Some Results on Coefficient Estimate Problems for Four-Leaf-Type Bounded Turning Functions

Abstract

Let $\mathcal{B}{T}_{4l}$ denote a subclass of bounded turning functions connected with a four-leaf-type domain. The goal of the study is to probe into the bounds of coefficients $|b_6|,|b_7|,|b_8|$, the bounds of the logarithmic coefficients, and the third-order determinants $|H_{3,1}|,|H_{3,2}|,|H_{3,3}|$ for the functions in this class.

1. Introduction and Definitions

Let $\mathcal{H}$ represent a family of mappings of the following kind, denoted by g

$$g\left(\xi \right)=\xi +\sum _{n=2}^{\infty }{b}_n{\xi }^n$$

(1)

in the unit disc $\mathcal{U}=\{\xi \in \mathbb{C}:|z|<1\}$. We refer to $S$ as a subfamily of $\mathcal{H}$, which consists of univalent functions in $\mathcal{U}$.

For each function $g\in \mathcal{H}$ and positive integers i and n, the Hankel determinant $H_{i,n}\left(g\right)$ is defined by

$$H_{i,n}\left(g\right):=\left|\begin{array}{cccc}{b}_n\hfill & b_{n+1}\hfill & \dots \hfill & b_{n+i-1}\hfill \\ b_{n+1}\hfill & b_{n+2}\hfill & \dots \hfill & b_{n+i}\hfill \\ ⋮\hfill & ⋮\hfill & \dots \hfill & ⋮\hfill \\ b_{n+i-1}\hfill & b_{n+i}\hfill & \dots \hfill & b_{n+2i-2}\hfill \end{array}\right|(b_1=1).$$

The Hankel determinant $H_{i,n}\left(g\right)$ was introduced by Pommerrenke [1,2]. The third Hankel determinants with $n\le 3$ are as follows:

$$H_{3,1}\left(g\right)=b_3{b}_5+2b_2{b}_3{b}_4-b_3^3-b_4^2-b_2^2{b}_5,$$

(2)

$$H_{3,2}\left(g\right)=b_4(b_3{b}_5-b_4^2)-b_5(b_2{b}_5-b_3{b}_4)+b_6(b_2{b}_4-b_3^2),$$

(3)

$$H_{3,3}\left(g\right)=b_5(b_4{b}_6-b_5^2)-b_6(b_3{b}_6-b_4{b}_5)+b_7(b_3{b}_5-b_4^2).$$

(4)

For the first time, the bounds of the third-order Hankel determinant $|H_{3,1}\left(g\right)|$ for the families of $S^{*}$ of starlike functions, $\mathcal{K}$ of convex functions and $\mathcal{R}$ of functions with bounded turning were investigated by Babalola [3]. Recently, the sharp bounds of the Hankel determinant $H_{3,1}\left(g\right)$ of subclasses of analytic functions were obtained by many authors [4,5,6,7,8,9,10].

For $g\in S$, let $F_g\left(\xi \right)$ be defined as the

$$F_g\left(\xi \right)=log\frac{g\left(\xi \right)}{\xi }=2\sum _{n=1}^{\infty }{\delta }_n{\xi }^n,$$

(5)

where ${\delta }_n={\delta }_n\left(g\right)$ is referred to as the logarithmic coefficients of g. In the theory of univalent functions, these coefficients play an important role for different estimates. The problem of the best upper bounds for ${\delta }_n$ is still open. In fact, even the proper order of magnitude is still not known. It is known, however, for the starlike functions that the best bound is $|{\delta }_n|\le \frac{1}{n}(n\ge 1)$, and that this is not true in general [11].

Differentiating (5), we have

$$\begin{array}{c}\hfill \frac{\xi g^{\prime }\left(\xi \right)-g\left(\xi \right)}{\xi g\left(\xi \right)}=2{\delta }_1+4{\delta }_2\xi +6{\delta }_3{\xi }^2+8{\delta }_4{\xi }^3+10{\delta }_5{\xi }^4+···.\end{array}$$

From (1), we obtain

$$\begin{array}{cc}\hfill \frac{\xi g^{\prime }\left(\xi \right)-g\left(\xi \right)}{\xi g\left(\xi \right)}& =b_2+(2b_3-b_2^2)\xi +(3b_4-3b_2{b}_3+b_2^3){\xi }^2+(4b_5-4b_2{b}_4-2b_3^2+4b_2^2{b}_3-b_2^4){\xi }^3\hfill \\ \hfill & +(5b_6-5b_2{b}_5-5b_3{b}_4+5b_4{b}_2^2+5b_2{b}_3^2-5b_2^3{b}_3+b_2^5){\xi }^4+···.\hfill \end{array}$$

By comparing the coefficients, we achieve

$$\left\{\begin{array}{c}{\delta }_1=\frac{1}{2}{b}_2,\hfill \\ {\delta }_2=\frac{1}{2}(b_3-\frac{1}{2}{b}_2^2),\hfill \\ {\delta }_3=\frac{1}{2}(b_4-b_2{b}_3+\frac{1}{3}{b}_2^3),\hfill \\ {\delta }_4=\frac{1}{2}(b_5-b_4{b}_2+b_3{b}_2^2-\frac{1}{2}{b}_3^2-\frac{1}{4}{b}_2^4),\hfill \\ {\delta }_5=\frac{1}{2}(b_6-b_2{b}_5-b_3{b}_4+b_4{b}_2^2+b_2{b}_3^2-b_2^3{b}_3+\frac{1}{5}{b}_2^5).\hfill \end{array}\right.$$

(6)

In 2022, Sunthrayuth et al. [12] introduced a subclass of bounded turning functions associated with a four-leaf function defined by

$$\mathcal{B}{T}_{4l}=\left\{g\in S:g^{\prime }\left(\xi \right)\prec 1+\frac{5}{6}\xi +\frac{1}{6}{\xi }^5\right\},$$

and obtained Kruskal inequality, the bounds of the coefficient inequalities and the two-order Hankel determinant of bounded turning class $\mathcal{B}{T}_{4l}$.

Utilizing the estimates of the coefficients of the Schwartz function, we study the third-order Hankel determinants $|H_{3,1}\left(g\right)|$, $|H_{3,2}\left(g\right)|$ and $|H_{3,3}\left(g\right)|$ for the class $\mathcal{B}{T}_{4l}$; also, we obtain the bounds for the logarithmic coefficients for $g\in \mathcal{B}{T}_{4l}$.

Let ${\omega }_0$ be the family of Schwarz functions. Thus, the functions $w\in {\omega }_0$ can be expressed as a power series:

$$w\left(\xi \right)=\sum _{n=1}^{\infty }{c}_n{\xi }^n.$$

(7)

Lemma 1

(see [13]). Suppose a function w is a member of ${\omega }_0$. Then,

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

$$|c_5|\le 1-|c_1{|}^2-|c_2{|}^2-\frac{|c_3{|}^2}{1+|c_1|},|c_6|\le 1-|c_1{|}^2-|c_2{|}^2-{|c_3|}^2,$$

$$|c_7|\le 1-|c_1{|}^2-|c_2{|}^2-{|c_3|}^2-\frac{|c_4{|}^2}{1+|c_1|}.$$

Lemma 2

(see [14,15]). Suppose a function w is a member of ${\omega }_0$. Then,

$$|c_1{c}_3-c_2^2|\le 1-|c_1{|}^2,|c_2{c}_4-c_3^2|\le {(1-c_1^2)}^2,|c_3{c}_5-c_4^2|\le 1-c_1^2.$$

Lemma 3

(see [12]). If $g\in \mathcal{B}{T}_{4l}$, then

$$|b_2|\le \frac{5}{12},|b_3|\le \frac{5}{18},|b_4|\le \frac{5}{24},|b_5|\le \frac{1}{6}.$$

Lemma 4

(see [11]). If $w\in {\omega }_0$, then $\left|c_n\right|\le 1(n\ge 1)$.

Lemma 5

(see [16]). Let $w\in {\omega }_0$; then, for $\gamma \in \mathbb{C}$, we have

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

Lemma 6

(see [17]). If $w\in {\omega }_0$. Then, we obtain

$$|c_3+\gamma c_1{c}_2+\varsigma c_1^3|\le 1,$$

where $(\gamma ,\varsigma )\in D_1\cup D_2$, with

$$D_1=\{(\gamma ,\varsigma )\in {\mathbb{R}}^2:|\gamma |\le \frac{1}{2},-1\le \varsigma \le 1\},$$

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

Lemma 7

(see [18]). Let $w\in {\omega }_0$ and $\gamma \in \mathbb{C}$; we obtain

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

Lemma 8

(see [19]). If $w\in {\omega }_0$, then for all $\gamma \in \mathbb{C}$, we have

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

2. The Bounds of the Third Hankel Determinant for $\mathit{g}\mathbf{\in }\mathcal{B}{\mathit{T}}_{\mathbf{4}\mathit{l}}$

Theorem 1.

If $g\in \mathcal{B}{T}_{4l}$, then

$$\begin{array}{c}\hfill |b_6|\le \frac{5}{36},|b_7|\le \frac{5}{42},|b_8|\le \frac{5}{48}.\end{array}$$

The bounds are sharp.

Proof. 

For a function $g\in \mathcal{B}{T}_{4l}$, there exists a Schwarz function $w\left(\xi \right)$, such that

$$g^{\prime }\left(\xi \right)=1+\frac{5}{6}w\left(\xi \right)+\frac{1}{6}{w}^5\left(\xi \right),$$

Comparing the coefficients, we yield

$$b_2=\frac{5}{12}{c}_1,b_3=\frac{5}{18}{c}_2,b_4=\frac{5}{24}{c}_3,b_5=\frac{1}{6}{c}_4,$$

(8)

$$b_6=\frac{5c_5+c_1^5}{36}$$

(9)

$$b_7=\frac{5c_6+5c_1^4{c}_2}{42},$$

(10)

$$b_8=\frac{5c_7+5c_1^4{c}_3+10c_1^3{c}_2^2}{48}.$$

(11)

From (9) and Lemma 1, we have

$$\begin{array}{ccc}\hfill \left|b_6\right|& & \le \frac{1}{36}\left[5|c_5|+|c_1{|}^5\right]\hfill \\ & & \le \frac{1}{36}[5(1-|c_1{|}^2-|c_2{|}^2-\frac{|c_3{|}^2}{1+|c_1|})+|c_1{|}^5]\hfill \\ & & =\frac{1}{36}(5-5|c_1{|}^2+|c_1{|}^5-5|c_2{|}^2-\frac{5|c_3{|}^2}{1+|c_1|})\hfill \\ & & \le \frac{1}{36}(5-5|c_1{|}^2+|c_1{|}^5)\hfill \\ & & \le \frac{5}{36}.\hfill \end{array}$$

From (10) and Lemma 1, we achieve

$$\begin{array}{ccc}\hfill \left|b_7\right|& & \le \frac{1}{42}\left[5|c_6|+5|c_1{|}^4|c_2|\right]\hfill \\ & & \le \frac{1}{42}\left[5(1-|c_1{|}^2-|c_2{|}^2-|c_3{|}^2)+5|c_1{|}^4|c_2|\right]\hfill \\ & & =\frac{1}{42}(5-5|c_1{|}^2+5|c_1{|}^4|c_2|-5|c_2{|}^2-5|c_3{|}^2)\hfill \\ & & \le \frac{1}{42}\left[5-5|c_1{|}^2+5|c_1{|}^4(1-|c_1{|}^2)\right]\hfill \\ & & =\frac{1}{42}\left(5-5|c_1{|}^2+5|c_1{|}^4-5{|c_1|}^6\right)\hfill \\ & & \le \frac{5}{42}.\hfill \end{array}$$

From (11) and Lemma 1, we obtain

$$\begin{array}{ccc}\hfill \left|b_8\right|& & \le \frac{1}{48}\left[5|c_7|+5|c_1{|}^4|c_3|+10|c_1{|}^3{|c_2|}^2\right]\hfill \\ & & \le \frac{1}{48}[5(1-|c_1{|}^2-|c_2{|}^2-|c_3{|}^2-\frac{|c_4{|}^2}{1+|c_1|})+5|c_1{|}^4|c_3|+10|c_1{|}^3{|c_2|}^2]\hfill \\ & & =\frac{1}{48}[5-5|c_1{|}^2+(-5+10|c_1{|}^3)|c_2{|}^2+5|c_1{|}^4|c_3|-5|c_3{|}^2-5\frac{|c_4{|}^2}{1+|c_1|}]\hfill \\ & & \le \frac{1}{48}[5-5|c_1{|}^2+(-5+10|c_1{|}^3)|c_2{|}^2+5|c_1{|}^4(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|})]\hfill \\ & & =\frac{1}{48}[5-5|c_1{|}^2+5|c_1{|}^4-5|c_1{|}^6+\frac{-5-5|c_1|+10|c_1{|}^3+5{|c_1|}^4}{1+|c_1|}|c_2{|}^2].\hfill \end{array}$$

Setting $c=|c_1|$ and $d=|c_2|$, we obtain

$$\begin{array}{c}\hfill \left|b_8\right|\le \frac{1}{48}{ϵ}_1(c,d)\end{array}$$

where

$$\begin{array}{c}\hfill {ϵ}_1(c,d)=5-5c^2+5c^4-5c^6+\frac{-5-5c+10c^3+5c^4}{1+c}{d}^2.\end{array}$$

The critical points of ${ϵ}_1$ satisfy

$$\left\{\begin{array}{c}\frac{\partial {ϵ}_1}{\partial c}=-10c+20c^3-30c^5+\frac{15c^4+40c^3+30c^2}{{(1+c)}^2}{d}^2,\hfill \\ \frac{\partial {ϵ}_1}{\partial d}=\frac{-10-10c+20c^3+10c^4}{1+c}d=0.\hfill \end{array}\right.$$

Applying numerical computations, we have

$$\left\{\begin{array}{c}{c}_0=0,\hfill \\ d_0=0,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_1=0.8668,\hfill \\ d_1=0.7940,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_2=0.8668,\hfill \\ d_2=-0.7940.\hfill \end{array}\right.$$

Thus, in $(0,1)\times (0,1-c^2)$, there is no critical points which satisfies $0\le d\le 1-c^2$.

For $c=0$,

$${ϵ}_1(0,d)=5-5d^2\le 5.$$

For $d=0$,

$${ϵ}_1(c,0)=5-5c^2+5c^4-5c^6\le 5.$$

For $d=1-c^2$,

$${ϵ}_1(c,1-c^2)=5c^2+10c^3-5c^4-15c^5+5c^7={\eta }_1\left(c\right)\le {\eta }_1\left(0.7202\right)=2.5799.$$

Thus, we have

$$|b_8|\le \frac{5}{48}.$$

These bounds can be determined by the following extremal functions:

$$g_1\left(\xi \right)={\int }_0^{\xi }(1+\frac{5}{6}{t}^5+\frac{1}{6}{t}^{25})dt=\xi +\frac{5}{36}{\xi }^6+\frac{1}{156}{\xi }^{26}+···,$$

$$g_2\left(\xi \right)={\int }_0^{\xi }(1+\frac{5}{6}{t}^6+\frac{1}{6}{t}^{30})dt=\xi +\frac{5}{42}{\xi }^7+\frac{1}{186}{\xi }^{31}+···,$$

$$g_3\left(\xi \right)={\int }_0^{\xi }(1+\frac{5}{6}{t}^7+\frac{1}{6}{t}^{35})dt=\xi +\frac{5}{48}{\xi }^8+\frac{1}{216}{\xi }^{36}+···.$$

□

Theorem 2.

If $g\in \mathcal{B}{T}_{4l}$, then

$$\begin{array}{ccc}\hfill \left|H_{2,3}\left(g\right)\right|& \le & 0.044516.\hfill \end{array}$$

Proof. 

Let $g\in \mathcal{B}{T}_{4l}$.

From (8), we receive

$$\begin{array}{c}\hfill \left|H_{2,3}\left(g\right)\right|=|b_3{b}_5-b_4^2|=\frac{5}{1728}\left|16c_2{c}_4-15c_3^2\right|=\frac{5}{1728}|15(c_2{c}_4-c_3^2)+c_2{c}_4|.\end{array}$$

(12)

Utilizing inequality, Lemma 1 and Lemma 2 in (12), we obtain

$$\begin{array}{ccc}\hfill \left|H_{2,3}\left(g\right)\right|& & \le \frac{5}{1728}[15|c_2{c}_4-c_3^2|+|c_2||c_4|]\hfill \\ & & \le \frac{5}{1728}[15(1-|c_1{|}^2{)}^2+|c_2|(1-|c_1{|}^2-|c_2{|}^2)]\hfill \\ & & =\frac{5}{1728}[15-30|c_1{|}^2+15|c_1{|}^4+(1-|c_1{|}^2)|c_2|-|c_2{|}^3)]\hfill \\ & & \le \frac{5}{1728}[15-30|c_1{|}^2+15|c_1{|}^4+(1-|c_1{|}^2)|c_2|-|c_2{|}^3].\hfill \end{array}$$

Let $|c_1|=c$ and $|c_2|=d$, we obtain

$$\begin{array}{cc}\hfill \left|H_{2,3}\left(g\right)\right|& \le \frac{5}{1728}{ϵ}_2(c,d).\end{array}$$

Consider

$$\left\{\begin{array}{c}\frac{\partial {ϵ}_2}{\partial c}=-60c+60c^3-2cd=0,\hfill \\ \frac{\partial {ϵ}_2}{\partial d}=1-c^2-3d^2=0.\hfill \end{array}\right.$$

Applying numerical computations, we obtain

$$\left\{\begin{array}{c}{c}_1=0,\hfill \\ d_1=-0.5774,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_2=0,\hfill \\ d_2=0.5774,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_3=-1,\hfill \\ d_3=0,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_4=1,\hfill \\ d_4=0,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_5=0.9998,\hfill \\ d_5=-0.0111,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_6=-0.9998,\hfill \\ d_6=-0.0111.\hfill \end{array}\right.$$

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

(1)

For $d=0$,

$$\begin{array}{c}\hfill {ϵ}_2(c,0)=15-30c^2+15c^4\le 15.\end{array}$$

(2)

For $c=0$,

$$\begin{array}{c}\hfill {ϵ}_2(0,d)=15+d-d^3\le {ϵ}_2(0,\frac{\sqrt{3}}{3})=\frac{2\sqrt{3}+135}{9}=15.384888.\end{array}$$

(3)

For $d=1-c^2$,

$$\begin{array}{c}\hfill {ϵ}_2(c,1-c^2)=15-29c^2+13c^4+c^6\le 15.\end{array}$$

Therefore, we yield

$$\begin{array}{c}\hfill \left|H_{2,3}(g)\right|\le \frac{10\sqrt{3}+675}{15552}=0.044516.\end{array}$$

The proof of Theorem 2 is completed.  □

Theorem 3.

If $g\in \mathcal{B}{T}_{4l}$, then

$$\begin{array}{ccc}\hfill \left|H_{3,1}(g)\right|& \le & \frac{3025}{46656}=0.064836.\hfill \end{array}$$

Proof. 

Assume that $g\in \mathcal{B}{T}_{4l}$. From (8) and (2), we achieve

$$\begin{array}{ccc}\hfill \left|H_{3,1}\left(g\right)\right|& \hfill & \hfill =\frac{1}{46656}\left|2250c_1{c}_2{c}_3-1000c_2^3-2025c_3^2+2160c_2{c}_4-1350c_1^2{c}_4\right|\\ \hfill & \hfill & \hfill =\frac{1}{46656}\left|1000c_2(c_1{c}_3-c_2^2)+1250c_1{c}_2{c}_3+2025(c_2{c}_4-c_3^2)+135c_2{c}_4-1350c_1^2{c}_4\right|\end{array}$$

By applying the triangle inequality, Lemma 1 and Lemma 2 in (13), we receive

$$\begin{array}{ccc}\hfill 46656\left|H_{3,1}(g)\right|& & \le 1000|c_2||c_1{c}_3-c_2^2|+1250|c_1||c_2||c_3|+2025|c_2{c}_4-c_3^2|+135|c_2||c_4|+1350|c_1{|}^2|c_4|\hfill \\ & & \le 1000|c_2|(1-|c_1{|}^2)+1250|c_1||c_2|(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|})+2025(1-|c_1{{|}^2)}^2\hfill \\ & & +135|c_2|(1-|c_1{|}^2-|c_2{|}^2)+1350|c_1{|}^2(1-|c_1{|}^2-|c_2{|}^2)\hfill \\ & & =2025-2700|c_1{|}^2+675|c_1{|}^4+(1135+1250|c_1|-1135|c_1{|}^2-1250|c_1{|}^3)|c_2|\hfill \\ & & -1350|c_1{|}^2|c_2{|}^2-\frac{135+1385|c_1|}{1+|c_1|}{|c_2|}^3.\hfill \end{array}$$

Setting $|c_1|=c$ and $|c_2|=d$, we have

$$\begin{array}{ccc}\hfill 46656\left|H_{3,1}\left(g\right)\right|& \le & 2025-2700c^2+675c^4+(1135+1250c-1135c^2-1250c^3)d\hfill \\ & & -1350c^2{d}^2-\frac{135+1385c}{1+c}{d}^3={ϵ}_3(c,d)\hfill \end{array}$$

where $c\in [0,1]$ and $d\in [0,1-c^2]$.

Taking the partial derivative with respect to c and y respectively, and we have

$$\begin{array}{cc}\hfill \frac{\partial {ϵ}_3}{\partial c}& =-5400c+2700c^3+(1250-2270c-3405c^2)d-2700c{d}^2-\frac{1250}{{(1+c)}^2}{d}^3\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \frac{\partial {ϵ}_3}{\partial d}& =1135+1250c-1135c^2-1250c^3-2700c^{2}d-\frac{405+4155c}{1+c}{d}^2.\hfill \end{array}$$

Setting $\frac{\partial {ϵ}_3}{\partial c}=\frac{\partial {ϵ}_3}{\partial d}=0$ and simplifying, we yield

$$\left\{\begin{array}{c}-5400c-10800c^2-2700c^3+5400c^4+2700c^5+(1250+230c-6695c^2-9080c^3-3405c^4)d\hfill \\ -2700c{(1+c)}^2{d}^2-1250d^3=0,\hfill \\ 1135+2358c+115c^2-2385c^3-1250c^4-2700c^2(1+c)d-(405+4155c)d^2=0.\hfill \end{array}\right.$$

Applying Newton’s methods, we yield

$$\left\{\begin{array}{c}{c}_1=-1,\hfill \\ d_1=0,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_2=0.1018,\hfill \\ d_2=-1.3080,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_3=1.0726,\hfill \\ d_3=-1.1909,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_4=1.7878\hfill \\ d_4=-0.3706,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_5=3.3060,\hfill \\ d_5=-6.5565,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_6=-1.3977,\hfill \\ d_6=0.0899.\hfill \end{array}\right.$$

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

For $c=0$,

$${ϵ}_3(0,d)=2025+1135d-135d^3\le {ϵ}_3(0,1)=3025.$$

For $d=0$,

$${ϵ}_3(c,0)=2025-2700c^2+675c^4\le 2025.$$

For $d=1-c^2$,

$${ϵ}_3(c,0)=3025-4665c^2+1605c^4+35c^6\le {ϵ}_3(0,0)=3025.$$

Thus, we obtain

$$\begin{array}{cc}\hfill \left|H_{3,1}(g)\right|& \le \frac{3025}{46656}=0.064836.\hfill \end{array}$$

The proof of Theorem 3 is completed.  □

Theorem 4.

If $g\in \mathcal{B}{T}_{4l}$, then

$$\begin{array}{c}\hfill |b_2{b}_5-b_3{b}_4|\le \frac{15.503407}{432}=0.035888.\end{array}$$

Proof. 

Let $g\in \mathcal{B}{T}_{4l}$. From (8), we obtain

$$|b_2{b}_5-b_3{b}_4|=\frac{1}{432}|30c_1{c}_4-25c_2{c}_3|.$$

(13)

Applying Lemma 1 and the triangle inequality in (14), we get

$$\begin{array}{cc}\hfill |b_2{b}_5-b_3{b}_4|& \le \frac{1}{432}[30|c_1|(1-|c_1{|}^2-|c_2{|}^2)+25|c_2|(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|})]\hfill \\ \hfill & =\frac{1}{432}[30|c_1|-30|c_1{|}^3+(25-25|c_1{|}^2)|c_2|-30|c_1||c_2{|}^2-\frac{25}{1+|c_1|}{|c_2|}^3].\hfill \end{array}$$

Setting $|c_1|=c$ and $|c_2|=d$, we yield

$$\begin{array}{c}\hfill |b_2{b}_5-b_3{b}_4|\le \frac{1}{432}{ϵ}_4(c,d).\end{array}$$

Taking the partial derivative with respect to c and d, respectively, we obtain

$$\begin{array}{cc}\hfill \frac{\partial {ϵ}_4}{\partial c}& =30-90c^2-50cd-30d^2+\frac{25}{{(1+c)}^2}{d}^3\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \frac{\partial {ϵ}_4}{\partial d}& =25-25c^2-60cd-\frac{75}{1+c}{d}^2.\hfill \end{array}$$

Setting $\frac{\partial {ϵ}_4}{\partial c}=\frac{\partial {ϵ}_4}{\partial d}=0$, and simplifying, we receive

$$\left\{\begin{array}{c}-90c^4-180c^3-60c^2+60c+30-50c{(1+c)}^{2}d-30{(1+c)}^2{d}^2+25d^3=0,\hfill \\ -25c^3-25c^2+25c+25-60(c^2+c)d-75d^2=0.\hfill \end{array}\right.$$

Applying Newton’s methods, we have

$$\left\{\begin{array}{c}{c}_1=-1,\hfill \\ d_1=0,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_2=0.4098,\hfill \\ d_2=-0.8978,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_3=0.4271,\hfill \\ d_3=0.4258,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_4=-0.4550\hfill \\ d_4=-0.2931,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_5=-0.7258,\hfill \\ d_5=0.3023.\hfill \end{array}\right.$$

Therefore, we obtain ${\epsilon }_4(c,d)\le {\epsilon }_4(0.4271,0.4258)=15.503407$.

(a)

For $c=0$,

$${\epsilon }_4(0,d)=25d-25d^3\le {\epsilon }_4(0,\frac{\sqrt{3}}{3})=\frac{50\sqrt{3}}{9}.$$

(b)

For $d=0$,

$${\epsilon }_4(c,0)=30c-30c^3\le {\epsilon }_4(\frac{\sqrt{3}}{3},0)=\frac{20\sqrt{3}}{3}=11.546666.$$

(c)

For $d=1-c^2$,

$${\epsilon }_4(c,1-c^2)=25c-20c^3-5c^5={\eta }_2(c)={\eta }_2(0.6017)=10.2913.$$

Thus, we obtain

$$|b_2{b}_5-b_3{b}_4|\le \frac{15.503407}{432}=0.035888.$$

□

Theorem 5.

If $g\in \mathcal{B}{T}_{4l}$, then

$$\begin{array}{c}\hfill |H_{2,2}(g)|=|b_2{b}_4-b_3^2|\le \frac{200.27}{2592}=0.077265.\end{array}$$

Proof. 

Let $g\in \mathcal{B}{T}_{4l}$. From (8), we obtain

$$\begin{array}{cc}\hfill |b_2{b}_4-b_3^2|& =\frac{1}{2592}|225c_1{c}_3-200c_2^2|\hfill \\ \hfill & =\frac{1}{2592}|200(c_1{c}_3-c_2^2)+25c_1{c}_3|.\hfill \end{array}$$

Applying Lemma 1, Lemma 2 and the triangle inequality, we receive

$$\begin{array}{cc}\hfill |b_2{b}_4-b_3^2|& \le \frac{1}{2592}[200(1-|c_1{|}^2)+25|c_1|(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|})]\hfill \\ \hfill & =\frac{1}{2592}[200+25|c_1|-200|c_1{|}^2-25|c_1{|}^3-\frac{|c_1|}{1+|c_1|}{|c_2|}^2]\hfill \\ \hfill & \le \frac{1}{2592}\left(200+25|c_1|-200|c_1{|}^2-25{|c_1|}^3\right)\hfill \\ \hfill & =\frac{1}{2592}{\eta }_3(|c_1|)\le \frac{1}{2592}{\eta }_3(0.0618)=\frac{200.2700}{2592}=0.077265.\hfill \end{array}$$

□

Theorem 6.

If $g\in \mathcal{B}{T}_{4l}$, then

$$\begin{array}{c}\hfill |H_{3,2}(g)|\le 0.025986.\end{array}$$

Proof. 

Let $g\in \mathcal{B}{T}_{4l}$. From Lemma 3, Theorems 1, 2, 4 and 5, we yield

$$\begin{array}{cc}\hfill |H_{3,2}(g)|& \le |b_4||b_3{b}_5-b_4^2|+|b_5||b_2{b}_5-b_3{b}_4|+|b_6||b_2{b}_4-b_3^2|\hfill \\ \hfill & =\frac{5}{24}\times 0.044516+\frac{1}{6}\times 0.035888+\frac{5}{36}\times 0.077265\hfill \\ \hfill & =0.025986.\hfill \end{array}$$

□

Theorem 7.

If $g\in \mathcal{B}{T}_{4l}$, then

$$\begin{array}{c}\hfill |H_{2,4}(g)|=|b_4{b}_6-b_5^2|\le \frac{24.385252}{864}=0.028224.\end{array}$$

Proof. 

Let $f\in \mathcal{B}{T}_{4l}$. From (8) and (9), we have

$$\begin{array}{c}\hfill |b_4{b}_6-b_5^2|=\frac{1}{864}|25c_3{c}_5-24c_4^2+5c_3{c}_1^5|=\frac{1}{864}|24(c_3{c}_5-c_4^2)+c_3{c}_5+5c_3{c}_1^5|.\end{array}$$

Using Lemmas 1 and 2, we have

$$\begin{array}{ccc}\hfill |b_4{b}_6-b_5^2|& & \le \frac{1}{864}(24|c_3{c}_5-c_4^2|+|c_3||c_5|+5|c_3||c_1^5|)\hfill \\ & & \le \frac{1}{864}[24(1-|c_1{|}^2)+|c_3|(1-|c_1{|}^2-|c_2{|}^2-\frac{|c_3{|}^2}{1+||c_1})+5|c_1{|}^5|c_3|]\hfill \\ & & =\frac{1}{864}[24-24|c_1{|}^2+(1-|c_1{|}^2+5|c_1{|}^5-|c_2{|}^2)|c_3|-\frac{1}{1+|c_1|}|c_3{|}^3].\hfill \end{array}$$

Setting $c=|c_1|,d=|c_2|$ and $e=|c_3|$, we yield

$$\begin{array}{ccc}\hfill |b_4{b}_6-b_5^2|& & \le \frac{1}{864}{ϵ}_5(c,d,e)\hfill \end{array}$$

where ${ϵ}_5(c,d,e)=24-24c^2+(1-c^2+5c^5-d^2)e-\frac{1}{1+c}{e}^3$ and $(c,d,e)\in \mathrm{\Omega }=\{(c,d,e):0\le c\le 1,0\le d\le 1-c^2,0\le e\le 1-c^2-\frac{d^2}{1+c}\}$. Consider

$$\frac{\partial {ϵ}_5}{\partial d}=-2de\le 0,$$

thus, there are no points in $\mathrm{\Omega }$.

(1)

For $c=0$,

$$\begin{array}{c}\hfill {ϵ}_5(0,d,e)=24+(1-d^2)e-e^3={\eta }_4(d,e).\end{array}$$

It is evident that there is on point in $(0,1)\times (0,1-d^2)$.

(2)

For $d=0$,

$$\begin{array}{c}\hfill {ϵ}_5(c,0,e)=24-24c^2+(1-c^2+5c^5)e-\frac{1}{1+c}{e}^3={\eta }_5(c,e).\end{array}$$

Partial derivative of ${\eta }_5$ with respect to c, and then with respect to e, we achieve

$$\begin{array}{c}\hfill \frac{\partial {\eta }_5}{\partial c}=-48c+(-2c+25c^4)e+\frac{1}{{(1+c)}^2}{e}^3,\end{array}$$

and

$$\begin{array}{c}\hfill \frac{\partial {\eta }_5}{\partial e}=1-c^2+5c^5-\frac{3}{1+c}{e}^2.\end{array}$$

Setting $\frac{\partial {\eta }_5}{\partial c}=\frac{\partial {\eta }_5}{\partial e}=0$ and simplifying, we yield

$$\left\{\begin{array}{c}-48c{(1+c)}^2+(25c^6+50c^5+25c^4-2c^3-4c^2-2c)e+e^3=0,\hfill \\ 5c^6+5c^5-c^3-c^2+c+1-3e^2=0.\hfill \end{array}\right.$$

We obtain a critical point $(0.0039,0.5785)$; thus, we have ${\eta }_5(c,e)\le {\eta }_5(0.0039,0.5785)=24.385252$.

(3)

For $e=0$,

$$\begin{array}{c}\hfill {ϵ}_5(c,d,0)=24-24c^2\le 24.\end{array}$$

(4)

For $e=1-c^2-\frac{d^2}{1+c}$,

$$\begin{array}{cc}{ϵ}_5(c,d,1-c^2-\frac{d^2}{1+c})\hfill & =24+c-24c^2-2c^3+6c^5-5c^7+\frac{-5c^5+4c^3-c^2-4c+1}{1+c}{d}^2\hfill \\ \hfill & +\frac{-2+4c}{{(1+c)}^2}{d}^4+\frac{1}{{(1+c)}^4}{d}^6={\eta }_6(c,d)\le {\eta }_6(0.0016,0.5767)=24.148169.\hfill \end{array}$$

(5)

For $c=d=0$

$$\begin{array}{c}\hfill {ϵ}_5(0,0,e)=24+e-e^3={\eta }_7(e)\le {\eta }_7(\frac{\sqrt{3}}{3})=24+\frac{2\sqrt{3}}{9}=24.384889.\end{array}$$

(6)

For $c=e=0$,

$$\begin{array}{c}\hfill {ϵ}_5(0,d,0)=24.\end{array}$$

(7)

For $d=e=0$,

$$\begin{array}{c}\hfill {ϵ}_5(c,0,0)=24+e-e^3\le 24.384889.\end{array}$$

(8)

For $e=0,d=1-c^2$,

$$\begin{array}{c}\hfill {ϵ}_5(c,1-c^2,0)=24-24c^2\le 24.\end{array}$$

(9)

For $d=0$ and $e=1-c^2$,

$$\begin{array}{c}\hfill {ϵ}_5(c,0,1-c^2)=24+c-24c^2-2c^3+6c^5-5c^7={\eta }_8(c)\le {\eta }_8(0.0206)=24.0104.\end{array}$$

(10)

For $c=0,e=1-d^2$

$$\begin{array}{c}\hfill {ϵ}_5(0,d,1-d^2)=24+d^2-2d^4+d^6={\eta }_9(d)\le {\eta }_9(\frac{\sqrt{3}}{3})=24.148148.\end{array}$$

Thus, we get

$$\begin{array}{c}\hfill |b_4{b}_6-b_5^2|\le \frac{24.385252}{864}=0.035888.\end{array}$$

□

Theorem 8.

If $g\in \mathcal{B}{T}_{4l}$, then

$$\begin{array}{c}\hfill |b_3{b}_6-b_4{b}_5|\le \frac{5}{144}.\end{array}$$

Proof. 

Let $g\in \mathcal{B}{T}_{4l}$. From (8) and (9), we receive

$$\begin{array}{c}\hfill |b_3{b}_6-b_4{b}_5|=\frac{1}{1296}|50c_2{c}_5-45c_3{c}_4+50c_2{c}_1^5|.\end{array}$$

Using Lemma 1, we obtain

$$\begin{array}{ccc}\hfill |b_3{b}_6-b_4{b}_5|& & \le \frac{1}{1296}(50|c_2||c_5|+45|c_3||c_4|+50|c_2||c_1^5|)\hfill \\ & & \le \frac{1}{1296}[50|c_2|(1-|c_1{|}^2-|c_2{|}^2-\frac{|c_3{|}^2}{1+|c_1|})+45|c_3|(1-|c_1{|}^2-|c_2{|}^2)+50|c_1{|}^5|c_2|]\hfill \\ & & =\frac{1}{1296}[(50-50|c_1{|}^2+50|c_1{|}^5)|c_2|-50|c_2{|}^3+(45-45|c_1{|}^2-45|c_2{|}^2)|c_3|-\frac{50|c_2|}{1+|c_1|}|c_3{|}^2].\hfill \end{array}$$

By setting $|c_1|=c,|c_2|=d$ and $|c_3|=e$, we have

$$\begin{array}{ccc}\hfill |b_3{b}_6-b_4{b}_5|& & \le \frac{1}{1296}\mathrm{\Psi }(c,d,e)\hfill \end{array}$$

where $\mathrm{\Psi }(c,d,e)=(50-50c^2+50c^5)d-50d^3+(45-45c^2-45d^2)e-\frac{50d}{1+c}{e}^2$, $(c,d,e)\in \mathrm{\Omega }$.

Differentiating partially with respect to c, d and e, respectively, we obtain

$$\begin{array}{cc}\hfill \frac{\partial \mathrm{\Psi }}{\partial c}& =(-100c+250c^4)d-90ce+\frac{50d}{{(1+c)}^2}{e}^2,\hfill \end{array}$$

$$\begin{array}{cc}\hfill \frac{\partial \mathrm{\Psi }}{\partial d}& =50-50c^2+50c^5-150d^2-90de-\frac{50}{1+c}{e}^2,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \frac{\partial \mathrm{\Psi }}{\partial e}& =45-45c^2-45d^2-\frac{100d}{1+c}e.\hfill \end{array}$$

By putting $\frac{\partial \mathrm{\Psi }}{\partial c}=\frac{\partial \mathrm{\Psi }}{\partial d}=\frac{\partial \mathrm{\Psi }}{\partial e}=0$, and simplifying, we obtain

$$\left\{\begin{array}{c}(250c^6+500c^5+250c^4-100c^3-200c^2-100c)d-900c{(1+c)}^{2}e+50d{e}^2=0,\hfill \\ 50c^6+50c^5-50c^3-50c^2+50c+50-150(1+c)d^2-90(1+c)de-50e^2=0,\hfill \\ -45c^3-45c^2+45c+45-45(1+c)d^2-100de=0.\hfill \end{array}\right.$$

By a numberical calculation, we obtain

$$\left\{\begin{array}{c}{c}_1=-1,\hfill \\ d_1=d,\hfill \\ e_1=0,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_2=0.8441,\hfill \\ d_2=0.4127,\hfill \\ e_2=0.2354,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_3=0.8441,\hfill \\ d_3=-0.4127,\hfill \\ e_3=-0.2354.\hfill \end{array}\right.$$

Thus, there is no critical point that satisfies $0\le c\le 1$ and $0\le d\le 1-c^2$.

(1)

For $c=0$,

$$\begin{array}{c}\hfill \mathrm{\Psi }(0,d,e)=50d-50d^3+(45-45d^2)e-50d{e}^2={\mathrm{\Lambda }}_1(d,e).\end{array}$$

Consider

$$\left\{\begin{array}{c}\frac{\partial {\mathrm{\Lambda }}_1}{\partial d}=50-150d^2-90de-50e^2=0,\hfill \\ \frac{\partial {\mathrm{\Lambda }}_1}{\partial e}=45-45d^2-100de=0.\hfill \end{array}\right.$$

A numerical calculation is that there is no critical point in $(0,1)\times (0,1-d^2)$.

(2)

For $d=0$,

$$\begin{array}{c}\hfill \mathrm{\Psi }(c,0,e)=(45-45c^2)e\le 45{(1-c^2)}^2\le 45.\end{array}$$

(3)

For $e=0$,

$$\begin{array}{c}\hfill \mathrm{\Psi }(c,d,0)=(50-50c^2+50c^5)d-50d^3={\mathrm{\Lambda }}_2(c,d)\le {\mathrm{\Lambda }}_2(0.7368,0.2828)=8.403278.\end{array}$$

(4)

For $e=1-c^2-\frac{d^2}{1+c}$,

$$\begin{array}{cc}\hfill \mathrm{\Psi }(c,d,1-c^2-\frac{d^2}{1+c})& =45-90c^2+45c^4+(50c-50c^3+50c^5)d+(-90+45c+45c^2)d^2\hfill \\ \hfill & +\frac{50-150c}{1+c}{d}^3+\frac{45}{1+c}{d}^4-\frac{50}{{(1+c)}^3}{d}^5={\mathrm{\Lambda }}_3(c,d).\hfill \end{array}$$

Differentiating ${\mathrm{\Lambda }}_3$ partially with respect to c and d, we yield

$$\begin{array}{cc}\hfill \frac{\partial {\mathrm{\Lambda }}_3}{\partial c}& =-180c+180c^3+(50-150c^2+250c^4)d+(45+90c)d^2-\frac{100d^3}{{(1+c)}^2}\hfill \\ \hfill & -\frac{45d^4}{{(1+c)}^2}+\frac{150d^5}{{(1+c)}^4}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \frac{\partial {\mathrm{\Lambda }}_3}{\partial d}& =50c-50c^3+50c^5+(-180+90c+90c^2)d+\frac{150-450c}{1+c}{d}^2+\frac{180d^3}{1+c}\hfill \\ \hfill & -\frac{250d^4}{{(1+c)}^3}.\hfill \end{array}$$

Setting $\frac{\partial {\mathrm{\Lambda }}_3}{\partial c}=\frac{\partial {\mathrm{\Lambda }}_3}{\partial d}=0$ and simplifying, we receive

$$\left\{\begin{array}{c}180c^7+720c^6+900c^5-900c^3-720c^2-180c+(250c^8+1000c^7+1350c^6+400c^5\hfill \\ -600c^4-400c^3+150c^2+200c+50)d+(90c^5+405c^4+720c^3+630c^2+270c+45)d^2\hfill \\ -100{(1+c)}^2{d}^3-45{(1+c)}^2{d}^4+150d^5=0,\hfill \\ 50c^8+150c^7+100c^6-100c^5-100c^4+100c^3+150c^2+50c+(90c^5+360c^4+360c^3-180c^2\hfill \\ -450c-180)d+(-450c^3-750c^2-150c+150)d^2+180{(1+c)}^2{d}^3-250d^4=0\hfill \end{array}\right.$$

Applying Newton’s methods, we recieve

$$\left\{\begin{array}{c}{c}_1=-1\hfill \\ d_1=0,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_2=0,\hfill \\ d_2=0,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_3=0.8336,\hfill \\ d_3=0.4141,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_4=-0.9194,\hfill \\ d_4=-0.0957.\hfill \end{array}\right.$$

Thus, there is no critical point satisfing $0\le c\le 1$ and $0\le d\le 1-c^2$.

(5)

For $d=c=0$,

$$\begin{array}{cc}\hfill \mathrm{\Psi }(0,0,e)& =45e\le 45.\hfill \end{array}$$

(6)

For $e=d=0$,

$$\begin{array}{cc}\hfill \mathrm{\Psi }(c,0,0)& =0.\hfill \end{array}$$

(7)

For $c=e=0$,

$$\begin{array}{cc}\hfill \mathrm{\Psi }(0,d,0)& =50d-50d^3={\mathrm{\Lambda }}_4(d)\le {\mathrm{\Lambda }}_4(\frac{\sqrt{3}}{3})=\frac{100\sqrt{3}}{9}=19.2444.\hfill \end{array}$$

(8)

For $d=1-c^2$ and $e=0$,

$$\begin{array}{cc}\hfill \mathrm{\Psi }(c,1-c^2,0)& =50c^2-100c^4+50c^5+50c^6-50c^7={\mathrm{\Lambda }}_5(c)\le {\mathrm{\Lambda }}_5(0.7145)=10.6734.\hfill \end{array}$$

(9)

For $e=1-c^2$ and $d=0$,

$$\begin{array}{cc}\hfill \mathrm{\Psi }(c,0,1-c^2)& =45{(1-c^2)}^2\le 45.\hfill \end{array}$$

(10)

For $e=1-d^2$ and $c=0$,

$$\begin{array}{cc}\hfill \mathrm{\Psi }(0,d,1-c^2)& =45-90d^2+50d^3+45d^4-50d^5={\mathrm{\Lambda }}_6(d)\le {\mathrm{\Lambda }}_6(0)=45.\hfill \end{array}$$

Hence, we obtain

$$\begin{array}{c}\hfill |b_3{b}_6-b_4{b}_5|\le \frac{45}{1296}=\frac{5}{144}.\end{array}$$

□

Theorem 9.

If $g\in \mathcal{B}{T}_{4l}$, then

$$\begin{array}{c}\hfill |H_{3,3}(g)|\le 0.016103.\end{array}$$

Proof. 

Let $g\in \mathcal{B}{T}_{4l}$. From Lemma 3 and Theorems 1–4, we receive

$$\begin{array}{ccc}\hfill |H_{3,3}(g)|& & \le |b_5||b_4{b}_6-b_5^2|+|b_6||b_3{b}_6-b_4{b}_5|+|b_7||b_3{b}_5-b_4^2|\hfill \\ \hfill & & \le \frac{1}{6}\times 0.035888+\frac{5}{36}\times \frac{5}{144}+\frac{5}{42}\times 0.044516\hfill \\ \hfill & & =0.016103.\hfill \end{array}$$

□

3. The Bounds of the Logarithmic Coefficients for $\mathit{g}\mathbf{\in }\mathcal{B}{\mathit{T}}_{\mathbf{4}\mathit{l}}$

Theorem 10.

If $g\in \mathcal{B}{T}_{4l}$, then

$$\begin{array}{c}\hfill |{\delta }_1|\le \frac{5}{24},|{\delta }_2|\le \frac{5}{36},|{\delta }_3|\le \frac{5}{48},|{\delta }_4|\le \frac{1}{12},|{\delta }_5|\le \frac{173400}{2488320}=0.069686.\end{array}$$

The first four bounds are the best possible.

Proof. 

Let $g\in \mathcal{B}{T}_{4l}$. From (6), (8), and (9), we have

$$\begin{array}{c}\hfill {\delta }_1=\frac{5c_1}{24},\end{array}$$

(14)

$$\begin{array}{c}\hfill {\delta }_2=\frac{5}{36}(c_2-\frac{5}{8}{c}_1^2),\end{array}$$

(15)

$$\begin{array}{c}\hfill {\delta }_3=\frac{5}{48}(c_3-\frac{5}{9}{c}_1{c}_2+\frac{25}{216}{c}_1^3),\end{array}$$

(16)

$$\begin{array}{c}\hfill {\delta }_4=\frac{1}{165888}(13824c_4-7200c_1{c}_3+4000c_1^2{c}_2-3200c_2^2-625c_1^4).\end{array}$$

(17)

$$\begin{array}{c}\hfill {\delta }_5=\frac{1}{2488320}(172800c_5-86400c_1{c}_4-72000c_2{c}_3+45000c_1^2{c}_3+40000c_1{c}_2^2-5000c_1^3{c}_2+37685c_1^5).\end{array}$$

(18)

Applying Lemma 4 to (15), we have

$$\begin{array}{c}\hfill |{\delta }_1|\le \frac{5}{24}\end{array}$$

Applying Lemma 5 to (16), we obtain

$$\begin{array}{c}\hfill |{\delta }_2|\le \frac{5}{36}.\end{array}$$

Utilizing the triangle inequality and Lemma 6 with $\gamma =-\frac{5}{9}$ and $\zeta =\frac{25}{216}$, we yield

$$\begin{array}{c}\hfill |{\delta }_3|\le \frac{5}{48}.\end{array}$$

Rearranging (18), we obtain

$$\begin{array}{ccc}\hfill |{\delta }_4|& & =\frac{1}{165888}|7200(c_4-c_1{c}_3-\frac{1}{2}{c}_2^2+\frac{3}{4}{c}_1^2{c}_2-\frac{1}{8}{c}_1^4)+6624c_4+400c_2^2-1400c_1^2{c}_2+275c_1^4|\hfill \\ \hfill & & \le \frac{1}{165888}|7200(c_4-c_1{c}_3-\frac{1}{2}{c}_2^2+\frac{3}{4}{c}_1^2{c}_2-\frac{1}{8}{c}_1^4)|+\frac{1}{165888}|6624c_4+400c_2^2-1400c_1^2{c}_2+275c_1^4|\hfill \\ \hfill & & =\frac{1}{165888}{\nabla }_1+\frac{1}{165888}{\nabla }_2,\hfill \end{array}$$

where ${\nabla }_1=|7200(c_4-c_1{c}_3-\frac{1}{2}{c}_2^2+\frac{3}{4}{c}_1^2{c}_2-\frac{1}{8}{c}_1^4)|$

$$\begin{array}{cc}\hfill {\nabla }_2& =|6624c_4+400c_2^2-1400c_1^2{c}_2+275c_1^4|.\end{array}$$

(19)

Using Lemma 7 with $\gamma =-\frac{1}{2}$, we obtain ${\nabla }_1\le 7200$.

Using Lemma 6 and the triangle inequality in (20), we receive

$$\begin{array}{ccc}\hfill {\nabla }_2& & \le 6624(1-|c_1{|}^2-|c_2{|}^2)+400|c_2{|}^2+1400|c_1{|}^2|c_2|+275|c_1{|}^4\hfill \\ \hfill & & =6624-6624|c_1{|}^2+275|c_1{|}^4+1400|c_1{|}^2|c_2|-6224|c_2{|}^2={\mathrm{{\rm Y}}}_1(c,d)\hfill \end{array}$$

where $|c_1|=c,|c_2|=d$.

Consider

$$\left\{\begin{array}{c}\frac{\partial {\mathrm{{\rm Y}}}_1}{\partial c}=-13248c+1100c^3+2800cd=0,\hfill \\ \frac{\partial {\mathrm{{\rm Y}}}_1}{\partial d}=1400c^2-12448d=0\hfill \end{array}\right.$$

Applying Newton’s methods, we have

$$\left\{\begin{array}{c}{c}_1=0\hfill \\ d_1=0,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_2=3.0599,\hfill \\ d_2=1.0531,\hfill \end{array}\right.\left\{\begin{array}{c}{c}_3=-3.0599,\hfill \\ d_3=1.0531.\hfill \end{array}\right.$$

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

(1)

For $c=0$,

$$\begin{array}{c}\hfill {\mathrm{{\rm Y}}}_1(0,d)=6624-6224d^2\le 6624.\end{array}$$

(2)

For $d=0$,

$$\begin{array}{c}\hfill {\mathrm{{\rm Y}}}_1(c,0)=6624-6624c^2+275c^4={\gamma }_1(c)\le {\gamma }_1(0)=6624.\end{array}$$

(3)

For $d=1-c^2$,

$$\begin{array}{c}\hfill {\mathrm{{\rm Y}}}_1(c,1-c^2)=400+7224c^2-7349c^4={\gamma }_2(c)\le {\gamma }_2(0.7011)=2175.3.\end{array}$$

Therefore, we have

$$\begin{array}{c}\hfill |{\gamma }_4|\le \frac{13824}{165888}=\frac{1}{12}.\end{array}$$

Rearranging (19), we obtain

$$\begin{array}{ccc}\hfill |{\delta }_5|& & =\frac{1}{2488320}|86400(c_5-c_1{c}_4-c_2{c}_3+\frac{3}{4}{c}_1{c}_2^2+\frac{3}{4}{c}_1^2{c}_3-\frac{1}{2}{c}_1^3{c}_2+\frac{1}{16}{c}_1^5)\hfill \\ \hfill & & +86400c_5+14400c_2{c}_3-19800c_1^2{c}_3-24800c_1{c}_2^2+38200c_1^3{c}_2+32285c_1^5|\hfill \\ \hfill & & \le \frac{1}{2488320}|86400(c_5-c_1{c}_4-c_2{c}_3+\frac{3}{4}{c}_1{c}_2^2+\frac{3}{4}{c}_1^2{c}_3-\frac{1}{2}{c}_1^3{c}_2+\frac{1}{16}{c}_1^5)|\hfill \\ \hfill & & +\frac{1}{2488320}|86400c_5+14400c_2{c}_3-19800c_1^2{c}_3-24800c_1{c}_2^2+38200c_1^3{c}_2+32285c_1^5|\hfill \\ \hfill & & =\frac{1}{2488320}{\nabla }_3+\frac{1}{2488320}{\nabla }_4,\hfill \end{array}$$

where ${\nabla }_3=|86400(c_5-c_1{c}_4-c_2{c}_3+\frac{3}{4}{c}_1{c}_2^2+\frac{3}{4}{c}_1^2{c}_3-\frac{1}{2}{c}_1^3{c}_2+\frac{1}{16}{c}_1^5)|$ and

$$\begin{array}{cc}\hfill {\nabla }_4& =|86400c_5+14400c_2{c}_3-19800c_1^2{c}_3-24800c_1{c}_2^2+38200c_1^3{c}_2+32285c_1^5|.\end{array}$$

(20)

Using Lemma 8 with $\gamma =-\frac{1}{2}$, we obtain ${\nabla }_3\le 86400$. Rearranging (21), we obtain

$$\begin{array}{cc}\hfill {\nabla }_4& =|86400c_5+14400c_2(c_3-2c_1{c}_2+c_1^3)-19800c_1^2(c_3-c_1{c}_2-c_1^3)+4000c_1{c}_2^2+4000c_1^3{c}_2+12485c_1^5|.\end{array}$$

Utilizing the triangle inequality, Lemma 1, Lemma 6 and 5, we obtain

$$\begin{array}{ccc}\hfill \nabla 4& & \le 86400|c_5|+14400|c_2||c_3-2c_1{c}_2+c_1^3|+19800|c_1{|}^2|c_3-c_1{c}_2-c_1^3|+4000|c_1||c_2{|}^2+4000|c_1{|}^3|c_2|+12485|c_1{|}^5\hfill \\ \hfill & & \le 86400(1-|c_1{|}^2-|c_2{|}^2-\frac{|c_3{|}^2}{1+|c_1|})+14400|c_2|+19800|c_1{|}^2+4000|c_1||c_2{|}^2+4000|c_1{|}^3|c_2|+12485|c_1{|}^5\hfill \\ \hfill & & =86400-666000|c_1{|}^2+12485|c_1{|}^5+(14400+4000|c_1{|}^3)|c_2|+(-86400+4000|c_1|)|c_2{|}^2-86400\frac{|c_3{|}^2}{1+|c_1|}\hfill \\ \hfill & & \le 86400-666000|c_1{|}^2+12485|c_1{|}^5+(14400+4000|c_1{|}^3)|c_2|+(-86400+4000|c_1|)|c_2{|}^2={\mathrm{{\rm Y}}}_2(c,d)\hfill \end{array}$$

where $|c_1|=c$, $|c_2|=d$. Consider

$$\left\{\begin{array}{c}\frac{\partial {\mathrm{{\rm Y}}}_2}{\partial c}=-133200c+62425c^4+12000c^{2}d+4000d^2=0,\hfill \\ \frac{\partial {\mathrm{{\rm Y}}}_2}{\partial d}=14400+4000c^3+(-172800+8000c)d=0.\hfill \end{array}\right.$$

We have a critical point $(0.000209,0.083334)$. Thus, we obtain ${\mathrm{{\rm Y}}}_2(c,d)\le {\mathrm{{\rm Y}}}_2$$(0.000209,0.083334)=86999.968$.

(1)

For $c=0$,

$$\begin{array}{c}\hfill {\mathrm{{\rm Y}}}_2(0,d)=86400+14400d-86400d^2={\gamma }_3(d)\le {\gamma }_3(\frac{1}{12})=87000.\end{array}$$

(2)

For $d=0$,

$$\begin{array}{c}\hfill {\mathrm{{\rm Y}}}_2(c,0)=86400-66600c^2+12485c^5\le 864000.\end{array}$$

(3)

For $d=1-c^2$,

$${\mathrm{{\rm Y}}}_2(c,1-c^2)=14400+4000c+91800c^2-4000c^3-86400c^4+12485c^5={\gamma }_4(c)\le {\gamma }_4(0.7771)=43098.$$

Therefore, we receive

$$\begin{array}{c}\hfill |{\delta }_5|\le \frac{173400}{2488320}=0.069686.\end{array}$$

These bounds can be determined by the following extremal functions:

$$g_4(\xi )={\int }_0^{\xi }(1+\frac{5}{6}t+\frac{1}{6}{t}^5)dt=\xi +\frac{5}{12}{\xi }^2+\frac{1}{36}{\xi }^6+···,$$

$$g_5(\xi )={\int }_0^{\xi }(1+\frac{5}{6}{t}^2+\frac{1}{6}{t}^{10})dt=\xi +\frac{5}{18}{\xi }^3+\frac{1}{66}{\xi }^{11}+···,$$

$$g_6(\xi )={\int }_0^{\xi }(1+\frac{5}{6}{t}^3+\frac{1}{6}{t}^{15})dt=\xi +\frac{5}{24}{\xi }^4+\frac{1}{96}{\xi }^{16}+···,$$

$$g_7(\xi )={\int }_0^{\xi }(1+\frac{5}{6}{t}^4+\frac{1}{6}{t}^{20})dt=\xi +\frac{1}{6}{\xi }^5+\frac{1}{126}{\xi }^{21}+···.$$

The proof of Theorem 10 is completed.  □

4. Conclusions

In this paper, we considered a subclass of bounded turning functions linked with a four-leaf-type domain. Utilizing the estimates of the coefficients of the Schwartz function, we obtained the estimates of coefficients of $|b_6|,|b_7|,|b_8|$, and the third-order determinants of $|H_{3,1}|,|H_{3,2}|,|H_{3,3}|$ for the class $\mathcal{B}{T}_{4l}$ for the first time. Also, one can easily use this new methodology to obtain the bounds the coefficients of $|b_6|,|b_7|,|b_8|$, and the third-order Hankel determinant of $|H_{3,1}|,|H_{3,2}|,|H_{3,3}|$ for other subclasses of univalent functions. The bounds provided in Theorems 2–9 lack sharpness and improve these bounds in further research.