On Certain Subclasses of Meromorphic Close to Convex Functions

Volume 2008, Article ID 246909,12pages doi:10.1155/2008/246909

Research Article

On Certain Subclasses of Meromorphic

Close-to-Convex Functions

Georgia Irina Oros, Adriana C ˘atas¸, and Gheorghe Oros

Department of Mathematics, University of Oradea, 1, Universit˘at¸ii street, 410087 Oradea, Romania

Correspondence should be addressed to Georgia Irina Oros,georgia oros [email protected]

Received 20 February 2008; Accepted 31 March 2008

Recommended by Narendra Kumar Govil

By using the operatorDn

λfz, z∈U, Definition 2.1, we introduce a class of meromorphic functions

denoted byΣα, λ, nand we obtain certain differential subordinations.

Copyrightq2008 Georgia Irina Oros 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. Introduction and preliminaries

Denote byUthe unit disc of the complex plane:

Uz∈C: |z|<1, U˙ U− {0}. 1.1

LetHUbe the space of holomorphic function inU. Let

An

f ∈ HU, fz zan1zn1· · · , z∈U

1.2

withA1A.

Fora∈Candn∈N, we let

Ha, n f ∈ HU, fz aanznan1zn1· · · , z∈U. 1.3

Let

K

f ∈A, Rezf

_z

fz 1>0, z∈U

1.4

If f and g are analytic functions inU, then we say thatf is subordinate to g, written

f ≺ g, if there is a function w analytic in U, with w0 0, |wz| < 1, for allz ∈ U such thatfz gwz for z ∈ U. If g is univalent, then f ≺ g if and only iff0 g0and

fU⊆gU.

A functionf, analytic inU, is said to be convex if it is univalent andfUis convex. Let ψ : C3_{×U → Cand let hbe univalent in U. Ifp is analytic in Uand satisfies the} second-orderdifferential subordination,

iψpz, zpz, z2_{pz;z≺hz, z∈U,}

thenpis called a solution of the differential subordination.

The univalent function q is called a dominant of the solutions of the differential subordination, or more simply a dominant, ifp≺qfor allpsatisfyingi.

A dominantqthat satisfiesq≺qfor all dominantsqofiis said to be the best dominant of i. Note that the best dominant is unique up to a rotation of U.In order to prove the original results, we use the following lemmas.

Lemma 1.1 see 1, Theorem 3.1.6, page 71, and the references therein . Let h be a convex function withh0 a, and letγ ∈C∗be a complex number withReγ ≥ 0. Ifp∈ Ha, n and

pz 1 γzp

_{z≺hz, z∈U, 1.5}

then

pz≺qz≺hz, z∈U, 1.6

where

qz γ nzγ/n

z

0

httγ/n−1dt, z∈U. 1.7

The functionqis convex and the best dominant.

Lemma 1.2 see 2, Lemma 13.5.1, page 375, and the references therein . Let g be a convex function inU,and let

hz gz nαzgz, z∈U, 1.8

whereα >0,andnis a positive integer. If

pz g0 pnznpn1zn1· · ·, z∈U 1.9

is holomorphic inU,and

pz αzpz≺hz, z∈U, 1.10

then

pz≺gz, 1.11

Lemma 1.3see1, Corollary 2.6.g.2, page 66 . Letf ∈Aand

Fz 2 z

z

0

ftdt, z∈U. 1.12

If

Re

zfz

fz 1 >−

1

2, 1.13

then

F ∈K. 1.14

Lemma 1.4see3, Lemma 1.5 . LetRec >0,and let

w k

2_|c|2_−k2_−c2

4kRec . 1.15

Lethbe an analytic function inUwithh0 1,and suppose that

Re

zhz

hz 1 >−w, z∈U. 1.16

Ifpz 1pkzk· · · (k ≥ 1integer) is analytic inUand

pz 1 czp

_{z≺hz, z∈U, 1.17}

then

pz≺qz, z∈U, 1.18

whereqis the solution of the differential equation:

qz k czq

_{z hz, qz 1, 1.19}

given by

qz c kzc/k

z

0

tc/k−1htdt. 1.20

Moreover,qis the best dominant.

Definition 1.5see4 . Forf ∈A,n∈N∗∪ {0}, the operatorSnf is defined bySn :A→A

S0fz fz, S1fz zfz,

. . .

Sn1fzzSnfz, z∈U.

Remark 1.6. Iff ∈A,

fz z ∞

j2

ajzj, 1.22

then

Snfz z ∞

j2

jnajzj, z∈U. 1.23

Definition 1.7see5 . Forf ∈A,n∈N∗∪ {0},the operatorRn_{fis defined byR}n_:A→A

R0fz fz, R1fz zfz,

. . .

n1Rn1fz zRnfznRnfz, z∈U.

1.24

Remark 1.8. Iff ∈A,

fz z ∞

j2

ajzj, 1.25

then

Rnfz z ∞

j2

Cn_nj−1ajzj, z∈U. 1.26

2. Main results

Definition 2.1. Letn∈N∗∪ {0}andλ ≥ 0. LetD_λnfdenote the operator defined byD_λn:A→A

D_λnfz 1−λSnfz λRnfz, z∈U, 2.1

where the operatorsSn_fandRn_{fare given by Definitions1.5and1.7, respectively.}

Remark 2.2. We observe thatD_λnis a linear operator and for

fz z ∞

j2

ajzj, 2.2

we have

Dn_λfz z ∞

j2

1−λjnλCn_nj−1ajzj. 2.3

Remark 2.3. Forn0,

D0_λfz 1−λS0fz λR0fz fz S0fz R0fz, 2.4

and forn1,

D1_λfz 1−λS1fz λR1fz zfz S1fz R1fz. 2.5

Remark 2.4. Iff ∈Σ,

fz 1

za0a1za2z

2_{· · ·, 2.6}

and we let

gz z2fz za0z2a1z3· · ·, z∈U. 2.7

Definition 2.5. If 0 ≤ α < 1, λ ≥ 0,andn ∈ N, letΣα, λ, n1denote the class of functions

f ∈Σwhich satisfy the inequality,

Re

Dn_λ1gzλzn

Rn_gz

n1

> α, 2.8

whereDn1

λ gis given byDefinition 2.1,gis given by2.7, andRngis given byDefinition 1.7.

Theorem 2.6. If0 ≤ α <1,λ ≥ 0, andn∈N, then

Σα, λ, n1⊂Σδ, λ, n1, 2.9

where

δδα 2α−121−αln 2. 2.10

Proof. Letf ∈Σα, λ, n1,

gz z2fz za0z2a1z3· · · , g∈A. 2.11

Sincef ∈Σα, λ, n1by usingDefinition 2.5, we deduce

Re

Dn_λ1gzλnz

Rn_gz

n1

> α, z∈U, 2.12

which is equivalent to

Dn_λ1gzλnz

Rngz n1 ≺

1 2α−1z

By using the properties of the operatorsDn

λg,Sng,andRng,we have

1−λSn1gz λRn1gzλnz

Rn_gz

n1

1−λzSngzλ

zRngznRngz n1

λnzRngz n1

1−λSngzzSngzλ

Rn_gzzRn_gznRn_gz

n1

λnzRn_gz

n1 1−λSngzλRngzz1−λSngzλRngz, z∈U.

2.14

Using2.14in2.13, we obtain

1−λSngzλRngzz1−λSngzλRngz≺ 1 2α−1z

1z , z∈U.

2.15

Let

pz Dn_λgz

1−λSngzλRngz

1−λ

z ∞

j2

jnajzj λ z ∞ j2

Cn_nj−1ajzj

1−λ

1

∞

j2

jn1ajzj−1 λ

1

∞

j2

jCn_nj−1ajzj−1

1

∞

j2

1−λjn1λjC_nn_j−1ajzj−1

1b1zb2z2· · · , z∈U.

2.16

We have thatp∈ H1,1 . From2.16, we have

pz 1−λSngzλRngz. 2.17

Using2.16and2.17in2.15, we obtain

pz zpz≺ 1 2α−1z

1z hz, z∈U. 2.18

By usingLemma 1.1, we have

where

qz 1 z

z

0

htdt 1 z

z

0

1 2α−1t

1t dt2α−121−α

ln1z

z , z∈U. 2.20

The functionqis convex and best dominant.

Sinceqis convex andqUis symmetric with respect to the real axis, we deduce

Repz>Req1 δδα 2α−121−αln 2, 2.21

from which we deduce thatΣα, λ, n1⊂Σδ, λ, n1.

Example 2.7. Ifn0,α1/2,λ ≥ 0, thenδ1/2 ln 2,and we deduce forf ∈Σthat

Re4zfz 5z2fz z3fz> 1

2, z∈U 2.22

implies

Re2zfz z2fz>ln 2, z∈U. 2.23

Theorem 2.8. Letrbe a convex function,r0 1, and lethbe a function such that

hz rz zrz, z∈U. 2.24

Iff ∈Σ,gis given by2.7, and the following differential subordination holds

Dn_λ1gzλnz

Rn_gz

n1 ≺hz rz zr

_{z, z∈U, 2.25}

then

D_λngz ≺rz, z∈U, 2.26

and this result is sharp.

Proof. By using the properties of the operatorDn_λg,we have

D_λn1gz 1−λSn1gz λRn1gz. 2.27

By using the properties of operators Sn_{gz, R}n_{gz, and by differentiating2.27, we}

obtain

D_λn1gz1−λSn1gz λRn1gz

1−λSngzzSngzλn1

Rn_gzzRn_gz

n1 .

Using2.28in2.25and relations2.16and2.17, after a simple calculation, Subordi-nation2.25becomes

pz zpz≺rz zrz, z∈U. 2.29 By usingLemma 1.2, we have

pz≺rz, z∈U, 2.30 that is,

Dn_λgz≺rz, z∈U. 2.31

Example 2.9. Ifn0,λ ≥ 0,rz 1z/1−z, fromTheorem 2.8, we deduce that iff ∈Σ

and

4zfz 5z2fz z3fz≺ 12z−z

2

1−z2 , z∈U, 2.32

then

2zfz z2fz≺ 1z

1−z, z∈U. 2.33 Theorem 2.10. Letrbe a convex function,r0 1, and

hz rz zrz, z∈U. 2.34

Iff ∈Σ,gis given by2.7, and the following differential subordination holds

D_λngz≺hz rz zrz, z∈U, 2.35

then

D_λngz

z ≺rz, z∈U, 2.36

and this result is sharp.

Proof. We let

pz D

n λgz

z , z∈U. 2.37

By differentiating2.37, we obtain

Dn_λgzpz zpz, z∈U. 2.38 Using2.38, Subordination2.35becomes

pz zpz≺rz zrz hz, z∈U. 2.39 By usingLemma 1.2, we have

pz≺rz, 2.40 that is,

D_λngz

z ≺rz, z∈U, 2.41

Example 2.11. If we letrz 1/1−z,n1,λ ≥ 0, then

hz 1

1−z2, 2.42

and fromTheorem 2.10, we deduce that iff ∈Σ,and

4zfz 5z2fz z3fz≺ 1

1−z2, z∈U, 2.43

then

2fz zfz≺ 1

1−z, z∈U, 2.44

and this result is sharp.

Theorem 2.12. Leth∈ HU, withh0 1,h0/0which verifies the inequality:

Re

1zh

_z

hz

>−1

2, z∈U. 2.45

Iff ∈Σ,gis given by2.7and the following differential subordination holds

Dn_λgz≺hz, z∈U, 2.46

then

Dn_λgz

z ≺qz, z∈U, 2.47

where

qz 1 z

_z

0

htdt, z∈U. 2.48

Functionqis convex and the best dominant.

Proof. In order to proveTheorem 2.12, we will use Lemmas1.3and1.4. We deduce the value ofwfromLemma 1.4by using the conditions ofTheorem 2.12. From2.37,Definition 2.1and

Remark 2.2, we have

pz D

n λgz

z

z

_∞

j2

1−λjn_λCn nj−1

ajzj

z

1

∞

j2

1−λjnλCn_nj−1ajzj−1

1b1zb2z· · ·, z∈U.

UsingLemma 1.4, we deduce from2.49thatk 1. Using2.38in Subordination2.46, we have

pz zpz≺hz, z∈U. 2.50

From Subordination2.50, by usingLemma 1.4, we deduce thatc1. Then,

w k

2_c2_{− |k}2_−c2_|

4kRec

11− |1−1|

4

1

2. 2.51 ApplyingLemma 1.4, from Subordination2.50, we obtain

pz≺qz 1 z

_z

0

htdt, z∈U, 2.52

that is,

Dn λgz

z ≺qz

1

z

_z

0

htdt, z∈U, 2.53

whereqis the best dominant.

Since the functionhverifies the relation2.45, fromLemma 1.3, we deduce thatqis a convex function.

Example 2.13. Ifn0,λ ≥ 0,hz e3/2z−1, fromTheorem 2.12, we deduce forf ∈Σthat if

4zfz 5z2fz z3fz≺e3/2z−1, z∈U, 2.54

then

2fz zfz≺ 2

3ze

3/2z₋ 2

3z−1, z∈U. 2.55 Theorem 2.14. Leth∈ HU, withh0 1,h0/0which verifies the inequality:

Re

1zh

_z

hz >−

1

2, z∈U. 2.56

Iff ∈Σ,gis given by2.7, and the following differential subordination holds

Dn_λ1gzλnz

Rngz

n1 ≺hz, z∈U, 2.57

then

Dn_λgz≺qz, z∈U, 2.58

where

qz 1 z

z

0

htdt 2.59

Proof. In order to prove Theorem 2.14, we will use Lemmas 1.3 and 1.4. The value of w is obtained using the conditions ofTheorem 2.14.

Using2.16and2.17, Subordination2.49becomes

pz zpz≺hz, z∈U. 2.60

From Subordination 2.60, by using Lemma 1.4, we deduce thatc 1; and from the relation2.16,Definition 2.1, andRemark 2.2, we obtain

pz Dn_λgz

z ∞

j2

1−λjnλC_nn_j−1ajzj

1

∞

j2

1−λjnλCn_nj−1·j·aj·zj−1

1c1zc2z2· · ·, z∈U.

2.61

From2.61, by usingLemma 1.4, we deduce thatk1, then

w k

2_|c|2_{− |k−c|}2

4kRec

11− |1−1|2

4

1

2. 2.62

ApplyingLemma 1.4, from Subordination2.60, we obtain

pz≺qz 1 z

z

0

htdt, z∈U, 2.63

that is,

Dn_λgz≺qz 1 z

_z

0

htdt, z∈U, 2.64

whereqis the best dominant.

Since the functionhverifies the inequality2.45, fromLemma 1.3, we deduce thatqis a convex function.

Example 2.15. Ifn 0,λ ≥ 0,f ∈Σ,hz 2zz2_/1z2_{, fromTheorem 2.14, we deduce}

that if

4zfz 5z2fz z3fz≺ 2zz

2

21z2, z∈U, 2.65

then

2fz zfz≺ 1

2z 1 2

1

References

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