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Inverse logarithmic coefficient bounds for starlike functions subordinated to the exponential functions

Abstract

In recent years, many subclasses of univalent functions, directly or not directly related to the exponential functions, have been introduced and studied. In this paper, we consider the class of \(\mathcal{S}^{\ast}_{e}\) for which \(zf^{\prime}(z)/f(z)\) is subordinate to \(e^{z}\) in the open unit disk. The classic concept of Hankel determinant is generalized by replacing the inverse logarithmic coefficient of functions belonging to certain subclasses of univalent functions. In particular, we obtain the best possible bounds for the second Hankel determinant of logarithmic coefficients of inverse starlike functions subordinated to exponential functions. This work may inspire to pay more attention to the coefficient properties with respect to the inverse functions of various classes of univalent functions.

1 Introduction and definitions

Let \(\mathcal{H} ( \mathbb{D} ) \) represent the family of analytic functions defined in the region of unit disc \(\mathbb{D}:= \{ z\in \mathbb{C}: \vert z \vert <1 \} \). For \(f\in \mathcal{H} (\mathbb{D} ) \), the normalized functions taking the form of

$$ f(z)=z+\sum_{s=2}^{\infty }b_{s}z^{s}, \quad z\in \mathbb{D} , $$
(1)

are belonging to the class \(\mathcal{A}\). Assuming also that \(\mathcal{S}\subset \mathcal{A}\) be the set of all univalent functions in \(\mathbb{D}\). In the theory of univalent functions, the Carathéodory function [1] is well studied. It is holomorphic in \(\mathbb{D}\) with positive real part, i.e., \(\Re ( p(z) ) >0\) and taking the series representation of

$$ p ( z ) =1+\sum_{s=1}^{\infty }\mu _{s}z^{s}, \quad z\in \mathbb{D}. $$
(2)

We denote by \(\mathcal{P}\) the set of these functions.

A basic relationship in geometry function theory is subordination. We write \(g\prec \widetilde{g}\) to illustrate that g is subordinate to . It is explained that for given two functions \(g, \widetilde{g}\in \mathcal{H} ( \mathbb{D} ) \), a Schwarz function ω is existing such that \(g(z)=\widetilde{g} (\omega (z) ) \) for \(z\in \mathbb{D}\). Once is univalent in \(\mathbb{D}\), then this relation is equivalent to saying that

$$ g ( z ) \prec \widetilde{g} ( z ) , \quad (z\in \mathbb{D} ) \quad \Longleftrightarrow \quad g(0)=\widetilde{g}(0)\quad \text{and}\quad g(\mathbb{D})\subset \widetilde{g}(\mathbb{D}). $$

The three classic classes of univalent functions are \(\mathcal{S}^{\ast }\), \(\mathcal{C}\) and \(\mathcal{R}\), which are known as starlike functions, convex functions and the bounded turning functions. These classes are characterized as

$$\begin{aligned} &\mathcal{C} \bigl( q^{*} \bigr)= \biggl\{ f\in \mathcal{A}: \frac{ ( zf^{\prime }(z) ) ^{\prime }}{f^{\prime }(z)}\prec q^{*}(z) \biggr\} , \\ &\mathcal{S}^{\ast } ( q{*} )= \biggl\{ f\in \mathcal{A}: \frac{zf^{\prime }(z)}{f(z)}\prec q^{*}(z) \biggr\} , \\ &\mathcal{R} \bigl( q^{*} \bigr)= \bigl\{ f\in \mathcal{A}:f^{ \prime }(z)\prec q^{*}(z) \bigr\} , \end{aligned}$$

with \(q^{*}(z)=\frac{1+z}{1-z}\), which maps \(\mathbb{D}\) to the right half plane. By choosing \(q^{*}\) as some other special functions, various interesting subfamilies of \(\mathcal{C}\), \(\mathcal{S}^{\ast }\), and \(\mathcal{R}\) were studied, the interested readers can refer to [2].

Define

$$\begin{aligned} &\mathcal{C}_{e}:= \biggl\{ f\in \mathcal{A}: \frac{ ( zf^{\prime }(z) ) ^{\prime }}{f^{\prime }(z)}\prec e^{z} \biggr\} , \\ &\mathcal{S}^{\ast }_{e}:= \biggl\{ f\in \mathcal{A}: \frac{zf^{\prime }(z)}{f(z)}\prec e^{z} \biggr\} , \\ &\mathcal{R}_{e}:= \bigl\{ f\in \mathcal{A}:f^{\prime }(z) \prec e^{z} \bigr\} . \end{aligned}$$

The class \(\mathcal{S}^{\ast}_{e}\) was introduced and studied by Mendiratta et al. [3]. Later, many interesting classes of univalent functions associated the exponential functions were intensively investigated. Cho et al. [4] introduced a class of starlike functions connected with sin function by letting that \(q^{*}=1+\sin z\). In [5], the authors considered a subclass of starlike function given by choosing \(q^{*}=\cos z\). Kumar et al. [6] defined a new class of starlike function by taking \(q^{*}=1+ ze^{z}\). For univalent functions defined by modified sigmoid functions [7], it uses the functions \(q^{*}=\frac{2}{1+e^{-z}}\). As it can be seen, all these specially chosen functions are closely related with the exponential function.

Let \(\iota , n\in \mathbb{N}= \{ 1,2,\ldots \} \). Then, the Hankel determinant \(H_{\iota ,n} ( f ) \), introduced by Pommerenke [8, 9], for \(f\in \mathcal{S}\) in the form of

$$ H_{\iota ,n} ( f ) :=\left \vert \textstyle\begin{array}{l@{\quad}l@{\quad}l@{\quad}l} b_{n} & b_{n+1} & \ldots & b_{n+\iota -1} \\ b_{n+1} & b_{n+2} & \ldots & b_{n+\iota} \\ \vdots & \vdots & \ldots & \vdots \\ b_{n+\iota -1} & b_{n+q} & \ldots & b_{n+2\iota -2}\end{array}\displaystyle \right \vert . $$
(3)

When investigating power series with integral coefficients and singularities, this method has proven to be effective by considering Hankel determinant, see [10]. In recent years, the bounds of \(H_{\iota ,n} ( f )\) for various types of univalent functions has been studied. For example, the absolute bounds of the second Hankel determinant \(H_{2,2} ( f ) =b_{2}b_{4}-b_{3}^{2}\) were calculated in [11, 12] for various subsets of univalent functions. However, there are still many unsolved problems in the exact estimation of this determinant, like the family of close-to-convex functions [13]. For the third Hankel determinant, the sharp bound of \(\vert H_{3,1} ( f ) \vert \) for convex functions \(\mathcal{C}\) was obtained in [14]. For \(f\in \mathcal{S}^{\ast}\), it is proved that \(\vert H_{3,1} (f ) \vert \leq \frac{4}{9}\) by Kowalczyk et al. [15]. For the bounded turning functions \(\mathcal{R}\), the upper bound was calculated to be \(\frac{1}{4}\) in [16]. For some subclasses of univalent or non-univalent functions, some interesting results on bounds on the Hankel determinant were also found in the studies like [1722].

The Bieberbach conjecture is based on logarithmic coefficients \(\gamma _{s}\) of f, where \(\gamma _{s}\) is defined by

$$ \log \biggl(\frac{f(z)}{z} \biggr)=2\sum_{s=1}^{\infty} \gamma _{s}z^{s}, \qquad \log 1=0. $$
(4)

These coefficients are well studied in the theory of univalent functions and it has been proven that

$$ \sum_{s=1}^{\infty} \vert \gamma _{s} \vert ^{2}\leq \frac{\pi ^{2}}{6}, $$
(5)

the bound is sharp for the Koebe function, see [23]. Recently, many authors have investigated the logarithmic coefficients related problems for various classes of univalent functions, e.g., [2428]. But the best upper bounds for \(\vert \gamma _{s} \vert \) \((s\geq 3)\) of univalent functions and some of their subfamilies are still open. In 2021, Kowalczyk et al. [29, 30] introduced the Hankel determinant using logarithmic coefficients for the first time, i.e., it replaces \(b_{n}\) as \(\gamma _{n}\) as entry.

According to 1/4-theorem proposed by Koebe, we know that the inverse function F of f defined in a neighborhood of origin exists with certainty. We may write as

$$ F(w):=f^{-1}(w)=w+B_{2}w^{2}+B_{3}w^{3}+ \cdots , \quad \vert w \vert < \frac{1}{4}. $$
(6)

Then the logarithmic coefficient \(\Gamma _{n}\) of F is given by

$$ \log \biggl(\frac{F(z)}{z} \biggr)=2\sum_{s=1}^{\infty} \Gamma _{s}w^{s}, \quad \vert w \vert < \frac{1}{4}. $$
(7)

The logarithmic coefficients of the inverses of univalent functions was studied by Ponnusamy et al. [31].

Motivated by the above works, it seems natural to consider the Hankel determinant with \(\Gamma _{n}\) replacing \(b_{n}\), see [32]. Using this idea, we have

$$ \mathcal{H}_{2,1} (F_{f^{-1}}/2 )=\Gamma _{1} \Gamma _{3}- \Gamma _{2}^{2} $$
(8)

and

$$ \mathcal{H}_{2,2} (F_{f^{-1}}/2 )=\Gamma _{2} \Gamma _{4}- \Gamma _{3}^{2}. $$
(9)

In [33], it was calculated that

$$\begin{aligned} \Gamma _{1}&=-\frac{1}{2} b_{2}, \end{aligned}$$
(10)
$$\begin{aligned} \Gamma _{2}&=-\frac{1}{2}b_{3}+ \frac{3}{4}b_{2}^{2}, \end{aligned}$$
(11)
$$\begin{aligned} \Gamma _{3}&=-\frac{1}{2}b_{4}+2b_{2}b_{3}- \frac{5}{3}b_{2}^{3}, \end{aligned}$$
(12)
$$\begin{aligned} \ \Gamma _{4}&=-\frac{1}{2}b_{5}+ \frac{5}{2}b_{2}b_{4}-\frac{15}{2}b_{2}^{2}b_{3}+ \frac{5}{4}b_{3}^{2}+\frac{35}{8}b_{2}^{4}. \end{aligned}$$
(13)

Thus, we have

$$ \mathcal{H}_{2,1} (F_{f^{-1}}/2 )= \frac{1}{4} \biggl( b_{2}b_{4}- b_{2}^{2}b_{3} -b_{3}^{2}+ \frac{13}{12}b_{2}^{4} \biggr) $$
(14)

and

$$\begin{aligned} \mathcal{H}_{2,2} (F_{f^{-1}}/2 ) =& \frac{145}{288}b_{2}^{6}- \frac{55}{48}b_{2}^{4}b_{3}+ \frac{5}{24}b_{2}^{3}b_{4}+ \frac{11}{16}b_{2}^{2}b_{3}^{2}- \frac{5}{8}b_{2}^{2}b_{5} \\ &{}+\frac{3}{4}b_{2}b_{3}b_{4}- \frac{1}{4}b_{2}^{4}+ \frac{1}{4}b_{3}b_{5}- \frac{5}{8}b_{3}^{3}. \end{aligned}$$
(15)

Let \(f_{\alpha}=e^{-i\alpha}f (e^{i\alpha}z )\). It is noted that

$$ \mathcal{H}_{2,1} (F_{f_{\alpha}^{-1}}/2 )=e^{4i\alpha} \mathcal{H}_{2,1} (F_{f^{-1}}/2 ) $$
(16)

and

$$ \mathcal{H}_{2,2} (F_{f_{\alpha}^{-1}}/2 )=e^{6i\alpha} \mathcal{H}_{2,2} (F_{f^{-1}}/2 ). $$
(17)

Hence, the functional \(\phi _{f}= \vert \mathcal{H}_{2,1} (F_{f^{-1}}/2 ) \vert \) and \(\varphi _{f}= \vert \mathcal{H}_{2,2} (F_{f^{-1}}/2 ) \vert \) are all rotation invariant. Recently, the upper bound of Hankel determinant for the class \(\mathcal{S}^{\ast }_{e}\) and \(\mathcal{C}_{e}\) were studied, including [34, 35]. On the inverse coefficient problem for the class \(\mathcal{R}_{e}\), it was investigated in [36]. In this article, we aim to calculate the sharp bounds on \(\vert \mathcal{H}_{2,1} (F_{f^{-1}}/2 ) \vert \) and \(\vert \mathcal{H}_{2,2} (F_{f^{-1}}/2 ) \vert \) for the class \(\mathcal{S}_{e}^{\ast}\).

2 A set of lemmas

We use the following lemmas to obtain our main results.

Lemma 2.1

([37]) Assume \(p\in \mathcal{P}\) be the form of (2). Then

$$\begin{aligned} 2\mu _{2} ={}&\mu _{1}^{2}+\kappa \bigl( 4-\mu _{1}^{2} \bigr) , \end{aligned}$$
(18)
$$\begin{aligned} 4\mu _{3} ={}&\mu _{1}^{3}+2 \bigl( 4-\mu _{1}^{2} \bigr) \mu _{1} \kappa -\mu _{1} \bigl( 4-\mu _{1}^{2} \bigr) \kappa ^{2}+2 \bigl( 4- \mu _{1}^{2} \bigr) \bigl( 1- \vert \kappa \vert ^{2} \bigr) \delta , \end{aligned}$$
(19)
$$\begin{aligned} 8\mu _{4} ={}&\mu _{1}^{4}+ \bigl( 4-\mu _{1}^{2} \bigr) \kappa \bigl[ c_{1}^{2} \bigl( \kappa ^{2}-3\kappa +3 \bigr) +4\kappa \bigr] -4 \bigl( 4-\mu _{1}^{2} \bigr) \bigl( 1- \vert \kappa \vert ^{2} \bigr) \\ &{}\times \bigl[ \mu _{1} ( \kappa -1 ) \delta + \overline{ \kappa}\delta ^{2}- \bigl( 1- \vert \delta \vert ^{2} \bigr) \varrho \bigr] , \end{aligned}$$
(20)

for some \(\kappa , \delta , \varrho \in \mathbb{\overline{D}}:= \{z\in \mathbb{C}: \vert z \vert \leq 1 \}\).

Lemma 2.2

(See [38]) For given real numbers A, B, C, let

$$ Y (A,B,C )=\max_{z\in \overline{{\mathbb{D}}}} \bigl\{ \bigl\vert A+Bz+Cz^{2} \bigr\vert +1- \vert z \vert ^{2} \bigr\} . $$

If \(A>0\) and \(C<0\), then

$$ Y (A,B,C )=\textstyle\begin{cases} 1-A+\frac{B^{2}}{4 (1+C )}, & \textit{if } B^{2} \geq - \frac{4AC^{3}}{1-C^{2}},\ \vert B \vert < 2 (1+C ), \\ 1+A+\frac{B^{2}}{4 (1-C )}, & \textit{if } B^{2}< \min \{4 (1-C )^{2}, -\frac{4AC^{3}}{1-C^{2}} \}, \\ R (A,B,C ), & \textit{otherwise}, \end{cases} $$

where

$$ R (A,B,C )=\textstyle\begin{cases} A+ \vert B \vert +C, & \textit{if } -C (4A+ \vert B \vert )\leq A \vert B \vert , \\ -A+ \vert B \vert -C, & \textit{if } -C (-4A+ \vert B \vert )\geq A \vert B \vert , \\ (A-C )\sqrt{1-\frac{B^{2}}{4AC}}, & \textit{otherwise}. \end{cases} $$

Lemma 2.3

Define \(\rho : [0,4]\rightarrow \mathbb{R}\) by

$$ \rho (t):=h_{1}(t)\sqrt{h_{2}(t)}, $$

where

$$ h_{1}(t)=23t^{2}-96t+576,\qquad h_{2}(t)= \frac{18-t}{12+t}. $$

Then ρ is convex on \([0,4]\).

Proof

By observing that

$$\begin{aligned} h_{2}^{\frac{3}{2}}(t)\rho ^{\prime \prime}(t)&=h_{1}^{\prime \prime}(t) \bigl(h_{2}(t) \bigr)^{2}+h_{1}^{\prime}(t)h_{2}(t)h_{2}^{\prime}(t)- \frac{1}{4}h_{1}(t)h_{2}^{\prime \prime}(t)+ \frac{1}{2}h_{1}h_{2}(t)h_{2}^{ \prime \prime}(t) (t) \\ &= \frac{46t^{4} + 138t^{3} -19{,}251t^{2} -209{,}088t + 2{,}949{,}696}{ (t+12 )^{4}} \geq 0, \end{aligned}$$

we have \(\rho ^{\prime \prime}(t)\geq 0\) on \([0,4]\). The assertion in Lemma 2.3 thus follows. □

3 Main results

We first determine the bounds of \(\vert \mathcal{H}_{2,1} ( f^{-1}/2 ) \vert \) for \(f\in \mathcal{S}^{\ast}_{e}\).

Theorem 3.1

Let \(f\in \mathcal{S}^{\ast}_{e}\). Then

$$ \bigl\vert \mathcal{H}_{2,1} \bigl( f^{-1}/2 \bigr) \bigr\vert = \bigl\vert \Gamma _{1}\Gamma _{3}- \Gamma _{2}^{2} \bigr\vert \leq \frac{29}{428}. $$

The result is sharp.

Proof

Let \(f\in \mathcal{S}^{\ast}_{e}\). From [35], we know

$$\begin{aligned} b_{2} &=\frac{1}{4}\mu _{1}, \end{aligned}$$
(21)
$$\begin{aligned} b_{3} &=\frac{1}{3} \biggl( \frac{1}{2}\mu _{2}-\frac{1}{8}\mu _{1}^{2} \biggr) , \end{aligned}$$
(22)
$$\begin{aligned} b_{4} &=\frac{1}{4} \biggl( \frac{1}{2}\mu _{3}+\frac{1}{48}\mu _{1}^{3}- \frac{1}{4}\mu _{1}\mu _{2} \biggr) , \end{aligned}$$
(23)
$$\begin{aligned} b_{5} &=\frac{1}{5} \biggl( \frac{1}{2}\mu _{4}-\frac{1}{8}\mu _{2}^{2}+ \frac{1}{384}\mu _{1}^{4}+\frac{1}{16} \mu _{1}^{2}\mu _{2}-\frac{1}{4} \mu _{1} \mu _{3} \biggr) \end{aligned}$$
(24)

for some \(p\in \mathcal{P}\) in the form of

$$ p(z)=1+\mu _{1}z+\mu _{2}z^{2}+\mu _{3}z^{3}+\cdots ,\quad z\in \mathbb{D}. $$
(25)

Using (14), we get

$$ \mathcal{H}_{2,1} \bigl( f^{-1}/2 \bigr)= \frac{107}{9216}\mu _{1}^{4} - \frac{7}{384} \mu _{1}^{2}\mu _{2} + \frac{1}{48}\mu _{1}\mu _{3} - \frac{1}{64} \mu _{2}^{2}. $$
(26)

By the rotation invariant property for the class \(\mathcal{S}^{\ast}_{e}\) and the functional \(\vert \mathcal{H}_{2,1} ( f^{-1}/2 ) \vert \), we can assume that \(\mu _{1}=\mu \in [0, 2]\). Using (18) and (19) to express \(\mu _{2}\) and \(\mu _{3}\), we obtain

$$\begin{aligned} \bigl\vert \mathcal{H}_{2,1} \bigl( f^{-1}/2 \bigr) \bigr\vert =& \biggl\vert \frac{35}{9216}\mu ^{4}- \frac{5}{768}\mu ^{2} \bigl( 4- \mu ^{2} \bigr) \kappa -\frac{1}{768} \bigl( 4-\mu ^{2} \bigr) \bigl(12+\mu ^{2} \bigr)\kappa ^{2} \\ &+\frac{1}{96}\mu \bigl( 4-\mu ^{2} \bigr) \bigl( 1- \vert \kappa \vert ^{2} \bigr)\delta \biggr\vert \end{aligned}$$

for some \(\kappa , \delta \in \overline{\mathbb{D}}\).

When \(\mu =0\), it is clear that \(\vert \mathcal{H}_{2,1} ( f^{-1}/2 ) \vert \leq \frac{1}{16}\approx 0.0625\). If \(\mu =2\), we have \(\vert \mathcal{H}_{2,1} ( f^{-1}/2 ) \vert = \frac{35}{576}\approx 0.0608\). For the case of \(\mu \in (0,2)\), using \(\vert \delta \vert \leq 1 \), we get that

$$\begin{aligned} \bigl\vert \mathcal{H}_{2,1} \bigl( f^{-1}/2 \bigr) \bigr\vert &\leq\frac{\mu ( 4-\mu ^{2} ) }{96} \biggl( \biggl\vert \frac{35\mu ^{3}}{96 (4-\mu ^{2} )}-\frac{5}{8}\mu \kappa - \frac{12+\mu ^{2}}{8\mu} \kappa ^{2} \biggr\vert + 1- \vert \kappa \vert ^{2} \biggr) \\ &=:\frac{\mu ( 4-\mu ^{2} ) }{96}\Psi (A,B,C ), \end{aligned}$$

where

$$ \Psi (A,B,C )= \bigl\vert A+B\kappa +C\kappa ^{2} \bigr\vert +1- \vert \kappa \vert ^{2}, $$
(27)

with

$$ A= \frac{35\mu ^{3}}{96 (4-\mu ^{2} )},\qquad B=- \frac{5}{8}\mu ,\qquad C=- \frac{12+\mu ^{2}}{8\mu}. $$
(28)

Obviously, \(A>0\), \(C<0\) and we can apply Lemma 2.2 to find the maximum of Ψ.

It is easy to be verified that \(\vert B \vert \geq 2 (1+C )\) and \(B^{2}\geq 4 (1-C )^{2}\) and thus we only need to consider the cases of \(R (A, B, C )\).

Noting that \(-C (4A+ \vert B \vert )\leq A \vert B \vert \) is equivalent to

$$ \frac{5 (- 19\mu ^{4} + 240\mu ^{2} + 576 )}{768 (4-\mu ^{2} )} \leq 0, $$
(29)

which is impossible to hold for all \(\mu \in (0,2)\). Hence, it is left to check the condition \(-C (-4A+ \vert B \vert )\geq A \vert B \vert \).

Let \(\mu _{0}=\sqrt{\frac{16\sqrt{39}-72}{25}}\approx 1.0568\) be the only positive root of the equation \(-25\mu ^{4}-144\mu ^{2}+192=0\). For \(\mu \in (0,\mu _{0} ]\), we have

$$ -C \bigl(-4A+ \vert B \vert \bigr)- A \vert B \vert = \frac{5 (-25\mu ^{4}-144\mu ^{2}+192 )}{256 (4-\mu ^{2} )} \geq 0. $$
(30)

Then \(\Psi (A,B,C )\leq (-A+ \vert B \vert -C )\) and thus

$$\begin{aligned} \bigl\vert \mathcal{H}_{2,1} \bigl( f^{-1}/2 \bigr) \bigr\vert & \leq \frac{\mu ( 4-\mu ^{2} ) }{96} \bigl(-A+ \vert B \vert -C \bigr) \\ &=\frac{- 107\mu ^{4} + 144\mu ^{2} + 576}{9216} \\ &=:\varrho _{1}(\mu ). \end{aligned}$$

It is an elementary work to get that \(\varrho _{1}\) attains its maximum value \(\frac{29}{428}\) at \(\mu _{1}=\frac{6\sqrt{214}}{107}\approx 0.8203\). Therefore, we obtain \(\vert \mathcal{H}_{2,1} ( f^{-1}/2 ) \vert \leq \frac{29}{428}\approx 0.0678\) when \(\mu \in (0,\mu _{0} ]\).

If \(\mu \in (\mu _{0},2 )\), then \(\Psi (A,B,C )\leq (A-C )\sqrt{1- \frac{B^{2}}{4AC}}\) and we have

$$\begin{aligned} \bigl\vert \mathcal{H}_{2,1} \bigl( f^{-1}/2 \bigr) \bigr\vert \leq &\frac{\mu ( 4-\mu ^{2} ) }{96} (A-C ) \sqrt{1-\frac{B^{2}}{4AC}} \\ =&\frac{23\mu ^{4} -96\mu ^{2} + 576}{32{,}256}\sqrt{ \frac{14 (18-\mu ^{2} )}{12+\mu ^{2}}} \\ =&:\frac{1}{32{,}256}\varrho _{2} \bigl(\mu ^{2} \bigr), \end{aligned}$$

where

$$ \varrho _{2}(t)= \bigl(23t^{2} -96t + 576 \bigr)\sqrt{ \frac{14 (18-t )}{12+t}},\quad t\in \bigl(\mu _{0}^{2},4 \bigr). $$
(31)

From Lemma 2.3, we know \(\varrho _{2}\) is convex on \([\mu _{0}^{2},4 ]\). Hence, we obtain

$$ \varrho _{2}(t)\leq \max \bigl\{ \varrho _{2} \bigl(\mu _{0}^{2} \bigr),4 \bigr\} =\varrho _{2} \bigl(\mu _{0}^{2} \bigr). $$
(32)

It follows that

$$ \bigl\vert \mathcal{H}_{2,1} \bigl( f^{-1}/2 \bigr) \bigr\vert \leq \frac{1}{32{,}256}\varrho _{2} \bigl(\mu _{0}^{2} \bigr)\approx 0.0655. $$
(33)

Combining all the above, we conclude that

$$ \bigl\vert \mathcal{H}_{2,1} \bigl( f^{-1}/2 \bigr) \bigr\vert \leq \frac{29}{428}\approx 0.0678. $$

The equality is achieved by the extremal function given by

$$ f(z)=z\exp \biggl( \int _{0}^{z}\frac{e^{\omega (t)}-1}{t}\,dt \biggr), $$
(34)

with

$$ \omega (z)=\frac{q(z)-1}{q(z)+1}, $$
(35)

and

$$ q(z)=\frac{1+\frac{6\sqrt{214}}{107}z+ z^{2}}{1-z^{2}}. $$
(36)

This completes the proof of Theorem 3.1. □

Now we will calculate the upper bounds of \(\vert \mathcal{H}_{2,2} ( f^{-1}/2 ) \vert \) for the class \(\mathcal{S}^{\ast}_{e}\).

Theorem 3.2

Let \(f\in \mathcal{S}_{e}^{\ast}\). Then

$$ \bigl\vert \mathcal{H}_{2,2} \bigl( f^{-1}/2 \bigr) \bigr\vert = \bigl\vert \Gamma _{2}\Gamma _{4}- \Gamma _{3}^{2} \bigr\vert \leq \frac{9}{192}. $$

This result is the best possible.

Proof

By putting (22), (23), and (24) with \(\mu _{1}=\mu \) into (15), we obtain

$$\begin{aligned} \mathcal{H}_{2,2} \bigl( f^{-1}/2 \bigr) ={}& \frac{1}{2{,}654{,}208} \bigl(9733 \mu ^{6}-31{,}020 \mu ^{4} \mu _{2}+20{,}736\mu _{2}\mu _{4} \\ &{} -25{,}920\mu _{2}^{3}+16{,}560 \mu ^{2}\mu _{2}^{2}+35{,}712\mu \mu _{2}\mu _{3} \\ &{} -25{,}920\mu ^{2}\mu _{4}+18{,}336\mu ^{3} \mu _{3}-18{,}432\mu _{3}^{2} \bigr) . \end{aligned}$$
(37)

Using \(\lambda =4-\mu ^{2}\) in (18), (19), and (20) of Lemma 2.1, we obtain

$$\begin{aligned} \mathcal{H}_{2,2} \bigl( f^{-1}/2 \bigr) ={}& \frac{1}{2{,}654{,}208} \bigl\{ 1075 \mu ^{6}-1944\mu ^{4} \kappa ^{3}\lambda -912\mu ^{4}\kappa ^{2} \lambda \\ & {}+144\mu ^{2}\kappa ^{4}\lambda ^{2}-3744\mu ^{2}\kappa ^{3} \lambda ^{2} + 2628\mu ^{2}\kappa ^{2}\lambda ^{2}-7776 \mu ^{2}\kappa ^{2}\lambda \\ &{} -3534 \mu ^{4} \kappa \lambda +5184\kappa ^{3}\lambda ^{2}-3240\kappa ^{3} \lambda ^{3}+ 4896\mu \kappa \lambda ^{2} \bigl(1- \vert \kappa \vert ^{2} \bigr)\delta \\ & {}-576\mu \kappa ^{2}\lambda ^{2} \bigl( 1- \vert \kappa \vert ^{2} \bigr) \delta +7776\mu ^{3} \kappa \lambda \bigl( 1- \vert \kappa \vert ^{2} \bigr) \delta \\ & {}+ 5712\mu ^{3}\lambda \bigl( 1- \vert \kappa \vert ^{2} \bigr) \delta -4608\lambda ^{2} \bigl( 1- \vert \kappa \vert ^{2} \bigr) ^{2}\delta ^{2} \\ & {}-5184\lambda ^{2} \vert \kappa \vert ^{2} \bigl( 1- \vert \kappa \vert ^{2} \bigr) \delta ^{2}+ 7776\mu ^{2}\lambda \overline{\kappa} \bigl( 1- \vert \kappa \vert ^{2} \bigr) \delta ^{2}\\ & {}+5184\kappa \lambda ^{2} \bigl( 1- \vert \kappa \vert ^{2} \bigr) \bigl( 1- \vert \delta \vert ^{2} \bigr) \varrho \\ & {}- 7776\mu ^{2}\lambda \bigl( 1- \vert \kappa \vert ^{2} \bigr) \bigl( 1- \vert \delta \vert ^{2} \bigr) \varrho \bigr\} . \end{aligned}$$

We note that it can be written in the form of

$$ H_{2,3} ( f ) =\frac{1}{2{,}654{,}208} \bigl[ \zeta _{1} ( \mu ,\kappa ) +\zeta _{2} ( \mu ,\kappa ) \delta + \zeta _{3} ( \mu ,\kappa ) \delta ^{2}+\Phi ( \mu , \kappa , \delta ) \varrho \bigr] . $$

Here \(\rho , \kappa , \delta \in \overline{\mathbb{D}}\) and

$$\begin{aligned}& \begin{aligned} \zeta _{1} ( \mu ,\kappa )={}&1075\mu ^{6}+ \bigl( 4-\mu ^{2} \bigr) \bigl[ 144\mu ^{2} \bigl( 4-\mu ^{2} \bigr)\kappa ^{4}-288 \bigl(5\mu ^{4}-20\mu ^{2}+108 \bigr)\kappa ^{3} \\ &{}+12\mu ^{2} \bigl(228-295\mu ^{2} \bigr)\kappa ^{2}-3534\mu ^{4} \kappa \bigr] , \end{aligned} \\& \zeta _{2} ( \mu ,\kappa )=48\mu \bigl( 4-\mu ^{2} \bigr) \bigl( 1- \vert \kappa \vert ^{2} \bigr) \bigl[ -12 \bigl( 4-\mu ^{2} \bigr)\kappa ^{2}+ \bigl(60\mu ^{2}+408 \bigr)\kappa +119\mu ^{2} \bigr] , \\& \zeta _{3} ( \mu ,\kappa )=576 \bigl( 4-\mu ^{2} \bigr) \bigl( 1- \vert \kappa \vert ^{2} \bigr) \biggl[ \bigl( 4- \mu ^{2} \bigr) \bigl( - \vert \kappa \vert ^{2}-8 \bigr) +\frac{27}{2}\mu ^{2}\overline{\kappa} \biggr] , \\& \Phi ( \mu ,\kappa ,\delta )=2592 \bigl( 4-\mu ^{2} \bigr) \bigl( 1- \vert \kappa \vert ^{2} \bigr) \bigl( 1- \vert \delta \vert ^{2} \bigr) \bigl[ 3\mu ^{2}+2 \kappa \bigl( 4- \mu ^{2} \bigr) \bigr] . \end{aligned}$$

By taking \(\vert \kappa \vert =x\), \(\vert \delta \vert =y\) along with \(\vert \varrho \vert \leq 1\), we obtain

$$\begin{aligned} \bigl\vert \mathcal{H}_{2,2} \bigl( f^{-1}/2 \bigr) \bigr\vert & \leq \frac{1}{2{,}654{,}208} \bigl[ \bigl\vert \zeta _{1} ( \mu ,x ) \bigr\vert + \bigl\vert \zeta _{2} ( \mu ,x ) \bigr\vert y+ \bigl\vert \zeta _{3} ( \mu ,x ) \bigr\vert y^{2}+ \bigl\vert \Phi ( \mu ,x,\delta ) \bigr\vert \bigr] . \\ &\leq \frac{1}{2{,}654{,}208} \bigl[\Theta ( \mu ,x,y ) \bigr] , \end{aligned}$$
(38)

where we set

$$ \Theta ( \mu , x, y ) =r_{1} ( \mu , x ) +r_{2} ( \mu , x ) y+r_{3} ( \mu , x ) y^{2}+r_{4} ( \mu , x ) \bigl( 1-y^{2} \bigr) , $$

with

$$\begin{aligned}& \begin{aligned} r_{1} ( \mu , x )={}&1075\mu ^{6}+ \bigl( 4-\mu ^{2} \bigr) \bigl[ 144\mu ^{2} \bigl(4-\mu ^{2} \bigr)x^{4}+288 \bigl(5\mu ^{4}-20 \mu ^{2}+108 \bigr)\mu ^{2}x^{3} \\ &{}+12\mu ^{2} \bigl\vert 288-295\mu ^{2} \bigr\vert x^{2}+3534 \mu ^{4}x \bigr] , \end{aligned} \\& r_{2} ( \mu , x )=48\mu \bigl( 4-\mu ^{2} \bigr) \bigl( 1-x^{2} \bigr) \bigl[12 \bigl( 4-\mu ^{2} \bigr)x^{2}+ \bigl(60 \mu ^{2}+408 \bigr) x+119\mu ^{2} \bigr] , \\& r_{3} ( \mu , x )=576 \bigl( 4-\mu ^{2} \bigr) \bigl( 1-x^{2} \bigr) \biggl[ \bigl( 4-\mu ^{2} \bigr) \bigl( x^{2}+8 \bigr) + \frac{27}{2}\mu ^{2}x \biggr] , \\& r_{4} ( \mu , x )=2592 \bigl( 4-\mu ^{2} \bigr) \bigl( 1-x^{2} \bigr) \bigl[3\mu ^{2}+2x \bigl( 4-\mu ^{2} \bigr) \bigr] . \end{aligned}$$

Then all that remains for us is to find the maximum value of Θ in the closed domain defined by \(\Omega := [ 0,2 ] \times [ 0,1 ] \times [ 0,1 ] \). In light of \(\Theta (0,1,1)=124{,}416\), it is seen that

$$ \max_{(\mu , x, y)\in \Omega} \bigl\{ \Theta (\mu , x, y) \bigr\} \geq 124{,}416. $$
(39)

Now we aim to illustrate that the maximum value of Θ with \((\mu , x, y)\in \Omega \) is equal to \(1{,}224{,}416\).

When \(x=1\), it reduces to

$$\begin{aligned} \Theta (\mu ,1,y)=&1075\mu ^{6}+ \bigl( 4-\mu ^{2} \bigr) \bigl(4830 \mu ^{4} - 5184\mu ^{2} + 31{,}104+12\mu ^{2} \bigl\vert 288-295\mu ^{2} \bigr\vert \bigr) \\ =&:\varpi (\mu ). \end{aligned}$$

Let \(\epsilon =\sqrt{\frac{295}{288}}\approx 0.9881\). If \(\mu \geq \epsilon \), then

$$ \varpi (\mu )=- 7295\mu ^{6} + 42{,}120\mu ^{4} - 65{,}664 \mu ^{2} + 124{,}416, $$
(40)

which has a maximum about \(110{,}662.6\) at \(\mu \approx 1.6624\) on \([\epsilon ,2 ]\). When \(\mu \in [0,\epsilon ]\), we obtain

$$ \varpi (\mu )=- 215\mu ^{6} + 6888\mu ^{4} - 38{,}016\mu ^{2} + 124{,}416, $$
(41)

which achieves its maximum \(124{,}416\) at \(\mu =0\) on \([0,\epsilon ]\). Taking \(\mu =2\), we have \(\Theta (2, x, y )\equiv 68{,}800\). In these cases, Θ gains a maximum \(124{,}416\) and thus we may assume \(\mu <2\) and \(x<1\) in the next discussions.

Let \(( \mu , x, y ) \in [0,2 ) \times [ 0,1 ) \times ( 0,1 ) \). Then

$$ \frac{\partial \Theta}{\partial y}=r_{2}(\mu ,x)+2 \bigl[r_{3}(\mu ,x)-r_{4}( \mu ,x) \bigr]y. $$
(42)

Plugging \(\frac{\partial \Theta}{\partial y}=0\) yields

$$ \widehat{y}= \frac{\mu [ 12 ( 4-\mu ^{2} )x^{2}+ (60\mu ^{2}+408 )x+119\mu ^{2} ] }{12 (1-x ) [ 2 ( 4-\mu ^{2} )x+43\mu ^{2}-64 ] }. $$

If \(\widehat{y}\in ( 0,1 ) \), then we must have the following inequalities:

$$\begin{aligned}& 12 (2-\mu ) (\mu + 2 )^{2}x^{2}+ \bigl(60\mu ^{3} + 540\mu ^{2}+ 408\mu -864 \bigr)x \\& \quad {}+ 119\mu ^{3} - 516\mu ^{2} + 768< 0, \end{aligned}$$
(43)
$$\begin{aligned}& \mu ^{2}>\frac{8 ( 8-x ) }{43-2x}=:h(x). \end{aligned}$$
(44)

It is not difficult to prove that the inequality in Equation (43) is false for \(x\in [\frac{1}{3},1 )\). Therefore, for the existence of a critical point \((\widehat{\mu},\widehat{x},\widehat{y} )\) and \(\widehat{y}\in (0,1)\), we must have \(\widehat{x}<\frac{1}{3}\). By observing that h is decreasing on \((0,1)\), then \(\widehat{\mu}^{2}>\frac{46}{31}\). As \(\widehat{x}<\frac{1}{3}\), we know

$$ r_{1} (\widehat{\mu},\widehat{x} )\leq r_{1} \biggl( \widehat{\mu},\frac{1}{3} \biggr)=:\phi _{1} (\widehat{\mu} ). $$
(45)

Using \(1-\widehat{x}^{2}<1\) and \(\widehat{x}<\frac{1}{3}\), we obtain

$$ r_{j} (\widehat{\mu},\widehat{x} ) \leq \frac{9}{8}r_{j} \biggl(\widehat{\mu},\frac{1}{3} \biggr)=:\phi _{j} ( \widehat{\mu} )\quad j=2,3,4. $$
(46)

Therefore, we deduce that

$$ \Theta (\widehat{\mu},\widehat{x},\widehat{y} )\leq \phi _{1} ( \widehat{\mu} )+\phi _{4} (\widehat{\mu} )+ \phi _{2} ( \widehat{\mu} )\widehat{y}+ \bigl[ \phi _{3} (\widehat{ \mu} )-\phi _{4} ( \widehat{\mu} ) \bigr] \widehat{y}^{2}=: \Xi ( \widehat{\mu},\widehat{y} ). $$

As \(\phi _{3} (\widehat{\mu} )-\phi _{4} (\widehat{\mu} )=64 (4-\widehat{\mu}^{2} ) (184-127 \widehat{\mu}^{2} )\leq 0\), we see \(\frac{\partial ^{2}\Xi }{\partial \widehat{y}^{2}}\leq 0\). Then it is found that

$$ \frac{\partial \Theta }{\partial \widehat{y}}\geq \frac{\partial \Theta }{\partial \widehat{y}}\vert_{\widehat{y}=1}=\phi _{2} (\widehat{\mu} )+2 \bigl[\phi _{3} (\widehat{\mu} )-\phi _{4} (\widehat{\mu} ) \bigr] \geq 0,\quad \widehat{\mu}\in \biggl(\sqrt{\frac{46}{31}},2 \biggr). $$

This means that

$$ \Xi (\widehat{\mu},\widehat{y} )\leq \Xi ( \widehat{\mu},1 )=\phi _{1} (\widehat{\mu} )+\phi _{2} (\widehat{\mu} )+\phi _{3} (\widehat{\mu} )=: \chi _{0} (\widehat{\mu} ). $$

Because \(\chi _{0}\) takes a maximum value \(107{,}665.6\), we have \(\Theta (\widehat{\mu},\widehat{x},\widehat{y} )<124{,}416\). Therefore, we have to check the cases of \(y=0\) and \(y=1\) to get the maximum point of Θ.

Suppose that \(y=0\). As

$$ \Theta (\mu ,x,0 )=r_{1} (\mu ,x )+r_{4} ( \mu ,x ) $$
(47)

and

$$ \Theta (\mu ,x,1 )=r_{1} (\mu ,x )+r_{2} ( \mu ,x )+r_{3} (\mu ,x ), $$
(48)

we have

$$\begin{aligned} \Theta (\mu ,x,1 )-\Theta (\mu ,x,0 )&=r_{2} (\mu ,x )+r_{3} (\mu ,x )-r_{4} (\mu ,x ) \\ &=48 \bigl(4-c^{2} \bigr) \bigl(1-x^{2} \bigr)\Pi (\mu ,x), \end{aligned}$$

where

$$\begin{aligned} \Pi (\mu ,x)={}&12 (1+\mu ) \bigl(4-\mu ^{2} \bigr)x^{2}+6 \bigl(10\mu ^{3} +45\mu ^{2} + 68\mu -72 \bigr)x \\ &{}+ 119 \mu ^{3} - 258 \mu ^{2} + 384. \end{aligned}$$
(49)

It is a tedious basic work to show that \(\Pi (\mu ,x)\geq 0\) on \([0,2]\times [0,1]\). Thus, we know \(\Theta (\mu ,x,1)\geq \Theta (\mu ,x,0)\) for all \((\mu ,x)\in [0,2]\times [0,1]\). Hence, we can only discuss the maximum of Θ in the case of \(y=1\).

Now it is time to find to find the maximum value of Θ on \(y=1\). Actually, when \(\mu \in [\epsilon ,2 )\), we have

$$\begin{aligned} \Theta (\mu ,x,1 )&=1075\mu ^{6}+ \bigl(4-\mu ^{2} \bigr) \bigl(m_{4}x^{4}+m_{3}x^{3}+m_{2}x^{2}+m_{1}x+m_{0} \bigr) \\ &=1075\mu ^{6}+ \bigl(4-\mu ^{2} \bigr)Q(\mu ,x) \\ &=:W(\mu ,x), \end{aligned}$$

where

$$ Q(\mu ,x)=m_{4}x^{4}+m_{3}x^{3}+m_{2}x^{2}+m_{1}x+m_{0} $$
(50)

with

$$\begin{aligned}& m_{4} =144 \bigl(4-\mu ^{2} \bigr) \bigl(\mu ^{2}-4\mu -4 \bigr), \\& m_{3} =288 \bigl(5\mu ^{4} - 10\mu ^{3} - 47\mu ^{2} - 68\mu + 108 \bigr), \\& m_{2} =12 \bigl(295\mu ^{4} - 524\mu ^{3} + 48\mu ^{2} + 192\mu - 1344 \bigr), \\& m_{1} =6\mu (\bigl(589\mu ^{3} + 480\mu ^{2} + 1296\mu + 3264 \bigr), \\& m_{0} =48 \bigl(119\mu ^{3} - 96\mu ^{2} + 384 \bigr). \end{aligned}$$

In view of \([\epsilon ,2 )\subset [\frac{9}{10},2 ]\), we next prove that \(W<124{,}416\) on the rectangle \(R_{1}:= [\frac{9}{10},2 ]\times [0,1]\). When \(\mu =\frac{9}{10}\), we obtain

$$ W \biggl(\frac{9}{10},x \biggr)=:\varsigma _{1}(x). $$
(51)

It is calculated that \(\varsigma _{1}\) attains its maximum about \(98{,}883.6\) at \(x\approx 0.7837\). When \(\mu =2\), we have \(W (2,x )\equiv 68{,}800\) for all \(x\in [0,1]\). If \(x=0\), then

$$\begin{aligned} W (\mu ,0 )&=1075\mu ^{6} - 5712\mu ^{5} + 4608\mu ^{4} + 22{,}848 \mu ^{3} - 36{,}864\mu ^{2} + 73{,}728 \\ &=:\varsigma _{2}(\mu ). \end{aligned}$$
(52)

It is observed that \(\varsigma _{2}\) achieves its maximum \(68{,}800\) at \(\mu =2\) on \([\frac{9}{10},2 ]\). If \(x=1\), then

$$ W (\mu ,1 )=- 7295\mu ^{6} + 42{,}120\mu ^{4} - 65{,}664\mu ^{2} + 124{,}416=:\varsigma _{3}(\mu ). $$
(53)

We see \(\varsigma _{3}\) obtains its maximum about \(110{,}662.6\) at \(\mu \approx 1.6624\) on \([\frac{9}{10},2 ]\). By numerical calculation, it is noted that the system of the equations

$$ \frac{\partial W}{\partial \mu}=0\quad \text{and}\quad \frac{\partial W}{\partial x}=0 $$
(54)

has no positive roots in \((\frac{9}{10},2 )\times (0,1)\). Thus, there are no critical points of W in the interior of \(R_{1}\). Based on these facts, we conclude that \(W<124{,}416\) when \(\mu \in [\epsilon ,2 )\).

It remains to consider \(\mu \in [0,\epsilon )\). In this case, we obtain

$$\begin{aligned} \Theta (\mu ,x,1 )&=1075\mu ^{6}+ \bigl(4-\mu ^{2} \bigr) \bigl(l_{4}x^{4}+l_{3}x^{3}+l_{2}x^{2}+l_{1}x+l_{0} \bigr) \\ &=1075\mu ^{6}+ \bigl(4-\mu ^{2} \bigr)L(\mu ,x) \\ &=:K(\mu ,x), \end{aligned}$$

where

$$ L(\mu ,x)=l_{4}x^{4}+l_{3}x^{3}+l_{2}x^{2}+l_{1}x+l_{0} $$
(55)

with

$$\begin{aligned}& l_{4} =144 \bigl(4-\mu ^{2} \bigr) \bigl(\mu ^{2}-4\mu -4 \bigr), \\& l_{3} =288 \bigl(5\mu ^{4} - 10\mu ^{3} - 47\mu ^{2} - 68\mu + 108 \bigr), \\& l_{2} =12 \bigl(-295\mu ^{4} - 524\mu ^{3} +624\mu ^{2} + 192\mu - 1344 \bigr), \\& l_{1} =6\mu (\bigl(589\mu ^{3} + 480\mu ^{2} + 1296\mu + 3264 \bigr), \\& l_{0} =48 \bigl(119\mu ^{3} - 96\mu ^{2} + 384 \bigr). \end{aligned}$$

As \([0,\epsilon )\subset [0,1 ]\), we next prove that \(W\leq 124{,}416\) on the rectangle \(R_{2}:= [0,1 ]\times [0,1]\). When \(\mu =0\), we obtain

$$ K (0,x )=- 9216x^{4} + 124{,}416x^{3} - 64{,}512x^{2} + 73{,}728:= \sigma _{1}(x). $$
(56)

It is calculated that \(\sigma _{1}\) attains its maximum \(124{,}416\) at \(x=1\). When \(\mu =2\), we have

$$ K (1,x )=- 9072x^{4} - 10{,}368x^{3} - 48{,}492x^{2} + 101{,}322x + 59{,}683=: \sigma _{2}(x). $$
(57)

It is noted that \(\sigma _{2}\) has a maximum about \(11{,}197.1\) attained at \(x\approx 0.7292\). If \(x=0\), then

$$ K (\mu ,0 )=1075\mu ^{6} - 5712\mu ^{5} + 4608\mu ^{4} + 22{,}848 \mu ^{3} - 36{,}864\mu ^{2} + 73{,}728=:\sigma _{3}(\mu ). $$
(58)

It is observed that \(\sigma _{3}\) achieves its maximum \(73{,}728\) at \(\mu =0\) on \([0,1 ]\). If \(x=1\), then

$$ K (\mu ,1 )=- 215\mu ^{6} + 6888\mu ^{4} - 38{,}016\mu ^{2} + 124{,}416=: \sigma _{4}(\mu ). $$
(59)

We see \(\sigma _{4}\) obtains its maximum about \(124{,}416\) at \(\mu =0\) on \([0,1 ]\). By numerical calculation, we get that the system of the equations

$$ \frac{\partial K}{\partial \mu}=0\quad \text{and}\quad \frac{\partial K}{\partial x}=0 $$
(60)

has two positive roots in \((0,1 )\times (0,1)\), i.e., \((0.0903, 0.0721 )\) and \((0.9180, 0.7954 )\). The two critical points of K have the values about \(73{,}686.7\) and \(100{,}800.0\). From these discussions, we get that \(K(\mu ,x)\leq 124{,}416\) if \(\mu \in [0,\epsilon )\). Therefore, we obtain the inequality that \(\Theta ( \mu ,x,y ) \leq 124{,}416\) on the whole domain Ω, which leads to

$$ \bigl\vert \mathcal{H}_{2,2} \bigl( f^{-1}/2 \bigr) \bigr\vert \leq \frac{124{,}416}{2{,}654{,}208}=\frac{9}{192}\approx 0.0469. $$

The equality is obtained by the extremal function f given by

$$ f(z)=z\exp \biggl( \int _{0}^{z}\frac{e^{t^{2}}-1}{t}\,dt \biggr)=z+ \frac{1}{2}z^{3}+\frac{1}{4}z^{5}+ \cdots ,\quad z\in \mathbb{D}. $$
(61)

 □

4 Conclusion

In this paper, we consider the Hankel determinant by taking coefficients of logarithmic coefficients of inverse functions for certain subclasses of univalent functions. This is a natural generalization of the classic concept and may help to understand more properties of the inverse functions. We have obtained the sharp bounds for the second Hankel determinant of logarithmic coefficients of inverse functions with respect to the subclass of starlike functions defined by subordination to the exponential functions. Given the importance of the logarithmic coefficients and inverse coefficients of univalent functions, our work provides a new basis for the study of the Hankel determinant. It could also inspire further similar results investigating other subfamilies of univalent functions or taking the bounds of higher-order Hankel determinants.

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Acknowledgements

This work was supported by the Key Project of Natural Science Foundation of Educational Committee of Henan Province under Grant no. 24B110001 of the P. R. China. The authors would like to express their gratitude for the referees’ valuable suggestions, which really improved the present work.

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The idea of the present paper was proposed by Lei Shi, Muhammad Abbas and improved by Mohsan Raza. Lei Shi and Muhammad Abbas wrote and completed the calculations. Muhammad Arif and Poom Kumam checked all the results. All authors read and approved the final manuscript.

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Correspondence to Mohsan Raza.

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Shi, L., Abbas, M., Raza, M. et al. Inverse logarithmic coefficient bounds for starlike functions subordinated to the exponential functions. J Inequal Appl 2024, 17 (2024). https://doi.org/10.1186/s13660-024-03094-5

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