28
1
0<α< ,
2 1−β
2.29
Hence we obtain
where
If
we have 1 − 2α1 − β > 0. It follows that T x > 0.
Therefore we obtain ϕ x ≥ 0 for x > β2 if 0 < α ≤ α1 or α ≥ α2 . It follows that
⎧
1−β
⎪
⎪
,
⎪ ⎨ α n μ − μβ − 2μ
⎪
min ϕx ϕ β2 ⎪
x≥β2
⎪
1−β
⎪
⎪
⎩ ,
α n−μ 1−β
if α ≥ α2 ,
2.30
if 0 < α ≤ α1 .
If α1 < α < α2 , we have
lim T x T β2 nα − μ 1 − α 1 − β μ 1 − 2α 1 − β < 0
2
x → β 2.31
by 2.13 and 2.21 for 1/1 − β ≤ α < α2 and by 2.23 for α1 < α < 1/1 − β. Note that T x
is an continuous increasing function for x ≥ β2 , and
lim T x > 0.
x→∞
2.32
Then there exists a unique x0 ∈ β2 , ∞, such that
T x0 0,
or ϕ x0 0.
2.33
Thus, x0 is the global minimum point of ϕx on β2 , ∞. It follows from 2.33 that
nα − μ
2
α x0 2α1 − αβ 1 − α2 μ 2α 1 − β − 1 ,
2.34
Journal of Inequalities and Applications
7
or
1
x0 2
α
2
μ2 2α 1 − β − 1
2
− 2α1 − αβ − 1 − α .
2
nα − μ
2.35
By a simple calculation, we may obtain
2α 1 − β − 1
min ϕx ϕx0 x≥β2
α n2 α2 2 μ2 1 − β − nμ α
2.36
for α1 < α < α2 . It follows from 2.30 and 2.36 that that inequality 2.13 holds. This shows
that inequality 2.10 holds, which contradicts with 2.7. Hence we must have
Re pz > β,
z ∈ U.
2.37
Theorem 2.3. Let α, μ, β, λ and Mn α, β, μ be defined as in Theorem 2.2. If fz ∈ Hn satisfies
μ
μ
zf z
z
z
1 − α
< λ,
α
−
1
fz
fz fz
2.38
where 0 < λ ≤ Mn α, β, μ, then fz ∈ S∗ β.
Proof. If μ 0, Mn α, β, 0 α1 − β and the result is trivial. Now, assume μ > 0. If we put
qz z
fz
μ
,
2.39
then by some transformations and 2.38 we get
qz −
α zq z ≺ 1 λz.
μ
2.40
By Lemma 2.1, we obtain
qz ≺ 1 μλ
z.
nα − μ
2.41
Let
pz zf z
.
fz
2.42
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Journal of Inequalities and Applications
Then we have
qz 1 − α αpz ≺ 1 λz.
2.43
By Theorem 2.2, we get
Re pz > β,
z ∈ U.
2.44
It follows that fz ∈ S∗ β.
For β 0, we get the following corollary.
Corollary 2.4. Let 0 ≤ μ < 1, nα > μ, and let
⎧ α nα − μ
⎪
⎪
⎪
,
⎪
⎪
⎪
α n μ − 2μ
⎪
⎪
⎪
⎪
⎪
⎪
√
⎪
⎪ ⎨ nα − μ 2α − 1 ,
Mn α, μ ⎪
n2 α2 2 μ2 − nμ α
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
nα − μ
⎪
⎪
,
⎩
n−μ
if α ≥ α2 ,
if α1 ≤ α < α2 ,
2.45
if 0 < α < α1 ,
where
n−μ
,
n
2
n 3μ n 3μ − 8nμ
α1 α2 2n
2.46
.
If fz ∈ Hn satisfies
μ
μ
zf z
z
z
< λ,
1 − α
α
−
1
fz
fz fz
2.47
where 0 < λ ≤ Mn α, μ, then fz ∈ S∗ .
Corollary 2.5. Let 0 ≤ μ < 1, 0 ≤ β < 1, and let
⎧
n−μ 1−β
⎪
⎪
⎪
,
⎪
⎪
⎪ n−μ 1β
⎨
Mn β, μ ⎪
n − μ 1 − 2β
⎪
⎪
⎪
,
⎪
⎪
⎩ n2 2μ2 1 − β − nμ
if 1 > β ≥
μ
,
nμ
μ
.
if 0 ≤ β <
nμ
2.48
Journal of Inequalities and Applications
9
If fz ∈ Hn satisfies
μ
zf z
z
< λ,
−
1
fz fz
2.49
where 0 < λ ≤ Mn β, μ, then fz ∈ S∗ β.
Proof. Note that
n−μ 1−β
μ
,
α1 ≥ 1, for β ≥
nμ
n 1−β
n−μ 1−β
μ
,
α1 < 1, for β <
nμ
n 1−β
2
− 8nμ 1 − β
n 3μ 1 − β n 3μ 1 − β
α2 ≥ 1.
2n 1 − β
2.50
Putting α 1 in Theorem 2.3, we obtain the above corollary.
Remark 2.6. Our results extend the results given by Obradović 5, and Obradović and Owa
6.
Theorem 2.7. Let 0 < μ < 1, 0 ≤ β < 1, Re{c} > −μ, and let
⎧
n − μ 1 − β n c − μ
⎪
⎪
⎪
,
⎪
⎪
⎪ n − μ 1 − β c − μ
⎨
βn β, μ ⎪
n − μ 1 − 2βn c − μ
⎪
⎪
⎪
⎪
,
⎪
⎩ n2 2μ2 1 − β − nμc − μ
if β ≥
μ
,
nμ
μ
if β <
.
nμ
2.51
If fz ∈ Hn satisfies
μ
zf z
z
< λ,
−
1
fz fz
2.52
μ
−1/μ
c−μ z
t
tc−μ−1 dt
,
Fz z c−μ
z
ft
0
2.53
where 0 < λ ≤ βn β, μ, and
then Fz ∈ S∗ β.
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Journal of Inequalities and Applications
Proof. Let
z
Qz F z
Fz
1μ
2.54
.
Then from 2.52 and 2.53 we obtain
1μ
1
z
Q z f z
≺ 1 λz.
Qz c−μ
fz
2.55
Hence, by using Theorem 1 given by Hallenbeck and Ruscheweyh 7, we have that
Qz ≺ 1 λ1 z,
c − μλ
z,
λ1 n c − μ
2.56
and the desired result easily follows from Corollary 2.5.
For c μ 1, we have the following corollary.
Corollary 2.8. Let 0 < μ < 1, 0 ≤ β < 1, and let
⎧
n − μ 1 − β n 1
⎪
⎪
⎪
,
⎪
⎪
⎪
n−μ 1−β
⎨
βn β, μ ⎪
n − μ 1 − 2βn 1
⎪
⎪
⎪
,
⎪
⎪
⎩ n2 2μ2 1 − β − nμ
if 1 > β ≥
μ
,
nμ
μ
.
if 0 ≤ β <
nμ
2.57
If fz ∈ Hn satisfies
μ
zf z
z
− 1 < λ,
fz fz
2.58
μ −1/μ
z 1
t
Fz z
dt
,
z 0 ft
2.59
where 0 < λ ≤ βn β, μ, and
then Fz ∈ S∗ β.
Acknowledgment
The authors would like to thank the referee for giving them thoughtful suggestions which
greatly improved the presentation of the paper. Bao Gejun was supported by NSF of
P.R.China no. 11071048.
Journal of Inequalities and Applications
11
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