Fractional Calculus of a Class of Univalent Functions With Negative ...

EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 4, No. 2, 2011, 162-173 ISSN 1307-5543 – www.ejpam.com

Fractional Calculus of a Class of Univalent Functions With Negative Coefficients Defined By Hadamard Product With Rafid -Operator Waggas Galib Atshan∗ , Rafid Habib Buti Department of Mathematics, College of Computer Science and Mathematics, University of Al-Qadisiya, Diwaniya, Iraq

Abstract. In our paper, we study a class W R (λ, β , α, µ, θ ), which consists of analytic and univalent functions with negative coefficients in the open unit disk U = {z ∈ C : |z| < 1} defined by Hadamard product (or convolution) with Rafid - Operator, we obtain coefficient bounds and extreme points for this class. Also distortion theorem using fractional calculus techniques and some results for this class are obtained. 2000 Mathematics Subject Classifications: 30C45 Key Words and Phrases: Univalent Function, Fractional Calculus, Hadamard Product, Distortion Theorem, Rafid-Operator, Extreme Point.

1. Introduction Let R denote the class of functions of the form: f (z) = z −

∞ X

an z n , (an ≥ 0, n ∈ IN = {1, 2, 3, · · · })

(1)

n=2

which are analytic and univalent in the unit disk U = {z ∈ C : |z| < 1}. If f ∈ R is given by (1) and g ∈ R given by ∞ X g(z) = z − bn z n , bn ≥ 0 n=2

then the Hadamard product (or convolution) f ∗ g of f and g is defined by f ∗ g(z) = z −

∞ X

an bn z n = (g ∗ f )(z).

(2)

n=2 ∗

Corresponding author.

Email addresses: waggashndyahoo. om (W. Atshan), Rafidhbyahoo. om (R. Buti) http://www.ejpam.com

162

c 2011 EJPAM All rights reserved.

W. Atshan, R. Buti / Eur. J. Pure Appl. Math, 4 (2011), 162-173

163

Lemma 1. The Rafid -Operator of f ∈ R for 0 ≤ µ < 1, 0 ≤ θ ≤ 1 is denoted by Rθµ and defined as following: Z∞ t 1 θ θ −1 − 1−µ Rµ ( f (z)) = t e f (zt)d t (1 − µ)1+θ Γ(θ + 1) 0 ∞ X = z− K(n, µ, θ )an z n , (3) n=2

where K(n, µ, θ ) =

(1−µ)n−1 Γ(θ +n) . Γ(θ +1)

Proof. Rθµ ( f

(z)) = =

=

∞

Z

1 (1 − µ)1+θ Γ(θ + 1) 1

t θ −1 e 0 Z∞

1 (1 − µ)1+θ Γ(θ + 1)

−

t 1−µ

t 1−µ

f (zt)d t 

t θ −1 e

(1 − µ)1+θ Γ(θ + 1)

−

zt −

0

 Z z

∞ X

 an (zt)n  d t

n=2 ∞

θ −

t e 0

t 1−µ

dt −

∞ X

Z

∞

an z n

n=2

 t

θ −1+n −

e

t 1−µ

d t

0

t Let x = 1−µ , then if t = 0, we get x = 0, t = ∞, we get x = ∞ and t = (1 − µ)x, then d t = (1 − µ)d x. Thus Z∞ 1 θ z Rµ ( f (z)) = (1 − µ)1+θ e−x x θ d x (1 − µ)1+θ Γ(θ + 1) 0  Z∞ ∞ X an z n (1 − µ)θ +n e−x x θ −1+n d x  − 0

n=2

=



1 (1 − µ)1+θ Γ(θ + 1)

= z−

∞ X (1 − µ)n−1 Γ(θ + n) n=2

= z−

z(1 − µ)1+θ Γ(θ + 1) −

∞ X

Γ(θ + 1)

∞ X

 an z n (1 − µ)θ +nΓ(θ + n)

n=2

an z n

K(n, µ, θ )an z n .

n=2

Definition 1. A function f (z) ∈ R, z ∈ U is said to be in the class W R(λ, β, α, µ, θ ) if and only if satisfies the inequality: ( ) z(Rθµ (( f ∗ g)(z)))′ + λz 2 (Rθµ (( f ∗ g)(z)))′′ Re (1 − λ)Rθµ (( f ∗ g)(z)) + λz(Rθµ (( f ∗ g)(z)))′


z(Rθ (( f ∗ g)(z)))′ + λz 2 (Rθ (( f ∗ g)(z)))′′ µ µ ≥β − 1 + α, (1 − λ)Rθµ (( f ∗ g)(z)) + λz(Rθµ (( f ∗ g)(z)))′

164

(4)

where 0 ≤ α < 1, 0 ≤ λ ≤ 1, β ≥ 0, z ∈ U, 0 ≤ µ < 1, 0 ≤ θ ≤ 1 and g(z) ∈ R given by g(z) = z −

∞ X

bn z n , bn ≥ 0.

n=2

Lemma 2. [1] Let w = u + iv. Then Re w ≥ σ if and only if |w − (1 + σ)| ≤ |w + (1 − σ)|. Lemma 3. [1] Let w = u + iv and σ, γ are real numbers. Then Re w > σ|w − 1| + γ if and only if Re {w(1 + σe iφ ) − σe iφ } > γ. We aim to study the coefficient bounds, extreme points, application of fractional calculus and Hadamard product of the class W R(λ, β, α, µ, θ ).

2. Coefficient Bounds and Extreme Points We obtain here a necessary and sufficient condition and extreme points for the functions f (z) in the class W R(λ, β, α, µ, θ ). Theorem 1. The function f (z) defined by (1) is in the class W R(λ, β, α, µ, θ ) if and only if ∞ X (1 − λ + nλ)[n(1 + β) − (β + α)]K(n, µ, θ )an bn ≤ 1 − α, n=2

where 0 ≤ α < 1, β ≥ 0, 0 ≤ λ ≤ 1, 0 ≤ µ < 1 and 0 ≤ θ ≤ 1. Proof. By Definition 1, we get ( ) z(Rθµ (( f ∗ g)(z)))′ + λz 2 (Rθµ (( f ∗ g)(z)))′′ Re (1 − λ)Rθµ (( f ∗ g)(z)) + λz(Rθµ (( f ∗ g)(z)))′ z(Rθ (( f ∗ g)(z)))′ + λz 2 (Rθ (( f ∗ g)(z)))′′ µ µ ≥β − 1 + α. (1 − λ)Rθµ (( f ∗ g)(z)) + λz(Rθµ (( f ∗ g)(z)))′ Then by Lemma 3, we have ( ) z(Rθµ (( f ∗ g)(z)))′ + λz 2 (Rθµ (( f ∗ g)(z)))′′ Re (1 + β e iφ ) − β e iφ ≥ α, (1 − λ)Rθµ (( f ∗ g)(z)) + λz(Rθµ (( f ∗ g)(z)))′ −π < φ ≤ π, or equivalently, ( (z(Rθµ (( f ∗ g)(z)))′ + λz 2 (Rθµ (( f ∗ g)(z)))′′ )(1 + β e iφ ) Re (1 − λ)Rθµ (( f ∗ g)(z)) + λz(Rθµ (( f ∗ g)(z)))′

(5)


−

165

β e iφ ((1 − λ)(Rθµ (( f ∗ g)(z)) + λz 2 (Rθµ (( f ∗ g)(z)))′ ))

)

(1 − λ)Rθµ (( f ∗ g)(z)) + λz(Rθµ (( f ∗ g)(z)))′

≥ α.

Let F(z) = [z(Rθµ (( f ∗ g)(z)))′ + λz 2 (Rθµ (( f ∗ g)(z)))′′ ](1 + β e iφ ) −β e iφ [(1 − λ)(Rθµ (( f ∗ g)(z))) + λz(Rθm (( f ∗ g)(z)))′ ], and

E(z) = (1 − λ)Rθµ (( f ∗ g)(z)) + λz(Rθµ (( f ∗ g)(z)))′ .

By Lemma 2, (6) is equivalent to |F(z) + (1 − α)E(z)| ≥ |F(z) − (1 + α)E(z)| for 0 ≤ α < 1. But |F(z) + (1 − α)E(z)|   ∞ ∞ X X K(n, µ, θ )an bn z n − λ n(n − 1)K(n, µ, θ )an bn z n  (1 + β e iφ ) = z − n=2 n=2   ∞ ∞ X X −β e iφ (1 − λ)(z − K(n, µ, θ )an bn z n ) + λz − λ nK(n, µ, θ )an bn z n  n=2

n=2

 +(1 − α) z − (1 − λ + nλ)K(n, µ, θ )an bn z n  n=2 ∞ X = (2 − α)z − [(n + λn(n − 1)) + (1 − α)(1 − λ + nλ)]K(n, µ, θ )an bn z n n=2 ∞ X iφ n −β e [n + λn(n − 1) − (1 − λ + nλ)]K(n, µ, θ )an bn z n=2 

∞ X

∞ X ≥ (2 − α)|z| − [(n + λn(n − 1)) + (1 − α)(1 − λ + λn)]K(n, µ, θ )an bn |z|n n=2

−β

∞ X

[n + λn(n − 2) − 1 + λ]K(n, µ, θ )an bn |z|n .

n=2

Also |F(z) − (1 + α)E(z)|   ∞ ∞ X X  n n = z− nK(n, µ, θ )an bn z − λ n(n − 1)K(n, µ, θ )an bn z (1 + β e iφ ) n=2 n=2

(6)


 −β e iφ z − (1 − λ)

∞ X n=2

K(n, µ, θ )an bn z n − λ

166 ∞ X

 nK(n, µ, θ )an bn z n 

n=2

 −(1 + α) z − (1 − λ + nλ)K(n, µ, θ )an bn z n  n=2 ∞ X [(n + λn(n − 1)) − (1 + α)(1 − λ + nλ)]K(n, µ, θ )an bn z n = −az − n=2 ∞ X −β e iφ [n + nλ(n − 1) − (1 − λ + nλ)]K(n, µ, θ )an bn z n n=2 

≤ α|z| +

∞ X

∞ X

[(n + nλ(n − 1)) − (1 + α)(1 − λ + nλ)]K(n, µ, θ )an bn |z|n

n=2 ∞ X +β [n + nλ(n − 1) − (1 − λ + nλ)]K(n, µ, θ )an bn |z|n n=2

and so |F(z) + (1 − α)E(z)| − |F(z) − (1 + α)E(z)| ≥ 2(1 − α)|z| ∞ X − [(2n + 2nλ(n − 1)) − 2α(1 − λ + nλ) − β(2n + 2nλ(n − 1) n=2

−2(1 − λ + nλ))]K(n, µ, θ )an bn |z|n ≥ 0 or ∞ X

[n(1 + β) + nλ(n − 1)(1 + β) − (1 − λ + nλ)(α + β)]K(n, µ, θ )an bn ≤ 1 − α.

n=2

This is equivalent to ∞ X (1 − λ + nλ)[n(1 + β) − (β + α)]K(n, µ, θ )an bn ≤ 1 − α. n=2

Conversely, suppose that (5) holds. Then we must show ( (z(Rθµ (( f ∗ g)(z)))′ + λz 2 (Rθµ (( f ∗ g)(z)))′′ )(1 + β e iφ ) Re (1 − λ)Rθµ (( f ∗ g)(z)) + λz(Rθµ (( f ∗ g)(z)))′ ) β e iφ ((1 − λ)(Rθµ (( f ∗ g)(z)) + λz(Rθµ (( f ∗ g)(z)))′ )) − ≥ α. (1 − λ)Rθµ (( f ∗ g)(z)) + λz(Rθµ (( f ∗ g)(z)))′


167

Upon choosing the values of z on the positive real axis where 0 ≤ z = r < 1, the above inequality reduces to   ∞ P iφ iφ n−1 (1 − α) − [n(1 + β e )(1 − λ + λn) − (α + β e )(1 − λ + nλ)]K(n, µ, θ )an bn r     n=2 ≥ 0. Re ∞   P   n−1 1− (1 − λ + nλ)K(n, µ, θ )an bn r n=2

Since Re(−e iφ ) ≥ −|e iφ | = −1, the above inequality reduces to   ∞ P n−1 (1 − α) − [n(1 + β)(1 − λ + λn) − (α + β)(1 − λ + nλ)]K(n, µ, θ )a b r   n n   n=2 Re ≥ 0. ∞   P   1− (1 − λ + nλ)K(n, µ, θ )an bn r n−1 n=2

Letting r → 1− , we get desired conclusion. Corollary 1. Let f (z) ∈ W R(λ, β, α, µ, θ ). Then an ≤

1−α (1 − λ + nλ)(n(1 + β) − (β + α))K(n, µ, θ )bn

,

where 0 ≤ α < 1, β ≥ 0, 0 ≤ λ ≤ 1, 0 ≤ µ < 1, 0 ≤ θ ≤ 1. Theorem 2. Let f1 (z) = z and f n (z) = z −

1−α (1 − λ + nλ)(n(1 + β) − (β + α))K(n, µ, θ )bn

zn,

where n ≥ 2, n ∈ IN , 0 ≤ α < 1, β ≥ 0, 0 ≤ λ ≤ 1, 0 ≤ µ < 1 and 0 ≤ θ ≤ 1. Then f (z) is in the class W R(λ, β, α, µ, θ ) if and only if it can be expressed in the form f (z) =

∞ X

σn f n (z),

n=1

where σn ≥ 0 and

∞ P

σn = 1 or 1 = σ1 +

n=1

Proof. Let f (z) =

∞ P

σn .

n=2 ∞ P

σn f n (z), where σn ≥ 0 and

n=1

f (z) = z −

∞ P

σn = 1. Then

n=1 ∞ X

1−α

(1 − λ + nλ)(n(1 + β) − (β + α))K(n, µ, θ )bn n=2

σn z n ,

and we get ∞ X (1 − λ + nλ)(n(1 + β) − (β + α))K(n, µ, θ )b n

n=2

1−α

×


1−α

σn =

168

(1 − λ + nλ)(n(1 + β) − (β + α))K(n, µ, θ )bn

∞ X

σn = 1 − σ1 ≤ 1 (by Theorem 1).

n=2

By virtue of Theorem 1, we can show that f (z) ∈ W R(λ, β, α, µ, θ ). Conversely, assume that f (z) of the form (1) belongs to W R(λ, β, α, µ, θ ). Then an ≤

1−α (1 − λ + nλ)(n(1 + β) − (β + α))K(n, µ, θ )bn

Setting σn = and σ1 = 1 −

∞ P

, n ∈ IN , n ≥ 2.

(1 − λ + nλ)(n(1 + β) − (β + α))K(n, µ, θ )an bn 1−α

σn , we obtain

n=2

f (z) =

∞ X n=1

σn f n (z) = σ1 f1 () +

∞ X

σn f n (z).

n=2

This completes the proof.

3. Application of the Fractional Calculus Various operators of fractional calculus (that is, fractional derivative and fractional integral) have been rather extensively studied by many researcher (c.f. [3-5]). However, we try to restrict ourselves to the following definitions given by Owa [2] for convenience. Definition 2 (Fractional integral operator). The fractional integral of order δ is defined, for a function f (z) , by Zz 1 f (t) −δ Dz f (z) = d t (δ > 0), (7) Γ(δ) 0 (z − t)1−δ where f (z) is an analytic function in a simply - connected region of the z-plane containing the origin, and the multiplicity of (z − t)δ−1 is removed by requiring log(z − t) to be real, when (z − t) > 0. Definition 3 (Fractional derivative operator). The fractional derivative of order δ is defined, for a function f (z) by Zz 1 d f (t) δ Dz f (z) = d t‘ (0 ≤ δ < 1), (8) Γ(1 − δ) dz 0 (z − t)δ where f (z) is as in Definition 2.


169

Definition 4 (Under the condition of Definition 3). The fractional derivative of order k + δ (k = 0, 1, 2, · · · ) is defined by Dzk+δ f (z) =

dk dz k

Dzδ f (z), (0 ≤ δ < 1).

(9)

From Definition 2 and 3 by applying a simple calculation we get Dz−δ f (z) =

Dzδ f (z) =

1 Γ(2 + δ) 1 Γ(2 − δ)

z δ+1 −

z 1−δ −

∞ X

Γ(n + 1)

an z n+δ ,

(10)

an z n−δ .

(11)

Γ(n + 1 + δ) n=2

∞ X

Γ(n + 1)

Γ(n + 1 − δ) n=2

Now making use of above (10), (11), we state and prove the theorems : Theorem 3. Let f (z) ∈ W R(λ, β, α, µ, θ ). Then 2(1 − α) 1 −δ δ+1 |Dz f (z)| ≤ |z| 1+ |z| , Γ(2 + δ) (2 + δ)(1 + λ)(2 + β − α)(θ + 1)b2 and |Dz−δ f

(z)| ≥

1 Γ(2 + δ)

|z|

δ+1

1−

2(1 − α)

(12)

(2 + δ)(1 + λ)(2 + β − α)(θ + 1)b2

|z| ,

(13)

The inequalities in (12) and (13) are attained for the function given by f (z) = z −

1−α

z2

(14)

.

(15)

(1 + λ)(2 + β − α)(θ + 1)b2

Proof. By using Theorem 1, we have ∞ X n=2

an ≤

1−α (1 + λ)(2 + β − α)(θ + 1)b2

By (10), we have Γ(2 + δ)z −δ Dz−δ f (z) = z −

∞ X

ℓ(n, δ)an z n ,

n=2

such that ℓ(n, δ) =

Γ(n + 1)Γ(2 + δ) Γ(n + 1 + δ)

, n ≥ 2.

we know that ℓ(n, δ) is a decreasing function of n and 0 < ℓ(n, δ) ≤ ℓ(2, δ) =

2 2+δ

.

(16)


170

Using (15) and (16), we have |Γ(2 + δ)z −δ Dz−δ f (z)| ≤ |z| + ℓ(2, δ)|z|2

∞ X

an

n=2

≤ |z| +

2(1 − α) (2 + δ)(1 + λ)(2 + β − α)(θ + 1)b2

|z|2 ,

which gives (12); we also have |Γ(2 + δ)z

−δ

Dz−δ f

(z)| ≥ |z| − ℓ(2, δ)|z|

2

∞ X

an

n=2

≥ |z| −

2(1 − α) (2 + δ)(1 + λ)(2 + β − α)(θ + 1)b2

|z|2 ,

which gives (13). Theorem 4. Let f (z) ∈ W R(λ, β, α, µ, θ ). Then |Dzδ f

(z)| ≤

and |Dzδ f

(z)| ≥

|z|1−δ Γ(2 − δ) |z|1−δ Γ(2 − δ)

1+

1−

2(1 − α)

(2 − δ)(1 + λ)(2 + β − α)(θ + 1)b2 2(1 − α)

|z| ,

(2 − δ)(1 + λ)(2 + β − α)(θ + 1)b2

|z| .

The inequalities (17) and (18) are attained for the function f (z) given by (14). Proof. By (11), we have Γ(2 − δ)z δ Dzδ f (z) = z −

∞ X Γ(n + 1)Γ(2 − δ)

Γ(n + 1 − δ)

n=2

= z−

∞ X

an z n

Φ(n, δ)an z n ,

n=2

where Φ(n, δ) =

Γ(n+1)Γ(2−δ) . Γ(n+1−δ)

For n ≥ 2, Φ(n, δ) is a decreasing function of n, then

Φ(n, δ) ≤ Φ(2, δ) =

Γ(3)Γ(2 − δ) Γ(3 − δ)

=

2Γ(2)Γ(2 − δ) (2 − δ)Γ(2 − δ)

=

2 2−δ

.

Also by using (15), we have |Γ(2 − δ)z δ Dzδ f (z)| ≤ |z| + Φ(2, δ)|z|2

∞ X

an

n=2

≤ |z| +

(17)

2(1 − α) (2 − δ)(1 + λ)(2 − β + α)(θ + 1)b2

|z|2 .

(18)


Then |Dzδ f (z)| ≤

|z|1−δ Γ(2 − δ)

1+

171

2(1 − α) (2 − δ)(1 + λ)(2 − β + α)(θ + 1)b2

|z| ,

and by the same way, we obtain |Dzδ f

(z)| ≥

|z|1−δ Γ(2 − δ)

1−

2(1 − α)

(2 − δ)(1 + λ)(2 − β + α)(θ + 1)b2

|z| .

Corollary 2. For every f ∈ W R(λ, β, α, µ, θ ), we have |z|2

2(1 − α)

1− |z| 2 3(1 + λ)(2 − β + α)(θ + 1)b2 Z |z|2 z 2(1 − α) f (t)d t ≤ |z| , ≤ 1+ 0 2 3(1 + λ)(2 − β + α)(θ + 1)b2

(19) (20)

and (1 − α) |z| 1 − |z| ≤ | f (z)| (1 + λ)(2 − β + α)(θ + 1)b2 (1 − α) |z| , ≤ |z| 1 + (1 + λ)(2 − β + α)(θ + 1)b2

(21) (22)

Proof. (i) By Definition 2 and Theorem 3 for δ = 1 we have Dz−1 f (z) = true.

Rz 0

f (t)d t, the result is

(ii) By Definition 3 and Theorem 4 for δ = 0, we have Zz d 0 Dz f (z) = f (t)d t = f (z), dz 0 the result is true. Corollary 3. Dz−δ f (z) and Dzδ f (z) are included in the disk with center at the origin and radii 1 Γ(2 + δ) and

1 Γ(2 − δ)

1+

1+

2(1 − α)

,

(2 + δ)(1 + λ)(2 − β + α)(θ + 1)b2 2(1 − α) (2 − δ)(1 + λ)(2 − β + α)(θ + 1)b2

.


172

4. Hadamard Product Theorem 5. Let f (z) = z −

∞ X

n

an z , g(z) = z −

n=2

∞ X

bn z n

n=2

belong to W R(λ, β, α, µ, θ ). Then the Hadamard product of f and g given by ∞ P ( f ∗ g)(z) = z − an bn z n belongs to W R(λ, β, α, µ, θ ). n=2

Proof. Since f and g ∈ W R(λ, β, α, µ, θ ), we have ∞ X (1 − λ + nλ)[n(1 + β) − (α + β)]K(n, µ, θ )bn 1−α

n=2

and

an ≤ 1

∞ X (1 − λ + nλ)[n(1 + β) − (α + β)]K(n, µ, θ )a n

1−α

n=2

bn ≤ 1

and by applying the Cauchy- Schwarz inequality, we have   p ∞ X (1 − λ + nλ)[n(1 + β) − (α + β)]K(n, µ, θ ) an bn p   an bn 1 − α n=2 !1/2 ∞ X (1 − λ + nλ)[n(1 + β) − (α + β)]K(n, µ, θ )bn ≤ an 1−α n=2 !1/2 ∞ X (1 − λ + nλ)[n(1 + β) − (α + β)]K(n, µ, θ )an × bn . 1−α n=2 However, we obtain   p ∞ X (1 − λ + nλ)[n(1 + β) − (α + β)]K(n, µ, θ ) an bn p   an bn ≤ 1. 1 − α n=2 Now, we want to prove ∞ X (1 − λ + nλ)[n(1 + β) − (α + β)]K(n, µ, θ ) 1−α

n=2

an bn ≤ 1.

Since ∞ X (1 − λ + nλ)[n(1 + β) − (α + β)]K(n, µ, θ ) 1−α

n=2

=

∞ X



an bn

(1 − λ + nλ)[n(1 + β) − (α + β)]K(n, µ, θ )

 n=2

Hence, we get the required result.

1−α

p

an bn

 p 

an bn .

REFERENCES

173

References [1] E. S. Aqlan, Some Problems Connected with Geometric Function Theory, Ph.D. Thesis, Pune University, Pune (unpublished), (2004). [2] S. Owa, On the distortion theorems, Kyungpook Math. J., 18: 53-59, 1978. [3] H. M. Srivastava and R. G. Buschman, Convolution integral equation with special function kernels, John Wiley and Sons, New York, London, Sydney and Toronto, 1977. [4] H. M. Srivastava and S. Owa, An application of the fractional derivative, Math. Japon, 29:384-389, 1984. [5] H. M. Srivastava and S. Owa, (Editors), Univalent Functions, Fractional Calculus and Their Applications, Halsted press (Ellis Harwood Limited, Chichester), John Wiley and Sons, New York, Chichester, Brisbane and Toronto, 1989.