Coefficient Problems for a Class of Univalent Functions

Abstract

In this paper, a new subclass has been defined as $\Omega$ of the univalent function in $\mathbb{D}=\{z\in \mathbb{C}:|z|<1\}$. The central goal of this paper is to determine estimates for logarithmic coefficients, inverse logarithmic coefficients, some cases of the Hankel determinant and Zalcman functionals $J_{n,m}$ of inverse functions.

1. Introduction and Definitions

Let $\mathbb{D}$ be the unit disk, $\{z\in \mathbb{C}:|z|<1\}$ denotes the open unit disc and the symbol $\mathcal{A}$ indicates the class of analytic functions f normalized by $f\left(0\right)=0=f^{\prime }\left(0\right)-1.$ It signifies that $f\in \mathcal{A}$ has the following representation

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

(1)

Let $S$ denote the subclass of all univalent functions in $\mathcal{A}$. In [1,2], Peng and Zhong discuss the properties of the class $\Omega$. A function f is in $\Omega$ if

$$\Omega =\left\{f\in \mathcal{A}:|z{f}^{\prime }\left(z\right)-f\left(z\right)|<\frac{1}{2}(z\in \mathbb{D})\right\}.$$

The study of function coefficients has focused on the estimation of the Hankel determinants. In the 1960s, Pommerenke [3] defined the qth Hankel determinant for a function f of the form (1) as

$${\mathcal{H}}_{q,n}\left(f\right):=\left|\begin{array}{cccc}{a}_n\hfill & a_{n+1}\hfill & \dots \hfill & a_{n+q-1}\hfill \\ a_{n+1}\hfill & a_{n+2}\hfill & \dots \hfill & a_{n+q}\hfill \\ ⋮\hfill & ⋮\hfill & \dots \hfill & ⋮\hfill \\ a_{n+q-1}\hfill & a_{n+q}\hfill & \dots \hfill & a_{n+2q-2}\hfill \end{array}\right|,$$

where $q\ge 1$ and $n\ge 1$. The sharp bounds of the second Hankel determinant $\left|H_{2,2}\left(f\right)\right|$ were studied for almost all meaningful subclasses of the class S [4,5,6,7]. The best known second-order case is due to Hayman ([8]), saying that $\left|H_2\left(n\right)\right|\le B{\lambda }^{1/2}$, where B is an absolute constant. The estimate of the third Hankel determinant $\left|H_{3,1}\left(f\right)\right|=\left|a_3{a}_5+2a_2{a}_3{a}^4-a_3^3-a_4^2-a_2^2{a}_5\right|$ is more difficult to obtain than $|H_{2,2}\left(f\right)|$. Even for principal subclasses of the class S, the estimates of the third Hankel determinant are still not sharp [9,10,11,12,13,14,15].

The logarithmic coefficients of $f\in S$, denoted by ${\gamma }_n={\gamma }_n\left(f\right),$ are described as

$$log\frac{f\left(z\right)}{z}=2\sum _{n=1}^{\infty }{\gamma }_n{z}^n\left(z\in \mathbb{D}\right).$$

If f is given by (1), then its logarithmic coefficients are given as follows

$$\begin{array}{c}{\gamma }_1=\frac{1}{2}{a}_2,\hfill \\ {\gamma }_2=\frac{1}{2}\left(a_3-\frac{1}{2}{a}_2^2\right),\hfill \\ {\gamma }_3=\frac{1}{2}\left(a_4-a_2{a}_3+\frac{1}{3}{a}_2^3\right),\hfill \\ {\gamma }_4=\frac{1}{2}\left(a_5-a_4{a}_2+a_3{a}_2^2-\frac{1}{2}{a}_3^2-\frac{1}{4}{a}_2^4\right),\hfill \\ {\gamma }_5=\frac{1}{2}\left(a_6-a_2{a}_5-a_3{a}_4+a_4{a}_2^2+a_2{a}_3^2-a_2^3{a}_3+\frac{1}{5}{a}_2^5\right).\hfill \end{array}.$$

(2)

The famous Köebe 1/4-theorem ensures that, for each univalent function f defined in $\mathbb{D}$, its inverse $f^{-1}$ exists at least on a disc of radius 1/4 with Taylor’s series of the form representation

$$f^{-1}\left(\omega \right)=\omega +A_2{\omega }^2+A_3{\omega }^3+A_4{\omega }^4+\cdots .$$

(3)

Using the representation $f\left(f^{-1}\left(\omega \right)\right)=\omega ,$ we obtain

$$\begin{array}{c}{A}_2=-a_2,\hfill \\ A_3=2a_2^2-a_3,\hfill \\ A_4=-a_4+5a_2{a}_3-5a_2^3,\hfill \\ A_5=-a_5+6a_4{a}_2-21a_3{a}_2^2+3a_3^2+14a_2^4,\hfill \\ A_6=-a_6+7a_5{a}_2-28a_4{a}_2^2-28a_2{a}_3^2+84a_3{a}_2^3-42a_2^5+7a_3{a}_4.\hfill \end{array}.$$

(4)

The logarithmic coefficients of $f^{-1}\in S$, denoted by ${\Gamma }_n={\Gamma }_n\left(f^{-1}\right),$ are described as

$$log\frac{f^{-1}\left(\omega \right)}{\omega }=2\sum _{n=1}^{\infty }{\Gamma }_n{\omega }^n\left(|\omega |<1/4\right).$$

Differentiating and using (4), we obtain

$$\begin{array}{c}{\Gamma }_1=-\frac{1}{2}{a}_2,\hfill \\ {\Gamma }_2=-\frac{1}{2}\left(a_3-\frac{3}{2}{a}_2^2\right),\hfill \\ {\Gamma }_3=-\frac{1}{2}\left(a_4-4a_2{a}_3+\frac{10}{3}{a}_2^3\right),\hfill \\ {\Gamma }_4=-\frac{1}{2}\left(a_5-5a_4{a}_2+15a_3{a}_2^2-\frac{5}{2}{a}_3^2-\frac{35}{4}{a}_2^4\right),\hfill \\ {\Gamma }_5=-\frac{1}{2}\left(a_6-6a_2{a}_5-6a_3{a}_4+21a_4{a}_2^2+21a_2{a}_3^2-56a_2^3{a}_3+\frac{126}{5}{a}_2^5\right).\hfill \end{array}.$$

(5)

Recently, researchers [16,17] have studied the Hankel determinant for the inverse function. In this paper, we give the bounds of the logarithmic coefficients and the upper bound of the second and third Hankel determinant for the inverse functions of the functions in the class $\Omega$. Let ${\mathcal{B}}_0$ be the class of analytic functions $\omega \left(z\right)={\displaystyle \sum _{n=1}^{\infty }}{c}_n{z}^n(z\in \mathbb{D})$ and satisfy the condition $\left|\omega \left(z\right)\right|\le 1$ for $z\in \mathbb{D}$. Consider the functional $\Phi \left(\omega \right)=\left|c_3+\mu c_1{c}_2+\nu c_1^3\right|$ for $\omega \in {\mathcal{B}}_0$ and $\mu ,\nu \in \mathbb{R}$. Now, we define the sets $D_1,D_2$ and $D_3$ by:

$$D_1=\left\{\left(\mu ,\nu \right)\in {\mathbb{R}}^2:\left|\mu \right|\le \frac{1}{2},-1\le \nu \le 1\right\},$$

$$D_2=\left\{\left(\mu ,\nu \right)\in {\mathbb{R}}^2:\frac{1}{2}\le \left|\mu \right|\le 2,\frac{4}{27}{\left(\left|\mu \right|+1\right)}^3-\left(\left|\mu \right|+1\right)\le \nu \le 1\right\},$$

$$D_3=\left\{\left(\mu ,\nu \right)\in {\mathbb{R}}^2:\frac{1}{2}\le |\mu |\le 2,-\frac{2}{3}(|\mu |+1)\le \nu \le \frac{4}{27}{(|\mu |+1)}^3-(|\mu |+1)\right\},$$

$$D_4=\left\{\left(\mu ,\nu \right)\in {\mathbb{R}}^2:\left|\mu \right|\le \frac{1}{2},\nu \le -1\right\}.$$

In this paper, we introduced Schwarz functions and applied the properties of function class $B_0$ (see References [18,19,20,21,22]) and obtained the exact inequalities and corresponding extreme value functions. When studying Hankel determinants, we encountered the extreme value problem of multivariate functions. Currently, the literature mostly avoids discussing the possible internal extreme points, and some methods for finding the extreme value were incorrect. However, we have introduced a method for calculating the solutions of multivariate nonlinear equations and completed the calculation of possible internal extreme points.

To prove our results, we need the following lemmas for Schwarz functions.

2. A Set of Lemmas

Lemma 1

(see [18]). If $\omega \in {\mathcal{B}}_0$, then the sharp estimate $\left|c_n\right|\le 1$ holds for $n\ge 1$.

Lemma 2

(see [19]). Let $\omega \in {\mathcal{B}}_0$, the following sharp estimate $\Phi \left(\omega \right)\le \mathsf{\Psi }\left(\mu ,\nu \right)$ holds for $\mu ,\nu \in \mathbb{R}$, where

$$\mathsf{\Psi }(\mu ,\nu )\le \left\{\begin{array}{cc}1,\hfill & if(\mu ,\nu )\in D_1\bigcup D_2,\hfill \\ \frac{2}{3}(|\mu |+1){\left(\frac{|\mu |+1}{3(|\mu |+1+\nu )}\right)}^{1/2},\hfill & if(\mu ,\nu )\in D_3,\hfill \\ |\nu |,\hfill & if(\mu ,\nu )\in D_4.\hfill \end{array}\right.$$

Lemma 3

(see [20]). Let $\omega \in {\mathcal{B}}_0$, then

$$\left|c_2\right|\le 1-{\left|c_1\right|}^2,|c_3|\le 1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|},|c_4|\le 1-|c_1{|}^2-|c_2{|}^2,|c_5|\le 1-|c_1{|}^2-{|c_2|}^2-\frac{|c_3{|}^2}{1+|c_1|}.$$

Lemma 4

(see [21]). Let $\omega \in {\mathcal{B}}_0$, then for all $\mu \in \mathbb{C}$, we have

$$\left|c_2+\mu c_1^2\right|\le max\left\{1,\left|\mu \right|\right\}$$

Lemma 5

(see [22]). Let $\omega \in {\mathcal{B}}_0$, then for all $\mu \in \mathbb{C}$, $\left|\mu \right|\le 1,$ we have

$$\left|c_4+2\mu c_1{c}_3+\mu c_2^2+3{\mu }^2{c}_1^2{c}_2+{\mu }^3{c}_1^4\right|\le 1$$

Lemma 6

(see [22]). Let $\omega \in {\mathcal{B}}_0$, then for all $\mu \in \mathbb{C}$, $\left|\mu \right|\le 1,$ we have

$$\left|c_5+(1+\mu )c_1{c}_4+(1+\mu )c_2{c}_3+3\mu c_1{c}_2^2+(1+\mu +{\mu }^2)c_1^2{c}_3+2\mu (1+\mu )c_1^3{c}_2+{\mu }^2{c}_1^5\right|\le 1,$$

$$\left|c_5+2\mu c_1{c}_4+2\mu c_2{c}_3+3{\mu }^2{c}_1{c}_2^2+3{\mu }^2{c}_1^2{c}_3+4{\mu }^3{c}_1^3{c}_2+{\mu }^4{c}_1^5\right|\le 1.$$

3. Logarithmic Coefficients and Hankel Determinants

In the first theorem, we derive the sharp bound for the logarithmic coefficients.

Theorem 1.

If $f\in \Omega$ is given by (1), then

$$|{\gamma }_1|\le \frac{1}{4},|{\gamma }_2|\le \frac{1}{4},|{\gamma }_3|\le \frac{1}{4},|{\gamma }_4|\le \frac{1}{4},|{\gamma }_5|\le \frac{1}{4}.$$

All bounds are sharp.

Proof. 

Let $f\in \Omega$. From the formula

$$f\left(z\right)=z+\frac{1}{2}z\omega \left(z\right),$$

(6)

we obtain

$$a_2=\frac{c_1}{2},a_3=\frac{c_2}{2},a_4=\frac{c_3}{2},a_5=\frac{c_4}{2},a_6=\frac{c_5}{2},a_7=\frac{c_6}{2}.$$

(7)

From (2) and (7), we obtain

$$\begin{array}{c}{\gamma }_1=\frac{1}{4}{c}_1,\hfill \\ {\gamma }_2=\frac{1}{4}\left(c_2-\frac{1}{4}{c}_1^2\right),\hfill \\ {\gamma }_3=\frac{1}{4}\left(c_3-\frac{1}{2}{c}_1{c}_2+\frac{1}{12}{c}_1^3\right),\hfill \\ {\gamma }_4=\frac{1}{4}\left(c_4-\frac{1}{2}{c}_1{c}_3+\frac{1}{4}{c}_1^2{c}_2-\frac{1}{4}{c}_2^2-\frac{1}{32}{c}_1^4\right),\hfill \\ {\gamma }_5=\frac{1}{4}\left(c_5-\frac{1}{2}{c}_1{c}_4-\frac{1}{2}{c}_2{c}_3+\frac{1}{4}{c}_1{c}_2^2+\frac{1}{4}{c}_3{c}_1^2-\frac{1}{8}{c}_1^3{c}_2+\frac{1}{80}{c}_1^5\right).\hfill \end{array}.$$

(8)

From Lemmas 1 and 4, the bounds of $\left|{\gamma }_1\right|$ and $\left|{\gamma }_2\right|$ are obvious. By using Lemma 2 with $\mu =-\frac{1}{2}$ and $\nu =\frac{1}{12}$, the inequality for $\left|{\gamma }_3\right|$ can be obtained.

${\gamma }_4$ in Formula (8) can be written as

$$\begin{array}{ccc}\hfill \left|{\gamma }_4\right|& =& \left|2c_4-c_1{c}_3+\frac{1}{2}{c}_1^2{c}_2-\frac{1}{2}{c}_2^2-\frac{1}{16}{c}_1^4\right|\hfill \\ & & =\left|c_4-c_1{c}_3-\frac{1}{2}{c}_2^2+\frac{3}{4}{c}_1^2{c}_2-\frac{1}{8}{c}_1^4+c_4-\frac{1}{4}{c}_1^2{c}_2+\frac{1}{16}{c}_1^4\right|.\hfill \end{array}$$

Applying the triangle inequality, we have

$$8\left|{\gamma }_4\right|\le \left|c_4-c_1{c}_3-\frac{1}{2}{c}_2^2+\frac{3}{4}{c}_1^2{c}_2-\frac{1}{8}{c}_1^4\right|+\left|c_4-\frac{1}{4}{c}_1^2{c}_2+\frac{1}{16}{c}_1^4\right|.$$

From Lemma 5 for $\mu =-\frac{1}{2}$, we get $\left|c_4-c_1{c}_3-\frac{1}{2}{c}_2^2+\frac{3}{4}{c}_1^2{c}_2-\frac{1}{8}{c}_1^4\right|\le 1,$ thus we have

$$8\left|{\gamma }_4\right|\le 1+\Delta$$

where

$$\Delta =\left|c_4-\frac{1}{4}{c}_1^2{c}_2+\frac{1}{16}{c}_1^4\right|.$$

(9)

Now, we claim that $\Delta \le 1.$ Applying Lemma 3 and the triangle inequality in (9), we get

$$\Delta \le 1-{\left|c_1\right|}^2-{\left|c_2\right|}^2+\frac{1}{4}\left|c_2\right|{\left|c_1\right|}^2+\frac{1}{16}{\left|c_1\right|}^4={\phi }_1\left(x,y\right)=1-x^2-y^2+\frac{1}{4}{x}^{2}y+\frac{1}{16}{x}^4$$

where $\left|c_1\right|=x,\left|c_2\right|=y$.

Assume that ${\phi }_1\left(x,y\right)$ has a maximum value at an interior point of $\mathsf{\Lambda }=\left\{\left(x,y\right):0\le x\le 1,\right.$ $\left.0\le y\le 1-x^2\right\}$. Then $\frac{\partial {\phi }_1}{\partial x}=-2x+\frac{1}{2}xy+\frac{1}{4}{x}^3=\frac{\partial {\phi }_1}{\partial y}=-2y+\frac{1}{4}{x}^2=0$ implies $x=y=0$ which is a contradiction.

On the boundary of $\mathsf{\Lambda }$, we get

$${\phi }_1\left(0,y\right)=1-y^2\le 1,$$

$${\phi }_1\left(x,0\right)=1-x^2+\frac{1}{16}{x}^4\le {\phi }_1\left(0,0\right)=1,$$

$${\phi }_1\left(x,1-x^2\right)=\frac{5}{4}{x}^2-\frac{19}{16}{x}^4={\tau }_1\left(x\right)\le {\tau }_1\left(\sqrt{\frac{10}{19}}\right)=\frac{25}{76}<1.$$

Hence, $\Delta \le 1,$ and so $8\left|{\gamma }_4\right|\le 2.$ Thus, we get the fourth inequality in Theorem 1.

${\gamma }_5$ in Formula (8) can be written as

$$\begin{array}{ccc}\hfill 8\left|{\gamma }_5\right|& =& \left|2c_5-c_1{c}_4-c_2{c}_3+\frac{1}{2}{c}_1{c}_2^2+\frac{1}{2}{c}_3{c}_1^2-\frac{1}{4}{c}_2{c}_1^3+\frac{1}{40}{c}_1^5\right|\hfill \\ & & =\left|c_5-c_1{c}_4-c_2{c}_3+\frac{3}{4}{c}_1{c}_2^2+\frac{3}{4}{c}_3{c}_1^2-\frac{1}{2}{c}_2{c}_1^3+\frac{1}{16}{c}_1^5+c_5-\frac{1}{4}{c}_1{c}_2^2-\frac{1}{4}{c}_3{c}_1^2+\frac{1}{4}{c}_2{c}_1^3-\frac{3}{80}{c}_1^5\right|.\hfill \end{array}$$

From Lemma 6 for $\mu =-\frac{1}{2}$, we know that

$$\begin{array}{c}\hfill \left|c_5-c_1{c}_4-c_2{c}_3+\frac{3}{4}{c}_1{c}_2^2+\frac{3}{4}{c}_3{c}_1^2-\frac{1}{2}{c}_2{c}_1^3+\frac{1}{16}{c}_1^5\right|\le 1.\end{array}$$

Thus, we obtain

$$8\left|{\gamma }_5\right|\le 1+\Delta$$

where

$$\Delta =\left|c_5-\frac{1}{4}{c}_1{c}_2^2-\frac{1}{4}{c}_3{c}_1^2+\frac{1}{4}{c}_2{c}_1^3-\frac{3}{80}{c}_1^5\right|.$$

(10)

Next, we claim that $\Delta \le 1.$ Applying Lemma 3 and the triangle inequality in (10), we have

$$\Delta \le 1-|c_1{|}^2-|c_2{|}^2-\frac{|c_3{|}^2}{1+|c_1|}+\frac{1}{4}|c_1||c_2{|}^2+\frac{1}{4}|c_3||c_1{|}^2+\frac{1}{4}|c_2||c_1{|}^3+\frac{3}{80}{|c_1|}^5.$$

The expression on the right side of the above inequality takes its greatest value with respect to $\left|c_3\right|$ when $\left|c_3\right|=\frac{1}{8}{\left|c_1\right|}^2\left(1+\left|c_1\right|\right),$ so

$$\Delta \le {\phi }_2\left(x,y\right)=1-x^2-y^2+\frac{1}{64}{x}^4\left(1+x\right)+\frac{1}{4}x{y}^2+\frac{1}{4}y{x}^3+\frac{3}{80}{x}^5$$

where $\left|c_1\right|=x,\left|c_2\right|=y.$

Assume that ${\phi }_2\left(x,y\right)$ has a maximum value at an interior point of $\mathsf{\Lambda }=\left\{\left(x,y\right):0\le x\le 1,\right.$ $\left.0\le y\le 1-x^2\right\}$. Then $\frac{\partial {\phi }_2}{\partial x}=-2x+\frac{1}{64}\left(5x^4+4x^3\right)+\frac{1}{4}{y}^2+\frac{3}{4}y{x}^2+\frac{3}{16}{x}^4=\frac{\partial {\phi }_2}{\partial y}=-2y+\frac{1}{2}xy+\frac{1}{4}{x}^3=0$ implies $x=y=0$ which is a contradiction.

On the boundary of $\mathsf{\Lambda }$, we get

$${\phi }_2\left(0,y\right)=1-y^2\le 1,$$

$${\phi }_2\left(x,0\right)=1-x^2+\frac{1}{16}{x}^4+\frac{17}{320}{x}^5\le {\phi }_2(0,0)=1,$$

$${\phi }_2\left(x,1-x^2\right)=\frac{1}{4}x+x^2-\frac{1}{4}{x}^3-\frac{63}{64}{x}^4+\frac{17}{320}{x}^5.$$

Since the functions $f_1\left(x\right)=\frac{1}{4}x-\frac{1}{4}{x}^3+\frac{17}{320}{x}^5$ and $f_2\left(x\right)=x^2-\frac{63}{64}{x}^4$ reach their greatest values for $x=\sqrt{\frac{24-4\sqrt{19}}{17}}$ and $x=\frac{4\sqrt{2}}{\sqrt{63}}$, respectively. Thus, $f_1\left(x\right)\le f_1\left(\sqrt{\frac{24-4\sqrt{19}}{17}}\right)\approx 0.1002856\cdots$ and $f_2\left(x\right)\le f_1\left(\frac{4\sqrt{2}}{\sqrt{63}}\right)=\frac{16}{63},$ and it follows that

$${\phi }_2\left(x,1-x^2\right)\le 0.1002856\cdots +\frac{16}{63}<1.$$

Hence, $\Delta \le 1$, and so $8\left|{\gamma }_5\right|\le 2$. Thus, we obtain the fifth inequality in Theorem 1.

The equalities in Theorem 1 hold for functions f given by (6) with $\omega \left(z\right)=z,\omega \left(z\right)=z^2,\omega \left(z\right)=z^3,\omega \left(z\right)=z^4,\omega \left(z\right)=z^5$ respectively. □

Remark 1.

The estimates of the coefficients $\left|{\gamma }_1\right|$, $\left|{\gamma }_2\right|$ and $\left|{\gamma }_3\right|$ of Theorem 1 are the improvement of the estimates obtained in ([2], Theorem 2.1).

Theorem 2.

Let $(x_0,y_0)\approx (0.147952,0.356608)$ be the approximate root of the system of linear equations

$$\left\{\begin{array}{c}-32y^4+\left(-16x-16\right)y^3+\left(8x^4+56x^3+120x^2+104x+32\right)y^2+(24x^6+24x^5-104x^4-200x^3\hfill \\ -96x^2+16x+16)y+6x^8+18x^7+50x^6+102x^5+48x^4-112x^3-144x^2-48x=0,\hfill \\ 64y^3+\left(-48x^2-48x\right)y^2+\left(8x^4+80x^3+72x^2-64x-64\right)y+6x^6-4x^5-42x^4-32x^3+32x^2\hfill \\ +48x+16=0.\hfill \end{array}\right.$$

If $f\in \Omega$ be of form (1) and its inverse $f^{-1}$ is given by (3), then we have

$$\left|H_{2,2}\left(f^{-1}\right)\right|=\left|A_2{A}_4-A_3^2\right|\le \frac{1}{4}$$

and

$$\begin{array}{ccc}\hfill \left|H_{3,1}\left(f^{-1}\right)\right|& \le & \frac{1}{64}\left[\frac{16y_0^4}{{(1+x_0)}^2}-\frac{16x_0}{1+x_0}{y}_0^3+(4x_0^2+32x_0-32)y_0^2+(6x_0^4-16x_0^3-16x_0^2+16x_0+16)y_0\right.\hfill \\ & & \left.+x_0^6+8x_0^4-24x_0^2+16\right]\hfill \\ & & \approx 0.289558.\hfill \end{array}$$

The first inequality is sharp.

Proof. 

From (4) and (7), we get

$$\begin{array}{c}{A}_2=-\frac{1}{2}{c}_1,\hfill \\ A_3=\frac{1}{2}\left(c_1^2-c_2\right),\hfill \\ A_4=\frac{1}{8}\left(-4c_3+10c_1{c}_2-5c_1^3\right),\hfill \\ A_5=\frac{1}{8}\left(-4c_4+12c_1{c}_3-21c_2{c}_1^2+6c_2^2+7c_1^4\right).\hfill \end{array}.$$

(11)

From (11), we have

$$\left|H_{2,2}\left(f^{-1}\right)\right|=\left|A_2{A}_4-A_3^2\right|=\frac{1}{16}\left|c_1^4-2c_1^2{c}_2+4c_3{c}_1-4c_2^2\right|.$$

Applying the triangle inequality and Lemma 3 in the above equality, we obtain

$$\begin{array}{ccc}\hfill \left|H_{2,2}\left(f^{-1}\right)\right|& & \le \frac{1}{16}\left(|c_1{|}^4+2|c_1{|}^2|c_2|+4|c_3||c_1|+4|c_2{|}^2\right)\hfill \\ & & \le \frac{1}{16}\left[|c_1{|}^4+2|c_1{|}^2|c_2|+4|c_1|\left(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|}\right)+4|c_2{|}^2|\right]\hfill \\ & & =\frac{1}{16}\left[|c_1{|}^4+2|c_1{|}^2|c_2|+4|c_1|-4|c_1{|}^3+\frac{4}{1+|c_1|}|c_2{|}^2|\right]\hfill \\ & & \le \frac{1}{16}\left[|c_1{|}^4+2|c_1{|}^2\left(1-|c_1{|}^2\right)+4|c_1|-4|c_1{|}^3+\frac{4}{1+|c_1|}{\left(1-|c_1{|}^2\right)}^2\right]\hfill \\ & & =\frac{1}{16}\left(4-2|c_1{|}^2-{|c_1|}^4\right)\le \frac{1}{4}.\hfill \end{array}$$

From (11), we have

$$64\left|H_{3,1}\left(f^{-1}\right)\right|=\left|-16c_3^2+16c_1{c}_2{c}_3+16c_2{c}_4-8c_4{c}_1^2-16c_2^3+12c_1^2{c}_2^2+c_1^6-6c_2{c}_1^4\right|.$$

Hence, applying the triangle inequality and Lemma 3, we get

$$\begin{array}{ccc}\hfill 64\left|H_{3,1}\left(f^{-1}\right)\right|& \le & 16|c_3{|}^2+16|c_1||c_2||c_3|+16|c_2||c_4|+8|c_1{|}^2|c_4|+16|c_2{|}^3+12|c_1{|}^2|c_2{|}^2+|c_1{|}^6+6|c_2||c_1{|}^4\hfill \\ & & \le 16{\left(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|}\right)}^2+16|c_1||c_2|\left(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|}\right)+16|c_2|\left(1-|c_1{|}^2-{|c_2|}^2\right)\hfill \\ & & +8|c_1{|}^2\left(1-|c_1{|}^2-{|c_2|}^2\right)+16|c_2{|}^3+12|c_1{|}^2|c_2{|}^2+|c_1{|}^6+6|c_2||c_1{|}^4\hfill \\ & & =\frac{16|c_2{|}^4}{{\left(1+|c_1|\right)}^2}-\frac{16|c_1|}{1+|c_1|}|c_2{|}^3+(4|c_1{|}^2+32|c_1|-32)|c_2{|}^2+(6|c_1{|}^4-16|c_1{|}^3-16{|c_1|}^2\hfill \\ & & +16|c_1|+16)|c_2|+|c_1{|}^6+8|c_1{|}^4-24{|c_1|}^2+16.\hfill \end{array}$$

Setting $x=|c_1|$ and $y=|c_2|,$ we obtain

$$\begin{array}{ccc}\hfill 64\left|H_{3,1}\left(f^{-1}\right)\right|& \le & \frac{16y^4}{{\left(1+x\right)}^2}-\frac{16x}{1+x}{y}^3+\left(4x^2+32x-32\right)y^2+\left(6x^4-16x^3-16x^2+16x+16\right)y\hfill \\ & & +x^6+8x^4-24x^2+16={\phi }_3\left(x,y\right).\hfill \end{array}$$

We need to calculate the maximum value of ${\phi }_3\left(x,y\right)$ on $\mathsf{\Lambda }=\left\{\left(x,y\right):0\le x\le 1,\right.$ $\left.0\le y\le 1-x^2\right\}$. Differentiating ${\phi }_3\left(x,y\right)$ with respect to x and y, we get

$$\frac{\partial {\phi }_3}{\partial x}=-\frac{32y^4}{{\left(1+x\right)}^3}-\frac{16}{{\left(1+x\right)}^2}{y}^3+\left(8x+32\right)y^2+\left(24x^3-48x^2-32x+16\right)y+6x^5+32x^3-48x$$

and

$$\frac{\partial {\phi }_3}{\partial y}=\frac{64y^3}{{\left(1+x\right)}^2}-\frac{48x}{1+x}{y}^2+\left(8x^2+64x-64\right)y+6x^4-16x^3-16x^2+16x+16.$$

Setting $\frac{\partial {\phi }_3}{\partial x}=0,\frac{\partial {\phi }_3}{\partial y}=0$ and simplifying, we obtain

$$\left\{\begin{array}{c}-32y^4+(-16x-16)y^3+(8x^4+56x^3+120x^2+104x+32)y^2+(24x^6+24x^5-104x^4-200x^3\hfill \\ -96x^2+16x+16)y+6x^8+18x^7+50x^6+102x^5+48x^4-112x^3-144x^2-48x=0,\hfill \\ 64y^3+(-48x^2-48x)y^2+(8x^4+80x^3+72x^2-64x-64)y+6x^6-4x^5-42x^4-32x^3+32x^2\hfill \\ +48x+16=0.\hfill \end{array}\right.$$

By numerical computation we obtain the following:

$\left\{\begin{array}{c}{x}_0\approx 0.147952,\hfill \\ y_0\approx 0.356608,\hfill \end{array}\right.$$\left\{\begin{array}{c}{x}_1=1,\hfill \\ y_1=0,\hfill \end{array}\right.$$\left\{\begin{array}{c}{x}_2\approx 0.062301,\hfill \\ y_2\approx -1.125603\hfill \end{array}\right.$$\left\{\begin{array}{c}{x}_3\approx 0.300973,\hfill \\ y_3\approx 0.9613356,\hfill \end{array}\right.$$\left\{\begin{array}{c}{x}_4\approx -1.086663,\hfill \\ y_4\approx 0.068860,\hfill \end{array}\right.$$\left\{\begin{array}{c}{x}_5\approx -1.403759,\hfill \\ y_5\approx 0.757992,\hfill \end{array}\right.$$\left\{\begin{array}{c}{x}_6\approx -74.674540,\hfill \\ y_6\approx -1439.155590,\hfill \end{array}\right.$

Thus, in $\Omega$, there is a critical point $\left(x_0,y_0\right)\approx \left(0.147952,0.356608\right)$ satisfying $y\le 1-x^2$. For the point, we have ${\phi }_3\left(x_0,y_0\right)\approx 18.5317.$

On the boundary of $\mathsf{\Lambda }$, we get

$${\phi }_3\left(0,y\right)=16y^4-32y^2+16y+16\le {\phi }_3\left(0,0.2696\right)\approx 18.0722,$$

$${\phi }_3\left(x,0\right)=x^6+8x^4-24x^2+16\le {\phi }_3\left(0,0\right)=16,$$

$${\phi }_3\left(x,1-x^2\right)=16x^6+30x^5-43x^4-60x^3+27x^2+30x+1={\tau }_2\left(x\right)\le {\tau }_2\left(0.518961\right)\approx 12.433.$$

Thus, $\left|H_{3,1}\left(f^{-1}\right)\right|\le \frac{18.5317}{64}\approx 0.289558$, and we obtain the second inequality in Theorem 2.

Observe that, if $c_2=1$ and $c_k=0(k\ne 2)$, then $A_2=A_2=0,A_3=-\frac{1}{2}$. This means that the first inequality is sharp. □

Theorem 3.

Let $f\in \Omega$ be of form (1) and its inverse $f^{-1}$ is given by (3). Then we have

$$|{\Gamma }_1|\le \frac{1}{4},|{\Gamma }_2|\le \frac{1}{4},|{\Gamma }_3|\le 2\sqrt{\frac{6}{13}},|{\Gamma }_4|\le \frac{1507}{2512},|{\Gamma }_5|\le \frac{1251}{320}.$$

All bounds are sharp.

Proof. 

From (5) and (7), we get

$$\begin{array}{c}{\Gamma }_1=-\frac{1}{4}{c}_1,\hfill \\ {\Gamma }_2=-\frac{1}{4}(c_2-\frac{3}{4}{c}_1^2),\hfill \\ {\Gamma }_3=-\frac{1}{4}(c_3-2c_1{c}_2+\frac{5}{6}{c}_1^3),\hfill \\ {\Gamma }_4=-\frac{1}{4}(c_4-\frac{5}{2}{c}_1{c}_3+\frac{15}{4}{c}_2{c}_1^2-\frac{5}{4}{c}_2^2-\frac{35}{32}{c}_1^4),\hfill \\ {\Gamma }_5=-\frac{1}{4}(c_5-3c_1{c}_4-3c_2{c}_3+\frac{21}{4}{c}_1{c}_2^2+\frac{21}{4}{c}_3{c}_1^2-7c_1^3{c}_2+\frac{63}{40}{c}_1^5).\hfill \end{array}.$$

(12)

The bounds of $\left|{\Gamma }_1\right|,\left|{\Gamma }_2\right|$ follow from Lemma 1 and Lemma 4, respectively.

Formula (12) for ${\Gamma }_3$ can be written as

$$4\left|{\Gamma }_3\right|=\left|c_3-2c_1{c}_2+\frac{5}{6}{c}_1^3\right|.$$

(13)

Applying the triangle and Lemma 3 in (13), we obtain

$$4\left|{\Gamma }_3\right|\le 1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|}+2|c_1||c_2|+\frac{5}{6}{|c_1|}^3={\phi }_4\left(x,y\right)=1-x^2-\frac{y^2}{1+x}+2xy+\frac{5}{6}{x}^3$$

where $x=\left|c_1\right|,y=\left|c_2\right|.$

Assume that ${\phi }_4\left(x,y\right)$ has a maximum value at an interior point of $\mathsf{\Lambda }=\left\{\left(x,y\right):0\le x\le 1,\right.$ $\left.0\le y\le 1-x^2\right\}$. Then, $\frac{\partial {\phi }_4}{\partial x}=-2x+\frac{y^2}{{(1+x)}^2}+2y+\frac{5}{2}{x}^2=\frac{\partial {\phi }_4}{\partial y}=2x-\frac{2y}{1+x}=0$ implies $x=y=0$, which is a contradiction.

On the boundary of $\mathsf{\Lambda }$, we get

$${\phi }_4\left(0,y\right)=1-y^2\le 1,$$

$${\phi }_4\left(x,0\right)=1-x^2+\frac{5}{6}{x}^3\le {\phi }_4\left(0,0\right)=1,$$

$${\phi }_4\left(x,1-x^2\right)=3x-\frac{13}{6}{x}^3={\tau }_3\left(x\right)\le {\tau }_3\left(\sqrt{\frac{6}{13}}\right)=2\sqrt{\frac{6}{13}}=1.3587\dots >1.$$

We note that equality in the inequality in $\left|{\Gamma }_3\right|\le 2\sqrt{\frac{6}{13}}$ is attained when $c_1=\sqrt{\frac{6}{13}},c_2=-\frac{7}{13}$ and $c_3=\frac{7}{13}\sqrt{\frac{6}{13}}$.

Formula (12) for ${\gamma }_4$ can be written

$$128\left|{\Gamma }_4\right|=\left|32c_4-80c_1{c}_3+120c_2{c}_1^2-40c_2^2-35c_1^4\right|.$$

Applying the triangle and Lemma 3, we obtain

$$\begin{array}{ccc}\hfill 128\left|{\Gamma }_4\right|& \le & 32\left(1-|c_1{|}^2-{|c_2|}^2\right)+80|c_1|\left(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|}\right)+120|c_2||c_1{|}^2+40|c_2{|}^2+35{|c_1|}^4\hfill \\ & & =32-32|c_1{|}^2+8|c_2{|}^2+80|c_1|-80|c_1{|}^3-\frac{80|c_1||c_2{|}^2}{1+|c_1|}+120|c_2||c_1{|}^2+35{|c_1|}^4\hfill \\ & & ={\phi }_5\left(x,y\right)=32+80x-32x^2+8y^2-80x^3-\frac{80x{y}^2}{1+x}+120y{x}^2+35x^4\hfill \end{array}$$

where $\left|c_1\right|=x,\left|c_2\right|=y.$

Assume that ${\phi }_5\left(x,y\right)$ has a maximum value at an interior point of $\mathsf{\Lambda }=\left\{(x,y):0\le x\le 1,\right.$ $\left.0\le y\le 1-x^2\right\}$. The critical points of $\mathsf{\Lambda }$ satisfy the conditions

$$80-64x-240x^2-\frac{80y^2}{{(1+x)}^2}+240xy+140x^3=0andy=\frac{15(x^2+x^3)}{2(9x-1)}.$$

Solving this system leads to the equality

$$27540x^5-12060x^4-2524x^3+7392x^2-1504x+80=0,$$

which has two solutions: $x_1=0.1218\cdots$ and $x_2=0.1005\cdots$. Hence, we get two critical points: $\left(0.1218\cdots ,1.2973\cdots \right)$ and $\left(0.1005\cdots ,-0.87277\cdots \right).$ Both points do not belong to $\mathsf{\Lambda }$.

On the boundary of $\mathsf{\Lambda }$, we get

$${\phi }_5\left(0,y\right)=32+8y^2\le {\phi }_5\left(0,1\right)=40,$$

$${\phi }_5\left(x,0\right)=32+80x-32x^2-80x^3+35x^4\le {\phi }_5\left(0.5275\cdots ,0\right)\approx 56.2633\cdots ,$$

$${\phi }_5\left(x,1-x^2\right)=40+152x^2-157x^4={\tau }_4\left(x\right)\le {\tau }_4\left(\sqrt{\frac{76}{157}}\right)=\frac{12056}{157}\approx 76.7898\cdots .$$

We note that equality in the inequality in $\left|{\Gamma }_4\right|\le \frac{1507}{2512}$ is attained when $c_1=\sqrt{\frac{76}{157}},c_2=-\frac{81}{157},c_3=\frac{81}{157}\sqrt{\frac{76}{157}}$ and $c_4=-\frac{6156}{24649}$.

Formula (12) for ${\Gamma }_5$ can be written as

$$\begin{array}{ccc}\hfill 8\left|{\Gamma }_5\right|& & =|2c_5-6c_1{c}_4-6c_2{c}_3+\frac{21}{2}{c}_1{c}_2^2+\frac{21}{2}{c}_3{c}_1^2-14c_1^3{c}_2+\frac{63}{20}{c}_1^5|\hfill \\ & & =|(c_5-2c_1{c}_4-2c_2{c}_3+3c_1{c}_2^2+3c_3{c}_1^2-4c_1^3{c}_2+c_1^5)+c_5-4c_1{c}_4-4c_2{c}_3+\frac{15}{2}{c}_1{c}_2^2+\frac{15}{2}{c}_3{c}_1^2\hfill \\ & & -10c_1^3{c}_2+\frac{43}{20}{c}_1^5|.\hfill \end{array}$$

From Lemma 6 for $\mu =-1$, we know that

$$\left|c_5-2c_1{c}_4-2c_2{c}_3+3c_1{c}_2^2+3c_3{c}_1^2-4c_1^3{c}_2+c_1^5\right|\le 1.$$

Thus, we get

$$8\left|{\Gamma }_5\right|\le 1+\Delta$$

where

$$\Delta =\left|c_5-4c_1{c}_4-4c_2{c}_3+\frac{15}{2}{c}_1{c}_2^2+\frac{15}{2}{c}_3{c}_1^2-10c_1^3{c}_2+\frac{43}{20}{c}_1^5\right|.$$

(14)

Now, we show that $\Delta \le \frac{1211}{40}.$ Applying the triangle inequality and Lemma 3 in (14), we have

$$\begin{array}{ccc}\hfill \Delta & \le & 1-|c_1{|}^2-|c_2{|}^2-\frac{|c_3{|}^2}{1+|c_1|}+4|c_1\left(1-|c_1{|}^2-{|c_2|}^2\right)+4|c_2||c_3|+\frac{15}{2}|c_1||c_2{|}^2+\frac{15}{2}|c_3||c_1{|}^2\hfill \\ & & +10|c_1{|}^3|c_2|+\frac{43}{20}{|c_1|}^5.\hfill \end{array}$$

The expression on the right side of the above inequality takes its greatest value with respect to $\left|c_3\right|$ when $\left|c_3\right|=\frac{\left(1+|c_1|\right)(8|c_2|+15|c_1{|}^2)}{4},$ so

$$\Delta \le {\phi }_6(x,y)=1+4x-x^2-4x^3+\frac{225}{16}{x}^4+\frac{1297}{80}{x}^5+\left(\frac{15}{2}x+3\right)y^2+\left(25x^3+15x^2\right)y$$

where $\left|c_1\right|=x,\left|c_2\right|=y.$

We try to find the maximum value of ${\phi }_6\left(x,y\right)$ on $\left[0,1\right]\times \left[1,1-x^2\right]$. Differentiating ${\phi }_6\left(x,y\right)$ with respect to x and y, we get

$$\frac{\partial {\phi }_6}{\partial x}=4-2x-12x^2+\frac{225}{4}{x}^3+\frac{1297}{16}{x}^4+\frac{15}{2}{y}^2+\left(75x^2+30x\right)y$$

and

$$\frac{\partial {\phi }_6}{\partial y}=25x^3+15x^2+15x+6y.$$

Setting $\frac{\partial {\phi }_6}{\partial x}=0,\frac{\partial {\phi }_6}{\partial y}=0$, we obtain

$\left\{\begin{array}{c}{x}_1\approx 0.390648,\hfill \\ y_1\approx -0.318681,\hfill \end{array}\right.$$\left\{\begin{array}{c}{x}_2\approx -0.398223,\hfill \\ y_2\approx -30.013200,\hfill \end{array}\right.$$\left\{\begin{array}{c}{x}_3\approx -0.574696,\hfill \\ y_3\approx 0.079731.\hfill \end{array}\right.$

Thus, in $\mathsf{\Lambda }$, there is no critical point.

On the boundary, we get

$${\phi }_6\left(0,y\right)=3y^2\le {\phi }_6\left(0,1\right)\le 3$$

$${\phi }_6\left(x,0\right)=1+4x-x^2-4x^3+\frac{225}{16}{x}^4+\frac{1297}{80}{x}^5\le {\phi }_6(1,0)=\frac{1211}{40},$$

$${\phi }_6\left(x,1-x^2\right)=4+\frac{23}{2}x+8x^2+6x^3+\frac{33}{16}{x}^4-\frac{103}{80}{x}^5={\tau }_5\left(x\right)\le {\tau }_5\left(1\right)=\frac{1211}{40}.$$

We note that equality in the inequality in $|{\Gamma }_5|\le \frac{1251}{320}$ is attained when $c_1=1$. □

Remark 2.

From Lemma 2 and (12), we have $\left|{\Gamma }_3\right|\le \frac{1}{2}\sqrt{\frac{6}{23}}$, which contradicts the estimate of $\left|{\Gamma }_3\right|$ in Theorem 3. This means that the function $\mathsf{\Psi }\left(\mu ,\nu \right)$ does not hold for $(\mu ,\nu )\in D_3.$

4. Generalized Zalcman Functional of Inverse Function

Let us consider some cases of the generalized Zalcman functional $\left|J_{n,m}\left(f^{-1}\right)\right|=$ $\left|A_{n+m-1}-A_n{A}_m\right|$ for $f\in \Omega .$

Theorem 4.

Let $f\in \Omega$ be of form (1) and its inverse $f^{-1}$ be given by (3). Then we have

$$\left|J_{2,3}\left(f^{-1}\right)\right|=\left|A_4-A_2{A}_3\right|\le \sqrt{\frac{12}{27}},\left|J_{2,4}\left(f^{-1}\right)\right|=\left|A_5-A_2{A}_4\right|\le \frac{47}{20},\left|J_{3,3}\left(f^{-1}\right)\right|=\left|A_5-A_3^2\right|\le \frac{1009}{768}.$$

All bounds are sharp.

Proof. 

From (11) and Lemma 3, we get

$$\begin{array}{ccc}\hfill 2\left|J_{2,3}\left(f^{-1}\right)\right|& =& 2|A_4-A_2{A}_3|=|c_3-2c_1{c}_2+\frac{3}{4}{c}_1^3|\le 1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|}+2|c_1||c_2|+\frac{3}{4}{|c_1|}^3\hfill \\ & & ={\phi }_7\left(x,y\right)=1-x^2-\frac{y^2}{1+x}+2xy+\frac{3}{4}{x}^3\hfill \end{array}$$

where $\left|c_1\right|=x,\left|c_2\right|=y.$

Assume that ${\phi }_7\left(x,y\right)$ has a maximum value at an interior point of $\mathsf{\Lambda }=\left\{\left(x,y\right):0\le x\le 1,\right.$ $\left.0\le y\le 1-x^2\right\}$. Then $\frac{\partial {\phi }_7}{\partial x}=-2x+\frac{y^2}{{(1+x)}^2}+2y+\frac{9}{4}{x}^2=\frac{\partial {\phi }_7}{\partial y}=-\frac{y}{1+x}+x=0$, implies $(x,y)=(0,0)$ which is a contradiction.

On the boundary of $\mathsf{\Lambda }$, we get

$${\phi }_7\left(0,y\right)=1-y^2\le 1$$

$${\phi }_7\left(x,0\right)=1-x^2+\frac{3}{4}{x}^3\le {\phi }_6(0,0)=1,$$

$${\phi }_7\left(x,1-x^2\right)=3x-\frac{9}{4}{x}^3={\phi }_7\left(\sqrt{\frac{12}{27}},\frac{15}{27}\right)=2\sqrt{\frac{12}{27}}=1.3333\dots >1.$$

We note the equality in the inequality in $\left|J_{2,3}\left(f^{-1}\right)\right|\le \sqrt{\frac{12}{27}}$ is attained when $c_1=\sqrt{\frac{12}{27}},c_2=-\frac{15}{27}$ and $c_3=\frac{15}{27}\sqrt{\frac{12}{27}}$.

From (11) and Lemma 3, we get

$$\begin{array}{ccc}\hfill 2\left|J_{2,4}\left(f^{-1}\right)\right|& =& 2\left|A_5-A_2{A}_4\right|=\left|-c_4+\frac{5}{2}{c}_3{c}_1-4c_2{c}_1^2+\frac{3}{4}{c}_2^2+\frac{9}{8}{c}_1^4\right|\hfill \\ & & \le 1-|c_1{|}^2-|c_2{|}^2+\frac{5}{2}|c_1|\left(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|}\right)+4|c_1{|}^2|c_2|+\frac{3}{4}|c_2{|}^2+\frac{9}{8}{|c_1|}^4\hfill \\ & & =1-|c_1{|}^2-\frac{1}{4}|c_2{|}^2+\frac{5}{2}|c_1|\left(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|}\right)+4|c_1{|}^2|c_2|+\frac{9}{8}{|c_1|}^4\hfill \\ & & ={\phi }_8\left(x,y\right)=1-x^2-\frac{1}{4}{y}^2+\frac{5}{2}x-\frac{5}{2}{x}^3-\frac{5x{y}^2}{2(1+x)}+4x^{2}y+\frac{9}{8}{x}^4\hfill \end{array}$$

where $|c_1|=x,|c_2|=y.$

Assume that ${\phi }_8\left(x,y\right)$ has a maximum value at an interior point of $\mathsf{\Lambda }=\left\{\left(x,y\right):0\le x\le 1,\right.$ $\left.0\le y\le 1-x^2\right\}$. The critical points of $\mathsf{\Lambda }$ satisfy the conditions

$$\frac{5}{2}-2x-\frac{15}{2}{x}^2-\frac{5y^2}{2{(1+x)}^2}+8xy+\frac{9}{2}{x}^3=0andy=\frac{8x^2(1+x)}{6x+1}.$$

Solving this system leads to an equality

$$1092x^5+144x^4-187x^3+117x^2+56x+5=0$$

which has no solutions.

On the boundary of $\mathsf{\Lambda }$, we get

$${\phi }_8\left(0,y\right)=1-\frac{1}{4}{y}^2\le 1$$

$${\phi }_8\left(x,0\right)=1+\frac{5}{2}x-x^2-\frac{5}{2}{x}^3+\frac{9}{8}{x}^4\le {\phi }_8(0.5305,0)\approx 1.76067,$$

$${\phi }_8\left(x,1-x^2\right)=\frac{3}{4}+6x^2-\frac{45}{8}{x}^4={\tau }_6\left(x\right)\le {\tau }_6\left(\sqrt{\frac{24}{45}}\right)=\frac{47}{20}.$$

We note that equality in the inequality in $\left|J_{2,4}\left(f^{-1}\right)\right|\le \frac{47}{20}$ is attained when $c_1=\sqrt{\frac{24}{45}}$, $c_2=-\frac{21}{45}$, $c_3=\frac{21}{45}\sqrt{\frac{24}{45}}$ and $c_4=-\frac{504}{2025}$.

From (11) and Lemma 3, we get

$$\begin{array}{ccc}\hfill 2\left|J_{3,3}\left(f^{-1}\right)\right|& =& 2\left|A_5-A_3^2\right|=\left|-c_4+3c_3{c}_1-\frac{17}{4}{c}_2{c}_1^2+c_2^2+\frac{5}{4}{c}_1^4\right|\hfill \\ & & \le 1-|c_1{|}^2-|c_2{|}^2+3|c_1|\left(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|}\right)+\frac{17}{4}|c_1{|}^2|c_2|+|c_2{|}^2+\frac{5}{4}{|c_1|}^4\hfill \\ & & =1-|c_1{|}^2+3|c_1|-3|c_1{|}^3-\frac{3|c_1||c_2{|}^2}{1+|c_1|}+\frac{17}{4}|c_1{|}^2|c_2|+\frac{5}{4}{|c_1|}^4\hfill \\ & & ={\phi }_9\left(x,y\right)=1+3x-x^2-3x^3-\frac{3x{y}^2}{1+x}+\frac{17}{4}{x}^{2}y+\frac{5}{4}{x}^4\hfill \end{array}$$

where $|c_1|=x,|c_2|=y.$

Assume that ${\phi }_9\left(x,y\right)$ has a maximum value at an interior point of $\left(0,1\right)\times \left(1,1-x^2\right)$. The critical points satisfy the conditions

$$3-2x-9x^2-\frac{3y^2}{{(1+x)}^2}+\frac{17}{2}xy+5x^3=0andy=\frac{17x(1+x)}{24}.$$

Solving this system leads to an equality

$$6348x^3-2583x^2-1152x+1728=0$$

which has no solutions.

On the boundary of $\mathsf{\Lambda }$, we get

$${\phi }_9\left(0,y\right)=1$$

$${\phi }_9\left(x,0\right)=1+3x-x^2-3x^3+\frac{5}{4}{x}^4\le {\phi }_9(0.5513,0)\approx 1.96277,$$

$${\phi }_9\left(x,1-x^2\right)=1+\frac{25}{4}{x}^2-6x^4={\tau }_7\left(x\right)\le {\tau }_7\left(\sqrt{\frac{25}{448}}\right)=\frac{1009}{384}.$$

We note that equality in the inequality in $\left|J_{3,3}\left(f^{-1}\right)\right|\le \frac{1009}{768}$ is attained when $c_1=\sqrt{\frac{25}{48}}$, $c_2=-\frac{23}{48}$, $c_3=\frac{23}{48}\sqrt{\frac{25}{48}}$ and $c_4=-\frac{575}{2304}$. □

5. Conclusions

In Geometric Function Theory, many authors have studied and investigated various coefficient functionals of other classes of analytic functions. By using the class ${\mathcal{B}}_0$ of Schwarz functions, we achieved the bounds of logarithmic coefficients, inverse logarithmic coefficients, the Hankel determinant and Zalcman functionals $J_{n,m}$ of inverse functions in the class $\Omega$ presented in this paper.

It is worth noting that the class ${\mathcal{B}}_0$ of the Schwartz function is a good tool to study the coefficients for different subclasses of analytic functions.