Skip to content

Advertisement

  • Research
  • Open Access

Some differential subordinations using Ruscheweyh derivative and Sălăgean operator

Advances in Difference Equations20132013:150

https://doi.org/10.1186/1687-1847-2013-150

Received: 9 April 2013

Accepted: 10 May 2013

Published: 28 May 2013

Abstract

In the present paper we study the operator defined by using the Ruscheweyh derivative R n f ( z ) and the Sălăgean operator S n f ( z ) , denoted by L α n : A A , L α n f ( z ) = ( 1 α ) R n f ( z ) + α S n 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 with A 1 = A . We obtain several differential subordinations regarding the operator L α n .

MSC:30C45, 30A20, 34A40.

Keywords

  • differential subordination
  • convex function
  • best dominant
  • differential operator
  • Sălăgean operator
  • Ruscheweyh derivative

1 Introduction

Denote by U the unit disc of the complex plane, U = { z C : | z | < 1 } , and by 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 } with A 1 = A 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 : 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 a 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 (Sălăgean [1])

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

Remark 1.1 If f A , f ( z ) = z + j = 2 a j z j , then S n f ( z ) = z + j = 2 j n a j z j , z U .

Definition 1.2 (Ruscheweyh [2])

For f A , 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 1.2 If f A , f ( z ) = z + j = 2 a j z j , then R n f ( z ) = z + j = 2 C n + j 1 n a j z j , z U .

Definition 1.3 ([3])

Let α 0 , n N . Denote by L α n the operator given by L α n : A A ,
L α n f ( z ) = ( 1 α ) R n f ( z ) + α S n f ( z ) , z U .

Remark 1.3 If f A , f ( z ) = z + j = 2 a j z j , then L α n f ( z ) = z + j = 2 ( α j n + ( 1 α ) C n + j 1 n ) a j z j , z U .

This operator was studied also in [35].

Lemma 1.1 (Hallenbeck and Ruscheweyh [[6], Th. 3.1.6, p.71])

Let h be 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 [6])

Let g be a convex function in U and let h ( z ) = g ( z ) + n α z g ( z ) , for z U , where α > 0 and n is a positive integer.

If p ( z ) = g ( 0 ) + p n z n + p n + 1 z n + 1 +  , z U , is holomorphic in U and
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 2.1 Let g be a convex function, g ( 0 ) = 1 and let h be the function h ( z ) = g ( z ) + z δ g ( z ) , z U .

If α , δ 0 , n N , f A and satisfies the differential subordination
( L α n f ( z ) z ) δ 1 ( L α n f ( z ) ) h ( z ) , z U ,
(2.1)
then
( L α n f ( z ) z ) δ g ( z ) , z U ,

and this result is sharp.

Proof By using the properties of the operator L α n , we have
L α n f ( z ) = z + j = 2 ( α j n + ( 1 α ) C n + j 1 n ) a j z j , z U .

Consider p ( z ) = ( L α n f ( z ) z ) δ = ( z + j = 2 ( α j n + ( 1 α ) C n + j 1 n ) a j z j z ) δ = 1 + p δ z δ + p δ + 1 z δ + 1 +  , z U .

We deduce that p H [ 1 , δ ] .

Differentiating, we obtain ( L α n f ( z ) z ) δ 1 ( L α n 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 ) + z δ g ( z ) , z U .
By using Lemma 1.2, we have
p ( z ) g ( z ) , z U , i.e. , ( L α n f ( z ) z ) δ g ( z ) , z U .

 □

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

If α , δ 0 , n N , f A and satisfies the differential subordination
( L α n f ( z ) z ) δ 1 ( L α n f ( z ) ) h ( z ) , z U ,
(2.2)
then
( L α n f ( z ) z ) δ q ( z ) , z U ,

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

Proof Let
p ( z ) = ( L α n f ( z ) z ) δ = ( z + j = 2 ( α j n + ( 1 α ) C n + j 1 n ) a j z j z ) δ = ( 1 + j = 2 ( α j n + ( 1 α ) C n + j 1 n ) a j z j 1 ) δ = 1 + j = δ + 1 p j z j 1

for z U , p H [ 1 , δ ] .

Differentiating, we obtain ( L α n f ( z ) z ) δ 1 ( L α n 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. , ( L α n f ( z ) z ) δ q ( z ) = δ z δ 0 z h ( t ) t δ 1 d t , z U ,

and q is the best dominant. □

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

If α , δ 0 , n N , f A and satisfies the differential subordination
( L α n f ( z ) z ) δ 1 ( L α n f ( z ) ) h ( z ) , z U ,
(2.3)
then
( L α n f ( z ) z ) δ q ( z ) , z U ,

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

Proof Following the same steps as in the proof of Theorem 2.2 and considering p ( z ) = ( L α n 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.,
( L α n f ( z ) z ) δ q ( z ) = δ z δ 0 z h ( t ) t δ 1 d t = δ z δ 0 z t δ 1 1 + ( 2 β 1 ) t 1 + t d t = δ z δ 0 z [ ( 2 β 1 ) t δ 1 + 2 ( 1 β ) t δ 1 1 + t ] d t = ( 2 β 1 ) + 2 ( 1 β ) δ z δ 0 z t δ 1 1 + t d t , z U .

 □

Remark 2.1 For n = 1 , α = 2 , δ = 1 , we obtain the same example as in [[7], Example 2.2.1, p.26].

Theorem 2.4 Let g be a convex function such that g ( 0 ) = 1 and let h be the function h ( z ) = g ( z ) + z γ g ( z ) , z U , where γ > 0 .

If α 0 , n N , f A and the differential subordination
( γ + 1 ) z γ L α n f ( z ) ( L α n + 1 f ( z ) ) 2 + z 2 γ L α n f ( z ) ( L α n + 1 f ( z ) ) 2 [ ( L α n f ( z ) ) L α n f ( z ) 2 ( L α n + 1 f ( z ) ) L α n + 1 f ( z ) ] h ( z ) , z U ,
(2.4)
holds, then
z L α n f ( z ) ( L α n + 1 f ( z ) ) 2 g ( z ) , z U ,

and this result is sharp.

Proof For f A , f ( z ) = z + j = 2 a j z j , we have L α n f ( z ) = z + j = 2 ( α j n + ( 1 α ) C n + j 1 n ) a j z j , z U .

Consider p ( z ) = z L α n f ( z ) ( L α n + 1 f ( z ) ) 2 and we obtain p ( z ) + z γ p ( z ) = ( γ + 1 ) z γ L α n f ( z ) ( L α n + 1 f ( z ) ) 2 + z 2 γ L α n f ( z ) ( L α n + 1 f ( z ) ) 2 [ ( L α n f ( z ) ) L α n f ( z ) 2 ( L α n + 1 f ( z ) ) L α n + 1 f ( z ) ] .

Relation (2.4) becomes
p ( z ) + z γ p ( z ) h ( z ) = g ( z ) + γ g ( z ) , z U .
By using Lemma 1.2, we have
p ( z ) g ( z ) , z U , i.e. , z L α n f ( z ) ( L α n + 1 f ( z ) ) 2 g ( z ) , z U .

 □

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

If α 0 , γ C { 0 } is a complex number with Re γ 0 , n N , f A and satisfies the differential subordination
( γ + 1 ) z γ L α n f ( z ) ( L α n + 1 f ( z ) ) 2 + z 2 γ L α n f ( z ) ( L α n + 1 f ( z ) ) 2 [ ( L α n f ( z ) ) L α n f ( z ) 2 ( L α n + 1 f ( z ) ) L α n + 1 f ( z ) ] h ( z ) , z U ,
(2.5)
then
z L α n f ( z ) ( L α n + 1 f ( z ) ) 2 q ( z ) , z U ,

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

Proof Let p ( z ) = z L α n f ( z ) ( L α n + 1 f ( z ) ) 2 , z U , p H [ 1 , 1 ] . Differentiating, we obtain p ( z ) + z γ p ( z ) = ( γ + 1 ) z γ L α n f ( z ) ( L α n + 1 f ( z ) ) 2 + z 2 γ L α n f ( z ) ( L α n + 1 f ( z ) ) 2 [ ( L α n f ( z ) ) L α n f ( z ) 2 ( L α n + 1 f ( z ) ) L α n + 1 f ( z ) ] , z U , and (2.5) becomes
p ( z ) + z γ p ( z ) h ( z ) , z U .
Using Lemma 1.1, we have
p ( z ) q ( z ) , z U , i.e. , z L α n f ( z ) ( L α n + 1 f ( z ) ) 2 q ( z ) = γ z γ 0 z h ( t ) t γ 1 d t , z U ,

and q is the best dominant. □

Theorem 2.6 Let g be a convex function such that g ( 0 ) = 1 and let h be the function h ( z ) = g ( z ) + z γ g ( z ) , z U , where γ > 0 .

If α 0 , n N , f A and the differential subordination
( γ + 2 ) z 2 γ ( L α n f ( z ) ) L α n f ( z ) + z 3 γ [ ( L α n f ( z ) ) L α n f ( z ) ( ( L α n f ( z ) ) L α n f ( z ) ) 2 ] h ( z ) , z U ,
(2.6)
holds, then
z 2 ( L α n f ( z ) ) L α n f ( z ) g ( z ) , z U .

This result is sharp.

Proof Let p ( z ) = z 2 ( L α n f ( z ) ) L α n f ( z ) . We deduce that p H [ 0 , 1 ] .

Differentiating, we obtain p ( z ) + z γ p ( z ) = ( γ + 2 ) z 2 γ ( L α n f ( z ) ) L α n f ( z ) + z 3 γ [ ( L α n f ( z ) ) L α n f ( z ) ( ( L α n f ( z ) ) L α n f ( z ) ) 2 ] , z U .

Using the notation in (2.6), the differential subordination becomes
p ( z ) + 1 γ z p ( z ) h ( z ) = g ( z ) + z γ g ( z ) .
By using Lemma 1.2, we have
p ( z ) g ( z ) , z U , i.e. , z 2 ( L α n f ( z ) ) L α n f ( z ) g ( z ) , z U ,

and this result is sharp. □

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

If α 0 , γ C { 0 } is a complex number with Re γ 0 , n N , f A and satisfies the differential subordination
( γ + 2 ) z 2 γ ( L α n f ( z ) ) L α n f ( z ) + z 3 γ [ ( L α n f ( z ) ) L α n f ( z ) ( ( L α n f ( z ) ) L α n f ( z ) ) 2 ] h ( z ) , z U ,
(2.7)
then
z 2 ( L α n f ( z ) ) L α n f ( z ) q ( z ) , z U ,

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

Proof Let p ( z ) = z 2 ( L α n f ( z ) ) L α n f ( z ) , z U , p H [ 0 , 1 ] .

Differentiating, we obtain p ( z ) + z γ p ( z ) = ( γ + 2 ) z 2 γ ( L α n f ( z ) ) L α n f ( z ) + z 3 γ [ ( L α n f ( z ) ) L α n f ( z ) ( ( L α n f ( z ) ) L α n f ( z ) ) 2 ] , z U , and (2.7) becomes
p ( z ) + 1 γ z p ( z ) h ( z ) , z U .
Using Lemma 1.1, we have
p ( z ) q ( z ) , z U , i.e. , z 2 ( L α n f ( z ) ) L α n f ( z ) q ( z ) = γ z γ 0 z h ( t ) t γ 1 d t , z U ,

and q is the best dominant. □

Theorem 2.8 Let g be a convex function such that g ( 0 ) = 1 and let h be the function h ( z ) = g ( z ) + z g ( z ) , z U .

If α 0 , n N , f A and the differential subordination
1 L α n f ( z ) ( L α n f ( z ) ) [ ( L α n f ( z ) ) ] 2 h ( z ) , z U ,
(2.8)
holds, then
L α n f ( z ) z ( L α n f ( z ) ) g ( z ) , z U .

This result is sharp.

Proof Let p ( z ) = L α n f ( z ) z ( L α n f ( z ) ) . We deduce that p H [ 1 , 1 ] .

Differentiating, we obtain 1 L α n f ( z ) ( L α n f ( z ) ) [ ( L α n 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 ) + z g ( z ) .
By using Lemma 1.2, we have
p ( z ) g ( z ) , z U , i.e. , L α n f ( z ) z ( L α n f ( z ) ) g ( z ) , z U ,

and this result is sharp. □

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

If α 0 , n N , f A and satisfies the differential subordination
1 L α n f ( z ) ( L α n f ( z ) ) [ ( L α n f ( z ) ) ] 2 h ( z ) , z U ,
(2.9)
then
L α n f ( z ) z ( L α n f ( z ) ) q ( z ) , z U ,

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

Proof Let p ( z ) = L α n f ( z ) z ( L α n f ( z ) ) , z U , p H [ 0 , 1 ] .

Differentiating, we obtain 1 L α n f ( z ) ( L α n f ( z ) ) [ ( L α n 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. , L α n f ( z ) z ( L α n f ( z ) ) q ( z ) = 1 z 0 z h ( t ) d t , z U ,

and q is the best dominant. □

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

If α 0 , n N , f A and satisfies the differential subordination
1 L α n f ( z ) ( L α n f ( z ) ) [ ( L α n f ( z ) ) ] 2 h ( z ) , z U ,
(2.10)
then
L α n f ( z ) z ( L α n f ( z ) ) q ( z ) , z U ,

where q is given by q ( z ) = ( 2 β 1 ) + 2 ( 1 β ) ln ( 1 + z ) z , z U . The function q is convex and it is the best dominant.

Proof Following the same steps as in the proof of Theorem 2.9 and considering p ( z ) = L α n f ( z ) z ( L α n 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.,
L α n f ( z ) z ( L α n f ( z ) ) q ( z ) = 1 z 0 z h ( t ) d t = 1 z 0 z 1 + ( 2 β 1 ) t 1 + t d t = 1 z 0 z [ ( 2 β 1 ) + 2 ( 1 β ) 1 + t ] d t = ( 2 β 1 ) + 2 ( 1 β ) ln ( 1 + z ) z , 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 , α = 2 , we obtain L 2 1 f ( z ) = R 1 f ( z ) + 2 S 1 f ( z ) = z f ( z ) + 2 z f ( z ) = z f ( z ) = z + 2 z 2 .

Then ( L 2 1 f ( z ) ) = 1 + 4 z ,
L 2 1 f ( z ) z ( L 2 1 f ( z ) ) = z + 2 z 2 z ( 1 + 4 z ) = 1 + 2 z 1 + 4 z , 1 L 2 1 f ( z ) ( L 2 1 f ( z ) ) [ ( L 2 1 f ( z ) ) ] 2 = 1 ( z + 2 z 2 ) 4 ( 1 + 4 z ) 2 = 8 z 2 + 4 z + 1 ( 1 + 4 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
8 z 2 + 4 z + 1 ( 1 + 4 z ) 2 1 z 1 + z , z U ,
induce
1 + 2 z 1 + 4 z 1 + 2 ln ( 1 + z ) z , z U .

Theorem 2.11 Let g be a convex function such that g ( 0 ) = 0 and let h be the function h ( z ) = g ( z ) + z g ( z ) , z U .

If α 0 , n N , f A and the differential subordination
[ ( L α n f ( z ) ) ] 2 + L α n f ( z ) ( L α n f ( z ) ) h ( z ) , z U ,
(2.11)
holds, then
L α n f ( z ) ( L α n f ( z ) ) z g ( z ) , z U .

This result is sharp.

Proof Let p ( z ) = L α n f ( z ) ( L α n f ( z ) ) z . We deduce that p H [ 0 , 1 ] .

Differentiating, we obtain [ ( L α n f ( z ) ) ] 2 + L α n f ( z ) ( L α n 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 ) + z g ( z ) .
By using Lemma 1.2, we have
p ( z ) g ( z ) , z U , i.e. , L α n f ( z ) ( L α n f ( z ) ) z g ( z ) , z U ,

and this result is sharp. □

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

If α 0 , n N , f A and satisfies the differential subordination
[ ( L α n f ( z ) ) ] 2 + L α n f ( z ) ( L α n f ( z ) ) h ( z ) , z U ,
(2.12)
then
L α n f ( z ) ( L α n f ( z ) ) z q ( z ) , z U ,

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

Proof Let p ( z ) = L α n f ( z ) ( L α n f ( z ) ) z , z U , p H [ 0 , 1 ] .

Differentiating, we obtain [ ( L α n f ( z ) ) ] 2 + L α n f ( z ) ( L α n 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. , L α n f ( z ) ( L α n f ( z ) ) z q ( z ) = 1 z 0 z h ( t ) d t , z U ,

and q is the best dominant. □

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

If α 0 , n N , f A and satisfies the differential subordination
[ ( L α n f ( z ) ) ] 2 + L α n f ( z ) ( L α n f ( z ) ) h ( z ) , z U ,
(2.13)
then
L α n f ( z ) ( L α n f ( z ) ) z q ( z ) , z U ,

where q is given by q ( z ) = ( 2 β 1 ) + 2 ( 1 β ) ln ( 1 + z ) z , z U . The function q is convex and it is the best dominant.

Proof Following the same steps as in the proof of Theorem 2.12 and considering p ( z ) = L α n f ( z ) ( L α n 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.,
L α n f ( z ) ( L α n f ( z ) ) z q ( z ) = 1 z 0 z h ( t ) d t = 1 z 0 z 1 + ( 2 β 1 ) t 1 + t d t = 1 z 0 z [ ( 2 β 1 ) + 2 ( 1 β ) 1 + t ] d t = ( 2 β 1 ) + 2 ( 1 β ) ln ( 1 + z ) z , 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 , α = 2 , we obtain L 2 1 f ( z ) = R 1 f ( z ) + 2 S 1 f ( z ) = z f ( z ) + 2 z f ( z ) = z f ( z ) = z + 2 z 2 , z U .

Then ( L 2 1 f ( z ) ) = 1 + 4 z ,
L 2 1 f ( z ) ( L 2 1 f ( z ) ) z = ( z + 2 z 2 ) ( 1 + 4 z ) z = 8 z 2 + 6 z + 1 , [ ( L 2 1 f ( z ) ) ] 2 + L 2 1 f ( z ) ( L 2 1 f ( z ) ) = ( 1 + 4 z ) 2 + ( z + 2 z 2 ) 4 = 24 z 2 + 12 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
24 z 2 + 12 z + 1 1 z 1 + z , z U ,
induce
8 z 2 + 6 z + 1 1 + 2 ln ( 1 + z ) z , z U .

Theorem 2.14 Let g be a convex function such that g ( 0 ) = 0 and let h be the function h ( z ) = g ( z ) + z 1 δ g ( z ) , z U .

If α 0 , δ ( 0 , 1 ) , n N , f A and the differential subordination
( z L α n f ( z ) ) δ L α n + 1 f ( z ) 1 δ ( ( L α n + 1 f ( z ) ) L α n + 1 f ( z ) δ ( L α n f ( z ) ) L α n f ( z ) ) h ( z ) , z U ,
(2.14)
holds, then
L α n + 1 f ( z ) z ( z L α n f ( z ) ) δ g ( z ) , z U .

This result is sharp.

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

Differentiating, we obtain ( z L α n f ( z ) ) δ L α n + 1 f ( z ) 1 δ ( ( L α n + 1 f ( z ) ) L α n + 1 f ( z ) δ ( L α n f ( z ) ) L α n 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 ) + z 1 δ g ( z ) .
By using Lemma 1.2, we have
p ( z ) g ( z ) , z U , i.e. , L α n + 1 f ( z ) z ( z L α n f ( z ) ) δ g ( z ) , z U ,

and this result is sharp. □

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

If α 0 , δ ( 0 , 1 ) , n N , f A and satisfies the differential subordination
( z L α n f ( z ) ) δ L α n + 1 f ( z ) 1 δ ( ( L α n + 1 f ( z ) ) L α n + 1 f ( z ) δ ( L α n f ( z ) ) L α n f ( z ) ) h ( z ) , z U ,
(2.15)
then
L α n + 1 f ( z ) z ( z L α n f ( z ) ) δ q ( z ) , z U ,

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

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

Differentiating, we obtain ( z L α n f ( z ) ) δ L α n + 1 f ( z ) 1 δ ( ( L α n + 1 f ( z ) ) L α n + 1 f ( z ) δ ( L α n f ( z ) ) L α n 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. , L α n + 1 f ( z ) z ( z L α n f ( z ) ) δ q ( z ) = 1 δ z 1 δ 0 z h ( t ) t δ d t , z U ,

and q is the best dominant. □

Declarations

Acknowledgements

Dedicated to Professor Hari M Srivastava.

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

Authors’ Affiliations

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

References

  1. Sălăgean GS Lecture Notes in Math. 1013. In Subclasses of Univalent Functions. Springer, Berlin; 1983:362–372.Google Scholar
  2. Ruscheweyh S: New criteria for univalent functions. Proc. Am. Math. Soc. 1975, 49: 109–115. 10.1090/S0002-9939-1975-0367176-1MathSciNetView ArticleGoogle Scholar
  3. Alb Lupaş A: On special differential subordinations using Sălăgean and Ruscheweyh operators. Math. Inequal. Appl. 2009, 12(4):781–790.MathSciNetGoogle Scholar
  4. Alb Lupaş A: On a certain subclass of analytic functions defined by Salagean and Ruscheweyh operators. J. Math. Appl. 2009, 31: 67–76.MathSciNetGoogle Scholar
  5. Alb Lupaş A, Breaz D: On special differential superordinations using Sălăgean and Ruscheweyh operators. Geometric Function Theory and Applications (Proc. of International Symposium, Sofia, 27-31 August 2010), pp. 98–103.Google Scholar
  6. Miller SS, Mocanu PT: Differential Subordinations. Theory and Applications. Dekker, New York; 2000.Google Scholar
  7. Alb Lupaş, DA: Subordinations and Superordinations. Lambert Academic Publishing (2011)Google Scholar

Copyright

© Alb Lupaş; licensee Springer 2013

This article is published under license to BioMed Central Ltd. 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.

Advertisement