Volume 2010, Article ID 175369, 11 pagesdoi:10.1155/2010/175369 Research Article Some Starlikeness Criterions for Analytic Functions Gejun Bao,1 Lifeng Guo,1 and Yi Ling2 1 Department of
Trang 1Volume 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 Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China
2 Department of Mathematical Science, Delaware State University, Dover, DE 19901, USA
Correspondence should be addressed to Gejun Bao,baogj@hit.edu.cnand
Yi Ling,yiling@desu.edu
Received 26 October 2010; Accepted 16 December 2010
Academic Editor: Vijay Gupta
Copyrightq 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 onα, μ, β, and λ for which |1−αz/fz μ αzfz/fzz/fz μ− 1| < λ implies fz ∈ S∗β, where S∗β is the class of starlike functions of order β Some results of
Obradovi´c and Owa are extended We also obtain some new results on starlikeness criterions
1 Introduction
Letn be a positive integer, and let H ndenote the class of function
fz z ∞
kn
that are analytic in the unit diskU {z : |z| < 1} For 0 ≤ β < 1, let
S∗
β
f ∈ H1: Rezfz
fz > β, z ∈ U
1.2
denote the class of starlike function of orderβ and S∗0 S∗
Letfz and Fz be analytic in U; then we say that the function fz is subordinate
toFz 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 tof0 F0 and fU ⊂ FU 1
Trang 2S λf ∈ H1:fz − 1 < λ,z ∈ U 1.3
Singh2 proved that S λ ⊂ S∗if 0< λ ≤ 2/√5 More recently, Fournier3,4 proved that
S λ ⊂ S∗⇐⇒ 0 ≤ λ ≤ √2
5,
ρ λ
⎧
⎪
⎪
⎪
⎪
1 − λ1 − λ/2
1− λ2/4 , if 0≤ λ ≤
2
3,
1/21− 5/4λ2
1− λ2/4 , if
2
3 ≤ λ ≤ 1,
1.4
is the order of starlikeness ofS λ Now, we define
Uλ, μ, n
f ∈ H n:
zf fzzfz z μ− 1
< λ, z ∈ U. 1.5 Clearly,Uλ, −1, 1 S λ In 1998, Obradovi´c5 proved that
Uλ, μ, 1⊂ S∗ 1.6
if 0< μ < 1 and 0 < λ ≤ 1 − μ/1 − μ2 μ2 Recently, Obradovi´c and Owa6 proved that
Uλ, μ, n⊂ S∗ 1.7
if 0< μ < 1 and 0 < λ ≤ n − μ/n − μ2 μ2
In this paper we find a condition onα, μ, β, and λ for which
1 − αfz z μ α zfz
fz
z
fz
μ
− 1
impliesfz ∈ S∗β and extend some results of Obradovi´c and Owa 5,6 Also, we obtain some new results on starlikeness criterions
Trang 32 Main Results
For our results we need the following lemma
Lemma 2.1 see 6 Let pz 1 p n z n p n1 z n1 · · · be analytic in U, n ≥ 1, and satisfy the
condition
pz −1μ zpz ≺ 1 λz, 0 < μ < 1, 0 < λ ≤ 1. 2.1
Then
pz ≺ 1 λ1z, 2.2
where
Theorem 2.2 Let 0 ≤ μ < 1, nα > μ, 0 ≤ β < 1, and
M n
α, β, μ
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
αnα − μ1− β
αn μ − μβ− 2μ , if α ≥ α2,
nα − μ2α1− β− 1
n2α2 2μ2
1− β− nμα , if α1< α < α2,
nα − μ1− β
n − μ1− β , if 0 < α ≤ α1,
2.4
where
α1 n − μ
1− β
n1− β ,
α2 n 3μ
1− β
n 3μ1− β2− 8nμ1− β
2.5
If pz 1 p n z n p n1 z n1 · · · and qz 1 q n z n q n1 z n1 · · · are analytic in U, satisfy
qz ≺ 1 nα − μ μλ z, 2.6
Trang 4where 0 < λ ≤ M n α, β, μ, then
Repz > β, for z ∈ U. 2.8
Proof If μ 0, it is easy to see the result is true Now, assume μ > 0 Let
If there existsz0∈ U, such that Re pz0 β, then we will show that
qz01− α αpz0− 1 ≥ λ 2.10 for 0< λ ≤ M n α, β, μ Note that |qz0 − 1| ≤ N for z ∈ U; it is sufficient to show that
α pz0 − 1 − N1 − α αpz0 ≥ λ 2.11 for 0< λ ≤ M n α, β, μ Let pz0 β iy, y ∈ R; then, the left-hand side of 2.11 is
α
β − 12 y2− N
αβ 1 − α2 α2y2
αβ2 y2 1 − 2β − Nα2β2 α2y2 2α1 − αβ 1 − α2.
2.12
Suppose thatx β2 y2and note thatnα − μN μλ; then inequality 2.11 is equivalent to
x 1 − 2β
nα − μ μα2x 2α1 − αβ 1 − α2 2.13 for allx ≥ β2and 0< λ ≤ M n α, β, μ Now, if we define
ϕx
x 1 − 2β
nα − μ μα2x 2α1 − αβ 1 − α2, x ≥ β2, 2.14
then we have
ϕx
nα − μψx μ1− 2α1− β
2ψxx 1 − 2βnα − μ μψx2, x > β2, 2.15
where
ψx α2x 2α1 − αβ 1 − α2. 2.16
Trang 5ψx α2x2 2α1 − αβ 1 − α2>1− α
1− β, for x > β2, 2.17
the denominator ofϕx is positive Further, let
Tx nα − μψx μ1− 2α1− β, x ≥ β2. 2.18
We have
Tx ≥nα − μ1− α
1− β μ1 − 2α1 − β. 2.19 If
1
we get
Tx ≥ nα2
1− β−n 3μ1− βα 2μ
n1− βα − r1α − r2, 2.21
where
r1 n 3μ
1− β−
n 3μ1− β2− 8nμ1− β
r2 n 3μ
1− β
n 3μ1− β2− 8nμ1− β
2.22
Note that
r1< 1
1− β < r2. 2.23
We obtain
Tx ≥ 0 for α ≥ α2 r2. 2.24 If
1
2
1− β ≤ α <
1
Trang 6we have
Tx ≥ αn − μ1− β− n1− βα. 2.26 Hence we obtain
Tx ≥ 0 for 1
2
where
α1 n − μ
1− β
n1− β <
1
If
0< α < 1
2
we have 1− 2α1 − β > 0 It follows that Tx > 0.
Therefore we obtainϕx ≥ 0 for x > β2if 0< α ≤ α1orα ≥ α2 It follows that
min
x≥β2ϕx ϕβ2
⎧
⎪
⎪
⎪
⎪
1− β
αn μ − μβ− 2μ , if α ≥ α2,
1− β
αn − μ1− β , if 0< α ≤ α1.
2.30
Ifα1< α < α2, we have
lim
x →β2Tx Tβ2
nα − μ1− α
1− β μ1 − 2α1 − β < 0 2.31
by2.13 and 2.21 for 1/1 − β ≤ α < α2and by2.23 for α1< α < 1/1 − β Note that Tx
is an continuous increasing function forx ≥ β2, and
lim
Then there exists a uniquex0 ∈ β2, ∞, such that
Thus,x0is the global minimum point ofϕx on β2, ∞ It follows from 2.33 that
nα − μα2x0 2α1 − αβ 1 − α2 μ2α1− β− 1, 2.34
Trang 7x0 α12
μ2
2α1− β− 12
nα − μ2 − 2α1 − αβ − 1 − α2
By a simple calculation, we may obtain
min
x≥β2ϕx ϕx0
2α1− β− 1
αn2α2 2μ2
forα1< α < α2 It follows from2.30 and 2.36 that that inequality 2.13 holds This shows that inequality2.10 holds, which contradicts with 2.7 Hence we must have
Repz > β, z ∈ U. 2.37
Theorem 2.3 Let α, μ, β, λ and M n α, β, μ be defined as in Theorem 2.2 If fz ∈ H n satisfies
1 − αfz z μ α zfz
fz
z
fz
μ
− 1
where 0 < λ ≤ M n α, β, μ, then fz ∈ S∗β.
Proof If μ 0, M n α, β, 0 α1 − β and the result is trivial Now, assume μ > 0 If we put
qz
fz
μ
then by some transformations and2.38 we get
qz − α μ zqz ≺ 1 λz. 2.40
qz ≺ 1 nα − μ μλ z. 2.41 Let
pz zf fzz 2.42
Trang 8Then we have
Repz > β, z ∈ U. 2.44
It follows thatfz ∈ S∗β.
Forβ 0, we get the following corollary.
Corollary 2.4 Let 0 ≤ μ < 1, nα > μ, and let
M nα, μ
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
αnα − μ
αn μ− 2μ , if α ≥ α2,
nα − μ√2α − 1
n2α2 2μ2− nμα , if α1 ≤ α < α2,
nα − μ
n − μ , if 0 < α < α1,
2.45
where
α1 n − μ n ,
α2 n 3μ
n 3μ2− 8nμ
2.46
If fz ∈ H n satisfies
1 − αfz z μ α zfz
fz
z
fz
μ
− 1
where 0 < λ ≤ M n α, μ, then fz ∈ S∗.
Corollary 2.5 Let 0 ≤ μ < 1, 0 ≤ β < 1, and let
M n
β, μ
⎧
⎪
⎪
⎨
⎪
⎪
⎩
n − μ1− β
n − μ1 β , if 1 > β ≥
μ
n μ ,
n − μ1− 2β
n2 2μ2
1− β− nμ , if 0 ≤ β <
μ
n μ .
2.48
Trang 9If fz ∈ H n satisfies
zf fzzfz z μ− 1
where 0 < λ ≤ M n β, μ, then fz ∈ S∗β.
Proof Note that
α1 n − μ
1− β
n1− β ≥ 1, for β ≥
μ
n μ ,
α1 n − μ
1− β
n1− β < 1, for β <
μ
n μ ,
α2 n 3μ
1− β
n 3μ1− β2− 8nμ1− β
2.50
Puttingα 1 inTheorem 2.3, we obtain the above corollary
Remark 2.6 Our results extend the results given by Obradovi´c5, and Obradovi´c and Owa
6
Theorem 2.7 Let 0 < μ < 1, 0 ≤ β < 1, Re{c} > −μ, and let
β nβ, μ
⎧
⎪
⎪
⎨
⎪
⎪
⎩
n − μ1− βn c − μ
n − μ1− βc − μ , if β ≥ n μ μ ,
n − μ1− 2βn c − μ
n2 2μ2
1− β− nμc − μ , if β < n μ μ .
2.51
If fz ∈ H n satisfies
zf fzzfz z μ− 1
where 0 < λ ≤ β n β, μ, and
Fz z c − μ
z c−μ
z
0
t
ft
μ
t c−μ−1 dt
−1/μ
then Fz ∈ S∗β.
Trang 10Proof Let
Qz Fz
Fz
1μ
Then from2.52 and 2.53 we obtain
Qz c − μ1 Qz fz
fz
1μ
Hence, by using Theorem 1 given by Hallenbeck and Ruscheweyh7, we have that
Qz ≺ 1 λ1z, λ1 c − μλ
and the desired result easily follows fromCorollary 2.5
Forc μ 1, we have the following corollary.
Corollary 2.8 Let 0 < μ < 1, 0 ≤ β < 1, and let
β nβ, μ
⎧
⎪
⎪
⎨
⎪
⎪
⎩
n − μ1− βn 1
n − μ1− β , if 1 > β ≥
μ
n μ ,
n − μ1− 2βn 1
n2 2μ2
1− β− nμ , if 0 ≤ β <
μ
n μ .
2.57
If fz ∈ H n satisfies
zf fzzfz z μ− 1
where 0 < λ ≤ β n β, μ, and
Fz z
1
z
z
0
t
ft
μ
dt
−1/μ
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.Chinano 11071048
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