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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;qzpq

k1

λkp;qakpzkpq, 1.2

where

λp;q: p!

pq!

pq;p∈N;q∈N∪ {0}

. 1.3

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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 asfzgz 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 classesSpφandCpφ consist, respectively, ofp-valent functionsf starlikewith respect toφandp-valent functions fconvexwith respect toφinUgiven by

fSpφ⇐⇒ 1 p

zfz

fzφz, fCpφ⇐⇒ 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 inSpφandCpφ. Ma and Minda14have introduced and investigated the classesSφ:S1φand:C1φ. For−1≤B < A≤1, the class SA, B S1Az/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;qzpqQz, zfq1z

fqzpq1≺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.

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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

PQ, 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

PQ, 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;qzpqQz 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

fqzzQz

Qz pq, 2.2

then

fqz

λp;qzpqQz, 2.3

andQis the best dominant.

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Proof. Define the analytic functionPzby

Pz: fqz

λp;qzpq. 2.4

Then a computation shows that

zfq1z

fqz zPz

Pz pq. 2.5

The subordination2.2yields zPz

Pz pqzQz

Qz pq, 2.6

or equivalently

zPz

PzzQz

Qz . 2.7

Define the function ϕ by ϕw : 1/w. Then 2.7 can be written as zPz·ϕPzzQz·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 < pqfqz< λp;q|z|pq−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

pq. 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

fqzzAB

1Az1Bzpq, 2.10

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then

fqz

λp;qzpq ≺ 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 ABz

1Az1Bz, hz: zFz

Fz 1−ABz2 1Az1Bz.

2.13

Withzre, note that R

h re

R 1−ABr2e2 1Are1Bre

1−ABr21ABr2 ABrcosθ

|1Are1Bre|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, chooseandB0 inCorollary 2.3. Sincewβz/1βz is equivalent to|w| ≤β|1−w|,it follows that iff∈ Apsatisfies

zfq1z

fqzpq β2 1−β2

< β

1−β2, 2.15

then

fqz λp;qzpq −1

< β. 2.16

2WithA1 andB0,it follows fromCorollary 2.3that wheneverf ∈ Apsatisfies

R

zfq1z fqzpq

< 1

2, 2.17

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then

fqz λp;qzpq −1

<1. 2.18

Takingq0 andQz hφ,p/zp,Theorem 2.1yields the following corollary.

Corollary 2.5see13. IffSpφ,then

fz zphφ,p

zp . 2.19

Similarly, choosing q 1 and Qz kφ,p /pzp−1,Theorem 2.1yields the following corollary.

Corollary 2.6see13. IffCpφ,then

fz zp−1kφ,p

zp−1. 2.20

Theorem 2.7. LetQzbe convex univalent inUandQ0 1. Iff∈ Apsatisfies fqz

λp;qzpq·

zfq1z fqzpq

zQz, 2.21

then

fqz

λp;qzpqQz, 2.22

andQis the best dominant.

Proof. Define the analytic functionPzbyPz :fqz/λp;qzpq.Then it follows from 2.5that

fqz λp;qzpq·

zfq1z fqzpq

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.

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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

fqzpq

≤ |z|pqfqzλp;qzpq≤ |z|pq. 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;qzpqzQz pqQz, 2.27

then

fqz

λp;qzpqQz, 2.28

andQis the best dominant.

Proof. Define the functionPzbyPz fqz/λp;qzpq.It follows from2.5that zPz pqPzzQz pqQz, 2.29

that is,

zPz θ Pz

zQz θ Qz

, 2.30

where θw pqw. The conditions in Lemma 1.2are clearly satisfied. Thus fqz/

λp;qzpqQz,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−1zQz pQz, 2.31

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then

fz

zpQz. 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

fzzQz Qz, 2.33

then

fz

zQz. 2.34

Theorem 2.12. LetQzbe univalent and nonzero inU,Q0 1, andzQz/Q2zbe starlike.

Iff∈ Apsatisfies

λp;qzpq fqz ·

zfq1z fqzpq

zQz

Q2z, 2.35

then

fqz

λp;qzpqQz, 2.36

andQis the best dominant.

Proof. Define the functionPzbyPz fqz/λp;qzpq.It follows from2.5that λp;qzpq

fqz ·

zfq1z

fqzpq

1

Pz·zPz

Pz zPz

P2z. 2.37

By assumption,

zPz

P2zzQz

Q2z. 2.38

Withϕw:1/w2,2.38can be written aszPz·ϕPzzQz·ϕQz.The function ϕwis analytic inC− {0}.SincezQzϕQzis starlike, it follows fromLemma 1.1that PzQz,andQzis the best dominant.

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The next four theorems give sufficient conditions for the following differential subor- dination

zfq1z

fqzpq1≺Qz 2.39

to hold.

Theorem 2.13. Let Qz be univalent and nonzero in U, Q0 1, Qz/qp 1, and zQz/QzQz pq−1be starlike inU. Iff∈ Apsatisfies

1 zfq2z/fq1zpq1

zfq1z/fqzpq1 ≺1 zQz

QzQz pq−1, 2.40

then

zfq1z

fqzpq1≺Qz, 2.41

andQis the best dominant.

Proof. Let the functionPzbe defined by

Pz zfq1z

fqzpq1. 2.42

Upon differentiating logarithmically both sides of2.42, it follows that zPz

Pz pq−1 1zfq2z

fq1zzfq1z

fqz . 2.43

Thus

1zfq2z

fq1zpq1 zPz

Pz pq−1 Pz. 2.44

The equations2.42and2.44yield

1 zfq2z/fq1zpq1

zfq1z/fqzpq−1 zPz

PzPz pq−11. 2.45

Iff∈ Apsatisfies the subordination2.40,2.45gives zPz

PzPz pq−1 ≺ zQz

QzQz pq−1, 2.46

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that is,

zPz·ϕ Pz

zQz·ϕ Qz

2.47 withϕw:1/wwpq−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 ABz

1Az2, 2.48

thenfSA, 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/qp1, and let zQz/Qz pq−1be starlike inU. Iff ∈ Apsatisfies

1zfq2z

fq1zzfq1z

fqzzQz

Qz pq−1, 2.49

then

zfq1z

fqzpq1≺Qz, 2.50

andQis the best dominant.

Proof. Let the functionPzbe defined by 2.42. It follows from2.43and the hypothesis that

zPz

Pz pq−1 ≺ zQz

Qz pq−1. 2.51

Define the functionϕbyϕw:1/wpq−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.

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Theorem 2.16. LetQzbe a convex function inU,andQ0 1. Iff ∈ Apsatisfies zfq1z

fqz 2zfq2z

fq1zzfq1z fqz

zQz Qz pq−1, 2.53

then

zfq1z

fqzpq1≺Qz, 2.54

andQis the best dominant.

Proof. Let the functionPzbe defined by2.42. Using2.43, it follows that zfq1z

fqz

1zfq2z

fq1zzfq1z fqz

zPz, 2.55

and, therefore,

zfq1z fqz

2zfq2z

fq1zzfq1z fqz

zPz Pz pq−1. 2.56

By assumption,

zPz Pz pq−1≺zQz Qz pq−1, 2.57

or

zPz θ Pz

zQz θ Qz

, 2.58

where the functionθw wpq1.The proof is completed by applyingLemma 1.2.

Theorem 2.17. LetQzbe a convex function inU, withQ0 1. Iff ∈ Apsatisfies zfq1z

fqz 1zfq2z

fq1zzfq1z fqz

zQz, 2.59

then

zfq1z

fqzpq1≺Qz, 2.60

andQis the best dominant.

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Proof. Let the functionPzbe defined by2.42. It follows from2.43thatzPz·ϕPzzQz·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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