Open Access

Differential subordinations using the Ruscheweyh derivative and the generalized Sălăgean operator

Advances in Difference Equations20132013:252

DOI: 10.1186/1687-1847-2013-252

Received: 13 June 2013

Accepted: 6 August 2013

Published: 20 August 2013

Abstract

In the present paper, we study the operator, using the Ruscheweyh derivative R m f ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq1_HTML.gif and the generalized Sălăgean operator D λ m f ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq2_HTML.gif, denote by R D λ , α m : A n A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq3_HTML.gif, R D λ , α m f ( z ) = ( 1 α ) R m f ( z ) + α D λ m f ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq4_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, where A n = { f H ( U ) : f ( z ) = z + a n + 1 z n + 1 + , z U } https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq6_HTML.gif is the class of normalized analytic functions. We obtain several differential subordinations regarding the operator R D λ , α m https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq7_HTML.gif.

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 } https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq8_HTML.gif and H ( U ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq9_HTML.gif the space of holomorphic functions in U.

Let A n = { f H ( U ) : f ( z ) = z + a n + 1 z n + 1 + , z U } https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq6_HTML.gif and H [ a , n ] = { f H ( U ) : f ( z ) = a + a n z n + a n + 1 z n + 1 + , z U } https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq10_HTML.gif for a C https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq11_HTML.gif and n N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq12_HTML.gif.

Denote by K = { f A n : Re z f ( z ) f ( z ) + 1 > 0 , z U } https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq13_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq14_HTML.gif, if there is a function w analytic in U, with w ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq15_HTML.gif, | w ( z ) | < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq16_HTML.gif, for all z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif such that f ( z ) = g ( w ( z ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq17_HTML.gif for all z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif. If g is univalent, then f g https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq14_HTML.gif if and only if f ( 0 ) = g ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq18_HTML.gif and f ( U ) g ( U ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq19_HTML.gif.

Let ψ : C 3 × U C https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq20_HTML.gif, 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equ1_HTML.gif
(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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq21_HTML.gif for all p satisfying (1.1).

A dominant q ˜ https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq22_HTML.gif that satisfies q ˜ q https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq23_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif, λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq25_HTML.gif and n , m N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq26_HTML.gif, the operator D λ m https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq27_HTML.gif is defined by D λ m : A n A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq28_HTML.gif,
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equa_HTML.gif

Remark 1.1 If f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif and f ( z ) = z + j = n + 1 a j z j https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq29_HTML.gif, then D λ m f ( z ) = z + j = n + 1 [ 1 + ( j 1 ) λ ] m a j z j https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq30_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

Remark 1.2 For λ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq31_HTML.gif, in the definition above, we obtain the Sălăgean differential operator [2].

Definition 1.2 (Ruscheweyh [3])

For f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif, n , m N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq26_HTML.gif, the operator R m https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq32_HTML.gif is defined by R m : A n A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq33_HTML.gif,
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equb_HTML.gif

Remark 1.3 If f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif, f ( z ) = z + j = n + 1 a j z j https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq29_HTML.gif, then R m f ( z ) = z + j = n + 1 C m + j 1 m a j z j https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq34_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

Definition 1.3 [4]

Let α , λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq35_HTML.gif, n , m N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq26_HTML.gif. Denote by R D λ , α m https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq7_HTML.gif the operator given by R D λ , α m : A n A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq3_HTML.gif,
R D λ , α m f ( z ) = ( 1 α ) R m f ( z ) + α D λ m f ( z ) , z U . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equc_HTML.gif

Remark 1.4 If f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif, f ( z ) = z + j = n + 1 a j z j https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq29_HTML.gif, 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq36_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

This operator was studied also in [46] and [7].

Remark 1.5 For α = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq37_HTML.gif, R D λ , 0 m f ( z ) = R m f ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq38_HTML.gif, where z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif and for α = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq39_HTML.gif, R D λ , 1 m f ( z ) = D λ m f ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq40_HTML.gif, where z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

For λ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq31_HTML.gif, we obtain R D 1 , α m f ( z ) = L α m f ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq41_HTML.gif, which was studied in [811].

For m = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq42_HTML.gif, 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq43_HTML.gif, where z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

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

Let h be a convex function with h ( 0 ) = a https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq44_HTML.gif, and let γ C { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq45_HTML.gif be a complex number with Re γ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq46_HTML.gif. If p H [ a , n ] https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq47_HTML.gif and
p ( z ) + 1 γ z p ( z ) h ( z ) , z U , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equd_HTML.gif
then
p ( z ) g ( z ) h ( z ) , z U , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Eque_HTML.gif

where g ( z ) = γ n z γ / n 0 z h ( t ) t γ / n 1 d t https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq48_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

Lemma 1.2 (Miller and Mocanu [12])

Let g be a convex function in U, and let h ( z ) = g ( z ) + n α z g ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq49_HTML.gif, for z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, where α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq50_HTML.gif and n is a positive integer.

If p ( z ) = g ( 0 ) + p n z n + p n + 1 z n + 1 + https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq51_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, is holomorphic in U and
p ( z ) + α z p ( z ) h ( z ) , z U , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equf_HTML.gif
then
p ( z ) g ( z ) , z U , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equg_HTML.gif

and this result is sharp.

2 Main results

Theorem 2.1 Let g be a convex function, g ( 0 ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq52_HTML.gif, and let h be the function h ( z ) = g ( z ) + n z δ g ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq53_HTML.gif, for z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

If α , λ , δ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq54_HTML.gif, n , m N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq26_HTML.gif, f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equ2_HTML.gif
(2.1)
then
( R D λ , α m f ( z ) z ) δ g ( z ) , z U , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equh_HTML.gif

and this result is sharp.

Proof By using the properties of operator R D λ , α m https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq7_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equi_HTML.gif

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 + https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq55_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

We deduce that p H [ 1 , n δ ] https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq56_HTML.gif.

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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq57_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

Then (2.1) becomes
p ( z ) + 1 δ z p ( z ) h ( z ) = g ( z ) + n z δ g ( z ) for  z U . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equj_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equk_HTML.gif

 □

Theorem 2.2 Let h be a holomorphic function, which satisfies the inequality Re ( 1 + z h ( z ) h ( z ) ) > 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq58_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, and h ( 0 ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq59_HTML.gif.

If α , λ , δ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq54_HTML.gif, n , m N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq26_HTML.gif, f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equ3_HTML.gif
(2.2)
then
( R D λ , α m f ( z ) z ) δ q ( z ) , z U , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equl_HTML.gif

where q ( z ) = δ n z δ n 0 z h ( t ) t δ n 1 d t https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq60_HTML.gif. The function q is 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equm_HTML.gif

for z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, p H [ 1 , n δ ] https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq56_HTML.gif.

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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq57_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, and (2.2) becomes
p ( z ) + 1 δ z p ( z ) h ( z ) , z U . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equn_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equo_HTML.gif

and q is the best dominant. □

Corollary 2.3 Let h ( z ) = 1 + ( 2 β 1 ) z 1 + z https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq61_HTML.gif be a convex function in U, where 0 β < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq62_HTML.gif.

If α , δ , λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq63_HTML.gif, n , m N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq26_HTML.gif, f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equ4_HTML.gif
(2.3)
then
( R D λ , α m f ( z ) z ) δ q ( z ) , z U , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equp_HTML.gif

where q is 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq64_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif. 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 ) = ( R D λ , α m f ( z ) z ) δ https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq65_HTML.gif, the differential subordination (2.3) becomes
p ( z ) + z δ p ( z ) h ( z ) = 1 + ( 2 β 1 ) z 1 + z , z U . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equq_HTML.gif
By using Lemma 1.1, for γ = δ https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq66_HTML.gif, we have p ( z ) q ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq67_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equr_HTML.gif

 □

Remark 2.1 For n = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq68_HTML.gif, λ = 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq69_HTML.gif, α = 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq70_HTML.gif, δ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq71_HTML.gif, we obtain the same example as in [[13], Example 4.2.1, p.125].

Theorem 2.4 Let g be a convex function such that g ( 0 ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq72_HTML.gif, and let h be the function h ( z ) = g ( z ) + n z δ g ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq73_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

If α , λ , δ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq54_HTML.gif, n , m N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq26_HTML.gif, f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equ5_HTML.gif
(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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equs_HTML.gif

and this result is sharp.

Proof For f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif, f ( z ) = z + j = n + 1 a j z j https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq29_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equt_HTML.gif
Consider p ( z ) = z R D λ , α m f ( z ) ( R D λ , α m + 1 f ( z ) ) 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq74_HTML.gif, 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 ) ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equu_HTML.gif
Relation (2.4) becomes
p ( z ) + z δ p ( z ) h ( z ) = g ( z ) + n z δ g ( z ) , z U . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equv_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equw_HTML.gif

 □

Theorem 2.5 Let h be a holomorphic function, which satisfies the inequality Re ( 1 + z h ( z ) h ( z ) ) > 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq58_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, and h ( 0 ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq59_HTML.gif.

If α , λ , δ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq54_HTML.gif, n , m N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq26_HTML.gif, f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equ6_HTML.gif
(2.5)
then
z R D λ , α m f ( z ) ( R D λ , α m + 1 f ( z ) ) 2 q ( z ) , z U , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equx_HTML.gif

where q ( z ) = δ n z δ n 0 z h ( t ) t δ n 1 d t https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq60_HTML.gif. The function q is 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq74_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, p H [ 1 , n ] https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq75_HTML.gif.

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 ) ] https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq76_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, and (2.5) becomes
p ( z ) + z δ p ( z ) h ( z ) , z U . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equy_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equz_HTML.gif

and q is the best dominant. □

Theorem 2.6 Let g be a convex function such that g ( 0 ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq52_HTML.gif, and let h be the function h ( z ) = g ( z ) + n z δ g ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq53_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

If α , λ , δ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq54_HTML.gif, n , m N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq26_HTML.gif, f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equ7_HTML.gif
(2.6)
holds, then
z 2 ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) g ( z ) , z U . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equaa_HTML.gif

This result is sharp.

Proof Let p ( z ) = z 2 ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq77_HTML.gif. We deduce that p H [ 0 , n ] https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq78_HTML.gif.

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 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq79_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equab_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equac_HTML.gif

and this result is sharp. □

Theorem 2.7 Let h be an holomorphic function, which satisfies the inequality Re ( 1 + z h ( z ) h ( z ) ) > 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq58_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, and h ( 0 ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq59_HTML.gif.

If α , λ , δ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq54_HTML.gif, n , m N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq26_HTML.gif, f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equ8_HTML.gif
(2.7)
then
z 2 ( R D λ , α m f ( z ) ) R D λ , α m f ( z ) q ( z ) , z U , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equad_HTML.gif

where q ( z ) = δ n z δ n 0 z h ( t ) t δ n 1 d t https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq60_HTML.gif. The function q is 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq77_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, p H [ 0 , n ] https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq78_HTML.gif.

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 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq79_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, and (2.7) becomes
p ( z ) + 1 δ z p ( z ) h ( z ) , z U . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equae_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equaf_HTML.gif

and q is the best dominant. □

Theorem 2.8 Let g be a convex function such that g ( 0 ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq52_HTML.gif, and let h be the function h ( z ) = g ( z ) + n z g ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq80_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

If α , λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq35_HTML.gif, n , m N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq26_HTML.gif, f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equ9_HTML.gif
(2.8)
holds, then
R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) g ( z ) , z U . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equag_HTML.gif

This result is sharp.

Proof Let p ( z ) = R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq81_HTML.gif. We deduce that p H [ 1 , n ] https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq75_HTML.gif.

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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq82_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

Using the notation in (2.8), the differential subordination becomes
p ( z ) + z p ( z ) h ( z ) = g ( z ) + n z g ( z ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equah_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equai_HTML.gif

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq58_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, and h ( 0 ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq59_HTML.gif.

If α , λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq35_HTML.gif, n , m N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq26_HTML.gif, f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equ10_HTML.gif
(2.9)
then
R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) q ( z ) , z U , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equaj_HTML.gif

where q ( z ) = 1 n z 1 n 0 z h ( t ) t 1 n 1 d t https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq83_HTML.gif. The function q is 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 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq81_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, p H [ 0 , n ] https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq78_HTML.gif.

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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq82_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, and (2.9) becomes
p ( z ) + z p ( z ) h ( z ) , z U . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equak_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equal_HTML.gif

and q is the best dominant. □

Corollary 2.10 Let h ( z ) = 1 + ( 2 β 1 ) z 1 + z https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq61_HTML.gif be a convex function in U, where 0 β < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq62_HTML.gif.

If α , λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq35_HTML.gif, n , m N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq26_HTML.gif, f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equ11_HTML.gif
(2.10)
then
R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) q ( z ) , z U , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equam_HTML.gif

where q is given by q ( z ) = ( 2 β 1 ) + 2 ( 1 β ) n z 1 n 0 z t 1 n 1 1 + t d t https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq84_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif. 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 ) = R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq85_HTML.gif, the differential subordination (2.10) becomes
p ( z ) + z p ( z ) h ( z ) = 1 + ( 2 β 1 ) z 1 + z , z U . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equan_HTML.gif
By using Lemma 1.1 for γ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq86_HTML.gif, we have p ( z ) q ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq67_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equao_HTML.gif

 □

Example 2.1 Let h ( z ) = 1 z 1 + z https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq87_HTML.gif be a convex function in U with h ( 0 ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq88_HTML.gif and Re ( z h ( z ) h ( z ) + 1 ) > 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq89_HTML.gif.

Let f ( z ) = z + z 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq90_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif. For n = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq68_HTML.gif, m = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq91_HTML.gif, λ = 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq69_HTML.gif, α = 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq70_HTML.gif, 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq92_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

Then ( R D 1 2 , 2 1 f ( z ) ) = f ( z ) = 1 + 2 z https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq93_HTML.gif,
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equap_HTML.gif

We have q ( z ) = 1 z 0 z 1 t 1 + t d t = 1 + 2 ln ( 1 + z ) z https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq94_HTML.gif.

Using Theorem 2.9, we obtain
2 z 2 + 2 z + 1 ( 1 + 2 z ) 2 1 z 1 + z , z U , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equaq_HTML.gif
induce
1 + z 1 + 2 z 1 + 2 ln ( 1 + z ) z , z U . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equar_HTML.gif

Theorem 2.11 Let g be a convex function such that g ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq95_HTML.gif, and let h be the function h ( z ) = g ( z ) + n z g ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq80_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

If α , λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq35_HTML.gif, n , m N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq26_HTML.gif, f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equ12_HTML.gif
(2.11)
holds, then
R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z g ( z ) , z U . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equas_HTML.gif

This result is sharp.

Proof Let p ( z ) = R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq96_HTML.gif. We deduce that p H [ 0 , n ] https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq78_HTML.gif.

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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq97_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

Using the notation in (2.11), the differential subordination becomes
p ( z ) + z p ( z ) h ( z ) = g ( z ) + n z g ( z ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equat_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equau_HTML.gif

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq98_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, and h ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq99_HTML.gif.

If α , λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq35_HTML.gif, n , m N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq26_HTML.gif, f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equ13_HTML.gif
(2.12)
then
R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z q ( z ) , z U , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equav_HTML.gif

where q ( z ) = 1 n z 1 n 0 z h ( t ) t 1 n 1 d t https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq83_HTML.gif. The function q is 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq96_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, p H [ 0 , n ] https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq78_HTML.gif.

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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq97_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, and (2.12) becomes
p ( z ) + z p ( z ) h ( z ) , z U . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equaw_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equax_HTML.gif

and q is the best dominant. □

Corollary 2.13 Let h ( z ) = 1 + ( 2 β 1 ) z 1 + z https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq61_HTML.gif be a convex function in U, where 0 β < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq62_HTML.gif.

If α , λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq35_HTML.gif, n , m N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq26_HTML.gif, f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equ14_HTML.gif
(2.13)
then
R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z q ( z ) , z U , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equay_HTML.gif

where q is given by q ( z ) = ( 2 β 1 ) + 2 ( 1 β ) n z 1 n 0 z t 1 n 1 1 + t d t https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq84_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif. 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 ) = R D λ , α m f ( z ) ( R D λ , α m f ( z ) ) z https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq100_HTML.gif, the differential subordination (2.13) becomes
p ( z ) + z p ( z ) h ( z ) = 1 + ( 2 β 1 ) z 1 + z , z U . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equaz_HTML.gif
By using Lemma 1.1 for γ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq86_HTML.gif, we have p ( z ) q ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq67_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equba_HTML.gif

 □

Example 2.2 Let h ( z ) = 1 z 1 + z https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq87_HTML.gif be a convex function in U with h ( 0 ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq88_HTML.gif and Re ( z h ( z ) h ( z ) + 1 ) > 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq89_HTML.gif.

Let f ( z ) = z + z 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq90_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif. For n = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq68_HTML.gif, m = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq91_HTML.gif, λ = 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq69_HTML.gif, α = 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq70_HTML.gif, 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq92_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

Then ( R D 1 2 , 2 1 f ( z ) ) = f ( z ) = 1 + 2 z https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq93_HTML.gif,
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equbb_HTML.gif

We have q ( z ) = 1 z 0 z 1 t 1 + t d t = 1 + 2 ln ( 1 + z ) z https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq94_HTML.gif.

Using Theorem 2.12, we obtain
6 z 2 + 6 z + 1 1 z 1 + z , z U , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equbc_HTML.gif
induce
2 z 2 + 3 z + 1 1 + 2 ln ( 1 + z ) z , z U . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equbd_HTML.gif

Theorem 2.14 Let g be a convex function such that g ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq95_HTML.gif, and let h be the function h ( z ) = g ( z ) + n z 1 δ g ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq101_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

If α , λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq35_HTML.gif, δ ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq102_HTML.gif, n , m N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq26_HTML.gif, f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equ15_HTML.gif
(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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Eqube_HTML.gif

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq103_HTML.gif. We deduce that p H [ 1 , n ] https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq75_HTML.gif.

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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq104_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif.

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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equbf_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equbg_HTML.gif

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq98_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, and h ( 0 ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq59_HTML.gif.

If α , λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq35_HTML.gif, δ ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq102_HTML.gif, n , m N https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq26_HTML.gif, f A n https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq24_HTML.gif 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equ16_HTML.gif
(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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equbh_HTML.gif

where q ( z ) = 1 δ n z 1 δ n 0 z h ( t ) t 1 δ n 1 d t https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq105_HTML.gif. The function q is 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq103_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, p H [ 0 , n ] https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq78_HTML.gif.

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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq106_HTML.gif, z U https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_IEq5_HTML.gif, and (2.15) becomes
p ( z ) + 1 1 δ z p ( z ) h ( z ) , z U . https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equbi_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-1847-2013-252/MediaObjects/13662_2013_Article_1369_Equbj_HTML.gif

and q is the best dominant. □

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 the present article.

Authors’ Affiliations

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

References

  1. Al-Oboudi FM: On univalent functions defined by a generalized Sălăgean operator. Int. J. Math. Math. Sci. 2004, 27: 1429–1436.MathSciNetView ArticleMATH
  2. Sălăgean GS Lecture Notes in Math. 1013. In Subclasses of Univalent Functions. Springer, Berlin; 1983:362–372.
  3. Ruscheweyh S: New criteria for univalent functions. Proc. Am. Math. Soc. 1975, 49: 109–115. 10.1090/S0002-9939-1975-0367176-1MathSciNetView ArticleMATH
  4. Lupaş AA: On special differential subordinations using a generalized Sălăgean operator and Ruscheweyh derivative. J. Comput. Anal. Appl. 2011, 13(1):98–107.MathSciNetMATH
  5. Lupaş AA: On a certain subclass of analytic functions defined by a generalized Sălăgean operator and Ruscheweyh derivative. Carpath. J. Math. 2012, 28(2):183–190.MATH
  6. Lupaş AA: On special differential superordinations using a generalized Sălăgean operator and Ruscheweyh derivative. Comput. Math. Appl. 2011, 61: 1048–1058. 10.1016/j.camwa.2010.12.055MathSciNetView ArticleMATH
  7. Lupaş AA: Certain special differential superordinations using a generalized Sălăgean operator and Ruscheweyh derivative. An. Univ. Oradea, Fasc. Mat. 2011, XVIII: 167–178.MATH
  8. Lupaş AA: On special differential subordinations using Sălăgean and Ruscheweyh operators. Math. Inequal. Appl. 2009, 12(4):781–790.MathSciNetMATH
  9. Lupaş AA: On a certain subclass of analytic functions defined by Sălăgean and Ruscheweyh operators. J. Math. Appl. 2009, 31: 67–76.MathSciNet
  10. Lupaş AA, Breaz D: On special differential superordinations using Sălăgean and Ruscheweyh operators. Geometric Function Theory and Applications 2010, 98–103. (Proc. of International Symposium, Sofia, 27–31 August 2010)
  11. Lupaş AA: Some differential subordinations using Ruscheweyh derivative and Sălăgean operator. Adv. Differ. Equ. 2013., 2013: Article ID 150 10.1186/1687-1847-2013-150
  12. Miller SS, Mocanu PT: Differential Subordinations: Theory and Applications. Dekker, New York; 2000.MATH
  13. Lupaş DAA: Subordinations and Superordinations. Lap Lambert Academic Publishing, Saarbrücken; 2011.

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© Andrei; licensee Springer. 2013

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