# Some properties of certain subclasses of analytic functions involving adifferential operator

## Abstract

In the present paper, we introduce and study certain subclasses of analyticfunctions in the open unit disk U which is defined by the differentialoperator $D R λ m , n$. We study and investigate some inclusionproperties of these classes. Furthermore, a generalizedBernardi-Libera-Livington integral operator is shown to be preserved for theseclasses.

MSC: 30C45.

## 1 Introduction

Let be aclass of functions f in the open unit disk $U={z∈C:|z|<1}$ normalized by $f(0)= f ′ (0)−1=0$. Thus each $f∈A$ has a Taylor series representation

$f(z)=z+ ∑ j = 2 ∞ a j z j .$
(1.1)

We denote by $S(ξ)$ the well-known subclass of consisting of allanalytic functions which are, respectively, starlike of order ξ[1, 2]

$S(ξ)= { f ∈ A : Re ( z f ′ ( z ) f ( z ) ) > ξ , z ∈ U } ,0≤ξ<1.$

Let be a class of all functions ϕ which are analytic andunivalent in U and for which $ϕ(U)$ is convex with $ϕ(0)=1$ and $Reϕ(z)>0$, $z∈U$.

For two functions f and g analytic in U, we say that thefunction f is subordinate to g in U and write$f(z)≺g(z)$, $z∈U$, if there exists a Schwarz function$w(z)$ which is analytic in U with$w(0)=0$ and $|w(z)|<1$ such that $f(z)=g(w(z))$, $z∈U$.

Making use of the principle of subordination between analytic functions, denote by$S(ξ,ϕ)$ a subclass of the class for$0≤ξ<1$ and $ϕ∈R$ which are defined by

$S(ξ,ϕ)= { f ∈ A : 1 1 − ξ ( z f ′ ( z ) f ( z ) − ζ ) ≺ ϕ ( z ) , z ∈ U } .$

Let $f,g∈A$, where f and g are defined by$f(z)=z+ ∑ j = 2 ∞ a j z j$ and $g(z)=z+ ∑ j = 2 ∞ b j z j$. Then the Hadamard product (or convolution)$f∗g$ of the functions f and g is definedby

$(f∗g)(z)=z+ ∑ j = 2 ∞ a j b j z j .$

Definition 1.1 (Al-Oboudi )

For $f∈A$, $λ≥0$ and $m∈N$, the operator $D λ m$ is defined by $D λ m :A→A$,

$D λ 0 f ( z ) = f ( z ) , D λ 1 f ( z ) = ( 1 − λ ) f ( z ) + λ z f ′ ( z ) = D λ f ( z ) , … , D λ m f ( z ) = ( 1 − λ ) D λ m − 1 f ( z ) + λ z ( D λ m f ( z ) ) ′ = D λ ( D λ m − 1 f ( z ) ) , z ∈ U .$

Remark 1.1 If $f∈A$ and $f(z)=z+ ∑ j = 2 ∞ a j z j$, then $D λ m f(z)=z+ ∑ j = 2 ∞ [ 1 + ( j − 1 ) λ ] m a j z j$, $z∈U$.

Remark 1.2 For $λ=1$ in the above definition, we obtain theSălăgean differential operator .

Definition 1.2 (Ruscheweyh )

For $f∈A$ and $n∈N$, 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 1.3 If $f∈A$, $f(z)=z+ ∑ j = 2 ∞ a j z j$, then $R n f(z)=z+ ∑ j = 2 ∞ ( n + j − 1 ) ! n ! ( j − 1 ) ! a j z j$, $z∈U$.

Definition 1.3 ()

Let $λ≥0$ and $n,m∈N$. Denote by $D R λ m , n :A→A$ the operator given by the Hadamard product of thegeneralized Sălăgean operator $D λ m$ and the Ruscheweyh operator $R n$,

$D R λ m , n f(z)= ( D λ m ∗ R n ) f(z),$

for any $z∈U$ and each nonnegative integer m,n.

Remark 1.4 If $f∈A$ and $f(z)=z+ ∑ j = 2 ∞ a j z j$, then $D R λ m , n f(z)=z+ ∑ j = 2 ∞ [ 1 + ( j − 1 ) λ ] m ( n + j − 1 ) ! n ! ( j − 1 ) ! a j 2 z j$, $z∈U$.

Remark 1.5 The operator $D R λ m , n$ was studied also in .

For $λ=1$, $m=n$, we obtain the Hadamard product$S R n$ of the Sălăgean operator $S n$ and the Ruscheweyh derivative $R n$, which was studied in [12, 13].

For $m=n$, we obtain the Hadamard product$D R λ n$ of the generalized Sălăgean operator $D λ n$ and the Ruscheweyh derivative $R n$, which was studied in .

Using a simple computation, one obtains the next result.

Proposition 1.1 ()

For$m,n∈N$and$λ≥0$, we have

$D R λ m + 1 , n f(z)=(1−λ)D R λ m , n f(z)+λz ( D R λ m , n f ( z ) ) ′$
(1.2)

and

$z ( D R λ m , n f ( z ) ) ′ =(n+1)D R λ m , n + 1 f(z)−nD R λ m , n f(z).$
(1.3)

By using the operator $D R λ m , n f(z)$, we define the following subclasses of analyticfunctions for $0≤ζ<1$ and $ϕ∈R$:

$S λ m , n ( ξ ) = { f ∈ A : D R λ m , n f ∈ S ( ξ ) } , S λ m , n ( ξ , ϕ ) = { f ∈ A : D R λ m , n f ∈ S ( ξ , ϕ ) } .$

In particular, we set

$S λ m , n ( ξ , 1 + A z 1 + B z ) = S λ m , n (ξ,A,B),−1

Next, we will investigate various inclusion relationships for the subclasses ofanalytic functions introduced above. Furthermore, we study the results of Faisalet al., Darus and Faisal .

## 2 Inclusion relationship associated with the operator $D R λ m , n$

First, we start with the following lemmas which we need for our main results.

Lemma 2.1 ([22, 23])

Let$φ(μ,v)$be a complex function such that$φ:D→C$, $D⊆C×C$, and let$μ= μ 1 +i μ 2$, $v= v 1 +i v 2$. Suppose that$φ(μ,v)$satisfies the following conditions:

1. 1.

$φ(μ,v)$ is continuous in D,

2. 2.

$(1,0)∈D$ and $Reφ(1,0)>0$,

3. 3.

$Reφ(i μ 2 , v 1 )≤0$ for all $(i μ 2 , v 1 )∈D$ such that $v 1 ≤− 1 2 (1+ μ 2 2 )$.

Let$h(z)=1+ c 1 z+ c 2 z 2 +⋯$be analytic in U, such that$(h(z),z h ′ (z))∈D$for all$z∈U$. If$Re{φh(z),z h ′ (z)}>0$, $z∈U$, then$Re{h(z)}>0$.

Lemma 2.2 ()

Let ϕ be convex univalent in U with$ϕ(0)=1$and$Re{kϕ(z)+ν}>0$, $k,ν∈C$. If p is analytic in U with$p(0)=1$, then

$p(z)+ z p ′ ( z ) k p ( z ) + ν ≺ϕ(z),z∈U,$

implies$p(z)≺ϕ(z)$, $z∈U$.

Theorem 2.1 Let$f∈A$, $0≤ξ<1$, $m,n∈N$, $λ>0$, then

$S λ m , n + 1 (ξ)⊆ S λ m , n (ξ)⊆ S λ m , n − 1 (ξ).$

Proof Let $f∈ S λ m , n + 1 (ξ)$ and suppose that

$z ( D R λ m , n f ( z ) ) ′ D R λ m , n f ( z ) =ξ+(1−ξ)h(z).$
(2.1)

Since from (1.3)

$(n+1) D R λ m , n + 1 f ( z ) D R λ m , n f ( z ) =n+ξ+(1−ξ)h(z),$

we obtain

$( 1 − ξ ) h ′ ( z ) = ( n + 1 ) [ ( D R λ m , n + 1 f ( z ) ) ′ D R λ m , n f ( z ) − D R λ m , n + 1 f ( z ) D R λ m , n f ( z ) ⋅ ( D R λ m , n f ( z ) ) ′ D R λ m , n f ( z ) ] , ( 1 − ξ ) z h ′ ( z ) = ( n + 1 ) D R λ m , n + 1 f ( z ) D R λ m , n f ( z ) [ z ( D R λ m , n + 1 f ( z ) ) ′ D R λ m , n + 1 f ( z ) − ξ − ( 1 − ξ ) h ( z ) ] , ( 1 − ξ ) h ′ ( z ) z n + ξ + ( 1 − ξ ) h ( z ) = z ( D R λ m , n + 1 f ( z ) ) ′ D R λ m , n + 1 f ( z ) − ξ − ( 1 − ξ ) h ( z ) , z ( D R λ m , n + 1 f ( z ) ) ′ D R λ m , n + 1 f ( z ) − ξ = ( 1 − ξ ) h ( z ) + ( 1 − ξ ) h ′ ( z ) z n + ξ + ( 1 − ξ ) h ( z ) .$

Taking $h(z)=μ= μ 1 +i μ 2$ and $z h ′ (z)=v= v 1 +i v 2$, we define $φ(μ,v)$ by

$φ(μ,v)=(1−ξ)μ+ ( 1 − ξ ) v n + ξ + ( 1 − ξ ) μ$

and

$Re { φ ( i μ 2 , v 1 ) } = ( 1 − ξ ) ( n + ξ ) v 1 ( n + ξ ) 2 + ( 1 − ξ ) 2 μ 2 2 , Re { φ ( i μ 2 , v 1 ) } ≤ − ( 1 − ξ ) ( n + ξ ) ( 1 + μ 2 2 ) 2 [ ( n + ξ ) 2 + ( 1 − ξ ) 2 μ 2 2 ] < 0 .$

Clearly, $φ(μ,v)$ satisfies the conditions of Lemma 2.1. Hence$Re{h(z)}>0$, $z∈U$, implies $f∈ S λ m , n (ξ)$. □

Remark 2.1 Using relation (1.2) and the same techniques as to prove theearlier results, we can obtain a new similar result.

Theorem 2.2 Let $f∈A$ and $ϕ∈R$ with

$Re { ϕ ( z ) } < ξ − 1 + 1 λ 1 − ξ .$

Then

$S λ m + 1 , n (ξ,ϕ)⊂ S λ m , n (ξ,ϕ)⊂ S λ m − 1 , n (ξ,ϕ).$

Proof Let $f(z)∈ S λ m + 1 , n (ξ,ϕ)$ and set

$p(z)= 1 1 − ξ ( z ( D R λ m , n f ( z ) ) ′ D R λ m , n f ( z ) − ξ ) ,$
(2.2)

where p is analytic in U with $p(0)=1$.

By using (1.2) we have

$z ( D R λ m , n f ( z ) ) ′ D R λ m , n f ( z ) = 1 λ D R λ m + 1 , n f ( z ) D R λ m , n f ( z ) − 1 − λ λ .$

Now, by using (2.2) we get

$p ′ ( z ) = 1 1 − ξ ( 1 λ D R λ m + 1 , n f ( z ) D R λ m , n f ( z ) − 1 − λ λ − ξ ) , 1 λ D R λ m + 1 , n f ( z ) D R λ m , n f ( z ) = ξ + 1 − λ λ + ( 1 − ξ ) p ( z ) .$
(2.3)

By using (2.2) and (2.3), we obtain

$z p ′ ( z ) = 1 1 − ξ ⋅ 1 λ [ z ( D R λ m + 1 , n f ( z ) ) ′ D R λ m , n f ( z ) − D R λ m + 1 , n f ( z ) D R λ m , n f ( z ) ⋅ z ( D R λ m , n f ( z ) ) ′ D R λ m , n f ( z ) ] , ( 1 − ξ ) z p ′ ( z ) = 1 λ ⋅ D R λ m + 1 , n f ( z ) D R λ m , n f ( z ) [ z ( D R λ m + 1 , n f ( z ) ) ′ D R λ m + 1 , n f ( z ) − z ( D R λ m , n f ( z ) ) ′ D R λ m , n f ( z ) ] , ( 1 − ξ ) z p ′ ( z ) = [ ζ − 1 + 1 λ + ( 1 − ξ ) p ( z ) ] [ z ( D R λ m + 1 , n f ( z ) ) ′ D R λ m + 1 , n f ( z ) − ( 1 − ξ ) p ( z ) − ξ ] , ( 1 − ξ ) z p ′ ( z ) ( 1 − ξ ) p ( z ) + ζ − 1 + 1 λ = z ( D R λ m + 1 , n f ( z ) ) ′ D R λ m + 1 , n f ( z ) − ξ − ( 1 − ξ ) p ( z ) .$

Hence,

$1 1 − ξ [ z ( D R λ m + 1 , n f ( z ) ) ′ D R λ m + 1 , n f ( z ) − ξ ] =p(z)+ z p ′ ( z ) ( 1 − ζ ) p ( z ) + ζ − 1 + 1 λ .$
(2.4)

Since $Re{ϕ(z)}< ξ − 1 + 1 λ 1 − ξ$ implies $Re{(1−ξ)p(z)+ξ−1+ 1 λ }>0$, applying Lemma 2.2 to (2.4) we have that$f(z)∈ S λ m , n (ξ,ϕ)$, as required. □

Remark 2.2 By using relation (1.3) and the same techniques as to prove theearlier results, we can obtain a new similar result.

Corollary 2.3 Let$1 + A 1 + B < ξ − 1 + 1 λ 1 − ξ$for$−1, then

$S λ m + 1 , n (ξ,A,B)⊂ S λ m , n (ξ,A,B)⊂ S λ m − 1 , n (ξ,A,B).$

Proof Taking $ϕ(z)= 1 + A z 1 + B z$, $−1 in Theorem 2.2, we get thecorollary. □

## 3 Integral-preserving properties

In this section, we present several integral-preserving properties for the subclassesof analytic functions defined above. We recall the generalizedBernardi-Libera-Livington integral operator  defined by

$F c [ f ( z ) ] = c + 1 z c ∫ 0 z t c − 1 f(t)dt=z+ ∑ j = 2 ∞ c + 1 j + c a j z c ,f∈A,c>−1,$
(3.1)

which satisfies the following equality:

$cD R λ m , n F c [ f ( z ) ] +z [ D R λ m , n F c ( f ( z ) ) ] ′ =(c+1)D R λ m , n f(z).$
(3.2)

Theorem 3.1 Let$c>−1$, $0≤ξ<1$. If$f∈ S λ m , n (ξ)$, then$F c f∈ S λ m , n (ξ)$.

Proof Let $f∈ S λ m , n (ξ)$. By using (3.2), we get

$z [ D R λ m , n F c [ f ( z ) ] ] ′ D R λ m , n F c [ f ( z ) ] =(c+1) D R λ m , n f ( z ) D R λ m , n F c [ f ( z ) ] −c.$

Let

$z [ D R λ m , n F c [ f ( z ) ] ] ′ D R λ m , n F c [ f ( z ) ] =ξ+(1−ξ)h(z),h(z)=1+ c 1 z+ c 2 z 2 +⋯.$

We obtain

$z [ D R λ m , n f ( z ) ] ′ D R λ m , n f ( z ) −ξ=(1−ξ)h(z)+ ( 1 − ξ ) z h ′ ( z ) ξ + ( 1 − ξ ) h ( z ) + c .$

This implies

$φ(μ,v)=(1−ξ)μ+ ( 1 − ξ ) v c + ξ + ( 1 − ξ ) μ$

(same as Theorem 2.1) and

$Re { φ ( i μ 2 , v 1 ) } = ( 1 − ξ ) ( c + ξ ) v 1 ( c + ξ ) 2 + ( 1 − ξ ) 2 μ 2 2 , Re { φ ( i μ 2 , v 1 ) } ≤ − ( 1 − ξ ) ( c + ξ ) ( 1 + μ 2 ) 2 2 [ ( c + ξ ) 2 + ( 1 − ξ ) 2 μ 2 2 ] < 0 .$

After using Lemma 2.1 and Theorem 2.1, we have

$F c f∈ S λ m , n (ξ).$

□

Theorem 3.2 Let $c>−1$ and $ϕ∈R$ with

$Re { ϕ ( z ) } < c + ξ 1 − ξ .$

If$f∈ S λ m , n (ξ,ϕ)$, then$F c f∈ S λ m , n (ξ,ϕ)$.

Proof Let $f(z)∈ S λ m , n (ξ,ϕ)$ and set

$p(z)= 1 1 − ξ ( z [ D R λ m , n F c [ f ( z ) ] ] ′ D R λ m , n F c [ f ( z ) ] − ξ ) ,$
(3.3)

where p is analytic in U with $p(0)=1$.

Using (3.2) and (3.3), we have

$(c+1) z [ D R λ m , n f ( z ) ] D R λ m , n F c [ f ( z ) ] =c+ξ+(1−ξ)p(z).$
(3.4)

Then, using (3.2), (3.3) and (3.4), we obtain

$1 1 − ξ ( z [ D R λ m , n f ( z ) ] ′ D R λ m , n f ( z ) − ξ ) =p(z)+ z p ′ ( z ) ( 1 − ξ ) p ( z ) + c + ξ .$
(3.5)

Applying Lemma 2.2 to (3.5), we conclude that

$F c f∈ S λ m , n (ξ,ϕ).$

□

## 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 thepresent article.

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Correspondence to Loriana Andrei. 