Certain Sharp Coefficient Results on a Subclass of Starlike Functions Defined by the Quotient of Analytic Functions

Abstract

In the present paper, we consider a subclass of starlike functions ${\mathcal{G}}_{3/2}$ defined by the ratio of analytic representations of convex and starlike functions. The main aim is to determine the bounds of Fekete–Szegö-type inequalities and Hankel determinants for functions in this class. It is proved that $max\left\{\left|{\mathcal{H}}_{3,1}\left(f\right)\right|:f\in {\mathcal{G}}_{3/2}\right\}$ is equal to $\frac{1}{81}$. The bounds for $f\in {\mathcal{G}}_{3/2}$ are sharp.

1. Introduction and Definitions

Let $\mathcal{A}$ denote the normalized analytic functions defined in the open unit disc $\mathbb{D}=\left\{z\in \mathbb{C}:\left|z\right|<1\right\}$ with the series expansion of the form

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

(1)

Let $\mathcal{S}\subseteq \mathcal{A}$ be the class of univalent functions in $\mathbb{D}$. Suppose that $\mathcal{P}$ is the Carathéodory class (see [1]) of all functions p that are analytic in $\mathbb{D}$ with $\Re \left(p\left(z\right)\right)>0$ and normalized by

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

(2)

A function $f\in \mathcal{A}$ is said to be in the class ${\mathcal{S}}^{*}$ of starlike functions, if, and only if,

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

(3)

For the class $\mathcal{C}$ of convex functions, the necessary and sufficient condition is

$$\Re \left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)>0,z\in \mathbb{D}.$$

(4)

It is known that if $f\in \mathcal{C},$ then $z{f}^{\prime }\left(z\right)\in {\mathcal{S}}^{*}$—see [2].

Nonlinear functional analysis is an active topic and has its applications in the natural sciences, economics, and numerical analysis—see, for example, [3,4,5,6,7]. In [8], Silverman introduced and studied a new subclass of $\mathcal{A}$ using the quotient of $\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)/\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}$. For $b\in (0,1]$, the class ${\mathcal{G}}_b$ was defined by

$${\mathcal{G}}_b=\left\{f\in \mathcal{A}:\left|\frac{1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}-1\right|<b,z\in \mathbb{D}\right\}.$$

(5)

In [9], Obradović and Tuneski found that, for $f\in \mathcal{A}$, if

$$\Re \left(\frac{1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\right)<\frac{3}{2},z\in \mathbb{D},$$

(6)

then $f\in {\mathcal{S}}^{*}$.

We say $g_1$ is subordinate to $g_2$ in $\mathbb{D}$, written as $g_1\prec g_2$, if $g_1\left(z\right)=g_2\left(\omega \left(z\right)\right)$ for some analytic function $\omega \left(z\right)$ with $\omega \left(0\right)=0$ and $\left|\omega \left(z\right)\right|<1$. If $g_2$ in $\mathbb{D}$ is univalent, then $g_1\prec g_2$ is equivalent to

$$g_1\left(0\right)=g_2\left(0\right)and{g}_1\left(\mathbb{D}\right)\subset g_2\left(\mathbb{D}\right).$$

(7)

Using subordination, we can also write

$${\mathcal{G}}_b=\left\{f\in \mathcal{A}:\frac{1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\prec 1+bz,z\in \mathbb{D}\right\}.$$

(8)

In [10], the upper bounds on the initial coefficients and Hankel determinants for $f\in {\mathcal{G}}_b$ were derived. Motivated by the above results, Rǎducanu [11] investigated the class ${\mathcal{G}}_{3/2}$ defined by

$${\mathcal{G}}_{3/2}:=\left\{f\in \mathcal{A}:\Re \left(\frac{1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\right)<\frac{3}{2},z\in \mathbb{D}\right\}.$$

(9)

Clearly, ${\mathcal{G}}_{3/2}$ is a subclass of starlike functions. With the additional restriction of $f^{″}\left(0\right)=p$ for $p\in [0,2]$, Rǎducanu obtained the upper bounds of some initial coefficients and the difference of moduli of successive coefficients $\left|a_3-a_2\right|$ and $\left|a_4-a_3\right|$ for $f\in {\mathcal{G}}_{3/2}$. For other investigations on the analytic functions associated with the ratio of analytic representations of convex and starlike functions, we refer to [12,13,14].

The Hankel determinant ${\mathcal{H}}_{q,n}\left(f\right)$ with $q,n\in \mathbb{N}$ and $a_1=1$ for a function $f\in \mathcal{S}$ of the series form (1) was given by Pommerenke [15,16] as

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

(10)

This is widely used in the theory of Markov processes, of non-stationary signals, in the Hamburger moment problem, and many other issues in both pure mathematics and technical applications—see, for example, [17,18]. In recent years, many results on the upper bounds of Hankel determinant for various subclasses of univalent functions were obtained—interested readers may refer to [19,20,21,22,23,24,25] and the references.

In the present paper, we aim to investigate coefficient problems related to Fekete–Szegö-type functionals and Hankel determinants for the functions $f\in {\mathcal{G}}_{3/2}$.

2. A Set of Lemmas

To prove our main results, we require the following lemmas.

Lemma 1.

(see [26]). Let $p\in \mathcal{P}$ be the form of (2). Then,

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

(11)

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

(12)

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

(13)

for some $x,\delta ,\rho \in \overline{\mathbb{D}}:=\left\{z\in \mathbb{C}:\left|z\right|\le 1\right\}$.

Lemma 2.

(see [27]). Let $\mu \in \mathbb{C}$. If $p\in \mathcal{P}$ is represented as (2), then

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

(14)

Lemma 3.

For all $(c,t)\in [0,2)\times \left[\frac{1}{3},1\right]$, we have

$$17{\left(2+c\right)}^2\left(2-c\right)t^2+2\left(2+c\right)\left(23c^2+116c-162\right)t+29c^3-290c^2+512\ge 0.$$

(15)

Proof. 

Let

$$F_1(c,t):=17{\left(2+c\right)}^2\left(2-c\right)t^2+2\left(2+c\right)\left(23c^2+116c-162\right)t+29c^3-290c^2+512.$$

(16)

It is noted that

$$F_1(c,t)=\left(2+c\right)F_2(c,t)+29c^3-290c^2+512,$$

(17)

where

$$F_2(c,t)=17\left(4-c^2\right)t^2+2\left(23c^2+116c-162\right)t.$$

(18)

Let ${\widehat{c}}_0=\frac{\sqrt{7090}-58}{23}\approx 1.1392$ be the only positive root of the equation $23c^2+116c-162=0$. If $c\ge {\widehat{c}}_0$, we have $23c^2+116c-162\ge 0$ and thus $F_2(c,t)\ge F_2\left(c,\frac{1}{3}\right)$. Then

$$F_1(c,t)\ge \left(2+c\right)F_2\left(c,\frac{1}{3}\right)+29c^3-290c^2+512=:s_1\left(c\right),c\in \left[{\widehat{c}}_0,2\right).$$

(19)

In light of

$$s_1\left(c\right)=\frac{2}{9}\left(191c^3-836c^2+244c+1400\right)$$

(20)

and $s_1^{\prime }\left(c\right)\le 0$ on $\left[{\widehat{c}}_0,2\right)$, we get that $s_1\left(c\right)\ge s_1\left(2\right)=16$. Hence, we obtain that $F_1(c,t)\ge 0$ on $\left[{\widehat{c}}_0,2\right]\times [0,1]$.

For $c<{\widehat{c}}_0$, we note that the symmetric axis of $F_2$ is defined by

$$t_0=\frac{-23c^2-116c+162}{17\left(4-c^2\right)}>0.$$

(21)

Let ${\widehat{c}}_1=\frac{\sqrt{982}-29}{3}\approx 0.7790$ be the only positive root of the equation $6c^2+116c-94=0$. If $c\le {\widehat{c}}_1$, we have $t_0\ge 1$. Then $F_2(c,t)\ge F_2(c,1)$, which leads to

$$F_1(c,t)\ge \left(2+c\right)F_1(c,1)+29c^3-290c^2+512=2c\left(104+29c^2\right)\ge 0.$$

(22)

If $c\in \left({\widehat{c}}_1,{\widehat{c}}_0\right)$, it is simple to observe that

$$\begin{array}{cc}\hfill F_1(c,t)& \ge 17{\left(2+c\right)}^2\left(2-c\right)·\frac{1}{9}+2\left(2+c\right)\left(23c^2+116c-162\right)·1+29c^3-290c^2+512\hfill \\ \hfill & =\frac{2}{9}\left(329c^3+136c^2+664c-544\right)>0\hfill \end{array}$$

on $\left({\widehat{c}}_1,{\widehat{c}}_0\right)$. Hence, $F_1(c,t)\ge 0$ for all $(c,t)\in \left[0,{\widehat{c}}_0\right]\times \left[\frac{1}{3},1\right]$. Therefore, we deduce that $F_1(c,t)\ge 0$ on $[0,2)\times \left[\frac{1}{3},1\right]$. The assertion in Lemma 3 thus follows. □

Lemma 4.

For all $(c,t)\in [0,2)\times \left[\frac{1}{3},1\right]$, we have

$$17\left(1+c\right)\left(4-c^2\right)t^2+\left(46c^3+162c^2+140c-324\right)t+29c^3-145c^2+256\ge 0.$$

(23)

Proof. 

Define

$$F_3(c,t):=17\left(1+c\right)\left(4-c^2\right)t^2+\left(46c^3+162c^2+140c-324\right)t+29c^3-145c^2+256.$$

(24)

It is observed that

$$\begin{array}{cc}\hfill F_3(c,t)-F_1(c,t)& =-17\left(4-c^2\right)t^2+162\left(2-c^2\right)t+145c^2-256\hfill \\ \hfill & \ge -17\left(4-c^2\right)t+162\left(2-c^2\right)t+145c^2-256\hfill \\ \hfill & =(256-145c^2)\left(1-t\right).\hfill \end{array}$$

For $c^2\le \frac{256}{145}$, it is clear that $F_3(c,t)\ge F_1(c,t)\ge 0$. If $c^2>\frac{256}{145}$, then $46c^3+162c^2+140c-324>0$ and it follows that $F_3(c,t)\ge F_3\left(c,\frac{1}{3}\right)>0$. Hence, we have $F_3(c,t)\ge 0$ on $\left[0,2\right)\times \left[\frac{1}{3},1\right]$, which leads to the assertion in Lemma 4. □

3. Main Results

For the class of analytic univalent functions, Fekete Szegö studied the maximum value of $\left|a_3-\zeta a_2\right|$ when $\zeta$ is real. After that, the upper bound for the functional $\left|a_3-\zeta a_2\right|$ were investigated extensively for various subclasses of univalent functions. For some recent works, see, for example, [28,29,30,31]. We begin by finding the Fekete–Szegö-type inequalities for $f\in {\mathcal{G}}_{3/2}.$

Theorem 1.

Let $\gamma \in \mathbb{C}.$ If $f\in {\mathcal{G}}_{3/2}$ is of the form (1), then

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

(25)

The inequality is sharp.

Proof. 

Let $f\in {\mathcal{G}}_{3/2}$. Then there exists a function $p\in \mathcal{P}$ such that

$$3\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-2\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)=p\left(z\right)\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)},z\in \mathbb{D}.$$

(26)

Assuming that

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

(27)

Using (1), it follows that

$$\begin{array}{cc}\hfill \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}& =1+a_{2}z+\left(2a_3-a_2^2\right)z^2+\left(a_2^3-3a_2{a}_3+3a_4\right)z^3\hfill \\ \hfill & +\left(4a_5-a_2^4+4a_2^2{a}_3-4a_2{a}_4-2a_3^2\right)z^4+\cdots \hfill \end{array}$$

(28)

and

$$\begin{array}{cc}\hfill 1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}& =1+2a_{2}z+\left(6a_3-4a_2^2\right)z^2+\left(8a_2^3-18a_2{a}_3+12a_4\right)z^3\hfill \\ \hfill & +\left(20a_5-16a_2^4+48a_2^2{a}_3-32a_2{a}_4-18a_3^2\right)z^4+\cdots .\hfill \end{array}$$

(29)

Combining (26)–(29), we obtain

$$\begin{array}{cc}\hfill a_2& =-\frac{1}{2}{c}_1,\hfill \end{array}$$

(30)

$$\begin{array}{cc}\hfill a_3& =-\frac{1}{8}\left(c_2-2c_1^2\right),\hfill \end{array}$$

(31)

$$\begin{array}{cc}\hfill a_4& =-\frac{1}{144}\left(8c_3-21c_1{c}_2+18c_1^3\right),\hfill \end{array}$$

(32)

$$\begin{array}{cc}\hfill a_5& =-\frac{1}{1152}\left(36c_4-27c_2^2-80c_1{c}_3+138c_1^2{c}_2-72c_1^4\right).\hfill \end{array}$$

(33)

Employing (30) and (31), we have

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

(34)

An application of Lemma 2 leads to the assertion in Theorem 1. The equality can be attained by the functions $f_1$ and $f_2$ given, respectively, by the equations

$$3\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-2\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)=\frac{1-z}{1+z}·\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)},z\in \mathbb{D},$$

(35)

$$3\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-2\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)=\frac{1-z^2}{1+z^2}·\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)},z\in \mathbb{D}.$$

(36)

Using series expansions, $f_1$ and $f_2$ are in the form of

$$f_1\left(z\right)=z+z^2+\frac{3}{4}{z}^3+\frac{19}{36}{z}^4+\frac{101}{288}{z}^5+\cdots ,z\in \mathbb{D},$$

(37)

$$f_2\left(z\right)=z+\frac{1}{4}{z}^3+\frac{1}{32}{z}^5+\cdots ,z\in \mathbb{D}.$$

(38)

□

The second Hankel determinant is known as the functional ${\mathcal{H}}_{2,2}\left(f\right)=a_2{a}_4-a_3^2$. In recent years, many results on the upper bounds of second Hankel determinant for certain subclass of univalent functions were obtained, see [32,33,34,35]. Now, we provide the sharp bound of ${\mathcal{H}}_{2,2}\left(f\right)$ for $f\in {\mathcal{G}}_{3/2}$.

Theorem 2.

If $f\in$ then

$$\left|H_{2,2}\left(f\right)\right|=\left|a_2{a}_4-a_3^2\right|\le \frac{1}{16}.$$

(39)

The bound is sharp.

Proof. 

From (30) to (32), we have

$${\mathcal{H}}_{2,2}\left(f\right)=a_2{a}_4-a_3^2=-\frac{1}{64}{c}_2^2-\frac{1}{96}{c}_1^2{c}_2+\frac{1}{36}{c}_1{c}_3.$$

(40)

Let $f\in {\mathcal{G}}_{3/2}$ and $f_{\theta }\left(z\right)=e^{-i\theta }f\left(e^{i\theta }z\right)$, $\theta \in \mathbb{R}$. We know $f_{\theta }\left(z\right)\in {\mathcal{G}}_{3/2}$ for all $\theta \in \mathbb{R}$ and $\left|{\mathcal{H}}_{2,2}\left(f_{\theta }\right)\right|=\left|{\mathcal{H}}_{2,2}\left(f\right)\right|$. Since the class ${\mathcal{G}}_{3/2}$ and the functional $\left|{\mathcal{H}}_{2,2}\left(f\right)\right|$ are rotation-invariant, we can assume that $a_2$ of f is real and, thus, $c_1:=c\in \left[0,2\right]$. Using (11) and (12) to express $c_2$ and $c_3$, we obtain

$$\begin{array}{cc}\hfill {\mathcal{H}}_{2,2}\left(f\right)& =-\frac{5}{2304}{c}^4+\frac{1}{1152}{c}^2\left(4-c^2\right)x-\frac{1}{2304}\left(36+7c^2\right)\left(4-c^2\right)x^2\hfill \\ \hfill & +\frac{1}{72}c\left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\delta .\hfill \end{array}$$

Let $\left|x\right|:=t$. From $\left|\delta \right|\le 1$, we get

$$\begin{array}{cc}\hfill \left|{\mathcal{H}}_{2,2}\left(f\right)\right|& \le \frac{5}{2304}{c}^4+\frac{1}{1152}{c}^2\left(4-c^2\right)t+\frac{1}{2304}\left(36+7c^2\right)\left(4-c^2\right)t^2\hfill \\ \hfill & +\frac{1}{72}c\left(4-c^2\right)\left(1-t^2\right)=:G\left(c,t\right).\hfill \end{array}$$

As $7c^2-32c+36\ge 0$ on $[0,2]$, it is noted that

$$\frac{\partial G}{\partial t}=\frac{1}{1152}\left(4-c^2\right)\left[\left(7c^2-32c+36\right)t+c^2\right]\ge 0.$$

(41)

Thus we have $G\left(c,t\right)\le G\left(c,1\right).$ Putting $t=1$ leads to

$$\left|{\mathcal{H}}_{2,2}\left(f\right)\right|\le \frac{1}{16}-\frac{1}{576}{c}^4\le \frac{1}{16}.$$

(42)

The bound is achieved by the function $f_2$ given by Equation (36). □

It is not easy to find the sharp bound of the third Hankel determinant for a certain subclass of univalent functions. For instance, Kowalczyk et al. [36] proved that $\left|{\mathcal{H}}_{3,1}\left(f\right)\right|\le \frac{4}{9}$ for $f\in {\mathcal{S}}^{*}$. Before it was solved, there are many works investigated this problem—for details, we refer to [37,38]. Now, we give the sharp bound of ${\mathcal{H}}_{3,1}\left(f\right)$ for $f\in {\mathcal{G}}_{3/2}.$

Theorem 3.

Suppose that $f\in {\mathcal{G}}_{3/2}$. Then,

$$\left|{\mathcal{H}}_{3,1}\left(f\right)\right|\le \frac{1}{81}.$$

(43)

The result is the best possible.

Proof. 

From the definition, we know

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

(44)

Let $f\in {\mathcal{G}}_{3/2}$ and $f_{\theta }\left(z\right)=e^{-i\theta }f\left(e^{i\theta }z\right)$, $\theta \in \mathbb{R}$. It is observed that $f_{\theta }\left(z\right)\in {\mathcal{G}}_{3/2}$ for all $\theta \in \mathbb{R}$ and $\left|{\mathcal{H}}_{3,1}\left(f_{\theta }\right)\right|=\left|{\mathcal{H}}_{3,1}\left(f\right)\right|$. As the class ${\mathcal{G}}_{3/2}$ and the functional $\left|{\mathcal{H}}_{3,1}\left(f\right)\right|$ are invariant under the rotations, we may assume that $a_2$ of f is real and, thus, $c_1:=c\in \left[0,2\right]$. Substituting (30)–(33) into (44), we have

$${\mathcal{H}}_{3,1}\left(f\right)=\frac{1}{82944}\left(-81c_2^3+48c{c}_2{c}_3+324c_2{c}_4-256c_3^2+18c^2{c}_2^2\right).$$

(45)

Let $b=4-c^2$. By applying Lemma 1 and inserting the formulae of $c_2$, $c_3$ and $c_4$ into (45), we obtain

$$\begin{array}{cc}\hfill {\mathcal{H}}_{3,1}\left(f\right)& =\frac{1}{663552}\left\{37c^6-81x^3{b}^3+648x^3{b}^2+648c^2{x}^{2}b-278c^4{x}^{2}b\right.\hfill \\ \hfill & +162c^4{x}^{3}b-137c^2{x}^2{b}^2-22c^2{x}^3{b}^2+34c^2{x}^4{b}^2+109c^{4}xb\hfill \\ \hfill & +232c^{3}b\left(1-{\left|x\right|}^2\right)\delta -648{\left|x\right|}^2{b}^2\left(1-{\left|x\right|}^2\right){\delta }^2-512b^2{\left(1-{\left|x\right|}^2\right)}^2{\delta }^2\hfill \\ \hfill & -136c{x}^2{b}^2(1-{\left|x\right|}^2)\delta -648c^2\overline{x}b(1-{\left|x\right|}^2){\delta }^2-280cx{b}^2\left(1-{\left|x\right|}^2\right)\delta \hfill \\ \hfill & -648c^{3}xb(1-{\left|x\right|}^2)\delta +648c^{2}b\left(1-{\left|x\right|}^2\right)\left(1-{\left|\delta \right|}^2\right)\rho \hfill \\ \hfill & \left.+648x{b}^2\left(1-{\left|x\right|}^2\right)\left(1-{\left|\delta \right|}^2\right)\rho \right\},\hfill \end{array}$$

where $x,\delta ,\rho \in \overline{\mathbb{D}}$. By suitable rearrangements, it is observed that ${\mathcal{H}}_{3,1}\left(f\right)$ can be written as

$$\begin{array}{c}\hfill {\mathcal{H}}_{3,1}\left(f\right)=\frac{1}{663552}\left[u_1\left(c,x\right)+u_2\left(c,x\right)\delta +u_3\left(c,x\right){\delta }^2+\Phi \left(c,x,\delta \right)\rho \right],\end{array}$$

(46)

where

$$\begin{array}{cc}\hfill u_1\left(c,x\right)& =37c^6+\left(4-c^2\right)x\left[34c^2\left(4-c^2\right)x^3+\left(103c^4-88c^2+1296\right)x^2\right.\hfill \\ \hfill & \left.+c^2\left(100-141c^2\right)x+109c^4\right],\hfill \\ \hfill u_2\left(c,x\right)& =8(4-c^2)\left(1-{\left|x\right|}^2\right)c\left[-17\left(4-c^2\right)x^2-\left(46c^2+140\right)x+29c^2\right],\hfill \\ \hfill u_3\left(c,x\right)& =8(4-c^2)\left(1-{\left|x\right|}^2\right)\left[\left(4-c^2\right)\left(-17{\left|x\right|}^2-64\right)-81c^2\overline{x}\right],\hfill \\ \hfill \Phi \left(c,x,\delta \right)& =648(4-c^2)\left(1-{\left|x\right|}^2\right)\left(1-{\left|\delta \right|}^2\right)\left[\left(4-c^2\right)x+c^2\right].\hfill \end{array}$$

Let $\left|x\right|=:t$ and $\left|\delta \right|=:y.$ By observing $\left|\rho \right|\le 1,$ it follows that

$$\begin{array}{cc}\hfill \left|{\mathcal{H}}_{3,1}\left(f\right)\right|& \le \frac{1}{663552}\left[\left|u_1\left(c,x\right)\right|+\left|u_2\left(c,x\right)\right|y+\left|u_3\left(c,x\right)\right|y^2+\left|\Phi \left(c,x,\delta \right)\right|\right]\hfill \\ \hfill & \le \frac{1}{663552}\left[\mathrm{\Gamma }\left(c,t,y\right)\right],\hfill \end{array}$$

(47)

where

$$\mathrm{\Gamma }\left(c,t,y\right)={\sigma }_1\left(c,t\right)+{\sigma }_2\left(c,t\right)y+{\sigma }_3\left(c,t\right)y^2+{\sigma }_4\left(c,t\right)\left(1-y^2\right),$$

(48)

with

$$\begin{array}{cc}\hfill {\sigma }_1\left(c,t\right)& =37c^6+(4-c^2)\left[34c^2\left(4-c^2\right)t^4+\left(103c^4-88c^2+1296\right)t^3\right.\hfill \\ \hfill & \left.+c^2\left(100+141c^2\right)t^2+109c^{4}t\right],\hfill \\ \hfill {\sigma }_2\left(c,t\right)& =8(4-c^2)\left(1-t^2\right)c\left[17\left(4-c^2\right)t^2+\left(46c^2+140\right)t+29c^2\right],\hfill \\ \hfill {\sigma }_3\left(c,t\right)& =8(4-c^2)\left(1-t^2\right)\left[\left(4-c^2\right)\left(17t^2+64\right)+81c^{2}t\right],\hfill \\ \hfill {\sigma }_4\left(c,t\right)& =648(4-c^2)\left(1-t^2\right)\left[\left(4-c^2\right)t+c^2\right].\hfill \end{array}$$

The inequality (47) comes from the fact that $\left|u_j(c,x)\right|\le {\sigma }_j(c,t)$ for $j=1,2,3$ and $\left|\Phi \left(c,x,\delta \right)\right|\le {\sigma }_4(c,t)\left(1-{\left|\delta \right|}^2\right)$. Here, $\left|u_1(c,x)\right|\le {\sigma }_1(c,t)$ follows from $4-c^2\ge 0$ and $103c^4-88c^2+1296\ge 0$ on $[0,2]$.

Now the problem reduces to find the maximum value of $\mathrm{\Gamma }$ in the closed cuboid $\mathrm{\Omega }:=[0,2]\times [0,1]\times [0,1].$ By noting that $\mathrm{\Gamma }\left(0,0,1\right)=8192$, we have ${\max }_{(c,t,y)\in \mathrm{\Omega }}\mathrm{\Gamma }(c,t,y)\ge 8192.$ In the following, we show that ${max}_{(c,t,y)\in \mathrm{\Omega }}\mathrm{\Gamma }(c,t,y)=8192$.

On the face, $c=2,$ $\mathrm{\Gamma }(2,t,y)\equiv 2368.$ For $t=1,$ we have

$$\mathrm{\Gamma }(c,1,y)=-282c^6+1128c^4-704c^2+5184=:r_1\left(c\right).$$

(49)

Since $r_1^{\prime }\left(c\right)=0$ has two positive roots ${\tilde{c}}_1=\sqrt{\frac{4}{3}-\frac{20\sqrt{141}}{141}}\approx 0.6007$ and ${\tilde{c}}_2=\sqrt{\frac{4}{3}+\frac{20\sqrt{141}}{141}}\approx 1.5185$, $r_1$ attains its maximum value about 6100.85 on $c={\tilde{c}}_2$ for $c\in [0,2]$. Thus, we may only consider the cases of $c<2$ and $t<1$.

First we suppose that $\left(c,t,y\right)\in \left[0,2\right)\times \left[0,1\right)\times \left(0,1\right).$ By differentiating partially $\mathrm{\Gamma }$ with respect to y, we obtain

$$\frac{\partial \mathrm{\Gamma }}{\partial y}={\sigma }_2(c,t)+2\left[{\sigma }_3(c,t)-{\sigma }_4(c,t)\right]y.$$

(50)

Taking $\frac{\partial \mathrm{\Gamma }}{\partial y}=0,$ we obtain

$$y=\frac{17c\left(4-c^2\right)t^2+\left(46c^3+140c\right)t+29c^3}{2(1-t)\left[17\left(4-c^2\right)t+145c^2-256\right]}=:y_0.$$

(51)

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

$$17{\left(2+c\right)}^2\left(2-c\right)t^2+2\left(2+c\right)\left(23c^2+116c-162\right)t+29c^3-290c^2+512<0$$

(52)

and

$$c^2>\frac{4\left(64-17t\right)}{145-17t}.$$

(53)

Now we have to obtain the solutions that satisfy both inequalities (52) and (53) for the existence of critical points with $y\in (0,1)$. From Lemma 3, we know (52) is impossible to hold for $t\in \left[\frac{1}{3},1\right]$.

Let $\left(\tilde{c},\tilde{t},\tilde{y}\right)$ be a critical point of $\mathrm{\Gamma }$ lies in $[0,2)\times [0,1)\times (0,1)$. From the above discussions, we know $\tilde{t}<\frac{1}{3}$. Assume that $g\left(t\right)=\frac{4\left(64-17t\right)}{145-17t}.$ As $g^{\prime }\left(t\right)<0$ in $\left[0,1\right],$ $g\left(t\right)$ is decreasing over $\left[0,1\right]$. Hence, ${\tilde{c}}^2\ge \frac{350}{209}.$ Under the conditions that $t\in \left[0,\frac{1}{3}\right)$, $c\in \left(\sqrt{\frac{350}{209}},2\right)$ and $y\in (0,1)$, we find that

$${\sigma }_1\left(c,t\right)\le {\sigma }_1\left(c,\frac{1}{3}\right)=:{\tau }_1\left(c\right)$$

(54)

and

$${\sigma }_j\left(c,x\right)\le \frac{9}{8}{\sigma }_j\left(c,\frac{1}{3}\right)=:{\tau }_j\left(c\right),j=2,3,4.$$

(55)

Therefore, we have

$$\mathrm{\Gamma }\left(c,t,y\right)\le {\tau }_1\left(c\right)+{\tau }_2\left(c\right)y+{\tau }_3\left(c\right)y^2+{\tau }_4\left(c\right)\left(1-y^2\right)=:\mathrm{\Xi }(c,y).$$

(56)

Since $\frac{\partial \mathrm{\Xi }}{\partial y}={\tau }_2\left(c\right)+2\left[{\tau }_3\left(c\right)-{\tau }_4\left(c\right)\right]y$, we have

$$\begin{array}{c}\hfill \frac{\partial \mathrm{\Xi }}{\partial y}{|}_{y=1}={\tau }_2\left(c\right)+2\left[{\tau }_3\left(c\right)-{\tau }_4\left(c\right)\right]=\frac{16}{9}(4-c^2)r_2\left(c\right),\end{array}$$

where $r_2\left(c\right)=191c^3-836c^2+244c+1400.$ It is easy to get that $r_2\left(c\right)\ge 0$ for $c\in [0,2]$. Thus, we have ${\frac{\partial \mathrm{\Xi }}{\partial y}|}_{y=1}\ge 0$. As ${\frac{\partial \mathrm{\Xi }}{\partial y}|}_{y=0}={\tau }_2\left(c\right)\ge 0$ and $\frac{\partial \mathrm{\Xi }}{\partial y}$ is a continuous linear function with respect to y, we deduce that

$$\frac{\partial \mathrm{\Xi }}{\partial y}\ge min\left\{\frac{\partial \mathrm{\Xi }}{\partial y}{|}_{y=0},\frac{\partial \mathrm{\Xi }}{\partial y}{|}_{y=1}\right\}\ge 0,y\in (0,1).$$

(57)

Thus,

$$\mathrm{\Xi }(c,y)\le \mathrm{\Xi }(c,1)={\tau }_1\left(c\right)+{\tau }_2\left(c\right)+{\tau }_3\left(c\right)=:r_3\left(c\right).$$

(58)

By numerical calculation, we know $r_3$ has a maximum value about $7391.58$ on $c=\sqrt{\frac{350}{209}}\approx 1.2941$ for $c\in \left(\sqrt{\frac{350}{209}},2\right)$. Therefore, we have $\mathrm{\Gamma }(c,t,y)<8192$ on $(c,t,y)\in \left(\sqrt{\frac{350}{209}},2\right)\times \left[0,\frac{1}{3}\right)\times (0,1)$. Hence, $\mathrm{\Gamma }\left(\tilde{c},\tilde{t},\tilde{y}\right)<8192$. Thus, any critical point which lies in $\mathrm{\Omega }$ with $y\in (0,1)$, has a maximum value less than $\mathrm{\Gamma }(0,0,1)$. Then it is sufficient to consider the face $y=0$ and $y=1$ of $\Omega$ to get the maximum value of $\mathrm{\Gamma }$.

Since

$$\mathrm{\Gamma }(c,t,0)={\sigma }_1(c,t)+{\sigma }_4(c,t)$$

(59)

and

$$\mathrm{\Gamma }(c,t,1)={\sigma }_1(c,t)+{\sigma }_2(c,t)+{\sigma }_3(c,t),$$

(60)

we note that

$$\begin{array}{cc}\hfill \mathrm{\Gamma }(c,t,1)-\mathrm{\Gamma }(c,t,0)& ={\sigma }_2(c,t)+{\sigma }_3(c,t)-{\sigma }_4(c,t)\hfill \\ \hfill & =:8\left(4-c^2\right)\left(1-t^2\right)\mathrm{\Theta }(c,t),\hfill \end{array}$$

where

$$\mathrm{\Theta }(c,t)=17\left(1+c\right)\left(4-c^2\right)t^2+\left(46c^3+162c^2+140c-324\right)t+29c^3-145c^2+256.$$

(61)

From Lemma 4, we know $\mathrm{\Theta }(c,t)\ge 0$ on $\left[0,2\right)\times \left[\frac{1}{3},1\right]$; thus, we have

$$\mathrm{\Gamma }(c,t,0)\le \mathrm{\Gamma }(c,t,1),(c,t)\in \left[0,2\right)\times \left[\frac{1}{3},1\right).$$

(62)

If $t<\frac{1}{3}$, from (54) and (55), we obtain

$$\mathrm{\Gamma }(c,t,0)\le {\tau }_1\left(c\right)+{\tau }_4\left(c\right)=:r_4\left(c\right).$$

(63)

Since

$$r_4\left(c\right)=\frac{2}{81}\left(-745c^6-8908c^4+34592c^2+147744\right)$$

(64)

has a maximum value about 4376.24 attained on $c\approx 1.2707$ and $\mathrm{\Gamma }(0,0,1)=8192$, we have

$$\mathrm{\Gamma }(c,t,0)<\mathrm{\Gamma }(0,0,1),(c,t)\in \left[0,2\right)\times \left[0,\frac{1}{3}\right).$$

(65)

Combining (62) and (65), it remains to find the maximum value of $\mathrm{\Gamma }$ on the face of $y=1$.

On the face $y=1,$ it is noted that

$$\begin{array}{cc}\hfill \mathrm{\Gamma }(c,t,1)& =37c^6+\left(4-c^2\right)\left({\lambda }_4{t}^4+{\lambda }_3{t}^3+{\lambda }_2{t}^2+{\lambda }_{1}t+{\lambda }_0\right)\hfill \\ \hfill & =:Q(c,t),\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill {\lambda }_4& =34(4-c^2)\left(c^2-4c-4\right),\hfill \\ \hfill {\lambda }_3& =103c^4-368c^3-736c^2-1120c+1296,\hfill \\ \hfill {\lambda }_2& =141c^4-368c^3+476c^2+544c-1504,\hfill \\ \hfill {\lambda }_1& =109c^4+368c^3+648c^2+1120c,\hfill \\ \hfill {\lambda }_0& =232c^3-512c^2+2048.\hfill \end{array}$$

In virtue of $c^2-4c-4\le 0$ on $[0,2]$, we have ${\lambda }_4\le 0$ and thus

$$Q(c,t)\le 37c^6+\left(4-c^2\right)\left({\lambda }_3{t}^3+{\lambda }_2{t}^2+{\lambda }_{1}t+{\lambda }_0\right)=:W(c,t).$$

(66)

Now we need to find the maximum value of W on $[0,2]\times [0,1]$. On the vertices $(0,0)$, $(2,0)$, $(0,1)$ and $(2,1)$, we have $W(0,0)=8192$, $W(2,0)=W(2,1)=2368$ and $W(0,1)=7360$. On the sides of $[0,2]\times [0,1]$, if $c=0$, we have

$$W(0,t)=5184t^3-6016t^2+8192=:r_5\left(t\right).$$

(67)

Since $r_5^{\prime }\left(t\right)=0$ has only one positive root ${\tilde{t}}_1=\frac{188}{243}\approx 0.7737$, we see the maximum value of $r_5$ is 8192 attained on $t=0$. If $c=2$, then $W(2,t)\equiv 2368$ for all $t\in [0,1]$. It is left to discuss the case of $(c,t)\in (0,2)\times (0,1)$. Solving the system of equations

$$\frac{\partial W}{\partial t}=(4-c^2)\left(3{\lambda }_3{t}^2+2{\lambda }_{2}t+{\lambda }_1\right)=0$$

(68)

and

$$\frac{\partial W}{\partial c}={\mu }_3\left(c\right)t^3+{\mu }_2\left(c\right)t^2+{\mu }_1\left(c\right)t+{\mu }_0=0,$$

(69)

where

$$\begin{array}{cc}\hfill {\mu }_3& =-618c^5+1840c^4+4592c^3-1056c^2-8480c-4480,\hfill \\ \hfill {\mu }_2& =-846c^5+1840c^4+352c^3-6048c^2+6816c+2176,\hfill \\ \hfill {\mu }_1& =-654c^5-1840c^4-848c^3+1056c^2+5184c+4480,\hfill \\ \hfill {\mu }_0& =2c\left(111c^4-580c^3+1024c^2+1392c-4096\right),\hfill \end{array}$$

it is found that all the three critical points lie in $(0,2)\times (0,1)$ are $\left({\overline{c}}_1,{\overline{t}}_1\right)\approx \left(0.4480,0.4137\right)$, $\left({\overline{c}}_2,{\overline{t}}_2\right)\approx \left(1.0708,0.7129\right)$ and $\left({\overline{c}}_3,{\overline{t}}_3\right)\approx \left(0.4973,0.9759\right)$. As $W\left({\overline{c}}_1,{\overline{t}}_1\right)\approx 7913.01$, $W\left({\overline{c}}_2,{\overline{t}}_2\right)\approx 8167.95$ and $W\left({\overline{c}}_3,{\overline{t}}_3\right)\approx 7823.08$, we see the maximum value of W is less than 8192 in $(0,2)\times (0,1)$. In conclusion, we obtain that $W(c,t)\le 8192$ on $[0,2]\times [0,1]$, which leads to $\mathrm{\Gamma }\left(c,t,y\right)\le 8192$ for all $(c,t,y)\in \left[0,2\right]\times \left[0,1\right]\times \left[0,1\right]$. Hence, we obtain

$$\left|{\mathcal{H}}_{3,1}\left(f\right)\right|\le \frac{8192}{663552}=\frac{1}{81}.$$

(70)

For the sharpness, we consider the function $f_3$ satisfying the equation

$$3\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-2\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)=\frac{1-z^3}{1+z^3}·\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)},z\in \mathbb{D}.$$

(71)

The series expansion of $f_3$ is given by

$$f_3\left(z\right)=z-\frac{1}{2}{z}^2+\frac{1}{4}{z}^3-\frac{1}{72}{z}^4-\frac{11}{144}{z}^5+\cdots ,z\in \mathbb{D}.$$

(72)

This completes the proof Theorem 3. □

4. Conclusions

In this paper, we investigated some coefficient-related problems for the class ${\mathcal{G}}_{3/2}$, which is defined by the ratio of analytic representations of convex and starlike functions. Fekete–Szegö-type inequalities and upper bounds for the second and third Hankel determinant were obtained. The calculation is based on the relationship between the coefficients of functions in the considered class and the coefficients for the corresponding Carathéodory class. The results may inspire further investigations on getting more sharp bounds related with functional such as $a_3{a}_5-a_4^2$, $a_4-a_2{a}_3$ and $a_5-a_3^2$. By making use of the concept of q-calculus, it is also interesting to investigate a more general class of this type.