International Journal of Scientific & Engineering Research, Volume 5, Issue 2, February-2014 8

ISSN 2229-5518

SANDWICH-TYPE THEOREMS FOR MEROMORPHIC MULTIVALENT FUNCTIONS ASSOCIATED WITH A LINEAR OPERATOR

H. E. Darwish, A. Y. Lashin, and B. H. Rizqan

Abstract— the purpose of this article is to obtain some subordination and superordination preserving properties of meromorphic

p

multivalent functions in the punctured open unit disk associated with the linear operator meromorphic multivalent functions are also considered.

Qα , β , γ

the sandwish- type results for these

2010 Mathematics Subject Classification: Primary 30C45.

Keywords— Meromorphic function, Hadamard product, Subordination, Superordination, Meromorphic multivalent function, Linear operators,

Sandwich result.

## 1 INTRODUCTION

—————————— ——————————
open unit disk

U = {z ∈ C :

z < 1}.

a dominant, if

p q

for all p satisfying (1.1). A domnant

ET H (U)

denote the class of analytic functions in the
solutions of the differential subordination, or more simply
For n ∈ N = {1, 2,...}
and

a ∈ C,

let
that satisfies for all dominants q of (1.1) is said

H [a, n] = { f H :

f ( z) = a + an

z n + a

n+1

z n+1 + ...}.

to be the best dominant.
Let f and F be members of

H . the function f is

Definition 2 [10]. Let ϕ

: C 2 → C

and let h be analytic
said to be subordinate to

F , or F is said to be superordi-

in U . If p and

ϕ ( p( z), zp′( z))

are univalent in U
nate to f , if there exists a function w analytic in U ,
and satisfy the differential superordination:
with w(0) = 0
and

w( z) < 1

( z U),

such that

h( z)  ϕ ( p( z), zp′( z))

( z U),

(1.2)

f ( z) = F (w( z))

( z U).

then p is called a solution of the differential superordination.
An analytic function q is called a subordinant of the
In such a case, we write
soltions of the differential superordination, or more simply

f F

( z U) or

f ( z)  F ( z)

( z U).

a subordinant if

q p

for all p satisfying (1.2). A uni-
If the function F is univalent in U , then we have (cf. [11])

f ( z)  F ( z)

( z U) ⇔

f (0) = F (0)

vlent subordinant that for all subordinants q of
(1.2) is said to be the best subordinant.

and

f (U) ⊂ F (U).

Definition 3 [10]. Denote by Q the class of functions f

Definition 1 [9]. Let φ

: C 2 → C

and let h be univalent
that are analytic and injective on

U \ E( f ),

where
in U . If p is analytic in U and satisfies the differential
subordination:

E( f ) = {ζ ∈ ∂U

: lim f ( z) = ∞},

z ζ

φ ( p( z), zp′( z))  h( z)

( z U),

(1.1)
and are such that
then p is called a solution of the differential subordination.
The univalent function q is called a dominant of the

f (ζ ) ≠ 0(ζ ∈ ∂U \ E( f )).

For any integer

m > − p,

Let

Σ p , m

denote the class of

————————————————

Department of Mathematics, Faculty of Science,

all meromorphic functions f of the form

H. E. Darwish, Mansoura University, Mansoura, Egypt

E-mail: Darwish333@yahoo.com

Y. Lashin, Mansoura University, Mansoura, Egypt

f ( z) = z p + a z k

k =m

( p ∈ N = {1, 2,...}),

B. H. Rizqan, Mansoura University, Mansoura, Egypt E-mail :

which are analytic and p -valent in the punctured unit disk
bakeralhaaiti@yahoo.com

U = {z ∈ C :

0 < z

< 1} = U \ {0}.

For convenience, we
write

Σ p ,1− p = Σ p .

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For functions

f ∈ Σ p , m

given by (1.3), and

g ∈ Σ p , m

If is not subordinate to , then there exist points

iθ

defined by

z0 = r0 e

U and ζ 0 ∈ ∂U \ E( f ),

g ( z) = z p + b z k

=

(m > − p; p ∈ N),

(1.4)
for which

q(U

) ⊂ p(U),

q( z

) = p(ζ )

k m r0 0 0

then the Hadamard product (or convolution) of f and g
is

and

z0 q′( z0

) = mζ 0

p′(ζ 0 )

(m n).

A function

L( z, t )

defined on

U × [0, ∞),

is the subordi-

( f g )( z) = z p + a b z k

= ( g f )( z)

nation chain (or Löwner chain) if

L( z,.)

is analytic and uni-

k k

k =m

(1.5)

valent in U for all

t ∈ [0, ∞),

L( z,.)

is continuously

(m > − p; p ∈ N).

Differentiable on

[0, ∞)

for all

z U

and
For

f ∈ Σ p , m

we now define the integral operator

L( z, s)  L( z, t )

( z U; 0 ≤ s < t ).

p

α , β , γ

: Σ p , m

→ Σ p , m

which was introduced and studied
by El-Ashwah et al. [3] as follows:

#### Lemma 2 [10]. Let

q H[a,1]

and ϕ

: C 2 → C.

Q p f ( z) =

Γ(α + β γ + 1) 1 ⋅

Also set

α , β , γ

Γ(β )Γ(α γ + 1) z β + p

ϕ (q( z), zq′( z)) ≡ h( z)

( z U).

z α −γ

If L( z, t ) = ϕ (q( z), tzq′( z))

is a subordination chain and

 − t t

f t dt

p H[a,1] ∩ Q,

1 

0 z

β + p −1 ( )

then.

= 1 + Γ(α + β γ + 1) ⋅

z p Γ(β )

Γ(β + p + k )

a z k

(1.6)

implies that

h( z)  ϕ ( p( z), zp′( z))

( z U).

k =m

Γ(α + β + p + k γ + 1)

Furthermore, if

ϕ (q( z), zp′( z)) = h( z)

has a univalent
and

p

(β > 0; α > γ − 1; γ

> 0;

p ∈ N;

z U),

solution

q Q,

then q is the best subordinate.

2

Qα −1, β , γ f ( z) =

f ( z)

(β > 0; γ

> 0;

p ∈ N;

z U ).

#### Lemma 3 [7]. Suppose that the function H : C → C

From (1.6), it is easy to verify that

p p

satisfies the following condition:

Re{H (is, t )} ≤ 0

z(Qα +1, β , γ f ( z))′ = (α + β γ + 1)Qα , β , γ f ( z) −

(α + β + p γ + 1)Q p f ( z).

(1.7)

for all real s and

t ≤ −n(1 + s 2 ) / 2

(n ∈ N).

If the function

p( z) = 1 + pn

z n + ....

is analytic in U and

Remark:

p p

p Re{H ( p( z), zp′( z))} > 0

( z U),

(i) For

γ = 1, Qα , β ,1 = Qα , β ,

where the operator

Qα , β

then
was introduced and studied by Aqlan et al. [2] (see also [1]);

Re{ p( z)} > 0

( z U).

1 p

(ii) For

p = γ

= 1, Qα , β ,1 = Qα , β ,

where the operator

Qα , β ,

was introduced and studied by Lashin [6].

β , γ ∈ C

with

β ≠ 0

and

h H(U)

## 2. A SET OF LEMMS

with

h(0) = c.

If

Re{βh( z) + γ } > 0

( z U),

The following lemmas will be required in our present
then, the solution of the differential equation
investigation.

#### Lemma 1 [9]. Let

p Q

with

p(0) = a

and let

q( z) +

zq′( z)

βq( z) + γ

= h( z)

( z U;

q(0) = c),

q( z) = a + an

z n + .....

is analytic in U satisfies the inequality
be analytic in U with

Re{βq( z) + γ } > 0

( z U).

q( z) ≠ a

and

n ≥ 1.

#### Lemma 5 [11]. The function

L( z, t ) = a1 (t ) z + ...

with

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a1 (t ) ≠ 0

and

lima1 (t ) = ∞

t →∞

is a subordination chain if
then,

Re{q( z)} > 0

( z U).

and only if

zL( z, t ) 

Taking the logarithmic differentiation on both sides of the se- cond equation in (3.5) and using the equation (1.7) we obtain

 

Re > 0

( z U; 0 ≤ t < ∞),

p(α + β γ )φ ( z) = p(α + β γ )G( z) + ( p η ) zG′( z).

=L( z, t ) 

(3.7)
and

 ∂t 

Now, by differentiating both sides of (3.7), We obtain the rela- tionship:

L( z; t ) ≤ K 0 a1 (t ) ,

z < r0 < 1, t ≥ 0,

1 + zφ ′ ( z) = 1 + zG′ ( z) +

zq′( z)

φ ′( z)

G′( z)

q( z) + p(α + β γ ) /( p η )

For some positive constants
a subordination chain.

K 0 and

r0 , then

L( z, t ) is

= q( z) +

zq′( z)

q( z) + p(α + β γ ) /( p η )

h( z).

(3.8)

## 3. MAIN RESULTS

We also note from (3.1) that
We begin with proving the following subordination theorem

p

Reh( z)

+ p(α + β γ

p η

) 

 > 0

( z U),

involving the operator

Qα , β , γ f ( z)

defined by (1.6).  
and so by Lemma 4, we conclude that the differential equation

#### Theorem 1. Let

(3.8) has a solution

q H(U)

with

f , g ∈ Σ p , m , α > γ ,

β > 0, γ

> 0,

p ∈ N,

0 ≤ η < p

q(0) = h(0) = 1.

and

z U.

Suppose that

zφ

Re1 +

( z) 

 > −δ ,

(3.1)
Let us put

H (u, v) = u +

v + δ

u + p(α + β γ ) /( p η )

(3.9)
setting
where

φ ′( z) 

where δ is given by (3.2).
From (3.1), (3.8) and (3.9), we obtain
Now, we proceed to show that

( p η ) 2 + p 2 (α + β γ ) 2

Re{

( , )} 0  ;

1 (1

2 ) .

δ = −

4 p( p η )(α + β γ )

H is t

s R

t ≤ −

2

+ s

(3.10)

( p η ) 2 p 2 (α + β γ ) 2

(0 < δ ≤ 1/ 2).

(3.2)

Indeed, From (3.9), we have

t

Re{H (is, t )} = Reis +

+ δ

4 p( p η )(α + β γ )

is + p(α + β γ ) /( p η ) 

Then, the following subordination relation

= tp(α + β γ ) / p η + δ

p(α + β γ ) /( p η ) + is 2

implies that
(3.3)

≤ − Eδ (s) ,

2 p(α + β γ ) /( p η ) + is 2

(3.11)

z p Q p f ( z)  z

p p

α , β , γ

( z U).

(3.4)

where

p p p α + β γ

p α β γ

Moreover, the function

z Qα , β , γ g ( z)

is the best dominant.

E (s) :=  (

) − 2δ s 2 − ( +

− ) ⋅

Proof. Let us define the functions F and G,

δ

respectively,

p η

p η

by

F ( z) := z

Qα , β , γ f ( z)

and

G( z) := z

Qα , β , γ g ( z). (3.5)

 2δ

p(α + β γ ) 

.

p η

(3.12)

We first show. If the function is defined by

q( z) := 1 + zG ′ ( z)

( z U),

(3.6)
For δ given by (3.2), we can prove easily that the expression

G ′( z)

Eδ (s)

given by (3.12) is greater than or equal to zero. Hence,

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from (3.9), we see that (3.10) holds true. Thus using Lemma 3, we conclude that.

L(ζ

0 , t ) = G(ζ

) + ( p η )(1 + t ) ζ

0 p(α + β γ )

0 G′(ζ 0 )

Re{q( z)} > 0

( z U).

= + p η

Moreover, we see that the condition:

G′(0) ≠ 0

F ( z0 )

p(α + β γ )

z0 F ′( z0 )

is satisfied. Hence, the function defined by (3.5) is convex

( p η ) p p

( ) η p p

( ) ( ),

in U.

= z0 Qα , β , γ f z0

p

+ z0 Qα , β , γ f p

z0 φ U

Next, we prove that the subordination condition (3.3) implies
by virtue of the subordination condition (3.3). This contra-
that
dicts the above observation

L(ζ 0 , t ) ∉ φ (U). Therefore, the

F ( z)  G( z)

( z U)

(3.13)
subordination condition (3.3) must imply the subordination
for the functions F and G defined by (3.5). Without loss
given by (3.13). Considering

F ( z) = G( z), we see that the

of generality, we can assume that G is analytic and univa-

lent on U and
function G is the best dominant. This evidently completes the proof of Theorem 1.

G′(ζ ) ≠ 0

(ζ ∈ ∂U).

We next provide a dual problem of Theorem 1, in the
For this purpose, we consider the function

L( z, t )

given by
sense that the subordinations are replaced by superordina-
tions.

L( z, t ) := G( z) + ( p η )(1 + t ) zG′( z)

p(α + β γ )

#### Theorem 2. Let

(0 ≤ t < ∞; 0 ≤ η < p;

z U).

f , g ∈ Σ p , m , α > γ ,

β > 0, γ

> 0,

p ∈ N,

0 ≤ η < p

We note that

L( z, t )

= G′(0) ⋅

and

z U.

Suppose that

Re1 +

zφ

( z) 

 > −δ ,

z z =0

φ ′( z) 

p α + β γ

+ p η

+ t

setting

 ( ) (

)(1

)  ≠ 0.

p η η

 (α + β γ ) 

φ ( z) :=

z Qα

p

, β , γ g ( z) +

z Qα , β , γ g ( z),

p

This shows that the function

L( z, t ) = α1 (t ) z + ...

where δ is given by (3.2), and

p η η

satisfies the condition

α1 (t ) ≠ 0

(0 ≤ t < ∞).

p p

p α −1

, β , γ f ( z) +

z p Q p f ( z),

p

p p

α β γ

H

Then,
Furthermore, we have
is univalent in U and

z Q , ,

f ( z)

[1,1] Q.

Re zL(

z, t) /

z

 =

the following superordination relation

 ∂L( z, t) / ∂t

φ ( z) 

p η z p Q p

f ( z) + η z p Q p

f ( z) ( z U),

p α + β γ

zGz 

p α −1 , β , γ

p α , β , γ

( Re

) + (1 + t )1 +

( )  > 0.

(3.14)

p η

G′( z) 

implies that

p p p p

Therefore, by virtue of Lemma 5,

L( z, t )

is a subordination

z Qα , β , γ g ( z)  z

p

Qα , β , γ f ( z)

p

( z U).

chain. We observe from the definition of subordination chain
that
Moreover, the function

z Qα , β , γ g ( z)

is the best subordi-

L(ζ , t ) ∉ L(U, 0) = φ (U)

(ζ ∈ ∂U; 0 ≤ t < ∞).

nant.

Proof. The first part of the proof is similar to that of Theorem 1

Now, suppose that F is not subordinate to G,
then by
and so we will use the same notation as in the proof of
Lemma 1, there exists points
that

z0 U

and

ζ 0 ∈ ∂U,

such
Therem1.
Now, let us define the functions F and G by (3.5). We

q is defined by (3.6), using

F ( z0 ) = G(ζ 0 )

(0 ≤ t < ∞).

and

z0 F ( z0 ) = (1 + t )ζ 0 G′(ζ 0 )

first note that, if the function
(3.7), then we obtain
Hence, we have

φ ( z) = G( z) +

p η zG

p(α + β γ )

:= ϕ (G( z), zG′( z)).

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(3.15) Then using the same method as in the proof of Theorem 1.

p p

α , β , γ 2

are the best subordinate and the best
We can prove that

Re{q( z)} > 0

( z U),

dominant, respectively.
The assumption of Theorem 3, that the functions
that is, G defined by (3.5) is convex (univalent) in U . Next, we prove that the subordination condition (3.14) implies that
need to be univalent in U , may be replaced by another

G( z)  F ( z)

(3.16)

( z U)

condition in the following result.
for the functions F and G defined by (3.5). Now, consider

Corollary 1. Let

the function

L( z, t )

defined by

f , g k ∈ Σ p , m

(k = 1, 2), α > γ ,

β > 0, γ

> 0,

p ∈ N,

L( z, t ) := G( z) +

( p η )t

zG′( z)(0 ≤ t < ∞;

z U).

0 ≤ η < p

p(α + β γ )

and

z U.

Suppose that condition (3.17) is satisfied and
As G is convex and

p(α + β γ ) /( p η ) > 0,

we can

Re1 +

zψ

( z) 

 > −δ ,

(3.18)
prove easily that

L( z, t )

is a subordination chain as in the

ψ ′( z) 

proof of Theorem 1. Therefore, according to Lemma 2, we conclude that the superordination condition (3.14) must
setting

ψ ( z) :=

p η z p Q p

f ( z) + η z p Q p

f ( z),

imply the superordination given by (3.16). Furthermore, as
the differential equation (3.15) has the univalent solution

p α −1 , β , γ

p α , β , γ

G , it is the best subordinate of the given differential su-

where δ is given by (3.2). Then, the following relation
perordination. Therefore, we complete the proof of Theo- rem 2. If we combine Theorems 1 and 2, then we obtain the

φ1 ( z) 

p η

p

p p

α −1

, β , γ f ( z) +

η z p

p

p

α , β , γ

following sandwich -type theorem.

#### Theorem 3. Let

φ2 ( z)

implies that

( z U),

f , g k ∈ Σ p , m

(k = 1, 2), α > γ ,

β > 0, γ

> 0,

p ∈ N,

0 ≤ η < p

Moreover, the function

z Qα , β , γ g1 ( z)

and
and

z U.

Suppose that

′′

p p

α , β , γ

g 2 ( z)

are the best subordinate and the best

zφ ( z)

Re1 + k  > −δ ,

(3.17)
dominant, respectively.

φ ( z) 

Proof. To prove Corollary 1, we have to show that condi-

ψ and

setting
tion (3.18) implies univalent of

p p

( z)

F ( z) := z

Qα , β , γ g1 ( z).

As 0 < δ ≤ 1 / 2

from Theorem 1, condition (3.18) means
where δ is given by (3.2), and
that ψ is a close- to-convex function in U (see [4]) and

p η

p

p p

α −1

, β , γ f ( z) +

η z p

p

Q p f ( z),

hence ψ is univalent in U . Furthermore, using the same techniques as in the proof of Theorem 1, we can prove the
is univalent in U and the following relation

z Qα , β , γ f ( z) ∈ H[1,1] ∩ Q.

Then,
convexity (univalent) of F and so the details may be omit- ted here. Therefore, by applying Theorem 3, we obtain
Corollary1.

φ1 ( z) 

p η

p

p p

α −1

, β , γ f ( z) +

η z p

p

p

α , β , γ

#### Theorem 4. Let

φ2 ( z)

( z U)

f , g k ∈ Σ p , m

(k = 1, 2), α > ( p η ) / p,

0 ≤ η < p

Implies that
and

z U .

Suppose that

zφ ′′ ( z) 

p p Re1 +

 > −δ ,

(3.19)
Moreover, the function

z Qα , β , γ g1 ( z)

and

φk ( z) 

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setting

z p Q p F ( z) ∈ H[1,1] ∩ Q.

Then, the following relation

φk ( z) :=

p η

p +1 p

α

, β , γ

g k ( z) +

η z p +1

p

α , β , γ

g k ( z),

φ ( z)  z

p Q p

f ( z)  φ

( z)

( z U)

p −1 p

where

1

Implies that

p p

α , β , γ 2

p p

( p η ) 2 + ( p(α + β γ − 1) + η ) 2

z Qα , β , γ Fµ ( g1 )( z)  z

Qα , β , γ Fµ ( f )( z)

δ = −

4( p η )( p(α + β γ − 1) + η )

z Q p F ( g )( z)

( z U).

( p η ) 2 − ( p(α + β γ − 1) + η ) 2

Moreover, the functions

z p Q p F ( g )( z)

and

,

4( p η )( p(α + β γ − 1) + η )

z p Q p F ( g )( z)

are the best subordinate and the best
and

p η

p +1 p

α

, β , γ f ( z) +

η z p +1

Q p f ( z),

dominant, respectively.
Proof. Let us define the functions F and Gk

(k = 1, 2) by

p −1

p

p +1 p

F ( z) := z p Q p

Fµ ( f )( z)

Is univalent in U and z

Qα , β , γ f ( z) ∈ H[0,1] ∩ Q.

and

G ( z) := z p Q p

Fµ ( g k

)( z),

Then, the following relation
respectively. Without loss of generality, as in the proof of

φ1 ( z) 

p η

p

z p +1

p

α −1

, β , γ f ( z) +

η z p +1

p

p

α , β , γ

Theorem 1, we can assume that Gk

univalent on U , and
is analytic and

φ2 ( z)

( z U),

Implies that

G(ζ ) ≠ 0

(ζ ∈ ∂U).

z p +1

Q p g ( z)  z

p +1

p

α , β , γ

From the definition of the integral operator

Fµ defined by

p +1

p

α , β , γ 2

( z U).

(3.20), we obtain

z(Q p

F ( f )( z))′ = µQ p

f ( z) −

Moreover, the functions z

p +1

p

α , β , γ 1

and dominant,

α , β , γ µ

p

α −1 , β , γ

respectively.
Next, we consider the integral operator by (ef. [5],[12])

z

µ + p −1

Fµ (µ > 0)

defined

(µ + p)Qα , β , γ Fµ ( f )( z).

Then, from (3.21) and (3.23), we have

µφ ( z) = µG ( z) + zG( z).

Setting

(3.23)

(3.24)

Fµ ( f )( z) := µ + p t

0

f ( z)dt

( f ∈ Σ p , m ;

µ > 0)

(3.20)

zG′′ ( z)

q ( z) = 1 + k

k G( z)

(k = 1, 2;

z U),

(3.25)
Now, we obtain the following result involving the integral
operator defined by (3.20).

k

And differentating both sides of (3.24), we obtain

#### Theorem 5. Let

f , g k ∈ Σ p , m

(k = 1, 2).

Suppose also

zφ ′′

1 + k

( z) = q

( z) +

zq′ ( z)

k .

q ( z) + µ

that

φk ( z) k



Re1

zφ ′′

+ k

( z) 

 > −δ

The remaining part of the proof is similar to that of
Theorem 1 and so is omitted the proof involved.
Setting



(3.21)

p p

φ ( z) 

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4µ

p p

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ISSN 2229-5518

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