# Some differential subordinations using Ruscheweyh derivative and Sălăgean operator

## 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.

## 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 )

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 )

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 ()

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 .

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 $h(z)=g(z)+ z δ g ′ (z)$, $z∈U$.

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 $q(z)= δ z δ ∫ 0 z h(t) t δ − 1 dt$. 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

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

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 $α,δ≥0$, $n∈N$, $f∈A$ 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 $q(z)=(2β−1)+ 2 ( 1 − β ) δ z δ ∫ 0 z t δ − 1 1 + t dt$, $z∈U$. 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 $p(z)= ( L α n f ( z ) z ) δ$, 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 [, 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 $Reγ≥0$, $n∈N$, $f∈A$ 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 $q(z)= γ z γ ∫ 0 z h(t) t γ − 1 dt$. 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

$p(z)≺q(z),z∈U, i.e. , z L α n f ( z ) ( L α n + 1 f ( z ) ) 2 ≺q(z)= γ z γ ∫ 0 z h(t) t γ − 1 dt,z∈U,$

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 $Reγ≥0$, $n∈N$, $f∈A$ 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 $q(z)= γ z γ ∫ 0 z h(t) t γ − 1 dt$. 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

$p(z)≺q(z),z∈U, i.e. , z 2 ( L α n f ( z ) ) ′ L α n f ( z ) ≺q(z)= γ z γ ∫ 0 z h(t) t γ − 1 dt,z∈U,$

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 $h(z)=g(z)+z g ′ (z)$, $z∈U$.

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 $q(z)= 1 z ∫ 0 z h(t)dt$. 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

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

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 $α≥0$, $n∈N$, $f∈A$ 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 $q(z)=(2β−1)+2(1−β) ln ( 1 + z ) z$, $z∈U$. 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 $L 2 1 f(z)=− R 1 f(z)+2 S 1 f(z)=−z f ′ (z)+2z f ′ (z)=z f ′ (z)=z+2 z 2$.

Then $( L 2 1 f ( z ) ) ′ =1+4z$,

$L 2 1 f ( z ) z ( L 2 1 f ( z ) ) ′ = z + 2 z 2 z ( 1 + 4 z ) = 1 + 2 z 1 + 4 z , 1 − L 2 1 f ( z ) ⋅ ( L 2 1 f ( z ) ) ′ ′ [ ( L 2 1 f ( z ) ) ′ ] 2 = 1 − ( z + 2 z 2 ) ⋅ 4 ( 1 + 4 z ) 2 = 8 z 2 + 4 z + 1 ( 1 + 4 z ) 2 .$

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

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 $q(z)= 1 z ∫ 0 z h(t)dt$. 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

$p(z)≺q(z),z∈U, i.e. , L α n f ( z ) ⋅ ( L α n f ( z ) ) ′ z ≺q(z)= 1 z ∫ 0 z h(t)dt,z∈U,$

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 $q(z)=(2β−1)+2(1−β) ln ( 1 + z ) z$, $z∈U$. 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 $L 2 1 f(z)=− R 1 f(z)+2 S 1 f(z)=−z f ′ (z)+2z f ′ (z)=z f ′ (z)=z+2 z 2$, $z∈U$.

Then $( L 2 1 f ( z ) ) ′ =1+4z$,

$L 2 1 f ( z ) ⋅ ( L 2 1 f ( z ) ) ′ z = ( z + 2 z 2 ) ( 1 + 4 z ) z = 8 z 2 + 6 z + 1 , [ ( L 2 1 f ( z ) ) ′ ] 2 + L 2 1 f ( z ) ⋅ ( L 2 1 f ( z ) ) ′ ′ = ( 1 + 4 z ) 2 + ( z + 2 z 2 ) ⋅ 4 = 24 z 2 + 12 z + 1 .$

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

Using Theorem 2.12, we obtain

$24 z 2 +12z+1≺ 1 − z 1 + z ,z∈U,$

induce

$8 z 2 +6z+1≺−1+ 2 ln ( 1 + z ) z ,z∈U.$

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 $q(z)= 1 − δ z 1 − δ ∫ 0 z h(t) t − δ dt$. 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. □

## References

1. 1.

Sălăgean GS Lecture Notes in Math. 1013. In Subclasses of Univalent Functions. Springer, Berlin; 1983:362–372.

2. 2.

Ruscheweyh S: New criteria for univalent functions. Proc. Am. Math. Soc. 1975, 49: 109–115. 10.1090/S0002-9939-1975-0367176-1

3. 3.

Alb Lupaş A: On special differential subordinations using Sălăgean and Ruscheweyh operators. Math. Inequal. Appl. 2009, 12(4):781–790.

4. 4.

Alb Lupaş A: On a certain subclass of analytic functions defined by Salagean and Ruscheweyh operators. J. Math. Appl. 2009, 31: 67–76.

5. 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.

6. 6.

Miller SS, Mocanu PT: Differential Subordinations. Theory and Applications. Dekker, New York; 2000.

7. 7.

Alb Lupaş, DA: Subordinations and Superordinations. Lambert Academic Publishing (2011)

## Acknowledgements

Dedicated to Professor Hari M Srivastava.

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

## Author information

Authors

### Corresponding author

Correspondence to Alina Alb Lupaş.

### Competing interests

The author declares that she has no competing interests.

### Authors’ contributions

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

## Rights and permissions

Reprints and Permissions

Alb Lupaş, A. Some differential subordinations using Ruscheweyh derivative and Sălăgean operator. Adv Differ Equ 2013, 150 (2013). https://doi.org/10.1186/1687-1847-2013-150

• Accepted:

• Published:

### Keywords

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