Upper Bounds of the Third Hankel Determinant for Close-to-Convex Functions

Abstract

In this paper, the third Hankel determinant for the class $\mathcal{N}$ of functions convex in one direction is estimated. An analogous problem is solved for a subclass of $\mathcal{N}$ consisting of functions with real coefficients. Additionally, this determinant for odd functions in $\mathcal{N}$ is estimated. Moreover, similar results are obtained in the relative class $\mathcal{M}$ consisting of functions $z{f}^{\prime }\left(z\right)$, where $f\in \mathcal{N}$. The majority of bounds is sharp.

1. Introduction

Let $\mathcal{A}$ be the family of all functions analytic in the open unit disk $\mathbb{D}=\{z\in \mathbb{C}:|z|<1\}$ having the power series expansion

$$f\left(z\right)=z+\sum _{n=2}^{\infty }{a}_n{z}^n$$

(1)

and let $\mathcal{S}$ denote the class of univalent functions in $\mathcal{A}$.

Since the early twentieth centry, coefficient problems have been in the mainstream of the geometric theory of analytic functions. The main motivation for studying these problems was the famous Bieberbach conjecture that $|a_n| \le n$ for $f\in \mathcal{S}$ (first proposed in 1916). The problem was finally solved by de Branges in 1985 (see [1,2] for the proof). There are many papers in which the n-th coefficient $a_n$ was estimated for various subclasses of analytic functions.

Recent studies in this area have focused on estimating so-called Hankel determinants. In the 1960s, Pommerenke defined the q-th Hankel determinant for a function f of the form (1) as

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

where $q,n\in \{1,2,\cdots \}$ (see [3,4]). The bound of $H_{q,n}\left(f\right)$ was investigated for various subfamilies of $\mathcal{A}$. The third Hankel determinant

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

has been discussed in many recent papers, which is quite complicated. Babalola [5] tried to estimate $|H_{3,1}\left(f\right)|$ for classes of starlike functions ${\mathcal{S}}^{*}={\mathcal{S}}^{*}\left(0\right)$, convex functions $\mathcal{K}$ and functions with bounded turning $\mathcal{R}$, where

$$\begin{array}{cc}\hfill {\mathcal{S}}^{*}\left(\alpha \right)& =\left\{f\in \mathcal{S}:\mathfrak{Re}\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}>\alpha \right\},\hfill \\ \hfill \mathcal{K}& =\left\{f\in \mathcal{S}:\mathfrak{Re}\left\{1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right\}>0\right\},\hfill \\ \hfill \mathcal{R}& =\left\{f\in \mathcal{S}:\mathfrak{Re}\left\{f^{\prime }\left(z\right)\right\}>0\right\}.\hfill \end{array}$$

For the classes $\mathcal{K}$ and ${\mathcal{S}}^{*}(1/2)$, results obtained by Kowalczyk et al. in [6] and Lecko et al. in [7], respectively, are sharp, i.e., $|H_{3,1}\left(f\right)| \le 4/135$ for $f\in \mathcal{K}$ and $|H_{3,1}\left(f\right)| \le 1/9$ for $f\in {\mathcal{S}}^{*}(1/2)$. In [8], it was proved that $|H_{3,1}\left(f\right)| \le 5/9$ for $f\in {\mathcal{S}}^{*}$ and in [9] it was proved that $|H_{3,1}\left(f\right)| \le 41/60$ for $f\in \mathcal{R}$. Moreover, in [9] sharp bounds for 2-fold and 3-fold symmetric starlike functions, convex functions and functions with bounded turning were obtained. However, some of the results obtained even for the most important subclasses of the class $\mathcal{S}$ are still not sharp (see, for example, [10,11,12,13,14,15,16,17,18]).

In this paper, we find estimates of the third Hankel determinant $H_{3,1}\left(f\right)$ for the following classes

$$\mathcal{M}=\left\{f\in \mathcal{A}:\mathfrak{Re}\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}<\frac{3}{2}\right\},$$

(2)

$$\mathcal{N}=\left\{f\in \mathcal{A}:\mathfrak{Re}\left\{1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right\}<\frac{3}{2}\right\}.$$

(3)

This problem was first studied by Prajapat et al. in [19].

The class $\mathcal{N}$ was introduced and discussed in 1941 by Ozaki (see [20]). He proved that each function f in $\mathcal{N}$ is univalent. After that, Umezawa (1952, [21]) showed that functions in $\mathcal{N}$ are convex in one direction. In fact, he proved much more. Namely, if for f the following inequality

$$-\frac{\alpha }{2\alpha -3}<\mathfrak{Re}\left\{1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right\}<\alpha$$

holds for any $\alpha \ge 3/2$, then f is univalent and convex in one direction. This observation was a background for defining two classes

$$\mathcal{F}\left(\lambda \right)=\left\{f\in \mathcal{A}:\mathfrak{Re}\left\{1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right\}>\lambda \right\},$$

(4)

and

$$\mathcal{N}\left(\lambda \right)=\left\{f\in \mathcal{A}:\mathfrak{Re}\left\{1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right\}<\lambda \right\}.$$

(5)

The class $\mathcal{N}\left(\lambda \right)$ was first discussed by Singh and Singh [22] in 1982. Here, it is worth recalling that Nunokawa and Saitoh showed that every $f\in \mathcal{N}\left(\lambda \right)$, $\lambda \in (1,3/2]$ is strongly starlike of order $2(\lambda -1)$.

Coefficient problems for the class $\mathcal{N}$ have been considered in a few papers. Obradovic found [23] the estimate of $|a_n| \le 1/n$ in the class $\mathcal{N}$ and later improved this result and showed [24] that

$$|a_n| \le \frac{1}{n(n-1)}.$$

Ponnusamy in [25] derived the bounds of initial logarithmic coefficients for $f\in \mathcal{N}$ and posed a conjecture on the general form of n-th logarithmic coefficient of $f\in \mathcal{N}$.

Taking into account the Alexander relation, it is interesting to consider a class of functions $z{f}^{\prime }\left(z\right)$ such that $f\in \mathcal{N}$ or $f\in \mathcal{N}\left(\lambda \right)$. For this reason, one can study the class $\mathcal{M}$ and the relative class $\mathcal{M}\left(\lambda \right)$. Unfortunatelly, $\mathcal{M}$ is not a subset of $\mathcal{S}$. It is enough to consider the function $f\left(z\right)=z-z^2$. It is not univalent in the unit disk. The example given below showes that this function is in $\mathcal{M}$.

Example 1.

Let

$$f_n\left(z\right)=z{(1-z^n)}^{\frac{1}{n}}=z-\frac{1}{n}{z}^{n+1}-\frac{n-1}{2n^2}{z}^{2n+1}-\cdots$$

Then

$$\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}=\frac{1-2z^n}{1-z^n}=\frac{3}{2}-\frac{1}{2}p\left(z\right),$$

where $p\left(z\right)=\frac{1+z^n}{1-z^n}$. Consequently, $\mathfrak{Re}\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}=\frac{3}{2}-\frac{1}{2}\mathfrak{Re}\left\{p\left(z\right)\right\}<\frac{3}{2}$, so $f\in \mathcal{M}$. In particular, for $n=1$ and $n=2$ we get

$$f_1\left(z\right)=z-z^2\mathit{and}{f}_2\left(z\right)=z\sqrt{1-z^2}=z-\frac{1}{2}{z}^3-\frac{1}{8}{z}^5-\cdots .$$

(6)

By the Alexander relation,

$$g_n\left(z\right)={\int }_0^z{(1-{\zeta }^n)}^{\frac{1}{n}}\mathrm{d}\zeta =z-\frac{1}{n(n+1)}{z}^{n+1}-\frac{n-1}{2n^2(2n+1)}{z}^{2n+1}-\cdots$$

belongs to $\mathcal{N}$. Thus, for $n=1$ and $n=2$ we obtain

$$f_1\left(z\right)=z-\frac{1}{2}{z}^2\mathit{and}{f}_2\left(z\right)=\frac{1}{2}(z\sqrt{1-z^2}+arcsinz)=z-\frac{1}{6}{z}^3-\frac{1}{40}{z}^5-\cdots .$$

(7)

2. Preliminary Results

Let $\mathcal{P}$ be the class of analytic functions p with a positive real part in $\mathbb{D}$, having the Taylor series expansion

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

(8)

By ${\mathcal{B}}_0$ we denote the class of Schwarz functions, i.e., analytic functions $w:\mathbb{D}\to \mathbb{D}$, $w\left(0\right)=0$. The function $w\in {\mathcal{B}}_0$ has the Taylor series expansion

$$w\left(z\right)=\sum _{n=1}^{\infty }{c}_n{z}^n.$$

(9)

It is known that for p and w given by (8) and (9), respectively, we have

$$\begin{array}{cc}\hfill p_1& =2c_1,p_2=2({c_1}^2+c_2),p_3=2({c_1}^3+2c_1{c}_2+c_3),\hfill \\ \hfill p_4& =2({c_1}^4+3{c_1}^2{c}_2+2c_1{c}_3+{c_2}^2+c_4).\hfill \end{array}$$

(10)

In order to prove our results, we need four lemmas concerning functions in the classes $\mathcal{P}$ and ${\mathcal{B}}_0$. The first one is known as Caratheodory’s lemma (see, for example, [26]) and the second is due to Zaprawa [27]. The third lemma was proved by Carlson [28].

Lemma 1

([26]). Let $p\in \mathcal{P}$ be given by (8). Then $|p_n| \le 2$ for $n\ge 1$.

Lemma 2

([27]). Let $p\in \mathcal{P}$ be given by (8). Then

$$|p_2{p}_4-{p_3}^2| \le 4.$$

Lemma 3

([28]). Let $w\in {\mathcal{B}}_0$ be given by (9). Then

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

Additonaly, we can prove what follows.

Lemma 4.

Let $w\in {\mathcal{B}}_0$ be given by (9). Then

$$|c_2{c}_4-{c_3}^2| \le 1 - |c_1{|}^2.$$

Proof. 

From Gutzmer’s inequality for functions in ${\mathcal{B}}_0$

$$|c_1{|}^2 + |c_2{|}^2 + |c_3{|}^2 + {|c_4|}^2 \le 1$$

and the triangle inequality we obtain

$$|c_2{c}_4 - {c_3}^2| \le |c_2|·|c_4| + \left(1-|c_1{|}^2 - |c_2{|}^2 - {|c_4|}^2\right) = 1 - |c_1{|}^2 - {\left(|c_2{|}^2 - |c_4|\right)}^2 - |c_2|·|c_4|.$$

The omission of the last two components which are non-positive completes the proof. □

The coefficients of $f\in \mathcal{M}$ can be expressed by coefficients of a relative function p from the class $\mathcal{P}$. For f and p given by (1) and (8), respectively, we have

$$\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}=\frac{3}{2}-\frac{1}{2}p\left(z\right).$$

(11)

Hence, comparing the coefficients of functions in (11), we get

$$\begin{array}{cc}\hfill a_2& =-\frac{1}{2}{p}_1,a_3=\frac{1}{8}({p_1}^2-2p_2),a_4=\frac{1}{48}(6p_1{p}_2-8p_3-{p_1}^3),\hfill \\ \hfill a_5& =\frac{1}{384}(32p_1{p}_3-48p_4+12{p_2}^2-12{p_1}^2{p}_2+{p_1}^4).\hfill \end{array}$$

(12)

From (10) and (12) we obtain

$$\begin{array}{cc}\hfill a_2& =-c_1,a_3=-\frac{1}{2}{c}_2,a_4=-\frac{1}{6}(c_1{c}_2+2c_3),\hfill \\ \hfill a_5& =-\frac{1}{24}(2{c_1}^2{c}_2+4c_1{c}_3+3{c_2}^2+6c_4).\hfill \end{array}$$

(13)

Analogously, the coefficients of $f\in \mathcal{N}$ can be expressed in terms of the coefficients of $p\in \mathcal{P}$ or $\omega \in {\mathcal{B}}_0$. From the Alexander relation $f\in \mathcal{N}$ if and only if $z{f}^{\prime }\left(z\right)\in \mathcal{M}$. Hence, for $f\in \mathcal{N}$,

$$\begin{array}{cc}\hfill a_2& =-\frac{1}{4}{p}_1,a_3=\frac{1}{24}({p_1}^2-2p_2),a_4=\frac{1}{192}(6p_1{p}_2-8p_3-{p_1}^3),\hfill \\ \hfill a_5& =\frac{1}{1920}(32p_1{p}_3-48p_4+12{p_2}^2-12{p_1}^2{p}_2+{p_1}^4)\hfill \end{array}$$

(14)

or

$$\begin{array}{cc}\hfill a_2& =-\frac{1}{2}{c}_1,a_3=-\frac{1}{6}{c}_2,a_4=-\frac{1}{24}(c_1{c}_2+2c_3)\hfill \\ \hfill a_5& =-\frac{1}{120}(2{c_1}^2{c}_2+4c_1{c}_3+3{c_2}^2+6c_4).\hfill \end{array}$$

(15)

3. Main Results

Theorem 1.

If $f\in \mathcal{M}$ is given by (1), then

$$|H_{3,1}\left(f\right)| \le \frac{43}{144}\approx 0.2986\cdots .$$

Proof. 

From (13) we get

$$\begin{array}{cc}\hfill 9|H_{3,1}\left(f\right)|& = |\frac{9}{8}{c}_2{c}_4 - {c_3}^2 + \frac{9}{4}{c_1}^2{c}_4 + \frac{3}{4}{c_1}^4{c}_2 - \frac{1}{4}{c_1}^2{c_2}^2 + \frac{27}{16}{c_2}^3 + \frac{3}{2}{c_1}^3{c}_3 - \frac{13}{4}{c}_1{c}_2{c}_3|\hfill \\ \hfill & = |c_2{c}_4 - {c_3}^2| + \frac{1}{8}(c_2 + 18{c_1}^2)c_4 + \frac{3}{4}{c_1}^4{c}_2 - \frac{1}{4}{c_1}^2{c_2}^2\hfill \\ \hfill & + \frac{27}{16}{c_2}^3 + \frac{1}{4}(6{c_1}^3 - 13c_1{c}_2)c_3|.\hfill \end{array}$$

Using the triangle inequality we get

$$\begin{array}{cc}\hfill 9|H_{3,1}\left(f\right)|& \le |c_2{c}_4 - {c_3}^2| + \frac{1}{8}(|c_2| + 18|c_1{|}^2)|c_4| + \frac{3}{4}|c_1{|}^4|c_2| + \frac{1}{4}|c_1{|}^2{|c_2|}^2\hfill \\ \hfill & + \frac{27}{16}|c_2{|}^3 + \frac{1}{4}(6|c_1{|}^3 + 13|c_1||c_2|)|c_3|.\hfill \end{array}$$

Now, applying Lemmas 3 and 4 we obtain

$$\begin{array}{cc}\hfill 9|H_{3,1}\left(f\right)|& \le 1 - |c_1{|}^2 + \frac{1}{8}(|c_2| + 18|c_1{|}^2\left)\right(1 - |c_1{|}^2 - |c_2{|}^2) + \frac{3}{4}|c_1{|}^4|c_2| + \frac{1}{4}|c_1{|}^2{|c_2|}^2\hfill \\ \hfill & + \frac{27}{16}|c_2{|}^3 + \frac{1}{4}(6|c_1{|}^3 + 13|c_1||c_2|)\left(1 - |c_1{|}^2 - \frac{|c_2{|}^2}{1+|c_1|}\right).\hfill \end{array}$$

Thus,

$$9|H_{3,1}\left(f\right)| \le h(|c_1|,|c_2|),$$

where

$$\begin{array}{cc}\hfill h(c,d)& =1-c^2+\frac{1}{8}(d+18c^2)(1-c^2-d^2)+\frac{3}{4}{c}^{4}d+\frac{1}{4}{c}^2{d}^2\hfill \\ \hfill & +\frac{27}{16}{d}^3+\frac{1}{4}(6c^3+13cd)\left(1-c^2-\frac{d^2}{1+c}\right).\hfill \end{array}$$

Arranging h with respect to d we have

$$\begin{array}{cc}\hfill h(c,d)& =\frac{25-27c}{16(1+c)}{d}^3-\frac{c^2(4+7c)}{2(1+c)}{d}^2+\frac{1}{8}(1+26c-c^2-26c^3+6c^4)d\hfill \\ & +\frac{1}{4}(1-c^2)(4+9c^2+6c^3).\hfill \end{array}$$

Since $\frac{c^2(4+7c)}{2(1+c)}>0$ for $c\in [0,1]$, so we can omit the second component. From $d \le 1-c^2$, we get $h(c,d) \le h_1(c,d)$, where

$$\begin{array}{cc}\hfill h_1(c,d)& =\frac{25-27c}{16(1+c)}{d}^3+\frac{1}{8}(1-c^2)(1+26c)d+\frac{1}{4}(1-c^2)(4+9c^2+6c^3)+\frac{3}{4}{c}^4(1-c^2).\hfill \end{array}$$

Since

$$\frac{\partial h_1(c,d)}{\partial d}=\frac{3(25-27c)}{16(1+c)}{d}^2+\frac{1}{8}(1-c^2)(1+26c)$$

and

$$\frac{\partial h_1(c,d)}{\partial d}{|}_{d=1-c^2}=\frac{1}{16}(1-c^2)(77-104c+81c^2)>0,$$

so $h_1$ is an increasing function of variable d. Thus,

$$h_1(c,d) \le h_1(c,1-c^2)=\frac{1}{16}(1-c^2)(43+36c^2+24c^3-15c^4).$$

Moreover, we have

$$\frac{\partial h_1(c,1-c^2)}{\partial c}=\frac{1}{8}c(-7+36c-102c^2-60c^3+45c^4)<0,$$

so $h_1(c,1-c^2)$ is an decreasing function of variable c, thus $h_1(c,1-c^2) \le h_1(0,1)=\frac{43}{16}$. Hence, $h(c,d) \le \frac{43}{16}$, so $|H_{3,1}\left(f\right)| \le \frac{43}{144}$. □

The result is not sharp. It is expected that the sharp bound of $|H_{3,1}\left(f\right)|$ is equal to $\frac{3}{16}$.

Sharp results can be obtained on additional assumptions. First, we can assume that the coefficients of f are real (see Theorem 2). Second, we can assume that $a_2=0$ (see Theorem 3).

Theorem 2.

If $f\in \mathcal{M}$ is given by (1) and the coefficients of f are real, then

$$H_{3,1}\left(f\right) \le \frac{3}{16}\approx 0.1875.$$

The result is sharp.

Proof. 

From

$$9H_{3,1}\left(f\right)=-{\left(c_3-\frac{3}{4}{c_1}^3\right)}^2-\frac{1}{4}{c_1}^2{c_2}^2+\frac{9}{16}{c_1}^6+\frac{9}{8}{c}_2{c}_4+\frac{9}{4}{c_1}^2{c}_4+\frac{3}{4}{c_1}^4{c}_2+\frac{27}{16}{c_2}^3-\frac{13}{4}{c}_1{c}_2{c}_3,$$

by omitting the first two non-positive components and applying Lemma 3 we get

$$H_{3,1}\left(f\right) \le \frac{1}{9}h(|c_1|,|c_2|),$$

where

$$h(c,d)=\frac{9}{16}{c}^6+\frac{9}{8}(d+2c^2)(1-c^2-d^2)+\frac{3}{4}{c}^{4}d+\frac{27}{16}{d}^3+\frac{13}{4}cd\left(1-c^2-\frac{d^2}{1+c}\right).$$

Since

$$\frac{\partial h}{\partial d}=\frac{9}{8}(1-c^2)+\frac{13}{4}c(1-c^2)+\frac{3}{4}{c}^4-\frac{9}{2}{c}^{2}d+\left(\frac{27}{16}-\frac{39c}{4(1+c)}\right)d^2,$$

we obtain

$$\frac{\partial h}{\partial d}{|}_{d=0}=\frac{9}{8}(1-c^2)+\frac{13}{4}c(1-c^2)+\frac{3}{4}{c}^4>0$$

and

$$\frac{\partial h}{\partial d}{|}_{d=1-c^2}=\frac{1}{16}\left(45-104c+12c^2+104c^3-45c^4\right)>0.$$

Hence

$$\begin{array}{cc}\hfill h(c,d)& \le h(c,1-c^2)=\frac{1}{16}\left(27-11c^2-11c^4+4c^6\right)\hfill \\ \hfill & =\frac{1}{16}\left[27-11c^2-7c^4-4c^4(1-c^2)\right] \le \frac{27}{16}.\hfill \end{array}$$

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

For the sharpness of the result it is enough to consider a function $\omega \left(z\right)$ with $c_2=1$ and $c_k=0$ for $k\ne 1$, i.e., $\omega \left(z\right)=z^2$. Consequently, the extremal function $f\in \mathcal{M}$ is given by

$$f\left(z\right)=z\sqrt{1-z^2}=z-\frac{1}{2}{z}^3-\frac{1}{8}{z}^5+\dots .$$

□

Theorem 3.

If $f\in \mathcal{M}$ is given by (1) and $a_2=0$, then

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

The result is sharp.

Proof. 

If $a_2=0$, then from (12) we have $p_1=0$, $a_3=-\frac{1}{4}{p}_2$, $a_4=-\frac{1}{6}{p}_3$, $a_5=-\frac{1}{8}{p}_4+\frac{1}{32}{p_2}^2$. Now, from Lemmas 1 and 2, we obtain

$$\begin{array}{cc}\hfill |H_{3,1}\left(f\right)|& = \frac{1}{1152}|36p_2{p}_4-32{p_3}^2+9{p_2}^3|\hfill \\ \hfill & \le \frac{1}{1152}(32|p_2{p}_4-{p_3}^2|+4|p_2||p_4|+9|p_2{|}^3)\hfill \\ \hfill & \le \frac{1}{1152}(32·4+4·2·2+9·8)=\frac{3}{16}.\hfill \end{array}$$

□

Corollary 1.

If $f\in \mathcal{M}$ is odd, then $|H_{3,1}\left(f\right)| \le \frac{3}{16}$.

Let us now turn to the class $\mathcal{N}$.

Theorem 4.

If $f\in \mathcal{N}$ is given by (1), then

$$|H_{3,1}\left(f\right)| \le \frac{17}{1080}\approx 0.0157\cdots .$$

Proof. 

Formula (15) results in

$$\begin{array}{cc}\hfill 144|H_{3,1}\left(f\right)|& = |\frac{6}{5}{c}_2{c}_4-{c_3}^2+\frac{9}{5}{c_1}^2{c}_4+\frac{3}{5}{c_1}^4{c}_2+\frac{1}{20}{c_1}^2{c_2}^2+\frac{19}{15}{c_2}^3+\frac{6}{5}{c_1}^3{c}_3-\frac{11}{5}{c}_1{c}_2{c}_3|\hfill \\ \hfill & = |c_2{c}_4-{c_3}^2+\frac{1}{5}(c_2+9{c_1}^2)c_4+\frac{3}{5}{c_1}^4{c}_2+\frac{1}{20}{c_1}^2{c_2}^2\hfill \\ \hfill & + \frac{19}{15}{c_2}^3+\frac{1}{5}(6{c_1}^3-11c_1{c}_2)c_3|.\hfill \end{array}$$

The triangle inequality leads to

$$\begin{array}{cc}\hfill 144|H_{3,1}\left(f\right)|& \le |c_2{c}_4 - {c_3}^2| + \frac{1}{5}(|c_2| + 9|c_1{|}^2)|c_4| + \frac{3}{5}|c_1{|}^4|c_2| + \frac{1}{20}|c_1{|}^2{|c_2|}^2\hfill \\ \hfill & + \frac{19}{15}|c_2{|}^3 + \frac{1}{5}(6|c_1{|}^3 + 11|c_1||c_2|)|c_3|.\hfill \end{array}$$

By Lemmas 3 and 4,

$$\begin{array}{cc}\hfill 144|H_{3,1}\left(f\right)|& \le 1 - |c_1{|}^2 + \frac{1}{5}(|c_2|+9|c_1{|}^2\left)\right(1 - |c_1{|}^2 - |c_2{|}^2) + \frac{3}{5}|c_1{|}^4|c_2| + \frac{1}{20}|c_1{|}^2{|c_2|}^2\hfill \\ \hfill & + \frac{19}{15}|c_2{|}^3 + \frac{1}{5}(6|c_1{|}^3 + 11|c_1||c_2|)\left(1 - |c_1{|}^2 - \frac{|c_2{|}^2}{1 + |c_1|}\right).\hfill \end{array}$$

Thus,

$$144|H_{3,1}\left(f\right)| \le h(|c_1|,|c_2|),$$

where

$$\begin{array}{cc}\hfill h(c,d)& = 1 - c^2 + \frac{1}{5}(d + 9c^2)(1 - c^2 - d^2) + \frac{3}{5}{c}^{4}d + \frac{1}{20}{c}^2{d}^2\hfill \\ \hfill & + \frac{19}{15}{d}^3 + \frac{1}{5}(6c^3 + 11cd)\left(1 - c^2 - \frac{d^2}{1 + c}\right),\hfill \end{array}$$

or equivalently,

$$\begin{array}{cc}\hfill h(c,d)& = \frac{16-17c}{15(1+c)}{d}^3-\frac{c^2(35+59c)}{20(1+c)}{d}^2+\frac{1}{5}(1+11c-c^2-11c^3+3c^4)d\hfill \\ \hfill & + \frac{1}{5}(1-c^2)(5+9c^2+6c^3).\hfill \end{array}$$

Note that $\frac{c^2(35+59c)}{20(1+c)}>0$ for $c\in [0,1]$. Hence, we can omit the second component. Applying $d \le 1-c^2$, we get $h(c,d) \le h_1(c,d)$, where

$$\begin{array}{cc}\hfill h_1(c,d)& =\frac{16-17c}{15(1+c)}{d}^3+\frac{1}{5}(1-c^2)(1+11c)d+\frac{1}{5}(1-c^2)(5+9c^2+6c^3)+\frac{3}{5}{c}^4(1-c^2).\hfill \end{array}$$

The function $h_1$ increases with respect to variable d because

$$\frac{\partial h_1(c,d)}{\partial d}=\frac{16-17c}{5(1+c)}{d}^2+\frac{1}{5}(1-c^2)(1+11c)$$

and

$$\frac{\partial h_1(c,d)}{\partial d}{|}_{d=1-c^2}=\frac{1}{5}(1-c^2)(17-22c+17c^2)>0.$$

Thus,

$$h_1(c,d) \le h_1(c,1-c^2)=\frac{1}{15}(1-c^2)(34+25c^2+18c^3-8c^4).$$

Moreover,

$$\frac{\partial h_1(c,1-c^2)}{\partial c}=\frac{2}{5}c(-3+9c-22c^2-15c^3+8c^4)<0,$$

so $h_1(c,1-c^2)$ is a decreasing function of variable c, thus $h_1(c,1-c^2) \le h_1(0,1)=\frac{34}{15}$. This means that $h(c,d) \le \frac{34}{15}$ and $|H_{3,1}\left(f\right)| \le \frac{17}{1080}$. □

Similarly as in Theorem 1, the result is not sharp. The sharp bound of $|H_{3,1}\left(f\right)|$ is expected to be $\frac{19}{2160}$.

Theorem 5.

If $f\in \mathcal{N}$ is given by (1) and the coefficients of f are real, then

$$H_{3,1}\left(f\right) \le \frac{19}{2160}\approx 0.0087\cdots .$$

The result is sharp.

Proof. 

Since

$$\begin{array}{cc}\hfill 144H_{3,1}\left(f\right)& = \frac{6}{5}{c}_2{c}_4+\frac{9}{5}{c_1}^2{c}_4-{c_3}^2+\frac{19}{15}{c_2}^3-\frac{11}{5}{c}_1{c}_2{c}_3+\frac{6}{5}{c_1}^3{c}_3+\frac{1}{20}{c_1}^2{c_2}^2+\frac{3}{5}{c_1}^4{c}_2\hfill \\ \hfill & = -\frac{3}{5}{\left(c_3-{c_1}^3\right)}^2-\frac{2}{5}{\left(c_3+2c_1{c}_2\right)}^2-\frac{3}{5}{c}_1{c}_2{c}_3\hfill \\ \hfill & + \frac{3}{5}{c_1}^6+\frac{6}{5}{c}_2{c}_4+\frac{9}{5}{c_1}^2{c}_4+\frac{19}{15}{c_2}^3+\frac{33}{20}{c_1}^2{c_2}^2+\frac{3}{5}{c_1}^4{c}_2,\hfill \end{array}$$

omitting the first two non-positive components and applying Lemma 3 yields

$$H_{3,1}\left(f\right) \le \frac{1}{144}h(|c_1|,|c_2|),$$

where

$$\begin{array}{cc}\hfill h(c,d)& =\frac{3}{5}{c}^6+(\frac{6}{5}d+\frac{9}{5}{c}^2)(1-c^2-d^2)+\frac{19}{15}{d}^3+\frac{33}{20}{c}^2{d}^2+\frac{3}{5}{c}^{4}d+\frac{3}{5}cd\left(1-c^2-\frac{d^2}{1+c}\right)\hfill \\ \hfill & =\frac{9}{5}{c}^2(1-c^2)+\frac{3}{5}{c}^4+\frac{3}{5}\left[(1-c^2)(2+c)+c^4\right]d-\frac{3}{20}{c}^2{d}^2+\frac{1}{15}\left(1-\frac{9c}{1+c}\right)d^3.\hfill \end{array}$$

A direct computation shows that

$$\frac{\partial h}{\partial d}{|}_{d=0}=\frac{3}{5}\left[(1-c^2)(2+c)+c^4\right]>0$$

and

$$\frac{\partial h}{\partial d}{|}_{d=1-c^2}=\frac{1}{10}\left(14-12c-c^2+12c^3-7c^4\right)>0.$$

Hence,

$$h(c,d) \le h(c,1-c^2)=\frac{19}{15}-\frac{7}{20}{c}^2-\frac{7}{10}{c}^4+\frac{23}{60}{c}^6=\frac{19}{15}-\frac{7}{20}{c}^2-\frac{19}{60}{c}^4-\frac{23}{60}{c}^4(1-c^2) \le \frac{19}{15},$$

so $H_{3,1}\left(f\right) \le \frac{19}{2160}$.

Observe that taking $\omega \left(z\right)=z^2$, i.e., $c_2=1$ and $c_k=0$ for $k\ne 1$, results in $144H_{3,1}\left(f\right)=\frac{19}{15}$. For this reason, the equality in Theorem 5 holds for $f_2$ given by (7). □

Theorem 6.

If $f\in \mathcal{N}$ is given by (1) and $a_2=0$, then

$$|H_{3,1}\left(f\right)| \le \frac{19}{2160}.$$

The result is sharp.

Proof. 

Assume that $a_2=0$. Then $p_1=0$, $a_3=-\frac{1}{12}{p}_2$, $a_4=-\frac{1}{24}{p}_3$, $a_5=-\frac{1}{40}{p}_4+\frac{1}{160}{p_2}^2$. By Lemmas 1 and 2,

$$\begin{array}{cc}\hfill |H_{3,1}\left(f\right)|& = \frac{1}{17280}|36p_2{p}_4-30{p_3}^2+{p_2}^3|\hfill \\ \hfill & \le \frac{1}{17280}(30|p_2{p}_4-{p_3}^2|+6|p_2||p_4|+|p_2{|}^3)\hfill \\ \hfill & \le \frac{1}{17280}(30·4+6·2·2+8)=\frac{19}{2160}.\hfill \end{array}$$

□

Corollary 2.

If $f\in \mathcal{N}$ is odd, then $|H_{3,1}\left(f\right)| \le \frac{19}{2160}$.

4. Conclusions

In this paper, a new approach to calculating the third Hankel determinant for close-to-convex functions was proposed. The bounds of the third Hankel determinant for the classes $\mathcal{M}$ and $\mathcal{N}$ derived with the new approach are better than those obtained by Prajapat in [19]

$$\begin{array}{cc}\hfill |H_{3,1}\left(f\right)|& \le \frac{81+16\sqrt{3}}{216}\approx 0.5033\cdots \mathrm{for}f\in \mathcal{M},\hfill \\ \hfill |H_{3,1}\left(f\right)|& \le \frac{139}{5760}\approx 0.0241\cdots \mathrm{for}f\in \mathcal{N}.\hfill \end{array}$$

The new method involves determining $H_{3,1}\left(f\right)$ based on relationships between the classes $\mathcal{M}$ and $\mathcal{N}$ with the class ${\mathcal{B}}_0$, as well as appropriate coefficient relations. The advantage of this approach is the possibility of determining estimates of these functionals when the function coefficients are real.