Volume 2009, Article ID 601597,10pages doi:10.1155/2009/601597
Research Article
Conditions for Carath ´eodory Functions
Nak Eun Cho and In Hwa Kim
Department of Applied Mathematics, Pukyong National University, Busan 608-737, South Korea
Correspondence should be addressed to Nak Eun Cho,[email protected]
Received 12 April 2009; Accepted 13 October 2009
Recommended by Yong Zhou
The purpose of the present paper is to derive some sufficient conditions for Carath´eodory functions in the open unit disk. Our results include several interesting corollaries as special cases.
Copyrightq2009 N. E. Cho and I. H. Kim. 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.
1. Introduction
LetPbe the class of functionspof the form
pz 1
∞
n1
pnzn, 1.1
which are analytic in the open unit diskU{z∈C:|z|<1}. IfpinPsatisfies
Repz>0 z∈U, 1.2
then we say thatpis the Catath´eodory function.
LetAdenote the class of all functionsfanalytic in the open unit diskU{z:|z|<1} with the usual normalizationf0 f0−10. Iffandgare analytic inU, we say thatfis subordinate tog, writtenf≺gorfz≺gz, ifgis univalent,f0 g0andfU⊂gU. For 0< α≤1, letSTCαandSTSαdenote the classes of functionsf∈ Awhich are strongly convex and starlike of orderα; that is, which satisfy
1zf
z
fz ≺
1z
1−z α
z∈U, 1.3
zfz fz ≺
1z
1−z α
respectively. We note that1.3and 1.4can be expressed, equivalently, by the argument functions. The classesSTCαandSTSαwere introduced by Brannan and Kirwan1and studied by Mocanu 2 and Nunokawa 3,4. Also, we note that ifα 1, thenSTSα
coincides with S∗, the well-known class of starlikeunivalent functions with respect to origin, and if 0 < α < 1, thenSTSαconsists only of bounded starlike functions1, and hence the inclusion relationSTSα ⊂ S∗ is proper. Furthermore, Nunokawa and Thomas
4 see also5found the valueβαsuch thatSTCβα⊂ STSα.
In the present paper, we consider general forms which cover the results by Mocanu
6 and Nunokawa and Thomas 4. An application of a certain integral operator is also considered. Moreover, we give some sufficient conditions for univalentclose-to-convexand
stronglystarlike functionsof orderβas special cases of main results.
2. Main Results
To prove our results, we need the following lemma due to Nunokawa3.
Lemma 2.1. Letpbe analytic inU, p0 1 andpz/0in U. Suppose that there exists a point
z0∈Usuch that
argpz< π
2α for|z|<|z0|,
argpz0 π
2α 0< α≤1.
2.1
Then we have
z0pz0
pz0 iαk,
2.2
where
k≥ 1
2
x1
x
when argpz0
π
2α,
k≤ −1
2
x1
x
when argpz0 −
π
2α,
pz0
1/α±
ix x >0.
2.3
With the help ofLemma 2.1, we now derive the following theorem.
Theorem 2.2. Letpbe nonzero analytic inUwithp0 1and letpsatisfy the differential equation
ηzpz Bzpz aibAz, 2.4
whereη > 0,a∈R,0≤b≤atanπ/2α,0< α <1,Az signImpzandBzis analytic inUwithB0 a. If
argBz< π
2β
where
βη, α, a, b 2 πtan
−1SαTαasinπ/2α−bcosπ/2α ηα
SαTαacosπ/2αbsinπ/2α
, 2.6
Sα 1α1α/2, Tα 1−α1−α/2, 2.7
then
argpz< π
2α z∈U. 2.8
Proof. If there exists a point z0 ∈ Usuch that the conditions 2.1 are satisfied, then by
Lemma 2.1we obtain2.2under the restrictions2.3. Then we obtain
Az0
⎧ ⎨ ⎩
1, ifpz0 ixα,
−1, ifpz0 −ixα,
Bz0
aibAz0
pz0 −η
z0pz0
pz0
aibAz0±ix−α−iηαk
a xαcos
π
2α
b
xαAz0sin
±π
2α
i
b
xαAz0cos
π
2α−
a xαsin
±π
2α
−ηαk
.
2.9
Now we suppose that
pz0
1/α
ix x >0. 2.10
Then we have
argBz0 −tan−1
asinπ/2α−bcosπ/2αηαxαk
acosπ/2αbsinπ/2α
, 2.11
where
kxα≥ 1
2
Then, by a simple calculation, we see that the functiongx takes the minimum value at
x1−α/1α. Hence, we have
argBz0≤ −tan−1
1α1α/21−α1−α/2asinπ/2α−bcosπ/2α ηα
1α1α/21−α1−α/2acosπ/2αbsinπ/2α
−π
2β
η, α, a, b ,
2.13
whereβη, α, a, bis given by2.6. This evidently contradicts the assumption ofTheorem 2.2. Next, we suppose that
pz0
1/α−
ix x >0. 2.14
Applying the same method as the above, we have
argBz0≥tan−1
1α1α/21−α1−α/2asinπ/2α−bcosπ/2α ηα
1α1α/21−α1−α/2acosπ/2αbsinπ/2α
π
2β
η, α, a, b ,
2.15
whereβη, α, a, bis given by2.6, which is a contradiction to the assumption ofTheorem 2.2. Therefore, we complete the proof ofTheorem 2.2.
Corollary 2.3. Letf ∈ Aandη >0,0< α <1. If
arg1−η zf
z
fz η
1zf
z
fz
< π
2β
η, α z∈U, 2.16
whereβη, αis given by2.6witha1andb0, thenf ∈ STSα.
Proof. Taking
pz fz
zfz, Bz
1−η zf
z
fz η
1zf
z
fz
2.17
in Theorem 2.2, we can see that 2.4 is satisfied. Therefore, the result follows from
Theorem 2.2.
By a similar method of the proof inTheorem 2.2, we have the following theorem.
Theorem 2.5. Letpbe nonzero analytic inUwithp0 1and letpsatisfy the differential equation
zpz
pz Bz aibAz, 2.18
wherea∈R,b∈R−∪ {0},Az signImpz, andBzis analytic inUwithB0 a. If
argBz< π
2αδ, a, b z∈U, 2.19
where
αδ:αδ, a, b 2 πtan
−1δ−b
a δ >0, 2.20
then
argpz< π
2δ z∈U. 2.21
Corollary 2.6. Letf ∈ STSαδ, whereαδis given by2.20witha1andb0. Then
argfz
z < π
2δ z∈U. 2.22
Proof. Letting
pz z
fz, Bz
zfz
fz 2.23
inTheorem 2.5, we haveCorollary 2.6immediately.
If we combine Corollaries2.4and2.6, then we obtain the following result obtained by Nunokawa and Thomas4.
Corollary 2.7. Letf ∈ STCβδ, where
βδ 2 πtan
−1
tanπ 2αδ
αδ
1αδ1αδ/21−αδ1−αδ/2cosπ/2αδ
2.24
andαδis given by2.20. Then
argfz
z < π
Corollary 2.8. Letf ∈ A,0< α <1andβ,γbe real numbers withβ /0andβγ >0. If
arg
βzf
z
fz γ
< π
2δ
α, β, γ z∈U, 2.26
where
δα, β, γ 2 πtan
−1
tanπ 2α
α
βγ 1α1α/21−α1−α/2cosπ/2α
, 2.27
then
arg
βzF
z
Fz γ
< π
2α z∈U, 2.28
whereFis the integral operator defined by
Fz
βγ
zγ
z
0
fβttγ−1dt 1/β
z∈U. 2.29
Proof. Let
Bz 1
βγ
βzf
z
fz γ
, 2.30
pz βγ
zγfβz
z
0
fβttγ−1dt. 2.31
ThenBzandpzare analytic inUwithB0 p0 1. By a simple calculation, we have
1
βγzp
z Bzpz 1. 2.32
Using a similar method of the proof inTheorem 2.2, we can obtain that
argpz< π
2α z∈U. 2.33
From2.29and2.31, we easily see that
Since
βzF
z
Fz γ βγ
pz, 2.35
the conclusion ofCorollary 2.8immediately follows.
Remark 2.9. Lettingα → 1 inCorollary 2.8, we have the result obtained by Miller and Mocanu
7.
The proof of the following theorem below is much akin to that ofTheorem 2.2and so we omit for details involved.
Theorem 2.10. Letpbe nonzero analytic inUwithp0 1and letpsatisfy the differential equation
zpz
pz Bzpz aibAz, 2.36
wherea∈R,b∈R−∪ {0},Az signImpzandBzis analytic inUwithB0 a. If
argBz< π
2βα, a, b z∈U, 2.37
where
βα, a, b α 2 πtan
−1α−b
a 0< α≤1, 2.38
then
argpz< π
2α z∈U. 2.39
Corollary 2.11. Letf∈ Awithfz/0inUand0< α≤1. If
argfz zfz < π
2βα z∈U, 2.40
whereβαis given by2.38witha1andb0, then
argfz< π
2α z∈U, 2.41
Proof. Let
pz 1
fz, Bz f
z zfz 2.42
inTheorem 2.10. Then2.36is satisfied and so the result follows.
By applyingTheorem 2.10, we have the following result obtained by Mocanu6.
Corollary 2.12. Letf∈ Awithfz/z /0andα0be the solution of the equation given by
2α 2 πtan
−1α1 0< α <1. 2.43
If
argfz< π
21−α0 z∈U, 2.44
thenf ∈ S∗.
Proof. Let
pz z
fz, Bz f
z. 2.45
Then, byTheorem 2.10, condition2.44implies that
arg z
fz < π
2α0. 2.46
Therefore, we have
argzf
z
fz
≤argfzarg z
fz < π
2, 2.47
which completes the proof ofCorollary 2.12.
Corollary 2.13. Letf∈ Awithfzfz/z /0inUand0< α≤1. If
argzf
z
fz
2zf
z
fz − zfz
fz
< π
2βα z∈U, 2.48
Finally, we have the following result.
Theorem 2.14. Letpbe nonzero analytic inUwithp0 1. If
arg1−λpz λzpz < π
2βλ, α, 2.49
βλ, α α 2 πtan
−1 λα
1−λ 0≤λ <1; 0< α <1, 2.50
then
argpz< π
2α z∈U. 2.51
Proof. If there exists a pointz0 ∈Usatisfying the conditions ofLemma 2.1, then we have
1−λpz0 λz0pz0 ±ixα1−λiλαk. 2.52
Now we suppose that
pz0
1/α
ix x >0. 2.53
Then we have
arg1−λpz0 λz0pz0
π
2αtan
−1 λαk
1−λ
≥ π
2
α 2
πtan
−1 λα
1−λ
π
2βλ, α,
2.54
whereβλ, αis given by2.50. Also, for the case
pz0
1/α−
ix x >0, 2.55
we obtain
arg1−λpz0 λz0pz0 ≤ −
π
2
α 2
πtan
−1 λα
1−λ
−π
2βλ, α,
2.56
Corollary 2.15. Letf∈ Awithfzfz/z /0inUand0< α <1. If
arg
zfz
fz
1zf
z
fz − zfz
fz
< π
2α1 z∈U, 2.57
thenf ∈ STSα.
Acknowledgment
This research was supported by Basic Science Research Program through the National Research Foundation of Korea NRF funded by the Ministry of Education, Science and TechnologyNo. 2009-0066192.
References
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