Open Access

Applications of differential subordinations for certain classes of p-valent functions associated with generalized Srivastava-Attiya operator

Journal of Inequalities and Applications20122012:153

https://doi.org/10.1186/1029-242X-2012-153

Received: 19 December 2011

Accepted: 21 June 2012

Published: 5 July 2012

Abstract

The object of the present paper is to investigate some inclusion relations and other interesting properties for certain classes of p-valent functions involving generalized Srivastava-Attiya operator by using the principle of differential subordination.

MSC:30C45.

Keywords

differential subordinationintegral operatorp-valent functions

1 Introduction

Let A ( p ) be the class of functions which are analytic and p-valent in the unit disc U = { z C : | z | < 1 } of the form
f ( z ) = z p + k = 1 a k + p z k + p ( p N = { 1 , 2 , } ) .
(1.1)
Let also A ( 1 ) = A 1 . For g ( z ) A ( p ) , given by g ( z ) = z p + k = 1 b k + p z k + p , the Hadamard product (or convolution) of f ( z ) and g ( z ) is defined by
( f g ) ( z ) = z p + k = 1 a k + p b k + p z k + p = ( g f ) ( z ) .
(1.2)
Next, in the usual notation, let Φ ( z , s , a ) denote the Hurwitz-Lerch Zeta function defined as follows:
(1.3)

For further interesting properties and characteristics of the Hurwitz-Lerch Zeta function Φ ( z , s , a ) see [2, 5, 8, 9, 11], and [21].

Recently, Srivastava and Attiya [20] have introduced the linear operator L s , b : A 1 A 1 , defined in terms of the Hadamard product by
L s , b ( f ) ( z ) = G s , b ( z ) f ( z ) ( z U ; b C Z 0 ; s C ) ,
(1.4)
where
G s , b = ( 1 + b ) s [ Φ ( z , s , b ) b s ] ( z U ) .
(1.5)

The Srivastava-Attiya operator L s , b contains, among its special cases, the integral operators introduced and investigated by Alexander [1], Libera [7] and Jung et al. [6].

Analogous to L s , b , Liu [10] defined the operator J p , s , b : A ( p ) A ( p ) by
J p , s , b ( f ) ( z ) = G p , s , b ( z ) f ( z ) ( z U ; b C Z 0 ; s C ; p N ) ,
(1.6)
where
G p , s , b = ( 1 + b ) s [ Φ p ( z , s , b ) b s ]
and
Φ p ( z , s , b ) = 1 b s + k = 0 z k + p ( k + 1 + b ) s .
(1.7)
It is easy to observe from (1.6) and (1.7) that
J p , s , b ( f ) ( z ) = z p + k = 1 ( 1 + b k + 1 + b ) s a k + p z k + p .
(1.8)
We note that
  1. (i)

    J p , 0 , b ( f ) ( z ) = f ( z ) ;

     
  2. (ii)

    J 1 , 1 , 0 ( f ) ( z ) = L f ( z ) = 0 z f ( t ) t d t ( f A 1 ), where the operator L was introduced by Alexander [1];

     
  3. (iii)

    J 1 , s , b ( f ) ( z ) = L s , b f ( z ) ( s C b C Z 0 ), where the operator L s , b was introduced by Srivastava-Attiya [20];

     
  4. (iv)

    J p , 1 , μ + p 1 ( f ) ( z ) = F μ , p ( f ) ( z ) ( μ > p p N ), where the operator F μ , p was introduced by Choi et al. [3];

     
  5. (v)

    J p , α , p ( f ) ( z ) = I p α f ( z ) ( α > 0 p N ), where the operator I p α was introduced by Shams et al. [18];

     
  6. (vi)

    J p , γ , p 1 ( f ) ( z ) = J p γ f ( z ) ( γ N 0 = N { 0 } p N ), where the operator J p γ was introduced by El-Ashwah and Aouf [4];

     
  7. (vii)

    J p , γ , p + l 1 ( f ) ( z ) = J p γ ( l ) f ( z ) ( γ N 0 p N l 0 ), where the operator J p γ ( l ) was introduced by El-Ashwah and Aouf [4].

     
It follows from (1.8) that
z ( J p , s , b ( f ) ( z ) ) = ( b + 1 ) J p , s 1 , b ( f ) ( z ) ( b + 1 p ) J p , s , b ( f ) ( z ) .
(1.9)
For two analytic functions f , g A ( p ) , we say that f is subordinate to g, written f ( z ) g ( z ) if there exists a Schwarz function w ( z ) , 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 ( z ) is univalent in U, then we have the following equivalence (see [14]):
f ( z ) g ( z ) f ( 0 ) = g ( 0 ) and f ( U ) g ( U ) .
Definition 1 For fixed parameters A and B, with 1 B < A 1 , we say that f A ( p ) is in the class S p s , b ( A , B ) if it satisfies the following subordination condition:
( J p , s , b ( f ) ( z ) ) p z p 1 1 + A z 1 + B z ( p N ) .
(1.10)
In view of the definition of subordination (1.10) is equivalent to the following condition:
| ( J p , s , b ( f ) ( z ) ) p z p 1 1 B ( J p , s , b ( f ) ( z ) ) p z p 1 A | < 1 ( z U ) .
For convenience, we write S p s , b ( 1 2 η p , 1 ) = S p s , b ( η ) , where S p s , b ( η ) denotes the class of functions in A ( p ) satisfying the inequality
Re ( ( J p , s , b ( f ) ( z ) ) p z p 1 ) > η ( 0 η < 1 ; p N ; z U ) .

In the present paper, we investigate some inclusion relations and other interesting properties for certain classes of p-valent functions involving an integral operator.

2 Preliminaries

To establish our main results, we need the following lemmas.

Lemma 1 ([13, 14])

Let h be analytic and convex (univalent) in U with h ( 0 ) = 1 . Suppose also that the function φ given by
φ ( z ) = 1 + c m z m + c m + 1 z m + 1 + ,
(2.1)
is analytic in U, where m is a positive integer. If
φ ( z ) + z φ ( z ) ϱ h ( z ) ( Re { ϱ } 0 ; ϱ 0 ) ,
(2.2)
then
φ ( z ) ψ ( z ) = ϱ m z ϱ m 0 z t ϱ m 1 h ( t ) d t h ( z )
(2.3)

and ψ ( z ) is the best dominant of (2.2).

We denote by H ( ϱ ) the class of functions Φ ( z ) given by
Φ ( z ) = 1 + c 1 z + c 2 z 2 + ,
(2.4)
which are analytic in U and satisfy the following inequality:
Re { Φ ( z ) } > ϱ ( 0 ϱ < 1 ; z U ) .

Lemma 2 ([17])

Let the function Φ ( z ) H ( ϱ ) , where Φ ( z ) given by (2.4). Then
Re { Φ ( ϱ ) } 2 ϱ 1 + 2 ( 1 ϱ ) 1 + | z | ( 0 ϱ < 1 ; z U ) .

Lemma 3 ([22])

For 0 ϱ 1 , ϱ 2 < 1 ,
H ( ϱ 1 ) H ( ϱ 2 ) H ( ϱ 3 ) , ϱ 3 = 1 2 ( 1 ϱ 1 ) ( 1 ϱ 2 ) .

The result is best possible.

Lemma 4 ([24])

Let μ be a positive measure on the unit interval [ 0 , 1 ] . Let g ( z , t ) be a complex valued function defined on U × [ 0 , 1 ] such that g ( 0 , t ) is analytic in U for each t [ 0 , 1 ] and such that g ( z , 0 ) is μ integrable on [ 0 , 1 ] for all z U . In addition, suppose that Re { g ( z , t ) } > 0 , g ( r , t ) is real and

Re { 1 g ( z , t ) } 1 g ( r , t ) ( | z | r < 1 ; t [ 0 , 1 ] ) .
If G is defined by
G ( z ) = 0 1 g ( z , t ) d μ ( t ) ,
then
Re { 1 G ( z ) } 1 G ( r ) ( | z | r < 1 ) .

Lemma 5 ([19])

Let the function g be analytic in U with g ( 0 ) = 1 and Re { g ( z ) } > 1 2 ( z U ). Then, for any function F analytic in U, ( g F ) ( U ) is contained in the convex hull of F ( U ) .

Lemma 6 ([16])

Let φ be analytic in U with φ ( 0 ) = 1 and φ ( z ) = 0 for 0 < | z | < 1 and let A , B C with A B , | B | 1 .
  1. (i)
    Let B 0 and γ C = C { 0 } satisfy either | γ ( A B ) B 1 | 1 or | γ ( A B ) B + 1 | 1 . If φ satisfies
    1 + z φ ( z ) γ φ ( z ) 1 + A z 1 + B z ,
    (2.5)
     
then
φ ( z ) ( 1 + B z ) γ ( A B B ) ,
and this is best dominant.
  1. (ii)
    Let B = 0 and γ C be such that | γ A | < π . If φ satisfies (2.5), then
    φ ( z ) e γ A z
     

and this is the best dominant.

For real or complex numbers a n and c ( c Z 0 ) and z U , the Gaussian hypergeometric function defined by
F 1 2 ( a , n ; c ; z ) = 1 + a n c z 1 ! + a ( a + 1 ) n ( n + 1 ) c ( c + 1 ) z 2 2 ! + = k = 0 ( a ) k ( n ) k ( c ) k z k k ! ,
(2.6)

where ( d ) k = d ( d + 1 ) ( d + k 1 ) and ( d ) 0 = 1 . We note that the series defined by (2.6) converges absolutely for z U , and hence, F 1 2 represents an analytic function in U (see, for details, [23], Ch.14]).

Lemma 7 ([23])

For real or complex numbers a, n and c ( c Z 0 )
(2.7)
(2.8)
and
F 1 2 ( a , n ; c ; z ) = 2 F 1 ( n , a ; c ; z ) .
(2.9)

3 Main results

Unless otherwise mentioned, we assume throughout this paper that 1 B < A 1 , s C , b C Z 0 , p N { 1 } , 0 < α 1 , m is a positive integer and the powers are understood as principle values.

Theorem 1 Let f given by (1.1) satisfy the following subordination condition:
( 1 α ) ( J p , s , b ( f ) ( z ) ) p z p 1 + α ( J p , s , b ( f ) ( z ) ) p ( p 1 ) z p 2 1 + A z 1 + B z .
(3.1)
Then
( J p , s , b ( f ) ( z ) ) p z p 1 Ψ ( z ) 1 + A z 1 + B z ,
(3.2)
where
Ψ ( z ) = { A B + ( 1 A B ) ( 1 + B z ) 1 2 F 1 ( 1 , 1 ; p 1 m α + 1 ; B z 1 + B z ) for B 0 , 1 + p 1 m α + p 1 A z for B = 0 ,
(3.3)
is the best dominant of (3.2). Furthermore,
f S p s , b ( β ) ,
(3.4)
where
β = { A B + ( 1 A B ) ( 1 B ) 1 2 F 1 ( 1 , 1 ; p 1 m α + 1 ; B B 1 ) for B 0 , 1 p 1 m α + p 1 A for B = 0 .
(3.5)

The estimate (3.4) is best possible.

Proof Let
θ ( z ) = ( J p , s , b ( f ) ( z ) ) p z p 1 ( z U ) ,
(3.6)
where θ is of the form (2.1) and is analytic in U. Differentiating (3.6) with respect to z, we get
( 1 α ) ( J p , s , b ( f ) ( z ) ) p z p 1 + α ( J p , s , b ( f ) ( z ) ) p ( p 1 ) z p 2 = θ ( z ) + α p 1 z θ ( z ) 1 + A z 1 + B z .
Applying Lemma 1 for ϱ = p 1 α and Lemma 7, we have
( J p , s , b ( f ) ( z ) ) p z p 1 Ψ ( z ) = p 1 m α z p 1 m α 0 z t p 1 m α 1 ( 1 + A t 1 + B t ) d t = { A B + ( 1 A B ) ( 1 + B z ) 1 2 F 1 ( 1 , 1 ; p 1 m α + 1 ; B z 1 + B z ) for B 0 , 1 + p 1 m α + p 1 A z for B = 0 .
This proves the assertion (3.2) of Theorem 1. Next, in order to prove the assertion (3.4) of Theorem 1, it suffices to show that
inf | z | < 1 { Re ( Ψ ( z ) ) } = Ψ ( 1 ) .
Indeed, we have
Re { 1 + A z 1 + B z } 1 A r 1 B r ( | z | r < 1 ) .
Setting
G ( z , ζ ) = 1 + A ζ z 1 + B ζ z and d ν ( ζ ) = p 1 m α ζ p 1 m α 1 d ζ ( 0 ζ 1 ) ,
which is a positive measure on the closed interval [ 0 , 1 ] , we get
Ψ ( z ) = 0 1 G ( z , ζ ) d ν ( ζ ) .
Then
Re { Ψ ( z ) } 0 1 1 A ζ r 1 B ζ r d ν ( ζ ) = Ψ ( r ) ( | z | r < 1 ) .

Letting r 1 in the above inequality, we obtain the assertion (3.4). Finally, the estimate (3.4) is best possible as Ψ is the best dominant of (3.2). This completes the proof of Theorem 1. □

Theorem 2 If f S p s , b ( η ) ( 0 η < 1 ), then
Re { ( 1 α ) ( J p , s , b ( f ) ( z ) ) p z p 1 + α ( J p , s , b ( f ) ( z ) ) p ( p 1 ) z p 2 } > η ( | z | < R ) ,
where
R = { ( p 1 ) 2 + ( m α ) 2 m α p 1 } 1 m .
(3.7)

The result is best possible.

Proof Let f S p s , b ( η ) , then we write
( J p , s , b ( f ) ( z ) ) p z p 1 = η + ( 1 η ) u ( z ) ( z U ) ,
(3.8)
where u is of the form (2.1), is analytic in U and has a positive real part in U. Differentiating (3.8) with respect to z, we have
1 1 η { ( 1 α ) ( J p , s , b ( f ) ( z ) ) p z p 1 + α ( J p , s , b ( f ) ( z ) ) p ( p 1 ) z p 2 η } = u ( z ) + α p 1 z u ( z ) .
(3.9)
Applying the following well-known estimate [12]:
| z u ( z ) | Re { u ( z ) } 2 m r m 1 r 2 m ( | z | = r < 1 ) ,
in (3.9), we have
(3.10)

such that the right-hand side of (3.10) is positive, if r < R , where R is given by (3.7).

In order to show that the bound R is best possible, we consider the function f A ( p ) defined by
( J p , s , b ( f ) ( z ) ) p z p 1 = η + ( 1 η ) 1 + z m 1 z m ( 0 η < 1 ; z U ) .
Note that
1 1 η { ( 1 α ) ( J p , s , b ( f ) ( z ) ) p z p 1 + α ( J p , s , b ( f ) ( z ) ) p ( p 1 ) z p 2 η } = ( p 1 ) ( 1 z 2 m ) 2 α m z m ( p 1 ) ( 1 z m ) 2 = 0 ,

for z = R exp { i π m } . This completes the proof of Theorem 2. □

For a function f A ( p ) , the generalized Bernardi-Libera-Livingston integral operator F μ , p is defined by
F μ , p ( f ) ( z ) = μ + p z μ 0 z t μ 1 f ( t ) d t = ( z p + k = 1 μ + p μ + p + k z k + p ) f ( z ) = z p 2 F 1 ( 1 , μ + p ; μ + p + 1 ; z ) f ( z ) ( μ > p ; z U ) .
(3.11)
From (1.8) and (3.11), we have
z ( J p , s , b F μ , p ( f ( z ) ) ) = ( μ + p ) J p , s , b ( f ) ( z ) μ J p , s , b F μ , p ( f ( z ) ) ( μ > p ; z U )
(3.12)
and
J p , s , b F μ , p ( f ( z ) ) = F μ , p ( J p , s , b ( f ) ( z ) ) .
Theorem 3 Let f S p s , b ( A , B ) and F μ , p be defined by (3.11). Then
( J p , s , b F μ , p ( f ( z ) ) ) p z p 1 Φ ( z ) 1 + A z 1 + B z ,
(3.13)
where
Φ ( z ) = { A B + ( 1 A B ) ( 1 + B z ) 1 2 F 1 ( 1 , 1 ; μ + p m + 1 ; B z 1 + B z ) for B 0 , 1 + μ + p μ + p + m A z for B = 0 ,
(3.14)
is the best dominant of (3.13). Furthermore,
Re { ( J p , s , b F μ , p ( f ( z ) ) ) p z p 1 } > ψ ( z U ) ,
where
ψ = { A B + ( 1 A B ) ( 1 B ) 1 2 F 1 ( 1 , 1 ; μ + p m + 1 ; B B 1 ) for B 0 , 1 μ + p μ + p + m A for B = 0 .

The result is best possible.

Proof Let
K ( z ) = ( J p , s , b F μ , p ( f ( z ) ) ) p z p 1 ( z U ) ,
(3.15)
where K is of the form (2.1) and is analytic in U. Using (3.12) in (3.15) and differentiating the resulting equation with respect to z, we have
( J p , s , b ( f ) ( z ) ) p z p 1 = K ( z ) + z K ( z ) p + μ 1 + A z 1 + B z .

The remaining part of the proof is similar to that of Theorem 1, and so we omit it. □

We note that
( J p , s , b F μ , p ( f ( z ) ) ) p z p 1 = p + μ p z p + μ 0 z t μ ( J p , s , b ( f ) ( t ) ) d t ( f A ( p ) ; z U ) .
(3.16)

Putting A = 1 2 δ p ( 0 δ < 1 ) and B = 1 in Theorem 3 and using (3.16), we obtain the following corollary.

Corollary 1 If f A ( p ) satisfies the following inequality:
Re { ( J p , s , b f ( z ) ) p z p 1 } > δ ( 0 δ < 1 ; z U ) ,
then
Re { p + μ p z p + μ 0 z t μ ( J p , s , b ( f ) ( t ) ) d t } > δ p + ( 1 δ p ) [ 2 F 1 ( 1 , 1 ; μ + p m + 1 ; 1 2 ) 1 ] ( z U ) .

The result is best possible.

Theorem 4 Let f , g A ( p ) satisfy the following inequality:

Re { J p , s , b ( g ) ( z ) z p } > 0 ( z U ) .
If
| J p , s , b ( f ) ( z ) J p , s , b ( g ) ( z ) 1 | < 1 ( z U ) ,
then
where
(3.17)
Proof Let
q ( z ) = J p , s , b ( f ) ( z ) J p , s , b ( g ) ( z ) 1 = c m z m + c m + 1 z m + 1 + ,
(3.18)
where q ( z ) is analytic in U with q ( 0 ) = 0 and | q ( z ) | | z | m . Then, by applying the familiar Schwartz Lemma [15], we have q ( z ) = z m X ( z ) , where X is analytic in U and | X ( z ) | 1 . Therefore (3.18) leads to
J p , s , b ( f ) ( z ) = J p , s , b ( g ) ( z ) ( 1 + z m X ( z ) ) ( z U ) .
(3.19)
Differentiating (3.19) logarithmically with respect to z, we have
z ( J p , s , b ( f ) ( z ) ) J p , s , b ( f ) ( z ) = z ( J p , s , b ( g ) ( z ) ) J p , s , b ( g ) ( z ) + z m { m X ( z ) + z X ( z ) } 1 + z m X ( z ) .
(3.20)
Letting
ω ( z ) = J p , s , b ( g ) ( z ) z p ( z U ) ,
where ω is in the form (2.1), is analytic in U Re { ω ( z ) } > 0 and
z ( J p , s , b ( g ) ( z ) ) J p , s , b ( g ) ( z ) = z ω ( z ) ω ( z ) + p ,
then we have
Re { z ( J p , s , b ( f ) ( z ) ) J p , s , b ( f ) ( z ) } p | z ω ( z ) ω ( z ) | | z m { m X ( z ) + z X ( z ) } 1 + z m X ( z ) | .
(3.21)
Using the following known estimates [12] (see also [15]):
| ω ( z ) ω ( z ) | 2 m r m 1 1 r 2 m and | m X ( z ) + z X ( z ) 1 + z m X ( z ) | m 1 r m ( | z | = r < 1 ) ,
in (3.21), we have
Re { z ( J p , s , b ( f ) ( z ) ) J p , s , b ( f ) ( z ) } p 3 m r m ( p + m ) r 2 m 1 r 2 m ( | z | = r < 1 ) ,

which is certainly positive, provided that , where is given by (3.17). This completes the proof of Theorem 4. □

Theorem 5 Let 1 B i < A i 1 ( i = 1 , 2 ) and τ < p . If each of the functions f i A ( p ) satisfies the following subordination condition:
( 1 α ) ( J p , s , b ( f i ) ( z ) ) p z p 1 + α ( J p , s , b ( f i ) ( z ) ) p ( p 1 ) z p 2 1 + A i z 1 + B i z ( i = 1 , 2 ) ,
(3.22)
then
( 1 α ) ( J p , s , b ( F ) ( z ) ) p z p 1 + α ( J p , s , b ( F ) ( z ) ) p ( p 1 ) z p 2 1 + ( 1 2 τ p ) z 1 z ,
(3.23)
where
F ( z ) = J p , s , b ( f 1 f 2 ) ( z )
(3.24)
and
τ = p 4 p ( A 1 B 1 ) ( A 2 B 2 ) ( 1 B 1 ) ( 1 B 2 ) [ 1 1 2 2 F 1 ( 1 , 1 ; p 1 α + 1 ; 1 2 ) ] .
(3.25)

The result is best possible when B 1 = B 2 = 1 .

Proof Suppose that the functions f i A ( p ) ( i = 1 , 2 ) satisfy the condition (3.22). Then by setting
h i ( z ) = ( 1 α ) ( J p , s , b ( f i ) ( z ) ) p z p 1 + α ( J p , s , b ( f i ) ( z ) ) p ( p 1 ) z p 2 ( i = 1 , 2 ) ,
(3.26)
we have
h i H ( ϱ i ) , ϱ i = 1 A i 1 B i ( i = 1 , 2 ) .
And
( J p , s , b ( f i ) ( z ) ) = p ( p 1 ) α z ( 1 p ) ( 1 α ) α 0 z t p 1 α 1 h i ( t ) d t ( i = 1 , 2 ) ,
(3.27)
from (3.24), (3.26) and (3.27), we have
( J p , s , b ( F ) ( z ) ) = p ( p 1 ) α z ( 1 p ) ( 1 α ) α 0 z t p 1 α 1 H ( t ) d t ( i = 1 , 2 ) .
(3.28)
For convenience,
H ( z ) = ( 1 α ) ( J p , s , b ( F ) ( z ) ) p z p 1 + α ( J p , s , b ( F ) ( z ) ) p ( p 1 ) z p 2 = p ( p 1 ) α z ( 1 p ) α 0 z t p 1 α 1 ( h 1 h 2 ) ( t ) d t .
(3.29)
Since h i H ( ϱ i ) ( i = 1 , 2 ), it follows from Lemma 3 that
( h 1 h 2 ) ( z ) H ( ϱ 3 ) , ϱ 3 = 1 2 ( 1 ϱ 1 ) ( 1 ϱ 2 ) .
(3.30)
By using (3.30) in (3.29) and applying Lemmas 2 and 3, we have
Re { H ( z ) } = p ( p 1 ) α 0 1 s p 1 α 1 Re { ( h 1 h 2 ) ( s z ) } d s p ( p 1 ) α 0 1 s p 1 α 1 ( 2 ϱ 3 1 + 2 ( 1 ϱ 3 ) 1 + s | z | ) d s > p ( p 1 ) α 0 1 s p 1 α 1 ( 2 ϱ 3 1 + 2 ( 1 ϱ 3 ) 1 + s ) d s = p 4 p ( A 1 B 1 ) ( A 2 B 2 ) ( 1 B 1 ) ( 1 B 2 ) [ 1 p 1 α 0 1 s p 1 α 1 ( 1 + s ) 1 d s ] = p 4 p ( A 1 B 1 ) ( A 2 B 2 ) ( 1 B 1 ) ( 1 B 2 ) [ 1 1 2 2 F 1 ( 1 , 1 ; p 1 α + 1 ; 1 2 ) ] ( z 1 ) = τ ( z U ) .
When B 1 = B 2 = 1 , we consider f i A ( p ) ( i = 1 , 2 ) satisfy the condition (3.22) and are defined by
( J p , s , b ( f i ) ( z ) ) = p ( p 1 ) α z ( 1 p ) ( 1 α ) α 0 z t p 1 α 1 ( 1 + A i t 1 t ) d t ( i = 1 , 2 ) .
By using (3.29) and applying Lemma 3, we have
H ( z ) = p ( p 1 ) α 0 1 s p 1 α 1 [ 1 ( 1 + A 1 ) ( 1 + A 2 ) + ( 1 + A 1 ) ( 1 + A 2 ) 1 s z ] d s = p p ( 1 + A 1 ) ( 1 + A 2 ) + p ( 1 + A 1 ) ( 1 + A 2 ) ( 1 z ) 1 2 F 1 ( 1 , 1 ; p 1 α + 1 ; z z 1 ) p p ( 1 + A 1 ) ( 1 + A 2 ) + p 2 ( 1 + A 1 ) ( 1 + A 2 ) 2 F 1 ( 1 , 1 ; p 1 α + 1 ; 1 2 ) ( z 1 ) .

This completes the proof of Theorem 5. □

Remark 1 Putting A i = 1 2 θ i ( 0 θ i < 1 ) and B i = 1 ( i = 1 , 2 ) in Theorem 5, we obtain the result obtained by Liu [10], Theorem 5].

Putting A i = 1 2 θ i ( 0 θ i < 1 ), B i = 1 ( i = 1 , 2 ) and s = 0 in Theorem 5, we obtain the following corollary.

Corollary 2 Let χ < p and f i A ( p ) satisfy the following inequality:
Re { ( 1 α ) f i ( z ) p z p 1 + α f i ( z ) p ( p 1 ) z p 2 } > θ i ( 0 θ i < 1 ; i = 1 , 2 ) ,
then
Re { ( 1 α ) ( f 1 f 2 ) ( z ) p z p 1 + α ( f 1 f 2 ) ( z ) p ( p 1 ) z p 2 } > χ p ,
where
χ = p 4 p ( 1 θ 1 ) ( 1 θ 2 ) [ 1 1 2 2 F 1 ( 1 , 1 ; p 1 α + 1 ; 1 2 ) ] .

The result is best possible.

Theorem 6 Let f S p s , b ( A , B ) and g A ( p ) satisfy the following inequality:

Re { g ( z ) z p } > 1 2 ( z U ) ,
(3.31)
then
( f g ) ( z ) S p s , b ( A , B ) .
Proof We have
( J p , s , b ( f g ) ( z ) ) p z p 1 = ( J p , s , b ( f ) ( z ) ) p z p 1 g ( z ) z p ( z U ) ,

where g ( z ) satisfies (3.31) and 1 + A z 1 + B z is convex (univalent) in U. By using (1.10) and applying Lemma 5, we complete the proof of Theorem 6. □

Theorem 7 Let σ > 0 and f A ( p ) satisfy the following subordination condition:
( 1 α ) J p , s , b ( f ) ( z ) z p + α ( J p , s , b ( f ) ( z ) ) p z p 1 1 + A z 1 + B z .
(3.32)
Then
Re { J p , s , b ( f ) ( z ) z p } 1 σ > γ 1 σ ,
where
γ = { A B + ( 1 A B ) ( 1 B ) 1 2 F 1 ( 1 , 1 ; p m α + 1 ; B B 1 ) for B 0 , 1 p m α + p A for B = 0 .

The result is best possible.

Proof Let
M ( z ) = J p , s , b ( f ) ( z ) z p ( z U ) ,
(3.33)
where M is of the form (2.1) and is analytic in U. Differentiating (3.33) with respect to z, we have
( 1 α ) J p , s , b ( f ) ( z ) z p + α ( J p , s , b ( f ) ( z ) ) p z p 1 = M ( z ) + α p z M ( z ) 1 + A z 1 + B z .
Now, by following steps similar to the proof of Theorem 1 and using the elementary inequality
Re { ϒ 1 / ϰ } { Re ϒ } 1 / ϰ ( Re { ϒ } > 0 ; ϰ N ) ,

we obtain the result asserted by Theorem 7. □

Theorem 8 Let ν C and A , B C with A B and | B | 1 . Suppose that
If f A ( p ) with J p , s , b ( f ) ( z ) 0 for all z U = U { 0 } , then
J p , s 1 , b ( f ) ( z ) J p , s , b ( f ) ( z ) 1 + A z 1 + B z ,
implies
( J p , s , b ( f ) ( z ) z p ) ν q 1 ( z ) ,
where
q 1 ( z ) = { ( 1 + B z ) ν ( b + 1 ) ( A B ) / B , if B 0 , e ν ( b + 1 ) A z , if B = 0 ,

is the best dominant.

Proof Let us put
φ ( z ) = ( J p , s , b ( f ) ( z ) z p ) ν ( z U ) .
(3.34)
Then φ is analytic in U, φ ( 0 ) = 1 and φ ( z ) 0 for all z U . Taking the logarithmic derivatives in both sides of (3.34) and using the identity (1.9), we have
1 + z φ ( z ) ν ( b + 1 ) φ ( z ) = J p , s 1 , b ( f ) ( z ) J p , s , b ( f ) ( z ) 1 + A z 1 + B z .

Now the assertions of Theorem 8 follow by using Lemma 6 for γ = ν ( b + 1 ) . □

Putting B = 1 and A = 1 2 σ , 0 σ < 1 , in Theorem 8, we obtain the following corollary.

Corollary 3 Assume that ν C satisfies either | 2 ν ( b + 1 ) ( 1 σ ) 1 | 1 or | 2 ν ( b + 1 ) ( 1 σ ) + 1 | 1 . If f A ( p ) with J p , s , b ( f ) ( z ) 0 for z U , then
Re { J p , s 1 , b ( f ) ( z ) J p , s , b ( f ) ( z ) } > σ ( z U ) ,
implies
( J p , s , b ( f ) ( z ) z p ) ν q 2 ( z ) = ( 1 z ) 2 ν ( b + 1 ) ( 1 σ ) ,

and q 2 is the best dominant.

Remark 2 Specializing the parameters s and b in the above results of this paper, we obtain the results for the corresponding operators F μ , p , I p α , J p γ and J p γ ( l ) which are defined in the introduction.

Declarations

Acknowledgements

The authors would like to thank the referees of the paper for their helpful suggestions.

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, Mansoura University

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