Open Access

Some properties of certain subclasses of analytic functions involving adifferential operator

Advances in Difference Equations20142014:142

https://doi.org/10.1186/1687-1847-2014-142

Received: 16 March 2014

Accepted: 9 April 2014

Published: 13 May 2014

Abstract

In the present paper, we introduce and study certain subclasses of analyticfunctions in the open unit disk U which is defined by the differentialoperator D R λ m , n . We study and investigate some inclusionproperties of these classes. Furthermore, a generalizedBernardi-Libera-Livington integral operator is shown to be preserved for theseclasses.

MSC: 30C45.

Keywords

analytic functions differential operator differential subordination differential superordination

1 Introduction

Let be aclass of functions f in the open unit disk U = { z C : | z | < 1 } normalized by f ( 0 ) = f ( 0 ) 1 = 0 . Thus each f A has a Taylor series representation
f ( z ) = z + j = 2 a j z j .
(1.1)
We denote by S ( ξ ) the well-known subclass of consisting of allanalytic functions which are, respectively, starlike of order ξ[1, 2]
S ( ξ ) = { f A : Re ( z f ( z ) f ( z ) ) > ξ , z U } , 0 ξ < 1 .

Let be a class of all functions ϕ which are analytic andunivalent in U and for which ϕ ( U ) is convex with ϕ ( 0 ) = 1 and Re ϕ ( z ) > 0 , z U .

For two functions f and g analytic in U, we say that thefunction f is subordinate to g in U and write f ( z ) g ( z ) , z U , if there exists a Schwarz function w ( z ) which is analytic in U with w ( 0 ) = 0 and | w ( z ) | < 1 such that f ( z ) = g ( w ( z ) ) , z U .

Making use of the principle of subordination between analytic functions, denote by S ( ξ , ϕ ) [3] a subclass of the class for 0 ξ < 1 and ϕ R which are defined by
S ( ξ , ϕ ) = { f A : 1 1 ξ ( z f ( z ) f ( z ) ζ ) ϕ ( z ) , z U } .
Let f , g A , where f and g are defined by f ( z ) = z + j = 2 a j z j and g ( z ) = z + j = 2 b j z j . Then the Hadamard product (or convolution) f g of the functions f and g is definedby
( f g ) ( z ) = z + j = 2 a j b j z j .

Definition 1.1 (Al-Oboudi [4])

For f A , λ 0 and m N , the operator D λ m is defined by D λ m : A A ,
D λ 0 f ( z ) = f ( z ) , D λ 1 f ( z ) = ( 1 λ ) f ( z ) + λ z f ( z ) = D λ f ( z ) , , D λ m f ( z ) = ( 1 λ ) D λ m 1 f ( z ) + λ z ( D λ m f ( z ) ) = D λ ( D λ m 1 f ( z ) ) , z U .

Remark 1.1 If f A and f ( z ) = z + j = 2 a j z j , then D λ m f ( z ) = z + j = 2 [ 1 + ( j 1 ) λ ] m a j z j , z U .

Remark 1.2 For λ = 1 in the above definition, we obtain theSălăgean differential operator [5].

Definition 1.2 (Ruscheweyh [6])

For f A and n N , the operator R n is defined by R n : A A ,
R 0 f ( z ) = f ( z ) , R 1 f ( z ) = z f ( z ) , , ( n + 1 ) R n + 1 f ( z ) = z ( R n f ( z ) ) + n R n f ( z ) , z U .

Remark 1.3 If f A , f ( z ) = z + j = 2 a j z j , then R n f ( z ) = z + j = 2 ( n + j 1 ) ! n ! ( j 1 ) ! a j z j , z U .

Definition 1.3 ([7])

Let λ 0 and n , m N . Denote by D R λ m , n : A A the operator given by the Hadamard product of thegeneralized Sălăgean operator D λ m and the Ruscheweyh operator R n ,
D R λ m , n f ( z ) = ( D λ m R n ) f ( z ) ,

for any z U and each nonnegative integer m,n.

Remark 1.4 If f A and f ( z ) = z + j = 2 a j z j , then D R λ m , n f ( z ) = z + j = 2 [ 1 + ( j 1 ) λ ] m ( n + j 1 ) ! n ! ( j 1 ) ! a j 2 z j , z U .

Remark 1.5 The operator D R λ m , n was studied also in [810].

For λ = 1 , m = n , we obtain the Hadamard product S R n [11] of the Sălăgean operator S n and the Ruscheweyh derivative R n , which was studied in [12, 13].

For m = n , we obtain the Hadamard product D R λ n [14] of the generalized Sălăgean operator D λ n and the Ruscheweyh derivative R n , which was studied in [1520].

Using a simple computation, one obtains the next result.

Proposition 1.1 ([7])

For m , n N and λ 0 , we have
D R λ m + 1 , n f ( z ) = ( 1 λ ) D R λ m , n f ( z ) + λ z ( D R λ m , n f ( z ) )
(1.2)
and
z ( D R λ m , n f ( z ) ) = ( n + 1 ) D R λ m , n + 1 f ( z ) n D R λ m , n f ( z ) .
(1.3)
By using the operator D R λ m , n f ( z ) , we define the following subclasses of analyticfunctions for 0 ζ < 1 and ϕ R :
S λ m , n ( ξ ) = { f A : D R λ m , n f S ( ξ ) } , S λ m , n ( ξ , ϕ ) = { f A : D R λ m , n f S ( ξ , ϕ ) } .
In particular, we set
S λ m , n ( ξ , 1 + A z 1 + B z ) = S λ m , n ( ξ , A , B ) , 1 < B < A 1 .

Next, we will investigate various inclusion relationships for the subclasses ofanalytic functions introduced above. Furthermore, we study the results of Faisalet al.[21], Darus and Faisal [3].

2 Inclusion relationship associated with the operator D R λ m , n

First, we start with the following lemmas which we need for our main results.

Lemma 2.1 ([22, 23])

Let φ ( μ , v ) be a complex function such that φ : D C , D C × C , and let μ = μ 1 + i μ 2 , v = v 1 + i v 2 . Suppose that φ ( μ , v ) satisfies the following conditions:
  1. 1.

    φ ( μ , v ) is continuous in D,

     
  2. 2.

    ( 1 , 0 ) D and Re φ ( 1 , 0 ) > 0 ,

     
  3. 3.

    Re φ ( i μ 2 , v 1 ) 0 for all ( i μ 2 , v 1 ) D such that v 1 1 2 ( 1 + μ 2 2 ) .

     

Let h ( z ) = 1 + c 1 z + c 2 z 2 + be analytic in U, such that ( h ( z ) , z h ( z ) ) D for all z U . If Re { φ h ( z ) , z h ( z ) } > 0 , z U , then Re { h ( z ) } > 0 .

Lemma 2.2 ([24])

Let ϕ be convex univalent in U with ϕ ( 0 ) = 1 and Re { k ϕ ( z ) + ν } > 0 , k , ν C . If p is analytic in U with p ( 0 ) = 1 , then
p ( z ) + z p ( z ) k p ( z ) + ν ϕ ( z ) , z U ,

implies p ( z ) ϕ ( z ) , z U .

Theorem 2.1 Let f A , 0 ξ < 1 , m , n N , λ > 0 , then
S λ m , n + 1 ( ξ ) S λ m , n ( ξ ) S λ m , n 1 ( ξ ) .
Proof Let f S λ m , n + 1 ( ξ ) and suppose that
z ( D R λ m , n f ( z ) ) D R λ m , n f ( z ) = ξ + ( 1 ξ ) h ( z ) .
(2.1)
Since from (1.3)
( n + 1 ) D R λ m , n + 1 f ( z ) D R λ m , n f ( z ) = n + ξ + ( 1 ξ ) h ( z ) ,
we obtain
( 1 ξ ) h ( z ) = ( n + 1 ) [ ( D R λ m , n + 1 f ( z ) ) D R λ m , n f ( z ) D R λ m , n + 1 f ( z ) D R λ m , n f ( z ) ( D R λ m , n f ( z ) ) D R λ m , n f ( z ) ] , ( 1 ξ ) z h ( z ) = ( n + 1 ) D R λ m , n + 1 f ( z ) D R λ m , n f ( z ) [ z ( D R λ m , n + 1 f ( z ) ) D R λ m , n + 1 f ( z ) ξ ( 1 ξ ) h ( z ) ] , ( 1 ξ ) h ( z ) z n + ξ + ( 1 ξ ) h ( z ) = z ( D R λ m , n + 1 f ( z ) ) D R λ m , n + 1 f ( z ) ξ ( 1 ξ ) h ( z ) , z ( D R λ m , n + 1 f ( z ) ) D R λ m , n + 1 f ( z ) ξ = ( 1 ξ ) h ( z ) + ( 1 ξ ) h ( z ) z n + ξ + ( 1 ξ ) h ( z ) .
Taking h ( z ) = μ = μ 1 + i μ 2 and z h ( z ) = v = v 1 + i v 2 , we define φ ( μ , v ) by
φ ( μ , v ) = ( 1 ξ ) μ + ( 1 ξ ) v n + ξ + ( 1 ξ ) μ
and
Re { φ ( i μ 2 , v 1 ) } = ( 1 ξ ) ( n + ξ ) v 1 ( n + ξ ) 2 + ( 1 ξ ) 2 μ 2 2 , Re { φ ( i μ 2 , v 1 ) } ( 1 ξ ) ( n + ξ ) ( 1 + μ 2 2 ) 2 [ ( n + ξ ) 2 + ( 1 ξ ) 2 μ 2 2 ] < 0 .

Clearly, φ ( μ , v ) satisfies the conditions of Lemma 2.1. Hence Re { h ( z ) } > 0 , z U , implies f S λ m , n ( ξ ) . □

Remark 2.1 Using relation (1.2) and the same techniques as to prove theearlier results, we can obtain a new similar result.

Theorem 2.2 Let f A and ϕ R with
Re { ϕ ( z ) } < ξ 1 + 1 λ 1 ξ .
Then
S λ m + 1 , n ( ξ , ϕ ) S λ m , n ( ξ , ϕ ) S λ m 1 , n ( ξ , ϕ ) .
Proof Let f ( z ) S λ m + 1 , n ( ξ , ϕ ) and set
p ( z ) = 1 1 ξ ( z ( D R λ m , n f ( z ) ) D R λ m , n f ( z ) ξ ) ,
(2.2)

where p is analytic in U with p ( 0 ) = 1 .

By using (1.2) we have
z ( D R λ m , n f ( z ) ) D R λ m , n f ( z ) = 1 λ D R λ m + 1 , n f ( z ) D R λ m , n f ( z ) 1 λ λ .
Now, by using (2.2) we get
p ( z ) = 1 1 ξ ( 1 λ D R λ m + 1 , n f ( z ) D R λ m , n f ( z ) 1 λ λ ξ ) , 1 λ D R λ m + 1 , n f ( z ) D R λ m , n f ( z ) = ξ + 1 λ λ + ( 1 ξ ) p ( z ) .
(2.3)
By using (2.2) and (2.3), we obtain
z p ( z ) = 1 1 ξ 1 λ [ z ( D R λ m + 1 , n f ( z ) ) D R λ m , n f ( z ) D R λ m + 1 , n f ( z ) D R λ m , n f ( z ) z ( D R λ m , n f ( z ) ) D R λ m , n f ( z ) ] , ( 1 ξ ) z p ( z ) = 1 λ D R λ m + 1 , n f ( z ) D R λ m , n f ( z ) [ z ( D R λ m + 1 , n f ( z ) ) D R λ m + 1 , n f ( z ) z ( D R λ m , n f ( z ) ) D R λ m , n f ( z ) ] , ( 1 ξ ) z p ( z ) = [ ζ 1 + 1 λ + ( 1 ξ ) p ( z ) ] [ z ( D R λ m + 1 , n f ( z ) ) D R λ m + 1 , n f ( z ) ( 1 ξ ) p ( z ) ξ ] , ( 1 ξ ) z p ( z ) ( 1 ξ ) p ( z ) + ζ 1 + 1 λ = z ( D R λ m + 1 , n f ( z ) ) D R λ m + 1 , n f ( z ) ξ ( 1 ξ ) p ( z ) .
Hence,
1 1 ξ [ z ( D R λ m + 1 , n f ( z ) ) D R λ m + 1 , n f ( z ) ξ ] = p ( z ) + z p ( z ) ( 1 ζ ) p ( z ) + ζ 1 + 1 λ .
(2.4)

Since Re { ϕ ( z ) } < ξ 1 + 1 λ 1 ξ implies Re { ( 1 ξ ) p ( z ) + ξ 1 + 1 λ } > 0 , applying Lemma 2.2 to (2.4) we have that f ( z ) S λ m , n ( ξ , ϕ ) , as required. □

Remark 2.2 By using relation (1.3) and the same techniques as to prove theearlier results, we can obtain a new similar result.

Corollary 2.3 Let 1 + A 1 + B < ξ 1 + 1 λ 1 ξ for 1 < B < A 1 , then
S λ m + 1 , n ( ξ , A , B ) S λ m , n ( ξ , A , B ) S λ m 1 , n ( ξ , A , B ) .

Proof Taking ϕ ( z ) = 1 + A z 1 + B z , 1 < B < A 1 in Theorem 2.2, we get thecorollary. □

3 Integral-preserving properties

In this section, we present several integral-preserving properties for the subclassesof analytic functions defined above. We recall the generalizedBernardi-Libera-Livington integral operator [25] defined by
F c [ f ( z ) ] = c + 1 z c 0 z t c 1 f ( t ) d t = z + j = 2 c + 1 j + c a j z c , f A , c > 1 ,
(3.1)
which satisfies the following equality:
c D R λ m , n F c [ f ( z ) ] + z [ D R λ m , n F c ( f ( z ) ) ] = ( c + 1 ) D R λ m , n f ( z ) .
(3.2)

Theorem 3.1 Let c > 1 , 0 ξ < 1 . If f S λ m , n ( ξ ) , then F c f S λ m , n ( ξ ) .

Proof Let f S λ m , n ( ξ ) . By using (3.2), we get
z [ D R λ m , n F c [ f ( z ) ] ] D R λ m , n F c [ f ( z ) ] = ( c + 1 ) D R λ m , n f ( z ) D R λ m , n F c [ f ( z ) ] c .
Let
z [ D R λ m , n F c [ f ( z ) ] ] D R λ m , n F c [ f ( z ) ] = ξ + ( 1 ξ ) h ( z ) , h ( z ) = 1 + c 1 z + c 2 z 2 + .
We obtain
z [ D R λ m , n f ( z ) ] D R λ m , n f ( z ) ξ = ( 1 ξ ) h ( z ) + ( 1 ξ ) z h ( z ) ξ + ( 1 ξ ) h ( z ) + c .
This implies
φ ( μ , v ) = ( 1 ξ ) μ + ( 1 ξ ) v c + ξ + ( 1 ξ ) μ
(same as Theorem 2.1) and
Re { φ ( i μ 2 , v 1 ) } = ( 1 ξ ) ( c + ξ ) v 1 ( c + ξ ) 2 + ( 1 ξ ) 2 μ 2 2 , Re { φ ( i μ 2 , v 1 ) } ( 1 ξ ) ( c + ξ ) ( 1 + μ 2 ) 2 2 [ ( c + ξ ) 2 + ( 1 ξ ) 2 μ 2 2 ] < 0 .
After using Lemma 2.1 and Theorem 2.1, we have
F c f S λ m , n ( ξ ) .

 □

Theorem 3.2 Let c > 1 and ϕ R with
Re { ϕ ( z ) } < c + ξ 1 ξ .

If f S λ m , n ( ξ , ϕ ) , then F c f S λ m , n ( ξ , ϕ ) .

Proof Let f ( z ) S λ m , n ( ξ , ϕ ) and set
p ( z ) = 1 1 ξ ( z [ D R λ m , n F c [ f ( z ) ] ] D R λ m , n F c [ f ( z ) ] ξ ) ,
(3.3)

where p is analytic in U with p ( 0 ) = 1 .

Using (3.2) and (3.3), we have
( c + 1 ) z [ D R λ m , n f ( z ) ] D R λ m , n F c [ f ( z ) ] = c + ξ + ( 1 ξ ) p ( z ) .
(3.4)
Then, using (3.2), (3.3) and (3.4), we obtain
1 1 ξ ( z [ D R λ m , n f ( z ) ] D R λ m , n f ( z ) ξ ) = p ( z ) + z p ( z ) ( 1 ξ ) p ( z ) + c + ξ .
(3.5)
Applying Lemma 2.2 to (3.5), we conclude that
F c f S λ m , n ( ξ , ϕ ) .

 □

Author’s contributions

The author drafted the manuscript, read and approved the final manuscript.

Declarations

Acknowledgements

The author thanks the referee for his/her valuable suggestions to improve thepresent article.

Authors’ Affiliations

(1)
Department of Mathematics and Computer Science, University of Oradea

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© Andrei; licensee Springer. 2014

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