Kyungpook Mathematical Journal 2018; 58(2): 243-255
Certain Subclasses of k–uniformly Functions Involving the Generalized Fractional Differintegral Operator
Tamer Mohamed Seoudy
Department of Mathematics, Faculty of Science, Fayoum University, Fayoum 63514, Egypt, e-mail : tms00@fayoum.edu.eg
Received: April 25, 2013; Accepted: April 12, 2016; Published online: June 23, 2018.

This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0/) which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract

We introduce several k–uniformly subclasses of p–valent functions defined by the generalized fractional differintegral operator and investigate various inclusion relationships for these subclasses. Some interesting applications involving certain classes of integral operators are also considered.

Keywords: analytic functions, k-uniformly starlike functions, k-uniformly convex functions, k-uniformly close-to-convex functions, k-uniformly quasi-convex functions, integral operator, Hadamard product, subordination
1. Introduction

Let denote the class of functions of the form:

$f (z)=zp+∑n=1∞an+p zn+p (p∈ℕ={1,2,3,…}),$

which are analytic and p–valent in the open unit disk = {z ∈ ℂ : |z| < 1}. If f and g are analytic in , we say that f is subordinate to g, written fg or f(z) ≺ g(z), if there exists a Schwarz function ω, analytic in with ω (0) = 0 and (z)| < 1 (z), such that f(z) = g(ω(z)) (z). In particular, if the function g is univalent in , the above subordination is equivalent to (see [8] and [9]).

For 0 ≤ γ, η < p, k ≥ 0 and z, we define $USp* (k;γ)$, UCp (k; γ), UKp (k; γ, η) and $UKp* (k;γ,η)$ the k–uniformly subclasses of consisting of all analytic functions which are, respectively, p–valent starlike of order γ, p–valent convex of order γ, p–valent close-to-convex of order γ, and type η and p–valent quasi-convex of order γ, and type η as follows:

$USp* (k;γ)={f∈Ap:ℜ (zf′(z)f (z)-γ)>k|zf′ (z)f (z)-p|},$$UCp (k;γ)={f∈Ap:ℜ (zf″ (z)f′ (z)-γ)>k|zf″ (z)f′ (z)-p|},$$UKp (k;γ,η)={f∈Ap:∃g∈USp* (k;η),ℜ (xf″ (z)g(z)-γ)>k|zf″ (z)g(z)-p|},$$UKp* (k;γ,η)={f∈Ap:∃g∈UCp (k;η),ℜ ((zf′ (z))′g′ (z)-γ)>k|(zf′ (z))′g′ (z)-p|}.$

These subclasses were introduced and studied by Al-Kharsani [1]. We note that

(i) $US1* (k;γ)=US*(k;γ)$ and UC1 (k; γ) = UC (k; γ) (0 ≤ γ <1) (see [6] and [20]);

(ii) $USp* (0;γ)=Sp* (γ) (0≤γ (see [12] and [15]);

(iii) UCp (0; γ) = Cp (γ) (0 ≤ γ < p) (see [12]);

(iv) UKp (0; γ, η) = Kp (γ, η) (0 ≤ γ, η < p) (see [2]);

(v) $UKp* (0;γ,η)=Kp* (γ,η) (0≤γ,η (see [10]).

Corresponding to a conic domain Ωp,k,γ defined by

$Ωp,k,γ={u+iv:u>k(u-p)2+v2+γ},$

we define the function qp,k,γ (z) which maps onto the conic domain Ωp,k,γ such that 1 ∈ Ωp,k,γ as the following (see [1]):

$qp,k,γ (z)={p+(p-2γ) z1-z(k=0),p-γ1-k2cos {2π (cos-1 k) i log1+z1-z}-k2p-γ1-k2(01).$

where $u (z)=z-x1-xz$, x ∈ (0, 1) and ζ (k) is such that $k=coshπζ′ (z)4ζ(z)$. By virtue of the properties of the conic domain Ωp,k,γ, we have

$ℜ {qp,k,γ (z)}>kp+γk+1.$

Making use of the principal of subordination between analytic functions and the definition of qp,k,γ (z), we may rewrite the subclasses $USp* (k;γ)$, UCp (k; γ), UKp (k; γ, β) and $UKp* (k;γ,β)$ as the following:

$USp* (k;γ)={f∈Ap:zf′ (z)f (z)≺qp,k,γ (z)},$$UCp (k;γ)={f∈Ap:1+zf″ (z)f′ (z)≺qp,k,γ (z)},$$UKp (k;γ,η)={f∈Ap:∃g∈USp* (k;η),zf′ (z)g(z)≺qp,k,γ (z)},$$UKp* (k;γ,η)={f∈Ap:∃g∈UCp (k;η),(zf′ (z))′g′ (z)≺qp,k,γ (z)}.$

Srivastava et al. [23] introduced the following generalized fractional integral and generalized fractional derivative operators as follows(see also [16] and [19]):

### Definition 1.1

([23]) For real numbers λ > 0, μ and η, the Saigo hypergeometric fractional integral operator $I0,zλ,μ,η:Ap→Ap$ is defined by

$I0,zλ,μ,ηf (z)=z-λ-μΓ (λ)∫0z(z-t)λ-1 F21 (λ+μ,-η;λ;1-tz) f (t) dt,$

where the function f(z) is analytic in a simply-connected region of the complex z–plane containing the origin, with the order

$f (z)=O(∣z∣ɛ) (z→0;ɛ>max{0,μ-λ}-1),$

and the multiplying of (zt)λ−1 is removed by requiring log (zt) to be real when (zt) > 0.

### Definition 1.2

([23]) Under the hypotheses of Definition 1.1, Saigo hypergeometric fractional derivative operator $J0,zλ,μ,η:Ap→Ap$ is defined by

$J0,zλ,μ,ηf (z)={1Γ (1-λ)ddz {zλ-μ∫0z(z-t)-λF21 (μ-λ,1-η;1-λ;1-tz) f (t) dt}(0≤λ<1),dndznJ0,zλ-μ,μ,ηf (z)(n≤λ

where the multiplying of (zt)−λ is removed as in Definition 1.1.

We note that

$I0,zλ-λ,ηf (z)=Dz-λf (z) (λ>0) and J0,zλ,λ,ηf (z)=Dzλf (z) (0≤λ<1),$

where $Dz-λ$ denotes fractional integral operator and $Dzλ$ denotes fractional derivative operator studied by Owa [11].

Recently, Goyal and Prajapat [7] (see also [17] and [18]) introduced the generalized fractional differintegral operator $S0,zλ,μ,η:Ap→Ap (p∈ℕ,η∈ℝ,μ by

$S0,zλ,μ,ηf (z)={Γ (1+p-μ) Γ (1+p+η-λ)Γ (1+p) Γ (1+p+η-μ)zμJ0,zλ,μ,η(0≤λ<η+p+1),Γ (1+p-μ) Γ (1+p+η-λ)Γ (1+p) Γ (1+p+η-μ)zμI0,z-λ,μ,η(-∞<λ<0).$

It is easily seen from a function f of the form (1.1), we have

$S0,zλ,μ,ηf (z)=zpF32 (1,1+p,1+p+η-μ;1+p-μ,1+p+η-λ;z)*f (z)=zp+∑n=1∞(1+p)n (1+p+η-μ)n (1+p-μ)n (1+p+η-λ)n an+p zn+p (z∈U;p∈ℕ;μ,η∈ℝ;μ

where qFs (qs + 1; q, s ∈ ℕ0 = ℕ∪{0}) is well known generalized hypergeometric function (see, for details, [13, 22]) and (v)n is the Pochhammer symbol defined, in terms of Gamma function, by

$(v)n =Γ (v+n)Γ (v)={1(n=0)v(v+1)…(v+n-1)(n∈ℕ).$

We note that

$S0,z0,0,0f (z)=f (z),S0,z1,0,0f (z)=zf′ (z)p$

and

$S0,zλ,λ,0f (z)=Ωz(λ,p)f (z)=Γ (p+1-λ)Γ (p+1)zλDzλf (z) (-∞≤λ

where the extended fractional differintegral operator $Ωz(λ,p)$ was introduced and studied by Patel and Mishra [14]. The fractional differential operator $Ωz(λ,p)$ with 0 ≤ λ < 1 was investigated by Srivastava and Aouf [21]. The operator $Ωz(λ,1)=Ωzλ$ was introduced by Owa and Srivastava [13];

Upon setting

$Gp,η,μλ (z)=zp+∑n=1∞(1+p)n (1+p+η-μ)n (1+p-μ)n (1+p+η-λ)n zn+p (z∈U;p∈ℕ;μ,η∈ℝ;μ

we define a new function $[Gp,μ,ηλ (z)]-1$ by means of the Hadamard product (or convolution):

$Gp,η,μλ (z)*[Gp,η,μλ (z)]-1=zp(1-z)δ+p (δ>-p;z∈U).$

Tang et al. [24] introduced the linear operator $Hp,η,μλ,δ:Ap→Ap$ as follows:

$Hp,η,μλ,δf (z)=[Gp,η,μλ (z)]-1*f (z).$

For f given by (1.1), then from (1.19), we have

$Hp,η,μλ,δf (z)=zp+∑n=1∞(δ+p)n (1+p-μ)n (1+p+η-λ)n n!(1+p)n (1+p+η-μ)n an+p zn+p$

by using (1.20), we get

$z (Hp,η,μλ+1,δf (z))′=(p+η-λ) Hp,η,μλ,δf (z)-(η-λ) Hp,η,μλ+1,δf (z)$

and

$z (Hp,η,μλ,δf (z))′=(δ+p) Hp,η,μλ,δ+1f (z)-δHp,η,μλ,δf (z).$

Next, using the operator $Hp,η,μλ,δ$, we introduce the following k–uniformly subclasses of p–valent functions for η ∈ ℝ, μ < p+1,−∞ < λ < η+ p+1, δ >p, p ∈ ℕ, k ≥ 0 and 0 ≤ γ, ρ < p:

$USp,η,μλ,δ (k;γ)={f∈Ap:Hp,η,μλ,δf (z)∈USp* (k;γ);z∈U},$$UCp,η,μλ,δ (k;γ)={f∈Ap:Hp,η,μλ,δf (z)∈UCp (k;γ);z∈U},$$UKp,η,μλ,δ (k;γ,ρ)={f∈Ap:Hp,η,μλ,δf (z)∈UKp (k;γ,ρ);z∈U},$$UQp,η,μλ,δ (k;γ,ρ)={f∈Ap:Hp,η,μλ,δf (z)∈UKp* (k;γ,ρ);z∈U}.$

We also note that

$f∈USp,η,μλ,δ (k;γ)⇔zfp∈UCp,η,μλ,δ (k;γ),$

and

$f∈UKp,η,μλ,δ (k;γ,ρ)⇔zf′p∈UQp,η,μλ,δ (k;γ,ρ).$

In this paper, we investigate several inclusion properties of the classes $USp,η,μλ,δ (k;γ),UCp,η,μλ,δ (k;γ),UKp,η,μλ,δ (k;γ,ρ)$ and $UQp,η,μλ,δ (k;γ,ρ)$ associated with the operator $Hp,η,μλ,δ$. Some applications involving integral operators are also considered.

2. Inclusion Properties Involving the Operator Hp,η,μλ,δ

In order to prove the main results, we shall need The following lemmas.

### Lemma 2.1

([5]) Let h (z) be convex univalent inwith ℜ{αh (z) + β} > 0 (α, β ∈ ℂ). If p (z) is analytic inwith p (0) = h (0), then

$p (z)+zp′ (z)αp (z)+β≺h(z)$

implies

$p (z)≺h(z).$

### Lemma 2.2

([8]) Let h (z) be convex univalent inand let w be analytic inwith ℜ{w (z)} ≥ 0. If p (z) is analytic inand p (0) = h (0), then

$p (z)+w(z) zp′ (z)≺h(z)$

implies

$p (z)≺h(z).$

### Theorem 2.3

Let δ (k + 1)+kp +γ > 0 and (ηλ) (k + 1)+kp +γ > 0. Then,

$USp,η,μλ,δ+1 (k;γ)⊂USp,η,μλ,δ (k;γ)⊂USp,η,μλ+1,δ (k;γ).$
Proof

We first prove that $USp,η,μλ,δ+1 (k;γ)⊂USp,η,μλ,δ (k;γ)$. Let $f∈USp,η,μλ,δ+1 (k;γ)$ and set

$p (z)=z(Hp,η,μλ,δf (z))′Hp,η,μλ,δf (z) (z∈U),$

where the function p (z) is analytic in with p (0) = p. Using (1.22), (2.5) and (2.6), we have

$z (Hp,η,μλ,δ+1f (z))′Hp,η,μλ,δ+1f (z)=p (z)+zp′(z)p (z)+δ≺qp,k,γ (z).$

Since δ (k + 1) + kp +γ > 0, we see that

$ℜ {qp,k,γ (z)+δ}>0 (z∈U).$

Applying Lemma 2.1 to (2.7), it follows that p (z) ≺ qp,k,γ (z), that is, $f∈USp,η,μλ,δ (k;γ)$. To prove the right part, let $f∈USp,η,μλ,δ (k;γ)$ and consider

$h(z)=z (Hp,η,μλ+1,δf (z))′Hp,η,μλ+1,δf (z) (z∈U),$

where the function h (z) is analytic in with h (0) = p. Then, by using the arguments similar to those detailed above, together with (1.21), it follows that p (z) ≺ qp,k,γ (z), which implies that $f∈USp,η,μλ+1,δ (k;γ)$. Therefore, we complete the proof of Theorem 2.3.

### Theorem 2.4

Let δ (k + 1)+kp +γ > 0 and (ηλ) (k + 1)+kp +γ > 0. Then,

$UCp,η,μλ,δ+1 (k;γ)⊂UCp,η,μλ,δ (k;γ)⊂UCp,η,μλ+1,δ (k;γ).$
Proof

Applying (1.27) and Theorem 2.3, we observe that

$f∈UCp,η,μλ,δ+1 (k;γ)⇔zf′p∈USp,η,μλ,δ+1 (k;γ)⇒zf′p∈USp,η,μλ,δ (k;γ) (by Theorem 2.3),⇔f∈UCp,η,μλ,δ (k;γ)$

and

$f∈UCp,η,μλ,δ (k;γ)⇔zf′p∈USp,η,μλ,δ (k;γ)⇒zf′p∈USp,η,μλ+1,δ (k;γ) (by Theorem 2.3),⇔f∈UCp,η,μλ+1,δ (k;γ),$

which evidently proves Theorem 2.4.

Next, by using Lemma 2.2, we obtain the following inclusion relation for the class $UKp,η,μλ,δ (k;γ,ρ)$.

### Theorem 2.5

Let δ (k + 1)+kp +ρ > 0 and (ηλ) (k + 1)+kp +ρ > 0. Then,

$UKp,η,μλ,δ+1 (k;γ,ρ)⊂UKp,η,μλ,δ (k;γ,ρ)⊂UKp,η,μλ+1,δ (k;γ,ρ).$
Proof

We begin by proving that $UKp,η,μλ,δ+1 (k;γ,ρ)⊂UKp,η,μλ,δ (k;γ,ρ)$. Let $f∈UKp,η,μλ,δ+1 (k;γ,ρ)$. Then, from the definition of $UKp,η,μλ,δ+1 (k;γ,ρ)$, there exists a function r (z) ∈ USp (k; γ) such that

$z (Hp,η,μλ,δ+1f (z))′r(z)≺qp,k,γ (z).$

Choose the function g such that $Hp,η,μλ,δ+1g(z)=r(z)$. Then, $g∈USp,η,μλ,δ+1 (k;γ)$ and

$z (Hp,η,μλ,δ+1f (z))′Hp,η,μλ,δ+1g(z)≺qp,k,γ (z).$

Now let

$p (z)=z (Hp,η,μλ,δf (z))′Hp,η,μλ,δ (z) (z∈U),$

where p (z) is analytic in with p (0) = p. Since $g∈USp,η,μλ,δ+1 (k;γ)$, by Theorem 2.3, we know that $g∈USp,η,μλ,δ (k;γ)$. Let

$t(z)=z (Hp,η,μλ,δg(z))′Hp,η,μλ,δg(z) (z∈U),$

where t (z) is analytic in with $ℜ {t(z)}>kp+ρk+1$. Also, from (2.13), we note that

$Hp,η,μλ,δzf′ (z)=Hp,η,μλ,δg(z) p (z).$

Differentiating both sides of (2.15) with respect to z, we obtain

$z (Hp,η,μλ,δzf′ (z))′Hp,η,μλ,δg(z)=z (Hp,η,μλ,δg(z))′Hp,η,μλ,δg(z)p (z)+zp′ (z)=t(z) p (z)+zp′ (z).$

Now using the identity (1.22) and (2.14), we obtain

$z (Hp,η,μλ,δ+1f (z))′Hp,η,μλ,δ+1g(z)=Hp,η,μλ,δ+1zf′ (z)Hp,η,μλ,δ+1g(z)=z (Hp,η,μλ,δzf′ (z))′+δHp,η,μλ,δzf′ (z)z (Hp,η,μλ,δg(z))′+δHp,η,μλ,δg(z)=z (Hp,η,μλ,δzf′ (z))′Hp,η,μλ,δg(z)+δz (Hp,η,μλ,δf (z))′Hp,η,μλ,δg(z)z (Hp,η,μλ,δg(z))′Hp,η,μλ,δg(z)+δ=t(z) p (z)+zp′ (z)+δp (z)t(z)+δ=p (z)+zp′ (z)t(z)+δ.$

Since δ (k + 1) + kp +ρ > 0 and $ℜ {t(z)}>kp+ρk+1$, we see that

$ℜ{t(z)+δ}>0 (z∈U).$

Hence, applying Lemma 2.2, we can show that p (z) ≺ qp,k,γ (z) so that $f∈UKp,η,μλ,δ (k;γ,ρ)$. For the second part, by using the arguments similar to those detailed above with (1:15), we obtain

$UKp,η,μλ,δ (k;γ,ρ)⊂UKp,η,μλ+1,δ (k;γ,ρ).$

Therefore, we complete the proof of Theorem 2.5.

### Theorem 2.6

Let δ (k + 1) + kp +ρ > 0and (ηλ) (k + 1) + kp +ρ > 0Then,

$UQp,η,μλ,δ+1 (k;γ,ρ)⊂UQp,η,μλ,δ (k;γ,ρ)⊂UQp,η,μλ+1,δ (k;γ,ρ).$
Proof

Just as we derived Theorem 2.4 as consequence of Theorem 2.3 by using the equivalence (1.27), we can also prove Theorem 2.6 by using Theorem 2.5 and the equivalence (1.28).

3. Inclusion Properties Involving the Integral Operator Fc,p

In this section, we present several integral-preserving properties of the p-valent function classes introduced here. We consider the generalized Libera integral operator Fc,p (f) (see [4] and [3]) defined by

$Fc,p (f) (z)=c+pzc∫tc-1f (z) dt (c>-p).$

### Theorem 3.1

Let c (k + 1) + kp + γ ≥ 0. If $f∈USp,η,μλ,δ (k;γ)$, then $Fc,p (f)∈USp,η,μλ,δ (k;γ)$.

Proof

Let $f∈USp,η,μλ,δ (k;γ)$ and set

$p (z)=z (Hp,η,μλ,δFc,p (f) (z))′Hp,η,μλ,δFc,p (f) (z) (z∈U),$

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

From (3.1), we have

$z (Hp,η,μλ,δFc,p (f) (z))′=(c+p) Hp,η,μλ,δf (z)-cHp,η,μλ,δFc,p (f) (z).$

Then, by using (3.2) and (3.3), we obtain

$(c+p)Hp,η,μλ,δf (z)Hp,η,μλ,δFc (f) (z)=p (z)+c.$

Taking the logarithmic differentiation on both sides of (3.4) and multiplying by z, we have

$z (Hp,η,μλ,δf (z))′Hp,η,μλ,δf (z)=p (z)+zp′ (z)p (z)+c≺qk,γ (z) (z∈U).$

Hence, by virtue of Lemma 2.1, we conclude that p (z) ≺ qk,γ (z) in , which implies that $Fc,p (f)∈USp,η,μλ,δ (k;γ)$.

Next, we derive an inclusion property involving Fc,p (f), which is given by the following.

### Theorem 3.2

Let c (k + 1) + kp + γ ≥ 0. If $f∈UCp,η,μλ,δ (k;γ)$, then $Fc,p (f)∈UCp,η,μλ,δ (k;γ)$.

Proof

By applying Theorem 2.5, it follows that

$f∈UCp,η,μλ,δ (k;γ)⇔zf′p∈USp,η,μλ,δ (k;γ)⇒Fc,p(zf′p)∈USp,η,μλ,δ (k;γ) (by Theorem 3.1)⇔z (Fc,p (f))′p∈UCp,η,μλ,δ (k;γ)⇔Fc,p (f)∈UCp,η,μλ,δ (k;γ),$

which proves Theorem 3.2.

### Theorem 3.3

Let c (k + 1)+kp + ρ ≥ 0. If $f∈UKp,η,μλ,δ (k;γ,ρ)$, then $Fc,p (f)∈UKp,η,μλ,δ (k;γ,ρ)$.

Proof

Let $f∈UKp,η,μλ,δ (k;γ,ρ)$. Then, in view of the definition of the class $UKp,η,μλ,δ (k;γ,ρ)$, there exists a function $g∈USp,η,μλ,δ (k;γ)$ such that

$z (Hp,η,μλ,δf (z))′Hp,η,μλ,δg(z)≺qk,γ (z).$

Thus, we set

$p (z)=z (Hp,η,μλ,δFc,p (f) (z))′Hp,η,μλ,δFc,p (g) (z) (z∈U),$

where p (z) is analytic in with p (0) = p. Since $g∈USp,η,μλ,δ (k;γ)$, we see from Theorem 3.1 that $Fc,p (g)∈USp,η,μλ,δ (k;γ)$. Let

$t(z)=z (Hp,η,μλ,δFc,p (g) (z))′Hp,η,μλ,δFc,p (g) (z) (z∈U),$

where t (z) is analytic in with $ℜ {t(z)}>kp+ηk+1$ Also, from (3.7), we note that

$Hp,η,μλ,δzFc,p′ (f) (z)=Hp,η,μλ,δFc,p (g) (z).p (z).$

Differentiating both sides of (3.9) with respect to z, we obtain

$z (Hp,η,μλ,δzFc,p′ (f) (z))′Hp,η,μλ,δFc,pg(z)=z (Hp,η,μλ,δFc,p (g) (z))′Hp,η,μλ,δFc,p (g) (z)p (z)+zp′ (z)=t(z) p (z)+zp′ (z).$

Now using the identity (3.3) and (3.10), we obtain

$z (Hp,η,μλ,δf (z))′Hp,η,μλ,δg(z)=z (Hp,η,μλ,δzFc,p′ (f) (z))′+cHp,η,μλ,δzFc,p′ (f) (z)z (Hp,η,μλ,δFc,p (g) (z))′+cHp,η,μλ,δFc,p (g) (z)=z (Hp,η,μλ,δzFc,p′ (f) (z))′Hp,η,μλ,δFc,p (g) (z)+cz (Hp,η,μλ,δFc,p (f) (z))′Hp,η,μλ,δFc,p (g) (z)z (Hp,η,μλ,δFc,p (g) (z))′Hp,η,μλ,δFc,p (g) (z)+c=t(z) p (z)+zp′ (z)+cp (z)t(z)+c=p (z)+zp′ (z)t(z)+c.$

Since c (k + 1) + kp + ρ ≥ 0 and $ℜ{t(z)}>kp+ηk+1$, we see that

$ℜ{t(z)+c}>0 (z∈U).$

Hence, applying Lemma 2.2 to (3.11), we can show that p (z) ≺ qp,k,γ (z) so that $Fc,p (f)∈UKp,η,μλ,δ (k;γ,ρ)$.

### Theorem 3.4

Let c (k + 1)+kp + η ≥ 0. If $f∈UQp,η,μλ,δ (k;γ,ρ)$, then $Fc,p (f)∈UQp,η,μλ,δ (k;γ,ρ)$.

Proof

Just as we derived Theorem 3.2 as consequence of Theorem 3.1, we easily deduce the integral-preserving property asserted by Theorem 3.4 by using Theorem 3.3.

Acknowledgements

The author is grateful to the referees for their valuable suggestions.

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