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### Article

Kyungpook Mathematical Journal 2024; 64(1): 57-68

Published online March 31, 2024

### Polylogarithms and Subordination of Some Cubic Polynomials

Manju Yadav, Sushma Gupta and Sukhjit Singh

Department of Mathematics, Sant Longowal Institute of Engineering and Technology, Longowal 148106, India
e-mail : manjuyadav1197@gmail.com, sushmagupta1@yahoo.com and sukhjit_d@yahoo.com

Received: January 20, 2023; Accepted: September 15, 2023

### Abstract

Let V3(z, f) and σ3(1)(z,f) be the cubic polynomials representing, respectively, the 3rd de la Vallée Poussin mean and the 3rd Cesàro mean of order 1 of a power series f(z). If K denotes the usual class of convex univalent functions in the open unit disk centered at the origin, we show that, in general, V3(z, f) ⊀ σ3(1)(z,f), for all fK. Making use of polylogarithms, we identify a transformation, Λ : KK, such that V3(z, Λ(f)) ≺ σ3(1)(z,(f)) for all fK. Here ‘≺’ stands for subordination between two analytic functions.

Keywords: Convex functions, Cesà,ro means, De la Vallé,e Poussin means, Subordination, Hadamard product

### 1. Introduction

For a real number r, 0<r<1 let Dr={zC:|z|<r} denote the open disc in the complex plane with center at the origin and radius r and D:=D1. We denote by A the class of all analytic functions defined in D that are normalized by the conditions f(0)=f'(0)-1=0, and by the class S of univalent functions in A. Further, we let K denote the subclass of those functions in S that map D onto convex domains and

S2=fK:zf'K.

Analytically, fK if and only if

1+zf''(z)f'(z)>0,zD.

If C denotes the class of close-to-convex functions in D, then S2KCS.

In 1983, Lewis [7], in an extremely involved proof, showed that the polylogarithmic functions

Liβ(z)=n=11nβzn,β0

are convex univalent in D. For β>0, the series in (1.1) has the integral representation given by the formula (see [13]):

Liβ(z)=1Γ(β)t01z(log(1/t))β-111-tzdt,|z|<1,β>0.

Here Γ stands for Euler gamma function. Note that

Li1(z)=t01z1-tzdt=-log(1-z),|z|<1

and

Li2(z)=01zlog(1/t)11tzdt=0 z log(1ζ)ζ dζ, z<1.

The dilogarithm, Li2, is of particular importance, since it arises in many integrals that cannot be expressed in terms of elementary functions. Due to this, it is used in many computer algebra systems, such as Maple (see [1]), where it takes the form dilog(z)=Li2(1-z).

The de la Vallée Poussin means are defined by

vn(f,t)=12πt02πwn(t-u)f(u)du,nN,

where f is periodic real-valued function and

wn(t)=2n(n!)2(2n)!(1+cost)n=12nnk=-nn2nn+keikt

are the de la Vallée Poussin kernels. In 1958, Pólya and Schoenberg [8] cast these means in terms of complex-valued analytic functions using the Hadamard product.

The Hadamard product or convolution of two power series f(z)=n=0anzn and g(z)=n=0bnzn is defined as the power series n=0anbnzn and is denoted by (f*g)(z). If f and g are analytic in D, then f*g is analytic in D as well.

Define

Vn(z)=2nn-1k=1n2nn+kzk,nN,zD.

For a given function f(z)=n=1anzn,

Vn(z,f)=(Vn*f)(z)=2nn-1k=1n2nn+kakzk

is the nth de la Vallée Poussin mean of f. Pólya and Schoenberg ([8, Theorem 2, p. 298]) proved that for all nN the de la Vallée Poussin means Vn(z,f) are convex if and only if f is convex.

Similarly, for a real number α0, if

σn(α)(z)=n+α-1n-1-1k=1nn+α-kn-kzk,nN,zD,

then the polynomial,

σn(α)(z,f)=(σn(α)*f)(z)=n+α-1n-1-1k=1nn+α-kn-kakzk,

is called the nth Cesàro mean of f of order α. Note that σn(α)(z)=σn(α)(z,z/(1-z)). Several authors (see [3, 4, 6, 10]) studied univalence properties of the polynomials σn(α)(z,z/(1-z)) and it is now known (see [6]) that for α1 and nN, σn(α)(z,z/(1-z))C. Using the fact that the class C is closed under convolution with convex functions (see [11, Theorem 2.2]), we immediately get that for α1 and nN, σn(α)(z,f)C for all fK. For some more geometric and subordination properties of Cesàro means we refer to [12, 14].

Let two functions f and g be analytic in Dr. Then f is called subordinate to g, written fg, in Dr if there exists an analytic function w satisfying the inequality |w(z)||z|<r such that f(z)=g(w(z)) in Dr. If g is univalent in Dr, then fg in Dr is equivalent to f(0)=g(0) and f(Dr)g(Dr).

In 1981, Singh and Singh [15] proved the following result:

Theorem 1.1. If fK, then V2(z,f)σ2(1)(z,f) in D.

Recently, the authors [18] proved that, infact, V2(z,f)σ2(α)(z,f) holds in D for all fK and for all α1.

A natural question which arises is: Is V3(z,f)σ3(α)(z,f) in D, for all fK?

The primary objective of this paper is to answer this question in the case that α=1. The paper is arranged as follows. In Section 2, we collect some known results which we shall use in the sequel. Section 3 starts with an example showing that answer to above question is `no'. Then, using polylogarithms, a subclass of K is identified such that for all f in this subclass, V3(z,f)σ3(1)(z,f) in D.

### 2. Preliminaries

In this section we recall the following definition and results which shall be needed to prove our results in this paper.

Definition 2.1. A sequence {bn}1 of complex numbers is said to be a subordinating factor sequence if, whenever f(z)=n=1anzn, a1=1, is univalent and convex in D, we have

n=1anbnznf(z).

Lemma 2.2. [16] A sequence {bn}1 of complex numbers is a subordinating factor sequence if and only if

1+2n=1bnzn>0,zD.

Lemma 2.3. [11] Let φ and ψ be convex functions in D and suppose that f is subordinate to φ. Then f*ψ is subordinate to φ*ψ in D.

Lemma 2.4. [2] Let Pn be the set of all polynomials of degree n,n2. Assume that QPn has all its critical points ζj in D¯,j=1,2,,n-1. Let PPn satisfy P(0)=Q(0) and

Q(ζj)P(D),j=1,2,,n-1.

Then PQ in D.

Lemma 2.5. [9] Given a polynomial,

r(z)=a0+a1z+a2z2+...+anzn

of degree n, let

Mk=detB¯kTAk A¯kTBk  (k=1,2,...,n),

where Ak and Bk are triangular matrices

Ak=a0 a1 ak1 0a0 ak2 00a0 ,   Bk= a¯n a¯n1 a¯nk+1 0 a¯n a¯nk+2 00 a¯n .

Then the polynomial r(z) has all its zeros inside the unit circle |z|=1 if and only if the determinants M1,M2,...,Mn are all positive.

### 3. Main Results

We begin with the following example which shows that the subscript 2 in Theorem 1.1 can not be replaced with 3, in general.

Example 3.1. Let I(z)=z/(1-z) be the right half plane mapping which maps D onto {wC:(w)>-1/2}. This function is a distinguished member of the class K. We claim that

V3(z,I)σ3(1)(z,I)

in D. We note that

σ3(1)(z,I)=z+23z2+13z3

and

V3(z,I)=34z+310z2+120z3.

Now ζ1=(-2+i5)/3,ζ2=(-2-i5)/3 are the only critical points of σ3(1)(z,I) and both lie on |z|=1. As V3(0,f)=σ3(1)(0,f), therefore, in view of Lemma 2.4, if V3(z,I)σ3(1)(z,I), we must have σ3(1)(ζj,I)V3(D,I),j=1,2. But this will be the case if all the zeros of the polynomials φj(z)=V3(z,I)-σ3(1)(ζj,I), j=1,2, lie outside the circle |z|=1. First consider

φ1(z)=120z3+310z2+34z+38-i10581.

Then all the zeros of φ1(z) will lie outside the circle |z|=1 if and only if all the zeros of the polynomial

φ11z=120+310z+34z2+38-i10581z3

lie inside the circle |z|=1. Now comparing φ1(1/z) with the polynomial r(z) (with n=3) in (2.1) we note that

a0=120,a1=310,a2=34,a3=38-i10581.

It is easy to verify that M1 and M2 as given by (2.2) are positive. We now calculate

M3=det B3 ¯TA3 A3 ¯TB3,

where

A3=12031034012031000120,B3=38+i1058134310038+i10581340038+i10581.

Using Mathematica [17], we get

M3=det38i1058100120310343438i10581001203103103438i10581001201200038+i10581343103101200038+i1058134343101200038+i105810.00145985.

Since M3<0, by Lemma 2.5 we get, φ1(1/z) does not have all its zeros inside the circle |z|=1. This implies that σ3(1)(ζ1,I)V3(D,I). Similarly, we can prove that σ3(1)(ζ2,I)V3(D,I). Hence, by Lemma 2.4, V3(z,I)σ3(1)(z,I), zD.

For geometric illustration of this example, we have drawn boundaries of the domains V3(D,I) and σ3(1)(D,I) in Figure 1. The zoomed portion of this figure clearly shows that V3(z,I)σ3(1)(z,I).

Figure 1. V3(D,I)σ3(1)(D,I)

Definition 3.2. We define a class of functions, Kβ, as under:

Kβ:={g:g(z)=(Liβ*f)(z),fK,β0}.

For β0, as Liβ is convex and convolution of two convex functions is convex, so, KβK. Obviously, in view of (1.2) and (1.3), we have

K1=0z f(ζ)ζdζ:fK,z<1

and

K2=0z logzζ f(ζ)ζdζ:fK,z<1.

Lemma 3.3. Let Liβ be given by (1.1). Then the cubic polynomial σ3(1)(z,Liβ) is convex univalent in D for all ββ0, where β0(1.039) is the positive root of the transcendental equation

22β+1(3β+2-27)+3β(25-3β+2)=0.

Proof. We have σ3(1)(z,Liβ)=z+23z22β+z331+β=g(z) (say).

Then

1+zg''(z)g'(z)=z23β-1+83z2β+1z23β+43z2β+1=(z23β-1+83z2β+1)(z¯23β+43z¯2β+1)|z23β+43z2β+1|2.

Thus g will be convex univalent in D provided (t(z))0 in D, where

t(z)=z23β-1+83z2β+1z¯23β+43z¯2β+1.

Putting z=eiθ,0θ<2π , we get

T(θ)=[t(eiθ)]=1+32914β+39β-43β+4cosθ2β1+53β+1+83βcos2θ.

We minimize T(θ), for θ[0,2π). Note that

Tθ(θ)=-4sinθ1356β+12β+4cosθ3β,

and, so, critical points of T(θ) are given by

sinθ=0,4cosθ=-32β1+53β+1.

It is now easy to verify that T(θ) has global minimum value

22β+19β+1(3β-1)22β+1(3β+2-27)+3β(25-3β+2)

at the critical point given by

4cosθ=-32β1+53β+1.

Using Mathematica [17], we conclude that minT(θ)0 for ββ0(1.039), where β0 is the positive root of (3.2). As, 1+zg''(z)/g'(z)=1 at z=0, using minimum principle for harmonic functions, we conclude that g(z)=σ3(1)(z,Liβ) maps D univalently onto a convex domain for ββ0.

Graph of transcendental function in (3.2) is shown in Figure 2 below.

Figure 2. Graph of transcendental function in eq(3.2)

Lemma 3.4. If Liβ is given by (1.1), then

V3(z,Liβ)σ3(1)(z,Liβ),

in D for all β>β0, where β0 is same as in Lemma 3.3.

Proof. As Liβ(z)=z+ n=2(zn/nβ), we have

σ3(1)(z,Liβ)=z+23z22β+13z33β,

and

V3(z,Liβ)=34z+310z22β+120z33β.

In view of Lemma 3.3 and Definition 2.1, it suffices to show that the sequence {3/4,9/20,3/20,0,0,...} is a subordinating factor sequence. Applying Lemma 2.2, this will be true if

(H(z))>0,zD,

where

H(z)=1+234z+920z2+320z3.

Putting z=eiθ, 0θ<2π, in H(z) and then taking the real part we get

(H(eiθ))=1+32cosθ+35cos2θ+15cos3θ=1101+32cosθ+6cos2θ+4cos3θ.

On calculating the critical points, and by simple calculations, we can easily verify that (3.3) has minimum value at cosθ=-1/(3+3) and this minimum value is equal to (3-3)/30=0.042..., which is certainly greater than zero. As (H(z))=1 at z=0, so we conclude that (H(z))>0 in D, by minimum principle for harmonic functions.

In Figure 3, we have drawn boundaries of the domains V3(D,Liβ) and σ3(1)(D,Liβ), taking β=2.

Figure 3. V3(D,Li2)σ3(1)(D,Li2)

Theorem 3.5. Let β0 be the number as in Lemma 3.3. Then for all real numbers β,ββ0, and for all fK, we have,

V3(z,Liβ*f)σ3(1)(z,Liβ*f),

in D; or, equivalently,

V3(z,g)σ3(1)(z,g),

in D for all gKβ(ββ0).

Proof. By Lemma 3.3, σ3(1)(z,Liβ) is convex univalent in D for all ββ0. The proof, therefore, follows from Lemma 2.3 and Lemma 3.4 and observing that σ3(1)(z,Liβ*f)=σ3(1)(z,Liβ)*f.

If we choose β=2 in above Theorem 3.5, we get, in view of (3.1), the following result:

Corollary 3.6. For all fK,

V3z,t0zlogzζf(ζ)ζdζσ3(1)z,t0zlogzζf(ζ)ζdζ,

in D.

If fS2, then zf'K and Liβ*zf'=Liβ-1*f. Thus we get:

Corollary 3.7. For all real numbers β,ββ0, and for all fS2, we have,

V3(z,Liβ-1*f)σ3(1)(z,Liβ-1*f),

in D. Here, the number β0 is same as in Lemma 3.3.

Taking β=2 in Corollary 3.7 and noting that Li1*f=-log(1-z)*f=t0zf(ζ)/ζdζ, we obtain:

Corollary 3.8. For all fS2, we have,

V3z,t0zf(ζ)ζdζσ3(1)z,t0zf(ζ)ζdζ,

in D.

Remark 3.9. In 1990, Komatu [5] introduced the integral operator, Lcβ:AA, by

Lcβ[f](z):=(1+c)βΓ(β)t01log1tβ-1tc-1f(tz)dt(c>-1,β0).

It is easy to verify that if f(z)=z+n=2anzn, then Lcβ[f] defined above can be expressed by the series expansion as follows:

Lcβ[f](z)=z+n=2c+1c+nβanzn.

In view of (1.1), we obtain:

L0β[f](z)=(Liβ*f)(z).

Therefore, Theorem 3.5 can be restated as under:

Theorem 3.10. For all real numbers β,ββ0, and for all fK, we have

V3z,L0β[f]σ3(1)z,L0β[f],

in D, where the number β0 is as in Lemma 3.3.

### Acknowledgements

The first author, Manju Yadav, acknowledges financial support from CSIR-UGC, Govt. of India, in the form of JRF vide Award Letter No. 211610111581.

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