The Bound of the Hankel Determinant of the Third Kind for Starlike Functions

In the present paper, the estimate of the third Hankel determinant H3,1(f)=a1a2a3a2a3a4a3a4a5\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \begin{aligned} H_{3,1}(f)&= \begin{vmatrix} a_{1}&a_{2}&a_{3} \\ a_{2}&a_{3}&a_{4} \\ a_{3}&a_{4}&a_{5} \end{vmatrix} \end{aligned} \end{aligned}$$\end{document}for the class of starlike functions, i.e., for the class of analytic functions f standardly normalized such that Re(zf′(z)/f(z))>0,z∈D:={z∈C:|z|<1},\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{\mathrm{Re}}}(zf'(z)/f(z)) > 0,\ z\in {{\mathbb {D}}}:=\{z \in {\mathbb {C}} : |z|<1\},$$\end{document} is improved.

Particularly, the third Hankel determinant H 3,1 ( f ) is given by To find the growth of the Hankel determinant H q,n ( f ) dependent on q and n for the whole class S ⊂ A of univalent functions as well as for its subclasses is an interesting problem to study. For the class S some important result was shown by Pommerenke [13]. For fixed q and n the growth problem can be reduced to an estimate of the Hankel determinant for the selected subclasses of A. Recently many authors examined the Hankel determinant H 2,2 ( f ) = a 2 a 4 − a 2 3 of order 2 (see, e.g., [3,4,6,8,12]). Note also that H 2,1 ( f ) = a 3 − a 2 2 . Thus the Hankel determinant H 2,1 ( f ) reduces to the well-known coefficient functional which for S was estimated in 1916 by Bieberbach (see, e.g., [5,Vol. I,p. 35]).
The problem to find the upper bound of the Hankel determinant H 3,1 ( f ) of order 3 is more sophisticated if we expect to get sharp result. From (1.3) by using the triangle inequality we get at once the following inequality This simple observation allowed to estimate of |H 3,1 ( f )| for compact subclasses F of A by various authors (see, e.g., [2,[15][16][17][18]). However, these results are far from sharpness. If case when a given subclass F of A has a representation with using the Carathéodory class P, i.e., the class of functions p ∈ H of the form having a positive real part in D, the coefficients of functions in F have a suitable representation expressed by the coefficients of functions in P. Therefore to get the upper bound of each term in (1.4) cited authors based their computing on the wellknown formulas on coefficient c 2 (e.g., [14, p. 166]) and on the formula c 3 due to Libera and Zlotkiewicz [9]. In order to improve the bound of |H 3,1 ( f )| we have to use directly formula (1.3), where we need to apply a formula for c 4 , similar to the formulas (2.1) and (2.2). In a recent paper [7] the authors found such a formula for c 4 . According to the authors' knowledge, formulas for the coefficients c n for n ≥ 5 analogous to the formulas (2.1) and (2.2) are not known.

Main Result
The basis for proof of the main result is the following lemma. It contains the wellknown formula for c 2 (e.g., [14, p. 166]), the formula for c 3 due to Libera and Zlotkiewicz [9,10] and the formula for c 4 found by the authors [7]. and for some ζ, η, ξ ∈ D := {z ∈ C : |z| ≤ 1}. Now, we will estimate the third-order Hankel determinant H 3,1 ( f ) for f ∈ S * . To this end, the following propositions are required.
Proof At first, note that the polynomial θ 3 has a unique zero x =: x 1 ≈ 0.9314 in (0, 1). Since x 1 ∈ (0.92, 0.95) and for x ∈ (0.92, 0.95), We have The polynomial θ 4 has exactly two zeros in (0, 1), namely, x =: x 2 ≈ 0.533701 and x =: x 3 ≈ 0.811327. We will now show that (2.5) Since x 1 > 0.9, so θ 3 (x) > 0, for x ∈ [0.5, 0.9] and the inequality (2.5) is equivalent to The above one can be equivalently written as As the polynomial on the left hand of the above inequality has a unique zero x ≈ 0.40928 in [0, 1], the above inequality is true, so is the inequality (2.5). Thus the function has no critical point in (0, 3) × (0, 1). Hence it is sufficient to show that > 0 on the boundary of [0, 3] × [0, 1]. We can easily check that the following inequalities hold: Thus the proof of the proposition is completed.

and
(1, Thus the proof of the proposition is completed.

Now using the equalities (2.1)-(2.3), by straightforward algebraic computation we have
By (2.15) we get A simple computing yields Hence and by Proposition 2.2 we have where the function is defined by (2.4). Thus the function [0, 3] t →F 1 (t, ·) is increasing, and therefore we havẽ (2.16) Indeed, the last inequality is true since, as easy to verify the inequality holds. Thus the inequality (2.16) confirms the inequality (2.14). ( and repeating the argumentation of Part (a) we get the inequality (2.14).

By (2.18) we get
A simple computing yields Hence and by Proposition 2.4 we have where the function is defined by (2.8). Thus the function (3, 4) t →F 1 (t, ·) is increasing, and therefore we havẽ Indeed, the last inequality is true since so is the following one Thus the inequality (2.19) confirms the inequality (2.14).

Remark 2.6
Although the constant 8/9 improves essentially the estimates found in [1] and [19], it is not the best possible. To find the sharp estimate of the Hankel determinant H 3,1 ( f ) for starlike functions is still an open problem.
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