Fractional Calculus of a Class of Univalent Functions With Negative Coefficients Defined By Hadamard Product With Rafid -Operator

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1 EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 4, No. 2, 2, ISSN 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 WR(λ,β,α,,θ), which consists of analytic and univalent functions with negative coefficients in the open unit disk U={ C : <} 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. 2 Mathematics Subject Classifications: 3C45 Key Words and Phrases: Univalent Function, Fractional Calculus, Hadamard Product, Distortion Theorem, Rafid-Operator, Extreme Point.. Introduction Let R denote the class of functions of the form: f()= a n n, (a n, n IN={,2,3, }) () which are analytic and univalent in the unit disk U={ C : <}. If f R is given by () and g R given by g()= b n n, b n then the Hadamard product (or convolution) f g of f and g is defined by f g()= a n b n n =(g f)(). (2) Corresponding author. addresses: Û ÒÝÓÓºÓÑ (W. Atshan), ÊÝÓÓºÓÑ (R. Buti) 62 c 2 EJPAM All rights reserved.

2 W. Atshan, R. Buti/Eur. J. Pure Appl. Math, 4 (2), Lemma. The Rafid -Operator of f R for <, θ is denoted by R θ and defined as following: R θ (f()) = ( ) +θ t θ e t f(t)d t Γ(θ+ ) = K(n,,θ)a n n, (3) where K(n,,θ)= ( )n Γ(θ+n). Γ(θ+) Proof. R θ (f()) = ( ) +θ Γ(θ+ ) = = ( ) +θ Γ(θ+ ) ( ) +θ Γ(θ+ ) t θ e t f(t)d t t θ e t t t θ e t d t a n (t) n d t a n n t θ +n e t d t Let x= t, then if t=, we get x=, t=, we get x= and t=( )x, then d t=( )d x. Thus R θ (f()) = ( ) +θ ( ) +θ e x x θ d x Γ(θ+ ) a n n ( ) θ+n e x x θ +n d x = ( ) +θ ( ) +θ Γ(θ+ ) a n n ( ) θ+n Γ(θ+ n) Γ(θ+ ) ( ) n Γ(θ+ n) = a n n Γ(θ+ ) = K(n,,θ)a n n. Definition. A function f() R, U is said to be in the class WR(λ,β,α,,θ) if and only if satisfies the inequality: (R θ ((f g)())) +λ 2 (R θ ((f g)())) ( λ)r θ ((f g)())+λ(rθ ((f g)()))

3 W. Atshan, R. Buti/Eur. J. Pure Appl. Math, 4 (2), (R θ β ((f g)())) +λ 2 (R θ ((f g)())) ( λ)r θ ((f +α, (4) g)())+λ(rθ ((f g)())) where α<, λ,β, U, <, θ and g() R given by g()= b n n, b n. Lemma 2.[] Let w= u+ iv. Then w σ if and only if w (+σ) w+( σ). Lemma 3.[] Let w= u+ iv andσ,γ are real numbers. Then w>σ w +γ if and only if {w(+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,, θ). 2. Coefficient Bounds and Extreme Points We obtain here a necessary and sufficient condition and extreme points for the functions f() in the class WR(λ,β,α,,θ). Theorem. The function f() defined by () is in the class WR(λ,β,α,,θ) if and only if ( λ+ nλ)[n(+β) (β+α)]k(n,,θ)a n b n, (5) where α<,β, λ, < and θ. Proof. By Definition, we get (R θ ((f g)())) +λ 2 (R θ ((f g)())) ( λ)r θ ((f g)())+λ(rθ ((f g)())) (R θ β ((f g)())) +λ 2 (R θ ((f g)())) ( λ)r θ ((f g)())+λ(rθ ((f g)())) +α. Then by Lemma 3, we have (R θ ((f g)())) +λ 2 (R θ ((f g)())) ( λ)r θ ((f g)())+λ(rθ ((f ) βe iφ α, g)())) (+βeiφ π < φ π, or equivalently, ((R θ ((f g)())) +λ 2 (R θ ((f g)())) )(+βe iφ ) ( λ)r θ ((f g)())+λ(rθ ((f g)()))

4 W. Atshan, R. Buti/Eur. J. Pure Appl. Math, 4 (2), Let and βeiφ (( λ)(r θ ((f g)())+λ2 (R θ ((f g)())) )) ( λ)r θ ((f α. (6) g)())+λ(rθ ((f g)())) F() = [(R θ ((f g)())) +λ 2 (R θ ((f g)())) ](+βe iφ ) By Lemma 2, (6) is equivalent to But Also βe iφ [( λ)(r θ ((f g)()))+λ(rθ m ((f g)())) ], E()=( λ)r θ ((f g)())+λ(rθ ((f g)())). F()+()E() F() (+α)e() for α<. F()+()E() = K(n,,θ)a n b n n λ n(n )K(n,,θ)a n b n n (+βe iφ ) βe iφ ( λ)( K(n,,θ)a n b n n )+λ λ nk(n,,θ)a n b n n +() ( λ+ nλ)k(n,,θ)a n b n n = (2 α) [(n+λn(n ))+()( λ+ nλ)]k(n,,θ)a n b n n βe iφ [n+λn(n ) ( λ+ nλ)]k(n,,θ)a n b n n (2 α) β [(n+λn(n ))+()( λ+λn)]k(n,,θ)a n b n n [n+λn(n 2) +λ]k(n,,θ)a n b n n. F() (+α)e() = nk(n,,θ)a n b n n λ n(n )K(n,,θ)a n b n n (+βe iφ )

5 W. Atshan, R. Buti/Eur. J. Pure Appl. Math, 4 (2), and so βe iφ ( λ) K(n,,θ)a n b n n λ nk(n,,θ)a n b n n (+α) ( λ+ nλ)k(n,,θ)a n b n n = a [(n+λn(n )) (+α)( λ+ nλ)]k(n,,θ)a n b n n βe iφ [n+ nλ(n ) ( λ+ nλ)]k(n,,θ)a n b n n α + +β [(n+ nλ(n )) (+α)( λ+ nλ)]k(n,,θ)a n b n n [n+nλ(n ) ( λ+ nλ)]k(n,,θ)a n b n n F()+()E() F() (+α)e() 2() [(2n+2nλ(n )) 2α( λ+ nλ) β(2n+2nλ(n ) 2( λ+ nλ))]k(n,,θ)a n b n n or [n(+β)+ nλ(n )(+β) ( λ+ nλ)(α+β)]k(n,,θ)a n b n. This is equivalent to ( λ+ nλ)[n(+β) (β+α)]k(n,,θ)a n b n. Conversely, suppose that (5) holds. Then we must show ((R θ ((f g)())) +λ 2 (R θ ((f g)())) )(+βe iφ ) ( λ)r θ ((f g)())+λ(rθ ((f g)())) βeiφ (( λ)(r θ ((f g)())+λ(rθ ((f g)())) )) ( λ)r θ ((f α. g)())+λ(rθ ((f g)()))

6 W. Atshan, R. Buti/Eur. J. Pure Appl. Math, 4 (2), Upon choosing the values of on the positive real axis where =r<, the above inequality reduces to () [n(+βe iφ )( λ+λn) (α+βe iφ )( λ+ nλ)]k(n,,θ)a n b n r n ( λ+ nλ)k(n,,θ)a n b n r n. Since ( e iφ ) e iφ =, the above inequality reduces to () [n(+β)( λ+λn) (α+β)( λ+ nλ)]k(n,,θ)a n b n r n. ( λ+ nλ)k(n,,θ)a n b n r n Letting r, we get desired conclusion. Corollary. Let f() WR(λ,β,α,,θ). Then a n ( λ+ nλ)(n(+β) (β+α))k(n,,θ)b n, where α<,β, λ, <, θ. Theorem 2. Let f ()= and f n ()= ( λ+ nλ)(n(+β) (β+α))k(n,,θ)b n n, where n 2, n IN, α<,β, λ, < and θ. Then f() is in the class WR(λ,β,α,,θ) if and only if it can be expressed in the form n= f()= σ n f n (), n= whereσ n and σ n = or =σ + σ n. Proof. Let f()= σ n f n (), whereσ n and σ n =. Then and we get n= f()= n= ( λ+nλ)(n(+β) (β+α))k(n,,θ)b n σ n n, ( λ+ nλ)(n(+β) (β+α))k(n,,θ)bn

7 W. Atshan, R. Buti/Eur. J. Pure Appl. Math, 4 (2), σ n ( λ+ nλ)(n(+β) (β+α))k(n,,θ)b n = σ n = σ (by Theorem ). By virtue of Theorem, we can show that f() WR(λ,β,α,,θ). Conversely, assume that f() of the form () belongs to WR(λ,β,α,,θ). Then Setting andσ = a n ( λ+nλ)(n(+β) (β+α))k(n,,θ)b n, n IN, n 2. σ n = ( λ+ nλ)(n(+β) (β+α))k(n,,θ)a nb n σ n, we obtain This completes the proof. f()= σ n f n ()=σ f ()+ σ n f n (). n= 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(), by D δ f()= f(t) d t(δ>), (7) Γ(δ) ( t) δ where f() is an analytic function in a simply - connected region of the -plane containing the origin, and the multiplicity of( t) δ is removed by requiring log( t) to be real, when ( t)>. Definition 3 (Fractional derivative operator). The fractional derivative of order δ is defined, for a function f() by D δ f()= Γ( δ) where f() is as in Definition 2. d d f(t) d t ( δ<), (8) ( t) δ

8 W. Atshan, R. Buti/Eur. J. Pure Appl. Math, 4 (2), Definition 4 (Under the condition of Definition 3). The fractional derivative of order k+δ(k=,,2, ) is defined by D k+δ f()= d k d k Dδ f(),( δ<). (9) From Definition 2 and 3 by applying a simple calculation we get D δ f()= Γ(2+δ) δ+ D δ f()= Γ(2 δ) δ Γ(n+) Γ(n++δ) a n n+δ, () Γ(n+) Γ(n+ δ) a n n δ. () Now making use of above (), (), we state and prove the theorems : Theorem 3. Let f() WR(λ,β,α,,θ). Then and D δ f() D δ f() Γ(2+δ) δ+ + Γ(2+δ) δ+ 2(), (2) (2+δ)(+λ)(2+β α)(θ+ )b 2 2(), (3) (2+δ)(+λ)(2+β α)(θ+ )b 2 The inequalities in (2) and (3) are attained for the function given by f()= Proof. By using Theorem, we have By (), we have such that (+λ)(2+β α)(θ+ )b 2 2 (4) a n. (5) (+λ)(2+β α)(θ+ )b 2 Γ(2+δ) δ D δ f()= l(n,δ)a n n, (6) l(n,δ)= Γ(n+)Γ(2+δ), n 2. Γ(n++δ) we know thatl(n,δ) is a decreasing function of n and <l(n,δ) l(2,δ)= 2 2+δ.

9 W. Atshan, R. Buti/Eur. J. Pure Appl. Math, 4 (2), Using (5) and (6), we have Γ(2+δ) δ D δ f() +l(2,δ) 2 which gives (2); we also have which gives (3). + Γ(2+δ) δ D δ f() l(2,δ) 2 Theorem 4. Let f() WR(λ,β,α,,θ). Then and D δ D δ a n 2() (2+δ)(+λ)(2+β α)(θ+ )b 2 2, a n 2() (2+δ)(+λ)(2+β α)(θ+ )b 2 2, δ 2() f() +, (7) Γ(2 δ) (2 δ)(+λ)(2+β α)(θ+ )b 2 δ 2() f(). (8) Γ(2 δ) (2 δ)(+λ)(2+β α)(θ+ )b 2 The inequalities (7) and (8) are attained for the function f() given by (4). Proof. By (), we have Γ(2 δ) δ D δ f() = Γ(n+)Γ(2 δ) a n n Γ(n+ δ) = Φ(n,δ)a n n, whereφ(n,δ)= Γ(n+)Γ(2 δ). For n 2, Φ(n, δ) is a decreasing function of n, then Γ(n+ δ) Also by using (5), we have Φ(n,δ) Φ(2,δ)= Γ(3)Γ(2 δ) Γ(3 δ) Γ(2 δ) δ D δ f() +Φ(2,δ) 2 + = 2Γ(2)Γ(2 δ) (2 δ)γ(2 δ) = 2 2 δ. a n 2() (2 δ)(+λ)(2 β+α)(θ+ )b 2 2.

10 W. Atshan, R. Buti/Eur. J. Pure Appl. Math, 4 (2), Then D δ and by the same way, we obtain D δ δ 2() f() +, Γ(2 δ) (2 δ)(+λ)(2 β+α)(θ+ )b 2 δ 2() f(). Γ(2 δ) (2 δ)(+λ)(2 β+α)(θ+ )b 2 Corollary 2. For every f WR(λ,β,α,,θ), we have 2 2() (9) 2 3(+λ)(2 β+α)(θ+ )b 2 2() f(t)d t 2 +, (2) 2 3(+λ)(2 β+α)(θ+ )b 2 and () f() (2) (+λ)(2 β+α)(θ+ )b 2 + () (+λ)(2 β+α)(θ+ )b 2, (22) Proof. (i) By Definition 2 and Theorem 3 forδ=we have D f()= true. (ii) By Definition 3 and Theorem 4 forδ=, we have the result is true. D f()= d d f(t)d t= f(), f(t)d t, the result is Corollary 3. D δ f() and D δ f() are included in the disk with center at the origin and radii 2() +, Γ(2+δ) (2+δ)(+λ)(2 β+α)(θ+ )b 2 and 2() +. Γ(2 δ) (2 δ)(+λ)(2 β+α)(θ+ )b 2

11 W. Atshan, R. Buti/Eur. J. Pure Appl. Math, 4 (2), Theorem 5. Let 4. Hadamard Product f()= a n n, g()= b n n belong to W R(λ, β, α,, θ). Then the Hadamard product of f and g given by (f g)()= a n b n n belongs to WR(λ,β,α,,θ). and Proof. Since f and g WR(λ,β,α,,θ), we have ( λ+ nλ)[n(+β) (α+β)]k(n,,θ)bn a n ( λ+ nλ)[n(+β) (α+β)]k(n,,θ)an b n and by applying the Cauchy- Schwar inequality, we have ( λ+ nλ)[n(+β) (α+β)]k(n,,θ) a n b n a n b n /2 ( λ+ nλ)[n(+β) (α+β)]k(n,,θ)bn a n /2 ( λ+ nλ)[n(+β) (α+β)]k(n,,θ)an b n. However, we obtain ( λ+ nλ)[n(+β) (α+β)]k(n,,θ) a n b n a n b n. Now, we want to prove ( λ+ nλ)[n(+β) (α+β)]k(n,,θ) a n b n. Since ( λ+ nλ)[n(+β) (α+β)]k(n,,θ) a n b n = ( λ+nλ)[n(+β) (α+β)]k(n,,θ) a n b n a n b n. Hence, we get the required result.

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