The Second Hankel Determinant Problem for a Class of Bi-Close-to-Convex Functions

Abstract

The purpose of the present work is to determine a bound for the functional $H_2\left(2\right)=a_2{a}_4-a_3^2$ for functions belonging to the class ${\mathcal{C}}_{\Sigma }$ of bi-close-to-convex functions. The main result presented here provides much improved estimation compared with the previous result by means of different proof methods than those used by others.

1. Introduction

Let $\mathcal{A}$ be a class of analytic functions in the open unit disk $\mathbb{D}=\{z\in \mathbb{C}:\left|z\right|<1\}$, of the form

$$f\left(z\right)=z+\sum _{n=2}^{\infty }{a}_n{z}^n\left(z\in \mathbb{D}\right).$$

(1)

Let $\mathcal{S}$ be the class of functions $f\in \mathcal{A}$ which are univalent in $\mathbb{D}$. A function $f\in \mathcal{A}$ is said to be starlike, if it satisfies the inequality

$$\mathrm{Re}\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)>0\left(z\in \mathbb{D}\right).$$

(2)

We denote by ${\mathcal{S}}^{*}$ the class which consists of all functions $f\in \mathcal{A}$ that are starlike. A function $f\in \mathcal{A}$ is said to be close-to-convex if there exits a function $g\in {\mathcal{S}}^{*}$ such that it satisfies the inequality

$$\mathrm{Re}\left(\frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}\right)>0\left(z\in \mathbb{D}\right).$$

(3)

We denote by $\mathcal{C}$ the class which consists of all functions $f\in \mathcal{A}$ that are close-to-convex. We note that ${\mathcal{S}}^{*}\subset \mathcal{C}\subset \mathcal{S}$ and that $\left|a_n\right|\le n$ for $f\in {\mathcal{S}}^{*}$.

For two functions f and g which are analytic in $\mathbb{D}$, we say that the function f is subordinate to g, and write $f\left(z\right)\prec g\left(z\right)$, if there exists a Schwarz function w, that is a function w analytic in $\mathbb{D}$ with $w\left(0\right)=0$ and $\left|w\left(z\right)\right|<1$ in $\mathbb{D}$, such that $f\left(z\right)=g\left(w\right(z\left)\right)$ for all $z\in \mathbb{D}$. In particular, if the function g is univalent in $\mathbb{D}$, then $f\prec g$ if and only if $f\left(0\right)=g\left(0\right)$ and $f\left(\mathbb{D}\right)\subseteq g\left(\mathbb{D}\right)$ [1].

In 1976, Noonan and Thomas [2] defined the q-th Hankel determinant for integers $n\ge 1$ and $q\ge 1$ by

$$H_q\left(n\right)=\left|\begin{array}{cccc}{a}_n& a_{n+1}& \cdots & a_{n+q-1}\\ a_{n+1}& a_{n+2}& \cdots & a_{n+q}\\ ⋮& ⋮& ⋮& ⋮\\ a_{n+q-1}& a_{n+q}& \cdots & a_{n+2q-2}\end{array}\right|\left(a_1=1\right).$$

In general, one of the important tools in the theory of univalent functions is the Hankel determinant. It is used, for example, in showing that a function of bounded characteristic in $\mathbb{D}$, that is, a function which is a ratio of two bounded analytic functions with its Laurent series around the origin having integral coefficients, is rational [3]. For the use of Hankel determinant in the study of meromorphic functions, see [4]. For detailed information, the readers are encouraged [5,6]. Various properties of these determinants can be found in [7] (Chapter 4). The investigations of Hankel determinants for different classes of analytic functions started in the 1960s. Pommerenke [8] proved that the Hankel determinants of univalent functions satisfy $\left|H_q\left(n\right)\right|\le K{n}^{-(\frac{1}{2}+\beta )q+\frac{3}{2}}$ where $n,q\in \mathbb{N},q\ge 2,\beta >1/4000$ and K depends only on q. Later, Hayman [9] proved that $\left|H_q\left(n\right)\right|\le A{n}^{\frac{1}{2}}$ where $n\in \mathbb{N}$ and A is an absolute constant for areally mean univalent functions. Pommerenke [10] investigated the Hankel determinant of areally mean p-valent functions, univalent functions as well as of starlike functions. For results related to these determinants, see also [11,12].

Note that

$$H_2\left(1\right)=\left|\begin{array}{cc}{a}_1& a_2\\ a_2& a_3\end{array}\right|\mathrm{and}{H}_2\left(2\right)=\left|\begin{array}{cc}{a}_2& a_3\\ a_3& a_4\end{array}\right|,$$

where the Hankel determinants $H_2\left(1\right)=a_3-a_2^2$ and $H_2\left(2\right)=a_2{a}_4-a_3^2$ are well-known as Fekete-Szegö and second Hankel determinant functionals, respectively. Further, Fekete and Szegö [13] introduced the generalized functional $a_3-\lambda a_2^2$, where $\lambda$ is some real number. In recent years, the research on Hankel determinants has focused on the estimation of $\left|H_2\left(2\right)\right|$. Problems in this field has also been argued by several authors for various classes of univalent functions [14,15,16,17,18,19,20,21,22,23,24].

The Koebe one-quarter theorem [1] ensures that the image of $\mathbb{D}$ under every univalent function $f\in \mathcal{S}$ contains a disk of radius $1/4.$ Thus every function $f\in \mathcal{S}$ has an inverse $f^{-1}$, such that

$$f^{-1}\left(f\left(z\right)\right)=z\left(z\in \mathbb{D}\right),\mathrm{and}f\left(f^{-1}\left(w\right)\right)=w\left(\left|w\right|<r_0\left(f\right);r_0\left(f\right)\ge \frac{1}{4}\right),$$

where the inverse $f^{-1}$ has the power series expansion (see [25])

$$f^{-1}\left(w\right)=w-a_2{w}^2+\left(2a_2^2-a_3\right)w^3-\left(5a_2^3-5a_2{a}_3+a_4\right)w^4+\cdots .$$

(4)

A function $f\in \mathcal{A}$ is said to be bi-univalent in $\mathbb{D}$ if both f and $f^{-1}$ are univalent in $\mathbb{D}$, in the sense that $f^{-1}$ has a univalent analytic continuation to $\mathbb{D}$. Let $\Sigma$ denote the class of bi-univalent functions in $\mathbb{D}$. For a brief history of functions in the class $\Sigma$ and also other different characteristics of these functions and the coefficient problems, see [25,26,27,28,29,30,31,32] and the references therein.

In 2014, Hamidi and Jahangiri [33] defined the class of bi-close-to-convex functions of order $\alpha (0\le \alpha <1)$ that this class is denoted by ${\mathcal{C}}_{\Sigma }\left(\alpha \right)$ and in particular, ${\mathcal{C}}_{\Sigma }\left(0\right)={\mathcal{C}}_{\Sigma }$.

Definition 1.

A function $f\in \Sigma$ is in the class of bi-close-to-convex functions of order α if the following conditions are satisfied:

$$\mathrm{Re}\left(\frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}\right)>\alpha \left(z\in \mathbb{D}\right)$$

(5)

and

$$\mathrm{Re}\left(\frac{w{F}^{\prime }\left(w\right)}{G\left(w\right)}\right)>\alpha \left(w\in \mathbb{D}\right),$$

(6)

where the function $F\left(w\right)=f^{-1}\left(w\right)$ is defined by (4), $g\left(z\right)=z+{\displaystyle \sum _{n=2}^{\infty }}{b}_n{z}^n\in {\mathcal{S}}^{*}$ and

$$G\left(w\right)=w+\sum _{n=2}^{\infty }{B}_n{z}^n\in {\mathcal{S}}^{*}.$$

(7)

Recently, Güney et al. [34] obtained the bound for the second Hankel determinant $H_2\left(2\right)$ for the class ${\mathcal{C}}_{\Sigma }$ of bi-close-to-convex functions as follows:

Theorem 1.

Let the function f given by (1) be in the class ${\mathcal{C}}_{\Sigma }$ and $G\left(w\right)=g^{-1}\left(w\right)$. Then

$$\left|H_2\left(2\right)\right|:=\left|a_2{a}_4-a_3^2\right|\le \frac{353}{36}.$$

Remark 1.

By means of the subordination, the conditions (5) and (6) are, respectively, equivalent to

$$\frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}\prec \frac{1+z}{1-z}\mathit{and}\frac{w{F}^{\prime }\left(z\right)}{G^{(}w)}\prec \frac{1+w}{1-w}.$$

The main purpose of this paper is to determine bounds for the functional $H_2\left(2\right)=a_2{a}_4-a_3^2$ for functions belonging to the subclass ${\mathcal{C}}_{\Sigma }$ of bi-close-to-convex functions, which is a much improved estimation than the previous result given by Güney et al. [34]. We note that our proof method is by means of the subordination and more direct than those used by others and so we get a smaller upper bound and more accurate estimation for the functional $|H_2\left(2\right)|$ for functions in the class ${\mathcal{C}}_{\Sigma }$.

2. Main Results

Theorem 2.

Let the function f given by (1) be in the class ${\mathcal{C}}_{\Sigma }$ and $G\left(w\right)=g^{-1}\left(w\right)$. Then

$$\left|H_2\left(2\right)\right|:=\left|a_2{a}_4-a_3^2\right|\le \frac{227}{36}.$$

In order to prove our main result, we need the following lemmas.

Lemma 1.

[1] (p. 190) Let u be analytic function in the unit disk $\mathbb{D}$, with $u\left(0\right)=0$, and $\left|u\left(z\right)\right|<1$ for all $z\in \mathbb{D}$, with the power series expansion

$$u\left(z\right)=\sum _{n=1}^{\infty }{c}_n{z}^n.$$

Then, $\left|c_n\right|\le 1$ for all $n\in \mathbb{N}$. Furthermore, $\left|c_n\right|=1$ for some $n\in \mathbb{N}$ if and only if $u\left(z\right)=e^{i\theta }{z}^n$, $\theta \in \mathbb{R}$.

Lemma 2.

[20] If $\psi \left(z\right)={\displaystyle \sum _{n=1}^{\infty }}{\psi }_n{z}^n$, $z\in \mathbb{D}$, is a Schwarz function with ${\psi }_1\in \mathbb{R}$, then

$$\begin{array}{cc}\hfill {\psi }_2& =x\left(1-{\psi }_1^2\right),\hfill \\ \hfill {\psi }_3& =\left(1-{\psi }_1^2\right)\left(1-{\left|x\right|}^2\right)s-{\psi }_1\left(1-{\psi }_1^2\right)x^2,\hfill \end{array}$$

for some $x,s$, with $\left|x\right|\le 1$ and $\left|s\right|\le 1$.

Lemma 3.

[35] Let the function $f\in {\mathcal{S}}^{*}$ be given by (1). Then, for any real number μ,

$$\left|a_3-\mu a_2^2\right|\left\{\begin{array}{ccc}3-4\mu \hfill & \mathrm{if}\hfill & \mu \le \frac{1}{2}\hfill \\ \hfill & \hfill \\ 1\hfill & \mathrm{if}\hfill & \frac{1}{2}\le \mu \le 1\hfill \\ \hfill & \hfill \\ 4\mu -3\hfill & \mathrm{if}\hfill & \mu \ge 1.\hfill \end{array}\right.$$

Lemma 4.

[19] Let the function $f\in {\mathcal{S}}^{*}$ be given by (1). Then

$$\left|H_2\left(2\right)\right|:=\left|a_2{a}_4-a_3^2\right|\le 1.$$

Equality holds true for the Koebe function $k\left(z\right)={\displaystyle \frac{z}{{(1-z)}^2}}$.

Lemma 5.

[36] Let the function $f\in {\mathcal{S}}^{*}$ be given by (1). Then

$$\left|a_2{a}_3-a_4\right|\le 2.$$

Equality holds true for the Koebe function $k\left(z\right)={\displaystyle \frac{z}{{(1-z)}^2}}$.

Proof of Theorem 2.

As noted in Remark 1, if $f\in {\mathcal{C}}_{\Sigma }$, then by definition of subordination, there exist two Schwarz functions u and v, of the form $u\left(z\right)={\displaystyle \sum _{n=1}^{\infty }}{c}_n{z}^n$ and $v\left(z\right)={\displaystyle \sum _{n=1}^{\infty }}{d}_n{z}^n$, $z\in \mathbb{D}$ that we can write

$$\frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}=\frac{1+u\left(z\right)}{1-u\left(z\right)}=1+2c_{1}z+(2c_2+2c_1^2)z^2+(2c_3+4c_1{c}_2+2c_1^3)z^3+\cdots$$

and

$$\frac{z{F}^{\prime }\left(z\right)}{G\left(z\right)}=\frac{1+v\left(w\right)}{1-v\left(w\right)}=1+2d_{1}w+(2d_2+2d_1^2)w^2+(2d_3+4d_1{d}_2+2d_1^3)w^3+\cdots .$$

Equating coefficients in two above relations then gives

$$\begin{array}{cc}\hfill & 2a_2-b_2=2c_1,\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill & 3a_3-b_3-2a_2{b}_2+b_2^2=2c_2+2c_1^2,\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill & 4a_4-b_4-2a_2{b}_3+2b_2{b}_3-3a_3{b}_2+2a_2{b}_2^2-b_2^3=2c_3+4c_1{c}_2+2c_1^3,\hfill \end{array}$$

(10)

and

$$\begin{array}{cc}\hfill & -2a_2+b_2=2d_1,\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill & -3a_3+b_3-2a_2{b}_2-b_2^2+6a_2^2=2d_2+2d_1^2,\hfill \\ \hfill & -4a_4+b_4-2a_2{b}_3-3b_2{b}_3-3a_3{b}_2+2a_2{b}_2^2+2b_2^3-20a_2^3\hfill \end{array}$$

(12)

$$\begin{array}{cc}\hfill & +20a_2{a}_3+6a_2^2{b}_2=2d_3+4d_1{d}_2+2d_1^3,\hfill \end{array}$$

(13)

respectively. From (8) and (11), we get that

$$c_1=-d_1,$$

(14)

Also, according to the proof of [34] (Theorem), it is enough that we set $2c_1,2c_2+2c_1^2,2c_3+4c_1{c}_2+2c_1^3$ instead of $c_1,c_2,c_3$, and $2d_1,2d_2+2d_1^2,2d_3+4d_1{d}_2+2d_1^3$ instead of $d_1,d_2,d_3$ in relations (2.5)–(2.10) in [34], respectively. Thus we can write (2.20) in [34], as given below:

$$\begin{array}{cc}\left|a_2{a}_4-a_3^2\right|=\hfill & |\frac{1}{8}(b_2{b}_4-b_3^2)+\frac{1}{72}{b}_3^2+\frac{2}{8}(b_4-b_2{b}_3)c_1-\frac{10}{48}{b}_2{b}_3{c}_1\hfill \\ \hfill & +\frac{4}{24}\left(b_3-\frac{13}{4}{b}_2^2\right)c_1^2\hfill \\ \hfill & -\frac{7}{144}\left(b_3-\frac{19}{14}{b}_2^2\right)b_2^2-\frac{2}{9}\left(b_3-\frac{19}{16}{b}_2^2\right)(c_2-d_2)\hfill \\ \hfill & -\frac{10}{32}{b}_2{c}_1\left[4c_1^2-\frac{8}{15}(c_2-d_2)-\frac{13}{15}{b}_2^2\right]\hfill \\ \hfill & -\frac{2}{16}{c}_1\left[8c_1^3-\frac{4}{3}{c}_1(c_2-d_2)-2(c_3-d_3)-4c_1^3-4c_1(c_2+d_2)\right]\hfill \\ \hfill & +\frac{1}{16}{b}_2\left[2(c_3-d_3)+4c_1^3+4c_1(c_2+d_2)\right]-\frac{4}{36}{(c_2-d_2)}^2|.\hfill \end{array}$$

(15)

According to Lemma 2 and (14), we find that

$$c_2-d_2=\left(1-c_1^2\right)(x-y)\mathrm{and}{c}_2+d_2=\left(1-c_1^2\right)(x+y)$$

(16)

and

$$c_3=\left(1-c_1^2\right)\left(1-{\left|x\right|}^2\right)s-c_1\left(1-c_1^2\right)x^2\mathrm{and}{d}_3=\left(1-d_1^2\right)\left(1-{\left|y\right|}^2\right)t-d_1\left(1-d_1^2\right)y^2,$$

where

$$c_3-d_3=(1-c_1^2)\left[(1-|x{|}^2)s-(1-|y{|}^2)t\right]-c_1(1-c_1^2)(x^2+y^2)$$

(17)

for some $x,y,s,t$ with $\left|x\right|\le 1,\left|y\right|\le 1,\left|s\right|\le 1$ and $\left|t\right|\le 1$. Applying (16) and (17) in (15), it follows that

$$\begin{array}{cc}\hfill \left|a_2{a}_4-a_3^2\right|=& |\frac{1}{8}(b_2{b}_4-b_3^2)+\frac{1}{72}{b}_3^2+\frac{2}{8}(b_4-b_2{b}_3)c_1-\frac{10}{48}{b}_2{b}_3{c}_1\hfill \\ \hfill & +\frac{4}{24}\left(b_3-\frac{13}{4}{b}_2^2\right)c_1^2-\frac{7}{144}\left(b_3-\frac{19}{14}{b}_2^2\right)b_2^2\hfill \\ \hfill & -\frac{40}{32}{b}_2{c}_1^3+\frac{26}{96}{b}_2^3{c}_1-\frac{16}{16}{c}_1^4+\frac{8}{16}{c}_1^4+\frac{4}{16}{b}_2{c}_1^3\hfill \\ \hfill & +2\left(1-c_1^2\right)(x-y)\left[-\frac{1}{9}\left(b_3-\frac{19}{16}{b}_2^2\right)+\frac{8}{96}{b}_2{c}_1+\frac{4}{48}{c}_1^2\right]\hfill \\ \hfill & +2c_1\left(1-c_1^2\right)(x+y)\left[\frac{4}{16}{c}_1+\frac{2}{16}{b}_2\right]\hfill \\ \hfill & +2(1-c_1^2)\left[(1-|x{|}^2)s-(1-|y{|}^2)t\right]\left[\frac{2}{16}{c}_1+\frac{1}{16}{b}_2\right]\hfill \\ \hfill & -2c_1(1-c_1^2)(x^2+y^2)\left[\frac{2}{16}{c}_1+\frac{1}{16}{b}_2\right]-\frac{4}{36}{\left(1-c_1^2\right)}^2{(x-y)}^2|.\hfill \end{array}$$

Since by Lemma 1, $\left|c_1\right|\le 1$, we assume that $c_1=c\in [0,1]$. So by utilizing the triangle inequality we have

$$\begin{array}{cc}\hfill & \left|a_2{a}_4-a_3^2\right|\hfill \\ \hfill \le & \frac{1}{8}\left|b_2{b}_4-b_3^2\right|+\frac{1}{72}\left|b_3^2\right|+\frac{2}{8}\left|b_4-b_2{b}_3\right|c+\frac{10}{48}\left|b_3-\frac{13}{10}{b}_2^2\right|c\left|b_2\right|\hfill \\ \hfill & +\frac{4}{24}\left|b_3-\frac{13}{4}{b}_2^2\right|c^2+\frac{7}{144}\left|b_3-\frac{19}{14}{b}_2^2\right|\left|b_2^2\right|\hfill \\ \hfill & +\left|-\frac{40}{32}+\frac{4}{16}\right|\left|b_2\right|c^3+\left|-\frac{16}{16}+\frac{8}{16}\right|c^4\hfill \\ \hfill & +2\left(1-c^2\right)\left[\frac{1}{9}\left|b_3-\frac{19}{16}{b}_2^2\right|+\frac{8}{96}\left|b_2\right|c+\frac{4}{48}{c}^2\right]\left(\left|x\right|+\left|y\right|\right)\hfill \\ \hfill & +2c\left(1-c^2\right)\left[\frac{4}{16}c+\frac{2}{16}\left|b_2\right|\right]\left(\left|x\right|+\left|y\right|\right)\hfill \\ \hfill & +2(1-c^2)\left[\frac{2}{16}c+\frac{1}{16}\left|b_2\right|\right]\left[(1-|x{|}^2)+(1-|y{|}^2)\right]\hfill \\ \hfill & +2c(1-c^2)\left[\frac{2}{16}c+\frac{1}{16}\left|b_2\right|\right]\left({\left|x\right|}^2+{\left|y\right|}^2\right)+\frac{4}{36}{\left(1-c^2\right)}^2({\left(\left|x\right|+\left|y\right|\right)}^2\hfill \\ \hfill =& \frac{1}{8}\left|b_2{b}_4-b_3^2\right|+\frac{1}{72}\left|b_3^2\right|+\frac{2}{8}\left|b_4-b_2{b}_3\right|c+\frac{10}{48}\left|b_3-\frac{13}{10}{b}_2^2\right|c\left|b_2\right|\hfill \\ \hfill & +\frac{4}{24}\left|b_3-\frac{13}{4}{b}_2^2\right|c^2+\frac{7}{144}\left|b_3-\frac{19}{14}{b}_2^2\right|\left|b_2^2\right|+\left|b_2\right|c^3+\frac{1}{2}{c}^4\hfill \\ \hfill & +4(1-c^2)\left[\frac{2}{16}c+\frac{1}{16}\left|b_2\right|\right]\hfill \\ \hfill & +\left(2\left[\frac{1}{9}\left|b_3-\frac{19}{16}{b}_2^2\right|+\frac{8}{96}\left|b_2\right|c+\frac{4}{48}{c}^2\right]+2c\left[\frac{4}{16}c+\frac{2}{16}\left|b_2\right|\right]\right)\left(1-c^2\right)\left(\left|x\right|+\left|y\right|\right)\hfill \\ \hfill & +2\left(\frac{2}{16}c+\frac{1}{16}\left|b_2\right|\right)\left(c-1\right)(1-c^2)\left({\left|x\right|}^2+{\left|y\right|}^2\right)+\frac{4}{36}{\left(1-c^2\right)}^2{\left(\left|x\right|+\left|y\right|\right)}^2.\hfill \end{array}$$

We now apply Lemmas 3–5 in order to deduce that

$$\begin{array}{cc}\hfill & \left|a_2{a}_4-a_3^2\right|\hfill \\ \hfill \le & \frac{1}{8}+\frac{1}{8}+\frac{2}{4}c+\frac{22}{24}c+\frac{40}{24}{c}^2+\frac{17}{36}+2c^3+\frac{1}{2}{c}^4+\frac{8}{16}(1-c^2)(c+1)\hfill \\ \hfill & +\left(2\left[\frac{7}{36}+\frac{16}{96}c+\frac{4}{48}{c}^2\right]+2c\left[\frac{4}{16}c+\frac{4}{16}\right]\right)\left(1-c^2\right)\left(\left|x\right|+\left|y\right|\right)\hfill \\ \hfill & +\frac{4}{16}\left(c+1\right)\left(c-1\right)(1-c^2)\left({\left|x\right|}^2+{\left|y\right|}^2\right)\hfill \\ \hfill & +\frac{4}{36}{\left(1-c^2\right)}^2{\left(\left|x\right|+\left|y\right|\right)}^2.\hfill \end{array}$$

Now, for $\lambda =\left|x\right|\le 1$ and $\mu =\left|y\right|\le 1$, we obtain

$$\left|a_2{a}_4-a_3^2\right|\le J_1+(\lambda +\mu )J_2+({\lambda }^2+{\mu }^2)J_3+{(\lambda +\mu )}^2{J}_4=L(\lambda ,\mu ),$$

where

$$\begin{array}{c}{J}_1=J_1\left(c\right)=\frac{26}{36}+\frac{34}{24}c+\frac{40}{24}{c}^2+2c^3+\frac{1}{2}{c}^4+\frac{8}{16}(1-c^2)(c+1)\ge 0\hfill \\ J_2=J_2\left(c\right)=\left(\frac{7}{18}+\frac{5}{6}c+\frac{4}{6}{c}^2\right)\left(1-c^2\right)\ge 0\hfill \\ J_3=J_3\left(c\right)=-\frac{1}{4}{(1-c^2)}^2\le 0\hfill \\ J_4=J_4\left(c\right)=\frac{1}{9}{\left(1-c^2\right)}^2\ge 0.\hfill \end{array}$$

We now need to maximize the function $L(\lambda ,\mu )$ on the closed square $[0,1]\times [0,1]$ for $c\in [0,1]$. With regards to $L(\lambda ,\mu )=L(\mu ,\lambda )$, it is sufficient to show that there exists the maximum of

$$H\left(\lambda \right)=L(\lambda ,\lambda )=J_1+2\lambda J_2+2{\lambda }^2(J_3+2J_4),$$

(18)

on $\lambda \in [0,1]$ according to $c\in [0,1]$. We let $c\in [0,1]$. Considering Equation (18) for $0<\lambda <1$ and $J_3+2J_4<0$, we consider for critical point

$${\lambda }_0={\displaystyle \frac{-J_2}{2(J_3+2J_4)}}={\displaystyle \frac{J_2}{2k}}=\frac{18\left(\frac{7}{18}+\frac{5}{6}c+\frac{4}{6}{c}^2\right)\left(1-c^2\right)}{{\left(1-c^2\right)}^2}=\frac{18\left(\frac{7}{18}+\frac{5}{6}c+\frac{4}{6}{c}^2\right)}{\left(1-c^2\right)}>1$$

for any fixed $c\in [0,1]$, where $k=-(J_3+2J_4)>0$. Therefore, for ${\lambda }_0={\displaystyle \frac{J_2}{2k}}>1$, it follows that $k<{\displaystyle \frac{J_2}{2}}\le J_2$, and so $J_2+J_3+2J_4\ge 0$. So,

$$H\left(0\right)=J_1\le J_1+2(J_2+J_3+2J_4)=H\left(1\right).$$

Therefore, it follows that

$$max\left\{H\left(\lambda \right):\lambda \in [0,1]\right\}=H\left(1\right)=J_1+2J_2+2J_3+4J_4.$$

Therefore, $maxL(\lambda ,\mu )=L(1,1)$ on the boundary of the square.

We define the real function W on $(0,1)$ by

$$W\left(c\right)=L(1,1)=J_1+2J_2+2J_3+4J_4.$$

Now putting $J_1,J_2,J_3$ and $J_4$ in the function W, we have

$$\begin{array}{c}\hfill W\left(c\right)=-\frac{8}{9}{c}^4-\frac{1}{6}{c}^3+\frac{11}{6}{c}^2+\frac{43}{12}c+\frac{70}{36}.\end{array}$$

By elementary calculations, we get that $W\left(c\right)$ is an increasing function of c. Therefore, we obtain the maximum of $W\left(c\right)$ on $c=1$ and

$$maxW\left(c\right)=W\left(1\right)=\frac{227}{36}.$$

This completes the proof. □

Example 1.

If we choose the functions

$$f\left(z\right)=z+{\displaystyle \frac{z^3}{2}},g\left(z\right)=z-{\displaystyle \frac{z^3}{3}},$$

then will have

$$F\left(w\right)=f^{-1}\left(w\right)=w-{\displaystyle \frac{w^3}{2}},G\left(w\right)=g^{-1}\left(w\right)=w+{\displaystyle \frac{w^3}{3}}$$

and so these functions satisfy in Definition 1. Thus function $f\in \Sigma$ is bi-close-to-convex, that is, $f\in {\mathcal{C}}_{\Sigma }$ (see for more details, [33]). Therefore, Theorem 2 holds for $f\left(z\right)=z+{\displaystyle \frac{z^3}{2}}$.

Remark 2.

The obtained bound for $\left|a_2{a}_4-a_3^2\right|$ in Theorem 2 is smaller than and more accurate the estimation given in Theorem 1.

3. Conclusions

In the present paper, we find a smaller upper bound and more accurate estimation for the functional $|H_2\left(2\right)|$ for functions in the class ${\mathcal{C}}_{\Sigma }$ with $G\left(w\right)=g^{-1}\left(w\right)$ which is an improvement of the result obtained by Guney et al. [34]. Obtaining a sharp estimate for $|H_2\left(2\right)|$ of the class ${\mathcal{C}}_{\Sigma }$ with $G\left(w\right)=g^{-1}\left(w\right)$ is still an open problem.