# Univalent Functions by Means of Chebyshev Polynomials

Authors: Sh. Najafzadeh; Z. Salleh
Journal: Journal of Function Spaces
 (2022)

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## Abstract

The primary motivation of the paper is to define a new classChδα,β,γ which consists of univalent functions associated with Chebyshev polynomials. For this class, we determine the coefficient bound and convolution preserving property. Furthermore, by using subordination structure, two new subclasses of Chδα,β,γ are introduced and denoted by Mλ1,λ2,s and Nλ1,λ2,s, respectively. For these subclasses, we obtain coefficient estimate, extreme points, integral representation, convexity, geometric interpretation, and inclusion results. Moreover, we prove that, under some restrictions on parameters, Chδα,β,γ=Nλ1,λ2,s.


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## Body

## 1. Introduction

LetA be the class of analytic univalent functions in the open unit disk:(1)D=z∈ℂ:z<1,with Taylor expansion series of the form(2)fz=z+∑k=2∞akzk.Also, denote byS the class of univalent functions which are normalized by f0=f′0−1=0, see [1, 2]. Furthermore, suppose that N be the subclass of A consisting of functions with negative coefficients of the type:(3)fz=z−∑k=2∞akzk,ak≥0.The significance of Chebyshev polynomials in numerical analysis is very important in both practical and theoretical points of view. There are four kinds of such polynomials. Many researchers consider orthogonal polynomials of Chebyshev and obtain many interest results.The Chebyshev polynomials of the first and second kinds are well known and introduced by(4)Tnt=cosθandUnt=sinn+1θsinθ,−1<t<1,where t=cosθ and n is the degree of polynomial. For more details, one may refer to [1–6]. The polynomials in (4) are connected by the following relations:(5)dTntdt=nUn−1t,Tnt=Unt−tUn−1t,(6)2Tnt=Unt−Un−2t.We note that ift=cosθ, where θ∈−π/3,π/3, then(7)Hz,t=11−2cosθz+z2=1+∑k=2∞sink+1θsinθzk,z∈D.Also, we can write(8)Hz,t=1+U1tz+U2tz2+⋯,z∈D,−1<t<1,where(9)Un−1t=sinnarccost1−t2,n∈ℕ,are the Chebyshev polynomials of the second kind, see [7–9].The Hadamard product (convolution) for functions(10)fz=z−∑k=2∞akzk,gz=z−∑k=2∞bkzk,is denoted by f∗g and defined as follows:(11)f∗gz=z−∑k=2∞akbkzk=g∗fz,z∈D.The generating function of the first kind of Chebyshev polynomialTnt, t∈−1,1 is given by(12)∑n=0∞Tntzn=1−tz1−2tz+z2,z∈D,see [10].Now, we consider the functions:(13)H1z=1+2cosθ+1z−Hz,t,H2z=1+1+cosθz−1−tz1−2tz+z2,(14)Vz=H1∗H1∗H2∗H2∗fz,where fz=z−∑k=2∞akzk∈N and “∗” denotes the convolution. By a simple calculation, we conclude that Vz∈N and in the form(15)Vz=z−∑k=2∞Tk2tsin2k+1θsin2θakzk,where θ∈−π/3,π/3 and t=cosθ.LetChδα,β,γ denote the subclass of N consisting of functions of form (15) satisfying the condition:(16)V′z+zV″z−12γV′z−α1+βγ<δ,where 0<α≤1, 0<β≤1, 0≤γ≤1, 0≤δ<1, and Vz is given by (15), see [11].


## 2. Main Results

In this section, we introduce a sharp coefficient bound forVz∈Chδα,β,γ. Also, convolution preserving property under parameters α and β is proved.Theorem 1.
Vz∈Chδα,β,γif and only if(17)∑k=2∞kk+2γδTk2tsin2k+1θsin2θak≤δγ2−α1+β.Proof.
Let inequality (17) hold true, and suppose z∈∂D=z∈D:z=1. Then, we obtain(18)V′z+zV″z−1−δ2γV′z−α1+βγ=−∑k=2∞kTktsink+1θsinθ2akzk−δ2γ−∑k=2∞2γkTk2tsin2k+1θsin2θakzk−1−α1+βγ=∑k=2∞kk+2γδTk2tsin2k+1θsin2θ−δγ2−α1+β≤0.
Hence, by maximum modulus theorem, we conclude thatVz∈Chδα,β,γ.
Conversely, letVz, defined by (15), be in the class Chδα,β,γ, so condition (16) yields(19)V′z+2V″z−12γV′z−α1+βγ=∑k=2∞k2Tk2tsin2k+1θ/sin2θakzk−12γ−∑k=2∞2γkTk2tsin2k+1θ/sin2θakzk−1−α1+βγ<δ.
Since, for anyz, Rez<z, then(20)Re∑k=2∞Tk2tsin2k+1θ/sin2θk2akzk−1γ2−α1+β−∑k=2∞Tk2tsin2k+1θ/sin2θ2γkakzk−1<δ.
By lettingz⟶1 through real values, we obtain(21)∑k=2∞Tk2tsin2k+1θsin2θk2ak≤δγ2−α1+β−∑k=2∞Tk2tsin2k+1θsin2θαγδkak,and this completes the proof.Remark 1.
We note that the function,(22)Wz=z−sin2θγδ2−α1+β41+γδcos22θsin23θz2,shows that inequality (17) is sharp.Theorem 2.
Iffz=z−∑k=2∞akzkandgz=z−∑k=2∞bkzkare in the classChδα,β,γ, then(i)
f∗gzbelongs toChδα,β,γα∗, where(23)α∗≤21+β−sinθ2−α1+βTktsink+1θ2γδkk+2γδ1+β.(ii)
f∗gzbelongs toChδα,β∗,γ, where(24)β∗≤2α−1+sinθ2−α1+βTktsink+1θ2γδkk+2γδ1+β.Proof.
(i)
It is sufficient to show that(25)∑k=2∞kk+2γδTk2tsin2k+1θsin2θδγ2−α1+βα∗akbk≤1.By using the Cauchy–Schwarz inequality from (31), we obtain(26)∑k=2∞kk+2γδsin2k+1θsin2θδγ2−α1+βα∗akbk≤1.Hence, we find the largestα∗ such that(27)∑k=2∞kk+2γδsin2k+1θsin2θδγ2−α∗1+βakbk≤∑k=2∞kk+2γδsin2k+1θsin2θδγ2−α1+βakbk≤1.This inequality holds if(28)sin2θδγ2−α1+βkk+2γδTk2tsin2k+1θ≤2−α∗1+β2−α1+β,or equivalently(29)α∗≤21+β+sinθ2−α1+βTktsink+1θ2γδkk+2γδ1+β.(ii)
By using the same techniques as in the part (i), we can easily prove the part (ii), so the proof is complete.


## 3. Subclass ofChδα,β,γ and Their Geometric Properties

In this section, we introduce two new subclasses ofChδα,β,γ and conclude their geometric properties.For analytic functionsfz and Fz in D, we say f is subordinate to F, written f≺F, if there exists a function w analytic in D, with w0=0 and wz<1 such that fz=Fwz, see [3, 12]. If F is univalent, then(30)f≺F⇔f0=F0,fD⊂FD.LetMλ1,λ2,s consist of all analytic functions gz∈D for which g0=1 and(31)gz≺1+λ2+λ1−λ21−sz1+λ2z,where −1≤λ1<λ2≤1, 0<λ2≤1, and 0≤s<1. Let Nλ1,λ2,s denote the class of all functions Vz∈Chδα,β,γ for which(32)zV′zVz∈Mλ1,λ2,s,where Vz is given by (15).Theorem 3.
Vz∈Nλ1,λ2,sif and only if(33)∑k=2∞1+k−1λ2+1λ2−λ11−sTktsink+1θsinθ2ak<1.Proof.
LetVz∈Nλ1,λ2,s; then, by (16), (31), and (32), we have(34)∑k=2∞Tk2tsin2k+1θ/sin2θk−1akzkλ2−λ11−s−∑k=2∞Tk2tsin2k+1θ/sin2θλ2k−1+λ2−λ11−sakzk<1,which implies that(35)Re∑k=2∞Tk2tsin2k+1θ/sin2θakzkλ2−λ11−s−∑k=2∞Tk2tsin2k+1θ/sin2θλ2k−1+λ2−λ11−sakzk<1.
Now, we choose the values ofz on the real axis, and letting z⟶1−, we get the required result.
Conversely, assume that condition (33) holds true. We must show that Vz∈Nλ1,λ2,s or equivalently(36)Y=Vz−zV′zλ2zV′z−λ2+λ2−λ11−sVs<1.However, we haveY=∑k=2∞Tk2tsin2k+1θ/sin2θakλ2−λ11−s−∑k=2∞Tk2tsin2k+1θ/sin2θλ2k−1+λ2−λ11−s.
By using (33), we get Y<1, so the proof is complete.Corollary 1.
LetV∈Nλ1,λ2,s; then,(37)ak<λ2−λ11−sTktsink+1θ/sinθ2λ2−λ11−s+λ2+1k−1.Theorem 4.
Letλ2≠1,Vz∈Nλ1,λ2,s, and(38)zV′zVz=a+ib=X.
Then, the values ofX lie in the circle.Proof.
By (31) and (32), we have(39)X=a+ib=1+λ2+λ2−λ11−sWz1+λ2Wz,Wz<1.
Then,(40)a+ib1+λ2Wz=1+λ2+λ2−λ11−sWzora−1+ib=λ2+λ2−λ11−s−aλ2−ibλ2Wz.
After a simple calculation, we obtain(41)a−1−λ2λ2+λ2−λ11−s1−λ222+b2<λ2−λ11−s1−λ222.
Hence, the value ofX lies in the circle with center at(42)1−λ2λ2+λ2−λ11−s1−λ22,0and radius λ2−λ11−s/1−λ22.Theorem 5.
If(43)λ2+1λ2−λ11−s≤k2+2γδk−1+γαδ1+βδγ2−α1+β,then Chδα,β,γ=Nλ1,λ2,s.Proof.
By equation (32), we have(44)Nλ1,λ2,s⊆Chδα,β,γ.
Now, assume thatV∈Chδα,β,γ; then, by Theorem 1, we have(45)∑k=2∞kk+2γδTktsink+1θsinθ2ak≤δγ2−α1+β.
By Theorem3, it is enough to show that (33) holds true, which is possible when(46)1+k−1λ2+1λ2−λ11−s≤kk+2γδδγ2−α1+β,or equivalently(47)k−1λ2+1λ2−λ11−s≤k2+2γδk−1+δγα1+βδγ2−α1+β.
Sincek starts from 2, then k−1≥1, and hence, from the last inequality, we obtain the required result.In the next theorems, we prove the inclusion property and convex combination concept. Also, extreme points and integral representation are introduced.Theorem 6.
Let0≤s2<s1<1; then,(48)Nλ1,λ2,ss1⊂Nλ1,λ2,ss2.Proof.
Supposef∈Nλ1,λ2,ss1; then, by Theorem 3, we have(49)∑k=2∞1+k−1λ2+1λ2−λ11−ss1Tktsink+1θsinθ2ak≤δγ2−α1+β.
We have to prove(50)∑k=2∞1+k−1λ2+1λ2−λ11−sTktsink+1θsinθ2ak≤δγ2−α1+β.
However, the last inequality holds true if(51)1+k−1λ2+1λ2−λ11−s≤1+k−1λ2+1λ2−λ11−s,and this inequality by hypothesis s2<s1 definitely holds true, so the proof is complete.Theorem 7.
Nλ1,λ2,sis a convex set.Proof.
We have to prove that if(52)Vjz=z−∑k=2∞Tktsink+1θsinθ2ak,jzk,j=1,2,…,m,is in the class Nλ1,λ2,s, then the function,(53)Lz=∑j=1mdjVjz,is also in Nλ1,λ2,s, where ∑j=1mdj=1. However, we have(54)Lz=z−∑k=2∞∑j=1mTktsink+1θsinθ2djak,jzk.
We have to prove that iffjzj=1,2,…,m is in the class Nλ1,λ2,s, then the function Lz=∑j=1mdjfjz is also in Nλ1,λ2,s, where ∑j=1mdj=1. We have(55)Lz=z−∑k=2∞∑j=1mdjak,jzk.
Since, by Theorem3,(56)∑k=2∞1+λ2+1k−1lsTktsink+1θsinθ2∑j=1mdjak,j=∑j=1m∑k=2∞1+λ2+1k−1λ2−λ11−sTktsink+1θsinθ2ak,jdj<∑j=1mdj=1so L∈Nλ1,λ2,s and the proof is complete.Theorem 8.
The functionV1z=zand(57)Vkz=z−λ2−λ11−sλ2−λ11−s+λ2+1k−1sinθTktsink+1θ2akzk,k≥2,are the extreme points of Nλ1,λ2,s.Proof.
We have to prove thatL∈Nλ1,λ2,s if and only if(58)Lz=∑k=2∞1dkVkz,where dk≥0k≥1 and ∑k=2∞1dk=1.Proof.
LetL∈Nλ1,λ2,s. If we set(59)dk=λ2−λ11−s+λ2+1k−1λ2−λ11−ssinθTktsink+1θ2ak,k≥2,we get dk≥0, and if we put d1=1−∑k=2∞dk, then we obtain(60)Lz=z−∑k=2∞λ2−λ11−sλ2−λ11−s+λ2+1k−1sinθTktsink+1θ2dkzk=z−∑k=2∞dkz−Vkz=∑k=2∞dkVkz.
Conversely, suppose(61)Lz=∑k=2∞1dkVkz.
Then, we have(62)Lz=d1V1z+∑k=2∞dkVkzd1z+∑k=2∞z−λ2−λ11−sλ2−λ11−s+λ2+1k−1sinθTktsink+1θ2dkzk=z−∑k=2∞λ2−λ11−sλ2−λ11−s+λ2+1k−1sinθTktsink+1θ2dkzk.
Since(63)∑k=2∞dk1+λ2+1k−1λ2−λ11−sλ2−λ11−sλ2−λ11−s+λ2+1k−1××sinθTktsink+1θ2Tktsink+1θsinθ2=∑k=2∞dk=1−d1<1,therefore, by Theorem 3, we conclude the result.Theorem 9.
Letf∈Nλ1,λ2,s; then,(64)Vz=exp∫0z1−λ2+λ2−λ11−sWtt1−λ2Wtdt,where Wz<1.Proof.
By lettingUz=zV′/V, since f∈Nλ1,λ2,s, so(65)Uz≺1+λ2+λ1−λ21−sz1+λ2z,or equivalently(66)Uz−1Uzλ2−λ2+λ1−λ21−s<1.
Therefore,(67)Uz−1Uzλ2−λ2+λ1−λ21−s=Wz,wz<1.
Hence, we can write(68)V′zVz=1−λ2+λ1−λ21−sWzz1−λ2Wz.
After integration, we get the required result.


## 4. Conclusion

Univalent functions have always been the main interests of many researchers in geometric function theory. Many studies recently related to Chebyshev polynomials revolved around classes of analytic normalized univalent functions. In this particular work, the geometric properties are obtained for functions in more general class denoted byChδα,β,γ using the Chebyshev polynomials associated with the convolution structure. Some other geometric results are introduced for the subclasses of Chδα,β,γ.



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