(1)http://jipam.vu.edu.au/ Volume 7, Issue 4, Article 134, 2006 SUBORDINATION RESULTS FOR A CLASS OF ANALYTIC FUNCTIONS DEFINED BY A LINEAR OPERATOR B.A

http://jipam.vu.edu.au/

Volume 7, Issue 4, Article 134, 2006

SUBORDINATION RESULTS FOR A CLASS OF ANALYTIC FUNCTIONS DEFINED BY A LINEAR OPERATOR

B.A. FRASIN

DEPARTMENT OFMATHEMATICS

AL AL-BAYTUNIVERSITY

P.O. BOX: 130095 MAFRAQ, JORDAN

[email protected]

Received 21 June, 2005; accepted 01 June, 2006 Communicated by A. Sofo

ABSTRACT. In this paper, we derive several interesting subordination results for certain class of analytic functions defined by the linear operatorL(a, c)f(z)which introduced and studied by Carlson and Shaffer [2].

Key words and phrases: Analytic functions, Hadamard product, Subordinating factor sequence.

2000 Mathematics Subject Classification. Primary 30C45; Secondary 30A10, 30C80.

1. INTRODUCTION ANDDEFINITIONS

LetAdenote the class of functions of the form:

(1.1) f(z) =z+

∞

X

n=2

anzⁿ

which are analytic in the open unit disc ∆ = {z :|z|<1}. For two functionsf(z)and g(z) given by

(1.2) f(z) =z+

∞

X

n=2

anzⁿ and g(z) = z+

∞

X

n=2

cnzⁿ

their Hadamard product (or convolution) is defined by

(1.3) (f∗g)(z) :=z+

∞

X

n=2

a_nc_nzⁿ.

ISSN (electronic): 1443-5756

c 2006 Victoria University. All rights reserved.

189-05

Define the functionφ(a, c;z)by (1.4) φ(a, c;z) :=

∞

X

n=0

(a)_n

(c)_nzⁿ⁺¹ (c /∈Z⁻0 :={0,−1,−2, . . .}, z∈∆), where(λ)nis the Pochhammer symbol given, in terms of Gamma functions,

(λ)_n:= Γ(λ+n) (1.5) Γ(λ)

=

( 1, n= 0,

λ(λ+ 1)(λ+ 2). . .(λ+n−1), n∈N:{1,2, . . .}.

Corresponding to the functionφ(a, c;z),Carlson and Shaffer [2] introduced a linear operator L(a, c) :A → Aby

(1.6) L(a, c)f(z) :=φ(a, c;z)∗f(z),

or, equivalently, by

L(a, c)f(z) :=z+

∞

X

n=1

(a)_n

(c)_na_n+1zⁿ⁺¹ (z ∈∆).

Note thatL(1,1)f(z) =f(z),L(2,1)f(z) = zf⁰(z)andL(3,1)f(z) =zf⁰(z) + ¹₂z²f⁰⁰(z).

For −1 ≤ α < 1, β ≥ 0, we let L(a, c;α, β) consist of functions f in A satisfying the condition

(1.7) Re

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

L(a, c)f(z) −(a−1)

> β

aL(a+ 1, c)f(z) L(a, c)f(z) −a

+α, (z ∈∆) The family L(a, c;α, β) is of special interest for it contains many well-known as well as many new classes of analytic univalent functions. For L(1,1;α,0), we obtain the family of starlike functions of order α(0 ≤ α < 1)and L(2,1;α,0) is the family of convex functions of order α (0 ≤ α < 1). ForL(1,1; 0, β) andL(2,1; 0, β), we obtain the class of uniformly β- starlike functions and uniformly β- convex functions, respectively, introduced by Kanas and Winsiowska ([3],[4]) (see also the work of Kanas and Srivastava [5], Goodman ([7],[8]), Rønning ([10],[11]), Ma and Minda [9] and Gangadharan et al. [6]).

Before we state and prove our main result we need the following definitions and lemmas.

Definition 1.1 (Subordination Principle). Let g(z)be analytic and univalent in ∆. If f(z) is analytic in∆, f(0) =g(0),andf(∆)⊂g(∆),then we see that the functionf(z)is subordinate tog(z)in∆,and we writef(z)≺g(z).

Definition 1.2 (Subordinating Factor Sequence). A sequence {b_n}^∞_n=1 of complex numbers is called a subordinating factor sequence if, wheneverf(z)is analytic , univalent and convex in

∆, we have the subordination given by (1.8)

∞

X

n=2

b_na_nzⁿ≺f(z) (z ∈∆, a₁ = 1).

Lemma 1.1 ([14]). The sequence{b_n}^∞_n=1 is a subordinating factor sequence if and only if

(1.9) Re

( 1 + 2

∞

X

n=1

b_nzⁿ )

>0 (z ∈∆).

Lemma 1.2. If (1.10)

∞

X

n=2

σ_n(a, c;α, β)|a_n| ≤1−α where, for convenience,

σn(a, c;α, β) := (1 +β)(a)_n+ [1−α−a(1 +β)](a)_n−1 (c)n−1

(1.11)

(−1≤α <1; β ≥0, n≥2), thenf(z)∈ L(a, c;α, β).

Proof. It suffices to show that

β

aL(a+ 1, c)f(z) L(a, c)f(z) −a

−Re

aL(a+ 1, c)f(z) L(a, c)f(z) −a

≤1−α.

We have β

aL(a+ 1, c)f(z) L(a, c)f(z) −a

−Re

aL(a+ 1, c)f(z) L(a, c)f(z) −a

≤(1 +β)

aL(a+ 1, c)f(z) L(a, c)f(z) −a

≤

(1 +β)P∞ n=2

a(a+1)n−1−a(a)n−1

(c)n−1

|a_n| |z|ⁿ⁻¹

1−P∞ n=2

(a)n−1

(c)n−1 |an| |z|ⁿ⁻¹

≤

(1 +β)P∞ n=2

(a)n−a(a)n−1

(c)n−1

|a_n| 1−P∞

n=2 (a)n−1

(c)n−1 |a_n| . The last expression is bounded above by1−αif

∞

X

n=2

(1 +β)(a)_n+ [1−α−a(1 +β)](a)_n−1 (c)n−1

|an| ≤1−α

and the proof is complete.

LetL^?(a, c;α, β)denote the class of functionsf(z)∈ Awhose coefficients satisfy the con- dition (1.10). We note thatL^?(a, c;α, β)⊆ L(a, c;α, β).

2. MAIN THEOREM

Employing the techniques used earlier by Srivastava and Attiya [13], Attiya [1] and Singh [12], we state and prove the following theorem.

Theorem 2.1. Let the functionf(z)defined by (1.1) be in the classL^?(a, c;α, β)where−1≤ α <1 ; β ≥0;a > 0;c >0.Also letKdenote the familiar class of functionsf(z)∈ Awhich are also univalent and convex in∆. Then

(2.1) σ₂(a, c;α, β)

2[1−α+σ₂(a, c;α, β)](f ∗g)(z)≺g(z) (z ∈∆; g ∈ K), and

(2.2) Re(f(z))>−1−α+σ₂(a, c;α, β)

σ₂(a, c;α, β) , (z ∈∆).

The constant _2[1−α+σ^{σ2(a,c;α,β)}

2(a,c;α,β)] is the best estimate.

Proof. Letf(z)∈ L^?(a, c;α, β)and letg(z) =z+P∞

n=2c_nzⁿ∈ K. Then (2.3) σ₂(a, c;α, β)

2[1−α+σ₂(a, c;α, β)](f ∗g)(z) = σ₂(a, c;α, β)

2[1−α+σ₂(a, c;α, β)] z+

∞

X

n=2

ancnzⁿ

! .

Thus, by Definition 1.2, the assertion of our theorem will hold if the sequence (2.4)

σ2(a, c;α, β)

2[1−α+σ₂(a, c;α, β)]a_n ∞

n=1

is a subordinating factor sequence, witha₁ = 1.In view of Lemma 1.1, this will be the case if and only if

(2.5) Re

( 1 + 2

∞

X

n=1

σ₂(a, c;α, β)

2[1−α+σ₂(a, c;α, β)]a_nzⁿ )

>0 (z ∈∆).

Now

Re (

1 + σ₂(a, c;α, β) 1−α+σ₂(a, c;α, β)

∞

X

n=1

a_nzⁿ )

= Re

1 + σ₂(a, c;α, β) 1−α+σ₂(a, c;α, β)z

+ 1

1−α+σ2(a, c;α, β)

∞

X

n=1

σ₂(a, c;α, β)a_nzⁿ )

≥1−

σ₂(a, c;α, β) 1−α+σ₂(a, c;α, β)r

− 1

1−α+σ₂(a, c;α, β)

∞

X

n=1

σn(a, c;α, β)anrⁿ )

.

Sinceσ_n(a, c;α, β)is an increasing function ofn(n≥2) 1−

σ₂(a, c;α, β) 1−α+σ2(a, c;α, β)r

− 1

1−α+σ₂(a, c;α, β)

∞

X

n=1

σ_n(a, c;α, β)a_nrⁿ )

>1− σ₂(a, c;α, β)

1−α+σ₂(a, c;α, β)r− 1−α

1−α+σ₂(a, c;α, β)r (|z|=r)

>0.

Thus (2.5) holds true in ∆. This proves the inequality (2.1). The inequality (2.2) follows by taking the convex functiong(z) = _1−z^z =z+P∞

n=2zⁿ in (2.1). To prove the sharpness of the constant _2[1−α+σ^{σ2(a,c;α,β)}

2(a,c;α,β)],we consider the functionf₀(z)∈ L^?(a, c;α, β)given by

(2.6) f₀(z) = z− 1−α

σ₂(a, c;α, β)z² (−1≤α <1; β ≥0).

Thus from (2.1), we have

(2.7) σ₂(a, c;α, β)

2[1−α+σ₂(a, c;α, β)]f₀(z)≺ z 1−z.

It can easily verified that

(2.8) min

Re

σ₂(a, c;α, β)

2[1−α+σ2(a, c;α, β)]f₀(z)

=−1

2 (z ∈∆), This shows that the constant _2[1−α+σ^{σ2(a,c;α,β)}

2(a,c;α,β)] is best possible.

Corollary 2.2. Let the functionf(z)defined by (1.1) be in the class L^?(1,1;α, β)and satisfy the condition

(2.9)

∞

X

n=2

[n(1 +β)−(α+β)]|a_n| ≤1−α then

β+ 2−α

2(β+ 3−2α)(f ∗g)(z)≺g(z) (2.10)

(−1≤α <1; β ≥0; z ∈∆; g ∈ K) and

(2.11) Re(f(z))>−β+ 3−2α

β+ 2−α , (z ∈∆).

The constant _{2(β+3−2α)}^β+2−α is the best estimate.

Corollary 2.3. Let the functionf(z)defined by (1.1) be in the classL^?(1,1;α,0)and satisfy the condition

(2.12)

∞

X

n=2

(n−α)|a_n| ≤1−α, then

(2.13) 2−α

6−4α(f∗g)(z)≺g(z) (z ∈∆; g ∈ K) and

(2.14) Re(f(z))>−3−2α

2−α , (z ∈∆).

The constant _6−4α^2−α is the best estimate.

Puttingα = 0in Corollary 2.3, we obtain

Corollary 2.4 ([12]). Let the function f(z) defined by (1.1) be in the class L^?(1,1; 0,0)and satisfy the condition

(2.15)

∞

X

n=2

n|an| ≤1 then

(2.16) 1

3(f∗g)(z)≺g(z) (z ∈∆; g ∈ K) and

(2.17) Re(f(z))>−3

2, (z ∈∆).

The constant1/3 is the best estimate.

Corollary 2.5. Let the functionf(z)defined by (1.1) be in the class L^?(2,1;α, β)and satisfy the condition

(2.18)

∞

X

n=2

n[n(1 +β)−(α+β)]|a_n| ≤1−α,

then

β+ 2−α

2β+ 5−3α(f∗g)(z)≺g(z) (2.19)

(−1≤α <1; β ≥0; z ∈∆; g ∈ K) and

(2.20) Re(f(z))>−2β+ 5−3α

2(β+ 2−α), (z ∈∆).

The constant _2β+5−3α^β+2−α is the best estimate.

Corollary 2.6. Let the functionf(z)defined by (1.1) be in the classL^?(2,1;α,0)and satisfy the condition

(2.21)

∞

X

n=2

n(n−α)|an| ≤1−α,

then

(2.22) 2−α

5−3α(f∗g)(z)≺g(z) (z ∈∆; g ∈ K) and

(2.23) Re(f(z))>− 5−3α

2(2−α), (z ∈∆).

The constant _5−3α^2−α is the best estimate.

Puttingα = 0in Corollary 2.6, we obtain

Corollary 2.7. Let the functionf(z)defined by (1.1) be in the class L^?(2,1; 0,0)and satisfy the condition

(2.24)

∞

X

n=2

n²|a_n| ≤1

then

(2.25) 2

5(f∗g)(z)≺g(z) (z ∈∆; g ∈ K) and

(2.26) Re(f(z))> −5

4 , (z ∈∆).

The constant2/5is the best estimate.

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