Hankel Determinant Containing Logarithmic Coefficients for Bounded Turning Functions Connected to a Three-Leaf-Shaped Domain

Abstract

The purpose of this study was to obtain the sharp Hankel determinant $H_{2,1}\left(F_f/2\right)$ and $H_{2,2}\left(F_f/2\right)$ with a logarithmic coefficient as entry for the class ${\mathcal{BT}}_{3\mathcal{L}}$ of bounded turning functions connected with a three-leaf-shaped domain. In this study, we developed a novel method to prove the bound sharpness. Although the calculations are much easier using numerical analysis, all the proofs of our results can be checked with a basic knowledge of calculus.

1. Introduction and Definitions

Let $\mathbb{D}=\left\{z\in \mathbb{C}:\left|z\right|<1\right\}$ be a subset of the complex numbers set $\mathbb{C}$, and $\mathcal{S}$ be the class of univalent functions in $\mathbb{D}$ with the normalization form of

$$f\left(z\right)=z+\sum _{l=2}^{\infty }{a}_l{z}^l.$$

(1)

Suppose that $\mathcal{P}$ represent the class of all functions p that are holomorphic in $\mathbb{D}$ with $\Re \left(p\left(z\right)\right)>0$ and have series representation in the form of

$$p\left(z\right)=1+\sum _{n=1}^{\infty }{c}_n{z}^n,z\in \mathbb{D}.$$

(2)

Recently, Aleman and Constantin [1] describes an interaction between univalent function theory and fluid dynamics. The logarithmic coefficients ${\beta }_n$ of $f\in \mathcal{S}$ are given by

$$F_f\left(z\right):=log\left(\frac{f\left(z\right)}{z}\right)=2\sum _{n=1}^{\infty }{\beta }_n{z}^n,z\in \mathbb{D}.$$

(3)

These coefficients significantly contribute, in many estimations, to the theory of univalent functions. In 1985, de Branges [2] obtained for $n\ge 1,$

$$\sum _{l=1}^{n}l\left(n-l+1\right){\left|{\beta }_n\right|}^2\le \sum _{l=1}^n\frac{n-l+1}{l}.$$

(4)

This inequality provides the famous Bieberbach–Robertson–Milin conjectures about the Taylor coefficients of f belonging to $\mathcal{S}$ in its most general form. In 2005, Kayumov [3] solved Brennan’s conjecture for conformal mappings by considering the logarithmic coefficients. We list a few studies that have produced important work on logarithmic coefficients [4,5,6,7].

For given parameters $q,n\in \mathbb{N}=\left\{1,2,\dots \right\}$, the Hankel determinant $H_{q,n}\left(f\right)$ was defined by Pommerenke [8,9] for a function $f\in \mathcal{S}$ of the form (1) as

$$H_{q,n}\left(f\right)=\left|\begin{array}{cccc}{a}_n\hfill & a_{n+1}\hfill & \dots \hfill & a_{n+q-1}\hfill \\ a_{n+1}\hfill & a_{n+2}\hfill & \dots \hfill & a_{n+q}\hfill \\ ⋮\hfill & ⋮\hfill & \dots \hfill & ⋮\hfill \\ a_{n+q-1}\hfill & a_{n+q}\hfill & \dots \hfill & a_{n+2q-2}\hfill \end{array}\right|.$$

(5)

Later, many results regarding the upper bounds of the Hankel determinant for different sub-collections of univalent functions were found [10,11,12,13,14,15,16,17].

According to the definition, for $f\in \mathcal{S}$, its logarithmic coefficients can be easily calculated by

$$\begin{array}{ccc}\hfill {\gamma }_1& =& \frac{1}{2}{a}_2,\hfill \end{array}$$

(6)

$$\begin{array}{ccc}\hfill {\gamma }_2& =& \frac{1}{2}\left(a_3-\frac{1}{2}{a}_2^2\right),\hfill \end{array}$$

(7)

$$\begin{array}{ccc}\hfill {\gamma }_3& =& \frac{1}{2}\left(a_4-a_2{a}_3+\frac{1}{3}{a}_2^3\right),\hfill \end{array}$$

(8)

$$\begin{array}{ccc}\hfill {\gamma }_4& =& \frac{1}{2}\left(a_5-a_2{a}_4+a_2^2{a}_3-\frac{1}{2}{a}_3^2-\frac{1}{4}{a}_2^4\right).\hfill \end{array}$$

(9)

Recently, Kowalczyk and Lecko [18,19] studied the Hankel determinant $H_{q,n}\left(F_f/2\right)$, whose elements are logarithmic coefficients of f, that is,

$$H_{q,n}\left(F_f/2\right)=\left|\begin{array}{cccc}{\gamma }_n\hfill & {\gamma }_{n+1}\hfill & \dots \hfill & {\gamma }_{n+q-1}\hfill \\ {\gamma }_{n+1}\hfill & {\gamma }_{n+2}\hfill & \dots \hfill & {\gamma }_{n+q}\hfill \\ ⋮\hfill & ⋮\hfill & \dots \hfill & ⋮\hfill \\ {\gamma }_{n+q-1}\hfill & {\gamma }_{n+q}\hfill & \dots \hfill & {\gamma }_{n+2q-2}\hfill \end{array}\right|.$$

(10)

We observe that $H_{2,1}\left(F_f/2\right)={\gamma }_1{\gamma }_3-{\gamma }_2^2$ corresponds to the well-known functional $H_{2,1}\left(f\right)=a_3-a_2^2$ over the class $\mathcal{S}$ or its subclasses. Similar results can also be found in [20].

In [21], Gandhi introduced a family of starlike functions connected with a three-leaf-type function defined by

$${\mathcal{S}}_{3\mathcal{L}}^{*}:=\left\{f\in \mathcal{S}:\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\prec 1+\frac{4}{5}z+\frac{1}{5}{z}^4,z\in \mathbb{D}\right\}.$$

(11)

The radius problems for this class were characterized. Later, some other interesting properties of this function were discussed [22,23].

The subclass of bounded turning function is a class of univalent function satisfying

$$\Re f^{\prime }\left(z\right)>0,z\in \mathbb{D}.$$

(12)

This condition implies that

$$\left|arg{f}^{\prime }\left(z\right)\right|<\frac{1}{2}\pi ,$$

(13)

where arg denotes the angle of rotation. The bounded turning function class is of special importance because it is not a part of starlike functions. It belongs to a wider subclass of univalent functions, namely close-to-convex functions, which fulfil the condition

$$\Re \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}>0,z\in \mathbb{D},$$

(14)

with $g\left(z\right)\in {\mathcal{S}}^{*}$. Taking $g\left(z\right)=z$, it reduces to the class of bounded turning functions. Using the three-leaf function, we introduce the subclass ${\mathcal{BT}}_{3\mathcal{L}}$, defined by

$${\mathcal{BT}}_{3\mathcal{L}}^{*}:=\left\{f\in \mathcal{S}:f^{\prime }\left(z\right)\prec 1+\frac{4}{5}z+\frac{1}{5}{z}^4,z\in \mathbb{D}\right\}.$$

(15)

In the present study, our main aim is to find the sharp bounds of $H_{2,1}\left(F_f/2\right)$ and $H_{2,2}\left(F_f/2\right)$ for the class ${\mathcal{BT}}_{3\mathcal{L}}.$

2. Main Results

Lemma 1

([24]). Let $p\in \mathcal{P}$. Then, for $x,\zeta ,\rho \in \overline{\mathbb{D}}$,

$$\begin{array}{ccc}\hfill 2c_2& =& c_1^2+x\left(4-c_1^2\right),\hfill \end{array}$$

(16)

$$\begin{array}{ccc}\hfill 4c_3& =& c_1^3+2\left(4-c_1^2\right)c_{1}x-c_1\left(4-c_1^2\right)x^2+2\left(4-c_1^2\right)\left(1-{\left|x\right|}^2\right)\zeta ,\hfill \end{array}$$

(17)

$$\begin{array}{ccc}\hfill 8c_4& =& c_1^4+\left(4-c_1^2\right)x\left[c_1^2\left(x^2-3x+3\right)+4x\right]-4\left(4-c_1^2\right)\left(1-{\left|x\right|}^2\right)\hfill \\ & & \left[c\left(x-1\right)\zeta +\overline{x}{\zeta }^2-\left(1-{\left|\zeta \right|}^2\right)\rho \right].\hfill \end{array}$$

(18)

Theorem 1.

If the function $f\in {\mathcal{BT}}_{3\mathcal{L}}$, then

$$\left|H_{2,1}\left(F_f/2\right)\right|=\left|{\gamma }_1{\gamma }_3-{\gamma }_2^2\right|\le \frac{4}{225}.$$

(19)

The result is the best possible and can be achieved by

$$f_1\left(z\right)={\int }_0^z\left(1+\frac{4}{5}\left(t^2\right)+\frac{1}{5}\left(t^8\right)\right)dt=z+\frac{4}{15}{z}^3+\frac{1}{45}{z}^9+\cdots .$$

(20)

Proof. 

From the definition of the class ${\mathcal{BT}}_{3\mathcal{L}}$ along with the subordination principal, there is a Schwarz function $\omega \left(z\right)$ such that

$$f^{\prime }\left(z\right)=1+\frac{4}{5}\left(\omega \left(z\right)\right)+\frac{1}{5}\left(\omega \left(z\right)\right):=\mathsf{\Theta }\left(z\right),z\in \mathbb{D}.$$

(21)

Assuming that $p\in \mathcal{P},$ by writing p in terms of Schwarz function w, we have

$$p\left(z\right)=\frac{1+\omega \left(z\right)}{1-\omega \left(z\right)}=1+c_{1}z+c_2{z}^2+c_3{z}^3+\cdots .$$

(22)

This is equivalent to

$$\omega \left(z\right)=\frac{p\left(z\right)-1}{p\left(z\right)+1}=\frac{c_{1}z+c_2{z}^2+c_3{z}^3+c_4{z}^4+\cdots }{2+c_{1}z+c_2{z}^2+c_3{z}^3+c_4{z}^4+\cdots }.$$

(23)

Using (1), we can easily obtain

$$f^{\prime }\left(z\right)=1+2a_{2}z+3a_3{z}^2+4a_4{z}^3+5a_5{z}^4+\cdots .$$

(24)

From the series expansion of $,$ we have

$$\begin{array}{ccc}\hfill \mathsf{\Theta }\left(z\right)& =& 1+\frac{2}{5}{c}_{1}z+\left(-\frac{1}{5}{c}_1^2+\frac{2}{5}{c}_2\right)z^2+\left(-\frac{2}{5}{c}_1{c}_2+\frac{1}{10}{c}_1^3+\frac{2}{5}{c}_3\right)z^3\hfill \\ & & +\left(-\frac{1}{5}{c}_2^2-\frac{3}{80}{c}_1^4+\frac{2}{5}{c}_4-\frac{2}{5}{c}_1{c}_3+\frac{3}{10}{c}_1^2{c}_2\right)z^4+\cdots .\hfill \end{array}$$

(25)

By comparing (24) and (25) it follows that

$$\begin{array}{ccc}\hfill a_2& =& \frac{1}{5}{c}_1,\hfill \\ \hfill a_3& =& \frac{1}{3}\left(-\frac{1}{5}{c}_1^2+\frac{2}{5}{c}_2\right),\hfill \\ \hfill a_4& =& \frac{1}{4}\left(-\frac{2}{5}{c}_1{c}_2+\frac{1}{10}{c}_1^3+\frac{2}{5}{c}_3\right),\hfill \\ \hfill a_5& =& \frac{1}{5}\left(-\frac{1}{5}{c}_2^2-\frac{3}{80}{c}_1^4+\frac{2}{5}{c}_4-\frac{2}{5}{c}_1{c}_3+\frac{3}{10}{c}_1^2{c}_2\right).\hfill \end{array}$$

(26)

Putting (26) in (6)–(9), we obtain

$$\begin{array}{ccc}\hfill {\gamma }_1& =& \frac{1}{10}{c}_1,\hfill \end{array}$$

(27)

$$\begin{array}{ccc}\hfill {\gamma }_2& =& -\frac{13}{300}{c}_1^2+\frac{1}{15}{c}_2,\hfill \end{array}$$

(28)

$$\begin{array}{ccc}\hfill {\gamma }_3& =& -\frac{19}{300}{c}_1{c}_2+\frac{41}{2000}{c}_1^3+\frac{1}{20}{c}_3,\hfill \end{array}$$

(29)

$$\begin{array}{ccc}\hfill {\gamma }_4& =& -\frac{11}{450}{c}_2^2-\frac{16101}{180000}{c}_1^4+\frac{1}{25}{c}_4-\frac{1}{20}{c}_1{c}_3+\frac{53}{1125}{c}_1^2{c}_2.\hfill \end{array}$$

(30)

From (27)–(29), we have

$$H_{2,1}\left(F_f/2\right)={\gamma }_1{\gamma }_3-{\gamma }_2^2=\frac{31}{180000}{c}_1^4-\frac{1}{1800}{c}_1^2{c}_2+\frac{1}{200}{c}_1{c}_3-\frac{1}{225}{c}_2^2.$$

(31)

Using Lemma 1 and $c_1=c$, we obtain

$$H_{2,1}\left(F_f/2\right)=\frac{1}{30000}{c}^4-\frac{1}{7200}\left(4-c^2\right)\left(32+c^2\right)x^2+\frac{1}{400}c\left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\zeta ,$$

(32)

Let $\left|x\right|=\xi$. By noting that $\left|\zeta \right|\le 1$ and $\xi \le 1$, it follows that

$$\begin{array}{ccc}\hfill \left|H_{2,1}\left(F_f/2\right)\right|& \le & \frac{1}{30000}{c}^4+\frac{1}{400}c\left(4-c^2\right)+\frac{1}{7200}(4-c^2)(c^2-18c+32){\xi }^2\hfill \\ & \le & \frac{1}{30000}{c}^4+\frac{1}{800}{c}^2\left(4-c^2\right)+\frac{1}{900}{\left(4-c^2\right)}^2:=\varpi \left(c\right).\hfill \end{array}$$

As $\varpi \left(c\right)$ attains its maximum $\frac{4}{225}$ at $c=0$, the assertion of (19) follows. □

Theorem 2.

If the function $f\in {\mathcal{BT}}_{3\mathcal{L}}$, then

$$\left|H_{2,2}\left(F_f/2\right)\right|=\left|{\gamma }_2{\gamma }_4-{\gamma }_3^2\right|\le \frac{1}{100}.$$

(33)

The result is sharp, with the extremal function given by

$$f_2\left(z\right)={\int }_0^z\left(1+\frac{4}{5}\left(t^3\right)+\frac{1}{5}\left(t^{12}\right)\right)dt=z+\frac{1}{5}{z}^4+\frac{1}{65}{z}^{13}+\cdots .$$

(34)

Proof. 

By virtue of (28)–(30) and $c_1=c,$ we show that

$$\begin{array}{cc}\hfill H_{2,2}\left(F_f/2\right)& =\frac{1}{M_0}\left(A_1{c}^3{c}_3+A_2{c}^6+A_3{c}^4{c}_2+A_4{c}_2^3+A_{5}c{c}_2{c}_3\right.\\ & \left.+A_6{c}_2{c}_4+A_7{c}^2{c}_4+A_8{c}_3^2+A_9{c}^2{c}_2^2\right),\end{array}$$

(35)

where $A_1$ = 12,600, $A_2=-3761$, $A_3=-4080$, $A_4=-$176,000, $A_5=$ 32,400, $A_6=$ 288,000, $A_7=-$187,200, $A_8=-$270,000, $A_9=$ 20,400, and $M_0=1.08\times {10}^8$.

Applying Lemma 1 yields to

$$H_{2,2}\left(F_f/2\right)=\frac{1}{M_0}\left({\nu }_1\left(c,x\right)+{\nu }_2\left(c,x\right)\zeta +{\nu }_3\left(c,x\right){\zeta }^2+\psi \left(c,x,\zeta \right)\rho \right),$$

(36)

where $\rho ,x,\zeta \in \overline{\mathbb{D}},$ and

$$\begin{array}{ccc}\hfill {\nu }_1\left(c,x\right)& =& -1326c^6+\left(4-c^2\right)\left[\left(4-c^2\right)\left(-5000c^2{x}^3+1125c^2{x}^4-16000x^3\right.\right.\hfill \\ & & \left.\left.+6600c^2{x}^2\right)+4260c^{4}x-21600c^2{x}^2+6300c^4{x}^2-5400c^4{x}^3\right],\hfill \\ \hfill {\nu }_2\left(c,x\right)& =& 100\left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\left[\left(4-c^2\right)\left(180cx-45c{x}^2\right)-18c^3+216c^{3}x\right],\hfill \\ \hfill {\nu }_3\left(c,x\right)& =& 100\left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\left[\left(4-c^2\right)\left(-675-45{\left|x\right|}^2\right)+216c^2\overline{x}\right],\hfill \\ \hfill \psi \left(c,x,\zeta \right)& =& 100\left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\left(1-{\left|\zeta \right|}^2\right)\left[720x\left(4-c^2\right)-216c^2\right].\hfill \end{array}$$

Let $\left|x\right|=x$ and $\left|\zeta \right|=y.$ By taking $\left|\rho \right|\le 1,$ we obtain

$$\begin{array}{ccc}\hfill \left|H_{2,2}\left(F_f/2\right)\right|& \le & \frac{1}{M_0}\left(\left|{\nu }_1\left(c,x\right)\right|+\left|{\nu }_2\left(c,x\right)\right|y+\left|{\nu }_3\left(c,x\right)\right|y^2+\left|\psi \left(c,x,\zeta \right)\right|\right).\hfill \\ & \le & \frac{1}{M_0}\left(\gamma \left(c,x,y\right)\right),\hfill \end{array}$$

(37)

where

$$\gamma \left(c,x,y\right)=h_1\left(c,x\right)+h_2\left(c,x\right)y+h_3\left(c,x\right)y^2+h_4\left(c,x\right)\left(1-y^2\right),$$

(38)

with

$$\begin{array}{ccc}\hfill h_1\left(c,x\right)& =& 1326c^6+\left(4-c^2\right)\left[\left(4-c^2\right)\left(5000c^2{x}^3+1125c^2{x}^4+16000x^3\right.\right.\hfill \\ & & \left.\left.+6600c^2{x}^2\right)+4260c^{4}x+21600c^2{x}^2+6300c^4{x}^2+5400c^4{x}^3\right],\hfill \\ \hfill h_2\left(c,x\right)& =& 100\left(4-c^2\right)\left(1-x^2\right)\left[\left(4-c^2\right)\left(180cx+45c{x}^2\right)+18c^3+216c^{3}x\right],\hfill \\ \hfill h_3\left(c,x\right)& =& 100\left(4-c^2\right)\left(1-x^2\right)\left[\left(4-c^2\right)\left(675+45x^2\right)+216c^{2}x\right],\hfill \\ \hfill h_4\left(c,x\right)& =& 100\left(4-c^2\right)\left(1-x^2\right)\left[720x\left(4-c^2\right)+216c^2\right].\hfill \end{array}$$

We can easily observe that $\gamma \left(0,0,1\right)=$ 1,080,000. Let $\mathsf{\Gamma }=\left[0,2\right]\times \left[0,1\right]\times \left[0,1\right]$ and $m_0=1.08\times {10}^6$. Thus, we know

$$max\gamma (c,x,y)\ge m_0,(c,x,y)\in \mathsf{\Gamma }.$$

(39)

We now prove that $\gamma \left(c,x,y\right)$ will attain its maximum value on the face $y=1$ of $\mathsf{\Gamma }$.

For $x=1$, $\gamma (c,x,y)$ is independent of y; it is seen that its maximum value is 396,165.779 achieved at $c\approx 1.294240$. This means that $\gamma (c,x,y)$ can not achieve its maxima on $x=1$. Thus, we only need to discuss $x\in [0,1)$.

First, we prove that the global maximum value of $\gamma (c,x,y)$ can not be obtained on $y=0$. $p_2(c,x)=\gamma (c,x,0)=h_1(c,x)+h_4(c,x)$. For $\frac{3}{5}<x\le 1$, we can easily check that

$$p_2\left(c,x\right)\le h_1(c,1)+(4-c^2)\left[46080(4-c^2)+13824c^2\right]:={\tau }_1\left(c\right).$$

(40)

Because ${\tau }_1\left(c\right)$ attains its maximum value 993,280.0 at $c=0$, we know that $p_2(c,x)<m_0$ for $x\in \left(\frac{3}{5},1\right]$. For $0\le x\le \frac{3}{5}$, we observe that

$$p_2\left(c,x\right)\le h_1\left(c,\frac{3}{5}\right)+(4-c^2)\left[43200(4-c^2)+21600c^2\right]:={\tau }_2\left(c\right).$$

(41)

Since ${\tau }_2\left(c\right)$ attains its maximum value 746,496.0 at $c=0$, we conclude that $\gamma (c,x,y)$ cannot achieve its global maximum value on $y=0$.

Second, we aim to show that all the critical points of $\gamma (c,x,y)$ with $y\in (0,1)$ must satisfy the inequality $\gamma (c,x,y)<$ 1,080,000. Let $\left(c,x,y\right)\in \left[0,2\right]\times \left[0,1\right)\times \left(0,1\right).$ Assuming that $\frac{\partial \gamma }{\partial y}=0$, we deduce that

$$y=\frac{c\left[5\left(x+4\right)x\left(4-c^2\right)+2c^2\left(12x+1\right)\right]}{10(x-1)\left[\left(15-x\right)\left(4-c^2\right)-\frac{24}{5}{c}^2\right]}=y_0.$$

(42)

Notably, $y_0\in \left(0,1\right)$ is possible only if

$$2c^3\left(12x+1\right)+5cx\left(x+4\right)\left(4-c^2\right)+10(1-x)\left(15-x\right)\left(4-c^2\right)<48(1-x)c^2$$

(43)

and

$$c^2>\frac{20\left(15-x\right)}{99-5x}.$$

(44)

For the existence of the critical points, we must obtain the solutions that satisfy both inequalities (43) and (44). Define $m\left(x\right)=\frac{20\left(15-x\right)}{99-5x}.$ As $m^{\prime }\left(x\right)<0$ in $\left(0,1\right),$ it follows that $m\left(x\right)$ is decreasing over $\left(0,1\right).$ Hence, $c^2>\frac{140}{47}.$ Let

$$\begin{array}{ccc}\hfill \mathsf{\Xi }(c,x)& & =2c^3\left(12x+1\right)+5cx\left(x+4\right)\left(4-c^2\right)\hfill \\ & & +10(1-x)\left(15-x\right)\left(4-c^2\right)-48(1-x)c^2\hfill \end{array}$$

with $(c,x)\in \left(\sqrt{\frac{140}{47}},2\right]\times [\frac{1}{2},1)$. Using $\frac{1}{2}\le x<1$ and $c^2>\frac{470}{14}$, it is easily seen that

$$\begin{array}{ccc}\hfill \mathsf{\Xi }(c,x)& \ge & 14c^3+\frac{45}{4}c\left(4-c^2\right)+140(1-x)\left(4-c^2\right)-48(1-x)c^2\hfill \\ & =& 14c^3+\frac{45}{4}c\left(4-c^2\right)+(560-188c^2)(1-x)\hfill \\ & \ge & 14c^3+\frac{45}{4}c\left(4-c^2\right)+280-94c^2\hfill \\ & =& 280+45c-94c^2+\frac{11}{4}{c}^3:=\varrho \left(c\right).\hfill \end{array}$$

Because $min\left\{\varrho \left(c\right)\right\}=16>0$ for $c\in \left(\sqrt{\frac{140}{47}},2\right]$, we know that (43) does not hold true in this case for all values of $x\in \left[\frac{1}{2},1\right).$ Thus, there is no critical point of $\gamma \left(c,x,y\right)$ in $\left[0,2\right]\times \left[\frac{1}{2},1\right)\times \left(0,1\right).$

If there is a critical point $(c_0,x_0,y_0)$ of $\gamma$ satisfying $y_0\in (0,1)$, it means that $x_0<\frac{1}{2}$. Additionally, using (44), it follows that $c_0^2>\frac{580}{193}$. For $(c,x)\in \left(\frac{580}{193},2\right]\times \left[0,\frac{1}{2}\right)$, we easily see that $h_1\left(c,x\right)\le h_1\left(c,\frac{1}{2}\right):={\varphi }_1\left(c\right)$ and

$$h_j\left(c,x\right)\le \frac{4}{3}{h}_j\left(c,\frac{1}{2}\right):={\varphi }_j\left(c\right),j=2,3,4.$$

(45)

Hence, we obtain that

$$\gamma (c,x,y)\le {\varphi }_1\left(c\right)+{\varphi }_2\left(c\right)y+{\varphi }_3\left(c\right)y^2+{\varphi }_4\left(c\right)\left(1-y^2\right):=\mathsf{{\rm Y}}(c,y).$$

(46)

Since $\frac{{\partial }^2\mathsf{{\rm Y}}}{\partial y^2}=(4-c^2)($130,500−72,225$c^2)\le 0$ for $c\in (\sqrt{\frac{580}{193}},2)$, it leads to

$$\frac{\partial \mathsf{{\rm Y}}}{\partial y}\ge \frac{\partial \mathsf{{\rm Y}}}{\partial y}{|}_{y=1}=(4-c^2)(130500+40500c-72225c^2+275c^3)\ge 0.$$

(47)

Then, it follows that

$$\mathsf{{\rm Y}}(c,y)\le \mathsf{{\rm Y}}(c,1)={\varphi }_1\left(c\right)+{\varphi }_2\left(c\right)+{\varphi }_3\left(c\right):=\mathsf{\Theta }\left(c\right).$$

(48)

As $\mathsf{\Theta }\left(c\right)$ attains its maximum value 283,312.9 at $c\approx 1.733546$, for all the critical points $(\tilde{c},\tilde{x},\tilde{y})$ with $\tilde{y}\in (0,1)$, we have $\gamma (\tilde{c},\tilde{x},\tilde{y})<m_0$.

The global maximum value of $\gamma$ in $\mathsf{\Gamma }$ certainly exists and can only be obtained at the critical points of $\mathsf{\Gamma }$ or the boundary $\partial \mathsf{\Gamma }$. Additionally, as we proved that for $y\in [0,1)$, all the critical points or boundary points of $\gamma$ are strictly less than $m_0$, we conclude that ${max}_{(c,x,y)\in \mathsf{\Gamma }}\gamma (c,x,y)$ can only be attained in $y=1$. Therefore, we obtain

$$\gamma (c,x,y)\le {max}_{(c,x,y)\in \mathsf{\Gamma }}\gamma (c,x,y)={max}_{(c,x)\in [0,2]\times [0,1]}\gamma (c,x,1).$$

(49)

Suppose

$$p_3(c,x):=\gamma (c,x,1).$$

(50)

Now, we prove that

$$p_3(c,x)\le m_0,(c,x)\in [0,2]\times [0,1].$$

(51)

A basic calculation produces

$$\begin{array}{ccc}\hfill p_3(c,x)& =& h_1(c,x)+h_2(c,x)+h_3(c,x)\hfill \\ & =& 1326c^6+\left(4-c^2\right)\left\{\left(4-c^2\right)\left[67500+18000cx+\right.\right.\hfill \\ & & 300\left(22c^2+15c-210\right)x^2+1000\left(5c^2-18c+16\right)x^3+\hfill \\ & & \left.1125\left(c^2-4c-4\right)x^4\right]+1800c^3+\hfill \\ & & 20\left(213c^2+1080c+1080\right)c^{2}x+900\left(7c^2-2c+24\right)c^2{x}^2\hfill \\ & & \left.+5400\left(c^2-4c-4\right)c^2{x}^4\right\}.\hfill \end{array}$$

As $c^2-4c-4\le 0$ for $c\in [0,2]$, we have

$$\begin{array}{ccc}\hfill p_3(c,x)& \le & 1326c^6+\left(4-c^2\right)\left\{\left(4-c^2\right)\left[67500+18000cx\right.\right.\hfill \\ & & \left.+300\left(22c^2+15c-210\right)x^2+1000\left(5c^2-18c+16\right)x^3\right]\hfill \\ & & +1800c^3+20\left(213c^2+1080c+1080\right)c^{2}x\hfill \\ & & \left.+900\left(7c^2-2c+24\right)c^2{x}^2\right\}:=\mathsf{\Lambda }(c,x).\hfill \end{array}$$

Assuming that $0\le c<1$, because $7c^2-2c+24\ge 0$ and $x^2\le x\le 1$, we obtain

$$\begin{array}{ccc}& & 1800c^3+20\left(213c^2+1080c+1080\right)c^{2}x+900\left(7c^2-2c+24\right)c^2{x}^2\hfill \\ \hfill \le & & 1800c^3+20\left(213c^2+1080c+1080\right)c^2+900\left(7c^2-2c+24\right)c^2\hfill \\ \hfill =& & 10560c^4+21600c^3+43200c^2.\hfill \end{array}$$

As $5c^2-18c+16\ge 0$ for $c\in [0,1]$ and $x^3\le x^2$, we have

$$\begin{array}{ccc}& & 67500+18000cx+300\left(22c^2+15c-210\right)x^2+1000\left(5c^2-18c+16\right)x^3\hfill \\ \hfill \le & & 67500+18000cx+300\left(22c^2+15c-210\right)x^2+1000\left(5c^2-18c+16\right)x^2\hfill \\ \hfill =& & 67500+18000cx+100\left(116c^2-135c-470\right)x^2\hfill \\ \hfill \le & & 100\left(675+180cx-470x^2\right).\hfill \end{array}$$

Define

$$W(c,x):=675+180cx-470x^2.$$

(52)

In virtue of

$$x_0=-\frac{180c}{2\times (-470)}=\frac{9}{47}c\in [0,1],$$

(53)

we see

$$W(c,x)\le \frac{4\times (-470)\times 675-32400c^2}{4\times (-470)}=675+\frac{810}{47}{c}^2\le 675+18c^2.$$

(54)

Then, we deduce that

$$\begin{array}{ccc}\hfill \mathsf{\Lambda }(c,x)& \le & 1326c^6+100{\left(4-c^2\right)}^2\left(675+18c^2\right)\hfill \\ & & +\left(4-c^2\right)\left(10560c^4+21600c^3+43200c^2\right):=\iota \left(c\right).\hfill \end{array}$$

A basic calculation shows that

$$\iota \left(c\right)=-7434c^6-21600c^5+52140c^4+86400c^3-338400c^2+1080000.$$

(55)

By observing that 52,140$c^4+$ 86,400$c^3-$ 338,400$c^2\le 0$ for $c\in [0,1]$, $\iota \left(c\right)$ achieved its maximum value 1,080,000 at $c=0$. Therefore, we obtain $p_3(c,x)\le \mathsf{\Lambda }(c,x)\le \iota \left(c\right)\le m_0$ with $(c,x)\in \left[0,1\right)\times [0,1].$

For $c\in [1,2)$, we easily observe $22c^2+15c-210\le 0$. Thus, we have

$$\begin{array}{ccc}\hfill \mathsf{\Lambda }(c,x)& \le & 1326c^6+\left(4-c^2\right)\left\{\left(4-c^2\right)\left[67500+18000cx\right.\right.\hfill \\ & & \left.+1000\left(5c^2-18c+16\right)x^3\right]\hfill \\ & & +1800c^3+20\left(213c^2+1080c+1080\right)c^{2}x\hfill \\ & & \left.+900\left(7c^2-2c+24\right)c^2{x}^2\right\}:=T(c,x).\hfill \end{array}$$

By noting that $5c^2-18c+16\le 3$ and $7c^2-2c+24\le 48$ for $c\in [1,2]$, we see

$$\begin{array}{ccc}\hfill T(c,x)& \le & 1326c^6+\left(4-c^2\right)\left\{\left(4-c^2\right)\left[67500+18000cx+3000x^3\right]\right.\hfill \\ & & \left.+1800c^3+20\left(213c^2+1080c+1080\right)c^{2}x+43200c^2{x}^2\right\}\hfill \\ & \le & 1326c^6+\left(4-c^2\right)\left[\left(4-c^2\right)\left(67500+18000c+3000\right)\right.\hfill \\ & & \left.+1800c^3+20\left(213c^2+1080c+1080\right)c^2+43200c^2\right]\hfill \\ & =& -2934c^6-5400c^5+22740c^4-50400c^3-304800c^2\hfill \\ & & +288000c+1128000:=\eta \left(c\right).\hfill \end{array}$$

From $1\le c<2$, we obtain that

$$\begin{array}{ccc}\hfill {\eta }^{\prime }\left(c\right)& =& -17604c^5-27000c^4+68220c^3-151200c^2-609600c+288000\hfill \\ & \le & -17604-27000+136440c^2-151200c^2-609600+288000\hfill \\ & =& -366204-14760c^2\le 0.\hfill \end{array}$$

So, $\eta \left(c\right)$ attains its maximum value 1,075,206 at $c=1$. Thus, we get $p_3(c,x)\le \mathsf{\Lambda }(c,x)\le \eta \left(c\right)\le \eta \left(1\right)<m_0$ for $(c,x)\in [1,2)\times [0,1)$.

From the above discussion, we conclude that

$$\gamma \left(c,x,y\right)\le m_0,\left[0,2\right]\times \left[0,1\right]\times \left[0,1\right].$$

(56)

Using (37), we have

$$\left|H_{2,2}\left(F_f/2\right)\right|\le \frac{m_0}{M_0}\le \frac{1}{100}=0.01.$$

(57)

Hence, the proof of Theorem 2 is complete. □