Sharp Bounds on Coefficients Functionals of Hankel Determinants for Ozaki Close-to-Convex Functions

Abstract

We determine the sharp bounds on the second Hankel determinants of logarithmic coefficients and the third Hankel determinants for Ozaki close-to-convex functions.

1 Introduction

Let ${\mathcal {A}}$ be the class of functions analytic in the unit disk $\mathbb {D}:=\{z\in \mathbb {C}:\, |z|<1\}$ of the form

$$\begin{aligned} f(z)=z+\sum _{k=2}^{\infty }a_{k}z^{k}. \end{aligned}$$

(1.1)

We denote ${\mathcal {S}}$ by the subclass of ${\mathcal {A}}$ consisting of univalent functions.

In 1941, Ozaki [33] introduced the classes of functions ${\mathcal {F}}$ and ${\mathcal {G}}$, and defined as

$$\begin{aligned} {\mathcal {F}}=\bigg \{f\in {\mathcal {A}}:\quad \Re \bigg (1+\frac{zf''(z)}{f'(z)}\bigg )>-\frac{1}{2}, \qquad z\in \mathbb {D}\bigg \}, \end{aligned}$$

(1.2)

and

$$\begin{aligned} {\mathcal {G}}=\bigg \{f\in {\mathcal {A}}:\quad \Re \bigg (1+\frac{zf''(z)}{f'(z)}\bigg )<\frac{3}{2}, \qquad z\in \mathbb {D}\bigg \}, \end{aligned}$$

(1.3)

respectively. The author proved the inclusion relation ${\mathcal {G}}\subset {\mathcal {S}}$. We also note that ${\mathcal {F}}$ follows from the original definition of Kaplan [15], and that Umezawa [39] subsequently proved that functions in ${\mathcal {F}}$ are not necessarily starlike, but are convex in one direction. The functions in the classes ${\mathcal {F}}$ and ${\mathcal {G}}$ are known as Ozaki close-to-convex functions, which have nice geometric properties and are used to understand the shape and behavior of various subclasses of univalent functions.

Given $q,n\in \mathbb {N}$, the Hankel determinant $H_{q,n}(f)$ of $f\in {\mathcal {A}}$ of the form (1.1) is defined by

$$\begin{aligned} H_{q,n}(f)= \left| \begin{array}{cccc} a_{n} &{} a_{n+1} &{} \cdots &{} a_{n+q-1}\\ a_{n+1} &{} a_{n+2} &{} \cdots &{} a_{n+q} \\ \vdots &{} \vdots &{} \vdots &{} \vdots \\ a_{n+q-1} &{} a_{n+q} &{} \cdots &{} a_{n+2(q-1)}\\ \end{array}\right| . \end{aligned}$$

In recent years, many papers have been devoted to finding bounds of determinants, whose elements are coefficients of functions in ${\mathcal {A}}$, or its subclasses. The sharp bounds on the second Hankel determinants $|H_{2,1}(f)|$ and $|H_{2,2}(f)|$ were obtained by [6, 9, 13, 14, 26, 32], for various classes of analytic functions. We refer to [4, 7, 8, 34, 36, 37, 43] for discussions on the upper bounds of the third Hankel determinants $|H_{3,1}(f)|$ for various classes of univalent functions. However, these results are far from sharpness. In a recent paper, Kwon et al. [23] found such a formula of expressing $c_{4}$ for Carathéodory functions, the sharp results of the third Hankel determinants are found for some classes of univalent functions (see e.g., [5, 19,20,21, 24, 25, 35, 40,41,42]).

Note that for $f\in {\mathcal {A}}$, $a_{1}=1$, $H_{3,1}(f)$ reduces to

$$\begin{aligned} H_{3,1}(f)=a_{3}\big (a_{2}a_{4}-a_{3}^{2}\big )-a_{4}\big (a_{4}-a_{2}a_{3}\big )+a_{5}\big (a_{3}-a_{2}^{2}\big ). \end{aligned}$$

(1.4)

Recently, Kowalczyk et al. [21] proved the sharp inequality $|H_{3,1}(f)|\le 1/16$ for $f\in {\mathcal {F}}$. In this paper, we prove that $|H_{3,1}(f)|\le 19/2160$ for $f\in {\mathcal {G}}$, and so giving the sharp bound for $|H_{3,1}(f)|$ for a significant subclass of ${\mathcal {G}}$.

For $f\in {\mathcal {S}}$, let

$$\begin{aligned} F_{f}(z):=\log \bigg (\frac{f(z)}{z}\bigg )=2\sum _{n=1}^{\infty }\gamma _{n}(f)z^{n}, \qquad z\in \mathbb {D}. \end{aligned}$$

(1.5)

The numbers $\gamma _{n}:=\gamma _{n}(f)$ are called logarithmic coefficients of f. It is well known that the logarithmic coefficients play a crucial role in Milin conjecture [29]. Sharp logarithmic coefficient estimates for the class ${\mathcal {S}}$ are already known for $n=1$ and $n=2$, given by $|\gamma _{1}|\le 1$ and $|\gamma _{2}|\le 1/2 + 1/e^{2}$, respectively. However, the bound of $\gamma _{n}$ for $n\ge 3$, is still an open problem. We refer to [1, 2, 10, 12, 22, 38] for discussions on the logarithmic coefficient for various classes of univalent functions.

Given $q,n\in \mathbb {N}$, the Hankel determinant $H_{q,n}(F_{f}/2)$ which entries are logarithmic coefficients of $f\in {\mathcal {A}}$ of the form (1.1) is defined by

$$\begin{aligned} H_{q,n}(F_{f}/2)= \left| \begin{array}{cccc} \gamma _{n} &{} \gamma _{n+1} &{} \cdots &{} \gamma _{n+q-1}\\ \gamma _{n+1} &{} \gamma _{n+2} &{} \cdots &{} \gamma _{n+q} \\ \vdots &{} \vdots &{} \vdots &{} \vdots \\ \gamma _{n+q-1} &{} \gamma _{n+q} &{} \cdots &{} \gamma _{n+2(q-1)}\\ \end{array}\right| . \end{aligned}$$

A study of Hankel determinant with entries as logarithmic coefficients was initiated by Kowalczyk and Lecko [16]. Due to the great importance of logarithmic coefficients, the proposed topic seems reasonable and interesting.

By differentiating (1.5) and using (1.1) we get

$$\begin{aligned} \begin{aligned}&\gamma _{1}=\frac{1}{2}a_{2}, \qquad \qquad \qquad \qquad \qquad \gamma _{2}=\frac{1}{2}\left( a_{3}-\frac{1}{2}a_{2}^{2}\right) , \\&\gamma _{3}=\frac{1}{2}\left( a_{4}-a_{2}a_{3}+\frac{1}{3}a_{2}^{3}\right) , \qquad \gamma _{4}=\frac{1}{2}\left( a_{5}-a_{2}a_{4}+a_{2}^{2}a_{3}-\frac{1}{2}a_{3}^{2}-\frac{1}{4}a_{2}^{4}\right) . \end{aligned}\nonumber \\ \end{aligned}$$

(1.6)

Therefore,

$$\begin{aligned} \begin{aligned} H_{2,2}(F_{f}/2)=\gamma _{2}\gamma _{4}-\gamma _{3}^{2}=&\frac{1}{4}\bigg [\bigg (a_{3}-\frac{1}{2}a_{2}^{2}\bigg )a_{5}+\bigg (a_{2}a_{3}-a_{4}-\frac{1}{6}a_{2}^{3}\bigg )a_{4}\\&\quad +\bigg (\frac{1}{4}a_{2}^{2}a_{3}-\frac{1}{2}a_{3}^{2}-\frac{1}{12}a_{2}^{4}\bigg )a_{3}+\frac{1}{72}a_{2}^{6}\bigg ]. \end{aligned}\nonumber \\ \end{aligned}$$

(1.7)

By observing that when $f\in {\mathcal {S}}$, for $f_{\theta }(z):=e^{-i\theta }f(e^{i\theta }z)$ with $\theta \in \mathbb {R}$, we have

$$\begin{aligned} H_{2,2}(F_{f_{\theta }}/2)=e^{6i\theta }H_{2,2}(F_{f}/2). \end{aligned}$$

Thus, the coefficients functional $|H_{2,2}(F_{f}/2)|$ is a rotationally invariant.

Recently, Kowalczyk and Lecko [16] obtained the sharp bounds of $|H_{2,1}(F_{f}/2)|$ for the classes of starlike and convex functions. Moreover, the sharp bounds of $|H_{2,1}(F_{f}/2)|$ for the classes of starlike and convex functions of order $\alpha (0\le \alpha <1)$ were found in [17]. Very recently, Kowalczyk and Lecko [18] examined the sharp bounds of $|H_{2,1}(F_{f}/2)|$ for the classes of strongly starlike and strongly convex functions. Allu et al. [3] (see also [31]) examined the sharp bounds of $|H_{2,1}(F_{f}/2)|$ for the classes of starlike and convex functions with respect to symmetric points. Moreover, the sharp bounds of $|H_{2,1}(F_{f}/2)|$ for the class of starlike functions of order $\alpha \,(0\le \alpha <1)$ with respect to symmetric points were investigated in [30]. Very recently, Eker et al. [11] obtained the sharp bounds of $|H_{2,1}(F_{f}/2)|$ and $|H_{2,1}(F_{f^{-1}}/2)|$ for the classes of strongly Ozaki close-to-convex functions and inverse functions, respectively.

The problem of finding sharp bounds of $|H_{2,2}(F_{f}/2)|$ is technically much more difficult. The purpose of this paper is to prove the sharp bounds of $|H_{2,2}(F_{f}/2)|$ for the classes ${\mathcal {F}}$ and ${\mathcal {G}}$, respectively.

Denote ${\mathcal {P}}$ by the class of Carathéodory functions p normalized by

$$\begin{aligned} p(z)=1+\sum _{n=1}^{\infty }c_{n}z^{n}, \qquad z\in \mathbb {D}, \end{aligned}$$

(1.8)

and satisfy the condition $\Re \big (p(z)\big )>0$.

The following results are known for functions belonging to the class ${\mathcal {P}}$, which will be required in the proof of our main results.

Lemma 1.1

(See [23, 27, 28]) If $p\in {\mathcal {P}}$ and is given by (1.8) with $c_{1}\ge 0$, then

$$\begin{aligned}{} & {} c_{1}=2\zeta _{1}, \end{aligned}$$

(1.9)

$$\begin{aligned}{} & {} c_{2}=2\zeta _{1}^{2}+2(1-\zeta _{1}^{2})\zeta _{2}, \end{aligned}$$

(1.10)

$$\begin{aligned}{} & {} c_{3}=2\zeta _{1}^{3}+4(1-\zeta _{1}^{2})\zeta _{1}\zeta _{2} -2(1-\zeta _{1}^{2})\zeta _{1}\zeta _{2}^{2}+2(1-\zeta _{1}^{2})(1-|\zeta _{2}|^{2})\zeta _{3}\qquad \qquad \quad \end{aligned}$$

(1.11)

and

$$\begin{aligned} \begin{aligned} c_{4}=\,&2\zeta _{1}^{4}+2(1-\zeta _{1}^{2})(3\zeta _{1}^{2}+\zeta _{2} -3\zeta _{1}^{2}\zeta _{2}+\zeta _{1}^{2}\zeta _{2}^{2})\zeta _{2}\\&+2(1-\zeta _{1}^{2})(1-|\zeta _{2}|^{2})\big (2\zeta _{1}-2\zeta _{1} \zeta _{2}-\overline{\zeta _{2}}\zeta _{3}\big )\zeta _{3}\\&+2(1-\zeta _{1}^{2})(1-|\zeta _{2}|^{2})(1-|\zeta _{3}|^{2})\zeta _{4}, \end{aligned} \end{aligned}$$

(1.12)

for some $\zeta _{1}\in [0,1]$ and $\zeta _{2}, \zeta _{3}, \zeta _{4}\in \overline{\mathbb {D}}:=\{z\in \mathbb {C}:\, |z|\le 1\}$.

2 Main Results

In this section, we will prove the sharp bounds on the second Hankel determinants of logarithmic coefficients for the classes ${\mathcal {F}}$ and ${\mathcal {G}}$, and the sharp bounds on the third Hankel determinants for the class ${\mathcal {G}}$, respectively.

We begin by deriving the sharp bounds of $|H_{2,2}(F_{f}/2)|$ for the class ${\mathcal {F}}$.

Theorem 2.1

If $f\in {\mathcal {F}}$ be of the form (1.1), then

$$\begin{aligned} \big |H_{2,2}(F_{f}/2)\big |\le \frac{1}{32}. \end{aligned}$$

(2.1)

The result is sharp for

$$\begin{aligned} 1+\frac{zf''(z)}{f'(z)}=\frac{1+2z^{2}}{1-z^{2}}, \qquad z\in \mathbb {D}, \end{aligned}$$

(2.2)

that is, $f(z)=z+z^{3}/2+3z^{5}/8+\cdots$.

Proof

For the function $f\in {\mathcal {F}}$ given by (1.1), there exists an analytic function $p\in {\mathcal {P}}$ in the unit disk $\mathbb {D}$ with $p(0)=1$ and $\Re \big (p(z)\big )>0$ such that

$$\begin{aligned} 1+\frac{zf''(z)}{f'(z)}=\frac{3}{2}p(z)-\frac{1}{2}, \qquad z\in \mathbb {D}. \end{aligned}$$

(2.3)

By elementary calculations, we have

$$\begin{aligned} \begin{aligned}&a_{2}=\frac{3}{4}c_{1}, \qquad a_{3}=\frac{1}{8}\big (2c_{2}+3c^{2}_{1}\big ), \qquad a_{4}=\frac{1}{64}\big (8c_{3}+18c_{1}c_{2}+9c^{3}_{1}\big ), \\&a_{5}=\frac{3}{640}\big (16c_{4}+32c_{1}c_{3}+36c^{2}_{1}c_{2}+12c^{2}_{2}+9c^{4}_{1}\big ). \end{aligned} \end{aligned}$$

(2.4)

Thus, (1.7) and (2.4) give

$$\begin{aligned} \begin{aligned} 655360\cdot H_{2,2}(F_{f}/2)=\,&384\big (8c_{2}+3c^{2}_{1}\big )c_{4}-1536c_{1}c_{2}c_{3}-2560c^{2}_{3}\\&+864c^{3}_{1}c_{3}-864c^{2}_{1}c^{2}_{2}+1024c^{3}_{2}-27c^{6}_{1}. \end{aligned}\nonumber \\ \end{aligned}$$

(2.5)

Since the class ${\mathcal {F}}$ and $\big |H_{2,2}(F_{f}/2)\big |$ are rotationally invariant, we may assume that $c_{1}\in [0,2]$. Thus, in view of (1.9) we assume that $\zeta _{1}\in [0,1]$. Using (2.5) and (1.9)-(1.12), we obtain

$$\begin{aligned} \begin{aligned}&655360\cdot H_{2,2}(F_{f}/2)\\&\quad = \big [5440\zeta _{1}^{6}+23552(1-\zeta _{1}^{2})\zeta _{1}^{4}\zeta _{2} +512(1-\zeta _{1}^{2})(7-54\zeta _{1}^{2})\zeta _{1}^{2}\zeta _{2}^{2}\\&\qquad +1024(1-\zeta _{1}^{2})(20-12\zeta _{1}^{2}+13\zeta _{1}^{4})\zeta _{2}^{3} +2048(1-\zeta _{1}^{2})^{2}\zeta _{1}^{2}\zeta _{2}^{4}\big ]\\&\qquad +512(1-\zeta _{1}^{2})(1-|\zeta _{2}|^{2})\big [47\zeta _{1}^{3}-28(2+\zeta _{1}^{2})\zeta _{1}\zeta _{2} -8(1-\zeta _{1}^{2})\zeta _{1}\zeta _{2}^{2}\big ]\zeta _{3}\\&\qquad -1024(1-\zeta _{1}^{2})(1-|\zeta _{2}|^{2})\big [(1-\zeta _{1}^{2})(10+2|\zeta _{2}|^{2})+21\zeta _{1}^{2}\overline{\zeta _{2}}\big ]\zeta _{3}^{2}\\&\qquad +3072(1-\zeta _{1}^{2})(1-|\zeta _{2}|^{2})(1-|\zeta _{3}|^{2})\big [7\zeta _{1}^{2}+4(1-\zeta _{1}^{2})\zeta _{2}\big ]\zeta _{4}, \end{aligned} \end{aligned}$$

for some $\zeta _{1}\in [0,1]$ and $\zeta _{2}, \zeta _{3}, \zeta _{4}\in \overline{\mathbb {D}}$. Since $|\zeta _{4}|\le 1$, we have

$$\begin{aligned} \begin{aligned}&10240\cdot \big |H_{2,2}(F_{f}/2)\big |\\&\quad \le \big |85\zeta _{1}^{6}+368(1-\zeta _{1}^{2})\zeta _{1}^{4}\zeta _{2} +8(1-\zeta _{1}^{2})(7-54\zeta _{1}^{2})\zeta _{1}^{2}\zeta _{2}^{2}\\&\qquad +16(1-\zeta _{1}^{2})(20-12\zeta _{1}^{2}+13\zeta _{1}^{4})\zeta _{2}^{3} +32(1-\zeta _{1}^{2})^{2}\zeta _{1}^{2}\zeta _{2}^{4}\big |\\&\qquad +8(1-\zeta _{1}^{2})(1-|\zeta _{2}|^{2})\big |47\zeta _{1}^{3}-28(2+\zeta _{1}^{2})\zeta _{1}\zeta _{2} -8(1-\zeta _{1}^{2})\zeta _{1}\zeta _{2}^{2}\big | \cdot |\zeta _{3}|\\&\qquad +16(1-\zeta _{1}^{2})(1-|\zeta _{2}|^{2})\big [\big |(1-\zeta _{1}^{2})(10+2|\zeta _{2}|^{2})+21\zeta _{1}^{2}\overline{\zeta _{2}}\big |\\&\qquad -3\big |7\zeta _{1}^{2}+4(1-\zeta _{1}^{2})\zeta _{2}\big |\big ]\cdot |\zeta _{3}|^{2}\\&\qquad +48(1-\zeta _{1}^{2})(1-|\zeta _{2}|^{2})\big |7\zeta _{1}^{2}+4(1-\zeta _{1}^{2})\zeta _{2}\big |. \end{aligned} \end{aligned}$$

$\textbf{A}$. Suppose that

$$\begin{aligned} \big |(1-\zeta _{1}^{2})(10+2|\zeta _{2}|^{2})+21\zeta _{1}^{2}\overline{\zeta _{2}}\big | -3\big |7\zeta _{1}^{2}+4(1-\zeta _{1}^{2})\zeta _{2}\big |\ge 0. \end{aligned}$$

Then

$$\begin{aligned} 10240\cdot \big |H_{2,2}(F_{f}/2)\big |\le u(\zeta _{1},|\zeta _{2}|), \end{aligned}$$

where $u:\ \mathbb {R}^{2}\rightarrow \mathbb {R}$ is defined by

$$\begin{aligned} \begin{aligned} u(x,y):=\,&85x^{6}+8(1-x^{2})(20-20x^{2}+47x^{3})\\&+8(1-x^{2})(56x+42x^{2}+28x^{3}+46x^{4})y\\&+8(1-x^{2})(-16+8x+16x^{2}-55x^{3}+x^{2}\cdot |7-54x^{2}|)y^{2}\\&+8(1-x^{2})(40-56x-66x^{2}-28x^{3}+26x^{4})y^{3}\\&+32(1-x^{2})^{2}(-1-2x+x^{2})y^{4}. \end{aligned} \end{aligned}$$

We show that $u(x,y)\le 320$ for $(x,y)\in [0, 1]\times [0, 1]$.

$\textbf{I}$. On the vertices of $[0, 1]\times [0, 1]$, we have

$$\begin{aligned} u(0,0)=160, \quad u(0,1)=320, \quad u(1,0)=u(1,1)=85. \end{aligned}$$

$\textbf{II}$. On the sides of $[0, 1]\times [0, 1]$, we get

$$\begin{aligned} \begin{aligned} u(0,y)&=32(5-4y^{2}+10y^{3}-y^{4})\le u(0,1)=320, \quad y\in (0, 1),\\ u(x,0)&=85x^{6}+8(1-x^{2})(20-20x^{2}+47x^{3})\le u(0,0)=160, \quad x\in (0, 1),\\ u(1,y)&=85, \quad y\in (0, 1),\\ u(x,1)&=85x^{6}+8(1-x^{2})(40-20x^{2}+68x^{4}+|7x^{2}-54x^{4}|)\\&\le u(0,1)=320, \ x\in (0, 1). \end{aligned} \end{aligned}$$

$\textbf{III}$. It remains to consider the set $(0, 1)\times (0, 1)$.

If $7-54x^{2}\ge 0$. Then all the real solutions ($x\ne 0,\pm 1$) of the system of equations

$$\begin{aligned} \begin{aligned} \frac{\partial u}{\partial x}=\,&(-640x+1128x^{2}+640x^{3}-1880x^{4}+510x^{5})\\&+8(56+84x-84x^{2}+16x^{3}-140x^{4}-276x^{5})y\\&+8(8+78x-189x^{2}-308x^{3}+275x^{4}+324x^{5})y^{2}\\&+8(-56-212x+84x^{2}+368x^{3}+140x^{4}-156x^{5})y^{3}\\&+64(-1+3x+6x^{2}-6x^{3}-5x^{4}+3x^{5})y^{4}=0, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \frac{\partial u}{\partial y}=\,&8(1-x^{2})\big [(56x+42x^{2}+28x^{3}+46x^{4})\\&+2(-16+8x+23x^{2}-55x^{3}-54x^{4})y\\&+3(40-56x-66x^{2}-28x^{3}+26x^{4})y^{2}\\&+16(-1-2x+2x^{2}+2x^{3}-x^{4})y^{3}\big ]=0, \end{aligned} \end{aligned}$$

by a numerical computation are the following

$$\begin{aligned} \begin{aligned}&{\left\{ \begin{array}{ll} x_{1}\approx -0.751907,\\ y_{1}\approx -26.2555,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{2}\approx 1.38642,\\ y_{2}\approx 16.9888,\\ \end{array}\right. } \qquad \ {\left\{ \begin{array}{ll} x_{3}\approx -1.62503,\\ y_{3}\approx 5.7912,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{4}\approx -0.808447,\\ y_{4}\approx -0.246107,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{5}\approx -0.764429,\\ y_{5}\approx 0.288228,\\ \end{array}\right. } \ \ \ {\left\{ \begin{array}{ll} x_{6}\approx -0.0197963,\\ y_{6}\approx -0.0303034.\\ \end{array}\right. } \end{aligned} \end{aligned}$$

Thus the function u has no critical point in $(0, \sqrt{7}/\sqrt{54}]\times (0, 1)$.

If $7-54x^{2}<0$. Then all the real solutions ($x\ne 0,\pm 1$) of the system of equations

$$\begin{aligned} \frac{\partial u}{\partial x}=\,{} & {} (-640x+1128x^{2}+640x^{3}-1880x^{4}+510x^{5})\\{} & {} +8(56+84x-84x^{2}+16x^{3}-140x^{4}-276x^{5})y\\{} & {} +8(8+50x-189x^{2}+180x^{3}+275x^{4}-324x^{5})y^{2}\\{} & {} +8(-56-212x+84x^{2}+368x^{3}+140x^{4}-156x^{5})y^{3}\\{} & {} +64(-1+3x+6x^{2}-6x^{3}-5x^{4}+3x^{5})y^{4}=0, \end{aligned}$$

and

$$\begin{aligned} \frac{\partial u}{\partial y}=\,{} & {} 8(1-x^{2})\big [(56x+42x^{2}+28x^{3}+46x^{4})\\{} & {} +2(-16+8x+9x^{2}-55x^{3}+54x^{4})y\\{} & {} +3(40-56x-66x^{2}-28x^{3}+26x^{4})y^{2}\\{} & {} +16(-1-2x+2x^{2}+2x^{3}-x^{4})y^{3}\big ]=0, \end{aligned}$$

by a numerical computation are the following

$$\begin{aligned} \begin{aligned}&{\left\{ \begin{array}{ll} x_{1}\approx -0.756068,\\ y_{1}\approx -26.0806,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{2}\approx 2.57483,\\ y_{2}\approx 13.8819,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{3}\approx -1.40321,\\ y_{3}\approx 9.79431,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{4}\approx 0.690931,\\ y_{4}\approx 0.735877,\\ \end{array}\right. } \qquad \ \ {\left\{ \begin{array}{ll} x_{5}\approx 0.452718,\\ y_{5}\approx 0.871419,\\ \end{array}\right. } \ \ \ \ \ {\left\{ \begin{array}{ll} x_{6}\approx -0.799775,\\ y_{6}\approx -0.478217,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{7}\approx -0.773773,\\ y_{7}\approx 0.164858,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{8}\approx -0.0197346,\\ y_{8}\approx -0.0302118.\\ \end{array}\right. } \end{aligned} \end{aligned}$$

Thus $(x_{4},y_{4})$ and $(x_{5},y_{5})$ are the critical points of u in $(\sqrt{7}/\sqrt{54}, 1)\times (0, 1)$ with

$$\begin{aligned} u(x_{4},y_{4})\approx 271.2564<320, \qquad u(x_{5},y_{5})\approx 256.9165<320. \end{aligned}$$

$\textbf{B}$. Suppose that

$$\begin{aligned} \big |(1-\zeta _{1}^{2})(10+2|\zeta _{2}|^{2})+21\zeta _{1}^{2}\overline{\zeta _{2}}\big | -3\big |7\zeta _{1}^{2}+4(1-\zeta _{1}^{2})\zeta _{2}\big |< 0. \end{aligned}$$

Then

$$\begin{aligned} 10240\cdot \big |H_{2,2}(F_{f}/2)\big |\le v(\zeta _{1},|\zeta _{2}|), \end{aligned}$$

where $v:\ \mathbb {R}^{2}\rightarrow \mathbb {R}$ is defined by

$$\begin{aligned} v(x,y):=\,{} & {} 85x^{6}+8(1-x^{2})(42x^{2}+47x^{3})\\{} & {} +8(1-x^{2})(24+56x-24x^{2}+28x^{3}+46x^{4})y\\{} & {} +8(1-x^{2})(8x-42x^{2}-55x^{3}+x^{2}\cdot |7-54x^{2}|)y^{2}\\{} & {} +8(1-x^{2})(16-56x-28x^{3}+26x^{4})y^{3}\\{} & {} +32(1-x^{2})^{2}(-2x+x^{2})y^{4}. \end{aligned}$$

We show now that $v(x,y)\le 320$ for $(x,y)\in [0, 1]\times [0, 1]$.

$\textbf{I}$. On the vertices of $[0, 1]\times [0, 1]$, we have

$$\begin{aligned} v(0,0)=0, \quad v(0,1)=320, \quad v(1,0)=v(1,1)=85. \end{aligned}$$

$\textbf{II}$. On the sides of $[0, 1]\times [0, 1]$, we get

$$\begin{aligned} v(0,y)= & {} 64(3y+2y^{3})\le v(0, 1)=320, \quad y\in (0, 1),\\ v(x,0)= & {} 85x^{6}+8(1-x^{2})(42x^{2}+47x^{3})\le v(x_{0}, 0)\approx 169.17, \\ x_{0}\approx & {} 0.788876,\ x\in (0, 1),\\ v(1,y)= & {} 85, \quad y\in (0, 1),\\ v(x,1)= & {} 85x^{6}+8(1-x^{2})\big (40-20x^{2}+68x^{4}+|7x^{2}-54x^{4}|\big )\\\le & {} v(0,1)=320, \ x\in (0, 1). \end{aligned}$$

$\textbf{III}$. It remains to consider the set $(0, 1)\times (0, 1)$.

If $7-54x^{2}\ge 0$. Then all the real solutions ($x\ne 0,\pm 1$) of the system of equations

$$\begin{aligned} \frac{\partial v}{\partial x}=\,{} & {} (672x+1128x^{2}-1344x^{3}-1880x^{4}+510x^{5})\\{} & {} +8(56-96x-84x^{2}+280x^{3}-140x^{4}-276x^{5})y\\{} & {} +8(8-70x-189x^{2}-76x^{3}+275x^{4}+324x^{5})y^{2}\\{} & {} +8(-56-32x+84x^{2}+104x^{3}+140x^{4}-156x^{5})y^{3}\\{} & {} +8(-8+8x+48x^{2}-32x^{3}-40x^{4}+24x^{5})y^{4}=0, \end{aligned}$$

and

$$\begin{aligned} \frac{\partial v}{\partial y}=\,{} & {} 8(1-x^{2})\big [(24+56x-24x^{2}+28x^{3}+46x^{4})\\{} & {} +2(8x-35x^{2}-55x^{3}-54x^{4})y\\{} & {} +3(16-56x-28x^{3}+26x^{4})y^{2}\\{} & {} +4(-8x+4x^{2}+8x^{3}-4x^{4})y^{3}\big ]=0, \end{aligned}$$

by a numerical computation are the following

$$\begin{aligned}{} & {} {\left\{ \begin{array}{ll} x_{1}\approx -2.3661,\\ y_{1}\approx -0.367626,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{2}\approx -0.93242,\\ y_{2}\approx 0.473722,\\ \end{array}\right. } \quad \ \ {\left\{ \begin{array}{ll} x_{3}\approx -0.7798,\\ y_{3}\approx -0.329245,\\ \end{array}\right. }\\{} & {} {\left\{ \begin{array}{ll} x_{4}\approx -0.62236,\\ y_{4}\approx 0.0596674,\\ \end{array}\right. } \qquad \ {\left\{ \begin{array}{ll} x_{5}\approx -0.325247,\\ y_{5}\approx 1.55834,\\ \end{array}\right. } \ \ \ {\left\{ \begin{array}{ll} x_{6}\approx -0.311599,\\ y_{6}\approx -1.54511,\\ \end{array}\right. }\\{} & {} {\left\{ \begin{array}{ll} x_{7}\approx 0.164715,\\ y_{7}\approx -0.77519,\\ \end{array}\right. } \qquad \ \ {\left\{ \begin{array}{ll} x_{8}\approx 0.273348,\\ y_{8}\approx 0.745703,\\ \end{array}\right. } \quad \ \ {\left\{ \begin{array}{ll} x_{9}\approx 1.46314,\\ y_{9}\approx 5.14397.\\ \end{array}\right. } \end{aligned}$$

Thus $(x_{8},y_{8})$ is the unique critical point of v in $(0, \sqrt{7}/\sqrt{54}]\times (0, 1)$ with

$$\begin{aligned} v(x_{8},y_{8})\approx 231.2083<320. \end{aligned}$$

If $7-54x^{2}<0$. Then all the real solutions ($x\ne 0, \pm 1$) of the system of equations

$$\begin{aligned} \frac{\partial v}{\partial x}=\,{} & {} (672x+1128x^{2}-1344x^{3}-1880x^{4}+510x^{5})\\{} & {} +8(56-96x-84x^{2}+280x^{3}-140x^{4}-276x^{5})y\\{} & {} +8(8-98x-189x^{2}+412x^{3}+275x^{4}-324x^{5})y^{2}\\{} & {} +8(-56-32x+84x^{2}+104x^{3}+140x^{4}-156x^{5})y^{3}\\{} & {} +8(-8+8x+48x^{2}-32x^{3}-40x^{4}+24x^{5})y^{4}=0, \end{aligned}$$

and

$$\begin{aligned} \frac{\partial v}{\partial y}=\,{} & {} 8(1-x^{2})\big [(24+56x-24x^{2}+28x^{3}+46x^{4})\\{} & {} +2(8x-49x^{2}-55x^{3}+54x^{4})y\\{} & {} +3(16-56x-28x^{3}+26x^{4})y^{2}\\{} & {} +4(-8x+4x^{2}+8x^{3}-4x^{4})y^{3}\big ]=0, \end{aligned}$$

by a numerical computation are the following

$$\begin{aligned}{} & {} {\left\{ \begin{array}{ll} x_{1}\approx -2.3661,\\ y_{1}\approx -0.367626,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{2}\approx -0.93242,\\ y_{2}\approx 0.473722,\\ \end{array}\right. } \quad {\left\{ \begin{array}{ll} x_{3}\approx -0.7798,\\ y_{3}\approx -0.329245,\\ \end{array}\right. }\\{} & {} {\left\{ \begin{array}{ll} x_{4}\approx -0.62236,\\ y_{4}\approx 0.0596674,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{5}\approx -0.325247,\\ y_{5}\approx 1.55834,\\ \end{array}\right. } \ \ \ {\left\{ \begin{array}{ll} x_{6}\approx -0.311599,\\ y_{6}\approx -1.54511,\\ \end{array}\right. }\\{} & {} {\left\{ \begin{array}{ll} x_{7}\approx 0.164715,\\ y_{7}\approx -0.77519,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{8}\approx 0.273348,\\ y_{8}\approx 0.745703,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{9}\approx 1.46314,\\ y_{9}\approx 5.14397.\\ \end{array}\right. } \end{aligned}$$

Thus the function v has no critical point in $(\sqrt{7}/\sqrt{54}, 1)\times (0, 1)$.

Summarizing, we see that the bounds obtained in Parts A and B give

$$\begin{aligned} \big |H_{2,2}(F_{f}/2)\big |\le \frac{1}{10240}\cdot 320=\frac{1}{32}. \end{aligned}$$

We finally note that equality in (2.1) holds for the function $f\in {\mathcal {F}}$ defined by (1.1), and satisfying (2.3) with

$$\begin{aligned} p(z):=\frac{1+z^{2}}{1-z^{2}}, \qquad z\in \mathbb {D}, \end{aligned}$$

for which $a_{2}=a_{4}=0$, $a_{3}=1/2$ and $a_{5}=3/8$. This completes the proof of the theorem. $\square$

We next consider the sharp bounds of $|H_{2,2}(F_{f}/2)|$ for the class ${\mathcal {G}}$.

Theorem 2.2

If $f\in {\mathcal {G}}$ be of the form (1.1), then

$$\begin{aligned} \big |H_{2,2}(F_{f}/2)\big |\le \frac{1}{576}. \end{aligned}$$

(2.6)

The result is sharp for

$$\begin{aligned} 1+\frac{zf''(z)}{f'(z)}=\frac{1-2z^{3}}{1-z^{3}}, \qquad z\in \mathbb {D}, \end{aligned}$$

(2.7)

that is, $f(z)=z-z^{4}/12+\cdots$.

Proof

For the function $f\in {\mathcal {G}}$ given by (1.1), there exists an analytic function $p\in {\mathcal {P}}$ in the unit disk $\mathbb {D}$ with $p(0)=1$ and $\Re \big (p(z)\big )>0$ such that

$$\begin{aligned} 1+\frac{zf''(z)}{f'(z)}=\frac{3}{2}-\frac{1}{2}p(z), \qquad z\in \mathbb {D}. \end{aligned}$$

(2.8)

By elementary calculations, we have

$$\begin{aligned} \begin{aligned}&a_{2}=-\frac{1}{4}c_{1}, \qquad a_{3}=\frac{1}{24}\big (-2c_{2}+c_{1}^{2}\big ), \qquad a_{4}=\frac{1}{192}\big (-8c_{3}+6c_{1}c_{2}-c_{1}^{3}\big ), \\&a_{5}=\frac{1}{1920}\big (-48c_{4}+32c_{1}c_{3}-12c_{1}^{2}c_{2}+12c^{2}_{2}+c_{1}^{4}\big ). \end{aligned} \end{aligned}$$

(2.9)

Thus, (1.7) and (2.9) give

$$\begin{aligned} 17694720\cdot H_{2,2}(F_{f}/2)=\,&1152\big (8c_{2}-c_{1}^{2}\big )c_{4}+1536c_{1}c_{2}c_{3}-7680c_{3}^{2}\nonumber \\&+288c_{1}^{3}c_{3}-288c_{1}^{2}c_{2}^{2}-1024c_{2}^{3}-c_{1}^{6}. \end{aligned}$$

(2.10)

Since the class ${\mathcal {G}}$ and $\big |H_{2,2}(F_{f}/2)\big |$ are rotationally invariant, we may assume that $c_{1}\in [0,2]$. Thus, in view of (1.9) we assume that $\zeta _{1}\in [0,1]$. Using (2.10) and (1.9)-(1.12), we obtain

$$\begin{aligned} \begin{aligned} 276480\cdot H_{2,2}(F_{f}/2)=\,&\big [15\zeta _{1}^{6}+144(1-\zeta _{1}^{2})\zeta _{1}^{4}\zeta _{2} +168(1-\zeta _{1}^{2})(1-2\zeta _{1}^{2})\zeta _{1}^{2}\zeta _{2}^{2}\\&+16(1-\zeta _{1}^{2})(28-20\zeta _{1}^{2}+19\zeta _{1}^{4})\zeta _{2}^{3} +96(1-\zeta _{1}^{2})^{2}\zeta _{1}^{2}\zeta _{2}^{4}\big ]\\&+24(1-\zeta _{1}^{2})(1-|\zeta _{2}|^{2})\big [7\zeta _{1}^{3}-12(2+\zeta _{1}^{2})\zeta _{1}\zeta _{2}\\&-8(1-\zeta _{1}^{2})\zeta _{1}\zeta _{2}^{2}\big ]\zeta _{3}-48(1-\zeta _{1}^{2})(1-|\zeta _{2}|^{2})\\&\big [(1-\zeta _{1}^{2})(10+2|\zeta _{2}|^{2})+9\zeta _{1}^{2}\overline{\zeta _{2}}\big ]\zeta _{3}^{2}+144(1-\zeta _{1}^{2})\\&(1-|\zeta _{2}|^{2})(1-|\zeta _{3}|^{2})\big [3\zeta _{1}^{2}+4(1-\zeta _{1}^{2})\zeta _{2}\big ]\zeta _{4}, \end{aligned} \end{aligned}$$

for some $\zeta _{1}\in [0,1]$ and $\zeta _{2}, \zeta _{3}, \zeta _{4}\in \overline{\mathbb {D}}$. Since $|\zeta _{4}|\le 1$, we have

$$\begin{aligned} \begin{aligned}&276480\cdot \big |H_{2,2}(F_{f}/2)\big |\\&\quad \le \big |15\zeta _{1}^{6}+144(1-\zeta _{1}^{2})\zeta _{1}^{4}\zeta _{2} +168(1-\zeta _{1}^{2})(1-2\zeta _{1}^{2})\zeta _{1}^{2}\zeta _{2}^{2}\\&+16(1-\zeta _{1}^{2})(28-20\zeta _{1}^{2}+19\zeta _{1}^{4})\zeta _{2}^{3} +96(1-\zeta _{1}^{2})^{2}\zeta _{1}^{2}\zeta _{2}^{4}\big |\\&+24(1-\zeta _{1}^{2})(1-|\zeta _{2}|^{2})\big |7\zeta _{1}^{3}-12(2+\zeta _{1}^{2})\zeta _{1}\zeta _{2} -8(1-\zeta _{1}^{2})\zeta _{1}\zeta _{2}^{2}\big | \cdot |\zeta _{3}|\\&+48(1-\zeta _{1}^{2})(1-|\zeta _{2}|^{2})\big [\big |(1-\zeta _{1}^{2})(10+2|\zeta _{2}|^{2})+9\zeta _{1}^{2}\overline{\zeta _{2}}\big | -3\big |3\zeta _{1}^{2}\\&+4(1-\zeta _{1}^{2})\zeta _{2}\big |\big ]\cdot |\zeta _{3}|^{2}+144(1-\zeta _{1}^{2})(1-|\zeta _{2}|^{2})\big |3\zeta _{1}^{2}+4(1-\zeta _{1}^{2})\zeta _{2}\big |. \end{aligned} \end{aligned}$$

$\textbf{A}$. Suppose that

$$\begin{aligned} \big |(1-\zeta _{1}^{2})(10+2|\zeta _{2}|^{2})+9\zeta _{1}^{2}\overline{\zeta _{2}}\big | -3\big |3\zeta _{1}^{2}+4(1-\zeta _{1}^{2})\zeta _{2}\big |\ge 0. \end{aligned}$$

Then

$$\begin{aligned} 276480\cdot \big |H_{2,2}(F_{f}/2)\big |\le \varphi (\zeta _{1},|\zeta _{2}|), \end{aligned}$$

where $\varphi :\ \mathbb {R}^{2}\rightarrow \mathbb {R}$ is defined by

$$\begin{aligned} \begin{aligned} \varphi (x,y):=\,&15x^{6}+24(1-x^{2})(20-20x^{2}+7x^{3})\\&+144(1-x^{2})(4x+3x^{2}+2x^{3}+x^{4})y\\&+24(1-x^{2})(-16+8x+16x^{2}-15x^{3}+7x^{2}\cdot |1-2x^{2}|)y^{2}\\&+16(1-x^{2})(28-36x-47x^{2}-18x^{3}+19x^{4})y^{3}\\&+96(1-x^{2})^{2}(-1-2x+x^{2})y^{4}. \end{aligned} \end{aligned}$$

We show that $\varphi (x,y)\le 480$ for $(x,y)\in [0, 1]\times [0, 1]$.

$\textbf{I}$. On the vertices of $[0, 1]\times [0, 1]$, we have

$$\begin{aligned} \varphi (0,0)=480, \quad \varphi (0,1)=448, \quad \varphi (1,0)=\varphi (1,1)=15. \end{aligned}$$

$\textbf{II}$. On the sides of $[0, 1]\times [0, 1]$, we get

$$\begin{aligned} \begin{aligned} \varphi (0,y)&= 32(15-12y^{2}+14y^{3}-3y^{4})\le \varphi (0,0)=480, \quad y\in (0, 1),\\ \varphi (x,0)&= 15x^{6}+24(1-x^{2})(20-20x^{2}+7x^{3})\le \varphi (0,0)=480, \quad x\in (0, 1),\\ \varphi (1,y)&=15, \quad y\in (0, 1),\\ \varphi (x,1)&=15x^{6}+8(1-x^{2})(56-28x^{2}+44x^{4}+21|x^{2}-2x^{4}|)\\&\le \varphi (0,1)=448, \ x\in (0, 1). \end{aligned} \end{aligned}$$

$\textbf{III}$. It remains to consider the set $(0, 1)\times (0, 1)$.

If $1-2x^{2}\ge 0$. Then all the real solutions ($x\ne 0,\pm 1$) of the system of equations

$$\begin{aligned} \begin{aligned} \frac{\partial \varphi }{\partial x}=\,&(-1920x+504x^{2}+1920x^{3}-840x^{4}+90x^{5})\\&+144(4+6x-6x^{2}-8x^{3}-10x^{4}-6x^{5})y\\&+24(8+78x-69x^{2}-148x^{3}+75x^{4}+84x^{5})y^{2}\\&+16(-36-150x+54x^{2}+264x^{3}+90x^{4}-114x^{5})y^{3}\\&+192(-1+3x+6x^{2}-6x^{3}-5x^{4}+3x^{5})y^{4}=0, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \frac{\partial \varphi }{\partial y}=\,&48(1-x^{2})\big [3(4x+3x^{2}+2x^{3}+x^{4})\\&+(-16+8x+23x^{2}-15x^{3}-14x^{4})y\\&+(28-36x-47x^{2}-18x^{3}+19x^{4})y^{2}\\&+8(-1-2x+2x^{2}+2x^{3}-x^{4})y^{3}\big ]=0, \end{aligned} \end{aligned}$$

by a numerical computation are the following

$$\begin{aligned} \begin{aligned}&{\left\{ \begin{array}{ll} x_{1}\approx -55.9185,\\ y_{1}\approx 0.774424,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{2}\approx -31.3008,\\ y_{2}\approx 0.399471,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{3}\approx -0.761847,\\ y_{3}\approx -11.6168,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{4}\approx 1.45524,\\ y_{4}\approx 6.40123,\\ \end{array}\right. } \qquad \ \ {\left\{ \begin{array}{ll} x_{5}\approx -1.62457,\\ y_{5}\approx 2.67252,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{6}\approx 0.116028,\\ y_{6}\approx 0.909375,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{7}\approx -0.94848,\\ y_{7}\approx -0.331374,\\ \end{array}\right. } \quad \ \ {\left\{ \begin{array}{ll} x_{8}\approx -0.894531,\\ y_{8}\approx 0.380566. \end{array}\right. } \end{aligned} \end{aligned}$$

Thus $(x_{6},y_{6})$ is the unique critical point of $\varphi$ in $(0, \sqrt{2}/2]\times (0, 1)$ with

$$\begin{aligned} \varphi (x_{6},y_{6})\approx 383.3639<480. \end{aligned}$$

If $1-2x^{2}<0$. Then all the real solutions ($x\ne 0,\pm 1$) of the system of equations

$$\begin{aligned} \begin{aligned} \frac{\partial \varphi }{\partial x}=\,&(-1920x+504x^{2}+1920x^{3}-840x^{4}+90x^{5})\\&+144(4+6x-6x^{2}-8x^{3}-10x^{4}-6x^{5})y\\&+24(8+50x-69x^{2}+20x^{3}+75x^{4}-84x^{5})y^{2}\\&+16(-36-150x+54x^{2}+264x^{3}+90x^{4}-114x^{5})y^{3}\\&+192(-1+3x+6x^{2}-6x^{3}-5x^{4}+3x^{5})y^{4}=0, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \frac{\partial \varphi }{\partial y}=\,&48(1-x^{2})\big [3(4x+3x^{2}+2x^{3}+x^{4})\\&+(-16+8x+9x^{2}-15x^{3}+14x^{4})y\\&+(28-36x-47x^{2}-18x^{3}+19x^{4})y^{2}\\&+8(-1-2x+2x^{2}+2x^{3}-x^{4})y^{3}\big ]=0, \end{aligned} \end{aligned}$$

by a numerical computation are the following

$$\begin{aligned} \begin{aligned}&{\left\{ \begin{array}{ll} x_{1}\approx -0.766314,\\ y_{1}\approx -11.6419,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{2}\approx 2.70716,\\ y_{2}\approx 7.14005,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{3}\approx -1.36535,\\ y_{3}\approx 4.72423,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{4}\approx -0.957006,\\ y_{4}\approx -0.450247,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{5}\approx 0.10236,\\ y_{5}\approx 0.87329,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{6}\approx -0.899079,\\ y_{6}\approx 0.308825.\\ \end{array}\right. } \end{aligned} \end{aligned}$$

Thus the function $\varphi$ has no critical point in $(\sqrt{2}/2, 1)\times (0, 1)$.

$\textbf{B}$. Suppose that

$$\begin{aligned} \big |(1-\zeta _{1}^{2})(10+2|\zeta _{2}|^{2})+9\zeta _{1}^{2}\overline{\zeta _{2}}\big | -3\big |3\zeta _{1}^{2}+4(1-\zeta _{1}^{2})\zeta _{2}\big |< 0. \end{aligned}$$

Then

$$\begin{aligned} 276480\cdot \big |H_{2,2}(F_{f}/2)\big |\le \psi (\zeta _{1},|\zeta _{2}|), \end{aligned}$$

where $\psi :\ \mathbb {R}^{2}\rightarrow \mathbb {R}$ is defined by

$$\begin{aligned} \begin{aligned} \psi (x,y):=\,&15x^{6}+24(1-x^{2})(18x^{2}+7x^{3})\\&+144(1-x^{2})(4+4x-4x^{2}+2x^{3}+x^{4})y\\&+24(1-x^{2})(8x-18x^{2}-15x^{3}+7x^{2}\cdot |1-2x^{2}|)y^{2}\\&+16(1-x^{2})(-8-36x+16x^{2}-18x^{3}+19x^{4})y^{3}\\&+96(1-x^{2})^{2}(-2x+x^{2})y^{4}. \end{aligned} \end{aligned}$$

We show that $\psi (x,y)\le 448$ for $(x,y)\in [0, 1]\times [0, 1]$.

$\textbf{I}$. On the vertices of $[0, 1]\times [0, 1]$, we have

$$\begin{aligned} \psi (0,0)=0, \quad \psi (0,1)=448, \quad \psi (1,0)=\psi (1,1)=15. \end{aligned}$$

$\textbf{II}$. On the sides of $[0, 1]\times [0, 1]$, we get

$$\begin{aligned} \psi (0,y)= & {} 64(9y-2y^{3})\le \psi (0,1)=448, \quad y\in (0, 1),\\ \psi (x,0)= & {} 15x^{6}+24(1-x^{2})(18x^{2}+7x^{3}) \le \psi (x_{0},0)\approx 140.341,\\ x_{0}\approx & {} 0.733049,\ x\in (0, 1),\\ \psi (1,y)= & {} 15, \quad y\in (0, 1),\\ \psi (x,1)= & {} 15x^{6}\!+\!8(1\!-\!x^{2})(56\!-\!28x^{2}\!+\!44x^{4}\!+\!21|x^{2}-2x^{4}|)\!\le \! 448, \quad \!\!x\!\in \!(0, 1). \end{aligned}$$

$\textbf{III}$. It remains to consider the set $(0, 1)\times (0, 1)$.

If $1-2x^{2}\ge 0$. Then all the real solutions ($x\ne 0,\pm 1$) of the system of equations

$$\begin{aligned} \begin{aligned} \frac{\partial \psi }{\partial x}=\,&(864x+504x^{2}-1728x^{3}-840x^{4}+90x^{5})\\&+144(4-16x-6x^{2}+20x^{3}-10x^{4}-6x^{5})y\\&+24(8-22x-69x^{2}-12x^{3}+75x^{4}+84x^{5})y^{2}\\&+16(-36+48x+54x^{2}+12x^{3}+90x^{4}-114x^{5})y^{3}\\&+192(-1+x+6x^{2}-4x^{3}-5x^{4}+3x^{5})y^{4}=0, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \frac{\partial \psi }{\partial y}=\,&48(1-x^{2})\big [3(4+4x-4x^{2}+2x^{3}+x^{4})\\&+(8x-11x^{2}-15x^{3}-14x^{4})y\\&+(-8-36x+16x^{2}-18x^{3}+19x^{4})y^{2}\\&+8(-2x+x^{2}+2x^{3}-x^{4})y^{3}\big ]=0, \end{aligned} \end{aligned}$$

by a numerical computation are the following

$$\begin{aligned} \begin{aligned}&{\left\{ \begin{array}{ll} x_{1}\approx -55.4838,\\ y_{1}\approx 0.776748,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{2}\approx -29.6139,\\ y_{2}\approx 0.387681,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{3}\approx 0.321635,\\ y_{3}\approx -4.5939,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{4}\approx 1.55353,\\ y_{4}\approx 3.22173,\\ \end{array}\right. } \qquad \ \ {\left\{ \begin{array}{ll} x_{5}\approx -1.73153,\\ y_{5}\approx 3.21781,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{6}\approx -1.90529,\\ y_{6}\approx -0.222475,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{7}\approx -0.761626,\\ y_{7}\approx -0.266647,\\ \end{array}\right. } \quad \ {\left\{ \begin{array}{ll} x_{8}\approx -0.570421,\\ y_{8}\approx 0.104924,\\ \end{array}\right. } \quad \ \ {\left\{ \begin{array}{ll} x_{9}\approx -0.925824,\\ y_{9}\approx 0.541028. \end{array}\right. } \end{aligned} \end{aligned}$$

Thus the function $\psi$ has no critical point in $(0, \sqrt{2}/2]\times (0, 1)$.

If $1-2x^{2}<0$. Then all the real solutions ($x\ne 0,\pm 1$) of the system of equations

$$\begin{aligned} \begin{aligned} \frac{\partial \psi }{\partial x}=\,&(864x+504x^{2}-1728x^{3}-840x^{4}+90x^{5})\\&+144(4-16x-6x^{2}+20x^{3}-10x^{4}-6x^{5})y\\&+24(8-50x-69x^{2}+156x^{3}+75x^{4}-84x^{5})y^{2}\\&+16(-36+48x+54x^{2}+12x^{3}+90x^{4}-114x^{5})y^{3}\\&+192(-1+x+6x^{2}-4x^{3}-5x^{4}+3x^{5})y^{4}=0, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \frac{\partial \psi }{\partial y}=\,&48(1-x^{2})\big [3(4+4x-4x^{2}+2x^{3}+x^{4})\\&+(8x-25x^{2}-15x^{3}+14x^{4})y\\&+(-8-36x+16x^{2}-18x^{3}+19x^{4})y^{2}\\&+8(-2x+x^{2}+2x^{3}-x^{4})y^{3}\big ]=0, \end{aligned} \end{aligned}$$

by a numerical computation are the following

$$\begin{aligned} \begin{aligned}&{\left\{ \begin{array}{ll} x_{1}\approx 2.62485,\\ y_{1}\approx 7.99341,\\ \end{array}\right. } \qquad \qquad {\left\{ \begin{array}{ll} x_{2}\approx -1.41281,\\ y_{2}\approx 5.38615,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{3}\approx 0.337397,\\ y_{3}\approx -4.50435,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{4}\approx -3.00552,\\ y_{4}\approx -0.50500,\\ \end{array}\right. } \qquad \quad \ {\left\{ \begin{array}{ll} x_{5}\approx -0.772503,\\ y_{5}\approx -0.286793,\\ \end{array}\right. } \quad \ {\left\{ \begin{array}{ll} x_{6}\approx -0.56427,\\ y_{6}\approx 0.109389,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{7}\approx -0.980877,\\ y_{7}\approx 0.459454.\\ \end{array}\right. } \end{aligned} \end{aligned}$$

Thus the function $\psi$ has no critical point in $(\sqrt{2}/2, 1)\times (0, 1)$.

Summarizing, we see that the bounds obtained in Parts A and B give

$$\begin{aligned} \big |H_{2,2}(F_{f}/2)\big |\le \frac{1}{276480}\cdot 480=\frac{1}{576}. \end{aligned}$$

We finally note that equality in (2.6) holds for the function $f\in {\mathcal {G}}$ defined by (1.1), and satisfying (2.8) with

$$\begin{aligned} p(z):=\frac{1+z^{3}}{1-z^{3}}, \qquad z\in \mathbb {D}, \end{aligned}$$

for which $a_{2}=a_{3}=a_{5}=0$ and $a_{4}=-1/12$. This completes the proof of the theorem. $\square$

We finally prove the sharp bounds of $|H_{3,1}(f)|$ for functions $f\in {\mathcal {G}}$.

Theorem 2.3

If $f\in {\mathcal {G}}$ be of the form (1.1), then

$$\begin{aligned} \big |H_{3,1}(f)\big |\le \frac{19}{2160}. \end{aligned}$$

(2.11)

The result is sharp for

$$\begin{aligned} 1+\frac{zf''(z)}{f'(z)}=\frac{1-2z^{2}}{1-z^{2}}, \qquad z\in \mathbb {D}, \end{aligned}$$

(2.12)

that is, $f(z)=z-z^{3}/6-z^{5}/40+\cdots$.

Proof

Let the function $f\in {\mathcal {G}}$ given by (1.1). Thus, (1.4) and (2.9) give

$$\begin{aligned} \begin{aligned} 552960\cdot H_{3,1}(f)=\,&288\big (4c_{2}+c_{1}^{2}\big )c_{4}-288c_{1}c_{2}c_{3}-960c_{3}^{2}+48c_{1}^{3}c_{3}\\&-84c_{1}^{2}c_{2}^{2}+32c_{2}^{3}-12c_{1}^{4}c_{2}-c_{1}^{6}. \end{aligned} \end{aligned}$$

(2.13)

Since the class ${\mathcal {G}}$ and $\big |H_{3,1}(f)\big |$ are rotationally invariant, we may assume that $c_{1}\in [0,2]$. Thus, in view of (1.9) we assume that $\zeta _{1}\in [0,1]$. Using (2.13) and (1.9)-(1.12), we obtain

$$\begin{aligned} \begin{aligned} 8640\cdot H_{3,1}(f)=\,&(1-\zeta _{1}^{2})\big \{\big [36\zeta _{1}^{4}\zeta _{2} +3(1-25\zeta _{1}^{2})\zeta _{1}^{2}\zeta _{2}^{2}\\&+4(19-5\zeta _{1}^{2}+13\zeta _{1}^{4})\zeta _{2}^{3} +12(1-\zeta _{1}^{2})\zeta _{1}^{2}\zeta _{2}^{4}\big ]\\&+12(1-|\zeta _{2}|^{2})\big [6\zeta _{1}^{3}-(11+7\zeta _{1}^{2})\zeta _{1}\zeta _{2} -2(1-\zeta _{1}^{2})\zeta _{1}\zeta _{2}^{2}\big ]\zeta _{3}\\&-12(1-|\zeta _{2}|^{2})\big [(1-\zeta _{1}^{2})(5+|\zeta _{2}|^{2})+9\zeta _{1}^{2}\overline{\zeta _{2}}\big ]\zeta _{3}^{2}\\&+36(1-|\zeta _{2}|^{2})(1-|\zeta _{3}|^{2})\big [3\zeta _{1}^{2}+2(1-\zeta _{1}^{2})\zeta _{2}\big ]\zeta _{4}\big \}, \end{aligned} \end{aligned}$$

for some $\zeta _{1}\in [0, 1]$ and $\zeta _{2}, \zeta _{3}, \zeta _{4}\in \overline{\mathbb {D}}$. Since $|\zeta _{4}|\le 1$, we have

$$\begin{aligned} \begin{aligned} 8640\cdot \big |H_{3,1}(f)\big |&\le (1-\zeta _{1}^{2})\big \{\big |36\zeta _{1}^{4}\zeta _{2} +3(1-25\zeta _{1}^{2})\zeta _{1}^{2}\zeta _{2}^{2} +4(19-5\zeta _{1}^{2}+13\zeta _{1}^{4})\zeta _{2}^{3}\\ {}&\quad +12(1-\zeta _{1}^{2})\zeta _{1}^{2}\zeta _{2}^{4}\big | +12(1-|\zeta _{2}|^{2})\big |6\zeta _{1}^{3}-(11+7\zeta _{1}^{2})\zeta _{1}\zeta _{2}\\ {}&\quad -2(1-\zeta _{1}^{2})\zeta _{1}\zeta _{2}^{2}\big |\cdot |\zeta _{3}| +12(1-|\zeta _{2}|^{2})\big [\big |(1-\zeta _{1}^{2})(5+|\zeta _{2}|^{2})\\ {}&\quad +9\zeta _{1}^{2}\overline{\zeta _{2}}\big | -3\big |3\zeta _{1}^{2}+2(1-\zeta _{1}^{2})\zeta _{2}\big |\big ]\cdot |\zeta _{3}|^{2}\\&\quad +36(1-|\zeta _{2}|^{2})\big |3\zeta _{1}^{2}+2(1-\zeta _{1}^{2})\zeta _{2}\big |\big \}. \end{aligned} \end{aligned}$$

$\textbf{A}$. Suppose that

$$\begin{aligned} \big |(1-\zeta _{1}^{2})(5+|\zeta _{2}|^{2})+9\zeta _{1}^{2}\overline{\zeta _{2}}\big | -3\big |3\zeta _{1}^{2}+2(1-\zeta _{1}^{2})\zeta _{2}\big |\ge 0. \end{aligned}$$

Then

$$\begin{aligned} 8640\cdot \big |H_{3,1}(f)\big |\le h(\zeta _{1},|\zeta _{2}|), \end{aligned}$$

where $h:\ \mathbb {R}^{2}\rightarrow \mathbb {R}$ is defined by

$$\begin{aligned} \begin{aligned} h(x,y):=\,&12(1-x^{2})(5-5x^{2}+6x^{3})\\&+12(1-x^{2})(11x+9x^{2}+7x^{3}+3x^{4})y\\&+3(1-x^{2})(-16+8x+16x^{2}-32x^{3}+x^{2}\cdot |1-25x^{2}|)y^{2}\\&+4(1-x^{2})(19-33x-32x^{2}-21x^{3}+13x^{4})y^{3}\\&+12(1-x^{2})^{2}(-1-2x+x^{2})y^{4}. \end{aligned} \end{aligned}$$

We show that $h(x,y)\le 76$ for $(x,y)\in [0, 1]\times [0, 1]$.

$\textbf{I}$. On the vertices of $[0, 1]\times [0, 1]$, we have

$$\begin{aligned} h(0,0)=60, \quad h(0,1)=76, \quad h(1,0)=h(1,1)=0. \end{aligned}$$

$\textbf{II}$. On the sides of $[0, 1]\times [0, 1]$, we get

$$\begin{aligned} \begin{aligned}&h(0,y)=76y^{3}+12(1-y^{2})(5+y^{2})\le h(0,1)=76, \quad y\in (0, 1),\\&h(x,0)=12(1-x^{2})(5-5x^{2}+6x^{3})\le h(0,0)=60, \quad x\in (0, 1),\\&h(1,y)=0, \quad y\in (0, 1),\\&h(x,1)=(1-x^{2})(76-8x^{2}+76x^{4}+3|x^{2}-25x^{4}|)\le h(0,1)=76, \quad x\in (0, 1). \end{aligned} \end{aligned}$$

$\textbf{III}$. It remains to consider the set $(0, 1)\times (0, 1)$.

If $1-25x^{2}\ge 0$. Then all the real solutions ($x\ne 0,\pm 1$) of the system of equations

$$\begin{aligned} \begin{aligned} \frac{\partial h}{\partial x}=\,&12(-20x+18x^{2}+20x^{3}-30x^{4})\\&+12(11+18x-12x^{2}-24x^{3}-35x^{4}-18x^{5})y\\&+3(8+66x-120x^{2}-168x^{3}+160x^{4}+150x^{5})y^{2}\\&+4(-33-102x+36x^{2}+180x^{3}+105x^{4}-78x^{5})y^{3}\\&+24(-1+3x+6x^{2}-6x^{3}-5x^{4}+3x^{5})y^{4}=0, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \frac{\partial h}{\partial y}=\,&6(1-x^{2})\big [2(11x+9x^{2}+7x^{3}+3x^{4})\\&+(-16+8x+17x^{2}-32x^{3}-25x^{4})y\\&+2(19-33x-32x^{2}-21x^{3}+13x^{4})y^{2}\\&+8(-1-2x+2x^{2}+2x^{3}-x^{4})y^{3}\big ]=0, \end{aligned} \end{aligned}$$

by a numerical computation are the following

$$\begin{aligned} \begin{aligned}&{\left\{ \begin{array}{ll} x_{1}\approx -9502.13,\\ y_{1}\approx 0.845903,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{2}\approx -0.731928,\\ y_{2}\approx -20.5476,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{3}\approx 1.35957,\\ y_{3}\approx 15.3823,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{4}\approx -14.6979,\\ y_{4}\approx 0.668684,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{5}\approx -13.3345,\\ y_{5}\approx 0.419256,\\ \end{array}\right. } \qquad \ {\left\{ \begin{array}{ll} x_{6}\approx -1.63900,\\ y_{6}\approx 5.10765,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{7}\approx 0.276393,\\ y_{7}\approx 0.848735,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{8}\approx 0.429674,\\ y_{8}\approx 0.674311,\\ \end{array}\right. } \qquad \quad {\left\{ \begin{array}{ll} x_{9}\approx -0.905519,\\ y_{9}\approx -0.339299,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{10}\approx -0.840874,\\ y_{10}\approx 0.380896,\\ \end{array}\right. } \quad {\left\{ \begin{array}{ll} x_{11}\approx 0.0345448,\\ y_{11}\approx 0.0570942. \end{array}\right. } \end{aligned} \end{aligned}$$

Thus $(x_{11},y_{11})$ is the unique critical point of h in $(0, 1/5]\times (0, 1)$ with

$$\begin{aligned} h(x_{11},y_{11})\approx 60.0155<76. \end{aligned}$$

If $1-25x^{2}< 0$. Then all the real solutions ($x\ne 0,\pm 1$) of the system of equations

$$\begin{aligned} \begin{aligned} \frac{\partial h}{\partial x}=\,&12(-20x+18x^{2}+20x^{3}-30x^{4})\\&+12(11+18x-12x^{2}-24x^{3}-35x^{4}-18x^{5})y\\&+3(8+62x-120x^{2}+40x^{3}+160x^{4}-150x^{5})y^{2}\\&+4(-33-102x+36x^{2}+180x^{3}+105x^{4}-78x^{5})y^{3}\\&+24(-1+3x+6x^{2}-6x^{3}-5x^{4}+3x^{5})y^{4}=0, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \frac{\partial h}{\partial y}=\,&6(1-x^{2})\big [2(11x+9x^{2}+7x^{3}+3x^{4})\\&+(-16+8x+15x^{2}-32x^{3}+25x^{4})y\\&+2(19-33x-32x^{2}-21x^{3}+13x^{4})y^{2}\\&+8(-1-2x+2x^{2}+2x^{3}-x^{4})y^{3}\big ]=0, \end{aligned} \end{aligned}$$

by a numerical computation are the following

$$\begin{aligned} \begin{aligned}&{\left\{ \begin{array}{ll} x_{1}\approx -0.735039,\\ y_{1}\approx -20.4190,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{2}\approx 1.54150,\\ y_{2}\approx 11.5368,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{3}\approx -1.44859,\\ y_{3}\approx 7.65791,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{4}\approx -0.926545,\\ y_{4}\approx -0.564185,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{5}\approx 0.26956,\\ y_{5}\approx 0.879153,\\ \end{array}\right. } \quad \ \ {\left\{ \begin{array}{ll} x_{6}\approx 0.524719,\\ y_{6}\approx 0.683744,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{7}\approx -0.85181,\\ y_{7}\approx 0.256192,\\ \end{array}\right. } \qquad \ \ {\left\{ \begin{array}{ll} x_{8}\approx 0.0345845,\\ y_{8}\approx 0.0571628. \end{array}\right. } \end{aligned} \end{aligned}$$

Thus $(x_{5},y_{5})$ and $(x_{6},y_{6})$ are the critical points of h in $(1/5, 1)\times (0, 1)$ with

$$\begin{aligned} h(x_{5},y_{5})\approx 71.02494<76, \qquad h(x_{6},y_{6})\approx 73.94978<76. \end{aligned}$$

$\textbf{B}$. Suppose that

$$\begin{aligned} \big |(1-\zeta _{1}^{2})(5+|\zeta _{2}|^{2})+9\zeta _{1}^{2}\overline{\zeta _{2}}\big | -3\big |3\zeta _{1}^{2}+2(1-\zeta _{1}^{2})\zeta _{2}\big |<0. \end{aligned}$$

Then

$$\begin{aligned} 8640\cdot \big |H_{3,1}(f)\big |\le g(\zeta _{1},|\zeta _{2}|), \end{aligned}$$

where $g:\ \mathbb {R}^{2}\rightarrow \mathbb {R}$ is defined by

$$\begin{aligned} \begin{aligned} g(x,y):=\,&36(1-x^{2})(3x^{2}+2x^{3})\\&+12(1-x^{2})(6+11x-6x^{2}+7x^{3}+3x^{4})y\\&+3(1-x^{2})(8x-36x^{2}-32x^{3}+x^{2}\cdot |1-25x^{2}|)y^{2}\\&+4(1-x^{2})(1-33x+13x^{2}-21x^{3}+13x^{4})y^{3}\\&+12(1-x^{2})^{2}(-2x+x^{2})y^{4}. \end{aligned} \end{aligned}$$

We show that $g(x,y)\le 76$ for $(x,y)\in [0, 1]\times [0, 1]$.

$\textbf{I}$. On the vertices of $[0, 1]\times [0, 1]$, we have

$$\begin{aligned} g(0,0)=0, \quad g(0,1)=76, \quad g(1,0)=g(1,1)=0. \end{aligned}$$

$\textbf{II}$. On the sides of $[0, 1]\times [0, 1]$, we get

$$\begin{aligned} \begin{aligned}&g(0,y)=72y+4y^{3}\le g(0,1)=76, \quad y\in (0, 1),\\&g(x,0)=36x^{2}(1-x^{2})(3+2x)\le g(x_{0},0)\approx 39.9705, \quad x_{0}\approx 0.733044,\ x\in (0, 1),\\&g(1,y)=0, \quad y\in (0, 1),\\&g(x,1)=(1-x^{2})(76-8x^{2}+76x^{4}+3|x^{2}-25x^{4}|)\le g(0,1)=76, \quad x\in (0, 1). \end{aligned} \end{aligned}$$

$\textbf{III}$. It remains to consider the set $(0, 1)\times (0, 1)$.

If $1-25x^{2}\ge 0$. Then all the real solutions ($x\ne 0,\pm 1$) of the system of equations

$$\begin{aligned} \begin{aligned} \frac{\partial g}{\partial x}=\,&36(6x+6x^{2}-12x^{3}-10x^{4})\\&+12(11-24x-12x^{2}+36x^{3}-35x^{4}-18x^{5})y\\&+3(8-70x-120x^{2}+40x^{3}+160x^{4}+150x^{5})y^{2}\\&+4(-33+24x+36x^{2}+105x^{4}-78x^{5})y^{3}\\&+24(-1+x+6x^{2}-4x^{3}-5x^{4}+3x^{5})y^{4}=0, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \frac{\partial g}{\partial y}=\,&6(1-x^{2})\big [2(6+11x-6x^{2}+7x^{3}+3x^{4})\\&+(8x-35x^{2}-32x^{3}-25x^{4})y\\&+2(1-33x+13x^{2}-21x^{3}+13x^{4})y^{2}\\&+8(-2x+x^{2}+2x^{3}-x^{4})y^{3}\big ]=0, \end{aligned} \end{aligned}$$

by a numerical computation are the following

$$\begin{aligned} \begin{aligned}&{\left\{ \begin{array}{ll} x_{1}\approx -9589.92,\\ y_{1}\approx 1.38296,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{2}\approx 1.41932,\\ y_{2}\approx 11.2938,\\ \end{array}\right. } \qquad \quad {\left\{ \begin{array}{ll} x_{3}\approx -14.5113,\\ y_{3}\approx 0.695679,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{4}\approx -12.4904,\\ y_{4}\approx 0.366752,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{5}\approx -0.0961383,\\ y_{5}\approx -5.70471,\\ \end{array}\right. } \quad \ {\left\{ \begin{array}{ll} x_{6}\approx -1.68318,\\ y_{6}\approx 6.24783,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{7}\approx -1.51857,\\ y_{7}\approx -0.274771,\\ \end{array}\right. } \quad \ \ {\left\{ \begin{array}{ll} x_{8}\approx 0.380253,\\ y_{8}\approx 0.809067,\\ \end{array}\right. } \qquad \ \ {\left\{ \begin{array}{ll} x_{9}\approx 0.539943,\\ y_{9}\approx 0.658839,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{10}\approx 0.776016,\\ y_{10}\approx -1.50891,\\ \end{array}\right. } \quad \ \ {\left\{ \begin{array}{ll} x_{11}\approx 0.572503,\\ y_{11}\approx -1.28199,\\ \end{array}\right. } \qquad \ {\left\{ \begin{array}{ll} x_{12}\approx -0.699973,\\ y_{12}\approx -0.30267,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{13}\approx -0.409285,\\ y_{13}\approx 0.170843,\\ \end{array}\right. } \quad {\left\{ \begin{array}{ll} x_{14}\approx -0.861525,\\ y_{14}\approx 0.540418. \end{array}\right. } \end{aligned} \end{aligned}$$

Thus the function g has no critical point in $(0, 1/5]\times (0, 1)$.

If $1-25x^{2}<0$. Then all the real solutions ($x\ne 0,\pm 1$) of the system of equations

$$\begin{aligned} \begin{aligned} \frac{\partial g}{\partial x}=\,&36(6x+6x^{2}-12x^{3}-10x^{4})\\&+12(11-24x-12x^{2}+36x^{3}-35x^{4}-18x^{5})y\\&+3(8-74x-120x^{2}+248x^{3}+160x^{4}-150x^{5})y^{2}\\&+4(-33+24x+36x^{2}+105x^{4}-78x^{5})y^{3}\\&+24(-1+x+6x^{2}-4x^{3}-5x^{4}+3x^{5})y^{4}=0, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \frac{\partial g}{\partial y}=\,&6(1-x^{2})\big [2(6+11x-6x^{2}+7x^{3}+3x^{4})\\&+(8x-37x^{2}-32x^{3}+25x^{4})y\\&+2(1-33x+13x^{2}-21x^{3}+13x^{4})y^{2}\\&+8(-2x+x^{2}+2x^{3}-x^{4})y^{3}\big ]=0, \end{aligned} \end{aligned}$$

by a numerical computation are the following

$$\begin{aligned} \begin{aligned}&{\left\{ \begin{array}{ll} x_{1}\approx 2.27466,\\ y_{1}\approx 9.76748,\\ \end{array}\right. } \qquad \quad {\left\{ \begin{array}{ll} x_{2}\approx -1.49035,\\ y_{2}\approx 8.66161,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{3}\approx -0.0959782,\\ y_{3}\approx -5.70831,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{4}\approx -2.76974,\\ y_{4}\approx -0.612726,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{5}\approx 0.328097,\\ y_{5}\approx 0.878105,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{6}\approx 0.605801,\\ y_{6}\approx 0.667028,\\ \end{array}\right. }\\&{\left\{ \begin{array}{ll} x_{7}\approx -0.723959,\\ y_{7}\approx -0.372024,\\ \end{array}\right. } \qquad {\left\{ \begin{array}{ll} x_{8}\approx -0.91188,\\ y_{8}\approx 0.413182,\\ \end{array}\right. } \quad \ \ {\left\{ \begin{array}{ll} x_{9}\approx -0.414826,\\ y_{9}\approx 0.173701.\\ \end{array}\right. } \end{aligned} \end{aligned}$$

Thus $(x_{5},y_{5})$ is the unique critical point of g in $(1/5, 1)\times (0, 1)$ with

$$\begin{aligned} g(x_{5},y_{5})\approx 70.12399<76, \qquad g(x_{6},y_{6})\approx 73.93581<76. \end{aligned}$$

Summarizing, we see that the bounds obtained in Parts A and B give

$$\begin{aligned} \big |H_{3,1}(f)\big |\le \frac{1}{8640}\cdot 76=\frac{19}{2160}. \end{aligned}$$

We finally note that equality in (2.6) holds for the function $f\in {\mathcal {G}}$ defined by (1.1), and satisfying (2.8) with

$$\begin{aligned} p(z):=\frac{1+z^{2}}{1-z^{2}}, \qquad z\in \mathbb {D}, \end{aligned}$$

for which $a_{2}=a_{4}=0$, $a_{3}=-1/6$ and $a_{5}=-1/40$. This completes the proof of the theorem. $\square$