# A Subclass of Analytic Functions Related tok-Uniformly Convex and Starlike Functions

Authors: Saqib Hussain; Akhter Rasheed; Muhammad Asad Zaighum; Maslina Darus
Journal: Journal of Function Spaces
 (2017)

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## Abstract

We investigate some subclasses ofk-uniformly convex and k-uniformly starlike functions in open unit disc, which is generalization of class of convex and starlike functions. Some coefficient inequalities, a distortion theorem, the radii of close-to-convexity, and starlikeness and convexity for these classes of functions are studied. The behavior of these classes under a certain modified convolution operator is also discussed.


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## Body

## 1. Introduction

LetA be the class of all analytic functions f in open unit disc Δ={z:z<1}, normalized by f0=0 and f′0=1. Thus, any f∈A has the following Maclaurin’s series:(1)fz=z+∑n=2∞anzn.A functionf is said to be univalent if it never takes same value twice. By S we mean the subclass of A which is composed of univalent functions. By ST and CV we mean the well-known subclasses of A that are, respectively, starlike and convex.In 1991, Goodman [1, 2] introduced the classes UCV and UST of uniformly convex and uniformly starlike functions, respectively. A function f∈S is uniformly convex if fz maps every circular arc γ contained in Δ with center ζ∈Δ onto a convex arc. The function f∈S is uniformly starlike if fz maps every circular arc γ contained in Δ with center ζ∈Δ onto a starlike arc with respect to f(ζ). A more useful representation of UCV and UST was given in [3–6] as(2)f∈UCV⟺f∈A,Re⁡zf′z′f′z>zf′′zf′z,z∈Δ.f∈UST⟺f∈A,Re⁡zf′zfz>zf′zfz-1,z∈Δ.In 1999, fork≥0, Kanas and Wisniowska [7] introduced the class k-UCV and k-UST as(3)f∈k-UCV⟺zf′∈k-UST⟺f∈A,Re⁡zf′z′f′z>kzf′′zf′z,z∈Δ.Observe that0-UCV≡CV, 0-UST≡UST and 1-UCV≡UCV, 1-UST≡UST.For fixedk≥0, these classes have a nice geometrical representation; for detail see [7–9].A lot of authors obtain very useful properties ofUCV and UST and their generalization in several direction; for example, see [1, 2, 7, 8, 10, 11] and reference cited therein.For0≤α<1, in [4] (see also [12]), Ronning introduced the following two important subclasses k-USTα and k-UCVα as(4)f∈k-USTα⟺Re⁡zf′zfz-α>kzf′zfz-1,z∈Δ.f∈k-UCVα⟺zf′∈k-USTα.Recently in [13] El-Ashwah et al. introduced two important subclass k-UCVα,β and k-USTα,β of k-uniformly convex starlike functions as(5)f∈k-UCVα,β⟺Re⁡zf′z′f′z-α>kzf′z′f′z-β,z∈Δ,f∈k-USTα,β⟺Re⁡zf′zfz-α>kzf′zfz-β,z∈Δ,where 0≤α<β≤1 and k1-β<1-α.Letfjj=1,2,… be defined by(6)fjz=z+∑n=2∞an,jzn,an,j≥0;j∈N;then the modified Hadmard product of f1z and f2z is defined by(7)f1∗f2z=z-∑n=2∞an,1an,2zn.We denoteT by subclass of S consisting of functions having all negative coefficients in their Maclaurin’s series expansions, so any f∈T has a series of the form:(8)fz=z-∑n=2∞anzn,z∈Δ.LetVη be the class of functions f∈S given in (1) for which arg⁡an=π+n-1η, n≥2. Note that V0=T [11].In recent years, more and more researchers are interested in the above defined classes (see [9, 11, 14–22]).In this paper, by taking inspiration from the above cited paper, we introduce some new subclasses of analytic functions and obtain some interesting results.Definition 1.
For0≤α<β≤1, 0≤δ<1, k1-β<1-α, and 0≤λ<1, a function f∈S is in class k-Uα,β,λ,δ if and only if(9)Re⁡1-δzf′+δzf′+1+2λz2f′′+λz3f′′′1-δf+δzf′+λz2f′′-α≥k1-δzf′+δzf′+1+2λz2f′′+λz3f′′′1-δf+δzf′+λz2f′′-β,z∈Δ.Also(10)k-VUηα,β,λ,δ=k-Uα,β,λ,δ∩Vη.It is worth mentioning that, for special values of parameters, these classes were extensively studied by many authors; here we mention few of them.(1)
k-Uα,β,λ,1=k-Uλ,β,α [21].(2)
k-VU0α,β,λ,1=k-VUηλ,β,α [21].(3)
0-VU0α,1,0,1=CVα [11].(4)
k-VU0α,1,0,1=k-UCVα [23].(5)
1-Uα,1,0,1=UCVα [4].(6)
k-Uα,β,0,0=k-USTα,β [13].Throughout the paper-1≤α≤β≤1, 0≤λ<1, 1-α>k1-β, and z∈Δ, unless otherwise stated.


## 2. Main Results

Theorem 2.
A functionf(z) given by (1) is in class k-Uα,β,λ,δ if(11)∑n=2∞Πn1+k-kβ+αΩnan≤1-α-k1-β,where(12)Πn=n+δnn-11+λn,Ωn=1-δ+δn1+λn-1.Proof.
It is sufficient to prove that inequality (9) holds. As we know(13)Re⁡w>kw-β+αiffRe⁡1+keiθw-βkeiθ≥α;then inequality (9) can be written as (14)Re⁡1+keiθ1-δzf′+δzf′+1+2λz2f′′+λz3f′′′1-δf+δzf′+λz2f′′-βkeiθ≥α.This is, (15)Re⁡AzBz≥α,where (16)Az=1+keiθzf′+δ1+2λz2f′′+λz3f′′′-βkeiθ1-δf+δzf′+λz2f′′,Bz=1-δf+δzf′+λz2f′′;then we have(17)Az+1-αBz-Az-1+αBz≥0.Now(18)Az+1-αBz=1-βkeiθ+2-αz-∑n=2∞βΩn-Πnkeiθ-1-αΩn-Πnanzn≥-1-βk+2-αz-∑n=2∞βΩn-Πnk+1-αΩn+Πnanzn,and also(19)Az-1+αBz=1-βkeiθ-αz+∑n=2∞Πn-βΩnkeiθ-1+αΩn+Πnanzn≤1-βk+αz+∑n=2∞Πn-βΩnk-1+αΩn+Πnanzn.From (18) and (19), we have (20)Az+1-αBz-Az-1+αBz≥21-α-k1-βz-2∑n=2∞Πn-βΩnk+Πn-αΩnanzn=21-α-k1-βz-∑n=2∞Πn1+k-kβ+αΩnanzn.The last expression is bounded below by 0 if (21)∑n=2∞Πn1+k-kβ+αΩnan≤1-α-k1-β,which completes the proof.In the next theorem, we prove that condition (11) is also necessary for function f∈k-Uα,β,λ,δ.Theorem 3.
Letfz be given by (1) and in Vη; then f∈k-VUηα,β,λ,δ if and only if(22)∑n=2∞Πn1+k-kβ+αΩnan≤1-α-k1-β.Proof.
From Theorem2, we need only to show that f∈k-VUηα,β,λ,δ satisfies inequality (22). If f∈k-VUηα,β,λ,δ, then by definition, we have (23)Re⁡1-α+∑n=2∞Πn-αΩnanzn-11+∑n=2∞Ωnanzn-1≥k1-β+∑n=2∞Πn-βΩnanzn-11+∑n=2∞Ωnanzn-1.
Sincef is function of form (1) with the argument property given in class Vη and letting z=reθ in the above inequality, we have(24)1-α-∑n=2∞Πn-αΩnanrn-11-∑n=2∞Ωnanrn-1≥k1-β-∑n=2∞Πn-βΩnanrn-11-∑n=2∞Ωnanrn-1for r→1, and (24) leads to require inequality (25)∑n=2∞Πn1+k-kβ+αΩnan≤1-α-k1-β.
The function(26)fn,ηz=z-1-α-k1-βei1-nηΠn1+k-kβ+αΩnzn,0≤η≤2π,n≥2is extremal function.Corollary 4.
Letfz given in (1) be in class k-VUηα,β,λ,δ. Then(27)an≤1-α-k1-βΠn1+k-kβ+αΩn,n≥2.Inequality (27) is attained for the function given in (26).Theorem 5.
Let the functionfz given in (1) be in class k-VUηα,β,λ,δ. Then for z<r=1(28)fz≥r-1-α-k1-βΠ21+k-kβ+αΩ2r2,fz≤r+1-α-k1-βΠ21+k-kβ+αΩ2r2.The results in (28) are attained for the function given in (26) for z=±r.Proof.
As we know from Theorem3(29)Π21+k-kβ+αΩ2∑n=2∞an≤∑n=2∞Πn1+k-kβ+αΩnan≤1-α-k1-β.As(30)fz≥z-∑n=2∞anzn≥r-r2∑n=2∞an≥r-1-α-k1-βΠ21+k-kβ+αΩ2r2,similarly (31)fz≤z+∑n=2∞anzn≤r+r2∑n=2∞an≤r+1-α-k1-βΠ21+k-kβ+αΩ2r2.
This completes the proof.Theorem 6.
Let the functionfz given in (1) be in class k-VUηα,β,λ,δ. Then for z<r=1(32)1-21-α-k1-βΠ21+k-kβ+αΩ2r≤fz≤1+21-α-k1-βΠ21+k-kβ+αΩ2r.Proof.
Forfz given by (1), we have(33)f′z≥1-∑n=2∞nanzn-1≥1-r∑n=2∞nan,f′z≤1+∑n=2∞nanzn-1≤1+r∑n=2∞nan.
In view of Theorem3,(34)Π21+k-kβ+αΩ22∑n=2∞nan≤∑n=2∞Πn1+k-kβ+αΩnan≤1-α-k1-β,or, equivalently,(35)∑n=2∞nan≤21-α-k1-βΠ21+k-kβ+αΩ2.
A substitution from (35) into (33) yields inequality (32), which is required.Theorem 7.
Letf∈k-VUηα,β,λ,δ with argument property as in class Vη. Define fjz=z and(36)fn,η=z-1-α-k1-βei1-nηΠn1+k-kβ+αΩnzn,where 0≤η≤2π, n≥2.
Then functionfz is in class k-VUηα,β,λ,δ if and only if it can be expressed as(37)fz=∑n=1∞μnfn,η,where μn≥0n≥1 and ∑n=1∞μn=1.Proof.
Assume that(38)fz=μ1f1z+∑n=2∞μnz-1-α-k1-βei1-nηΠn1+k-kβ+αΩnzn=∑n=1∞μnz-∑n=2∞1-α-k1-βei1-nηΠn1+k-kβ+αΩnμnzn.
Then it follows that(39)∑n=2∞1-α-k1-βei1-nηΠn1+k-kβ+αΩnμnΠn1+k-kβ+αΩn=∑n=2∞μn1-α-k1-β≤1-μ11-α-k1-β≤1-α-k1-β,by Theorem 3, f∈k-VUηα,β,λ,δ. Conversely, assume that the function fz defined by (1) belongs to class k-VUηα,β,λ,δ, and then (40)an≤1-α-k1-βΠn1+k-kβ+αΩn,n≥2.
Set(41)μn=Πn1+k-kβ+αΩn1-α-k1-βan,n≥2,and μ1=1-∑n=2∞μn, n≥2. Then fz=∑n=1∞μnfn,η and this completes the proof.Theorem 8.
Letf∈k-VUηα,β,λ,δ. Then fz is close to convex of order σ0≤σ<1 in the disc z<r1, where(42)r1=inf⁡1-σΠn1+k-kβ+αΩnn1-α-k1-β1/n-1,n≥2.Proof.
Asf∈Vη, where f is close to convex of order σ, we have(43)f′z-1<1-σ,as (44)f′z-1≤∑n=2∞nanzn-1,this expression is less than 1-σ if (45)∑n=2∞n1-σanzn-1<1.
By the fact thatf∈k-VUηα,β,λ,δ if and only if (46)∑n=2∞Πn1+k-kβ+αΩn1-α-k1-βan≤1;inequality (43) is true if (47)n1-σzn-1≤Πn1+k-kβ+αΩn1-α-k1-β;or, equivalently, (48)zn-1=1-σΠn1+k-kβ+αΩnn1-α-k1-β.Theorem 9.
Letf∈k-VUηα,β,λ,δ. Then fz is close to convex of order σ0≤σ<1 in the disc⁡z<r2, where(49)r2=inf⁡1-σΠn1+k-kβ+αΩnn-σ1-α-k1-β1/n-1,n≥2.Proof.
Asf∈Vη and f is starlike of order σ, then we have(50)zf′zfz-1<1-σ,
as(51)zf′zfz-1≤∑n=2∞n-1anzn-11-∑n=2∞anzn-1.
The last expression is less than1-σ if (52)∑n=2∞n-σ1-σanzn-1<1.
Using the fact thatf∈k-VUηα,β,λ,δ if and only if (53)∑n=2∞Πn1+k-kβ+αΩn1-α-k1-βan≤1(50) is true if (54)n-σ1-σzn-1<Πn1+k-kβ+αΩn1-α-k1-β.
Or equivalently(55)zn-1=1-σΠn1+k-kβ+αΩnn-σ1-α-k1-β,which is required.Theorem 10.
Letf∈k-VUηα,β,λ,δ. Then fz is convex of order σ0≤σ<1 in the disc⁡z<r3, where(56)r3=inf⁡1-σΠn1+k-kβ+αΩnnn-σ1-α-k1-β1/n-1,n≥2.Proof.
Using the fact thatf is convex if and only if zf′ is starlike, following the lines of Theorem 9, we have the required results.Theorem 11.
Letfjzj=1,2,… given by (6) be in class k-VUηα,β,λ,δ. Then f1∗f2∈k-VUηϕ1,λ,δ, for(57)ϕ1=Πn1+k-kβ+αΩn2-Πn1+k-kβΩn1-α-k1-β2Πn1+k-kβ+αΩn2-Ωn1-α-k1-β2.Proof.
We need to prove the largestϕ1 such that (58)Πn1+k-kβ+αΩn1-ϕ1-k1-βan,1an,2≤1.
From Theorem3, we have (59)∑n=2∞Πn1+k-kβ+αΩn1-α-k1-βan,1≤1,∑n=2∞Πn1+k-kβ+αΩn1-α-k1-βan,2≤1.
By Cauchy-Schwarz inequality, we have(60)∑n=2∞Πn1+k-kβ+αΩn1-α-k1-βan,1an,2≤1.Thus, it is sufficient to show(61)Πn1+k-kβ+ϕ1Ωn1-ϕ1-k1-βan,1an,2≤Πn1+k-kβ+αΩn1-α-k1-βan,1an,2,n≥2.
Forn≥2(62)an,1an,2≤Πn1+k-kβ+αΩn1-ϕ1-k1-βΠn1+k-kβ+ϕ1Ωn1-α-k1-β.
Note that(63)an,1an,2≤1-α-k1-βΠn1+k-kβ+αΩn.We need to show(64)1-α-k1-βΠn1+k-kβ+αΩn≤Πn1+k-kβ+αΩn1-ϕ1-k1-βΠn1+k-kβ+ϕ1Ωn1-α-k1-β,or equivalently(65)ϕ1≤Πn1+k-kβ+αΩn2-Πn1+k-kβΩn1-α-k1-β2Πn1+k-kβ+αΩn2-Ωn1-α-k1-β2=ωn.
ω
n is an increasing function for n≥2. For n=2 in (65),(66)ϕ1≤ω2=Π21+k-kβ+αΩ22-Π21+k-kβΩ21-α-k1-β2Π21+k-kβ+αΩ22-Ω21-α-k1-β2,which proves main assertion of Theorem 11.



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