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

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.

1 Introduction

Let A(p) be the class of functions which are analytic and p-valent in the unit disc U={zC:|z|<1} of the form

f(z)= z p + k = 1 a k + p z k + p (pN={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

(fg)(z)= z p + k = 1 a k + p b k + p z k + p =(gf)(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 ] (zU).
(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)=Lf(z)= 0 z f ( t ) t dt (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) (sCbC 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>ppN), 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>0pN), 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}pN), 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 pNl0), 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+1p) J p , s , b (f)(z).
(1.9)

For two analytic functions f,gA(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 zU, such that f(z)=g(w(z))zU. 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)andf(U)g(U).

Definition 1 For fixed parameters A and B, with 1B<A1, we say that fA(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 (pN).
(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(zU).

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(0η<1;pN;zU).

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 withh(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)dth(z)
(2.3)

andψ(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(0ϱ<1;zU).

Lemma 2 ([17])

Let the functionΦ(z)H(ϱ), whereΦ(z)given by (2.4). Then

Re { Φ ( ϱ ) } 2ϱ1+ 2 ( 1 ϱ ) 1 + | z | (0ϱ<1;zU).

Lemma 3 ([22])

For0 ϱ 1 , ϱ 2 <1,

H( ϱ 1 )H( ϱ 2 )H( ϱ 3 ), ϱ 3 =12(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]. Letg(z,t)be a complex valued function defined onU×[0,1]such thatg(0,t)is analytic in U for eacht[0,1]and such thatg(z,0)is μ integrable on[0,1]for allzU. In addition, suppose thatRe{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)dμ(t),

then

Re { 1 G ( z ) } 1 G ( r ) ( | z | r < 1 ) .

Lemma 5 ([19])

Let the function g be analytic in U withg(0)=1andRe{g(z)}> 1 2 (zU). Then, for any function F analytic in U, (gF)(U)is contained in the convex hull ofF(U).

Lemma 6 ([16])

Let φ be analytic in U withφ(0)=1andφ(z)=0for0<|z|<1and letA,BCwithAB, |B|1.

  1. (i)

    LetB0andγ C =C{0}satisfy either| γ ( A B ) B 1|1or| 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)

    LetB=0andγ C be such that|γA|<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 zU, 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+k1) and ( d ) 0 =1. We note that the series defined by (2.6) converges absolutely for zU, 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 1B<A1, sC, bC Z 0 , pN{1}, 0<α1, 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:

(1α) ( 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 (zU),
(3.6)

where θ is of the form (2.1) and is analytic in U. Differentiating (3.6) with respect to z, we get

(1α) ( 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 anddν(ζ)= p 1 m α ζ p 1 m α 1 dζ(0ζ1),

which is a positive measure on the closed interval [0,1], we get

Ψ(z)= 0 1 G(z,ζ)dν(ζ).

Then

Re { Ψ ( z ) } 0 1 1 A ζ r 1 B ζ r dν(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 Iff S p s , b (η) (0η<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+(1η)u(z)(zU),
(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 fA(p) defined by

( J p , s , b ( f ) ( z ) ) p z p 1 =η+(1η) 1 + z m 1 z m (0η<1;zU).

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 fA(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;zU)
(3.12)

and

J p , s , b F μ , p ( f ( z ) ) = F μ , p ( J p , s , b ( f ) ( z ) ) .

Theorem 3 Letf 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(zU),

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 (zU),
(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 ) ) dt ( f A ( p ) ; z U ) .
(3.16)

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

Corollary 1 IffA(p)satisfies the following inequality:

Re { ( J p , s , b f ( z ) ) p z p 1 } >δ(0δ<1;zU),

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](zU).

The result is best possible.

Theorem 4 Letf,gA(p)satisfy the following inequality:

Re { J p , s , b ( g ) ( z ) z p } >0(zU).

If

| J p , s , b ( f ) ( z ) J p , s , b ( g ) ( z ) 1|<1(zU),

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 ) ) (zU).
(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 (zU),

where ω is in the form (2.1), is analytic in URe{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 Let1 B i < A i 1 (i=1,2) andτ<p. If each of the functions f i A(p)satisfies the following subordination condition:

(1α) ( 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

(1α) ( 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=p4p ( 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)=(1α) ( 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)dt(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)dt(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 =12(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 ) dt(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 =12 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 =12 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 Letχ<pand 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=p4p(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 Letf S p s , b (A,B)andgA(p)satisfy the following inequality:

Re { g ( z ) z p } > 1 2 (zU),
(3.31)

then

(fg)(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 (zU),

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 Letσ>0andfA(p)satisfy the following subordination condition:

(1α) 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 (zU),
(3.33)

where M is of the form (2.1) and is analytic in U. Differentiating (3.33) with respect to z, we have

(1α) 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 Letν C andA,BCwithABand|B|1. Suppose that

IffA(p)with J p , s , b (f)(z)0for allz 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 (zU).
(3.34)

Then φ is analytic in U, φ(0)=1 and φ(z)0 for all zU. 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=12σ, 0σ<1, in Theorem 8, we obtain the following corollary.

Corollary 3 Assume thatν C satisfies either|2ν(b+1)(1σ)1|1or|2ν(b+1)(1σ)+1|1. IffA(p)with J p , s , b (f)(z)0forz U , then

Re { J p , s 1 , b ( f ) ( z ) J p , s , b ( f ) ( z ) } >σ(zU),

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.

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Aouf, M., Mostafa, A., Shahin, A. et al. Applications of differential subordinations for certain classes of p-valent functions associated with generalized Srivastava-Attiya operator. J Inequal Appl 2012, 153 (2012). https://doi.org/10.1186/1029-242X-2012-153

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Keywords

  • differential subordination
  • integral operator
  • p-valent functions