For a real number r, 0<r<1 let Dr={z∈C:|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

Analytically, f∈K if and only if

If C denotes the class of close-to-convex functions in D, then S2⊂K⊂C⊂S.

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

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

Here Γ stands for Euler gamma function. Note that

and

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

where f is periodic real-valued function and

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=0∞anzn and g(z)=∑n=0∞bnzn is defined as the power series ∑n=0∞anbnzn 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

For a given function f(z)=∑n=1∞anzn,

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

Similarly, for a real number α≥0, if

then the polynomial,

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 α≥1 and n∈N, 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 α≥1 and n∈N, σn(α)(z,f)∈C for all f∈K. 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 f≺g, 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 f≺g 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 f∈K, 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 f∈K and for all α≥1.

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

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.

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=1​∞anzn, a1=1, is univalent and convex in D, we have

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

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,n≥2. Assume that Q∈Pn has all its critical points ζj in D¯,j=1,2,…,n-1. Let P∈Pn satisfy P(0)=Q(0) and

Then P≺Q in D.

Lemma 2.5. [9] Given a polynomial,

of degree n, let

where Ak and Bk are triangular matrices

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.

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 {w∈C:ℜ(w)>-1/2}. This function is a distinguished member of the class K. We claim that

in D. We note that

and

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

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

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

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

where

Using Mathematica [17], we get

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), z∈D.

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:

For β≥0, as Liβ is convex and convolution of two convex functions is convex, so, Kβ⊂K. Obviously, in view of (1.2) and (1.3), we have

and

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 β≥β0, where β0 (≈1.039) is the positive root of the transcendental equation

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

Then

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

Putting z=eiθ,0≤θ<2π , we get

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

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

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

at the critical point given by

Using Mathematica [17], we conclude that min​T(θ)≥0 for β≥β0 (≈1.039), where β0 is the positive root of (3.2). As, ℜ1+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 β≥β0.

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

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

and

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

where

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

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 ℜ(H(z))=1 at z=0, so we conclude that ℜ(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, β≥β0, and for all f∈K, we have,

in D; or, equivalently,

in D for all g∈Kβ (β≥β0).

Proof. By Lemma 3.3, σ3(1)(z,Liβ) is convex univalent in D for all β≥β0. 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 f∈K,

in D.

If f∈S2, then zf'∈K and Liβ*zf'=Liβ-1*f. Thus we get:

Corollary 3.7. For all real numbers β, β≥β0, and for all f∈S2, we have,

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 f∈S2, we have,

in D.

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

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

In view of (1.1), we obtain:

Therefore, Theorem 3.5 can be restated as under:

Theorem 3.10. For all real numbers β, β≥β0, and for all f∈K, we have

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

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