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# Aspects of univalent holomorphic functions involving multiplier transformation and Ruscheweyh derivative

## Abstract

Making use multiplier transformation and Ruscheweyh derivative,we introduce a new class of analytic functions $RI(γ,λ,l,α,β)$ defined on the open unit disc, and investigate its various characteristics. Further we obtain distortion bounds, extreme points and radii of close-to-convexity, starlikeness and convexity and neighborhood property for functions belonging to the class $RI(γ,λ,l,α,β)$.

MSC:30C45, 30A20, 34A40.

## 1 Introduction

Let denote the class of functions of the form $f(z)=z+ ∑ j = 2 ∞ a j z j$, which are analytic and univalent in the open unit disc $U={z:z∈C:|z|<1}$. is a subclass of consisting of the functions of the form $f(z)=z− ∑ j = 2 ∞ | a j | z j$. For functions $f,g∈A$ given by $f(z)=z+ ∑ j = 2 ∞ a j z j$, $g(z)=z+ ∑ j = 2 ∞ b j z j$, we define the Hadamard product (or convolution) of f and g by $(f∗g)(z)=z+ ∑ j = 2 ∞ a j b j z j$, $z∈U$.

Definition 1.1 (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 1.1 If $f∈A$, $f(z)=z+ ∑ j = 2 ∞ a j z j$, then $R n f(z)=z+ ∑ j = 2 ∞ ( n + j − 1 ) ! n ! ( j − 1 ) ! a j z j$, $z∈U$.

If $f∈T$, $f(z)=z− ∑ j = 2 ∞ a j z j$, then $R n f(z)=z− ∑ j = t + 1 ∞ ( n + j − 1 ) ! n ! ( j − 1 ) ! a j z j$, $z∈U$.

Definition 1.2 [2, 3]

For $f∈A$, $n∈N∪{0}$, $λ,l≥0$, the operator $I(n,λ,l)f(z)$ is defined by the following infinite series:

$I(n,λ,l)f(z):=z+ ∑ j = 2 ∞ ( λ ( j − 1 ) + l + 1 l + 1 ) n a j z j .$

Remark 1.2 

It follows from the above definition that

$I ( 0 , λ , l ) f ( z ) = f ( z ) , ( l + 1 ) I ( n + 1 , λ , l ) f ( z ) = ( l + 1 − λ ) I ( n , λ , l ) f ( z ) + λ z ( I ( n , λ , l ) f ( z ) ) ′ , z ∈ U .$

Remark 1.3 The operator $I(n,λ,0)= D λ n$ is the generalized Sălăgean operator introduced by Al-Oboudi , and $I(n,1,0)= S n$ is the Sălăgean differential operator .

Definition 1.3 [7, 8]

Let $γ,λ,l≥0$, $n∈N$. Denote by $R I n , λ , l γ$ the operator given by $R I n , λ , l γ :A→A$,

$R I n , λ , l γ f(z)=(1−γ) R n f(z)+γI(n,λ,l)f(z),z∈U.$

Remark 1.4 If $f∈A$, $f(z)=z+ ∑ j = 2 ∞ a j z j$, then

$R I n , λ , l γ f(z)=z+ ∑ j = 2 ∞ { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } a j z j ,z∈U.$

If $f∈T$, $f(z)=z− ∑ j = 2 ∞ a j z j$, then

$R I n , λ , l γ f(z)=z− ∑ j = 2 ∞ { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } a j z j ,z∈U.$

Remark 1.5 The operator $R I n , λ , 0 γ f(z)=R D λ , γ n f(z)$ which was introduced in  and the operator $R I n , 1 , 0 γ f(z)= L γ n f(z)$ which was introduced in .

Following the work of Najafzadeh and Pezeshki  we can define the class $RI(γ,λ,l,α,β)$ as follows.

Definition 1.4 For $γ,λ,l≥0$, $0≤α<1$ and $0<β≤1$, let $RI(γ,λ,l,α,β)$ be the subclass of consisting of functions that satisfying the inequality

$| R I n , λ , l μ , γ f ( z ) − 1 2 ν ( R I n , λ , l μ , γ f ( z ) − α ) − ( R I n , λ , l μ , γ f ( z ) − 1 ) |<β,$
(1.1)

where

$R I n , λ , l μ , γ f(z)=(1−μ) R I n , λ , l γ f ( z ) z +μ ( R I n , λ , l γ f ( z ) ) ′ ,$
(1.2)

$0<ν≤1$.

Remark 1.6 If $f∈T$, $f(z)=z− ∑ j = 2 ∞ a j z j$, then

$R I n , λ , l μ , γ f ( z ) = 1 − ∑ j = t + 1 ∞ [ 1 + μ ( j − 1 ) ] × { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } a j z j − 1 , z ∈ U .$

Remark 1.7 The class $RI(γ,λ,0,α,β)=RD(γ,λ,α,β)$ defined and studied in  and $RI(γ,1,0,α,β)=L(γ,α,β)$ defined and studied in .

## 2 Coefficient bounds

In this section we obtain coefficient bounds and extreme points for functions in $RI(γ,λ,l,α,β)$.

Theorem 2.1 Let the function $f∈T$. Then $f∈RI(γ,λ,l,α,β)$ if and only if

$∑ j = t + 1 ∞ ( 1 + μ ( j − 1 ) ) [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } a j < 2 β ν ( 1 − α ) .$
(2.1)

The result is sharp for the function $F(z)$ defined by

$F(z)=z− 2 β ν ( 1 − α ) ( 1 + μ ( j − 1 ) ) [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } z j ,j≥t+1.$

Proof Suppose f satisfies (2.1). Then for $|z|<1$, we have

$| R I n , λ , l μ , γ f ( z ) − 1 | − β | 2 ν ( R I n , λ , l μ , γ f ( z ) − α ) − ( R I n , λ , l μ , γ f ( z ) − 1 ) | = | − ∑ j = t + 1 ∞ ( 1 + μ ( j − 1 ) ) { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } a j z j − 1 | − β | 2 ν ( 1 − α ) − ( 2 ν − 1 ) ∑ j = t + 1 ∞ { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } × [ 1 + μ ( j − 1 ) ] a j z j − 1 | ≤ ∑ j = t + 1 ∞ [ 1 + μ ( j − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } a k − 2 β ν ( 1 − α ) + ∑ j = t + 1 ∞ β ( 2 ν − 1 ) ( 1 + μ ( j − 1 ) ) { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } a j = ∑ j = t + 1 ∞ [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } a j − 2 β ν ( 1 − α ) < 0 .$

Hence, by using the maximum modulus Theorem and (1.1), $f∈RI(γ,λ,l,α,β)$. Conversely, assume that

$| R I n , λ , l μ , γ f ( z ) − 1 2 ν ( R I n , λ , l μ , γ f ( z ) − α ) − ( R I n , λ , l μ , γ f ( z ) − 1 ) | = | − ∑ j = t + 1 ∞ [ 1 + μ ( j − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } a j z j − 1 2 ν ( 1 − α ) − ∑ j = t + 1 ∞ [ 1 + μ ( j − 1 ) ] ( 2 ν − 1 ) { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } a j z j − 1 | < β , z ∈ U .$

Since $Re(z)≤|z|$ for all $z∈U$, we have

$Re { ∑ j = t + 1 ∞ [ 1 + μ ( j − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } a j z j − 1 2 ν ( 1 − α ) − ∑ j = t + 1 ∞ [ 1 + μ ( j − 1 ) ] ( 2 ν − 1 ) { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } a j z j − 1 } < β .$
(2.2)

By choosing choose values of z on the real axis so that $R I n , λ , l μ , γ f(z)$ is real and letting $z→1$ through real values, we obtain the desired inequality (2.1). □

Corollary 2.2 If $f∈T$ is in $RI(γ,λ,l,α,β)$, then

$a j ≤ 2 β ν ( 1 − α ) [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } ,j≥t+1,$
(2.3)

with equality only for functions of the form $F(z)$.

Theorem 2.3 Let $f 1 (z)=z$ and

$f j ( z ) = z − 2 β ν ( 1 − α ) [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } z j , j ≥ t + 1 ,$
(2.4)

for $0≤α<1$, $0<β≤1$, $γ,λ,l≥0$ and $0<ν≤1$. Then $f(z)$ is in the class $RI(γ,λ,l,α,β)$ if and only if it can be expressed in the form

$f(z)= ∑ j = t ∞ ω j f j (z),$
(2.5)

where $ω j ≥0$ and $∑ j = 1 ∞ ω j =1$.

Proof Suppose $f(z)$ can be written as in (2.5). Then

$f(z)=z− ∑ j = t + 1 ∞ ω j 2 β ν ( 1 − α ) [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } z j .$

Now,

$∑ j = t + 1 ∞ [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } 2 β ν ( 1 − α ) ω j × 2 β ν ( 1 − α ) [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } = ∑ j = t + 1 ∞ ω j = 1 − ω 1 ≤ 1 .$

Thus $f∈RI(γ,λ,l,α,β)$.

Conversely, let $f∈RI(γ,λ,l,α,β)$. Then by using (2.3), setting

$ω j = [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } 2 β ν ( 1 − α ) a j ,j≥t+1,$

and $ω 1 =1− ∑ j = 2 ∞ ω j$, we have $f(z)= ∑ j = t ∞ ω j f j (z)$. This completes the proof of Theorem 2.3. □

## 3 Distortion bounds

In this section we obtain distortion bounds for the class $RI(γ,λ,l,α,β)$.

Theorem 3.1 If $f∈RI(γ,λ,l,α,β)$, then

$r − 2 β ν ( 1 − α ) ( 1 + μ t ) [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ t + l l + 1 ) n + ( 1 − γ ) ( n + t ) ! n ! t ! } r t + 1 ≤ | f ( z ) | ≤ r + 2 β ν ( 1 − α ) ( 1 + μ t ) [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ t + l l + 1 ) n + ( 1 − γ ) ( n + t ) ! n ! t ! } r t + 1$
(3.1)

holds if the sequence ${ σ j ( γ , λ , l , β , ν ) } j = t + 1 ∞$ is non-decreasing, and

$1 − 2 β ν ( 1 − α ) ( t + 1 ) ( 1 + μ t ) [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ t + l l + 1 ) n + ( 1 − γ ) ( n + t ) ! n ! t ! } r t ≤ | f ′ ( z ) | ≤ 1 + 2 β ν ( 1 − α ) ( t + 1 ) ( 1 + μ t ) [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ t + l l + 1 ) n + ( 1 − γ ) ( n + t ) ! n ! t ! } r t$
(3.2)

holds if the sequence ${ σ j ( γ , λ , l , β , ν ) j } j = t + 1 ∞$ is non-decreasing, where

$σ j (γ,β,ν)= [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } .$

The bounds in (3.1) and (3.2) are sharp, for $f(z)$ given by

$f(z)=z− 2 β ν ( 1 − α ) ( 1 + μ t ) [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ t + l l + 1 ) n + ( 1 − γ ) ( n + t ) ! n ! t ! } z t + 1 ,z=±r.$
(3.3)

Proof In view of Theorem 2.1, we have

$∑ j = t + 1 ∞ a j ≤ 2 β ν ( 1 − α ) ( 1 + μ t ) [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ t + l l + 1 ) n + ( 1 − γ ) ( n + t ) ! n ! t ! } .$
(3.4)

We obtain

$|z|−|z | t + 1 ∑ j = t + 1 ∞ a j ≤|f(z)|≤|z|+|z | t + 1 ∑ j = t + 1 ∞ a j .$

Thus

$r − 2 β ν ( 1 − α ) ( 1 + μ t ) [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ t + l l + 1 ) n + ( 1 − γ ) ( n + t ) ! n ! t ! } r t + 1 ≤ | f ( z ) | ≤ r + 2 β ν ( 1 − α ) ( 1 + μ t ) [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ t + l l + 1 ) n + ( 1 − γ ) ( n + t ) ! n ! t ! } r t + 1 .$
(3.5)

Hence (3.1) follows from (3.5). Further,

$∑ j = t + 1 ∞ j a j ≤ 2 β ν ( 1 − α ) ( 1 + μ t ) [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ t + l l + 1 ) n + ( 1 − γ ) ( n + t ) ! n ! t ! } .$

Hence (3.2) follows from

$1− r t ∑ j = t + 1 ∞ j a j ≤| f ′ (z)|≤1+ r t ∑ j = t + 1 ∞ j a j .$

□

## 4 Radius of starlikeness and convexity

The radii of close-to-convexity, starlikeness, and convexity for the class $RI(γ,λ,l,α,β)$ are given in this section.

Theorem 4.1 Let the function $f∈T$ belong to the class $RI(γ,λ,l,α,β)$, Then $f(z)$ is close-to-convex of order δ, $0≤δ<1$, in the disc $|z|, where

$r:= inf j ≥ t + 1 [ ( 1 − δ ) [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } 2 β ν j ( 1 − α ) ] 1 t .$
(4.1)

The result is sharp, with extremal function $f(z)$ given by (2.3).

Proof For given $f∈T$ we must show that

$| f ′ (z)−1|<1−δ.$
(4.2)

By a simple calculation we have

$| f ′ (z)−1|≤ ∑ j = t + 1 ∞ j a j |z | t .$

The last expression is less than $1−δ$ if

$∑ j = t + 1 ∞ j 1 − δ a j |z | t <1.$

We use the fact that $f∈RI(γ,λ,l,α,β)$ if and only if

$∑ j = t + 1 ∞ [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } 2 β ν ( 1 − α ) a j ≤1.$

Equation (4.2) holds true if

$j 1 − δ |z | t ≤ ∑ j = t + 1 ∞ [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } 2 β ν ( 1 − α ) .$

Or, equivalently,

$|z | t ≤ ∑ j = t + 1 ∞ ( 1 − δ ) [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } 2 β ν j ( 1 − α ) ,$

which completes the proof. □

Theorem 4.2 Let $f∈RI(γ,λ,l,α,β)$. Then

1. 1.

f is starlike of order δ, $0≤δ<1$, in the disc $|z|< r 1$, where

$r 1 = inf j ≥ t + 1 { ( 1 − δ ) [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } 2 β ν ( 1 − α ) ( j − δ ) } 1 t .$
2. 2.

f is convex of order δ, $0≤δ<1$, in the disc $|z|< r 2$ where,

$r 2 = inf j ≥ t + 1 { ( 1 − δ ) [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } 2 β ν j ( j − 1 ) ( 1 − α ) } 1 t .$

Each of these results is sharp for the extremal function $f(z)$ given by (2.5).

Proof 1. For $0≤δ<1$ we need to show that

$| z f ′ ( z ) f ( z ) −1|<1−δ.$
(4.3)

We have

$| z f ′ ( z ) f ( z ) −1|≤| ∑ j = t + 1 ∞ ( j − 1 ) a j | z | t 1 − ∑ j = t + 1 ∞ a j | z | t |.$

The last expression is less than $1−δ$ if

$∑ j = t + 1 ∞ ( j − δ ) 1 − δ a j |z | t <1.$

We use the fact that $f∈RI(γ,λ,l,α,β)$ if and only if

$∑ j = t + 1 ∞ [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } 2 β ν ( 1 − α ) a j <1.$

Equation (4.3) holds true if

$j − δ 1 − δ |z | t < [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } 2 β ν ( 1 − α ) .$

Or, equivalently,

$|z | t < ( 1 − δ ) [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } 2 β ν ( 1 − α ) ( j − δ ) ,$

which yields the starlikeness of the family.

1. 2.

Using the fact that f is convex if and only $z f ′$ is starlike, we can prove (2) with a similar way of the proof of (1). The function f is convex if and only if

$|z f ″ (z)|<1−δ.$
(4.4)

We have

$| z f ″ ( z ) | ≤ | ∑ j = t + 1 ∞ j ( j − 1 ) a j | z | t − 1 | < 1 − δ , ∑ j = t + 1 ∞ j ( j − 1 ) 1 − δ a j | z | t − 1 < 1 .$

We use the fact that $f∈RI(γ,λ,l,α,β)$ if and only if

$∑ j = t + 1 ∞ [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } 2 β ν ( 1 − α ) a j <1.$

Equation (4.4) holds true if

$j ( j − 1 ) 1 − δ |z | t − 1 < [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } 2 β ν ( 1 − α ) ,$

or, equivalently,

$|z | t − 1 < ( 1 − δ ) [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } 2 β ν j ( j − 1 ) ( 1 − α ) ,$

which yields the convexity of the family. □

## 5 Neighborhood property

In this section we study neighborhood property for functions in the class $RI(γ,λ,l,α,β)$.

Definition 5.1 For functions f belong to of the form and $ε≥0$, we define $(η−ε)$-neighborhood of f by

$N ε η (f)= { g ( z ) ∈ A : g ( z ) = z + ∑ j = 2 ∞ b j z j , ∑ j = 2 ∞ j η + 1 | a j − b j | ≤ ε } ,$

where η is a fixed positive integer.

By using the following lemmas we will investigate the $(η−ε)$-neighborhood of function in $RI(γ,λ,l,α,β)$.

Lemma 5.1 Let $−1≤β<1$, if $g(z)=z+ ∑ j = 2 ∞ b j z j$ satisfies

$∑ j = 2 ∞ j ρ + 1 | b j |≤ 2 β ν ( 1 − α ) 1 + β ( 2 ν − 1 )$

then $g(z)∈RI(γ,λ,l,α,β)$.

Proof By using of Theorem 2.1, it is sufficient to show that

$[ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) ρ + ( 1 − γ ) ( ρ + j − 1 ) ! ρ ! ( j − 1 ) ! } 2 β ν ( 1 − α ) = j ρ + 1 2 β ν ( 1 − α ) [ 1 + β ( 2 ν − 1 ) ] .$

But

$[ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) ρ + ( 1 − γ ) ( ρ + j − 1 ) ! ρ ! ( j − 1 ) ! } 2 β ν ( 1 − α ) ≤ j ρ + 1 2 β ν ( 1 − α ) [ 1 + β ( 2 ν − 1 ) ] .$

Therefore it is enough to prove that

$Q(j,ρ)= γ ( 1 + λ ( j − 1 ) + l l + 1 ) ρ + ( 1 − γ ) ( ρ + j − 1 ) ! ρ ! ( j − 1 ) ! j ρ + 1 ≤1,$

the result follows because the last inequality holds for all $j≥t+1$. □

Lemma 5.2 Let $f(z)=z− ∑ k = 2 ∞ a k z k ∈T$, $γ,λ,l≥0$, $0≤α<1$, $0<β≤1$ and $ε≥0$. If $f ( z ) + ϵ z 1 + ϵ ∈RI(γ,λ,l,α,β)$, then

$∑ j = t + 1 ∞ j ρ + 1 a j ≤ 2 β ν ( 1 − α ) ( 1 + ϵ ) ( t + 1 ) ρ + 1 [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ t + l l + 1 ) n + ( 1 − γ ) ( n + t ) ! n ! t ! } ,$

where either $ρ=0$ or $ρ=1$. The result is sharp with the extremal function

$f(z)=z− 2 β ν ( 1 − α ) ( 1 + ϵ ) [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ t + l l + 1 ) n + ( 1 − γ ) ( n + t ) ! n ! t ! } z t + 1 ,z∈U.$

Proof Letting $g(z)= f ( z ) + ϵ z 1 + ϵ$ we have

$g(z)=z− ∑ j = t + 1 ∞ a j 1 + ϵ z j ,z∈U.$

In view of Theorem 2.3, $g(z)= ∑ j = 1 ∞ η j g j (z)$ where $η j ≥0$, $∑ j = 1 ∞ η j =1$,

$g 1 (z)=z$

and

$g j (z)=z− 2 β ν ( 1 − α ) ( 1 + ϵ ) [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } z j ,j≥t+1.$

So we obtain

$g ( z ) = η 1 z + ∑ j = t + 1 ∞ η j [ z − 2 β ν ( 1 − α ) ( 1 + ϵ ) [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } z j ] = z − ∑ j = t + 1 ∞ η k 2 β ν ( 1 − α ) ( 1 + ϵ ) [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } z j .$

Since $η j ≥0$ and $∑ j = 2 ∞ η j ≤1$, it follows that

$∑ j = t + 1 ∞ a k ≤ sup j ≥ t + 1 j ρ + 1 2 β ν ( 1 − α ) ( 1 + ϵ ) [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! } .$

Since whenever $ρ=0$ or $ρ=1$ we conclude that

$W(j,ρ,γ,α,β,ϵ)= j ρ + 1 2 β ν ( 1 − α ) ( 1 + ϵ ) [ 1 + μ ( j − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ ( j − 1 ) + l l + 1 ) n + ( 1 − γ ) ( n + j − 1 ) ! n ! ( j − 1 ) ! }$

is a decreasing function of j, the result will follow. The proof is complete. □

Theorem 5.1 Let $ρ=0$ or $ρ=1$ and suppose $0≤β<1$ and

$− 1 ≤ θ < [ 1 + μ ( t − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ t + l l + 1 ) n + ( 1 − γ ) ( n + t ) ! n ! t ! } − 2 β ν ( 1 − α ) ( 1 + ϵ ) ( t + 1 ) η + 1 [ 1 + μ ( t − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ t + l l + 1 ) n + ( 1 − γ ) ( n + t ) ! n ! t ! } ,$

$f(z)∈T$ and $f ( z ) + ϵ z 1 + ϵ ∈RI(γ,λ,l,α,β)$, then the $(η−ε)$-neighborhood of f is the subset of $RI(λ,λ,l,α,β)$, where

$ε ≤ 2 ( 1 − α ) { θ γ [ 1 + μ ( t − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ t + l l + 1 ) n + ( 1 − γ ) ( n + t ) ! n ! t ! } − β γ [ 1 + θ ( 2 ν − 1 ) ] ( 1 + ϵ ) ( t + 1 ) η + 1 } / ( [ 1 + θ ( 2 ν − 1 ) ] [ 1 + μ ( t − 1 ) ] × [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ t + l l + 1 ) n + ( 1 − γ ) ( n + t ) ! n ! t ! } ) .$

The result is sharp.

Proof For $f(z)=z− ∑ j = 2 ∞ | a j | z j$, let $g(z)=z+ ∑ j = 2 ∞ b j z j$ be in $N ε η (f)$. So by Lemma 5.2, we have

$∑ j = 2 ∞ j η + 1 | b j | = ∑ j = 2 ∞ j η + 1 | a j − b j − a j | ≤ ε + 2 β ν ( 1 − α ) ( 1 + ϵ ) ( t + 1 ) η + 1 [ 1 + μ ( t − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ t + l l + 1 ) n + ( 1 − γ ) ( n + t ) ! n ! t ! } .$

By using Lemma 5.1, $g(z)∈L(γ,α,β)$ if

$ε+ 2 β ν ( 1 − α ) ( 1 + ϵ ) ( t + 1 ) η + 1 [ 1 + μ ( t − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ t + l l + 1 ) n + ( 1 − γ ) ( n + t ) ! n ! t ! } ≤ 2 θ ν ( 1 − α ) 1 + θ ( 2 ν − 1 ) ,$

that is, $ε≤ 2 ( 1 − α ) { θ γ [ 1 + μ ( t − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ t + l l + 1 ) n + ( 1 − γ ) ( n + t ) ! n ! t ! } − β γ [ 1 + θ ( 2 ν − 1 ) ] ( 1 + ϵ ) ( t + 1 ) η + 1 } [ 1 + θ ( 2 ν − 1 ) ] [ 1 + μ ( t − 1 ) ] [ 1 + β ( 2 ν − 1 ) ] { γ ( 1 + λ t + l l + 1 ) n + ( 1 − γ ) ( n + t ) ! n ! t ! }$, and the proof is complete. □

## Author’s contributions

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

## References

1. 1.

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

2. 2.

Alb Lupaş A: A special comprehensive class of analytic functions defined by multiplier transformation. J. Comput. Anal. Appl. 2010, 12(2):387–395.

3. 3.

Alb Lupaş A: A new comprehensive class of analytic functions defined by multiplier transformation. Math. Comput. Model. 2011, 54: 2355–2362. 10.1016/j.mcm.2011.05.044

4. 4.

Alb Lupaş A: On special differential superordinations using multiplier transformation. J. Comput. Anal. Appl. 2011, 13(1):121–126.

5. 5.

Al-Oboudi FM: On univalent functions defined by a generalized Sălăgean operator. Int. J. Math. Math. Sci. 2004, 27: 1429–1436.

6. 6.

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

7. 7.

Alb Lupaş A: On special differential subordinations using multiplier transformation and Ruscheweyh derivative. Romai J. 2010, 6(2):1–14.

8. 8.

Alb Lupaş A: Certain special differential superordinations using multiplier transformation and Ruscheweyh derivative. J. Comput. Anal. Appl. 2011, 13(1):108–115.

9. 9.

Alb Lupaş A: On special differential subordinations using a generalized Sălăgean operator and Ruscheweyh derivative. J. Comput. Anal. Appl. 2011, 13(1):98–107.

10. 10.

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

11. 11.

Najafzadeh S, Pezeshki E: Some aspects of univalent holomorphic functions involving Ruscheweyh and Salagean operator. An. Univ. Oradea, Fasc. Mat. 2013, XX(1):61–70.

12. 12.

Alb Lupas, A, Andrei, L: Aspects of univalent holomorphic functions involving Ruscheweyh derivative and generalized Sălăgean operator. J. Comput. Anal. Appl. (to appear)

13. 13.

Alb Lupas, A: Aspects of univalent holomorphic functions involving Sălăgean operator and Ruscheweyh derivative. J. Concr. Appl. Math. (to appear)

## Author information

Correspondence to Alina Alb Lupaş. 