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Differential Subordination Results Rabha W. Ibrahim and

Maslina Darus vol. 10, iss. 1, art. 8, 2009

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DIFFERENTIAL SUBORDINATION RESULTS FOR NEW CLASSES OF THE FAMILY E (Φ, Ψ)

RABHA W. IBRAHIM AND MASLINA DARUS

School of Mathematical Sciences Faculty of Science and Technology Universiti Kebangsaan Malaysia Bangi 43600, Selangor Darul Ehsan Malaysia

EMail:rabhaibrahim@yahoo.com maslina@ukm.my

Received: 05 July, 2008

Accepted: 02 December, 2008

Communicated by: S.S. Dragomir 2000 AMS Sub. Class.: 34G10, 26A33, 30C45.

Key words: Fractional calculus; Subordination; Hadamard product.

Abstract: We define new classes of the family E(Φ,Ψ), in a unit diskU := {z C,|z| < 1},as follows: for analytic functionsF(z),Φ(z)and Ψ(z)so that

<{F(z)∗Φ(z)F(z)∗Ψ(z)}>0, z U, F(z)Ψ(z)6= 0where the operatordenotes the convolution or Hadamard product. Moreover, we establish some subordination results for these new classes.

Acknowledgement: The work presented here was supported by SAGA: STGL-012-2006, Academy of Sciences, Malaysia.

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Contents

1 Introduction and preliminaries. 3

2 Main Results 7

3 Applications 14

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1. Introduction and preliminaries.

Let Bα+ be the class of all analytic functions F(z) in the open disk U := {z ∈ C, |z|<1},of the form

F(z) = 1 +

X

n=1

anzn+α−1, 0< α≤1,

satisfyingF(0) = 1. And let Bα be the class of all analytic functions F(z)in the open diskU of the form

F(z) = 1−

X

n=1

anzn+α−1, 0< α≤1, an≥0; n= 1,2,3, . . . , satisfyingF(0) = 1.With a view to recalling the principle of subordination between analytic functions, let the functionsf and g be analytic in U.Then we say that the functionf is subordinate tog if there exists a Schwarz functionw(z),analytic inU such that

f(z) =g(w(z)), z ∈U.

We denote this subordination byf ≺ g orf(z) ≺g(z), z ∈ U.If the functiong is univalent inU the above subordination is equivalent to

f(0) =g(0) and f(U)⊂g(U).

Letφ : C3 ×U → Cand let hbe univalent inU.Assume thatp, φ are analytic and univalent inU andpsatisfies the differential superordination

(1.1) h(z)≺φ(p(z)), zp0(z), z2p00(z);z), thenpis called a solution of the differential superordination.

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An analytic functionq is called a subordinant ifq ≺ pfor allpsatisfying (1.1).

A univalent functionqsuch thatp≺qfor all subordinantspof (1.1) is said to be the best subordinant.

LetB+be the class of analytic functions of the form f(z) = 1 +

X

n=1

anzn, an ≥0.

Given two functionsf, g ∈ B+, f(z) = 1 +

X

n=1

anzn and g(z) = 1 +

X

n=1

bnzn their convolution or Hadamard productf(z)∗g(z)is defined by

f(z)∗g(z) = 1 +

X

n=1

anbnzn, an≥0, bn ≥0, z ∈U.

Juneja et al. [1] define the familyE(Φ,Ψ),so that

<

f(z)∗Φ(z) f(z)∗Ψ(z)

>0, z ∈U where

Φ(z) =z+

X

n=2

ϕnzn and Ψ(z) = z+

X

n=2

ψnzn

are analytic in U with the conditions ϕn ≥ 0, ψn ≥ 0, ϕn ≥ ψn for n ≥ 2 and f(z)∗Ψ(z)6= 0.

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Definition 1.1. LetF(z) ∈ B+α,we define the familyEα+(Φ,Ψ)so that

(1.2) <

F(z)∗Φ(z) F(z)∗Ψ(z)

>0, z ∈U, where

Φ(z) = 1 +

X

n=1

ϕnzn+α−1 and Ψ(z) = 1 +

X

n=1

ψnzn+α−1

are analytic in U under the conditions ϕn ≥ 0, ψn ≥ 0, ϕn ≥ ψn forn ≥ 1 and F(z)∗Ψ(z)6= 0.

Definition 1.2. LettingF(z) ∈ Bα,we define the familyEα(Φ,Ψ) which satisfies (1.2) where

Φ(z) = 1−

X

n=1

ϕnzn+α−1 and Ψ(z) = 1−

X

n=1

ψnzn+α−1

are analytic in U under the conditions ϕn ≥ 0, ψn ≥ 0, ϕn ≥ ψn forn ≥ 1 and F(z)∗Ψ(z)6= 0.

In the present paper, we establish some sufficient conditions for functions F ∈ B+α andF ∈ Bαto satisfy

(1.3) F(z)∗Φ(z)

F(z)∗Ψ(z) ≺q(z), z ∈U,

where q(z) is a given univalent function in U such that q(0) = 1. Moreover, we give applications for these results in fractional calculus. We shall need the following known results.

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Lemma 1.3 ([2]). Letq(z)be convex in the unit disk U withq(0) = 1and<{q} >

1

2, z ∈U.If0≤µ <1, pis an analytic function inU withp(0) = 1and if (1−µ)p2(z) + (2µ−1)p(z)−µ+ (1−µ)zp0(z)

≺(1−µ)q2(z) + (2µ−1)q(z)−µ+ (1−µ)zq0(z), thenp(z)≺q(z)andq(z)is the best dominant.

Lemma 1.4 ([3]). Letq(z)be univalent in the unit diskU and letθ(z)be analytic in a domainDcontainingq(U).Ifzq0(z)θ(q)is starlike inU,and

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

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2. Main Results

In this section, we verify some sufficient conditions of subordination for analytic functions in the classesB+α andBα.

Theorem 2.1. Let the functionq(z)be convex in the unit diskU such thatq(0) = 1 and <{q} > 12. If F ∈ B+α and FF(z)∗Ψ(z)(z)∗Φ(z) an analytic function in U satisfies the subordination

(1−µ)

F(z)∗Φ(z) F(z)∗Ψ(z)

2

+ (2µ−1)

F(z)∗Φ(z) F(z)∗Ψ(z)

−µ + (1−µ)

F(z)∗Φ(z) F(z)∗Ψ(z)

z(F(z)∗Φ(z))0

F(z)∗Φ(z) − z(F(z)∗Ψ(z))0 F(z)∗Ψ(z)

≺(1−µ)q2(z) + (2µ−1)q(z)−µ+ (1−µ)zq0(z),

then F(z)∗Φ(z)

F(z)∗Ψ(z) ≺q(z) andq(z)is the best dominant.

Proof. Let the functionp(z)be defined by p(z) := F(z)∗Φ(z)

F(z)∗Ψ(z), z ∈U.

It is clear thatp(0) = 1.Then straightforward computation gives us (1−µ)p2(z) + (2µ−1)p(z)−µ+ (1−µ)zp0(z)

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= (1−µ)

F(z)∗Φ(z) F(z)∗Ψ(z)

2

+ (2µ−1)

F(z)∗Φ(z) F(z)∗Ψ(z)

−µ + (1−µ)z

F(z)∗Φ(z) F(z)∗Ψ(z)

0

= (1−µ)

F(z)∗Φ(z) F(z)∗Ψ(z)

2

+ (2µ−1)

F(z)∗Φ(z) F(z)∗Ψ(z)

−µ + (1−µ)

F(z)∗Φ(z) F(z)∗Ψ(z)

z(F(z)∗Φ(z))0

F(z)∗Φ(z) − z(F(z)∗Ψ(z))0 F(z)∗Ψ(z)

≺(1−µ)q2(z) + (2µ−1)q(z)−µ+ (1−µ)zq0(z).

By the assumption of the theorem we have that the assertion of the theorem follows by an application of Lemma1.3.

Corollary 2.2. IfF ∈ Bα+ and FF(z)∗Φ(z)(z)∗Ψ(z) is an analytic function in U satisfying the subordination

(1−µ)

F(z)∗Φ(z) F(z)∗Ψ(z)

2

+ (2µ−1)

F(z)∗Φ(z) F(z)∗Ψ(z)

−µ + (1−µ)

F(z)∗Φ(z) F(z)∗Ψ(z)

z(F(z)∗Φ(z))0

F(z)∗Φ(z) − z(F(z)∗Ψ(z))0 F(z)∗Ψ(z)

≺(1−µ)

1 +Az 1 +Bz

2

+ (2µ−1)

1 +Az 1 +Bz

−µ+ (1−µ)

1 +Az 1 +Bz

(A−B)z (1 +Az)(1 +Bz),

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then F(z)∗Φ(z)

F(z)∗Ψ(z) ≺

1 +Az 1 +Bz

, −1≤B < A≤1 and(1+Az1+Bz)is the best dominant.

Proof. Let the functionq(z)be defined by q(z) :=

1 +Az 1 +Bz

, z ∈U.

It is clear thatq(0) = 1and<{q} > 12 for arbitraryA, B, z ∈ U,then in view of Theorem2.1we obtain the result.

Corollary 2.3. IfF ∈ Bα+ and FF(z)∗Φ(z)(z)∗Ψ(z) is an analytic function in U satisfying the subordination

(1−µ)

F(z)∗Φ(z) F(z)∗Ψ(z)

2

+ (2µ−1)

F(z)∗Φ(z) F(z)∗Ψ(z)

−µ + (1−µ)

F(z)∗Φ(z) F(z)∗Ψ(z)

z(F(z)∗Φ(z))0

F(z)∗Φ(z) − z(F(z)∗Ψ(z))0 F(z)∗Ψ(z)

≺(1−µ)

1 +z 1−z

2

+ (2µ−1)

1 +z 1−z

−µ + (1−µ)

1 +z 1−z

2z 1−z2

,

then F(z)∗Φ(z)

F(z)∗Ψ(z) ≺

1 +z 1−z

, and(1+z1−z)is the best dominant.

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Define the functionϕα(a, c;z)by ϕα(a, c;z) := 1 +

X

n=1

(a)n

(c)nzn+α−1, (z ∈U; a ∈R, c ∈R\{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).

Corresponding to the functionϕα(a, c;z),define a linear operatorLα(a, c)by Lα(a, c)F(z) := ϕα(a, c;z)∗F(z), F(z) ∈ B+α

or equivalently by

Lα(a, c)F(z) := 1 +

X

n=1

(a)n

(c)nanzn+α−1. For details see [4]. Hence we have the following result:

Corollary 2.4. Let the functionq(z)be convex in the unit diskU such thatq(0) = 1 and<{q}> 12.IfLLα(a,c)Φ(z)

α(a,c)Ψ(z) is an analytic function inU satisfying the subordination (1−µ)

Lα(a, c)Φ(z) Lα(a, c)Ψ(z)

2

+ (2µ−1)

Lα(a, c)Φ(z) Lα(a, c)Ψ(z)

−µ + (1−µ)

Lα(a, c)Φ(z) Lα(a, c)Ψ(z)

z(Lα(a, c)Φ(z))0

Lα(a, c)Φ(z) − z(Lα(a, c)Ψ(z))0 Lα(a, c)Ψ(z)

≺(1−µ)q2(z) + (2µ−1)q(z)−µ+ (1−µ)zq0(z),

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then Lα(a, c)Φ(z)

Lα(a, c)Ψ(z) ≺q(z) andq(z)is the best dominant.

Theorem 2.5. Let the functionq(z)be univalent in the unit diskU such thatq0(z)6=

0and zqq(z)0(z) is starlike inU.IfF ∈ Bαsatisfies the subordination a

z(F(z)∗Φ(z))0

F(z)∗Φ(z) − z(F(z)∗Ψ(z))0 F(z)∗Ψ(z)

≺azq0(z) q(z) ,

then F(z)∗Φ(z)

F(z)∗Ψ(z) ≺q(z), z ∈U, andq(z)is the best dominant.

Proof. Let the functionp(z)be defined by p(z) :=

F(z)∗Φ(z) F(z)∗Ψ(z)

, z ∈U.

By setting

θ(ω) := a

ω, a6= 0,

it can easily observed thatθ(ω)is analytic inC− {0}.Then we obtain azp0(z)

p(z) =a

z(F(z)∗Φ(z))0

F(z)∗Φ(z) − z(F(z)∗Ψ(z))0 F(z)∗Ψ(z)

≺azq0(z) q(z) .

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By the assumption of the theorem we have that the assertion of the theorem follows by an application of Lemma1.4.

Corollary 2.6. IfF ∈ Bαsatisfies the subordination a

z(F(z)∗Φ(z))0

F(z)∗Φ(z) − z(F(z)∗Ψ(z))0 F(z)∗Ψ(z)

≺a (A−B)z (1 +Az)(1 +Bz)

then F(z)∗Φ(z)

F(z)∗Ψ(z) ≺

1 +Az 1 +Bz

, −1≤B < A≤1 and(1+Az1+Bz)is the best dominant.

Corollary 2.7. IfF ∈ Bαsatisfies the subordination a

z(F(z)∗Φ(z))0

F(z)∗Φ(z) −z(F(z)∗Ψ(z))0 F(z)∗Ψ(z)

≺a 2z

1−z2

, then

F(z)∗Φ(z) F(z)∗Ψ(z) ≺

1 +z 1−z

, and(1+z1−z)is the best dominant.

Define the functionφα(a, c;z)by φα(a, c;z) := 1−

X

n=1

(a)n

(c)nzn+α−1, (z ∈U; a ∈R, c ∈R\{0,−1,−2, . . .}), where(a)n is the Pochhammer symbol. Corresponding to the functionφα(a, c;z), define a linear operatorLα(a, c)by

Lα(a, c)F(z) := φα(a, c;z)∗F(z), F(z) ∈ Bα

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or equivalently by

Lα(a, c)F(z) := 1−

X

n=1

(a)n

(c)nanzn+α−1. Hence we obtain the next result.

Corollary 2.8. Let the functionq(z)be univalent in the unit diskU such thatq0(z)6=

0and zqq(z)0(z) is starlike inU.IfF ∈ Bαsatisfies the subordination a

z(Lα(a, c)Φ(z))0

Lα(a, c)Φ(z) − z(Lα(a, c)Ψ(z))0 Lα(a, c)Ψ(z)

≺azq0(z) q(z) ,

then Lα(a, c)Φ(z)

Lα(a, c)Ψ(z) ≺q(z), z ∈U, andq(z)is the best dominant.

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3. Applications

In this section, we introduce some applications of Section2containing fractional in- tegral operators. Assume thatf(z) =P

n=1ϕnznand let us begin with the following definitions

Definition 3.1 ([5]). The fractional integral of orderα for the function f(z)is de- fined by

Izαf(z) := 1 Γ(α)

Z z 0

f(ζ)(z−ζ)α−1dζ; 0≤α <1,

where the function f(z) is analytic in a simply-connected region of the complex z−plane (C) containing the origin. The multiplicity of (z −ζ)α−1 is removed by requiringlog(z−ζ)to be real when(z−ζ)>0.Note that,Izαf(z) =h

zα−1 Γ(α)

i f(z), forz >0and0forz ≤0(see [6]).

From Definition3.1, we have Izαf(z) =

zα−1 Γ(α)

f(z) = zα−1 Γ(α)

X

n=1

ϕnzn=

X

n=1

anzn+α−1

wherean := Γ(α)ϕn for alln = 1,2,3, . . . ,thus1 +Izαf(z) ∈ Bα+and1−Izαf(z) ∈ Bαn≥0).Then we have the following results.

Theorem 3.2. Let the assumptions of Theorem2.1hold. Then (1 +Izαf(z))∗Φ(z)

(1 +Izαf(z))∗Ψ(z)

≺q(z), z 6= 0, z ∈U andq(z)is the best dominant.

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Proof. Let the functionF(z)be defined by

F(z) := 1 +Izαf(z), z ∈U.

Theorem 3.3. Let the assumptions of Theorem2.5hold. Then (1−Izαf(z))∗Φ(z)

(1−Izαf(z))∗Ψ(z)

≺q(z), z ∈U andq(z)is the best dominant.

Proof. Let the functionF(z)be defined by

F(z) := 1−Izαf(z), z ∈U.

LetF(a, b;c;z)be the Gauss hypergeometric function (see [7]) defined, forz ∈ U,by

F(a, b;c;z) =

X

n=0

(a)n(b)n (c)n(1)nzn.

We need the following definitions of fractional operators in Saigo type fractional calculus (see [8,9]).

Definition 3.4. Forα > 0andβ, η ∈ R,the fractional integral operator I0,zα,β,η is defined by

I0,zα,β,ηf(z) = z−α−β Γ(α)

Z z 0

(z−ζ)α−1F

α+β,−η;α; 1− ζ z

f(ζ)dζ,

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where the function f(z) is analytic in a simply-connected region of the z−plane containing the origin, with order

f(z) = O(|z|)(z →0), >max{0, β−η} −1

and the multiplicity of(z−ζ)α−1 is removed by requiringlog(z−ζ)to be real when z−ζ >0.

From Definition3.4, withβ <0,we have I0,zα,β,ηf(z) = z−α−β

Γ(α) Z z

0

(z−ζ)α−1F(α+β,−η;α; 1− ζ

z)f(ζ)dζ

=

X

n=0

(α+β)n(−η)n (α)n(1)n

z−α−β Γ(α)

Z z 0

(z−ζ)α−1

1− ζ z

n

f(ζ)dζ

=

X

n=0

Bnz−α−β−n Γ(α)

Z z 0

(z−ζ)n+α−1f(ζ)dζ

=

X

n=0

Bn

z−β−1 Γ(α) f(ζ)

= B

Γ(α)

X

n=1

ϕnzn−β−1, where

Bn:= (α+β)n(−η)n (α)n(1)n

and B :=

X

n=0

Bn. Denotean := Γ(α)n for all n = 1,2,3, . . . ,and letα=−β.Thus,

1 +I0,zα,β,ηf(z) ∈ Bα+ and 1−I0,zα,β,ηf(z) ∈ Bαn≥0),

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andn we have the following results

Theorem 3.5. Let the assumptions of Theorem2.1hold. Then

"

(1 +I0,zα,β,ηf(z))∗Φ(z) (1 +I0,zα,β,ηf(z))∗Ψ(z)

#

≺q(z), z ∈U andq(z)is the best dominant.

Proof. Let the functionF(z)be defined by

F(z) := 1 +I0,zα,β,ηf(z), z ∈U.

Theorem 3.6. Let the assumptions of Theorem2.5hold. Then

"

(1−I0,zα,β,ηf(z))∗Φ(z) (1−I0,zα,β,ηf(z))∗Ψ(z)

#

≺q(z), z ∈U andq(z)is the best dominant.

Proof. Let the functionF(z)be defined by

F(z) := 1−I0,zα,β,ηf(z), z ∈U.

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