# Multivalent Functions Related with an Integral Operator

Authors: Syed Ghoos Ali Shah; Saqib Hussain; Saima Noor; Maslina Darus; Ibrar Ahmad
Journal: International Journal of Mathematics and Mathematical Sciences
 (2021)

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## Abstract

In this present paper, we introduce and explore certain new classes of uniformly convex and starlike functions related to the Liu–Owa integral operator. We explore various properties and characteristics, such as coefficient estimates, rate of growth, distortion result, radii of close-to-convexity, starlikeness, convexity, and Hadamard product. It is important to mention that our results are a generalization of the number of existing results in the literature.


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## Body

## 1. Introduction

Letℂ denote the complex plane and assume that Ap denotes the class of p-valent function of the form(1)λω=ωp+∑t=1∞at+pωt+pp∈ℕ=1,2,…,which are analytic in the open unit disc ⋓=ω:ω∈ℂ and ω<1. Specially, for p=1, we denote A=A1.ByU, K, and S, the subclasses of Ap consist of all univalent, convex, and starlike functions Sα. We also denote Sα, the class of starlike function of order α, α∈01. In 1991, Goodman [1, 2] introduced the classes UST and UCV of uniformly starlike and uniformly convex functions, respectively. A function λ is uniformly starlike (uniformly convex) in ⋓ if λ is in USTUCV and has the property that, for every circular arc γ contained in ⋓, with center ζ also in ⋓, the arc λγ is starlike (convex) with respect to λζ. A more useful representation of UST and UCV was given in [3]; see [4, 5], for details:(2)λ∈UST⇔ωλ′ωλω−1≤ℜωλ′ωλω,ω∈⋓,and(3)λ∈UCV⇔ωλ′′ωλ′ω≤ℜ1+ωλ′′ωλ′ω,ω∈⋓.In 1999, fork≥0, Kanas and Wisniowska [6] introduced the classes k−UST and k−UCV of k-uniformly convex and k−uniformly starlike functions, respectively, see also [7–10].Letk−USTα,β denote the subclass of Ap consisting of functions of form (1) and satisfy the following inequality:(4)ℜωλ′ωλω−α>kωλ′ωλω−β,0≤α<β≤1;k1−β<1−α;ω∈⋓.Also, letk−UCVα,β denote the subclass of Ap consisting of functions of form (1) and satisfy the following inequality:(5)ℜ1+ωλ′′ωλ′ω−α>k1+ωλ′′ωλ′ω−β,0≤α<β≤1;k1−β<1−α;ω∈⋓.It follows from (4) and (5) that(6)k−UCVα,β⇔ωλ′∈k−USTα,β.Notice that,k−USTα,0=Sα and k−UCVα,0=Kα, for k=0. The convolution (Hadamard product) for two functions λ, δ∈Ap, is defined by(7)λω∗δω=ωp+∑t=0∞at+pbt+pωt+p,where λ is given in (1) and δω=ωp+∑t=0∞bt+pωt+p.Taking from the above cited work and using Liu–Owa integral operator, we introduce the following class ofp-valent analytic function. In 2004, Liu and Owa [11] (see also [12–14]) introduced the integral operator Qb,pa: Ap⟶Ap as follows:(8)Qb,paλω=p+a+b−1p+b−1aωb∫0ω1−xωa−1xb−1λxdxa>0;b>−1;p∈ℕand(9)Qb,p0λω=λωa=0;b>−1.Forλ∈Ap, given by (1), and using properties of gamma function, we have(10)Qb,paλω=ωp+Γa+b+pΓb+p∑t=1∞Γb+p+tΓa+b+p+tat+pωt+pa≥0;b>−1;p∈ℕ.Definition 1.
For0≤α<β≤1, 0≤ν<1, k1−β<1−α, and 0≤μ<1, a function λ∈Ap is in class k−Ua,b,p,α,β,μ,ν if and only if(11)ℜ1−νωQb,paλω′+νωQb,paλω′+1+2μω2Qb,paλω′′+μω3Qb,paλω′′′1−νQb,paλω+νωQb,paλω′+μω2Qb,paλω′′−α≥k1−νωQb,paλω′+νωQb,paλω′+1+2μω2Qb,paλω′′+μω3Qb,paλω′′′1−νQb,paλω+νωQb,paλω′+μω2Qb,paλω′′−β.
We also denotek−£Uηa,b,p,α,β,μ,ν=k−Ua,b,p,α,β,μ,ν∩£η, where £η the class of functions λ∈Ap of form (1) for which argat=π+n−1η. For more details, see [15–20].


### 1.1. Special Cases

Specializing parameters,a,b,p,α,β,μ, and ν, we obtain the following subclasses studied by various authors:(1)k−U0,b,p,α,β,μ,1=k−Uα,β,μ [21](2)k−£U00,b,p,α,β,μ,1=k−£Uηα,β,μ,ν [21](3)0−£U00,b,p,α,1,0,1=CVα [22](4)k−£U00,b,p,α,1,0,1=k−UCVα [23](5)1−U0,b,p,α,1,0,1=UCVα [24](6)k−U0,b,p,α,β,0,0=k−USTα,β [25]


## 1.1. Special Cases

Specializing parameters,a,b,p,α,β,μ, and ν, we obtain the following subclasses studied by various authors:(1)k−U0,b,p,α,β,μ,1=k−Uα,β,μ [21](2)k−£U00,b,p,α,β,μ,1=k−£Uηα,β,μ,ν [21](3)0−£U00,b,p,α,1,0,1=CVα [22](4)k−£U00,b,p,α,1,0,1=k−UCVα [23](5)1−U0,b,p,α,1,0,1=UCVα [24](6)k−U0,b,p,α,β,0,0=k−USTα,β [25]


## 2. Main Results for the Classk−Ua,b,p,α,β,μ,ν

### 2.1. Coefficient Estimates

In this section, we obtain a necessary and sufficient condition for functionsλω in the classes k−£Uηa,b,p,α,β,μ,ν.Theorem 1.
A functionλω given by (1) is in the class k−Ua,b,p,α,β,μ,ν if(12)Γa+b+pΓb+p∑t=1∞Dt+p1+k−Ct+pα+kβΓb+p+tΓa+b+p+tat+p≤Cp1+k−Dpα+kβ,where(13)Ct+p=1−ν+νt+p1+μt+p−1,(14)Dt+p=t+p+νt+pt+p−11+μt+p,(15)Cp=p+νpp−11+μp,(16)Dp=1−ν+νp1+μp−1,and(17)−1≤α<β≤1,0≤μ<1,k1−β<1−α,a≥0,b>−1,p∈ℕandω∈⋓.Proof.
It suffices to show that inequality (11) holds true. As we know,(18)ℜϖ>kϖ−β+α⇔ℜ1+keiθϖ−βkeiθ>α.
Then, inequality (11) may be written as(19)ℜ1+keiθ1−νωQb,paλω′+νωQb,paλω′+1+2μω2Qb,paλω″+μω3Qb,paλω″′1−νQb,paλω+νωQb,paλω′+μω2Qb,paλω″−βkeiθ≥α,which can be written as ℜAω/Bω≥α, where(20)Aω=1+keiθ1−νωQb,paλω′+νωQb,paλω′+1+2μω2Qb,paλω′′+μω3Qb,paλω′′′−βkeiθ1−νQb,paλω+νωQb,paλω′+μω2Qb,paλω′′and(21)Bω=1−νQb,paλω+νωQb,paλω′+μω2Qb,paλω′′.
Then, we have(22)Aω+1−αBω−Aω−1+αBω≥0.
Now,(23)Aω+1−αBωCp+1−αDp+keiθCp−βDpωp−Γa+b+pΓb+p×∑t=1∞keiθβCt+p−Dt+p−Dt+p−1−αCt+p×Γb+p+tΓa+b+p+tat+pωt+p≥keiθβDp−Cp−Cp−1−αDpωp−Γa+b+pΓb+p×∑t=1∞keiθβCt+p−Dt+p−Dt+p−1−αCt+p×Γb+p+tΓa+b+p+tat+pωt+p.
Also,(24)Aω−1+αBω=Cp−1+αDp+keiθCp−βDpωp+Γa+b+pΓb+p×∑t=1∞keiθβCt+p−Dt+p−Dt+p+1+αCt+p×Γb+p+tΓa+b+p+tat+pωt+p≤keiθCp−βDp+Cp−1+αDpωp−Γa+b+pΓb+p×∑t=1∞keiθβCt+p−Dt+p−Dt+p+1+αCt+p×Γb+p+tΓa+b+p+tat+pωt+p.
Using (23) and (24), then we can obtain the following inequality:(25)Aω+1−αBω−Aω−1+αBω≥keiθβDp−Cp−Cp−βDp−Cp−1−αDp−Cp+1+αDpωp−Γa+b+pΓb+p×∑t=1∞keiθβCt+p−Dt+p−βCt+p−Dt+p−Dt+p−1−αCt+p+Dt+p−1+αCt+p×Γb+p+tΓa+b+p+tat+pωt+p.
The last expression is bounded below by 0 if(26)Γa+b+pΓb+p∑t=1∞Dt+p1+k−Ct+pα+kβΓb+p+tΓa+b+p+tat+p≤Cp1+k−Dpα+kβ,which complete the proof.Theorem 2.
Letλω be given by (1) and in £η; then, λ∈k−£Uηa,b,p,α,β,μ,ν if and only if(27)Γa+b+pΓb+p∑t=1∞Dt+p1+k−Ct+pα+kβΓb+p+tΓa+b+p+tat+p≤Cp1+k−Dpα+kβ,where Ct+p, Dt+p, Cp, and Dp are given by (13)–(16), respectively.Proof.
In view of Theorem2, we need only to show that λ∈k−£Uηa,b,p,α,β,μ,ν satisfies coefficient inequality (27). If λ∈k−£Uηa,b,p,α,β,μ,ν, then, by definition, we have(28)ℜCp−αDp+Γa+b+p/Γb+p∑t=1∞Dt+p−αCt+pΓb+p+t/Γa+b+p+tat+pωt+p−11+Γa+b+p/Γb+p∑t=1∞Ct+pΓb+p+t/Γa+b+p+tat+pωt+p−1≥kCp−βDp+Γa+b+p/Γb+p∑t=1∞Dt+p−βCt+pΓb+p+t/Γa+b+p+tat+pωt+p−11+Γa+b+p/Γb+p∑t=1∞Ct+pΓb+p+t/Γa+b+p+tat+pωt+p−1.
Sinceλ is a function of form (1) with the argument properties given in the class £η and setting ω=reiη in the above inequality, we have(29)ℜCp−αDp−Γa+b+p/Γb+p∑t=1∞Dt+p−αCt+pΓb+p+t/Γa+b+p+tat+pωt+p−11−Γa+b+p/Γb+p∑t=1∞Ct+pΓb+p+t/Γa+b+p+tat+pωt+p−1≥kCp−βDp+Γa+b+p/Γb+p∑t=1∞Dt+p−βCt+pΓb+p+t/Γa+b+p+tat+pωt+p−11−Γa+b+p/Γb+p∑t=1∞Ct+pΓb+p+t/Γa+b+p+tat+pωt+p−1.
Lettingr⟶1− (29) leads to the desired inequality:(30)Γa+b+pΓb+p∑t=1∞Dt+p1+k−Ct+pα+kβΓb+p+tΓa+b+p+tat+p≤Cp1+k−Dpα+kβ.
The function,(31)λt.ηω=ωp−Cp1+k−Dpα+kβei1−tηΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tωt+p,0≤η≤2π,t≥1,is an external function for (27).Corollary 1.
Let the functionλω defined by (1) be in the class k−£Uηa,b,p,α,β,μ,ν; then,(32)at+p≤Cp1+k−Dpα+kβΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tt∈N,with equality in (32), is attained for the function λt.ηω given by (31).Theorem 3.
Let the functionλ∈k−£Uηa,b,p,α,β,μ,ν with argument property as in class £η. Define λjω=ω and(33)λt.ηω=ωp−Cp1+k−Dpα+kβei1−tηΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tωt+p,where 0≤η≤2π and t≥1. Then, λω is in the class k−£Uηa,b,p,α,β,μ,ν if and only if it can be expressed as(34)λω=∑t=0∞ϑtλt.η,where ϑt≥0t≥0 and ∑t=0∞ϑt=1.Proof.
Assume that(35)λω=ϑ0λ0ω+∑t=1∞ϑtωp−Cp1+k−Dpα+kβei1−tηΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tωt+p=∑t=1∞ϑtωp−∑t=1∞Cp1+k−Dpα+kβei1−tηΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tϑtωt+p.
Then, by Theorem2, λ∈k−£Uηa,b,p,α,β,μ,ν. It follows that(36)λω=∑t=1∞Cp1+k−Dpα+kβei1−tηΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tϑt×Γa+b+pΓb+pDt+p1+k−Ct+pα+kβΓb+p+tΓa+b+p+t=∑t=1∞Cp1+k−Dpα+kβϑt≤1−ϑ1Cp1+k−Dpα+kβ≤Cp1+k−Dpα+kβ.
Conversely, assume that the functionλω defined by (1) belongs to the class k−£Uηa,b,p,α,β,μ,ν; then,(37)at+p≤Cp1+k−Dpα+kβΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t,t∈N.
Settingϑt=Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t/Cp1+k−Dpα+kβat+p, t≥1 and ϑ1=1−∑t=1∞ϑt, then λω=∑t=0∞ϑtλt.η and this completes the proof.


### 2.2. Growth and Distortion Result

In this section, we find a growth and distortion bound for functions in the classesk−£Uηa,b,p,α,β,μ,ν.Theorem 4.
Let the functionλω be defined by (1) in the class k−£Uηa,b,p,α,β,μ,ν; then, for ω=r<1,(38)rp−Cp1+k−Dpα+kβb+p/a+b+pDp+11+k−Cp+1α+kβrp+1≤λω≤rp+Cp1+k−Dpα+kβb+p/a+b+pDp+11+k−Cp+1α+kβrp+1ω=rand(39)prp−1−p+1Cp1+k−Dpα+kβb+p/a+b+pDp+11+k−Cp+1α+kβrp≤λ′ω≤prp−1+p+1Cp1+k−Dpα+kβb+p/a+b+pDp+11+k−Cp+1α+kβrpω=r,where equalities (38) and (39) hold for the function λω given by (27), for ω=±r.Proof.
From Theorem2, we have(40)b+pa+b+pDp+11+k−Cp+1α+kβ∑t=1∞at+p≤Γa+b+pΓb+p∑t=1∞Dt+p1+k−Ct+pα+kβΓb+p+tΓa+b+p+tat+p≤Cp1+k−Dpα+kβ.
The last inequality follows from Theorem2 Thus,(41)λω≤ωp+∑t=1∞at+pωt+p≤rp+rp+1∑t=1∞at+p≤rp+Cp1+k−Dpα+kβb+p/a+b+pDp+11+k−Cp+1α+kβrp+1.
Similarly,(42)λω≥ωp−∑t=1∞at+pωt+p≥rp−rp+1∑t=1∞at+p≥rp−Cp1+k−Dpα+kβb+p/a+b+pDp+11+k−Cp+1α+kβrp+1.
Now, by differentiating (1), we obtain(43)λ′ω≤pωp−1+∑t=1∞t+pat+pωt+p−1≤prp−1+rp∑t=1∞t+pat+p≤prp−1+p+1Cp1+k−Dpα+kβb+p/a+b+pDp+11+k−Cp+1α+kβrpand(44)λ′ω≥ωp−∑t=1∞at+pωt+p≥prp−1−rp∑t=1∞t+pat+p≥prp−1−p+1Cp1+k−Dpα+kβb+p/a+b+pDp+11+k−Cp+1α+kβrp.
Using Theorem2 in (44), we have(45)b+pa+b+pDp+11+k−Cp+1α+kβ2∑t=1∞t+pat+p≤Γa+b+pΓb+p∑t=1∞Dt+p1+k−Ct+pα+kβΓb+p+tΓa+b+p+tat+p≤Cp1+k−Dpα+kβ,or, equivalently(46)∑t=1∞t+pat+p≤2a+b+pCp1+k−Dpα+kβb+pDp+11+k−Cp+1α+kβ.
Using (46) into (43) and (44) yields inequality (39).


### 2.3. Radii of Close-to-Convexity, Starlikeness, and Convexity

In this section, we obtain the radii of close-to-convexity, starlikeness, and convexity for functionsλω in the classes k−£Uηa,b,p,α,β,μ,ν.Theorem 5.
Letλ∈k−£Uηa,b,p,α,β,μ,ν; then,
(i)λω is starlike of order ϰ0≤ϰ<1 in the disc ω<r1, where(47)r1=inf2−p−ϰt+p−ϰΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβ1/t,t≥1.
(ii)λω is convex of order ϰ0≤ϰ<1 in the disc ω<r2, where(48)r2=infp2−p−ϰt+pt+p−ϰΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβ1/t,t≥1.These results are sharp for the extremal functionλω given by (31).Proof.
(i) Givenλ∈£η and λ is starlike of order ϰ, we have(49)ωλ′ωλω−1<1−ϰ.
For the left-hand side of (49), we have(50)ωλ′ωλω−1≤p−1+∑t=1∞k+p−1at+pωt1−∑t=1∞at+pωt.
The last expression is less than1−ϰ if(51)∑t=1∞k+p−ϰ2−p−ϰat+pωt<1.
Use the fact thatλ∈k−£Uηa,b,p,α,β,μ,ν if and only if(52)∑t=1∞Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβat+p≤1.
We can say (49) is true if(53)k+p−ϰ2−p−ϰωt=Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβ.or, equivalently,(54)ωt=Γa+b+p/Γb+p2−p−ϰDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tk+p−ϰCp1+k−Dpα+kβ,which is required.
(ii) Using the fact thatλ is convex if and only if ωλ′ω is starlike, we can prove (ii) on similar lines to the proof of (i).


### 2.4. Modified Hadamard Product

Let the functionλjωj=1,2 be defined by(55)λjω=ωp+∑∞t=1at+p,iωt+p,at+p,i≥0,i∈ℕ.Then, we define the modified Hadamard product ofλ1ω and λ2ω by(56)λ1∗λ2ω=ωp−∑∞t=0at+p,1at+p,2ωt+p.Now, we prove the following.Theorem 6.
Letλjωj=1,2,… given by (55) be in the class k−£Uηa,b,p,α,β,μ,ν; then, λ1∗λ2∈k−£Uηa,b,p,Φ1,μ,ν for(57)Φ1=Cp1+k−DpkβΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t2−Dt+p1+k−Ct+pkβCp1+k−Dpα+kβ2DpΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t2−Ct+pCp1+k−Dpα+kβ2.Proof.
We need to prove the largestΦ1, such that(58)Dt+p1+k−Ct+pΦ1+kβCp1+k−DpΦ1+kβat+p,1at+p,2≤1.
From Theorem2, we have(59)∑t=1∞Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβat+p,1≤1,∑t=1∞Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβat+p,2≤1.
By Cauchy–Schwarz inequality, we have(60)∑t=1∞Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβat+p,1at+p,2≤1.
Thus, it is sufficient to show that(61)Dt+p1+k−Ct+pΦ1+kβCp1+k−DpΦ1+kβat+p,1at+p,2≤Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβat+p,1at+p,2.
That is,(62)at+p,1at+p,2≤Cp1+k−DpΦ1+kβΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβDt+p1+k−Ct+pΦ1+kβ.
Note that(63)at+p,1at+p,2≤Cp1+k−Dpα+kβΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t.
Consequently, from (62) and (63), we obtain(64)Cp1+k−Dpα+kβΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t≤Cp1+k−DpΦ1+kβΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tDt+p1+k−Ct+pΦ1+kβCp1+k−Dpα+kβ,or, equivalently,(65)Φ1≤Cp1+k−DpkβΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t2−Dt+p1+k−Ct+pkβCp1+k−Dpα+kβ2DpΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t2−Ct+pCp1+k−Dpα+kβ2=χt.
Sinceχ1t is an increasing function for t≥1, letting t=1 in (65), we obtain(66)Φ1≤χ11=Cp1+k−Dpkβb+p/a+b+pDp+11+k−Cp+1α+kβ2−Dp+11+k−Cp+1kβCp1+k−Dpα+kβ2Dpb+p/a+b+pDp+11+k−Cp+1α+kβ2−CP+1Cp1+k−Dpα+kβ2.
The proof of our theorem is now completed.Theorem 7.
Letλjωj=1,2 given by (55) be in the class k−£Uηa,b,p,α,β,μ,ν. If the sequence Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t is nondecreasing, then function(67)hω=ωp−∑t=1∞at+p,12+at+p,22ωt+pbelongs to the class k−£Uηa,b,p,Φ2,μ,ν, where(68)Φ2≤Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t2Cp1+k−Dpkβ−2Cp1+k−Dpα+kβ2Dt+p1+k−Ct+pkβDpΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t2−2Ct+pCp1+k−Dpα+kβ2.Proof.
We need to prove the largestΦ2.
From Theorem2, we have(69)∑t=1∞Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβ2at+p,12≤∑t=1∞Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβat+p,12≤1and(70)∑t=1∞Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβ2at+p,22≤∑t=1∞Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβat+p,22≤1.
It follows from (69) and (70) that(71)∑t=1∞12Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβ2at+p,12+at+p,22≤1.
Therefore, we need to find the largestΦ2, such that(72)Dt+p1+k−Ct+pΦ2+kβCp1+k−DpΦ2+kβ≤12Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβ2,
That is,(73)Φ2≤Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t2Cp1+k−Dpkβ−2Cp1+k−Dpα+kβ2Dt+p1+k−Ct+pkβDpΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t2−2Ct+pCp1+k−Dpα+kβ2.
Sinceχ2t is an increasing function for t≥1, letting t=1 in (73), we readily have(74)Φ2≤χ21=Γa+b+p/Γb+pDp+11+k−Cp+1α+kβΓb+p+t/Γa+b+p+t2Cp1+k−Dpkβ−2Cp1+k−Dpα+kβ2Dp+11+k−Cp+1kβDpΓa+b+p/Γb+pDp+11+k−Cp+1α+kβΓb+p+t/Γa+b+p+t2−2Cp+1Cp1+k−Dpα+kβ2.
The proof of our theorem is now completed.


## 2.1. Coefficient Estimates

In this section, we obtain a necessary and sufficient condition for functionsλω in the classes k−£Uηa,b,p,α,β,μ,ν.Theorem 1.
A functionλω given by (1) is in the class k−Ua,b,p,α,β,μ,ν if(12)Γa+b+pΓb+p∑t=1∞Dt+p1+k−Ct+pα+kβΓb+p+tΓa+b+p+tat+p≤Cp1+k−Dpα+kβ,where(13)Ct+p=1−ν+νt+p1+μt+p−1,(14)Dt+p=t+p+νt+pt+p−11+μt+p,(15)Cp=p+νpp−11+μp,(16)Dp=1−ν+νp1+μp−1,and(17)−1≤α<β≤1,0≤μ<1,k1−β<1−α,a≥0,b>−1,p∈ℕandω∈⋓.Proof.
It suffices to show that inequality (11) holds true. As we know,(18)ℜϖ>kϖ−β+α⇔ℜ1+keiθϖ−βkeiθ>α.
Then, inequality (11) may be written as(19)ℜ1+keiθ1−νωQb,paλω′+νωQb,paλω′+1+2μω2Qb,paλω″+μω3Qb,paλω″′1−νQb,paλω+νωQb,paλω′+μω2Qb,paλω″−βkeiθ≥α,which can be written as ℜAω/Bω≥α, where(20)Aω=1+keiθ1−νωQb,paλω′+νωQb,paλω′+1+2μω2Qb,paλω′′+μω3Qb,paλω′′′−βkeiθ1−νQb,paλω+νωQb,paλω′+μω2Qb,paλω′′and(21)Bω=1−νQb,paλω+νωQb,paλω′+μω2Qb,paλω′′.
Then, we have(22)Aω+1−αBω−Aω−1+αBω≥0.
Now,(23)Aω+1−αBωCp+1−αDp+keiθCp−βDpωp−Γa+b+pΓb+p×∑t=1∞keiθβCt+p−Dt+p−Dt+p−1−αCt+p×Γb+p+tΓa+b+p+tat+pωt+p≥keiθβDp−Cp−Cp−1−αDpωp−Γa+b+pΓb+p×∑t=1∞keiθβCt+p−Dt+p−Dt+p−1−αCt+p×Γb+p+tΓa+b+p+tat+pωt+p.
Also,(24)Aω−1+αBω=Cp−1+αDp+keiθCp−βDpωp+Γa+b+pΓb+p×∑t=1∞keiθβCt+p−Dt+p−Dt+p+1+αCt+p×Γb+p+tΓa+b+p+tat+pωt+p≤keiθCp−βDp+Cp−1+αDpωp−Γa+b+pΓb+p×∑t=1∞keiθβCt+p−Dt+p−Dt+p+1+αCt+p×Γb+p+tΓa+b+p+tat+pωt+p.
Using (23) and (24), then we can obtain the following inequality:(25)Aω+1−αBω−Aω−1+αBω≥keiθβDp−Cp−Cp−βDp−Cp−1−αDp−Cp+1+αDpωp−Γa+b+pΓb+p×∑t=1∞keiθβCt+p−Dt+p−βCt+p−Dt+p−Dt+p−1−αCt+p+Dt+p−1+αCt+p×Γb+p+tΓa+b+p+tat+pωt+p.
The last expression is bounded below by 0 if(26)Γa+b+pΓb+p∑t=1∞Dt+p1+k−Ct+pα+kβΓb+p+tΓa+b+p+tat+p≤Cp1+k−Dpα+kβ,which complete the proof.Theorem 2.
Letλω be given by (1) and in £η; then, λ∈k−£Uηa,b,p,α,β,μ,ν if and only if(27)Γa+b+pΓb+p∑t=1∞Dt+p1+k−Ct+pα+kβΓb+p+tΓa+b+p+tat+p≤Cp1+k−Dpα+kβ,where Ct+p, Dt+p, Cp, and Dp are given by (13)–(16), respectively.Proof.
In view of Theorem2, we need only to show that λ∈k−£Uηa,b,p,α,β,μ,ν satisfies coefficient inequality (27). If λ∈k−£Uηa,b,p,α,β,μ,ν, then, by definition, we have(28)ℜCp−αDp+Γa+b+p/Γb+p∑t=1∞Dt+p−αCt+pΓb+p+t/Γa+b+p+tat+pωt+p−11+Γa+b+p/Γb+p∑t=1∞Ct+pΓb+p+t/Γa+b+p+tat+pωt+p−1≥kCp−βDp+Γa+b+p/Γb+p∑t=1∞Dt+p−βCt+pΓb+p+t/Γa+b+p+tat+pωt+p−11+Γa+b+p/Γb+p∑t=1∞Ct+pΓb+p+t/Γa+b+p+tat+pωt+p−1.
Sinceλ is a function of form (1) with the argument properties given in the class £η and setting ω=reiη in the above inequality, we have(29)ℜCp−αDp−Γa+b+p/Γb+p∑t=1∞Dt+p−αCt+pΓb+p+t/Γa+b+p+tat+pωt+p−11−Γa+b+p/Γb+p∑t=1∞Ct+pΓb+p+t/Γa+b+p+tat+pωt+p−1≥kCp−βDp+Γa+b+p/Γb+p∑t=1∞Dt+p−βCt+pΓb+p+t/Γa+b+p+tat+pωt+p−11−Γa+b+p/Γb+p∑t=1∞Ct+pΓb+p+t/Γa+b+p+tat+pωt+p−1.
Lettingr⟶1− (29) leads to the desired inequality:(30)Γa+b+pΓb+p∑t=1∞Dt+p1+k−Ct+pα+kβΓb+p+tΓa+b+p+tat+p≤Cp1+k−Dpα+kβ.
The function,(31)λt.ηω=ωp−Cp1+k−Dpα+kβei1−tηΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tωt+p,0≤η≤2π,t≥1,is an external function for (27).Corollary 1.
Let the functionλω defined by (1) be in the class k−£Uηa,b,p,α,β,μ,ν; then,(32)at+p≤Cp1+k−Dpα+kβΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tt∈N,with equality in (32), is attained for the function λt.ηω given by (31).Theorem 3.
Let the functionλ∈k−£Uηa,b,p,α,β,μ,ν with argument property as in class £η. Define λjω=ω and(33)λt.ηω=ωp−Cp1+k−Dpα+kβei1−tηΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tωt+p,where 0≤η≤2π and t≥1. Then, λω is in the class k−£Uηa,b,p,α,β,μ,ν if and only if it can be expressed as(34)λω=∑t=0∞ϑtλt.η,where ϑt≥0t≥0 and ∑t=0∞ϑt=1.Proof.
Assume that(35)λω=ϑ0λ0ω+∑t=1∞ϑtωp−Cp1+k−Dpα+kβei1−tηΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tωt+p=∑t=1∞ϑtωp−∑t=1∞Cp1+k−Dpα+kβei1−tηΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tϑtωt+p.
Then, by Theorem2, λ∈k−£Uηa,b,p,α,β,μ,ν. It follows that(36)λω=∑t=1∞Cp1+k−Dpα+kβei1−tηΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tϑt×Γa+b+pΓb+pDt+p1+k−Ct+pα+kβΓb+p+tΓa+b+p+t=∑t=1∞Cp1+k−Dpα+kβϑt≤1−ϑ1Cp1+k−Dpα+kβ≤Cp1+k−Dpα+kβ.
Conversely, assume that the functionλω defined by (1) belongs to the class k−£Uηa,b,p,α,β,μ,ν; then,(37)at+p≤Cp1+k−Dpα+kβΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t,t∈N.
Settingϑt=Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t/Cp1+k−Dpα+kβat+p, t≥1 and ϑ1=1−∑t=1∞ϑt, then λω=∑t=0∞ϑtλt.η and this completes the proof.


## 2.2. Growth and Distortion Result

In this section, we find a growth and distortion bound for functions in the classesk−£Uηa,b,p,α,β,μ,ν.Theorem 4.
Let the functionλω be defined by (1) in the class k−£Uηa,b,p,α,β,μ,ν; then, for ω=r<1,(38)rp−Cp1+k−Dpα+kβb+p/a+b+pDp+11+k−Cp+1α+kβrp+1≤λω≤rp+Cp1+k−Dpα+kβb+p/a+b+pDp+11+k−Cp+1α+kβrp+1ω=rand(39)prp−1−p+1Cp1+k−Dpα+kβb+p/a+b+pDp+11+k−Cp+1α+kβrp≤λ′ω≤prp−1+p+1Cp1+k−Dpα+kβb+p/a+b+pDp+11+k−Cp+1α+kβrpω=r,where equalities (38) and (39) hold for the function λω given by (27), for ω=±r.Proof.
From Theorem2, we have(40)b+pa+b+pDp+11+k−Cp+1α+kβ∑t=1∞at+p≤Γa+b+pΓb+p∑t=1∞Dt+p1+k−Ct+pα+kβΓb+p+tΓa+b+p+tat+p≤Cp1+k−Dpα+kβ.
The last inequality follows from Theorem2 Thus,(41)λω≤ωp+∑t=1∞at+pωt+p≤rp+rp+1∑t=1∞at+p≤rp+Cp1+k−Dpα+kβb+p/a+b+pDp+11+k−Cp+1α+kβrp+1.
Similarly,(42)λω≥ωp−∑t=1∞at+pωt+p≥rp−rp+1∑t=1∞at+p≥rp−Cp1+k−Dpα+kβb+p/a+b+pDp+11+k−Cp+1α+kβrp+1.
Now, by differentiating (1), we obtain(43)λ′ω≤pωp−1+∑t=1∞t+pat+pωt+p−1≤prp−1+rp∑t=1∞t+pat+p≤prp−1+p+1Cp1+k−Dpα+kβb+p/a+b+pDp+11+k−Cp+1α+kβrpand(44)λ′ω≥ωp−∑t=1∞at+pωt+p≥prp−1−rp∑t=1∞t+pat+p≥prp−1−p+1Cp1+k−Dpα+kβb+p/a+b+pDp+11+k−Cp+1α+kβrp.
Using Theorem2 in (44), we have(45)b+pa+b+pDp+11+k−Cp+1α+kβ2∑t=1∞t+pat+p≤Γa+b+pΓb+p∑t=1∞Dt+p1+k−Ct+pα+kβΓb+p+tΓa+b+p+tat+p≤Cp1+k−Dpα+kβ,or, equivalently(46)∑t=1∞t+pat+p≤2a+b+pCp1+k−Dpα+kβb+pDp+11+k−Cp+1α+kβ.
Using (46) into (43) and (44) yields inequality (39).


## 2.3. Radii of Close-to-Convexity, Starlikeness, and Convexity

In this section, we obtain the radii of close-to-convexity, starlikeness, and convexity for functionsλω in the classes k−£Uηa,b,p,α,β,μ,ν.Theorem 5.
Letλ∈k−£Uηa,b,p,α,β,μ,ν; then,
(i)λω is starlike of order ϰ0≤ϰ<1 in the disc ω<r1, where(47)r1=inf2−p−ϰt+p−ϰΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβ1/t,t≥1.
(ii)λω is convex of order ϰ0≤ϰ<1 in the disc ω<r2, where(48)r2=infp2−p−ϰt+pt+p−ϰΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβ1/t,t≥1.These results are sharp for the extremal functionλω given by (31).Proof.
(i) Givenλ∈£η and λ is starlike of order ϰ, we have(49)ωλ′ωλω−1<1−ϰ.
For the left-hand side of (49), we have(50)ωλ′ωλω−1≤p−1+∑t=1∞k+p−1at+pωt1−∑t=1∞at+pωt.
The last expression is less than1−ϰ if(51)∑t=1∞k+p−ϰ2−p−ϰat+pωt<1.
Use the fact thatλ∈k−£Uηa,b,p,α,β,μ,ν if and only if(52)∑t=1∞Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβat+p≤1.
We can say (49) is true if(53)k+p−ϰ2−p−ϰωt=Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβ.or, equivalently,(54)ωt=Γa+b+p/Γb+p2−p−ϰDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tk+p−ϰCp1+k−Dpα+kβ,which is required.
(ii) Using the fact thatλ is convex if and only if ωλ′ω is starlike, we can prove (ii) on similar lines to the proof of (i).


## 2.4. Modified Hadamard Product

Let the functionλjωj=1,2 be defined by(55)λjω=ωp+∑∞t=1at+p,iωt+p,at+p,i≥0,i∈ℕ.Then, we define the modified Hadamard product ofλ1ω and λ2ω by(56)λ1∗λ2ω=ωp−∑∞t=0at+p,1at+p,2ωt+p.Now, we prove the following.Theorem 6.
Letλjωj=1,2,… given by (55) be in the class k−£Uηa,b,p,α,β,μ,ν; then, λ1∗λ2∈k−£Uηa,b,p,Φ1,μ,ν for(57)Φ1=Cp1+k−DpkβΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t2−Dt+p1+k−Ct+pkβCp1+k−Dpα+kβ2DpΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t2−Ct+pCp1+k−Dpα+kβ2.Proof.
We need to prove the largestΦ1, such that(58)Dt+p1+k−Ct+pΦ1+kβCp1+k−DpΦ1+kβat+p,1at+p,2≤1.
From Theorem2, we have(59)∑t=1∞Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβat+p,1≤1,∑t=1∞Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβat+p,2≤1.
By Cauchy–Schwarz inequality, we have(60)∑t=1∞Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβat+p,1at+p,2≤1.
Thus, it is sufficient to show that(61)Dt+p1+k−Ct+pΦ1+kβCp1+k−DpΦ1+kβat+p,1at+p,2≤Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβat+p,1at+p,2.
That is,(62)at+p,1at+p,2≤Cp1+k−DpΦ1+kβΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβDt+p1+k−Ct+pΦ1+kβ.
Note that(63)at+p,1at+p,2≤Cp1+k−Dpα+kβΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t.
Consequently, from (62) and (63), we obtain(64)Cp1+k−Dpα+kβΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t≤Cp1+k−DpΦ1+kβΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tDt+p1+k−Ct+pΦ1+kβCp1+k−Dpα+kβ,or, equivalently,(65)Φ1≤Cp1+k−DpkβΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t2−Dt+p1+k−Ct+pkβCp1+k−Dpα+kβ2DpΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t2−Ct+pCp1+k−Dpα+kβ2=χt.
Sinceχ1t is an increasing function for t≥1, letting t=1 in (65), we obtain(66)Φ1≤χ11=Cp1+k−Dpkβb+p/a+b+pDp+11+k−Cp+1α+kβ2−Dp+11+k−Cp+1kβCp1+k−Dpα+kβ2Dpb+p/a+b+pDp+11+k−Cp+1α+kβ2−CP+1Cp1+k−Dpα+kβ2.
The proof of our theorem is now completed.Theorem 7.
Letλjωj=1,2 given by (55) be in the class k−£Uηa,b,p,α,β,μ,ν. If the sequence Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t is nondecreasing, then function(67)hω=ωp−∑t=1∞at+p,12+at+p,22ωt+pbelongs to the class k−£Uηa,b,p,Φ2,μ,ν, where(68)Φ2≤Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t2Cp1+k−Dpkβ−2Cp1+k−Dpα+kβ2Dt+p1+k−Ct+pkβDpΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t2−2Ct+pCp1+k−Dpα+kβ2.Proof.
We need to prove the largestΦ2.
From Theorem2, we have(69)∑t=1∞Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβ2at+p,12≤∑t=1∞Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβat+p,12≤1and(70)∑t=1∞Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβ2at+p,22≤∑t=1∞Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβat+p,22≤1.
It follows from (69) and (70) that(71)∑t=1∞12Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβ2at+p,12+at+p,22≤1.
Therefore, we need to find the largestΦ2, such that(72)Dt+p1+k−Ct+pΦ2+kβCp1+k−DpΦ2+kβ≤12Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+tCp1+k−Dpα+kβ2,
That is,(73)Φ2≤Γa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t2Cp1+k−Dpkβ−2Cp1+k−Dpα+kβ2Dt+p1+k−Ct+pkβDpΓa+b+p/Γb+pDt+p1+k−Ct+pα+kβΓb+p+t/Γa+b+p+t2−2Ct+pCp1+k−Dpα+kβ2.
Sinceχ2t is an increasing function for t≥1, letting t=1 in (73), we readily have(74)Φ2≤χ21=Γa+b+p/Γb+pDp+11+k−Cp+1α+kβΓb+p+t/Γa+b+p+t2Cp1+k−Dpkβ−2Cp1+k−Dpα+kβ2Dp+11+k−Cp+1kβDpΓa+b+p/Γb+pDp+11+k−Cp+1α+kβΓb+p+t/Γa+b+p+t2−2Cp+1Cp1+k−Dpα+kβ2.
The proof of our theorem is now completed.


## 3. Conclusion

In our current investigation, we have presented and studied thoroughly some new subclasses ofp−valent functions related with uniformly convex and starlike functions, in connection with the Liu–Owa integral operator Qb,paλω given by (8). We have obtained sufficient and necessary conditions in relation to these classes, including growth, distortion theorem, and radius problem. Some special cases have been discussed as applications of our main results. The techniques and ideas of this paper may stimulate for further research in this area of knowledge.



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*Source: 5882343-2021-12-06.xml*