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Volume 7, Issue 4, Article 152, 2006

DIFFERENTIAL SUBORDINATIONS AND SUPERORDINATIONS FOR ANALYTIC FUNCTIONS DEFINED BY THE DZIOK-SRIVASTAVA LINEAR

OPERATOR

G. MURUGUSUNDARAMOORTHY AND N. MAGESH SCHOOL OFSCIENCE ANDHUMANITIES

VELLOREINSTITUTE OFTECHNOLOGY

DEEMEDUNIVERSITY, VELLORE- 632014, INDIA. gmsmoorthy@yahoo.com

DEPARTMENT OFMATHEMATICS

ADHIYAMAANCOLLEGE OFENGINEERING

HOSUR- 635109, INDIA. nmagi_2000@yahoo.co.in

Received 27 March, 2006; accepted 19 September, 2006 Communicated by N.K. Govil

ABSTRACT. In the present investigation, we obtain some subordination and superordination results involving Dziok-Srivastava linear operatorHml 1]for certain normalized analytic func- tions in the open unit disk. Our results extend corresponding previously known results.

Key words and phrases: Univalent functions, Starlike functions, Convex functions, Differential subordination, Convolution, Dziok-Srivastava linear operator.

2000 Mathematics Subject Classification. Primary 30C45; Secondary 30C80.

1. INTRODUCTION

LetHbe the class of functions analytic in ∆ := {z : |z| < 1} andH(a, n)be the subclass of H consisting of functions of the form f(z) = a+anzn +an+1zn+1 +· · ·. LetA be the subclass ofHconsisting of functions of the formf(z) =z+a2z2+· · ·. Letp, h∈ Hand let φ(r, s, t;z) : C3 ×∆ → C. Ifpandφ(p(z), zp0(z), z2p00(z);z)are univalent and if psatisfies the second order superordination

(1.1) h(z)≺φ(p(z), zp0(z), z2p00(z);z),

thenpis a solution of the differential superordination (1.1). (Iff is subordinate toF, thenF is superordinate tof.) An analytic functionqis called a subordinant ifq ≺pfor allpsatisfying (1.1). A univalent subordinantqethat satisfiesq ≺ eqfor all subordinantsqof (1.1) is said to be

ISSN (electronic): 1443-5756

c 2006 Victoria University. All rights reserved.

092-06

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the best subordinant. Recently Miller and Mocanu [14] obtained conditions on h, qand φfor which the following implication holds:

h(z)≺φ(p(z), zp0(z), z2p00(z);z)⇒q(z)≺p(z).

Using the results of Miller and Mocanu [14], Bulboac˘a [5] considered certain classes of first order differential superordinations as well as superordination-preserving integral operators [4].

Ali et al. [1] have used the results of Bulboac˘a [5] and obtained sufficient conditions for certain normalized analytic functionsf(z)to satisfy

q1(z)≺ zf0(z)

f(z) ≺q2(z),

whereq1 andq2 are given univalent functions in∆withq1(0) = 1andq2(0) = 1.Shanmugam et al. [19] obtained sufficient conditions for a normalized analytic functionf(z)to satisfy

q1(z)≺ f(z)

zf0(z) ≺q2(z) and q1(z)≺ z2f0(z)

{f(z)}2 ≺q2(z) whereq1 andq2are given univalent functions in∆withq1(0) = 1andq2(0) = 1.

In [2], for functionsf ∈ Asuch thatδ >0,

<

(zf0(z) f(z)

f(z) z

δ)

>0, z ∈∆,

a class of Bazilevic type functions was considered and certain properties were studied. In this paper motivated by Liu [11], we define a class

B(λ, δ, A, B) :=

(

f ∈ A: (1−λ)

f(z) z

δ

+λzf0(z) f(z)

f(z) z

δ

≺ 1 +Az 1 +Bz

) , where δ > 0, λ ≥ 0, −1 ≤ B < A ≤ 1and studied certain interesting properties based on subordination. Further we obtained a sandwich result for functions in the classB(λ, δ, A, B).

2. PRELIMINARIES

For our present investigation, we shall need the following definition and results.

Definition 2.1 ([14, Definition 2, p. 817]). Denote byQ, the set of all functionsf(z)that are analytic and injective on∆−E(f), where

E(f) =

ζ ∈∂∆ : lim

z→ζf(z) =∞

and are such thatf0(ζ)6= 0forζ ∈∂∆−E(f).

Lemma 2.1 ([13, Theorem 3.4h, p. 132]). Let q(z) be univalent in the unit diskand θ and φ be analytic in a domain D containing q(∆) with φ(w) 6= 0 when w ∈ q(∆). Set Q(z) =zq0(z)φ(q(z)),h(z) = θ(q(z)) +Q(z). Suppose that

(1) Q(z)is starlike univalent in∆, and (2) <nzh0(z)

Q(z)

o

>0forz ∈∆.

If

θ(p(z)) +zp0(z)φ(p(z))≺θ(q(z)) +zq0(z)φ(q(z)), thenp(z)≺q(z)andq(z)is the best dominant.

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Lemma 2.2 ([19]). Letqbe a convex univalent function inandψ, γ ∈Cwith

<

1 + zq00(z) q0(z) +ψ

γ

>0.

Ifp(z)is analytic inand

ψp(z) +γzp0(z)≺ψq(z) +γzq0(z) thenp(z)≺q(z)andq(z)is the best dominant.

Lemma 2.3 ([5]). Letq(z)be convex univalent in the unit diskandϑandϕbe analytic in a domainDcontainingq(∆). Suppose that

(1) <[ϑ0(q(z))/ϕ(q(z))]>0forz ∈∆, (2) zq0(z)ϕ(q(z))is starlike univalent in∆.

Ifp(z) ∈ H[q(0),1]∩Q, withp(∆) ⊆ D, andϑ(p(z)) +zp0(z)ϕ(p(z))is univalent in ∆, and

(2.1) ϑ(q(z)) +zq0(z)ϕ(q(z))≺ϑ(p(z)) +zp0(z)ϕ(p(z)), thenq(z)≺p(z)andq(z)is the best subordinant.

Lemma 2.4 ([14, Theorem 8, p. 822]). Letq be convex univalent inandγ ∈ C. Further assume that<[γ]>0.Ifp(z)∈ H[q(0),1]∩Q,p(z) +γzp0(z)is univalent in∆, then

q(z) +γzq0(z)≺p(z) +γzp0(z) impliesq(z)≺p(z)andq(z)is the best subordinant.

For two functionsf(z) =z+P

n=2anznandg(z) =z+P

n=2bnzn, the Hadamard product (or convolution) off andg is defined by

(f ∗g)(z) :=z+

X

n=2

anbnzn=: (g∗f)(z).

Forαj ∈C (j = 1,2, . . . , l)andβj ∈C\ {0,−1,−2, . . .}(j = 1,2, . . . , m),the generalized hypergeometric functionlFm1, . . . , αl1, . . . , βm;z)is defined by the infinite series

lFm1, . . . , αl1, . . . , βm;z) :=

X

n=0

1)n· · ·(αl)n1)n· · ·(βm)n

zn n!

(l≤m+ 1;l, m∈N0 :={0,1,2, . . .}) where(a)nis the Pochhammer symbol defined by

(a)n:= Γ(a+n) Γ(a) =

1, (n= 0);

a(a+ 1)(a+ 2)· · ·(a+n−1), (n∈N:={1,2,3. . .}).

Corresponding to the function

h(α1, . . . , αl1, . . . , βm;z) :=z lFm1, . . . , αl1, . . . , βm;z),

the Dziok-Srivastava operator [7] (see also [8, 20]) Hml1, . . . , αl1, . . . , βm) is defined by the Hadamard product

Hml1, . . . , αl1, . . . , βm)f(z) (2.2)

:=h(α1, . . . , αl1, . . . , βm;z)∗f(z)

=z+

X

n=2

1)n−1· · ·(αl)n−1

1)n−1· · ·(βm)n−1

anzn (n−1)!.

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For brevity, we write

Hml1]f(z) :=Hml1, . . . , αl1, . . . , βm)f(z).

It is easy to verify from (2.2) that

(2.3) z(Hml1]f(z))01Hml1+ 1]f(z)−(α1−1)Hml1]f(z).

Special cases of the Dziok-Srivastava linear operator include the Hohlov linear operator [9], the Carlson-Shaffer linear operator L(a, c) [6], the Ruscheweyh derivative operator Dn [18], the generalized Bernardi-Libera-Livingston linear integral operator (cf. [3], [10], [12]) and the Srivastava-Owa fractional derivative operators (cf. [16], [17]).

The main object of the present paper is to find sufficient conditions for certain normalized analytic functionsf(z)to satisfy

q1(z)≺

Hml1]f(z) z

δ

≺q2(z)

whereq1andq2are given univalent functions in∆.Also, we obtain the number of known results as special cases.

3. MAINRESULTS

We begin with the following:

Theorem 3.1. Letq(z)be univalent in∆, λ∈C andα1 >0,δ >0.Supposeq(z)satisfies

(3.1) <

1 + zq00(z) q0(z) +λ

δ

>0.

Iff ∈ Asatisfies the subordination,

(3.2) (1−λα1)

Hml1]f(z) z

δ +λα1

Hml1]f(z) z

δ

Hml1+ 1]f(z) Hml1]f(z)

≺q(z) + λ

δzq0(z), then

Hml1]f(z) z

δ

≺q(z)

andq(z)is the best dominant.

Proof. Define the functionp(z)by

(3.3) p(z) :=

Hml1]f(z) z

δ . Then

zp0(z) δ :=α1

Hml1]f(z) z

δ

Hml1+ 1]f(z) Hml1]f(z) −1

, hence the hypothesis (3.2) of Theorem 3.1 yields the subordination:

p(z) + λzp0(z)

δ ≺q(z) + λzq0(z) δ .

Now Theorem 3.1 follows by applying Lemma 2.2 withψ = 1andγ = λδ. Whenl = 2, m = 1, α1 = a, α2 = 1,and β1 = cin Theorem 3.1, we have the following corollary.

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Corollary 3.2. Let q(z)be univalent in ∆, λ ∈ C andα1 > 0, δ > 0.Supposeq(z) satisfies (3.1). Iff ∈ Aand satisfies the subordination,

(3.4) (1−λa)

L(a, c)f(z) z

δ

+λa

L(a, c)f(z) z

δ

L(a+ 1, c)f(z) L(a, c)f(z)

≺q(z) + λ

δzq0(z), then

L(a, c)f(z) z

δ

≺q(z)

andq(z)is the best dominant.

By takingl= 1, m= 0andα1 = 1in Theorem 3.1, we get the following corollary.

Corollary 3.3. Let q(z)be univalent in ∆, λ ∈ C andα1 > 0, δ > 0.Supposeq(z) satisfies (3.1). Iff ∈ Aand satisfies the subordination,

(3.5) (1−λ)

f(z) z

δ

f(z) z

δ zf0(z)

f(z)

≺q(z) + λ

δzq0(z), then

f(z) z

δ

≺q(z)

andq(z)is the best dominant.

Corollary 3.4. Let−1≤B < A≤1and (3.1) hold. Iff ∈ Aand (1−λα1)

Hml1]f(z) z

δ

+λα1

Hml1]f(z) z

δ

Hml1 + 1]f(z) Hml1]f(z)

≺ λ(A−B)z

δ(1 +Bz)2 + 1 +Az 1 +Bz, then

Hml1]f(z) z

δ

≺ 1 +Az 1 +Bz and 1+Bz1+Az is the best dominant.

Theorem 3.5. Letq(z)be univalent in∆,λ, δ∈C.Supposeq(z)satisfies

(3.6) <

1 + zq00(z)

q0(z) − zq0(z) q(z)

>0.

Iff ∈ Asatisfies the subordination:

(3.7) 1 +γδα1

Hml1+ 1]f(z) Hml1]f(z) −1

≺1 +γzq0(z) q(z) , then

Hml1]f(z) z

δ

≺q(z) andq(z)is the best dominant.

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Proof. Define the functionp(z)by p(z) =

Hml1]f(z) z

δ

.

It is clear thatp(0) = 1andp(z)is analytic in∆. By using the identity (2.3), from (3.3) we get,

(3.8) zp0(z)

p(z) =α1δ

Hml1+ 1]f(z) Hml1]f(z) −1

.

Using (3.8) in (3.7), we see that the subordination becomes 1 +γzp0(z)

p(z) ≺1 +γzq0(z) q(z) . By setting

θ(w) = 1 and ϕ(w) = γ w, we observe thatϕandθare analytic inC\ {0}. Also we see that

Q(z) :=zq0(z)ϕ(q(z)) = γzq0(z) q(z) , and

h(z) := ϑ(q(z)) +Q(z) = 1 +γzq0(z) q(z) . It is clear thatQ(z)is starlike univalent in∆and

<zh0(z) Q(z) =<

1 + zq00(z)

q0(z) − zq0(z) q(z)

≥0.

By the hypothesis of Theorem 3.5, the result now follows by an application of Lemma 2.1.

Specializing the values ofl = 1, m= 0, α1 = 1andq(z) = (1−z)1 2b (b ∈C− {0}), γ= 1b andδ = 1in Theorem 3.5 above, we have the following corollary as stated in [21].

Corollary 3.6. Letbbe a non zero complex number. Iff ∈ Aand 1 + 1

b

zf0(z) f(z) −1

≺ 1 +z 1−z, then

f(z)

z ≺ 1

(1−z)2b and (1−z)1 2b is the best dominant.

Choosing the values ofl = 1, m = 0, α1 = 1 andq(z) = (1−z)12ab (b ∈ C − {0}), γ = 1b andδ =a 6= 0in Theorem 3.5 above, we have the following corollary as stated in [15].

Corollary 3.7. Letbbe a non zero complex number. Iff ∈ Aand 1 + 1

b

zf0(z) f(z) −1

≺ 1 +z 1−z,

then

f(z) z

a

≺ 1

(1−z)2ab

wherea6= 0is a complex number and (1−z)12ab is the best dominant.

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Similarly forl = 2, m = 1, α1 = 1, α2 = 1, β1 = 1andq(z) = (1−z)1 2b (b ∈ C− {0}), γ = 1b andδ= 1in Theorem 3.5 above, we get the following result as stated in [21].

Corollary 3.8. Letbbe a non zero complex number. Iff ∈ Aand

1 + 1 b

zf00(z) f0(z) −1

≺ 1 +z 1−z, then

f0(z)≺ 1 (1−z)2b and (1−z)1 2b is the best dominant.

Next, applying Lemma 2.3, we have the following theorem.

Theorem 3.9. Let q(z) be convex univalent in ∆, λ ∈ C and 0 < δ < 1. Suppose f ∈ A satisfies

(3.9) Re

δ λ

>0

and

Hml 1]f(z) z

δ

∈H[q(0),1]∩Q.Let

(1−λα1)

Hml1]f(z) z

δ +λα1

Hml1]f(z) z

δ

Hml1+ 1]f(z) Hml1]f(z)

be univalent in∆.Iff ∈ Asatisfies the superordination, (3.10) q(z) + λ

δzq0(z)≺(1−λα1)

Hml1]f(z) z

δ

+λα1

Hml1]f(z) z

δ

Hml1+ 1]f(z) Hml1]f(z)

then

q(z)≺

Hml1]f(z) z

δ

andq(z)is the best subordinant.

Proof. Define the functionp(z)by

(3.11) p(z) :=

Hml1]f(z) z

δ

. Using (3.11), simple computation produces

zp0(z) δ :=α1

Hml1]f(z) z

δ

Hml1+ 1]f(z) Hml1]f(z) −1

, then

q(z) + λ

δzq0(z)≺p(z) + λ

δzp0(z).

By settingϑ(w) = w andφ(w) = λδ,it is easily observed thatϑ(w)is analytic in C.Also, φ(w)is analytic inC\{0}andφ(w)6= 0,(w∈C\{0}).

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Sinceq(z)is a convex univalent function, it follows that

<

ϑ0(q(z)) φ(q(z))

=<

δ λ

>0, z ∈∆, δ, λ∈C, δ, λ6= 0.

Now Theorem 3.9 follows by applying Lemma 2.3.

Concluding the results of differential subordination and superordination, we state the follow- ing sandwich result.

Theorem 3.10. Let q1 and q2 be convex univalent in ∆, λ ∈ C and 0 < δ < 1. Supposeq2 satisfies (3.1) andq1 satisfies (3.9). If

Hml1]f(z) z

δ

∈ H[q(0),1]∩Q,

(1−λα1)

Hml1]f(z) z

δ

+λα1

Hml1]f(z) z

δ

Hml1+ 1]f(z) Hml1]f(z)

is univalent in∆.Iff ∈ Asatisfies q1(z) + λ

δzq10(z) (3.12)

≺(1−λα1)

Hml1]f(z) z

δ +λα1

Hml1]f(z) z

δ

Hml1+ 1]f(z) Hml1]f(z)

≺q2(z) + λ

δzq20(z), then

q1(z)≺

Hml1]f(z) z

δ

≺q2(z) andq1, q2 are respectively the best subordinant and best dominant.

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