Research

# Differential subordinations using the Ruscheweyh derivative and the generalized Sălăgean operator

Loriana Andrei

Author Affiliations

Department of Mathematics and Computer Science, University of Oradea, 1 Universitatii street, Oradea, 410087, Romania

Advances in Difference Equations 2013, 2013:252  doi:10.1186/1687-1847-2013-252

 Received: 13 June 2013 Accepted: 6 August 2013 Published: 20 August 2013

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### Abstract

In the present paper, we study the operator, using the Ruscheweyh derivative R m f ( z ) and the generalized Sălăgean operator D λ m f ( z ) , denote by R D λ , α m : A n A n , R D λ , α m f ( z ) = ( 1 α ) R m f ( z ) + α D λ m f ( z ) , z U , where A n = { f H ( U ) : f ( z ) = z + a n + 1 z n + 1 + , z U } is the class of normalized analytic functions. We obtain several differential subordinations regarding the operator R D λ , α m .

MSC: 30C45, 30A20, 34A40.

##### Keywords:
differential subordination; convex function; best dominant; differential operator; generalized Sălăgean operator; Ruscheweyh derivative

### 1 Introduction

Denote by U the unit disc of the complex plane, U = { z C : | z | < 1 } and H ( U ) the space of holomorphic functions in U.

Let A n = { f H ( U ) : f ( z ) = z + a n + 1 z n + 1 + , z U } and H [ a , n ] = { f H ( U ) : f ( z ) = a + a n z n + a n + 1 z n + 1 + , z U } for a C and n N .

Denote by K = { f A n : Re z f ( z ) f ( z ) + 1 > 0 , z U } the class of normalized convex functions in U.

If f and g are analytic functions in U, we say that f is subordinate to g, written f g , if there is a function w analytic in U, with w ( 0 ) = 0 , | w ( z ) | < 1 , for all z U such that f ( z ) = g ( w ( z ) ) for all z U . If g is univalent, then f g if and only if f ( 0 ) = g ( 0 ) and f ( U ) g ( U ) .

Let ψ : C 3 × U C , and let h be an univalent function in U. If p is analytic in U and satisfies the (second-order) differential subordination

ψ ( p ( z ) , z p ( z ) , z 2 p ( z ) ; z ) h ( z ) , z U , (1.1)

then p is called a solution of the differential subordination. The univalent function q is called a dominant of the solutions of the differential subordination, or more simply a dominant, if p q for all p satisfying (1.1).

A dominant q ˜ that satisfies q ˜ q for all dominants q of (1.1) is said to be the best dominant of (1.1). The best dominant is unique up to a rotation of U.

Definition 1.1 (Al-Oboudi [1])

For f A n , λ 0 and n , m N , the operator D λ m is defined by D λ m : A n A n ,

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 + 1 f ( z ) = ( 1 λ ) D λ m f ( z ) + λ z ( D λ m f ( z ) ) = D λ ( D λ m f ( z ) ) , z U .

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

Remark 1.2 For λ = 1 , in the definition above, we obtain the Sălăgean differential operator [2].

Definition 1.2 (Ruscheweyh [3])

For f A n , n , m N , the operator R m is defined by R m : A n A n ,

R 0 f ( z ) = f ( z ) , R 1 f ( z ) = z f ( z ) , , ( m + 1 ) R m + 1 f ( z ) = z ( R m f ( z ) ) + m R m f ( z ) , z U .

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

Definition 1.3[4]

Let α , λ 0 , n , m N . Denote by R D λ , α m the operator given by R D λ , α m : A n A n ,

R D λ , α m f ( z ) = ( 1 α ) R m f ( z ) + α D λ m f ( z ) , z U .

Remark 1.4 If f A n , f ( z ) = z + j = n + 1 a j z j , then R D λ , α m f ( z ) = z + j = n + 1 { α [ 1 + ( j 1 ) λ ] m + ( 1 α ) C m + j 1 m } a j z j , z U .

This operator was studied also in [4-6] and [7].

Remark 1.5 For α = 0 , R D λ , 0 m f ( z ) = R m f ( z ) , where z U and for α = 1 , R D λ , 1 m f ( z ) = D λ m f ( z ) , where z U .

For λ = 1 , we obtain R D 1 , α m f ( z ) = L α m f ( z ) , which was studied in [8-11].

For m = 0 , R D λ , α 0 f ( z ) = ( 1 α ) R 0 f ( z ) + α D λ 0 f ( z ) = f ( z ) = R 0 f ( z ) = D λ 0 f ( z ) , where z U .

Lemma 1.1 (Hallenbeck and Ruscheweyh [[12], Th. 3.1.6, p.71])

Lethbe a convex function with h ( 0 ) = a , and let γ C { 0 } be a complex number with Re γ 0 . If p H [ a , n ] and

p ( z ) + 1 γ z p ( z ) h ( z ) , z U ,

then

p ( z ) g ( z ) h ( z ) , z U ,

where g ( z ) = γ n z γ / n 0 z h ( t ) t γ / n 1 d t , z U .

Lemma 1.2 (Miller and Mocanu [12])

Letgbe a convex function inU, and let h ( z ) = g ( z ) + n α z g ( z ) , for z U , where α > 0 andnis a positive integer.

If p ( z ) = g ( 0 ) + p n z n + p n + 1 z n + 1 + , z U , is holomorphic inUand

p ( z ) + α z p ( z ) h ( z ) , z U ,

then

p ( z ) g ( z ) , z U ,

and this result is sharp.

### 2 Main results

Theorem 2.1Letgbe a convex function, g ( 0 ) = 1 , and lethbe the function h ( z ) = g ( z ) + n z δ g ( z ) , for z U .

If α , λ , δ 0 , n , m N , f A n and satisfies the differential subordination

( R D λ , α m f ( z ) z ) δ 1 ( R D λ , α m f ( z ) ) h ( z ) , z U , (2.1)

then

( R D λ , α m f ( z ) z ) δ g ( z ) , z U ,

and this result is sharp.

Proof By using the properties of operator R D λ , α m , we have

R D λ , α m f ( z ) = z + j = n + 1 { α [ 1 + ( j 1 ) λ ] m + ( 1 α ) C m + j 1 m } a j z j , z U .

Consider p ( z ) = ( R D λ , α m f ( z ) z ) δ = ( z + j = n + 1 { α [ 1 + ( j 1 ) λ ] m + ( 1 α ) C m + j 1 m } a j z j z ) δ = 1 + p n δ z n δ + p n δ + 1 z n δ + 1 + , z U .

We deduce that p H [ 1 , n δ ] .

Differentiating we obtain ( R D λ , α m f ( z ) z ) δ 1 ( R D λ , α m f ( z ) ) = p ( z ) + 1 δ z p ( z ) , z U .

Then (2.1) becomes

p ( z ) + 1 δ z p ( z ) h ( z ) = g ( z ) + n z δ g ( z ) for  z U .

By using Lemma 1.2, we have

p ( z ) g ( z ) , z U , i.e. , ( R D λ , α m f ( z ) z ) δ g ( z ) , z U .

□

Theorem 2.2Lethbe a holomorphic function, which satisfies the inequality Re ( 1 + z h ( z ) h ( z ) ) > 1 2 , z U , and h ( 0 ) = 1 .

If α , λ , δ 0 , n , m N , f A n and satisfies the differential subordination

( R D λ , α m f ( z ) z ) δ 1 ( R D λ , α m f ( z ) ) h ( z ) , z U , (2.2)

then

( R D λ , α m f ( z ) z ) δ q ( z ) , z U ,

where q ( z ) = δ n z δ n 0 z h ( t ) t δ n 1 d t . The functionqis convex, and it is the best dominant.

Proof Let

p ( z ) = ( R D λ , α m f ( z ) z ) δ = ( z + j = n + 1 { α [ 1 + ( j 1 ) λ ] m + ( 1 α ) C m + j 1 m } a j z j z ) δ = ( 1 + j = n + 1 { α [ 1 + ( j 1 ) λ ] m + ( 1 α ) C m + j 1 m } a j z j 1 ) δ = 1 + j = n δ p j z j

for z U , p H [ 1 , n δ ] .

Differentiating, we obtain ( R D λ , α m f ( z ) z ) δ 1 ( R D λ , α m f ( z ) ) = p ( z ) + 1 δ z p ( z ) , z U , and (2.2) becomes

p ( z ) + 1 δ z p ( z ) h ( z ) , z U .

Using Lemma 1.1, we have

p ( z ) q ( z ) , z U , i.e. , ( R D λ , α m f ( z ) z ) δ q ( z ) = δ n z δ n 0 z h ( t ) t δ n 1 d t , z U ,

and q is the best dominant. □

Corollary 2.3Let h ( z ) = 1 + ( 2 β 1 ) z 1 + z be a convex function inU, where 0 β < 1 .

If α , δ , λ 0 , n , m N , f A n and satisfies the differential subordination

( R D λ , α m f ( z ) z ) δ 1 ( R D λ , α m f ( z ) ) h ( z ) , z U , (2.3)

then

( R D λ , α m f ( z ) z ) δ q ( z ) , z U ,

whereqis given by q ( z ) = ( 2 β 1 ) + 2 ( 1 β ) δ n z δ n 0 z t δ n 1 1 + t d t , z U . The functionqis convex, and it is the best dominant.

Proof Following the same steps as in the proof of Theorem 2.2 and considering p ( z ) = ( R D λ , α m f ( z ) z ) δ , the differential subordination (2.3) becomes

p ( z ) + z δ p ( z ) h ( z ) = 1 + ( 2 β 1 ) z 1 + z , z U .

By using Lemma 1.1, for γ = δ , we have p ( z ) q ( z ) , i.e.,

( R D λ , α m f ( z ) z ) δ q ( z ) = δ n z δ n 0 z h ( t ) t δ n 1 d t = δ n z δ n 0 z t δ n 1 1 + ( 2 β 1 ) t 1 + t d t = δ n z δ n 0 z [ ( 2 β 1 ) t δ n 1 + 2 ( 1 β ) t δ n 1 1 + t ] d t = ( 2 β 1 ) + 2 ( 1 β ) δ n z δ n 0 z t δ n 1 1 + t d t , z U .

□

Remark 2.1 For n = 1 , λ = 1 2 , α = 2 , δ = 1 , we obtain the same example as in [[13], Example 4.2.1, p.125].

Theorem 2.4Letgbe a convex function such that g ( 0 ) = 1 , and lethbe the function h ( z ) = g ( z ) + n z δ g ( z ) , z U .

If α , λ , δ 0 , n , m N , f A n and the differential subordination

z δ + 1 δ R D λ , α n f ( z ) ( R D λ , α n + 1 f ( z ) ) 2 + z 2 δ R D λ , α n f ( z ) ( R D λ , α n + 1 f ( z ) ) 2 [ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) 2 ( R D λ , α n + 1 f ( z ) ) R D λ , α n + 1 f ( z ) ] h ( z ) , z U (2.4)

holds, then

z R D λ , α m f ( z ) ( R D λ , α m + 1 f ( z ) ) 2 g ( z ) , z U ,

and this result is sharp.

Proof For f A n , f ( z ) = z + j = n + 1 a j z j , we have

R D λ , α m f ( z ) = z + j = n + 1 { α [ 1 + ( j 1 ) λ ] m + ( 1 α ) C m + j 1 m } a j z j , z U .

Consider p ( z ) = z R D λ , α m f ( z ) ( R D λ , α m + 1 f ( z ) ) 2 , and we obtain

p ( z ) + z δ p ( z ) = z δ + 1 δ R D λ , α n f ( z ) ( R D λ , α n + 1 f ( z ) ) 2 + z 2 δ R D λ , α n f ( z ) ( R D λ , α n + 1 f ( z ) ) 2 [ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) 2 ( R D λ , α n + 1 f ( z ) ) R D λ , α n + 1 f ( z ) ] .

Relation (2.4) becomes

p ( z ) + z δ p ( z ) h ( z ) = g ( z ) + n z δ g ( z ) , z U .

By using Lemma 1.2, we have

p ( z ) g ( z ) , z U , i.e. , z R D λ , α m f ( z ) ( R D λ , α m + 1 f ( z ) ) 2 g ( z ) , z U .

□

Theorem 2.5Lethbe a holomorphic function, which satisfies the inequality Re ( 1 + z h ( z ) h ( z ) ) > 1 2 , z U , and h ( 0 ) = 1 .

If α , λ , δ 0 , n , m N , f A n and satisfies the differential subordination

z δ + 1 δ R D λ , α n f ( z ) ( R D λ , α n + 1 f ( z ) ) 2 + z 2 δ R D λ , α n f ( z ) ( R D λ , α n + 1 f ( z ) ) 2 [ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) 2 ( R D λ , α n + 1 f ( z ) ) R D λ , α n + 1 f ( z ) ] h ( z ) , z U , (2.5)

then

z R D λ , α m f ( z ) ( R D λ , α m + 1 f ( z ) ) 2 q ( z ) , z U ,

where q ( z ) = δ n z δ n 0 z h ( t ) t δ n 1 d t . The functionqis convex, and it is the best dominant.

Proof Let p ( z ) = z R D λ , α m f ( z ) ( R D λ , α m + 1 f ( z ) ) 2 , z U , p H [ 1 , n ] .

Differentiating, we obtain p ( z ) + z δ p ( z ) = z δ + 1 δ R D λ , α n f ( z ) ( R D λ , α n + 1 f ( z ) ) 2 + z 2 δ R D λ , α n f ( z ) ( R D λ , α n + 1 f ( z ) ) 2 [ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) 2 ( R D λ , α n + 1 f ( z ) ) R D λ , α n + 1 f ( z ) ] , z U , and (2.5) becomes

p ( z ) + z δ p ( z ) h ( z ) , z U .

Using Lemma 1.1, we have

p ( z ) q ( z ) , z U , i.e. , z R D λ , α m f ( z ) ( R D λ , α m + 1 f ( z ) ) 2 q ( z ) = δ n z δ n 0 z h ( t ) t δ n 1 d t , z U ,

and q is the best dominant. □

Theorem 2.6Letgbe a convex function such that g ( 0 ) = 1 , and lethbe the function h ( z ) = g ( z ) + n z δ g ( z ) , z U .

If α , λ , δ 0 , n , m N , f A n and the differential subordination

z 2 δ + 2 δ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) + z 3 δ [ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) ( ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) ) 2 ] h ( z ) , z U (2.6)

holds, then

z 2 ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) g ( z ) , z U .

This result is sharp.

Proof Let p ( z ) = z 2 ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) . We deduce that p H [ 0 , n ] .

Differentiating, we obtain p ( z ) + z δ p ( z ) = z 2 δ + 2 δ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) + z 3 δ [ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) ( ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) ) 2 ] , z U .

Using the notation in (2.6), the differential subordination becomes

p ( z ) + 1 δ z p ( z ) h ( z ) = g ( z ) + n z δ g ( z ) .

By using Lemma 1.2, we have

p ( z ) g ( z ) , z U , i.e. , z 2 ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) g ( z ) , z U ,

and this result is sharp. □

Theorem 2.7Lethbe an holomorphic function, which satisfies the inequality Re ( 1 + z h ( z ) h ( z ) ) > 1 2 , z U , and h ( 0 ) = 1 .

If α , λ , δ 0 , n , m N , f A n and satisfies the differential subordination

z 2 δ + 2 δ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) + z 3 δ [ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) ( ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) ) 2 ] h ( z ) , z U , (2.7)

then

z 2 ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) q ( z ) , z U ,

where q ( z ) = δ n z δ n 0 z h ( t ) t δ n 1 d t . The functionqis convex, and it is the best dominant.

Proof Let p ( z ) = z 2 ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) , z U , p H [ 0 , n ] .

Differentiating, we obtain p ( z ) + z δ p ( z ) = z 2 δ + 2 δ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) + z 3 δ [ ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) ( ( R D λ , α n f ( z ) ) R D λ , α n f ( z ) ) 2 ] , z U , and (2.7) becomes

p ( z ) + 1 δ z p ( z ) h ( z ) , z U .

Using Lemma 1.1, we have

p ( z ) q ( z ) , z U , i.e. , z 2 ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) q ( z ) = δ n z δ n 0 z h ( t ) t δ n 1 d t , z U ,

and q is the best dominant. □

Theorem 2.8Letgbe a convex function such that g ( 0 ) = 1 , and lethbe the function h ( z ) = g ( z ) + n z g ( z ) , z U .

If α , λ 0 , n , m N , f A n and the differential subordination

1 R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) [ ( R D λ , α m f ( z ) ) ] 2 h ( z ) , z U (2.8)

holds, then

R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) g ( z ) , z U .

This result is sharp.

Proof Let p ( z ) = R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) . We deduce that p H [ 1 , n ] .

Differentiating, we obtain 1 R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) [ ( R D λ , α m f ( z ) ) ] 2 = p ( z ) + z p ( z ) , z U .

Using the notation in (2.8), the differential subordination becomes

p ( z ) + z p ( z ) h ( z ) = g ( z ) + n z g ( z ) .

By using Lemma 1.2, we have

p ( z ) g ( z ) , z U , i.e. , R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) g ( z ) , z U ,

and this result is sharp. □

Theorem 2.9Lethbe a holomorphic function, which satisfies the inequality Re ( 1 + z h ( z ) h ( z ) ) > 1 2 , z U , and h ( 0 ) = 1 .

If α , λ 0 , n , m N , f A n and satisfies the differential subordination

1 R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) [ ( R D λ , α m f ( z ) ) ] 2 h ( z ) , z U , (2.9)

then

R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) q ( z ) , z U ,

where q ( z ) = 1 n z 1 n 0 z h ( t ) t 1 n 1 d t . The functionqis convex, and it is the best dominant.

Proof Let p ( z ) = R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) , z U , p H [ 0 , n ] .

Differentiating, we obtain 1 R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) [ ( R D λ , α m f ( z ) ) ] 2 = p ( z ) + z p ( z ) , z U , and (2.9) becomes

p ( z ) + z p ( z ) h ( z ) , z U .

Using Lemma 1.1, we have

p ( z ) q ( z ) , z U , i.e. , R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) q ( z ) = 1 n z 1 n 0 z h ( t ) t 1 n 1 d t , z U ,

and q is the best dominant. □

Corollary 2.10Let h ( z ) = 1 + ( 2 β 1 ) z 1 + z be a convex function inU, where 0 β < 1 .

If α , λ 0 , n , m N , f A n and satisfies the differential subordination

1 R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) [ ( R D λ , α m f ( z ) ) ] 2 h ( z ) , z U , (2.10)

then

R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) q ( z ) , z U ,

whereqis given by q ( z ) = ( 2 β 1 ) + 2 ( 1 β ) n z 1 n 0 z t 1 n 1 1 + t d t , z U . The functionqis convex, and it is the best dominant.

Proof Following the same steps as in the proof of Theorem 2.9 and considering p ( z ) = R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) , the differential subordination (2.10) becomes

p ( z ) + z p ( z ) h ( z ) = 1 + ( 2 β 1 ) z 1 + z , z U .

By using Lemma 1.1 for γ = 1 , we have p ( z ) q ( z ) , i.e.,

R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) q ( z ) = 1 n z 1 n 0 z h ( t ) t 1 n 1 d t = 1 n z 1 n 0 z 1 + ( 2 β 1 ) t 1 + t t 1 n 1 d t = 1 n z 1 n 0 z t 1 n 1 [ ( 2 β 1 ) + 2 ( 1 β ) 1 + t ] d t = ( 2 β 1 ) + 2 ( 1 β ) n z 1 n 0 z t 1 n 1 1 + t d t , z U .

□

Example 2.1 Let h ( z ) = 1 z 1 + z be a convex function in U with h ( 0 ) = 1 and Re ( z h ( z ) h ( z ) + 1 ) > 1 2 .

Let f ( z ) = z + z 2 , z U . For n = 1 , m = 1 , λ = 1 2 , α = 2 , we obtain R D 1 2 , 2 1 f ( z ) = R 1 f ( z ) + 2 D 1 2 1 f ( z ) = z f ( z ) + 2 ( 1 2 f ( z ) + 1 2 z f ( z ) ) = f ( z ) = z + z 2 , z U .

Then ( R D 1 2 , 2 1 f ( z ) ) = f ( z ) = 1 + 2 z ,

R D 1 2 , 2 1 f ( z ) z ( R D 1 2 , 2 1 f ( z ) ) = z + z 2 z ( 1 + 2 z ) = 1 + z 1 + 2 z , 1 R D 1 2 , 2 1 f ( z ) ( R D 1 2 , 2 1 f ( z ) ) [ ( R D 1 2 , 2 1 f ( z ) ) ] 2 = 1 ( z + z 2 ) 2 ( 1 + 2 z ) 2 = 2 z 2 + 2 z + 1 ( 1 + 2 z ) 2 .

We have q ( z ) = 1 z 0 z 1 t 1 + t d t = 1 + 2 ln ( 1 + z ) z .

Using Theorem 2.9, we obtain

2 z 2 + 2 z + 1 ( 1 + 2 z ) 2 1 z 1 + z , z U ,

induce

1 + z 1 + 2 z 1 + 2 ln ( 1 + z ) z , z U .

Theorem 2.11Letgbe a convex function such that g ( 0 ) = 0 , and lethbe the function h ( z ) = g ( z ) + n z g ( z ) , z U .

If α , λ 0 , n , m N , f A n and the differential subordination

[ ( R D λ , α m f ( z ) ) ] 2 + R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) h ( z ) , z U (2.11)

holds, then

R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z g ( z ) , z U .

This result is sharp.

Proof Let p ( z ) = R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z . We deduce that p H [ 0 , n ] .

Differentiating, we obtain [ ( R D λ , α m f ( z ) ) ] 2 + R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) = p ( z ) + z p ( z ) , z U .

Using the notation in (2.11), the differential subordination becomes

p ( z ) + z p ( z ) h ( z ) = g ( z ) + n z g ( z ) .

By using Lemma 1.2, we have

p ( z ) g ( z ) , z U , i.e. , R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z g ( z ) , z U ,

and this result is sharp. □

Theorem 2.12Lethbe a holomorphic function, which satisfies the inequality Re ( 1 + z h ( z ) h ( z ) ) > 1 2 , z U , and h ( 0 ) = 0 .

If α , λ 0 , n , m N , f A n and satisfies the differential subordination

[ ( R D λ , α m f ( z ) ) ] 2 + R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) h ( z ) , z U , (2.12)

then

R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z q ( z ) , z U ,

where q ( z ) = 1 n z 1 n 0 z h ( t ) t 1 n 1 d t . The functionqis convex, and it is the best dominant.

Proof Let p ( z ) = R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z , z U , p H [ 0 , n ] .

Differentiating, we obtain [ ( R D λ , α m f ( z ) ) ] 2 + R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) = p ( z ) + z p ( z ) , z U , and (2.12) becomes

p ( z ) + z p ( z ) h ( z ) , z U .

Using Lemma 1.1, we have

p ( z ) q ( z ) , z U , i.e. , R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z q ( z ) = 1 n z 1 n 0 z h ( t ) t 1 n 1 d t , z U ,

and q is the best dominant. □

Corollary 2.13Let h ( z ) = 1 + ( 2 β 1 ) z 1 + z be a convex function inU, where 0 β < 1 .

If α , λ 0 , n , m N , f A n and satisfies the differential subordination

[ ( R D λ , α m f ( z ) ) ] 2 + R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) h ( z ) , z U , (2.13)

then

R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z q ( z ) , z U ,

whereqis given by q ( z ) = ( 2 β 1 ) + 2 ( 1 β ) n z 1 n 0 z t 1 n 1 1 + t d t , z U . The functionqis convex, and it is the best dominant.

Proof Following the same steps as in the proof of Theorem 2.12 and considering p ( z ) = R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z , the differential subordination (2.13) becomes

p ( z ) + z p ( z ) h ( z ) = 1 + ( 2 β 1 ) z 1 + z , z U .

By using Lemma 1.1 for γ = 1 , we have p ( z ) q ( z ) , i.e.,

R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z q ( z ) = 1 n z 1 n 0 z h ( t ) t 1 n 1 d t = 1 n z 1 n 0 z 1 + ( 2 β 1 ) t 1 + t t 1 n 1 d t = 1 n z 1 n 0 z t 1 n 1 [ ( 2 β 1 ) + 2 ( 1 β ) 1 + t ] d t = ( 2 β 1 ) + 2 ( 1 β ) n z 1 n 0 z t 1 n 1 1 + t d t , z U .

□

Example 2.2 Let h ( z ) = 1 z 1 + z be a convex function in U with h ( 0 ) = 1 and Re ( z h ( z ) h ( z ) + 1 ) > 1 2 .

Let f ( z ) = z + z 2 , z U . For n = 1 , m = 1 , λ = 1 2 , α = 2 , we obtain R D 1 2 , 2 1 f ( z ) = R 1 f ( z ) + 2 D 1 2 1 f ( z ) = z f ( z ) + 2 ( 1 2 f ( z ) + 1 2 z f ( z ) ) = f ( z ) = z + z 2 , z U .

Then ( R D 1 2 , 2 1 f ( z ) ) = f ( z ) = 1 + 2 z ,

R D 1 2 , 2 1 f ( z ) ( R D 1 2 , 2 1 f ( z ) ) z = ( z + z 2 ) ( 1 + 2 z ) z = 2 z 2 + 3 z + 1 , [ ( R D 1 2 , 2 1 f ( z ) ) ] 2 + R D 1 2 , 2 1 f ( z ) ( R D 1 2 , 2 1 f ( z ) ) = ( 1 + 2 z ) 2 + ( z + z 2 ) 2 = 6 z 2 + 6 z + 1 .

We have q ( z ) = 1 z 0 z 1 t 1 + t d t = 1 + 2 ln ( 1 + z ) z .

Using Theorem 2.12, we obtain

6 z 2 + 6 z + 1 1 z 1 + z , z U ,

induce

2 z 2 + 3 z + 1 1 + 2 ln ( 1 + z ) z , z U .

Theorem 2.14Letgbe a convex function such that g ( 0 ) = 0 , and lethbe the function h ( z ) = g ( z ) + n z 1 δ g ( z ) , z U .

If α , λ 0 , δ ( 0 , 1 ) , n , m N , f A n and the differential subordination

( z R D λ , α m f ( z ) ) δ R D λ , α m + 1 f ( z ) 1 δ ( ( R D λ , α m + 1 f ( z ) ) R D λ , α m + 1 f ( z ) δ ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) ) h ( z ) , z U (2.14)

holds, then

R D λ , α m + 1 f ( z ) z ( z R D λ , α m f ( z ) ) δ g ( z ) , z U .

This result is sharp.

Proof Let p ( z ) = R D λ , α m + 1 f ( z ) z ( z R D λ , α m f ( z ) ) δ . We deduce that p H [ 1 , n ] .

Differentiating, we obtain ( z R D λ , α m f ( z ) ) δ R D λ , α m + 1 f ( z ) 1 δ ( ( R D λ , α m + 1 f ( z ) ) R D λ , α m + 1 f ( z ) δ ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) ) = p ( z ) + 1 1 δ z p ( z ) , z U .

Using the notation in (2.14), the differential subordination becomes

p ( z ) + 1 1 δ z p ( z ) h ( z ) = g ( z ) + n z 1 δ g ( z ) .

By using Lemma 1.2, we have

p ( z ) g ( z ) , z U , i.e. , R D λ , α m + 1 f ( z ) z ( z R D λ , α m f ( z ) ) δ g ( z ) , z U ,

and this result is sharp. □

Theorem 2.15Lethbe a holomorphic function, which satisfies the inequality Re ( 1 + z h ( z ) h ( z ) ) > 1 2 , z U , and h ( 0 ) = 1 .

If α , λ 0 , δ ( 0 , 1 ) , n , m N , f A n and satisfies the differential subordination

( z R D λ , α m f ( z ) ) δ R D λ , α m + 1 f ( z ) 1 δ ( ( R D λ , α m + 1 f ( z ) ) R D λ , α m + 1 f ( z ) δ ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) ) h ( z ) , z U , (2.15)

then

R D λ , α m + 1 f ( z ) z ( z R D λ , α m f ( z ) ) δ q ( z ) , z U ,

where q ( z ) = 1 δ n z 1 δ n 0 z h ( t ) t 1 δ n 1 d t . The functionqis convex, and it is the best dominant.

Proof Let p ( z ) = R D λ , α m + 1 f ( z ) z ( z R D λ , α m f ( z ) ) δ , z U , p H [ 0 , n ] .

Differentiating, we obtain ( z R D λ , α m f ( z ) ) δ R D λ , α m + 1 f ( z ) 1 δ ( ( R D λ , α m + 1 f ( z ) ) R D λ , α m + 1 f ( z ) δ ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) ) = p ( z ) + 1 1 δ z p ( z ) , z U , and (2.15) becomes

p ( z ) + 1 1 δ z p ( z ) h ( z ) , z U .

Using Lemma 1.1, we have

p ( z ) q ( z ) , z U , i.e. , R D λ , α m + 1 f ( z ) z ( z R D λ , α m f ( z ) ) δ q ( z ) = 1 δ n z 1 δ n 0 z h ( t ) t 1 δ n 1 d t , z U ,

and q is the best dominant. □

### Competing interests

The author declares that she has no competing interests.

### Author’s contributions

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

### Acknowledgements

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

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