1_483_bulboaca.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 3, No. 1, 2010, 1-12

ISSN 1307-5543 – www.ejpam.com

Sandwich Theorems for Some Analytic Functions Defined by

Convolution

A. O. Mostafa1, T. Bulboacă2∗ , and M. K. Aouf1

1 Department of Mathematics, Faculty of Science, Mansoura University, Mansoura 35516, Egypt
2 Faculty of Mathematics and Computer Science, Babeş-Bolyai University, 400084 Cluj-Napoca,

Romania

Abstract. For certain analytic functions defined by convolution products, we obtain several applica-

tions of first order differential subordination and superordination, that generalize some previous results

obtained by different authors.

2000 Mathematics Subject Classifications: 30C80, 30C45

Key Words and Phrases: Analytic functions, differential subordination, differential superordination,

sandwich theorems, convolution product.

1. Introduction

LetA denote the class of functions of the form

f (z) = z +

∞∑

k=2

akzk, (1)

which are analytic in the unit disc U = {z ∈ C : |z| < 1}. If f and g are analytic functions in

U, we say that f is subordinate to g, written f (z)≺ g(z), if there exists a Schwarz function w,

which (by definition) is analytic in U, with w(0) = 0, and |w(z)| < 1 for all z ∈ U, such that

f (z) = g(w(z)), z ∈ U. Furthermore, if the function g is univalent in U, then we have the

equivalence

f (z) ≺ g(z)⇔ f (0) = g(0) and f (U)⊂ g(U).

Let H(U) denote the class of analytic functions in U, and let H[a, n] denote the subclass

of the functions f ∈ H(U) of the form

f (z) = a+ anzn + an+1zn+1 + . . . (a ∈ C, n ∈ N) .

∗Corresponding author.

Email addresses: adelaeg254�yahoo.
om (A. Mostafa), bulboa
a�math.ubb
luj.ro, (T. Bulboacă),mkaouf127�yahoo.
om, (M. Aouf)

http://www.ejpam.com 1 c© 2009 EJPAM All rights reserved.



A. Mostafa, T. Bulboacă, and M. Aouf / Eur. J. Pure Appl. Math, 3 (2010), 1-12 2

Supposing that h and g are two analytic functions in U, let

ϕ(r, s, t; z) : C3×U→ C.

If h and ϕ(h(z), zh′(z), z2h′′(z); z) are univalent functions in U, and if h satisfies the second-

order superordination

g(z) ≺ ϕ
�

h(z), zh′(z), z2h′′(z); z
�

, (2)

a function q ∈ H(U) is called a subordinant of (2), if q(z) ≺ h(z) for all the functions h

satisfying (2). A univalent subordinant eq that satisfies q(z) ≺ eq(z) for all of the subordinants

q of (2), is said to be the best subordinant.

Recently, Miller and Mocanu [14] obtained sufficient conditions for the functions g, h and

ϕ, such that the following implication holds:

g(z)≺ ϕ
�

h(z), zh′(z), z2h′′(z); z
�
⇒ g(z)≺ h(z).

Using the results of [14], [4] investigated certain classes of first order differential super-

ordinations, as well as superordination-preserving integral operators [5]. Ali et al. [1] used

the results of [4] to obtain sufficient conditions for normalized analytic functions to satisfy

q1(z) ≺
z f ′(z)

f (z)
≺ q2(z),

where q1 and q2 are given univalent normalized functions in U.

Very recently, Shanmugam et al. [21] obtained sufficient conditions for a normalized

analytic function f to satisfy

q1(z) ≺
f (z)

z f ′(z)
≺ q2(z) and q1(z) ≺

z2 f ′(z)
�

f (z)
�2 ≺ q2(z),

where q1 and q2 are given univalent functions in U, with q1(0) = q2(0) = 1.

For the functions f given by (1), and g ∈ A given by g(z) = z +
∞∑

k=2

bkzk, the Hadamard

(or convolution) product of f and g is defined by

( f ∗ g)(z) = z +

∞∑

k=2

ak bkzk, z ∈ U.

In this paper we obtained several interesting subordination results for the function

�
( f ∗ g)(z)

z

�α
,

α ∈ C∗, that generalize some previous results obtained by different authors.

Remark 1. (i) For different choices of the function g, the convolution product f ∗ g reduces to

several interesting functions. For example, if

g(z) = z +

∞∑

k=2

(α1)k−1 · . . . · (αl)k−1

(β1)k−1 · . . . · (βs)k−1(1)k−1

zk, z ∈ U, (3)



A. Mostafa, T. Bulboacă, and M. Aouf / Eur. J. Pure Appl. Math, 3 (2010), 1-12 3

where, αi > 0 (i = 1,2, . . . l), β j > 0 ( j = 1,2, . . . s), l ≤ s + 1, l, s ∈ N0 = N ∪ {0}, where

N = {1,2, . . . }, we see that f ∗ g = Hl ,s(α1) f , where Hl ,s(α1) is the Dziok-Srivastava operator,

introduced and studied in [8] (see also [9], [10]).

The operator Hl ,s(α1), contains many interesting operators, such as Hohlov linear opera-

tor (see [11], [19]), the Bernardi-Libera-Livingston operator (see [12]), and Owa-Srivastava

fractional derivative operator (see [17]).

(ii) Also, if

g(z) = z +

∞∑

k=2

�
1+ l +λ(k− 1)

1+ l

�m
zk, z ∈ U, (4)

where λ ≥ 0, l ≥ 0, m ∈ N0, we see that f ∗ g = I(m,λ, l) f , where I(m,λ, l) is the generalized

multiplier transformation introduced and studied by Cătaş et. al. [6].

The operator I(m,λ, l) contains, as special cases, the multiplier transformation (see [7]), the

generalized Sălăgean operator introduced and studied by Al-Oboudi [2] (see also [20]).

2. Definitions and Preliminaries

To prove our results we shall need the following definition and lemmas.

Lemma 1. [13] Let q be univalent in the unit disc U and let θ and ϕ be analytic in a domain D

containing q(U), with ϕ(w) 6= 0 when w ∈ q(U). Set Q(z) = zq′(z)ϕ(q(z)), h(z) = θ(q(z)) +

Q(z) and suppose that

(i) Q is a starlike function in U,

(ii) Re
zh′(z)

Q(z)
> 0, z ∈ U.

If p is analytic in U, with p(0) = q(0), p(U)⊂ D and

θ(p(z)) + zp′(z)ϕ(p(z)) ≺ θ(q(z)) + zq′(z)ϕ(q(z)), (5)

then p(z) ≺ q(z), and q is the best dominant of (5).

Lemma 2. [21] Let µ ∈ C, γ ∈ C∗ = C \ {0} and let q be a convex function in U, with

Re

�
1+

zq′′(z)

q′(z)
+
µ

γ

�
> 0, z ∈ U.

If p is analytic in U and

µp(z) + γzp′(z) ≺ µq(z) + γzq′(z), (6)

then p(z) ≺ q(z), and q is the best dominant of (6).



A. Mostafa, T. Bulboacă, and M. Aouf / Eur. J. Pure Appl. Math, 3 (2010), 1-12 4

Definition 1. [14] LetQ be the set of all functions f that are analytic and injective on U\ E( f ),

where

E( f ) =

�
ζ ∈ ∂U : lim

z→ζ
f (z) =∞
�

,

and are such that f ′(ζ) 6= 0 for ζ ∈ ∂U \ E( f ).

Lemma 3. [5] Let q be univalent in the unit disc U and let θ and ϕ be analytic in a domain D

containing q(U). Suppose that

(i) Re
θ ′(q(z))

ϕ(q(z))
> 0, z ∈ U,

(ii) h(z) = zq′(z)ϕ(q(z)) is starlike in U.

If p ∈ H[q(0), 1]∩Q, with p(U) ⊂ D, the function θ(p(z)) + zp′(z)ϕ(p(z)) is univalent in U

and

θ(q(z)) + zq′(z)ϕ(q(z))≺ θ(p(z)) + zp′(z)ϕ(p(z)), (7)

then q(z)≺ p(z), and q is the best subordinant of (7).

Lemma 4. [18] The function q(z) = (1− z)−2ab is univalent in U if and only if |2ab− 1| ≤ 1

or |2ab+ 1| ≤ 1.

3. Main Results

Theorem 1. Let q be convex in U, and let α,η ∈ C∗ such that

Re

�
1+

zq′′(z)

q′(z)
+
α

η

�
> 0, z ∈ U. (8)

Let g ∈ A , and for all functions f ∈A with ( f ∗ g)(z) 6= 0, z ∈ U̇= U \ {0}, set

χg(α,η; f )(z) = (1−η)
�
( f ∗ g)(z)

z

�α
+η

z( f ∗ g)′(z)

( f ∗ g)(z)

�
( f ∗ g)(z)

z

�α
. (9)

Then,

χg(α,η; f )≺ q(z) +
η

α
zq′(z) (10)

implies �
( f ∗ g)(z)

z

�α
≺ q(z),

and q is the best dominant of (10). (All the powers are the principal ones)



A. Mostafa, T. Bulboacă, and M. Aouf / Eur. J. Pure Appl. Math, 3 (2010), 1-12 5

Proof. If we define the function ψ by

ψ(z) =

�
( f ∗ g)(z)

z

�α
, z ∈ U, (11)

then ψ is analytic in U and ψ(0) = 1. Therefore, by differentiating (11) logarithmically with

respect to z, we have

ψ(z) +
η

α
zψ′(z) = (1−η)

�
( f ∗ g)(z)

z

�α
+η

z( f ∗ g)′(z)

( f ∗ g)(z)

�
( f ∗ g)(z)

z

�α
.

From the assumption (10) and the above relation we deduce

ψ(z) +
η

α
zψ′(z)≺ q(z) +

η

α
zq′(z),

hence, the assertion of our theorem follows by using Lemma 2 with µ = 1 and γ= η/α.

Taking q(z) = (1+Az)/(1+Bz) (−1≤ B < A≤ 1) in Theorem 1, the condition (8) becomes

Re

�
1− Bz

1+ Bz
+
η

α

�
> 0, z ∈ U. (12)

It is easy to check that the function φ(z) = (1− ζ)/(1 + ζ), |ζ| < |B| ≤ 1, is convex in U,

and since φ(ζ) = φ(ζ) for all |ζ| < |B|, it follows that the image φ(U) is a convex domain

symmetric with respect to the real axis, hence

inf

�
Re

1− Bz

1+ Bz
: z ∈ U

�
=

1− |B|

1+ |B|
≥ 0.

Then, the inequality (12) is equivalent to

Re
α

η
≥
|B| − 1

1+ |B|
, (13)

hence, we have the following corollary:

Corollary 1. Let −1 ≤ B < A≤ 1, let α,η ∈ C∗, and suppose that the condition (13) holds. Let

g ∈ A , and for all functions f ∈A with ( f ∗ g)(z) 6= 0, z ∈ U̇, suppose that

χg(α,η; f )≺
1+ Az

1+ Bz
+
η

α

(A− B)z

(1+ Bz)2
, (14)

where χg(α,η; f ) is given by (9).

Then �
( f ∗ g)(z)

z

�α
≺

1+ Az

1+ Bz
,

and (1+ Az)/(1+ Bz) is the best dominant of (14). (All the powers are the principal ones)



A. Mostafa, T. Bulboacă, and M. Aouf / Eur. J. Pure Appl. Math, 3 (2010), 1-12 6

Letting g be of the form (3), and using the identity [8]

z
�

Hl ,s(α1) f (z)
�′
= α1Hl ,s(α1 + 1) f (z)− (α1− 1)Hl ,s(α1) f (z), (15)

we obtain the next result:

Corollary 2. Let q be convex in U, let α,η ∈ C∗, and suppose that q satisfies the condition (8).

For all functions f ∈ A with Hl ,s(α1) f (z)(z) 6= 0, z ∈ U̇, set

χ1(α1;α,η; f )(z) = (1−ηα1)

�
Hl ,s(α1) f (z)

z

�α
+

η
α1Hl ,s(α1 + 1) f (z)

Hl ,s(α1) f (z)

�
Hl ,s(α1) f (z)

z

�α
. (16)

Then,

χ1(α1;α,η; f )(z)≺ q(z) +
η

α
zq′(z), (17)

implies �
Hl ,s(α1) f (z)

z

�α
≺ q(z),

and q is the best dominant of (17). (All the powers are the principal ones)

Remark 2. The Corollary 2 was also obtained by Murugusundaramoorthy and Magesh [15,

Theorem 3.1].

Letting g be of the form (4), and using the identity [6]

λz
�
I(m,λ, l) f (z)
�′
= (l + 1) I(m+ 1,λ, l) f (z)− (1+ l −λ) I(m,λ, l) f (z), (18)

where λ > 0, l ≥ 0, m ∈ N0, we deduce:

Corollary 3. Let q be convex in U, let α,η ∈ C∗, and suppose that q satisfies the condition (8).

For all functions f ∈ A with I(m,λ, l) f (z)(z) 6= 0, z ∈ U̇
�
λ > 0, l ≥ 0, m ∈ N0

�
, set

χ2(m,λ, l;α,η; f )(z) =

�
1−

η(l + 1)

λ

��
I(m,λ, l) f (z)

z

�α
+

η(l + 1)

λ

I(m+ 1,λ, l) f (z)

I(m,λ, l) f (z)

�
I(m,λ, l) f (z)

z

�α
, (19)

Then,

χ2(m,λ, l;α,η; f )(z)≺ q(z) +
η

α
zq′(z), (20)

implies �
I(m,λ, l) f (z)

z

�α
≺ q(z),

and q is the best dominant of (20). (All the powers are the principal ones)



A. Mostafa, T. Bulboacă, and M. Aouf / Eur. J. Pure Appl. Math, 3 (2010), 1-12 7

Theorem 2. Let α,γ ∈ C∗, and let q be univalent in U, with q(0) = 1 and q(z) 6= 0 for all z ∈ U,

such that q satisfies

Re

�
1+

zq′′(z)

q′(z)
−

zq′(z)

q(z)

�
> 0, z ∈ U. (21)

Let g ∈ A , and for all functions f ∈A with ( f ∗ g)(z) 6= 0, z ∈ U̇, suppose that

1+ γα

�
z( f ∗ g)′(z)

( f ∗ g)(z)
− 1

�
≺ 1+ γ

zq′(z)

q(z)
. (22)

Then, �
( f ∗ g)(z)

z

�α
≺ q(z),

and q is the best dominant of (22). (The power is the principal one)

Proof. If we define the function φ by

φ(z) =

�
( f ∗ g)(z)

z

�α
, (23)

then φ is analytic in U and φ(0) = 1. Differentiating (23) logarithmically with respect to z,

we get

zφ′(z)

φ(z)
= α

�
z( f ∗ g)′(z)

( f ∗ g)(z)
− 1

�
.

Using the above relation in (22), we have

1+ γ
zφ′(z)

φ(z)
≺ 1+ γ

zq′(z)

q(z)
.

Setting θ(w) = 1 and ϕ(w) = γ/w, then ϕ and θ are analytic in C∗. A simple computation

shows that

Q(z) = zq′(z)ϕ(q(z)) = γ
zq′(z)

q(z)
,

h(z) = θ(q(z)) +Q(z) = 1+ γ
zq′(z)

q(z)
,

and it is easily to see that the conditions of Lemma 1 are satisfied whenever (21) holds. Then,

by applying Lemma 1, our conclusion follows.

Putting q(z) = (1+ Az)/(1+ Bz) (−1≤ B < A≤ 1) in Theorem 2, it is easy to check that

the condition (21) holds whenever −1≤ B < A≤ 1, hence we obtain:



A. Mostafa, T. Bulboacă, and M. Aouf / Eur. J. Pure Appl. Math, 3 (2010), 1-12 8

Corollary 4. Let −1≤ B < A≤ 1. Let g ∈ A , and for all functions f ∈ A with ( f ∗ g)(z) 6= 0,

z ∈ U̇, suppose that

1+α

�
z( f ∗ g)′(z)

( f ∗ g)(z)
− 1

�
≺ 1+

(A− B)z

(1+ Az)(1+ Bz)
. (24)

Then, �
( f ∗ g)(z)

z

�α
≺

1+ Az

1+ Bz
,

and (1+ Az)/(1+ Bz) is the best dominant of (24). (The power is the principal one)

Putting q(z) = (1+ Bz)α(A−B)/B (−1≤ B < A≤ 1, B 6= 0) and γ = 1 in Theorem 2, and

according to Lemma 4, we have the following result:

Corollary 5. Let −1≤ B < A≤ 1, with B 6= 0, such that

����
α(A− B)

B
− 1

����≤ 1 or

����
α(A− B)

B
+ 1

����≤ 1.

Let g ∈ A , and for all functions f ∈A with ( f ∗ g)(z) 6= 0, z ∈ U̇, suppose that

1+α

�
z( f ∗ g)′(z)

( f ∗ g)(z)
− 1

�
≺

1+ [B+α(A− B)] z

1+ Bz
. (25)

Then, �
( f ∗ g)(z)

z

�α
≺ (1+ Bz)α(A−B)/B ,

and (1+ Bz)α(A−B)/B is the best dominant of (25). (The power is the principal one)

Taking γ = 1/ab, (a, b ∈ C∗), α = a and q(z) = (1− z)−2ab in Theorem 2 and combining

this together with Lemma 4, we obtain the next corollary:

Corollary 6. Let a, b ∈ C∗ such that

|2ab− 1| ≤ 1 or |2ab+ 1| ≤ 1.

Let g ∈ A , and for all functions f ∈A with ( f ∗ g)(z) 6= 0, z ∈ U̇, suppose that

1+
1

b

�
z( f ∗ g)′(z)

( f ∗ g)(z)
− 1

�
≺

1+ z

1− z
. (26)

Then, �
( f ∗ g)(z)

z

�a
≺ (1− z)−2ab,

and (1− z)−2ab is the best dominant of (26). (The power is the principal one)



A. Mostafa, T. Bulboacă, and M. Aouf / Eur. J. Pure Appl. Math, 3 (2010), 1-12 9

Remark 3. (i) Taking g(z) = z/(1− z) in Corollary 6, we obtain the result of Obradovíc et al.

[16, Theorem 1].

(ii) For g(z) = z/(1− z) and a = 1, Corollary 6 reduces to the recent result of Srivastava

and Lashin [22, Theorem 3].

(iii) The special case of Corollary 6, when g(z) = z/(1−z), γ= eiλ/(ab cosλ) (a, b ∈ C∗, |λ| < π/2),
and q(z) = (1− z)−2ab cosλe−iλ

, is due to Aouf et al. [3, Theorem 1].

Theorem 3. Let q be convex in U, and let α,η ∈ C∗ with

Re
α

η
> 0. (27)

Let g ∈ A , and for all functions f ∈ A with ( f ∗ g)(z) 6= 0, z ∈ U̇, suppose that�
( f ∗ g)(z)

z

�α
∈ H[q(0), 1] ∩ Q, and that χg(α,η; f ) is univalent in U, where χg(α,η; f )

is given by (9).

Then,

q(z) +
η

α
zq′(z)≺ χg(α,η; f )(z), (28)

implies

q(z) ≺
�
( f ∗ g)(z)

z

�α
,

and q is the best subordinant of (28). (All the powers are the principal ones)

Proof. If we let the function ψ be given by (11), a simple computation shows that

ψ(z) +
η

α
zψ′(z) = χg(α,η; f )(z).

Setting θ(w) = w and ϕ(w) = η/α, then θ and ϕ are analytic in C, and from (27) we have

Re
θ ′(q(z))

ϕ(q(z))
= Re

α

η
> 0, z ∈ U.

Since q is a convex function, it follows that h(z) = zq′(z)ϕ(q(z)) =
�
ηzq′(z)
�
/α is starlike

in U, and using Lemma 3 we obtain our result.

Letting g be of the form (3) in Theorem 3 and using the identity (15), we get the following

result obtained by Murugusundaramoorthy and Magesh [15, Theorem 3.9]:

Corollary 7. Let q be convex in U, and suppose that α,η ∈ C∗ satisfies the condition (27). For

all functions f ∈ A with Hl ,s(α1) f (z)(z) 6= 0, z ∈ U̇, suppose that

�
Hl ,s(α1) f (z)(z)

z

�α
∈

H[q(0), 1] ∩ Q, and that χ1(α1;α,η; f ) is univalent in U, where χ1(α1;α,η; f ) is given by

(16).

Then,

q(z) +
η

α
zq′(z)≺ χ1(α1;α,η; f )(z), (29)



A. Mostafa, T. Bulboacă, and M. Aouf / Eur. J. Pure Appl. Math, 3 (2010), 1-12 10

implies

q(z) ≺

�
Hl ,s(α1) f (z)

z

�α
,

and q is the best subordinant of (29). (All the powers are the principal ones)

Letting g be of the form (4) in Theorem 3 and using the identity (19), we have:

Corollary 8. Let q be convex in U, and suppose that α,η ∈ C∗ satisfies the condition (27).

For all functions f ∈ A with I(m,λ, l) f (z) 6= 0, z ∈ U̇
�
λ > 0, l ≥ 0, m ∈ N0

�
, suppose that�

I(m,λ, l) f (z)

z

�α
∈ H[q(0), 1] ∩ Q, and that χ2(m,λ, l;α,η; f ) is univalent in U, where

χ2(m,λ, l;α,η; f ) is given by (19).

Then,

q(z) +
η

α
zq′(z) ≺ χ2(m,λ, l;α,η; f )(z), (30)

implies

q(z)≺
�

I(m,λ, l) f (z)

z

�α
,

and q is the best subordinant of (30). (All the powers are the principal ones)

Combining Theorem 1 and Theorem 3, we deduce the following sandwich theorem:

Theorem 4. Let q1 and q2 be convex functions in U. Suppose that α,η ∈ C∗ satisfies (27) and

q2 satisfies (8).

Let g ∈ A , and for all functions f ∈ A with ( f ∗ g)(z) 6= 0, z ∈ U̇, suppose that�
( f ∗ g)(z)

z

�α
∈ H[q(0), 1] ∩ Q, and that χg(α,η; f ) is univalent in U, where χg(α,η; f )

is given by (9).

Then,

q1(z) +
η

α
zq′1(z)≺ χg(α,η; f )(z)≺ q2(z) +

η

α
zq′2(z), (31)

implies

q1(z) ≺
�
( f ∗ g)(z)

z

�α
≺ q2(z),

and, moreover, q1 and q2 are respectively, the best subordinant and the best dominant of (31).

(All the powers are the principal ones)

Remark 4. Combining Corollary 2 and Corollary 7, we get the sandwich result obtained by

Murugusundaramoorthy and Magesh [15, Theorem 3.10].

From Corollary 3 and Corollary 8, we get the next sandwich theorem:



REFERENCES 11

Theorem 5. Let q1 and q2 be convex functions in U. Suppose that α,η ∈ C∗ satisfies (27) and

q2 satisfies (8). For all functions f ∈ A with I(m,λ, l) f (z) 6= 0, z ∈ U̇
�
λ > 0, l ≥ 0, m ∈ N0

�
,

suppose that

�
I(m,λ, l) f (z)

z

�α
∈ H[q(0), 1] ∩Q, and that χ2(m,λ, l;α,η; f ) is univalent in

U, where χ2(m,λ, l;α,η; f ) is given by (19).

Then,

q1(z) +
η

α
zq′1(z) ≺ χ2(m,λ, l;α,η; f )(z)≺ q2(z) +

η

α
zq′2(z), (32)

implies

q1(z)≺
�

I(m,λ, l) f (z)

z

�α
≺ q2(z),

and, moreover, q1 and q2 are respectively, the best subordinant and the best dominant of (32).

(All the powers are the principal ones)

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