Coefficient Estimates for the Class \( \mathcal CL \)

1. Introduction

Let $\Delta =\{ϵ:|ϵ|<1\}$ be the unit disk in complex plane. Let $\mathcal{A}$ be the class of analytic functions of the form:

$$\begin{array}{c}\hfill f\left(ϵ\right)=ϵ+{\displaystyle \sum _{i=2}^{\infty }}{a}_i{ϵ}^i(ϵ\in \Delta ).\end{array}$$

(1)

which are defined in the unit disk $\Delta$ and normalized by $f\left(0\right)=f^{\prime }\left(0\right)-1=0$. Let $\mathcal{S}$ denote the class of all functions of $\mathcal{A}$ that are univalent. A function $f\in \mathcal{A}$ is said to belong to the class $\mathcal{C}$ of convex functions in $\Delta$, if it satisfies the following inequality:

$$\begin{array}{c}\hfill Re[1+{\displaystyle \frac{ϵf^{″}\left(ϵ\right)}{f^{\prime }\left(ϵ\right)}}]>0.\end{array}$$

(2)

In [1], Noonan and Thomas studied the $j^{th}$ Hankel determinants $H_{j,i}$ of functions $f\in \mathcal{A}$ of the form (1) for $j\ge 1$, which is defined by

$$H_{j,i}\left(f\right)=\left|\begin{array}{cccc}{a}_i& a_{i+1}& ···& a_{i+j-1}\\ a_{i+1}& a_{i+2}& ···& a_{i+j}\\ ⋮& ⋮& ⋮& ⋮& \\ a_{i+j-1}& a_{i+j}& ···& a_{i+2j-2}\end{array}\right|.$$

For $j=2,i=2,$$j=2,i=3$ and $j=1,i=3$, we yield

$$\begin{array}{c}\hfill H_{2,2}\left(f\right)=-a_3^2+a_2{a}_4,\end{array}$$

(3)

$$\begin{array}{c}\hfill H_{2,3}\left(f\right)=-a_4^2+a_3{a}_5,\end{array}$$

(4)

$$\begin{array}{c}\hfill H_{3,1}\left(f\right)=-a_3^3+2a_4{a}_3{a}_2-a_5{a}_2^2+a_5{a}_3-a_4^2\end{array}$$

(5)

The estimation of $|H_{2,3}\left(f\right)|$and $|H_{3,1}\left(f\right)|$ is tedious. Recently, many authors [2,3,4,5,6,7,8,9,10,11,12,13] obtained the sharp bounds of $|H_{2,3}\left(f\right)|$ and $|H_{3,1}\left(f\right)|$ for subclasses of analytic functions. At the end of 1960’s, Zalcman made a conjecture that each $f\in \mathcal{S}$ satisfies the inequality $|a_i^2-a_{2i-1}{|\le (i-1)}^2(i\ge 2)$ with equality for the Koebe function $K\left(ϵ\right)={\displaystyle \frac{ϵ}{{(1-ϵ)}^2}}$ and its rotations. In [14], Ma proposed a generalized version of the Zalcman conjecture as follows: for $f\in \mathcal{S},$$|b_i{b}_j-b_{i+j-1}|\le (i-1)(j-1)(i\ge 2,j\ge 2)$ and proved that this holds for starlike functions and univalent functions with real coefficients.

The logarithmic coefficients of function $f\in \mathcal{S}$, can be written as

$$F_f\left(ϵ\right):=log{\displaystyle \frac{f\left(ϵ\right)}{ϵ}}=2{\displaystyle \sum _{i=1}^{\infty }}{\gamma }_i{ϵ}^i,(ϵ\in \Delta ).$$

(6)

Differentiating (6), we achieve

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

(7)

Milin’s conjecture relies heavily on the logarithmic coefficients.

For $f\in \mathcal{S},$ has an inverse $f^{-1}$ given by

$$\begin{array}{cc}\hfill f^{-1}\left(w\right)& =w+{\displaystyle \sum _{m=2}^{\infty }}{D}_m{w}^m|w|<r_0\left(f\right);(r_0\left(f\right)\ge {\displaystyle \frac{1}{4}})\hfill \\ \hfill & =w-a_2{w}^2+(2a_2^2-a_3)w^3-(5a_2^3-5a_2{a}_3+a_4)w^4+(14a_2^4+3a_3^2-21a_2^2{a}_3\hfill \\ \hfill & +6a_2{a}_4-a_5)w^5+....\hfill \end{array}$$

(8)

From (8) and (5), we get

$$\begin{array}{cc}\hfill {\mathcal{H}}_{3,1}\left(f^{-1}\right)& =2D_4{D}_3{D}_2-D_3^3+D_5{D}_3-D_5{D}_2^2-D_4^2\hfill \\ \hfill & =-3a_2^4{a}_3+2a_2{a}_3{a}_4+3a_2^2{a}_3^2-a_2^2{a}_5+a_2^6-a_4^2-2a_3^3+a_3{a}_5\hfill \end{array}$$

(9)

In 2022, Wang and his coauthors [13] introduced the following class $\mathcal{R}L$ of analytic functions:

$$\begin{array}{c}\hfill \mathcal{R}L=\left\{f:f\in \mathcal{S}and{f}^{\prime }\left(ϵ\right)\prec \sqrt{1+ϵ}(ϵ\in \Delta )\right\}.\end{array}$$

In view of the class $\mathcal{R}L$, we introduced and studied the following subclass of convex functions in $\mathcal{A}$

$$\begin{array}{c}\hfill \mathcal{C}L=\{f:f\in \mathcal{S}and1+{\displaystyle \frac{f^{″}\left(ϵ\right)}{f^{\prime }\left(ϵ\right)}}\prec \sqrt{1+ϵ}(ϵ\in \Delta )\}.\end{array}$$

The sharp bounds of initial coefficients, the second and third Hankel determinants for the class $\mathcal{R}L$ were studied by Wang et al. [13]. The aim of this paper is to prove the sharp bounds of initial coefficients $|a_i|(i\le 6)$, the generalized Zalcman conjecture, the second and third Hankel determinants, the second Hankel determinant of logarithmic coefficients, and the third Hankel determinant of inverse functions for the class $\mathcal{C}L$. All bounds are sharp.

Let ${\mathcal{B}}_0$ denotes the class of Schwarz functions $\varpi$ which are analytic in $\Delta$ given by

$$\varpi \left(ϵ\right)=e_1ϵ+e_2{ϵ}^2+e_3{ϵ}^3+...,(ϵ\in \Delta )$$

(10)

and satisfying $\varpi \left(0\right)=0$ and $|\varpi (ϵ)|<1$. To derive our results, we shall need the following Lemmas:

Lemma 1.1 ([15]) If $\varpi \in {\mathcal{B}}_0$, then $|e_i|\le 1(i\ge 1)$.

Lemma 1.2 ([16]) Let $\varpi \in {\mathcal{B}}_0$ is given by (10). Then

$$|e_2|\le 1-|e_1{|}^2,|e_3|\le 1-|e_1{|}^2-{\displaystyle \frac{|e_2{|}^2}{1+|e_1|}},|e_4|\le 1-|e_1{|}^2-{|e_2|}^2,$$

$$|e_5|\le 1-|e_1{|}^2-|e_2{|}^2-{\displaystyle \frac{|e_3{|}^2}{1+|e_1|}},|e_6|\le 1-|e_1{|}^2-|e_2{|}^2-|e_3{|}^2,|e_7|\le 1-|e_1{|}^2-|e_2{|}^2-{|e_3|}^2-{\displaystyle \frac{|e_4{|}^2}{1+|e_1|}}.$$

Lemma 1.3 ([17]) If $\varpi \in {\mathcal{B}}_0$ and it is of the form (10), $\tau \in \mathbb{C}$ and $|\tau |\le 1$, then

$$\begin{array}{c}\hfill |e_3+2\tau e_1{e}_2+{\tau }^2{e}_1^3|\le 1.\end{array}$$

Lemma 1.4 ([17]) Let $\varpi \in {\mathcal{B}}_0$, $\tau \in \mathbb{C}$ and $|\tau |<1.$ Then

$$\begin{array}{c}\hfill |e_4+2\tau e_1{e}_3+3{\tau }^2{e}_1^2{e}_2+\tau e_2^2+{\tau }^3{e}_1^4|\le 1.\end{array}$$

Lemma 1.5 ([18]) Let $\varpi \in {\mathcal{B}}_0$, $\tau \in \mathbb{C}$ and $|\tau |\le 1$, then

$$|e_5+2\tau e_1{e}_4+2\tau e_2{e}_3+3{\tau }^2{e}_1{e}_2^2+3{\tau }^2{e}_1^2{e}_3+4{\tau }^3{e}_1^3{e}_2+{\tau }^4{e}_1^5|\le 1,$$

$$|e_5+(1+\tau )e_1{e}_4+2\tau e_3{e}_2+\tau (1+2\tau )e_1{e}_2^2+\tau (2+\tau )e_1^2{e}_3+{\tau }^2(3+\tau )e_1^3{e}_2+{\tau }^3{e}_1^5|\le 1.$$

Lemma 1.6 ([19]) If $\varpi \in {\mathcal{B}}_0$ is of the form (10), and $\tau \in \mathbb{C}$, then

$$\begin{array}{c}\hfill |e_2+\tau e_1^2|\le max\{1,|\tau |\}.\end{array}$$

Lemma 1.7 ([20]) $\varpi \in {\mathcal{B}}_0$. Then

$$\begin{array}{ccc}\hfill e_2& =& (1-|e_1{|}^2){\xi }_1,\hfill \\ \hfill e_3& =& -\overline{e_1}{\xi }_1^2(1-|e_1{|}^2)+(1-|e_1{|}^2\left)\right(1-|{\xi }_1{|}^2){\xi }_2\hfill \\ \hfill e_4& =& {\overline{e_1}}^2{\xi }_1^3(1-|e_1{|}^2)-(1-|e_1{|}^2\left)\right(1-|{\xi }_1{|}^2)(\overline{{\xi }_1}{\xi }_2^2+2{\xi }_2\overline{e_1}{\xi }_1)\hfill \\ & +& (1-|e_1{|}^2\left)\right(1-|{\xi }_1{|}^2\left)\right(1-|{\xi }_2{|}^2){\xi }_3.\hfill \end{array}$$

for some ${\xi }_i\in \mathbb{C}$ with $|{\xi }_i|\le 1(i=1,2,3).$

2. Zalcman Functionals and Hankel Determinant for the Class $\mathcal{C}L$

Theorem 2.1 If $f\in \mathcal{C}L$, then

$$|a_2|\le {\displaystyle \frac{1}{4}},|a_3|\le {\displaystyle \frac{1}{12}},|a_4|\le {\displaystyle \frac{1}{24}},|a_5|\le {\displaystyle \frac{1}{40}},|a_6|\le {\displaystyle \frac{1}{60}}.$$

The results are the best possible.

Proof Let $f\in \mathcal{C}L$. Then there exists $\varpi \in {\mathcal{B}}_0$ of the form (10) such that

$$\begin{array}{cc}\hfill 1+{\displaystyle \frac{ϵf^{″}\left(ϵ\right)}{f^{\prime }\left(ϵ\right)}}=& \sqrt{1+\varpi \left(ϵ\right)}.\hfill \end{array}$$

(11)

From (1) we yield

$$\begin{array}{cc}\hfill 1+{\displaystyle \frac{ϵf^{″}\left(ϵ\right)}{f^{\prime }\left(ϵ\right)}}& =1+2a_2ϵ+(6a_3-4a_2^2){ϵ}^2+(-18a_2{a}_3+12b_4+8a_2^3){ϵ}^3+(48a_2^2{a}_3+20a_5\hfill \\ \hfill & -32a_2{a}_4-16a_2^4-18a_3^2){ϵ}^4+(-120a_2^3{a}_3-50a_2{a}_5+30a_6-60a_3{a}_4+80a_2^2{a}_4\hfill \\ \hfill & +32a_2^5+90a_2{a}_3^2){ϵ}^5+···.\hfill \end{array}$$

(12)

Let

$$\varpi \left(ϵ\right)=e_1ϵ+e_2{ϵ}^2+e_3{ϵ}^3+...(ϵ\in \Delta ).$$

After computation, we yield

$$\begin{array}{cc}\hfill \sqrt{1+\varpi \left(ϵ\right)}& =1+{\displaystyle \frac{1}{2}}{e}_1ϵ+({\displaystyle \frac{1}{2}}{e}_2-{\displaystyle \frac{1}{8}}{e}_1^2){ϵ}^2+({\displaystyle \frac{1}{2}}{e}_3-{\displaystyle \frac{1}{4}}{e}_1{e}_2+{\displaystyle \frac{1}{16}}{e}_1^3){ϵ}^3+({\displaystyle \frac{1}{2}}{e}_4-{\displaystyle \frac{1}{4}}{e}_1{e}_3-{\displaystyle \frac{1}{8}}{e}_2^2\hfill \\ \hfill & +{\displaystyle \frac{3}{16}}{e}_1^2{e}_2-{\displaystyle \frac{5}{128}}{e}_1^4){ϵ}^4+({\displaystyle \frac{1}{2}}{e}_5-{\displaystyle \frac{1}{4}}{e}_2{e}_3-{\displaystyle \frac{1}{4}}{e}_1{e}_4+{\displaystyle \frac{3}{16}}{e}_1{e}_2^2+{\displaystyle \frac{3}{16}}{e}_1^2{e}_3-{\displaystyle \frac{5}{32}}{e}_1^3{e}_2\hfill \\ \hfill & +{\displaystyle \frac{7}{256}}{e}_1^5){ϵ}^5+···.\hfill \end{array}$$

(13)

By (12) and (13), we have

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

(14)

$$\begin{array}{c}\hfill a_3={\displaystyle \frac{4e_2+e_1^2}{48}},\end{array}$$

(15)

$$\begin{array}{c}\hfill a_4={\displaystyle \frac{16e_3+4e_1{e}_2+e_1^3}{384}},\end{array}$$

(16)

$$\begin{array}{c}\hfill a_5={\displaystyle \frac{48e_4+8e_1{e}_3+8e_1^2{e}_2-e_1^4}{1920}},\end{array}$$

(17)

$$\begin{array}{c}\hfill a_6={\displaystyle \frac{1}{23040}}(384e_5+48e_1{e}_4-32e_2{e}_3+64e_1^2{e}_3-40e_1^3{e}_2+64e_1{e}_2^2+7e_1^5).\end{array}$$

(18)

From (14) and Lemma 1.1, we have $|a_2|\le {\displaystyle \frac{1}{4}}.$

From (15) and Lemma 1.2, we yield

$$\begin{array}{c}\hfill |a_3|\le {\displaystyle \frac{1}{48}}\left[4\right(1-|e_1{|}^2)+|e_1{|}^2]={\displaystyle \frac{1}{48}}(4-3|e_1{|}^2)\le {\displaystyle \frac{4}{48}}={\displaystyle \frac{1}{12}}.\end{array}$$

From (16), Lemma 1.3 and 1.2, we achieve

$$\begin{array}{cc}\hfill |a_4|& ={\displaystyle \frac{1}{384}}|2(2e_1{e}_2+e_3+e_1^3)+14e_3-e_1^3|\hfill \\ \hfill & \le {\displaystyle \frac{1}{384}}[2|e_1^3+2e_1{e}_2+e_3|+14(1-|e_1{|}^2-{\displaystyle \frac{|e_2{|}^2}{1+|e_1|}})+|e_1{|}^3]\hfill \\ \hfill & \le {\displaystyle \frac{1}{384}}(16-14|e_1{|}^2+|e_1{|}^3-{\displaystyle \frac{14|e_2{|}^2}{1+|e_1|}})\hfill \\ \hfill & \le {\displaystyle \frac{1}{384}}(16-14|e_1{|}^2+|e_1{|}^3)\le {\displaystyle \frac{16}{384}}={\displaystyle \frac{1}{24}}.\hfill \end{array}$$

From (17) Lemma 1.4 and 1.2, we have

$$\begin{array}{cc}\hfill |a_5|& ={\displaystyle \frac{1}{1920}}|44e_4+4(e_4+2e_1{e}_3+e_2^2+3e_1^2{e}_2+e_1^4)-4e_2^2-4e_1^2{e}_2-5e_1^4|\hfill \\ \hfill & \le {\displaystyle \frac{1}{1920}}\left[44\right(1-|e_1{|}^2-|e_2{|}^2)+4+4|e_2{|}^2+4|e_1{|}^2(1-|e_1{|}^2)+5|e_1{|}^4]\hfill \\ \hfill & ={\displaystyle \frac{1}{1920}}(48-40|e_1{|}^2+|e_1{|}^4-40|e_2{|}^2)\hfill \\ \hfill & \le {\displaystyle \frac{1}{1920}}(48-40|e_1{|}^2+|e_1{|}^4)\le {\displaystyle \frac{48}{1920}}={\displaystyle \frac{1}{40}}.\hfill \end{array}$$

From (18) Lemma 1.5, Lemma 1.3 and 1.2, we achieve

$$\begin{array}{cc}\hfill |a_6|& ={\displaystyle \frac{1}{23040}}|48(e_5+e_1{e}_4+e_2{e}_3+{\displaystyle \frac{3}{4}}{e}_1{e}_2^2+{\displaystyle \frac{3}{4}}{e}_1^2{e}_3+{\displaystyle \frac{1}{2}}{e}_1^3{e}_2+{\displaystyle \frac{1}{16}}{e}_1^5)+40(-2e_3{e}_2+e_5+e_1{c}_2^2\hfill \\ \hfill & -e_1^2{e}_3+2e_1^3{e}_2-e_1^5)+296e_5-12e_1{e}_2^2+68e_1^2(e_1^3-2e_1{e}_2+e_3)-8e_1^3{e}_2-24e_1^5|\hfill \\ \hfill & \le {\displaystyle \frac{1}{23040}}[48+40+296(1-|e_1{|}^2-|e_2{|}^2-{\displaystyle \frac{|e_3{|}^2}{1+|e_1|}})+12|e_1||e_2{|}^2+68|e_1{|}^2+8|e_1{|}^3|e_2|+24|e_1{|}^5]\hfill \\ \hfill & ={\displaystyle \frac{1}{23040}}[384-228|e_1{|}^2+24|e_1{|}^5+8|e_1{|}^3|e_2|+(-296+12|e_1|)|e_2{|}^2-{\displaystyle \frac{296|e_3{|}^2}{1+|e_1|}}]\hfill \\ \hfill & \le {\displaystyle \frac{1}{23040}}[24|e_1{|}^5-228|e_1{|}^2+384+8|e_2||e_1{|}^3+(12|e_1|-296)|e_2{|}^2].\hfill \end{array}$$

By putting $|e_1|=x,y=|e_2|$, we have

$$\begin{array}{cc}\hfill |a_6|& \le {\displaystyle \frac{{\alpha }_1(x,y)}{23040}},\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill {\alpha }_1(x,y)& =24x^5-228x^2+384+8x^{3}y+(12x-296)y^2,\hfill \end{array}$$

with $x\in [0,1]$ and $y\in [0,1-x^2].$

The optimal points of ${\alpha }_1$ satisfy the system of equations

$$\left\{\begin{array}{c}{\displaystyle \frac{\partial {\alpha }_1}{\partial x}}=-456x+120x^4+24x^{2}y+12y^2=0,\hfill \\ {\displaystyle \frac{\partial {\alpha }_1}{\partial y}}=8x^3+(-592+24x)y=0.\hfill \end{array}\right.$$

Using numerical computations, we yield

$\left\{\begin{array}{c}{x}_1=0,\hfill \\ y_1=0,\hfill \end{array}\right.$$\left\{\begin{array}{c}{x}_2\approx 1.5581,\hfill \\ y_2\approx 0.0546.\hfill \end{array}\right.$

Thus, there is no critical point in the interior of $(0,1)\times (0,1-x^2).$

(1) For $y=0,$

$$\begin{array}{cc}\hfill {\alpha }_1(x,0)& =24x^5-228x^2+384\le 384.\hfill \end{array}$$

(2) For $x=0,$

$$\begin{array}{cc}\hfill {\alpha }_1(0,y)& =384-296y^2\le 384.\hfill \end{array}$$

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

$$\begin{array}{cc}\hfill {\alpha }_1(x,1-x^2)& =88+12x+364x^2-16x^3-296x^4+28x^5={\delta }_1\left(x\right)\le {\delta }_1\left(0.8109\right)\approx 210.3826.\hfill \end{array}$$

Therefore, we have

$$\begin{array}{c}\hfill |a_6|\le {\displaystyle \frac{384}{23040}}={\displaystyle \frac{1}{60}}.\end{array}$$

The bounds are sharp for the functions $f_i:\Delta \to \mathbb{C}$ given by

$$\begin{array}{c}\hfill f_1\left(ϵ\right)={\int }_0^{ϵ}\left(exp\left({\int }_0^x{\displaystyle \frac{\sqrt{1+t}-1}{t}}dt\right)\right)dx=ϵ+{\displaystyle \frac{1}{4}}{ϵ}^2+···,\end{array}$$

(19)

$$\begin{array}{c}\hfill f_2\left(ϵ\right)={\int }_0^{ϵ}\left(exp\left({\int }_0^x{\displaystyle \frac{\sqrt{1+t^2}-1}{t}}dt\right)\right)dx=ϵ+{\displaystyle \frac{1}{12}}{ϵ}^3+···,\end{array}$$

(20)

$$\begin{array}{c}\hfill f_3\left(ϵ\right)={\int }_0^{ϵ}\left(exp\left({\int }_0^x{\displaystyle \frac{\sqrt{1+t^3}-1}{t}}dt\right)\right)dx=ϵ+{\displaystyle \frac{1}{24}}{ϵ}^4+···,\end{array}$$

(21)

$$\begin{array}{c}\hfill f_4\left(ϵ\right)={\int }_0^{ϵ}\left(exp\left({\int }_0^x{\displaystyle \frac{\sqrt{1+t^4}-1}{t}}dt\right)\right)dx=ϵ+{\displaystyle \frac{1}{40}}{ϵ}^5+···,\end{array}$$

(22)

$$\begin{array}{c}\hfill f_5\left(ϵ\right)={\int }_0^{ϵ}\left(exp\left({\int }_0^x{\displaystyle \frac{\sqrt{1+t^5}-1}{t}}dt\right)\right)dx=ϵ+{\displaystyle \frac{1}{60}}{ϵ}^6+···.\end{array}$$

(23)

□

Conjecture 2.2 If $f\in \mathcal{C}L$, then

$$|a_i|\le {\displaystyle \frac{1}{2i(i-1)}}(i\ge 2).$$

Theorem 2.3 If $f\in \mathcal{C}L$, and $\tau \in \mathcal{C}$, then

$$|a_3-\tau a_2^2|\le {\displaystyle \frac{1}{12}}max\left\{1,|{\displaystyle \frac{1-3\tau }{4}}|\right\}.$$

Proof From (14)–(15), Lemma 1.6, we achieve

$$\begin{array}{cc}\hfill |a_3-\tau a_2^2|& ={\displaystyle \frac{1}{12}}|e_2+{\displaystyle \frac{1-3\tau }{4}}{e}_1^2|\le {\displaystyle \frac{1}{12}}max\left\{1,|{\displaystyle \frac{1-3\tau }{4}}|\right\}.\hfill \end{array}$$

□

Setting $\tau =1,$ we acquire the following corollary.

Corollary 2.4 If $f\in \mathcal{C}L$, then

$$|a_3-a_2^2|\le {\displaystyle \frac{1}{12}}.$$

□

Theorem 2.5 If $f\in \mathcal{C}L$, then

$$|a_4-a_2{a}_3|\le {\displaystyle \frac{1}{24}}.$$

The bound is sharp and is achieved by $f_3.$

Proof From From (14)–(16), Lemma 1.2 and 1.3, we get

$$\begin{array}{cc}\hfill |a_4-a_2{a}_3|& ={\displaystyle \frac{1}{384}}|16e_3-4e_1{e}_2-e_1^3|={\displaystyle \frac{1}{384}}|4(e_3-e_1{e}_2+{\displaystyle \frac{1}{4}}{e}_1^3)+12e_3-2e_1^3|\hfill \\ \hfill & \le {\displaystyle \frac{1}{384}}[4+12(1-|e_1{|}^2-{\displaystyle \frac{|e_2{|}^2}{1+|e_1|}})+2|e_1{|}^3]\hfill \\ \hfill & ={\displaystyle \frac{1}{384}}[16-12|e_1{|}^2+2|e_1{|}^3-{\displaystyle \frac{12|e_2{|}^2}{1+|e_1|}}]\hfill \\ \hfill & \le {\displaystyle \frac{1}{384}}[16-12|e_1{|}^2+2|e_1{|}^3]\le {\displaystyle \frac{16}{384}}={\displaystyle \frac{1}{24}}.\hfill \end{array}$$

□Theorem 2.6 If $f\in \mathcal{C}L$, then

$$|a_5-a_2{a}_4|\le {\displaystyle \frac{1}{40}}.$$

The result is sharp and is achieved by $f_4$.

Proof From (14), (16), (17), Lemma 1.4 and 1.2, we yield

$$\begin{array}{cc}\hfill |a_5-a_2{a}_4|& ={\displaystyle \frac{1}{7680}}|192e_4-48e_1{e}_3+12e_1^2{e}_2-9e_1^4|={\displaystyle \frac{1}{7680}}|48(-e_1{e}_3+e_4-{\displaystyle \frac{1}{2}}{e}_2^2+{\displaystyle \frac{3}{4}}{e}_1^2{e}_2-{\displaystyle \frac{1}{8}}{e}_1^4)\hfill \\ \hfill & +144e_4+24e_2^2-24e_1^2{e}_2-3e_1^4|\hfill \\ \hfill & \le {\displaystyle \frac{1}{7680}}[48+144(1-|e_2{|}^2-|e_1{|}^2)+24|e_2{|}^2+24|e_1{|}^2(-|e_1{|}^2+1)+3|e_1{|}^4]\hfill \\ \hfill & ={\displaystyle \frac{1}{7680}}(192-120|e_1{|}^2-21|e_1{|}^4-120|e_2{|}^2)\hfill \\ \hfill & \le {\displaystyle \frac{1}{7680}}(192-120|e_1{|}^2-21|e_1{|}^4)\le {\displaystyle \frac{192}{7680}}={\displaystyle \frac{1}{40}}.\hfill \end{array}$$

□

Theorem 2.7 If $f\in \mathcal{C}L$, then

$$|a_5-a_3^2|\le {\displaystyle \frac{1}{40}}.$$

The bound is sharp for the function $f_4$.

Proof From (15), (17), Lemma 1.2 and 1.4, we achieve

$$\begin{array}{cc}\hfill |a_5-a_3^2|& ={\displaystyle \frac{1}{11520}}|288e_4-80e_2^2+48e_1{e}_3+8e_1^2{e}_2-11e_1^4|\hfill \\ \hfill & ={\displaystyle \frac{1}{11520}}|48({\displaystyle \frac{1}{8}}{e}_1^4+e_1{e}_3+e_4+{\displaystyle \frac{1}{2}}{e}_2^2+{\displaystyle \frac{3}{4}}{e}_1^2{e}_2)-28e_1^2{e}_2-104e_2^2+240e_4-17e_1^4|\hfill \\ \hfill & \le {\displaystyle \frac{1}{11520}}[48+28|e_1{|}^2(1-|e_1{|}^2)+104|e_2{|}^2+240(1-|e_1{|}^2-|e_2{|}^2)+17|e_1{|}^4]\hfill \\ \hfill & ={\displaystyle \frac{1}{11520}}(288-212|e_1{|}^2-11|e_1{|}^4-136|e_2{|}^2)\hfill \\ \hfill & \le {\displaystyle \frac{1}{11520}}(288-212|e_1{|}^2-11|e_1{|}^4)\le {\displaystyle \frac{288}{11520}}={\displaystyle \frac{1}{40}}.\hfill \end{array}$$

□

Theorem 2.8 If $f\in \mathcal{C}L$, then

$$|a_6-a_2{a}_5|\le {\displaystyle \frac{1}{60}}.$$

The bound is sharp and is achieved by $f_5$.

Proof From (14), (17) (18), Lemma 1.2 and 1.5, we achieve

$$\begin{array}{cc}\hfill |a_6-a_2{a}_5|& ={\displaystyle \frac{1}{11520}}|192e_5-48e_4{e}_1-16e_2{e}_3+20e_1^2{e}_3-32e_1^3{e}_2+5e_1^5+32e_1{e}_2^2|\hfill \\ \hfill & ={\displaystyle \frac{1}{11520}}|48(-e_1{e}_4-e_2{e}_3+e_5+{\displaystyle \frac{1}{16}}{e}_1^5-{\displaystyle \frac{1}{2}}{e}_1^3{e}_2+{\displaystyle \frac{3}{4}}{e}_1^2{e}_3+{\displaystyle \frac{3}{4}}{e}_1{e}_2^2)+16(2e_2{e}_3-e_1^5\hfill \\ \hfill & +2e_1^3{e}_2-e_3{e}_1^2+e_2^2{e}_1+e_5)+128e_5-20e_1{e}_2^2+40e_1^3{e}_2+18e_1^5|\hfill \\ \hfill & \le {\displaystyle \frac{1}{11520}}[48+16+128(1-{\displaystyle \frac{|e_3{|}^3}{1+|e_1|}}-|e_1{|}^2-|e_2{|}^2)+20|e_1||e_2{|}^2+40|a_1{|}^3|e_2|+18|e_1{|}^5]\hfill \\ \hfill & ={\displaystyle \frac{1}{11520}}[192-128|e_1{|}^2+18|e_1{|}^5+40|e_1{|}^3|e_2|+(-128+20|e_1|)|e_2{|}^2-{\displaystyle \frac{128|e_3{|}^2}{1+|e_1|}}]\hfill \\ \hfill & \le {\displaystyle \frac{1}{11520}}[18|e_1{|}^5-128|e_1{|}^2+192+40|e_2||e_1{|}^3+(20|e_1|-128)|e_2{|}^2].\hfill \end{array}$$

Setting $x=|e_1|,|e_2|=y,$ we yield

$$\begin{array}{c}\hfill |a_6-a_2{a}_5|\le {\displaystyle \frac{{\alpha }_2(x,y)}{11520}},\end{array}$$

where

$$\begin{array}{c}\hfill {\alpha }_2(x,y)=18x^5-128x^2+192+40x^{3}y+(20x-128)y^2.\end{array}$$

The optimal points of ${\alpha }_2(x,y)$ satisfy the system of equations

$$\left\{\begin{array}{c}{\displaystyle \frac{\partial {\alpha }_2}{\partial x}}=-256x+90x^4+120x^{2}y+20y^2=0,\hfill \\ {\displaystyle \frac{\partial {\alpha }_2}{\partial y}}=40y^3+(-256+40x)y=0.\hfill \end{array}\right.$$

By utilizing numerical computations, we yield

$\left\{\begin{array}{c}{x}_1=0,\hfill \\ y_1=0,\hfill \end{array}\right.$$\left\{\begin{array}{c}{x}_2=1.2826,\hfill \\ y_2=0.4123,\hfill \end{array}\right.$$\left\{\begin{array}{c}{x}_3=7.9506,\hfill \\ y_3=-324.1117.\hfill \end{array}\right.$

Therefore, there is no optimal point in $(0,1)\times (0,1-x^2).$

(4) For $y=0,$

$$\begin{array}{c}\hfill {\alpha }_2(x,0)=192-128x^2+18x^5\le 192.\end{array}$$

(5) For $x=0,$

$$\begin{array}{c}\hfill {\alpha }_2(0,y)=192-128y^2\le 192.\end{array}$$

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

$$\begin{array}{c}\hfill {\alpha }_2(x,1-x^2)=64+20x+128x^2-128x^4-2x^5={\delta }_2\left(x\right)\le {\delta }_2\left(0.7383\right)\approx 110.0671.\end{array}$$

Thus, we get

$$\begin{array}{c}\hfill |a_6-a_2{a}_5|\le {\displaystyle \frac{192}{11520}}={\displaystyle \frac{1}{60}}.\end{array}$$

□

Theorem 2.9 If $f\in \mathcal{C}L$, then

$$|a_6-a_3{a}_4|\le {\displaystyle \frac{1}{60}}.$$

The result is sharp for $f_5$.

Proof From (15), (16), (18), Lemma 1.1, 1.3 and 1.5, we obtain

$$\begin{array}{cc}\hfill |a_6-a_3{a}_4|& ={\displaystyle \frac{1}{92160}}|1536e_5+192e_1{e}_4-448e_2{e}_3+176e_1^2{e}_3-200e_1^3{e}_2+176e_2^2{e}_1+23e_1^5|\hfill \\ \hfill & ={\displaystyle \frac{1}{92160}}|192(e_1{e}_4+{\displaystyle \frac{1}{16}}{e}_1^5+{\displaystyle \frac{1}{2}}{e}_1^3{e}_2+{\displaystyle \frac{3}{4}}{e}_1^2{e}_3+e_2{e}_3+{\displaystyle \frac{3}{4}}{e}_1{e}_2^2+e_5)+320(-2e_2{e}_3+2e_1^3{e}_2\hfill \\ \hfill & -e_1^5-e_1^2{e}_3+e_1{e}_2^2+e_5)+1024e_5-288e_1{e}_2^2+352e_1^2(-2e_1{e}_2+e_1^3+e_3)-232e_1^3{e}_2-21e_1^5|\hfill \\ \hfill & \le {\displaystyle \frac{1}{92160}}[192+320+1024(-|e_1{|}^2+1-|e_2{|}^2-{\displaystyle \frac{|e_3{|}^3}{1+|e_1|}})+288|e_1||e_2{|}^2+352{|e_1|}^2\hfill \\ \hfill & +232|e_1{|}^3|e_2|+21|e_1{|}^5]\hfill \\ \hfill & ={\displaystyle \frac{1}{92160}}[1536-672|e_1{|}^2+21|e_1{|}^5+232|e_2||e_1{|}^3+(288|e_1|-1024)|e_2{|}^2-{\displaystyle \frac{1024|e_3{|}^2}{1+|e_1|}}]\hfill \\ \hfill & \le {\displaystyle \frac{1}{2560}}[21|e_1{|}^5-672|e_1{|}^2+1536+232|e_1{|}^3|e_2|+(-1024+288|e_1|)|e_2{|}^2].\hfill \end{array}$$

Let $|e_2|=y,|e_1|=x,$ then we have

$$\begin{array}{c}\hfill |a_6-a_3{a}_4|\le {\displaystyle \frac{{\alpha }_3(x,y)}{2560}},\end{array}$$

where

$$\begin{array}{c}\hfill {\alpha }_3(x,y)=21x^5-672x^2+1536+232x^{3}y+(-1024+288x)y^2.\end{array}$$

The optimal points of ${\alpha }_3$ satisfy the system of equations

$$\left\{\begin{array}{c}{\displaystyle \frac{\partial {\alpha }_3}{\partial x}}=105x^4-1344x+696y{x}^2+288y^2=0,\hfill \\ {\displaystyle \frac{\partial {\alpha }_3}{\partial y}}=232x^3+(-2048+576x)y=0.\hfill \end{array}\right.$$

By applying numerical computations, we yield

$\left\{\begin{array}{c}{x}_1=0,\hfill \\ y_1=0,\hfill \end{array}\right.$$\left\{\begin{array}{c}{x}_2\approx 1.5613,\hfill \\ y_2\approx 0.7687,\hfill \end{array}\right.$$\left\{\begin{array}{c}{x}_3\approx 4.3175,\hfill \\ y_3\approx -42.5443,\hfill \end{array}\right.$$\left\{\begin{array}{c}{x}_4=-3.7197,\hfill \\ y_4=-2.8492.\hfill \end{array}\right.$

Thus, there is no optimal point in $(0,1)\times (0,1-x^2).$

(7) For $y=0,$

$$\begin{array}{c}\hfill {\alpha }_3(x,0)=1536-672x^2+21x^5\le 1536.\end{array}$$

(8) For $x=0,$

$$\begin{array}{c}\hfill {\alpha }_3(0,y)=1536-1024y^2\le 1536.\end{array}$$

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

$$\begin{array}{c}\hfill {\alpha }_3(x,1-x^2)=512+288x+1376x^2-344x^3-1024x^4+77x^5={\delta }_3\left(x\right)\le {\delta }_3\left(0.7810\right)\approx 1053.8.\end{array}$$

Therefore, we achieve

$$\begin{array}{c}\hfill |a_6-a_3{a}_4|\le {\displaystyle \frac{1536}{92160}}={\displaystyle \frac{1}{60}}.\end{array}$$

□

Conjecture 2.10 If $f\in \mathcal{C}L$, then

$$|a_i{a}_j-a_{i+j-1}|\le {\displaystyle \frac{1}{2(i+j-2)(i+j-1)}}.$$

Theorem 2.11 If $f\in \mathcal{C}L$, then

$$|H_{2,2}\left(f\right)|=|a_2{a}_4-a_3^2|\le {\displaystyle \frac{1}{144}}.$$

The bound is sharp for functions $f_2$ given by (20).

Proof Let $f\in \mathcal{C}L.$ From (14)–(16) and Lemma 1.2, we have

$$\begin{array}{cc}\hfill |a_2{a}_4-a_3^2|& ={\displaystyle \frac{1}{4608}}|32e_2^2-48e_1{e}_3+4e_1^2{e}_2-e_1^4|\hfill \\ \hfill & \le {\displaystyle \frac{1}{4608}}[32|e_2{|}^2+48|e_1|(1-{\displaystyle \frac{|e_2{|}^2}{1+|e_1|}}-|e_1{|}^2)+4|e_1{|}^2|e_2|+|e_1{|}^4]\hfill \\ \hfill & ={\displaystyle \frac{1}{4608}}[48|e_1|-48|e_1{|}^3+|e_1{|}^4+{\displaystyle \frac{32-16|e_1|}{1+|e_1|}}|e_2{|}^2+4|e_1{|}^2|e_2|]\hfill \\ \hfill & \le {\displaystyle \frac{1}{4608}}[48|e_1|-48|e_1{|}^3+|e_1{|}^4+{\displaystyle \frac{32-16|e_1|}{1+|e_1|}}(1-|e_1{|}^2{)}^2+4|e_1{|}^2(1-|e_1{|}^2\left)\right]\hfill \\ \hfill & ={\displaystyle \frac{1}{4608}}(32-12|e_1{|}^2-19|e_1{|}^4)\le {\displaystyle \frac{32}{4608}}={\displaystyle \frac{1}{144}}.\hfill \end{array}$$

□

Theorem 2.12 If $f\in \mathcal{C}L$, then

$$|H_{2,3}\left(f\right)|=|a_5{a}_3-a_4^2|\le {\displaystyle \frac{1}{576}}.$$

The bound is sharp for functions $f_3$ given by (21).

Proof Let $f\in \mathcal{C}L.$ From (15)–(17) and (4), we achieve

$$\begin{array}{ccc}\hfill a_3{a}_5-a_4^2& =& {\displaystyle \frac{1}{737280}}(-8e_1^4{e}_2-96e_1^3{e}_3-13e_1^6+176e_1^2{e}_2^2+1536e_2{e}_4-384e_1{e}_2{e}_3\hfill \\ & +& 384e_1^2{e}_4-1280e_3^2).\hfill \end{array}$$

(24)

Using (24) and Lemma 1.7, we yield

$$\begin{array}{ccc}\hfill H_{2,3}\left(f\right)& =& {\displaystyle \frac{1}{737280}}[256{\xi }_1^4(1-|e_1{|}^2{)}^2{\overline{e_1}}^2-1536|{\xi }_1{|}^2{\xi }_2^2(1-|{\xi }_1{|}^2\left)\right(1-|e_1{|}^2{)}^2+384{\xi }_1^3(1-|e_1{|}^2{)}^2{|e_1|}^2\hfill \\ & -& 512\overline{e_1}{\xi }_1^2(1-|{\xi }_1{|}^2\left)\right(1-|e_1{|}^2{)}^2{\xi }_2-384{\xi }_2{e}_1{\xi }_1(1-|e_1{|}^2{)}^2(1-|{\xi }_1{|}^2)+1536{\xi }_1{\xi }_3(1-|{\xi }_1{|}^2)\hfill \\ & \times & (1-|e_1{|}^2{)}^2(1-|{\xi }_2{|}^2)-13e_1^6+176e_1^2(1-|e_1{|}^2{)}^2{\xi }_1^2-8{\xi }_1(1-|e_1{|}^2)e_1^4+384{\xi }_1^3(1-|e_1{|}^2){\overline{e_1}}^2{e}_1^2\hfill \\ & +& 96{\xi }_1^2\overline{e_1}{e}_1^3(1-|e_1{|}^2)-1280{\xi }_2^2(1-|e_1{|}^2{)}^2(1-|{\xi }_1{|}^2{)}^2-96{\xi }_2{e}_1^3(1-|e_1{|}^2\left)\right(1-|{\xi }_1{|}^2)+384{\xi }_3{e}_1^2\hfill \\ & \times & (1-|e_1{|}^2\left)\right(1-|{\xi }_2{|}^2\left)\right(1-|{\xi }_1{|}^2)-384{\xi }_2^2{e}_1^2\overline{{\xi }_1}(1-|e_1{|}^2\left)\right(1-|{\xi }_1{|}^2)-768{\xi }_2{\xi }_1{e}_1^2\overline{e_1}(1-|e_1{|}^2\left)\right(1-|{\xi }_1{|}^2)].\hfill \end{array}$$

Therefore, we yield

$$H_{2,3}\left(f\right)={\displaystyle \frac{1}{737280}}\left[{\varrho }_1\left(e_1,{\xi }_1\right)+{\varrho }_2\left(e_1,{\xi }_1\right){\xi }_2+{\varrho }_3\left(e_1,{\xi }_1\right){\xi }_2^2+{\varrho }_4\left(e_1,{\xi }_1,{\xi }_2\right){\xi }_3\right],$$

where

$$\begin{array}{ccc}\hfill {\varrho }_1\left(e_1,{\xi }_1\right)& =& -13e_1^6+\left(1-|e_1{|}^2\right)\left[\left(1-|e_1{|}^2\right)\left(176e_1^2{\xi }_1^2++384{|e_1|}^2{\xi }_1^3+256{\overline{e_1}}^2{\xi }_1^4\right)+96e_1^3\overline{e_1}{\xi }_1^2\right.\hfill \\ & & \left.+384e_1^2{\overline{e_1}}^2{\xi }_1^3-8e_1^4{\xi }_1\right],\hfill \\ \hfill {\varrho }_2\left(e_1,{\xi }_1\right)& =& (1-|e_1{|}^2)\left(1-|{\xi }_1{|}^2\right)\left[(-384e_1{\xi }_1-512{\xi }_1^2\overline{e_1})(1-|e_1{|}^2)-768\overline{e_1}{e}_1^2{\xi }_1-96e_1^3\right],\hfill \\ \hfill {\varrho }_3\left(e_1,{\xi }_1\right)& =& \left(1-|e_1{|}^2\right)\left(1-|{\xi }_1{|}^2\right)\left[(-256{\left|{\xi }_1\right|}^2-1280)\left(1-|e_1{|}^2\right)-384\overline{{\xi }_1}{e}_1^2\right],\hfill \\ \hfill {\varrho }_4\left(e_1,{\xi }_1,{\xi }_2\right)& =& \left(1-|e_1{|}^2\right)\left(1-|{\xi }_1{|}^2\right)\left(1-|{\xi }_2{|}^2\right)\left[1536{\xi }_1\left(1-|e_1{|}^2\right)+384e_1^2\right].\hfill \end{array}$$

By setting $|e_1|=e,|{\xi }_1|=\aleph ,$$|{\xi }_2|=\nu$ and upon taking $\left|{\xi }_3\right|\le 1,$ we achieve

$$\begin{array}{ccc}\hfill \left|H_{2,3}\left(f\right)\right|& \le & {\displaystyle \frac{1}{737280}}\left(\left|{\varrho }_1\left(e_1,{\xi }_1\right)\right|+\left|{\varrho }_2\left(e_1,{\xi }_1\right)\right|\nu +\left|{\varrho }_3\left(e_1,{\xi }_1\right)\right|{\nu }^2+\left|{\varrho }_4\left(e_1,{\xi }_1,{\xi }_2\right)\right|\right)\hfill \\ & \le & {\displaystyle \frac{1}{737280}}{\mathsf{ϝ}}_1\left(e,\aleph ,\nu \right),\hfill \end{array}$$

where

$$\begin{array}{c}\hfill {\mathsf{ϝ}}_1\left(e,\aleph ,\nu \right)={\iota }_1\left(e,\aleph \right)+{\iota }_2\left(e,\aleph \right)\nu +{\iota }_3\left(e,\aleph \right){\nu }^2+{\iota }_4\left(e,\aleph \right)(1-{\nu }^2),\end{array}$$

with

$$\begin{array}{ccc}\hfill {\iota }_1\left(e,\aleph \right)& =& 13e^6+\left(1-e^2\right)\left[\left(1-e^2\right)\left(176e^2{\aleph }^2+384e^2{\aleph }^3+256e^2{\aleph }^4\right)+8e^4\aleph +384e^4{\aleph }^3+96e^4{\aleph }^2\right],\hfill \\ \hfill {\iota }_2\left(e,\aleph \right)& =& \left(1-e^2\right)\left(1-{\aleph }^2\right)\left[\left(384e\aleph +512e{\aleph }^2\right)\left(1-e^2\right)+768e^3\aleph +96e^3\right],\hfill \\ \hfill {\iota }_3(e,\aleph )& =& (1-e^2)(1-{\aleph }^2)[(1280+256{\aleph }^2)(1-e^2)+384\aleph e^2],\hfill \\ \hfill {\iota }_4\left(e,\aleph \right)& =& 384\left(1-e^2\right)\left(1-{\aleph }^2\right)\left[4\aleph \left(1-e^2\right)+e^2\right].\hfill \end{array}$$

We will maximize ${\mathsf{ϝ}}_1$ in the closed cuboid $\Xi :[0,1]\times [0,1]\times [0,1].$

Upon $(e,\aleph )\in [0,1]\times [0,1],$ we yield

$$\begin{array}{ccc}\hfill {\iota }_3(e,\aleph )& \le & (1-{\aleph }^2)(1-e^2)[(256{\aleph }^2+1280)(1-e^2)+384e^2]={\tau }_3(e,\aleph ).\hfill \end{array}$$

Taking ${\tau }_j={\iota }_j$ for $j=1,2,4$ and

$${\Gamma }_1\left(e,\aleph ,\nu \right)={\tau }_1\left(e,\aleph \right)+{\tau }_2\left(e,\aleph \right)\nu +{\tau }_3\left(e,\aleph \right){\nu }^2+{\tau }_4\left(e,\aleph \right)(1-{\nu }^2),$$

It is not difficult to see that ${\mathsf{ϝ}}_1\left(e,\aleph ,\nu \right)\le {\Gamma }_1\left(e,\aleph ,\nu \right)$ in the cuboid $\Xi .$

Differentiating ${\Gamma }_1$ with respect to $\nu$, we achieve

$${\displaystyle \frac{\partial {\Gamma }_1}{\partial \nu }}={\tau }_2(e,\aleph )+2\nu [{\tau }_3(e,\aleph )-{\tau }_4(e,\aleph )].$$

In light of ${\tau }_2(e,\aleph )\ge 0$ and

$${\tau }_3(e,\aleph )-{\tau }_4(e,\aleph )=256{(1-e^2)}^2(1-{\aleph }^2)({\aleph }^2-6\aleph +5)\ge 0$$

on $[0,1]\times [0,1],$ we yield ${\displaystyle \frac{\partial {\Gamma }_1}{\partial \nu }}\ge 0$ for all $\nu \in [0,1].$ Thus, we have

$$\begin{array}{ccc}\hfill {\mathsf{ϝ}}_1\left(e,\aleph ,\nu \right)& \le & {\Gamma }_1\left(e,\aleph ,y\right)\le {\Gamma }_1\left(e,\aleph ,1\right)={\tau }_1(e,\aleph )+{\tau }_2(e,\aleph )+{\tau }_3(e,\aleph )\hfill \\ & =& 13e^6-96e^5+896e^4+96e^3-2176e^2+1280+(-8e^6-384e^5+8e^4+384e)\aleph \hfill \\ & +& (80e^6+608e^5-896e^4-1120e^3+1840e^2+512e-1024){\aleph }^2+(384e^5-384e^4\hfill \\ & +& 384e^2-384e){\aleph }^3+(256e^6-512e^5-768e^4+1024e^3+768e^2-512e-256){\aleph }^4\hfill \\ & & ={\vartheta }_1(e,\aleph ).\hfill \end{array}$$

Consider

$$\left\{\begin{array}{c}{\displaystyle \frac{\partial {\vartheta }_1}{\partial e}}=78e^5-480e^4+3584e^3+288e^2-4352e+(-48e^5-1920e^4+32e^3+384)\aleph +(480e^5+3040e^4\hfill \\ -3584e^3-3360e^2+3680e+512){\aleph }^2+(1920e^4-1536e^3+768e-384){\aleph }^3+(1536e^5-2560e^4\hfill \\ -3072e^3+3072e^2+1536e-512){\aleph }^4=0\hfill \\ {\displaystyle \frac{\partial {\vartheta }_1}{\partial \aleph }}=384e+8e^4-384e^5-8e^6+(160e^6+1216e^5-1792e^4-2240e^3+3680e^2+1024e-2048)\aleph \hfill \\ +(1152e^5-1152e^4+1152e^2-1152e){\aleph }^2+(1024e^6-2048e^5-3072e^4+4096e^3+3072e^2-2048e\hfill \\ -1024){\aleph }^3=0\hfill \end{array}\right.$$

By calculating, we have

$\left\{\begin{array}{c}{e}_{1,1}\approx 1,\hfill \\ {\aleph }_{1,1}\approx 1.2799,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,2}\approx 1,\hfill \\ {\aleph }_{1,2}\approx -1.7742,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,3}\approx 1,\hfill \\ {\aleph }_{1,3}\approx -0.5057,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,4}\approx -1,\hfill \\ {\aleph }_{1,4}\approx 0.6698,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,5}\approx -1,\hfill \\ {\aleph }_{1,5}\approx -0.8729,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,6}\approx -1,\hfill \\ {\aleph }_{1,6}\approx 0.3697,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,7}=0,\hfill \\ {\aleph }_{1,7}=0,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,8}\approx 1.3544,\hfill \\ {\aleph }_{1,8}\approx 1.5451,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,9}\approx 1.3593,\hfill \\ {\aleph }_{1,9}\approx 0.7288,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,10}\approx -0.6646,\hfill \\ {\aleph }_{1,10}\approx 0.9982,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,11}\approx -0.7003,\hfill \\ {\aleph }_{1,11}\approx -4.5145,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,12}\approx -0.9906,\hfill \\ {\aleph }_{1,12}\approx 0.5353,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,13}\approx -1.1841,\hfill \\ {\aleph }_{1,13}\approx -0.5150.\hfill \end{array}\right.$

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

(10) For $e=1,$

$$\begin{array}{cc}\hfill {\vartheta }_1(1,\aleph )& =13.\end{array}$$

(11) For $e=0,$

$$\begin{array}{ccc}\hfill {\vartheta }_1(0,\aleph )& =& 1280-1024{\aleph }^2-256{\aleph }^4\le 1280.\hfill \end{array}$$

(12) For $\aleph =0$

$$\begin{array}{c}\hfill {\vartheta }_1(e,0)=1280-2176e^2+96e^3+896e^4-96e^5+13e^6\le 1280.\end{array}$$

(13) For $\aleph =1,$

$$\begin{array}{c}\hfill {\vartheta }_1(e,1)=341e^6-1144e^4+816e^2={\delta }_4\left(e\right)\le {\delta }_4\left(0.6673\right)\approx 166.6288.\end{array}$$

Thus, we achieve

$$\begin{array}{c}\hfill |a_5{a}_3-a_4^2|\le {\displaystyle \frac{1280}{737280}}={\displaystyle \frac{1}{576}}.\end{array}$$

□

Theorem 2.13 If $f\in \mathcal{C}L$, then

$$|H_{3,1}\left(f\right)|\le {\displaystyle \frac{1}{576}}.$$

The outcome is sharp for the function $f_3$.

Proof Let $f\in \mathcal{C}L$. From (14)–(17) and (5), we yield

$$\begin{array}{ccc}\hfill H_{3,1}\left(f\right)& =& {\displaystyle \frac{1}{2211840}}(-360e_2{e}_1^4+96e_3{e}_1^3+528e_2^2{e}_1^2+4608e_2{e}_4+2688e_1{e}_3{e}_2\hfill \\ & -& 2304e_4{e}_1^2-1280e_2^3-3840e_3^2+73e_1^6).\hfill \end{array}$$

(25)

From Lemma 1.7 and (25), we obtain

$$\begin{array}{ccc}\hfill H_{3,1}\left(f\right)& =& {\displaystyle \frac{1}{2211840}}\left[768\right(1-|e_1{|}^2{)}^2{\xi }_1^4{\overline{e_1}}^2-4608|{\xi }_1{|}^2{\xi }_2^2(1-|{\xi }_1{|}^2\left)\right(1-|e_1{|}^2{)}^2-2688{\xi }_1^3(1-|e_1{|}^2{)}^2{|e_1|}^2\hfill \\ & -& 1536\overline{e_1}{\xi }_1^2(1-|{\xi }_1{|}^2\left)\right(1-|e_1{|}^2{)}^2{\xi }_2+2688{\xi }_2{e}_1{\xi }_1(1-|e_1{|}^2{)}^2(1-|{\xi }_1{|}^2)+4608{\xi }_1{\xi }_3(1-|{\xi }_1{|}^2)\hfill \\ & \times & (1-|e_1{|}^2{)}^2(1-|{\xi }_2{|}^2)+73e_1^6+528e_1^2(1-|e_1{|}^2{)}^2{\xi }_1^2-1280(1-|e_1{|}^2{)}^3{\xi }_1^3-360{\xi }_1(1-|e_1{|}^2)e_1^4\hfill \\ & -& 2304{\xi }_1^3(1-|e_1{|}^2){\overline{e_1}}^2{e}_1^2-96{\xi }_1^2\overline{e_1}{e}_1^3(1-|e_1{|}^2)-3840{\xi }_2^2(1-|e_1{|}^2{)}^2(1-|{\xi }_1{|}^2{)}^2+96{\xi }_2{e}_1^3(1-|e_1{|}^2)\hfill \\ & \times & (1-|{\xi }_1{|}^2)-2304{\xi }_3{e}_1^2(1-|e_1{|}^2\left)\right(1-|{\xi }_2{|}^2\left)\right(1-|{\xi }_1{|}^2)+2304{\xi }_2^2{e}_1^2\overline{{\xi }_1}(1-|e_1{|}^2\left)\right(1-|{\xi }_1{|}^2)\hfill \\ & +& 4608{\xi }_2{\xi }_1{e}_1^2\overline{e_1}(1-|e_1{|}^2\left)\right(1-|{\xi }_1{|}^2)].\hfill \end{array}$$

Thus, we have

$$H_{3,1}\left(f\right)={\displaystyle \frac{1}{2211840}}\left[{\varrho }_5\left(e_1,{\xi }_1\right)+{\varrho }_6\left(e_1,{\xi }_1\right){\xi }_2+{\varrho }_7\left(e_1,{\xi }_1\right){\xi }_2^2+{\varrho }_8\left(e_1,{\xi }_1,{\xi }_2\right){\xi }_3\right],$$

where

$$\begin{array}{ccc}\hfill {\varrho }_5\left(e_1,{\xi }_1\right)& =& 73e_1^6+\left(1-|e_1{|}^2\right)\left[\left(1-|e_1{|}^2\right)\left(768{\overline{e_1}}^2{\xi }_1^4+528e_1^2{\xi }_1^2-1408{|e_1|}^2{\xi }_1^3-1280{\xi }_1^3\right)-96e_1^3\overline{e_1}{\xi }_1^2\right.\hfill \\ & & \left.-2304e_1^2{\overline{e_1}}^2{\xi }_1^3-360e_1^4{\xi }_1\right],\hfill \\ \hfill {\varrho }_6\left(e_1,{\xi }_1\right)& =& (1-|e_1{|}^2)\left(1-|{\xi }_1{|}^2\right)\left[(2688e_1{\xi }_1-1536{\xi }_1^2\overline{e_1})(1-|e_1{|}^2)+4608e_1^2\overline{e_1}{\xi }_1+96e_1^3\right],\hfill \\ \hfill {\varrho }_7\left(e_1,{\xi }_1\right)& =& \left(1-|e_1{|}^2\right)\left(1-|{\xi }_1{|}^2\right)\left[(-768{\left|{\xi }_1\right|}^2-3840)\left(1-|e_1{|}^2\right)+2304e_1^2\overline{{\xi }_1}\right],\hfill \\ \hfill {\varrho }_8\left(e_1,{\xi }_1,{\xi }_2\right)& =& \left(1-|e_1{|}^2\right)\left(1-|{\xi }_2{|}^2\right)\left(1-|{\xi }_1{|}^2\right)\left[4608{\xi }_1\left(1-|e_1{|}^2\right)-2304e_1^2\right].\hfill \end{array}$$

Setting $\aleph =|{\xi }_1|,$$\nu =|{\xi }_2|,|e_1|=e$, and upon taking $\left|{\xi }_3\right|\le 1,$ we yield

$$\begin{array}{ccc}\hfill \left|H_{3,1}\left(f\right)\right|& \le & {\displaystyle \frac{1}{2211840}}\left(\left|{\varrho }_5\left(e_1,{\xi }_1\right)\right|+\left|{\varrho }_6\left(e_1,{\xi }_1\right)\right|\nu +\left|{\varrho }_7\left(e_1,{\xi }_1\right)\right|{\nu }^2+\left|{\varrho }_8\left(e_1,{\xi }_1,{\xi }_2\right)\right|\right)\hfill \\ & \le & {\displaystyle \frac{1}{2211840}}{\mathsf{ϝ}}_2\left(e,\aleph ,\nu \right),\hfill \end{array}$$

where

$$\begin{array}{c}\hfill {\mathsf{ϝ}}_2\left(c,\aleph ,\nu \right)={\iota }_5\left(e,\aleph \right)+{\iota }_6\left(e,\aleph \right)\nu +{\iota }_7\left(e,\aleph \right){\nu }^2+{\iota }_8\left(e,\aleph \right)(1-{\nu }^2),\end{array}$$

$$\begin{array}{ccc}\hfill {\iota }_5\left(e,\aleph \right)& =& 73e^6+\left(1-e^2\right)\left[\left(528e^2{\aleph }^2+1408e^2{\aleph }^3+1280{\aleph }^3+768e^2{\aleph }^4\right)\left(1-e^2\right)+360e^4\aleph +96e^4{\aleph }^2+2304e^4{\aleph }^3\right],\hfill \\ \hfill {\iota }_6\left(e,\aleph \right)& =& \left(1-{\aleph }^2\right)\left(1-e^2\right)\left[\left(2688\aleph e+1536{\aleph }^{2}e\right)\left(1-e^2\right)+96e^3+4608\aleph e^3\right],\hfill \\ \hfill {\iota }_7(e,\aleph )& =& (1-e^2)(1-{\aleph }^2)[(3840+768{\aleph }^2)(1-e^2)+2304\aleph e^2],\hfill \\ \hfill {\iota }_8\left(e,\aleph \right)& =& \left(1-{\aleph }^2\right)\left(1-e^2\right)\left[2304e^2+4608\left(1-e^2\right)\aleph \right].\hfill \end{array}$$

In view of $(e,\aleph )\in [0,1]\times [0,1],$ we have

$$\begin{array}{ccc}\hfill {\iota }_7(e,\aleph ))& \le & (1-e^2)(1-{\aleph }^2)[(768{\aleph }^2+3840)(1-e^2)+2304e^2]={\tau }_7(e,\aleph ).\hfill \end{array}$$

Setting ${\tau }_j={\iota }_j$ for $j=5,6,8$ and

$${\Gamma }_2\left(e,\aleph ,\nu \right)={\tau }_5\left(e,\aleph \right)+{\tau }_6\left(e,\aleph \right)\nu +{\tau }_7\left(e,\aleph \right){\nu }^2+{\tau }_8\left(e,\aleph \right)(1-{\nu }^2).$$

Differentiating ${\Gamma }_2$ with respect to $\nu$, we yield

$${\displaystyle \frac{\partial {\Gamma }_2}{\partial \nu }}={\tau }_6(e,\aleph )+2[{\tau }_7(e,\aleph )-{\tau }_8(e,\aleph )]\nu .$$

Upon ${\tau }_6(e,\aleph )\ge 0$ and

$${\tau }_7(e,\aleph )-{\tau }_8(e,\aleph )=768(1-{\aleph }^2){(1-e^2)}^2(\aleph -1)(\aleph -5)\ge 0,$$

we have ${\displaystyle \frac{\partial {\Gamma }_2}{\partial \nu }}\ge 0.$ Therefore, we obtain

$$\begin{array}{ccc}\hfill {\mathsf{ϝ}}_2\left(e,\aleph ,\nu \right)& \le & {\Gamma }_2\left(e,\aleph ,\nu \right)\le {\Gamma }_2\left(e,\aleph ,1\right)={\tau }_5(e,\aleph )+{\tau }_6(e\aleph )+{\tau }_7(e,\aleph )\hfill \\ & =& 73e^6-96e^5+1536e^4+96e^3-5376e^2+3840+(768e^6-1536e^5-2304e^4+3072e^3\hfill \\ & +& 2304e^2-1536e-768){\aleph }^4+(-896e^6+1920e^5+768e^4+768e^3-1152e^2-2688e+1280){\aleph }^3\hfill \\ & +& (432e^6+1632e^5-1728e^4-3168e^3+4368e^2+1536e-3072){\aleph }^2+(-360e^6-1920e^5+360e^4\hfill \\ & -& 768e^3+2688e)\aleph ={\vartheta }_2(e,\aleph ).\hfill \end{array}$$

Consider

$$\left\{\begin{array}{c}{\displaystyle \frac{\partial {\vartheta }_2}{\partial e}}=438e^5-480e^4+6144e^3+288e^2-10752e+(4608e^5-7680e^4-9216e^3+9216e^2+4608e-1536){\aleph }^4\hfill \\ +(-5376e^5+9600e^4+3072e^3+2304e^2-2304e-2688){\aleph }^3+(2592e^5+8160e^4-6912e^3-9504e^2+8736e\hfill \\ +1536){\aleph }^2+(-2160e^5-9600e^4+1440e^3-2304e^2+2688)\aleph =0\hfill \\ {\displaystyle \frac{\partial {\vartheta }_2}{\partial \aleph }}=2688e-768e^3+360e^4-1920e^5-360e^6+(3072e^6-6144e^5-9216e^4+12288e^3+9216e^2-6144e\hfill \\ -3072){\aleph }^3+(-2688e^6+5760e^5+2304e^4+2304e^3-3456e^2-8064e+3840){\aleph }^2+(864e^6+3264e^5-3456e^4\hfill \\ -6336e^3+8736e^2+3072e-6144)\aleph =0\hfill \end{array}\right.$$

By calculating, we achieve

$\left\{\begin{array}{c}{e}_{1,1}\approx 1,\hfill \\ {\aleph }_{1,1}\approx 1.2745,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,2}\approx 1,\hfill \\ {\aleph }_{1,2}\approx -1.8793,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,3}\approx 1,\hfill \\ {\aleph }_{1,3}\approx -0.3952\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,4}\approx -1,\hfill \\ {\aleph }_{1,4}\approx -0.8379,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,5}=0,\hfill \\ {\aleph }_{1,5}=0,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,6}\approx 1.6312,\hfill \\ {\aleph }_{1,6}\approx 2.1674,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,7}\approx -0.4195,\hfill \\ {\aleph }_{1,7}\approx 1.0202,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,8}\approx -0.7005,\hfill \\ {\aleph }_{1,8}\approx -9.2713,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,9}\approx -1.0177,\hfill \\ {\aleph }_{1,9}\approx 0.5591,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,10}\approx -1.5719,\hfill \\ {\aleph }_{1,10}\approx -0.4054,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,11}\approx -1.9249,\hfill \\ {\aleph }_{1,11}\approx 1.8266.\hfill \end{array}\right.$

Therefore, there is no critical point in $(0,1)\times (0,1)$.

(14) For $e=1,$

$$\begin{array}{c}\hfill {\vartheta }_2(1,\aleph )=73.\end{array}$$

(15) For $e=0,$

$$\begin{array}{c}\hfill {\vartheta }_2(0,\aleph )=3840-3072{\aleph }^2+1280{\aleph }^3-768{\aleph }^4\le 3840.\end{array}$$

(16) For $\aleph =0,$

$$\begin{array}{c}\hfill {\vartheta }_2(e,0)=3840-5376e^2+96e^3+1536e^4-96e^5+73e^6\le 3840.\end{array}$$

(17) For $\aleph =1$

$$\begin{array}{c}\hfill {\vartheta }_2(e,1)=1280+144e^2-1368e^4+17e^6={\delta }_5\left(e\right)\le {\delta }_5\left(0.2295\right)\approx 1283.8.\end{array}$$

Thus, we have

$$\begin{array}{c}\hfill |H_{3,1}\left(f\right)|\le {\displaystyle \frac{3840}{2211840}}={\displaystyle \frac{1}{576}}.\end{array}$$

□

3. The Second-Order Hankel Determinant of Logarithmic Coefficients for $\mathcal{C}L$

Theorem 3.1 If $f\in \mathcal{C}L$, then

$$|{\gamma }_1|\le {\displaystyle \frac{1}{8}},|{\gamma }_2|\le {\displaystyle \frac{1}{24}},|{\gamma }_3|\le {\displaystyle \frac{1}{48}},|{\gamma }_4|\le {\displaystyle \frac{1}{80}},|{\gamma }_5|\le {\displaystyle \frac{1}{120}}.$$

These sharpness are given by (19), (20), (21), (22) and (23), respectively.

Proof Let $f\in \mathcal{C}L$. From (14)–(18) and (7), we have

$$\begin{array}{c}\hfill {\gamma }_1={\displaystyle \frac{1}{8}}{e}_1,\end{array}$$

(26)

$$\begin{array}{c}\hfill {\gamma }_2={\displaystyle \frac{8e_2-e_1^2}{192}},\end{array}$$

(27)

$$\begin{array}{c}\hfill {\gamma }_3={\displaystyle \frac{16e_3-4e_2{e}_1+e_1^3}{768}},\end{array}$$

(28)

$$\begin{array}{c}\hfill {\gamma }_4={\displaystyle \frac{1152e_4-288e_1{e}_3+232e_1^2{e}_2-160e_2^2-49e_1^4}{92160}},\end{array}$$

(29)

$$\begin{array}{c}\hfill {\gamma }_5={\displaystyle \frac{1}{921600}}(7680e_5-1920e_1{e}_4-2240e_2{e}_3+1600e_1^2{e}_3-1380e_1^3{e}_2+1680e_1{e}_2^2+240e_1^5).\end{array}$$

(30)

From (26), (27), Lemma 1.1 and 1.6, we have

$$|{\gamma }_1|\le {\displaystyle \frac{1}{8}},|{\gamma }_2|={\displaystyle \frac{1}{24}}|e_2-{\displaystyle \frac{1}{8}}{e}_1^2|\le {\displaystyle \frac{1}{24}}max\{1,|-{\displaystyle \frac{1}{8}}|\}={\displaystyle \frac{1}{24}}.$$

From Lemma 1.2, 1.3 and (28), we obtain

$$\begin{array}{cc}\hfill |{\gamma }_3|& ={\displaystyle \frac{1}{768}}|2(e_1^3-2e_1{e}_2+e_3)-e_1^3+14e_3|\hfill \\ \hfill & \le {\displaystyle \frac{1}{768}}[2|e_1^3-2e_1{e}_2+e_3|+|e_1{|}^3+14(1-|e_1{|}^2-{\displaystyle \frac{|e_2{|}^2}{1+|e_1|}}\left)\right]\hfill \\ \hfill & \le {\displaystyle \frac{1}{768}}(-{\displaystyle \frac{14|e_2{|}^2}{1+|e_1|}}+16+|e_1{|}^3-14|e_1{|}^2)\hfill \\ \hfill & \le {\displaystyle \frac{1}{768}}(16-14|e_1{|}^2+|e_1{|}^3)\le {\displaystyle \frac{16}{768}}={\displaystyle \frac{1}{48}}.\hfill \end{array}$$

From (29), Lemma 1.4 and 1.2, we yield

$$\begin{array}{cc}\hfill |{\gamma }_4|& ={\displaystyle \frac{1}{92160}}|1008e_4+144(e_4+3e_2{e}_1^2-2e_3{e}_1-e_1^4-e_2^2)-16e_2^2-200e_1^2{e}_2+95e_1^4|\hfill \\ \hfill & \le {\displaystyle \frac{1}{92160}}\left[1008\right(1-|e_2{|}^2-|e_1{|}^2)+144+16|e_2{|}^2+200(1-|e_1{|}^2)|e_1{|}^2+95|e_1{|}^4]\hfill \\ \hfill & ={\displaystyle \frac{1}{92160}}(1152-808|e_1{|}^2-105|e_1{|}^4-992|e_2{|}^2)\hfill \\ \hfill & \le {\displaystyle \frac{1}{92160}}(1152-808|e_1{|}^2-105|e_1{|}^4)\le {\displaystyle \frac{1152}{96120}}={\displaystyle \frac{1}{80}}.\hfill \end{array}$$

From (30)

$$\begin{array}{cc}\hfill |{\gamma }_5|& ={\displaystyle \frac{1}{921600}}|1920(-e_1{e}_4+e_5-e_2{e}_3+{\displaystyle \frac{3}{4}}{e}_2^2{e}_1-{\displaystyle \frac{1}{2}}{e}_2{e}_1^3+{\displaystyle \frac{3}{4}}{e}_3{e}_1^2+{\displaystyle \frac{1}{16}}{e}_1^5)+160(-2e_2{e}_3+e_5-e_3{e}_1^2\hfill \\ \hfill & +e_2^2{e}_1+2e_2{e}_1^3-e_1^5)+5600e_5+80e_1{e}_2^2+320(e_3+e_1^3-2e_2{e}_1)e_1^2-100e_2{e}_1^3-40e_1^5|\hfill \\ \hfill & \le {\displaystyle \frac{1}{921600}}[1920+160+5600(1-{\displaystyle \frac{|e_3{|}^2}{1+|e_1|}}-|e_1{|}^2-|e_2{|}^2)+80|e_1||e_2{|}^2+320|e_1{|}^2+100|e_1{|}^3|e_2|+40e_1{|}^5]\hfill \\ \hfill & ={\displaystyle \frac{1}{961200}}[7680-5280|e_1{|}^2+40|e_1{|}^5+100|e_1{|}^3|e_2|+(-5600+80|e_1|)|e_2{|}^2-{\displaystyle \frac{5600|e_3{|}^2}{1+|e_1|}}]\hfill \\ \hfill & \le {\displaystyle \frac{1}{921600}}[40|e_1{|}^5-5280|e_1{|}^2+100|e_2||e_1{|}^3+(80|e_1|-5600)|e_2{|}^2+7680].\hfill \end{array}$$

Setting $|e_1|=x,y=|e_2|$, we yield

$$\begin{array}{cc}\hfill |{\gamma }_5|& \le {\displaystyle \frac{{\alpha }_4(x,y)}{921600}},\hfill \end{array}$$

where

$$\begin{array}{c}\hfill {\alpha }_4(x,y)=40x^5-5280x^2+100y{x}^3+(80x-5600)y^2+7680.\end{array}$$

Consider

$$\left\{\begin{array}{c}{\displaystyle \frac{\partial {\alpha }_4}{\partial x}}=-10560x+200x^4+300x^{2}y+80y^2=0,\hfill \\ {\displaystyle \frac{\partial {\alpha }_4}{\partial y}}=100x^3+(-11200+160x)y=0.\hfill \end{array}\right.$$

Using numerical computations, we have

$\left\{\begin{array}{c}{x}_1=0,\hfill \\ y_1=0,\hfill \end{array}\right.$$\left\{\begin{array}{c}{x}_2\approx 3.687967,\hfill \\ y_2\approx 0.472769,\hfill \end{array}\right.$$\left\{\begin{array}{c}{x}_3\approx 89.376635,\hfill \\ y_3\approx -23028.924455.\hfill \end{array}\right.$

Thus, there are no critical points in the interior of $(0,1)\times (0,1-x^2).$

(18) For $y=0,$

$$\begin{array}{cc}\hfill {\alpha }_4(x,0)& =40x^5-5280x^2+7680\le 7680.\hfill \end{array}$$

(19) For $x=0,$

$$\begin{array}{cc}\hfill {\alpha }_4(0,y)& =7680-5600y^2\le 7680.\hfill \end{array}$$

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

$$\begin{array}{cc}\hfill {\alpha }_4(x,1-x^2)& =2080+5920x^2-60x^3-5600x^4+20x^5={\delta }_6\left(x\right)\le {\delta }_6\left(1\right)=2360.\hfill \end{array}$$

Therefore, we get

$$\begin{array}{c}\hfill |{\gamma }_5|\le {\displaystyle \frac{7680}{921600}}={\displaystyle \frac{1}{120}}.\end{array}$$

□Theorem 3.2 If $f\in \mathcal{C}L$, then

$$|{\gamma }_1{\gamma }_3-{\gamma }_2^2|\le {\displaystyle \frac{1}{576}}.$$

The result is the best possible for the function $f_2$.

Proof Let $f\in \mathcal{C}L$. From (26), (27) and (28), we have

$$\begin{array}{ccc}\hfill |{\gamma }_1{\gamma }_3-{\gamma }_2^2|& =& {\displaystyle \frac{1}{36864}}|96e_1{e}_3-8e_1^2{e}_2-64e_2^2+5e_1^4|.\hfill \end{array}$$

(31)

From (31) and Lemma 1.7, we yield

$$\begin{array}{ccc}\hfill |{\gamma }_1{\gamma }_3-{\gamma }_2^2|& =& {\displaystyle \frac{1}{36864}}|-96(1-|e_1{|}^2)|e_1{|}^2{\xi }_1^2+96e_1(1-|e_1{|}^2\left)\right(1-|{\xi }_1{|}^2){\xi }_2-8(1-|e_1{|}^2)e_1^2{\xi }_1\hfill \\ & +& 5e_1^4-64(1-|e_1{|}^2{)}^2{\xi }_1^2|.\hfill \end{array}$$

By putting $e=|e_1|,|{\xi }_1|=\aleph$, and upon utilizing $|{\xi }_2|\le 1$, we obtain

$$\begin{array}{ccc}\hfill |{\gamma }_1{\gamma }_3-{\gamma }_2^2|& \le & {\displaystyle \frac{1}{36864}}[96(e^2-e^4){\aleph }^2+96(e-e^3)(1-{\aleph }^2)+8(e^2-e^4)\aleph +5e^4+64{(1-e^2)}^2{\aleph }^2]\hfill \\ & =& {\vartheta }_3(e,\aleph ).\hfill \end{array}$$

Differentiating ${\vartheta }_3(e,\aleph )$ with respect to ℵ, we yield

$${\displaystyle \frac{\partial {\vartheta }_3}{\partial \aleph }}={\displaystyle \frac{1}{36864}}[8(e^2-e^4)+64\aleph (1-e^2)(2-3e+e^2)]\ge 0.$$

Thus, ${\vartheta }_3(e,\aleph )$ attains its maximum value at $\aleph =1$. Therefore, we achieve

$$\begin{array}{ccc}\hfill |{\gamma }_1{\gamma }_3-{\gamma }_2^2|& \le & {\vartheta }_3(e,1)={\displaystyle \frac{1}{36864}}[64-24e^2-35e^4]\le {\displaystyle \frac{64}{36864}}={\displaystyle \frac{1}{576}}.\hfill \end{array}$$

□Theorem 3.3 If $f\in \mathcal{C}L$, then

$$|{\gamma }_2{\gamma }_4-{\gamma }_3^2|\le {\displaystyle \frac{1}{2304}}.$$

The bound is sharp for the function $f_3$.

Proof Let $f\in \mathcal{C}L$. From From (27), (28) and (29), we have

$$\begin{array}{ccc}\hfill {\gamma }_2{\gamma }_4-{\gamma }_3^2& =& {\displaystyle \frac{1}{17694720}}(-7680e_3^2+1536e_3{e}_2{e}_1-672e_1^3{e}_3+19e_1^6-1280e_2^3-384e_1^4{e}_2+1536e_1^2{e}_2^2\hfill \\ & -& 1152e_1^2{e}_4+9216e_2{e}_4).\hfill \end{array}$$

(32)

From (32) and Lemma 1.7, we have

$$\begin{array}{ccc}\hfill {\gamma }_2{\gamma }_4-{\gamma }_3^2& =& {\displaystyle \frac{1}{17694720}}[1536{\overline{e_1}}^2(1-|e_1{|}^2{)}^2{\xi }_1^4-9216{\xi }_2^2|{\xi }_1{|}^2(1-|{\xi }_1{|}^2\left)\right(1-|e_1{|}^2{)}^2-1536{\xi }_1^3(1-|e_1{|}^2{)}^2{|e_1|}^2\hfill \\ & -& 3072{\xi }_2\overline{e_1}{\xi }_1^2(1-|{\xi }_1{|}^2\left)\right(1-|e_1{|}^2{)}^2+1536{\xi }_2{e}_1{\xi }_1(1-|{\xi }_1{|}^2\left)\right(1-|e_1{|}^2{)}^2+9216{\xi }_1{\xi }_3(1-|{\xi }_1{|}^2)\hfill \\ & \times & (1-|e_1{|}^2{)}^2(1-|{\xi }_2{|}^2)+19e_1^6+1536e_1^2(1-|e_1{|}^2{)}^2{\xi }_1^2-1280(1-|e_1{|}^2{)}^3{\xi }_1^3-384{\xi }_1(1-|e_1{|}^2)e_1^4\hfill \\ & -& 1152{\xi }_1^3(1-|e_1{|}^2){\overline{e_1}}^2{e}_1^2+672{\xi }_1^2\overline{e_1}{e}_1^3(1-|e_1{|}^2)-7680{\xi }_2^2(1-|e_1{|}^2{)}^2(1-|{\xi }_1{{|}^2)}^2-672{\xi }_2{e}_1^3\hfill \\ & \times & (1-|e_1{|}^2\left)\right(1-|{\xi }_1{|}^2)-1152{\xi }_3{e}_1^2(1-|e_1{|}^2\left)\right(1-|{\xi }_2{|}^2\left)\right(1-|{\xi }_1{|}^2)+1152{\xi }_2^2{e}_1^2\overline{{\xi }_1}(1-|e_1{|}^2\left)\right(1-|{\xi }_1{|}^2)\hfill \\ & +& 2304{\xi }_2{\xi }_1{e}_1^2\overline{e_1}(1-|e_1{|}^2\left)\right(1-|{\xi }_1{|}^2)].\hfill \end{array}$$

Therefore, we achieve

$${\gamma }_2{\gamma }_4-{\gamma }_3^2={\displaystyle \frac{1}{17694720}}\left[{\varrho }_9\left(e_1,{\xi }_1\right)+{\varrho }_{10}\left(e_1,{\xi }_1\right){\xi }_2+{\varrho }_{11}\left(e_1,{\xi }_1\right){\xi }_2^2+{\varrho }_{12}\left(e_1,{\xi }_1,{\xi }_2\right){\xi }_3\right],$$

where

$$\begin{array}{ccc}\hfill {\varrho }_9\left(e_1,{\xi }_1\right)& =& 19e_1^6+\left(1-|e_1{|}^2\right)\left[\left(1536{\overline{e_1}}^2{\xi }_1^4+1536e_1^2{\xi }_1^2-1280{\xi }_1^3-256{|e_1|}^2{\xi }_1^3\right)\left(1-|e_1{|}^2\right)+672\overline{e_1}{e}_1^3{\xi }_1^2\right.\hfill \\ & & \left.-1152{\overline{e_1}}^2{e}_1^2{\xi }_1^3-384e_1^4{\xi }_1\right],\hfill \\ \hfill {\varrho }_{10}\left(e_1,{\xi }_1\right)& =& \left(1-|{\xi }_1{|}^2\right)(1-|e_1{|}^2)\left[(1536{\xi }_1{e}_1-3072\overline{e_1}{\xi }_1^2)(1-|e_1{|}^2)+2304{\xi }_1\overline{e_1}{e}_1^2-672e_1^3\right],\hfill \\ \hfill {\varrho }_{11}\left(e_1,{\xi }_1\right)& =& \left(1-|e_1{|}^2\right)\left(1-|{\xi }_1{|}^2\right)\left[(-1536{\left|{\xi }_1\right|}^2-7680)\left(1-|e_1{|}^2\right)+1152\overline{{\xi }_1}{e}_1^2\right],\hfill \\ \hfill {\varrho }_{12}\left(e_1,{\xi }_1,{\xi }_2\right)& =& \left(1-|{\xi }_2{|}^2\right)\left(1-|{\xi }_1{|}^2\right)\left(1-|e_1{|}^2\right)\left[9216{\xi }_1\left(1-|e_1{|}^2\right)-1152e_1^2\right].\hfill \end{array}$$

Putting $\nu =|{\xi }_2|,|{\xi }_1|=\aleph$, $e=|e_1|$ and upon taking $\left|{\xi }_3\right|\le 1,$ we have

$$\begin{array}{ccc}\hfill \left|{\gamma }_2{\gamma }_4-{\gamma }_3^2\right|& \le & {\displaystyle \frac{1}{17694720}}\left(\left|{\varrho }_9\left(e_1,{\xi }_1\right)\right|+\left|{\varrho }_{10}\left(e_1,{\xi }_1\right)\right|\nu +\left|{\varrho }_{11}\left(e_1,{\xi }_1\right)\right|{\nu }^2+\left|{\varrho }_{12}\left(e_1,{\xi }_1,{\xi }_2\right)\right|\right)\hfill \\ & \le & {\displaystyle \frac{1}{17694720}}{\mathsf{ϝ}}_3\left(e,\aleph ,\nu \right),\hfill \end{array}$$

where

$$\begin{array}{c}\hfill {\mathsf{ϝ}}_3\left(c,\aleph ,\nu \right)={\iota }_9\left(e,\aleph \right)+{\iota }_{10}\left(e,\aleph \right)\nu +{\iota }_{11}\left(e,\aleph \right){\nu }^2+{\iota }_{12}\left(e,\aleph \right)(1-{\nu }^2),\end{array}$$

$$\begin{array}{ccc}\hfill {\iota }_9\left(e,\aleph \right)& =& 19e^6+\left(1-e^2\right)\left[\left(1536{\aleph }^2{e}^2+1280{\aleph }^3+256{\aleph }^3{e}^2+1536{\aleph }^4{e}^2\right)\left(1-e^2\right)+384\aleph e^4\right.\hfill \\ & & \left.+672{\aleph }^2{e}^4+1152{\aleph }^3{e}^4\right],\hfill \\ \hfill {\iota }_{10}\left(e,\aleph \right)& =& \left(1-{\aleph }^2\right)\left(1-e^2\right)\left[\left(1536\aleph e+3072{\aleph }^{2}e\right)\left(1-e^2\right)+672e^3+2304\aleph e^3\right],\hfill \\ \hfill {\iota }_{11}(e,\aleph )& =& (1-e^2)(1-{\aleph }^2)[(7680+1536{\aleph }^2)(1-e^2)+1152\aleph e^2],\hfill \\ \hfill {\iota }_{12}\left(e,\aleph \right)& =& \left(1-{\aleph }^2\right)\left(1-e^2\right)\left[1152e^2+9216\left(1-e^2\right)\aleph \right].\hfill \end{array}$$

In view of $\aleph \in [0,1],$ we yield

$$\begin{array}{ccc}\hfill {\iota }_{11}(e,\aleph ))& \le & (1-{\aleph }^2)(1-e^2)[(1536{\aleph }^2+7680)(1-e^2)+1152e^2]={\tau }_{11}(e,\aleph ).\hfill \end{array}$$

Setting ${\tau }_j={\iota }_j$ for $j=9,10,12$ and

$${\Gamma }_3\left(e,\aleph ,\nu \right)={\tau }_9\left(e,\aleph \right)+{\tau }_{10}\left(e,\aleph \right)\nu +{\tau }_{11}\left(e,\aleph \right){\nu }^2+{\tau }_{12}\left(e,\aleph \right)(1-{\nu }^2).$$

Differentiating ${\Gamma }_3$ with respect to $\nu$, we achieve

$${\displaystyle \frac{\partial {\Gamma }_3}{\partial \nu }}={\tau }_{10}(e,\aleph )+2[{\tau }_{11}(e,\aleph )-{\tau }_{12}(e,\aleph )]\nu .$$

Upon ${\tau }_{10}(e,\aleph )\ge 0$ and

$${\tau }_{11}(e,\aleph )-{\tau }_{12}(e,\aleph )=1536{(1-e^2)}^2(1-{\aleph }^2)({\aleph }^2-6\aleph +5)\ge 0,$$

we have ${\displaystyle \frac{\partial {\Gamma }_3}{\partial \nu }}\ge 0.$ Thus, we get

$$\begin{array}{ccc}\hfill {\mathsf{ϝ}}_3\left(e,\aleph ,\nu \right)& \le & {\Gamma }_3\left(e,\aleph ,\nu \right)\le {\Gamma }_3\left(e,\aleph ,1\right)={\tau }_9(e,\aleph )+{\tau }_{10}(e\aleph )+{\tau }_{11}(e,\aleph )\hfill \\ & =& 19e^6-672e^5+6528e^4+672e^3-14208e^2+7680+(1536e^6-3072e^5-4608e^4+6144e^3\hfill \\ & +& 4608e^2-3072e-1536){\aleph }^4+(-896e^6+768e^5+1920e^4+768e^3-2304e^2-1536e+1280){\aleph }^3\hfill \\ & +& (864e^6+3744e^5-7392e^4-6816e^3+12672e^2+3072e-6144){\aleph }^2+(-384e^6-768e^5+384e^4\hfill \\ & -& 768e^3+1536e)\aleph ={\vartheta }_4(e,\aleph ).\hfill \end{array}$$

Consider

$$\left\{\begin{array}{c}{\displaystyle \frac{\partial {\vartheta }_4}{\partial e}}=114e^5-3360e^4+26112e^3+2016e^2-28416e+(9216e^5-15360e^4-18432e^3+18432e^2+9216e-3072){\aleph }^4\hfill \\ +(-5376e^5+3840e^4+7680e^3+2304e^2-4608e-1536){\aleph }^3+(5184e^5+18720e^4-29568e^3-20448e^2+25344e\hfill \\ +3072){\aleph }^2+(-2304e^5-3840e^4+1536e^3-2304e^2+1536)\aleph =0\hfill \\ {\displaystyle \frac{\partial {\vartheta }_4}{\partial \aleph }}=1536e-768e^3+384e^4-768e^5-384e^6+(6144e^6-12288e^5-18432e^4+24576e^3+18432e^2-12288e\hfill \\ -6144){\aleph }^3+(-2688e^6+2304e^5+5760e^4+2304e^3-6912e^2-4608e+3840){\aleph }^2+(1728e^6+7488e^5-14784e^4\hfill \\ -13632e^3+25344e^2+6144e-12288)\aleph =0\hfill \end{array}\right.$$

By calculating, we achieve

$\left\{\begin{array}{c}{e}_{1,1}\approx 1,\hfill \\ {\aleph }_{1,1}\approx 1.4150,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,2}\approx 1,\hfill \\ {\aleph }_{1,2}\approx -1.8192,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,3}\approx 1,\hfill \\ {\aleph }_{1,3}\approx -0.5959\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,4}\approx -1,\hfill \\ {\aleph }_{1,4}\approx 0.5499,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,5}\approx -1,\hfill \\ {\aleph }_{1,5}\approx -0.8633,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,6}\approx -1,\hfill \\ {\aleph }_{1,6}\approx 0.2578,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,7}=0,\hfill \\ {\aleph }_{1,7}=0,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,8}\approx 5.6564,\hfill \\ {\aleph }_{1,8}\approx 0.2278,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,9}\approx -0.6438,\hfill \\ {\aleph }_{1,9}\approx 1.0868,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,10}\approx -0.8122,\hfill \\ {\aleph }_{1,10}\approx -3.4952,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,11}\approx -0.9963,\hfill \\ {\aleph }_{1,11}\approx 0.4177,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,12}\approx -1.0456,\hfill \\ {\aleph }_{1,12}\approx -0.5108,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,13}\approx -21.8175,\hfill \\ {\aleph }_{1,13}\approx 0.3078.\hfill \end{array}\right.$

Therefore, there is no critical point in $(0,1)\times (0,1)$.

(21) For $e=1,$

$$\begin{array}{c}\hfill {\vartheta }_4(1,\aleph )=19.\end{array}$$

(22) For $e=0,$

$$\begin{array}{c}\hfill {\vartheta }_4(0,\aleph )=7680-6144{\aleph }^2+1280{\aleph }^3-1536{\aleph }^4\le 7680.\end{array}$$

(23) For $\aleph =0,$

$$\begin{array}{c}\hfill {\vartheta }_4(e,0)=7680-14208e^2+672e^3+6528e^4-672e^5+19e^6\le 7680.\end{array}$$

(24) For $\aleph =1$

$$\begin{array}{c}\hfill {\vartheta }_4(e,1)=1280+768e^2-3168e^4+1139e^6={\delta }_7\left(e\right)\le {\delta }_7\left(0.3611\right)\approx 1328.8.\end{array}$$

Thus, we yield

$$\begin{array}{c}\hfill |{\gamma }_2{\gamma }_4-{\gamma }_3^2|\le {\displaystyle \frac{7680}{17694720}}={\displaystyle \frac{1}{2304}}.\end{array}$$

□

4. The Third-Order Hankel Determinant of Inverse Functions for the Class $\mathcal{C}L$

Theorem 4.1 If $f\in \mathcal{C}L$, then

$$|H_{3,1}\left(f^{-1}\right)|\le {\displaystyle \frac{1}{576}}.$$

The bound is sharp for the function $f_3$.

Proof Let $f\in \mathcal{C}L$. From (14)–(17) and (9), we yield

$$\begin{array}{ccc}\hfill H_{3,1}\left(f^{-1}\right)& =& {\displaystyle \frac{1}{2211840}}(-1320e_2{e}_1^4+96e_1^3{e}_3+2448e_2^2{e}_1^2+4608e_4{e}_2+2688e_3{e}_2{e}_1\hfill \\ & -& 2304e_1^2{e}_4-2560e_2^3-3840e_3^2+233e_1^6).\hfill \end{array}$$

(33)

From Lemma 1.7 and (33), we achieve

$$\begin{array}{ccc}\hfill {\mathcal{H}}_{3,1}\left(f^{-1}\right)& =& {\displaystyle \frac{1}{2211840}}[768{\overline{e_1}}^2(1-|e_1{|}^2{)}^2{\xi }_1^4-4608(1-|e_1{|}^2{)}^2(1-|{\xi }_1{|}^2){\xi }_2^2|{\xi }_1{|}^2-2688|e_1{|}^2{\xi }_1^3(1-|e_1{{|}^2)}^2\hfill \\ & -& 1536\overline{e_1}{\xi }_1^2(1-|{\xi }_1{|}^2\left)\right(1-|e_1{|}^2{)}^2{\xi }_2+2688{\xi }_2{\xi }_1{e}_1(1-|{\xi }_1{|}^2\left)\right(1-|e_1{|}^2{)}^2+4608{\xi }_3{\xi }_1(1-|{\xi }_2{|}^2)\hfill \\ & \times & (1-|{\xi }_1{|}^2\left)\right(1-|e_1{|}^2{)}^2+233e_1^6+2448e_1^2(1-|e_1{|}^2{)}^2{\xi }_1^2-2560(1-|e_1{|}^2{)}^3{\xi }_1^3-1340e_1^4(1-|e_1{|}^2){\xi }_1\hfill \\ & -& 2304e_1^2{\overline{e_1}}^2(1-|e_1{|}^2){\xi }_1^3-3840(1-|{\xi }_1{|}^2{)}^2(1-|e_1{|}^2{)}^2{\xi }_2^2+96{\xi }_2(1-|e_1{|}^2\left)\right(1-|{\xi }_1{|}^2)e_1^3\hfill \\ & -& 2304e_1^2{\xi }_3(1-|e_1{|}^2\left)\right(1-|{\xi }_2{|}^2\left)\right(1-|{\xi }_1{|}^2)+2304e_1^2(1-|e_1{|}^2)\overline{{\xi }_1}{\xi }_2^2(1-|{\xi }_1{|}^2)+4608{\xi }_2{\xi }_1\hfill \\ & \times & (1-|{\xi }_1{|}^2\left)\right(1-|e_1{|}^2)e_1^2\overline{e_1}-96(1-e_1{|}^2)\overline{e_1}{e}_1^3{\xi }_1^2].\hfill \end{array}$$

Thus, we achieve

$${\mathcal{H}}_{3,1}\left(f^{-1}\right)={\displaystyle \frac{1}{2211840}}\left[{\varrho }_{13}\left(e_1,{\xi }_1\right)+{\varrho }_{14}\left(e_1,{\xi }_1\right){\xi }_2+{\varrho }_{15}\left(e_1,{\xi }_1\right){\xi }_2^2+{\varrho }_{16}\left(e_1,{\xi }_1,{\xi }_2\right){\xi }_3\right],$$

where

$$\begin{array}{ccc}\hfill {\varrho }_{13}\left(e_1,{\xi }_1\right)& =& 233e_1^6+\left(1-|e_1{|}^2\right)\left[\left(768{\overline{e_1}}^2{\xi }_1^4+2448e_1^2{\xi }_1^2-2560{\xi }_1^3-128{|e_1|}^2{\xi }_1^3\right)\left(1-|e_1{|}^2\right)-96\overline{e_1}{e}_1^3{\xi }_1^2\right.\hfill \\ & & \left.-2304{\overline{e_1}}^2{e}_1^2{\xi }_1^3-1340e_1^4{\xi }_1\right],\hfill \\ \hfill {\varrho }_{14}\left(e_1,{\xi }_1\right)& =& \left(1-|{\xi }_1{|}^2\right)(1-|e_1{|}^2)\left[(2688{\xi }_1{e}_1-1536\overline{e_1}{\xi }_1^2)(1-|e_1{|}^2)+4608{\xi }_1\overline{e_1}{e}_1^2+96e_1^3\right],\hfill \\ \hfill {\varrho }_{15}\left(e_1,{\xi }_1\right)& =& \left(1-|e_1{|}^2\right)\left(1-|{\xi }_1{|}^2\right)\left[(-768{\left|{\xi }_1\right|}^2-3840)\left(1-|e_1{|}^2\right)+2304\overline{{\xi }_1}{e}_1^2\right],\hfill \\ \hfill {\varrho }_{16}\left(e_1,{\xi }_1,{\xi }_2\right)& =& \left(1-|{\xi }_1{|}^2\right)\left(1-|e_1{|}^2\right)\left(1-|{\xi }_2{|}^2\right)\left[4608\left(1-|e_1{|}^2\right){\xi }_1-2304e_1^2\right].\hfill \end{array}$$

By putting $|{\xi }_1|=\aleph ,\nu =|{\xi }_2|$, $|e_1|=2$ and upon taking $\left|{\xi }_3\right|\le 1,$ we yield

$$\begin{array}{ccc}\hfill \left|{\mathcal{H}}_{3,1}\left(f^{-1}\right)\right|& \le & {\displaystyle \frac{1}{2211840}}\left(\left|{\varrho }_{13}\left(e_1,{\xi }_1\right)\right|+\left|{\varrho }_{14}\left(e_1,{\xi }_1\right)\right|\nu +\left|{\varrho }_{15}\left(e_1,{\xi }_1\right)\right|{\nu }^2+\left|{\varrho }_{16}\left(e_1,{\xi }_1,{\xi }_2\right)\right|\right)\hfill \\ & \le & {\displaystyle \frac{1}{2211840}}{\mathsf{ϝ}}_4\left(e,\aleph ,\nu \right),\hfill \end{array}$$

where

$$\begin{array}{c}\hfill {\mathsf{ϝ}}_4\left(c,\aleph ,\nu \right)={\iota }_{13}\left(e,\aleph \right)+{\iota }_{14}\left(e,\aleph \right)\nu +{\iota }_{15}\left(e,\aleph \right){\nu }^2+{\iota }_{16}\left(e,\aleph \right)(1-{\nu }^2),\end{array}$$

$$\begin{array}{ccc}\hfill {\iota }_{13}\left(e,\aleph \right)& =& 233e^6+\left(1-e^2\right)\left[\left(2448{\aleph }^2{e}^2+2560{\aleph }^3+128{\aleph }^3{e}^2+768{\aleph }^4{e}^2\right)\left(1-e^2\right)+1340\aleph e^4\right.\hfill \\ & & \left.+96{\aleph }^2{e}^4+2304{\aleph }^3{e}^4\right],\hfill \\ \hfill {\iota }_{14}\left(e,\aleph \right)& =& \left(1-{\aleph }^2\right)\left(1-e^2\right)\left[\left(2688\aleph e+1536{\aleph }^{2}e\right)\left(1-e^2\right)+96e^3+4608\aleph e^3\right],\hfill \\ \hfill {\iota }_{15}(e,\aleph )& =& (1-e^2)(1-{\aleph }^2)[(3840+768{\aleph }^2)(1-e^2)+2304\aleph e^2],\hfill \\ \hfill {\iota }_{16}\left(e,\aleph \right)& =& \left(1-{\aleph }^2\right)\left(1-e^2\right)\left[2304e^2+4608\left(1-e^2\right)\aleph \right].\hfill \end{array}$$

In view of $\aleph \in [0,1],$ we have

$$\begin{array}{ccc}\hfill {\iota }_{15}(e,\aleph ))& \le & (1-{\aleph }^2)(1-e^2)[(768{\aleph }^2+3840)(1-e^2)+2304e^2]={\tau }_{15}(e,\aleph ).\hfill \end{array}$$

Setting ${\tau }_j={\iota }_j$ for $j=13,14,16$ and

$${\Gamma }_4\left(e,\aleph ,\nu \right)={\tau }_{13}\left(e,\aleph \right)+{\tau }_{14}\left(e,\aleph \right)\nu +{\tau }_{15}\left(e,\aleph \right){\nu }^2+{\tau }_{16}\left(e,\aleph \right)(1-{\nu }^2).$$

Differentiating ${\Gamma }_4$ with respect to $\nu$, we achieve

$${\displaystyle \frac{\partial {\Gamma }_4}{\partial \nu }}={\tau }_{14}(e,\aleph )+2[{\tau }_{15}(e,\aleph )-{\tau }_{16}(e,\aleph )]\nu .$$

Upon ${\tau }_{14}(e,\aleph )\ge 0$ and

$${\tau }_{15}(e,\aleph )-{\tau }_{16}(e,\aleph )=768{(1-e^2)}^2(5-6\aleph +{\aleph }^2)(1-{\aleph }^2)\ge 0,$$

we get ${\displaystyle \frac{\partial {\Gamma }_4}{\partial \nu }}\ge 0.$ Therefore, we obtain

$$\begin{array}{ccc}\hfill {\mathsf{ϝ}}_4\left(e,\aleph ,\nu \right)& \le & {\Gamma }_4\left(e,\aleph ,\nu \right)\le {\Gamma }_4\left(e,\aleph ,1\right)={\tau }_{13}(e,\aleph )+{\tau }_{14}(e\aleph )+{\tau }_{15}(e,\aleph )\hfill \\ & =& 3840-5376e^2+96e^3+1536e^4-96e^5+233e^6+(-768-1536e+2304e^2+3072e^3\hfill \\ & -& 2304e^4-1536e^5+768e^6){\aleph }^4+(2560-2688e-4992e^2+768e^3+4608e^4+1920e^5\hfill \\ & -& 2176e^6){\aleph }^3+(-3072+1536e+6288e^2-3168e^3-5568e^4+1632e^5+2352e^6){\aleph }^2\hfill \\ & +& (2688e-768e^3+1340e^4-1920e^5-1340e^6)\aleph ={\vartheta }_5(e,\aleph ).\hfill \end{array}$$

Consider

$$\left\{\begin{array}{c}{\displaystyle \frac{\partial {\vartheta }_5}{\partial e}}=-10752e+288e^2+6144e^3-480e^4+1398e^5+(-1536+4608e+9216e^2-9216e^3-7680e^4\hfill \\ +4608e^5){\aleph }^4+(-2688-9984e+2304e^2+18432e^3+9600e^4-13056e^5){\aleph }^3+(1536+12576e-9504e^2\hfill \\ -22272e^3+8160e^4+14112e^5){\aleph }^2+(2688-2304e^2+5360e^3-9600e^4-8040e^5)\aleph =0\hfill \\ {\displaystyle \frac{\partial {\vartheta }_5}{\partial \aleph }}=-1340e^6-1920e^5+1340e^4-768e^3+2688e+(-3072-6144e+9216e^2+12288e^3-9216e^4\hfill \\ -6144e^5+3072e^6){\aleph }^3+(7680-8064e-14976e^2+2304e^3+13824e^4+5760e^5-6528e^6){\aleph }^2\hfill \\ +(-6144+3072e+12576e^2-6336e^3-11136e^4+3264e^5+4704e^6)\aleph =0\hfill \end{array}\right.$$

By calculating, we achieve

$\left\{\begin{array}{c}{e}_{1,1}\approx 1,\hfill \\ {\aleph }_{1,1}\approx 1.3393,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,2}\approx 1,\hfill \\ {\aleph }_{1,2}\approx -2.0734,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,3}\approx 1,\hfill \\ {\aleph }_{1,3}\approx -0.2659,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,4}\approx -1,\hfill \\ {\aleph }_{1,4}\approx -0.7373,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,5}=0,\hfill \\ {\aleph }_{1,5}=0,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,6}\approx 3.7699,\hfill \\ {\aleph }_{1,6}\approx 2.0674,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,7}\approx -0.1419,\hfill \\ {\aleph }_{1,7}\approx 1.0477,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,8}\approx -0.7281,\hfill \\ {\aleph }_{1,8}\approx -9.9896,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,9}\approx -1.0215,\hfill \\ {\aleph }_{1,9}\approx 0.5050,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,10}\approx -1.1901,\hfill \\ {\aleph }_{1,10}\approx -0.2819,\hfill \end{array}\right.$$\left\{\begin{array}{c}{e}_{1,11}\approx -1.7088,\hfill \\ {\aleph }_{1,11}\approx 2.4591.\hfill \end{array}\right.$

Therefore, there is no critical point in $(0,1)\times (0,1)$.

(25) For $e=1,$

$$\begin{array}{c}\hfill {\vartheta }_5(1,\aleph )=233.\end{array}$$

(26) For $e=0,$

$$\begin{array}{c}\hfill {\vartheta }_5(0,\aleph )=-738{\aleph }^4+2560{\aleph }^3-3072{\aleph }^2+3840\le 3840.\end{array}$$

(27) For $\aleph =0,$

$$\begin{array}{c}\hfill {\vartheta }_5(e,0)=233e^6-96e^5+1536e^4+96e^3-5376e^2+3840\le 3840.\end{array}$$

(28) For $\aleph =1$

$$\begin{array}{c}\hfill {\vartheta }_5(e,1)=2560-1776e^2-388e^4-163e^6\le 2560.\end{array}$$

Thus, we yield

$$\begin{array}{c}\hfill |H_{3,1}\left(f^{-1}\right)|\le {\displaystyle \frac{3840}{2211840}}={\displaystyle \frac{1}{576}}.\end{array}$$

□