On new class of analytic functions defined by using differintegral operator

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ISSN: 2347-2529

Available online atwww.ijaamm.com

International Journal of Advances in

Applied Mathematics and Mechanics

On new class of analytic functions defined by using

differintegral operator

Research Article

Jumana Hekma Salman

1

_{, Ahmed Sallal Joudah}

2,∗

1_{Department of Statistical and Informatics,College of Computer Science and Mathematics, University of Al-Qadisiya,}

Diwiniya-Iraq

2_{Department of Medical Mathematics,College of Computer Science and Mathematics, University of Al-Qadisiya, Diwiniya- Iraq}

Received 22 May 2014; accepted (in revised version) 23 August 2014

Abstract:

Making use a fractional differintegral operator .We introduce a new class of univalent analytic functions in the unit diskU.Among the results investigated for this class of functions include the coefficient inequalities ,star-likenees,convexity,close -to-convexity,extreme points and Integral means inequalities.

MSC:

30C45• 30C50• 26A33

Keywords:

Fractional Calculus(fractional integral and fractional derivative)•Univalent functions•Principle of subordina-tion

c

1. Introduction

Letℑdenote the class of analytic functions in the open unit diskU ={z∈C :|z|<1},and of the form:

f(z) =z+

∞

X

n=2

anzn, (n∈N ={1, 2, 3, . . .}) (1)

LetHdenote the subclass of the classℑwhich is the functions defined by:

f(z) =z−

∞

X

n=2

anzn, (an≥0,n∈N) (2)

A function f ∈H is said to be in the class of starlike functions of orderβ(0≤β <1)inU, denoted byδ∗(β),if

R e

§_{z f}′_(z)

f(z)

ª

> β (z∈U) (3)

Forβ=0 the classδ∗(0) =δ∗is the class of starlike functions inU. (for details, see[1,2])

Iff andgare two analytic inU , then we say that the functionf(z)is subordinate tog(z)inUand writef(z)≺g(z)

,if there exits a Schwarz functionw(z)(analytic inU withw(0) =0,a n d|w(z)|<1),such thatf(z) =g(w(z)),z∈U. In particular .if the function g(z)is univalent inU , the above subordination is equivalent to f(0) = g(0) and

f(U)⊂g(U).[3–5]

We recall here the following definition of fractional calculus (that is fractional integral and fractional derivative of an arbitrary order)consider by Owa[6]. See also[7–10]

∗ _{Corresponding author.}

E-mail address:ahmedhiq@yahoo.com

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Definition 1.1.

The fractional integral of orderλ(λ >0)is defined for a function f(z)analytic in a simply-connected region of the complex plane containing the origin by

D_z−λf(z) = 1

Γ(λ)

Zz

0

f(ξ)

(z−ξ)1−λdξ, (4)

where the multiplicity(z−ξ)λ−1_{is removed by requiring}_{l o g}_(z₋_ξ₎_{to be real when}_(z₋_{ξ >}₀₎_.

Definition 1.2.

Under the hypothesis of definition1.2the fractional derivative operator of orderλ(λ≥0)is defined by :

D_zλf(z) =

   

  

1

Γ(λ) d d z

Z z

0

f(ξ)

(z−ξ)λdξ (0

¶_{λ <}₁₎

dn

d znD

λ−n

z f(z) (n−1¶λ <n;n∈N0=N∪ {0})

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where the multiplicity(z−ξ)−λ_{is removed as in definition}_1.2

For the purpose of this paper, we define here a fractional differintegral operator

Ωλz:H→H (−∞< λ <2;n∈N),

for a functionf(z)of the form(2)by:

Ωλ_zf(z) =z−

∞

X

n=2

Γ(2−λ)_Γ(n+1)

Γ(n−λ+1) anz

n₌

Γ(2−λ)zλD_zλf(z) (−∞< λ <2;z∈U), (6)

Definition 1.3.

A functionf(z)in H is in the classH(λ,γ,β,α,µ)if it satisfies the condition

R e

¨

(1−γ)−γ βΩ

λ

zf(z) + (1−β)z

+ (1−γ)z2 Ωλ

zf(z) ′′

z Ωλ

zf(z) ′

« >1

2(α+µ), (7)

where 0≤αµ <1; 0≤β,γ≤1;−∞< λ <2;z∈U.

Lemma 1.1.

see[11]Ifζis any complex number, then R e(ζ)> τif and only if|ζ−(1+τ)|<|ζ+ (1−τ)|whereτ≥0.

Some of the following properties have been found on other classes in[12–14].

2. Main results

Theorem 2.1.

Let the functions f ∈H be given by(2).Then f ∈H(λ,γ,β,α,µ)if and only if

∞

X

n=2

n

(1−γ)n−1

2(α+µ)

+γβ

Γ(2−λ)_Γ(n+1)

Γ(n−λ+1) an≤1−

1

2(α+µ). (8)

The result is sharp for the function

f(z) =z− 1−

1 2(α+µ)

n (1−γ)n−1₂(α+µ)+γβΓ(2_Γ−_(nλ₋)Γ_λ(n+₊₁₎1)z

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Proof. Suppose thatf ∈H(λ,γ,β,α,µ)by Using Lemma1.1and letting

ζ=(1−γ)z Ω

λ

zf(z) ′

−γ β(_Ωλ

zf(z)) + (1−β)z

+ (1−γ)z2 Ωλzf(z) ′′

z Ωλ

zf(z)

′ =

A(z) B(z).

Then it is sufficient prove that

A(z)−

1+1

2(α+µ) B(z) < A(z) +

1−1

2(α+µ) B(z) = −1

2(α+µ)z−

∞

X

n=2

anzn

Γ(2−λ)_Γ(n+1)

Γ(n−λ+1)

n2(1−γ) +γβ−n

1+1

2(α+µ) <

2−1 2(α+µ)

z−

∞

X

n=2

anzn

Γ(2−λ)Γ(n+1)

Γ(n−λ+1)

n2(1−γ) +γβ+n

1−1

2(α+µ)

which is equivalent to

∞ X n=2 n

(1−γ)n−1

2(α+µ)

+γβ

Γ(2−λ)_Γ(n+1)

Γ(n−λ+1) an−

1−1

2(α+µ)

≤0.

Conversely, assume that

R e

¨

(1−γ)−γ βΩ

λ

zf(z) + (1−β)z

+ (1−γ)z2 Ωλzf(z) ′′

z Ωλ

zf(z) ′

«

=R e

¨

(1−γ)z Ωλ_zf(z)′−γ β(_Ωλ

zf(z)) + (1−β)z

+ (1−γ)z2Ωλ_zf(z)′′ z Ωλ

zf(z) ′

«

=R e

       z− ∞ P n=2

Γ(2−λ)Γ(n+1)

Γ(n−λ+1) anzn

n2(1−γ) +γβ

z−

∞

P n=2

Γ(2−λ)Γ(n+1)

Γ(n−λ+1) n anzn

       (10)

We can choose the value 0fzon the real axis , so thatz Ωλ

zf(z) ′

is real .Letz→1. through real value , so we can write (10)as ∞ X n=2 n

(1−γ)n−1

2(α+µ)

+γβ

_Γ₍₂₋_λ₎_Γ_(n₊₁₎

Γ(n−λ+1) an≤1−

1 2(α+µ).

Finally, sharpness follows if we take

f(z) =z− 1−

1 2(α+µ)

n_,_n_≥_2. ₍₁₁₎

The proof is complete.

Corollary 2.1.

Let the function f ∈H be given by(2). If f ∈H(λ,γ,β,α,µ), then

an≤

1−1 2(α+µ)

n_,_n_≥_2. _(n_∈_N₎ ₍₁₂₎

The result is sharp for the function given by(9).

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3. Radius of starlikeness and convexity and close-to-convexity

Theorem 3.1.

Let f ∈H(λ,γ,β,α,µ).Then f is starlike of orderρ, 0≤ρ <1in|z|<r=r1(λ,γ,β,α,µ,ρ),where

r1(λ,γ,β,α,µ,ρ) =i n fn ¨

(1−ρ)n (1−γ)n−1₂(α+µ)+γβΓ(_Γ2_(n−λ₋)Γ_λ(n+₊₁₎1) 1−1

2(α+µ)

(n−ρ)

«n1−1

,n=2, 3, . . . .

The estimate is sharp for the function

f(z) =z− 1−

1 2(α+µ)

n (1−γ)n−₂1(α+µ)+γβΓ(2_Γ−_(nλ₋)Γ_λ(n+₊₁₎1)z n_,

n≥2 (13)

Proof. f is starlike of orderρ, 0≤ρ <1 if

R e

§_{z f}′_(z)

f(z)

ª

> ρ (14)

that is if

z f′(z) f(z) −1

≤1−ρ, (15)

which simplies to

∞

X

n=2

(n−ρ)an|z|n−1

(1−ρ) ≤1. (16)

By Theorem2.1, we have

an≤

1−1 2(α+µ)

n (1−γ)n−1₂(α+µ)+γβΓ(_Γ2−_(nλ₋)Γ_λ(n+₊₁₎1),n≥2. (17)

Using(16)and(17), we get

|z|n−1≤(1−ρ)

n (1−γ)n−₂1(α+µ)+γβΓ(_Γ2−_(nλ₋)Γ_λ(n+₊₁₎1) 1−1

2(α+µ)

(n−ρ) (18)

thus

|z|<r1(λ,γ,β,α,µ,ρ) =i n fn ¨

(1−ρ)n (1−γ)n−1₂(α+µ)+γβΓ(2_Γ−_(nλ₋)Γ_λ(₊n₁+₎1) 1−1

2(α+µ)

(n−ρ)

«n1−1

,n=2, 3, . . . .

Theorem 3.2.

Let f ∈H(λ,γ,β,α,µ).Then f is convex of orderρ, 0≤ρ <1in|z|<r=r2(λ,γ,β,α,µ,ρ),where

r2(λ,γ,β,α,µ,ρ) =i n fn ¨

2(α+µ)

n(n−ρ)

«n1−1

,n=2, 3, . . . .

f(z) =z− 1−

1 2(α+µ)

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Proof. f ∈H(λ,γ,β,α,µ)is convex of orderρ, 0≤ρ <1 if

R e

§ 1+z f

′′_(z₎

f′_(z₎

ª

> ρ (20)

that is if

z f′′(z) f′_(z₎

≤1−ρ, (21)

which simplies to

∞

X

n=2

n(n−ρ)an|z|n−1

(1−ρ) ≤1. (22)

By Theorem2.1, we have

an≤

1−1 2(α+µ)

n (1−γ)n−₂1(α+µ)+γβΓ(2_Γ−_(nλ₋)Γ_λ(n++1)1)

,n≥2. (23)

Using(22)and(23), we get

|z|n−1≤

2(α+µ)

n(n−ρ) (24)

thus

|z|<r2(λ,γ,β,α,µ,ρ) =i n fn ¨

(1−ρ)n (1−γ)n−1₂(α+µ)+γβΓ(2_Γ_(n−λ₋)Γ_λ(n+₊₁₎1) 1−1

2(α+µ)

n(n−ρ)

«n1−1

,n=2, 3, . . . .

Theorem 3.3.

Let f ∈H(λ,γ,β,α,µ).Then f is close -to-convex of orderρ, 0≤ρ <1in|z|<r=r3(λ,γ,β,α,µ,ρ),where

r3(λ,γ,β,α,µ,ρ) =i n fn ¨

2(α+µ)

«n1−1

,n=2, 3, . . . . (25)

f(z) =z− 1−

1 2(α+µ)

n (1−γ)n−1₂(α+µ)+γβΓ(2_Γ−(nλ−)Γλ(n++1)1)

zn,n≥2 (26)

Proof. Letf ∈H(λ,γ,β,α,µ).Then by Theorem2.1,

∞

X

n=2

n (1−γ)n−1₂(α+µ)+γβΓ(2_Γ−_(nλ₋)Γ_λ(₊n₁+₎1) 1−1

2(α+µ)

an≤1, (27)

for 0≤ρ <1, we need to show thatf′(z)−1

≤1−ρfor|z|<r=r₃(λ,γ,β,α,µ,ρ), when is given by(25). Now

f′(z)−1 =

∞

X

n=2

n anzn−1

≤

∞

X

n=2

n an|z|n−1. (28)

Thusf′(z)−1

≤1−ρif

∞

X

n=2

n

1−ρan|z|

n−1_≤_1, ₍₂₉₎

but by Theorem2.1above inequality holds true if

|z|n−1≤

¨

(1−ρ)n (1−γ)n−1₂(α+µ)+γβΓ(2_Γ_(n−λ₋)Γ_λ(n+₊₁₎1) 1−1

2(α+µ)

«

,n=2, 3, . . . ,

and this completes the proof.

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4. Extreme points for the function class

H

(

λ

,

γ

,

β

,

α

,

µ

)

Theorem 4.1. Let

f1(z) =z (30)

and

fn(z) =z−

1−1 2(α+µ)

n (1−γ)n−1 2(α+µ)

+γβΓ(2_Γ_(n−λ₋)Γ_λ(n+₊₁₎1)z

n_,_(n_∈_N_\{1}₎_. ₍₃₁₎

Then f ∈H(λ,γ,β,α,µ)if and only if it can be expressed in the following form :

f(z) =

∞

X

n=2 χnfn(z)

whereχn≥0and P∞

n=2χn=1.

Proof. Suppose that

f(z) =

∞

X

n=2

χnfn(z) =z−

∞

X

n=2 χn

1−1 2(α+µ)

n (1−γ)n−1₂(α+µ)+γβΓ(_Γ2−_(nλ₋)Γ_λ(₊n₁+₎1)z n_.

The from Theorem2.1, we have

∞

X

n=2

n

(1−γ)n−1

2(α+µ)

+γβ

_Γ₍₂₋_λ₎_Γ_(n₊₁₎

Γ(n−λ+1)

1−1 2(α+µ)

n (1−γ)n−1₂(α+µ)+γβΓ(_Γ2−_(nλ₋)Γ_λ(₊n₁+₎1)χn

=

1−1

2(α+µ) _X∞

n=2 χn=

1−1

2(α+µ)

(1−χn)≤1− 1 2(α+µ).

Thus ,in view of Theorem2.1, we find thatf ∈H(λ,γ,β,α,µ).

Conversely,let us suppose thatf ∈H(λ,γ,β,α,µ). Then, since

an≤

1−1 2(α+µ)

n (1−γ)n−1₂(α+µ)+γβΓ(2_Γ_(n−λ₋)Γ_λ(₊n₁+₎1) (n∈N\{1}), we may set

χn=

n (1−γ)n−₂1(α+µ)+γβΓ(2_Γ−_(nλ₋)Γ_λ(n+₊₁₎1) 1−1

2(α+µ)

an, (n∈N\{1})

andχn=1− P∞

n=2χn. Thus clearly we have

f(z) =z−

∞

X

n=2

χnfn(z).

This completed the proof of Theorem4.1

Corollary 4.1.

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5. Integral means inequalities for the function class

H

(

λ

,

γ

,

β

,

α

,

µ

)

In the year 1925, Littlewood[15]prove the following subordination theorem.

Theorem 5.1.

If the functions f and g are analytic in U with f(z)≺g(z),(z∈U),

then ,forδ >0and z=r eiϑ_,₍₀_<_r_<₁₎_,

Z2Π

0 f(z)

δdϑ≤

Z 2Π

0 g(z)

δdϑ.

We now make use of Theorem5.1to prove Theorem5.2below.

Theorem 5.2.

Let f ∈H(λ,γ,β,α,µ).Suppose also that fnis defined be equation(31). If there exists an analytic function w(z)given

by

[w(z)]n−1=

n (1−γ)n−₂1(α+µ)+γβΓ(2_Γ−_(nλ₋)Γ_λ(n++1)1) 1−1

2(α+µ)

∞

X

n=2

anzn−1,

there for z=r eiϑ_and₍₀_<_r_<₁_),

Z2Π

0 f(r eiϑ)

δdϑ≤

Z2Π

0

f_n(r eiϑ)

δdϑ (δ >0).

Proof. We must show that

Z2Π

0

1−

∞

X

n=2

anzn−1

δ

dϑ≤ Z 2Π

0

1− 1− 1 2(α+µ)

n (1−γ)n−₂1(α+µ)+γβΓ(2_Γ−_(nλ₋)Γ_λ(n+₊₁₎1)z n−1

δ

dϑ.

By applying Littlewood´s subordination theorem (Theorem5.1above), it would suffice to show that

1−

∞

X

n=2

anzn−1≺1−

1−1 2(α+µ)

n−1 _(z_∈_U₎_.

By setting

1−

∞

X

n=2

anzn−1=1−

1−1 2(α+µ)

n (1−γ)n−1₂(α+µ)+γβΓ(2_Γ−_(nλ₋)Γ_λ(n+₊₁₎1)[w(z)] n−1_,

we find that

[w(z)]n−1=

n (1−γ)n−₂1(α+µ)+γβΓ(2_Γ−_(nλ₋)Γ_λ(n++1)1) 1−1

2(α+µ)

∞

X

n=2

anzn−1,

which readily yieldsw(0) =0.

Next by using equation(9), we obtain

|w(z)|n−1≤

n (1−γ)n−₂1(α+µ)+γβΓ(2_Γ−_(nλ₋)Γ_λ(n+₊₁₎1) 1−1

2(α+µ)

∞

X

n=2

anzn−1

≤

n (1−γ)n−₂1(α+µ)+γβΓ(2_Γ−_(nλ₋)Γ_λ(n++1)1) 1−1

2(α+µ)

∞

X

n=2

an zn−1

≤ |z|n−1<1.

This completes the proof of Theorem5.2.

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