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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


The electronic version of this article is the complete one and can be found online at: http://www.advancesindifferenceequations.com/content/2013/1/252


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

© 2013 Andrei; licensee Springer

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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