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

1. Introduction and Definitions

Let $\mathcal{H}$ be a representation for 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 $\mathcal{S}$ as a subfamily of $\mathcal{H}$, which consists of univalent functions in $\mathcal{U}.$

For functions $g\in \mathcal{H}$ of the form $g\left(\xi \right)=\xi +b_2{\xi }^2+b_3{\xi }^3+···$ and positive integers i and n, the Hankel determinant $H_{i,n}\left(g\right)$ is defined by

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

The Hankel determinant $H_{i,n}\left(g\right)$ was introduced by Pommerrenke [1,2]. The third Hankel determinants are 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 ${\mathcal{S}}^{*}$, $\mathcal{K}$ and $\mathcal{R}$ 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 $g\in \mathcal{S}$, Let $F_g\left(\xi \right)$ be defined as the logarithmic coefficients of $g\left(\xi \right)$,

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

(5)

The ${\delta }_n={\delta }_n\left(g\right)$ are referred to as the 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].

Using (1) and differentiating (5), we have

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

(6)

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

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

Sunthrayuth et al. [12] 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 of the logarithmic coefficients for $g\in \mathcal{B}{T}_{4l}$.

Let ${\omega }_0$ be the family of Schwarz functions. Thus, the function $w\in {\omega }_0$ may 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 member of ${\omega }_0$. Then

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

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

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

Lemma 2

(see [14,15]). Suppose a function w is 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

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

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 $w\in {\omega }_0$ is in the form (7). Then, we get

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

where $(\gamma ,\varsigma )\in D_0\cup D_1,$ with

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

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

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 [18]). 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 $g\in \mathcal{B}{T}_{4l}$

Theorem 1.

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

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

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

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

From (10) and Lemma 1, we achieve

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

From (11) and Lemma 1, we get

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

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

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

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

$$\begin{array}{cc}\hfill |b_8|& \le \frac{5}{48}.\hfill \end{array}$$

The bounds hold for $c=0.$ The proof of Theorem 1 is completed. □

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 get

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

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 get

$\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

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

The proof of Theorem 2 is completed. ☐

Theorem 3.

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

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

Proof. 

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

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

(13)

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

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

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

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

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

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

(14)

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

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

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 get

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

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

(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 $d=0$,

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

(b)For $d=1-c^2$,

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

Thus, we get

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

☐

Theorem 5.

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

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

Proof. 

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

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

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

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

☐

Theorem 6.

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

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

Proof. 

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

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

☐

Theorem 7.

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

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

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 Lemma 1 and 2, we have

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

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 $(c,d,e)\in \Omega =\{(c,d,e):0\le 1,0\le d\le 1-c^2,0\le e\le 1-c^2-\frac{d^2}{1+c}\}$. Consider

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

thus there are no points in $\Omega$.

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

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

(5)For $c=d=0$

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

(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 $e=1-c^2$,

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

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

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

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

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

The bound is sharp.

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 get

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

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

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

where $\Psi (c,d,e)=(50-50c^2+50c^5)d-50d^3+(45-45c^2-45d^2)e-\frac{50d}{1+c}{e}^2$ , $(c,d,e)\in \Omega$. Differentiating partially with respect to c, d and e, respectively, we get

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

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

and

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

By putting $\frac{\partial \Psi }{\partial c}=\frac{\partial \Psi }{\partial d}=\frac{\partial \Psi }{\partial e}=0$, 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 caculation, we get

$\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’s no critical point which satisfies $0\le c\le 1$ and $0\le d\le 1-c^2.$

(1)For $c=0,$

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

Consider

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

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

(2)For $d=0,$

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

(3)For $e=0,$

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

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

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

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

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

and

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

Setting $\frac{\partial {\Lambda }_3}{\partial c}=\frac{\partial {\Lambda }_3}{\partial d}=0$ 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 $d=c=0,$

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

(6)For $e=d=0,$

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

(7)For $c=e=0,$

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

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

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

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

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

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

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

Hence, we get

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

The equality holds for $c_1=c_2=0$ and $c_3=c_4=1.$ ☐

Theorem 9.

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

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

Proof. 

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

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

☐

3. The bounds of the logarithmic coefficients for $g\in \mathcal{B}{T}_{4l}$

Theorem 10.

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

$$\begin{array}{c}\hfill |{\delta }_1|\le \frac{5}{24},|{\delta }_2|\le \frac{5}{36},|{\delta }_3|\le \frac{5}{48},|{\delta }_4|\le \frac{13827}{165888}=0.083351,\left|{\delta }_5\right|\le \frac{173400}{2488320}=0.069686.\end{array}$$

The first three bounds are the best possible.

Proof. 

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

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

(15)

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

(16)

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

(17)

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

(18)

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

(19)

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 get

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

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

$$\begin{array}{ccc}\hfill |{\delta }_4|& & =\frac{1}{165888}|6912(c_4-c_1{c}_3-\frac{1}{2}{c}_2^2+\frac{3}{4}{c}_1^2{c}_2-\frac{1}{8}{c}_1^4)+6912c_4-288c_1{c}_3+256c_2^2-1184c_1^2{c}_2+239c_1^4|\hfill \\ & & \le \frac{1}{165888}|6912(c_4-c_1{c}_3-\frac{1}{2}{c}_2^2+\frac{3}{4}{c}_1^2{c}_2-\frac{1}{8}{c}_1^4)|+\frac{1}{165888}|6912c_4-288c_1{c}_3+256c_2^2-1184c_1^2{c}_2+239c_1^4|\hfill \\ & & =\frac{1}{165888}{D}_1+\frac{1}{165888}{D}_2,\hfill \end{array}$$

where $D_1=\left|6912(c_4-c_1{c}_3-\frac{1}{2}{c}_2^2+\frac{3}{4}{c}_1^2{c}_2-\frac{1}{8}{c}_1^4)\right|$

$$\begin{array}{cc}\hfill D_2& =|6912c_4-288c_1{c}_3+256c_2^2-1184c_1^2{c}_2+239c_1^4|.\end{array}$$

(20)

Using Lemma 7 with $\gamma =-\frac{1}{2}$, we get $D_1\le 6912.$ Rearranging (20), we get

$$\begin{array}{cc}\hfill D_2& =|6912c_4-288c_1(c_3+2c_1^2{c}_2+c_1^4)+256c_2^2-608c_1^2{c}_2+527c_1^4|.\end{array}$$

Using Lemma 1, Lemma 6 and the triangle inequality, we receive

$$\begin{array}{ccc}\hfill D_2& & \le 6912(1-|c_1{|}^2-|c_2{|}^2)+288|c_1|+256|c_2{|}^2+608|c_1{|}^2|c_2|+527|c_1{|}^4\hfill \\ & & =6912+288|c_1|-6912|c_1{|}^2+527|c_1{|}^4+608|c_1{|}^2|c_2|-6656|c_2{|}^2={{\rm Y}}_1(c,d)\hfill \end{array}$$

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

Consider

$$\left\{\begin{array}{c}\frac{\partial {{\rm Y}}_1}{\partial c}=288-13824c+2108c^3+1216cd=0,\hfill \\ \frac{\partial {{\rm Y}}_1}{\partial d}=608c^2-13312d=0\hfill \end{array}\right.$$

Applying Newton’s methods, we have

$\left\{\begin{array}{c}{c}_1=2.5173\hfill \\ d_1=0.2894,\hfill \end{array}\right.$$\left\{\begin{array}{c}{c}_2=-2.5173,\hfill \\ d_2=0.2942,\hfill \end{array}\right.$$\left\{\begin{array}{c}{c}_3=0.0208,\hfill \\ d_3=0.\hfill \end{array}\right.$ Thus, in $(0,1)\times (0,1-c^2),$ there is no critical point.

(1)For $c=0,$

$$\begin{array}{c}\hfill {{\rm Y}}_1(0,d)=6912-6912d^2\le 6912.\end{array}$$

(2)For $d=0,$

$$\begin{array}{c}\hfill {{\rm Y}}_1(c,0)=6912+288c-6912c^2+527c^4={\gamma }_1\left(c\right)\le {\gamma }_1\left(0.0208\right)=6915.\end{array}$$

(3)For $d=1-c^2,$

$$\begin{array}{c}\hfill {{\rm Y}}_1(c,1-c^2)=256+288c+7008c^2-6737c^4={\gamma }_2\left(d\right)\le {\gamma }_2\left(0.7313\right)=2287.6.\end{array}$$

Therefore, we have

$$\begin{array}{c}\hfill |{\gamma }_4|\le \frac{13827}{165888}=0.083351.\end{array}$$

Rearranging (19), we obtain

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

where $D_3=\left|86400(c_5-c_1{c}_4-c_2{c}_3+\frac{3}{4}{c}_1{c}_2^2+\frac{3}{4}{c}_1^2{c}_3-\frac{1}{2}{c}_1^3{c}_2+\frac{1}{16}{c}_1^5)\right|$ and

$$\begin{array}{cc}\hfill D_4& =|86400c_5+14400c_2{c}_3-19800c_1^2{c}_3-24800c_1{c}_2^2+38200c_1^3{c}_2+32285c_1^5|.\end{array}$$

(21)

Using Lemma 8 with $\gamma =-\frac{1}{2}$, we get $D_3\le 86400.$ Rearranging (21), we get

$$\begin{array}{cc}\hfill D_4& =|86400c_5+14400c_2(c_3-2c_1{c}_2+c_1^3)-19800c_1^2(c_3-c_1{c}_2-c_1^3)+4000c_1{c}_2^2+4000c_1^3{c}_2+12485c_1^5|.\end{array}$$

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

$$\begin{array}{ccc}\hfill D_4& & \le 86400|c_5|+14400|c_2\left|\right|c_3-2c_1{c}_2+c_1^3|+19800|c_1{|}^2|c_3-c_1{c}_2-c_1^3|+4000|c_1\left|\right|c_2{|}^2+4000|c_1{|}^3|c_2|+12485|c_1{|}^5\hfill \\ & & \le 86400(1-|c_1{|}^2-|c_2{|}^2-\frac{|c_3{|}^2}{1+|c_1|})+14400|c_2|+19800|c_1{|}^2+4000|c_1\left|\right|c_2{|}^2+4000|c_1{|}^3|c_2|+12485|c_1{|}^5\hfill \\ & & =86400-666000|c_1{|}^2+12485|c_1{|}^5+(14400+4000|c_1{|}^3\left)\right|c_2|+(-86400+4000|c_1\left|\right)|c_2{|}^2-86400\frac{|c_3{|}^2}{1+|c_1|}\hfill \\ & & \le 86400-666000|c_1{|}^2+12485|c_1{|}^5+(14400+4000|c_1{|}^3\left)\right|c_2|+(-86400+4000|c_1\left|\right)|c_2{|}^2={{\rm Y}}_2(c,d)\hfill \end{array}$$

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

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

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

(1)For $c=0,$

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

(2)For $d=0,$

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

(3)For $d=1-c^2,$

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

Therefore, we receive

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

The proof of Theorem 10 is completed. ☐

4. Conclusion

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 coefficients of $|b_6|,|b_7|,|b_8|,$ and the third-order determinants of $H_{3,2},H_{3,3}$ of 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 $H_{3,2},H_{3,3}$ for other subclasses of univalent functions.