Paper-1-Acta23.2010. 147KB Jun 04 2011 12:03:09 AM

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SUBORDINATION AND SUPERORDINATION PROPERTIES OF MULTIVALENT FUNCTIONS DEFINED BY EXTENDED

MULTIPLIER TRANSFORMATION

M. K. Aouf, A. Shamandy, R. M. El-Ashwah and E. E. Ali

Abstract. In this paper, we study different applications of the theory of dif-ferential subordination and superordination results for certain normalized analytic functions in the open unit disc, which are acted upon by a class of extended multi-plier transformation.

2000Mathematics Subject Classification: 30C45. 1. Introduction

LetH(U) denotes the class of analytic functions in the open unit discU ={z∈

C _:_|z|<1} and let H[a, p] denotes the subclass of the functions f ∈ H(U) of the form

f(z) =a+apzp+ap+1zp+1+. . . (a∈C, p∈N={1,2, . . .}). (1.1) Also, let A(p) be the subclass of the functionsf ∈H(U) of the form

f(z) =zp+ ∞

X

k=p+1

akzk (p∈N), (1.2)

and set A≡A(1).

Forf, g ∈H(U), we say that the function f (z)is subordinate to g(z), written symbolically as follows:

f ≺g or f(z)≺g(z) ,

if there exists a Schwarz function w(z), which (by definition) is analytic in U with

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f(z)≺g(z) ⇔f(0)≺g(0) and f(U)⊂g(U).

Supposing thatp and h are two analytic functions inU, let

ϕ(r, s, t;z) :C3_×_U _→C_.

If p andϕ(p(z), zp′(z), z2p′′(z);z) are univalent functions in U and if p satisfies the second-order superordination

h(z)≺ϕ(p(z), zp′(z), z2p′′(z);z), (1.3) then p is called to be a solution of the differential superordination (1.3). (If f is subordinate to F, thenF is superordination tof). An analytic function q is called a subordinant of (1.3), if q(z) ≺ p(z) for all the functions p satisfying (1.3). A univalent subordinant _eq that satisfies q ≺q_efor all of the subordinants q of (1.3), is called the best subordinant (cf., e.g.,[11], see also [12]).

Recently, Miller and Mocanu [13] obtained sufficient conditions on the functions

h,q and ϕfor which the following implication holds:

k(z)≺ϕ(p(z), zp′(z), z2p′′(z);z)⇒q(z)≺p(z). (1.4) Using the results Miller and Mocanu [13], Bulboaca [5] considered certain classes of first-order differential superordinations as well as superordination preserving inte-gral operators [4]. Ali et al. [1], have used the results of Bulboaca [5] and obtained sufficient conditions for certain normalized analytic functions f(z)to satisfy

q1(z)≺

zf′ (z)

f(z) ≺q2(z), (1.5)

whereq1 andq2are given univalent functions inU withq1(0) = 1.Shanmugam et al. [17] obtained sufficient conditions for normalized analytic functions f(z) to satisfy

q1(z)≺

f(z)

zf′

(z) ≺q2(z), and

q₁(z)≺ z

2_f′ (z)

{f(z)}2 ≺q2(z),

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Definition 1[6]. Let the function f(z) ∈ A(p). For m ∈ N0 = N ∪ {0}, λ ≥ 0, ℓ≥0.The extended multiplier transformationI_pm(λ, ℓ) on A(p) is defined by the following infinite series:

I_pm(λ, ℓ)f(z) =zp+ ∞

X

k=p+1

p+λ(k−p) +ℓ p+ℓ

m

akzk. (1.6) (λ≥0;ℓ≥0;p∈N;m∈N0;z∈U).

We can write (1.6) as follows:

I_pm(λ, ℓ)f(z) = (Φp,m_λ,ℓ ∗f)(z),

where

Φp,m_λ,ℓ (z) =zp+ ∞

X

k=p+1

p+λ(k−p) +ℓ p+ℓ

m

zk. (1.7)

It is easily verified from (1.6), that

λz(I_pm(λ, ℓ)f(z))′ = (p+ℓ)I_pm+1(λ, ℓ)f(z)−[p(1−λ) +ℓ]I_pm(λ, ℓ)f(z) (λ >0). (1.8) We note that:

I_p0(λ, ℓ)f(z) =f(z) , I_p1(1,0)f(z) = zf ′

(z)

p and I

2

p(1,0)f(z) =

z(zf′(z))′

p2 .

Also by specilizing the parameters λ, ℓ, m and p, we obtain the following operators studied by various authors:

(i) I_pm(1, ℓ) =Ip(m, ℓ)f(z) (see Kumar et al. [10] and Srivastava et al. [18]); (ii)I_pm(1,0)f(z) =D_pmf(z) (see [3], [9] and [15]);

(iii) Im

1 (1, ℓ)f(z) =Iℓmf(z) (see Cho and Kim [7] and Cho and Srivastava [8]); (iv)I₁m(1,0) =Dmf(z) (m∈N0) (see Salagean [16]);

(v) I₁m(λ,0) =D_λm (see Al-Aboudi [2]); (vi)Im

1 (1,1) =Imf(z) (see Uralegaddi and Somanatha [19]); (vii) I_pm(λ,0) =D_λ,pm f(z), where D_λ,pm f(z) is defined by

Dm_λ,pf(z) =zp+ ∞

X

k=p+1

p+λ(k−p)

p

m

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2.Preliminaries

In order to prove our subordination and superordination results, we make use of the following known definition and lemmas.

Definition 2[13]. Denote by Q the set of all functions f (z)that are analytic and injective on U \E(f), where

E(f) ={ζ :ζ ∈∂U and lim

z→ζf(z) =∞}, (2.1) and are such that f′

(ζ)6= 0 forζ ∈∂U\E(f).

Lemma 1 [12]. Let the function q(z) be univalent in the unit disc U, and let θ

and ϕ be analytic in a domain D containing q(U), with ϕ(w)6= 0 when w∈q(U). SetQ(z) =zq′

(z)ϕ(q(z)), h(z) =θ(q(z)) +Q(z) and suppose that (i) Qis a starlike function in U,

(ii) Re

zh′

(z)

Q(z)

>0f or z∈U.

If pis analytic in U withp(0) =q(0), p(U)⊆D and

θ(p(z)) +zp′

(z)ϕ(p(z))≺θ(q(z)) +zq′

(z)ϕ(q(z)), (2.2) then p(z)≺q(z), and q is the best dominant .

Lemma 2 [17]. Let q be a convex function in U and let ψ ∈C with δ ∈ C∗ =

C\{0} with

Re 1 +zq ′′

(z)

q′₍_z₎

!

>max

0;−Reψ

δ

, z∈U.

If p(z) is analytic inU, and

ψp(z) +δzp′

(z)≺ψq(z) +δzq′

(z), (2.3)

then p(z)≺q(z), and q is the best dominant .

Lemma 3 [4]. Let q(z) be a convex univalent function in the unit disc U and let θ

and ϕbe analytic in a domain D containing q(U). Suppose that (i) Re

θ′

(q(z))

ϕ(q(z))

>0 forz∈U; (ii)zq′

(z)ϕ(q(z)) is starlike inU.

If p∈H[q(0),1]∩Q withp(U)⊆D, andθ(p(z)) +zp′

(z)ϕ(p(z)) is univalent inU, and

θ(q(z)) +zq′

(z)ϕ(q(z))≺θ(p(z)) +zp′

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then q(z)≺p(z), andq is the best subordinant.

Lemma 4[13]. Let q be convex univalent in U and let δ ∈C, with Re (δ) >0. If

p∈H[q(0),1]∩Q and p(z) +δzp′

(z) is univalent in U, then

q(z) +δzq′

(z)≺p(z) +δzp′

(z), (2.4)

implies

q(z)≺p(z) (z∈U) and q is the best subordinant .

3.Subordination results for analytic functions

Unless otherwise mentioned we shall assume throught this paper that λ > o, ℓ≥0, p∈N and m∈N0.

Theorem 1. Let q be univalent in U, with q(0) = 1,β ∈C∗

. Suppose q satisfies

Re 1 +zq ′′

(z)

q′₍_z₎

!

>max

0;−Re 1

β

. (3.1)

If f ∈A(p), I_pm+1(λ, ℓ)f(z)6= 0 (z∈U∗

=U − {0}) and satisfies the subordination Φ(f, β, p, m, λ, ℓ)≺q(z) +βzq′

(z), (3.2)

where

Φ(f, β, p, m, λ, ℓ) = I m

p (λ, ℓ)f(z)

Ipm+1(λ, ℓ)f(z) +β

p+ℓ

λ (

1−I

m+2

p (λ, ℓ)f(z)Ipm(λ, ℓ)f(z)

Ipm+1(λ, ℓ)f(z)2

) ,

(3.3) then

Im

p (λ, ℓ)f(z)

Ipm+1(λ, ℓ)f(z)

≺q(z), (3.4)

and q is the best dominant of (3.2). Proof Define the functionp(z) by

p(z) = I m

p (λ, ℓ)f(z)

Ipm+1(λ, ℓ)f(z)

(z∈U). (3.5)

Then the function p is analytic in U andp(0) = 1.Therefore, differentiating (3.5) logarithmically with respect to z, and using the identity (1.8) in the resulting equa-tion, we have

zp′ (z)

p(z) =β

p+ℓ

λ (

1

p(z) −

I_pm+2(λ, ℓ)f(z)

Ipm+1(λ, ℓ)f(z)

)

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and

p(z) +βzp′ (z) =

I_pm(λ, ℓ)f(z)

Ipm+1(λ, ℓ)f(z) +β

p+ℓ

λ (

1−I

m+2

p (λ, ℓ)f(z)Ipm(λ, ℓ)f(z)

Ipm+1(λ, ℓ)f(z)

2

)

. (3.7)

The subordination (3.1) from hypothesis becomes

p(z) +βzp′

(z)≺q(z) +βzq′

(z). (3.8)

The assertion(3.4) of Theorem 1 now follows by an application of Lemma 2. Puttingm=ℓ= 0 in Theorem 1, we obtain the following corollary. Corollary 1. Assume that (3.1) holds. If f ∈A(p), β∈C∗

, and

Ψ(f, β, p, λ)≺q(z) +βzq′

(z), (3.9)

where

Ψ(f, β, p, λ) =

1−β

λ(2−λ)p

f(z)

h

(1−λ)f(z) +λ_pzf′ (z)i

+

βp λ

  



1−

λ p

2

z2f′′(z)f(z)−(1−λ_p)f2(z)

h

(1−λ)f(z) + λ_pzf′ (z)i2

  



 (3.10)

then

f(z)

h

≺q(z), (3.11)

and q is the best dominant .

Remark 1. Putting p= 1 in Corollary 1, we obtain the result obtained by Nechita [14, Corollary 6].

Putting λ= 1 andℓ= 0 in Theorem 1 we obtain the following corollary. Corollary 2. Assume that (3.1) holds. If f ∈A(p), β∈C∗

, and

D_pmf(z)

Dmp +1f(z) +βp

(

1−D

m+2

p f(z).Dmp f(z)

Dpm+1f(z)

2

)

≺q(z) +βzq′

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then

Dm p f(z)

Dpm+1f(z)

≺q(z), (3.15)

Remark 2. Putting p= 1 in Corollary 2, we obtain the result obtained by Nechita [14, Corollary 7] and correct the result obtained by Shanmugam et al. [17, Theorem 5.1].

Putting m=ℓ= 0 and λ= 1 in Theorem 1, we obtain the following corollary. Corollary 3. Assume that (3.1) holds. If f ∈A(p), β∈C∗

,and

(1−βp)pf(z)

zf′₍_z₎ +βp

(

1− z

2_f′′

(z)f(z)−p(p−1)f2(z) [zf′₍_z_)]2

)

≺q(z) +βzq′

(z), (3.16)

then

pf(z)

zf′₍_z₎ ≺q(z) (3.17)

Remark 3. Putting p = 1 in Corollary 3, we obtain the result obtained by Shan-mugam et al. [17, Theorem 3.2].

Puttingq(z) = 1 +Az

1 +Bz (−1≤B < A≤1) in Theorem 1, we obtain the following

corollary.

Corollary 4. Let −1≤B < A≤1, β∈C∗

,and suppose that

Re

1−Bz

1 +Bz

>max

0,−Re 1

β

. (3.18)

If f ∈A(p),and

Φ(f, β, p, m, λ, ℓ)≺ 1 +Az

1 +Bz +β

(A−B)z

(1 +Bz)2, where Φ(f, β, p, m, λ, ℓ) is given by (3.3), then

Im

p (λ, ℓ)f(z)

Ipm+1(λ, ℓ)f(z)

≺ 1 +Az

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and q(z) = 1 +Az

1 +Bz is the best dominant.

Putting A= 1 and B=−1 in Corollary 4, we obtain the following corollary. Corollary 5. If f ∈A(p) and β ∈C∗

satisfy Φ(f, β, p, m, λ, ℓ)≺ 1 +z

1−z +

2βz

(1−z)2, where Φ(f, β, p, m, λ, ℓ) is given by (3.3), then

I_pm(λ, ℓ)f(z)

Ipm+1(λ, ℓ)f(z)

≺ 1 +z

1−z

and q(z) = 1 +z

1−z is the best dominant.

4.Superordination and sandwich results

Theorem 2. Let q be convex univalent in U, β ∈C.Suppose

Reβ >0. (4.1)

If f ∈A(p), Ipm(λ,ℓ)f(z)

Ipm+1(λ,ℓ)f(z) ∈

H[q(0),1]∩Q, Φ(f, β, p, m, λ, ℓ) is univalent in the unit disc U, where Φ(f, β, p, m, λ, ℓ) is defined by (3.3). and

q(z) +βzq′

(z)≺Φ(f, β, p, m, λ, ℓ), (4.2) then

q(z)≺ I

m

p (λ, ℓ)f(z)

Ipm+1(λ, ℓ)f(z) and q is the best subordinant of (4.1).

Proof. Define the functionp(z) by (3.5). Differentiating (3.5) logarithmically with respect toz, and using the identity (1.8) in the resulting equation, we have

p(z) +βzp′

(z)≺Φ(f, β, p, m, λ, ℓ). (4.3) Theorem 2 follows by an applying of Lemma 4.

Putting m=ℓ= 0 in Theorem 2, we obtain the following corollary.

Corollary 6. Let q be convex in U with q(0) = 1, and β ∈C, Reβ > 0. If f ∈ A(p), h f(z)

(1−λ)f(z)+λ pzf

′₍_z₎i ∈ H[q(0),1]∩Q, Ψ(f, β, p, λ) is univalent in the unit disc

U, where Ψ(f, β, p, λ) is defined by (3.10), and

q(z) +βzq′

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then

q(z)≺ h f(z)

(4.5)

and q is the best subordinant.

Putting λ= 1 andℓ= 0 in Theorem 2, we obtain the following corollary. Corollary 7. Let qbe convex in U with q(0) = 1, and β∈C,Reβ >0.If f ∈A(p)

Dm pf(z)

Dm+1p f(z) ∈ H[q(0),1]∩Q,

Dm pf(z)

Dpm+1f(z) +βp

1− D

m+2

p f(z).Dpmf(z)

[Dm+1p f(z)] 2

is univalent in the unit disc U, and

q(z) +βzq′

(z)≺ D

m p f(z)

Dmp +1f(z) +βp

(

1−D

m+2

p f(z).Dpmf(z)

Dmp +1f(z)

2

)

, (4.6)

then

q(z)≺ D

m p f(z)

Dpm+1f(z)

,

and q is the best subordinant .

Remark 4. Putting p= 1 in Corollary 7, we obtain the result obtained by Nechita [14, Corollary 12] and correct the result obtained by Shanmugam et al. [17, Theorem 5.3].

Combining Theorem 1 and Theorem 2, we get the following sandawich theorem. Theorem 3. Let q1, q2 be convex in U with q1(0) =q2(0) = 1,β ∈C, Reβ >0and

q₂(z)satisfies (3.1). If f ∈A(p), Ipm(λ,ℓ)f(z)

Ipm+1(λ,ℓ)f(z) ∈

H[q(0),1]∩Q, Φ(f, β, p, m, λ, ℓ) is univalent in the unit disc U, where Φ(f, β, p, m, λ, ℓ) is defined by (3.3) and

q1(z) +βzq1′(z)≺Φ(f, β, m, λ, ℓ)≺q2(z) +βzq′2(z), (4.7) then

q₁(z)≺ I

m

p (λ, ℓ)f(z)

Ipm+1(λ, ℓ)f(z)

≺q₂(z)

and the functionsq1, q2 are respectively the best subordinant and the best dominant. Theorem 4. Let q(z) be univalent inU with q(0) = 1, β∈C∗

. Assume that (3.1) holds. If f ∈A(p),

ζ(f, β, m, p, λ, ℓ)≺q(z) +βzq′

(z) (4.8)

where

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1 +β

p+ℓ λ

zpI

m+1

p (λ, ℓ)f(z)

Im

p (λ, ℓ)f(z)

2 +β

p+ℓ

λ

(4.9)

zpI_pm+2(λ, ℓ)f(z)

Im

p (λ, ℓ)f(z)

2 −2β

p+ℓ

λ

zp

I_pm+1(λ, ℓ)f(z)2

Im

p (λ, ℓ)f(z)

3 ,

then

zpI

m+1

p (λ, ℓ)f(z)

Im

p (λ, ℓ)f(z)

2 ≺q(z) (4.10)

Proof. Define the functionp(z) by

p(z) =zpI

m+1

p (λ, ℓ)f(z)

Im

p (λ, ℓ)f(z)

2 (z∈U). (4.11)

Then, simple computations show that

zp′ (z)

p(z) =p+

zIm+1

p (λ, ℓ)f(z)

′

Ipm+1(λ, ℓ)f(z)

−2z

Im

p (λ, ℓ)f(z)

′

Im

p (λ, ℓ)f(z)

. (4.12)

We use the identity (1.8) in (4.12) we obtain

zp′ (z)

p(z) =

p+ℓ λ

(

1 +I m+2

p (λ, ℓ)f(z)

Im

p (λ, ℓ)f(z)

−2I m+1

p (λ, ℓ)f(z)

Im

p (λ, ℓ)f(z)

)

and

p(z) +βzp′ (z) =

1 +β

p+ℓ λ

zpI

m+1

p (λ, ℓ)f(z)

Im

p (λ, ℓ)f(z)

2 +β

p+ℓ

λ

.

.z

p_Im+2

p (λ, ℓ)f(z)

Im

p (λ, ℓ)f(z)

2 −2β

p+ℓ

λ

zp

I_pm+1(λ, ℓ)f(z)2

Im

p (λ, ℓ)f(z)

3 .

The subordination (4.9) becomes

p(z) +βzp′

(z)≺q(z) +βzq′ (z).

Theorem 4 follows by an applying of Lemma 2.

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Corollary 8. Letq(z) be univalent in U withq(0) = 1, β∈C∗

(1 +βp)(1−λ)z p

f(z) +

λ

p + (2λ−1)β

zp+1_f′ (z) [f(z)]2 +

βλ p

zp+2f′′(z) [f(z)]2 −2β

λ p

zp+2hf′(z)i2

[f(z)]3 ≺q(z) +βzq ′

(z) (4.13) then

(1−λ) z p

f(z)+

λ p

zp+1f′(z)

[f(z)]2 ≺q(z), (4.14) and q is the best dominant .

Putting λ= 1 andℓ= 0 in Theorem 4, we obtain the following corollary. Corollary 9. Let q(z) be univalent in U with q(0) = 1, β∈C∗

(1 +βp)z p_Dm+1

p f(z)

Dm

p f(z)

2 +βp

zp_Dm+2 p f(z)

Dm

p f(z)

2 −2βp

zp_Dm+1 p f(z)

2

Dm

p f(z)

3 ≺q(z) +βzq

′ (z) (4.15) then

zpD_pm+1f(z)

Dm

p f(z)

2 ≺q(z), and q is the best dominant .

Remark 6. Putting p= 1 in Corollary 9, we obtain the result obtained by Shan-mugam et al. [17, Theorem 5.4].

Puttingm=ℓ= 0 andλ= 1 in Theorem 4, we obtain the following corollary. Corollary 10. Let q(z) be univalent in U with q(0) = 1, β ∈ C∗

zp+1f′(z)

p[f(z)]2 −β

zp+1 p

zp f(z)

′′

≺q(z) +βzq′

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zp+1_f′ (z)

p[f(z)]2 ≺q(z) and q is the best dominant.

Remark 7. Putting p = 1 in Corollary 10, we obtain the result obtained by Shanmugam et al. [17, Theorem 3.4].

Puttingq(z) = 1 +Az

1 +Bz (−1≤B < A≤1) in Theorem 4, we obtain the following

corollary.

Corollary 11. Letq(z) be univalent inU withq(0) = 1, β∈C∗

ζ(f, β, p, m, λ, ℓ)≺ 1 +Az

1 +Bz +β

(A−B)z

(1 +Bz)2, (4.17)

where ζ(f, β, p, m, λ, ℓ) is given by (4.9), then

zpI_pm+1(λ, ℓ)f(z)

Im

p (λ, ℓ)f(z)

2 ≺

1 +Az

1 +Bz (4.18)

and q(z) = 1 +Az

1 +Bz is the best dominant .

Next, applying Lemma 4, we have the following theorem.

Theorem 5. Let q be convex in U with q(0) = 1, and β ∈ C, Reβ >0. If f ∈ A(p), zpIpm+1(λ,ℓ)f(z)

[Im

p (λ,ℓ)f(z)]

2 ∈ H[q(0),1]∩ Q, ζ(f, β, p, m, λ, ℓ) is univalent in U, where

ζ(f, β, p, m, λ, ℓ) is defined by (4.9), and

q(z) +βzq′

(z)≺ζ(f, β, p, m, λ, ℓ), (4.19) then

q(z)≺ z

p_Im+1

p (λ, ℓ)f(z)

Im

p (λ, ℓ)f(z)

2

and q is the best subordinant .

Puttingm=ℓ= 0 in Theorem 5, we obtain the following corollary.

Corollary 12. Let q be convex in U with q(0) = 1, and β ∈ C, Reβ > 0. If f ∈A(p) , (1−λ)_fz₍p_z₎ +λ_pzp+1f

′ (z)

[f(z)]2 ∈H[q(0),1]∩Q. ζ(f, β, p, λ) is univalent in

U, where ζ(f, β, p, λ) is defined by (4.9), and

q(z) +βzq′

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then

q(z)≺(1−λ) z p

f(z) +

λ p

zp+1_f′ (z) [f(z)]2 and q is the best subordinant .

Combining Theorem 4 and Theorem 5, we get the following sandawich theorem. Theorem 6. Let q1, q2 be convex in U with q1(0) =q2(0) = 1,β ∈C, Reβ >0and

q2(z)satisfies (3.1). Iff ∈A(p), z

p_Im+1

p (λ,ℓ)f(z)

[Im

p (λ,ℓ)f(z)]

2 ∈H[q(0),1]∩Q,ζ(f, β, p, m, λ, ℓ) is

univalent in disc U, where ζ(f, β, p, m, λ, ℓ) is defined by (4.9) and

q1(z) +βzq′1(z)≺ζ(f, β, m, λ, ℓ)≺q2(z) +βzq′2(z), (4.21) then

q1(z)≺

zp_Im+1

p (λ, ℓ)f(z)

Im

p (λ, ℓ)f(z)

2 ≺q2(z)

and the functions q1, q2 are, respectively, the best subordinant and the best domi-nant.

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M. K. Aouf, A. Shamandy, R. M. EL-Ashwah and E. E. Ali Department of Mathematics

Faculty of Science Mansoura University Mansoura 35516, Egypt.