On Coefficient Problems for Starlike and Convex Functions Associated with a Strip Domain

1. Introduction

Let $\mathbb{B}$ denote the class of functions ℏ which are analytic in the unit disk $\aleph =\{\upsilon \in \mathbb{C}:|\upsilon |<1\}$ normalized by $\hslash \left(0\right)={\hslash }^{\prime }\left(0\right)-1=0$. Then for $\upsilon \in \mathbb{C}$, $\hslash \in \mathbb{B}$ has the following representation

$$\begin{array}{c}\hfill \hslash \left(\upsilon \right)=\upsilon +{\displaystyle \sum _{k=2}^{\infty }}{t}_k{\upsilon }^k.\end{array}$$

(1)

Let $\mathcal{S}$ represent the collection of functions in $\mathbb{B}$ that are univalent (one-to-one) in ℵ. The well-known and important subclasses of $\mathcal{S}$ are the class ${\mathcal{S}}^{*}$ consists of all starlike functions, the class $\mathcal{C}$ comprises convex functions.

For $\hslash \in \mathbb{B}$ and $k,p\ge 0$, Pommerenke [1] introduced the Hankel determinant ${\mathcal{H}}_{p,k}$ of $p^{th}$ order as follows:

$${\mathcal{H}}_{p,k}(\hslash )=\left|\begin{array}{cccc}{t}_k& t_{k+1}& ···& t_{k+p-1}\\ t_{k+1}& t_{k+2}& ···& t_{k+p}\\ ⋮& ⋮& ⋮& ⋮& \\ t_{k+p-1}& t_{k+p}& ···& t_{k+2p-2}\end{array}\right|.$$

For $k=1$ and $p=2$, we get

$$\begin{array}{c}\hfill {\mathcal{H}}_{2,1}(\hslash )=t_3-t_2^2.\end{array}$$

For specific values of k and p the second and third-order Hankel determinants ${\mathcal{H}}_{2,2}(\hslash ),{\mathcal{H}}_{2,3}(\hslash )$ and ${\mathcal{H}}_{3,1}(\hslash )$ are given by

$$\begin{array}{c}\hfill {\mathcal{H}}_{2,2}(\hslash )=t_2{t}_4-t_3^2,\end{array}$$

(2)

$$\begin{array}{c}\hfill {\mathcal{H}}_{2,3}(\Gamma )=t_3{t}_5-t_4^2\end{array}$$

(3)

and

$$\begin{array}{c}\hfill {\mathcal{H}}_{3,1}(\Gamma )=2t_4{t}_3{t}_2-t_3^3+t_5{t}_3-t_5{t}_2^2-t_4^2.\end{array}$$

(4)

The sharp bounds of Hankel determinant of ${\mathcal{H}}_{3,1}(\Gamma )$, we refer to the articles [2,3,4,5,6,7,8,9,10,11].

For $\hslash \in \mathcal{S}$, the logarithmic coefficients ${\lambda }_k$ are defined by

$$log{\displaystyle \frac{\hslash \left(\upsilon \right)}{\upsilon }}=2{\displaystyle \sum _{k=1}^{\infty }}{\lambda }_k{\upsilon }^k,(\upsilon \in \aleph ).$$

(5)

Differentiating (5), we have

$$\left\{\begin{array}{c}{\lambda }_1={\displaystyle \frac{1}{2}}{t}_2,\hfill \\ {\lambda }_2={\displaystyle \frac{1}{2}}(t_3-{\displaystyle \frac{1}{2}}{t}_2^2),\hfill \\ {\lambda }_3={\displaystyle \frac{1}{2}}(t_4-t_2{t}_3+{\displaystyle \frac{1}{3}}{t}_2^3),\hfill \\ {\lambda }_4={\displaystyle \frac{1}{2}}(t_5-{\displaystyle \frac{1}{2}}{t}_3^2+t_3{t}_2^2-t_4{t}_2-{\displaystyle \frac{1}{4}}{t}_2^4).\hfill \end{array}\right.$$

(6)

Therefore, we obtain

$$\begin{array}{c}\hfill {\lambda }_3{\lambda }_1-{\lambda }_2^2={\displaystyle \frac{1}{48}}(-12t_3^2+12t_2{t}_4+t_2^4),\end{array}$$

(7)

$$\begin{array}{ccc}\hfill {\lambda }_2{\lambda }_4-{\lambda }_3^2& =& {\displaystyle \frac{1}{288}}(72t_3{t}_5-6t_2^4{t}_3+72t_2{t}_3{t}_4+18t_2^2{t}_3^2-36t_3^3\hfill \\ & -& 36t_2^2{t}_5-72t_4^2-12t_2^3{t}_4+t_2^6).\hfill \end{array}$$

(8)

Let G be the inverse function of $\hslash \in \mathcal{S}$ defined in a neighborhood of the origin with the Taylor series expansion

$$\begin{array}{c}\hfill G\left(\omega \right)={\hslash }^{-1}\left(\omega \right)=\omega +{\displaystyle \sum _{k=2}^{\infty }}{B}_k{\omega }^k.\end{array}$$

(9)

where we may choose $\left|\omega \right|<1/4$ from Koebe’s 1/4-theorem. Since $\hslash \left({\hslash }^{-1}\left(\omega \right)\right)=\omega$, it follows from (9) that

$$\left\{\begin{array}{c}{B}_2=-t_2,\hfill \\ B_3=-t_3+2t_2^2,\hfill \\ B_4=-t_4+5t_2{t}_3-5t_2^3,\hfill \\ B_5=-t_5+3t_3^2-21t_3{t}_2^2+6t_4{t}_2+14t_2^4.\hfill \end{array}\right.$$

(10)

From (4) and (10), we have

$$\begin{array}{ccc}\hfill {\mathcal{H}}_{3,1}\left({\hslash }^{-1}\right)& =\hfill & \hfill 2B_4{B}_3{B}_2-B_3^3+B_5{B}_3-B_5{B}_2^2-A_4^2\\ \hfill & =\hfill & \hfill -3t_2^4{t}_3+t_2^6+3t_2^2{t}_3^2+2t_2{t}_3{t}_4-2t_3^3-t_4^2-t_2^2{t}_5+t_3{t}_5\end{array}$$

(11)

In 2025, Kumar and Verma [11] introduced a subclass of starlike functions linked with a strip domain domain, defined by

$$\begin{array}{c}\hfill {\mathcal{S}}_{⊺}^{*}=\left\{\hslash :\hslash \in \mathbb{B}and{\displaystyle \frac{\upsilon {\hslash }^{\prime }\left(\upsilon \right)}{\hslash \left(\upsilon \right)}}\prec ϝ\left(\upsilon \right)=1+arctan\upsilon (\upsilon \in \aleph )\right\},\end{array}$$

where $ϝ\left(\upsilon \right)=1+arctan\left(\upsilon \right)$ is univalent in $\aleph .$ we introduce and study the class of convex functions connected with a strip domain, which is given by

$$\begin{array}{c}\hfill {\mathcal{C}}_{⊺}=\left\{\hslash :\hslash \in \mathbb{B}and1+{\displaystyle \frac{\upsilon {\hslash }^{\prime \prime }\left(\upsilon \right)}{{\hslash }^{\prime }\left(\upsilon \right)}}\prec ϝ\left(\upsilon \right)=1+arctan\upsilon (\upsilon \in \aleph )\right\}.\end{array}$$

Kumar and Verma [11] obtained the sharp bounds of the coefficients $t_k$, $k=1$ to 4 and established the following:$|t_k|\le {\displaystyle \frac{1}{k-1}}(k=2,3,4),|{\mathcal{H}}_{2,2}(\hslash )|\le {\displaystyle \frac{1}{4}},\left|{\mathcal{H}}_{3,1}(\hslash )\right|\le {\displaystyle \frac{1}{9}}.$

Motivated by the results obtained by the authors mentioned above, in this paper, we are making an attempt to estimate sharp bound for initial coefficients and coefficient functionals namely $|t_k|(2\le k\le 4),|{\mathcal{H}}_{2,1}(\hslash )|,|{\mathcal{H}}_{2,2}(\hslash )|,|{\mathcal{H}}_{2,3}(\hslash )\left|and\right|{\mathcal{H}}_{3,1}(\hslash )|$, when ℏ belongs to the class ${\mathcal{C}}_{⊺}.$ Also, we obtain the sharp bounds of the third Hankel determinant of inverse functions and the second order Hankel determinant of logarithmic inverse functions for the classes ${\mathcal{C}}_{⊺}$ and ${\mathcal{S}}_{⊺}^{*}.$

Let $\Xi$ be the collection of regular functions in ℵ of the form

$$\begin{array}{c}\hfill \zeta \left(\upsilon \right)=\upsilon +{\displaystyle \sum _{k=2}^{\infty }}{\ell }_k{\upsilon }^k.\end{array}$$

(12)

satisfying $\left|\zeta \right(\upsilon \left)\right|<1,\upsilon \in \aleph$. The foundation for proofs of our main results are the following lemmas.

Lemma 1.1 

If $\zeta \in \Xi$ is of the form (12), then

$$\begin{array}{c}\hfill |{\ell }_k|\le 1fork\ge 1,\end{array}$$

(13)

and for $\gamma \in \mathcal{C}$, we have

$$\begin{array}{c}\hfill |{\ell }_2+\gamma {\ell }_1^2|\le max\{1,\left|\gamma \right|\}.\end{array}$$

(14)

For the inequality in (13), we refer to [12]. Also, see [13] for the inequality (14).

Lemma 1.2 

([14]) If $\zeta \in \Xi$, then for any real numbers α and β such that

$$\begin{array}{c}\hfill (\alpha ,\beta )\in \{(\alpha ,\beta ):|\alpha |\le {\displaystyle \frac{1}{2}},-1\le \beta \le 1\}\cup \{(\alpha ,\beta ):{\displaystyle \frac{1}{2}}\le \left|\alpha \right|\le 2,{\displaystyle \frac{4}{27}}{\left(\right|\alpha |+1)}^3-\left(\right|\alpha |+1)\le \beta \le 1\},\end{array}$$

the following sharp estimates holds

$$\begin{array}{c}\hfill |{\ell }_3+\alpha {\ell }_1{\ell }_2+\beta {\ell }_1^3|\le 1.\end{array}$$

Lemma 1.3 

([15]) Let $\zeta \in \Xi$, $\tau \in \mathbb{C}$ and $\left|\tau \right|\le 1.$ Then

$$\begin{array}{c}\hfill |{\ell }_4+2\tau {\ell }_1{\ell }_3+3{\tau }^2{\ell }_1^2{\ell }_2+\tau {\ell }_2^2+{\tau }^3{\ell }_1^4|\le 1.\end{array}$$

Lemma 1.4 

([16]) If $\zeta \in \Xi$, then

$$|{\ell }_2|\le 1-|{\ell }_1{|}^2,|{\ell }_4|\le 1-|{\ell }_1{|}^2-{\left|{\ell }_2\right|}^2.$$

Lemma 1.5 

([17]) Let $\zeta \in \Xi$, then

$$\begin{array}{ccc}\hfill {\ell }_1& =& {\sigma }_1,\hfill \\ \hfill {\ell }_2& =& (1-|{\sigma }_1{|}^2){\sigma }_2,\hfill \\ \hfill {\ell }_3& =& -(1-|{\sigma }_1{|}^2)\overline{{\sigma }_1}{\sigma }_2^2+(1-|{\sigma }_1{|}^2\left)\right(1-|{\sigma }_2{|}^2){\sigma }_3\hfill \\ \hfill {\ell }_4& =& (1-|{\sigma }_1{|}^2){\overline{{\sigma }_1}}^2{\sigma }_2^3-(1-|{\sigma }_1{|}^2\left)\right(1-|{\sigma }_2{|}^2)(\overline{{\sigma }_2}{\sigma }_3^2+2\overline{{\sigma }_1}{\sigma }_2{\sigma }_3)\hfill \\ & +& (1-|{\sigma }_1{|}^2\left)\right(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_3{|}^2){\sigma }_4.\hfill \end{array}$$

for some ${\sigma }_k\in \mathbb{C}$ with $|{\sigma }_k|\le 1(k=1,2,3,4).$

2. The Second and Third Order Hankel Determinants for the Class ${\mathcal{C}}_{⊺}$

Theorem 2.1 

If $\hslash \in {\mathcal{C}}_{⊺}$, then

$$|t_2|\le {\displaystyle \frac{1}{2}},|t_3|\le {\displaystyle \frac{1}{6}},|t_4|\le {\displaystyle \frac{1}{12}},\left|t_5\right|\le 0.082576.$$

The first three bounds are sharp and are given by

$$\begin{array}{c}\hfill {\hslash }_1\left(\upsilon \right)={\int }_0^{\upsilon }\left(exp\left({\int }_0^x{\displaystyle \frac{arctant}{t}}dt\right)\right)dx=\upsilon +{\displaystyle \frac{1}{2}}{\upsilon }^2+···,\end{array}$$

(15)

$$\begin{array}{c}\hfill {\hslash }_2\left(\upsilon \right)={\int }_0^{\upsilon }\left(exp\left({\int }_0^x{\displaystyle \frac{arctan{t}^2}{t}}dt\right)\right)dx=\upsilon +{\displaystyle \frac{1}{6}}{\upsilon }^3+···,\end{array}$$

(16)

$$\begin{array}{c}\hfill {\hslash }_3\left(\upsilon \right)={\int }_0^{\upsilon }\left(exp\left({\int }_0^x{\displaystyle \frac{arctan{t}^3}{t}}dt\right)\right)dx=\upsilon +{\displaystyle \frac{1}{12}}{\upsilon }^4+···.\end{array}$$

(17)

Proof 

For any function ℏ belonging to the set ${\mathcal{C}}_{⊺}$, there exists an analytic function $\zeta \left(\upsilon \right)$, $\left|\zeta \right(\upsilon \left)\right|<1$ and $\zeta \left(0\right)=0$ such that

$$\begin{array}{c}\hfill 1+{\displaystyle \frac{\upsilon {\hslash }^{\prime \prime }\left(\upsilon \right)}{{\hslash }^{\prime }\left(\upsilon \right)}}=1+arctan\left(\zeta \left(\upsilon \right)\right).\end{array}$$

(18)

Utilizing (1), we have

$$\begin{array}{cc}\hfill 1+{\displaystyle \frac{\upsilon {\hslash }^{\prime \prime }\left(\upsilon \right)}{{\hslash }^{\prime }\left(\upsilon \right)}}& =1+2t_2\upsilon +(6t_3-4t_2^2){\upsilon }^2+(8t_2^3-18t_2{t}_3+12t_4){\upsilon }^3\hfill \\ \hfill & +(-16t_2^4-18t_3^2+48t_2^2{t}_3-32t_2{t}_4+20t_5){\upsilon }^4+···.\hfill \end{array}$$

(19)

Using (12) together with $1+arctan\left(\zeta \right(\upsilon \left)\right)$, we receive

$$\begin{array}{cc}\hfill 1+arctan\left(\zeta \right(\upsilon \left)\right)& =1+{\ell }_1\upsilon +{\ell }_2{\upsilon }^2+(-{\displaystyle \frac{1}{3}}{\ell }_1^3+{\ell }_3){\upsilon }^3+(-{\ell }_2{\ell }_1^2+{\ell }_4){\upsilon }^4+···.\hfill \end{array}$$

(20)

Comparing (19) and (20), we achieve

$$\begin{array}{c}\hfill t_2={\displaystyle \frac{1}{2}}{\ell }_1,\end{array}$$

(21)

$$\begin{array}{c}\hfill t_3={\displaystyle \frac{1}{6}}{\ell }_2+{\displaystyle \frac{1}{6}}{\ell }_1^2,\end{array}$$

(22)

$$\begin{array}{c}\hfill t_4={\displaystyle \frac{1}{72}}{\ell }_1^3+{\displaystyle \frac{1}{8}}{\ell }_1{\ell }_2+{\displaystyle \frac{1}{12}}{\ell }_3,\end{array}$$

(23)

$$\begin{array}{c}\hfill t_5={\displaystyle \frac{1}{360}}(18{\ell }_4+24{\ell }_1{\ell }_3+9{\ell }_2^2-5{\ell }_1^4).\end{array}$$

(24)

Applying (13) to (21), we get

$$\begin{array}{c}\hfill |t_2|\le {\displaystyle \frac{1}{2}}.\end{array}$$

From (22) and using (14), we have

$$\begin{array}{c}\hfill |t_3|={\displaystyle \frac{1}{6}}|{\ell }_2+{\ell }_1^2|\le {\displaystyle \frac{1}{6}}max\{1,|1\left|\right\}={\displaystyle \frac{1}{6}}.\end{array}$$

Applying Lemma 1.2 to the equation (23), we get

$$\begin{array}{c}\hfill |t_4|={\displaystyle \frac{1}{12}}|{\ell }_1^3+{\displaystyle \frac{3}{2}}{\ell }_1{\ell }_2+{\displaystyle \frac{1}{6}}{\ell }_3|\le {\displaystyle \frac{1}{12}}.\end{array}$$

From (24), Lemma 1.3 and 1.4, we have

$$\begin{array}{ccc}\hfill |t_5|& =& {\displaystyle \frac{1}{360}}|12({\ell }_1^4+{\ell }_2^2+3{\ell }_2{\ell }_1^2+2{\ell }_3{\ell }_1+{\ell }_4)+6{\ell }_4-36{\ell }_1^2{\ell }_2-3{\ell }_2^2-17{\ell }_1^4|\hfill \\ & & \le {\displaystyle \frac{1}{360}}[12+6(1-|{\ell }_1{|}^2-|{\ell }_2{|}^2)+36|{\ell }_1{|}^2|{\ell }_2|+3|{\ell }_2{|}^2+17|{\ell }_1{|}^4]\hfill \\ & & ={\displaystyle \frac{1}{360}}[18-6|{\ell }_1{|}^2+36|{\ell }_1{|}^2|{\ell }_2|-3|{\ell }_2{|}^2+17|{\ell }_1{|}^4]\hfill \\ & & ={\displaystyle \frac{1}{360}}{\eta }_1(x,y),\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill {\eta }_1(x,y)& =& 18-6x^2+36x^{2}y-3y^2+17x^4\hfill \end{array}$$

with $x=|{\ell }_1|\in [0,1]$ and $y=|{\ell }_2|\in [0,1].$ The optimal points of ${\eta }_1(x,y)$ satisfy the conditions

$$\left\{\begin{array}{c}{\displaystyle \frac{\partial {\eta }_1}{\partial x}}=-12x+72yx+68x^3=0\hfill \\ {\displaystyle \frac{\partial {\eta }_1}{\partial y}}=36x^2-6y=0\hfill \end{array}\right.$$

By calculating, we have $\left\{\begin{array}{c}{x}_1=0,\hfill \\ y_1=0,\hfill \end{array}\right.$ $\left\{\begin{array}{c}{x}_2=0.1549,\hfill \\ y_2=0.1440,\hfill \end{array}\right.$ $\left\{\begin{array}{c}{x}_3=-0.1549,\hfill \\ x_3=0.1440.\hfill \end{array}\right.$

Hence, there is an optimal point $(x_2,y_2)$ in $(0,1)\times (0,1)$. so, we have $max\left\{{\eta }_1(x,y)\right\}\le {\eta }_1(x_2,y_2)=17.928016.$

(1) For $x=0,$

$$\begin{array}{cc}\hfill {\eta }_1(0,x)& =18-3y^2\le 18..\end{array}$$

(2) For $y=0,$

$$\begin{array}{ccc}\hfill {\eta }_1(0,x)& =& 18-6x^2+17x^4\le 29.\hfill \end{array}$$

(3) For $y=1-x^2$

$$\begin{array}{c}\hfill {\eta }_1(x,1-x^2)=15+36x^2-22x^4={ϵ}_1\left(x\right)\le {ϵ}_1\left(0.9045\right)=29.7273.\end{array}$$

Therefore, we have

$$\begin{array}{c}\hfill |t_5|\le {\displaystyle \frac{29.7273}{360}}=0.082576.\end{array}$$

Theorem 2.2 

If $\hslash \in {\mathcal{C}}_T$, then

$$|t_3-\gamma t_2^2|\le {\displaystyle \frac{1}{6}}\{1,{\displaystyle \frac{|2-3\gamma |}{2}}\}.$$

The sharpness is given by (15) and (16).

Proof 

Let $\hslash \in {\mathcal{C}}_T.$ From (21) , (22) and (14), we yield

$$\begin{array}{c}\hfill |t_3-\gamma t_2^2|={\displaystyle \frac{1}{6}}|{\ell }_2+{\displaystyle \frac{2-3\gamma }{2}}{\ell }_1^2|\le {\displaystyle \frac{1}{6}}\{1,{\displaystyle \frac{|2-3\gamma |}{2}}\}.\end{array}$$

□

Theorem 2.3 

If $\hslash \in {\mathcal{C}}_T$, then

$$|t_2{t}_4-t_3^2|\le {\displaystyle \frac{1}{36}}.$$

The bound is sharp for the function ${\hslash }_2\left(\upsilon \right)$ given by (16).

Proof 

Let $\hslash \in {\mathcal{C}}_T.$ From (21) , (22) , (23) and (2), we have

$$\begin{array}{c}\hfill |t_2{t}_4-t_3^2|={\displaystyle \frac{1}{144}}|6{\ell }_1{\ell }_3+{\ell }_1^2{\ell }_2-4{\ell }_2^2-3{\ell }_1^4|.\end{array}$$

(25)

Applying Lemma 1.5 to the equation (25), we achieve

$$\begin{array}{c}\hfill |t_2{t}_4-t_3^2|={\displaystyle \frac{1}{144}}|-6(1-|{\sigma }_1{|}^2\left)\right|{\sigma }_1{|}^2{\sigma }_2^2+6{\sigma }_1(1-|{\sigma }_1{|}^2\left)\right(1-|{\sigma }_2{|}^2){\sigma }_3+{\sigma }_1^2(1-|{\sigma }_1{|}^2){\sigma }_2-4{\sigma }_2^2(1-|{\sigma }_1{|}^2{)}^2-3{\sigma }_1^4|.\end{array}$$

By setting $|{\sigma }_1|=\sigma ,|{\sigma }_2|=\rho$, and upon taking $|{\sigma }_3|\le 1$, we have

$$\begin{array}{c}\hfill |t_2{t}_4-t_3^2|\le {\displaystyle \frac{1}{144}}[6(1-{\sigma }^2){\sigma }^2{\rho }^2+6(\sigma -{\sigma }^3)(1-{\rho }^2)+{\sigma }^2(1-{\sigma }^2)\rho +4{(1-{\sigma }^2)}^2{\rho }^2+3{\sigma }^4]={\displaystyle \frac{1}{144}}{\xi }_1(\sigma ,\rho ).\end{array}$$

Differentiating ${\xi }_1(\sigma ,\rho )$ with respect to $\rho$, we yield

$${\displaystyle \frac{\partial {\xi }_1}{\partial \rho }}={\sigma }^2-{\sigma }^4+4\rho (1-{\sigma }^2)({\sigma }^2-3\sigma +2)\ge 0.$$

Hence, ${\xi }_1$ attains its maximum value at $\rho =1$. Therefore, we get

$$\begin{array}{c}\hfill |t_2{t}_4-t_3^2|\le {\displaystyle \frac{1}{144}}{\xi }_1(\sigma ,1)={\displaystyle \frac{1}{144}}(4-{\sigma }^2)\le {\displaystyle \frac{4}{144}}={\displaystyle \frac{1}{36}}.\end{array}$$

Theorem 2.4 

If $\hslash \in {\mathcal{C}}_T$, then

$$|{\mathcal{H}}_{2,3}(\hslash )|\le {\displaystyle \frac{210.88628}{25920}}=0.008136.$$

Proof 

Let $\hslash \in {\mathcal{C}}_T.$ From $\left(22\right)-\left(24\right)$ and $\left(3\right)$, we yield

$$\begin{array}{ccc}\hfill {\mathcal{H}}_{2,3}(\hslash )& =& {\displaystyle \frac{1}{25920}}(216{\ell }_2{\ell }_4-252{\ell }_1{\ell }_2{\ell }_3-150{\ell }_1^4{\ell }_2+216{\ell }_1^2{\ell }_4+228{\ell }_1^3{\ell }_3-180{\ell }_3^2+108{\ell }_2^3\hfill \\ & & -297{\ell }_1^2{\ell }_2^2-65{\ell }_1^6).\hfill \end{array}$$

(26)

From $\left(26\right)$ and Lemma 1.5 , we have

$$\begin{array}{ccc}\hfill {\mathcal{H}}_{2,3}(\hslash )& =& {\displaystyle \frac{1}{25920}}[-297(1-|{\sigma }_1{|}^2{)}^2{\sigma }_1^2{\sigma }_2^2+252|{\sigma }_1{|}^2{\sigma }_2^3(1-|{\sigma }_1{|}^2{)}^2-252{\sigma }_1{\sigma }_2(1-|{\sigma }_1{|}^2{)}^2(1-|{\sigma }_2{|}^2){\sigma }_3\hfill \\ & +& 36{\overline{{\sigma }_1}}^2(1-|{\sigma }_1{|}^2{)}^2{\sigma }_2^4-216|{\sigma }_2{|}^2(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2{)}^2{\sigma }_3^2-72\overline{{\sigma }_1}{\sigma }_2^2(1-|{\sigma }_1{|}^2{)}^2(1-|{\sigma }_2{|}^2){\sigma }_3\hfill \\ & +& 216{\sigma }_2(1-|{\sigma }_3{|}^2\left)\right(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2{)}^2{\sigma }_4-180(1-|{\sigma }_2{|}^2{)}^2(1-|{\sigma }_1{|}^2{)}^2{\sigma }_3^2-228{\sigma }_1^3\overline{{\sigma }_1}{\sigma }_2^2(1-|{\sigma }_1{|}^2)\hfill \\ & +& 288(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2){\sigma }_1^3{\sigma }_3+216{\sigma }_1^2{\overline{{\sigma }_1}}^2{\sigma }_2^3(1-|{\sigma }_1{|}^2)-65{\sigma }_1^6-216\overline{{\sigma }_2}{\sigma }_1^2(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2){\sigma }_3^2\hfill \\ & -& 432\overline{{\sigma }_1}{\sigma }_1^2{\sigma }_2(1-|{\sigma }_1{|}^2\left)\right(1-|{\sigma }_2{|}^2){\sigma }_3+216{\sigma }_1^2(1-|{\sigma }_3{|}^2\left)\right(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2){\sigma }_4-150(1-|{\sigma }_1{|}^2){\sigma }_1^4{\sigma }_2\hfill \\ & +& 108(1-|{\sigma }_1{{|}^2)}^3{\sigma }_2^3].\hfill \end{array}$$

Therefore, we achieve

$${\mathcal{H}}_{2,3}(\hslash )={\displaystyle \frac{1}{25920}}\left[{\zeta }_1\left({\sigma }_1,{\sigma }_2\right)+{\zeta }_2\left({\sigma }_1,{\sigma }_2\right){\sigma }_3+{\zeta }_3\left({\sigma }_1,{\sigma }_2\right){\sigma }_3^2+{\Psi }_1\left({\sigma }_1,{\sigma }_2,{\sigma }_3\right){\sigma }_4\right],$$

where

$$\begin{array}{ccc}\hfill {\zeta }_1\left({\sigma }_1,{\sigma }_2\right)& =& \left(1-|{\sigma }_1{|}^2\right)\left[(-297{\sigma }_1^2{\sigma }_2^2+144|{\sigma }_1{|}^2{\sigma }_2^3+108{\sigma }_2^3+36{\overline{{\sigma }_1}}^2{\sigma }_2^4\left)\right(1-|{\sigma }_1{|}^2)+216{\sigma }_1^2{\overline{{\sigma }_1}}^2{\sigma }_2^3\right.\hfill \\ & & \left.-228{\sigma }_1^3\overline{{\sigma }_1}{\sigma }_2^2-150{\sigma }_1^4{\sigma }_2\right]-65{\sigma }_1^6,\hfill \\ \hfill {\zeta }_2\left({\sigma }_1,{\sigma }_2\right)& =& (1-|{\sigma }_1{|}^2)\left(1-|{\sigma }_2{|}^2\right)\left[(-252{\sigma }_1{\sigma }_2-72\overline{{\sigma }_1}{\sigma }_2^2)(1-|{\sigma }_1{|}^2)-432{\sigma }_1^2\overline{{\sigma }_1}{\sigma }_2+288{\sigma }_1^3\right],\hfill \\ \hfill {\zeta }_3\left({\sigma }_1,{\sigma }_2\right)& =& (1-|{\sigma }_2{|}^2)\left(1-|{\sigma }_1{|}^2\right)\left[(-36|{\sigma }_2{|}^2-180\left)\right(1-|{\sigma }_1{|}^2)-216{\sigma }_1^2\overline{{\sigma }_2}\right],\hfill \\ \hfill {\Psi }_1\left({\sigma }_1,{\sigma }_2,{\sigma }_3\right)& =& (1-|{\sigma }_1{|}^2\left)\right(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_3{|}^2)\left[216(1-|{\sigma }_1{|}^2){\sigma }_2+216{\sigma }_1^2\right].\hfill \end{array}$$

Setting $|{\sigma }_1|=\sigma ,$$|{\sigma }_2|=\rho$, $|{\sigma }_3|=\iota$, and utilizing $\left|{\sigma }_4\right|\le 1,$ we achieve

$$\begin{array}{ccc}\hfill \left|{\mathcal{H}}_{2,3}(\hslash )\right|& \le & {\displaystyle \frac{1}{25920}}\left(\left|{\zeta }_1\left({\sigma }_1,{\sigma }_2\right)\right|+\left|{\zeta }_2\left({\sigma }_1,{\sigma }_2\right)\right|\iota +\left|{\zeta }_3\left({\sigma }_1,{\sigma }_2\right)\right|{\iota }^2+\left|{\Psi }_1\left({\sigma }_1,{\sigma }_2,{\sigma }_3\right)\right|\right)\hfill \\ & \le & {\displaystyle \frac{1}{25920}}{\phi }_1\left(\sigma ,\rho ,\iota \right),\hfill \end{array}$$

where

$$\begin{array}{c}\hfill {\phi }_1\left(\sigma ,\rho ,\iota \right)={\kappa }_1\left(\sigma ,\rho \right)+{\kappa }_2\left(\sigma ,\rho \right)\iota +{\kappa }_3\left(\sigma ,\rho \right){\iota }^2+{\kappa }_4\left(\sigma ,\rho \right)(1-{\iota }^2),\end{array}$$

with

$$\begin{array}{ccc}\hfill {\kappa }_1\left(\sigma ,\rho \right)& =& \left(1-{\sigma }^2\right)\left[(1-{\sigma }^2)(36{\sigma }^2{\rho }^4+144{\sigma }^2{\rho }^3+108{\rho }^3+297{\sigma }^2{\rho }^2)+216{\sigma }^4{\rho }^3+228{\sigma }^4{\rho }^2+150{\sigma }^4\rho \right]+65{\sigma }^6,\hfill \\ \hfill {\kappa }_2\left(\sigma ,\rho \right)& =& \left(1-{\sigma }^2\right)(1-{\rho }^2)\left[(225\sigma \rho +72\sigma {\rho }^2)(1-{\sigma }^2)+432{\sigma }^3\rho +288{\sigma }^3\right],\hfill \\ \hfill {\kappa }_3\left(\sigma ,\rho \right)& =& \left(1-{\sigma }^2\right)(1-{\rho }^2)\left[(36{\rho }^2+180)(1-{\sigma }^2)+216{\sigma }^2\rho \right],\hfill \\ \hfill {\kappa }_4\left(\sigma ,\rho \right)& =& (1-{\rho }^2)(1-{\sigma }^2)\left[216(1-{\sigma }^2)\rho +216{\sigma }^2\right].\hfill \end{array}$$

Note that

$$\begin{array}{ccc}\hfill {\kappa }_3\left(\sigma ,\rho \right)& =& (1-{\rho }^2)\left(1-{\sigma }^2\right)\left[(180+36{\rho }^2)(1-{\sigma }^2)+216{\sigma }^2\right]={\pi }_1(\sigma ,\rho ).\hfill \end{array}$$

Hence, we get

$$\begin{array}{c}\hfill {\phi }_1\left(\sigma ,\rho ,\iota \right)\le {\kappa }_1\left(\sigma ,\rho \right)+{\kappa }_2\left(\sigma ,\rho \right)\iota +{\pi }_1\left(\sigma ,\rho \right){\iota }^2+{\kappa }_4\left(\sigma ,\rho \right)(1-{\iota }^2)={\mu }_1\left(\sigma ,\rho ,\iota \right),\end{array}$$

Differentiating ${\mu }_1\left(\sigma ,\rho ,\iota \right)$ with respect to $\iota$, we obtain

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\partial {\mu }_1}{\partial \iota }}& =& {\kappa }_2(\sigma ,\rho )+2\iota \left[{\pi }_1(\sigma ,\rho )-{\kappa }_4(\sigma ,\rho )\right]\hfill \\ & =& {\kappa }_2(\sigma ,\rho )+72{(1-{\sigma }^2)}^2(1-{\rho }^2)({\rho }^2-6\rho +5)\iota \ge 0.\hfill \end{array}$$

So, ${\mu }_1\left(\sigma ,\rho ,\iota \right)$ attains its maximum value at $\iota =1$. . Therefore, we have

$$\begin{array}{ccc}\hfill {\phi }_1\left(\sigma ,\rho ,\iota \right)& \le & {\mu }_1\left(\sigma ,\rho ,\iota \right)\le {\mu }_1\left(\sigma ,\rho ,1\right)={\kappa }_1\left(\sigma ,\rho \right)+{\kappa }_2\left(\sigma ,\rho \right)+{\pi }_1\left(\sigma ,\rho \right)\hfill \\ & =& 65{\sigma }^6-288{\sigma }^5-36{\sigma }^4+288{\sigma }^3-144{\sigma }^2+180+(36{\sigma }^6-72{\sigma }^5-108{\sigma }^4+144{\sigma }^3+108{\sigma }^2\hfill \\ & -& 72\sigma -36){\rho }^4+(-72{\sigma }^6+180{\sigma }^5+36{\sigma }^4+72{\sigma }^3-72{\sigma }^2-252\sigma +108){\rho }^3+(69{\sigma }^6+360{\sigma }^5\hfill \\ & -& 294{\sigma }^4-432{\sigma }^3+369{\sigma }^2+72\sigma -144){\rho }^2+(-150{\sigma }^6-180{\sigma }^5+150{\sigma }^4-72{\sigma }^3+252\sigma )\rho ={\xi }_2(\sigma ,\rho ).\hfill \end{array}$$

The critical points of ${\xi }_2(\sigma ,\rho )$ satisfy the conditions

$$\left\{\begin{array}{c}{\displaystyle \frac{\partial {\xi }_2}{\partial \sigma }}=390{\sigma }^5-1440{\sigma }^4-144{\sigma }^3+864{\sigma }^2-288\sigma +(216{\sigma }^5-360{\sigma }^4-432{\sigma }^3+432{\sigma }^2+216\sigma -72){\rho }^4\hfill \\ +(-432{\sigma }^5+900{\sigma }^4+144{\sigma }^3+216{\sigma }^2-144\sigma -252){\rho }^3+(414{\sigma }^5+1800{\sigma }^4-1176{\sigma }^3-1296{\sigma }^2+738\sigma \hfill \\ +72){\rho }^2+(-900{\sigma }^5-900{\sigma }^4+600{\sigma }^3-216{\sigma }^2+252)\rho =0\hfill \\ {\displaystyle \frac{\partial {\xi }_2}{\partial \rho }}=252\sigma -72{\sigma }^3+150{\sigma }^4-180{\sigma }^5-150{\sigma }^6+(144{\sigma }^6-288{\sigma }^5-432{\sigma }^4+576{\sigma }^3+432{\sigma }^2-288\sigma -144){\rho }^3\hfill \\ +(-216{\sigma }^6+540{\sigma }^5+108{\sigma }^4+216{\sigma }^3-216{\sigma }^2-756\sigma +324){\rho }^2+(138{\sigma }^6+720{\sigma }^5-588{\sigma }^4-864{\sigma }^3\hfill \\ +738{\sigma }^2+144\sigma -288)\rho =0\hfill \end{array}\right.$$

By calculating, we achieve $\left\{\begin{array}{c}{\sigma }_{1,1}=1,\hfill \\ {\rho }_{1,1}=1.3932,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,2}=1,\hfill \\ {\rho }_{1,2}=-2.2054,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,3}=1,\hfill \\ {\rho }_{1,3}=-0.4656,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,4}=-1,\hfill \\ {\rho }_{1,4}=0.7778,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,5}=0,\hfill \\ {\rho }_{1,5}=0,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,6}=0.0522,\hfill \\ {\rho }_{1,6}=0.0496,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,7}=0.6209,\hfill \\ {\rho }_{1,7}=0.5731,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,8}=1.2857,\hfill \\ {\rho }_{1,8}=6.6123.\hfill \end{array}\right.$

Therefore, there are two critical points of $({\sigma }_{1,6},{\rho }_{1,6})$ and $({\sigma }_{1,7},{\rho }_{1,7})$ in $(0,1)\times (0,1)$. Thus, we have $max\left\{{\xi }_2(\sigma ,\rho )\right\}\le max\{{\xi }_2({\sigma }_{1,6},{\rho }_{1,6}),{\xi }_2({\sigma }_{1,7},{\rho }_{1,7})\}=max\{179.963264,210.88628\}=210.88628.$

(4) For $\sigma =1,$

$$\begin{array}{cc}\hfill {\xi }_2(1,\rho )& =65.\end{array}$$

(5) For $\sigma =0,$

$$\begin{array}{ccc}\hfill {\xi }_2(0,\rho )& =& 180-144{\rho }^2+108{\rho }^3-36{\rho }^4\le 180.\hfill \end{array}$$

(6) For $\rho =0$

$$\begin{array}{c}\hfill {\xi }_2(\sigma ,0)=180-144{\sigma }^2+288{\sigma }^3-36{\sigma }^4-288{\sigma }^5+65{\sigma }^6\le 180.\end{array}$$

(7) For $\rho =1,$

$$\begin{array}{c}\hfill {\xi }_2(\sigma ,1)=108+261{\sigma }^2-252{\sigma }^4-52{\sigma }^6={ϵ}_2\left(\sigma \right)\le {ϵ}_2\left(0.6738\right)=169.6868.\end{array}$$

Thus, we achieve

$$\begin{array}{c}\hfill |{\mathcal{H}}_{2,3}(\hslash )|\le {\displaystyle \frac{210.88628}{25920}}=0.008136.\end{array}$$

Theorem 2.5 

If $\hslash \in {\mathcal{C}}_T$, then

$$|{\mathcal{H}}_{3,1}(\hslash )|\le {\displaystyle \frac{1}{144}}.$$

The outcome is the best possible and is obtained by (17).

Proof 

Let $\hslash \in {\mathcal{C}}_T.$ From $\left(21\right)-\left(24\right)$ and $\left(4\right)$, we achieve

$$\begin{array}{ccc}\hfill {\mathcal{H}}_{3,1}(\hslash )& =& {\displaystyle \frac{1}{25920}}(216{\ell }_4{\ell }_2+108{\ell }_1{\ell }_2{\ell }_3+90{\ell }_1^4{\ell }_2-108{\ell }_1^2{\ell }_4+156{\ell }_1^3{\ell }_3-180{\ell }_3^2-12{\ell }_2^3\hfill \\ & & -279{\ell }_1^2{\ell }_2^2-35{\ell }_1^6).\hfill \end{array}$$

(27)

From Lemma 1.5 and $\left(27\right)$, we yield

$$\begin{array}{ccc}\hfill {\mathcal{H}}_{3,1}(\hslash )& =& {\displaystyle \frac{1}{25920}}[-35{\sigma }_1^6-216|{\sigma }_2{|}^2(1-|{\sigma }_1{|}^2{)}^2(1-|{\sigma }_2{|}^2){\sigma }_3^2+36{\overline{{\sigma }_1}}^2(1-|{\sigma }_1{|}^2{)}^2{\sigma }_2^4+216(1-|{\sigma }_1{{|}^2)}^2\hfill \\ & \times & (1-|{\sigma }_3{|}^2\left)\right(1-|{\sigma }_2{|}^2){\sigma }_2{\sigma }_4-72\overline{{\sigma }_1}(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2{)}^2{\sigma }_2^2{\sigma }_3+90{\sigma }_1^4(1-|{\sigma }_1{|}^2){\sigma }_2-108{\overline{{\sigma }_1}}^2{\sigma }_1^2{\sigma }_2^3\hfill \\ & \times & (1-|{\sigma }_1{|}^2)+216{\sigma }_1^2\overline{{\sigma }_1}{\sigma }_2(1-|{\sigma }_1{|}^2\left)\right(1-|{\sigma }_2{|}^2){\sigma }_3-12(1-|{\sigma }_1{|}^2{)}^3{\sigma }_2^3-108{\sigma }_1^2(1-|{\sigma }_1{|}^2\left)\right(1-|{\sigma }_3{|}^2)\hfill \\ & \times & (1-|{\sigma }_2{|}^2){\sigma }_4-279{\sigma }_1^2{\sigma }_2^2(1-|{\sigma }_1{|}^2{)}^2-108|{\sigma }_1{|}^2{\sigma }_2^3(1-|{\sigma }_1{|}^2{)}^2-180(1-|{\sigma }_2{|}^2{)}^2{\sigma }_3^2(1-|{\sigma }_1{{|}^2)}^2\hfill \\ & +& 156{\sigma }_1^3(1-|{\sigma }_1{|}^2\left)\right(1-|{\sigma }_2{|}^2){\sigma }_3+108{\sigma }_1{\sigma }_2(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2{)}^2{\sigma }_3-156{\sigma }_1^3(1-|{\sigma }_1{|}^2)\overline{{\sigma }_1}{\sigma }_2^2\hfill \\ & +& 108{\sigma }_1^2(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2)\overline{{\sigma }_2}{\sigma }_3^2].\hfill \end{array}$$

Hence, we have

$${\mathcal{H}}_{3,1}(\hslash )={\displaystyle \frac{1}{25920}}\left[{\zeta }_4\left({\sigma }_1,{\sigma }_2\right)+{\zeta }_5\left({\sigma }_1,{\sigma }_2\right){\sigma }_3+{\zeta }_6\left({\sigma }_1,{\sigma }_2\right){\sigma }_3^2+{\Psi }_2\left({\sigma }_1,{\sigma }_2,{\sigma }_3\right){\sigma }_4\right],$$

where

$$\begin{array}{ccc}\hfill {\zeta }_4\left({\sigma }_1,{\sigma }_2\right)& =& \left(1-|{\sigma }_1{|}^2\right)\left[(-279{\sigma }_1^2{\sigma }_2^2-96|{\sigma }_1{|}^2{\sigma }_2^3-12{\sigma }_2^3+36{\sigma }_2^4{\overline{{\sigma }_1}}^2\left)\right(1-|{\sigma }_1{|}^2)-108{\sigma }_1^2{\overline{{\sigma }_1}}^2{\sigma }_2^3\right.\hfill \\ & & \left.-156\overline{{\sigma }_1}{\sigma }_1^3{\sigma }_2^2+90{\sigma }_1^4{\sigma }_2\right]-35{\sigma }_1^6,\hfill \\ \hfill {\zeta }_5\left({\sigma }_1,{\sigma }_2\right)& =& (1-|{\sigma }_1{|}^2)\left(1-|{\sigma }_2{|}^2\right)\left[(108{\sigma }_1{\sigma }_2-72{\sigma }_2^2\overline{{\sigma }_1})(1-|{\sigma }_1{|}^2)+216{\sigma }_1^2\overline{{\sigma }_1}{\sigma }_2+156{\sigma }_1^3\right],\hfill \\ \hfill {\zeta }_6\left({\sigma }_1,{\sigma }_2\right)& =& (1-|{\sigma }_2{|}^2)\left(1-|{\sigma }_1{|}^2\right)\left[(-108-36|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2)+108{\sigma }_1^2\overline{{\sigma }_2}\right],\hfill \\ \hfill {\Psi }_2\left({\sigma }_1,{\sigma }_2,{\sigma }_3\right)& =& (1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_3{|}^2\left)\right(1-|{\sigma }_1{|}^2)\left[216{\sigma }_2(1-|{\sigma }_1{|}^2)-108{\sigma }_1^2\right].\hfill \end{array}$$

Let $\sigma =|{\sigma }_1|,\rho =|{\sigma }_2|,\iota =|{\sigma }_3|.$ By noting that $|{\sigma }_4|\le 1$, we achieve

$$\begin{array}{ccc}\hfill \left|{\mathcal{H}}_{3,1}(\hslash )\right|& \le & {\displaystyle \frac{1}{25920}}\left(\left|{\zeta }_4\left({\sigma }_1,{\sigma }_2\right)\right|+\left|{\zeta }_5\left({\sigma }_1,{\sigma }_2\right)\right|\iota +\left|{\zeta }_6\left({\sigma }_1,{\sigma }_2\right)\right|{\iota }^2+\left|{\Psi }_2\left({\sigma }_1,{\sigma }_2,{\sigma }_3\right)\right|\right)\hfill \\ & \le & {\displaystyle \frac{1}{25920}}{\phi }_2\left(\sigma ,\rho ,\iota \right),\hfill \end{array}$$

where

$$\begin{array}{c}\hfill {\phi }_2\left(\sigma ,\rho ,\iota \right)={\kappa }_5\left(\sigma ,\rho \right)+{\kappa }_6\left(\sigma ,\rho \right)\iota +{\kappa }_7\left(\sigma ,\rho \right){\iota }^2+{\kappa }_8\left(\sigma ,\rho \right)(1-{\iota }^2),\end{array}$$

with

$$\begin{array}{ccc}\hfill {\kappa }_5\left(\sigma ,\rho \right)& =& \left(1-{\sigma }^2\right)\left[(36{\rho }^4{\sigma }^2+96{\sigma }^2{\rho }^3+12{\rho }^3+279{\sigma }^2{\rho }^2)(1-{\sigma }^2)+108{\sigma }^4{\rho }^3+156{\sigma }^4{\rho }^2+90{\sigma }^4\rho \right]+35{\sigma }^6,\hfill \\ \hfill {\kappa }_6\left(\sigma ,\rho \right)& =& \left(1-{\sigma }^2\right)(1-{\rho }^2)\left[(108\sigma \rho +72{\rho }^2\sigma )(1-{\sigma }^2)+216{\sigma }^3\rho +156{\sigma }^3\right],\hfill \\ \hfill {\kappa }_7\left(\sigma ,\rho \right)& =& \left(1-{\sigma }^2\right)(1-{\rho }^2)\left[(1-{\sigma }^2)(180+36{\rho }^2)+108{\sigma }^2\rho \right],\hfill \\ \hfill {\kappa }_8\left(\sigma ,\rho \right)& =& (1-{\sigma }^2)(1-{\rho }^2)\left[216\rho (1-{\sigma }^2)+108{\sigma }^2\right].\hfill \end{array}$$

Note that

$$\begin{array}{ccc}\hfill {\kappa }_7(\sigma ,\rho ))& \le & (1-{\rho }^2)(1-{\sigma }^2)[(180+36{\rho }^2)(1-{\sigma }^2)+108{\sigma }^2]={\pi }_2(\sigma ,\rho ).\hfill \end{array}$$

Hence, we have

$$\begin{array}{c}\hfill {\phi }_2\left(\sigma ,\rho ,\iota \right)\le {\kappa }_5\left(\sigma ,\rho \right)+{\kappa }_6\left(\sigma ,\rho \right)\iota +{\pi }_2\left(\sigma ,\rho \right){\iota }^2+{\kappa }_8\left(\sigma ,\rho \right)(1-{\iota }^2)={\mu }_2\left(\sigma ,\rho ,\iota \right),\end{array}$$

Differentiating ${\mu }_2\left(\sigma ,\rho ,\iota \right)$ with respect to $\iota$, we yield

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\partial {\mu }_2}{\partial \iota }}& =& {\kappa }_6(\sigma ,\rho )+2\iota \left[{\pi }_2(\sigma ,\rho )-{\kappa }_8(\sigma ,\rho )\right]\hfill \\ & =& {\kappa }_6(\sigma ,\rho )+72(1-{\rho }^2)(5-6\rho +{\rho }^2){(1-{\sigma }^2)}^2\iota \ge 0.\hfill \end{array}$$

Hence, ${\mu }_2\left(\sigma ,\rho ,\iota \right)$ attains its maximum value at $\iota =1$. So, we get

$$\begin{array}{ccc}\hfill {\phi }_2\left(\sigma ,\rho ,\iota \right)& \le & {\mu }_2\left(\sigma ,\rho ,\iota \right)\le {\mu }_2\left(\sigma ,\rho ,1\right)={\kappa }_5(\sigma ,\rho )+{\kappa }_6(\sigma ,\rho )+{\pi }_2(\sigma ,\rho )\hfill \\ & =& 180-252{\sigma }^2+156{\sigma }^3+72{\sigma }^4-156{\sigma }^5+35{\sigma }^6+(-36-72\sigma +108{\sigma }^2+144{\sigma }^3-108{\sigma }^4-72{\sigma }^5\hfill \\ & +& 36{\sigma }^6){\rho }^4+(12-108\sigma +72{\sigma }^2-72{\sigma }^4+108{\sigma }^5-12{\sigma }^6){\rho }^3+(-144+72\sigma +459{\sigma }^2-300{\sigma }^3\hfill \\ & -& 438{\sigma }^4+228{\sigma }^5+123{\sigma }^6){\rho }^2+(108\sigma +90{\sigma }^4-108{\sigma }^5-90{\sigma }^6)\rho ={\xi }_3(\sigma ,\rho ).\hfill \end{array}$$

Consider

$$\left\{\begin{array}{c}{\displaystyle \frac{\partial {\xi }_3}{\partial \sigma }}=-504\sigma +468{\sigma }^2+288{\sigma }^3-780{\sigma }^4+210{\sigma }^5+(-72+216\sigma +432{\sigma }^2-432{\sigma }^3-360{\sigma }^4+216{\sigma }^5){\rho }^4\hfill \\ +(-108+144\sigma -288{\sigma }^3+540{\sigma }^4-72{\sigma }^5){\rho }^3+(72+918\sigma -900{\sigma }^2-1752{\sigma }^3+1140{\sigma }^4+738{\sigma }^5){\rho }^2\hfill \\ +(108+300{\sigma }^3-540{\sigma }^4-540{\sigma }^5)\rho =0\hfill \\ {\displaystyle \frac{\partial {\xi }_3}{\partial \rho }}=-90{\sigma }^6-108{\sigma }^5+90{\sigma }^4+108\sigma +(-144-288\sigma +432{\sigma }^2+576{\sigma }^3-432{\sigma }^4-288{\sigma }^5+144{\sigma }^6){\iota }^3+(36\hfill \\ -324\sigma +216{\sigma }^2-216{\sigma }^4+324{\sigma }^5-36{\sigma }^6){\rho }^2+(-288+144\sigma +918{\sigma }^2-600{\sigma }^3-876{\sigma }^4+456{\sigma }^5+246{\sigma }^6)\rho =0\hfill \end{array}\right.$$

By calculating, we acquire $\left\{\begin{array}{c}{\sigma }_{1,1}=1,\hfill \\ {\rho }_{1,1}=1.5139,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,2}=1,\hfill \\ {\rho }_{1,2}=-2.0363,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,3}=1,\hfill \\ {\rho }_{1,3}=-0.4776,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,4}=-1,\hfill \\ {\rho }_{1,4}=0.7445,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,5}=0,\hfill \\ {\rho }_{1,5}=0,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,6}=1.2123,\hfill \\ {\rho }_{1,6}=4.3877,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,7}=6.9938,\hfill \\ {\rho }_{1,7}=0.3396,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,8}=-0.4782,\hfill \\ {\rho }_{1,8}=0.8180,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,9}=-0.6679,\hfill \\ {\rho }_{1,9}=-7.5224,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,10}=-0.8774,\hfill \\ {\rho }_{1,10}=0.2630,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,11}=-0.9372,\hfill \\ {\rho }_{1,11}=-0.6130.\hfill \end{array}\right.$

Thus, there is no optimal point in $(0,1)\times (0,1).$

(8) For $\sigma =0,$

$$\begin{array}{c}\hfill {\xi }_3(0,\rho )=180-144{\rho }^2+12{\rho }^3-36{\rho }^4\le 180.\end{array}$$

(9) For $\sigma =1,$

$$\begin{array}{c}\hfill {\xi }_3(1,\rho )=35.\end{array}$$

(10)For $\rho =0,$

$$\begin{array}{c}\hfill {\xi }_3(\sigma ,0)=35{\sigma }^6-156{\sigma }^5+72{\sigma }^4+156{\sigma }^3-252{\sigma }^2+180\le 180.\end{array}$$

(11) For $\rho =1,$

$$\begin{array}{c}\hfill {\xi }_3(\sigma ,1)=92{\sigma }^6-456{\sigma }^4+387{\sigma }^2+12={ϵ}_3\left(\sigma \right)\le {ϵ}_3\left(0.7071\right)=103.\end{array}$$

Therefore, we acquire

$$\begin{array}{c}\hfill |{\mathcal{H}}_{3,1}(\hslash )|\le {\displaystyle \frac{180}{25920}}={\displaystyle \frac{1}{144}}.\end{array}$$

3. The Second-Order Hankel Determinants of Logarithmic Coefficients for the Class ${\mathcal{C}}_T$

Theorem 3.1 

If $\hslash \in {\mathcal{C}}_T$, then

$$|{\lambda }_1|\le {\displaystyle \frac{1}{4}}|{\lambda }_2|\le {\displaystyle \frac{1}{12}},\left|{\lambda }_3\right|\le {\displaystyle \frac{1}{24}}.$$

The sharpness is given by $\left(15\right)$, $\left(16\right)$ and $\left(17\right).$

Proof 

Let $\hslash \in {\mathcal{C}}_T.$ From $\left(21\right)-\left(24\right)$ and $\left(6\right)$, we have

$$\begin{array}{c}\hfill {\lambda }_1={\displaystyle \frac{1}{4}}{\ell }_1,\end{array}$$

(28)

$$\begin{array}{c}\hfill {\lambda }_2={\displaystyle \frac{1}{12}}{\ell }_2+{\displaystyle \frac{1}{48}}{\ell }_1^2,\end{array}$$

(29)

$$\begin{array}{c}\hfill {\lambda }_3=-{\displaystyle \frac{1}{72}}{\ell }_1^3+{\displaystyle \frac{1}{48}}{\ell }_1{\ell }_2+{\displaystyle \frac{1}{24}}{\ell }_3,\end{array}$$

(30)

$$\begin{array}{c}\hfill {\lambda }_4={\displaystyle \frac{1}{5760}}(144{\ell }_4+72{\ell }_1{\ell }_3+32{\ell }_2^2-140{\ell }_1^2{\ell }_2-25{\ell }_1^4).\end{array}$$

(31)

Applying $\left(13\right)$ to $\left(28\right)$, we get

$$\begin{array}{c}\hfill |{\lambda }_1|\le {\displaystyle \frac{1}{4}}.\end{array}$$

From $\left(29\right)$ and using $\left(14\right)$, we achieve

$$\begin{array}{c}\hfill |{\lambda }_2|={\displaystyle \frac{1}{12}}|{\ell }_2+{\displaystyle \frac{1}{4}}{\ell }_1^2|\le {\displaystyle \frac{1}{12}}max\{1,|{\displaystyle \frac{1}{4}}\left|\right\}={\displaystyle \frac{1}{12}}.\end{array}$$

Applying Lemma 1.2 to the equation $\left(30\right)$, we get

$$\begin{array}{c}\hfill |{\lambda }_3|={\displaystyle \frac{1}{24}}|-{\displaystyle \frac{1}{3}}{\ell }_1^3+{\displaystyle \frac{1}{2}}{\ell }_1{\ell }_2+{\ell }_3|\le {\displaystyle \frac{1}{24}}.\end{array}$$

Theorem 3.2 

If $\hslash \in {\mathcal{C}}_T$, then

$$|{\lambda }_1{\lambda }_3-{\lambda }_2^2|\le {\displaystyle \frac{1}{144}}.$$

The result is sharp, and the equality is achieved from the function ${\hslash }_2\left(\upsilon \right).$

Proof 

Let $\hslash \in {\mathcal{C}}_T.$ From $\left(28\right)$, $\left(29\right)$ and $\left(30\right)$, we yield

$$\begin{array}{ccc}\hfill |{\lambda }_1{\lambda }_3-{\lambda }_2^2|& =& {\displaystyle \frac{1}{2304}}(24{\ell }_1{\ell }_3+4{\ell }_1^2{\ell }_2-16{\ell }_2^2-9{\ell }_1^4).\hfill \end{array}$$

(32)

From $\left(32\right)$ and Lemma 1.5, we yield

$$\begin{array}{ccc}\hfill |{\lambda }_1{\lambda }_3-{\lambda }_2^2|& =& {\displaystyle \frac{1}{2304}}|-24(1-|{\sigma }_1{|}^2\left)\right|{\sigma }_1{|}^2{\sigma }_2^2+24{\sigma }_1(1-|{\sigma }_1{|}^2\left)\right(1-|{\sigma }_2{|}^2){\sigma }_3+4(1-|{\sigma }_1{|}^2){\sigma }_1^2{\sigma }_2-9{\sigma }_1^4-16(1-|{\sigma }_1{|}^2{)}^2{\sigma }_2^2|.\hfill \end{array}$$

By putting $\sigma =|e_1|,|{\sigma }_2|=\rho$, and upon utilizing $|{\sigma }_3|\le 1$, we get

$$\begin{array}{ccc}\hfill |{\lambda }_1{\lambda }_3-{\lambda }_2^2|& \le & {\displaystyle \frac{1}{2304}}[24({\sigma }^2-{\sigma }^4){\rho }^2+24(\sigma -{\sigma }^3)(1-{\rho }^2)+4({\sigma }^2-{\sigma }^4)\rho +9{\sigma }^4+16{(1-{\sigma }^2)}^2{\rho }^2]={\xi }_4(\sigma ,\rho ).\hfill \end{array}$$

Differentiating ${\xi }_4(\sigma ,\rho )$ with respect to $\rho$, we have

$${\displaystyle \frac{\partial {\xi }_4}{\partial \rho }}={\displaystyle \frac{1}{2304}}[4{\sigma }^2(1-{\sigma }^2)+16\rho (1-{\sigma }^2)(2-3\sigma +{\sigma }^2)]\ge 0.$$

Therefore, ${\xi }_4(\sigma ,\rho )$ attains its maximum value at $\rho =1$. Thus, we have

$$\begin{array}{ccc}\hfill |{\lambda }_1{\lambda }_3-{\lambda }_2^2|& \le & {\xi }_4(\sigma ,1)={\displaystyle \frac{1}{2304}}[16-4{\sigma }^2-3{\sigma }^4]\le {\displaystyle \frac{16}{2304}}={\displaystyle \frac{1}{144}}.\hfill \end{array}$$

Theorem 3.3 

If $\hslash \in {\mathcal{C}}_T$, then

$$|{\lambda }_2{\lambda }_4-{\lambda }_3^2|\le {\displaystyle \frac{1}{576}}.$$

The bound is sharp and is achieved from the function ${\hslash }_3\left(\upsilon \right)$.

Proof 

Let $\hslash \in {\mathcal{C}}_T.$ By using $\left(29\right)$,$\left(30\right)$ and $\left(31\right)$, we obtain

$$\begin{array}{ccc}\hfill {\lambda }_2{\lambda }_4-{\lambda }_3^2& =& {\displaystyle \frac{1}{829440}}(-1440{\ell }_3^2-576{\ell }_3{\ell }_2{\ell }_1+1176{\ell }_1^3{\ell }_3-235{\ell }_1^6+384{\ell }_2^3-240{\ell }_1^4{\ell }_2-1944{\ell }_1^2{\ell }_2^2\hfill \\ & & +432{\ell }_1^2{\ell }_4+1728{\ell }_2{\ell }_4).\hfill \end{array}$$

(33)

From $\left(33\right)$ and Lemma 1.5 , we obtain

$$\begin{array}{ccc}\hfill {\lambda }_2{\lambda }_4-{\lambda }_3^2& =& {\displaystyle \frac{1}{829440}}[288{\overline{{\sigma }_1}}^2(1-|{\sigma }_1{|}^2{)}^2{\sigma }_2^4-576\overline{{\sigma }_1}{\sigma }_2^2(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2{)}^2{\sigma }_3-1440(1-|{\sigma }_2{|}^2{)}^2(1-|{\sigma }_1{{|}^2)}^2\hfill \\ & \times & {\sigma }_3^2-235{\sigma }_1^6+432{\sigma }_1^2{\overline{{\sigma }_1}}^2(1-|{\sigma }_1{|}^2){\sigma }_2^3-864{\sigma }_1^2\overline{{\sigma }_1}{\sigma }_2(1-|{\sigma }_1{|}^2\left)\right(1-|{\sigma }_2{|}^2){\sigma }_3-432{\sigma }_1^2\overline{{\sigma }_2}(1-|{\sigma }_1{|}^2)\hfill \\ & \times & (1-|{\sigma }_2{|}^2){\sigma }_3^2+432(1-|{\sigma }_1{|}^2\left)\right(1-|{\sigma }_3{|}^2\left)\right(1-|{\sigma }_2{|}^2){\sigma }_1^2{\sigma }_4-1728(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2{)}^2{\left|{\sigma }_2\right|}^2{\sigma }_3^2\hfill \\ & +& 384{\sigma }_2^3(1-|{\sigma }_1{|}^2{)}^3+1728{\sigma }_2(1-|{\sigma }_3{|}^2\left)\right(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2{)}^2{\sigma }_4-1176{\sigma }_1^3\overline{{\sigma }_1}(1-|{\sigma }_1{|}^2){\sigma }_2^2\hfill \\ & -& 240{\sigma }_1^4(1-|{\sigma }_1{|}^2){\sigma }_2-576{\sigma }_1{\sigma }_2(1-|{\sigma }_1{|}^2{)}^2(1-|{\sigma }_2{|}^2){\sigma }_3+1176{\sigma }_1^3(1-|{\sigma }_1{|}^2\left)\right(1-|{\sigma }_2{|}^2){\sigma }_3\hfill \\ & +& 576|{\sigma }_1{|}^2{\sigma }_2^3(1-|{\sigma }_1{|}^2{)}^2-1944{\sigma }_1^2{\sigma }_2^2(1-|{\sigma }_1{{|}^2)}^2].\hfill \end{array}$$

Hence, we have

$${\lambda }_2{\lambda }_4-{\lambda }_3^2={\displaystyle \frac{1}{829440}}\left[{\zeta }_7\left({\sigma }_1,{\sigma }_2\right)+{\zeta }_8\left({\sigma }_1,{\sigma }_2\right){\sigma }_3+{\zeta }_9\left({\sigma }_1,{\sigma }_2\right){\sigma }_3^2+{\Psi }_3\left({\sigma }_1,{\sigma }_2,{\sigma }_3\right){\sigma }_4\right],$$

where

$$\begin{array}{ccc}\hfill {\zeta }_7\left({\sigma }_1,{\sigma }_2\right)& =& \left(1-|{\sigma }_1{|}^2\right)\left[(-1944{\sigma }_1^2{\sigma }_2^2+192|{\sigma }_1{|}^2{\sigma }_2^3+384{\sigma }_2^3+288{\sigma }_2^4{\overline{{\sigma }_1}}^2\left)\right(1-|{\sigma }_1{|}^2)+432{\sigma }_1^2{\overline{{\sigma }_1}}^2{\sigma }_2^3\right.\hfill \\ & & \left.-1176\overline{{\sigma }_1}{\sigma }_1^3{\sigma }_2^2-240{\sigma }_1^4{\sigma }_2\right]-235{\sigma }_1^6,\hfill \\ \hfill {\zeta }_8\left({\sigma }_1,{\sigma }_2\right)& =& (1-|{\sigma }_1{|}^2)\left(1-|{\sigma }_2{|}^2\right)\left[(-576{\sigma }_1{\sigma }_2-576{\sigma }_2^2\overline{{\sigma }_1})(1-|{\sigma }_1{|}^2)-864{\sigma }_1^2\overline{{\sigma }_1}{\sigma }_2+1176{\sigma }_1^3\right],\hfill \\ \hfill {\zeta }_9\left({\sigma }_1,{\sigma }_2\right)& =& (1-|{\sigma }_2{|}^2)\left(1-|{\sigma }_1{|}^2\right)\left[(-1440-288|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2)-432{\sigma }_1^2\overline{{\sigma }_2}\right],\hfill \\ \hfill {\Psi }_3\left({\sigma }_1,{\sigma }_2,{\sigma }_3\right)& =& (1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_3{|}^2\left)\right(1-|{\sigma }_1{|}^2)\left[1728{\sigma }_2(1-|{\sigma }_1{|}^2)+432{\sigma }_1^2\right].\hfill \end{array}$$

Let $\iota =|{\sigma }_3|,|{\sigma }_1|=\sigma ,\rho =|{\sigma }_2|.$ By noting that $|{\sigma }_4|\le 1$, we have

$$\begin{array}{ccc}\hfill \left|{\lambda }_2{\lambda }_4-{\lambda }_3^2\right|& \le & {\displaystyle \frac{1}{829440}}\left(\left|{\zeta }_7\left({\sigma }_1,{\sigma }_2\right)\right|+\left|{\zeta }_8\left({\sigma }_1,{\sigma }_2\right)\right|\iota +\left|{\zeta }_9\left({\sigma }_1,{\sigma }_2\right)\right|{\iota }^2+\left|{\Psi }_3\left({\sigma }_1,{\sigma }_2,{\sigma }_3\right)\right|\right)\hfill \\ & \le & {\displaystyle \frac{1}{829440}}{\phi }_3\left(\sigma ,\rho ,\iota \right),\hfill \end{array}$$

where

$$\begin{array}{c}\hfill {\phi }_3\left(\sigma ,\rho ,\iota \right)={\kappa }_9\left(\sigma ,\rho \right)+{\kappa }_{10}\left(\sigma ,\rho \right)\iota +{\kappa }_{11}\left(\sigma ,\rho \right){\iota }^2+{\kappa }_{12}\left(\sigma ,\rho \right)(1-{\iota }^2),\end{array}$$

with

$$\begin{array}{ccc}\hfill {\kappa }_9\left(\sigma ,\rho \right)& =& \left(1-{\sigma }^2\right)\left[(288{\rho }^4{\sigma }^2+192{\sigma }^2{\rho }^3+384{\rho }^3+1944{\sigma }^2{\rho }^2)(1-{\sigma }^2)+432{\sigma }^4{\rho }^3+1176{\sigma }^4{\rho }^2+240{\sigma }^4\rho \right]+235{\sigma }^6,\hfill \\ \hfill {\kappa }_{10}\left(\sigma ,\rho \right)& =& \left(1-{\sigma }^2\right)(1-{\rho }^2)\left[(576\sigma \rho +576{\rho }^2\sigma )(1-{\sigma }^2)+864{\sigma }^3\rho +1176{\sigma }^3\right],\hfill \\ \hfill {\kappa }_{11}\left(\sigma ,\rho \right)& =& \left(1-{\sigma }^2\right)(1-{\rho }^2)\left[(1-{\sigma }^2)(1440+288{\rho }^2)+432{\sigma }^2\rho \right],\hfill \\ \hfill {\kappa }_{12}\left(\sigma ,\rho \right)& =& (1-{\sigma }^2)(1-{\rho }^2)\left[1728\rho (1-{\sigma }^2)+432{\sigma }^2\right].\hfill \end{array}$$

It is clear that

$$\begin{array}{ccc}\hfill {\kappa }_{11}(\sigma ,\rho ))& \le & (1-{\rho }^2)(1-{\sigma }^2)[(1440+288{\rho }^2)(1-{\sigma }^2)+432{\sigma }^2]={\pi }_3(\sigma ,\rho ).\hfill \end{array}$$

Hence, we achieve

$$\begin{array}{c}\hfill {\phi }_3\left(\sigma ,\rho ,\iota \right)\le {\kappa }_9\left(\sigma ,\rho \right)+{\kappa }_{10}\left(\sigma ,\rho \right)\iota +{\pi }_3\left(\sigma ,\rho \right){\iota }^2+{\kappa }_{12}\left(\sigma ,\rho \right)(1-{\iota }^2)={\mu }_3\left(\sigma ,\rho ,\iota \right),\end{array}$$

Differentiating ${\mu }_3\left(\sigma ,\rho ,\iota \right)$ with respect to $\iota$, we have

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\partial {\mu }_3}{\partial \iota }}& =& {\kappa }_{10}(\sigma ,\rho )+2\iota \left[{\pi }_3(\sigma ,\rho )-{\kappa }_{12}(\sigma ,\rho )\right]\hfill \\ & =& {\kappa }_{10}(\sigma ,\rho )+576{(1-{\sigma }^2)}^2(5-6\rho +{\rho }^2)(1-{\rho }^2)\iota \ge 0.\hfill \end{array}$$

So, ${\mu }_3\left(\sigma ,\rho ,\iota \right)$ attains its maximum value at $\iota =1$. Therefore, we have

$$\begin{array}{ccc}\hfill {\phi }_3\left(\sigma ,\rho ,\iota \right)& \le & {\mu }_3\left(\sigma ,\rho ,\iota \right)\le {\mu }_3\left(\sigma ,\rho ,1\right)={\kappa }_9(\sigma ,\rho )+{\kappa }_{10}(\sigma ,\rho )+{\pi }_3(\sigma ,\rho )\hfill \\ & =& 1440-2448{\sigma }^2+1176{\sigma }^3+1008{\sigma }^4-1176{\sigma }^5+235{\sigma }^6+(-288-576\sigma +864{\sigma }^2+1152{\sigma }^3\hfill \\ & -& 864{\sigma }^4-576{\sigma }^5+288{\sigma }^6){\rho }^4+(384-576\sigma -576{\sigma }^2+288{\sigma }^3+432{\sigma }^4+288{\sigma }^5-240{\sigma }^6){\rho }^3\hfill \\ & +& (-1152+576\sigma +3816{\sigma }^2-2328{\sigma }^3-3432{\sigma }^4+1752{\sigma }^5+768{\sigma }^6){\rho }^2+(576\sigma -288{\sigma }^3+240{\sigma }^4\hfill \\ & -& 288{\sigma }^5-240{\sigma }^6)\rho ={\xi }_5(\sigma ,\rho ).\hfill \end{array}$$

Consider

$$\left\{\begin{array}{c}{\displaystyle \frac{\partial {\xi }_5}{\partial \sigma }}=-4896\sigma +3528{\sigma }^2+4032{\sigma }^3-5880{\sigma }^4+1410{\sigma }^5+(-576+1728\sigma +3456{\sigma }^2-3456{\sigma }^3-2880{\sigma }^4\hfill \\ +1728{\sigma }^5){\rho }^4+(-576-1152\sigma +864{\sigma }^2+1728{\sigma }^3+1440{\sigma }^4-1440{\sigma }^5){\rho }^3+(576+7632\sigma -6984{\sigma }^2-13728{\sigma }^3\hfill \\ +8760{\sigma }^4+4608{\sigma }^5){\rho }^2+(576-864{\sigma }^2+960{\sigma }^3-1440{\sigma }^4-1440{\sigma }^5)\rho =0\hfill \\ {\displaystyle \frac{\partial {\xi }_5}{\partial \rho }}=-240{\sigma }^6-288{\sigma }^5+240{\sigma }^4-288{\sigma }^3+576\sigma +(-1152-2304\sigma +3456{\sigma }^2+4608{\sigma }^3-3456{\sigma }^4-2304{\sigma }^5\hfill \\ +1152{\sigma }^6){\rho }^3+(1152-1728\sigma -1728{\sigma }^2+864{\sigma }^3+1296{\sigma }^4+864{\sigma }^5-720{\sigma }^6){\rho }^2+(-2304+1152\sigma +7632{\sigma }^2\hfill \\ -4656{\sigma }^3-6864{\sigma }^4+3504{\sigma }^5+1536{\sigma }^6)\rho =0\hfill \end{array}\right.$$

By calculating, we achieve $\left\{\begin{array}{c}{\sigma }_{1,1}=1,\hfill \\ {\rho }_{1,1}=1.5403,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,2}=1,\hfill \\ {\rho }_{1,2}=-1.7762,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,3}=1,\hfill \\ {\rho }_{1,3}=-0.7640,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,4}=-1,\hfill \\ {\rho }_{1,4}=0.8112,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,5}=0,\hfill \\ {\rho }_{1,5}=0,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,6}=1.2116,\hfill \\ {\rho }_{1,6}=2.4301,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,7}=3.6137,\hfill \\ {\rho }_{1,7}=0.1634,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,8}=-0.4476,\hfill \\ {\rho }_{1,8}=0.9032,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,9}=-0.7728,\hfill \\ {\rho }_{1,9}=-4.8254,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,10}=-0.8740,\hfill \\ {\rho }_{1,10}=0.1952,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,11}=-0.9536,\hfill \\ {\rho }_{1,11}=-1.1010.\hfill \end{array}\right.$

Thus, there are no optimal points in $(0,1)\times (0,1).$

(12) For $\sigma =0,$

$$\begin{array}{c}\hfill {\xi }_5(0,\rho )=1440-1152{\rho }^2+384{\rho }^3-288{\rho }^4\le 1440.\end{array}$$

(13) For $\sigma =1,$

$$\begin{array}{c}\hfill {\xi }_5(1,\rho )=235.\end{array}$$

(14)For $\rho =0,$

$$\begin{array}{c}\hfill {\xi }_5(\sigma ,0)=235{\sigma }^6-1176{\sigma }^5+1008{\sigma }^4+1176{\sigma }^3-2448{\sigma }^2+1440\le 1440.\end{array}$$

(15) For $\rho =1,$

$$\begin{array}{c}\hfill {\xi }_5(\sigma ,1)=811{\sigma }^6-2616{\sigma }^4+1656{\sigma }^2+384={ϵ}_4\left(\sigma \right)\le {ϵ}_4\left(0.6210\right)=680.0852.\end{array}$$

Therefore, we have

$$\begin{array}{c}\hfill |{\lambda }_2{\lambda }_4-{\lambda }_3^2|\le {\displaystyle \frac{1440}{829440}}={\displaystyle \frac{1}{576}}.\end{array}$$

4. The Third-Order Hankel Determinants of Inverse Functions for the Classes ${\mathcal{C}}_T$ and ${\mathcal{S}}_T^{*}$

Theorem 4.1 

If $\hslash \in {\mathcal{C}}_T$, then

$$|{\mathcal{H}}_{3,1}\left({\hslash }^{-1}\right)|\le {\displaystyle \frac{1}{144}}.$$

The outcome is the best possible and is obtained by $\left(17\right)$.

Proof 

Let $\hslash \in {\mathcal{C}}_T.$ From $\left(21\right)-\left(24\right)$ and $\left(11\right)$, we have

$$\begin{array}{ccc}\hfill {\mathcal{H}}_{3,1}\left({\hslash }^{-1}\right)& =& {\displaystyle \frac{1}{25920}}(216{\ell }_2{\ell }_4+108{\ell }_3{\ell }_2{\ell }_1-108{\ell }_4{\ell }_1^2+156{\ell }_3{\ell }_1^3-132{\ell }_2^3-180{\ell }_3^2\hfill \\ & & -99{\ell }_1^2{\ell }_2^2-20{\ell }_1^6).\hfill \end{array}$$

(34)

From $\left(34\right)$ and Lemma 1.5, we get

$$\begin{array}{ccc}\hfill {\mathcal{H}}_{3,1}\left({\hslash }^{-1}\right)& =& {\displaystyle \frac{1}{25920}}[-20{\sigma }_1^6-216{\sigma }_3^2|{\sigma }_2{|}^2(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2{)}^2+36{\sigma }_2^4{\overline{{\sigma }_1}}^2(1-|{\sigma }_1{|}^2{)}^2+216(1-|{\sigma }_3{|}^2)\hfill \\ & \times & (1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2{)}^2{\sigma }_2{\sigma }_4-72{\sigma }_3{\sigma }_2^2\overline{{\sigma }_1}(1-|{\sigma }_1{|}^2{)}^2(1-|{\sigma }_2{|}^2)-108{\sigma }_1^2{\overline{{\sigma }_1}}^2(1-|{\sigma }_1{|}^2){\sigma }_2^3\hfill \\ & +& 216{\sigma }_3{\sigma }_2{\sigma }_1^2\overline{{\sigma }_1}(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2)-132{\sigma }_2^3(1-|{\sigma }_1{|}^2{)}^3-108(1-|{\sigma }_3{|}^2\left)\right(1-|{\sigma }_1{|}^2\left)\right(1-|{\sigma }_2{|}^2){\sigma }_1^2{\sigma }_4\hfill \\ & -& 99{\sigma }_2^2{\sigma }_1^2(1-|{\sigma }_1{|}^2{)}^2-108(1-|{\sigma }_1{|}^2{)}^2|{\sigma }_1{|}^2{\sigma }_2^3-180{\sigma }_3^2(1-|{\sigma }_1{|}^2{)}^2(1-|{\sigma }_2{|}^2{)}^2-156{\sigma }_2^2\overline{{\sigma }_1}{\sigma }_1^3(1-|{\sigma }_1{|}^2)\hfill \\ & +& 156{\sigma }_3{\sigma }_1^3(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2)+108(1-|{\sigma }_1{|}^2{)}^2(1-|{\sigma }_2{|}^2){\sigma }_1{\sigma }_2{\sigma }_3+108{\sigma }_3^2\overline{{\sigma }_2}{\sigma }_1^2(1-|{\sigma }_1{|}^2\left)\right(1-|{\sigma }_2{|}^2)].\hfill \end{array}$$

Thus, we get

$${\mathcal{H}}_{3,1}\left({\hslash }^{-1}\right)={\displaystyle \frac{1}{25920}}\left[{\zeta }_{10}\left({\sigma }_1,{\sigma }_2\right)+{\zeta }_{11}\left({\sigma }_1,{\sigma }_2\right){\sigma }_3+{\zeta }_{12}\left({\sigma }_1,{\sigma }_2\right){\sigma }_3^2+{\Psi }_4\left({\sigma }_1,{\sigma }_2,{\sigma }_3\right){\sigma }_4\right],$$

where

$$\begin{array}{ccc}\hfill {\zeta }_{10}\left({\sigma }_1,{\sigma }_2\right)& =& \left(1-|{\sigma }_1{|}^2\right)\left[(-99{\sigma }_1^2{\sigma }_2^2+24|{\sigma }_1{|}^2{\sigma }_2^3-132{\sigma }_2^3+36{\overline{{\sigma }_1}}^2{\sigma }_2^4\left)\right(1-|{\sigma }_1{|}^2)-108{\sigma }_2^3{\overline{{\sigma }_1}}^2{\sigma }_1^2\right.\hfill \\ & & \left.-156{\sigma }_2^2{\sigma }_1^3\overline{{\sigma }_1}\right]-20{\sigma }_1^6,\hfill \\ \hfill {\zeta }_{11}\left({\sigma }_1,{\sigma }_2\right)& =& (1-|{\sigma }_1{|}^2)\left(1-|{\sigma }_2{|}^2\right)\left[(-72\overline{{\sigma }_1}{\sigma }_2^2+108{\sigma }_2{\sigma }_1)(1-|{\sigma }_1{|}^2)+156{\sigma }_1^3+216{\sigma }_2\overline{{\sigma }_1}{\sigma }_1^2\right],\hfill \\ \hfill {\zeta }_{12}\left({\sigma }_1,{\sigma }_2\right)& =& (1-|{\sigma }_2{|}^2)\left(1-|{\sigma }_1{|}^2\right)\left[(-36|{\sigma }_2{|}^2-108\left)\right(1-|{\sigma }_1{|}^2)+108\overline{{\sigma }_2}{\sigma }_1^2\right],\hfill \\ \hfill {\Psi }_4\left({\sigma }_1,{\sigma }_2,{\sigma }_3\right)& =& (1-|{\sigma }_3{|}^2\left)\right(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2)\left[-108{\sigma }_1^2+216(1-|{\sigma }_1{|}^2){\sigma }_2\right].\hfill \end{array}$$

Using the triangle inequality and let $|{\sigma }_1|=\sigma ,\rho =|{\sigma }_2|,$$|{\sigma }_3|=\iota$, and using $|{\sigma }_4|\le 1$, we yield

$$\begin{array}{ccc}\hfill \left|{\mathcal{H}}_{3,1}\left({\hslash }^{-1}\right)\right|& \le & {\displaystyle \frac{1}{25920}}\left(\left|{\zeta }_{10}\left({\sigma }_1,{\sigma }_2\right)\right|+\left|{\zeta }_{11}\left({\sigma }_1,{\sigma }_2\right)\right|\iota +\left|{\zeta }_{12}\left({\sigma }_1,{\sigma }_2\right)\right|{\iota }^2+\left|{\Psi }_4\left({\sigma }_1,{\sigma }_2,{\sigma }_3\right)\right|\right)\hfill \\ & \le & {\displaystyle \frac{1}{25920}}{\phi }_4\left(\sigma ,\rho ,\iota \right),\hfill \end{array}$$

where

$$\begin{array}{c}\hfill {\phi }_4\left(\sigma ,\rho ,\iota \right)={\kappa }_{13}\left(\sigma ,\rho \right)+{\kappa }_{14}\left(\sigma ,\rho \right)\iota +{\kappa }_{15}\left(\sigma ,\rho \right){\iota }^2+{\kappa }_{16}\left(\sigma ,\rho \right)(1-{\iota }^2),\end{array}$$

with

$$\begin{array}{ccc}\hfill {\kappa }_{13}\left(\sigma ,\rho \right)& =& \left(1-{\sigma }^2\right)\left[(36{\sigma }^2{\rho }^4+24{\rho }^3{\sigma }^2+132{\rho }^3+99{\rho }^2{\sigma }^2)(1-{\sigma }^2)+108{\rho }^3{\sigma }^4+156{\rho }^2{\sigma }^4\right]+20{\sigma }^6,\hfill \\ \hfill {\kappa }_{14}\left(\sigma ,\rho \right)& =& \left(1-{\sigma }^2\right)(1-{\rho }^2)\left[(108\rho \sigma +72\sigma {\rho }^2)(1-{\sigma }^2)+156{\sigma }^3+216\rho {\sigma }^3\right],\hfill \\ \hfill {\kappa }_{15}\left(\sigma ,\rho \right)& =& \left(1-{\sigma }^2\right)(1-{\rho }^2)\left[108\rho {\sigma }^2+(36{\rho }^2+180)(1-{\sigma }^2)\right],\hfill \\ \hfill {\kappa }_{16}\left(\sigma ,\rho \right)& =& (1-{\sigma }^2)(1-{\rho }^2)\left[108{\sigma }^2+216(1-{\sigma }^2)\rho \right].\hfill \end{array}$$

Note that

$$\begin{array}{ccc}\hfill {\kappa }_{15}(\sigma ,\rho ))& \le & (1-{\rho }^2)(1-{\sigma }^2)[(180+36{\rho }^2)(1-{\sigma }^2)+108{\sigma }^2]={\pi }_4(\sigma ,\rho ).\hfill \end{array}$$

Hence, we yield

$$\begin{array}{c}\hfill {\phi }_4\left(\sigma ,\rho ,\iota \right)\le {\kappa }_{13}\left(\sigma ,\rho \right)+{\kappa }_{14}\left(\sigma ,\rho \right)\iota +{\pi }_4\left(\sigma ,\rho \right){\iota }^2+{\kappa }_{16}\left(\sigma ,\rho \right)(1-{\iota }^2)={\mu }_4\left(\sigma ,\rho ,\iota \right),\end{array}$$

Differentiating ${\mu }_4\left(\sigma ,\rho ,\iota \right)$ with respect to $\iota$, we achieve

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\partial {\mu }_4}{\partial \iota }}& =& {\kappa }_{14}(\sigma ,\rho )+2\iota \left[{\pi }_4(\sigma ,\rho )-{\kappa }_{16}(\sigma ,\rho )\right]\hfill \\ & =& {\kappa }_{14}(\sigma ,\rho )+72\iota (5-6\rho +{\rho }^2)(1-{\rho }^2){(1-{\sigma }^2)}^2\ge 0.\hfill \end{array}$$

Therefore, ${\mu }_4\left(\sigma ,\rho ,\iota \right)$ attains its maximum value at $\iota =1$. Thus, we have

$$\begin{array}{ccc}\hfill {\phi }_4\left(\sigma ,\rho ,\iota \right)& \le & {\mu }_4\left(\sigma ,\rho ,\iota \right)\le {\mu }_4\left(\sigma ,\rho ,1\right)={\kappa }_{13}\left(\sigma ,\rho \right)+{\kappa }_{14}\left(\sigma ,\rho \right)+{\pi }_4\left(\sigma ,\rho \right)\hfill \\ & =& 20{\sigma }^6-156{\sigma }^5+72{\sigma }^4+156{\sigma }^3-252{\sigma }^2+180+(36{\sigma }^6-72{\sigma }^5-108{\sigma }^4+144{\sigma }^3+108{\sigma }^2\hfill \\ & -& 72\sigma -36){\rho }^4+(-84{\sigma }^6+108{\sigma }^5+192{\sigma }^4-240{\sigma }^2-108\sigma +132){\rho }^3+(-57{\sigma }^6+228{\sigma }^5\hfill \\ & -& 78{\sigma }^4-300{\sigma }^3+279{\sigma }^2+72\sigma -144){\rho }^2+(-108{\sigma }^5+108\sigma )\rho ={\xi }_6(\sigma ,\rho ).\hfill \end{array}$$

The critical points of ${\xi }_6(\sigma ,\rho )$ satisfy the conditions

$$\left\{\begin{array}{c}{\displaystyle \frac{\partial {\xi }_6}{\partial \sigma }}=120{\sigma }^5-780{\sigma }^4+288{\sigma }^3+468{\sigma }^2-504\sigma +(216{\sigma }^5-360{\sigma }^4-432{\sigma }^3+432{\sigma }^2+216\sigma -72){\rho }^4+(-504{\sigma }^5\hfill \\ +540{\sigma }^4+768{\sigma }^3-480\sigma -108){\rho }^3+(-342{\sigma }^5+1140{\sigma }^4-312{\sigma }^3-900{\sigma }^2+558\sigma +72){\rho }^2+(-540{\sigma }^4+108)\rho =0\hfill \\ {\displaystyle \frac{\partial {\xi }_6}{\partial \rho }}=108\sigma -108{\sigma }^5+(144{\sigma }^6-288{\sigma }^5-432{\sigma }^4+576{\sigma }^3+432{\sigma }^2-288\sigma -144){\rho }^3+(-252{\sigma }^6+324{\sigma }^5+576{\sigma }^4\hfill \\ -720{\sigma }^2-324\sigma +396){\rho }^2+(-114{\sigma }^6+456{\sigma }^5-156{\sigma }^4-600{\sigma }^3+558{\sigma }^2+144\sigma -288)\rho =0\hfill \end{array}\right.$$

By calculating, we acquire $\left\{\begin{array}{c}{\sigma }_{1,1}=1,\hfill \\ {\rho }_{1,1}=1.3977,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,2}=1,\hfill \\ {\rho }_{1,2}=-1.4919,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,3}=1,\hfill \\ {\rho }_{1,3}=-0.9058,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,4}=-1,\hfill \\ {\rho }_{1,4}=0.7780,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,5}=-1,\hfill \\ {\rho }_{1,5}=-0.9625,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,6}=-1,\hfill \\ {\rho }_{1,6}=-0.4452,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,7}=0,\hfill \\ {\rho }_{1,7}=0,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,8}=1.0582,\hfill \\ {\rho }_{1,8}=-1.0906,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,9}=1.1417,\hfill \\ {\rho }_{1,9}=-1.0043,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,10}=2.8187,\hfill \\ {\rho }_{1,10}=4.2318,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,11}=-0.0768,\hfill \\ {\rho }_{1,11}=1.0373,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,12}=-0.7400,\hfill \\ {\rho }_{1,12}=-9.9597,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,13}=-0.8874,\hfill \\ {\rho }_{1,13}=0.3561,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,14}=-1.2181,\hfill \\ {\rho }_{1,14}=-0.8576,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,15}=-1.4078,\hfill \\ {\rho }_{1,15}=4.5937.\hfill \end{array}\right.$

Thus, there is no solution of this system in $(0,1)\times (0,1).$

(16) For $\sigma =1,$

$$\begin{array}{c}\hfill {\xi }_6(1,\rho )=20.\end{array}$$

(17) For $\sigma =0,$

$$\begin{array}{c}\hfill {\xi }_6(0,\rho )=180-144{\rho }^2+132{\rho }^3-36{\rho }^4\le 180.\end{array}$$

(18) For $\rho =0,$

$$\begin{array}{c}\hfill {\xi }_6(\sigma ,0)=180-252{\sigma }^2+156{\sigma }^3+72{\sigma }^4-156{\sigma }^5+20{\sigma }^6\le 180.\end{array}$$

(19) For $\rho =1$

$$\begin{array}{c}\hfill {\xi }_6(\sigma ,1)=132-105{\sigma }^2+78{\sigma }^4-85{\sigma }^6\le 132.\end{array}$$

Thus, we yield

$$\begin{array}{c}\hfill |{\mathcal{H}}_{3,1}\left({\hslash }^{-1}\right)|\le {\displaystyle \frac{180}{25920}}={\displaystyle \frac{1}{144}}.\end{array}$$

Theorem 4.2 

If $\hslash \in {\mathcal{S}}_T^{*}$, then

$$|{\mathcal{H}}_{3,1}\left({\hslash }^{-1}\right)|\le {\displaystyle \frac{3}{16}}.$$

The bound is sharp for the function ${\hslash }_5\left(\upsilon \right)$.

Proof 

Let $\hslash \in {\mathcal{S}}_T^{*},$ then there exists an analytic function $\zeta \left(\upsilon \right)$, such that

$$\begin{array}{c}\hfill {\displaystyle \frac{\upsilon {\hslash }^{\prime }\left(\upsilon \right)}{\hslash \left(\upsilon \right)}}=1+arctan\left(\zeta \left(\upsilon \right)\right).\end{array}$$

(35)

From $\left(1\right)$, we have

$$\begin{array}{cc}\hfill {\displaystyle \frac{\upsilon {\hslash }^{\prime }\left(\upsilon \right)}{\hslash \left(\upsilon \right)}}& =1+t_2\upsilon +(2t_3-t_2^2){\upsilon }^2+(t_2^3-3t_2{t}_3+3t_4){\upsilon }^3+(-t_2^4-2t_3^2+4t_2^2{t}_3\hfill \\ \hfill & -4t_2{t}_4+4t_5){\upsilon }^4+···.\hfill \end{array}$$

(36)

From $\left(36\right)$ and $\left(20\right)$, we have

$$\left\{\begin{array}{c}{t}_2={\ell }_1,\hfill \\ t_3={\displaystyle \frac{1}{2}}({\ell }_2+{\ell }_1^2),\hfill \\ t_4={\displaystyle \frac{1}{18}}(6{\ell }_3+9{\ell }_1{\ell }_2+{\ell }_1^3),\hfill \\ t_5={\displaystyle \frac{1}{72}}(18{\ell }_4+24{\ell }_3{\ell }_1+9{\ell }_2^2-5{\ell }_1^4).\hfill \end{array}\right.$$

(37)

From $\left(11\right)$ and $\left(37\right)$, we have

$$\begin{array}{ccc}\hfill {\mathcal{H}}_{3,1}\left({\hslash }^{-1}\right)& =& {\displaystyle \frac{1}{1296}}(-243{\ell }_2^3-144{\ell }_3^2+216{\ell }_3{\ell }_2{\ell }_1+168{\ell }_1^3{\ell }_3+113{\ell }_1^6-369{\ell }_1^4{\ell }_2+81{\ell }_1^2{\ell }_2^2\hfill \\ & & -162{\ell }_1^2{\ell }_4+162{\ell }_2{\ell }_4).\hfill \end{array}$$

(38)

From $\left(38\right)$ and Lemma 1.5 , we obtain

$$\begin{array}{ccc}\hfill H_{3,1}\left({\hslash }^{-1}\right)& =& {\displaystyle \frac{1}{1296}}[18{\sigma }_2^4{\overline{{\sigma }_1}}^2(1-|{\sigma }_1{|}^2{)}^2-36{\sigma }_3\overline{{\sigma }_1}{\sigma }_2^2(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2{)}^2-144{\sigma }_3^2(1-|{\sigma }_2{|}^2{)}^2(1-|{\sigma }_1{{|}^2)}^2\hfill \\ & +& 113{\sigma }_1^6-162{\sigma }_2^3{\overline{{\sigma }_1}}^2{\sigma }_1^2(1-|{\sigma }_1{|}^2)+324{\sigma }_3\overline{{\sigma }_1}{\sigma }_2{\sigma }_1^2(1-|{\sigma }_1{|}^2\left)\right(1-|{\sigma }_2{|}^2)+162{\sigma }_3^2\overline{{\sigma }_2}{\sigma }_1^2(1-|{\sigma }_1{|}^2)\hfill \\ & \times & (1-|{\sigma }_2{|}^2)-162{\sigma }_1^2{\sigma }_4(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2\left)\right(1-|{\sigma }_3{|}^2)-162|{\sigma }_2{|}^2{\sigma }_3^2(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{{|}^2)}^2\hfill \\ & -& 243{\sigma }_2^3(1-|{\sigma }_1{|}^2{)}^3+162{\sigma }_4{\sigma }_2(1-|{\sigma }_3{|}^2\left)\right(1-|{\sigma }_1{|}^2{)}^2(1-|{\sigma }_2{|}^2)-168{\sigma }_2^2\overline{{\sigma }_1}{\sigma }_1^3(1-|{\sigma }_1{|}^2)\hfill \\ & -& 369{\sigma }_2{\sigma }_1^4(1-|{\sigma }_1{|}^2)+216{\sigma }_3{\sigma }_2{\sigma }_1(1-|{\sigma }_1{|}^2{)}^2(1-|{\sigma }_2{|}^2)+168{\sigma }_3{\sigma }_1^3(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2)\hfill \\ & -& 216{\sigma }_2^3(1-|{\sigma }_1{|}^2{)}^2|{\sigma }_1{|}^2+81{\sigma }_2^2{\sigma }_1^2(1-|{\sigma }_1{{|}^2)}^2].\hfill \end{array}$$

Hence, we acquire

$$H_{3,1}\left({\hslash }^{-1}\right)={\displaystyle \frac{1}{1296}}\left[{\zeta }_{13}\left({\sigma }_1,{\sigma }_2\right)+{\zeta }_{14}\left({\sigma }_1,{\sigma }_2\right){\sigma }_3+{\zeta }_{15}\left({\sigma }_1,{\sigma }_2\right){\sigma }_3^2+{\Psi }_5\left({\sigma }_1,{\sigma }_2,{\sigma }_3\right){\sigma }_4\right],$$

where

$$\begin{array}{ccc}\hfill {\zeta }_{13}\left({\sigma }_1,{\sigma }_2\right)& =& \left(1-|{\sigma }_1{|}^2\right)\left\{[81{\sigma }_1^2{\sigma }_2^2-(243-27|{\sigma }_1{|}^2){\sigma }_2^3+18{\sigma }_2^4{\overline{{\sigma }_1}}^2\left]\right(1-|{\sigma }_1{|}^2)-162{\sigma }_1^2{\overline{{\sigma }_1}}^2{\sigma }_2^3\right.\hfill \\ & & \left.-168\overline{{\sigma }_1}{\sigma }_1^3{\sigma }_2^2-369{\sigma }_1^4{\sigma }_2\right\}+113{\sigma }_1^6,\hfill \\ \hfill {\zeta }_{14}\left({\sigma }_1,{\sigma }_2\right)& =& (1-|{\sigma }_1{|}^2)\left(1-|{\sigma }_2{|}^2\right)\left[(216{\sigma }_1{\sigma }_2-36{\sigma }_2^2\overline{{\sigma }_1})(1-|{\sigma }_1{|}^2)+324{\sigma }_1^2\overline{{\sigma }_1}{\sigma }_2+168{\sigma }_1^3\right],\hfill \\ \hfill {\zeta }_{15}\left({\sigma }_1,{\sigma }_2\right)& =& (1-|{\sigma }_2{|}^2)\left(1-|{\sigma }_1{|}^2\right)\left[(-144-18|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2)+162{\sigma }_1^2\overline{{\sigma }_2}\right],\hfill \\ \hfill {\Psi }_5\left({\sigma }_1,{\sigma }_2,{\sigma }_3\right)& =& (1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2\left)\right(1-|{\sigma }_3{|}^2)\left[162(1-|{\sigma }_1{|}^2){\sigma }_2-162{\sigma }_1^2\right].\hfill \end{array}$$

By writing $|{\sigma }_3|=\iota ,\rho =|{\sigma }_2|,$$|{\sigma }_1|=\sigma$ and using $|{\sigma }_4|\le 1$, we yield

$$\begin{array}{ccc}\hfill \left|{\mathcal{H}}_{3,1}\left({\hslash }^{-1}\right)\right|& \le & {\displaystyle \frac{1}{1296}}\left(\left|{\zeta }_{13}\left({\sigma }_1,{\sigma }_2\right)\right|+\left|{\zeta }_{14}\left({\sigma }_1,{\sigma }_2\right)\right|\iota +\left|{\zeta }_{15}\left({\sigma }_1,{\sigma }_2\right)\right|{\iota }^2+\left|{\Psi }_5\left({\sigma }_1,{\sigma }_2,{\sigma }_3\right)\right|\right)\hfill \\ & \le & {\displaystyle \frac{1}{1296}}{\phi }_5\left(\sigma ,\rho ,\iota \right),\hfill \end{array}$$

where

$$\begin{array}{c}\hfill {\phi }_5\left(\sigma ,\rho ,\iota \right)={\kappa }_{17}\left(\sigma ,\rho \right)+{\kappa }_{18}\left(\sigma ,\rho \right)\iota +{\kappa }_{19}\left(\sigma ,\rho \right){\iota }^2+{\kappa }_{20}\left(\sigma ,\rho \right)(1-{\iota }^2),\end{array}$$

with

$$\begin{array}{ccc}\hfill {\kappa }_{17}\left(\sigma ,\rho \right)& =& \left(1-{\sigma }^2\right)\left[(18{\sigma }^2{\rho }^4-27{\rho }^3{\sigma }^2+243{\rho }^3+81{\rho }^2{\sigma }^2)(1-{\sigma }^2)+162{\rho }^3{\sigma }^4+168{\rho }^2{\sigma }^4+369{\sigma }^4\rho \right]+113{\sigma }^6,\hfill \\ \hfill {\kappa }_{18}\left(\sigma ,\rho \right)& =& \left(1-{\sigma }^2\right)(1-{\rho }^2)\left[(216\rho \sigma +36\sigma {\rho }^2)(1-{\sigma }^2)+168{\sigma }^3+324\rho {\sigma }^3\right],\hfill \\ \hfill {\kappa }_{19}\left(\sigma ,\rho \right)& =& \left(1-{\sigma }^2\right)(1-{\rho }^2)\left[162\rho {\sigma }^2+(18{\rho }^2+144)(1-{\sigma }^2)\right],\hfill \\ \hfill {\kappa }_{20}\left(\sigma ,\rho \right)& =& (1-{\sigma }^2)(1-{\rho }^2)\left[162{\sigma }^2+162\rho (1-{\sigma }^2)\right].\hfill \end{array}$$

It is clear that for $\rho \in [0,1],$

$$\begin{array}{ccc}\hfill {\kappa }_{19}\left(\sigma ,\rho \right)& \le & (1-{\rho }^2)(1-{\sigma }^2)[(144+18{\rho }^2)(1-{\sigma }^2)+162{\sigma }^2]={\pi }_5(\sigma ,\rho ).\hfill \end{array}$$

Thus, we have

$$\begin{array}{c}\hfill {\phi }_5\left(\sigma ,\rho ,\iota \right)\le {\kappa }_{17}\left(\sigma ,\rho \right)+{\kappa }_{18}\left(\sigma ,\rho \right)\iota +{\pi }_5\left(\sigma ,\rho \right){\iota }^2+{\kappa }_{20}\left(\sigma ,\rho \right)(1-{\iota }^2)={\mu }_5\left(\sigma ,\rho ,\iota \right),\end{array}$$

Differentiating ${\mu }_5\left(\sigma ,\rho ,\iota \right)$ with respect to $\iota$, we acquire

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\partial {\mu }_5}{\partial \iota }}& =& {\kappa }_{18}(\sigma ,\rho )+2\iota \left[{\pi }_5(\sigma ,\rho )-{\kappa }_{20}(\sigma ,\rho )\right]\hfill \\ & =& {\kappa }_{18}(\sigma ,\rho )+36\iota {(1-{\sigma }^2)}^2(8-9\rho +{\rho }^2)(1-{\rho }^2)\ge 0.\hfill \end{array}$$

Therefore, ${\mu }_5\left(\sigma ,\rho ,\iota \right)$ attains its maximum value at $\iota =1$. Thus, we yield

$$\begin{array}{ccc}\hfill {\phi }_5\left(\sigma ,\rho ,\iota \right)& \le & {\mu }_5\left(\sigma ,\rho ,\iota \right)\le {\mu }_5\left(\sigma ,\rho ,1\right)={\kappa }_{17}\left(\sigma ,\rho \right)+{\kappa }_{18}\left(\sigma ,\rho \right)+{\pi }_5\left(\sigma ,\rho \right)\hfill \\ & =& 113{\sigma }^6-168{\sigma }^5-18{\sigma }^4+168{\sigma }^3-126{\sigma }^2+144+(18{\sigma }^6-36{\sigma }^5-54{\sigma }^4+72{\sigma }^3+54{\sigma }^2-36\sigma \hfill \\ & -& 18){\rho }^4+(-189{\sigma }^6+108{\sigma }^5+459{\sigma }^4+108{\sigma }^3-513{\sigma }^2-216\sigma +243){\rho }^3+(-87{\sigma }^6+204{\sigma }^5+42{\sigma }^4\hfill \\ & -& 240{\sigma }^3+171{\sigma }^2+36\sigma -126){\rho }^2+(-369{\sigma }^6-108{\sigma }^5+369{\sigma }^4-108{\sigma }^3+216\sigma )\rho ={\xi }_7(\sigma ,\rho ).\hfill \end{array}$$

The critical points of ${\xi }_7(\sigma ,\rho )$ satisfy the conditions

$$\left\{\begin{array}{c}{\displaystyle \frac{\partial {\xi }_7}{\partial \sigma }}=678{\sigma }^5-840{\sigma }^4-72{\sigma }^3+504{\sigma }^2-252\sigma +(108{\sigma }^5-180{\sigma }^4-216{\sigma }^3+216{\sigma }^2+108\sigma -36){\rho }^4+(-1134{\sigma }^5\hfill \\ +540{\sigma }^4+1836{\sigma }^3+324{\sigma }^2-1026\sigma -216){\rho }^3+(-522{\sigma }^5+1020{\sigma }^4+168{\sigma }^3-720{\sigma }^2+342\sigma +36){\rho }^2+(-2214{\sigma }^5\hfill \\ -540{\sigma }^4+1476{\sigma }^3-324{\sigma }^2+216)\rho =0\hfill \\ {\displaystyle \frac{\partial {\xi }_7}{\partial \rho }}=216\sigma -108{\sigma }^3+369{\sigma }^4-108{\sigma }^5-369{\sigma }^6+(72{\sigma }^6-144{\sigma }^5-216{\sigma }^4+288{\sigma }^3+216{\sigma }^2-144\sigma -72){\rho }^3+(-567{\sigma }^6\hfill \\ +324{\sigma }^5+1377{\sigma }^4+324{\sigma }^3-1539{\sigma }^2-648\sigma +729){\rho }^2+(-174{\sigma }^6+408{\sigma }^5+84{\sigma }^4-480{\sigma }^3+342{\sigma }^2+72\sigma -252)\rho =0\hfill \end{array}\right.$$

By calculating, we get $\left\{\begin{array}{c}{\sigma }_{1,1}=1,\hfill \\ {\rho }_{1,1}=1.6198,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,2}=1,\hfill \\ {\rho }_{1,2}=-2.6328,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,3}=1,\hfill \\ {\rho }_{1,3}=0.0130,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,4}=-1,\hfill \\ {\rho }_{1,4}=0.7573,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,5}=0,\hfill \\ {\rho }_{1,5}=0,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,6}=0.0545,\hfill \\ {\rho }_{1,6}=0.0564,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,7}=0.5663,\hfill \\ {\rho }_{1,7}=0.8352,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,8}=0.7188,\hfill \\ {\rho }_{1,8}=0.7530,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,9}=3.1931,\hfill \\ {\rho }_{1,9}=21.0511,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,10}=-0.7441,\hfill \\ {\rho }_{1,10}=0.3083,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,11}=-0.7461,\hfill \\ {\rho }_{1,11}=-32.0416,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,12}=-0.8186,\hfill \\ {\rho }_{1,12}=-0.2719,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,13}=-1.4409,\hfill \\ {\rho }_{1,13}=11.5977.\hfill \end{array}\right.$

Thus, there are three critical points of $({\sigma }_{1,6},{\rho }_{1,6})$, $({\sigma }_{1,7},{\rho }_{1,7})$ and $({\sigma }_{1,8},{\rho }_{1,8})$ in $(0,1)\times (0,1)$. Therefore, we have $max\left\{{\xi }_2(\sigma ,\rho )\right\}\le max\{{\xi }_2({\sigma }_{1,6},{\rho }_{1,6}),{\xi }_2({\sigma }_{1,7},{\rho }_{1,7,{\xi }_2({\sigma }_{1,8},{\rho }_{1,8})})\}=max\{143.961357,182.386461,187.761713\}=187.761713.$

(20) For $\sigma =1,$

$$\begin{array}{c}\hfill {\xi }_7(1,\rho )=113.\end{array}$$

(21) For $\sigma =0,$

$$\begin{array}{c}\hfill {\xi }_7(0,\rho )=144-126{\rho }^2+243{\rho }^3-18{\rho }^4\le 243.\end{array}$$

(22) For $\rho =0,$

$$\begin{array}{c}\hfill {\xi }_7(\sigma ,0)=144-126{\sigma }^2+168{\sigma }^3-18{\sigma }^4-168{\sigma }^5+113{\sigma }^6\le 144.\end{array}$$

(23) For $\rho =1$

$$\begin{array}{c}\hfill {\xi }_7(\sigma ,1)=243-414{\sigma }^2+798{\sigma }^4-514{\sigma }^6\le 243.\end{array}$$

Thus, we acquire

$$\begin{array}{c}\hfill |{\mathcal{H}}_{3,1}\left({\hslash }^{-1}\right)|\le {\displaystyle \frac{243}{1296}}={\displaystyle \frac{3}{16}}.\end{array}$$

5. The Second-Order Hankel Determinants of Logarithmic Coefficients for the Class ${\mathcal{S}}_T^{*}$

Theorem 5.1 

If $\hslash \in {\mathcal{S}}_T^{*}$, then

$$|{\lambda }_1|\le {\displaystyle \frac{1}{2}},|{\lambda }_2|\le {\displaystyle \frac{1}{4}},\left|{\lambda }_3\right|\le {\displaystyle \frac{1}{6}}.$$

The bounds are sharp with the extremal function:

$$\begin{array}{c}\hfill {\hslash }_4\left(\upsilon \right)=exp{\int }_0^{\upsilon }{\displaystyle \frac{1+arctant}{t}}dt=\upsilon +{\upsilon }^2++···,\end{array}$$

(39)

$$\begin{array}{c}\hfill {\hslash }_5\left(\upsilon \right)=exp{\int }_0^{\upsilon }{\displaystyle \frac{1+arctan{t}^2}{t}}dt=\upsilon +{\displaystyle \frac{1}{2}}{\upsilon }^3++···,\end{array}$$

(40)

$$\begin{array}{c}\hfill {\hslash }_6\left(\upsilon \right)=exp{\int }_0^{\upsilon }{\displaystyle \frac{1+arctan{t}^3}{t}}dt=\upsilon +{\displaystyle \frac{1}{3}}{\upsilon }^4++···.\end{array}$$

(41)

Proof 

Let $\hslash \in {\mathcal{S}}_T^{*}.$ From $\left(37\right)$ and $\left(6\right)$, we have

$$\begin{array}{c}\hfill {\lambda }_1={\displaystyle \frac{1}{2}}{\ell }_1,\end{array}$$

(42)

$$\begin{array}{c}\hfill {\lambda }_2={\displaystyle \frac{1}{4}}{\ell }_2,\end{array}$$

(43)

$$\begin{array}{c}\hfill {\lambda }_3=-{\displaystyle \frac{1}{18}}{\ell }_1^3+{\displaystyle \frac{1}{6}}{\ell }_3,\end{array}$$

(44)

Applying $\left(13\right)$ to $\left(42\right)$ and $\left(43\right)$ , we have

$$\begin{array}{c}\hfill |{\lambda }_1|\le {\displaystyle \frac{1}{2}},\left|{\lambda }_2\right|\le {\displaystyle \frac{1}{4}}.\end{array}$$

From Lemma 1.2 and $\left(44\right)$, we get

$$\begin{array}{c}\hfill |{\lambda }_3|={\displaystyle \frac{1}{6}}|-{\displaystyle \frac{1}{3}}{\ell }_1^3+{\ell }_3|\le {\displaystyle \frac{1}{6}}.\end{array}$$

Theorem 5.2 

If $\hslash \in {\mathcal{S}}_T^{*}$, then

$$|{\lambda }_1{\lambda }_3-{\lambda }_2^2|\le {\displaystyle \frac{1}{16}}.$$

The result is sharp for the function ${\hslash }_5\left(\upsilon \right)$.

Proof 

Let $\hslash \in {\mathcal{S}}_T^{*}.$ From $\left(42\right)$, $\left(43\right)$ and $\left(44\right)$, we acquire

$$\begin{array}{ccc}\hfill |{\lambda }_1{\lambda }_3-{\lambda }_2^2|& =& {\displaystyle \frac{1}{144}}|12{\ell }_1{\ell }_3-9{\ell }_2^2-4{\ell }_1^4|.\hfill \end{array}$$

(45)

From $\left(45\right)$ and Lemma 1.5, we have

$$\begin{array}{ccc}\hfill |{\lambda }_1{\lambda }_3-{\lambda }_2^2|& =& {\displaystyle \frac{1}{144}}|-12({\sigma }_1{|}^2-|{\sigma }_1{|}^4\left)\right|{\sigma }_2^2+12{\sigma }_3{\sigma }_1(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2)-9(1-|{\sigma }_1{|}^2{)}^2{\sigma }_2^2-4{\sigma }_1^4|.\hfill \end{array}$$

Setting $|{\sigma }_2|=\rho .|{\sigma }_1|=\sigma$ and upon taking $|{\sigma }_3|\le 1$, we acquire

$$\begin{array}{c}\hfill |{\lambda }_1{\lambda }_3-{\lambda }_2^2|\le {\displaystyle \frac{1}{144}}[12({\sigma }^2-{\sigma }^4){\rho }^2+12(1-{\rho }^2)(\sigma -{\sigma }^3)+9{(1-{\sigma }^2)}^2{\rho }^2+4{\sigma }^4]={\displaystyle \frac{1}{144}}{\xi }_8(\sigma ,\rho ).\end{array}$$

By differentiating ${\xi }_8(\sigma ,\rho )$ with respect to $\rho$, we yield

$${\displaystyle \frac{\partial {\xi }_8}{\partial \rho }}=6\rho (1-{\sigma }^2)({\sigma }^2-4\sigma +3)]\ge 0.$$

Therefore, ${\xi }_8(\sigma ,\rho )$ is an creasing function on $\rho \in [0,1]$. Hence, we get

$$\begin{array}{c}\hfill |{\lambda }_1{\lambda }_3-{\lambda }_2^2|\le {\displaystyle \frac{1}{144}}{\xi }_8(\sigma ,1)={\displaystyle \frac{1}{144}}[9-6{\sigma }^2+{\sigma }^4]\le {\displaystyle \frac{9}{144}}={\displaystyle \frac{1}{16}}.\end{array}$$

□

Theorem 5.3 

If $\hslash \in {\mathcal{S}}_T^{*}$, then

$$|{\lambda }_2{\lambda }_4-{\lambda }_3^2|\le {\displaystyle \frac{1}{36}}.$$

The inequality is best possible for the function ${\hslash }_6\left(\upsilon \right)$.

Proof 

Let $\hslash \in {\mathcal{S}}_T^{*}.$ From $\left(37\right)$ and $\left(8\right)$, we have

$$\begin{array}{ccc}\hfill {\lambda }_2{\lambda }_4-{\lambda }_3^2& =& {\displaystyle \frac{1}{5184}}(-144{\ell }_3^2+96{\ell }_1^3{\ell }_3-16{\ell }_1^6+162{\ell }_1^2{\ell }_2^2+162{\ell }_2{\ell }_4).\hfill \end{array}$$

From $\left(46\right)$ and Lemma 1.5 , we achieve

$$\begin{array}{ccc}\hfill {\lambda }_2{\lambda }_4-{\lambda }_3^2& =& {\displaystyle \frac{1}{5184}}[-162{\sigma }_1^2{\sigma }_2^2(1-|{\sigma }_1{|}^2{)}^2+18{\overline{{\sigma }_1}}^2{\sigma }_2^4(1-|{\sigma }_1{|}^2{)}^2-162|{\sigma }_2{|}^2(1-|{\sigma }_1{|}^2{)}^2(1-|{\sigma }_2{|}^2){\sigma }_3^2-16{\sigma }_1^6\hfill \\ & -& 36\overline{{\sigma }_1}{\sigma }_2^2(1-|{\sigma }_1{|}^2{)}^2(1-|{\sigma }_2{|}^2){\sigma }_3+162{\sigma }_2(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_3{|}^2\left)\right(1-|{\sigma }_1{|}^2{)}^2{\sigma }_4-144{\sigma }_3^2(1-|{\sigma }_2{{|}^2)}^2\hfill \\ & \times & (1-|{\sigma }_1{|}^2{)}^2-96{\sigma }_2^2{\sigma }_1^3\overline{{\sigma }_1}(1-|{\sigma }_1{|}^2)+96{\sigma }_3{\sigma }_1^3(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2)].\hfill \end{array}$$

So, we yield

$${\lambda }_2{\lambda }_4-{\lambda }_3^2={\displaystyle \frac{1}{5184}}\left[{\zeta }_{16}\left({\sigma }_1,{\sigma }_2\right)+{\zeta }_{17}\left({\sigma }_1,{\sigma }_2\right){\sigma }_3+{\zeta }_{18}\left({\sigma }_1,{\sigma }_2\right){\sigma }_3^2+{\Psi }_6\left({\sigma }_1,{\sigma }_2,{\sigma }_3\right){\sigma }_4\right],$$

where

$$\begin{array}{ccc}\hfill {\zeta }_{16}\left({\sigma }_1,{\sigma }_2\right)& =& -162{\left(1-|{\sigma }_1{|}^2\right)}^2{\sigma }_1^2{\sigma }_2^2+18{\sigma }_2^4{\overline{{\sigma }_1}}^2(1-|{\sigma }_1{|}^2{)}^2-96(1-|{\sigma }_1{|}^2)\overline{{\sigma }_1}{\sigma }_1^3{\sigma }_2^2-16{\sigma }_1^6,\hfill \\ \hfill {\zeta }_{17}\left({\sigma }_1,{\sigma }_2\right)& =& -36{\sigma }_2^2\overline{{\sigma }_1}(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2{)}^2+96(1-|{\sigma }_1{|}^2)\left(1-|{\sigma }_2{|}^2\right){\sigma }_1^3],\hfill \\ \hfill {\zeta }_{18}\left({\sigma }_1,{\sigma }_2\right)& =& (-18|{\sigma }_2{|}^2-144\left)\right(1-|{\sigma }_2{|}^2){\left(1-|{\sigma }_1{|}^2\right)}^2,\hfill \\ \hfill {\Psi }_6\left({\sigma }_1,{\sigma }_2,{\sigma }_3\right)& =& 162{\sigma }_2(1-|{\sigma }_2{|}^2\left)\right(1-|{\sigma }_1{|}^2{)}^2(1-|{\sigma }_3{|}^2).\hfill \end{array}$$

Setting $\iota =|{\sigma }_3|,\rho =|{\sigma }_2|,$$\sigma =|{\sigma }_1|$, and upon taking $|{\sigma }_1|\le 1$, we achieve

$$\begin{array}{ccc}\hfill {\lambda }_2{\lambda }_4-{\lambda }_3^2& \le & {\displaystyle \frac{1}{5184}}\left(\left|{\zeta }_{16}\left({\sigma }_1,{\sigma }_2\right)\right|+\left|{\zeta }_{17}\left({\sigma }_1,{\sigma }_2\right)\right|\iota +\left|{\zeta }_{18}\left({\sigma }_1,{\sigma }_2\right)\right|{\iota }^2+\left|{\Psi }_6\left({\sigma }_1,{\sigma }_2,{\sigma }_3\right)\right|\right)\hfill \\ & \le & {\displaystyle \frac{1}{5184}}{\phi }_6\left(\sigma ,\rho ,\iota \right),\hfill \end{array}$$

where

$$\begin{array}{c}\hfill {\phi }_6\left(\sigma ,\rho ,\iota \right)={\kappa }_{21}\left(\sigma ,\rho \right)+{\kappa }_{22}\left(\sigma ,\rho \right)\iota +{\kappa }_{23}\left(\sigma ,\rho \right){\iota }^2+{\kappa }_{24}\left(\sigma ,\rho \right)(1-{\iota }^2),\end{array}$$

with

$$\begin{array}{ccc}\hfill {\kappa }_{21}(\sigma ,\rho )& =& 18{\sigma }^2{(1-{\sigma }^2)}^2{\rho }^4+162{\sigma }^2{(1-{\sigma }^2)}^2{\rho }^2+96({\sigma }^4-{\sigma }^6){\rho }^2+16{\sigma }^6,\hfill \\ \hfill {\kappa }_{22}(\sigma ,\rho )& =& 36\sigma {(1-{\sigma }^2)}^2({\rho }^2-{\rho }^4)+96({\sigma }^3-{\sigma }^5)(1-{\rho }^2),\hfill \\ \hfill {\kappa }_{23}(\sigma ,\rho )& =& {(1-{\sigma }^2)}^2(1-{\rho }^2)(18{\rho }^2+144),\hfill \\ \hfill {\kappa }_{24}(\sigma ,\rho )& =& 162(\rho -{\rho }^3){(1-{\sigma }^2)}^2.\hfill \end{array}$$

Differentiating ${\phi }_6\left(\sigma ,\rho ,\iota \right)$ with respect to $\iota$, we yield

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\partial {\phi }_6}{\partial \iota }}& =& {\kappa }_{22}(\sigma ,\rho )+2\iota \left[{\kappa }_{23}(\sigma ,\rho )-{\kappa }_{24}(\sigma ,\rho )\right]\hfill \\ & =& {\kappa }_{22}(\sigma ,\rho )+36\iota ({\rho }^2-9\rho +8)(1-{\rho }^2){(1-{\sigma }^2)}^2\ge 0.\hfill \end{array}$$

Thus, ${\phi }_6\left(\sigma ,\rho ,\iota \right)$ is an creasing function on $\iota \in [0,1]$. Hence, we get

$$\begin{array}{ccc}\hfill {\phi }_6\left(\sigma ,\rho ,\iota \right)& \le & {\phi }_6\left(\sigma ,\rho ,1\right)={\kappa }_{21}\left(\sigma ,\rho \right)+{\kappa }_{22}\left(\sigma ,\rho \right)+{\kappa }_{23}\left(\sigma ,\rho \right)\hfill \\ & =& 144-288{\sigma }^2+96{\sigma }^3+144{\sigma }^4-96{\sigma }^5+16{\sigma }^6+(-18-36\sigma +54{\sigma }^2+72{\sigma }^3-54{\sigma }^4-36{\sigma }^5\hfill \\ & +& 18{\sigma }^6){\rho }^4+(-126+36\sigma +414{\sigma }^2-168{\sigma }^3-354{\sigma }^4+132{\sigma }^5+66{\sigma }^6){\rho }^2={\xi }_9(\sigma ,\rho ).\hfill \end{array}$$

Consider

$$\left\{\begin{array}{c}{\displaystyle \frac{\partial {\xi }_9}{\partial \sigma }}=-576\sigma +288{\sigma }^2+576{\sigma }^3-480{\sigma }^4+96{\sigma }^5+(-36+108\sigma +216{\sigma }^2-216{\sigma }^3-180{\sigma }^4+108{\sigma }^5){\rho }^4\hfill \\ +(36+828\sigma -504{\sigma }^2-1416{\sigma }^3+660{\sigma }^4+396{\sigma }^5){\rho }^2=0\hfill \\ {\displaystyle \frac{\partial {\xi }_9}{\partial \rho }}(-72-144\sigma +216{\sigma }^2+288{\sigma }^3-216{\sigma }^4-144{\sigma }^5+72{\sigma }^6){\rho }^3+(-252+72\sigma +828{\sigma }^2-336{\sigma }^3-708{\sigma }^4\hfill \\ +264{\sigma }^5+132{\sigma }^6)\rho =0\hfill \end{array}\right.$$

By calculating, we get $\left\{\begin{array}{c}{\sigma }_{1,1}=0,\hfill \\ {\rho }_{1,1}=0,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,2}=2,\hfill \\ {\rho }_{1,2}=0,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,3}=2.5321,\hfill \\ {\rho }_{1,3}=0,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,4}=-0.8794,\hfill \\ {\rho }_{1,4}=0,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,5}=1.3473,\hfill \\ {\rho }_{1,5}=0,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,6}=-1,\hfill \\ {\rho }_{1,6}=0.8660,\hfill \end{array}\right.$$\left\{\begin{array}{c}{\sigma }_{1,7}=-1,\hfill \\ {\rho }_{1,7}=-0.8660.\hfill \end{array}\right.$

Thus, there is no optimal point in in $(0,1)\times (0,1)$.

(24) For $\sigma =1,$

$$\begin{array}{c}\hfill {\xi }_9(1,\rho )=16\end{array}$$

(25) For $\sigma =0,$

$$\begin{array}{c}\hfill {\xi }_9(0,\rho )=144-126{\rho }^2-18{\rho }^4\le 144.\end{array}$$

(26) For $\rho =0,$

$$\begin{array}{c}\hfill {\xi }_9(\sigma ,0)=16{\sigma }^6-96{\rho }^5+144{\rho }^4+96{\rho }^3-288{\rho }^2+144\le 144.\end{array}$$

(27) For $\rho =1$

$$\begin{array}{c}\hfill {\xi }_9(\sigma ,1)=180{\sigma }^2-264{\sigma }^4+100{\sigma }^6={ϵ}_5\left(\sigma \right)\le {ϵ}_5\left(0.6800\right)=36.6719.\end{array}$$

Therefore, we yield

$$\begin{array}{c}\hfill |{\lambda }_2{\lambda }_4-{\lambda }_3^2|\le {\displaystyle \frac{144}{5184}}={\displaystyle \frac{1}{36}}.\end{array}$$