Volume 2008, Article ID 830138,12pages doi:10.1155/2008/830138
Research Article
Subordination for Higher-Order Derivatives of Multivalent Functions
Rosihan M. Ali,1Abeer O. Badghaish,1, 2 and V. Ravichandran3
1School of Mathematical Sciences, Universiti Sains Malaysia (USM), 11800 Penang, Malaysia
2Mathematics Department, King Abdul Aziz University, P.O. Box 581, Jeddah 21421, Saudi Arabia
3Department of Mathematics, University of Delhi, Delhi 110 007, India
Correspondence should be addressed to Rosihan M. Ali,[email protected] Received 18 July 2008; Accepted 24 November 2008
Recommended by Vijay Gupta
Differential subordination methods are used to obtain several interesting subordination results and best dominants for higher-order derivatives ofp-valent functions. These results are next applied to yield various known results as special cases.
Copyrightq2008 Rosihan M. Ali 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.
1. Motivation and preliminaries
For a fixedp∈N:{1,2, . . .}, letApdenote the class of all analytic functions of the form
fz zp∞
k1
akpzkp, 1.1
which are p-valent in the open unit disc U {z ∈ C : |z| < 1} and let A : A1. Upon differentiating both sides of1.1q-times with respect toz, the following differential operator is obtained:
fqz λp;qzp−q∞
k1
λkp;qakpzkp−q, 1.2
where
λp;q: p!
p−q!
p≥q;p∈N;q∈N∪ {0}
. 1.3
Several researchers have investigated higher-order derivatives of multivalent functions, see, for example,1–10. Recently, by the use of the well-known Jack’s lemma11,12, Irmak and Cho5obtained interesting results for certain classes of functions defined by higher-order derivatives.
Letfandgbe analytic inU. Thenfissubordinatetog, written asfz≺gz z∈U if there is an analytic functionwzwithw0 0 and|wz|<1,such thatfz gwz.
In particular, if g is univalent in U, then f subordinate to g is equivalent to f0 g0 and fU ⊆ gU. A p-valent function f ∈ Ap is starlike if it satisfies the condition 1/pRzfz/fz > 0z ∈ U. More generally, let φz be an analytic function with positive real part inU, φ0 1, φ0>0,andφzmaps the unit discUonto a region starlike with respect to 1 and symmetric with respect to the real axis. The classesS∗pφandCpφ consist, respectively, ofp-valent functionsf starlikewith respect toφandp-valent functions fconvexwith respect toφinUgiven by
f∈S∗pφ⇐⇒ 1 p
zfz
fz ≺φz, f∈Cpφ⇐⇒ 1 p
1zfz fz
≺φz. 1.4
These classes were introduced and investigated in 13, and the functions hφ,p and kφ,p, defined, respectively, by
1 p
zhφ,p
hφ,p φz
z∈U, hφ,p∈ Ap , 1
p
1zkφ,p kφ,p
φz
z∈U, kφ,p∈ Ap ,
1.5
are important examples of functions inS∗pφandCp∗φ. Ma and Minda14have introduced and investigated the classesS∗φ:S∗1φandCφ:C1φ. For−1≤B < A≤1, the class S∗A, B S∗1Az/1Bzis the class of Janowski starlike functionscf.15,16.
In this paper, corresponding to an appropriate subordinate functionQzdefined on the unit disk U, sufficient conditions are obtained for a p-valent function f to satisfy the subordination
fqz
λp;qzp−q ≺Qz, zfq1z
fqz −pq1≺Qz. 1.6
In the particular case when q 1 and p 1, and Qz is a function with positive real part, the first subordination gives a sufficient condition for univalence of analytic functions, while the second subordination implication gives conditions for convexity of functions. If q 0 and p 1, the second subordination gives conditions for starlikeness of functions.
Thus results obtained in this paper give important information on the geometric prop- erties of functions satisfying differential subordination conditions involving higher-order derivatives.
The following lemmas are needed to prove our main results.
Lemma 1.1see12, page 135, Corollary 3.4h.1. LetQbe univalent inU, andϕbe analytic in a domainDcontainingQU. IfzQz·ϕQzis starlike, andPis analytic inUwithP0 Q0 andPU⊂D, then
zPz·ϕ Pz
≺zQz·ϕ Qz
⇒P ≺Q, 1.7
andQis the best dominant.
Lemma 1.2see12, page 135, Corollary 3.4h.2. LetQbe convex univalent inU, and letθbe analytic in a domainDcontainingQU. Assume that
R θQz 1zQz Qz
>0. 1.8
IfPis analytic inUwithP0 Q0andPU⊂D, then zPz θ
Pz
≺zQz θ Qz
⇒P ≺Q, 1.9
andQis the best dominant.
2. Main results
The first four theorems below give sufficient conditions for a differential subordination of the form
fqz
λp;qzp−q ≺Qz 2.1
to hold.
Theorem 2.1. LetQzbe univalent and nonzero inU,Q0 1, and letzQz/Qzbe starlike inU. If a functionf ∈ Apsatisfies the subordination
zfq1z
fqz ≺ zQz
Qz p−q, 2.2
then
fqz
λp;qzp−q ≺Qz, 2.3
andQis the best dominant.
Proof. Define the analytic functionPzby
Pz: fqz
λp;qzp−q. 2.4
Then a computation shows that
zfq1z
fqz zPz
Pz p−q. 2.5
The subordination2.2yields zPz
Pz p−q≺ zQz
Qz p−q, 2.6
or equivalently
zPz
Pz ≺ zQz
Qz . 2.7
Define the function ϕ by ϕw : 1/w. Then 2.7 can be written as zPz·ϕPz ≺ zQz·ϕQz. Since Qz/0, ϕw is analytic in a domain containing QU. Also zQz·ϕQz zQz/Qzis starlike. The result now follows fromLemma 1.1.
Remark 2.2. Forf ∈ Ap, Irmak and Cho5, page 2, Theorem 2.1showed that
Rzfq1z
fqz < p−q⇒fqz< λp;q|z|p−q−1. 2.8 However, it should be noted that the hypothesis of this implication cannot be satisfied by any function inApas the quantity
zfq1z fqz
z0
p−q. 2.9
Theorem 2.1is the correct formulation of their result in a more general setting.
Corollary 2.3. Let−1≤B < A≤1. Iff∈ Apsatisfies zfq1z
fqz ≺ zA−B
1Az1Bzp−q, 2.10
then
fqz
λp;qzp−q ≺ 1Az
1Bz. 2.11
Proof. For−1≤B < A≤1, define the functionQby Qz 1Az
1Bz. 2.12
Then a computation shows that
Fz: zQz
Qz A−Bz
1Az1Bz, hz: zFz
Fz 1−ABz2 1Az1Bz.
2.13
Withzreiθ, note that R
h reiθ
R 1−ABr2e2iθ 1Areiθ1Breiθ
1−ABr21ABr2 ABrcosθ
|1Areiθ1Breiθ|2 .
2.14
Since 1ABr2 ABrcosθ≥1−Ar1−Br>0 forAB≥0, and similarly, 1ABr2 ABrcosθ ≥ 1Ar1Br> 0 forAB ≤0, it follows thatRhz > 0,and hence zQz/Qzis starlike. The desired result now follows fromTheorem 2.1.
Example 2.4. 1For 0< β <1, chooseAβandB0 inCorollary 2.3. Sincew≺βz/1βz is equivalent to|w| ≤β|1−w|,it follows that iff∈ Apsatisfies
zfq1z
fqz −pq β2 1−β2
< β
1−β2, 2.15
then
fqz λp;qzp−q −1
< β. 2.16
2WithA1 andB0,it follows fromCorollary 2.3that wheneverf ∈ Apsatisfies
R
zfq1z fqz −pq
< 1
2, 2.17
then
fqz λp;qzp−q −1
<1. 2.18
Takingq0 andQz hφ,p/zp,Theorem 2.1yields the following corollary.
Corollary 2.5see13. Iff ∈S∗pφ,then
fz zp ≺ hφ,p
zp . 2.19
Similarly, choosing q 1 and Qz kφ,p /pzp−1,Theorem 2.1yields the following corollary.
Corollary 2.6see13. Iff ∈Cp∗φ,then
fz zp−1 ≺ kφ,p
zp−1. 2.20
Theorem 2.7. LetQzbe convex univalent inUandQ0 1. Iff∈ Apsatisfies fqz
λp;qzp−q·
zfq1z fqz −pq
≺zQz, 2.21
then
fqz
λp;qzp−q ≺Qz, 2.22
andQis the best dominant.
Proof. Define the analytic functionPzbyPz :fqz/λp;qzp−q.Then it follows from 2.5that
fqz λp;qzp−q·
zfq1z fqz −pq
zPz. 2.23
By assumption, it follows that
zPz·ϕ Pz
≺zQz·ϕ Qz
, 2.24
whereϕw 1. SinceQzis convex, andzQz·ϕQz zQzis starlike,Lemma 1.1 gives the desired result.
Example 2.8. When
Qz:1 z
λp;q, 2.25
Theorem 2.7is reduced to the following result in5, page 4, Theorem 2.4. Forf∈ Ap, fqz·
zfq1z
fqz −pq
≤ |z|p−q⇒fqz−λp;qzp−q≤ |z|p−q. 2.26 In the special caseq 1, this result gives a sufficient condition for the multivalent function fzto be close-to-convex.
Theorem 2.9. LetQzbe convex univalent inUandQ0 1. Iff∈ Apsatisfies zfq1z
λp;qzp−q ≺zQz p−qQz, 2.27
then
fqz
λp;qzp−q ≺Qz, 2.28
andQis the best dominant.
Proof. Define the functionPzbyPz fqz/λp;qzp−q.It follows from2.5that zPz p−qPz≺zQz p−qQz, 2.29
that is,
zPz θ Pz
≺zQz θ Qz
, 2.30
where θw p−qw. The conditions in Lemma 1.2are clearly satisfied. Thus fqz/
λp;qzp−q≺Qz,andQis the best dominant.
Takingq0,Theorem 2.9yields the following corollary.
Corollary 2.10see17, Corollary 2.11. LetQzbe convex univalent inU,andQ0 1. If f∈ Apsatisfies
fz
zp−1 ≺zQz pQz, 2.31
then
fz
zp ≺Qz. 2.32
Withp1,Corollary 2.10yields the following corollary.
Corollary 2.11 see17, Corollary 2.9. LetQzbe convex univalent inU,and Q0 1. If f∈ Asatisfies
fz≺zQz Qz, 2.33
then
fz
z ≺Qz. 2.34
Theorem 2.12. LetQzbe univalent and nonzero inU,Q0 1, andzQz/Q2zbe starlike.
Iff∈ Apsatisfies
λp;qzp−q fqz ·
zfq1z fqz −pq
≺ zQz
Q2z, 2.35
then
fqz
λp;qzp−q ≺Qz, 2.36
andQis the best dominant.
Proof. Define the functionPzbyPz fqz/λp;qzp−q.It follows from2.5that λp;qzp−q
fqz ·
zfq1z
fqz −p−q
1
Pz·zPz
Pz zPz
P2z. 2.37
By assumption,
zPz
P2z ≺ zQz
Q2z. 2.38
Withϕw:1/w2,2.38can be written aszPz·ϕPz≺zQz·ϕQz.The function ϕwis analytic inC− {0}.SincezQzϕQzis starlike, it follows fromLemma 1.1that Pz≺Qz,andQzis the best dominant.
The next four theorems give sufficient conditions for the following differential subor- dination
zfq1z
fqz −pq1≺Qz 2.39
to hold.
Theorem 2.13. Let Qz be univalent and nonzero in U, Q0 1, Qz/q− p 1, and zQz/QzQz p−q−1be starlike inU. Iff∈ Apsatisfies
1 zfq2z/fq1z−pq1
zfq1z/fqz−pq1 ≺1 zQz
QzQz p−q−1, 2.40
then
zfq1z
fqz −pq1≺Qz, 2.41
andQis the best dominant.
Proof. Let the functionPzbe defined by
Pz zfq1z
fqz −pq1. 2.42
Upon differentiating logarithmically both sides of2.42, it follows that zPz
Pz p−q−1 1zfq2z
fq1z −zfq1z
fqz . 2.43
Thus
1zfq2z
fq1z −pq1 zPz
Pz p−q−1 Pz. 2.44
The equations2.42and2.44yield
1 zfq2z/fq1z−pq1
zfq1z/fqz−pq−1 zPz
PzPz p−q−11. 2.45
Iff∈ Apsatisfies the subordination2.40,2.45gives zPz
PzPz p−q−1 ≺ zQz
QzQz p−q−1, 2.46
that is,
zPz·ϕ Pz
≺zQz·ϕ Qz
2.47 withϕw:1/wwp−q−1.The desired result is now established by an application of Lemma 1.1.
Theorem 2.13 contains a result in 18, page 122, Corollary 4 as a special case. In particular, we note that Theorem 2.13with p 1, q 0, and Qz 1Az/1 Bz for−1≤B < A≤1 yields the following corollary.
Corollary 2.14see18, page 123, Corollary 6. Let−1≤B < A≤1. Iff∈ Asatisfies 1 zfz/fz
zfz/fz ≺1 A−Bz
1Az2, 2.48
thenf ∈S∗A, B.
ForA0, BbandA1, B−1,Corollary 2.14gives the results of Obradoviˇc and Tuneski19.
Theorem 2.15. Let Qz be univalent and nonzero in U, Q0 1, Qz/q−p1, and let zQz/Qz p−q−1be starlike inU. Iff ∈ Apsatisfies
1zfq2z
fq1z −zfq1z
fqz ≺ zQz
Qz p−q−1, 2.49
then
zfq1z
fqz −pq1≺Qz, 2.50
andQis the best dominant.
Proof. Let the functionPzbe defined by 2.42. It follows from2.43and the hypothesis that
zPz
Pz p−q−1 ≺ zQz
Qz p−q−1. 2.51
Define the functionϕbyϕw:1/wp−q−1.Then2.51can be written as zPz·ϕ
Pz
≺zQz·ϕ Qz
. 2.52
Sinceϕwis analytic in a domain containingQU, andzQz·ϕQzis starlike, the result follows fromLemma 1.1.
Theorem 2.16. LetQzbe a convex function inU,andQ0 1. Iff ∈ Apsatisfies zfq1z
fqz 2zfq2z
fq1z −zfq1z fqz
≺zQz Qz p−q−1, 2.53
then
zfq1z
fqz −pq1≺Qz, 2.54
andQis the best dominant.
Proof. Let the functionPzbe defined by2.42. Using2.43, it follows that zfq1z
fqz
1zfq2z
fq1z −zfq1z fqz
zPz, 2.55
and, therefore,
zfq1z fqz
2zfq2z
fq1z − zfq1z fqz
zPz Pz p−q−1. 2.56
By assumption,
zPz Pz p−q−1≺zQz Qz p−q−1, 2.57
or
zPz θ Pz
≺zQz θ Qz
, 2.58
where the functionθw wp−q1.The proof is completed by applyingLemma 1.2.
Theorem 2.17. LetQzbe a convex function inU, withQ0 1. Iff ∈ Apsatisfies zfq1z
fqz 1zfq2z
fq1z −zfq1z fqz
≺zQz, 2.59
then
zfq1z
fqz −pq1≺Qz, 2.60
andQis the best dominant.
Proof. Let the functionPzbe defined by2.42. It follows from2.43thatzPz·ϕPz≺ zQz·ϕQz,whereϕw 1.The result follows easily fromLemma 1.1.
Acknowledgment
This work was supported in part by the FRGS and Science Fund research grants, and was completed while the third author was visiting USM.
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