# On the Chebyshev Polynomial for a Certain Class of Analytic Univalent Functions

Authors: Shahram Najafzadeh; Mugur Acu
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this work, by considering the Chebyshev polynomial of the first and second kind, a new subclass of univalent functions is defined. We obtain the coefficient estimate, extreme points, and convolution preserving property. Also, we discuss the radii of starlikeness, convexity, and close-to-convexity.


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## Body

## 1. Introduction

LetΔ be the open unit disk z∈ℂ:z<1 and A be the class of analytic functions in Δ, satisfying the normalized conditions:
(1)f0=0,f′0=1.Thus, eachf∈A has the following Taylor expansion:
(2)fz=z+∑k=2∞akzk.Furthermore, byS, we shall denote the family of all functions in A that are univalent in Δ. Denote by N the subclass of A consisting of functions with negative coefficients of the type:
(3)fz=z−∑k=2∞akzk,ak≥0,z∈Δ,see [1].Many researchers deal with orthogonal polynomials of Chebyshev, see [2, 3] and [4]. The Chebyshev polynomials of first kind and the second kind are defined by
(4)Tkt=coskθ,Ukt=sink+1θsinθ,respectively, where−1<t<1, t=cosθ, and k is the degree of polynomial.The polynomial in (1) is connected by the following relations:
(5)dTktdt=kUk−1t,Tkt=Ukt−kUk−1t,(6)2Tkt=Ukt−Uk−2t.We note that ift=cosθ, −π/3<θ<π/3, then
(7)Hz,t=11−2zcosθ+z2=1+∑k=1∞sink+1θsinθzk,z∈Δ.Also, we have(8)Hz,t=1+U1tz+U2tz2+⋯,z∈Δ,−1<t<1,where
(9)Uk−1t=sinkarccost1−t2,k∈ℕ,are the Chebyshev polynomials of the second kind, see [5, 6] and [7].The generating function of the first kind of Chebyshev polynomialTkt, t∈−1,1 is given by
(10)∑k=0∞Tktzk=1−tz1−2tz+z2.For more details, see [8, 9] and [10].For two functionsf and g, analytic in Δ, we say that f is subordinate to g in Δ, written
(11)fz≺gz,z∈Δ,if there exists a Schwarz function w which is analytic in Δ with
(12)w0=0,wz<1,z∈Δ,such thatfz=gwz, (z∈Δ), see [11].Also, ifg is univalent in Δ, then
(13)fz≺gz,z∈Δ⟺f0=g0,fΔ⊂gΔ.Furthermore, iffz=z−∑k=2∞akzk and gz=z−∑k=2∞bkzk, then the Hadamard product (or covolution) of f and g is defined by
(14)f∗gz=z−∑k=2∞akbkzk=g∗fz.Now, we consider the following functions which are connected with the Chebyshev polynomial of the first and second kind:(15)C1z=1+1+cosθz−1−tz1−2tz+z2,(16)C2z=1+2cosθ+1z−Hz,t,(17)Qz=C1∗C1∗C2∗C2∗fz,where fz=z−∑k=2∞akzk∈N and “∗” denotes the Hadamard product.With a simple calculation we conclude thatQz belongs to N and it is of the form:
(18)Qz=z−∑k=2∞sink+1θsinθTkt2akzk,where−π/3<θ<π/3 and t=cosθ.Definition.
ForM=α+β−α1−γ, −1≤β<α≤1, 0<γ<1, and 0≤λ≤1, we say that Qz of the form (18) is a member of Eγλα,β if the following subordination relation holds
(19)zQ′zfλz≺1+Mz1+αz,
wherefλz=1−λz+λfz, fz∈N.Equation (19) is equivalent to the following inequality:
(20)zQ′z/fλz−1M−αzQ′z/fλz<1.


## 2. Main Results

In this section, we introduce a sharp coefficient bound for the classEγλα,β. Also, the convolution preserving property is investigated.Theorem 1.
The functionQz of form (18) belongs to Eγλα,β if and only if
(21)∑k=2∞ksink+1θsinθTkt2−λ1−α+λβ−α1−γak≤β−α1−γ.Proof.
Let the inequality (21) holds and z∈∂Δ=z∈ℂ:z=1. We have to prove that (19) or equivalently (20) holds true. But we have
(22)Y=zQ′z−fλz−Mfλz−αzQ′z=z−∑k=2∞sink+1θsinθTkt2kakzk−1−λz−λz−∑k=2∞akzk−M1−λz+λz−∑k=2∞akzk−αz+∑k=2∞sink+1θsinθTkt2αkakzk=−∑k=2∞sink+1θsinθTkt2k−λakzk−M−αz−∑k=2∞λM−αksink+1θsinθTkt2akzk.
By puttingz∈∂Δ and
(23)λM−αksink+1θsinθTkt2=λM−α−ksink+1θsinθTkθ2−λα,the above expression reduces to
(24)Y≤∑k=2∞ksink+1θsinθTkt2−λ1−α+λM−αak−M−α.
SinceH−α=β−α1−γ, by using inequality (21), we get Y≤0, so Q∈Eγλα,β.
To prove the converse, letQ∈Eγλα,β, thus
(25)zQ′z/fλz−1M−αzQ′z/fλz=z−∑k=2∞ksink+1θ/sinθTkt2akzk−1−λz+λfzM1−λz+λfz−αz1−∑k=2∞ksink+1θ/sinθTkt2akzk−1<1,
for allz∈Δ. By Rez≤z for all z∈Δ, we have
(26)Re∑k=2∞ksink+1θ/sinθTkt2−λakzkM−αz−∑k=2∞λM−αksink+1θ/sinθTkt2akzk<1.
By lettingz⟶1, through positive values and choose the values of z such that zQ′z/fλz is real, we have
(27)∑k=2∞ksink+1θsinθTkt2−λ1−α+tM−αak≤M−α,and this completes the proof.Remark.
We note that the function:(28)Vz=z−β−α1−γ2sin3θ/sinθcos2θ2−λ1−α+λβ−α1−γz2,shows that the inequality (21) is sharp.Theorem 2.
Let(29)Q1z=z−∑k=2∞sink+1θsinθTkt2akzk,(30)Q2z=z−∑k=2∞sink+1θsinθTkt2bkzk,
be in the classEγλα,β, then Q1∗Q2z belongs to Eγλα,β~, where
(31)β~≤α+β−α21−γX1−αX1−α+λβ−α1−γ2−λ1−γ2β−α2,(32)X=ksink+1θsinθTkt2−λ.Proof.
It is sufficient to show that(33)∑k=2∞ksink+1θsinθTkt2−λ1−αβ−α1−γ+λakbk≤1.
By using the Cauchy-Schwarz inequality, from (21), we obtain
(34)∑k=2∞ksink+1θsinθTkt2−λ1−αβ−α1−γ+λakbk≤1.
Here, we find the largestk such that
(35)∑k=2∞ksink+1θsinθTkt2−λ1−αβ−α1−γ−λakbk≤1,
or equivalently fork≥2,
(36)akbk≤X1−α+λβ−α1−γβ~−αX1−α+λβ−α1−γβ−α,
whereX is given by (32).
This inequality holds if(37)β−α1−γX1−α+λβ−α1−γ≤X1−α+λβ−α1−γβ~−αX1−α+λβ~−α1−γβ−α,or equivalently
(38)β~≤α+β−α21−γX1−αX1−α+λβ−α1−γ2−λ1−γ2β−α2,
whereX is given by (32), so the proof is complete.


## 3. Geometric Properties ofEγλα,β

In this section, we show that the classEγλα,β is a convex set. Also, the radii of starlikeness, convexity, and close-to-convexity are obtained.Theorem 3.
The classEγλα,β is a convex set.Proof.
It is enough to prove that if forj=1,2,⋯,m,
(39)Qjz=z−∑k=2∞sink+1θsinθkak,jzk,
be inEγλα,β, then the function
(40)Fz=∑j=1mdjQjz,
is also inEγλα,β, where ∑j=1mdj=1. But, we have
(41)Fz=z−∑k=2∞∑j=1msink+1θsinθTkt2djak,jzk,=z−∑k=2∞sink+1θsinθTkt2∑j=1mdjak,jzk.
Since by Theorem1,
(42)∑k=2∞ksink+1θsinθTkt2−λ1−α+λβ−α1−γ∑j=1mdjak,j=∑j=1m∑k=2∞ksink+1θsinθTkt2−λ1−α+λβ−α1−γak,jdj<∑j=1mβ−α1−γdj=β−α1−γ∑j=1mdj=β−α1−γ,
so,Fz∈Eγλα,β. Hence, the proof is complete.Theorem 4.
Letf∈Eγλα,β, then
(i)
f is a starlike of order θ1 (cosθ1<1) in z<R1 where(43)R1=infkksink+1θsinθTkt2−λ1−αβ−α1−γ+λ1−θ1k−θ11/k−1(ii)
f is convex of order θ2 (0≤θ2<1) in z<R2 where(44)R2=infkksink+1θsinθTkt2−λ1−αβ−α1−γ+λ1−θ2kk−θ21/k−1(iii)
f is close-to-convex of order θ3 (0θ3<1) in ∣z∣<R3, where(45)R3=infkksink+1θsinθTkt2−λ1−αβ−α1−γ+λ1−θ3k1/k−1Proof.
(i)
For0≤θ1<1, we need to show that(46)zf′f−1<1−θ1
In other words, it is sufficient to show that(47)zf′zfz−1=∑k=2∞k−1akzk−11−∑k=2∞akzk−1≤∑k=2∞k−1akzk−11−∑k=2∞akzk−1<1−θ1,(48)∑k=2∞k−θ11−θ1akzk−1<1.
By (21), it is easy to see that above inequality holds if
(49)zk−1≤ksink+1θsinθTkt2−λ1−αβ−α1−γ+λ1−θ1k−θ1.
This completes the proof of (i).(ii)
Sincef is convex if and only if zf′ is starlike, we get the required result (ii)(iii)
We must show thatf′z−1≤1−θ3. But(50)f′z−1=∑k=2∞kakzk−1≤∑k=2∞kakzk−1
Thus,f′z−1<1−θ3 if ∑k=2∞k/1−θ3akzk−1≤1. But by Theorem 1, the above inequality holds true, if
(51)zk−1≤ksink+1θsinθTkt2−λ1−αβ−α1−γ+λ1−θ3k.
Hence, the proof is complete.


## 4. Conclusions

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 using the Chebyshev polynomials associated with a convolution structure. In this paper, when the parameters being complex numbers could be subject to further investigation. Also, by changing the operator and extending, it may be for future studies.



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