Sharp Coefficient and Hankel Problems Related to a Symmetric Domain

Abstract

In the current article, we utilize the concept of subordination to establish a new subclass of analytic functions associated with a bounded domain that is symmetric about the real axis. By applying the convolution technique, we derive the necessary and sufficient condition, the radius of convexity for this recently introduced class. Furthermore, we prove the sharp upper bounds for the second-order Hankel determinants $|{\mathcal{H}}_{2,1}\xi |,|{\mathcal{H}}_{2,2}\xi |$ and third-order Hankel determinant $|{\mathcal{H}}_{3,1}\xi |$ for the functions $\xi$ belonging to the newly defined class.

1. Introduction

Let $\mathcal{A}$denote the class of all of analytic functions $\xi \left(\mu \right)$ defined in the open unit disk $\mathbf{U}=\{\mu :\mu \in \mathbb{C}$ and $\left|\mu \right|<1\},$whose normalized Maclaurin series representation is given as:

$$\xi \left(\mu \right)=\mu +\sum _{t=2}^{\infty }{a}_t{\mu }^t,\mu \in \mathbf{U},$$

(1)

with normalization conditions $\xi \left(0\right)=0$ and ${\xi }^{\prime }\left(0\right)=1$. A function is said to be univalent if it never takes the same value twice. By $\mathcal{S}$, we mean the subclass of $\mathcal{A}$ containing all the functions which are univalent in $\mathbf{U}$. Subordination is an important tool used to investigate various inclusion and radii problems for certain classes of analytic functions.

A function $\xi$ is subordinate to $\mathbf{g}$ in $\mathbf{U}$ written as $\xi \prec \mathbf{g}$ if there exists a Schwarz function $\chi$, which is analytic in $\mathbf{U}$ and $\chi \left(0\right)=0$ with $\left|\chi \left(\mu \right)\right|<1$, such that $\xi \left(\mu \right)=\mathbf{g}\left(\chi \left(\mu \right)\right)$. In addition, if the function $\mathbf{g}$ is univalent in $\mathbf{U}$, then we have

$$\xi \left(0\right)=\mathbf{g}\left(0\right)\mathrm{and}\xi \left(\mathbf{U}\right)\subset \mathbf{g}\left(\mathbf{U}\right).$$

By ${\mathcal{S}}^{\ast },\mathcal{C},\mathcal{K}$ and $\mathcal{R}$, we mean the well known subclasses of $\mathcal{S}$ consisting of starlike, convex, close-to-convex and functions with bounded turnings, respectively.

The convolution of two regular functions $\xi$ and $\varsigma$ in $\mathbf{U}$, with the series representation of $\xi$ given in (1) and $\varsigma =\mu +{\displaystyle \sum _{t=2}^{\infty }}{b}_t{\mu }^t$, is defined as

$$\left(\xi \ast \varsigma \right)\left(\mu \right)=\mu +\sum _{t=2}^{\infty }{a}_t{b}_t{\mu }^t,\mu \in \mathbf{U}.$$

(2)

In 1985, Padmanabhan et al. [1] defined the integrated families of starlike and convex functions by using the theory of convolution along with the function $\frac{\mu }{{\left(1-\mu \right)}^a}$, where $a\in \mathbb{R}$. Shanmugam [2] generalized the idea of [1] and introduced the general form of functions class ${\mathcal{S}}_h^{\ast }\left(\varphi \right)$ by considering an analytic function $\varphi \left(\mu \right)$ with $\varphi \left(0\right)=1,$ and $h\left(\mu \right)\in \mathcal{A},$ as:

$${\mathcal{S}}_h^{\ast }\left(\varphi \right)=\left\{\xi \in \mathcal{A}:\frac{\mu {(\xi \ast h)}^{\prime }}{(\xi \ast h)}\prec \varphi \left(\mu \right),\mu \in \mathbf{U}\right\}.$$

(3)

By taking $h\left(\mu \right)=\frac{\mu }{1-\mu }$ or $\frac{\mu }{{\left(1-\mu \right)}^2}$, we obtain the well known classes ${\mathcal{S}}^{\ast }\left(\varphi \right)$ and $\mathcal{C}\left(\varphi \right)$ of the Ma and Minda types of starlike and convex function defined in [3]. Further, by choosing $\varphi \left(\mu \right)=\frac{1+\mu }{1-\mu }$, the classes ${\mathcal{S}}^{\ast }\left(\varphi \right)$ and $\mathcal{C}\left(\varphi \right)$ reduce to ${\mathcal{S}}^{\ast }$ and $\mathcal{C}$ classes.

In the recent past, a large number of researchers have defined and studied several interesting subclasses of analytic and univalent functions by restricting $\varphi \left(\mu \right)$ in the general form of ${\mathcal{S}}^{\ast }\left(\varphi \right)$ and $\mathcal{C}\left(\varphi \right)$. Here we highlight some of them:

(i)

Let $\varphi \left(\mu \right)=\frac{1+F\mu }{1+G\mu },$ $-1\le G<F\le 1$. Geometrically, $\varphi \left(\mu \right)$ maps the open unit disc onto a domain whose center lies on the real axis and is also symmetric about the real axis. For function value $\varphi \left(\mu \right)$, we get ${\mathcal{S}}^{\ast }[F,G]={\mathcal{S}}^{\ast }\left(\frac{1+F\mu }{1+G\mu }\right)$, the class of Janowski starlike functions; see [4].

(ii)

For $\varphi \left(\mu \right)=cos\mu ,$ the class ${\mathcal{S}}_{cos\mu }^{\ast }$ was studied by Bano et al. [5], while for $\varphi \left(\mu \right)=cosh\mu ,$ the function class ${\mathcal{S}}_{cosh\mu }^{\ast }$ was introduced and studied by Alotaibi et al. [6].

(iii)

For $\varphi \left(\mu \right)=e^{\mu },$ the class ${\mathcal{S}}_e^{\ast }$ was defined and studied by Mendiratta [7].

(iv)

For $\varphi \left(\mu \right)=1+sin\mu ,$ the class ${\mathcal{S}}^{\ast }\left(\varphi \right)$ reduces to ${\mathcal{S}}_{sin}^{\ast }$, which was introduced and investigated by Cho et al. [8].

(v)

For $\varphi \left(\mu \right)=1+\mu -\frac{1}{3}{\mu }^3$, we obtain the class ${\mathcal{S}}_{nep}^{\ast }$ investigated by Wani and Swaminathan [9].

(vi)

For $\varphi \left(\mu \right)=1+{sinh}^{-1}\left(\mu \right)$, the class ${\mathcal{S}}^{\ast }\left(\varphi \right)$ was introduced and investigated by Kumar et al. [10].

(vii)

For $\varphi \left(\mu \right)=\frac{2}{1+e^{-\mu }},$ the class ${\mathcal{S}}^{\ast }\left(\varphi \right)$ reduces to ${\mathcal{S}}_{sig}^{\ast }$; see [11].

(viii)

For $\varphi \left(\mu \right)=\sqrt{1+\mu }$, we get ${\mathcal{S}}^{\ast }\sqrt{1+\mu }={\mathcal{S}}_L^{\ast }$ investigated by Sokol et al. [12].

(ix)

For $\varphi \left(\mu \right)=1+tanh\mu$, the class becomes ${\mathcal{S}}_{tanh\mu }^{\ast },$ which was introduced by Khalil et al. [13]; see also [14].

By taking inspiration and motivation from above cited work, we introduce the following subclass of analytic functions.

Definition 1. 

Let $\xi \in \mathcal{A}$be given in (1). Then, $\xi \in {\mathcal{CV}}_{\mathcal{T}}$ if the following condition holds true:

$$\frac{{\left(\mu {\xi }^{\prime }\left(\mu \right)\right)}^{\prime }}{{\xi }^{\prime }\left(\mu \right)}\prec 1+tanh\mu ,\mu \in \mathbf{U}.$$

(4)

Geometrically, the class ${\mathcal{CV}}_{\mathcal{T}}$ contains all the functions $\xi$ that lie within the image domain of $1+tanh\mu$ for a specified radius. It can be easily seen that the image of unit disc under $1+tanh\mu$ has an appealing geometry which is symmetric about the real axis.

Let $\xi \in \mathcal{A}$, and for given values of $n,$ $q\ge 1$ the $q^{th}$ Hankel determinant ${\mathcal{H}}_{q,n}\left(\xi \right)$ was defined in [15] as follows:

$${\mathcal{H}}_{q,n}\left(\xi \right)=\left|\begin{array}{cccccc}{a}_n& a_{n+1}& .& .& .& a_{n+q-1}\\ a_{n+1}& .& .& .& .& .\\ .& .& .& .& .& .\\ .& .& .& .& .& .\\ .& .& .& .& .& .\\ a_{n+q-1}& .& .& .& .& a_{n+2\left(q-1\right)}\end{array}\right|.$$

(5)

Finding the sharp upper bounds of Hankel determinants $\left|{\mathcal{H}}_{q,n}\right|$ for different subfamilies and certain values of n and q has become the most attractive and important problem in the theory of analytic functions.

In [16], Hayman proved that for $\xi \in \mathcal{S}$ the best sharp inequality is

$$\left|{\mathcal{H}}_{2,n}\left(\xi \right)\right|\le \lambda \sqrt{n},$$

where $\lambda >0$. Similarly, Janteng et al. [17] estimated the sharp bounds for the second Hankel determinant for the classes of $\mathcal{K}$, ${\mathcal{S}}^{\ast }$ and $\mathcal{R}$ as:

$$\left|{\mathcal{H}}_{2,2}\left(\xi \right)\right|\le \left\{\begin{array}{c}\frac{1}{8},\xi \in \mathcal{K},\\ 1,\xi \in {\mathcal{S}}^{\ast },\\ \frac{4}{9},\xi \in \mathcal{R}.\end{array}\right.$$

In $2010,$ Babalola [18] calculated the upper bounds of $\left|{\mathcal{H}}_{3,1}\left(\xi \right)\right|$ for ${\mathcal{S}}^{\ast }$ and $\mathcal{C}$. Later, many authors have estimated $\left|{\mathcal{H}}_{3,1}\left(\xi \right)\right|$ for different subfamilies of analytic functions, see [8,19] and the references cited therein. Zaprawa [20] amended Babalola’s [18] results and claimed that they are still not the best possible, and gives the estimated value as:

$$\left|{\mathcal{H}}_{3,1}\left(\xi \right)\right|\le \left\{\begin{array}{c}\frac{49}{540},\xi \in \mathcal{K},\\ 1,\xi \in {\mathcal{S}}^{\ast },\\ \frac{41}{60},\xi \in \mathcal{R}.\end{array}\right.$$

Furthermore, Kwon et al. [21] improved this bound as $\left|{\mathcal{H}}_{3,1}\left(\xi \right)\right|\le 8/9$ for $\xi \in {\mathcal{S}}^{\ast }$. Furthermore, Zaprawa et al. [22] extend their work by estimating $\left|{\mathcal{H}}_{3,1}\left(\xi \right)\right|\le 5/9$ for $\xi \in {\mathcal{S}}^{\ast }$. Working in a similar direction, the authors of [23,24,25,26] contributed by extending this work for various subfamilies of univalent functions. In the recent past, Arif et al. [27] calculated the sharpness of $\left|{\mathcal{H}}_{3,1}\left(\xi \right)\right|$ and the coefficient bounds for a subfamily of starlike functions related to sigmoid functions, while, Shi et al. [28] estimated the sharp third Hankel determinant for functions with bounded turning along with logarithmic coefficients connected to a petal-shaped domain, whereas Ponnusamy et al. [29] defined these logarithmic coefficients for the families of inverse, starlike and convex univalent functions. The coefficient estimates help us to study the area and arc length problems.

The main objective of this study is to determine the sharp bounds for the second and third Hankel determinants, as well as the sharp upper bounds of the coefficient inequalities for functions belonging to ${\mathcal{CV}}_{\mathcal{T}}$.

2. Set of Lemmas

Let $\mathcal{P}$ denote the class of analytic functions p such that $p\left(0\right)=1$ and

$\Re p\left(\mu \right)>0$, $\mu \in \mathbf{U}$ and having the series of the form:

$$p\left(\mu \right)=1+\underset{t=1}{\sum ^{\infty }}{c}_t{\mu }^t,\mu \in \mathbf{U}.$$

(6)

The following lemmas will be helpful in our subsequent work.

Lemma 1

([15]). Let $p\in \mathcal{P}$ given by (6). Then, the following hold

$$\begin{array}{ccc}\hfill \left|c_t\right|& \le & 2fort\ge 1,\hfill \end{array}$$

(7)

$$\begin{array}{ccc}\hfill \left|c_{t+k}-\lambda c_t{c}_k\right|& <& 2for0\le \lambda \le 1,\hfill \end{array}$$

(8)

$$\begin{array}{ccc}\hfill \left|c_{t+2k}-\lambda c_t{c}_k^2\right|& \le & 2(1+2\lambda ),for\lambda \in \mathbb{R},\hfill \end{array}$$

(9)

and

$$\left|c_2-\frac{c_1^2}{2}\right|\le 2-\frac{\left|c_1^2\right|}{2}.$$

(10)

Lemma 2. 

Let $p\in \mathcal{P},$ given by (6). Then there exist $k,\delta$and ρ with $\left|k\right|\le 1,\left|\delta \right|\le 1,$and $\left|\rho \right|\le 1$ such that

$$c_2=\frac{1}{2}\left(c_1^2+k\left(4-c_1^2\right)\right),$$

(11)

$$c_3=\frac{1}{4}\left(c_1^3+2c_{1}k\left(4-c_1^2\right)-\left(4-c_1^2\right)c_1{k}^2+2\left(4-c_1^2\right)\left(1-{\left|k\right|}^2\right)\delta \right),$$

(12)

$$c_4=\frac{1}{8}\left(\begin{array}{c}{c}_1^4+k\left(4-c_1^2\right)(4k+(k^2-3k+3)c_1^2)-4\left(4-c_1^2\right)\left(1-{\left|k\right|}^2\right)\\ (c(k-1)\delta -\rho \left(1-{\left|\delta \right|}^2\right)+{\delta }^2\tilde{k})\end{array}\right).$$

(13)

The inequalities given in (11), (12) and (13) are due to [15], [30] and [31], respectively.

Lemma 3

([30]). If $p\in \mathcal{P}$ is given by (6), then for $0\le B\le 1$ and

$B\left(2B-1\right)\le D\le B$ we have

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

(14)

3. Main Results

Theorem 1. 

Let $\xi \in {\mathcal{CV}}_{\mathcal{T}}$, given in (1). Then

$$\frac{1}{\mu }\left[\xi \left(\mu \right)\ast \left(\frac{M{\mu }^2-L\mu }{{\left(1-\mu \right)}^3}\right)\right]\ne 0,$$

(15)

where

$$M=2+tanh\left(e^{i\theta }\right)\mathit{and}L=tanh\left(e^{i\theta }\right).$$

(16)

Proof. 

As $\xi \in {\mathcal{CV}}_{\mathcal{T}}$ is analytic in $\mathbf{U},$$\frac{1}{\mu }\xi \left(\mu \right)\ne 0$ for all $\mu$ in $\mathbf{U}$. Then, by using the definition of subordination and (4), we have

$$\frac{{\left(\mu {\xi }^{\prime }\left(\mu \right)\right)}^{\prime }}{{\xi }^{\prime }\left(\mu \right)}=1+tanh\chi \left(\mu \right),$$

(17)

where, $\chi \left(\mu \right)$ is the Schwarz function. Let $\chi \left(\mu \right)=e^{i\theta },$$-\pi \le \theta \le \pi .$ Then (17) becomes

$$\frac{\mu {\xi }^{″}\left(\mu \right)}{{\xi }^{\prime }\left(\mu \right)}\ne tanh\left(e^{i\theta }\right),$$

which implies

$${\mu }^2{\xi }^{″}\left(\mu \right)-\mu {\xi }^{\prime }\left(\mu \right)tanh\left(e^{i\theta }\right)\ne 0.$$

(18)

It can be easily seen that

$${\mu }^2{\xi }^{″}\left(\mu \right)+\mu {\xi }^{\prime }\left(\mu \right)=\xi \left(\mu \right)\ast \frac{\mu \left(1+\mu \right)}{{\left(1-\mu \right)}^3}\mathrm{and}\mu {\xi }^{\prime }\left(\mu \right)=\xi \left(\mu \right)\ast \frac{\mu }{{\left(1-\mu \right)}^2}.$$

(19)

Using (19) in (18) we get

$$\xi \left(\mu \right)\ast \left(\frac{M{\mu }^2-L\mu }{{\left(1-\mu \right)}^3}\right)\ne 0.$$

(20)

From (20), we will obtain (15), where M and L are given in (16). □

Theorem 2. 

Let $\xi \in \mathcal{A}$. Then, $\xi \in {\mathcal{CV}}_{\mathcal{T}}$ if

$$\sum _{n=2}^{\infty }\left[\frac{M}{L}n\left(n-1\right)-\frac{n\left(n+1\right)}{2}\right]a_n{\mu }^{n-1}-1\ne 0,$$

(21)

where M and L are given in (16).

Proof. 

If $\xi \in {\mathcal{CV}}_{\mathcal{T}}$, then from Theorem 1, we have

$$\frac{1}{\mu }\left[\xi \left(\mu \right)\ast \left(\frac{M{\mu }^2-L\mu }{{\left(1-\mu \right)}^3}\right)\right]\ne 0,$$

which is equivalent to

$$\frac{1}{\mu }\left[\xi \left(\mu \right)\ast \frac{M{\mu }^2}{{\left(1-\mu \right)}^3}-\xi \left(\mu \right)\ast \frac{L\mu }{{\left(1-\mu \right)}^3}\right]\ne 0.$$

Since ${\mu }^2=\mu \left(1+\mu \right)-\mu ,$ we have

$$\frac{1}{\mu }\left[M\left(\xi \left(\mu \right)\ast \frac{\mu \left(1+\mu \right)}{{\left(1-\mu \right)}^3}-\xi \left(\mu \right)\ast \frac{\mu }{{\left(1-\mu \right)}^3}\right)-L\left(\xi \left(\mu \right)\ast \frac{\mu }{{\left(1-\mu \right)}^3}\right)\right]\ne 0.$$

(22)

Now, applying (19) and some properties of convolution, (22), reduces to

$$\frac{1}{\mu }\left[M\left({\mu }^2{\xi }^{″}\left(\mu \right)\right)-L\left(\frac{1}{2}{\mu }^2{\xi }^{″}\left(\mu \right)+\mu {\xi }^{\prime }\left(\mu \right)\right)\right]\ne 0.$$

Using (1) and after some simplification, we obtain (21). □

Theorem 3. 

Let $\xi \in \mathcal{A}$ be given in (1). Then, $\xi \in {\mathcal{CV}}_{\mathcal{T}}$ if

$$\sum _{n=2}^{\infty }\left|4+n\left(n-3\right)tanh\left(e^{i\theta }\right)\right|\left|a_n\right|<\left|2tanh\left(e^{i\theta }\right)\right|.$$

(23)

Proof. 

To prove our required result, we use relation (21) as

$$\left|1-\sum _{n=2}^{\infty }\frac{4+n\left(n-3\right)tanh\left(e^{i\theta }\right)}{2tanh\left(e^{i\theta }\right)}{a}_n{\mu }^{n-1}\right|>1-\sum _{n=2}^{\infty }\frac{\left|4+n\left(n-3\right)tanh\left(e^{i\theta }\right)\right|}{\left|2tanh\left(e^{i\theta }\right)\right|}\left|a_n\right|{\left|\mu \right|}^{n-1}.$$

(24)

From (23), we have

$$1-\sum _{n=2}^{\infty }\left|\frac{4+n\left(n-3\right)tanh\left(e^{i\theta }\right)}{2tanh\left(e^{i\theta }\right)}\right|\left|a_n\right|>0.$$

(25)

From (24), (25) and using Theorem 2, we have the desired result. □

Theorem 4. 

Let $\xi \in {\mathcal{CV}}_{\mathcal{T}}.$ Then, ξ is convex of order $\alpha ,$ $0\le \alpha <1$ and $\left|\mu \right|<r_1$, where

$$r_1=\underset{n\ge 2}{inf}{\left(\frac{\left|4+n\left(n-3\right)tanh\left(e^{i\theta }\right)\right|}{\left|2tanh\left(e^{i\theta }\right)\right|}\frac{\left(1-\alpha \right)}{n\left(n-\alpha \right)}\right)}^{\frac{1}{n-1}}.$$

(26)

Proof. 

It is sufficient to show that

$$\left|\frac{{\left(\mu {\xi }^{\prime }\left(\mu \right)\right)}^{\prime }}{{\xi }^{\prime }\left(\mu \right)}-1\right|\le 1-\alpha .$$

(27)

From (1), we have

$$\left|\frac{\mu {\xi }^{″}\left(\mu \right)}{{\xi }^{\prime }\left(\mu \right)}\right|\le \frac{{\displaystyle \sum _{n=2}^{\infty }}n\left(n-1\right)a_n{\left|\mu \right|}^{n-1}}{1-{\displaystyle \sum _{n=2}^{\infty }}n{a}_n{\left|\mu \right|}^{n-1}}.$$

(28)

(28) is bounded above by $1-\alpha$ if

$$\sum _{n=2}^{\infty }\left[\frac{n\left(n-1\right)+n\left(1-\alpha \right)}{1-\alpha }\right]\left|a_n\right|{\left|\mu \right|}^{n-1}\le 1.$$

(29)

As $\xi \in {\mathcal{CV}}_{\mathcal{T}}$, from Theorem 1 and for some fixed $n,$ we have

$$\left|a_n\right|\le \frac{\left|2tanh\left(e^{i\theta }\right)\right|}{\left|4+n\left(n-3\right)tanh\left(e^{i\theta }\right)\right|}.$$

(30)

From (29) and (30), we can obtain (26). □

Theorem 5. 

Let $\xi \in {\mathcal{CV}}_{\mathcal{T}}.$ Then

$$\left|a_2\right|\le \frac{1}{2},\left|a_3\right|\le \frac{1}{6}\mathit{and}\left|a_4\right|\le \frac{1}{12}.$$

The first two inequalities are sharp for ${\xi }_1$, and the third inequality is sharp for ${\xi }_3:$

$${\xi }_1=\underset{0}{\overset{\mu }{\int }}exp\left(\underset{0}{\overset{t}{\int }}\frac{tanhx}{x}dx\right)dt=\mu +\frac{1}{2}{\mu }^2+\frac{1}{6}{\mu }^3+\frac{1}{72}{\mu }^4....$$

(31)

$${\xi }_3=\underset{0}{\overset{\mu }{\int }}exp\left(\underset{0}{\overset{t}{\int }}\frac{tanh{x}^3}{x}dx\right)dt=\mu +\frac{{\mu }^4}{12}+\frac{{\mu }^7}{126}+....$$

(32)

Proof. 

As $\xi \in {\mathcal{CV}}_{\mathcal{T}},$

$$\frac{{\left(\mu {\xi }^{\prime }\left(\mu \right)\right)}^{\prime }}{{\xi }^{\prime }\left(\mu \right)}\prec 1+tanh\mu ,$$

which implies

$$\frac{{\left(\mu {\xi }^{\prime }\left(\mu \right)\right)}^{\prime }}{{\xi }^{\prime }\left(\mu \right)}=1+tanh\left(\chi \left(\mu \right)\right).$$

(33)

From (1), we have

$$\begin{array}{ccc}\hfill \frac{{\left(\mu {\xi }^{\prime }\left(\mu \right)\right)}^{\prime }}{{\xi }^{\prime }\left(\mu \right)}& =& 1+2a_2\mu +\left(6a_3-4a_2^2\right){\mu }^2+\left(12a_4-18a_2{a}_3+8a_2^3\right){\mu }^3\hfill \\ & & +\left(20a_5-32a_2{a}_4-18a_3^2+48a_3{a}_2^2-16a_2^4\right){\mu }^4+....\hfill \end{array}$$

(34)

Let$p\in \mathcal{P}$, then we have

$$p\left(\mu \right)=\frac{1+\chi \left(\mu \right)}{1-\chi \left(\mu \right)}=1+c_1\mu +c_2{\mu }^2+c_3{\mu }^3+c_4{\mu }^4+\dots .$$

This implies that

$$\begin{array}{ccc}\hfill \chi \left(\mu \right)& =& \frac{1}{2}{c}_1\mu +\left(\frac{1}{2}{c}_2-\frac{1}{4}{c}_1^2\right){\mu }^2+\left(\frac{1}{8}{c}_1^3-\frac{1}{2}{c}_1{c}_2+\frac{1}{2}{c}_3\right){\mu }^3\hfill \\ & & +\left(\frac{1}{2}{c}_4-\frac{1}{2}{c}_1{c}_3-\frac{1}{4}{c}_2^2-\frac{1}{16}{c}_1^4+\frac{3}{8}{c}_1^2{c}_2\right){\mu }^4+\dots .\hfill \end{array}$$

Then

$$\begin{array}{ccc}\hfill 1+tanh\chi \left(\mu \right)& =& 1+\frac{1}{2}{c}_1\mu +\left(\frac{1}{2}{c}_2-\frac{1}{4}{c}_1^2\right){\mu }^2+\left(\frac{1}{12}{c}_1^3-\frac{1}{2}{c}_1{c}_2+\frac{1}{2}{c}_3\right){\mu }^3\hfill \\ & & +\left(\frac{1}{2}{c}_4+\frac{1}{4}{c}_1^2{c}_2-\frac{1}{2}{c}_1{c}_3-\frac{1}{4}{c}_2^2\right){\mu }^4+\dots .\hfill \end{array}$$

(35)

From (34) and (35), we have

$$\begin{array}{ccc}\hfill a_2& =& \frac{1}{4}{c}_1,\hfill \end{array}$$

(36)

$$\begin{array}{ccc}\hfill a_3& =& \frac{1}{12}{c}_2,\hfill \end{array}$$

(37)

$$\begin{array}{ccc}\hfill a_4& =& \frac{1}{12}\left(-\frac{1}{24}{c}_1^3-\frac{1}{8}{c}_2{c}_1+\frac{1}{2}{c}_3\right),\hfill \end{array}$$

(38)

$$\begin{array}{ccc}\hfill a_5& =& \frac{1}{20}\left(\frac{5}{144}{c}_1^4-\frac{1}{12}{c}_1^2{c}_2-\frac{1}{6}{c}_3{c}_1-\frac{1}{8}{c}_2^2+\frac{1}{2}{c}_4\right).\hfill \end{array}$$

(39)

Then, from Lemma 1 we have

$$\left|a_2\right|\le \frac{1}{2},\left|a_3\right|\le \frac{1}{6}.$$

From (38), we have

$$\left|a_4\right|=\frac{1}{24}\left|c_3-\frac{1}{4}{c}_1{c}_2-\frac{1}{12}{c}_1^3\right|,$$

by using Lemma 3 we have

$$\left|a_4\right|\le \frac{1}{12}.$$

□

Theorem 6. 

Let $\xi \in {\mathcal{CV}}_{\mathcal{T}},$ then

$$\left|{\mathcal{H}}_{2,1}\xi \right|=\left|a_3-a_2^2\right|\le \frac{1}{6}.$$

(40)

The result is sharp for the function

$${\xi }_2=\underset{0}{\overset{\mu }{\int }}exp\left(\underset{0}{\overset{t}{\int }}\frac{tanh{x}^2}{x}dx\right)dt=\mu +\frac{{\mu }^3}{6}+\frac{{\mu }^5}{40}+....$$

(41)

Proof. 

As $\xi \in {\mathcal{CV}}_{\mathcal{T}},$ from (36) and (37), we have

$$\left|a_3-a_2^2\right|=\left|\frac{1}{12}{c}_2-\frac{1}{16}{c}_1^2\right|=\frac{1}{12}\left|c_2-\frac{3}{4}{c}_1^2\right|,$$

then from Lemma 1 we get

$$\left|{\mathcal{H}}_{2,1}\xi \right|\le \frac{1}{6}.$$

□

Theorem 7. 

Let $\xi \in {\mathcal{CV}}_{\mathcal{T}}.$ Then

$$\left|{\mathcal{H}}_{2,2}\xi \right|=\left|a_2{a}_4-a_3^2\right|\le \frac{1}{36}.$$

(42)

This inequality is sharp for the function given in (41).

Proof. 

As $\xi \in {\mathcal{CV}}_{\mathcal{T}},$ from (36)–(38), we have

$$\left|a_2{a}_4-a_3^2\right|=\left|-\frac{1}{1152}{c}_1^4-\frac{1}{384}{c}_1^2{c}_2+\frac{1}{96}{c}_3{c}_1-\frac{1}{144}{c}_2^2\right|.$$

Using Lemma 2 we obtain

$$\left|a_2{a}_4-a_3^2\right|=\left|-\frac{1}{576}{t}^2{k}^2-\frac{1}{384}t{k}^2{c}_1^2+\frac{1}{2304}tk{c}_1^2+\frac{1}{192}\left(1-{\left|k\right|}^2\right)\delta t{c}_1-\frac{1}{768}{c}_1^4\right|.$$

From $t=4-c_1^2$, $c_1=c,$ $\left|\delta \right|\le 1, \left|k\right|=b$ and using triangular inequality, we have

$$\left|a_2{a}_4-a_3^2\right|\le \frac{1}{2304}\left(4{(4-c^2)}^2{b}^2+6(4-c^2)b^2{c}^2+(4-c^2)b{c}^2+12(1-b^2)(4-c^2)c+3c^4\right).$$

Consider

$$\xi (c,b)=4{(4-c^2)}^2{b}^2+6(4-c^2)b^2{c}^2+(4-c^2)b{c}^2+12(1-b^2)(4-c^2)c+3c^4.$$

To find the maximum value of $\xi (c,b),$ let

$$\frac{\partial \xi }{\partial b}=\left(4-c^2\right)\left(32b+4b{c}^2-24bc+c^2\right)\ge 0,$$

which implies

$$\xi (c,b)\le \xi (c,1).$$

Putting $b=1,$ we get

$$\left|a_2{a}_4-a_3^2\right|\le \frac{1}{2304}\left(4{(4-c^2)}^2+6(4-c^2)c^2+(4-c^2)c^2+3c^4\right).$$

(43)

Now let

$$G\left(c\right)=\left(4{(4-c^2)}^2+7(4-c^2)c^2+3c^4\right),$$

which implies

$$\frac{\partial G}{\partial c}=-8c\le 0\mathrm{for}\mathrm{all}c\in \left[0,2\right],$$

which shows that $G\left(c\right)$ is the decreasing function, so the maximum value occurs at $c=0,$ so (43) becomes

$$\left|{\mathcal{H}}_{2,2}\xi \right|\le \frac{1}{36}.$$

□

Theorem 8. 

Let $\xi \in {\mathcal{CV}}_{\mathcal{T}}.$ Then

$$\left|{\mathcal{H}}_{3,1}\xi \right|\le \frac{1}{144}.$$

(44)

This result is sharp for the function given in (32).

Proof. 

From (5), we have

$${\mathcal{H}}_{3,1}\xi =2a_2{a}_3{a}_4-a_2^2{a}_5-a_3^3+a_3{a}_5-a_4^2.$$

Using (36)–(39) with $c_1=c,$ we have

$${\mathcal{H}}_{3,1}\xi =\frac{1}{414720}\left(\begin{array}{c}-50c^6-207c^2{c}_2^2-456c_2^3-720c_3^2+864c_2{c}_4-648c^2{c}_4+336c^3{c}_3+\\ 78c^4{c}_2+792c{c}_2{c}_3\end{array}\right).$$

(45)

From Lemma 3, with $4-c^2=t,$ we can simplify terms occurring in (45) as

$$\begin{array}{ccc}\hfill 207c^2{c}_2^2& =& \frac{207}{4}{c}^6+\frac{207}{2}{c}^{4}tk+\frac{207}{4}{c}^2{t}^2{k}^2,\hfill \\ \hfill 456c_2^3& =& 57t^3{k}^3+171t^2{k}^2{c}^2+171tk{c}^4+57c^6,\hfill \\ \hfill 720c_3^2& =& 45c^6-90c^{4}t{k}^2+180c^{4}tk+180c^3\left(1-{\left|k\right|}^2\right)t\delta +45c^2{t}^2{k}^4-180c^2{t}^2{k}^3+180c^2{t}^2{k}^2\hfill \\ & & -180c\left(1-{\left|k\right|}^2\right)t^2{k}^2\delta +360c\left(1-{\left|k\right|}^2\right)t^{2}k\delta +180{\left(1-{\left|k\right|}^2\right)}^2{t}^2{\delta }^2,\hfill \\ \hfill 864c_2{c}_4& =& 54c^6+54c^{4}t{k}^3-162c^{4}t{k}^2+216c^{4}tk-216\left(1-{\left|k\right|}^2\right)c^{3}tk\delta +216\left(1-{\left|k\right|}^2\right)c^{3}t\delta \hfill \\ & & +54c^2{t}^2{k}^4-162c^2{t}^2{k}^3+162c^2{t}^2{k}^2+216c^{2}t{k}^2-216\left(1-{\left|k\right|}^2\right)c^{2}tk{\delta }^2\hfill \\ & & +216\left(1-{\left|k\right|}^2\right)\left(1-{\left|\delta \right|}^2\right)\rho c^{2}t-216\left(1-{\left|k\right|}^2\right)c{t}^2{k}^2\delta +216\left(1-{\left|k\right|}^2\right)c{t}^{2}k\delta +216t^2{k}^3\hfill \\ & & -216\left(1-{\left|k\right|}^2\right)t^2{k}^2{\delta }^2+216\left(1-{\left|k\right|}^2\right)\left(1-{\left|\delta \right|}^2\right)\rho t^{2}k,\hfill \\ \hfill 648c^2{c}_4& =& 81c^6+81t{c}^4{k}^3-243t{c}^4{k}^2+243t{c}^{4}k-324\left(1-{\left|k\right|}^2\right)t{c}^{3}k\delta +324\left(1-{\left|k\right|}^2\right)t{c}^3\delta \hfill \\ & & +324t{c}^2{k}^2-324\left(1-{\left|k\right|}^2\right)t{c}^{2}k{\delta }^2+324\left(1-{\left|k\right|}^2\right)t\left(1-{\left|\delta \right|}^2\right)\rho c^2,\hfill \\ \hfill 336c^3{c}_3& =& 84c^6-84t{c}^4{k}^2+168t{c}^{4}k+168\left(1-{\left|k\right|}^2\right)t\delta c^3,\hfill \\ \hfill 78c^4{c}_2& =& 39c^6+39tk{c}^4,\hfill \\ \hfill 792c{c}_2{c}_3& =& 99c^6-99c^{4}t{k}^2+297c^{4}tk+198\left(1-{\left|k\right|}^2\right)\delta c^{3}t-99c^2{t}^2{k}^3+198c^2{t}^2{k}^2\hfill \\ & & +198\left(1-{\left|k\right|}^2\right)\delta c{t}^{2}k.\hfill \end{array}$$

Putting above values in (45), we have

$${\mathcal{H}}_{3,1}\xi =\frac{1}{414720}\left(\begin{array}{c}-\frac{35}{4}{c}^6-27c^{4}t{k}^3-12c^{4}t{k}^2+\frac{45}{2}{c}^{4}tk+108c^3\left(1-{\left|k\right|}^2\right)tk\delta \\ +78c^3\left(1-{\left|k\right|}^2\right)t\delta +108c^2\left(1-{\left|k\right|}^2\right)tk{\delta }^2-108\left(1-{\left|\delta \right|}^2\right)\rho c^2\left(1-{\left|k\right|}^2\right)t\\ +9c^2{t}^2{k}^4-81c^2{t}^2{k}^3-\frac{171}{4}{c}^2{t}^2{k}^2-108c^{2}t{k}^2-36c\left(1-{\left|k\right|}^2\right)t^2{k}^2\delta \\ +54c\left(1-{\left|k\right|}^2\right)t^{2}k\delta -180{\left(1-{\left|k\right|}^2\right)}^2{t}^2{\delta }^2-216\left(1-{\left|k\right|}^2\right)t^2{k}^2{\delta }^2\\ +216\left(1-{\left|\delta \right|}^2\right)\rho \left(1-{\left|k\right|}^2\right)t^{2}k-57t^3{k}^3+216t^2{k}^3\end{array}\right).$$

Since $t=4-c^2,$ it follows that

$${\mathcal{H}}_{3,1}\xi \left(\mu \right)=\frac{1}{414720}\left[u_1(c,k)+u_2(c,k)\delta +u_3(c,k){\delta }^2+\varphi (c,k,\delta )\rho \right],$$

where

$$\begin{array}{ccc}\hfill u_1(c,k)& =& -\frac{35}{4}{c}^6-\frac{3}{4}(4-c^2)k\left[\begin{array}{c}(4-c^2)k\left(57c^2-12c^2{k}^2+32c^{2}k+16k\right)\\ +144c^{2}k+16c^{4}k+36c^4{k}^2-30c^4\end{array}\right],\hfill \\ \hfill u_2(c,k)& =& -6c(1-{\left|k\right|}^2)(4-c^2)\left[\left(6k^2-9k\right)(4-c^2)-18c^{2}k-13c^2\right],\hfill \\ \hfill u_3(c,k)& =& -36(1-{\left|k\right|}^2)(4-c^2)\left[\left(k^2+5\right)(4-c^2)-3c^{2}k\right],\hfill \\ \hfill \varphi (c,k,\delta )& =& 108(1-{\left|k\right|}^2)(4-c^2)\left[2(4-c^2)k-c^2\right](1-{\left|\delta \right|}^2).\hfill \end{array}$$

$\left|k\right|=k$ and $\left|\delta \right|=y,$ are replaced with $\left|\rho \right|\le 1,$ such that

$$\begin{array}{ccc}\hfill \left|{\mathcal{H}}_{3,1}\xi \right|& =& \frac{1}{414720}\left[\left|u_1(c,k)\right|+\left|u_2(c,k)\right|y+\left|u_3(c,k)\right|y^2+\left|\varphi (c,k,\delta )\right|\right],\hfill \\ & \le & \frac{1}{414720}\psi (c,k,y).\hfill \end{array}$$

(46)

where

$$\psi (c,k,y)=v_1(c,k)+v_2(c,k)y+v_3(c,k)y^2+v_4(c,k)(1-y^2),$$

(47)

with

$$\begin{array}{ccc}\hfill v_1(c,k)& =& \frac{35}{4}{c}^6+\frac{3}{4}(4-c^2)k\left[\begin{array}{c}k\left(57c^2+12c^2{k}^2+32c^{2}k+16k\right)(4-c^2)\\ +144c^{2}k+16c^{4}k+36c^4{k}^2+30c^4\end{array}\right],\hfill \\ \hfill v_2(c,k)& =& 6c(1-k^2)(4-c^2)\left[\left(6k^2+9k\right)(4-c^2)+18c^{2}k+13c^2\right],\hfill \\ \hfill v_3(c,k)& =& 36(1-k^2)(4-c^2)\left[\left(k^2+5\right)(4-c^2)+3c^{2}k\right],\hfill \\ \hfill v_4(c,k)& =& 108(1-k^2)(4-c^2)\left[\left(2k(4-c^2)+c^2\right)\right].\hfill \end{array}$$

To maximize the function $\psi (c,k,y)$in the cuboid, $\mathbf{\Delta }=\left[0,2\right]\times \left[0,1\right]\times \left[0,1\right],$we need to discuss the maximum values of $\psi (c,k,y)$ inside on the six faces as well as on edges of $\mathbf{\Delta }$.

1. Interior points of the cuboid

First, we check the maximum value in the interior of the cuboid; for this, let $\psi \left(c,k,y\right)\in \left(0,2\right)\times \left(0,1\right)\times \left(0,1\right)$ and consider

$$\begin{array}{ccc}\hfill \frac{\partial \psi }{\partial y}& =& 6c(1-k^2)(4-c^2)\left[\left(6k^2+9k\right)(4-c^2)+18c^{2}k+13c^2\right]\hfill \\ & & +72(1-k^2)(4-c^2)\left[\left(4-c^2\right)\left(k-5\right)+3c^2\right]\left(k-1\right)y,\hfill \end{array}$$

for critical point in $\Delta$, $\frac{\partial \psi }{\partial y}=0,$ which gives

$$y=\frac{c\left[3k(4-c^2)\left(2k+3\right)+c^2\left(18k+13\right)\right]}{\left[12\left(4-c^2\right)\left(5-k\right)-36c^2\right]\left(k-1\right)}=y_1,$$

(48)

since $y_1\in \left(0,1\right),$

$$c\left[3k(4-c^2)\left(2k+3\right)+c^2\left(18k+13\right)\right]+12\left(4-c^2\right)\left(5-k\right)\left(1-k\right)\le 36c^2\left(1-k\right)$$

(49)

and

$$c^2\ge \frac{4\left(5-k\right)}{8-k}.$$

(50)

In order to show the existence of the critical points, we must find solutions that meet the inequalities in (49) and (50). For this let $h\left(k\right)=\frac{4\left(5-k\right)}{8-k},$ and calculations show that $h^{\prime }\left(k\right)<0$, that is a decreasing function on $\left(0,1\right)$, so

$$c^2\ge \frac{16}{7}.$$

Simple calculation shows that inequality (49) does not hold true for $k\in \left(0,1\right),$which means in the interior of the cuboid, there are no critical points.

2. On the six faces of the cuboid

We will now examine the interior of each of the six faces of $\mathbf{\Delta }$ to find the maximum of $\psi \left(c,k,y\right)$.

$\left(\mathrm{i}\right)$ On the face $c=0,$ we have

$$p_1\left(k,y\right)=\psi (0,k,y)=192k^3+\left(3456k+576y^2\left(k-1\right)\left(k-5\right)\left(1-k^2\right)\right),$$

for critical points, let

$$\frac{\partial p_1}{\partial y}=-1152y\left(k^2-1\right)\left(k-1\right)\left(k-5\right)\ne 0,$$

which indicates that $p_1$ has no optimum points in $\left(0,1\right)\times \left(0,1\right).$

$\left(\mathrm{ii}\right)$ On the face $c=2,$ we have

$$\psi (2,k,y)=560.$$

(51)

$\left(\mathrm{iii}\right)$ On the face $k=0$

$$p_2\left(c,y\right)=\psi (c,0,y)=\frac{35}{4}{c}^6+78c^{3}y\left(4-c^2\right)+288c^4{y}^2-108c^4-1872c^2{y}^2+432c^2+2880y^2.$$

For the critical point, let

$$\frac{\partial p_2}{\partial y}=6\left(4-c^2\right)\left(13c^3-96y{c}^2+240y\right),$$

putting $\frac{\partial p_2}{\partial y}=0$ gives

$$y=\frac{13c^3}{48\left(2c^2-5\right)}=y_0.$$

(52)

For the specified range of $y,$ $y_0\in \left(0,1\right),$ if $c>c_0\approx 1.5811388$.

Further $\frac{\partial p_2}{\partial c}=0$ gives

$$\frac{3}{2}c\left(35c^4-260c^{3}y+768c^2{y}^2-288c^2+624cy-2496y^2+576\right)=0.$$

(53)

Putting the value of (52) in (53), we get

$$333c^9-8070c^7+48564c^5-112320c^3+86400c=0.$$

Now solving for $c\in \left(0,2\right)$, we get $c\approx 1.2691,$ which shows that no optimal solution exists.

$\left(\mathrm{iv}\right)$ On the face $k=1$

$$p_3\left(c,y\right)=\psi (c,1,y)=23c^6-456c^4+1548c^2+192,$$

for critical points, let

$$\frac{\partial p_3}{\partial c}=138c^5-1824c^3+3096c.$$

Putting $\frac{\partial p_3}{\partial c}=0$, we obtained $c\approx 1.4142$, at which $p_3$ attains its maximum value, which is

$$p_3\left(c,y\right)\le 1648.$$

(54)

$\left(\mathrm{v}\right)$ On the face $y=0$

$$\begin{array}{ccc}\hfill p_4\left(c,k\right)& =& \psi (c,k,0)=9c^6{k}^4-3c^6{k}^3+\frac{123}{4}{c}^6{k}^2-\frac{45}{2}{c}^{6}k+\frac{35}{4}{c}^6-72c^4{k}^4-288c^4{k}^3\hfill \\ & & -294c^4{k}^2+306c^{4}k-108c^4+144c^2{k}^4+2016c^2{k}^3+684c^2{k}^2-1728c^{2}k+432c^2\hfill \\ & & -3264k^3+3456k,\hfill \end{array}$$

for critical points inside $\left(0,2\right)\times \left(0,1\right),$ let

$$\begin{array}{ccc}\hfill \frac{\partial p_4}{\partial c}& =& 54c^5{k}^4-18c^5{k}^3+\frac{369}{2}{c}^5{k}^2-135c^{5}k+\frac{105}{2}{c}^5-288c^3{k}^4-1152c^3{k}^3-1176c^3{k}^2\hfill \\ & & +1224c^{3}k-432c^3+288c{k}^4+4032c{k}^3+1368c{k}^2-3456ck+864c\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \frac{\partial p_4}{\partial k}& =& 36c^6{k}^3-9c^6{k}^2+\frac{123}{2}{c}^{6}k-\frac{45}{2}{c}^6-288c^4{k}^3-864c^4{k}^2-588c^{4}k+306c^4\hfill \\ & & +576c^2{k}^3+6048c^2{k}^2+1368c^{2}k-1728c^2-9792k^2+3456.\hfill \end{array}$$

Computation shows that these systems of equations $\frac{\partial p_4}{\partial c}=0$ and $\frac{\partial p_4}{\partial k}=0$ do not have a solution in $\left(0,2\right)\times \left(0,1\right)$.

$\left(\mathrm{vi}\right)$ On the face $y=1$

$$\begin{array}{ccc}\hfill p_5\left(c,k\right)& =& \psi (c,k,1)=9c^6{k}^4-3c^6{k}^3+\frac{123}{4}{c}^6{k}^2-\frac{45}{2}{c}^{6}k+\frac{35}{4}{c}^6-36c^5{k}^4+54c^5{k}^3+114c^5{k}^2\hfill \\ & & -54c^{5}k-78c^5-108c^4{k}^4+36c^4{k}^3-546c^4{k}^2-18c^{4}k+180c^4+288c^3{k}^4-600c^3{k}^2\hfill \\ & & +312c^3+432c^2{k}^4-144c^2{k}^3+2268c^2{k}^2+432c^{2}k-1440c^2-576c{k}^4-864c{k}^3\hfill \\ & & +576c{k}^2+864ck-576k^4+192k^3-2304k^2+2880.\hfill \end{array}$$

Now, for the critical point, let

$$\begin{array}{ccc}\hfill \frac{\partial p_5}{\partial c}& =& 54c^5{k}^4-18c^5{k}^3+\frac{369}{2}{c}^5{k}^2-135c^{5}k+\frac{105}{2}{c}^5-180c^4{k}^4+270c^4{k}^3+570c^4{k}^2\hfill \\ & & -270c^{4}k-390c^4-432c^3{k}^4+144c^3{k}^3-2184c^3{k}^2-72c^{3}k+720c^3+864c^2{k}^4-1800c^2{k}^2\hfill \\ & & +936c^2+864c{k}^4-288c{k}^3+4536c{k}^2+864ck-2880c-576k^4-864k^3+576k^2+864k\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \frac{\partial p_5}{\partial k}& =& 36c^6{k}^3-9c^6{k}^2+\frac{123}{2}{c}^{6}k-\frac{45}{2}{c}^6-144c^5{k}^3+162c^5{k}^2+228c^{5}k-54c^5-432c^4{k}^3\hfill \\ & & +108c^4{k}^2-1092c^{4}k-18c^4+1152c^3{k}^3-1200c^{3}k+1728c^2{k}^3-432c^2{k}^2+4536c^{2}k\hfill \\ & & +432c^2-2304c{k}^3-2592c{k}^2+1152ck+864c-2304k^3+576k^2-4608k.\hfill \end{array}$$

Computation indicates that the solution to the system of equation $\frac{\partial p_5}{\partial c}=0$ and $\frac{\partial p_5}{\partial k}=0$ does not exist in $\left(0,2\right)\times \left(0,1\right)$.

3. On the twelve edges of the cuboid

Now, to find the maximum of $\psi \left(c,k,y\right)$ on the twelve edges of the cuboid

$\left(\mathrm{i}\right)$On $k=0$ and $y=0$

$$p_6\left(c\right)=\psi (c,0,0)=\frac{35}{4}{c}^6-108c^4+432c^2,$$

putting $\frac{\partial p_6}{\partial c}=0,$ gives the critical value $c\approx 1.8516$, at which we have

$$\psi (c,0,0)\le 564.$$

(55)

$\left(\mathrm{ii}\right)$ On $k=0$ and $y=1$

$$p_7\left(c\right)=\psi (c,0,1)=\frac{35}{4}{c}^6-78c^5+180c^4+312c^3-1440c^2+2880.$$

Differentiating $w.r.t$ “c”, we have

$$\frac{\partial p_7}{\partial c}=\frac{105}{2}{c}^5-390c^4+720c^3+936c^2-2880c\le 0,$$

for all $c\in [0,2]$; thus, the maximum occurs at $c=0$

$$\psi (c,0,1)\le 2880.$$

(56)

$\left(\mathrm{iii}\right)$ On $k=0$ and $c=0,$ we obtain

$$p_8\left(y\right)=\psi (0,0,y)=2880y^2.$$

Clearly, $\frac{\partial p_8}{\partial y}>0$ for $y\in [0,1]$, which shows that the function is increasing, so the maximum value occurs at $y=1$

$$p_8\left(y\right)\le 2880.$$

(57)

As the terms $\psi (c,1,0)$ and $\psi (c,1,1)$ are free from k, that is

$$p_9\left(c\right)=\psi (c,1,0)=\psi (c,1,1)=23c^6-456c^4+1548c^2+192.$$

Putting $\frac{\partial p_9}{\partial c}=0$ and solving, we obtained $c\approx 1.4142$, at which $p_9$ attains its maximum value, which is

$$p_9\left(c\right)\le 1648.$$

(58)

$\left(\mathrm{iv}\right)$ On $c=0$ and $k=1$

$$p_{10}\left(y\right)=\psi \left(0,1,y\right)=192.$$

(59)

$\left(\mathrm{v}\right)$On $c=2$

$$\psi (2,k,0)=\psi (2,k,1)=\psi (2,0,y)=\psi (2,1,y)=560.$$

(60)

$\left(\mathrm{vi}\right)$On $c=0$ and $y=0$

$$p_{11}\left(k\right)=\psi (0,k,0)=3456k-3264k^3,$$

for critical points, $\frac{\partial p_{11}}{\partial k}=0,$ and we get $k\approx 0.59409$, at which the maximum value is

$$p_{11}\left(k\right)\le 1368.$$

(61)

$\left(\mathrm{vii}\right)$On $y=1$ and $c=0$, we have

$$p_{12}\left(k\right)=\psi (0,k,1)=-576k^4+192k^3-2304k^2+2880.$$

By checking $\frac{\partial p_{12}}{\partial k}$, the maximum value occurs at $k=0,$ which is

$$p_{12}\left(k\right)\le 2880.$$

(62)

From all the computations, we conclude that

$$\psi \left(c,k,y\right)\le 2880.$$

Hence, it follows from (46) that

$$\left|{\mathcal{H}}_{3,1}\xi \right|\le \frac{1}{144}.$$

□

4. Conclusions

In the present investigation, we studied a new subclass of analytic functions related to a bounded, symmetric domain. We focused our attention on deriving sharp coefficient bounds, second and third Hankel determinants and the radius problem for functions belonging to this class. This study can be extended to the classes of meromorphic and harmonic functions and can be related to q-calculus as well.