Investigation of the Hankel Determinant Sharp Bounds for a Specific Analytic Function Linked to a Cardioid-Shaped Domain

Abstract

One of the challenging tasks in the study of function theory is how to obtain sharp estimates of coefficients that appear in the Taylor–Maclaurin series of analytic univalent functions, and for obtaining these bounds, researchers used the concepts of Carathéodory functions. Among these coefficient-related problems, the problem of the third-order Hankel determinant sharp bound is the most difficult one. The aim of the present study is to determine the sharp bound of the Hankel determinant of third order by using the methodology of the aforementioned Carathéodory function family. Further, we also study some other coefficient-related problems, such as the Fekete–Szegő inequality and the second-order Hankel determinant. We examine these results for the family of bounded turning functions linked with a cardioid-shaped domain.

1. Introduction and Definitions

To fully comprehend the fundamental notions used in our new research, we must revisit certain introductory principles. For this purpose, we consider the open unit disc denoted by ${\mathbb{O}}_d=\left\{z\in \mathbb{C}:\left|z\right|<1\right\}$ and use the symbol $\mathcal{A}$ to represent the family of analytic functions normalized by $g\left(0\right)=g^{\prime }\left(0\right)-1=0.$ This normalization ensures that if $g\in \mathcal{A}$, then g possesses the Taylor’s series expansion

$$g\left(z\right)=z+\sum _{j=1}^{\infty }{d}_j{z}^j,z\in {\mathbb{O}}_d.$$

(1)

Further, recall that an analytic function is univalent in the region ${\mathbb{O}}_d$ if it takes no value more than once in ${\mathbb{O}}_d.$ That is, g being univalent in ${\mathbb{O}}_d$ means mathematically that $g\left(z_1\right)=g\left(z_2\right)$ if $z_1=z_2$ for $z_1,z_2\in {\mathbb{O}}_d.$ Therefore, using the notation $\mathcal{S}$, we refer to the family of univalent functions characterized by the series expansion given in Equation (1). This family was originally introduced by Köebe in 1907. In 2012, Aleman and Constantin [1] established an astonishing connection between fluid dynamics and univalent function theory. They actually provided a simple method that shows how to use a univalent harmonic map for finding explicit solutions to incompressible two-dimensional Euler equations. It has several implications in a variety of applied scientific disciplines, including modern mathematical physics, fluid dynamics, nonlinear integrable system theory, and the theory of partial differential equations.

In 1916, Bieberbach [2] proposed the famous “Bieberbach conjecture” in function theory, stating that if $g\in \mathcal{S}$ (the family of univalent functions with the series expansion as in Equation (1)), then $\left|d_n\right|\le n$ for all $n\ge 2$. He proved this for $n=2$, and subsequent researchers, including Löwner [3], Garabedian and Schiffer [4], Pederson and Schiffer [5], and Pederson [6], confirmed it for $n=3,$ $n=4,$ $n=5,$ and $n=6$, respectively. However, settling the conjecture for $n\ge 7$ remained elusive until 1985 when de-Branges [7] used hypergeometric functions to prove it for every $n\ge 2$. During the period between 1916 and 1985, various subfamilies of $\mathcal{S}$ were introduced, such as starlike (${\mathcal{S}}^{*}$), convex ($\mathcal{C}$), close-to-convex ($\mathcal{K}$), and bounded turning functions ($\mathcal{BT}$), and more subclasses were contributed by recent researchers [8,9,10,11,12,13]. Among these subfamilies, the family ${\mathcal{S}}_{car}^{*}$ of starlike functions was established in [14] and characterized by the condition $\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}\prec \tilde{Q}\left(z\right)$, where $\tilde{Q}\left(z\right)=1+\frac{4}{3}z+\frac{2}{3}{z}^2$, which maps the open unit disk ${\mathbb{O}}_d$ onto a region bounded by the cardioid equation given as ${\left(9x^2+9y^2-18x+5\right)}^2-16{\left(9x^2+9y^2-6x+1\right)}^2=0$.

It is simple to conclude from the above description of the family ${\mathcal{S}}_{car}^{*}$ that

$$g\in {\mathcal{S}}_{car}^{*}\iff g\left(z\right)=zexp\left(\underset{0}{\overset{z}{\int }}\frac{v\left(t\right)-1}{t}dt\right),$$

(2)

for some $v\left(z\right)\prec \tilde{Q}\left(z\right).$ By substituting $v\left(z\right)=\tilde{Q}\left(z\right)$ in (2), we achieve the function

$$G_0\left(z\right)=zexp\left(\frac{2}{3}{\int }_0^z\left(2+t\right)dt\right)=z+\frac{4}{3}{z}^2+\frac{11}{9}{z}^3+\frac{68}{81}{z}^4\cdots ,$$

(3)

and it serves as the extremal function in a number of ${\mathcal{S}}_{car}^{*}$-family problems. Applying the function $\tilde{Q}\left(z\right)$ given in (2)$,$ we now consider the below subfamily ${\mathcal{BT}}_{car}$ of analytic functions

$${\mathcal{BT}}_{car}=\left\{g\in \mathcal{S}:g^{\prime }\left(z\right)\prec \tilde{Q}\left(z\right)\left(z\in {\mathbb{O}}_d\right)\right\}.$$

(4)

The determinant ${\mathcal{H}}_{\lambda ,n}\left(g\right)$, where $n,\lambda \in \mathbb{N}=\{1,2,\dots \}$, is known as the Hankel determinant and was contributed by Pommerenke [15,16]. It is formed by the coefficients of the function $g\in \mathcal{S}$ and is defined as:

$${\mathcal{H}}_{\lambda ,n}\left(g\right)=\left|\begin{array}{cccc}{d}_n\hfill & d_{n+1}\hfill & \dots \hfill & d_{n+\lambda -1}\hfill \\ d_{n+1}\hfill & d_{n+2}\hfill & \dots \hfill & d_{n+\lambda }\hfill \\ ⋮\hfill & ⋮\hfill & \dots \hfill & ⋮\hfill \\ d_{n+\lambda -1}\hfill & d_{n+\lambda }\hfill & \dots \hfill & d_{n+2\lambda -2}\hfill \end{array}\right|.$$

Numerous applications, notably those on power series with integral coefficients by Polya ([17], p. 323) and Cantor [18] and singularities by Hadamard ([17], p. 329) and Edrei [19], highlight the significance of this determinant. There are relatively few publications in the literature that give the bounds of the Hankel determinant for functions of general class $\mathcal{S}$. The best estimate for $g\in \mathcal{S}$ was determined by Hayman in [20] and is $\left|{\mathcal{H}}_{2,n}\left(g\right)\right|\le \left|\eta \right|$, where $\eta$ is a constant. Additionally, for $g\in \mathcal{S}$, it was shown in [21] that the second-order Hankel determinant $\left|{\mathcal{H}}_{2,2}\left(g\right)\right|\le \eta$ for $0\le \eta \le 11/3$. The two determinants ${\mathcal{H}}_{2,1}\left(g\right)$ and ${\mathcal{H}}_{2,2}\left(g\right)$ have been extensively studied in the literature for various subfamilies of univalent functions. The work done by the authors [22,23,24,25,26,27,28], where they determined sharp bounds for the second determinant, is particularly noteworthy. For further exploration of this determinant, refer to the articles [29,30,31,32,33,34].

The most challenging problem to study is the below third-order determinant, especially in finding sharp bounds for this determinant.

$${\mathcal{H}}_{3,1}\left(g\right)=\left|\begin{array}{ccc}1\hfill & d_2\hfill & d_3\hfill \\ d_2\hfill & d_3\hfill & d_4\hfill \\ d_3\hfill & d_4\hfill & d_5\hfill \end{array}\right|=d_3\left(d_2{d}_4-d_3^2\right)-d_4\left(d_4-d_2{d}_3\right)+d_5\left(d_3-d_2^2\right).$$

(5)

Although there are several papers on the investigation of non-sharp bounds of this determinant, we cite here a few of them. See [35,36,37,38]. In fact, Babalola was the very first person to study the bounds of this third-order determinant for $\mathcal{K},$ ${\mathcal{S}}^{*}$ and $\mathcal{BT}$ families in a paper [39] that surfaced in 2010. Subsequently, employing an innovative methodology, Zaprawa [40] further advanced Babalola’s 2017 findings by establishing the subsequent bounds:

$$\left|{\mathcal{H}}_{3,1}\left(g\right)\right|\le \left\{\begin{array}{ccc}\frac{49}{540},\hfill & \mathrm{for}\hfill & g\in \mathcal{C},\hfill \\ 1,\hfill & \mathrm{for}\hfill & g\in {\mathcal{S}}^{*},\hfill \\ \frac{41}{60},\hfill & \mathrm{for}\hfill & g\in \mathcal{BT}.\hfill \end{array}\right.$$

Additionally, it is worth noting that Zaprawa emphasized the potential for enhancing the aforementioned bounds, as they may not represent the optimal limits, as stated by Babalola.

Following that, scientists worked hard to prove sharp bounds (which cannot be improved further) for these inequalities, and some of them [41,42] were successful in obtaining improved bounds for the class ${\mathcal{S}}^{*}$. The sharp bounds of this determinant were finally obtained for classes $\mathcal{C},$ ${\mathcal{S}}^{*},$ and $\mathcal{BT}$ in the articles [43,44,45], respectively. These sharp bounds are

$$\left|{\mathcal{H}}_{3,1}\left(g\right)\right|\le \left\{\begin{array}{ccc}\frac{4}{135},\hfill & \mathrm{for}\hfill & g\in \mathcal{C},\hfill \\ \frac{4}{9},\hfill & \mathrm{for}\hfill & g\in {\mathcal{S}}^{*},\hfill \\ \frac{1}{4},\hfill & \mathrm{for}\hfill & g\in \mathcal{BT}.\hfill \end{array}\right.$$

By employing similar techniques, Khalil Ullah et al. [46] and Lecko et al. [47] derived the sharp bounds for $\left|{\mathcal{H}}_{3,1}\left(g\right)\right|$ when considering functions belonging to the families ${\mathcal{S}}_{tanh}^{*}$ and ${\mathcal{S}}^{*}\left(1/2\right)$, respectively. Additionally, the works of authors [48,49,50,51,52,53] proved sharp bounds for the third-order Hankel determinant in various novel subfamilies of univalent functions. In the present work, we consider a family ${\mathcal{BT}}_{car}$ of bounded turning functions related to the cardioid-shaped domain. Using the approach of the aforementioned Carathéodory function family, we calculate the sharp bound of the third-order Hankel determinant for functions belonging to the family ${\mathcal{BT}}_{car}$. The Fekete–Szegő inequality and the second-order Hankel determinant are a couple of the other coefficient-related problems that we study here in this article.

2. A Set of Lemmas

Let $\mathcal{P}$ represent the class of all functions p that are analytic in ${\mathbb{O}}_d$ with $\mathfrak{Re}\left(p\left(z\right)\right)>0$ and also has the series representation

$$p\left(z\right)=1+\sum _{n=1}^{\infty }{c}_n{z}^{n}z\in {\mathbb{O}}_d.$$

(6)

The core of our proof lies in the following lemma, which encompasses the well-known formula for $c_2$ (see [54]), the formula for $c_3$ derived by Libera and Zlotkiewicz [55,56], and the formula for $c_4$ contributed by the authors [57].

Lemma 1. 

Let $p\in \mathcal{P}$ and be given by (6)$.$ Then, for some complex valued x with $\left|x\right|\le 1,$ and some complex valued $\zeta ,\rho$ with $\left|\zeta \right|\le 1,\left|\rho \right|\le 1$

$$\begin{array}{ccc}\hfill 2c_2& =& c_1^2+x\left(4-c_1^2\right),\hfill \end{array}$$

(7)

$$\begin{array}{ccc}\hfill 4c_3& =& c_1^3+2\left(4-c_1^2\right)c_{1}x-c_1\left(4-c_1^2\right)x^2+2\left(4-c_1^2\right)\left(1-{\left|x\right|}^2\right)\zeta ,\hfill \\ \hfill 8c_4& =& c_1^4+(4-c_1^2)x\left[c_1^2\left(x^2-3x+3\right)+4x\right]-4(4-c_1^2)(1-{\left|x\right|}^2)\hfill \end{array}$$

(8)

$$\begin{array}{ccc}& & \left[c(x-1)\zeta +\overline{x}{\zeta }^2-(1-{\left|\zeta \right|}^2)\rho \right].\hfill \end{array}$$

(9)

Lemma 2. 

For any $p\in \mathcal{P}$ expressed in the form (6), and considering $\mu \in \mathbb{C}$, we have the inequality

$$\left|c_{n+k}-\mu c_n{c}_k\right|\le 2max\left\{1,\left|2\mu -1\right|\right\}.$$

(10)

Furthermore, if $B\in [0,1]$ and satisfies $B(2B-1)\le D\le B$, then the following holds:

$$\left|c_3-2B{c}_1{c}_2+D{c}_1^3\right|\le 2$$

(11)

These inequalities, (10) and (11), are extracted from references [56,58].

3. Hankel Determinant Problem for ${\mathcal{BT}}_{\mathbf{car}}$

Theorem 1. 

If g has the form (1)$,$ and belongs to ${\mathcal{BT}}_{car}.$ Then

$$\left|d_3-\gamma d_2^2\right|\le \frac{4}{9}max\left\{1,\left|\frac{2\gamma -1}{2}\right|\right\},\mathit{for}\gamma \in \mathbb{C}.$$

This outcome is precise and cannot be further enhanced.

Proof. 

Consider $g\in \mathcal{BT}car$. Utilizing the Schwarz function $w\left(z\right)$, we readily express

$$g^{\prime }\left(z\right)=1+\frac{4}{3}w\left(z\right)+\frac{2}{3}{\left(w\left(z\right)\right)}^2,\left(z\in {\mathbb{O}}_d\right).$$

Furthermore, let $p\in \mathcal{P}$, which can be represented in terms of the Schwarz function $w\left(z\right)$ as follows:

$$p\left(z\right)=\frac{1+w\left(z\right)}{1-w\left(z\right)}=1+c_{1}z+c_2{z}^2+c_3{z}^3+\cdots .$$

Alternatively, the expression for $w\left(z\right)$ can be written as:

$$w\left(z\right)=\frac{p\left(z\right)-1}{p\left(z\right)+1}=\frac{c_{1}z+c_2{z}^2+c_3{z}^3+c_4{z}^4+\cdots }{2+c_{1}z+c_2{z}^2+c_3{z}^3+c_4{z}^4+\cdots }.$$

(12)

From (1)$,$ we have

$$g^{\prime }\left(z\right)=1+2d_{2}z+3d_3{z}^2+4d_4{z}^3+5d_5{z}^4+\cdots .$$

(13)

By employing the series expansion of (12) and performing straightforward calculations, we obtain the following expression:

$$\begin{array}{ccc}\hfill 1+\frac{4}{3}w\left(z\right)+\frac{2}{3}{\left(w\left(z\right)\right)}^2& =& 1+\frac{2}{3}{c}_{1}z+\left(\frac{2}{3}{c}_2-\frac{1}{6}{c}_1^2\right)z^2+\left(\frac{2}{3}{c}_3-\frac{1}{3}{c}_1{c}_2\right)z^3\hfill \\ & & +\left(\frac{2}{3}{c}_4+\frac{1}{24}{c}_1^4-\frac{1}{6}{c}_2^2-\frac{1}{3}{c}_1{c}_3\right)z^4+\cdots .\hfill \end{array}$$

(14)

Comparing (13) and (14), we obtain

$$\begin{array}{ccc}\hfill d_2& =& \frac{1}{3}{c}_1,\hfill \end{array}$$

(15)

$$\begin{array}{ccc}\hfill d_3& =& \frac{1}{3}\left(\frac{2}{3}{c}_2-\frac{1}{6}{c}_1^2\right),\hfill \end{array}$$

(16)

$$\begin{array}{ccc}\hfill d_4& =& \frac{1}{4}\left(\frac{2}{3}{c}_3-\frac{1}{3}{c}_1{c}_2\right),\hfill \end{array}$$

(17)

$$\begin{array}{ccc}\hfill d_5& =& \frac{1}{5}\left(\frac{2}{3}{c}_4-\frac{1}{6}{c}_2^2+\frac{1}{24}{c}_1^4-\frac{1}{3}{c}_1{c}_3\right).\hfill \end{array}$$

(18)

From (15) and (16), we easily get

$$\left|d_3-\gamma d_2^2\right|=\left|\frac{2}{9}{c}_2-\frac{1}{18}{c}_1^2-\gamma \frac{1}{9}{c}_1^2\right|.$$

By rearranging, it yields

$$\left|d_3-\gamma d_2^2\right|=\frac{2}{9}\left|\left(c_2-\left(\frac{2\gamma +1}{4}\right)c_1^2\right)\right|.$$

Application of (10)$,$ leads us to

$$\left|d_3-\gamma d_2^2\right|\le \frac{4}{9}max\left\{1,\left|\left(\frac{2\gamma +1}{2}\right)-1\right|\right\},$$

and after simplification, we get the required result.

Equality is obtained by using coefficients in the below function

$${\int }_0^z\left(1+\frac{4}{3}\left(t\right)+\frac{2}{3}\left(t^2\right)\right)dt=z+\frac{2}{3}{z}^2+\frac{2}{9}{z}^3+\cdots .$$

□

Theorem 2. 

If g has the form (1)$,$ and belongs to ${\mathcal{BT}}_{car}.$ Then

$$\left|d_2{d}_3-d_4\right|\le \frac{1}{3}.$$

This inequality stands as its optimal form, with no potential for further enhancement.

Proof. 

Employing (15)–(17)$,$ we obtain

$$\left|d_2{d}_3-d_4\right|=\frac{1}{6}\left|c_3-2\left(\frac{17}{36}\right)c_1{c}_2+\frac{1}{9}{c}_1^3\right|.$$

From (11)$,$ we have

$$0<B=\frac{17}{36}<1,B=\frac{17}{36}>D=\frac{1}{9}.$$

and

$$B\left(2B-1\right)=-\frac{17}{648}<D=\frac{1}{9}.$$

Then, by using $\left|c_3-2\left(\frac{17}{36}\right)c_1{c}_2+\frac{1}{9}{c}_1^3\right|$ $\le 2$ along with the triangle inequality, we obtain

$$\left|d_2{d}_3-d_4\right|\le \frac{1}{3}.$$

This inequality attains its optimal state and is realized by

$${\int }_0^z\left(1+\frac{4}{3}\left(t^3\right)+\frac{2}{3}\left(t^6\right)\right)dt=z+\frac{1}{3}{z}^4+\frac{2}{21}{z}^7+\cdots .$$

□

Theorem 3. 

For g in ${\mathcal{BT}}_{car}$, the second Hankel determinant is given by:

$$\left|{\mathcal{H}}_{2,2}\left(g\right)\right|=\left|d_2{d}_4-d_3^2\right|\le \frac{16}{81}.$$

The above bound is the best possible.

Proof. 

From (15)–(17)$,$ we have

$$d_2{d}_4-d_3^2=-\frac{1}{324}{c}_1^4-\frac{1}{324}{c}_1^2{c}_2+\frac{1}{18}{c}_1{c}_3-\frac{4}{81}{c}_2^2.$$

Using Equations (7) and (8) to express $c_2$ and $c_3$ in terms of $c_1$ (noting that we can assume without loss of generality that $c_1=c$ with $0\le c\le 2$), we obtain the following expression:

$$\begin{array}{ccc}\hfill \left|d_2{d}_4-d_3^2\right|& =& \left|-\frac{1}{324}{c}^4-\frac{1}{72}{c}^2\left(4-c^2\right)x^2-\frac{1}{81}{\left(4-c^2\right)}^2{x}^2+\frac{1}{648}{c}^2\left(4-c^2\right)x\right.\hfill \\ & & \left.+\frac{1}{36}c\left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\zeta \right|,\hfill \end{array}$$

where we used the triangle inequality and replaced $\left|\zeta \right|\le 1$ and $\left|x\right|=b,$ with $b\le 1$ and $c\in \left[0,2\right]$. Thus, we have:

$$\begin{array}{ccc}\hfill \left|d_2{d}_4-d_3^2\right|& \le & \left|\frac{1}{324}{c}^4+\frac{1}{72}{c}^2\left(4-c^2\right)b^2+\frac{1}{81}{\left(4-c^2\right)}^2{b}^2+\frac{1}{648}{c}^2\left(4-c^2\right)b\right.\hfill \\ & & \left.+\frac{1}{36}c\left(4-c^2\right)\left(1-b^2\right):=\varphi \left(c,b\right)\right|.\hfill \end{array}$$

With a simple exercise, we can show that ${\varphi }^{\prime }\left(c,b\right)\ge 0$ on $\left[0,1\right]$, so $\varphi \left(c,b\right)\le \varphi \left(c,1\right)$. Setting $b=1$ gives:

$$\left|d_2{d}_4-d_3^2\right|\le \frac{1}{324}{c}^4+\frac{5}{324}{c}^2\left(4-c^2\right)+\frac{1}{81}{\left(4-c^2\right)}^2:=\varphi \left(c,1\right).$$

Since ${\varphi }^{\prime }\left(c,1\right)<0,$ $\varphi \left(c,1\right)$ is a decreasing function and attains its maximum value at $c=0$:

$$\left|d_2{d}_4-d_3^2\right|\le \frac{16}{81}.$$

The bound of second Hankel determinant is best possible and is achieved by using the coefficients of the function

$${\int }_0^z\left(1+\frac{4}{3}\left(t^2\right)+\frac{2}{3}\left(t^4\right)\right)dt=z+\frac{4}{9}{z}^3+\frac{2}{15}{z}^5+\cdots .$$

□

Theorem 4. 

For an f that is an element of ${\mathcal{BT}}_{car}$ and follows the structure presented in (1), we have:

$$\left|{\mathcal{H}}_{2,3}\left(g\right)\right|=\left|d_3{d}_5-d_4^2\right|\le \frac{1}{9}.$$

This outcome represents the optimal condition for this result.

Proof. 

Using (16)–(18) along with $c_1=c,$ we get

$$\begin{array}{ccc}\hfill {\mathcal{H}}_{2,3}\left(g\right)& =& \frac{1}{2160}\left(-c^6+4c^4{c}_2+8c^3{c}_3-11c^2{c}_2^2-16c^2{c}_4+28c{c}_2{c}_3\right.\hfill \\ & & \left.-16c_2^3+64c_2{c}_4-60c_3^2\right).\hfill \end{array}$$

(19)

Let $n=4-c^2$ in Equations (7)–(9). Now, using these formulae, we obtain

$$\begin{array}{ccc}\hfill 4c^4{c}_2& =& 2\left(c^{4}nx+c^6\right),\hfill \\ \hfill 8c^3{c}_3& =& -2c^{4}n{x}^2+4c^{3}n\left(1-{\left|x\right|}^2\right)\zeta +4c^{4}nx+2c^6,\hfill \\ \hfill 11c^2{c}_2^2& =& \frac{11}{4}{c}^6+\frac{11}{2}{c}^{4}nx+\frac{11}{4}{c}^2{n}^2{x}^2,\hfill \\ \hfill 16c^2{c}_4& =& 2c^6+2c^{4}n{x}^3-6c^{4}n{x}^2+6c^{4}nx+8c^{2}n{x}^2-8c^{3}nx\left(1-{\left|x\right|}^2\right)\zeta \hfill \\ & & -8c^{2}n\overline{x}\left(1-{\left|x\right|}^2\right){\zeta }^2+8c^{2}n\left(1-{\left|x\right|}^2\right)\left(1-{\left|\zeta \right|}^2\right)\rho +8c^{3}n\left(1-{\left|x\right|}^2\right)\zeta ,\hfill \\ \hfill 28c{c}_2{c}_3& =& \frac{7}{2}{c}^6+\frac{21}{2}{c}^{4}nx-\frac{7}{2}{c}^{4}n{x}^2+7c^{3}n\left(1-{\left|x\right|}^2\right)\zeta +7c^2{n}^2{x}^2-\frac{7}{2}{c}^2{n}^2{x}^3\hfill \\ & & +7c{n}^{2}x\left(1-{\left|x\right|}^2\right)\zeta ,\hfill \\ \hfill 16c_2^3& =& 2c^6+6c^{4}nx+6c^2{n}^2{x}^2+2n^3{x}^3,\hfill \\ \hfill 64c_2{c}_4& =& 4c^6+4c^{4}n{x}^3-12c^{4}n{x}^2+16c^{4}nx+16c^{2}n{x}^2-16c^{3}nx\left(1-{\left|x\right|}^2\right)\zeta \hfill \\ & & -16c^{2}n\overline{x}\left(1-{\left|x\right|}^2\right){\zeta }^2+16c^{2}n\left(1-{\left|x\right|}^2\right)\left(1-{\left|\zeta \right|}^2\right)\rho +16c^{3}n\hfill \\ & & \left(1-{\left|x\right|}^2\right)\zeta +4c^2{n}^2{x}^4-12c^2{n}^2{x}^3+12c^2{n}^2{x}^2+16n^2{x}^3\hfill \\ & & -16c{n}^2{x}^2\left(1-{\left|x\right|}^2\right)\zeta -16n^{2}x\overline{x}\left(1-{\left|x\right|}^2\right){\zeta }^2+16n^{2}x\left(1-{\left|x\right|}^2\right)\hfill \\ & & \left(1-{\left|\zeta \right|}^2\right)\rho +16c{n}^{2}x\left(1-{\left|x\right|}^2\right)\zeta ,\hfill \\ \hfill 60c_3^2& =& \frac{15}{4}{c}^6+15c^{4}nx-\frac{15}{2}{c}^{4}n{x}^2+15c^{3}n\left(1-{\left|x\right|}^2\right)\zeta +15c^2{n}^2{x}^2\hfill \\ & & -15c^2{n}^2{x}^3+30c{n}^{2}x\left(1-{\left|x\right|}^2\right)\zeta +\frac{15}{4}{c}^2{n}^2{x}^4-15c{n}^2{x}^2\hfill \\ & & \left(1-{\left|x\right|}^2\right)\zeta +15n^2{\left(1-{\left|x\right|}^2\right)}^2{\zeta }^2.\hfill \end{array}$$

Substituting the above expressions in (19)$,$ we have

$$\begin{array}{ccc}\hfill {\mathcal{H}}_{2,3}\left(g\right)& =& \frac{1}{2160}\left\{-2x^3{n}^3+16x^3{n}^2-\frac{19}{4}{c}^2{x}^2{n}^2+8c^2{x}^{2}n+2c^4{x}^{3}n-\frac{1}{2}{c}^2{x}^3{n}^2\right.\hfill \\ & & +\frac{1}{4}{c}^2{x}^4{n}^2-15n^2{\left(1-{\left|x\right|}^2\right)}^2{\zeta }^2-4c^4{x}^{2}n-8c^{3}xn\left(1-{\left|x\right|}^2\right)\zeta \hfill \\ & & -8c^{2}n\overline{x}\left(1-{\left|x\right|}^2\right){\zeta }^2+8c^{2}n\left(1-{\left|x\right|}^2\right)\left(1-{\left|\zeta \right|}^2\right)\rho -7cx{n}^2\hfill \\ & & \left(1-{\left|x\right|}^2\right)\zeta -c{x}^2{n}^2\left(1-{\left|x\right|}^2\right)\zeta -16x{n}^2\overline{x}\left(1-{\left|x\right|}^2\right){\zeta }^2\hfill \\ & & \left.+16x{n}^2\left(1-{\left|x\right|}^2\right)\left(1-{\left|\zeta \right|}^2\right)\rho +4c^{3}n\left(1-{\left|x\right|}^2\right)\zeta \right\}.\hfill \end{array}$$

Since $n=4-c^2,$

$${\mathcal{H}}_{2,3}\left(g\right)=\frac{1}{2160}\left(w_1\left(c,x\right)+w_2\left(c,x\right)\zeta +w_3\left(c,x\right){\zeta }^2+w_4\left(c,x,\zeta \right)\rho \right),$$

where $\rho ,x,\zeta \in {\mathbb{O}}_d$, and

$$\begin{array}{ccc}\hfill w_1\left(c,x\right)& =& \left(4-c^2\right)[\left(4-c^2\right)\left(8x^3+\frac{3}{2}{c}^2{x}^3-\frac{19}{4}{c}^2{x}^2+\frac{1}{4}{c}^2{x}^4\right)\hfill \\ & & +8c^2{x}^2+2c^4{x}^3-4c^4{x}^2],\hfill \\ \hfill w_2\left(c,x\right)& =& \left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\left[\left(4-c^2\right)\left(-7cx-c{x}^2\right)-8c^{3}x+4c^3\right],\hfill \\ \hfill w_3\left(c,x\right)& =& \left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\left[\left(4-c^2\right)\left(-x^2-15\right)-8c^2\overline{x}\right],\hfill \\ \hfill w_4\left(c,x,\zeta \right)& =& \left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\left(1-{\left|\zeta \right|}^2\right)\left[8c^2+16x\left(4-c^2\right)\right].\hfill \end{array}$$

By using $\left|x\right|=x,\left|\zeta \right|=y$ and utilizing the fact $\left|\rho \right|\le 1,$ we obtain

$$\begin{array}{ccc}\hfill \left|{\mathcal{H}}_{2,3}\left(g\right)\right|& \le & \frac{1}{2160}\left(\left|w_1\left(c,x\right)\right|+\left|w_2\left(c,x\right)\right|y+\left|w_3\left(c,x\right)\right|y^2+\left|w_4\left(c,x,\zeta \right)\right|\right).\hfill \\ & \le & \frac{1}{2160}\left(W\left(c,x,y\right)\right),\hfill \end{array}$$

(20)

where

$$W\left(c,x,y\right)=l_1\left(c,x\right)+l_2\left(c,x\right)y+l_3\left(c,x\right)y^2+l_4\left(c,x\right)\left(1-y^2\right),$$

with

$$\begin{array}{ccc}\hfill l_1\left(c,x\right)& =& \left(4-c^2\right)[\left(4-c^2\right)\left(8x^3+\frac{3}{2}{c}^2{x}^3+\frac{19}{4}{c}^2{x}^2+\frac{1}{4}{c}^2{x}^4\right)\hfill \\ & & +8c^2{x}^2+2c^4{x}^3+4c^4{x}^2],\hfill \\ \hfill l_2\left(c,x\right)& =& \left(4-c^2\right)\left(1-x^2\right)\left[\left(4-c^2\right)\left(7cx+c{x}^2\right)+8c^{3}x+4c^3\right],\hfill \\ \hfill l_3\left(c,x\right)& =& \left(4-c^2\right)\left(1-x^2\right)\left[\left(4-c^2\right)\left(x^2+15\right)+8c^{2}x\right],\hfill \\ \hfill l_4\left(c,x\right)& =& \left(4-c^2\right)\left(1-x^2\right)\left[8c^2+16x\left(4-c^2\right)\right].\hfill \end{array}$$

We now aim to maximize $W\left(c,x,y\right)$ within the closed cuboid $\mathsf{\Theta }:\left[0,2\right]\times \left[0,1\right]\times \left[0,1\right].$

To achieve this, we must consider the maximum values of $W\left(c,x,y\right)$ in the interior of $\mathsf{\Theta }$, within the interior of its six faces, and on its twelve edges.

• Interior points of cuboid $\mathsf{\Theta }$:

Let $\left(c,x,y\right)\in \left(0,2\right)\times \left(0,1\right)\times \left(0,1\right).$ Differentiating $W\left(c,x,y\right)$ partially with respect to y, we obtain:

$$\begin{array}{ccc}\hfill \frac{\partial W}{\partial y}& =& \left(4-c^2\right)(1-x^2)\left[2y(x-1)\left(\left(4-c^2\right)\left(x-15\right)+8c^2\right)\right.\hfill \\ & & \left.+c\left(x\left(4-c^2\right)\left(x+7\right)+4c^2\left(2x+1\right)\right)\right].\hfill \end{array}$$

Setting $\frac{\partial W}{\partial y}=0,$ we find:

$$y=\frac{c\left(x\left(4-c^2\right)\left(x+7\right)+4c^2\left(2x+1\right)\right)}{2(x-1)\left(\left(4-c^2\right)\left(15-x\right)-8c^2\right)}=y_0.$$

If $y_0$ is a critical point inside $\mathsf{\Theta }$, then $y_0\in \left(0,1\right),$ which is only possible if:

$$4c^3\left(2x+1\right)+cx\left(4-c^2\right)\left(x+7\right)+2(1-x)\left(4-c^2\right)\left(15-x\right)<16c^2(1-x),$$

(21)

and

$$c^2>\frac{4(15-x)}{23-x}.$$

(22)

Now, we need to find the solutions that satisfy both inequalities (21) and (22) for the existence of critical points.

Let $g\left(x\right)=\frac{4(15 - x)}{23 - x}.$ Since $g^{\prime }\left(x\right)<0$ in $\left(0,1\right),$ $g\left(x\right)$ is decreasing over $\left(0,1\right).$ Hence, $c^2>\frac{28}{11},$ and a calculation shows that the Equation (21) is satisfied for $c>1.754523050$ and $x<\frac{1}{4}.$

Subsequently, we establish that $W(c,x,y)<\frac{1}{36}$ in the region $(1.754523050,2)\times (0,\frac{1}{4})\times (0,1)$. For values of x less than $\frac{13}{32}$, we have $1-x^2<1$, thereby allowing us to express the following inequalities:

$$\begin{array}{ccc}\hfill l_1(c,x)& \le & \frac{1}{9}{c}^6+(4-c^2)\left(\frac{762125}{2359296}{c}^4+\frac{3176017}{196608}{c}^2+\frac{2197}{192}\right):={\mathsf{\Phi }}_1\left(c\right),\hfill \\ \hfill l_2(c,x)& \le & c(4-c^2)\left(\frac{1027}{16}+\frac{1783}{192}{c}^2\right):={\mathsf{\Phi }}_2\left(c\right),\hfill \\ \hfill l_3(c,x)& \le & (4-c^2)\left(\frac{15529}{48}-\frac{4067}{64}{c}^2\right):={\mathsf{\Phi }}_3\left(c\right),\hfill \\ \hfill l_4(c,x)& \le & (4-c^2)\left(8c^2+\frac{416}{3}\right):={\mathsf{\Phi }}_4\left(c\right).\hfill \end{array}$$

Hence, we obtain:

$$W(c,x,y)\le \frac{1}{20480}\left[{\mathsf{\Phi }}_1\left(c\right)+{\mathsf{\Phi }}_4\left(c\right)+{\mathsf{\Phi }}_2\left(c\right)y+\left[{\mathsf{\Phi }}_3\left(c\right)-{\mathsf{\Phi }}_4\left(c\right)\right]y^2\right]:=\psi (c,y).$$

At this point, we examine the derivatives:

$$\frac{\partial \psi }{\partial y}=\frac{1}{20480}\left[{\mathsf{\Phi }}_2\left(c\right)+2\left[{\mathsf{\Phi }}_3\left(c\right)-{\mathsf{\Phi }}_4\left(c\right)\right]y\right]$$

and:

$$\frac{{\partial }^2\psi }{\partial y^2}=\frac{1}{10240}\left[{\mathsf{\Phi }}_3\left(c\right)-{\mathsf{\Phi }}_4\left(c\right)\right].$$

Given that ${\mathsf{\Phi }}_3\left(c\right)-{\mathsf{\Phi }}_4\left(c\right)\le 0$ for $c\in (1.754523050,2)$, it follows that $\frac{{\partial }^2\psi }{\partial y^2}\le 0$ for $y\in (0,1)$. Consequently, $\frac{\partial \psi }{\partial y}$ is a decreasing function, resulting in:

$$\frac{\partial \psi }{\partial y}\le \frac{\partial \psi }{\partial y}{|}_{y=0}={\mathsf{\Phi }}_2\left(c\right)\ge 0,y\in (0,1).$$

Thus, we deduce:

$$\psi (c,y)\le \psi (c,1)=\frac{1}{20480}\left[{\mathsf{\Phi }}_1\left(c\right)+{\mathsf{\Phi }}_2\left(c\right)+{\mathsf{\Phi }}_3\left(c\right)\right]:=\kappa \left(c\right).$$

It is easy to see that $\kappa$ takes its maximum value $0.02169278324$ at $c=1.754523050.$ Thus, we have:

$$W(c,x,y)<\frac{1}{9}\approx 0.0625,(c,x,y)\in \left(1.754523050,2\right)\times \left(0,\frac{1}{4}\right)\times \left(0,1\right).$$

Hence, $W(c,x,y)<\frac{1}{9}$. Therefore, W does not have an optimal solution in the interior of $\mathsf{\Theta }.$

2.

Interior of all the six faces of cuboid $\mathsf{\Theta }$:

(i)

On the face $c=0,$ $W(c,x,y)$ becomes:

$$I_1(x,y)=W(0,x,y)=16\left[(1-x^2)(y^2(x-15)(x-1)+16x)+8x^3\right],x,y\in \left(0,1\right).$$

Differentiating $I_1(x,y)$ partially with respect to $y,$ we have:

$$\frac{\partial I_1}{\partial y}=32y(1-x^2)(x-15)(x-1),x,y\in \left(0,1\right).$$

Thus, $I_1(x,y)$ has no critical point in the interval $\left(0,1\right)\times \left(0,1\right).$

(ii)

On the face $c=2,$ $W(c,x,y)$ takes the form:

$$W(2,x,y)=0.$$

(iii)

On the face $x=0,$ $W(c,x,y)$ yields:

$$I_2(c,y)=W(c,0,y)=(4-c^2)(4c^{3}y+(60-23c^2)y^2+8c^2).$$

Differentiating $I_2(c,y)$ partially with respect to $y,$ we have:

$$\frac{\partial I_2}{\partial y}=(4-c^2)(4c^3+(120-46c^2)y).$$

Setting $\frac{\partial I_2}{\partial y}=0,$ we obtain:

$$y=\frac{2c^3}{23c^2-60}=y_1.$$

For the given range of $y,$ $y_1$ should belong to $\left(0,1\right),$ which is possible only if $c>c_0, c_0\approx 1.75452304.$ In addition, the derivative of $I_2(c,y)$ partially with respect to c is:

$$\begin{array}{ccc}\hfill \frac{\partial I_2}{\partial c}& =& (4-c^2)(12c^{2}y-46c{y}^2+16c)-16c^3-8c^{4}y\hfill \\ & & -\left(120-46c^2\right)c{y}^2.\hfill \end{array}$$

(23)

Plugging the value of y in (23)$,$ and setting $\frac{\partial I_2}{\partial c}=0,$ we obtain:

$$\frac{\partial I_2}{\partial c}=-8c(69c^8+1692c^6-14552c^4+36480c^2-28800)=0.$$

(24)

A calculation gives the solution of (24) in the interval $\left(0,1\right)$, which is $c\approx 1.297674409.$ Thus, $I_2(c,y)$ has no optimal point in the interval $\left(0,2\right)\times \left(0,1\right).$

(iv)

On the face $x=1,$ $W(c,x,y)$ becomes:

$$I_3(c,y)=W(c,1,y)=(4-c^2)((4-c^2)(\frac{13}{2}{c}^2+8)+6c^4+8c^2).$$

Then:

$$\frac{\partial I_3}{\partial c}=3c^5-112c^3+144c.$$

Setting $\frac{\partial I_3}{\partial c}=0,$ we get $c\approx 1.1547005383,$ at which $I_3(c,y)$ achieves its maximum, which is $max{I}_3(c,y)=175.4074074.$ Hence:

$$\left|{\mathcal{H}}_{2,3}\left(g\right)\right|\le 0.081207133.$$

(v)

On the face $y=0,$ $W(c,x,y)$ reduces to

$$\begin{array}{ccc}\hfill I_4(c,x)& =& W(c,x,0)=-12c^4{x}^3+88c^2{x}^3-128x^3-\frac{1}{2}{c}^6{x}^3+\frac{1}{4}{c}^6{x}^4\hfill \\ & & -2c^4{x}^4+4c^2{x}^4+\frac{3}{4}{c}^6{x}^2-22c^4{x}^2+76c^2{x}^2\hfill \\ & & +16c^{4}x-8c^4-128c^{2}x+32c^2+256x.\hfill \end{array}$$

Now differentiating partially with respect to $c,$ then with respect to x, we have

$$\begin{array}{ccc}\hfill \frac{\partial I_4}{\partial c}& =& -48c^3{x}^3+176c{x}^3-3c^5{x}^3+\frac{3}{2}{c}^5{x}^4-8c^3{x}^4+8c{x}^4+\frac{9}{2}{c}^5{x}^2\hfill \\ & & -88c^3{x}^2+152c{x}^2+64c^{3}x-32c^3-256cx+64c.\hfill \end{array}$$

(25)

and

$$\begin{array}{ccc}\hfill \frac{\partial I_4}{\partial x}& =& -36c^4{x}^2+264c^2{x}^2-384x^2-\frac{3}{2}{c}^6{x}^2+c^6{x}^3-8c^4{x}^3+16c^2{x}^3\hfill \\ & & +\frac{3}{2}{c}^{6}x-44c^{4}x+152c^{2}x+16c^4-128c^2+256.\hfill \end{array}$$

(26)

A numerical calculation shows that the solution does not exist for the system of Equations (25) and (26) in $\left(0,2\right)\times \left(0,1\right).$

(vi)

On the face $y=1,$ $W(c,x,y)$ after some simplification becomes

$$\begin{array}{ccc}\hfill I_5(c,x)& =& W(c,x,1)=240-\frac{1}{2}{c}^6{x}^3+\frac{1}{4}{c}^6{x}^4+\frac{3}{4}{c}^6{x}^2-c^5{x}^4+c^5{x}^3+5c^5{x}^2\hfill \\ & & +8c^3{x}^4-c^{5}x+24c^3{x}^3-24c^3{x}^2-16c{x}^4-112c{x}^3-4c^5\hfill \\ & & -3c^4{x}^4-8c^{4}x-16x^4+15c^4-120c^2+128x^3-224x^2\hfill \\ & & +16c^3-72c^2{x}^3+12c^2{x}^4+220c^2{x}^2+12c^4{x}^3-44c^4{x}^2\hfill \\ & & +112cx+16c{x}^2-24c^{3}x+32c^{2}x.\hfill \end{array}$$

Partial derivative of $I_5(c,x)$ with respect to c and then with respect to x, we have

$$\begin{array}{ccc}\hfill \frac{\partial I_5}{\partial c}& =& \frac{3}{2}{c}^5{x}^4-3c^5{x}^3+\frac{9}{2}{c}^5{x}^2-12c^3{x}^4+48c^3{x}^3-176c^3{x}^2+24c{x}^4\hfill \\ & & -144c{x}^3-5c^4{x}^4-5c^{4}x-16x^4-20c^4+48c^2-112x^3+16x^2\hfill \\ & & +60c^3+72c^2{x}^3+24c^2{x}^4-72c^2{x}^2+5c^4{x}^3+25c^4{x}^2+64cx\hfill \\ & & +440c{x}^2-32c^{3}x-72c^{2}x-240c+112x.\hfill \end{array}$$

(27)

and

$$\begin{array}{ccc}\hfill \frac{\partial I_5}{\partial x}& =& \frac{3}{2}{c}^{6}x+c^6{x}^3-\frac{3}{2}{c}^6{x}^2-4c^5{x}^3+3c^5{x}^2+10c^{5}x+32c^3{x}^3+72c^3{x}^2\hfill \\ & & -64c{x}^3-c^5-88c^{4}x-8c^4+32c^2-64x^3+384x^2-24c^3\hfill \\ & & +48c^2{x}^3-216c^2{x}^2-12c^4{x}^3+36c^4{x}^2+32cx-336c{x}^2\hfill \\ & & -48c^{3}x+440c^{2}x+112c-448x.\hfill \end{array}$$

(28)

As in the above case, we deduce the same result for the face $y=0,$ that is, that the system of Equations (27) and (28) does not have a solution in $\left(0,2\right)\times \left(0,1\right).$

3.

On the Edges of Cuboid $\mathsf{\Theta }$:

(i)

Along the edges $x=0$ and $y=0$, the function $W(c,x,y)$ yields:

$$W(c,0,0)=-8c^4+32c^2=I_6\left(c\right).$$

Upon differentiating $I_6\left(c\right)$ with respect to $c,$ we have:

$$I_6^{\prime }\left(c\right)=-32c^3+64c.$$

Setting $I_6^{\prime }\left(c\right)=0$ gives $c_0\approx 1.4142135$ at which $W(c,0,0)=I_6\left(c\right)$ achieved its maximum which is $max{I}_6\left(c\right)=I_6\left(c_0\right)=32$. Thus,

$$\left|{\mathcal{H}}_{2,3}\left(g\right)\right|\le \frac{2}{135}.$$

(ii)

simplifies to $W(c,0,1)$ as shown below:

$$W(c,0,1)=-4c^5+15c^4+16c^3-120c^2+240=I_7\left(c\right).$$

Hence, the derivative is:

$$I_7^{\prime }\left(c\right)=-20c^4+60c^3+48c^2-240c.$$

Given that $I_7^{\prime }\left(c\right)<0$ in the interval $\left[0,2\right]$, $I_7\left(c\right)$ decreases within this interval. Thus, the maximum is attained at $c=0$, resulting in $max{I}_7\left(c\right)=I_7\left(0\right)=240$. Consequently,

$$\left|{\mathcal{H}}_{2,3}\left(g\right)\right|\le \frac{1}{9}.$$

(iii)

On the edge $c=0$ and $x=0,$ then $W(c,x,y)$ becomes:

$$W(0,0,y)=240y^2=I_8\left(y\right).$$

Since $I_8^{\prime }\left(y\right)>0$ in $\left[0,1\right].$ Thus, $I_8\left(y\right)$ is increasing in $\left[0,1\right]$ and hence maxima is achieved at $y=1.$ Therefore, $max{I}_8\left(y\right)=I_8\left(1\right)=240$. Thus, we have

$$\left|{\mathcal{H}}_{2,3}\left(g\right)\right|\le \frac{1}{9}.$$

(iv)

On the edges $W(c,1,0)$ and $W(c,1,1)$

Since $W(c,1,y)$ is independent of $y,$ we have:

$$W(c,1,0)=W(c,1,1)=\frac{1}{2}{c}^6-28c^4+72c^2+128=I_9\left(c\right).$$

The derivative is:

$$I_9^{\prime }\left(c\right)=3c^5-112c^3+144c.$$

Setting $I_9^{\prime }\left(c\right)=0$ gives $c_0\approx 1.15470053837$ at which $W(c,1,0)=W(c,1,1)=I_9\left(c\right)$ achieved its maximum. Therefore, $max{I}_9\left(c\right)=I_9\left(c_0\right)=175.40740$. Hence,

$$\left|{\mathcal{H}}_{2,3}\left(g\right)\right|\le 0.081207133.$$

(v)

At the boundary where $c=0$ and $x=1$, the function $W(c,x,y)$ assumes the expression:

$$W(0,1,y)=128.$$

This leads to the conclusion that:

$$\left|{\mathcal{H}}_{2,3}\left(g\right)\right|\le \frac{8}{135}.$$

(vi)

At the boundary where $c=2$, the function $W(c,x,y)$ evaluates to:

$$W(2,x,y)=0.$$

Considering that $W(2,x,y)$ remains unaffected by variations in x and y, it follows that:

$$W(2,0,y)=W(2,1,y)=W(2,x,0)=W(2,x,1)=0.$$

(vii)

On the edge $c=0$ and $y=1,$ then $W(c,x,y)$ reduces to

$$W(0,x,1)=-16x^4+128x^3-224x^2+240=I_{10}\left(x\right).$$

Then,

$$I_{10}^{\prime }\left(x\right)=-64x^3+384x^2-448x.$$

Since $I_{10}^{\prime }\left(x\right)<0$ in $\left[0,1\right].$ Thus, $I_{10}\left(x\right)$ is decreasing in $\left[0,1\right]$ and hence maxima is achieved at $x=0.$ Therefore, $max{I}_{10}\left(x\right)=I_{10}\left(0\right)=240$. Thus, we have

$$\left|{\mathcal{H}}_{2,3}\left(g\right)\right|\le \frac{1}{9}.$$

(viii)

When $c=0$ and $y=0$, the function $W(c,x,y)$ transforms into:

$$W(0,x,0)=-128x^3+256x=I_{11}\left(x\right).$$

The derivative is:

$$I_{11}^{\prime }\left(x\right)=-384x^2+256.$$

putting $I_{11}^{\prime }\left(x\right)=0$ we get the critical point $x_0\approx 0.81649658$ at which $I_{11}\left(x\right)$ achieved its maximum. Therefore, $max{I}_{11}\left(x\right)=I_{11}\left(x_0\right)=139.3487498$. Hence

$$\left|{\mathcal{H}}_{2,3}\left(g\right)\right|\le 0.06451331.$$

From the preceding cases, it is evident that:

$$W\left(c,x,y\right)\le 240\mathrm{on}\left[0,2\right]\times \left[0,1\right]\times \left[0,1\right].$$

Applying (20), we can express:

$$\left|{\mathcal{H}}_{2,3}\left(g\right)\right|\le \frac{1}{2160}\left(W\left(c,x,y\right)\right)\le \frac{1}{9}.$$

In the scenario where $g\in {\mathcal{BT}}_{car}$, the tight bound for this Hankel determinant is achieved by:

$$\left|{\mathcal{H}}_{2,3}\left(g\right)\right|=\frac{1}{9}$$

with the corresponding extremal function

$${\int }_0^z\left(1+\frac{4}{3}\left(t^3\right)+\frac{2}{3}\left(t^6\right)\right)dt=z+\frac{1}{3}{z}^4+\frac{2}{21}{z}^7+\cdots .$$

□

Next, we will enhance the precision of the third-order Hankel determinant ${\mathcal{H}}_{3,1}\left(g\right)$ for functions belonging to the class ${\mathcal{BT}}_{car}.$

Theorem 5. 

If g is a member of the ${\mathcal{BT}}_{car}$ class, then the magnitude of the third Hankel determinant is bounded by

$$\left|{\mathcal{H}}_{3,1}\left(g\right)\right|\le \frac{1}{9}.$$

This bound is achieved with optimal precision.

Proof. 

The third Hankel determinant can be written as

$${\mathcal{H}}_{3,1}\left(g\right)=2d_2{d}_3{d}_4-d_3^3-d_4^2+d_3{d}_5-d_2^2{d}_5.$$

Putting (15)–(18)$,$ with $c_1=c$, we get

$$\begin{array}{ccc}\hfill {\mathcal{H}}_{3,1}\left(g\right)& =& \frac{1}{58320}\left(-71c^6+168c^4{c}_2+288c^3{c}_3-321c^2{c}_2^2-1296c^2{c}_4\right.\hfill \\ & & \left.+2196c{c}_2{c}_3-1072c_2^3+1728c_2{c}_4-1620c_3^2\right).\hfill \end{array}$$

(29)

Let $n=4-c^2$ in Lemmas (7)–(9). Now using these formulae, we obtain

$$\begin{array}{ccc}\hfill 168c^4{c}_2& =& 84\left(c^6+c^{4}nx\right),\hfill \\ \hfill 288c^3{c}_3& =& 72c^6+144c^{4}nx-72c^{4}n{x}^2+144c^{3}n\left(1-{\left|x\right|}^2\right)\zeta ,\hfill \\ \hfill 321c^2{c}_2^2& =& \frac{321}{4}{c}^6+\frac{321}{2}{c}^{4}nx+\frac{321}{4}{c}^2{n}^2{x}^2,\hfill \\ \hfill 1296c^2{c}_4& =& 162c^{4}n{x}^3-648c^{3}nx\left(1-{\left|x\right|}^2\right)\zeta -648c^{2}n\overline{x}\left(1-{\left|x\right|}^2\right){\zeta }^2\hfill \\ & & -486c^{4}n{x}^2+648c^{2}n\left(1-{\left|x\right|}^2\right)\left(1-{\left|\zeta \right|}^2\right)\rho +648c^{3}n\left(1-{\left|x\right|}^2\right)\zeta \hfill \\ & & +486c^{4}nx+162c^6+648c^{2}n{x}^2,\hfill \\ \hfill 2196c{c}_2{c}_3& =& -\frac{549}{2}{c}^2{n}^2{x}^3-\frac{549}{2}{c}^{4}n{x}^2+549c{n}^{2}x\left(1-{\left|x\right|}^2\right)\zeta +549c^2{n}^2{x}^2\hfill \\ & & +549c^{3}n\left(1-{\left|x\right|}^2\right)\zeta +\frac{1647}{2}{c}^{4}nx+\frac{549}{2}{c}^6,\hfill \\ \hfill 1072c_2^3& =& 134n^3{x}^3+402c^2{n}^2{x}^2+402c^{4}nx+134c^6,\hfill \\ \hfill 1728c_2{c}_4& =& 432c^{2}n{x}^2+432n^2{x}^3+108c^6+432c^{4}nx+432c^{3}n\left(1-{\left|x\right|}^2\right)\zeta \hfill \\ & & +432c^{2}n\left(1-{\left|x\right|}^2\right)\left(1-{\left|\zeta \right|}^2\right)\rho +324c^2{n}^2{x}^2+432c{n}^{2}x\left(1-{\left|x\right|}^2\right)\zeta \hfill \\ & & +432n^{2}x\left(1-{\left|x\right|}^2\right)\left(1-{\left|\zeta \right|}^2\right)\rho -324c^{4}n{x}^2-432c^{2}n\overline{x}\left(1-{\left|x\right|}^2\right){\zeta }^2\hfill \\ & & -432c^{3}nx\left(1-{\left|x\right|}^2\right)\zeta -324c^2{n}^2{x}^3-432n^{2}x\overline{x}\left(1-{\left|x\right|}^2\right){\zeta }^2\hfill \\ & & +108c^{4}n{x}^3+108c^2{n}^2{x}^4-432c{n}^2{x}^2\left(1-{\left|x\right|}^2\right)\zeta ,\hfill \\ \hfill 1620c_3^2& =& \frac{405}{4}{c}^2{n}^2{x}^4-405c{n}^2{x}^2\left(1-{\left|x\right|}^2\right)\zeta -405c^2{n}^2{x}^3-\frac{405}{2}{c}^{4}n{x}^2\hfill \\ & & +405n^2{\left(1-{\left|x\right|}^2\right)}^2{\zeta }^2+810c{n}^{2}x\left(1-{\left|x\right|}^2\right)\zeta +405c^2{n}^2{x}^2\hfill \\ & & +405c^{3}n\left(1-{\left|x\right|}^2\right)\zeta +405c^{4}nx+\frac{405}{4}{c}^6.\hfill \end{array}$$

Substituting the above expressions in (29)$,$ we have

$$\begin{array}{ccc}\hfill {\mathcal{H}}_{3,1}\left(g\right)& =& \frac{1}{58320}\left\{-10c^6+432n^2{x}^3-134n^3{x}^3-216c^{2}n{x}^2-54c^{4}n{x}^3+18c^{4}n{x}^2\right.\hfill \\ & & +30c^{4}nx+\frac{27}{4}{c}^2{n}^2{x}^4-\frac{387}{2}{c}^2{n}^2{x}^3-\frac{57}{4}{c}^2{n}^2{x}^2-405n^2{\left(1-{\left|x\right|}^2\right)}^2{\zeta }^2\hfill \\ & & +72c^{3}n\left(1-{\left|x\right|}^2\right)\zeta +216c^{3}nx\left(1-{\left|x\right|}^2\right)\zeta +216c^{2}n\overline{x}\left(1-{\left|x\right|}^2\right){\zeta }^2\hfill \\ & & -216c^{2}n\left(1-{\left|x\right|}^2\right)\left(1-{\left|\zeta \right|}^2\right)\rho -27c{n}^2{x}^2\left(1-{\left|x\right|}^2\right)\zeta -432n^{2}x\overline{x}\left(1-{\left|x\right|}^2\right){\zeta }^2\hfill \\ & & \left.+171c{n}^{2}x\left(1-{\left|x\right|}^2\right)\zeta +432n^{2}x\left(1-{\left|x\right|}^2\right)\left(1-{\left|\zeta \right|}^2\right)\rho \right\}.\hfill \end{array}$$

Since $n=4-c^2,$

$${\mathcal{H}}_{3,1}\left(g\right)=\frac{1}{58320}\left(v_1\left(c,x\right)+v_2\left(c,x\right)\zeta +v_3\left(c,x\right){\zeta }^2+\psi \left(c,x,\zeta \right)\rho \right),$$

where $\rho ,x,\zeta \in {\mathbb{O}}_d$, and

$$\begin{array}{ccc}\hfill v_1\left(c,x\right)& =& -10c^6+\left(4-c^2\right)[\left(4-c^2\right)\left(-104x^3-\frac{119}{2}{c}^2{x}^3+\frac{27}{4}{c}^2{x}^4-\frac{57}{4}{c}^2{x}^2\right)\hfill \\ & & -216c^2{x}^2-54c^4{x}^3+18c^4{x}^2+30c^{4}x],\hfill \\ \hfill v_2\left(c,x\right)& =& \left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\left[\left(4-c^2\right)\left(171cx-27c{x}^2\right)+216c^{3}x+72c^3\right],\hfill \\ \hfill v_3\left(c,x\right)& =& \left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\left[\left(4-c^2\right)\left(-27x^2-405\right)+216c^2\overline{x}\right],\hfill \\ \hfill \psi \left(c,x,\zeta \right)& =& \left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\left(1-{\left|\zeta \right|}^2\right)\left[-216c^2+432x\left(4-c^2\right)\right].\hfill \end{array}$$

By using $\left|x\right|=x,\left|\zeta \right|=y$ and utilizing the fact $\left|\rho \right|\le 1,$ we obtain

$$\begin{array}{ccc}\hfill \left|{\mathcal{H}}_{3,1}\left(g\right)\right|& \le & \frac{1}{58320}\left(\left|v_1\left(c,x\right)\right|+\left|v_2\left(c,x\right)\right|y+\left|v_3\left(c,x\right)\right|y^2+\left|\psi \left(c,x,\zeta \right)\right|\right).\hfill \\ & \le & \frac{1}{58320}\left(G\left(c,x,y\right)\right),\hfill \end{array}$$

(30)

where

$$G\left(c,x,y\right)=g_1\left(c,x\right)+g_2\left(c,x\right)y+g_3\left(c,x\right)y^2+g_4\left(c,x\right)\left(1-y^2\right),$$

with

$$\begin{array}{ccc}\hfill g_1\left(c,x\right)& =& 10c^6+\left(4-c^2\right)[\left(4-c^2\right)\left(104x^3+\frac{119}{2}{c}^2{x}^3+\frac{27}{4}{c}^2{x}^4+\frac{57}{4}{c}^2{x}^2\right)\hfill \\ & & +216c^2{x}^2+54c^4{x}^3+18c^4{x}^2+30c^{4}x],\hfill \\ \hfill g_2\left(c,x\right)& =& \left(4-c^2\right)\left(1-x^2\right)\left[\left(4-c^2\right)\left(171cx+27c{x}^2\right)+216c^{3}x+72c^3\right],\hfill \\ \hfill g_3\left(c,x\right)& =& \left(4-c^2\right)\left(1-x^2\right)\left[\left(4-c^2\right)\left(27x^2+405\right)+216c^{2}x\right],\hfill \\ \hfill g_4\left(c,x\right)& =& \left(4-c^2\right)\left(1-x^2\right)\left[216c^2+432x\left(4-c^2\right)\right].\hfill \end{array}$$

Now, we have to maximize $G\left(c,x,y\right)$ in the closed cuboid $\mathsf{\Theta }:\left[0,2\right]\times \left[0,1\right]\times \left[0,1\right].$

To address this, we need to examine the maximum values of $G\left(c,x,y\right)$ within the confines of $\mathsf{\Theta }$, encompassing its interior, six faces, and twelve edges.

1.

Interior points of cuboid $\mathsf{\Theta }$:

Let $\left(c,x,y\right)\in \left(0,2\right)\times \left(0,1\right)\times \left(0,1\right),$ and differentiate partially $G\left(c,x,y\right)$ with respect to y, we obtain

$$\begin{array}{ccc}\hfill \frac{\partial G}{\partial y}& =& \left(4-c^2\right)(1-x^2)\left[54y(x-1)\left(\left(4-c^2\right)\left(x-15\right)+8c^2\right)\right.\hfill \\ & & \left.+c\left(x\left(4-c^2\right)\left(171+27x\right)+72c^2\left(3x+1\right)\right)\right].\hfill \end{array}$$

Putting $\frac{\partial G}{\partial y}=0,$ we get

$$y=\frac{c\left(x\left(4-c^2\right)\left(171+27x\right)+72c^2\left(3x+1\right)\right)}{54(x-1)\left(\left(4-c^2\right)\left(15-x\right)-8c^2\right)}=y_0.$$

If $y_0$ is a critical point inside $\mathsf{\Theta },$ then $y_0\in \left(0,1\right),$ which is possible only if

$$72c^3\left(3x+1\right)+cx\left(4-c^2\right)\left(171+27x\right)+54(1-x)\left(4-c^2\right)\left(15-x\right)<432c^2(1-x).$$

(31)

and

$$c^2>\frac{4(15-x)}{23-x}.$$

(32)

We must now determine the solutions that simultaneously satisfy the inequalities (31) and (32) to establish the existence of critical points.

Let $g\left(x\right)=\frac{4(15 - x)}{23 - x}.$ As $g^{\prime }\left(x\right)<0$ in $\left(0,1\right).$ so, $g\left(x\right)$ is decreasing over $\left(0,1\right).$ Hence, $c^2>\frac{28}{11}$ and a simple exercise shows that (31) does not hold in this case for all values of $x\in \left(0,1\right)$ and there is no critical point of $\mathsf{\Theta }$ in $\left(0,2\right)\times \left(0,1\right)\times \left(0,1\right).$

2.

Interior of all the six faces of cuboid $\mathsf{\Theta }$:

(i)

On the face $c=0,$ $G(c,x,y)$ becomes

$$J_1(x,y)=G(0,x,y)=16\left[27(1-x^2)(y^2(x-15)(x-1)+16x)+104x^3\right],x,y\in \left(0,1\right).$$

Differentiate $J_1(x,y)$ partially with respect to $y,$ we have

$$\frac{\partial J_1}{\partial y}=864y(1-x^2)(x-15)(x-1),x,y\in \left(0,1\right).$$

Thus, $J_1(x,y)$ has no critical point in the interval $\left(0,1\right)\times \left(0,1\right).$

(ii)

On the face $c=2,$ $G(c,x,y)$ takes the form

$$G(2,x,y)=640.$$

Thus

$$\left|{\mathcal{H}}_{3,1}\left(g\right)\right|\le \frac{8}{729}.$$

(iii)

On the face $x=0,$ $G(c,x,y)$ yields

$$J_2(c,y)=G(c,0,y)=10c^6+(4-c^2)(72c^{3}y+(1620-405c^2)y^2+216c^2(1-y^2)).$$

Differentiate $J_2(c,y)$ partially with respect to $y,$ we have

$$\frac{\partial J_2}{\partial y}=(4-c^2)(72c^3+2(1620-405c^2)y-432c^{2}y).$$

Putting $\frac{\partial J_2}{\partial y}=0$ and we obtain

$$y=\frac{4c^3}{3(23c^2-60)}=y_1.$$

For the given range of $y,$ $y_1$ should belong to $\left(0,1\right),$ which is possible only if $c>c_0, c_0\approx 1.70121012.$ In addition, the derivative of $J_2(c,y)$ partially with respect to c is

$$\begin{array}{ccc}\hfill \frac{\partial J_2}{\partial c}& =& 60c^5-144c^{4}y+(4-c^2)(216c^{2}y-810c{y}^2+432c(1-y^2))\hfill \\ & & -432c^3(1-y^2)-2c\left(-405c^2+1620\right)y^2.\hfill \end{array}$$

(33)

Plugging the value of y in (33)$,$ and setting $\frac{\partial J_2}{\partial c}=0$ we obtain

$$\frac{\partial J_2}{\partial c}=12c(2093c^8-48496c^6+287136c^4-656640c^2+518400)=0.$$

(34)

A calculation gives the solution of (34) in the interval $\left(0,1\right)$ that is $c\approx 1.402457566$. Thus $J_2(c,y)$ has no optimal point in the interval $\left(0,2\right)\times \left(0,1\right).$

(iv)

On the face $x=1,$ $G(c,x,y)$ becomes

$$J_3(c,y)=G(c,1,y)=10c^6+(4-c^2)((4-c^2)(104+\frac{161}{2}{c}^2)+216c^2+102c^4).$$

Then

$$\frac{\partial J_3}{\partial c}=-69c^5-1392c^3+2640c.$$

Setting $\frac{\partial J_3}{\partial c}=0,$ we get $c\approx 1.32118226$ at which $J_3(c,y)$ achieved its maximum which is $max{J}_3(c,y)=2846.62537$. Hence,

$$\left|{\mathcal{H}}_{3,1}\left(g\right)\right|\le 0.04881044.$$

(v)

On the face $y=0,$ $G(c,x,y)$ reduces to

$$\begin{array}{ccc}\hfill J_4(c,x)& =& G(c,x,0)=10c^6-588c^4{x}^3+3576c^2{x}^3-5248x^3+\frac{11}{2}{c}^6{x}^3\hfill \\ & & +\frac{27}{4}{c}^6{x}^4-54c^4{x}^4+108c^2{x}^4-\frac{15}{4}{c}^6{x}^2-42c^4{x}^2\hfill \\ & & +288c^2{x}^2-30c^{6}x+522c^{4}x-216c^4+864c^2\hfill \\ & & -3456c^{2}x+6912x.\hfill \end{array}$$

Now, differentiating partially with respect to $c,$ then with respect to x, we have

$$\begin{array}{ccc}\hfill \frac{\partial J_4}{\partial c}& =& 60c^5-2352c^3{x}^3+7152c{x}^3+33c^5{x}^3+\frac{81}{2}{c}^5{x}^4-216c^3{x}^4\hfill \\ & & +216c{x}^4-\frac{45}{2}{c}^5{x}^2-168c^3{x}^2+456c{x}^2-180c^{5}x+2208c^{3}x\hfill \\ & & -864c^3+1728c-6912cx.\hfill \end{array}$$

(35)

and

$$\begin{array}{ccc}\hfill \frac{\partial J_4}{\partial x}& =& -1764c^4{x}^2+10728c^2{x}^2-15744x^2+\frac{33}{2}{c}^6{x}^2+27c^6{x}^3\hfill \\ & & -216c^4{x}^3+432c^2{x}^3-\frac{15}{2}{c}^{6}x-84c^{4}x+456c^{2}x-30c^6\hfill \\ & & +552c^4-3456c^2+6912.\hfill \end{array}$$

(36)

A numerical computation shows that the solution does not exist for the system of Equations (35) and (36) in $\left(0,2\right)\times \left(0,1\right).$

(vi)

On the face $y=1,$ $G(c,x,y)$ after some simplification becomes

$$\begin{array}{ccc}\hfill J_5(c,x)& =& G(c,x,1)=6480-72c^5-\frac{15}{4}{c}^6{x}^2-30c^{6}x-27c^5{x}^4+45c^5{x}^3+99c^5{x}^2\hfill \\ & & +216c^3{x}^4-45c^{5}x+504c^3{x}^3-504c^3{x}^2-432c{x}^4-2736c{x}^3+\frac{11}{2}{c}^6{x}^3\hfill \\ & & +\frac{27}{4}{c}^6{x}^4-81c^4{x}^4-432x^4+405c^4+10c^6-3240c^2+1664x^3-6048x^2\hfill \\ & & +288c^3-744c^2{x}^3+324c^2{x}^4+4116c^2{x}^2+60c^4{x}^3-636c^4{x}^2-96c^{4}x\hfill \\ & & +2736cx+432c{x}^2-504c^{3}x+864c^{2}x.\hfill \end{array}$$

Partial derivative of $J_5(c,x)$ with respect to c and then with respect to x, we have

$$\begin{array}{ccc}\hfill \frac{\partial J_5}{\partial c}& =& 60c^5+\frac{81}{2}{c}^5{x}^4+33c^5{x}^3-\frac{45}{2}{c}^5{x}^2-324c^3{x}^4-180c^{5}x+240c^3{x}^3\hfill \\ & & -2544c^3{x}^2+648c{x}^4-1488c{x}^3-135c^4{x}^4-432x^4-360c^4\hfill \\ & & +864c^2-2736x^3+432x^2+1620c^3+1512c^2{x}^3+648c^2{x}^4\hfill \\ & & -1512c^2{x}^2+225c^4{x}^3+495c^4{x}^2-225c^{4}x+1728cx\hfill \\ & & +8232c{x}^2-384c^{3}x-1512c^{2}x-6480c+2736x.\hfill \end{array}$$

(37)

and

$$\begin{array}{ccc}\hfill \frac{\partial J_5}{\partial x}& =& -45c^5+\frac{33}{2}{c}^6{x}^2-\frac{15}{2}{c}^{6}x-108c^5{x}^3+135c^5{x}^2+198c^{5}x+864c^3{x}^3\hfill \\ & & +1512c^3{x}^2-1728c{x}^3+27c^6{x}^3-96c^4-30c^6+864c^2\hfill \\ & & -1728x^3+4992x^2-504c^3+1296c^2{x}^3-2232c^2{x}^2\hfill \\ & & -324c^4{x}^3+180c^4{x}^2-1272c^{4}x+864cx-8208c{x}^2\hfill \\ & & -1008c^{3}x+8232c^{2}x+2736c-12096x.\hfill \end{array}$$

(38)

As in the above case, we deduce the same result for the face $y=0,$ that is, that the system of Equations (37) and (38) has no solution in $\left(0,2\right)\times \left(0,1\right).$

3.

On the edges of cuboid $\mathsf{\Theta }$:

(i)

On the edge $x=0$ and $y=0,$ then $G(c,x,y)$ yields

$$G(c,0,0)=10c^6-216c^4+864c^2=J_6\left(c\right).$$

Differentiating $J_6\left(c\right)$ with respect to $c,$ we have

$$J_6^{\prime }\left(c\right)=60c^5-864c^3+1728c.$$

Setting $J_6^{\prime }\left(c\right)=0$ gives $c_0\approx 1.54919333$ at which $G(c,0,0)=J_6\left(c\right)$ gets its maximum which is $max{J}_6\left(c\right)=J_6\left(c_0\right)=967.6800$. Thus

$$\left|{\mathcal{H}}_{3,1}\left(g\right)\right|\le 0.01659259.$$

(ii)

On the edge $x=0$ and $y=1,$ then $G(c,x,y)$ reduce to $G(c,0,1)$ given below

$$G(c,0,1)=10c^6-72c^5+405c^4+288c^3-3240c^2+6480=J_7\left(c\right).$$

Then

$$J_7^{\prime }\left(c\right)=60c^5-360c^4+1620c^3+864c^2-6480c.$$

Since $J_7^{\prime }\left(c\right)<0$ in $\left[0,2\right].$ Thus, $J_7\left(c\right)$ is decreasing in $\left[0,2\right]$ and hence maxima is achieved at $c=0.$ Therefore $max{J}_7\left(c\right)=J_7\left(0\right)=6480$. Thus, we have

$$\left|{\mathcal{H}}_{3,1}\left(g\right)\right|\le \frac{1}{9}.$$

(iii)

On the edge $c=0$ and $x=0,$ then $G(c,x,y)$ becomes

$$G(0,0,y)=6480y^2=J_8\left(y\right).$$

Since $J_8^{\prime }\left(y\right)>0$ in $\left[0,1\right].$ Thus, $J_8\left(y\right)$ is increasing in $\left[0,1\right]$ and hence maxima is achieved at $y=1.$ Therefore, $max{J}_8\left(y\right)=J_8\left(1\right)=6480$. Thus, we have

$$\left|{\mathcal{H}}_{3,1}\left(g\right)\right|\le \frac{1}{9}.$$

(iv)

On the edges $G(c,1,0)$ and $G(c,1,1)$

Since $G(c,1,y)$ is free of $y,$ therefore

$$G(c,1,0)=G(c,1,1)=-\frac{23}{2}{c}^6-348c^4+1320c^2+1664=J_9\left(c\right).$$

Then

$$J_9^{\prime }\left(c\right)=-69c^5-1392c^3+2604c.$$

Setting $J_9^{\prime }\left(c\right)=0$ gives $c_0\approx 1.3211822619$ at which $G(c,1,0)=G(c,1,1)=J_9\left(c\right)$ achieved its maximum. Therefore $max{J}_9\left(c\right)=J_9\left(c_0\right)=2846.62537$. Hence

$$\left|{\mathcal{H}}_{3,1}\left(g\right)\right|\le 0.048810448.$$

(v)

On the edge $c=0$ and $x=1,$ then $G(c,x,y)$ takes the form

$$G(0,1,y)=1664.$$

Hence we have

$$\left|{\mathcal{H}}_{3,1}\left(g\right)\right|\le \frac{104}{3645}.$$

(vi)

On the edge $c=2,$ then $G(c,x,y)$ yields

$$G(2,x,y)=640.$$

$G(2,x,y)$ is independent of x and $y,$ therefore

$$G(2,0,y)=G(2,1,y)=G(2,x,0)=G(2,x,1)=640.$$

Thus

$$\left|{\mathcal{H}}_{3,1}\left(g\right)\right|\le \frac{8}{729}.$$

(vii)

On the edge $c=0$ and $y=1,$ then $G(c,x,y)$ reduces to

$$G(0,x,1)=-432x^4+1664x^3-6048x^2+6480=J_{10}\left(x\right).$$

Then

$$J_{10}^{\prime }\left(x\right)=-1728x^3+4992x^2-12096x.$$

Since $J_{10}^{\prime }\left(x\right)<0$ in $\left[0,1\right].$ Thus $J_{10}\left(x\right)$ is decreasing in $\left[0,1\right]$ and hence, the maximum is achieved at $x=0.$ Therefore $max{J}_{10}\left(x\right)=J_{10}\left(0\right)=6480$. Thus we have

$$\left|{\mathcal{H}}_{3,1}\left(g\right)\right|\le \frac{1}{9}.$$

(viii)

On the edge $c=0$ and $y=0,$ then $G(c,x,y)$ becomes

$$G(0,x,0)=-5248x^3+6912x=J_{11}\left(x\right).$$

Then,

$$J_{11}^{\prime }\left(x\right)=-15744x^2+6912.$$

Upon setting $J_{11}^{\prime }\left(x\right)=0$, we obtain the critical point $x_0\approx 0.66258915$ at which $J_{11}\left(x\right)$ achieved its maximum. Therefore, $max{J}_{11}\left(x\right)=J_{11}\left(x_0\right)=3053.210832$. Hence

$$\left|{\mathcal{H}}_{3,1}\left(g\right)\right|\le 0.05235272.$$

Thus, based on the aforementioned cases, we can conclude that

$$G\left(c,x,y\right)\le 6480\mathrm{on}\left[0,2\right]\times \left[0,1\right]\times \left[0,1\right].$$

From (30), we can write

$$\left|{\mathcal{H}}_{3,1}\left(g\right)\right|\le \frac{1}{58320}\left(G\left(c,x,y\right)\right)\le \frac{1}{9}.$$

For $g\in {\mathcal{BT}}_{car},$ the sharp bound for the Hankel determinant is obtained as follows:

$$\left|{\mathcal{H}}_{3,1}\left(g\right)\right|=\frac{1}{9},$$

and this bound is achieved by the extremal function:

$${\int }_0^z\left(1+\frac{4}{3}\left(t^3\right)+\frac{2}{3}\left(t^6\right)\right)dt=z+\frac{1}{3}{z}^4+\frac{2}{21}{z}^7+\cdots .$$

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4. Conclusions

The main idea behind investigating coefficient problems in various families of analytic functions is to express the coefficients of these functions using the well-known class of Carathéodory functions in the open unit disk. Recent applications of this approach have yielded several interesting results. However, many of the obtained bounds were non-sharp for analytic univalent functions associated with different image domains. In this study, we determined sharp estimates of coefficient-related problems for functions belonging to the family ${\mathcal{BT}}_{car}$ of bounded turning functions related to the cardioid domain. Most of the derived bounds are proven to be sharp. This study could act as an inspiration for further investigations into the sharp bounds of analytic functions associated with novel image domains. In addition, using the same methodology, one can investigate the bounds of higher-order Hankel determinants as studied in the articles [59,60,61,62,63,64,65,66].