# On Subclass ofk-Uniformly Convex Functions of Complex Order Involving Multiplier Transformations

Authors: Waggas Galib Atshan; Ali Hamza Abada
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We introduce a subclass ofk-uniformly convex functions of order α with negative coefficients by using the multiplier transformations in the open unit disk U={z∈ℂ:|z|<1}. We obtain coefficient estimates, radii of convexity and close-to-convexity, extreme points, and integral means inequalities for the function f that belongs to the class 𝒩mℓ(α,β,k,υ).


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## Body

## 1. Introduction

Let𝒩 denote the class of functions of the form:(1.1)f(z)β=zβ+∑n=2∞βanzβ+n-1,β>0,
which are analytic and univalent in the open unit disk U={z∈ℂ:|z|<1} (see [1]). Also denote by ℳ the subclass of 𝒩 consisting of functions of the form:(1.2)f(z)β=zβ-∑n=2∞βanzβ+n-1,(an≥0,β>0).
For any integer m, we define the multiplier transformations Imℓ (see [2, 3]) of functions f∈𝒩(n) by(1.3)Imlf(z)β=zβ-∑n=2∞β(β+lβ+l+n-1)manzβ+n-1=zβ-∑n=2∞βQ(n,β,l)anzβ+n-1,(l≥0,z∈U),
where 𝒬(n,β,ℓ)=((β+ℓ)/(β+ℓ+n-1))^m.A functionf∈ℳ is said to be in the class USL(α,k) (k-uniformly starlike Functions of order α) if it satisfies the condition:(1.4)Re{zf′(z)βf(z)β-α}>k|zf′(z)βf(z)β-1|,(0≤α<1,k≥0),z∈U
and is said to be in the class UCV(α,k) (k-uniformly convex Functions of order α) if it satisfies the condition:(1.5)Re{1+zf′′(z)βf′(z)β-α}>k|zf′′(z)βf′(z)β|,(0≤α<1,k≥0),z∈U.
Indeed it follows from (1.4) and (1.5) that(1.6)f∈UCV(α,k)⟺zf′∈USL(α,k).The interesting geometric properties of these function classes were extensively studied by Kanas et al., in [4, 5], motivated by Altintas et al. [6], Murugusundaramoorthy and Srivastava [7], and Murugusundaramoorthy and Magesh [8, 9], Atshan and Kulkarni [10] and Atshan and Buti [11].Now, we define a new subclass of uniformly convex functions of complex order.For0≤α<1, k≥0, υ∈ℂ∖{0}, we let 𝒩mℓ(α,β,k,υ) be the class of functions f satisfying (1.2) with the analytic criterion:(1.7)Re{1+1υ(1+z(Imlf(z)β)′′(Imlf(z)β)′-α)}>k|1+1υ(z(Imlf(z)β)′′(Imlf(z)β)′)|,z∈U,
where Imℓf(z)β is given by (1.3).


## 2. Main Results

First, we obtain the necessary and sufficient condition for functionsf in the class 𝒩mℓ(α,β,k,υ).Theorem 2.1.
The necessary and sufficient condition forf of the form of (1.2) to be in the class 𝒩mℓ(α,β,k,υ) is
(2.1)∑n=2∞(β+n-1)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l)an≤(k-α)+(1-k)(β+|υ|),
where 0≤α<1, k≥0, υ∈ℂ∖{0}.Proof.
Suppose that (2.1) is true for z∈U. Then
(2.2)Re{1+1υ(1+z(Imlf(z)β)′′(Imlf(z)β)′-α)}-k|1+1υ(z(Imlf(z)β)′′(Imlf(z)β)′)|>0,
if
(2.3)1+1|υ|((β-α)-∑n=2∞(β+n-1)(β+n-α-1)Q(n,β,l)an|z|n-11-∑n=2∞(β+n-1)Q(n,β,l)an|z|n-1)-k[1+1|υ|((β-1)-∑n=2∞(β+n-1)(β+n-2)Q(n,β,l)an|z|n-11-∑n=2∞(β+n-1)Q(n,β,l)an|z|n-1)]>0,
that is, if
(2.4)∑n=2∞(β+n-1)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l)an≤(k-α)+(1-k)(β+|υ|).
Conversely, assume that f∈𝒩mℓ(α,β,k,υ), then
(2.5)Re{1+1υ(1+z(Imlf(z)β)′′(Imlf(z)β)′-α)}>k|1+1υ(z(Imlf(z)β)′′(Imlf(z)β)′)|,Re{1+1υ((β-α)-∑n=2∞(β+n-1)(β+n-α-1)Q(n,β,l)anzn-11-∑n=2∞(β+n-1)Q(n,β,l)anzn-1)}>k|1+1υ((β-1)-∑n=2∞(β+n-1)(β+n-2)Q(n,β,l)anzn-11-∑n=2∞(β+n-1)Q(n,β,l)anzn-1)|.
Letting z→1- along the real axis, we have
(2.6)1+1|υ|((β-α)-∑n=2∞(β+n-1)(β+n-α-1)Q(n,β,l)an1-∑n=2∞(β+n-1)Q(n,β,l)an)>k[1+1|υ|((β-1)-∑n=2∞(β+n-1)(β+n-2)Q(n,β,l)an1-∑n=2∞(β+n-1)Q(n,β,l)an)].
Hence, by maximum modulus theorem, the simple computation leads to the desired inequality
(2.7)∑n=2∞(β+n-1)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l)an≤(k-α)+(1-k)(β+|υ|),
which completes the proof.Corollary 2.2.
Let the functionf defined by (1.2) belong to 𝒩mℓ(α,β,k,υ). Then,
(2.8)an≤(k-α)+(1-k)(β+|υ|)(β+n-1)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l),
where 0≤α<1, k≥0,υ∈ℂ∖{0}, with equality for
(2.9)f(z)β=zβ-β(k-α)+(1-k)(β+|υ|)(β+n-1)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l)zβ+n-1.


## 3. Radii of Convexity and Close-to-Convexity

We obtain the radii of convexity and close-to-convexity results forf functions in the class 𝒩mℓ(α,β,k,υ) in the following theorems.Theorem 3.1.
Letf∈𝒩mℓ(α,β,k,υ). Then f is convex of order δ(0≤δ<1) in the disk |z|<r=r1(α,β,k,υ,n,δ), where
(3.1)r1=inf⁡n≥2⁡[(2-δ-β)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l)(3-δ-β-n)[(k-α)+(1-k)(β+|υ|)]]1/n-1.Proof.
Letf∈𝒩mℓ(α,β,k,υ). Then by Theorem 2.1, we have
(3.2)∑n=2∞(β+n-1)[(β+n-1+|υ|)(1-k)+(k-α)](k-α)+(1-k)(β+|υ|)Q(n,β,l)an≤1.
For  0≤δ<1, we need to show that
(3.3)|zf′′(z)βf′(z)β|≤1-δ,
and we have to show that
(3.4)|zf′′(z)βf′(z)β|≤(β-1)-∑n=2∞(β+n-1)(β+n-2)an|z|n-11-∑n=2∞(β+n-1)an|z|n-1≤1-δ.
Hence,
(3.5)∑n=2∞(β+n-1)(3-δ-β-n)(2-δ-β)an|z|n-1≤1.
This is enough to consider
(3.6)|z|n-1≤(2-δ-β)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l)(3-δ-β-n)[(k-α)+(1-k)(β+|υ|)].
Therefore,
(3.7)|z|≤{(2-δ-β)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l)(3-δ-β-n)[(k-α)+(1-k)(β+|υ|)]}1/n-1.
Setting z=r1(α,β,k,υ,n,δ) in (3.7), we get the radius of convexity, which completes the proof of Theorem 3.1.Theorem 3.2.
Letf∈𝒩mℓ(α,β,k,υ). Then f is close-to-convex of order δ(0≤δ<1) in the disk |z|<r=r2(α,β,k,υ,n,δ), where
(3.8)r2=inf⁡n≥2⁡[(β+n-1)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l)(k-α)+(1-k)(β+|υ|)]1/n-1.Proof.
Letf∈𝒩mℓ(α,β,k,υ). Then by Theorem 2.1, we have
(3.9)∑n=2∞(β+n-1)[(β+n-1+|υ|)(1-k)+(k-α)](k-α)+(1-k)(β+|υ|)Q(n,β,l)an≤1.
For  0≤δ<1, we need to show that
(3.10)|f′(z)βzβ-1-1|≤1-δ,
and we have to show that
(3.11)|f′(z)βzβ-1-1|≤(β-1)+∑n=2∞β(β+n-1)an|z|n-1≤1-δ.
Hence,
(3.12)∑n=2∞β(β+n-1)(2-δ-β)an|z|n-1≤1.
This is enough to consider
(3.13)|z|n-1≤(2-δ-β)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l)β[(k-α)+(1-k)(β+|υ|)].
Therefore,
(3.14)|z|≤{(2-δ-β)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l)β[(k-α)+(1-k)(β+|υ|)]}1/n-1.
Setting z=r2(α,β,k,υ,n,δ) in (3.14), we get the radius of close-to-convexity, which completes the proof of Theorem 3.2.


## 4. Extreme Points

The extreme points of the class𝒩mℓ(α,β,k,υ) are given by the following theorem.Theorem 4.1.
Let(4.1)f1(z)β=zβfn(z)β=zβ-β(k-α)+(1-k)(β+|υ|)(β+n-1)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l)zβ+n-1,
for n=2,3,4,….
Then,f∈𝒩mℓ(α,β,k,υ) if and only if it can be expressed in the form:
(4.2)f(z)β=∑n=1∞Υnfn(z)β,
where Υn≥0 and
(4.3)∑n=1∞Υn=1.Proof.
Suppose thatf can be expressed as in (4.2). Our goal is to show that f∈𝒩mℓ(α,β,k,υ). By (4.2), we have that
(4.4)f(z)β=∑n=1∞Υnfn(z)β=Υ1f1(z)β+∑n=2∞Υnfn(z)β=Υ1f1(z)β+∑n=2∞Υn(zβ-β(k-α)+(1-k)(β+|υ|)(β+n-1)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l)zβ+n-1)=∑n=1∞Υnzβ-∑n=2∞βΥn(k-α)+(1-k)(β+|υ|)(β+n-1)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l)zβ+n-1=zβ-∑n=2∞βΥn[(k-α)+(1-k)(β+|υ|)](β+n-1)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l)zβ+n-1.
Now,
(4.5)∑n=2∞(β+n-1)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l)(k-α)+(1-k)(β+|υ|)×Υn[(k-α)+(1-k)(β+|υ|)](β+n-1)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l)=∑n=2∞Υn=1-Υ1≤1.
Thus, f∈𝒩mℓ(α,β,k,υ).
Conversely, assume thatf∈𝒩mℓ(α,β,k,υ). Since
(4.6)an≤(k-α)+(1-k)(β+|υ|)(β+n-1)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l)(n≥2),
we can set
(4.7)Υn=(β+n-1)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l)(k-α)+(1-k)(β+|υ|)an(n≥2),Υ1=1-∑n=2∞Υn.
Then,
(4.8)f(z)β=zβ-∑n=2∞βanzβ+n-1=zβ-∑n=2∞βΥn[(k-α)+(1-k)(β+|υ|)](β+n-1)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l)zβ+n-1=zβ-∑n=2∞Υn(zβ-fn(z)β)=zβ(1-∑n=2∞Υn)+∑n=2∞Υnfn(z)β=Υ1f1(z)β+∑n=2∞Υnfn(z)β=∑n=1∞Υnfn(z)β.
This completes the proof of Theorem 4.1.


## 5. Integral Means

In order to find the integral means inequality and to verify the Silverman Conjuncture [12] for f∈𝒩mℓ(α,β,k,υ), we need the following definition of subordination and subordination result according to Littlewood [13].Definition 5.1 (see [13]).
Letf and g be analytic in U. Then, we say that the function f is subordinate to g if there exists a Schwarz function w, analytic in U with w(0)=0, |w(z)|<1 such that f(z)=g(w(z))(z∈U). We denote this subordination f≺g or f(z)≺g(z)(z∈U). In particular, if the function g is univalent in U, the above subordination is equivalent to f(0)=g(0), f(U)⊂g(U).Lemma 5.2 (see [13]).
If the functionsf and g are analytic in Uwith g≺f, then
(5.1)∫02π|g(reiθ)|ηdθ≤∫02π|f(reiθ)|ηdθ,η>0,z=reiθ,0<r<1.Applying Theorem 2.1 with the extremal function and Lemma 5.2, we prove the following theorem.Theorem 5.3.
Letη>0. If f∈𝒩mℓ(α,β,k,υ) and {Φ(α,β,k,υ,n)}n=2∞ are nondecreasing sequences, then, for z=reiθ and 0<r<1, one has
(5.2)∫02π|f(reiθ)β|ηdθ≤∫02π|f2(reiθ)β|ηdθ,
where
(5.3)f2(z)β=zβ-β(k-α)+(1-k)(β+|υ|)Φ(α,β,k,υ,2)zβ+1,Φ(α,β,k,υ,n)=(β+n-1)[(β+n-1+|υ|)(1-k)+(k-α)]Q(n,β,l).Proof.
Letfof the form of (1.2) and
(5.4)f2(z)β=zβ-β(k-α)+(1-k)(β+|υ|)Φ(α,β,k,υ,2)zβ+1,
then we must show that
(5.5)∫02π|1-∑n=2∞βanzn-1|ηdθ≤∫02π|1-β(k-α)+(1-k)(β+|υ|)Φ(α,β,k,υ,2)z|ηdθ.
By Lemma 5.2, it suffices to show that
(5.6)1-∑n=2∞βanzn-1≺1-β(k-α)+(1-k)(β+|υ|)Φ(α,β,k,υ,2)z.
Setting
(5.7)1-∑n=2∞βanzn-1=1-β(k-α)+(1-k)(β+|υ|)Φ(α,β,k,υ,2)w(z),
from (5.7) and (2.1) we obtain
(5.8)|w(z)|=|∑n=2∞Φ(α,β,k,υ,2)(k-α)+(1-k)(β+|υ|)anzn-1|≤|z|∑n=2∞Φ(α,β,k,υ,n)(k-α)+(1-k)(β+|υ|)an≤|z|<1.
This completes the proof of Theorem 5.3.



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