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Coefficient inequalities for a subclass of Bazilevič functions

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Abstract

Let f be analytic in D={z:|z| < 1} with f(z)=z+n=2anzn, and for α ≥ 0 and 0 < λ ≤ 1, let 1(α,λ) denote the subclass of Bazilevič functions satisfying |f(z)(zf(z))1α1|<λ for 0 < λ ≤ 1. We give sharp bounds for various coefficient problems when f1(α,λ), thus extending recent work in the case λ = 1.

1 Definitions and preliminaries

Denote by A the class of analytic functions f defined for zD={z:|z|<1}, and normalized so that f(0) = 0 and f′(0) = 1, and by S the subclass of A consisting of functions that are univalent in D={z:|z|<1}. Let f be given by

(1)f(z)=z+n=2anzn.

Then, for α > 0, it was shown by Bazilevič [1] that if fA and is given by eq. (1), then there exists starlike functions g such that

Rezf(z)f(z)1αg(z)α>0,

it follows that fS. We denote this class of Bazilevič functions by (α), so that (α)S when α > 0.

The case α = 0 was subsequently considered by Sheil-Small [2], who showed that (α)S when α ≥ 0.

Taking g(z) ≡ z gives the class 1(α) of Bazilevič functions, which has been the subject of much recent research. We note that 1(0) is the class S of starlike functions, and 1(1) the well-known class of functions whose derivative has positive real part in D.

Thus, f1(α), if and only if, for zD

Ref(z)(zf(z))1α>0.

Various properties have been obtained for functions in 1(α). Among other results, Singh [3] found sharp estimates for the moduli of the first four coefficients and obtained the solution to the Fekete–Szegö problem. Sharp bounds for the second Hankel determinant, the initial coefficients of the function log(f(z)/z), and the initial coefficients of the inverse function f−1 were obtained in [4], and distortion theorems and some length–area results were also obtained in [5,6,7].

We now define the subclass 1(α,λ) of 1(α), which was introduced in 1996 by Ponnusamy and Singh [8, Theorem 3]. In this article, the authors determine condition on λ so that functions in 1(α,λ) are starlike in D (see also [9, Theorem 3] for an extension of this result). Later in [10, Theorems 1 and 2], the authors considered complex values of α and obtained condition on λ such as the functions in 1(α,λ) are spirallike in D. Two of the present authors in [11] studied the class 1(α,λ) in the case λ = 1. Therefore, it is natural to consider the investigation of the problems discussed in this article for the complex values of α in the context of the investigation from [10].

Definition 1.1

Let fS and be given by eq. (1). Then, for α ≥ 0 and 0 < λ ≤ 1, f1(α,λ), if and only if, for zD

(2)|f(z)(zf(z))1α1|<λ.

We note that f1(0,1) reduces to the class of bounded starlike functions considered by Singh [12].

Although the aforementioned definition requires that α ≥ 0, choosing α = −1 gives the class U(λ) of univalent functions defined for zD by

|f(z)(zf(z))21|<λ.

The class U(λ) has been the focus of a great deal of research in recent years (see e.g. [13,14], and for a summary of some known results, see [15]). Although the classes 1(α,λ) for α ≥ 0 and U(λ) have similar structural representations, they are fundamentally different in many ways, and we shall see in the following analysis that the methods used in this study cannot be applied to the class U(λ). It is also interesting to note that the only known negative value of α which gives a subset of S appears to be α = −1. See [16,17] and references therein for recent investigation, which also deals with the case α = −1 for meromorphic functions.

In this study, we give sharp bounds for the modulus of the coefficients an for f1(α,λ) when 2 ≤ n ≤ 5, together with other related results, noting that when fU(λ), sharp bounds have been found only for some initial coefficients.

First note that from eq. (2), we can write

(3)f(z)(zf(z))1α=1+λω(z),

for zD, where ω is the Schwarz function.

Next, recall the class P of functions with positive real part in D, so that hP, if, and only if, Re h(z) > 0 for zD.

We write

(4)h(z)=1+n=1cnzn.

Thus, as we can write

h(z)=1+ω(z)1ω(z),

eq. (3) can be written as

(5)f(z)(zf(z))1α=1+λh(z)1h(z)+1.

We shall use the following results concerning the coefficients of hP.

Lemma 1.1

[18] IfhP, then for some complex valued x with|x|1, and some complex valued ζ with |ζ| ≤ 1,

2c2=c12+x(4c12),4c3=c13+2(4c12)c1xc1(4c12)x2+2(4c12)(1|x|2)ζ.

Lemma 1.2

[19] IfhP, then |cn| ≤ 2 for n ≥ 1, and

|c2μ2c12|max{2,2|μ1|}={2,0μ2,2|μ1|,elsewhere.

Lemma 1.3

[19] IfhP, then

|c3(1+μ)c1c2+μc13|max{2,2|2μ1|}={2,0μ1,2|2μ1|,elsewhere.

Lemma 1.4

[19] LethP. If 0 ≤ B ≤ 1 and B(2B − 1) ≤ DB, then

|c32Bc1c2+Dc13|2.

Lemma 1.5

[20] IfhP, and α1, α2, β1and β2satisfy 0 < α1 < 1, 0 < α2 < 1 and

(6)8α1(1α1)((α2β22β1)2+(α2(α1+α2)β2)2)+α2(1α2)(β22α1α2)24α22(1α2)2α1(1α1),
then
|β1c14+α1c22+2α2c1c3(3/2)β2c12c2c4|2.

2 Initial coefficients

We first give sharp bounds for some initial coefficients for f1(α,λ), extending those given in [11].

Theorem 2.1

Letf1(α,λ)for α ≥ 0 and 0 < λ ≤ 1 and be given by eq. (1).

Then, for 2 ≤ n ≤ 5,

|an|λ(α+n1).
The inequalities are sharp.

Proof

Equating coefficients in eq. (5) gives

(7)a2=c1λ2(1+α),a3=λ2(2+α)(c2(2+4α+2α22λ+αλ+α2λ)4(1+α)2c12),a4=λ2(3+α)(c3(4+6α+2α23λ+2αλ+α2λ)2(1+α)(2+α)c1c2+(12+42α+54α2+30α3+6α418λ24αλ+12α2λ+24α3λ+6α4λ+6λ213αλ2+6α4λ+6λ213αλ22α2λ2+7α3λ2+2α4λ2)c1324(1+α)3(2+α)),a5=λ2(4+α)(Mc14+Nc22+2Qc1c332Rc12c2c4),

where

N=8+8α+2α24λ+3αλ+α2λ4(2+α)2,
Q=6+8α+2α24λ+3αλ+α2λ4(1+α)(3+α),
M=(288+1536α+3432α2+4152α3+2928α4+1200α5+264α6+24α7528λ1764αλ1704α2λ+444α3λ+1872α4λ+1284α5λ+360α6λ+36α7λ+288λ2+24αλ21068α2λ2744α3λ2+552α4λ2+696α5λ2+228α6λ2+24α7λ248λ3+212αλ3116α2λ3295α3λ3+37α4λ3+149α5λ3+55α6λ3+6α7λ3)/(192(1+α)4(2+α)2(3+α)),
R=(72+240α+306α2+186α3+54α4+6α588λ118αλ+40α2λ+112α3λ+48α4λ+6α5λ+24λ246αλ221α2λ2+26α3λ2+15α4λ2+2α5λ2)/(12(1+α)2(2+α)2(3+α)).

The inequality for |a2| is trivial.

For a3, we apply Lemma 1.2 with

μ=(2+4α+2α22λ+αλ+α2λ)2(1+α)2.

Since 0 ≤ μ ≤ 2 for α ≥ 0 and 0 < λ ≤ 1, the inequality for |a3| follows.

For a4, we use Lemma 1.4 with

B=(4+6α+2α23λ+2αλ+α2λ)4(1+α)(2+α),

and

D=(12+42α+54α2+30α3+6α418λ24αλ+12α2λ+24α3λ+6α4λ+6λ213αλ2+6α4λ+6λ213αλ22α2λ2+7α3λ2+2α4λ2)/(24(1+α)3(2+α)).

Since 0 ≤ B ≤ 1, and B(2B − 1) ≤ DB, when α ≥ 0 and 0 < λ ≤ 1, the inequality for |a4| follows.

For a5, we apply Lemma 1.5 with α1, α2, β1 and β2 the respective coefficients of a5 in eq. (7). Since 0 < α1 < 1 and 0 < α2 < 1, for α ≥ 0 and 0 < λ ≤ 1, then by expanding both sides and subtracting, it is easily seen that the conditions (6) of Lemma 1.5 are satisfied (the detailed proof of this step can be found in [21]), and so the inequality for |a5| follows. The inequality of |ai| is sharp on choosing ci = 2 when 2 ≤ i ≤ 5, and cj = 0 when ij.□

3 Inverse coefficients

Since 1(α,λ)S, inverse functions f−1 exist, and so we can write

(8)f1(w)=w+n=2Anwn,

valid in some disk |w| ≤ r0(f). It is an easy exercise to show that

(9)A2=a2,A3=2a22a3,A4=5a23+5a2a3a4.

We first prove the following.

Theorem 3.1

Iff1(α,λ), with inverse coefficients given by eq. (8), then for 0 < λ ≤ 1,

|A2|λ1+αwhenα0,|A3|λ2+αwhen0α12(1+17),and0<λ2(1+α)2(2+α)(3+α),λ2+αwhenα>12(1+17),and0<λ1,(3+α)λ22(1+α)2when0α12(1+17),and2(1+α)2(2+α)(3+α)<λ1.
All the inequalities are sharp.

Proof

Substituting eq. (7) into eq. (9) gives

(10)2(1+α)A2=λc1,2(2+α)A3=λ(c22(1+α)2+(2+α)(3+α)λ4(1+α)2c12).

The inequality for |A2| is trivial, since |c1| ≤ 2.

For A3, we use Lemma 1.2 with

μ=2(1+α)2+(2+α)(3+α)λ2(1+α)2,

and the inequalities for |A3| easily follow. The inequality for |A2| is sharp when c1 = 2. The first and second inequalities for |A3| are sharp on choosing c1 = 0 and c2 = 2. The third inequality for |A3| is sharp when c1 = c2 = 2.□

When λ = 1, obtaining sharp bounds for |A4| follows relatively easily from an application of Lemmas 1.3 and 1.4 [11]. However, finding sharp bounds when 0 < λ ≤ 1 appears to be a much more difficult problem, as the next theorem demonstrates.

The inequalities for |A4| for f1(α,λ) are complicated, and in the interest of brevity, we omit many of the detailed calculations. Also, to simplify the analysis and presentation of the results, we define Γi(α) for i = 1, 2, 3 as follows:

Γ1(α)=3(1+α)3(2+α)(3+α)(4+α),Γ2(α)=2(1+α)(2+α)(3+α)(4+α),Γ3(α)=6(1+α)3(2+α)2(3+α)(4+α)(20+33α+12α2+α3).

We also denote the positive real root of the equation 21 + 17α − 2α3 = 0 by α1=3.40366, α2=12(1+33)=3.37228, and the positive real root of the equation 4 + 9αα3 = 0 by α3=3.20147.

Theorem 3.2

Iff1(α,λ), with inverse coefficients given by eq. (8), then

(11)|A4|λ3+α,
when either (i)α>α1and 0 < λ ≤ 1; (ii)α2αα1and 0 < λΓ1(α); (iii)α3<α<α2and 0 < λΓ1(α); or (iv)0αα3, and 0 < λΓ3(α).

Also,

(12)|A4|(2+α)(4+α)λ33(1+α)3,
when either (v)α2αα1and Γ1(α) ≤ λ ≤ 1; (vi)α3<α<α2and Γ1(α) < λΓ2(α); or (vii)α3<α<α2and Γ2(α) < λ ≤ 1.

All the inequalities are sharp.

We note that using the aforementioned lemmas, Theorem 3.2 proves that sharp inequalities for |A4| are established for α ≥ 0 and 0 < λ ≤ 1, apart from the intervals 0αα3 and Γ3(α) < λ ≤ 1, where the methods used fail.

Proof

Again on substituting eq. (7) in eq. (9) we have

(13)A4=λ2(3+α)(c3(4+6α+2α2+12λ+7αλ+α2λ)2(1+α)(2+α)c1c2+c1312(1+α)3(2+α)(6+21α+27α2+15α3+3α4+36λ+93αλ+81α2λ+27α3λ+3α4λ+48λ2+76αλ2+44α2λ2+11α3λ2+α4λ2)).

To find the maximum of the modulus of eq. (13), we first use Lemmas 1.3 and 1.4.

Let

B=(4+6α+2α2+12λ+7αλ+α2λ)4(1+α)(2+α),

and

D=112(1+α)3(2+α)(6+21α+27α2+15α3+3α4+36λ+93αλ+81α2λ+27α3λ+3α4λ+48λ2+76αλ2+44α2λ2+11α3λ2+α4λ2)

To see that eq. (11) holds in cases (i)–(iv), we use Lemma 1.4, noting that a long computation shows that both 0 ≤ B ≤ 1 and B(2B − 1) ≤ DB are valid in all cases. This proves inequality (11).

For inequality (12), we write eq. (13) as

(14)λ2(3+α)(c32Bc1c2+Bc13)=λ2(3+α)(c32Bc1c2+Bc13+(DB)c13).

In this case, we can therefore apply Lemma 1.4, provided that both 0 ≤ B ≤ 1 and DB ≥ 0 are valid, and again a long computation shows that these inequalities are valid in cases (v) and (vi).

A simple calculation shows that

DB=112((2+α)(3+α)(4+α)λ2(1+α)33),

and so we obtain from Lemma 1.4 and the inequality |c1| ≤ 2 that

|A4|λ2(3+α)(2+23((2+α)(3+α)(4+α)λ2(1+α)33))=(2+α)(4+α)λ33(1+α)3.

We are therefore left to prove eq. (12) in case (vii), where we use Lemma 1.3, with

μ=(3+α)(4+α)λ2(1+α)(2+α).

Then, μ > 1 when 0α<α2, and Γ2(α) < λ ≤ 1, and Dμ ≥ 0 when α3α<α2 and Γ2(α) < λ ≤ 1, and so both inequalities are satisfied when α3α<α2 and Γ2(α) < λ ≤ 1.

Thus, writing

A4=λ2(3+α)(c3(μ+1)c1c2+μc13+(Dμ)c13),

Lemma 1.3, and the inequality |c1| ≤ 2, gives

|A4|λ2(3+α)(2|2μ1|+8(Dμ))=(2+α)(4+α)λ33(1+α)3,

provided α3α<α2 and Γ2(α) < λ ≤ 1, which gives inequality (12) in case (vii).□

4 The logarithmic coefficients

The logarithmic coefficients γn of f are defined in D by

(15)logf(z)z=2n=1γnzn.

Differentiating eq. (15) and equating coefficients give

(16)γ1=12a2,γ2=12(a312a22),γ3=12(a4a2a3+13a23),γ4=18(a24+4a22a32a324a2a4+4a5).

Using the same techniques as in the proof of Theorem 2.1, it is possible to prove the following (proofs can be found in [21]).

Theorem 4.1

Letf1(α,λ)for α ≥ 0 and 0 < λ ≤ 1 with logarithmic coefficients given by eq. (16). Then, for 1 ≤ n ≤ 4,

(17)|γn|λ2(n+α).
All the inequalities are sharp.

5 The second Hankel determinant

The qth Hankel determinant of f is defined for q ≥ 1 and n ≥ 1 as follows and has been extensively studied (see e.g. [22,23,24,25])

Hq(n)=|anan+1an+q+1an+1an+q1an+2q2|.

We prove the following, noting that the result is valid for α ≥ 0.

Theorem 5.1

Iff1(α,λ), then for α ≥ 0, and 0 < λ ≥ 1,

H2(2)=|a2a4a32|λ2(2+α)2.
The inequality is sharp.

Proof

We use the idea first developed in [23].

Equating coefficients in eq. (7) gives

(18)H2(2)=c14λ216(1+α)(2+α)2(3+α)c12c2λ24(1+α)(2+α)2(3+α)c22λ24(2+α)2+c1c3λ24(1+α)(3+α)(1α)c14λ4192(1+α).3

Next applying Lemma 1.1, noting that H2(2) is rotationally invariant, and again writing c1:=c, so that 0 ≤ c ≤ 2, it follows that

(19)H2(2)=c2(4c2)ζ2λ216(1+α)(3+α)(4c2)2ζ2λ216(2+α)2(1α)c4λ4192(1+α)3+c(4c2)λ2(1|ζ|2)η8(1+α)(3+α).

Taking the modulus in eq. (19), and noting that |η| ≤ 1, gives

(20)H2(2)c2(4c2)|ζ|2λ216(1+α)(3+α)+(4c2)2|ζ|2λ216(2+α)2+|1α|c4λ4192(1+α)3+c(4c2)λ2(1|ζ|2)8(1+α)(3+α):=ψ(α,λ,|ζ|,c).

Since the derivative of ψ(α, λ, |ζ|, c) with respect to |ζ| is positive, we deduce from eq. (20) that

(21)H2(2)c2(4c2)λ216(1+α)(3+α)+(4c2)2λ216(2+α)2+|1α|c4λ4192(1+α)3:=ψ1(α,λ,c).

Thus, we must find the maximum value of ψ1(α, λ, c), when 0 ≤ c ≤ 2.

Elementary calculus shows that ψ1(α,λ,c)=0 has three roots in 0 ≤ c ≤ 2, but the only valid root is at c = 0.

Since ψ1(α,λ,0)=λ2(2+α)2 and ψ1(α,λ,2)=λ4|1α|12(1+α)3, the proof of the theorem is complete on noting that ψ1(α, λ, 0) ≥ ψ1(α, λ, 2) when α ≥ 0 and 0 < λ ≤ 1.

The inequality is sharp on choosing c1 = 0 and c2 = c3 = 2 in eq. (18).□

6 A Fekete–Szegö theorem

We finally give a sharp Fekete–Szegö inequality for B1(α,λ) omitting the proof, which is a straightforward application of Lemma 1.2.

Theorem 6.1

Letf1(α,λ). Then, for α ≥ 0, 0 < λ ≤ 1, andν,

|a3νa22|{λ2(α1+2ν)2(1+α)2,ifν5α+3α24+2α,λ2+α,if5α+3α24+2αν4+3α+α24+2α,λ2(α1+2ν)2(1+α)2,ifν4+3α+α24+2α.
All the inequalities are sharp.

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Received: 2019-08-23
Revised: 2020-01-29
Accepted: 2020-01-30
Published Online: 2020-05-07

© 2020 Sa’adatul Fitri et al., published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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  30. Exponential spline method for singularly perturbed third-order boundary value problems
  31. Existence results of noninstantaneous impulsive fractional integro-differential equation
  32. Review Articles
  33. On a characterization of exponential, Pearson and Pareto distributions via covariance and pseudo-covariance
Heruntergeladen am 10.5.2026 von https://www.degruyterbrill.com/document/doi/10.1515/dema-2020-0040/html?lang=de
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