SECOND HANKEL DETERMINANT FOR LOGARITHMIC INVERSE COEFFICIENTS OF CONVEX AND STARLIKE FUNCTIONS

Abstract

We obtain sharp bounds for the second Hankel determinant of logarithmic inverse coefficients for starlike and convex functions.

1. Introduction

Let $\mathcal {H}$ denote the class of analytic functions in the unit disk $\mathbb {D}:=\{z\in \mathbb {C}:\, |z|<1\}$ . Here $\mathcal {H}$ is a locally convex topological vector space endowed with the topology of uniform convergence over compact subsets of $\mathbb {D}$ . Let $\mathcal {A}$ denote the class of functions $f\in \mathcal {H}$ such that $f(0)=0$ and $f'(0)=1$ . Let $\mathcal {S}$ denote the subclass of $\mathcal {A}$ consisting of functions which are univalent (that is, one-to-one) in $\mathbb {D}$ . If $f\in \mathcal {A}$ , then it has the series representation

(1.1) $$ \begin{align} f(z)= z+\sum_{n=2}^{\infty}a_n z^n, \quad z\in \mathbb{D}. \end{align} $$

For $q,n \in \mathbb {N}$ , the Hankel determinant $H_{q,n}(f)$ of the Taylor coefficients of the function $f \in \mathcal {A}$ of the form (1.1) is

$$ \begin{align*} H_{q,n}(f) = \begin{vmatrix} a_n & a_{n+1} & \cdots & a_{n+q-1} \\ a_{n+1} & a_{n+2} & \cdots & a_{n+q} \\ \vdots & \vdots & \ddots &\vdots \\ a_{n+q-1} & a_{n+q} & \cdots & a_{n+2(q-1)} \end{vmatrix}. \end{align*} $$

Hankel determinants of various orders have been studied in many contexts (see for instance [5]). The Fekete–Szegö functional is the second Hankel determinant $H_{2,1}(f)$ . Fekete–Szegö obtained estimates for $|a_3 - \mu a_2 ^2|$ with $\mu $ real (see [10, Theorem 3.8]).

Let g be the inverse function of $f\in \mathcal {S}$ defined in a neighbourhood of the origin with the Taylor series expansion

(1.2) $$ \begin{align} g(w)=f^{-1}(w)=w+\sum_{n=2}^{\infty}A_n w^n, \end{align} $$

where we may choose $|w| < 1/4$ from Koebe’s $1/4$ -theorem. Using variational methods, Löwner [16] obtained the sharp estimate

$$ \begin{align*} |A_n| \leq K_n \quad \text{for each}\ n \in \mathbb{N},\end{align*} $$

where $K_n = (2n)!/(n!(n + 1)!)$ and $K(w) = w + K_2w^2 + K_3w^3 + \cdots $ is the inverse of the Koebe function. Let $f(z)= z+\sum _{n=2}^{\infty }a_n z^n $ be a function in class $\mathcal {S}$ . Since $f(f^{-1})(w)=w$ , it follows from (1.2) that

$$ \begin{align*} & A_{2}=-a_2,\\ & A_{3}=-a_3+2a_2^2,\\ & A_{4}=-a_4+5a_2a_3-5a_2^3. \end{align*} $$

The logarithmic coefficients $\gamma _{n}$ of $f\in \mathcal {S}$ are defined by

(1.3) $$ \begin{align} F_{f}(z):= \log\frac{f(z)}{z}=2\sum\limits_{n=1}^{\infty}\gamma_{n}z^{n}, \quad z \in \mathbb{D}. \end{align} $$

Few exact upper bounds for $\gamma _{n}$ have been established. The significance of this problem in the context of the Bieberbach conjecture was pointed out by Milin [17]. Milin’s conjecture that for $f\in \mathcal {S}$ and $n\ge 2$ ,

$$ \begin{align*}\sum\limits_{m=1}^{n}\sum\limits_{k=1}^{m}\bigg(k|\gamma_{k}|^{2}-\frac{1}{k}\bigg)\le 0,\end{align*} $$

led De Branges, by proving this conjecture, to the proof of the Bieberbach conjecture [9]. For the Koebe function $k(z)=z/(1-z)^{2}$ , the logarithmic coefficients are $\gamma _{n}=1/n$ . Since the Koebe function k plays the role of extremal function for most of the extremal problems in the class $\mathcal {S}$ , it might be expected that $|\gamma _{n}|\le 1/n$ holds for functions in $\mathcal {S}$ . However, this is not true in general, even in order of magnitude. Indeed, there exists a bounded function f in the class $\mathcal {S}$ with logarithmic coefficients $\gamma _{n}\ne O(n^{-0.83})$ (see [10, Theorem 8.4]). By differentiating (1.3) and equating coefficients,

(1.4) $$ \begin{align} \nonumber & \gamma_{1}=\tfrac{1}{2}a_{2}, \\ & \gamma_{2}=\tfrac{1}{2}(a_{3}-\tfrac{1}{2}a_{2}^{2}),\\ \nonumber & \gamma_{3}=\tfrac{1}{2}(a_{4}-a_{2}a_{3}+\tfrac{1}{3}a_{2}^{3}). \end{align} $$

If $f\in \mathcal {S}$ , it is easy to see that $|\gamma _{1}|\le 1$ , because $|a_2| \leq 2$ . Using the Fekete–Szegö inequality [10, Theorem 3.8] for functions in $\mathcal {S}$ in (1.3), we obtain the sharp estimate

$$ \begin{align*}|\gamma_{2}|\le\tfrac{1}{2}(1+2e^{-2})=0.635\ldots.\end{align*} $$

For $n\ge 3$ , the problem seems much harder and no significant bound for $|\gamma _{n}|$ when $f\in \mathcal {S}$ appears to be known. In 2017, Ali and Allu [1] obtained initial logarithmic coefficient bounds for close-to-convex functions. For recent results on several subclasses of close-to-convex functions, see [2, 6, 21].

The notion of logarithmic inverse coefficients, that is, logarithmic coefficients of the inverse of f, was proposed by Ponnusamy et al. [20]. The logarithmic inverse coefficients $\Gamma _n$ , $n \in \mathbb {N}$ , of f are defined by the equation

$$ \begin{align*} F_{f^{-1}}(w):= \log\frac{f^{-1}(w)}{w}=2\sum\limits_{n=1}^{\infty}\Gamma_{n}w^{n}, \quad |w|<1/4. \end{align*} $$

In [20], Ponnusamy et al. found sharp upper bounds for the logarithmic inverse coefficients for the class $\mathcal {S}$ , namely

$$ \begin{align*}|\Gamma_n| \leq \frac{1}{2n} \bigg( \begin{matrix} 2n \\ n \end{matrix} \bigg), \quad n \in \mathbb{N}, \end{align*} $$

with equality only for the Koebe function or one of its rotations. Ponnusamy et al. [20] also obtained sharp bounds for the initial logarithmic inverse coefficients for some of the important geometric subclasses of $ \mathcal {S}$ .

Recently, Kowalczyk and Lecko [12] proposed the study of the Hankel determinant whose entries are logarithmic coefficients of $ f \in \mathcal {S}$ , given by

$$ \begin{align*} H_{q,n}(F_f/2) = \begin{vmatrix} \gamma_n & \gamma_{n+1} & \cdots & \gamma_{n+q-1} \\ \gamma_{n+1} & \gamma_{n+2} & \cdots & \gamma_{n+q} \\ \vdots & \vdots & \ddots &\vdots \\ \gamma_{n+q-1} & \gamma_{n+q} & \cdots & \gamma_{n+2(q-1)} \end{vmatrix}. \end{align*} $$

Kowalczyk and Lecko [12] obtained a sharp bound for the second Hankel determinant $H_{2,1}(F_f/2)$ for starlike and convex functions. Sharp bounds for $H_{2,1}(F_f/2)$ for various subclasses of $\mathcal {S}$ are considered in [3, 4, 11, 13, 18]).

In this paper, we consider the second Hankel determinant for logarithmic inverse coefficients. From (1.4), for $ f \in \mathcal {S}$ given by (1.1), the second Hankel determinant of $F_{f^{-1}}/2$ is given by

(1.5) $$ \begin{align}H_{2,1}(F_{f^{-1}}/2) =\Gamma_1\Gamma_3-\Gamma_{2}^2 &=\tfrac{1}{4}(A_2A_4-A_3^2+\tfrac{1}{4}A_2^4)\nonumber\\ &=\tfrac{1}{48}(13a_2^4-12a_2^2a3-12a_3^2+12a_2a_4).\end{align} $$

We note that |$H_{2,1}(F_{f^{-1}}/2)|$ is invariant under rotation, since for $f_{\theta }(z):=e^{-i \theta } f(e^{i \theta } z)$ , $\theta \in \mathbb {R}$ and $f \in \mathcal {S}$ ,

$$ \begin{align*} H_{2,1}(F_{f_{\theta}^{-1}}/2)=\frac{e^{4 i \theta}}{48}(13a_2^4-12a_2^2a3-12a_3^2+12a_2a_4)=e^{4 i \theta} H_{2,1}(F_{f^{-1}}/2). \end{align*} $$

The main aim of this paper is to find a sharp upper bound for $|H_{2,1}(F_{f^{-1}}/2)| $ when f belongs to the class of convex or starlike functions. A domain $\Omega \subseteq \mathbb {C}$ is said to be starlike with respect to a point $z_{0}\in \Omega $ if the line segment joining $z_{0}$ to any point in $\Omega $ lies entirely in $\Omega $ . If $z_0$ is the origin, then we say that $\Omega $ is a starlike domain. A function $f \in \mathcal {A}$ is said to be starlike if $f(\mathbb {D})$ is a starlike domain. We denote by $\mathcal {S}^*$ the class of starlike functions f in $\mathcal {S}$ . It is well known that a function $f \in \mathcal {A}$ is in $\mathcal {S}^*$ if and only if

(1.6) $$ \begin{align} \mathrm{Re\,}\bigg( \frac{zf'(z)}{f(z)} \bigg)> 0 \quad \text{for}\ z \in \mathbb{D}. \end{align} $$

Further, a domain $\Omega \subseteq \mathbb {C}$ is called convex if the line segment joining any two points of $\Omega $ lies entirely in $\Omega $ . A function $f\in \mathcal {A}$ is called convex if $f(\mathbb {D})$ is a convex domain. We denote by $\mathcal {C}$ the class of convex functions in $\mathcal {S}$ . A function $f \in \mathcal {A}$ is in $\mathcal {C}$ if and only if

(1.7) $$ \begin{align} \mathrm{Re\,}\bigg( 1+\frac{zf"(z)}{f'(z)} \bigg)> 0 \quad \text{for}\ z \in \mathbb{D}. \end{align} $$

2. Preliminary results

In this section, we present the key lemmas which will be used to prove the main results of this paper. Let $\mathcal {P}$ denote the class of all analytic functions p having positive real part in $\mathbb {D}$ , with the form

(2.1) $$ \begin{align} p(z)=1+c_{1} z+c_{2} z^{2}+c_{3} z^{3}+ \cdots. \end{align} $$

A member of $\mathcal {P}$ is called a Carathéodory function. It is known that $|c_{n}| \leq 2, n \geq 1$ , for $p \in \mathcal {P}$ . By using (1.6) and (1.7), functions in the classes $\mathcal {S}^*$ and $\mathcal {C}$ can be represented in terms of functions in the Carathéodory class $\mathcal {P}$ .

Parametric representations of the coefficients are often useful. In Lemma 2.1, (2.2) is due to Carathéodory [10]. Equation (2.3) can be found in [19]. In 1982, Libera and Zlotkiewicz [14, 15] derived (2.4) with the assumption that $c_1 \geq 0$ . Later, Cho et al. [7] derived (2.4) in the general case and gave the explicit form of the extremal function.

Lemma 2.1. If $p \in \mathcal {P}$ is of the form (2.1), then

(2.2) $$ \begin{align} c_{1}=2 p_{1}, \end{align} $$

(2.3) $$ \begin{align} c_{2}=2 p_{1}^{2}+2(1-p_{1}^{2}) p_{2} \end{align} $$

and

(2.4) $$ \begin{align} c_{3}=2 p_{1}^{3}+4\big(1-p_{1}^{2}\big) p_{1} p_{2}-2\big(1-p_{1}^{2}\big) p_{1} p_{2}^{2}+2\big(1-p_{1}^{2}\big)\big(1-|p_{2}|^{2}\big) p_{3} \end{align} $$

for some $p_{1}, p_{2}, p_{3} \in \overline {\mathbb {D}}:=\{z \in \mathbb {C}:|z| \leq 1\}$ .

For $p_{1} \in \mathbb {T}:=\{z \in \mathbb {C}:|z|=1\}$ , there is a unique function $p \in \mathcal {P}$ with $c_{1}$ as in (2.2), namely

$$ \begin{align*} p(z)=\frac{1+p_{1} z}{1-p_{1} z}, \quad z \in \mathbb{D}. \end{align*} $$

For $p_{1} \in \mathbb {D}$ and $p_{2} \in \mathbb {T}$ , there is a unique function $p \in \mathcal {P}$ with $c_{1}$ and $c_{2}$ as in (2.2) and (2.3), namely

(2.5) $$ \begin{align} p(z)=\frac{1+(p_{1}+\overline{p_{1}} p_{2}) z+p_{2} z^{2}}{1-(p_{1}-\overline{p_{1}} p_{2}) z-p_{2} z^{2}}. \end{align} $$

For $p_{1}, p_{2} \in \mathbb {D}$ and $p_{3} \in \mathbb {T}$ , there is unique function $p \in \mathcal {P}$ with $c_{1}, c_{2}$ and $c_{3}$ as in (2.2)–(2.4), namely

$$ \begin{align*} p(z)=\frac{1+(\overline{p_{2}} p_{3}+\overline{p_{1}} p_{2}+p_{1}) z+(\overline{p_{1}} p_{3}+p_{1} \overline{p_{2}} p_{3}+p_{2}) z^{2}+p_{3} z^{3}}{1+(\overline{p_{2}} p_{3}+\overline{p_{1}} p_{2}-p_{1}) z+(\overline{p_{1}} p_{3}-p_{1} \overline{p_{2}} p_{3}-p_{2}) z^{2}-p_{3} z^{3}}, \quad z \in \mathbb{D}. \end{align*} $$

Next we recall the following well-known result due to Choi et al. [8].

Lemma 2.2. Let $A, B, C$ be real numbers and

$$ \begin{align*} Y(A, B, C):=\max _{z \in \overline{\mathbb{D}}}(|A+B z+C z^{2}|+1-|z|^{2}). \end{align*} $$

1. (i) If $A C \geq 0$ , then

   $$ \begin{align*} Y(A, B, C)= \begin{cases}|A|+|B|+|C|, & |B| \geq 2(1-|C|), \\ 1+|A|+\displaystyle\frac{B^{2}}{4(1-|C|)}, & |B|<2(1-|C|) .\end{cases} \end{align*} $$

2. (ii) If $A C<0$ , then

   $$ \begin{align*} Y(A, B, C)= \begin{cases}1-|A|+\displaystyle\frac{B^{2}}{4(1-|C|)}, & -4 A C(C^{-2}-1) \leq B^{2} \wedge|B|<2(1-|C|), \\ 1+|A|+\displaystyle\frac{B^{2}}{4(1+|C|)}, & B^{2}<\min \{4(1+|C|)^{2},-4 A C(C^{-2}-1)\}, \\ R(A, B, C), & \text {otherwise, }\end{cases} \end{align*} $$

   where

   $$ \begin{align*} R(A, B, C)= \begin{cases}|A|+|B|+|C|, & |C|(|B|+4|A|) \leq|A B|, \\ -|A|+|B|+|C|, & |A B| \leq|C|(|B|-4|A|), \\ (|A|+|C|) \sqrt{1-\displaystyle\frac{B^{2}}{4 A C}}, & \! \ \text{otherwise.}\end{cases} \end{align*} $$

3. Main results

Now we will prove the first main result of this paper. We obtain the following sharp bound for $H_{2,1}(F_{f^{-1}}/2)$ for functions in the class $\mathcal {C}$ .

Theorem 3.1. For $f\in \mathcal {C}$ given by (1.1),

(3.1) $$ \begin{align} |H_{2,1}(F_{f^{-1}}/2)| \leq \tfrac{1}{33}. \end{align} $$

The inequality is sharp.

Proof. Let $f\in \mathcal {C}$ be of the form (1.1). Then by (1.7),

(3.2) $$ \begin{align} 1+\frac{zf"(z)}{f'(z)}=p(z) \end{align} $$

for some $p \in \mathcal {P}$ of the form (2.1). Since the class $\mathcal {C}$ is invariant under rotation and the function is also rotationally invariant, we can assume that $c_1 \in [0,2]$ . Comparing the coefficients on both sides of (3.2) yields

$$ \begin{align*} & a_2=\tfrac{1}{2}c_1, \\ & a_3=\tfrac{1}{6}(c_2+c_1^2), \\ & a_4=\tfrac{1}{24}(2c_3+3c_1c_2+c_1^3 ). \end{align*} $$

Hence, by (1.5),

$$ \begin{align*} H_{2,1}(F_{f^{-1}}/2)=\tfrac{1}{2304}(11c_1^4-20c_1^2c_2-16c_2^2+24c_1c_3). \end{align*} $$

By (2.2)–(2.4), after simplification,

(3.3) $$ \begin{align} \nonumber H_{2,1}(F_{f^{-1}}/2) &=\frac{p_1^4}{48} -\frac{1}{24}(1-p_1^2)p_1^2p_2-\frac{1}{72}(1-p_1^2)(2+p_1^2)p_2^2 \\ &\quad+\frac{1}{24}(1-p_1^2)(1-|p_1^2|)p_1p_3. \end{align} $$

We consider three cases according to the value of $p_1$ .

Case 1: $p_1=1$ . By (3.3),

$$ \begin{align*}|H_{2,1}(F_{f^{-1}}/2)|=\tfrac{1}{48}.\end{align*} $$

Case 2: $p_1=0$ . By (3.3),

$$ \begin{align*}|H_{2,1}(F_{f^{-1}}/2)|=\tfrac{1}{36}|p_2^2|\leq\tfrac{1}{36}.\end{align*} $$

Case 3: $p_1 \in (0,1)$ . Since $|p_3| \leq 1$ , applying the triangle inequality in (3.3) gives

(3.4) $$ \begin{align} \nonumber |H_{2,1}(F_{f^{-1}}/2)| &=\frac{1}{24}p_1(1-p_1^2)\bigg(\bigg| \frac{p_1^3}{2(1-p_1^2)}-p_1 p_2-\frac{2+p_1^2}{3p_1}p_2^2 \bigg|+1-|p_2^2|\bigg) \notag \\ &\leq \frac{1}{24}p_1(1-p_1^2)(| A+B p_2+C p_2^2 |+1-|p_2^2|), \end{align} $$

where

$$ \begin{align*} A:=\frac{p_1^3}{2(1-p_1^2)}, \quad B:=-p_1,\quad C:=-\frac{2+p_1^2}{3p_1}. \end{align*} $$

Since $AC < 0$ , we can apply Lemma 2.2(ii). The argument now divides into five parts.

3(a). For $p_{1} \in (0,1)$ ,

$$ \begin{align*} -4 A C\bigg(\frac{1}{C^{2}}-1\bigg)-B^{2}=-\frac{p_{1}^{2}(14+p_1^2)}{3(2+p_{1}^{2})} \leq 0. \end{align*} $$

The inequality $|B|<2(1-|C|)$ is equivalent to $p_1(4 - 6 p_1 + 5 p_1^2) < 0$ which is not true for $p_{1} \in (0,1)$ .

3(b). It is easy to check that

$$ \begin{align*} \min \bigg\{4(1+|C|)^{2},-4 A C\bigg(\frac{1}{C^{2}}-1\bigg)\bigg\}=-4 A C\bigg(\frac{1}{C^{2}}-1\bigg), \end{align*} $$

and from 3(a),

$$ \begin{align*} -4 A C\bigg(\frac{1}{C^{2}}-1\bigg) \leq B^{2}. \end{align*} $$

Therefore, the inequality $B^{2} < \min \{4(1+|C|)^{2},-4 A C({1}/{C^{2}}-1)\}$ does not hold for $0<p_1<1$ .

3(c). The inequality $|C|(|B|\hspace{-0.5pt}+\hspace{-0.5pt}4|A|)\hspace{-0.5pt}-\hspace{-0.5pt}|A B|\hspace{-0.5pt} \leq\hspace{-0.5pt} 0$ is equivalent to ${4\hspace{-0.5pt}+\hspace{-0.5pt}6p_1^2\hspace{-0.5pt}-\hspace{-0.5pt}p_1^4\hspace{-0.5pt}\leq\hspace{-0.5pt} 0}$ , which is false for $p_{1} \in (0,1)$ .

3(d). The inequality

$$ \begin{align*}|A B|-|C|(|B|-4|A|)=\frac{9 p_1^4+10 p_1^2-4}{1 - p_1^2} \leq 0 \end{align*} $$

is equivalent to $9 p_1^4+10 p_1^2-4\leq 0$ , which is true for

$$ \begin{align*} 0<p_1 \leq p_1'=\tfrac{1}{3}\sqrt{\sqrt{61}-5}\approx 0.5588. \end{align*} $$

It follows from Lemma 2.2 and (3.4) that

$$ \begin{align*} |H_{2,1}(F_{f^{-1}}/2)| \leq \tfrac{1}{24}p_1(1-p_1^2)(-|A|+|B|+|C|) = \tfrac{1}{144}(4+4p_1^2-11p_1^4)=h(p_1), \end{align*} $$

where $h(x)=4+4x^2-11x^4$ . By a simple calculation, the maximum of the function $h(x)$ for $0<x\leq p_1'$ occurs at the point $x_0=\sqrt {2/11}$ . We conclude that

$$ \begin{align*} |H_{2,1}(F_{f^{-1}}/2)|\leq h\Big(\sqrt{\tfrac{2}{11}}\Big)=\tfrac{1}{33}. \end{align*} $$

3(e). For $p_1'<p_1<1$ , we use the last case of Lemma 2.2 together with (3.4) to obtain

$$ \begin{align*} |H_{2,1}(F_{f^{-1}}/2)| &\leq \frac{1}{24}p_1(1-p_1^2)(|C|+|A|) \sqrt{1-\frac{B^{2}}{4 A C}} \notag \\ &= \frac{1}{144}(p_1^4-2p_1^2+4)\sqrt{\frac{7-p_1^2}{4+2p_1^2}}=k(p_1), \end{align*} $$

where

$$ \begin{align*}k(x)=\frac{1}{144}(x^4-2x^2+4)\sqrt{\frac{7-x^2}{4+2x^2}}. \end{align*} $$

We want to find the maximum of $k(x)$ over the interval $p_1'<x<1$ . Observe that

$$ \begin{align*} k'(x)=\frac{x}{144} \sqrt{\frac{7 - x^2}{ 4 + 2 x^2}}\bigg( \frac{92 - 54x^2 - 15 x^4 + 4 x^6}{(-7 + x^2) (2 + x^2)}\bigg)=0 \end{align*} $$

if and only if $92 - 54x^2 - 15 x^4 + 4 x^6=0$ . However, all the real roots of this equation lie outside the interval $p_1'<x<1$ and $k'(x)<0$ for $p_1'<x<1$ . So k is decreasing and hence $k(x)\leq k(p_1')$ for $p_1'<x<1$ . We conclude that, for $p_1'<x<1,$

$$ \begin{align*} |H_{2,1}(F_{f^{-1}}/2)|\leq k(p_1') \approx 0.0290035. \end{align*} $$

The desired inequality (3.1) follows from Cases 1–3. By tracking back in the proof, we see that equality in (3.1) holds when

$$ \begin{align*} p_1=\sqrt{\tfrac{2}{11}},\quad p_3=1, \end{align*} $$

and

(3.5) $$ \begin{align} |A+Bp_2+Cp_2^2|+1-|p_2^2|=-|A|+|B|+|C|, \end{align} $$

where

$$ \begin{align*} A=\tfrac{1}{9}\sqrt{\tfrac{2}{11}},\quad B=-\sqrt{\tfrac{2}{11}},\quad C=4\sqrt{\tfrac{2}{11}}. \end{align*} $$

Indeed, we can easily verify that one of the solutions of (3.5) is $p_2=1.$ In view of Lemma 2.2, we conclude that equality holds for the function $f\in \mathcal {A}$ given by (1.7), corresponding to the function $p \in \mathcal {P}$ of the form (2.5) with $p_1=\sqrt {2/11},p_2=1$ and $p_3=1$ , that is,

$$ \begin{align*} p(z)=\frac{1+2\sqrt{2/11}z+z^2}{1-z^2}. \end{align*} $$

This complete the proof.

Next, we obtain the sharp bound for $H_{2,1}(F_{f^{-1}}/2)$ for functions in the class $\mathcal {S}^*$ .

Theorem 3.2. For $f\in \mathcal {S}^*$ given by (1.1),

(3.6) $$ \begin{align} |H_{2,1}(F_{f^{-1}}/2)| \leq \tfrac{13}{12}. \end{align} $$

The inequality is sharp.

Proof. Let $f\in \mathcal {S}^*$ be of the form (1.1). By (1.6),

(3.7) $$ \begin{align} \frac{zf'(z)}{f(z)}=p(z) \end{align} $$

for some $p \in \mathcal {P}$ of the form (2.1). By comparing the coefficients on both sides of (3.7),

$$ \begin{align*} & a_2=c_1, \\ & a_3=\tfrac{1}{2}(c_2+c_1^2), \\ & a_4=\tfrac{1}{6}(2c_3+3c_1c_2+c_1^3). \end{align*} $$

Hence, by (1.5),

$$ \begin{align*} H_{2,1}(F_{f^{-1}}/2)=\tfrac{1}{48}\big(6c_1^4-6c_1^2c_2-3c_2^2+4 c_1c_3\big). \end{align*} $$

From (2.2)–(2.4), by straightforward computation,

(3.8) $$ \begin{align} \notag H_{2,1}(F_{f^{-1}}/2)& = \tfrac{13}{12}p_1^4-\tfrac{5}{2}(1-p_1^2)p_2-\tfrac{1}{12}(1-p_1^2)(3+p_1^2)p_2^2 \\[5pt] & \quad +\tfrac{1}{3}(1-p_1^2)(1-|p_1^2|)p_1p_3. \end{align} $$

Now we consider three cases according to the value of $p_1$ .

Case 1: $p_1=1$ . By (3.8),

$$ \begin{align*}|H_{2,1}(F_{f^{-1}}/2)|=\tfrac{13}{12}.\end{align*} $$

Case 2: $p_1=0$ . By (3.8),

$$ \begin{align*}|H_{2,1}(F_{f^{-1}}/2)|=\tfrac{1}{4}|p_2^2|\leq\tfrac{1}{4}.\end{align*} $$

Case 3: $p_1 \in (0,1)$ . Applying the triangle inequality in (3.8) and using the fact that $|p_3| \leq 1$ ,

$$ \begin{align*} |H_{2,1}(F_{f^{-1}}/2)| &=\frac{1}{3}p_1(1-p_1^2)\bigg(\bigg| \frac{13p_1^3}{4(1-p_1^2)}-\frac{5}{2}p_1 p_2-\frac{3+p_1^2}{4p_1}p_2^2 \bigg|+1-|p_2^2|\bigg) \notag \\[5pt] &\leq \frac{1}{24}p_1(1-p_1^2)(| A+B p_2+C p_2^2 |+1-|p_2^2|), \end{align*} $$

where

$$ \begin{align*} A:=\frac{13p_1^3}{4(1-p_1^2)}, \quad B:=-\frac{5}{2}p_1,\quad C:=-\frac{3+p_1^2}{4p_1}. \end{align*} $$

Since $AC < 0$ , we can apply Lemma 2.2(ii).

3(a). For $p_{1} \in (0,1)$ ,

$$ \begin{align*} -4 A C\bigg(\frac{1}{C^{2}}-1\bigg)-B^{2}=-\frac{3p_{1}^{2}(16+p_1^2)}{(3+p_{1}^{2})} \leq 0. \end{align*} $$

The inequality $|B|<2(1-|C|)$ is equivalent to $3 - 4 p_1 + 2 p_1^2 < 0$ which is not true for $p_{1} \in (0,1)$ .

3(b). It is easy to see that

$$ \begin{align*} \min \bigg\{4(1+|C|)^{2},-4 A C\bigg(\frac{1}{C^{2}}-1\bigg)\bigg\}=-4 A C\bigg(\frac{1}{C^{2}}-1\bigg), \end{align*} $$

and from $3(a),$

$$ \begin{align*} -4 A C\bigg(\frac{1}{C^{2}}-1\bigg) \leq B^{2}. \end{align*} $$

Therefore, the inequality $B^{2} < \min \{4(1+|C|)^{2},-4 A C({1}/{C^{2}}-1)\}$ does not hold for $0<p_1<1$ .

3(c). The inequality $|C|(|B|+4|A|)-|A B| \leq 0$ is equivalent to the inequality $44p_1^4-68p_1^2 -16-p_1^4\geq 0$ , which is false for $p_{1} \in (0,1)$ .

3(d). The inequality

$$ \begin{align*}|A B|-|C|(|B|-4|A|)=\frac{96 p_1^4+88 p_1^2-15}{1 - p1^2} \leq 0 \end{align*} $$

is equivalent to $96 p_1^4+88 p_1^2-15\leq 0$ , which is true for

$$ \begin{align*} 0<p_1\leq p_1"=\frac{1}{2}\sqrt{\frac{\sqrt{211}-11}{6}} \approx 0.38328. \end{align*} $$

From (3.7) and Lemma 2.2,

(3.9) $$ \begin{align}| H_{2,1}(F_{f^{-1}}/2)| \leq \tfrac{1}{3}p_1(1-p_1^2)(-|A|+|B|+|C|) = \tfrac{1}{12}(3+8p_1^2-24p_1^4)=h(p_1), \end{align} $$

where $h(x)=3+8x^2-24x^4$ . Since $h'(x)>0$ in $ 0< x\leq p_1"$ , we have $h(x)\leq h(p_1")$ for $ 0< x\leq p_1"$ . Therefore,

$$ \begin{align*} |H_{2,1}(F_{f^{-1}}/2)|\leq \tfrac{1}{48}(-58+5\sqrt{211}) \approx 0.304775. \end{align*} $$

3(e). Furthermore, for $p_1"<p_1<1$ , from (3.8) and Lemma 2.2,

$$ \begin{align*} |H_{2,1}(F_{f^{-1}}/2)| &\leq \frac{1}{24}p_1(1-p_1^2)(|C|+|A|) \sqrt{1-\frac{B^{2}}{4 A C}} \notag \\ &= \frac{1}{6}(12p_1^4 -2p_1^2+3)\sqrt{\frac{16-3p_1^2}{39+13p_1^2}}=k(p_1), \end{align*} $$

where

$$ \begin{align*}k(x)=\frac{1}{6}\sqrt{\frac{16-3x^2}{39+13x^2}}(12x^4 -2x^2+3). \end{align*} $$

As $k'(x)=0$ has no solution in $(p_1",1)$ and $k'(x)>0$ , the maximum occurs at $x=1$ and we conclude that

$$ \begin{align*} |H_{2,1}(F_{f^{-1}}/2)| \leq k(1)=\tfrac{13}{12} \quad \text{for} \ p_1"<x<1. \end{align*} $$

The desired inequality (3.6) follows from Cases 1–3. For the equality, consider the Koebe function

$$ \begin{align*} k(z)=\frac{z}{(1-z)^2}. \end{align*} $$

Clearly, $k \in \mathcal {S}^*$ and it is easy to show that

$$ \begin{align*} |H_{2,1}(F_{k^{-1}}/2)|=\tfrac{13}{12}. \end{align*} $$

This completes the proof.