International Journal of Analysis and Applications

Volume 19, Number 5 (2021), 784-793

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

DOI: 10.28924/2291-8639-19-2021-784

ON MOCANU-TYPE FUNCTIONS WITH GENERALIZED BOUNDED VARIATIONS

SHUJAAT ALI SHAH∗, MUHAMMAD AFZAL SOOMRO AND ASGHAR ALI MAITLO

Department of Mathematics and Statistics, Quaid-i-Awam University of Engineering, Science and

Technology, Nawabshah, 67480 Sindh, Pakistan

∗Corresponding author: shahglike@yahoo.com

Abstract. The main focus of this article is the study of classes Mδµ (ϕ, H) and Qδµ (ϕ, g1, H). We present

various inclusion relationships and some applications of our investigations are considered. Also, we include

radius problem.

1. Introduction

Let A be the class of analytic functions of the form

f(z) = z +

∞∑
n=2

anz
n, (1.1)

in the open unit disk U = {z : |z| < 1}. If f and g are analytic in U, we say that f is subordinate to g,

written f ≺ g or f(z) ≺ g(z), if there exists a Schwartz function w in U such that f(z) = g(w(z)).

The convolution or Hadamard product of two functions f,g ∈A is denoted by f ∗g and is defined as

(f ∗g)(z) = z +
∞∑
n=2

anbnz
n, z ∈U. (1.2)

Analytic functions p in the class P[A,B] can be defined by using subordination as follows [3].

Let p be analytic in U with p(0) = 1. Then p ∈P[A,B], if and only if,

p(z) ≺
1 + Az

1 + Bz
, − 1 ≤ B < A ≤ 1, z ∈U.

Received July 12th, 2021; accepted August 17th, 2021; published September 14th, 2021.

2010 Mathematics Subject Classification. 30C45, 30C55.

Key words and phrases. analytic functions; Janowski functions; conic regions; bounded boundary rotations.

©2021 Authors retain the copyrights

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

784

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-19-2021-784


Int. J. Anal. Appl. 19 (5) (2021) 785

For k > 0, the conic domains Ωk, defined as;

Ωk =

{
u + iv : u > k

√
(u− 1)2 + v2

}
.

The domains Ωk (k = 0) represents right half plane, Ωk (0 < k < 1) represents hyperbola, Ωk (k = 1)

represents a parabola and Ωk (k > 1) represents an ellipse. The extremal functions for these conic regions

are given as;

pk(z) =




1+z
1−z , k = 0

1 + 2
π2

(
log

1+
√
z

1−
√
z

)2
, k = 1

1 + 2
1−k2

[(
2
π

arccos k
)

arctan h
√
z
]
, 0 < k < 1

1 + 1
k2−1 sin

(
π

2R(t)

∫ u(z)√t
0

1√
1−x2
√

1−(tx)2
dx

)
+ 1

k2−1,k > 1,

(1.3)

where u(z) = z−
√
t

z−
√
tz
, t ∈ (0, 1), z ∈ U and z is chosen such that k = cosh

(
πR′(t)
4R(t)

)
, R(t) is Legendre’s

complete elliptic integral of the first kind and R′(t) is complementary integral of R(t). See [4, 5] for more

information. These conic regions are being studied by several authors, see [6, 9, 12].

In 2017, Dziok and Noor [2] introduced and studied the concepts of some general classes given as below.

Definition 1.1. Let µ ≥ 0, Φ = (φ1(z),φ2(z)) and H = (h1(z),h2(z)) where hi ∈ A with hi(0) = 1,

(i = 1, 2). Then

Pµ(H) = {µq1 + (1 −µ) q2 : q1 ∈P (h1) , q2 ∈P (h2)} ,

where

P (h) = {q ∈A : q ≺ h with q(0) = 1} .

Some special cases:

(i) Pµ(h) = Pµ((h,h)). If µ = m4 +
1
2
, (m ≥ 2), then Pµ(h) = Pm(h).

(ii) If µ = m
4

+ 1
2
, (m ≥ 2) , and h(z) = 1+(1−2ρ)z

1−z , then Pµ(h) = Pm(ρ), this class was introduced by

Padmanabhan et al. [13].

(iii) If µ = m
4

+ 1
2
, (m ≥ 2) and h(z) = 1+Az

1+Bz
(−1 ≤ B < A ≤ 1), then Pµ(h) = Pm [A,B], this class was

introduced by Noor [10]. Moreover, for A = 1 and B = −1 we have Pµ(h) = Pm; see [14].

(iv) If µ = m
4

+ 1
2
, (m ≥ 2) and h(z) = pκ(z) (κ ≥ 0), then Pµ(h) = Pm (pκ), this class was defined by

Noor et al. [11].

Definition 1.2. Let f ∈A and δ ≥ 0. Then f ∈ Mδµ (Φ,ξ,H) if and only if Jδ (f ((z))) ∈Pµ(H), where

Jδ (f ((z))) = (1 −δ)
(ξ ∗φ2) ∗f
(ξ ∗φ1) ∗f

+ δ
φ2 ∗f
φ1 ∗f

.

If ξ1(z) = z +
∞∑
n=2

1
n
zn, φ1(z) = zϕ

′(z) and φ2(z) = zφ
′
1(z), then we have the following special cases.

Mδ (Φ,ξ,h) = Mδ1 (Φ,ξ, (h,h)) , M
δ
µ (Φ,H) = M

δ
µ (Φ,ξ1,H) ,



Int. J. Anal. Appl. 19 (5) (2021) 786

Mδµ (ϕ,H) = M
δ
µ ((φ2,φ1) ,H) , (1.4)

S∗µ (ϕ,H) = M
0
µ (ϕ,H) , S

∗ (ϕ,h) = M01 (ϕ,h) . (1.5)

Definition 1.3. Let f ∈ A, G = (g1,g2), where gi ∈ A with gi(0) = 1 (i = 1, 2), and δ,ϑ ≥ 0. Then

f ∈Qδµ,ϑ (Φ,ξ,G,H) if there exists g ∈ S
∗
ϑ (ϕ,G) such that

(1 −δ)
(ξ ∗φ2) ∗f
(ξ ∗φ1) ∗g

+ δ
φ2 ∗f
φ1 ∗g

∈Pµ (H) .

If ξ1(z) = z +
∞∑
n=2

1
n
zn, φ1(z) = zϕ

′(z) and φ2(z) = zφ
′
1(z), then we have the following special cases.

Qδ (Φ,ξ,g1,h1) = Mδ1,1 (Φ,ξ, (g1,g2) , (h1,h2)) ,

Qδµ,ϑ (Φ,G,H) = M
δ
µ,ϑ (Φ,ξ1,G,H) ,

Qδµ (ϕ,g1,H) = Q
δ
µ,1 ((φ2,φ1) , (g1,g1) ,H) . (1.6)

From (1.4), we denote the class Mδµ (ϕ,H) of functions f ∈A satisfies Jδ (f(z)) ∈Pµ(H), where

Jδ (f(z)) = (1 − δ)
z (ϕ∗f)′

(ϕ∗f)
+ δ

(
z (ϕ∗f)′

)′
(ϕ∗f)′

,

and Pµ(H) is given by Definition 1.1.

Similarly, from (1.6), we denote the class Qδµ (ϕ,h,H) of functions f ∈A satisfies Jδ (f(z),g(z)) ∈Pµ(H),

where

Jδ (f(z),g(z)) = (1 −δ)
z (ϕ∗f)′

(ϕ∗g)
+ δ

(
z (ϕ∗f)′

)′
(ϕ∗g)′

,

for g ∈ S∗ (ϕ,h), the class S∗ (ϕ,h) is given by (1.5).

2. Preliminary Results

Lemma 2.1. [2] Let H = (h1,h2), where hi (i = 1, 2) are analytic, univalent convex functions with hi(0) = 1

(i = 1, 2) and let κ : U → C (set of complex numbers) with <(κ) > 0. If p(z) is analytic, with p(0) = 1 in

U, satisfies

p(z) + κzp′(z) ∈Pµ(H),

then p(z) ∈Pµ(H).



Int. J. Anal. Appl. 19 (5) (2021) 787

Lemma 2.2. [8] Let h be analytic, univalent convex function in U with h(0) = 1 and Re (γh(z) + σ) > 0,

σ,γ ∈ C and γ 6= 0. If p(z) is analytic in U and p(0) = h(0), then{
p(z) +

zp′(z)

γp(z) + σ

}
≺ h(z),

implies p(z) ≺ q(z) ≺ h(z), where q(z) is best dominant and is given as,

q(z) =

[{∫ 1
0

(
exp

∫ tz
t

h(u) − 1
u

du

)
dt

}−1
−
σ

γ

]
.

Lemma 2.3. [15] If f ∈ C,g ∈ S∗, then for each h analytic in U with h(0) = 1,

(f ∗hg) (U)
(f ∗g) (U)

⊂ Coh(U),

where Coh(U) denotes the convex hull of h(U).

3. Main Results

3.1. Inclusion Results.

Theorem 3.1. Let δ ≥ 0, ϕ ∈A and h be any convex univalent function in U. Then

Mδ1 (ϕ,h) ⊂ M
0
1 (ϕ,h) .

Proof. Let f ∈ Mδ1 (ϕ,h). Then, by definition,

(1 −δ)
z (ϕ∗f)′

(ϕ∗f)
+ δ

(
z (ϕ∗f)′

)′
(ϕ∗f)′

∈P(h),

or

(1 − δ)
z (ϕ∗f)′

(ϕ∗f)
+ δ

(
z (ϕ∗f)′

)′
(ϕ∗f)′

≺ h(z). (3.1)

Consider
z (ϕ∗f)′

(ϕ∗f)
= p(z). (3.2)

On logarithmic differentiation of (3.2), we have(
z (ϕ∗f)′

)′
(ϕ∗f)′

=
z (ϕ∗f)′

(ϕ∗f)
+
zp′(z)

p(z)
. (3.3)

From (3.2) and (3.3), we get (
z (ϕ∗f)′

)′
(ϕ∗f)′

= p(z) +
zp′(z)

p(z)
. (3.4)

On making use of (3.2) and (3.4) in (3.1), we obtain

(1 −δ) p(z) + δ
[
p(z) +

zp′(z)

p(z)

]
≺ h(z),

this implies

p(z) + δ
zp′(z)

p(z)
≺ h(z).

By using Lemma 2.2, we conclude p(z) ≺ h(z). Hence f ∈ M01 (ϕ,h). �



Int. J. Anal. Appl. 19 (5) (2021) 788

Remark 3.1. Following different choices of ϕ and h give certain inclusion results for the above theorem.

(i) ϕ ∈ A, h(z) = 1+Az
1+Bz

, where −1 ≤ B < A ≤ 1.

(ii) ϕ ∈ A, h(z) = pk(z), where pk(z) is given by (1.3).

Corollary 3.1. Let δ ≥ 1. Then

Mδ1 (ϕ,h) ⊂ M
1
1 (ϕ,h) .

Proof. Let f ∈ Mδ1 (ϕ,h). Then , by definition,

(1 − δ)
z (ϕ∗f)′

(ϕ∗f)
+ δ

(
z (ϕ∗f)′

)′
(ϕ∗f)′

= s1(z) ≺ h(z),

from previous theorem, we can write

z (ϕ∗f)′

(ϕ∗f)
= s2(z) ≺ h(z).

Now,

δ

(
z (ϕ∗f)′

)′
(ϕ∗f)′

=

[
(1 −δ)

z (ϕ∗f)′

(ϕ∗f)
+ δ

(
z (ϕ∗f)′

)′
(ϕ∗f)′

]
+ (δ − 1)

z (ϕ∗f)′

(ϕ∗f)

= s1(z) + (δ − 1) s2(z).

Implies that (
z (ϕ∗f)′

)′
(ϕ∗f)′

=

(
1 −

1

δ

)
s2(z) +

1

δ
s1(z). (3.5)

Since s1,s2 ≺ h(z), (3.5) gives us (
z (ϕ∗f)′

)′
(ϕ∗f)′

≺ h(z).

Hence f ∈ Mδ1 (ϕ,h). �

Remark 3.2. The different choices of ϕ and h given in Remark 3.1 hold the inclusion result proved in above

theorem.

Theorem 3.2. Let δ, µ ≥ 0, ϕ ∈A, H = (h1,h2) where hi,h ∈A with hi(0) = h(0) = 1 (i = 1, 2). Then

Qδµ (ϕ,h,H) ⊂Q
0
µ (ϕ,h,H) .

Proof. Let f ∈Qδµ (ϕ,h,H). Then, by definition,

(1 − δ)
z (ϕ∗f)′

(ϕ∗g)
+ δ

(
z (ϕ∗f)′

)′
(ϕ∗g)′

∈Pµ(H), (3.6)

for g ∈ S∗ (ϕ,h).

Consider

z (ϕ∗f)′

(ϕ∗g)
= p(z), (3.7)



Int. J. Anal. Appl. 19 (5) (2021) 789

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

On logarithmic differentiation of (3.7), we get(
z (ϕ∗f)′

)′
(ϕ∗f)′

=
z (ϕ∗g)′

(ϕ∗g)
+
zp′(z)

p(z)
,

(
z (ϕ∗f)′

)′
(ϕ∗g)′

=
z (ϕ∗f)′

(ϕ∗g)′


z (ϕ∗g)′

(ϕ∗g)
+

zp′(z)
z(ϕ∗f)′
(ϕ∗g)


 ,

this implies (
z (ϕ∗f)′

)′
(ϕ∗g)′

=
z (ϕ∗f)′

(ϕ∗g)
+
zp′(z)
z(ϕ∗g)′
(ϕ∗g)

. (3.8)

From (3.7) and (3.8), we have(
z (ϕ∗f)′

)′
(ϕ∗g)′

= p(z) +
zp′(z)

p0(z)
; with p0(z) =

z (ϕ∗g)′

(ϕ∗g)
. (3.9)

Now, from (3.6), (3.7) and (3.9), we obtain

(1 − δ) p(z) + δ
(
p(z) +

zp′(z)

p0(z)

)
∈Pµ(H),

or equivalently,

p(z) +
δ

p0(z)
zp′(z) ∈Pµ(H).

If g ∈ S∗ (ϕ,h), then z(ϕ∗g)
′

(ϕ∗g) ≺ h(z); h ∈ P. This implies <(p0(z)) > 0 in U. Thus, by Lemma 2.1, we

conclude p(z) ∈Pµ(H). Consequently,
z(ϕ∗f)′
(ϕ∗g) ∈Pµ(H). Hence, f ∈Q

0
µ (ϕ,h,H). �

Remark 3.3. It is easy to see that the inclusion in Theorem 3.2 is true for different choices of ϕ, h and

H = (h1,h2) given as following.

(i) ϕ ∈ A, h1(z) = 1+Az1+Bz = h2(z), where −1 ≤ B < A ≤ 1.

(ii) ϕ ∈ A, h1(z) = pk(z) = h2(z), where pk(z) is given by (1.3).

(iii) ϕ ∈ A, h1(z) = 1+Az1+Bz , h2(z) = pk(z).

(iv) ϕ ∈ A, h1(z) = pk(z), h2(z) = 1+Az1+Bz .

Corollary 3.2. Let δ ≥ 1. Then

Qδµ (ϕ,h,H) ⊂Q
1
µ (ϕ,h,H) .

Proof. Let f ∈Qδµ (ϕ,h,H). Then, by definition,

(1 −δ)
z (ϕ∗f)′

(ϕ∗g)
+ δ

(
z (ϕ∗f)′

)′
(ϕ∗g)′

= p1(z) ∈Pµ(H),

where g ∈ S∗ (ϕ,h).

From previous theorem, we can write

z (ϕ∗f)′

(ϕ∗g)
= p2(z) ∈Pµ(H).



Int. J. Anal. Appl. 19 (5) (2021) 790

Now,

δ

(
z (ϕ∗f)′

)′
(ϕ∗g)′

=

[
(1 − δ)

z (ϕ∗f)′

(ϕ∗g)
+ δ

(
z (ϕ∗f)′

)′
(ϕ∗g)′

]
+ (δ − 1)

z (ϕ∗f)′

(ϕ∗g)

= p1(z) + (δ − 1) p2(z).

This implies (
z (ϕ∗f)′

)′
(ϕ∗g)′

=

(
1 −

1

δ

)
p2(z) +

1

δ
p1(z).

Since p1, p2 ∈Pµ(H) and Pµ(H) is convex set, then(
z (ϕ∗f)′

)′
(ϕ∗g)′

∈Pµ(H).

Hence f ∈Q1µ (ϕ,h,H). �

Theorem 3.3. Let 0 ≤ δ1 < δ. Then

Qδµ (ϕ,h,H) ⊂Q
δ1
µ (ϕ,h,H) .

Proof. If δ1 = 0, then it is obvious from Theorem 3.2.

For δ1 > 0. Let f ∈Qδµ (ϕ,h,H). Then, from Theorem 3.2

z (ϕ∗f)′

(ϕ∗g)
= p2(z) ∈Pµ(H). (3.10)

As we can write

(1 − δ1)
z (ϕ∗f)′

(ϕ∗g)
+ δ1

(
z (ϕ∗f)′

)′
(ϕ∗g)′

=
δ1
δ

[(
δ

δ1
− 1
)
z (ϕ∗f)′

(ϕ∗g)
+ (1 − δ)

z (ϕ∗f)′

(ϕ∗g)
+ δ

(
z (ϕ∗f)′

)′
(ϕ∗g)′

]
. (3.11)

Since f ∈Qδµ (ϕ,h,H), from definition of Qδµ (ϕ,h,H), we have

(1 −δ)
z (ϕ∗f)′

(ϕ∗g)
+ δ

(
z (ϕ∗f)′

)′
(ϕ∗g)′

= p1(z) ∈Pµ(H). (3.12)

From (3.10-3.12) and the convexity of Pµ(H) implies

(1 − δ1)
z (ϕ∗f)′

(ϕ∗g)
+ δ1

(
z (ϕ∗f)′

)′
(ϕ∗g)′

∈Pµ(H).

Hence f ∈Qδ1µ (ϕ,h,H). �

Remark 3.4. It is easy to see that the inclusion in Theorem 3.3 is true for all choices given in Remark 3.3.

Theorem 3.4. The class Qδµ (ϕ,h,H) is closed under the convex convolution.



Int. J. Anal. Appl. 19 (5) (2021) 791

Proof. Let f ∈Qδµ (ϕ,h,H). Then, by definition,

(1 − δ)
z (ϕ∗f)′

(ϕ∗g)
+ δ

(
z (ϕ∗f)′

)′
(ϕ∗g)′

∈Pµ(H). (3.13)

First, we need to prove ς ∗f ∈Q0µ (ϕ,h,H) for ς ∈ C.

We take δ = 0, then (3.13) implies

z (ϕ∗f)′

(ϕ∗g)
∈Pµ(H). (3.14)

Let

z (ϕ∗ (ς ∗f))′ (z)
(ϕ∗ (ς ∗g)) (z)

=
ς ∗ z(ϕ∗f)

′

(ϕ∗g) ((ϕ∗g)) (z)
ς ∗ (ϕ∗g) (z)

=
ς ∗h0(z) ((ϕ∗g)) (z)

ς ∗ (ϕ∗g) (z)
,

where h0(z) =
z(ϕ∗f)′
(ϕ∗g) ∈ Pµ(H). Since g ∈ S

∗(ϕ,h) implies ϕ ∗ g ∈ S∗(h) ⊂ S∗; h ∈ P. Thus, by Lemma

2.3, we conclude

z (ϕ∗ (ς ∗f))′ (z)
(ϕ∗ (ς ∗g)) (z)

∈Pµ(H). (3.15)

Similarly, for δ = 1, we can easily prove

z
(
ϕ∗ (ς ∗f)′

)′
(z)

(ϕ∗ (ς ∗g))′ (z)
∈Pµ(H). (3.16)

Our required result follows from (3.15) and (3.16). �

Corollary 3.3. The class Qδµ (ϕ,h,H) is closed under the following operators.

(i) f1(z) =
∫ z
0
f(t)
t
dt.

(ii) f2(z) =
2
z

∫ z
0
f(t)dt, (Libera’s operator [7]).

(iii) f3(z) =
∫ z
0
f(t)−f(xt)

t−xt dt, |x| ≤ 1, x 6= 1.

(iv) f4(z) =
c+1
zc

∫ z
0
tc−1f(t), Re(c) ≥ 0, (Generalized Bernardi operator [1]).

Proof. We may write, fi(z) = f(z) ∗φi(z), where φi(z), i = 1, 2, 3, 4, are convex and given by

φ1(z) = − log (1 −z) =
∞∑
n=1

1
n
zn,

φ2(z) =
−2[z−log(1−z)]

z
=
∞∑
n=1

2
n+1

zn,

φ3(z) =
1

1−x log
(

1−xz
1−z

)
=
∞∑
n=1

1−xn
(1−x)n z

n, |x| ≤ 1, x 6= 1,

φ4(z) =
∞∑
n=1

1+c
n+c

zn, Re(c) ≥ 0.

The proof follows easily by using Theorem 3.4. �



Int. J. Anal. Appl. 19 (5) (2021) 792

3.2. Radius Problem.

Theorem 3.5. Let f ∈ M01
(
ϕ, 1+Az

1+Bz

)
. Then, f ∈ Mδ1

(
ϕ, 1+z

1−z

)
for |z| < rδ, where

rδ =
2A2

{δ (A−B) + 2A} +
√
δ2 (A−B)2 + 4Aδ (A−B)

.

Proof. Let f ∈ M01
(
ϕ, 1+Az

1+Bz

)
. Then, by definition,

z (ϕ∗f)′

(ϕ∗f)
= p(z) ≺

1 + Az

1 + Bz
. (3.17)

On logrithmic differentiation of (3.17), we get

(
z (ϕ∗f)′

)′
(ϕ∗f)′

=
z (ϕ∗f)′

(ϕ∗f)
+
zp′(z)

p(z)
. (3.18)

By (3.17) and (3.18), we obtain (
z (ϕ∗f)′

)′
(ϕ∗f)′

= p(z) +
zp′(z)

p(z)
. (3.19)

Now,

(1 − δ)
z (ϕ∗f)′

(ϕ∗g)
+ δ

(
z (ϕ∗f)′

)′
(ϕ∗g)′

= p(z) + δ
zp′(z)

p(z)
.

<(Jδ (f(z))) ≥
A2r2 −{δ (A−B) + 2A}r + 1

(1 −Ar) (1 −Br)
.

For <(Jδ (f(z))) > 0 in U, we get

rδ =
2A2

{δ (A−B) + 2A} +
√
δ2 (A−B)2 + 4Aδ (A−B)

.

�

Corollary 3.4. Let f ∈ M01
(

z
1−z ,

1+z
1−z

)
= S∗. Then

f ∈ Mδ1
(

z

1 −z
,

1 + z

1 −z

)
= M(δ),

for |z| < rδ = 1(1+δ)+√δ2+2δ . Moreover, for δ = 1, we have well known result

S∗ ⊂ C, for |z| < r1 =
1

2 +
√

3
.

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

of this paper.



Int. J. Anal. Appl. 19 (5) (2021) 793

References

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	1. Introduction
	2. Preliminary Results
	3. Main Results
	3.1. Inclusion Results
	3.2. Radius Problem

	References