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

## 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.

## 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 )

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 .

Definition 1.2 (Ruscheweyh )

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 

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  and .

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 .

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 [, 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 $Reγ≥0$. 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 $g(z)= γ n z γ / n ∫ 0 z h(t) t γ / n − 1 dt$, $z∈U$.

Lemma 1.2 (Miller and Mocanu )

Let g be a convex function in U, and let $h(z)=g(z)+nαz g ′ (z)$, 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

$p(z)+αz p ′ (z)≺h(z),z∈U,$

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 $h(z)=g(z)+ n z δ g ′ (z)$, 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 $p∈H[1,nδ]$.

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

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 2.2 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 $α,λ,δ≥0$, $n,m∈N$, $f∈ A n$ 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 $q(z)= δ n z δ n ∫ 0 z h(t) t δ n − 1 dt$. The function q is convex, and it is the best dominant.

Proof Let

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

for $z∈U$, $p∈H[1,nδ]$.

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

$p(z)≺q(z),z∈U,i.e., ( R D λ , α m f ( z ) z ) δ ≺q(z)= δ n z δ n ∫ 0 z h(t) t δ n − 1 dt,z∈U,$

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 $α,δ,λ≥0$, $n,m∈N$, $f∈ A n$ 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

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

where q is given by $q(z)=(2β−1)+ 2 ( 1 − β ) δ n z δ n ∫ 0 z t δ n − 1 1 + t dt$, $z∈U$. 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 $p(z)= ( R D λ , α m f ( z ) z ) δ$, 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 [, Example 4.2.1, p.125].

Theorem 2.4 Let g be a convex function such that $g(0)=1$, and let h be 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 2.5 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 $α,λ,δ≥0$, $n,m∈N$, $f∈ A n$ 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 $q(z)= δ n z δ n ∫ 0 z h(t) t δ n − 1 dt$. The function q is convex, and it is the best dominant.

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

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 2.6 Let g be a convex function such that $g(0)=1$, and let h be 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 2.7 Let h be 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 $q(z)= δ n z δ n ∫ 0 z h(t) t δ n − 1 dt$. The function q is convex, and it is the best dominant.

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

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

$p(z)≺q(z),z∈U,i.e., z 2 ( R D λ , α m f ( z ) ) ′ R D λ , α m f ( z ) ≺q(z)= δ n z δ n ∫ 0 z h(t) t δ n − 1 dt,z∈U,$

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 $h(z)=g(z)+nz g ′ (z)$, $z∈U$.

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

$p(z)+z p ′ (z)≺h(z)=g(z)+nz g ′ (z).$

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 2.9 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 $α,λ≥0$, $n,m∈N$, $f∈ A n$ 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 $q(z)= 1 n z 1 n ∫ 0 z h(t) t 1 n − 1 dt$. The function q is convex, and it is the best dominant.

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

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

$p(z)≺q(z),z∈U,i.e., R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) ′ ≺q(z)= 1 n z 1 n ∫ 0 z h(t) t 1 n − 1 dt,z∈U,$

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 $α,λ≥0$, $n,m∈N$, $f∈ A n$ 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

$R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) ′ ≺q(z),z∈U,$

where q is given by $q(z)=(2β−1)+ 2 ( 1 − β ) n z 1 n ∫ 0 z t 1 n − 1 1 + t dt$, $z∈U$. 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 $p(z)= R D λ , α m f ( z ) z ( R D λ , α m f ( z ) ) ′$, 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 $( R D 1 2 , 2 1 f ( z ) ) ′ = f ′ (z)=1+2z$,

$R D 1 2 , 2 1 f ( z ) z ( R D 1 2 , 2 1 f ( z ) ) ′ = z + z 2 z ( 1 + 2 z ) = 1 + z 1 + 2 z , 1 − R D 1 2 , 2 1 f ( z ) ⋅ ( R D 1 2 , 2 1 f ( z ) ) ″ [ ( R D 1 2 , 2 1 f ( z ) ) ′ ] 2 = 1 − ( z + z 2 ) ⋅ 2 ( 1 + 2 z ) 2 = 2 z 2 + 2 z + 1 ( 1 + 2 z ) 2 .$

We have $q(z)= 1 z ∫ 0 z 1 − t 1 + t dt=−1+ 2 ln ( 1 + z ) z$.

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

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

$p(z)+z p ′ (z)≺h(z)=g(z)+nz g ′ (z).$

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 2.12 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 $α,λ≥0$, $n,m∈N$, $f∈ A n$ 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 $q(z)= 1 n z 1 n ∫ 0 z h(t) t 1 n − 1 dt$. The function q is convex, and it is the best dominant.

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

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 2.13 Let $h(z)= 1 + ( 2 β − 1 ) z 1 + z$ be a convex function in U, 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

$R D λ , α m f ( z ) ⋅ ( R D λ , α m f ( z ) ) ′ z ≺q(z),z∈U,$

where q is given by $q(z)=(2β−1)+ 2 ( 1 − β ) n z 1 n ∫ 0 z t 1 n − 1 1 + t dt$, $z∈U$. 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 $p(z)= R D λ , α m f ( z ) ⋅ ( R D λ , α m f ( z ) ) ′ z$, 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 $( R D 1 2 , 2 1 f ( z ) ) ′ = f ′ (z)=1+2z$,

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

We have $q(z)= 1 z ∫ 0 z 1 − t 1 + t dt=−1+ 2 ln ( 1 + z ) z$.

Using Theorem 2.12, we obtain

$6 z 2 +6z+1≺ 1 − z 1 + z ,z∈U,$

induce

$2 z 2 +3z+1≺−1+ 2 ln ( 1 + z ) z ,z∈U.$

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

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 2.15 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 $α,λ≥0$, $δ∈(0,1)$, $n,m∈N$, $f∈ A n$ 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 $q(z)= 1 − δ n z 1 − δ n ∫ 0 z h(t) t 1 − δ n − 1 dt$. The function q is convex, and it is the best dominant.

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

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. □

## Author’s contributions

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

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## Acknowledgements

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

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Correspondence to Loriana Andrei.

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Andrei, L. Differential subordinations using the Ruscheweyh derivative and the generalized Sălăgean operator. Adv Differ Equ 2013, 252 (2013). https://doi.org/10.1186/1687-1847-2013-252

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### Keywords

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