International Journal of Analysis and Applications

Volume 18, Number 4 (2020), 550-558

URL: https://doi.org/10.28924/2291-8639

DOI: 10.28924/2291-8639-18-2020-550

ON q-MOCANU TYPE FUNCTIONS ASSOCIATED WITH q-RUSCHEWEYH

DERIVATIVE OPERATOR

KHALIDA INAYAT NOOR AND SHUJAAT ALI SHAH∗

COMSATS University Islamabad, Pakistan

∗Corresponding author: shaglike@yahoo.com

Abstract. In this paper, we introduce certain subclasses of analytic functions defined by using the q-

difference operator. Mainly we give several inclusion results for defined classes. Also, certain applications

due to q-Ruscheweyh derivative operator will be discussed.

1. Introduction

Let A denotes the class of analytic functions f(z) in the open unit disk E = {z : |z| < 1} such that

f(z) = z +

∞∑
n=2

anz
n. (1.1)

Subordination of two functions f and g is denoted by f ≺ g and defined as f(z) = g(w(z)), where w(z) is

Schwartz function in E (see [10]). Let S, S∗ and C denote the subclasses of A of univalent functions, starlike

functions and convex functions respectively. Mocanu [11] introduced the class M (α) of α−convex functions

f ∈ S satisfies; (
(1 −α)

zf′(z)

f(z)
+ α

(zf′(z))
′

f′(z)

)
≺

1 + z

1 −z
,

where α ∈ [0, 1], f(z)
z
f′(z) 6= 0 and z ∈ E. We see that M0 = S∗ and M1 = C. This class is vastly studied

by several authors, see [2, 14].

We recall here some basic definitions and concept details of q-calculus that are used in this paper.

Received February 4th, 2020; accepted March 31st, 2020; published May 11th, 2020.

2010 Mathematics Subject Classification. 30C45, 30C55.

Key words and phrases. Mocanu functions; q-difference operator; q-Ruscheweyh derivative operator.

©2020 Authors retain the copyrights

of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

550

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-18-2020-550


Int. J. Anal. Appl. 18 (4) (2020) 551

The q-difference operator, which was introduced by Jackson [7], defined by

Dqf(z) =
f(z) −f(qz)

(1 −q)z
; q 6= 1, z 6= 0,

for q ∈ (0, 1). It is clear that limq→1− Dqf(z) = f′(z), where f′(z) is the ordinary derivative of the function.

For more properties of Dq; see [3–5, 9, 18].

It can easily be seen that, for n ∈ N = {1, 2, 3, ..} and z ∈ E,

Dq

{
∞∑
n=1

anz
n

}
=

∞∑
n=1

[n]q z
n−1,

where

[n]q =
1 −qn

1 −q
= 1 + q + q2 + ... .

We have the following rules of Dq.

Dq (af (z) ± bg (z)) = aDqf (z) ± bDqg (z) .

Dq (f (z) g (z)) = f (qz) Dq (g (z)) + g(z)Dq (f (z)) .

Dq

(
f(z)

g(z)

)
=
Dq (f(z)) g(z) −f(z)Dq (g(z))

g(qz)g(z)
, g(qz)g(z) 6= 0.

Dq (log f(z)) =
Dq (f(z))

f(z)
.

Some properties related with function theory involving q-theory were first introduced by Ismail et al. [6].

Moreover, several authors studied in this matter such as [1, 12, 13, 15].

Now, by making use of the principle of subordination together with q-difference operator, we have the

following classes:

Let a function p ∈ A with p(0) = 1 is in the class P̃q(β) if and only if

p(z) ≺ pq,β(z), where pq,β(z) =
(

1 + z

1 −qz

)β
, (0 < β ≤ 1) . (1.2)

It is very easy to see that pq,β(z) is convex univalent in E for 0 < β ≤ 1. Aslo, pq,β(z) is symmetric with

respect to the real axis, that is,

0 < <(pq,β(z)) <
(

1

1 −q

)β
.



Int. J. Anal. Appl. 18 (4) (2020) 552

Definition 1.1. Let function f ∈ A and 0 ≤ α ≤ 1, q ∈ (0, 1). Then f ∈ Mβq (α) if and only if

Jq (α,f) ∈ P̃q(β),

where

Jq (α,f) = (1 −α)
zDqf

f
+ α

Dq (zDqf)

Dqf
.

Moreover, let us denote

Mβq (0) = S
∗
q (β) , M

β
q (1) = Cq (β) .

A function f ∈ A is said to be in S∗q (β) and Cq (β) if and only if

zDqf(z)

f(z)
≺ pq,β(z) and

Dq (zDqf(z))

Dqf(z)
≺ pq,β(z),

respectively.

Special cases:

(i) If q → 1−, then the class Mβq (α) reduces to the class Mβ (α).

(ii) If q → 1− and β = 1, then the class Mβq (α) reduces to the class M (α) introduced by Mocanu [11].

(iii) If q → 1−, α = 0 and β = 1, then the class Mβq (α) reduces to the well known class S∗ of starlike

functions.

(iv) If q → 1−, α = 1 and β = 1, then the class Mβq (α) reduces to the well known class C of convex

functions.

The authors in [8], introduced an operator Rλq : A → A defined as:

Rλqf(z) = zλ+1,q(z) ∗f(z) (1.3)

= z +

∞∑
n=2

[n + λ− 1]q!
[λ]q! [n− 1]q!

anz
n, (1.4)

where f ∈ A, zλ+1,q(z) = z +
∑∞
n=2

[n+λ−1]
q
!

[λ]
q
![n−1]

q
!
zn and * denotes convolution.

This series (1.4) is absolutely convergent in E. For q → 1−, we have the operator Rλ, called Ruscheweyh

derivative operator introduced in [16].

In this case

Rλf(z) = lim
q→1−

Rλqf(z) = z +

∞∑
n=2

(n + λ− 1)!
λ! (n− 1)!

anz
n

=
z

(1 −z)λ+1
∗f(z).

We note that R0qf(z) = f(z) and R
1
qf(z) = zDqf(z). Also

Rnq f(z) =
zDnq

(
zn−1f(z)

)
[n]q!

; n ∈ N = {1, 2, 3, ...} .



Int. J. Anal. Appl. 18 (4) (2020) 553

The following identity can be easily obtained from (1.4)

zDq
(
Rλqf(z)

)
=

(
1 +

[λ]q
qλ

)
Rλ+1q f(z) −

[λ]q
qλ

Rλqf(z). (1.5)

Now, we define

Definition 1.2. Let f ∈ A and n ∈ N, 0 ≤ α ≤ 1, q ∈ (0, 1) and β ∈ (0, 1]. Then

f ∈ Mβq (n,α) if and only if R
n
q f(z) ∈ M

β
q (α) .

Moreover, let us denote

Mβq (n, 0) = S
∗
q (n,β) and M

β
q (n, 1) = Cq (n,β) .

Note that

f ∈ Cq (n,β) ⇔ zDqf ∈ S∗q (n,β) . (1.6)

2. Main Results

We need the following basic result to prove our main results:

Lemma 2.1. [17] Let β and γ be complex numbers with β 6= 0 and let h(z) be analytic in E with h(0) = 1

and Re{βh(z) + γ} > 0. If p(z) = 1 + p1z + p2z2 + ... is analytic in E, then

p(z) +
zDqp(z)

βp(z) + γ
≺ h(z)

implies that p(z) ≺ h(z).

Theorem 2.1. Let 0 ≤ α ≤ 1, β ∈ (0, 1] and q ∈ (0, 1). Then

Mβq (α) ⊂ S
∗
q (β) .

Proof. Let f ∈ Mβq (α) and let
zDqf(z)

f(z)
= p(z). (2.1)

We note that p(z) is analytic in E with p(0) = 1.

The q-logarithmic differentiation of (2.1) yields

Dq (zDq (f(z)))

Dqf(z)
−
Dq (f(z))

f(z)
=
Dqp(z)

p(z)
.

Equivalently

Dq (zDq (f(z)))

Dqf(z)
= p(z) +

zDqp(z)

p(z)
.

Since f ∈ Mβq (α), so we get

Jq (α,f) = p(z) + α
zDqp(z)

p(z)
≺ pq,β(z). (2.2)



Int. J. Anal. Appl. 18 (4) (2020) 554

Since Re
{

1
α
pq,β(z)

}
> 0 in E, so by (2.2) together with Lemma 2.1, we obtain p(z) ≺ pq,β(z). Consequently

f ∈ S∗q (β). �

Corollary 2.1. For q → 1−, we have Mβ(α) ⊂ S∗ (β). Furthermore, for β = 1, M(α) ⊂ S∗.

Corollary 2.2. For q → 1−, α = 1 and β = 1, we have well known fundamental result C ⊂ S∗.

Theorem 2.2. Let α > 1, β ∈ (0, 1] and q ∈ (0, 1). Then

Mβq (α) ⊂ Cq (β) .

Proof. Let f ∈ Mβq (α). Then, by Definition 1.1,

(1 −α)
zDqf(z)

f(z)
+ α

Dq (zDqf(z))

Dqf(z)
= p1(z) ∈ P̃q(β).

Now,

α
Dq (zDqf(z))

Dqf(z)
= (1 −α)

zDqf(z)

f(z)
+ α

Dq (zDqf(z))

Dqf(z)
+ (α− 1)

zDqf(z)

f(z)

= (α− 1)
zDqf(z)

f(z)
+ p1(z).

This implies

Dq (zDqf)

Dqf
=

(
1

α
− 1
)
zDqf

f
+

1

α
p1(z)

=

(
1

α
− 1
)
p2(z) +

1

α
p1(z).

Since p1,p2 ∈ P̃q(β) and is P̃q(β) convex set, so
Dq(zDqf)
Dqf

∈ P̃q(β). Hence, proof is complete. �

Theorem 2.3. For 0 ≤ α1 < α2 < 1

Mβq (α2) ⊂ M
β
q (α1) .

Proof. For α1 = 0, this is obvious from Theorem 2.1.

Let f ∈ Mβq (α2). Then, by Definition 1.1,

(1 −α2)
zDqf(z)

f(z)
+ α2

Dq (zDqf(z))

Dqf(z)
= q1(z) ∈ P̃q(β). (2.3)

Now, we can easily write

Jq (α1,f(z)) =
α1
α2
q1(z) +

(
1 −

α1
α2

)
q2(z), (2.4)

where we have used (2.3) and
zDqf(z)
f(z)

= q2(z) ∈ P̃q(β). Since P̃q(β) is convex set, so (2.4) follows our

required result. �



Int. J. Anal. Appl. 18 (4) (2020) 555

Remark 2.1. If α2 = 1 and let f ∈ Mβq (1) = Cq(β). Then, from Theorem 2.3, we can write

f ∈ Mβq (α1) ,for 0 ≤ α1 < 1,

Now, by making use of Theorem 2.1, we obtain f ∈ S∗q (β). Thus we have, Cq(β) ⊂ S∗q (β).

We develop some applications in terms of q-linear operator, which we call q-Ruscheweyh derivative oper-

ator, given by (1.3).

Theorem 2.4. Let 0 ≤ α ≤ 1, β ∈ (0, 1], n ∈ N0 and q ∈ (0, 1). Then

Mβq (n + 1,α) ⊂ S
∗
q (n + 1,β) .

Proof. One can easily prove this result by using similar arguments as used in Theorem 2.1 and letting

zDqfn+1,q(z)

fn+1,q(z)
= p(z)

(
for fn+1,q(z) = R

n+1
q f(z)

)
,

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

Theorem 2.5. Let 0 ≤ α ≤ 1, β ∈ (0, 1], n ∈ N0 and q ∈ (0, 1). Then

S∗q (n + 1,β) ⊂ S
∗
q (n,β) .

Proof. Let f ∈ S∗q (n + 1,β) and let fn+1(z) = Rn+1q f(z). Then

zDqfn+1,q(z)

fn+1,q(z)
≺ pq,β(z),

where pq,β(z) is given by (1.2).

Now, let

zDqfn,q(z)

fn,q(z)
= H(z), (2.5)

where H(z) is analytic in E with H(0) = 1. Using identity (1.5) and (2.5), we get

zDq (fn,q(z))

fn,q(z)
= (1 + Nq)

fn+1,q(z)

fn,q(z)
−Nq,

equivalently

(1 + Nq)
fn+1,q(z)

fn,q(z)
= H(z) + Nq,

(
for Nq =

[n]q
qn

)
.

The q-logarithmic differentiation yields,

zDq (fn+1,q(z))

fn+1,q(z)
= p(z) +

zDqH(z)

H(z) + Nq
. (2.6)

Since f ∈ S∗q (n + 1,β), So (2.6) implies

p(z) +
zDqH(z)

H(z) + Nq
≺ pq,β(z). (2.7)



Int. J. Anal. Appl. 18 (4) (2020) 556

Since Re{pq,β(z) + Nq} > 0 in E, we use Lemma 2.1 along with (2.7), to get

H(z) ≺ pq,β(z). Consequently, f ∈ S∗q (n,β). �

Theorem 2.6. Let 0 ≤ α ≤ 1, β ∈ (0, 1], n ∈ N0 and q ∈ (0, 1). Then

Cq (n + 1,β) ⊂ Cq (n,β) .

Proof. Let

f ∈ Cq (n + 1,β)

⇔ zf′ ∈ S∗q (n + 1,β) (by (1.6))

⇒ zf′ ∈ S∗q (n,β) (by Theorem2.5)

⇔ f ∈ Cq (n,β) . (by (1.6))

�

Remark 2.2. From Theorem 2.4 and Theorem 2.5, we can extend the inclusions as following

Mβq (n + 1,α) ⊂ S
∗
q (n + 1,β) ⊂ S

∗
q (n,β) ⊂ ... ⊂ S

∗
q (β) .

Cq (n + 1,β) ⊂ Cq (n,β) ⊂ ... ⊂ Cq (β) .

Theorem 2.7. Let f ∈ A. Then f ∈ Mβq (n + 1,α), α 6= 0, if and only if there exists g ∈ S∗q (n + 1,β) such

that

f(z) =

[
1

α

]
q

[∫ t
0

t
1
α
−1
(
g(t)

t

) 1
α

dqt

]α
. (2.8)

Proof. Let f ∈ Mβq (n + 1,α). Then

Jq (α,f) = (1 −α)
zDqf(z)

f(z)
+ α

Dq (zDqf(z))

Dqf(z)
∈ P̃q(β). (2.9)

On some simple calculations of (2.8), we get

zDqf(z) (f(z))
1
α
−1

= (g(z))
1
α . (2.10)

The q-logarithmic differentiation of (2.10), gives

(1 −α)
zDqf(z)

f(z)
+ α

Dq (zDqf(z))

Dqf(z)
=
zDqg(z)

g(z)
. (2.11)

From (2.9) and (2.11), we conclude our required result. �



Int. J. Anal. Appl. 18 (4) (2020) 557

Theorem 2.8. Let f ∈ A and define, for f ∈ Mβq (n,α),

Fc,q(z) =
[c + 1]q
zc

∫ z
0

tb−1f(t)dqt. (2.12)

Then Fc,q ∈ S∗q (n,β).

Proof. Let f ∈ Mβq (n,α). If we set, for Fnc,q(z) = Rnq (Fc,q(z))

zDq
(
Fnc,q(z)

)
Fnc,q(z)

= Q(z), (2.13)

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

From (2.12), we can write

Dq (z
cFc,q(z))

[c + 1]q
= zc−1f(z).

Using product rule of the q-difference operator, we get

zDqFc,q(z) =

(
1 +

[c]q
qc

)
f(z) −

[c]q
qc
Fc,q(z). (2.14)

From (2.13), (2.14) and (1.3), we have

Q(z) =

(
1 +

[c]q
qc

)
z (fn,q(z))

Fnc,q(z)
−

[c]q
qc
,

where Fnc,q(z) = R
n
q (Fc,q(z)) and fn,q(z) = R

n
q (f(z))

On q-logarithmic differentiation, we get

zDq (fn,q(z))

fn,q(z)
= Q(z) +

zDqQ(z)

Q(z) + [N]q
,

(
for Nq =

[c]q
qc

)
. (2.15)

Since f ∈ Mβq (n,α) ⊂ S∗q (n,β), so (2.15) implies

Q(z) +
zDqQ(z)

Q(z) + [c]q
≺ pq,β(z).

Now, by applying Lemma 2.1, we conclude Q(z) ≺ pq,β(z). Consequently,
zDq(Fnc,q(z))

Fnc,q(z)
≺ pq,β(z). Hence

Fc,q ∈ S∗q (n,β). �

Conflicts of Interest: The author(s) declare that there are no conflicts of interest regarding the publication

of this paper.



Int. J. Anal. Appl. 18 (4) (2020) 558

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	1. Introduction
	2. Main Results
	References