Sharp Coefficient Bounds for a Subclass of Bounded Turning Functions with a Cardioid Domain

Abstract

In the present paper, we give a new simple proof on the sharp bounds of coefficient functionals related to the Carathéodory functions and make a correction on the extremal functions. The result is further used to investigate some initial coefficient bounds on a subclass of bounded turning functions ${\mathcal{R}}_{\wp }$ associated with a cardioid domain. For functions in this class, we calculate the bounds of the Fekete–Szegö-type inequality and the second- and third-order Hankel determinants. All the results are proved to be sharp.

1. Introduction and Definitions

Let $\mathcal{H}\left(\mathbb{D}\right)$ represent the family of functions which are analytic in the unit disc $\mathbb{D}=\left\{z\in \mathbb{C}:\left|z\right|<1\right\}$. Let $\mathcal{A}$ denote the subfamily of $\mathcal{H}\left(\mathbb{D}\right)$ consisting of functions in the form of

$$f\left(z\right)=z+\sum _{k=2}^{\infty }{a}_k{z}^k,z\in \mathbb{D}.$$

(1)

Suppose that $\mathcal{P}$ indicates the class of the class of all functions p that are analytic in $\mathbb{D}$ with $\Re \left(p\left(z\right)\right)>0$ and

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

(2)

If $p\in \mathcal{P}$, it is a Carathéodory function. Assume that the set $\mathcal{S}\subset \mathcal{A}$ contains all univalent functions in $\mathbb{D}$. Using the Koebe theorem, it is known that for each univalent function $f\in \mathcal{S}$, there exist an inverse function $f^{-1}$ defined at least on a disc of radius $1/4$ with the Taylor’s series of the form

$$f^{-1}\left(w\right):=w+\sum _{n=2}^{\infty }{B}_n{w}^n,\left|w\right|<\frac{1}{4}.$$

(3)

For two functions $F_1,F_2\in \mathcal{H}\left(\mathbb{D}\right),$ we say $F_1$ is subordinate to $F_2,$ written by $F_1\prec F_2,$ if there exists a function u which is analytic in $\mathbb{D}$ with $u\left(0\right)=0$ and $\left|u\left(z\right)\right|<1$, such that $F_1\left(z\right)=F_2\left(u\left(z\right)\right),$$z\in \mathbb{D}$. The function u is called a Schwarz function. In the case of $F_2$ being univalent in $\mathbb{D},$ then we have the relation

$$\begin{array}{cccccc}\hfill F_1\left(z\right)\prec F_2\left(z\right)& \left(z\in \mathbb{D}\right)& ⟺\hfill & F_1\left(0\right)=F_2\left(0\right)& \mathrm{and}& F_1\left(\mathbb{D}\right)\subset F_2\left(\mathbb{D}\right).\end{array}$$

In geometric function theory, the most basic and important subfamilies of the set $\mathcal{S}$ are the family ${\mathcal{S}}^{\ast }$ of starlike functions, the family $\mathcal{C}$ of convex functions and the family $\mathcal{R}$ of bounded turning functions. The interested readers are referred to [1], (Chapter II). In 1994, Ma and Minda [2] introduced a class of analytic univalent functions $\phi \left(z\right)$, which maps $\mathbb{D}$ onto the starlike domain with respect to $\phi \left(0\right)=1$ in the right half plane and is symmetric about the real axis. The Ma and Minda classes of $\mathcal{C}\left(\phi \right)$, ${\mathcal{S}}^{\ast }\left(\phi \right)$ and $\mathcal{R}\left(\phi \right)$ are characterized, respectively, as

$$\begin{array}{cc}\hfill \mathcal{C}\left(\phi \right)& :=\left\{f\in \mathcal{A}:\frac{{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{f^{\prime }\left(z\right)}\prec \phi \left(z\right)\right\},\hfill \\ \hfill {\mathcal{S}}^{\ast }\left(\phi \right)& :=\left\{f\in \mathcal{A}:\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\prec \phi \left(z\right)\right\},\hfill \\ \hfill \mathcal{R}\left(\phi \right)& :=\left\{f\in \mathcal{A}:f^{\prime }\left(z\right)\prec \phi \left(z\right)\right\},\hfill \end{array}$$

see [3,4]. By considering different image domains $\phi \left(\mathbb{D}\right)$, various classes $\mathcal{C}\left(\phi \right)$, ${\mathcal{S}}^{\ast }\left(\phi \right)$ and $\mathcal{R}\left(\phi \right)$ of univalent functions were considered in recent years. For example, setting $\phi \left(z\right)=\sqrt{1+z}$, we obtain the class ${\mathcal{S}}_L^{\ast }={\mathcal{S}}^{\ast }\left(\sqrt{1+z}\right)$, which represents the collection of functions in the class $\mathcal{A}$ that $\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}$ lies in the domain bounded by the lemniscate of Bernoulli $\left|w^2-1\right|=1$, see [5]. Choosing $\tilde{\phi }=1+\frac{2}{{\pi }^2}{\left(log\left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)\right)}^2$, ${\mathcal{S}}_p^{\ast }={\mathcal{S}}^{\ast }\left(\tilde{\phi }\right)$ is the class of parabolic starlike functions. For functions $f\in {\mathcal{S}}_p^{\ast }$, its image of $\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}$ under $\mathbb{D}$ is the parabolic domain given by $\left\{w\in \mathbb{C}:\Re \left(w\right)>\left|w-1\right|\right\}$, see [6]. The class ${\mathcal{S}}_c^{\ast }={\mathcal{S}}^{\ast }\left(1+\frac{4}{3}z+\frac{2}{3}{z}^2\right)$ is a collection of starlike functions $f\in \mathcal{A}$ where $\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}$ lies in the domain bounded by the cardiod ${\Omega }_c=\left\{u+iv:{\left(9u^2+9v^2-18u+5\right)}^2-16\left(9u^2+9v^2-6u+1\right)=0\right\}$; for further reading we refer to [7]. In [8], Wani and Swaminathan investigated the class ${\mathcal{S}}_{Ne}^{\ast }={\mathcal{S}}^{\ast }\left(1+z-\frac{1}{3}{z}^3\right)$, consisting of functions associated with a nephroid domain. For other related works, see, for instance, [9,10,11]. Recently, S. Sivaprasad Kumar et al. [12] introduced and studied a class of starlike functions defined by

$${\mathcal{S}}_{\mathcal{C}}^{\ast }:=\left\{f\in \mathcal{A}:\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\prec 1+z{e}^z=:\wp \left(z\right),z\in \mathbb{D}\right\},$$

(4)

where $\wp \left(z\right)$ maps the unit disk onto a cardioid domain.

Motivated by the above works, we now consider a subfamily ${\mathcal{R}}_{\wp }$ of bounded turning functions defined by

$${\mathcal{R}}_{\wp }:=\left\{f\in \mathcal{A}:f^{\prime }\left(z\right)\prec 1+z{e}^z,z\in \mathbb{D}\right\}.$$

(5)

For given parameters $q,n\in \mathbb{N}=\left\{1,2,\cdots \right\}$, the Hankel determinant $H_{q,n}\left(f\right)$ was defined by Pommerenke [13,14] for a function $f\in \mathcal{S}$ of the form (1) as

$$H_{q,n}\left(f\right)=\left|\begin{array}{cccc}{a}_n\hfill & a_{n+1}\hfill & \dots \hfill & a_{n+q-1}\hfill \\ a_{n+1}\hfill & a_{n+2}\hfill & \dots \hfill & a_{n+q}\hfill \\ ⋮\hfill & ⋮\hfill & \dots \hfill & ⋮\hfill \\ a_{n+q-1}\hfill & a_{n+q}\hfill & \dots \hfill & a_{n+2q-2}\hfill \end{array}\right|,a_1=1.$$

(6)

The upper bounds of $\left|H_{q,n}\left(f\right)\right|$ have been investigated for different subclasses of univalent functions. By applying Schwarz Lemma [15,16], Selin Aydinoğlua and Bülent Nafi Örnek [17] determined the sharp bounds of Hankel determinant ${\mathcal{H}}_{2,1}\left(f\right)=a_3-a_2^2$ for the class ${\mathcal{M}}_{\alpha }$, defined by the condition $f\in \mathcal{A}$ and $\left|{\left(\frac{z}{f\left(z\right)}\right)}^2{f}^{\prime }\left(z\right)-\alpha \right|<1$, where $\alpha \in \mathbb{C}$. Of note, the Hankel determinant ${\mathcal{H}}_{2,1}\left(f\right)$ is also known as Fekete–Szegö inequality. The absolute sharp bounds of the functional $H_{2,2}\left(f\right)=a_2{a}_4-a_3^2$ were found in [18,19] for each of the sets $\mathcal{C},$${\mathcal{S}}^{\ast }$ and $\mathcal{R}$. The Hankel determinant of order three is given as

$$H_{3,1}\left(f\right)=\left|\begin{array}{ccc}1& a_2& a_3\\ a_2& a_3& a_4\\ a_3& a_4& a_5\end{array}\right|=-a_5{a}_2^2+2a_2{a}_3{a}_4-a_3^3+a_5{a}_3-a_4^2.$$

(7)

The estimation of the determinant $\left|H_{3,1}\left(f\right)\right|$ seems a little harder compared to the bound of $\left|H_{2,2}\left(f\right)\right|$, see [20,21,22]. In 2010, Babalola [23] obtained the upper bound of $\left|H_{3,1}\left(f\right)\right|$ for the families of ${\mathcal{S}}^{\ast },$$\mathcal{C}$ and $\mathcal{R}$. Later on, many authors obtained non-sharp bounds on $\left|H_{3,1}\left(f\right)\right|$ for different subfamilies of univalent functions, see, for example, [24,25,26]. The sharp bound of the third Hankel determinant for convex functions $\mathcal{C}$ was obtained in [27]. For $f\in {\mathcal{S}}^{\ast }$, the upper bound of $\left|H_{3,1}\left(f\right)\right|$ was finally proved to be $\frac{4}{9}$ by Kowalczyk et al. [28]. For the bounded turning functions $\mathcal{R}$, the sharp upper bound of third Hankel determinant was calculated to be $\frac{1}{4}$ in [29]. For some subclasses of convex functions, starlike functions and bounded turning functions, some sharp bounds of third Hankel determinant were also obtained in [30,31,32,33].

In the current article, our main goal is to calculate the sharp bounds on some initial coefficients for the class ${\mathcal{R}}_{\wp }$ of bounded turning functions linked with a cardioid domain. We also obtain the Fekete–Szegö inequality, and the sharp bounds of the second- and third-order Hankel determinants for this class. In proof of our results, we give a new simple proof of an estimation for the Carathéodory function and correct an error on the extremal function in Lemma 2.1 of [34].

2. A Set of Lemmas

The key to the proof of our results is the following lemmas.

Lemma 1 

([35]). Let $p\in \mathcal{P}$ be given by (2). Then, we have

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

(8)

$$\begin{array}{cc}\hfill 4c_3& =c_1^3+2\left(4-c_1^2\right)c_{1}x-c_1\left(4-c_1^2\right)x^2+2\left(4-c_1^2\right)\left(1-{\left|x\right|}^2\right)\delta ,\hfill \\ \hfill 8c_4& =c_1^4+\left(4-c_1^2\right)x\left[c_1^2\left(x^2-3x+3\right)+4x\right]-4\left(4-c_1^2\right)\left(1-{\left|x\right|}^2\right)\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill & ·\left[c_1\left(x-1\right)\delta +\overline{x}{\delta }^2-\left(1-{\left|\delta \right|}^2\right)\rho \right]\hfill \end{array}$$

(10)

for some complex numbers x, δ and ρ, such that $\left|x\right|\le 1$, $\left|\delta \right|\le 1$ and $\left|\rho \right|\le 1.$

Lemma 2 

([36]). If $p\in \mathcal{P}$ has the form (2), then

$$\left|c_n\right|\le 2\mathrm{for}n\ge 1.$$

(11)

Lemma 3 

([37]). For any complex number μ and $p\in \mathcal{P}$,

$$\left|c_{n+k}-\mu c_n{c}_k\right|\le 2max\left\{1,\left|2\mu -1\right|\right\}.$$

(12)

Lemma 4 

([38]). Let $\omega \left(z\right)={\sum }_{n=1}^{\infty }{w}_k{z}^k$ be a Schwarz function. Then, for real numbers μ and ν, we have the following sharp estimate given by

$$\Psi \left(\omega \right)=\left|w_3+\mu w_1{w}_2+\nu w_1^3\right|\le \Phi (\mu ,\nu ),$$

(13)

where $\Phi (\mu ,\nu )$ is defined by

$$\Phi (\mu ,\nu )=\left\{\begin{array}{cc}1,\hfill & (\mu ,\nu )\in {\mathbb{D}}_1\bigcup {\mathbb{D}}_2\bigcup \left\{\left(2,1\right)\right\},\\ \\ \left|\nu \right|,\hfill & (\mu ,\nu )\in \bigcup _{k=3}^7{\mathbb{D}}_k,\\ \\ \frac{2}{3}\left(\left|\mu \right|+1\right)\sqrt{\frac{\left|\mu \right|+1}{3(\left|\mu \right|+1+\nu )}},\hfill & (\mu ,\nu )\in {\mathbb{D}}_8\bigcup {\mathbb{D}}_9,\\ \\ \frac{1}{3}\nu \left(\frac{{\mu }^2-4}{{\mu }^2-4\nu }\right)\sqrt{\frac{{\mu }^2-4}{3\left(\nu -1\right)}},\hfill & (\mu ,\nu )\in {\mathbb{D}}_{10}\bigcup {\mathbb{D}}_{11}\setminus \left\{\left(2,1\right)\right\},\\ \\ \frac{2}{3}\left(\left|\nu \right|-1\right)\sqrt{\frac{\left|\mu \right|-1}{3\left(\left|\mu \right|-1-\nu \right)}},\hfill & (\mu ,\nu )\in {\mathbb{D}}_{12},\end{array}\right.$$

(14)

and

$$\begin{array}{cc}\hfill {\mathbb{D}}_1& =\left\{(\mu ,\nu ):\left|\mu \right|\le \frac{1}{2},-1\le \nu \le 1\right\},\hfill \\ \hfill {\mathbb{D}}_2& =\left\{(\mu ,\nu ):\frac{1}{2}\le \left|\mu \right|\le 2,\frac{4}{27}{\left(\left|\mu \right|+1\right)}^3-\left(\left|\mu \right|+1\right)\le \nu \le 1\right\},\hfill \\ \hfill {\mathbb{D}}_3& =\left\{(\mu ,\nu ):\left|\mu \right|\le \frac{1}{2},\nu \le -1\right\},\hfill \\ \hfill {\mathbb{D}}_4& =\left\{(\mu ,\nu ):\left|\mu \right|\ge \frac{1}{2},\nu \le -\frac{2}{3}\left(\left|\mu \right|+1\right)\right\},\hfill \\ \hfill {\mathbb{D}}_5& =\left\{(\mu ,\nu ):\left|\mu \right|\le 2,\nu \ge 1\right\},\hfill \\ \hfill {\mathbb{D}}_6& =\left\{(\mu ,\nu ):2\le \left|\mu \right|\le 4,\nu \ge \frac{1}{12}\left({\mu }^2+8\right)\right\},\hfill \\ \hfill {\mathbb{D}}_7& =\left\{(\mu ,\nu ):\left|\mu \right|\ge 4,\nu \ge \frac{2}{3}\left(\left|\mu \right|-1\right)\right\},\hfill \\ \hfill {\mathbb{D}}_8& =\left\{(\mu ,\nu ):\frac{1}{2}\le \left|\mu \right|\le 2,-\frac{2}{3}\left(\left|\mu \right|+1\right)\le \nu \le \frac{4}{27}{\left(\left|\mu \right|+1\right)}^3-\left(\left|\mu \right|+1\right)\right\},\hfill \\ \hfill {\mathbb{D}}_9& =\left\{(\mu ,\nu ):\left|\mu \right|\ge 2,-\frac{2}{3}\left(\left|\mu \right|+1\right)\le \nu \le \frac{2\left|\mu \right|\left(\left|\mu \right|+1\right)}{{\mu }^2+2\left|\mu \right|+4}\right\},\hfill \\ \hfill {\mathbb{D}}_{10}& =\left\{(\mu ,\nu ):2\le \left|\mu \right|\le 4,\frac{2\left|\mu \right|\left(\left|\mu \right|+1\right)}{{\mu }^2+2\left|\mu \right|+4}\le \nu \le \frac{1}{12}\left({\mu }^2+8\right)\right\},\hfill \\ \hfill {\mathbb{D}}_{11}& =\left\{(\mu ,\nu ):\left|\mu \right|\ge 4,\frac{2\left|\mu \right|\left(\left|\mu \right|+1\right)}{{\mu }^2+2\left|\mu \right|+4}\le \nu \le \frac{2\left|\mu \right|\left(\left|\mu \right|-1\right)}{{\mu }^2-2\left|\mu \right|+4}\right\},\hfill \\ \hfill {\mathbb{D}}_{12}& =\left\{(\mu ,\nu ):\left|\mu \right|\ge 4,\frac{2\left|\mu \right|\left(\left|\mu \right|-1\right)}{{\mu }^2-2\left|\mu \right|+4}\le \nu \le \frac{2}{3}\left(\left|\mu \right|-1\right)\right\}.\hfill \end{array}$$

The following Lemma was obtained by Virendra Kumar et al. [34] in 2019. As the authors point out, it is of independent interest as well. Unfortunately, there are some minor mistakes on the extremal function. Next, we will give a new more simple proof of this result using Lemma 4.

Lemma 5. 

Let $p\left(z\right)=1+p_{1}z+p_2{z}^2+p_3{z}^3+\cdots \in \mathcal{P}$. Then, for any real number σ,

$$\left|\sigma p_3-p_1^3\right|\le \left\{\begin{array}{cc}2\left|\sigma -4\right|,\hfill & \sigma <\frac{4}{3},\\ 2\sigma \sqrt{\frac{\sigma }{\sigma -1}},\hfill & \sigma \ge \frac{4}{3}.\end{array}\right.$$

The above estimate is sharp.

Proof. 

Let $p\in \mathcal{P}$ and

$$p\left(z\right)=1+p_{1}z+p_2{z}^2+p_3{z}^3+\cdots ,z\in \mathbb{D}.$$

Suppose that $\omega \left(z\right)=\frac{p\left(z\right)-1}{p\left(z\right)+1}$. Clearly, $\omega \left(0\right)=0$. Since $p\left(z\right)$ lies in the right half plane and $\frac{z-1}{z+1}$ maps the right half plane to the unit disk, we know $\left|\omega \left(z\right)\right|<1$. Thus, $\omega$ is a Schwarz function. Assume that

$$\omega \left(z\right)=w_{1}z+w_2{z}^2+w_3{z}^3+w_4{z}^4+\cdots ,z\in \mathbb{D}.$$

From the fact that

$$p\left(z\right)=\frac{1+\omega \left(z\right)}{1-\omega \left(z\right)}=1+2w_{1}z+\left(2w_1^2+2w_2\right)z^2+\left(2w_1^3+4w_2{w}_1+2w_3\right)z^3+\cdots ,$$

we have $p_1=2w_1,$$p_2=2\left(w_2+w_1^2\right)$ and $p_3=2\left(w_3+2w_2{w}_1+w_1^3\right)$. It follows that

$$\sigma p_3-p_1^3=2\sigma w_3+4\sigma w_1{w}_2+\left(2\sigma -8\right)w_1^3.$$

As $\sigma =0$ the proof is trivial, we assume that $\sigma \ne 0$ in the following. Then, we obtain

$$\left|\sigma p_3-p_1^3\right|=2\left|\sigma \right|\left|w_3+2w_1{w}_2+\frac{\sigma -4}{\sigma }{w}_1^3\right|.$$

(15)

Let $\mu =2$, $\nu =\frac{\sigma -4}{\sigma }$. Clearly, $\left(\mu ,\nu \right)$ is only possible to lie in the disk ${\mathbb{D}}_4$, ${\mathbb{D}}_5$, ${\mathbb{D}}_6$, ${\mathbb{D}}_8$ and ${\mathbb{D}}_9$, which can be further specified as

$$\begin{array}{cc}\hfill {\mathbb{D}}_4& =\left\{(\mu ,\nu ):\left|\mu \right|\ge \frac{1}{2},\nu \le -2\right\},\hfill \\ \hfill {\mathbb{D}}_5& =\left\{(\mu ,\nu ):\left|\mu \right|\le 2,\nu \ge 1\right\},\hfill \\ \hfill {\mathbb{D}}_6& =\left\{(\mu ,\nu ):2\le \left|\mu \right|\le 4,\nu \ge 1\right\},\hfill \\ \hfill {\mathbb{D}}_8& =\left\{(\mu ,\nu ):\frac{1}{2}\le \left|\mu \right|\le 2,-2\le \nu \le 1\right\},\hfill \\ \hfill {\mathbb{D}}_9& =\left\{(\mu ,\nu ):\left|\mu \right|\ge 2,-2\le \nu \le 1\right\}.\hfill \end{array}$$

For $\sigma <0$, it is observed that $\nu >1$ and $(\mu ,\nu )\in {\mathbb{D}}_5\bigcup {\mathbb{D}}_6$. Thus, we have

$$\left|w_3+2w_1{w}_2+\frac{\sigma -4}{\sigma }{w}_1^3\right|\le \left|\nu \right|=\frac{\sigma -4}{\sigma }.$$

(16)

For $0<\sigma <\frac{4}{3}$, we see $\nu <-2$ and $(\mu ,\nu )\in {\mathbb{D}}_4$. Thus,

$$\left|w_3+2w_1{w}_2+\frac{\sigma -4}{\sigma }{w}_1^3\right|\le \left|\nu \right|=-\frac{\sigma -4}{\sigma }.$$

(17)

For $\sigma \ge \frac{4}{3}$, we know $-2\le \nu <1$ and $(\mu ,\nu )\in {\mathbb{D}}_8\bigcup {\mathbb{D}}_9$. Then we deduce that

$$\left|w_3+2w_1{w}_2+\frac{\sigma -4}{\sigma }{w}_1^3\right|\le \frac{2}{3}\left(\left|\mu \right|+1\right)\sqrt{\frac{\left|\mu \right|+1}{3(\left|\mu \right|+1+\nu )}}=\sqrt{\frac{\sigma }{\sigma -1}}.$$

(18)

Combining (15)–(18), the result of Lemma 5 follows. □

Remark 1. 

In [34], the authors gave an extremal function f given by

$$f\left(z\right)=\frac{1-z^2}{1-2\sqrt{\frac{\sigma }{\sigma -1}}z+z^2},\sigma >\frac{4}{3}.$$

Let $q=\sqrt{\frac{\sigma }{\sigma -1}}$. It is seen that

$$f\left(z\right)=1+2qz+2\left(2q^2-1\right)z^2+2q\left(4q^2-3\right)z^3+2\left(8q^4-8q^2+1\right)z^4+\cdots ,z\in \mathbb{D}.$$

We know $f\notin \mathcal{P}$ because $c_1=2q>2$. Hence, the extremal function is not correct, since it is not a Carathéodory function. Indeed, the extremal function $\hat{f}$ for $\sigma >\frac{4}{3}$ can be defined by taking

$$\hat{f}\left(z\right)=1+{\hat{p}}_{1}z+{\hat{p}}_2{z}^2+{\hat{p}}_3{z}^3+\cdots ,z\in \mathbb{D},$$

(19)

where ${\hat{p}}_1=\sqrt{\frac{\sigma }{\sigma -1}},$${\hat{p}}_2=-\frac{\sigma -2}{\sigma -1}$ and ${\hat{p}}_3=-\frac{2\sigma -3}{\sigma }\sqrt{{\left(\frac{\sigma }{\sigma -1}\right)}^3}.$

3. Coefficient Bounds for the Family ${\mathcal{R}}_{\wp }$

We begin this section by finding the bounds on some initial coefficients for functions in the class ${\mathcal{R}}_{\wp }.$

Theorem 1. 

If $f\in {\mathcal{R}}_{\wp }$ has the series representation of the form (1), then

$$\begin{array}{cc}\hfill \left|a_2\right|& \le \frac{1}{2},\hfill \\ \hfill \left|a_3\right|& \le \frac{1}{3},\hfill \\ \hfill \left|a_4\right|& \le \frac{\sqrt{14}}{14}\approx 0.2673.\hfill \end{array}$$

These bounds are best possible.

Proof. 

Let $f\in {\mathcal{R}}_{\wp }.$ Then (5) can be written by Schwarz function as

$$f^{\prime }\left(z\right)=1+\omega \left(z\right)e^{\omega \left(z\right)},z\in \mathbb{D}.$$

(20)

Assuming that

$$\omega \left(z\right)=w_{1}z+w_2{z}^2+w_3{z}^3+\cdots ,z\in \mathbb{D},$$

(21)

and

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

(22)

It is seen that $p\in \mathcal{P}$ and

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

(23)

From (1) we obtain

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

(24)

By simplifications and using the series expansion of (23), we obtain

$$\begin{array}{cc}\hfill 1+\omega \left(z\right)e^{\omega \left(z\right)}& =1+\frac{1}{2}{c}_{1}z+\frac{1}{2}{c}_2{z}^2+\left(-\frac{1}{16}{c}_1^3+\frac{1}{2}{c}_3\right)z^3\hfill \\ \hfill & +\left(\frac{1}{24}{c}_1^4-\frac{3}{16}{c}_1^2{c}_2+\frac{1}{2}{c}_4\right)z^4+\cdots .\hfill \end{array}$$

(25)

Comparing (24) and (25), we have

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

(26)

$$\begin{array}{cc}\hfill a_3& =\frac{1}{6}{c}_2,\hfill \end{array}$$

(27)

$$\begin{array}{cc}\hfill a_4& =\frac{1}{8}\left(-\frac{1}{8}{c}_1^3+c_3\right),\hfill \end{array}$$

(28)

$$\begin{array}{cc}\hfill a_5& =\frac{1}{10}\left(\frac{1}{12}{c}_1^4-\frac{3}{8}{c}_1^2{c}_2+c_4\right).\hfill \end{array}$$

(29)

For $a_2$ and $a_3$, implementing Lemma 2, we obtain $\left|a_2\right|\le \frac{1}{2}$ and $\left|a_3\right|\le \frac{1}{3}$. For $a_4,$ an application of Lemma 5 leads us to $\left|a_4\right|\le \frac{1}{4}\sqrt{\frac{8}{7}}=\frac{\sqrt{14}}{14}.$ The equality of $\left|a_2\right|$ and $\left|a_3\right|$ are achieved by the functions $f_1$ and $f_2$ given, respectively, by

$$\begin{array}{cc}\hfill f_1\left(z\right)& ={\int }_0^z\left(1+t{e}^t\right)dt=z+\frac{1}{2}{z}^2+\frac{1}{3}{z}^3+\frac{1}{8}{z}^4+\frac{1}{30}{z}^5+\cdots ,z\in \mathbb{D},\hfill \end{array}$$

(30)

$$\begin{array}{cc}\hfill f_2\left(z\right)& ={\int }_0^z\left(1+t^2{e}^{t^2}\right)dt=z+\frac{1}{3}{z}^3+\frac{1}{5}{z}^5+\frac{1}{14}{z}^7+\frac{1}{54}{z}^9+\cdots ,z\in \mathbb{D}.\hfill \end{array}$$

(31)

The equality on the bounds of $\left|a_4\right|$ is obtained by $f_3$ defined by

$$f_3\left(z\right)={\int }_0^z\left(1+\omega \left(t\right)e^{\omega \left(t\right)}\right)dt,z\in \mathbb{D},$$

(32)

where $\omega \left(z\right)=\frac{p\left(z\right)-1}{p\left(z\right)+1}$ and

$$p\left(z\right)=1+\frac{\sqrt{56}}{7}z-\frac{6}{7}{z}^2-\frac{26\sqrt{14}}{49}{z}^3+\cdots ,z\in \mathbb{D}.$$

(33)

It is verified that

$$f_3\left(z\right)=z+\frac{\sqrt{14}}{14}{z}^2-\frac{1}{7}{z}^3-\frac{\sqrt{14}}{14}{z}^4+\cdots ,z\in \mathbb{D}.$$

(34)

The proof of Theorem 1 is thus completed. □

Theorem 2. 

If f is of the form (1) belonging to ${\mathcal{R}}_{\wp },$ then

$$\left|a_3-\gamma a_2^2\right|\le max\left\{\frac{1}{3},\left|\frac{3\gamma -4}{12}\right|\right\},\gamma \in \mathbb{C}.$$

This inequality is sharp.

Proof. 

Employing (26) and (27), we may write

$$\left|a_3-\gamma a_2^2\right|=\frac{1}{6}\left|c_2-\frac{3}{8}\gamma c_1^2\right|.$$

An application of (12) leads us to

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

This result is sharp for the functions $f_1$ and $f_2$ given by (30) and (31). □

Theorem 3. 

Let $f\in {\mathcal{R}}_{\wp }$. Then

$$\left|a_2{a}_3-a_4\right|\le \frac{1}{4}.$$

This inequality is sharp with the extremal function $f_4$ given by

$$f_4\left(z\right)={\int }_0^z\left(1+t^3{e}^{t^3}\right)dt=z+\frac{1}{4}{z}^4+\frac{1}{7}{z}^7+\frac{1}{20}{z}^{10}+\frac{1}{78}{z}^{13}+\cdots ,z\in \mathbb{D}.$$

(35)

Proof. 

Using (26)–(28), we have

$$\left|a_2{a}_3-a_4\right|=\frac{1}{8}\left|c_3-\frac{1}{3}{c}_1{c}_2-\frac{1}{8}{c}_1^3\right|.$$

(36)

From (21) and (22), it is noted that

$$\begin{array}{cc}\hfill c_1& =2w_1,\hfill \end{array}$$

(37)

$$\begin{array}{cc}\hfill c_2& =2\left(w_2+w_1^2\right),\hfill \end{array}$$

(38)

$$\begin{array}{cc}\hfill c_3& =2\left(w_3+2w_1{w}_2+w_1^3\right).\hfill \end{array}$$

(39)

Hence, we obtain

$$\left|a_2{a}_3-a_4\right|=\frac{1}{4}\left|w_3+\frac{4}{3}{w}_1{w}_2-\frac{1}{6}{w}_1^3\right|.$$

(40)

Taking $\mu =\frac{4}{3}$ and $\nu =-\frac{1}{6}$, we know $(\mu ,\nu )\in {\mathbb{D}}_2$. Using Lemma 4, we easily obtain

$$\left|a_2{a}_3-a_4\right|\le \frac{1}{4}.$$

Clearly, the bound is sharp with the extremal function given by (35). □

Theorem 4. 

If $f\in {\mathcal{R}}_{\wp },$ then

$$\left|H_{2,2}\left(f\right)\right|=\left|a_2{a}_4-a_3^2\right|\le \frac{1}{9}.$$

The inequality is sharp with the extremal function given by (31).

Proof. 

From (26)–(28), we have

$$H_{2,2}\left(f\right)=-\frac{1}{256}{c}_1^4+\frac{1}{32}{c}_1{c}_3-\frac{1}{36}{c}_2^2.$$

Let $f\in {\mathcal{R}}_{\wp }$ and $f_{\theta }\left(z\right)=e^{-i\theta }f\left(e^{i\theta }z\right),\theta \in \mathbb{R}$. We have $\left|H_{2,2}\left(f_{\theta }\right)\right|=\left|H_{2,2}\left(f\right)\right|$ for all $\theta \in \mathbb{R}$. Hence, when estimating the upper bounds of $\left|H_{2,2}\left(f\right)\right|$, we may assume $a_2$ of f to be real, and thus $c_1:=c\in [0,2]$. Using (8) and (9) to express $c_2$ and $c_3$ in terms of $c_1=c$, we obtain

$$\begin{array}{cc}\hfill \left|H_{2,2}\left(f\right)\right|& =\left|-\frac{7}{2304}{c}^4+\frac{1}{576}{c}^2\left(4-c^2\right)x-\frac{1}{1152}\left(4-c^2\right)\left(c^2+32\right)x^2\right.\hfill \\ \hfill & \left.+\frac{1}{64}c\left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\delta \right|.\hfill \end{array}$$

With the aid of the triangle inequality, replacing $\left|\delta \right|\le 1,\left|x\right|=t\le 1$ and taking $c\in \left[0,2\right]$, we obtain

$$\begin{array}{cc}\hfill \left|H_{2,2}\left(f\right)\right|& \le \frac{7}{2304}{c}^4+\frac{1}{576}{c}^2\left(4-c^2\right)t+\frac{1}{1152}\left(4-c^2\right)\left(c^2+32\right)t^2\hfill \\ \hfill & +\frac{1}{64}c\left(4-c^2\right)\left(1-t^2\right)=:K\left(c,t\right).\hfill \end{array}$$

It is noted that

$$\frac{\partial K}{\partial t}=\frac{1}{576}{c}^2(4-c^2)+\frac{1}{576}\left(4-c^2\right)\left(c^2-18c+32\right)t\ge 0$$

for $t\in \left[0,1\right]$, thus $K\left(c,t\right)\le K\left(c,1\right).$ Putting $t=1$ gives

$$\left|H_{2,2}\left(f\right)\right|\le \frac{7}{2304}{c}^4+\frac{1}{576}{c}^2\left(4-c^2\right)+\frac{1}{1152}\left(4-c^2\right)\left(c^2+32\right)=:\chi \left(c\right).$$

Since $\chi \left(c\right)=\frac{1}{2304}\left(c^4-40c^2+256\right)$ and ${\chi }^{\prime }\left(c\right)\le 0$ on $[0,2]$, we know $\chi$ is decreasing for $c\in [0,2]$ and

$$\left|H_{2,2}\left(f\right)\right|\le \chi \left(0\right)=\frac{1}{9}.$$

The equality is obtained by the extremal function defined by (31). This completes the proof of Theorem 4. □

Theorem 5. 

If $f\in {\mathcal{R}}_{\wp }$ has the form (1), then

$$\left|H_{3,1}\left(f\right)\right|\le \frac{1}{16}.$$

This inequality is sharp with the extremal function $f_4$ given by (35).

Proof. 

From the definition, $H_{3,1}\left(f\right)$ can be written as

$$H_{3,1}\left(f\right)=2a_2{a}_3{a}_4-a_3^3-a_4^2+a_3{a}_5-a_2^2{a}_5.$$

(41)

Let $c_1=c$. By putting (26)–(29) into (41), we obtain

$$\begin{array}{cc}\hfill H_{3,1}\left(f\right)& =\frac{1}{552,960}\left(-423c^6+1344c^4{c}_2+2160c^3{c}_3-3456c^2{c}_2^2-3456c^2{c}_4\right.\hfill \\ \hfill & \left.+5760c{c}_2{c}_3-2560c_2^3+9216c_2{c}_4-8640c_3^2\right).\hfill \end{array}$$

(42)

Let $f\in {\mathcal{R}}_{\wp }$ and $f_{\theta }=e^{-i\theta }f\left(e^{i\theta }z\right),\theta \in \mathbb{R}$. Note that $\left|H_{3,1}\left(f_{\theta }\right)\right|=\left|H_{3,1}\left(f\right)\right|$ for all $\theta \in \mathbb{R}$, we may also assume that $c\in [0,2]$. Suppose that $b=4-c^2.$ Using (8)–(10), we obtain

$$\begin{array}{cc}\hfill H_{3,1}\left(f\right)& =\frac{1}{552,960}\left\{-71c^6+2304b^2{x}^3-320b^3{x}^3+576c^{2}b{x}^2+144c^{4}b{x}^3-612c^{4}b{x}^2\right.\hfill \\ \hfill & +72c^{4}bx+36c^2{b}^2{x}^4-288c^2{b}^2{x}^3-816c^2{b}^2{x}^2-2160b^2{\left(1-{\left|x\right|}^2\right)}^2{\delta }^2\hfill \\ \hfill & +576c^{2}b\left(1-{\left|x\right|}^2\right)\left(1-{\left|\delta \right|}^2\right)\rho -576c^{3}bx\left(1-{\left|x\right|}^2\right)\delta -576c^{2}b\overline{x}\left(1-{\left|x\right|}^2\right){\delta }^2\hfill \\ \hfill & +936c^{3}b\left(1-{\left|x\right|}^2\right)\delta -144c{b}^2{x}^2\left(1-{\left|x\right|}^2\right)\delta -2304b^2{\left|x\right|}^2\left(1-{\left|x\right|}^2\right){\delta }^2\hfill \\ \hfill & \left.-576c{b}^{2}x\left(1-{\left|x\right|}^2\right)\delta +2304b^{2}x\left(1-{\left|x\right|}^2\right)\left(1-{\left|\delta \right|}^2\right)\rho \right\},\hfill \end{array}$$

where $\rho ,x,\delta \in \overline{\mathbb{D}}:=\left\{z:\left|z\right|\le 1\right\}$. Observing that $H_{3,1}\left(f\right)$ can be written as

$$H_{3,1}\left(f\right)=\frac{1}{552,960}\left[d_1\left(c,x\right)+d_2\left(c,x\right)\delta +d_3\left(c,x\right){\delta }^2+\Phi \left(c,x,\delta \right)\rho \right],$$

with

$$\begin{array}{cc}\hfill d_1\left(c,x\right)& =-71c^6+\left(4-c^2\right)\left[\left(4-c^2\right)\left(1024x^3+32c^2{x}^3+36c^2{x}^4-816c^2{x}^2\right)\right.\hfill \\ \hfill & \left.+576c^2{x}^2-612c^4{x}^2+144c^4{x}^3+72c^{4}x\right],\hfill \\ \hfill d_2\left(c,x\right)& =72\left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\left[\left(4-c^2\right)\left(-2c{x}^2\right)-32cx+13c^3\right],\hfill \\ \hfill d_3\left(c,x\right)& =144\left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\left[\left(4-c^2\right)\left(-{\left|x\right|}^2-15\right)-4c^2\overline{x}\right],\hfill \\ \hfill \Phi \left(c,x,\delta \right)& =576\left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\left(1-{\left|\delta \right|}^2\right)\left[c^2+4x\left(4-c^2\right)\right].\hfill \end{array}$$

Taking $\left|x\right|=t,\left|\delta \right|=y$ and utilizing the fact $\left|\rho \right|\le 1,$ we obtain

$$\begin{array}{cc}\hfill \left|H_{3,1}\left(f\right)\right|& \le \frac{1}{552,960}\left[\left|d_1\left(c,x\right)\right|+\left|d_2\left(c,x\right)\right|y+\left|d_3\left(c,x\right)\right|y^2+\left|\Phi \left(c,x,\delta \right)\right|\right].\hfill \\ \hfill & \le \frac{1}{552,960}\left[\Gamma \left(c,t,y\right)\right],\hfill \end{array}$$

(43)

where

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

with

$$\begin{array}{cc}\hfill h_1\left(c,t\right)& =71c^6+\left(4-c^2\right)\left[\left(4-c^2\right)\left(1024t^3+32c^2{t}^3+36c^2{t}^4+816c^2{t}^2\right)\right.\hfill \\ \hfill & \left.+576c^2{t}^2+144c^4{t}^3+612c^4{t}^2+72c^{4}t\right],\hfill \\ \hfill h_2\left(c,t\right)& =72\left(4-c^2\right)\left(1-t^2\right)\left[\left(4-c^2\right)\left(2c{t}^2\right)+13c^3+32ct\right],\hfill \\ \hfill h_3\left(c,t\right)& =144\left(4-c^2\right)\left(1-t^2\right)\left[\left(4-c^2\right)\left(t^2+15\right)+4c^{2}t\right],\hfill \\ \hfill h_4\left(c,t\right)& =576\left(4-c^2\right)\left(1-t^2\right)\left[c^2+4t\left(4-c^2\right)\right].\hfill \end{array}$$

Now, we have to maximize $\Gamma$ in the closed cuboid $\Theta :=\left[0,2\right]\times \left[0,1\right]\times \left[0,1\right].$ It is not hard to see that $\Gamma (0,0,1)=34,560$. Thus, we have ${max}_{(c,t,y)\in \Theta }\left\{\Gamma (c,t,y)\right\}\ge 34,560.$ We aim to prove that the maximum values of $\Gamma$ with $(c,t,y)\in \Theta$ is simply equal to 34,560. For this, we first show that the maximum value of $\Gamma$ is obtained on the face $y=1$ of $\Theta$.

On the face $t=1,$ it reduces to $\Gamma (c,1,y)=r_1\left(c\right)=127c^6-3312c^4+8256c^2+16,384.$ Then,

$$\frac{\partial r_1}{\partial c}=6c\left(127c^4-2208c^2+2752c\right).$$

Putting $\frac{\partial r_1}{\partial c}=0$, we obtain the only critical point ${\hat{c}}_0=\sqrt{\frac{1104-136\sqrt{47}}{127}}\approx 1.1625$ for $c\in (0,2)$. Therefore, $max{r}_1\left(c\right)\approx 21,805.95$ with the maximum value attained on $c={\hat{c}}_0$. Thus, we assume that $t<1$. Furthermore, for the points on the face $c=2,$$\Gamma (2,t,y)\equiv 4544$ for all $(t,y)\in [0,1]\times [0,1]$. Hence, we further assume that $c<2$.

Let $\left(c,t,y\right)\in \left[0,2\right)\times \left[0,1\right)\times \left[0,1\right].$ By differentiating $\Gamma$ partially with respect to y, we obtain

$$\frac{\partial \Gamma }{\partial y}=h_2(c,t)+2\left[h_3(c,t)-h_4(c,t)\right]y.$$

Obviously, we have

$${\left.\frac{\partial H}{\partial y}\right|}_{y=0}=h_2(c,t)\ge 0.$$

Let

$${\left.\frac{\partial H}{\partial y}\right|}_{y=1}=h_2(c,t)+2\left[h_3(c,t)-h_4(c,t)\right]=:{\zeta }_1(c,t).$$

(44)

It is noted that

$${\zeta }_1(c,t)=72(4-c^2)(1-t^2){\zeta }_2(c,t),$$

where

$${\zeta }_2(c,t)=(4-c^2)(2c{t}^2+4t^2+60-64t)+13c^3+32ct+16c^{2}t-16c^2.$$

Clearly, we have

$${\zeta }_2(c,t)\ge (4-c^2)(4t^2+60-64t)+13c^3+16c^{2}t-16c^2=:\eta (c,t).$$

Suppose that $\eta (c,t)={\eta }_0+{\eta }_{1}t+{\eta }_2{t}^2,$ where ${\eta }_0=240-76c^2+13c^3,$${\eta }_1=80c^2-256$ and ${\eta }_2=16-4c^2.$ Taking $\eta$ as a polynomial of degree 2 with respect to t, we know ${\eta }_2>0$ and the symmetric axis $t_0$ is defined as

$$t_0=-\frac{{\eta }_1}{2{\eta }_2}=\frac{2(16-5c^2)}{4-c^2}.$$

Let ${\tilde{c}}_0=\frac{4}{\sqrt{5}}$. For $c\in [{\tilde{c}}_0,2)$, it is observed that $t_0\le 0$. Then, the minimum value of $\eta$ is achieved on $t=0$. We thus have

$$\eta (c,t)\ge \eta (c,0)={\eta }_0\ge 40>0,c\in [{\tilde{c}}_0,2).$$

(45)

Let ${\overline{c}}_0=\frac{2\sqrt{7}}{3}$. It is seen that $t_0\ge 1$ for $c\in [0,{\overline{c}}_0]$. It follows that

$$\eta (c,t)\ge \eta (c,1)={\eta }_0+{\eta }_1+{\eta }_2=13c^3\ge 0,c\in \left[0,{\overline{c}}_0\right].$$

(46)

Assume that $c\in ({\overline{c}}_0,{\tilde{c}}_0)$. Then $t_0\in (0,1)$. Hence, the minimum value of $\eta$ is obtained on $t=t_0$. This leads to

$$\eta (c,t)\ge \eta \left(c,t_0\right)={\eta }_0-\frac{{\eta }_1^2}{4{\eta }_2}=\frac{\iota \left(c\right)}{4-c^2},$$

where

$$\iota \left(c\right)=-13c^5-324c^4+52c^3+2016c^2-3136,c\in ({\overline{c}}_0,{\tilde{c}}_0).$$

It is calculated that $\iota$ achieves its minimum value of about $56.9731$ on $c={\tilde{c}}_0$, thus we know

$$\eta (c,t)>0,c\in ({\overline{c}}_0,{\tilde{c}}_0).$$

(47)

Combining (45)–(47), we have $\eta (c,t)\ge 0$ on $\left[0,2\right)\times [0,1)$, which leads to ${\zeta }_1(c,t)\ge 0$ for all $(c,t)\in [0,2)\times \left[0,1\right)$. Therefore, we have ${\left.\frac{\partial \Gamma }{\partial y}\right|}_{y=1}\ge 0$. As $\frac{\partial \Gamma }{\partial y}$ is a linear continuous function with respect to y, we have

$$\frac{\partial \Gamma }{\partial y}\ge min\left\{{\left.\frac{\partial \Gamma }{\partial y}\right|}_{y=0},{\left.\frac{\partial \Gamma }{\partial y}\right|}_{y=1}\right\}\ge 0,y\in [0,1].$$

Hence, $\Gamma (c,t,y)\le \Gamma (c,t,1)$ for all $(c,t,y)\in [0,2)\times [0,1)\times [0,1]$. Based on the above discussions, it reduces to find the global maximum value of $\Gamma$ on the face $y=1$ of $\Theta$. On the face $y=1,$ we have

$$\begin{array}{cc}\hfill \Gamma (c,t,1)& =71c^6+{\left(4-c^2\right)}^2\left[36(c^2-4c-4)t^4+32(c^2+32)t^3+48(17c^2+3c-42)t^2+2160\right]\hfill \\ \hfill & +(4-c^2)\left[144c(c^3-4c-16)t^3+36c^2(17c^2-26c+16)t^2+72c(c^3+8c+32)t+936c^3\right]\hfill \\ \hfill & =:\Lambda (c,t).\hfill \end{array}$$

By observing that $c^2-4c-4\le 0$ and $c^3-4c-16\le 0$ for $c\in [0,2)$, we have

$$\begin{array}{cc}\hfill \Lambda (c,t)& \le 71c^6+{\left(4-c^2\right)}^2\left[32(c^2+32)t^3+48(17c^2+3c-42)t^2+2160\right]\hfill \\ \hfill & +(4-c^2)\left[36c^2(17c^2-26c+16)t^2+72c(c^3+8c+32)t+936c^3\right]\hfill \\ \hfill & =:Q(c,t).\hfill \end{array}$$

Furthermore, using $17c^2-26c+16\ge 0$, $t^3\le t^2\le t$ leads to

$$\begin{array}{cc}\hfill Q(c,t)& \le 71c^6+{\left(4-c^2\right)}^2\left[32(c^2+32)t^2+48(17c^2+3c-42)t^2+2160\right]\hfill \\ \hfill & +(4-c^2)\left[36c^2(17c^2-26c+16)t+72c(c^3+8c+32)t+936c^3\right]\hfill \\ \hfill & =4(4-c^2)R(c,t)+71c^6+2160{(4-c^2)}^2+936(4-c^2)c^3\hfill \\ \hfill & =:W(c,t),\hfill \end{array}$$

where

$$R(c,t)=4(4-c^2)(53c^2+9c-62)t^2+9c(19c^3-26c^2+32c+64)t.$$

Clearly, if $c\ge 1$, we have $53c^2+9c-62\ge 0$ and $19c^3-26c^2+32c+64\ge 0$, which leads to

$$R(c,t)\le R(c,1),c\in [1,2).$$

Then, we obtain

$$W(c,t)\le 4(4-c^2)R(c,1)+71c^6+2160{(4-c^2)}^2+936(4-c^2)c^3=:{\varrho }_1\left(c\right),c\in [1,2).$$

In virtue of ${\varrho }_1\left(c\right)=235c^6+144c^5-4032c^4-3456c^3+8832c^2+11,520c+18,688$ obtaining its maximum value of about 32,192.46 on $c\approx 1.1053$ for $c\in [1,2)$, we have $\Lambda (c,t)<34,560$ on $[1,2)\times [0,1).$ Suppose that $c\in [0,1)$ and $m\left(c\right)=19c^3-26c^2+32c+64$. It is noted that $m^{\prime }\left(c\right)=54c^2-52c+32\ge 0$ for $c\in [0,1)$. Thus, we have $m\left(c\right)\in [64,89)$. Since $0<4-c^2\le 4$ and $c^2\le c$, it is not hard to see that

$$R(c,t)\le 992(c-1)t^2+801ct=:V(c,t).$$

Let $V(c,t)=v_{1}t+v_2{t}^2$, where $v_1=801c$ and $v_2=992(c-1).$ Obviously, we have $v_2<0$. Considering V as a polynomial of degree 2 with respect to t, we obtain the symmetric axis ${\overline{t}}_0$ defined by

$${\overline{t}}_0=-\frac{v_1}{2v_2}=\frac{801c}{1984(1-c)}.$$

(48)

For $c>{\dot{c}}_0=\frac{1984}{2785}\approx 0.7124$, we have ${\overline{t}}_0>1$. Then, the maximum value of V is attained on $t=1$, which implies that $V(c,t)\le V(c,1)=1793c-992.$ Then,

$$W(c,t)\le 4(4-c^2)V(c,1)+71c^6+2160{(4-c^2)}^2+936(4-c^2)c^3=:{\varrho }_2\left(c\right),c\in \left[{\dot{c}}_0,1\right).$$

It is calculated that

$${\varrho }_2\left(c\right)=71c^6-936c^5+2160c^4-3428c^3-13,312c^2+28,688c+18,688,c\in \left[{\dot{c}}_0,1\right),$$

which obtains its maximum value of about 32,127.89 on $c\approx 0.8966$. Hence, we obtain

$$\Gamma (c,t)<34560,(c,t)\in [{\dot{c}}_0,1)\times [0,1).$$

For $c\in [0,{\dot{c}}_0)$, we have $t_0\in [0,1)$. Then, we obtain

$$V(c,t)\le -\frac{v_1^2}{4v_2}=\frac{{801}^2}{3968}·\frac{c^2}{1-c}\le \frac{162c^2}{1-c}\le 162c^2,$$

which yields to

$$W(c,t)\le 648(4-c^2)c^2+71c^6+2160{(4-c^2)}^2+936(4-c^2)c^3=:{\varrho }_3\left(c\right),c\in [0,{\dot{c}}_0).$$

In light of

$${\varrho }_3\left(c\right)=71c^6-936c^5+1512c^4+3744c^3-14,688c^2+34,560,c\in [0,{\dot{c}}_0),$$

it is not hard to see that ${\varrho }_3$ achieves its maximum value 34,560 on $c=0$. Therefore, we conclude that

$$\Lambda (c,t)\le 34,560,(c,t)\in [0,2)\times [0,1).$$

From the above cases, we obtain

$$\Gamma \left(c,t,y\right)\le 34,560\mathrm{on}\left[0,2\right]\times \left[0,1\right]\times \left[0,1\right].$$

Using (43), it follows that

$$\left|H_{3,1}\left(f\right)\right|\le \frac{1}{552,960}\left[\Gamma \left(c,t,y\right)\right]\le \frac{1}{16}=0.0625.$$

The proof of Theorem 5 is thus completed. □