9_xiong.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 5, No. 3, 2012, 380-389
ISSN 1307-5543 – www.ejpam.com

Some Results for Certain Subclasses of Functions with

Differential Equation and Subordination

Liangpeng Xiong∗, Xiaoli Liu

The College of Engineering and Technical, ChengDu University of Technology University, Leshan,

Sichuan, 614000, P.R China

Abstract. By applying the differential subordination theorem, we further investigate the subclass
H

n,γ
λ
[α,β] of functions which are analytic in the unit disk. Several subordination results on a con-

vex function and a incomplete beta function are obtained. Moreover, the function that belongs to the
H

n,γ
λ
[α,β] with a Cauchy-Euler differential equation is also discussed on similar subject. Our results

extend some earlier works.
2010 Mathematics Subject Classifications: 30C45
Key Words and Phrases: Analytic functions, Convex function, Subordination, Differential equation

1. Introduction and Definition

Let A denote the class of all functions of the form f (z) = z+
∞
∑

k=2
akzk, which are analytic

in the open unit disk U = {z ∈ C, |z| < 1} and let S be the subclass of A consisting of
univalent functions. K denotes the usual class of convex functions.

Suppose that the functions f and g are analytic in U. We say that f is subordinate to g

in U if there exists a functions φ analytic in U such that φ(0) = 0, |φ(z)| < 1 (|z| < 1) and
f (z) = g(φ(z)) (|z| < 1), written f ≺ g.

Let be given two functions f (z) = z +
∞
∑

k=2
akzk and g(z) = z +

∞
∑

k=2
bkzk analytic in the

open unit disc U = {z ∈ C : |z| < 1}, then the Hadamard product(or convolution) f ∗ g of two

functions f , g is defined by f ∗ g(z) = z +
∞
∑

k=2
ak bkzk.

Let (x)k be the pochhammer symbol defined by
(

1, k = 0, x ∈C/{0},

x(x + 1)(x + 2) . . . (x + k− 1), k ∈ N = {1,2,3, . . .}, x ∈ C.

∗Corresponding author.

Email addresses: xlpwxf�163.
om (L. Xiong), travel-lxl�163.
om (X. Liu)

http://www.ejpam.com 380 c© 2012 EJPAM All rights reserved.



L. Xiong, X. Liu / Eur. J. Pure Appl. Math, 5 (2012), 380-389 381

In [7], Ruscheweyh defined the incomplete beta function

h(a, c; z) = z +

∞
∑

k=2

(a)k−1

(c)k−1
zk |z| < 1, (1)

where a is any real number and c 6= {0,−1,−2, . . .}.
Now we recall the linear multiplier fractional differential operator D

n,γ
λ

introduced and
studied by Al-Oboudi and Al-Amoudi [1] as follows:

D
0,0
λ

f (z) = f (z),

D
1,γ
λ

f (z) = λz(Ωγ f (z))′+ (1− γ)Ωγ f (z) = D
γ

λ
f (z),

D
2,γ
λ

f (z) = D
γ

λ
(D

1,γ
λ

f (z)),

. . . . . .

D
n,γ
λ

f (z) = D
γ

λ
(D

n−1,γ
λ

f (z)),

for n ∈ N,λ ¾ 0 and 0 ¶ γ < 1, where Ωγ f (z) = Γ(2− γ)zγD
γ
z f (z) is an extension of the

fractional derivative and fractional integral defined by Owa and Srivastava [6].

Suppose f (z) = z +
∞
∑

k=2
akzk, in the light of the above definitions, it is easy to conclude

that

D
n,γ
λ
= z +

∞
∑

k=2

[ψk(γ,λ)]nakzk, n ∈ N0 = N∪ {0},

where

ψk(γ,λ) =
Γ(k+ 1)Γ(2− γ)

Γ(k+ 1− γ)
[1+λ(k− 1)] (k = 2,3, . . .). (2)

Let T denote the subclass of S whose elements can be expressed in the form

f (z) = z +

∞
∑

k=2

akzk ak ¶ 0.

Using the differential operator D
n,γ
λ

, Marouf [5] introduced and studied the class H
n,γ
λ
[α,β].

As a function f (z) ∈ T is in the H
n,γ
λ
[α,β] if and only if it satisfies

ℜ
�

α
D

n+2,γ
λ

f (z)

D
n,γ
λ

f (z)
+ (1−α)

D
n+1,γ
λ

f (z)

D
n,γ
λ

f (z)

�

> β (α¾ 0; 0¶ β < 0).

In particular, the class H
0,0
1 [α,β] ≡ H̄[α,β] was studied by Lashin [3] and the classes

H
0,0
1 [0,β]≡ T ∗(β) and H

0,0
1 [1,β]≡ C(β) were studied by Silverman [8].

To prove our results we shall need the following Definition and Lemma:

Definition 1. [See 9] An infinite sequence {bn}
∞
n=1 of complex numbers will be called a subordi-

nating factor sequence if whenever f ∈K , we have the subordination given by

∞
∑

n=1

an bnzn ≺ f (z) (z ∈ U, a1 = 1).



L. Xiong, X. Liu / Eur. J. Pure Appl. Math, 5 (2012), 380-389 382

Lemma 1. [See 9] The sequence {bn}
∞
n=1 is subordinating factor sequence if and only if

ℜ
�

1+ 2
∞
∑

n=1

bnzn
	

> 0 (z ∈ U).

Lemma 2. [See 7] Let 0< a ¶ c. If c ¾ 2 or a+ c ¾ 3, then the function

h(a, c; z) = z +

∞
∑

k=2

(a)k−1

(c)k−1
zk (z ∈ U)

belongs to the class K of convex functions.

In [5], Marouf proved the sufficient and necessary condition on a function

f (z) = z +
∞
∑

k=2
akzk ∈ T to be H

n,γ
λ
[α,β], which is equivalent to the following Lemma:

Lemma 3. [See 5] A function f (z) ∈ T is in the H
n,γ
λ
[α,β] if and only if

∞
∑

k=2

[(αψk(γ,λ) + 1)(ψk(γ,λ)− 1)+ 1− β][ψk(γ,λ)]n|ak| ¶ 1− β (3)

which ψk(γ,λ) is defined as (2).

Lemma 4. [See 4] If the functions f (z) and g(z) are analytic in U with g(z) ≺ f (z), then for

s > 0 and z = reiθ (0< r < 1), we have

∫ 2π

0

| f (reiθ )|s ¶

∫ 2π

0

|g(reiθ )|s.

2. Some Results on the Class H
n,γ
λ
[α,β]

We begin with the following theorem:

Theorem 1. If f ∈ H
n,γ
λ
[α,β] in U and s > 0, 0< |z| = r < 1, then for function g ∈K

Φ(2)

Φ(2)+ 1− β
f ∗ g(z) ≺ 2g(z) (4)

and
Φ(2)

Φ(2)+ 1− β

∫ 2π

0

| f ∗ g(reiθ )|sdθ ¶ 2

∫ 2π

0

|g(reiθ )|sdθ (5)

where Φ(2) = [(αψ2(γ,λ) + 1)(ψ2(γ,λ)− 1) + 1− β][ψ2(γ,λ)]n.



L. Xiong, X. Liu / Eur. J. Pure Appl. Math, 5 (2012), 380-389 383

Proof. Suppose we take f (z) = z +
∞
∑

k=2
akzk ∈ H

n,γ
λ
[α,β] and g(z) = z +

∞
∑

k=2
bkzk ∈ K ,

then

Φ(2)

2Φ(2)+ 2(1−β)
f ∗ g(z) =

Φ(2)

2Φ(2)+ 2(1− β)
z +

∞
∑

k=2

Φ(2)

2Φ(2)+ 2(1− β)
ak bkzk.

If we can know

ℜ
�

1+ 2
∞
∑

k=2

Φ(2)

2Φ(2)+ 2(1− β)
akzk

�

> 0

From Lemma 1, it implies that the sequence

�

Φ(2)

2Φ(2)+ 2(1− β)
ak

�∞

1

is a subordination factor sequence, with a1 = 1. Now

ℜ
�

1+ 2
∞
∑

k=2

Φ(2)

2Φ(2)+ 2(1− β)
akzk

�

=ℜ
�

1+
∞
∑

k=2

Φ(2)

Φ(2)+ 1− β
akzk

�

=ℜ
�

1+
Φ(2)

Φ(2) + 1− β
z +

1

Φ(2)+ 1− β

∞
∑

k=2

Φ(2)akzk

�

¾ 1−
Φ(2)

Φ(2) + 1− β
r −

1

Φ(2) + 1− β

∞
∑

k=2

Φ(2)|ak|r
k. (6)

since

Φ(k) = [(αψk(γ,λ) + 1)(ψk(γ,λ)− 1) + 1− β][ψk(γ,λ)]n (k = 2,3, . . .)

and

ψk(γ,λ) =
Γ(k+ 1)Γ(2− γ)

Γ(k+ 1− γ)
[1+λ(k− 1)] (k = 2,3, . . .)

is a increasing function of k, so 0< Φ(2)¶ Φ(k) (k = 2,3, . . .).
Following (6), we can write

¾ 1−
Φ(2)

Φ(2)+ 1− β
r −

1

Φ(2) + 1−β

∞
∑

k=2

Φ(k)|ak|r
k.

As 0< r < 1, it can make sure

¾ 1−
Φ(2)

Φ(2)+ 1− β
r −

r

Φ(2) + 1− β

∞
∑

k=2

Φ(k)|ak|. (7)



L. Xiong, X. Liu / Eur. J. Pure Appl. Math, 5 (2012), 380-389 384

Using Lemma 3 in (3) and following (7), we obtain

ℜ
�

1+ 2
∞
∑

k=2

Φ(2)

2Φ(2)+ 2(1− β)
akzk

�

¾ 1−
Φ(2)

Φ(2)+ 1− β
r −

1− β

Φ(2) + 1− β
r = 1− r > 0,

In the light of Definition 1, we have

Φ(2)

2Φ(2)+ 2(1− β)
f ∗ g(z) =

∞
∑

k=1

Φ(2)

2Φ(2)+ 2(1− β)
bkckzk ≺ g(z),

Furthermore, it is easy to deduce the result in (5) by using (4) and Lemma 4.

Corollary 1. If f (z) = z +
∞
∑

k=2
akzk ∈ H

n,γ
λ
[α,β] and F(z) = z +

∞
∑

k=2

(a)k−1

(c)k−1
akzk, then

Φ(2)

Φ(2)+ 1− β
F(z) ≺ 2h(a, c; z) (8)

and

ℜ f (z) >
β − 1−Φ(2)

Φ(2)
, (9)

where Φ(2) = [(αψ2(γ,λ)+1)(ψ2(γ,λ)−1)+1−β][ψ2(γ,λ)]n, and h(a, c; z) is the incomplete

beta function defined in (1) with 0< a ¶ c, c ¾ 2 or a+ c ¾ 3.

Proof. Since 0< a ¶ c, c ¾ 2 or a+ c ¾ 3, using Lemma 2, we can know that

h(a, c; z) = z +

∞
∑

k=2

(a)k−1

(c)k−1
zk ∈K .

Taking g(z) = h(a, c; z) and g(z) = z

1−z
in Theorem 1, respectively, the results (8) and (9) are

obtained.

Corollary 2. If f ∈ H̄[α,β] in U and s > 0, 0< |z| = r < 1, then for function g ∈K

2(α+ 1)− β

2(α− β) + 3
f ∗ g(z) ≺ 2g(z)

and
[2(α+ 1)− β]

2(α− β) + 3

∫ 2π

0

| f ∗ g(reiθ )|sdθ ¶ 2

∫ 2π

0

|g(reiθ )|sdθ .

Proof. By taking n= 0, γ = 0 and λ= 1 in Theorem 1, Corollary 2 is given.



L. Xiong, X. Liu / Eur. J. Pure Appl. Math, 5 (2012), 380-389 385

Corollary 3. If f ∈ T ∗(β) in U and s > 0, 0< |z| = r < 1, then for function g ∈K

2− β

3− 2β
f ∗ g(z) ≺ 2g(z)

and
2− β

3− 2β

∫ 2π

0

| f ∗ g(reiθ )|sdθ ¶ 2

∫ 2π

0

|g(reiθ )|sdθ .

Proof. By taking α = 0 in Corollary 2, Corollary 3 is given.

Corollary 4. If f ∈ C(β) in U and s > 0, 0< |z| = r < 1, then for function g ∈K

4− β

5− 2β
f ∗ g(z) ≺ 2g(z)

and
4− β

5− 2β

∫ 2π

0

| f ∗ g(reiθ )|sdθ ¶ 2

∫ 2π

0

|g(reiθ )|sdθ .

Proof. By taking α = 1 in Corollary 2, Corollary 4 is given.

3. Some Results on the Class H
n,γ
λ
[α,β] with Fixed Equation

In this section, we shall obtain several interesting results on the functions which are de-
fined by the class H

n,γ
λ
[α,β] with the following nonhomogeneous Cauchy-Euler differential

equation:

z2 d2 L

dz2
+ 2(µ+ 1)z

d L

dz
+µ(µ+ 1)L = (1+µ)(2+µ) f (z) (10)

where L(z) ∈ T , f (z) ∈ H
n,γ
λ
[α,β], µ+ 1> 0, µ ∈ R.

The cauchy-Euler differential equation was introduced earlier to study the distortion in-
equalities and neighborhoods problems of the other class of functions by O. Altintaş et al.
[2].

Theorem 2. If the function L(z) = z +
∞
∑

k=2
ckzk ∈ T satisfy the equation (10) with

f (z) = z +
∞
∑

k=2
akzk ∈ H

n,γ
λ
[α,β], then for function g(z) ∈K ,

(µ+ 3)Φ(2)

(µ+ 3)Φ(2)+ (µ+ 1)(1− β)
L ∗ g(z) ≺ 2g(z) (11)

and
(µ+ 3)Φ(2)

(µ+ 3)Φ(2)+ (µ+ 1)(1− β)

∫ 2π

0

|L ∗ g(reiθ )|sdθ ¶ 2

∫ 2π

0

|g(reiθ )|sdθ , (12)

where Φ(2) = [(αψ2(γ,λ) + 1)(ψ2(γ,λ)− 1) + 1− β][ψ2(γ,λ)]n, 0< |z| = r < 1, s > 0.



L. Xiong, X. Liu / Eur. J. Pure Appl. Math, 5 (2012), 380-389 386

Proof. Suppose g(z) = z +
∞
∑

k=2
bkzk ∈K , then

(µ+ 3)Φ(2)

2(µ+ 3)Φ(2)+ 2(µ+ 1)(1− β)
L ∗ g(z) =

(µ+ 3)Φ(2)

2(µ+ 3)Φ(2)+ 2(µ+ 1)(1−β)
z

+

∞
∑

k=2

(µ+ 3)Φ(2)

2(µ+ 3)Φ(2)+ 2(µ+ 1)(1− β)
bkckzk.

If we show that

ℜ{1+ 2
∞
∑

k=2

(µ+ 3)Φ(2)

2(µ+ 3)Φ(2)+ 2(µ+ 1)(1− β)
ckzk}> 0

Then from Lemma 1, we say that the sequence
�

(µ+ 3)Φ(2)

2(µ+ 3)Φ(2)+ 2(µ+ 1)(1− β)
ck

�∞

1

is a subordination factor sequence, with c1 = 1. Now

ℜ
�

1+ 2
∞
∑

k=2

(µ+ 3)Φ(2)

2(µ+ 3)Φ(2)+ 2(µ+ 1)(1−β)
ckzk

�

=ℜ
�

1+
∞
∑

k=2

(µ+ 3)Φ(2)

(µ+ 3)Φ(2)+ (µ+ 1)(1− β)
ckzk

�

=ℜ
�

1+
(µ+ 3)Φ(2)

(µ+ 3)Φ(2)+ (µ+ 1)(1− β)
z +

(µ+ 3)

(µ+ 3)Φ(2)+ (µ+ 1)(1− β)

∞
∑

k=2

Φ(2)ckzk

�

¾ 1−
(µ+ 3)Φ(2)

(µ+ 3)Φ(2)+ (µ+ 1)(1− β)
r −

(µ+ 3)

(µ+ 3)Φ(2)+ (µ+ 1)(1− β)

∞
∑

k=2

Φ(2)|ck|r
k (13)

Because L(z) satisfies the differential equation with the f (z) ∈ H
n,γ
λ
[α,β], so

ck =
(µ+ 1)(µ+ 2)

(k+µ)(k+µ+ 1)
ak

Following (13), we have

¾1−
(µ+ 3)Φ(2)

(µ+ 3)Φ(2)+ (µ+ 1)(1− β)
r −

(µ+ 3)

(µ+ 3)Φ(2)+ (µ+ 1)(1−β)

∞
∑

k=2

Φ(2)
(µ+ 1)(µ+ 2)

(k+µ)(k+µ+ 1)
|ak|r

k

¾1−
(µ+ 3)Φ(2)

(µ+ 3)Φ(2)+ (µ+ 1)(1− β)
r −

(µ+ 3)

(µ+ 3)Φ(2)+ (µ+ 1)(1−β)

∞
∑

k=2

Φ(2)
(µ+ 1)(µ+ 2)

(2+µ)(µ+ 3)
|ak|r

k

¾1−
(µ+ 3)Φ(2)

(µ+ 3)Φ(2)+ (µ+ 1)(1− β)
r −

(µ+ 1)

(µ+ 3)Φ(2)+ (µ+ 1)(1−β)

∞
∑

k=2

Φ(2)|ak|r
k (14)



L. Xiong, X. Liu / Eur. J. Pure Appl. Math, 5 (2012), 380-389 387

Since

Φ(k) = [(αψk(γ,λ) + 1)(ψk(γ,λ)− 1) + 1− β][ψk(γ,λ)]n (k = 2,3, . . .)

and

ψk(γ,λ) =
Γ(k+ 1)Γ(2− γ)

Γ(k+ 1− γ)
[1+λ(k− 1)] (k = 2,3, . . .)

is a increasing function of k, so 0< Φ(2)¶ Φ(k) (k = 2,3, . . .).
Following (14), we can write

¾ 1−
(µ+ 3)Φ(2)

(µ+ 3)Φ(2)+ (µ+ 1)(1− β)
r −

(µ+ 1)

(µ+ 3)Φ(2)+ (µ+ 1)(1− β)

∞
∑

k=2

Φ(k)|ak|r
k.

As 0< r < 1, it can make sure

¾ 1−
(µ+ 3)Φ(2)

(µ+ 3)Φ(2)+ (µ+ 1)(1− β)
r −

(µ+ 1)r

(µ+ 3)Φ(2)+ (µ+ 1)(1− β)

∞
∑

k=2

Φ(k)|ak|. (15)

Since f (z) = z +
∞
∑

k=2
∈ H

n,γ
λ
[α,β], using Lemma 3 and following (15), we obtain

ℜ{1+ 2
∞
∑

k=2

(µ+ 3)Φ(2)

2(µ+ 3)Φ(2)+ 2(µ+ 1)(1− β)
ckzk}

¾ 1−
(µ+ 3)Φ(2)

(µ+ 3)Φ(2)+ (µ+ 1)(1− β)
r −

(1−β)(µ+ 1)

(µ+ 3)Φ(2)+ (µ+ 1)(1− β)
r = 1− r > 0.

In the light of Definition 1, we have

(µ+ 3)Φ(2)

2(µ+ 3)Φ(2)+ 2(µ+ 1)(1− β)
L ∗ g(z) =

∞
∑

k=1

(µ+ 3)Φ(2)

2(µ+ 3)Φ(2)+ 2(µ+ 1)(1−β)
bkckzk ≺ g(z).

Furthermore, it is easy to deduce the result in (12) by using (11) and Lemma 4.

Corollary 5. If the function L(z) = z +
∞
∑

k=2
ckzk ∈ T satisfy the equation (10) with

f (z) = z +
∞
∑

k=2
akzk ∈ H

n,γ
λ
[α,β] and F(z) = z +

∞
∑

k=2

(a)k−1

(c)k−1
ckzk, then

(µ+ 3)Φ(2)

(µ+ 3)Φ(2)+ (µ+ 1)(1− β)
F(z) ≺ 2h(a, c; z) (16)

and

ℜL(z) > −
(µ+ 3)Φ(2)+ (µ+ 1)(1− β)

(µ+ 3)Φ(2)
, (17)

where Φ(2) = [(αψ2(γ,λ)+1)(ψ2(γ,λ)−1)+1−β][ψ2(γ,λ)]n, and h(a, c; z) is the incomplete

beta function with 0< a ¶ c, c ¾ 2 or a+ c ¾ 3 and 0< |z| = r < 1, s > 0.



L. Xiong, X. Liu / Eur. J. Pure Appl. Math, 5 (2012), 380-389 388

Proof. Since 0< a ¶ c, c ¾ 2 or a+ c ¾ 3, using Lemma 2, we can know that

h(a, c; z) = z +

∞
∑

k=2

(a)k−1

(c)k−1
zk ∈K .

Taking g(z) = h(a, c; z) and g(z) = z

1−z
in Theorem 2, respectively, the results (16) and (17)

are obtained.

Corollary 6. If the function L(z) = z +
∞
∑

k=2
ckzk ∈ T satisfy the equation (10) with

f (z) = z +
∞
∑

k=2
akzk ∈ H̄[α,β], then for function g(z) ∈K ,

(µ+ 3)[2(α+ 1)− β]

(µ+ 3)[2(α+ 1)− β] + (µ+ 1)(1− β)
L ∗ g(z) ≺ 2g(z)

and

(µ+ 3)[2(α+ 1)− β]

(µ+ 3)[2(α+ 1)− β] + (µ+ 1)(1− β)

∫ 2π

0

|L ∗ g(reiθ )|sdθ ¶ 2

∫ 2π

0

|g(reiθ )|sdθ .

Proof. By taking n= 0, γ = 0 and λ= 1 in Theorem 2, Corollary 6 is given.

Corollary 7. If the function L(z) = z +
∞
∑

k=2
ckzk ∈ T satisfy the equation (10) with

f (z) = z +
∞
∑

k=2
akzk ∈ T ∗(β), then for function g(z) ∈K ,

(µ+ 3)(2− β)

(µ+ 3)(2− β) + (µ+ 1)(1− β)
L ∗ g(z) ≺ 2g(z)

and
(µ+ 3)(2− β)

(µ+ 3)(2− β) + (µ+ 1)(1− β)

∫ 2π

0

|L ∗ g(reiθ )|sdθ ¶ 2

∫ 2π

0

|g(reiθ )|sdθ .

Proof. By taking α = 0 in Corollary 6, Corollary 7 is given.

Corollary 8. If the function L(z) = z +
∞
∑

k=2
ckzk ∈ T satisfy the equation (10) with

f (z) = z +
∞
∑

k=2
akzk ∈ C(β), then for function g(z) ∈K ,

(µ+ 3)(4− β)

(µ+ 3)(4− β) + (µ+ 1)(1− β)
L ∗ g(z) ≺ 2g(z)

and
(µ+ 3)(4− β)

(µ+ 3)(4− β) + (µ+ 1)(1− β)

∫ 2π

0

|L ∗ g(reiθ )|sdθ ¶ 2

∫ 2π

0

|g(reiθ )|sdθ .

Proof. By taking α = 1 in Corollary 6, Corollary 8 is given.



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