# A New Subclass of Salagean-Type Harmonic Univalent Functions

Authors: Khalifa Al-Shaqsi; Maslina Darus; Olubunmi  Abidemi Fadipe-Joseph
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

We define and investigate a new subclass of Salagean-type
harmonic univalent functions. We obtain coefficient conditions, extreme points,
distortion bounds, convolution, and convex combination for the above subclass of
harmonic functions.


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## Body

## 1. Introduction

Let𝒜 denote the class of functions of the form:(1.1)f(z)=z+∑k=2∞akzk
which are analytic in the open unit disk 𝕌={z∈ℂ:|z|<1}.We denote the subclass ofA consisting of analytic and univalent functions f(z) in the unit disk 𝕌 by S.The following classes of functions and many others are well known and have been studied repeatedly by many authors, namely, Sălăgean [1], Abdul Halim [2], and Darus [3] to mention but a few. (i)
S0={f(z)∈𝒜:Re{f(z)/z}>0,z∈𝕌}.(ii)
B(α)={f(z)∈𝒜:Re{f(z)/z}>α,0≤α<1,z∈𝕌}.(iii)
δ(α)={f(z)∈𝒜:Re{f′(z)}>α,0≤α<1,z∈𝕌}.(iv)
Bn(β)={f(z)∈𝒜:Re{Dnf(z)β/zβ}>0,z∈U,n∈ℕ0=N∪{0},β>0}.In 1994, Opoola defined the classTnβ(α) to be a subclass of A consisting of analytic functions satisfying the condition(1.2)Re{Dnf(z)βzβ}>α,z∈U,n∈N0,0≤α<1,β>0,
where Dn is the Salagean differential operator defined as follows: (1.3)D0f(z)=f(z),D1f(z)=Df(z)=zf′(z),Dnf(z)=D(Dn-1f(z))=z(Dn-1f(z)).
We note that Tnβ(α) is a generalization of the classes of functions S0,B(α),δ(α), and Bn(β).Some properties of this class of functions were established by Opoola [4] namely,  (i)
Tnβ(α) is a subclass of univalent functions;(ii)
Tn+1β(α)⊂Tnβ(α);(iii)
iff(z)∈Tnβ(α), then the integral operator(1.4)Fc(z)β=β+czβ∫0ctβ-1f(z)βdt(c≥0)
is also in Tnβ(α).Now, by Binomial expansion, we have(1.5)f(z)β=zβ+βa2zβ+1+[βa3+β(β-1)2!a23]zβ+2+[βa4+β(β-1)2!2a2a3+β(β-1)(β-2)3!a23]zβ+3+⋯.
Hence, we define(1.6)f(z)β=zβ+∑k=2∞βakzβ+k-1,β>0,Dnf(z)β=zβ+∑k=2∞βknakzβ+k-1,n∈N0.


## 2. Preliminaries

A continuous functionf=u+iv is a complex-valued harmonic function in a domain D⊂ℂ if both u and v are real harmonic in D. In any simply connected domain, we can write(2.1)f=h+g¯,
where h and g are analytic in D. We call h the analytic part and g the coanalytic part of f. A necessary and sufficient condition for f to be locally univalent and sense-preserving in D is that |h′|>|g′| in D.Denote bySℋ the class of functions f of the form (2.1) that are harmonic univalent and sense-preserving in the unit disk 𝕌. The subclasses of harmonic univalent functions have been studied by some authors for different purposes and different properties (see examples [5–12]). In this work, we may express the analytic functions h and g as(2.2)h(z)β=zβ+∑k=2∞βakzβ+k-1,g(z)β=∑k=1∞βbkzβ+k-1,|b1|<1.
Therefore,(2.3)f(z)β=h(z)β+g(z)β¯.
We define the modified Salagean operator of f as (2.4)Dnf(z)β=Dnh(z)β+(-1)nDng(z)β¯,
where(2.5)Dnh(z)β=zβ+∑k=2∞βknakzβ+k-1,Dng(z)β=∑k=1∞βknbkzβ+k-1.
We let ℋ(n,β,α) be the family of harmonic functions of the form (2.3) such that(2.6)Re{Dn+1f(z)βDnf(z)β}>α,β≥1,0≤α<1,n∈N0,
where Dnf(z)β is defined by (2.4).It is clear that the classℋ(n,β,α) includes a variety of well-known subclasses of Sℋ. For example, ℋ(0,1,α)≡Sℋ*(α) is the class of sense-preserving, harmonic univalent functions f which are starlike of order α in 𝕌, that is, ∂/∂θ{arg(f(reiθ))}>α, and ℋ(1,1,α)≡ℋ𝒦(α) is the class of sense-preserving, harmonic univalent functions f which are convex of order α in 𝕌, that is ∂/∂θ{arg(∂/∂θf(reiθ))}>α. Note that the classes Sℋ*(α) and ℋ𝒦(α) were introduced and studied by Jahangiri [5]. Also note that the class ℋ(n,1,α)≡ℋ𝒦(α) is the class of Salagean-type harmonic univalent functions introduced by Jahangiri et al. [13].We let the subclassℋ¯(n,β,α) consist of harmonic functions fn=h+gn¯ in ℋ(n,β,α) so h and g are of the form(2.7)hβ(z)=zβ-∑k=2∞|ak|zβ+k-1,gnβ(z)=(-1)n∑k=1∞|bk|zβ+k-1.In 1984, Clunie and Sheil-Small [14] investigated the class Sℋ as well as its geometric subclasses and obtained some coefficient bounds. Since then, there have been several related papers on Sℋ and its subclasses such that Silverman [15], Silverman and Silvia [16], and Jahangiri [5, 17] studied the harmonic univalent functions. Jahangiri [5] proved the following theorem.Theorem 2.1.
Letf=h+g¯ where  h=z+∑k=2∞akzk and  g=∑k=1∞bkzk. If
(2.8)∑k=1∞k-α1-α|ak|+k+α1-α|bk|≤2,(0≤α<1),
then f is sense-preserving, harmonic, and univalent in 𝕌 and f∈Sℋ*(α). The condition (2.8) is also necessary if f∈𝒯H(α)≡ℋ¯(0,1,α).In this paper, we will give the sufficient condition for functionsfβ=hβ+gβ¯ where hβ and gβ are given by (2.2) to be in the class ℋ(n,β,α) and it is shown that these coefficient conditions are also necessary for functions in the class ℋ¯(n,β,α). Also, we obtain distortion theorems and characterize the extreme points for functions in ℋ¯(n,β,α). Convolution and convex combination are also obtained.


## 3. Main Results

In this section, we prove the main results.


### 3.1. Coefficient Estimates

Theorem 3.1.
Letfβ=hβ+gβ¯, where  hβ and  gβ are given by (2.2). If
(3.1)∑k=1∞[(k-α)|ak|+(k+α)|bk|]βkn≤(1+β)(1-α),
where a1=1, n∈ℕ0, β≥1, and 0≤α<1, then fβ is sense-preserving, harmonic univalent in U, and f∈ℋ(n,β,α).Proof.
Ifz1β≠z2β, then
(3.2)|f(z1)β-f(z2)βh(z1)β-h(z2)β|≥1-|g(z1)β-g(z2)βh(z1)β-h(z2)β|=1-|∑k=1∞βbk(z1k+β-1-z2k+β-1)(z1β-z2β)+∑k=2∞βak(z1k+β-1-z2k+β-1)|>1-∑k=1∞(k+β-1)bk1-∑k=2∞(k+β-1)ak≥1-∑k=1∞(k+α)βkn/(1-α)|bk|1-∑k=2∞(k-α)βkn/(1-α)|ak|≥0,
which proves univalence. Note that f is sense-preserving in 𝕌. This is because
(3.3)|h′(z)β|≥β(|z|β-1-∑k=2∞(k+β-1)|ak||z|k+β-2)>β(1-∑k=2∞(k-α)βkn1-α|ak|)≥β(∑k=1∞(k+α)βkn1-α|bk|)≥∑k=1∞β(k+β-1)|bk||z|k+β-2≥|g′(z)β|.
By (2.6),
(3.4)Re{Dn+1f(z)βDnf(z)β}=Re{Dn+1h(z)β+(-1)n+1Dn+1g(z)β¯Dnh(z)β+(-1)nDng(z)β¯}>α.
Using the fact that Re(w)>α if and only if |1-α+w|≥|1+α-w|, it suffices to show that
(3.5)|1-α+Dn+1f(z)βDnf(z)β|-|1+α-Dn+1f(z)βDnf(z)β|≥0,(3.6)|Dn+1f(z)β+(1-α)Dnf(z)β|-|Dn+1f(z)β-(1+α)Dnf(z)β|≥0.
Substituting for Dn+1f(z)β, Dnf(z)β in (3.6), we have
(3.7)|Dn+1h(z)β+(-1)n+1Dn+1g(z)β¯+(1-α)[Dnh(z)β+(-1)nDng(z)β¯]|-|Dn+1h(z)β+(-1)n+1Dn+1g(z)β¯-(1+α)[Dnh(z)β+(-1)nDng(z)β¯]|=|zβ+∑k=2∞βkn+1akzβ+k-1+(-1)n+1∑k=1∞βkn+1bkzβ+k-1¯+(1-α)[zβ+∑k=2∞βknakzβ+k-1+(-1)n∑k=1∞βknbkzβ+k-1¯]|-|zβ+∑k=2∞βkn+1akzβ+k-1+(-1)n+1∑k=1∞βkn+1bkzβ+k-1¯-(1+α)[zβ+∑k=2∞βknakzβ+k-1+(-1)n∑k=1∞βknbkzβ+k-1¯]|=|(2-α)zβ+∑k=2∞β(k+1-α)knakzβ+k-1-(-1)n∑k=1∞β(k-1+α)knbkzβ+k-1|-|(-α)zβ+∑k=2∞β(k-1-α)knakzβ+k-1-(-1)n∑k=1∞β(k+1+α)knbkzβ+k-1|≥2(1-α)|z|β-∑k=2∞2βkn(k-α)|ak||zβ+k-1|-∑k=1∞2βkn(k-α)|bk||zβ+k-1|=2(1-α)[1-∑k=2∞βkn(k-α)1-α|ak|-∑k=1∞βkn(k+α)1-α|bk|].
This last expression is nonnegative by (3.1), and so the proof is complete.The harmonic function(3.8)f(z)β=zβ+∑k=2∞β1-α(k-α)βknxkzk+β-1+∑k=1∞β1-α(k+α)βknykzk+β-1¯,
where n∈ℕ0, β≥1, 0≤α<1, and ∑k=2∞|xk|+∑k=1∞|yk|=1, shows that the coefficient bound given by (3.1) is sharp. The functions of the form (3.8) are in ℋ(n,β,α) because (3.9)∑k=1∞[k-α1-α|ak|+k+α1-α|bk|]βkn=β+∑k=2∞|xk|+∑k=1∞|yk|=β+1.In the following theorem, it is shown that the condition (3.1) is also necessary for functions fnβ=hβ+gnβ¯ where hβ and gnβ are of the form (2.7).Theorem 3.2.
Letfnβ=hβ+gnβ¯ be given by (2.7). Then  fnβ∈ℋ¯(n,β,α), if and only if
(3.10)∑k=1∞[(k-α)|ak|+(k+α)|bk|]βkn≤(1+β)(1-α),
where a1=1, n∈ℕ0, β≥1, and 0≤α<1.Proof.
Sinceℋ¯(n,β,α)⊂ℋ(n,β,α), we only need to prove the “only if” part of the theorem. To this end, for functions fnβ of the form (2.7), we notice that the condition (2.6) is equivalent to
(3.11)Re{(1-α)zβ-∑k=2∞(k-α)βknakzk+β-1-(-1)2n∑k=1∞(k+α)βknbkzk+β-1¯zβ-∑k=2∞βknakzk+β-1+(-1)n∑k=1∞βknbkzk+β-1¯}≥0.
The above required condition (3.11) must hold for all values of z in 𝕌. Upon choosing the values of z on the positive real axis where 0≤z=r<1, we must have
(3.12)1-α-∑k=2∞(k-α)βknakrk-1-∑k=1∞(k+α)βknbkrk-11-∑k=2∞βknakrk-1+∑k=1∞βknbkrk-1¯≥0.
If the condition (3.10) does not hold, then the numerator in (3.12) is negative for r sufficiently close to 1. Hence there exist z0=r0 in (0,1) for which the quotient in (3.12) is negative. This contradicts the required condition for fnβ∈ℋ¯(n,λ,α) and so the proof is complete.


### 3.2. Distortion Bounds and Extreme Points

In this section, first we will obtain distortion bounds for functions inℋ¯(n,β,α).Theorem 3.3.
Letfnβ∈ℋ¯(n,β,α). Then for  |z|=r<1, we have
(3.13)|fn(z)β|≤(1+|b1|)rβ+1β2n(1-α2-α-1+α2-αβ|b1|)rβ+1,|fn(z)β|≥(1-|b1|)rβ-1β2n(1-α2-α-1+α2-αβ|b1|)rβ+1.Proof.
We only prove the right-hand inequality. The proof for the left-hand inequality is similar and will be omitted. Letfnβ∈ℋ¯(n,β,α). Taking the absolute value of fnβ, we obtain
(3.14)|fn(z)β|=|zβ+∑k=2∞akzk+β-1+(-1)n∑k=1∞bkzk+β-1¯|≤(1+|b1|)rβ+∑k=2∞(|ak|+|bk|)rk+β-1≤(1+|b1|)rβ+rβ+1∑k=2∞(|ak|+|bk|)≤(1+|b1|)rβ+1-α(2-α)β2n(∑k=2∞(2-α)β2n1-α|ak|+(2-α)β2n1-α|bk|)rβ+1≤(1+|b1|)rβ+1-α(2-α)β2n(∑k=2∞(k-α)βkn1-α|ak|+(k+α)βkn1-α|bk|)rβ+1≤(1+|b1|)rβ+1-α(2-α)β2n(1-1+α1-αβ|b1|)rβ+1,
for |b1|<1. This shows that the bounds given in Theorem 3.3 are sharp.The following covering result follows from the left-hand inequality in Theorem3.3.Corollary 3.4.
If functionfnβ=hβ+gβ¯, where hβ and gβ are given by (2.7), is in ℋ¯(n,β,α), then
(3.15){w:|w|<β2n+1-1-(β2n-1)αβ2n(2-α)-2n+1+12n(2-α)|b1|}⊂fn(U).Next we determine the extreme points of closed convex hulls ofℋ̅(n,β,α) denoted by clcoℋ¯(n,β,α).Theorem 3.5.
Letfnβ=hβ+gβ¯, where  hβ and  gβ are given by (2.7). Then  fnβ∈ℋ̅(n,β,α) if and only if
(3.16)fn(z)β=∑k=1∞(Xkhk(z)+Ykgnk(z)),
where h1(z)β=zβ, hk(z)β=zβ-(1-α)/((k-α)kn)zk+β-1(k=2,3,…), gnk(z)β=zβ+(-1)n(1-α)/((k+α)kn)zk+β-1¯(k=1,2,3,..), and  ∑k=1∞(Xk+Yk)=1,  Xk≥0, Yk≥0. In particular, the extreme points of ℋ¯(n,β,α) are {hk} and {gnk}.Proof.
For functionsfnβ=hβ+gβ¯, where hβ and gβ are given by (3.16), we have
(3.17)fn(z)β=∑k=1∞(Xkhk(z)+Ykgnk(z))=∑k=1∞(Xk+Yk)zβ-∑k=2∞1-α(k-α)knXkzk+β-1+(-1)n∑k=1∞1-α(k+α)knYkzk+β-1¯.
Then
(3.18)∑k=2∞(k-α)βkn1-α|ak|+∑k=1∞(k+α)βkn1-α|bk|=∑k=2∞Xk+∑k=1∞Yk=1-X1≤1,
and so fnβ∈clcoℋ¯(n,β,α).
Conversely, suppose thatfnβ∈clcoℋ¯(n,β,α). Setting
(3.19)Xk=(k-α)βkn1-α|ak|0≤Xk≤1(k=2,3,…),Yk=(k+α)βkn1-α|bk|0≤Yk≤1(k=1,2,3,…),
and X1=1-∑k=2∞Xk-∑k=1∞Yk; therefore, fnβ can be written as
(3.20)fn(z)β=zβ-∑k=2∞β|ak|zk+β-1+(-1)n∑k=1∞β|bk|zk+β-1¯=zβ-∑k=2∞(1-α)Xk(k-α)knzk+β-1+(-1)n∑k=1∞(1-α)Yk(k+α)knzk+β-1¯=zβ+∑k=2∞(hk(z)β-zβ)Xk+∑k=1∞(gnk(z)β-zβ)Yk=∑k=2∞hk(z)βXk+∑k=1∞gnk(z)βYk+zβ(1-∑k=2∞Xk-∑k=1∞Yk)=∑k=1∞(hk(z)βXk+gnk(z)βYk),asrequired.


### 3.3. Convolution and Convex Combination

In this section, we show that the classℋ¯(n,β,α) is invariant under convolution and convex combination of its member.For harmonic functionsfn(z)β=zβ-∑k=2∞|ak|zk+β-1+(-1)n∑k=1∞|bk|zk+β-1¯ and Fn(z)β=zβ-∑k=2∞|Ak|zk+β-1+(-1)n∑k=1∞|Bk|zk+β-1¯.The convolution offnβ and Fnβ is given by(3.21)(fnβ*Fnβ)(z)=fn(z)β*Fn(z)β=zβ-∑k=2∞|ak||Ak|zk+β-1+(-1)n∑k=1∞|bk||Bk|zk+β-1¯.Theorem 3.6.
For0≤λ≤α<1, let  fnβ∈ℋ¯(n,β,α) and  Fnβ∈ℋ¯(n,β,β). Then  fnβ*Fnβ∈ℋ¯(n,β,α)⊂ℋ¯(n,β,λ).Proof.
Let the functionsfn(z)β=zβ-∑k=2∞|ak|zk+β-1+(-1)n∑k=1∞|bk|zk+β-1¯ be in the class ℋ¯(n,β,α) and let the functions Fn(z)β=zβ-∑k=2∞|Ak|zk+β-1+(-1)n∑k=1∞|Bk|zk+β-1¯ be in the class ℋ¯(n,β,λ). Then the convolution fnβ*Fnβ is given by (3.21). We wish to show that the coefficients of fnβ*Fnβ satisfy the required condition given in Theorem 3.2. For Fnβ∈ℋ¯(n,β,λ), we note that |Ak|≤1 and |Bk|≤1. Now, for the convolution function fnβ*Fnβ, we obtain
(3.22)∑k=2∞(k-β)βkn1-β|ak||Ak|+∑k=1∞(k+β)βkn1-β|bk||Bk|≤∑k=2∞(k-β)βkn1-β|ak|+∑k=1∞(k+β)βkn1-β|bk|≤∑k=2∞(k-α)βkn1-α|ak|+∑k=1∞(k+α)βkn1-α|bk|≤1,
since 0≤λ≤α<1 and fnβ∈ℋ¯(n,β,α). Therefore, fnβ*Fnβ∈ℋ¯(n,β,α)⊂ℋ¯(n,β,λ).We now examine the convex combination ofℋ¯(n,β,α).Let the functionsfnj(z)β be defined, for j=1,2,…, by(3.23)fnj(z)β=zβ-∑k=2∞|ak,j|zk+β-1+(-1)n∑k=1∞|bk,j|zk+β-1¯.Theorem 3.7.
Let the functionsfnj(z)β defined by (3.23) be in the class  ℋ¯(n,β,α) for every  j=1,2,…,m. Then the functions  tj(z)β defined by
(3.24)tj(z)β=∑j=1mcjfnj(z)(0≤cj≤1)
are also in the class ℋ¯(n,β,α) where ∑j=1mcj=1.Proof.
According to the definition oftβ, we can write
(3.25)t(z)β=zβ-∑k=2∞(∑j=1mcjak,j)zk+β-1+(-1)n∑k=1∞(∑j=1mcjbn,j)zk+β-1¯.
Further, since fnj(z)β are in ℋ¯(n,β,α) for every (j=1,2,…), then by (3.1) we have
(3.26)∑k=1∞{[(k-α)(∑j=1mcj|ak,j|)+(k+α)(∑j=1mcj|bk,j|)]βkn}=∑j=1mcj(∑k=1∞[(k-α)|an,j|+(k+α)|bn,j|]βkn)≤∑j=1mcj2(1-α)≤2(1-α).
Hence the theorem follows.Corollary 3.8.
The classℋ¯(n,β,α) is close under convex linear combination.Proof.
Let the functionsfnj(z)β(j=1,2) defined by (3.23) be in the class Mℋ¯(n,λ,α). Then the function Ψ(z)β defined by
(3.27)Ψ(z)β=μfn1(z)β+(1-μ)fn2(z)β(0≤μ≤1)
is in the class Mℋ¯(n,λ,α). Also, by taking m=2, t1=μ, and t2=(1-μ) in Theorem 3.7, we have the above corollary.


## 3.1. Coefficient Estimates

Theorem 3.1.
Letfβ=hβ+gβ¯, where  hβ and  gβ are given by (2.2). If
(3.1)∑k=1∞[(k-α)|ak|+(k+α)|bk|]βkn≤(1+β)(1-α),
where a1=1, n∈ℕ0, β≥1, and 0≤α<1, then fβ is sense-preserving, harmonic univalent in U, and f∈ℋ(n,β,α).Proof.
Ifz1β≠z2β, then
(3.2)|f(z1)β-f(z2)βh(z1)β-h(z2)β|≥1-|g(z1)β-g(z2)βh(z1)β-h(z2)β|=1-|∑k=1∞βbk(z1k+β-1-z2k+β-1)(z1β-z2β)+∑k=2∞βak(z1k+β-1-z2k+β-1)|>1-∑k=1∞(k+β-1)bk1-∑k=2∞(k+β-1)ak≥1-∑k=1∞(k+α)βkn/(1-α)|bk|1-∑k=2∞(k-α)βkn/(1-α)|ak|≥0,
which proves univalence. Note that f is sense-preserving in 𝕌. This is because
(3.3)|h′(z)β|≥β(|z|β-1-∑k=2∞(k+β-1)|ak||z|k+β-2)>β(1-∑k=2∞(k-α)βkn1-α|ak|)≥β(∑k=1∞(k+α)βkn1-α|bk|)≥∑k=1∞β(k+β-1)|bk||z|k+β-2≥|g′(z)β|.
By (2.6),
(3.4)Re{Dn+1f(z)βDnf(z)β}=Re{Dn+1h(z)β+(-1)n+1Dn+1g(z)β¯Dnh(z)β+(-1)nDng(z)β¯}>α.
Using the fact that Re(w)>α if and only if |1-α+w|≥|1+α-w|, it suffices to show that
(3.5)|1-α+Dn+1f(z)βDnf(z)β|-|1+α-Dn+1f(z)βDnf(z)β|≥0,(3.6)|Dn+1f(z)β+(1-α)Dnf(z)β|-|Dn+1f(z)β-(1+α)Dnf(z)β|≥0.
Substituting for Dn+1f(z)β, Dnf(z)β in (3.6), we have
(3.7)|Dn+1h(z)β+(-1)n+1Dn+1g(z)β¯+(1-α)[Dnh(z)β+(-1)nDng(z)β¯]|-|Dn+1h(z)β+(-1)n+1Dn+1g(z)β¯-(1+α)[Dnh(z)β+(-1)nDng(z)β¯]|=|zβ+∑k=2∞βkn+1akzβ+k-1+(-1)n+1∑k=1∞βkn+1bkzβ+k-1¯+(1-α)[zβ+∑k=2∞βknakzβ+k-1+(-1)n∑k=1∞βknbkzβ+k-1¯]|-|zβ+∑k=2∞βkn+1akzβ+k-1+(-1)n+1∑k=1∞βkn+1bkzβ+k-1¯-(1+α)[zβ+∑k=2∞βknakzβ+k-1+(-1)n∑k=1∞βknbkzβ+k-1¯]|=|(2-α)zβ+∑k=2∞β(k+1-α)knakzβ+k-1-(-1)n∑k=1∞β(k-1+α)knbkzβ+k-1|-|(-α)zβ+∑k=2∞β(k-1-α)knakzβ+k-1-(-1)n∑k=1∞β(k+1+α)knbkzβ+k-1|≥2(1-α)|z|β-∑k=2∞2βkn(k-α)|ak||zβ+k-1|-∑k=1∞2βkn(k-α)|bk||zβ+k-1|=2(1-α)[1-∑k=2∞βkn(k-α)1-α|ak|-∑k=1∞βkn(k+α)1-α|bk|].
This last expression is nonnegative by (3.1), and so the proof is complete.The harmonic function(3.8)f(z)β=zβ+∑k=2∞β1-α(k-α)βknxkzk+β-1+∑k=1∞β1-α(k+α)βknykzk+β-1¯,
where n∈ℕ0, β≥1, 0≤α<1, and ∑k=2∞|xk|+∑k=1∞|yk|=1, shows that the coefficient bound given by (3.1) is sharp. The functions of the form (3.8) are in ℋ(n,β,α) because (3.9)∑k=1∞[k-α1-α|ak|+k+α1-α|bk|]βkn=β+∑k=2∞|xk|+∑k=1∞|yk|=β+1.In the following theorem, it is shown that the condition (3.1) is also necessary for functions fnβ=hβ+gnβ¯ where hβ and gnβ are of the form (2.7).Theorem 3.2.
Letfnβ=hβ+gnβ¯ be given by (2.7). Then  fnβ∈ℋ¯(n,β,α), if and only if
(3.10)∑k=1∞[(k-α)|ak|+(k+α)|bk|]βkn≤(1+β)(1-α),
where a1=1, n∈ℕ0, β≥1, and 0≤α<1.Proof.
Sinceℋ¯(n,β,α)⊂ℋ(n,β,α), we only need to prove the “only if” part of the theorem. To this end, for functions fnβ of the form (2.7), we notice that the condition (2.6) is equivalent to
(3.11)Re{(1-α)zβ-∑k=2∞(k-α)βknakzk+β-1-(-1)2n∑k=1∞(k+α)βknbkzk+β-1¯zβ-∑k=2∞βknakzk+β-1+(-1)n∑k=1∞βknbkzk+β-1¯}≥0.
The above required condition (3.11) must hold for all values of z in 𝕌. Upon choosing the values of z on the positive real axis where 0≤z=r<1, we must have
(3.12)1-α-∑k=2∞(k-α)βknakrk-1-∑k=1∞(k+α)βknbkrk-11-∑k=2∞βknakrk-1+∑k=1∞βknbkrk-1¯≥0.
If the condition (3.10) does not hold, then the numerator in (3.12) is negative for r sufficiently close to 1. Hence there exist z0=r0 in (0,1) for which the quotient in (3.12) is negative. This contradicts the required condition for fnβ∈ℋ¯(n,λ,α) and so the proof is complete.


## 3.2. Distortion Bounds and Extreme Points

In this section, first we will obtain distortion bounds for functions inℋ¯(n,β,α).Theorem 3.3.
Letfnβ∈ℋ¯(n,β,α). Then for  |z|=r<1, we have
(3.13)|fn(z)β|≤(1+|b1|)rβ+1β2n(1-α2-α-1+α2-αβ|b1|)rβ+1,|fn(z)β|≥(1-|b1|)rβ-1β2n(1-α2-α-1+α2-αβ|b1|)rβ+1.Proof.
We only prove the right-hand inequality. The proof for the left-hand inequality is similar and will be omitted. Letfnβ∈ℋ¯(n,β,α). Taking the absolute value of fnβ, we obtain
(3.14)|fn(z)β|=|zβ+∑k=2∞akzk+β-1+(-1)n∑k=1∞bkzk+β-1¯|≤(1+|b1|)rβ+∑k=2∞(|ak|+|bk|)rk+β-1≤(1+|b1|)rβ+rβ+1∑k=2∞(|ak|+|bk|)≤(1+|b1|)rβ+1-α(2-α)β2n(∑k=2∞(2-α)β2n1-α|ak|+(2-α)β2n1-α|bk|)rβ+1≤(1+|b1|)rβ+1-α(2-α)β2n(∑k=2∞(k-α)βkn1-α|ak|+(k+α)βkn1-α|bk|)rβ+1≤(1+|b1|)rβ+1-α(2-α)β2n(1-1+α1-αβ|b1|)rβ+1,
for |b1|<1. This shows that the bounds given in Theorem 3.3 are sharp.The following covering result follows from the left-hand inequality in Theorem3.3.Corollary 3.4.
If functionfnβ=hβ+gβ¯, where hβ and gβ are given by (2.7), is in ℋ¯(n,β,α), then
(3.15){w:|w|<β2n+1-1-(β2n-1)αβ2n(2-α)-2n+1+12n(2-α)|b1|}⊂fn(U).Next we determine the extreme points of closed convex hulls ofℋ̅(n,β,α) denoted by clcoℋ¯(n,β,α).Theorem 3.5.
Letfnβ=hβ+gβ¯, where  hβ and  gβ are given by (2.7). Then  fnβ∈ℋ̅(n,β,α) if and only if
(3.16)fn(z)β=∑k=1∞(Xkhk(z)+Ykgnk(z)),
where h1(z)β=zβ, hk(z)β=zβ-(1-α)/((k-α)kn)zk+β-1(k=2,3,…), gnk(z)β=zβ+(-1)n(1-α)/((k+α)kn)zk+β-1¯(k=1,2,3,..), and  ∑k=1∞(Xk+Yk)=1,  Xk≥0, Yk≥0. In particular, the extreme points of ℋ¯(n,β,α) are {hk} and {gnk}.Proof.
For functionsfnβ=hβ+gβ¯, where hβ and gβ are given by (3.16), we have
(3.17)fn(z)β=∑k=1∞(Xkhk(z)+Ykgnk(z))=∑k=1∞(Xk+Yk)zβ-∑k=2∞1-α(k-α)knXkzk+β-1+(-1)n∑k=1∞1-α(k+α)knYkzk+β-1¯.
Then
(3.18)∑k=2∞(k-α)βkn1-α|ak|+∑k=1∞(k+α)βkn1-α|bk|=∑k=2∞Xk+∑k=1∞Yk=1-X1≤1,
and so fnβ∈clcoℋ¯(n,β,α).
Conversely, suppose thatfnβ∈clcoℋ¯(n,β,α). Setting
(3.19)Xk=(k-α)βkn1-α|ak|0≤Xk≤1(k=2,3,…),Yk=(k+α)βkn1-α|bk|0≤Yk≤1(k=1,2,3,…),
and X1=1-∑k=2∞Xk-∑k=1∞Yk; therefore, fnβ can be written as
(3.20)fn(z)β=zβ-∑k=2∞β|ak|zk+β-1+(-1)n∑k=1∞β|bk|zk+β-1¯=zβ-∑k=2∞(1-α)Xk(k-α)knzk+β-1+(-1)n∑k=1∞(1-α)Yk(k+α)knzk+β-1¯=zβ+∑k=2∞(hk(z)β-zβ)Xk+∑k=1∞(gnk(z)β-zβ)Yk=∑k=2∞hk(z)βXk+∑k=1∞gnk(z)βYk+zβ(1-∑k=2∞Xk-∑k=1∞Yk)=∑k=1∞(hk(z)βXk+gnk(z)βYk),asrequired.


## 3.3. Convolution and Convex Combination

In this section, we show that the classℋ¯(n,β,α) is invariant under convolution and convex combination of its member.For harmonic functionsfn(z)β=zβ-∑k=2∞|ak|zk+β-1+(-1)n∑k=1∞|bk|zk+β-1¯ and Fn(z)β=zβ-∑k=2∞|Ak|zk+β-1+(-1)n∑k=1∞|Bk|zk+β-1¯.The convolution offnβ and Fnβ is given by(3.21)(fnβ*Fnβ)(z)=fn(z)β*Fn(z)β=zβ-∑k=2∞|ak||Ak|zk+β-1+(-1)n∑k=1∞|bk||Bk|zk+β-1¯.Theorem 3.6.
For0≤λ≤α<1, let  fnβ∈ℋ¯(n,β,α) and  Fnβ∈ℋ¯(n,β,β). Then  fnβ*Fnβ∈ℋ¯(n,β,α)⊂ℋ¯(n,β,λ).Proof.
Let the functionsfn(z)β=zβ-∑k=2∞|ak|zk+β-1+(-1)n∑k=1∞|bk|zk+β-1¯ be in the class ℋ¯(n,β,α) and let the functions Fn(z)β=zβ-∑k=2∞|Ak|zk+β-1+(-1)n∑k=1∞|Bk|zk+β-1¯ be in the class ℋ¯(n,β,λ). Then the convolution fnβ*Fnβ is given by (3.21). We wish to show that the coefficients of fnβ*Fnβ satisfy the required condition given in Theorem 3.2. For Fnβ∈ℋ¯(n,β,λ), we note that |Ak|≤1 and |Bk|≤1. Now, for the convolution function fnβ*Fnβ, we obtain
(3.22)∑k=2∞(k-β)βkn1-β|ak||Ak|+∑k=1∞(k+β)βkn1-β|bk||Bk|≤∑k=2∞(k-β)βkn1-β|ak|+∑k=1∞(k+β)βkn1-β|bk|≤∑k=2∞(k-α)βkn1-α|ak|+∑k=1∞(k+α)βkn1-α|bk|≤1,
since 0≤λ≤α<1 and fnβ∈ℋ¯(n,β,α). Therefore, fnβ*Fnβ∈ℋ¯(n,β,α)⊂ℋ¯(n,β,λ).We now examine the convex combination ofℋ¯(n,β,α).Let the functionsfnj(z)β be defined, for j=1,2,…, by(3.23)fnj(z)β=zβ-∑k=2∞|ak,j|zk+β-1+(-1)n∑k=1∞|bk,j|zk+β-1¯.Theorem 3.7.
Let the functionsfnj(z)β defined by (3.23) be in the class  ℋ¯(n,β,α) for every  j=1,2,…,m. Then the functions  tj(z)β defined by
(3.24)tj(z)β=∑j=1mcjfnj(z)(0≤cj≤1)
are also in the class ℋ¯(n,β,α) where ∑j=1mcj=1.Proof.
According to the definition oftβ, we can write
(3.25)t(z)β=zβ-∑k=2∞(∑j=1mcjak,j)zk+β-1+(-1)n∑k=1∞(∑j=1mcjbn,j)zk+β-1¯.
Further, since fnj(z)β are in ℋ¯(n,β,α) for every (j=1,2,…), then by (3.1) we have
(3.26)∑k=1∞{[(k-α)(∑j=1mcj|ak,j|)+(k+α)(∑j=1mcj|bk,j|)]βkn}=∑j=1mcj(∑k=1∞[(k-α)|an,j|+(k+α)|bn,j|]βkn)≤∑j=1mcj2(1-α)≤2(1-α).
Hence the theorem follows.Corollary 3.8.
The classℋ¯(n,β,α) is close under convex linear combination.Proof.
Let the functionsfnj(z)β(j=1,2) defined by (3.23) be in the class Mℋ¯(n,λ,α). Then the function Ψ(z)β defined by
(3.27)Ψ(z)β=μfn1(z)β+(1-μ)fn2(z)β(0≤μ≤1)
is in the class Mℋ¯(n,λ,α). Also, by taking m=2, t1=μ, and t2=(1-μ) in Theorem 3.7, we have the above corollary.



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*Source: 821531-2011-02-06.xml*