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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2017; 57(1): 109-124

Published online March 23, 2017

### Some New Subclasses of Analytic Functions defined by Srivastava-Owa-Ruscheweyh Fractional Derivative Operator

Khalida Inayat Noor1
Rashid Murtaza1
Janusz Sokól2

Department of Mathematics, COMSATS Institute of Information Technology Park Road, Islamabad, Pakistan1
Faculty of Mathematics and Natural Sciences, University of Rzeszów, ul. Prof. Pigonia 1, 35-310 Rzeszów, Poland2

Received: January 5, 2016; Accepted: December 6, 2016

### Abstract

In this article the Srivastava-Owa-Ruscheweyh fractional derivative operator La,λα is applied for defining and studying some new subclasses of analytic functions in the unit disk E. Inclusion results, radius problem and other results related to Bernardi integral operator are also discussed. Some applications related to conic domains are given.

Keywords: analytic functions, convolution, subordination, Srivastava-Owa-Ruscheweyh  ,fractional derivative operator, multiplier linear fractional differential operator, gamma function, incomplete beta function

### 1. Introduction

Let denote the class of all normalized functions of the form

f(z)=z+j=2ajzj,zE,

which are analytic in the open unit disk E = {z : |z| < 1}.

Let , , , (β) and (β) denote the subclasses of consisting of functions that are univalent, convex, starlike, convex of order β and starlike of order β in E respectively, see [10]. For the functions

f(z)=z+j=2ajzj         and         g(z)=z+j=2bjzj,zE,

the convolution (Hadamard product) is defined as

(f*g)(z)=z+j=2ajbjzj=(g*f)(z),zE.

Let f and g be analytic functions in E. Then f is said to be subordinate to g written as fg and f(z) ≺ g(z), zE, if there exits a Schwarz function ω analytic in E, with ω (0) = 0 and |ω(z)| < 1 for zE, such that f(z) = g (ω(z)), zE.

Recently, the theory of fractional calculus has found interesting applications in the theory of analytic functions. The classical definitions of fractional operators and their generalizations have fruitfully been applied in obtaining, for example, the characterization properties, coefficient estimates, distortion inequalities and convolution structures for various subclasses of analytic functions.

The fractional derivative of order α, 0 ≤ α < 1 is defined in [23] as follows

Dzαf(z)=1Γ(1-α)ddz0zf(t)(z-t)αdt,   0α<1,zE,

where the function f(z) is analytic in a simply connected domain in the complex plane containing the origin and the multiplicity of (zt)α is removed by requiring log(zt) ∈ ℝ, whenever (zt) > 0. The gamma function Γ is defined as

Γ(z)=0e-xxz-1dx,   ez>0.

Note that

Dz0f(z)=f(z).

Owa and Srivastava [24, 30] introduced the operator Ωα : as follows

Ωαf(z)=Γ(2-α)zαDzαf(z),   0α<1,=z+j=2Γ(1+j)Γ(2-α)Γ(1+j-α)ajzj,   zE,=ϕ(2,2-α;z)*f(z),

where the incomplete beta function φ(a, c; z) is defined as follows

φ(a,c;z)=j=0Γ(a+j)Γ(c)Γ(c+j)Γ(a)zj,

where a, c are complex numbers different from 0, −1, −2, …. It can be seen that

Ω0f(z)=f(z).

Mishra and Gochhayat [19] have studied some properties of the operator Ωα and introduced new subclass of k-uniformly convex functions. In a recent paper Ibrahim and Darus [12] introduced the fractional differential subordination based on the operator Ωα. Srivastava and Mishra [29] studied the applications of fractional calculus to the parabolic starlike and uniformly convex functions using the operator Ωα.

Al-Oboudi [1, 2] defined the linear multiplier fractional differential operator of order 1 as follows

Dλαf(z)=(1-λ)Ωαf(z)+λz(Ωαf(z))=z+j=2Γ(1+j)Γ(2-α)Γ(1+j-α)(1+λ(j-1))ajzj,=φ(2,2-α;z)*gλ(z)*f(z),   zE,

where 0 ≤ λ ≤ 1, 0 ≤ α < 1 and

gλ(z)=z-(1-λ)z2(1-z)2,   0λ1.

Obviously

D0αf(z)=Ωαf(z),

where Ωα was defined by (1.4) and

D00f(z)=Ω0f(z)=f(z).

Bulut [5, 6, 7, 8] used the Al-Oboudi fractional differential operator of order n to define some new integral operators and obtain interesting results. Recently, Noor et al [22] used fractional derivative to define some new subclasses of analytic functions in the conic regions. Now for a > 0, let

ka(z)=z(1-z)a=z+j=2Γ(a+j-1)Γ(a)Γ(j)zj,   zE.

Ozkan [25] defined the convolution of f given by (1.1) and ka(z) such that

(ka*f)(z)=z+j=2Γ(a+j-1)Γ(a)Γ(j)ajzj,   zE.

Now extending the concept of [4] and [25, 30], we define the Srivastava-Owa-Ruscheweyh La,λα:AA as follows

La,λαf(z)=z(1-z)a*Dλαf(z)=z+j=2Γ(a+j-1)Γ(2-α)Γ(j+1)Γ(1+j-α)Γ(a)Γ(j){1+λ(j-1)}ajzj=z(1-z)a*φ(2,2-α;z)*gλ(z)*f(z),   zE,

which is called the Srivastava-Owa-Ruscheweyh fractional derivative operator.

It can easily be seen that

• For α = 0, λ = 0. La,00f(z)=(ka*f)(z),

see [25].

• For a = n + 1 > 0, α = 0, λ = 0. Ln+1,00f(z)=z(1-z)n+1*f(z),

see [28].

• For a = n + 1 > 0, α = 0. Ln+1,00f(z)=z(1-z)n+1*Dλ0f(z),

see [26].

From the definition of La,λα, we can establish the following identity as well

(a-1)(La,λαf(z))+z(La,λαf(z))=a(La+1,λαf(z)).

We assume that h is analytic, convex, univalent in E with h(0) = 1 and ℜ h(z) > 0, zE. Using the operator La,λα, we define the following

### Definition 1.1

Let Pa,λα(h,δ) be the class of functions f satisfying

z(La,λαf(z))+δz2(La,λαf(z))(1-δ)(La,λαf(z))+δz(La,λαf(z))h(z),

for a > 0, 0 ≤ δ ≤ 1, 0 ≤ λ ≤ 1, 0 ≤ α < 1 and for all zE.

We note that the class Pa,00(h,δ)=Pa(h,δ), was studied by Ozkan in [25] and the class Pa,00(h,0)=Sa(h), was studied by Padmanabhan and Parvatham in [26].

Obviously, for the special choices of function h and parameters α, a, λ, we have the following relationship.

P1,00(1+z1-z,0)=S*,P2,00(1+z1-z,0)=P1,00(1+z1-z,1)=C,P1,00(1+(1-2β)z1-z,0)=S*(β),         0β<1,P2,00(1+(1-2β)z1-z,0)=P1,00(1+(1-2β)z1-z,1)=C(β).

### Definition 1.2

Let Ja,λα(h,γ) denote the subclass of consisting of the functions f which satisfies the following condition

(1-γ)(La,λαf(z))z+γ(La,λαf(z))h(z),

for some a > 0, 0 ≤ λ ≤ 1, 0 ≤ α < 1, 0 ≤ γ ≤ 1 and for all zE. Note that

Ja,00(h,γ)=Ja(h,γ),

see [25].

### Definition 1.3

Let Ma,λα(h,γ) be the class of functions f which satisfies the following condition

(La,λαf(z))+γz(La,λαf(z))h(z),

for some a > 0, 0 ≤ λ ≤ 1, 0 ≤ α < 1, 0 ≤ γ ≤ 1 and for all zE.

### 2. Preliminaries

To prove our main results, we need the following Lemmas.

### Lemma 2.1.([18])

Let h be analytic, univalent, convex in E, with h(0) = 1 and[β1h(z)+γ1] > 0, β1, γ1 ∈ ℂ, zE. If p is analytic in E, with p(0) = h(0), then

p(z)+zp(z)β1p(z)+γ1h(z),

implies

p(z)h(z).

### Lemma 2.2.([11])

Let h be analytic, univalent, convex in E, with h(0) = 1. Let p be analytic in E, with p(0) = h(0). If γ2 ≠ 0, ℜ γ2 ≥ 0 and

p(z)+zp(z)γ2h(z),   γ20,   zE,

then

p(z)q(z)h(z),

where

q(z)=γ2zγ20ztγ2-1h(t)dt.

### Lemma 2.3.([9])

If ψ, gand F is analytic function with(F(z)) > 0 for zE, then

e((ψ*Fg)(z)(ψ*g)(z))>0,   zE.

### Lemma 2.4.([21])

Let p(z) and q(z) be analytic in E, with p(0) = q(0) = 1 and e(q(z))>12 for |z| < ρ, (0 < ρ ≤ 1). Then the image of Eρ = {z : |z| < ρ} under p * q is a subset of closed convex hull of p(E).

### 3. Main Results

In this section, we will prove our main results.

### Theorem 3.1

Assume that a ≥ 1, 0 ≤ δ ≤ 1, 0 ≤ λ ≤ 1, 0 ≤ α < 1 and

1z{(1-δ)(La,λαf(z))+δz(La,λαf(z))}0   zE.

Then

Pa+1,λα(h,δ)Pa,λα(h,δ),   forallzE.
Proof

We suppose that fPa+1,λα(h,δ) and let

p(z)=z(La,λαf(z))+δz2(La,λαf(z))(1-δ)(La,λαf(z))+δz(La,λαf(z)).

Then by (3.1) the function p(z) is analytic in E, with p(0) = 1. From (1.7) and (3.2), we obtain

p(z)+zp(z)p(z)+(a-1)=az(La+1,λαf(z))+δaz2(La+1,λαf(z))(1-δ)a(La+1,λαf(z))+δaz(La+1,λαf(z)).

Since fPa+1,λα(h,δ), therefore by using (3.2), we have

p(z)+zp(z)p(z)+(a-1)h(z)in E.

Thus it follows from the Lemma 2.1 that

p(z)h(z)   for a1in E.

Hence fPa,λα(h,δ) for a ≥ 1.

As special cases of Theorem 3.1, we have the following results.

### Corollary 3.2.([25])

If (3.1) is satisfied for α = 0 and λ = 0, then Pa+1,00(h,δ)Pa,00(h,δ).

### Corollary 3.3

If (3.1) is satisfied for δ = 0, α = 0, λ = 0, a = 1 and h(z)=1+z1-z,P2,00(1+z1-z,0)P1,00(1+z1-z,0). That is, zE.

### Theorem 3.4

Assume that

1z{(1-δ)(La,λαFc(z))+δz(La,λαFc(z))}zE.

If fPa,λα(h,δ) for a ≥ 1, then FcPa,λα(h,δ), where Fc is defined as

Fc(z)=1+czc0ztc-1f(t)dt,   c>-1,   zE.
Proof

Suppose that fPa,λα(h,δ) and let

p(z)=z(La,λαFc(z))+δz2(La,λαFc(z))(1-δ)(La,λαFc(z))+δz(La,λαFc(z)).

Then by (3.4) the function p(z) is analytic in E, with p(0) = 1. From (3.3), we can write

z(Fc(z))+cFc(z)=(1+c)f(z).

This means that

z(La,λαFc(z))+c(La,λαFc(z))=(1+c)(La,λαf(z)).

From (3.4) and (3.5), we have

p(z)+zp(z)p(z)+c=z(La,λαf(z))+δz2(La,λαf(z))(1-δ)(La,λαf(z))+δz(La,λαf(z)).

Since fPa,λα(h,δ), therefore from (3.6), it follows that

p(z)+zp(z)p(z)+ch(z)   in E.

This implies by using Lemma 2.1 with β1 = 1 and γ1 = c that p(z) ≺ h(z) in E. Thus FcPa,λα(h,δ) in E.

Taking a = 1, α = 0, λ = 0, h(z)=1+z1-z, zE in Theorem 3.4, we deduce Theorem 1 and Theorem 2 of Bernardi [3] with δ = 0 and δ = 1 respectively.

### Theorem 3.5

For a > 0, 0 ≤ γ ≤ 1, 0 ≤ λ ≤ 1, 0 ≤ α < 1, we have

Ja+1,λα(h,γ)Ja,λα(h,γ).
Proof

Suppose that fJa+1,λα(h,γ) and let

p(z)=(1-γ)(La,λαf(z))z+γ(La,λαf(z)),

where p(z) is analytic in E, with p(0) = 1. Taking δ = 1 in (1.7), we obtain

z(La,λαf(z))=a(La+1,λαf(z))-(a-1)(La,λαf(z)).

Differentiating (3.8) and using (3.7), we have

p(z)+zp(z)a=(1-γ)(La+1,λαf(z))z+γ(La+1,λαf(z)).

Now by applying Lemma 2.2 and from (3.9), we can write ph in E. Thus fJa,λα(h,γ) in E.

### Theorem 3.6

If fJa,λα(h,γ), then FcJa,λα(h,γ), where Fc is defined by (3.3).

Proof

We assume that fJa,λα(h,γ). Let

p(z)=(1-γ)(La,λαFc(z))z+γ(La,λαFc(z)),

where p(z) is analytic in E, with p(0) = 1. From (3.8) and (3.10), we obtain

p(z)+zp(z)1+c=(1-γ)(La,λαf(z))z+γ(La,λαf(z)).

Using Lemma 2.2 with (3.11), we have ph in E and hence FcJa,λα(h,γ).

For α = 0, λ = 0, we have the following result, see [25].

### Corollary 3.7

If fJa,00(h,γ)=Ja(h,γ), then Fc (h, γ), where Fc is defined by (3.3).

### Theorem 3.8

For a > 0, 0 ≤ γ ≤ 1, 0 ≤ λ ≤ 1, 0 ≤ α < 1, we have

Ma+1,λα(h,γ)Ma,λα(h,γ).
Proof

Let fMa+1,λα(h,γ) and

p(z)=(La,λαf(z))+γz(La,λαf(z)),

where p(z) is analytic in E, with p(0) = 1.

Differentiating (3.8) and using (3.12), we obtain

p(z)+zp(z)a=(La+1,λαf(z))+γz(La+1,λαf(z)).

Since fMa+1,λα(h,γ), therefore

(La+1,λαf(z))+γz(La+1,λαf(z))h(z)   for all zE.

By using (3.13), it follows that

p(z)+zp(z)ah(z)   in   E.

Now by using Lemma 2.2, we obtain ph in E and hence fMa,λα(h,γ).

### Corollary 3.9.([25])

For α = 0, λ = 0, we have

Ma+1,00(h,γ)Ma,00(h,γ).

### Theorem 3.10

• fMa,λα(h,γ)zfJa,λα(h,γ),

• Ma,λα(h,γ)Ja,λα(h,γ).

Proof

(i) Let fMa,λα(h,γ). Now using the fact

z(La,λαf(z))=(La,λα(zf(z)),

we can write

(1-γ)La,λα(zf(z))z+γ(La,λα(zf(z)))=(La,λαf(z))+γz((La,λαf(z)).

Since fMa,λα(h,γ), therefore

(La,λαf(z))+γz(Lζa,λαf(z))h(z)in E.

From (3.14), it follows that

(1-γ)La,λα(zf(z))z+γ(La,λα(zf(z)))in E.

Hence zfJa,λα(h,γ).

(ii) Let fMa,λα(h,γ) and let

p(z)=(1-γ)(La,λαf(z))z+γ(La,λαf(z)),

where p(z) is analytic in E, with p(0) = 1. From (3.15), we obtain

p(z)+zp(z)=(La,λαf(z))+γz(La,λαf(z)).

Thus from Lemma 2.2, we have ph in E. Hence fJa,λα(h,γ).

For α = 0, λ = 0, we have some well known results due to [25].

### Corollary 3.11

• fMa,00(h,γ)zfJa,00(h,γ),

• Ma,00(h,γ)Ja,00(h,γ).

### Theorem 3.12

For γ > η ≥ 0, 0 ≤ λ ≤ 1, 0 ≤ α < 1, we have

• Ja,λα(h,γ)Ja,λα(h,η),

• Ma,λα(h,γ)Ma,λα(h,η).

Proof

(i). The case η = 0 is trivial. Suppose that η > 0. Let fJa,λα(h,γ) and let z1 be any arbitrary point in E. Then

(1-γ)La,λαf(z1)z1+γ(La,λαf(z1))h(E).

Since (La,λαf(z1))zh(E), we can write the following equality

(1-η)(La,λαf(z))z+η(La,λαf(z))=(1-ηγ)(La,λαf(z))z+ηγ[(1-γ)(La,λαf(z))z+γ(La,λαf(z))].

Now as ηγ<1 and h(E) is convex. Therefore from (3.16), it follows that

(1-η)(La,λαf(z))z+η(La,λαf(z))h(E).

Thus fJa,λα(h,η).

The proof of part (ii) is similar to (i).

### Theorem 3.13

If Ψ ∈ and fPa,λα(h,δ), then fPa,λα(h,δ), for a ≥ 1.

Proof

Let Ψ ∈ , G(z)=(1-δ)(La,λαf(z))+δz(La,λαf(z))S*(h)S* and p(z)=zG(z)G(z). Consider

z[(1-δ){Ψ*La,λαf(z)}+δ{z(Ψ*La,λαf(z))}](1-δ){Ψ*La,λαf(z)}+δz{Ψ*La,λαf(z)}=Ψ*[(1-δ)z(La,λαf(z))+δz{(z(La,λαf(z))}]Ψ*[(1-δ)La,λαf(z)+δz(La,λαf(z))]=Ψ*zG(z)Ψ*G(z)=(Ψ*pG)(z)(Ψ*G)(z),

where p(z)=zG(z)G(z)h(z)in E. We now apply Lemma 2.3 and using (3.17), we have Ψ*fPa,λα(h,δ).

Note that the above result also holds for the classes Ja,λα(h,γ) and Ma,λα(h,γ).

### Theorem 3.14

For δ ≥ 0, 0 ≤ λ ≤ 1, 0 ≤ α < 1, we have

Pa,λα(h,δ)Pa,λα(h,0).
Proof

The case δ = 0 is trivial, so we suppose that δ > 0. Assume that Pa,λα(h,δ) and let

N(z)=(1-δ)(La,λαf(z))+δz(La,λαf(z)).

Then

zN(z)N(z)=z(La,λαf(z))+δz2(La,λαf(z))(1-δ)(La,λαf(z))+δz(La,λαf(z)).

This means that zN(z)N(z)h(z) in E. That is N(h). Let

p(z)=z(La,λαf(z))(La,λαf(z)),

where p(z) is analytic in E, with p(0) = 1. Now

zN(z)N(z)=z(La,λαf(z))+δz2(La,λαf(z))(1-δ)La,λαf(z)+δz(La,λαf(z))=z(La,λαf(z))+δz{(z(La,λαf(z))}-δz(La,λαf(z))(1-δ)La,λαf(z)+δz(La,λαf(z))=(1-δ)z(La,λαf(z))La,λαf(z)+δz(z(La,λαf(z)))La,λαf(z)(1-δ)+δz(La,λαf(z))La,λαf(z).

Therefore using (3.18), we obtain

zN(z)N(z)=(1-δ)p(z)+δ(p2(z)+zp(z))(1-δ)+δp(z)=p(z)+zp(z)p(z)+(1/δ-1).

Since zN(z)N(z)h(z) in E. This implies that

p(z)+zp(z)p(z)+(1/δ-1)h(z)in E.

Now by using Lemma 2.2, we have p(z) ≺ h(z) in E. Thus fPa,λα(h,δ).

### 4. Applications

For different choices of analytic function h, we have some applications of the main results. For k ∈ [0,∞), Kanas [13, 14, 15] defined the conic domain Ωk as follows

Ωk={u+iv:u>k(u-1)2+v2}.

For k = 0, this domain represents the whole right half plane. Also we obtain the right branch of hyperbola for 0 < k < 1, the parabola for k = 1 and the ellipse for k > 1. The following functions pk(z) are univalent with pk (0) = 1, pk(0)>0 and plays the role of extremal functions mapping E onto the conic domains Ωk

pk(z)={1+z1-z,k=01+2π2[log(1+z1-z)]2k=11+21-k2sinh2(2πcos-1)tanh-1z0<k<11+2k2-1sin{22R(t)0u(z)z11-x21-(tx)2dx}+1k2-1,k>1.

Here u(z)=z-t1-tz,t(0,1), t ∈ (0, 1), zE and z is chosen such that k=cosh (πR(t)4R(t)), R(t) is Legender’s elliptic integral of the first kind and R′ (t) is the complementary integral of R(t). For details, we refer to [13], [14], [15] and [20]. In [16] some linear operators associated with k-uniformly convex functions were considered. Now by choosing h(z) = pk(z) in Theorem 3.14, we can easily prove the following

### Corollary 4.1

For δ ≥ 0, 0 ≤ λ ≤ 1, 0 ≤ α < 1, we have

Pa,λα(pk,δ)Pa,λα(qk,0),

where

qk(z)=[01(expttzpk(u)-1udu)dt]-1.

Some of the special cases are given as follows.

• Let k = 0. Then fPa,λα(1+z1-z,δ)fPa,λα(11-z,0). That is e(z(La,λαf(z))La,λαf(z))>12in E.

• For k = 1, we have Pa,λα((1+2π2[log(1+z1-z)]2,δ))Pa,λα(q1,0)

and e(z(La,λαf(z))La,λαf(z))>q1(-1)=12.

• Let k > 1 and fPa,λα(pk,δ)fPa,λα(z(z-k)log(1-zk),0). That is e(z(La,λαf(z))La,λαf(z))>1(k+1)log(1+1k).

• For the case k = 2, we note that Pa,λα(p2,δ)Pa,λα(q2,0). This gives us e(z(La,λαf(z))La,λαf(z))>q2(-1)13log320.813.

Now we prove a radius result for the class Ma,λα(h,σ).

### Theorem 4.2

Let fMa,λα(h,0). Then fMa,λα(h,σ), for |z| < rσ, where

rσ=(1+σ2)12-σ.
Proof

Let

p(z)=(La,λαf(z)),

where p(z) is analytic in E, with p(0) = 1. Then

p(z)+σzp(z)=(La,λαf(z))*gσ(z),

where gσ(z)=z-(1-σ)z2(1-z)2. It is known [17] that e(gσ(z)z)>12 in |z| < rσ. This implies that gσ(z)zh2(z) and e(h2(z))>12 in |z| < rσ. Now by using Lemma 2.4 and from (4.2), we obtain

p(z)+σzp(z)h2(z)*h(z)h(z)   in   z   <rσ.

It follows that

p(z)+σzp(z)=(La,λαf(z))+σz(La,λαf(z))h(z)   in   z   <rσ.

Thus fMa,λα(h,σ) in |z| < rσ, where rσ is given by (4.1).

### Acknowledgements

The authors are grateful to Dr. S. M. Junaid Zaidi, Rector, COMSATS Institute of Information Technology, Pakistan for providing excellent research and academic environment. This research is supported by the HEC NPRU project No: 20-1966/R&D/11-2553, titled, Research unit of Academic Excellence in Geometric Functions Theory and Applications.

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