The Second Hankel Determinant of Logarithmic Coefficients for Strongly Ozaki Close-to-Convex Functions

Abstract

The aim of this paper is to determine sharp bound for the second Hankel determinant of logarithmic coefficients $H_{2,1}(F_{f}/2)$ of strongly Ozaki close-to-convex functions in the open unit disk. Furthermore, sharp bound of $H_{2,1}(F_{f^{-1}}/2)$, where $f^{-1}$ is the inverse function of f, is also computed. The results show an invariance property of the second Hankel determinants of logarithmic coefficients $H_{2,1}(F_{f}/2)$ and $H_{2,1}(F_{f^{-1}}/2)$ for strongly convex functions.

1 Introduction

Let $\mathscr {A}$ denote the class of analytic functions of the form

$$\begin{aligned} f(z) =z+\sum _{k=2}^{\infty }a_{k}z^{k}, \qquad (z\in \mathbb {U} :=\{ z\in \mathbb {C} : |z| <1\}), \end{aligned}$$

(1.1)

and let $\mathscr {S}$ be the class of functions in $\mathscr {A}$ which are univalent in $\mathbb {U}$.

A function f of the form (1.1) is said to be starlike of order $\alpha$, $(0\le \alpha <1)$, in $\mathbb {U}$ if

$$\begin{aligned} {{\,\textrm{Re}\,}}\frac{zf'(z)}{f(z)}>\alpha \qquad (z\in \mathbb {U}). \end{aligned}$$

The set of all such functions is denoted by $\mathscr {S}^{*}(\alpha )$.

By $\mathscr {K}(\alpha )$, we denote the class of convex functions of order $\alpha$ $(\alpha <1)$, in $\mathbb {U}$ that satisfy the following inequality:

$$\begin{aligned} {{\,\textrm{Re}\,}}\left\{ 1+\frac{zf''(z)}{f'(z)}\right\} >\alpha \qquad (z\in \mathbb {U}). \end{aligned}$$

For $\alpha :=0$, these classes reduce to the well-known classes $\mathscr {S}^{*}$ and $\mathscr {K}$, the class of starlike functions and the class of convex functions, respectively.

Moreover, a function f of the form (1.1) is said to be strongly convex of order $\alpha$, $(0<\alpha \le 1)$, in $\mathbb {U}$ if

$$\begin{aligned} \left| {{\,\textrm{arg}\,}}\left\{ 1+\frac{zf''(z)}{f'(z)}\right\} \right| <\frac{\pi \alpha }{2}\qquad (z\in \mathbb {U}). \end{aligned}$$

The set of all such functions is denoted by $\mathscr {K}_{c}(\alpha )$.

A function $f\in \mathscr {A}$ belongs to $\mathscr {C}$, the class of close-to-convex functions in $\mathbb {U},$ if and only if there exists $g\in \mathscr {S}^{*}$ and $\theta \in (-\pi /2,\pi /2)$ such that

$$\begin{aligned} {{\,\textrm{Re}\,}}\left\{ \textrm{e}^{\textrm{i}\theta }\frac{zf'(z)}{g(z)}\right\} >0 \qquad (z\in \mathbb {U}). \end{aligned}$$

Geometrically, f is close-to-convex if and only if the image of $C_R:=\{z\in \mathbb {C}: |z|=R\}$ for every $R\in (0,1),$ has no “hairpin turns”; that is, there are no sections of the curve $f(C_{R})$ in which the tangent vector turns backward through an angle $\ge \pi$.

Although the class of close-to-convex functions was introduced by Kaplan [12] in 1952, in 1935 Ozaki [21, 22] had already considered the functions in $\mathscr {A}$ satisfying the following condition:

$$\begin{aligned} {{\,\textrm{Re}\,}}\left\{ 1+\frac{zf''(z)}{f'(z)}\right\} >-\frac{1}{2}, \qquad (z\in \mathbb {U}). \end{aligned}$$

(1.2)

Functions satisfying the inequality (1.2) are close-to-convex, and therefore, they are in $\mathscr {S}$ by the definition of Kaplan [12].

Recently, Kargar and Ebadian [13] generalized Ozaki’s condition as follows:

Definition 1

[13] Let $\mathscr {F}(\lambda )$ for $-1/2<\lambda \le 1$, denote the class of locally univalent normalized analytic functions f in the unit disk satisfying the condition

$$\begin{aligned} {{\,\textrm{Re}\,}}\left\{ 1+\frac{zf''(z)}{f'(z)}\right\} >\frac{1}{2}-\lambda , \qquad (z\in \mathbb {U}). \end{aligned}$$

When $1/2\le \lambda \le 1$, the functions in $\mathscr {F}(\lambda )$ are called Ozaki close-to-convex. The class $\mathscr {F}(1)$ was studied by Ponnusamy et al. [23]. Also, $\mathscr {F}(1/2)=\mathscr {K}$. Clearly, $\mathscr {F}(\lambda )\subset \mathscr {K} \subset \mathscr {S}^{*}$ for all $\lambda \in (-1/2,1/2)$.

Recently, Allu et al. extended the class $\mathscr {F}(\lambda )$ as follows:

Definition 2

[3, 31] Let $0<\alpha \le 1$ and $1/2 \le \lambda \le 1.$ Then $f\in \mathscr {A}$ is called strongly Ozaki-close-to-convex if and only if

$$\begin{aligned} \left| {{\,\textrm{arg}\,}}\left\{ \frac{2\lambda -1}{2\lambda +1}+\frac{2}{2\lambda +1}\left( 1+\frac{zf''(z)}{f'(z)}\right) \right\} \right| <\frac{\alpha \pi }{2}, \qquad (z\in \mathbb {U}). \end{aligned}$$

(1.3)

This class is denoted by $\mathscr {F}_{O}(\lambda ,\alpha )$.

The class $\mathscr {F}_{O}(\lambda ,\alpha )$ is the subclass of $\mathscr {S}$, and it is obvious that $\mathscr {F}_{O}(1/2,\alpha )=\mathscr {K}_{c}(\alpha )$ (see [3]).

Associated with each $f \in \mathscr {S}$ is a function

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

(1.4)

The numbers $\gamma _{k}$ are called the logarithmic coefficients of f. It is well known that the logarithmic coefficients play a crucial role in Milin conjecture (cf. [20], see also [9, p. 155]). It is surprising that for the class $\mathscr {S}$ the sharp estimates of single logarithmic coefficients are known only for two initial ones, namely

$$\begin{aligned} |\gamma _{1}|\le 1 \qquad \text {and} \qquad |\gamma _{2}|\le \frac{1}{2}+\frac{1}{\textrm{e}^{2}}=0.6353\dots \end{aligned}$$

and are unknown for $k\ge 3.$ Recently, logarithmic coefficients have been studied by many researches and upper bounds of logarithmic coefficients of functions in various subclasses of $\mathscr {S}$ have been obtained (e.g., [1, 2, 6, 17, 30, 34]). For a summary of some of the significant results concerning the logarithmic coefficients for univalent functions, we refer to [32].

Since each class $\mathscr {F}_{O}(\lambda ,\alpha )$ is compact and $f(0)=f'(0)-1=0$ for every $f\in \mathscr {F}_{O}(\lambda ,\alpha )$, there exists $r_{0}\in (0,1)$ such that $\mathbb {U}_{r_{0}}:=\{z\in \mathbb {C}: |z|<r_{0}\}\subset f(\mathbb {U})$ for every $f\in \mathscr {F}_{O}(\lambda ,\alpha )$. Thus, every function in $\mathscr {F}_{O}(\lambda ,\alpha )$ is invertible and

$$\begin{aligned} \begin{aligned} f^{-1}(w)&=w+\sum _{k=2}^{\infty }\delta _{k}w^{k}\\&=w-a_{2}w^{2}+(2a_{2}^{2}-a_{3})w^{3}-(5a_{2}^{3}-5a_{2}a_{3}+a_{4})w^{4}+\cdots ,\\&\qquad (w\in \mathbb {U}_{r_0}), \end{aligned} \end{aligned}$$

(1.5)

in $\mathbb {U}_{r_0}$ (see, e.g., [10, pp. 56-57]). Therefore for each $f \in \mathscr {F}_{O}(\lambda ,\alpha )$ we can define

$$\begin{aligned} F_{f^{-1}}(w):=\log \frac{f^{-1}(w)}{w}=2\sum _{k=1}^{\infty }\Gamma _{k}w^{k}, \qquad (w\in \mathbb {U}_{r_0}). \end{aligned}$$

(1.6)

The numbers $\Gamma _{k}$ can be called as the logarithmic coefficients of the inverse function of f.

For $q,n\in \mathbb {N}$, the Hankel determinant $H_{q,n}(f)$ of $f\in \mathscr {A}$ of form (1.1) is defined as

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

(1.7)

The Hankel determinant $H_{2,1}(f)=a_{3}-a_{2}^{2}$ is the well-known Fekete–Szegö functional. The second Hankel determinant $H_{2,2}(f)$ is given by $H_{2,2}(f)=a_{2}a_{4}-a_{3}^{2}$.

The problem of computing the upper bound of $|H_{q,n}(f)|$ over various subfamilies of $\mathscr {A}$ is interesting and widely studied in Geometric Function Theory. Sharp upper bounds of $|H_{2,2}(f)|$ and $|H_{3,1}(f)|$ for subclasses of analytic functions were obtained by various authors [7, 11, 16, 18, 19, 25,26,27].

Very recently, Kowalczyk and Lecko [14] introduced the Hankel determinant $H_{q,n}(F_{f}/2)$, which entries are logarithmic coefficients of f, i.e., $H_{q,n}(F_{f}/2)$ is of the form (1.7) with $a_n$ replaced by $\gamma _n.$ Similarly, we can define the determinant $H_{q,n}(F_{f^{-1}}/2)$ by replacing $a_n$ by $\Gamma _n$ in (1.7).

For a function $f\in \mathscr {S}$ given in (1.1), by differentiating (1.4), one can obtain

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

Therefore,

$$\begin{aligned} H_{2,1}(F_{f}/2)=\gamma _{1}\gamma _{3}-\gamma _{2}^{2}=\frac{1}{4}\left( a_{2}a_{4}-a_{3}^{2}+\frac{1}{12}a_{2}^{4}\right) . \end{aligned}$$

(1.8)

Furthermore, if $f\in \mathscr {S}$, then for $f_\theta \in \mathscr {S},$ $\theta \in \mathbb {R},$ defined as

$$\begin{aligned} f_{\theta }(z):= \textrm{e}^{-\textrm{i}\theta } f(\textrm{e}^{\textrm{i}\theta }z) \qquad (z\in \mathbb {U}), \end{aligned}$$

we find that (see [15])

$$\begin{aligned} H_{2,1}\left( \frac{1}{2}F_{f_{\theta }}\right) =\textrm{e}^{4\textrm{i}\theta }H_{2,1}\left( \frac{1}{2}F_{f}\right) . \end{aligned}$$

Kowalczyk and Lecko [15] obtained sharp bounds for $|H_{2,1}(F_{f}/2)|$ for the classes of starlike and convex functions of order $\alpha$. The problem of computing the sharp bounds of $|H_{2,1}(F_{f}/2)|$ for strongly starlike and strongly convex functions has been considered by Sümer Eker et. al. [29]. Furthermore, upper bounds for the second Hankel determinant of logarithmic coefficients for some different subclasses of class $\mathscr {S}$ have been obtained by Srivastava et al. [28] and Allu and Arora [4].

For a function $f\in \mathscr {S}$ given in (1.1), by differentiating (1.6) together with (1.5), one can obtain

$$\begin{aligned} \Gamma _1=-\frac{1}{2}a_2,\quad \Gamma _2=-\frac{1}{2}a_3+\frac{3}{4}a_2^2,\quad \Gamma _3=-\frac{1}{2}a_4+2a_2a_3-\frac{5}{3}a_2^3. \end{aligned}$$

Therefore,

$$\begin{aligned} H_{2,1}(F_{f^{-1}}/2)=\Gamma _{1}\Gamma _{3}-\Gamma _{2}^{2}=\frac{1}{4}\left( a_{2}a_{4}-a_{3}^{2}-a_2^2a_3+\frac{13}{12}a_{2}^{4}\right) . \end{aligned}$$

(1.9)

The aim of this paper is to give the sharp bounds for $|H_{2,1}(F_{f}/2)|$ and $|H_{2,1}(F_{f^{-1}}/2)|$ for the class of strongly Ozaki close-to-convex functions.

Let $\mathscr {P}$ denote the class of analytic functions p in $\mathbb {U}$ satisfying $p(0)=1$ and ${{\,\textrm{Re}\,}}p(z)>0$ for $z\in \mathbb {U}.$ Thus, every $p\in \mathscr {P}$ can be represented as

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

(1.10)

Elements of $\mathscr {P}$ are called Carathéodory functions.

To establish our main results, we will require the following lemmas.

Lemma 1

([5] (see also [15])) If $p\in \mathscr {P}$ is of the form (1.10) with $c_{1}\ge 0$, then

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

(1.11)

for some $d_{1}\in [0,1]$ and $d_{2},d_{3}\in \overline{\mathbb {U}}:=\left\{ z\in \mathbb {C}: |z| \le 1\right\}$.

For $d_{1}\in \mathbb {U}$ and $d_{2}\in \partial {\mathbb {U}}:=\left\{ z\in \mathbb {C}: |z|=1\right\}$, there is a unique function $p\in \mathscr {P}$ with $c_{1}$ and $c_{2}$ as in (1.11), namely

$$\begin{aligned} p(z)=\frac{1+(\overline{d_{1}}d_{2}+d_{1})z+d_{2}z^{2}}{1+(\overline{d_{1}}d_{2}-d_{1})z-d_{2}z^{2}}, \qquad (z\in \mathbb {U}). \end{aligned}$$

Lemma 2

[8] Given real numbers A, B, C, let

$$\begin{aligned} Y(A,B,C):=\max \left\{ \big |A+Bz+Cz^{2}\big |+1-|z|^{2}: z\in \overline{\mathbb {U}}\right\} . \end{aligned}$$

I. If $AC\ge 0$, then

$$\begin{aligned} Y(A,B,C)= \left\{ \begin{array}{ll} |A|+|B|+|C|, &{}\quad |B|\ge 2(1-|C|),\\ 1+|A|+\dfrac{B^{2}}{4(1-|C|)}, &{}\quad |B|<2(1-|C|). \end{array} \right. \end{aligned}$$

II. If $AC<0$, then

$$\begin{aligned} Y(A,B,C)= \left\{ \begin{array}{ll} 1-|A|+\dfrac{B^{2}}{4(1-|C|)}, &{}\quad -4AC(C^{-2}-1)\le B^{2}\wedge |B|<2(1-|C|),\\ 1+|A|+\dfrac{B^{2}}{4(1+|C|)}, &{}\quad B^{2}<\min \{4(1+|C|)^{2},-4AC(C^{-2}-1)\}\\ R(A,B,C), &{}\quad \mathrm {otherwise. }\end{array} \right. \end{aligned}$$

where

$$\begin{aligned} R(A,B,C):= \left\{ \begin{array}{ll} |A|+|B|-|C|, &{}\quad |C|(|B|+4|A|)\le |AB|,\\ -|A|+|B|+|C|, &{}\quad |AB|\le |C|(|B|-4|A|),\\ (|A|+|C|)\sqrt{1-\dfrac{B^{2}}{4AC}}, &{}\quad \mathrm {otherwise.}\end{array} \right. \end{aligned}$$

2 Second Hankel Determinant of Logarithmic Coefficients for Strongly Ozaki Close-to-Convex Functions

Theorem 1

Let $\alpha \in (0,1]$ and $\lambda \in [1/2,1]$. If $f\in \mathscr {F}_{O}(\lambda ,\alpha )$, then

$$\begin{aligned} \left| H_{2,1}(F_{f}/2)\right| \le \left\{ \begin{array}{ll} \dfrac{\alpha ^{2}(1+2\lambda )^{2}}{144}, &{}\quad F\le 2, \\ \dfrac{\alpha ^{2}(1+2\lambda )^{2}}{576}\left( 4+\dfrac{(F-2)^{2}}{16+4F-E}\right) , &{}\quad F>2, \end{array} \right. \end{aligned}$$

(2.1)

where $E:=\alpha ^{2}(4\lambda ^{2}-4\lambda -3)$ and $F:=\alpha (5+2\lambda )$. The inequalities in (2.1) are sharp.

Proof

Let $\alpha \in (0,1],$ $\lambda \in [1/2,1]$ and $f\in \mathscr {F}_{O}(\lambda ,\alpha )$ be of the form (1.1). Then by (1.3), we have

$$\begin{aligned} \frac{2\lambda -1}{2\lambda +1}+\frac{2}{2\lambda +1}\left( 1+\frac{zf''(z)}{f'(z)}\right) =(p(z))^{\alpha }, \qquad (z\in \mathbb {U}), \end{aligned}$$

(2.2)

for some function $p\in \mathscr {P}$ of the form (1.10). So equating coefficients we obtain

$$\begin{aligned} \begin{aligned}&a_{2}=\frac{\alpha (1+2\lambda )}{4} c_{1}, \\ {}&a_{3}=\frac{\alpha (1+2\lambda )}{24}\bigg (2c_{2}+(2\alpha +2\alpha \lambda -1)c_{1}^{2}\bigg ),\\ {}&a_{4}=\frac{\alpha (1+2\lambda )}{576} \bigg ((8-21\alpha +16\alpha ^2-18\alpha \lambda +30\alpha ^{2}\lambda +12\alpha ^{2}\lambda ^{2})c_{1}^{3}\\&\qquad -6(4-7\alpha -6\alpha \lambda )c_{1}c_{2}+24c_{3}\bigg ). \end{aligned} \end{aligned}$$

(2.3)

Since the class $\mathscr {F}_{O}(\lambda ,\alpha )$ and $|H_{2,1}(F_{f}/2)|$ are rotationally invariant, without loss of generality we may assume that $a_2\ge 0$, so $c:=c_{1}\in [0,2]$ (i.e., in view of (1.11) that $d_{1}\in [0,1])$. By using (1.8), (2.3) and (1.11), we obtain

$$\begin{aligned} \begin{aligned} \gamma _{1}\gamma _{3}-\gamma _{2}^{2}&=\frac{1}{4}\left( a_{2}a_{4}-a_{3}^{2}+\frac{1}{12}a_{2}^{4}\right) \\&= \frac{\alpha ^{2}(1+2\lambda )^{2}}{2304}\left[ (8-E)d_{1}^{4}+4F (1-d_{1}^{2})d_{1}^{2}d_{2}\right. \\&\quad \left. -8(1-d_{1}^{2})(d_{1}^{2}+2)d_{2}^{2}+ 24(1-d_{1}^{2})(1-|d_{2}|^{2})d_{1}d_{3}\right] , \end{aligned} \end{aligned}$$

(2.4)

where $E=\alpha ^{2}(4\lambda ^{2}-4\lambda -3)$ and $F=\alpha (5+2\lambda )$.

Now, we may have the following cases on $d_{1}$.

Case 1. Suppose that $d_{1}=1$. Then by (2.4) we obtain

$$\begin{aligned} \left| \gamma _{1}\gamma _{3}-\gamma _{2}^{2}\right| =\frac{\alpha ^{2}(1+2\lambda )^{2}}{2304}(8-E) \end{aligned}$$

Case 2. Suppose that $d_{1}=0$. Then by (2.4) we obtain

$$\begin{aligned} \left| \gamma _{1}\gamma _{3}-\gamma _{2}^{2}\right| =\frac{\alpha ^{2}(1+2\lambda )^{2}}{144}|d_{2}|^{2}\le \frac{\alpha ^{2}(1+2\lambda )^{2}}{144}. \end{aligned}$$

Case 3. Suppose that $d_{1}\in (0,1)$. By the fact that $|d_{3}|\le 1$, applying the triangle inequality to (2.4) we can write

$$\begin{aligned} \left| \gamma _{1}\gamma _{3}-\gamma _{2}^{2}\right|&=\left| \frac{\alpha ^{2}(1+2\lambda )^{2}}{2304}\left[ (8-E)d_{1}^{4}+4F (1-d_{1}^{2})d_{1}^{2}d_{2}\right. \right. \nonumber \\&\left. \quad -8(1-d_{1}^{2})(d_{1}^{2}+2)d_{2}^{2}+ 24(1-d_{1}^{2})(1-|d_{2}|^{2})d_{1}d_{3}\right] \Big |\nonumber \\&\le \frac{\alpha ^{2}(1+2\lambda )^{2}d_{1}(1-d_{1}^{2})}{96}\left[ \bigg |\frac{8-E}{24(1-d_{1}^{2})}d_{1}^{3}+ \frac{F}{6} d_{1}d_{2}-\frac{d_{1}^{2}+2}{3d_{1}}d_{2}^{2}\bigg |+1-|d_{2}|^{2}\right] \nonumber \\&= \frac{\alpha ^{2}(1+2\lambda )^{2}d_{1}(1-d_{1}^{2})}{96}\left[ \left| A+Bd_{2}+Cd_{2}^{2}\right| +1-|d_{2}|^{2}\right] \end{aligned}$$

(2.5)

where

$$\begin{aligned} A:=\frac{8-E}{24(1-d_{1}^{2})}d_{1}^{3},\quad B:=\frac{F}{6} d_{1} \quad \text {and} \quad C:=-\frac{d_{1}^{2}+2}{3d_{1}}. \end{aligned}$$

Since $AC < 0$, we apply Lemma 2 only for the case II.

We consider the following sub-cases.

3(a) Note that

$$\begin{aligned} |B|-2(1-|C|)=\frac{1}{6d_{1}}\left[ 4(1-d_{1})(2-d_{1})+Fd_{1}^{2}\right] >0. \end{aligned}$$

Therefore, $|B|<2(1-|C|)$ does not hold for $d_{1}\in (0,1)$, $\lambda \in [1/2,1]$ and $\alpha \in (0,1]$.

3(b) We can easily see that

$$\begin{aligned} 4\big (1+|C|\big )^{2}>0. \end{aligned}$$

Furthermore, since $AC<0$ and

$$\begin{aligned} \frac{1}{C^{2}}-1=-\frac{(1-d_{1}^{2})(4-d_{1}^{2})}{(d_{1}^{2}+2)^{2}}<0, \end{aligned}$$

the inequality

$$\begin{aligned} B^2<\min \left\{ 4(1+|C|)^{2},-4AC\left( \frac{1}{C^{2}}-1\right) \right\} \end{aligned}$$

is false for $d_{1}\in (0,1)$, $\lambda \in [1/2,1]$ and $\alpha \in (0,1]$.

3(c) Since $0<F\le 7$, we obtain

$$\begin{aligned} 4|C|-|B|=\frac{1}{6d_{1}}\left( (8-F)d_{1}^{2}+16\right) >0, \end{aligned}$$

and this implies

$$\begin{aligned} |C|(|B|+4|A|)-|AB|=|BC|+|A|(4|C|-|B|)>0. \end{aligned}$$

Consequently, the inequality $|C|(|B|+4|A|)\le |AB|$ does not hold for $d_{1}\in (0,1)$, $\lambda \in [1/2,1]$ and $\alpha \in (0,1]$.

3(d) We can write

$$\begin{aligned} \begin{aligned}|AB|-|C|(|B|-4|A|)&= \frac{(8-E)F}{144(1-d_{1}^{2})}d_{1}^{4}-\frac{d_{1}^{2}+2}{3d_{1}}\left( \frac{F}{6}d_{1}-\frac{8-E}{6(1-d_{1}^{2})}d_{1}^{3}\right) \\&=\frac{1}{144(1-t)}( Kt^{2}+Lt+M), \end{aligned} \end{aligned}$$

where $t:=d_{1}^{2}\in (0,1)$ and

$$\begin{aligned} K := 64+16F-8E-EF,\quad L :=128+8F-16E ,\quad M :=-16F. \end{aligned}$$

Since $-4\le E<0$ and $0<F\le 7$, it is easy to see that $K>0$, $L>0$ and $M<0$ for $\lambda \in [1/2,1]$ and $\alpha \in (0,1]$.

For the equation $Kt^{2}+Lt+M=0$, we have $\Delta >0$. Since

$$\begin{aligned} \frac{M}{K}<0\quad \text {and}\quad K+L+M>0, \end{aligned}$$

for $\lambda \in [1/2,1]$ and $\alpha \in (0,1]$, the equation $Kt^{2}+Lt+M=0$ has a unique positive root $t_{1}<1$. Thus, the inequality $|AB|-|C|\left( |B|-4|A|\right) \le 0$ holds for $(0,d_{1}^{*}]$, where $d_{1}^{*}=\sqrt{t_{1}}$. So we can write from (2.5) and Lemma 2,

$$\begin{aligned} \begin{aligned} \left| \gamma _{1}\gamma _{3}-\gamma _{2}^{2}\right|&\le \frac{\alpha ^{2}(1+2\lambda )^{2}d_{1}(1-d_{1}^{2})}{96}\left( -|A|+|B|+|C|\right) \\&=\frac{\alpha ^{2}(1+2\lambda )^2}{2304}\Phi (d_{1}), \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \Phi (x):=(E-4F-16)x^{4}+4(F-2)x^{2}+16,\quad x\in [0,d_{1}^{*}]. \end{aligned}$$

(2.6)

We note that $\Phi '(x)=0$ for $x\in (0,d_{1}^{*})$ holds only for

$$\begin{aligned} x=\sqrt{\frac{2(F-2)}{16+4F-E}}=:\xi , \end{aligned}$$

(2.7)

in the case when $F-2>0$. Clearly $\xi >0$. Now, we will show that $0<\xi <d_{1}^{*}$. Since $-4\le E<0$ and $0<F\le 7$, we obtain

$$\begin{aligned} \begin{aligned} K\xi ^{4}+L\xi ^{2}+M=&-\frac{4}{(16+4F-E)^{2}}\left[ F^{3}(E+32)+8F^{2}(E+28)\right. \\&\left. +4F(208-9E-E^{2})+16E^{2}-352E+1792\right] <0, \end{aligned} \end{aligned}$$

which confirms that $0<\xi <d_{1}^{*}$. Moreover, the function $\Phi$ attains its maximum value at $\xi$ on $[0,d_{1}^{*}].$ Thus for $F-2>0$, we obtain

$$\begin{aligned} \begin{aligned} \left| \gamma _{1}\gamma _{3}-\gamma _{2}^{2}\right|&\le \frac{\alpha ^{2}(1+2\lambda )^{2}}{2304}\Phi (\xi )\\&= \frac{\alpha ^{2}(1+2\lambda )^{2}}{576}\left( 4+\frac{(F-2)^{2}}{16+4F-E}\right) . \end{aligned} \end{aligned}$$

Furthermore, if $F-2\le 0$, then the function $\Phi$ is decreasing on $[0,d_{1}^{*}]$. Thus, we have

$$\begin{aligned} \begin{aligned} \left| \gamma _{1}\gamma _{3}-\gamma _{2}^{2}\right|&\le \frac{\alpha ^{2}(1+2\lambda )^{2}}{2304}\Phi (d_{1})\\&\le \frac{\alpha ^{2}(1+2\lambda )^{2}}{2304}\Phi (0)=\frac{\alpha ^{2}(1+2\lambda )^{2}}{144}. \end{aligned} \end{aligned}$$

3(e) Next consider the case $d_{1}\in [d_{1}^{*},1)$. Using the last case of the Lemma 2,

$$\begin{aligned} \begin{aligned} \left| \gamma _{1}\gamma _{3}-\gamma _{2}^{2}\right|&\le \frac{\alpha ^{2}(1+2\lambda )^{2}d_{1}(1-d_{1}^{2})}{96}\left( (|A|+|C|)\sqrt{1-\frac{B^{2}}{4AC}}\right) \\&=\frac{\alpha ^{2}(1+2\lambda )^{2}}{2304}\Psi (d_{1}), \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \Psi (x):= (16-8x^{2}-Ex^{4})\sqrt{1+\frac{F^{2}(1-x^{2})}{2(8-E)(x^{2}+2)}},\quad x\in [d_{1}^{*},1]. \end{aligned}$$

For $x\in [d_{1}^{*},1]$, we have

$$\begin{aligned} \begin{aligned} \Psi ^{\prime }(x)=&(-16x-4Ex^{3})\sqrt{1+\frac{F^{2}(1-x^{2})}{2(8-E)(x^{2}+2)}}\\&+(Ex^{4}+8x^{2}-16)\dfrac{3F^{2}x}{2(8-E)(x^{2}+2)^{2}\sqrt{1+\dfrac{F^{2}(1-x^{2})}{2(8-E)(x^{2}+2)}}}. \end{aligned} \end{aligned}$$

Since for $-4\le E<0,$

$$\begin{aligned} -16x-4Ex^{3}<0 \end{aligned}$$

and

$$\begin{aligned} Ex^{4}+8x^{2}-16<8(x^{2}-2)<0 \end{aligned}$$

for $\alpha \in (0,1]$ and $x\in [d_{1}^{*},1]$, we deduce that $\Psi$ is a decreasing function. This implies that

$$\begin{aligned} \begin{aligned} \left| \gamma _{1}\gamma _{3}-\gamma _{2}^{2}\right|&\le \frac{\alpha ^{2}(1+2\lambda )^{2}}{2304}\Psi (d_{1})\\&\le \frac{\alpha ^{2}(1+2\lambda )^{2}}{2304}\Psi (d_{1}^{*})\\&= \frac{\alpha ^{2}(1+2\lambda )^{2}}{2304}\Phi (d_{1}^{*}), \end{aligned} \end{aligned}$$

where $\Phi$ is given in (2.6).

Summarizing parts from Case 1-3, it follows the inequalities (2.1).

To show the sharpness for the case $F-2\le 0$, consider the function

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

It is obvious that the function p is in $\mathscr {P}$ with $c_{1}=c_{3}=0$ and $c_{2}=-2$. The corresponding function $f\in \mathscr {F}_{O}(\lambda ,\alpha )$ is described by (2.2). Hence by (2.3) it follows that $a_{2}=a_{4}=0$ and $a_{3}=-\alpha (1+2\lambda )/6$. From (2.4), we obtain

$$\begin{aligned} \left| \gamma _{1}\gamma _{3}-\gamma _{2}^{2}\right| = \frac{\alpha ^{2}(1+2\lambda )^{2}}{144}. \end{aligned}$$

For the case $F-2>0$, consider the function

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

where $\xi$ is given by (2.7). From Lemma 1, it follows that $p\in \mathscr {P}$. The corresponding function $f\in \mathscr {F}_{O}(\lambda ,\alpha )$ is described by (2.2) and has the following coefficients

$$\begin{aligned} \begin{aligned} a_{2} =&\frac{1}{2}\alpha \xi (1+2\lambda ), \\ a_{3} =&\frac{1}{6}\alpha (1+2\lambda ) \big [-1+\xi ^{2}(1+2\alpha +2\alpha \lambda ) \big ], \\ a_{4} =&\frac{1}{72}\alpha (1+2\lambda ) \big [-3\xi (2+7\alpha +6\alpha \lambda )\\&+\xi ^{3}(8+21\alpha +18\alpha \lambda +16\alpha ^{2}+30\alpha ^{2}\lambda +12\alpha ^{2}\lambda ^{2}\big )\big ]. \end{aligned} \end{aligned}$$

Hence, from (2.4) we obtain

$$\begin{aligned} \left| \gamma _{1}\gamma _{3}-\gamma _{2}^{2}\right| =\frac{\alpha ^{2}(1+2\lambda )^{2}}{576}\left( 4+\frac{(F-2)^{2}}{16+4F-E}\right) . \end{aligned}$$

This completes the proof.

For $\lambda = 1/2$, we get the bounds for the class $\mathscr {K}_{c}(\alpha )$ given in [29].

Corollary 1

Let $\alpha \in (0,1]$. If $f\in \mathscr {K}_{c}(\alpha )$, then

$$\begin{aligned} \left| \gamma _{1}\gamma _{3}-\gamma _{2}^{2}\right| \le \left\{ \begin{array}{ll} \dfrac{\alpha ^{2}}{36}, &{}\quad 0<\alpha \le \dfrac{1}{3}, \\ \dfrac{\alpha ^{2}(13\alpha ^{2}+18\alpha +17)}{144(\alpha ^{2}+6\alpha +4)}, &{}\quad \dfrac{1}{3}<\alpha \le 1. \end{array} \right. \end{aligned}$$

The inequalities are sharp.

3 Second Hankel Determinant of Logarithmic Coefficients for Inverse Functions

The following lemma will be used in the proof of the main result of this section.

Lemma 3

Let $T\in (2,11]$ and $S\in (4,39]$. Define $H:[0,1]\rightarrow \mathbb {R}$ by

$$\begin{aligned} H(x):=h_{1}(x)\sqrt{h_{2}(x)}, \end{aligned}$$

where for $x\in [0,1],$

$$\begin{aligned} h_{1}(x):= Sx^{2}-8x+16 \quad \text {and}\quad h_{2}(x):=1+\frac{T^{2}(1-x)}{2(S+8)(x+2)}. \end{aligned}$$

Then H is a convex function.

Proof

To prove the lemma, we will use the same method as in [24, p. 2524]. By differentiating H twice, we obtain

$$\begin{aligned} \begin{aligned} (h_{2}(x))^{3/2}H''(x) =&h_{1}''(x)\big (h_{2}(x)\big )^{2}+h_{1}'(x)h_{2}(x)h_{2}'(x)\\&+\frac{1}{2}h_{1}(x)h_{2}(x)h_{2}''(x)-\frac{1}{4}h_{1}(x)\big (h_{2}'(x)\big )^{2} \\ =&\frac{G(x)}{16(S+8)^{2}(x+2)^{4}}, \end{aligned} \end{aligned}$$

where for $x\in [0,1],$

$$\begin{aligned} \begin{aligned} G(x):=&\big [(8{x}^{4}+28{x}^{3}+3{x}^{2}-80x+32)S-312x+240\big ] T^{4}\\&+\big [\big (-32x^{4}-184x^{3}-336x^{2}-64x+256\big )S^{2}\\&+ \big (-256x^{4}-1472x^{3}-2688x^{2}+256x+3584\big )S+6144x+12288\big ]T^{2}\\&+\big ( 32{x}^{4}+256{x}^{3}+768{x}^{2}+1024x+512 \big ){S}^{3}\\&+\big (512{x}^{4}+4096{x}^{3}+12288{x}^{2}+16384x+8192\big ){S}^{2}\\&+\big (2048{x}^{4}+16384{x}^{3}+49152{x}^{2}+65536x+32768\big )S. \end{aligned} \end{aligned}$$

We show that our assertion is true by proving that $G(x)\ge 0$ for $x\in [0,1]$. For $x\in [0,1]$ and $S\in (4,39]$, define

$$\begin{aligned} J(u):=A_{0}+A_{1}u+A_{2}u^{2} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} A_{0} :=&\left( 32{x}^{4}+256{x}^{3}+768{x}^{2}+1024x+512 \right) {S}^{3}\\&+ \left( 512{x}^{4}+4096{x}^{3}+12288{x}^{2}+16384x+8192\right) {S}^{2}\\&+\left( 2048{x}^{4}+16384{x}^{3}+49152{x}^{2}+ 65536x+32768 \right) S \\ A_{1} :=&\left( -32{x}^{4}-184{x}^{3}-336{x}^{2}-64x+256 \right) {S}^{2}\\&+\left( -256{x}^{4}-1472{x}^{3}-2688{x}^{2}+256x+3584 \right) S\\&+6144x+12288 \\ A_{2}:=&\left( 8{x}^{4}+28{x}^{3}+3{x}^{2}-80x+32 \right) S-312x+240. \end{aligned} \end{aligned}$$

I. Consider the first case $A_{2}\le 0$. Then

$$\begin{aligned} \begin{aligned} J(4) =&\left( 32{x}^{4}+256{x}^{3}+768{x}^{2}+1024x+512 \right) {S}^{3} \\&+ \left( 384{x}^{4}+3360{x}^{3}+10944{x}^{2}+16128x+9216 \right) {S}^{2}\\&+ \left( 1152{x}^{4}+10944{x}^{3}+38448{x}^{2}+65280x+47616 \right) S+19584x+52992>0 \end{aligned} \end{aligned}$$

for $x\in [0,1]$. Furthermore,

$$\begin{aligned} J(121)=k_{0}+k_{1}S+k_{2}S^{2}+k_{3}S^{3}, \end{aligned}$$

where for $x\in [0,1],$

$$\begin{aligned} \begin{aligned} k_{0}&:= -3824568x+5000688 \\ k_{1}&:= 88200{x}^{4}+248220{x}^{3}-232173{x}^{2}-1074768x+934944\\ k_{2}&:=-3360{x}^{4}-18168{x}^{3}-28368{x}^{2}+8640x+39168\\ k_{3}&:=32{x}^{4}+256{x}^{3}+768{x}^{2}+1024x+512. \end{aligned} \end{aligned}$$

Since $k_{3}>0$ for $x\in [0,1]$ and $S>4$, we see that

$$\begin{aligned} J(121)\ge k_{0}+k_{1}S+(k_{2}+4k_{3})S^{2}, \end{aligned}$$

and

$$\begin{aligned} k_{2}+ 4k_{3}=-3232{x}^{4}-17144{x}^{3}-25296{x}^{2}+12736x+41216>0. \end{aligned}$$

Hence and by the fact that $S>4$, we obtain

$$\begin{aligned} J(121)\ge k_{0}+\left( k_{1}+4k_{2}+16k_{3}\right) S. \end{aligned}$$

Thus, since $k_{0}>0$, if $k_{1}+4k_{2}+16k_{3}\ge 0$, then $J(121)>0$.

If $k_{1}+4k_{2}+16k_{3}<0$, then

$$\begin{aligned} \left( k_{1}+4k_{2}+16k_{3}\right) S> \left( k_{1}+4k_{2}+16k_{3}\right) 39, \end{aligned}$$

and therefore for $x\in [0,1],$

$$\begin{aligned} \begin{aligned} J(121)&\ge k_{0}+\left( k_{1}+4k_{2}+16k_{3}\right) 39 \\&=-504192{x}^{4}-2674464{x}^{3}-3946176{x}^{2}-147171336x+201456528>0. \end{aligned} \end{aligned}$$

Thus, since $A_{2}\le 0$, we deduce that

$$\begin{aligned} J(u)\ge \min \{J(4), J(121)\}>0, \quad u\in [4,121]. \end{aligned}$$

II. Next we consider the case $A_{2}>0$. Then

$$\begin{aligned} J(u)=A_{0}+A_{1}u+A_{2}u^{2}\ge A_{0}+\big (A_{1}+4A_{2}\big )u=:\tilde{J}(u). \end{aligned}$$

We can easily see that

$$\begin{aligned} \begin{aligned} \tilde{J}(4)&=\left( 32{x}^{4}+256{x}^{3}+768{x}^{2}+1024x+512 \right) {S}^{3}\\&\quad + \left( 384{x}^{4}+3360{x}^{3}+10944{x}^{2}+16128x+9216 \right) {S}^{2}\\&\quad + \left( 1152{x}^{4}+10944{x}^{3}+38448{x}^{2}+65280x+47616 \right) S\\&\quad +19584x+52992>0. \end{aligned} \end{aligned}$$

Furthermore,

$$\begin{aligned} \tilde{J}(121)=q_{0}+q_{1}S+q_{2}S^{2}+q_{3}S^{3}, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} q_{0}&:= 592416x+1603008 \\ q_{1}&:= -25056{x}^{4}-148176{x}^{3}-274644{x}^{2}+57792x+481920\\ q_{2}&:=-3360{x}^{4}-18168{x}^{3}-28368{x}^{2}+8640x+39168\\ q_{3}&:=32{x}^{4}+256{x}^{3}+768{x}^{2}+1024x+512. \end{aligned} \end{aligned}$$

Since $S>4$, we obtain

$$\begin{aligned} \tilde{J}(121)> q_{0}+ q_{1}S+(q_{2}+4q_{3})S^{2}. \end{aligned}$$

Since $q_{0}>0$, $q_{1}>0$ and

$$\begin{aligned} q_{2}+4q_{3}=-3232{x}^{4}-17144{x}^{3}-25296{x}^{2}+12736x+41216>0, \end{aligned}$$

it follows that $\tilde{J}(121)>0$.

Hence, since the function $\tilde{J}$ is linear with respect to u, $\tilde{J}(4)>0$ and $\tilde{J}(121)>0$, we deduce that

$$\begin{aligned} J(u)\ge \tilde{J}(u)\ge \min \{\tilde{J}(4), \tilde{J}(121)\}>0, \quad u\in [4,121]. \end{aligned}$$

Finally, note that the cases I and II imply that $J(u)>0$ for $u\in (4,121]$, $S\in [4,39]$ and $x\in [0,1]$, which shows that $G(x)\ge 0$. This completes the proof of Lemma 3.

Theorem 2

Let $\alpha \in (0,1]$ and $\lambda \in [1/2,1]$. If $f\in \mathscr {F}_{O}(\lambda ,\alpha )$, then

$$\begin{aligned} \begin{aligned}&\left| H_{2,1}(F_{f^{-1}}/2)\right| \\&\le {\left\{ \begin{array}{ll}\dfrac{\alpha ^{2}(1+2\lambda )^{2}}{144}, &{} T\le 2, \\ \dfrac{\alpha ^{2}(1+2\lambda )^{2}}{576}\left( 4+\dfrac{(T-2)^{2}}{16+4T+S}\right) , &{} T>2,\ S \le 2\left( \sqrt{2T^{2}+8T+40}-T-2\right) , \\ \dfrac{\alpha ^{2}(1+2\lambda )^{2}}{2304}(8+S), &{} T>2,\ S > 2\left( \sqrt{2T^{2}+8T+40}-T-2\right) , \end{array}\right. } \end{aligned} \end{aligned}$$

(3.1)

where $T:=\alpha (1+10\lambda )$ and $S:=\alpha ^{2}(44\lambda ^{2}+4\lambda -9)$.

The inequalities in (3.1) are sharp.

Proof

Let $\alpha \in (0,1],$ $\lambda \in [1/2,1]$ and $f\in \mathscr {F}_{O}(\lambda ,\alpha )$ be of the form (1.1). Then, (2.2) holds for some function $p\in \mathscr {P}$ of the form (1.10). Since the class $\mathscr {F}_{O}(\lambda ,\alpha )$ and $|H_{2,1}(F_{f^{-1}}/2)|$ are rotationally invariant, without loss of generality we may assume that $a_2\ge 0$. Thus, by (2.3) we assume that $c:=c_{1}\in [0,2],$ i.e., in view of (1.11) that $d_{1}\in [0,1]$. By (1.9), (2.3) and (1.11), we get

$$\begin{aligned} \begin{aligned} H_{2,1}(F_{f^{-1}}/2)&=\Gamma _{1}\Gamma _{3}-\Gamma _{2}^{2}\\&=\frac{1}{4}\left( a_{2}a_{4}-a_{3}^{2}-a_2^2a_3+\frac{13}{12}a_{2}^{4}\right) \\&=\frac{\alpha ^{2}(1+2\lambda )^{2}}{2304}\Theta , \end{aligned} \end{aligned}$$

(3.2)

where

$$\begin{aligned} \begin{aligned} \Theta :=&-8(2+d_{1}^{2})(1-d_{1}^{2})d_{2}^{2}-4T(1-d_{1}^{2})d_{1}^{2}d_{2}\\&+24(1-d_{1}^{2})(1-|d_{2}|^{2})d_{1}d_{3}+(8+S)d_{1}^{4} \end{aligned} \end{aligned}$$

(3.3)

for some $d_{1}\in [0,1]$ and $d_{2},d_{3}\in \overline{\mathbb {U}}$.

I. Assume first that $T=\alpha (1 + 10\lambda )\le 2$. Then, by applying the triangle inequality to (3.3) and by the fact that $|d_{2}|\le 1$ and $|d_{3}|\le 1$, we obtain

$$\begin{aligned} |\Theta |\le (S-4T)d_{1}^{4}+4(T-2)d_{1}^{2}+16. \end{aligned}$$

(3.4)

Since $S-4T<0$ and $T\le 2$, by (3.4) we have

$$\begin{aligned} |\Theta |\le 16, \end{aligned}$$

which together with (3.2) shows the first inequality in (3.1).

II. Next assume that $T=\alpha (1 + 10\lambda )> 2$.

Case 1. Suppose that $d_{1}=1$. Then by (3.2) and (3.3), we obtain

$$\begin{aligned} \left| \Gamma _{1}\Gamma _{3}-\Gamma _{2}^{2}\right| =\frac{\alpha ^{2}(1+2\lambda )^{2}}{2304}(8+S). \end{aligned}$$

Case 2. Suppose that $d_{1}=0$. Then by (3.2) and (3.3), we obtain

$$\begin{aligned} \left| \Gamma _{1}\Gamma _{3}-\Gamma _{2}^{2}\right| =\frac{\alpha ^{2}(1+2\lambda )^{2}}{144}|d_{2}|^{2}\le \frac{\alpha ^{2}(1+2\lambda )^{2}}{144}. \end{aligned}$$

Case 3. Suppose that $d_{1}\in (0,1)$. Since $|d_{3}|\le 1$, by applying the triangle inequality to (3.2) we can write

$$\begin{aligned} \left| \Gamma _{1}\Gamma _{3}-\Gamma _{2}^{2}\right| \le \frac{\alpha ^{2}(1+2\lambda )^{2}d_{1}(1-d_{1}^{2})}{96}\Big [\left| A+Bd_{2}+Cd_{2}^{2}\right| +1-|d_{2}|^{2}\Big ], \end{aligned}$$

(3.5)

where

$$\begin{aligned} A:=-\frac{8+S}{24(1-d_{1}^{2})}d_{1}^{3},\quad B:=\frac{T}{6} d_{1}, \quad C:=\frac{d_{1}^{2}+2}{3d_{1}}. \end{aligned}$$

Since $AC < 0$, we apply Lemma 2 only for the case II.

We have

$$\begin{aligned} 2<T=\alpha (1+10\lambda )\le 11, \end{aligned}$$

and therefore

$$\begin{aligned} \frac{4}{9}<S=\alpha ^{2}(44\lambda ^{2}+4\lambda -9)\le 39 \end{aligned}$$

for $\lambda \in [1/2,1]$ and $\alpha \in (0,1]$.

We consider the following sub-cases.

3(a)

Note that

$$\begin{aligned} \begin{aligned} |B|-2(1-|C|)&=\frac{T}{6}d_{1}-2\bigg (1-\frac{d_{1}^{2}+2}{3d_{1}^{2}}\bigg )\\&=\frac{1}{6d_{1}}\bigg (4d_{1}^{2}-12d_{1}+Td_{1}^{2}+8\bigg )\\&=\frac{1}{6d_{1}}\bigg (4(1-d_{1})(2-d_{1})+Td_{1}^{2}\bigg )>0. \end{aligned} \end{aligned}$$

Therefore, $|B|<2(1-|C|)$ does not hold for $d_{1}\in (0,1)$, $\lambda \in [1/2,1]$ and $\alpha \in (0,1]$.

3(b) We can easily see that

$$\begin{aligned} \begin{aligned} \frac{1}{C^{2}}-1&=\frac{9d_{1}^{2}}{(d_{1}^{2}+2)^{2}}-1\\&=-\frac{1}{(d_{1}^{2}+2)^{2}}(d_{1}^{4}-5d_{1}^{2}+4)\\&= -\frac{1}{(d_{1}^{2}+2)^{2}}(1-d_{1}^{2})(4-d_{1}^{2})<0, \end{aligned} \end{aligned}$$

which yields

$$\begin{aligned} -4AC\left( \frac{1}{C^{2}}-1\right) <0. \end{aligned}$$

Therefore, the inequality

$$\begin{aligned} B^2< \min \left\{ 4(1+|C|)^{2},-4AC\left( \frac{1}{C^{2}}-1\right) \right\} \end{aligned}$$

is false for $d_{1}\in (0,1)$, $\lambda \in [1/2,1]$ and $\alpha \in (0,1]$.

3(c) Since

$$\begin{aligned} \begin{aligned} 4|C|-|B|&= 4\frac{d_{1}^{2}+2}{3d_{1}}-\frac{T}{6}d_{1}\\&=\frac{1}{6}((8-T)d_{1}^{2}+16)\\&\ge \frac{1}{6}(16-3d_{1}^{2})>0, \end{aligned} \end{aligned}$$

we have

$$\begin{aligned} |C|(|B|+4|A|)-|AB|=|BC|+|A|(4|C|-|B|)>0. \end{aligned}$$

Consequently, the inequality $|C|(|B|+4|A|)\le |AB|$ does not hold for $d_{1}\in (0,1)$, $\lambda \in [1/2,1]$ and $\alpha \in (0,1]$.

3(d) We can write

$$\begin{aligned} \begin{aligned}|AB|-|C|(|B|-4|A|)&= \frac{(8+S)T}{144(1-d_{1}^{2})}d_{1}^{4}-\frac{d_{1}^{2}+2}{3d_{1}}\left( \frac{T}{6}d_{1}-\frac{4(8+S)}{24(1-d_{1}^{2})}d_{1}^{3}\right) \\&=\frac{1}{144(1-t)}( Pt^{2}+Qt+R), \end{aligned} \end{aligned}$$

where $t:=d_{1}^{2}\in (0,1)$ and

$$\begin{aligned} P := 64+8S+16T+ST,\quad Q := 128+8T+16S,\quad R :=-16T. \end{aligned}$$

It is easy to see that $P>0$, $Q>0$ and $R<0$ for $\lambda \in [1/2,1]$ and $\alpha \in (0,1]$.

For the equation $Pt^{2}+Qt+R=0$, we have $\Delta >0$. Since

$$\begin{aligned} \frac{R}{P}<0 \quad \text {and} \quad P+Q+R>0, \end{aligned}$$

for $\lambda \in [1/2,1]$ and $\alpha \in (0,1]$, the equation $Pt^{2}+Qt+R=0$ has a unique positive root $t_{1}<1$. Thus, the inequality $|AB|-|C|(|B|-4|A|)\le 0$ holds for $(0,d_{1}^{**}]$, where $d_{1}^{**}:=\sqrt{t_{1}}$. So we can write from (3.5) and Lemma 2,

$$\begin{aligned} \begin{aligned} \left| \Gamma _{1}\Gamma _{3}-\Gamma _{2}^{2}\right|&\le \frac{\alpha ^{2}(1+2\lambda )^{2}d_{1}(1-d_{1}^{2})}{96}(-|A|+|B|+|C|)\\&=\frac{\alpha ^{2}(1+2\lambda )^2}{2304}\Phi (d_{1}), \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \Phi (x):=(-16-4T-S)x^{4}+4(T-2)x^{2}+16,\quad x\in [0,1]. \end{aligned}$$

We note that $\Phi '(x)=4(-16-4T-S)x^3+8(T-2)x=0$ for $x\in (0,1)$ holds only for

$$\begin{aligned} x=\sqrt{\frac{2(T-2)}{16+4T+S}}=:\xi , \end{aligned}$$

(3.6)

in the case when $T>2$, i.e., for $\alpha (1+10\lambda )>2$. Clearly $0<\xi <1$ and the function $\Phi$ attains at $\xi$ its maximum value on [0, 1]. Therefore, in the case when $0<\xi \le d_{1}^{**}$ we have

$$\begin{aligned} \begin{aligned} \left| \Gamma _{1}\Gamma _{3}-\Gamma _{2}^{2}\right|&\le \frac{\alpha ^{2}(1+2\lambda )^{2}}{2304}\Phi (\xi )\\&= \frac{\alpha ^{2}(1+2\lambda )^{2}}{2304}\left[ (-16-4T-S)\xi ^{4}+4(T-2)\xi ^{2}+16\right] \\&=\frac{\alpha ^{2}(1+2\lambda )^{2}}{576}\left( 4+\frac{(T-2)^{2}}{16+4T+S}\right) . \end{aligned} \end{aligned}$$

3(e) Next consider the case $x\in [d_{1}^{**},1)$. Using the last case of the Lemma 2,

$$\begin{aligned} \begin{aligned} \left| \Gamma _{1}\Gamma _{3}-\Gamma _{2}^{2}\right|&\le \frac{\alpha ^{2}(1+2\lambda )^{2}d_{1}(1-d_{1}^{2})}{96}\left( (|A|+|C|)\sqrt{1-\frac{B^{2}}{4AC}}\right) \\&=\frac{\alpha ^{2}(1+2\lambda )^{2}}{2304}h_{1}(d_{1}^{2})\sqrt{h_{2}(d_{1}^{2})}, \end{aligned} \end{aligned}$$

where for $x \in [0,1]$,

$$\begin{aligned} h_{1}(x):= Sx^{2}-8x+16 \quad \text {and}\quad h_{2}(x):=1+\frac{T^{2}(1-x)}{2(S+8)(x+2)}. \end{aligned}$$

It is easy to see that $h_{2}$ is a positive decreasing function in $[d_{1}^{**}, 1)$.

i) If $S\le 4$, then $h_{1}$ is a positive decreasing function in $[d_{1}^{**}, 1)$. Hence,

$$\begin{aligned} \begin{aligned}P\xi ^{4}+Q\xi ^{2}+R=&P\left( \frac{2(T-2)}{16+4T+S}\right) ^{2}+Q\left( \frac{2(T-2)}{16+4T+S}\right) +R\\ =&\frac{4}{(16+4T+S)^{2}}\left[ T^{3}(S-32)+8T^{2}(S-28)+4T(S^{2}-9S-208)\right. \\&\left. -16S^{2}-352S-1792\right] <0. \end{aligned} \end{aligned}$$

Therefore by Part 3(d), it follows that $0<\xi <d_{1}^{**}$. Since $h_1\sqrt{h_2}$ is decreasing in $[d_{1}^{**}, 1)$ and as easy to check $h_{1}(d_{1}^{**})\sqrt{h_{2}(d_{1}^{**})}=\Phi (d_{1}^{**}),$ we get

$$\begin{aligned} \begin{aligned} \left| \Gamma _{1}\Gamma _{3}-\Gamma _{2}^{2}\right|&\le \frac{\alpha ^{2}(1+2\lambda )^{2}}{2304}h_{1}(d_{1}^{**})\sqrt{h_{2}(d_{1}^{**})}\\&=\frac{\alpha ^{2}(1+2\lambda )^{2}}{2304}\Phi (d_{1}^{**})\\&\le \frac{\alpha ^{2}(1+2\lambda )^{2}}{2304}\Phi (\xi ). \end{aligned} \end{aligned}$$

ii) When $4<S\le 39$ and $2<T\le 11$, we can write

$$\begin{aligned} H(d_{1}^{2})=h_{1}(d_{1}^{2})\sqrt{h_{2}(d_{1}^{2})}, \end{aligned}$$

where H is the function defined in Lemma 3. Since by Lemma 3 the function H is convex, we deduce that

$$\begin{aligned} H(d_{1}^{2})\le \max \{H(d_{1}^{**}),H(1)\}=\max \{H(d_{1}^{**}),8+S\}. \end{aligned}$$

Suppose that $S\le 2(\sqrt{2T^{2}+8T+40}-T-2),$ i.e., that $S^{2}\le -4ST-8S+4T^{2}+16T+144.$ Then

$$\begin{aligned} \begin{aligned} P\xi ^{4}+Q\xi ^{2}+R =&\frac{4}{(16+4T+S)^{2}}\left[ 4S^{2}T-16S^{2}+ST^{3}+8ST^{2}-36ST\right. \\&\left. -352S-32T^{3}-224T^{2}-832T-1792\right] \\ \le&\frac{4}{(16+4T+S)^{2}}\left[ 4T(-4ST-8S+4T^{2}+16T+144)\right. \\&-16S^{2}+ST^{3}+8ST^{2}-36ST-352S-32T^{3}\\&\left. -224T^{2}-832T-1792\right] \\ =&\frac{4}{(16+4T+S)^{2}}\left[ -16S^{2}+ST^{3}-8ST^{2}-68ST-352S\right. \\&\left. -16T^{3}- 160T^{2}-256T-1792\right] <0, \end{aligned} \end{aligned}$$

which by Part 3(d) yields $0<\xi <d_{1}^{**}$. Since then $\Phi (\xi )\ge 8+S$, we get

$$\begin{aligned} H(d_{1}^{2})\le \max \{H(d_{1}^{**}),H(1)\}=\max \{H(d_{1}^{**}),8+S \}\le \Phi (\xi ). \end{aligned}$$

Suppose that $S > 2(\sqrt{2T^{2}+8T+40}-T-2)$. Then,

$$H(d_{1}^{**})=\Phi (d_{1}^{**})\le \Phi (\xi )\le 8+S$$

and hence

$$\begin{aligned} H(d_{1}^{2})\le \max \{H(d_{1}^{**}),H(1)\}=\max \{H(d_{1}^{**}),8+S \}= 8+S. \end{aligned}$$

Summarizing parts from Case 1–3, it follows (3.1).

In order to show that the inequalities are sharp, first let $f\in \mathscr {F}_{O}(\lambda ,\alpha )$ be defined by (2.2) with

$$\begin{aligned} p(z):= \frac{1+z^{2}}{1-z^{2}}=1+\sum _{k=1}^{\infty }2z^{2k}, \qquad (z\in \mathbb {U}), \end{aligned}$$

Then, in view of (2.3) we have

$$\begin{aligned} a_{2}=0,\quad a_{3}=\frac{\alpha (1+2\lambda )}{6},\quad a_{4}=0. \end{aligned}$$

Thus, from (3.2) we get

$$\begin{aligned} \left| H_{2,1}(F_{f^{-1}}/2)\right|&=\left| \frac{13}{48}a_{2}^{4}-\frac{1}{4}a_{2}^{2}a_{3}+\frac{1}{4}a_{4}a_{2}-\frac{1}{4}a_{3}^{2}\right| \\&=\frac{\alpha ^{2}(1+2\lambda )^{2}}{144}, \end{aligned}$$

which shows that the first bound in (3.1) is sharp.

Next let $f\in \mathscr {F}_{O}(\lambda ,\alpha )$ be defined by (2.2) with

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

where $\xi$ given by (3.6). Then in view of (2.3) we have

$$\begin{aligned} \begin{aligned} a_{2}=&\frac{\alpha (1+2\lambda )}{2} \xi ,\\ a_{3}=&\frac{\alpha (1+2\lambda )}{6}\left( 1+(2\alpha +2\alpha \lambda -1)\xi ^{2}\right) ,\\ a_{4}=&\frac{\alpha (1+2\lambda )\xi }{72} \left[ (8-21\alpha +16\alpha ^2-18\alpha \lambda +30\alpha ^{2}\lambda +12\alpha ^{2}\lambda ^{2})\xi ^{2}\right. \\&\left. \qquad \qquad \quad +21\alpha +18\alpha \lambda -6\right] . \end{aligned} \end{aligned}$$

Thus, from (3.2) we get

$$\begin{aligned} \begin{aligned} \left| H_{2,1}(F_{f^{-1}}/2)\right|&=\left| \frac{13}{48}a_{2}^{4}-\frac{1}{4}a_{2}^{2}a_{3}+\frac{1}{4}a_{4}a_{2}-\frac{1}{4}a_{3}^{2}\right| \\&=\frac{\alpha ^{2}(1+2\lambda )^{2}}{576}\left( 4+\frac{(T-2)^{2}}{16+4T+S}\right) , \end{aligned} \end{aligned}$$

which shows that the second bound in (3.1) is sharp.

Finally, let $f\in \mathscr {F}_{O}(\lambda ,\alpha )$ be defined by (2.2) with

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

Then, in view of (2.3) we have

$$\begin{aligned} a_{2}=&\frac{\alpha (1+2\lambda )}{2},\quad a_{3}=\frac{\alpha ^{2}}{3}(1+\lambda )(1+2\lambda )\\ a_{4}=&\frac{\alpha (1+2\lambda )}{36}(6\alpha ^{2}\lambda ^{2}+15\alpha ^{2}\lambda +8\alpha ^{2}+1). \end{aligned}$$

Thus, from (3.2) we get

$$\begin{aligned} \left| H_{2,1}(F_{f^{-1}}/2)\right|&= \left| \frac{13}{48}a_{2}^{4}-\frac{1}{4}a_{2}^{2}a_{3}+\frac{1}{4}a_{4}a_{2}-\frac{1}{4}a_{3}^{2}\right| \\&=\frac{\alpha ^{2}(1+2\lambda )^{2}}{2304}(8+S), \end{aligned}$$

which shows that the third bound in (3.1) is sharp.

In [33], it was shown that the bounds of $H_{2,2}(f)$ and $H_{2,2}(f^{-1})$ for the convex functions of order alpha were the same, reflecting other invariant properties related to the coefficients f and $f^{-1}$. Sim et al. [24] improved these bounds to achieve sharp bounds. The following result shows that for $\lambda =1/2$, i.e., for the class $\mathscr {K}_{c}(\alpha )$, the sharp bounds for the second Hankel determinant of logarithmic coefficients $H_{2,1}(F_{f}/2)$, given in Corollary 1, and $H_{2,1}(F_{f^{-1}}/2)$ are also the same.

Corollary 2

Let $\alpha \in (0,1]$. If $f\in \mathscr {K}_{c}(\alpha )$, then

$$\begin{aligned} \left| \Gamma _{1}\Gamma _{3}-\Gamma _{2}^{2}\right| \le \left\{ \begin{array}{ll} \dfrac{\alpha ^{2}}{36}, &{}\quad 0<\alpha \le \dfrac{1}{3}, \\ \dfrac{\alpha ^{2}(13\alpha ^{2}+18\alpha +17)}{144(\alpha ^{2}+6\alpha +4)}, &{}\quad \dfrac{1}{3}<\alpha \le 1. \end{array} \right. \end{aligned}$$

The inequalities are sharp.