On coefficient problems for functions starlike with respect to symmetric points

Abstract

The main idea of the study on coefficient problems in various classes of analytic functions (univalent or nonunivalent) is to express the coefficients of functions in a given class by the coefficients of corresponding functions with positive real part. Thus, coefficient functionals can be studied using inequalities known for the class ${\mathcal {P}}$. Lemmas obtained by Libera and Złotkiewicz and by Prokhorov and Szynal play a special role in this approach. Recently, a new way leading to results on coefficient functionals has been pointed out. This approach is based on relating the coefficients of functions in a given class and the coefficients of corresponding Schwarz functions. In many cases, if we follow this approach, it is easy to predict the exact estimate of the functional and make the appropriate computations. In the proofs of these estimates are used not only classical results (the Schwarz–Pick Lemma or Wiener’s inequality), but also inequalities obtained either recently (e.g. by Efraimidis) or long ago yet almost forgotten (Carlson’s inequality). In this paper, a number of coefficient problems will be solved using the new approach described above. The object of our study is the class of starlike functions with respect to symmetric points associated with the exponential function.

1 Introduction

Let ${\mathbb {D}}$ be the unit disk $\{z\in {\mathbb {C}}:|z|<1\}$ and ${\mathcal {A}}$ be the family of all functions f analytic in ${\mathbb {D}}$, normalized by the condition $f(0)=f'(0)-1=0$. It means that f has the expansion

$$\begin{aligned} f(z)=z+\sum _{n=2}^{\infty } a_n z^n\ . \end{aligned}$$

(1.1)

Let also ${\mathcal {B}}_0$ be the class of Schwarz functions, i.e., analytic functions $\omega :{\mathbb {D}}\rightarrow {\mathbb {D}}$, $\omega (0)=0$. The function $\omega \in {\mathcal {B}}_0$ can be written as a power series

$$\begin{aligned} \omega (z)=\sum _{n=1}^\infty c_nz^n\ . \end{aligned}$$

(1.2)

For given analytic functions f and g in ${\mathbb {D}}$, we say that f is subordinate to g in ${\mathbb {D}}$ and write $f\prec g$ if there exists $\omega \in {\mathcal {B}}_0$ such that

$$\begin{aligned} f(z) = g(\omega (z)),\quad z\in {\mathbb {D}}. \end{aligned}$$

Moreover, if the function g is univalent in ${\mathbb {D}}$, then $f\prec g$ if and only if $f(0) = g(0)$ and $f({\mathbb {D}})\subset g({\mathbb {D}})$.

Let ${\mathcal {S}}_S^*$ denote the class of functions which are starlike with respect to the symmetric points. A function f is in ${\mathcal {S}}_S^*$ if

$$\begin{aligned} \frac{2zf'(z)}{f(z)-f(-z)} \prec \varphi _0(z) , \quad z \in {\mathbb {D}}, \end{aligned}$$

where

$$\begin{aligned} \varphi _0(z)=\frac{1+z}{1-z}\ . \end{aligned}$$

If the function $\varphi _0(z)$ is replaced by any analytic univalent function $\varphi$ with positive real part in ${\mathbb {D}}$ and symmetric with respect to the real axis, then we obtain the class ${\mathcal {S}}_S^*(\varphi (z))$.

In this paper, we consider the class ${\mathcal {S}}_S^*(\varphi (z))$ with $\varphi (z)=e^z$. Hence, we can write

$$\begin{aligned} {\mathcal {S}}_S^*(e^z) = \left\{ f\in {\mathcal {S}}: \frac{2zf'(z)}{f(z)-f(-z)} = e^{\omega (z)}, \quad \omega \in {\mathcal {B}}_0\ ,\ z \in {\mathbb {D}} \right\} \ . \end{aligned}$$

This class was first discussed by Ganesh et al. in [3], where some coefficients functionals were estimated. The majority of results were not sharp. The main tool used to obtain those results was a lemma proved by Libera and Złotkiewicz.

In this paper, we follow a new approach which is based on relating the coefficients of functions in a given class and the coefficients of corresponding Schwarz functions. In many cases, it is easy to predict the exact estimate of the functional and make the appropriate computations. It is the case for the class ${\mathcal {S}}_S^*(e^z)$. By applying the lemmas proved by Libera and Złotkiewicz and by Prokhorov and Szynal as well as some other tools, and by performing the calculus more precisely, we are able to derive better estimates, almost all of them being sharp.

To prove our results, we need the following lemmas for Schwarz functions. The first one is the above-mentioned result obtained by Prokhorov and Szynal.

Lemma 1

([5]) Let $\omega (z)=c_{1}z+c_{2}z^{2}+\cdots$ be a Schwarz function. Then, for any real numbers $\mu$ and $\nu$ such that

$$\begin{aligned} (\mu ,\nu ) \in \left\{ |\mu |\le \frac{1}{2}\ ,\ -1 \le \nu \le 1 \right\} \cup \left\{ \frac{1}{2}\le |\mu |\le 2\ ,\ \frac{4}{27} \left( |\mu |+1\right) ^3-\left( |\mu |+1\right) \le \nu \le 1 \right\} \ , \end{aligned}$$

the following sharp estimate holds

$$\begin{aligned} \left| c_{3}+\mu c_{1}c_{2}+\nu c_{1}^{3}\right| \le 1. \end{aligned}$$

From the Schwarz–Pick Lemma, it follows that for $\omega \in {\mathcal {B}}_0$ of the form (1.2),

$$\begin{aligned} |c_2|\le 1-|c_1|^2\ . \end{aligned}$$

This inequality can be improved (see, for example, [2]) as follows. For any $\lambda \in {\mathbb {C}}$,

$$\begin{aligned} \left| c_2+\lambda c_1^2\right| \le \max \{1,|\lambda |\}\ . \end{aligned}$$

(1.3)

Carlson in [1] obtained another generalization of the Schwarz–Pick Lemma. Here, we state only these inequalities which are useful for our purpose (for all details, see [7]).

Lemma 2

( [1]) Let $\omega (z)=c_{1}z+c_{2}z^{2}+\cdots$ be a Schwarz function. Then,

$$\begin{aligned}&|c_3|\le 1-|c_1|^2 -\tfrac{|c_2|^2}{1+|c_1|} \quad ,\quad |c_4|\le 1-|c_1|^2 -|c_2|^2\quad ,\\&\quad |c_5|\le 1-|c_1|^2 -|c_2|^2-\tfrac{|c_3|^2}{1+|c_1|} \ . \end{aligned}$$

The above lemma immediately results in the following fact.

Lemma 3

Let $\omega (z)=c_{1}z+c_{2}z^{2}+\cdots$ be a Schwarz function. Then

$$\begin{aligned} |c_1c_3-c_2^2| \le 1-|c_1|^2\ . \end{aligned}$$

We also need the results obtained by Efraimidis.

Lemma 4

([2]) Let $\omega (z)=c_{1}z+c_{2}z^{2}+\cdots$ be a Schwarz function and $\lambda \in {\mathbb {C}}$. Then

$$\begin{aligned} \left| c_4+(1+\lambda )c_1c_3+c_2^2+(1+2\lambda )c_1^2c_2+\lambda c_1^4\right| \le \max \left\{ 1,|\lambda |\right\} \end{aligned}$$

(1.4)

and

$$\begin{aligned} \left| c_4+2c_1c_3+\lambda c_2^2+(1+2\lambda )c_1^2c_2+\lambda c_1^4\right| \le \max \left\{ 1,|\lambda |\right\} \ . \end{aligned}$$

(1.5)

The method used by Efraimidis in the proof of his lemma has much greater potential. Based on this method, we can obtain some inequalities involving the fifth coefficient of $\omega \in {\mathcal {B}}_0$ (see also [6]).

Lemma 5

If $\omega \in {\mathcal {B}}_0$ is of the form (1.2) and $\mu \in {\mathbb {C}}$, $|\mu |\le 1$, then

$$\begin{aligned}&\left| c_5+(1+\mu ) c_1c_4 + (1+\mu ) c_2c_3 + 3\mu c_1c_2^2 + \left( 1+\mu +\mu ^2\right) c_1^2c_3\right. \nonumber \\&\quad \qquad \left. + 2\mu (1+\mu ) c_1^3c_2 + \mu ^2 c_1^5\right| \le 1 \ . \end{aligned}$$

(1.6)

2 Coefficient bounds

We start with the coefficients of $f\in {\mathcal {S}}_S^*(e^z)$. Applying in

$$\begin{aligned} \frac{2zf'(z)}{f(z)-f(-z)} = e^{\omega (z)}, \end{aligned}$$

(2.1)

the expansions of f and $\omega$ given by (1.1) and (1.2), we obtain

$$\begin{aligned} a_2 & = {} \tfrac{1}{2} c_1 \nonumber \\ a_3 & = {} \tfrac{1}{2} c_2 + \tfrac{1}{4} c_1^2 \nonumber \\ a_4 & = {} \tfrac{1}{4} c_3 + \tfrac{3}{8} c_1c_2 + \tfrac{5}{48} c_1^3 \nonumber \\ a_5 & = {} \tfrac{1}{4} c_4 + \tfrac{1}{4} c_1c_3 + \tfrac{1}{4} c_2^2 + \tfrac{1}{4} c_1^2c_2 + \tfrac{1}{24} c_1^4 \nonumber \\ a_6 & = {} \tfrac{1}{6} c_5 + \tfrac{5}{24} c_1c_4 + \tfrac{1}{4} c_2c_3 + \tfrac{5}{24} c_1c_2^2 + \tfrac{1}{6} c_1^2c_3 + \tfrac{1}{8} c_1^3c_2 + \tfrac{11}{720} c_1^5 \ . \end{aligned}$$

(2.2)

From [3], it is known that if $f\in {\mathcal {S}}_S^*(e^z)$ is of the form (1.1), then $|a_2|\le \tfrac{1}{2}$ and $|a_3|\le \tfrac{1}{2}$.

Theorem 1

If $f\in {\mathcal {S}}_S^*(e^z)$ is of the form (1.1), then

$$\begin{aligned}|a_4|\le \tfrac{1}{4}\quad \text {and}\quad |a_5|\le \tfrac{1}{4}\ .\end{aligned}$$

The bounds are sharp.

Proof

Lemma 1 with $\mu =\tfrac{3}{2}$ and $\nu =\tfrac{5}{12}$ applied to

$$\begin{aligned} |a_4| = \tfrac{1}{4} \left| c_3 + \tfrac{3}{2} c_1c_2 + \tfrac{5}{12} c_1^3\right| , \end{aligned}$$

results in the first inequality.

To prove the second inequality, we can write

$$\begin{aligned} 4|a_5| = \left| \tfrac{1}{2}\left( c_4 + 2c_1c_3 + c_2^2 + 3c_1^2c_2 + c_1^4\right) + \tfrac{1}{2} \left( c_4 + c_2^2 - c_1^2c_2 - \tfrac{2}{3} c_1^4\right) \right| \ . \end{aligned}$$

(2.3)

From (1.4) with $\lambda =1$, the first component is bounded by 1/2. By Lemma 2, the second component, can be estimated as follows

$$\begin{aligned} \tfrac{1}{2}\left| c_4 + c_2^2 - c_1^2c_2 - \tfrac{2}{3} c_1^4\right| & \le {} \tfrac{1}{2}\left[ 1-|c_1|^2-|c_2|^2+|c_2|^2+|c_1|^2\left( 1-|c_1|^2\right) +\tfrac{2}{3}|c_1|^4\right] \\ & = {} \tfrac{1}{2}-\tfrac{1}{6}|c_1|^4 \le \tfrac{1}{2}\ . \end{aligned}$$

Combining the estimates of both components of (2.3), we get $|a_5|\le \tfrac{1}{4}$.

If $c_3=1$ and $c_k=0$ for $k\ne 3$, then $a_4=\tfrac{1}{4}$. Similarly, if $c_4=1$ and $c_k=0$ for $k\ne 4$, then $a_5=\tfrac{1}{4}$. This means that the equalities in the assertion of this theorem hold for the functions given by (2.1) with $\omega (z)=z^3$ and $\omega (z)=z^4$, respectively. $\square$

The logarithmic coefficients of a given univalent function f, denoted by $\gamma _n=\gamma _n(f)$, are defined as

$$\begin{aligned} \tfrac{1}{2}\log \left( \frac{f(z)}{z}\right) = \sum _{n=1}^\infty \gamma _n z^n\ . \end{aligned}$$

(2.4)

If f is given by (1.1), then its logarithmic coefficients are given as follows

$$\begin{aligned} \gamma _1 & = {} \tfrac{1}{2} a_2 \nonumber \\ \gamma _2 & = {} \tfrac{1}{2} \left( a_3 - \tfrac{1}{2} a_2^2\right) \nonumber \\ \gamma _3 & = {} \tfrac{1}{2} \left( a_4 - a_2a_3 + \tfrac{1}{3} a_2^3\right) \nonumber \\ \gamma _4 & = {} \tfrac{1}{2} \left( a_5 - a_2a_4 + a_2^2a_3 - \tfrac{1}{2} a_3^2 - \tfrac{1}{4} a_2^4\right) \nonumber \\ \gamma _5 & = {} \tfrac{1}{2} \left( a_6 - a_2a_5 - a_3a_4 + a_2a_3^2 + a_2^2a_4 - a_2^3a_3 + \tfrac{1}{5} a_2^5\right) \ . \end{aligned}$$

(2.5)

The sharp bounds of $\gamma _k$, $k=1,2,3,4,5$ are established in the next theorem.

Theorem 2

If $f\in {\mathcal {S}}_S^*(e^z)$ is of the form (1.1), then

$$\begin{aligned} |\gamma _1|\le \tfrac{1}{4} \quad ,\quad |\gamma _2|\le \tfrac{1}{4} \quad ,\quad |\gamma _3|\le \tfrac{1}{8} \quad ,\quad |\gamma _4|\le \tfrac{1}{8} \quad ,\quad |\gamma _5|\le \tfrac{1}{12}\ . \end{aligned}$$

The bounds are sharp.

Proof

Applying (2.2) in (2.5), we have

$$\begin{aligned} \gamma _1 & = {} \tfrac{1}{4} c_1 \nonumber \\ \gamma _2 & = {} \tfrac{1}{4} \left( c_2 + \tfrac{1}{4} c_1^2\right) \nonumber \\ \gamma _3 & = {} \tfrac{1}{8} \left( c_3 + \tfrac{1}{2} c_1c_2 + \tfrac{1}{12} c_1^3\right) \nonumber \\ \gamma _4 & = {} \tfrac{1}{8} \left( c_4 + \tfrac{1}{2} c_1c_3 + \tfrac{1}{2} c_2^2 + \tfrac{1}{4} c_1^2c_2 + \tfrac{1}{48} c_1^4\right) \nonumber \\ \gamma _5 & = {} \tfrac{1}{12} \left( c_5 + \tfrac{1}{2} c_1c_4 + \tfrac{3}{4} c_2c_3 + \tfrac{1}{16} c_1^3c_2 + \tfrac{1}{4} c_1^2c_3 + \tfrac{1}{8} c_1c_2^2 + \tfrac{1}{240} c_1^5\right) \ . \end{aligned}$$

(2.6)

The bounds of $\gamma _1$ and $\gamma _2$ are clear. The result for $\gamma _3$ immediately follows from Lemma 1 with $\mu =\tfrac{1}{2}$ and $\nu =\tfrac{1}{12}$.

Observe that

$$\begin{aligned}&\left| c_4 + \tfrac{1}{2} c_1c_3 + \tfrac{1}{2} c_2^2 + \tfrac{1}{4} c_1^2c_2 + \tfrac{1}{48} c_1^4\right| \\&\quad = \left| \tfrac{1}{2}\left( c_4+c_1c_3+c_2^2+c_1^2c_2\right) +\tfrac{1}{2} \left( c_4-\tfrac{1}{2}c_1^2c_2+\tfrac{1}{24} c_1^4\right) \right| \le 1\ . \end{aligned}$$

Indeed, from (1.4) with $\lambda =0$,

$$\begin{aligned} \tfrac{1}{2}\left| c_4+c_1c_3+c_2^2+c_1^2c_2\right| \le \tfrac{1}{2}\ \end{aligned}$$

and, by Lemma 2 and (1.3),

$$\begin{aligned} \left| \tfrac{1}{2} c_4-\tfrac{1}{4}c_1^2c_2+\tfrac{1}{48} c_1^4\right| \le \tfrac{1}{2}\left[ 1-|c_1|^2-|c_2|^2+\tfrac{1}{2}|c_1|^2(1-|c_1|^2)+\tfrac{1}{24} |c_1|^4\right] \ , \end{aligned}$$

which is clearly less than or equal to 1/2.

For $\gamma _5$, we have

$$\begin{aligned} 12|\gamma _5| & = {} \tfrac{1}{2}\left| \left( c_5 + c_1c_4 + c_2c_3 + c_1^2c_3\right) \right. \\&\quad + \left. \left( c_5 + \tfrac{1}{2} c_2c_3 - \tfrac{1}{2} c_1^2c_3 + \tfrac{1}{8} c_1^3c_2 + \tfrac{1}{4} c_1c_2^2 + \tfrac{1}{120} c_1^5\right) \right| \le \tfrac{1}{2}(1+W)\ . \end{aligned}$$

The bound of the first component follows from (1.6) with $\mu =0$. The triangle inequality and Lemma 2 result in

$$\begin{aligned}W & \le 1-|c_1|^2 -|c_2|^2-\tfrac{|c_3|^2}{1+|c_1|} + \tfrac{1}{2}\left( |c_2|+|c_1|^2\right) |c_3|\\&\quad + \tfrac{1}{8}|c_1|^3|c_2| +\tfrac{1}{4}|c_1||c_2|^2 + \tfrac{1}{120}|c_1|^5\ . \end{aligned}$$

The above expression takes its greatest value with respect to $|c_3|$ when $|c_3|=\tfrac{1}{4}(|c_2|+|c_1|^2)(1+|c_1|)$, so

$$\begin{aligned}W &\le 1-|c_1|^2 +\tfrac{1}{16}|c_1|^4+\tfrac{17}{240}|c_1|^5-\tfrac{15}{16}|c_2|^2+ \tfrac{5}{16}|c_1||c_2|^2\\&\quad +\tfrac{1}{8}|c_1|^2|c_2|+\tfrac{1}{4}|c_1|^3|c_2|\ . \end{aligned}$$

Since $|c_1|^3|c_2|\le |c_1|^2|c_2|$, we can write

$$\begin{aligned} W \le h(c,d)\ , \end{aligned}$$

where $c=|c_1|$, $d=|c_2|$ and

$$\begin{aligned} h(c,d) = 1-c^2 +\tfrac{1}{16}c^4+\tfrac{17}{240}c^5-\tfrac{15}{16}d^2+ \tfrac{5}{16}cd^2+\tfrac{3}{8} c^2d\ . \end{aligned}$$

The shape of the region of variability of (c, d) is a simple consequence of the Schwarz–Pick Lemma. It coincides with

$$\begin{aligned} \varOmega =\left\{ (c,d): 0\le c\le 1, 0\le d\le 1-c^2\right\} \ . \end{aligned}$$

(2.7)

It is not difficult to show that (0, 0) is the only critical point of h in $\varOmega$. Consequently, it is enough to derive the greatest value of h on the boundary of $\varOmega$. But,

$$\begin{aligned} h(c,0) & = {} 1-c^2 +\tfrac{1}{16}c^4+\tfrac{17}{240}c^5 \le 1\ , \\ h(0,d) & = {} 1-\tfrac{15}{16}d^2\le 1\ ,\\ h(c,1-c^2) & = {} \tfrac{1}{16}+\tfrac{5}{16}c+\tfrac{5}{4} c^2-\tfrac{5}{8} c^3-\tfrac{5}{4} c^4+\tfrac{23}{60}c^5\ . \end{aligned}$$

Since $\tfrac{5}{4} (c^2-c^4)\le \tfrac{5}{16}$ and $\tfrac{5}{16}(c-2c^3+c^5)\le \tfrac{\sqrt{5}}{25}$ we conclude that

$$\begin{aligned} h\left( c,1-c^2\right) \le \,\tfrac{1}{16}+\tfrac{5}{16}+\tfrac{\sqrt{5}}{25}+\left( \tfrac{23}{60}-\tfrac{5}{16}\right) c^5 <1\ . \end{aligned}$$

Combining all these inequalities, we get

$$\begin{aligned} h(c,d) \le h(0,0) =1\ , \end{aligned}$$

which results in the desired bound.

Observe that we obtain equalities in each bound of $\gamma _k$, $k=1,2,3,4,5$ when $\omega (z)=z^k$. This means that the obtained estimates are sharp. $\square$

3 Estimates of Zalcman functionals and Hankel determinants

It is known ([3]) that if $f\in {\mathcal {S}}_S^*(e^z)$ is of the form (1.1), then $|a_3-a_2^2|\le \tfrac{1}{2}$. This functional, known as the Fekete–Szegö functional, is a particular case of the Zalcman functional $a_{n+m-1}-a_na_m$. Let us consider other cases of the Zalcman functional.

The estimate of $a_{4}-a_2a_3$ is a simple consequence of Lemma 1. Namely,

$$\begin{aligned} \left| a_{4}-a_2a_3\right| = \tfrac{1}{4}\left| c_3+\tfrac{1}{2} c_1c_2-\tfrac{1}{12} c_1^3\right| \ , \end{aligned}$$

so taking $\mu =\tfrac{1}{2}$ and $\nu =-\tfrac{1}{12}$ in Lemma 1 yields

$$\begin{aligned} |a_4-a_2a_3|\le \tfrac{1}{4}\ . \end{aligned}$$

To estimate $a_{5}-a_3^2$, we write this expression as follows

$$\begin{aligned} \left| a_5-a_3^2\right| & = {} \tfrac{1}{4}\left| c_4+c_1c_3-\tfrac{1}{12} c_1^4\right| \\ & = {} \tfrac{1}{4}\left| \tfrac{1}{2}\left( c_4+2c_1c_3-\tfrac{1}{2} c_2^2-\tfrac{1}{2} c_1^4\right) +\tfrac{1}{2}\left( c_4+\tfrac{1}{2} c_2^2+\tfrac{1}{3} c_1^4\right) \right| \ . \end{aligned}$$

Applying (1.5) with $\lambda =-\tfrac{1}{2}$, the first component is bounded by 1/2. To estimate the other, we use the triangle inequality and Lemma 2. Hence

$$\begin{aligned} \tfrac{1}{2}\left| c_4+\tfrac{1}{2} c_2^2+\tfrac{1}{3} c_1^4\right|\le & {} \tfrac{1}{2}\left( 1-|c_1|^2-|c_2|^2+\tfrac{1}{2} |c_2|^2+\tfrac{1}{3} |c_1|^4\right) \\\le & {} \tfrac{1}{2}\left[ 1-\tfrac{1}{3} |c_1|^2\left( 3-|c_1|^2\right) -\tfrac{1}{2} |c_2|^2\right] \le \tfrac{1}{2}\ , \end{aligned}$$

so

$$\begin{aligned} |a_5-a_3^2|\le \tfrac{1}{4}\ . \end{aligned}$$

Considering $\omega (z)=z^3$ and $\omega (z)=z^4$, we can observe that the estimates of the two cases of the Zalcman functional are sharp. We have proved what follows.

Theorem 3

If $f\in {\mathcal {S}}_S^*(e^z)$ is of the form (1.1), then the following sharp bounds hold

$$\begin{aligned} |a_4-a_2a_3|\le \tfrac{1}{4}\quad \text {and}\quad |a_5-a_3^2|\le \tfrac{1}{4}\ . \end{aligned}$$

Let us turn to Hankel determinants for the class ${\mathcal {S}}_S^*(e^z)$. The first result is easy to obtain.

Theorem 4

If $f\in {\mathcal {S}}_S^*(e^z)$ is of the form (1.1), then

$$\begin{aligned}|H_{2,2}|\le \tfrac{1}{4}\ .\end{aligned}$$

Proof

If $f\in {\mathcal {S}}_S^*(e^z)$, then

$$\begin{aligned} |H_{2,2}| = \left| a_2a_4-a_3^2\right| = \tfrac{1}{4}\left| \tfrac{1}{2}\left( c_2^2- c_1c_3\right) +\tfrac{1}{2}\left( c_2^2+\tfrac{1}{2} c_1^2c_2+\tfrac{1}{12} c_1^4\right) \right| \ . \end{aligned}$$

Since

$$\begin{aligned} \left| c_2^2+\tfrac{1}{2} c_1^2c_2+\tfrac{1}{12} c_1^4\right| \le \left( 1-|c_1|^2\right) ^2 + \tfrac{1}{2} c_1^2\left( 1-|c_1|^2\right) + \tfrac{1}{12} c_1^4 = 1 - \tfrac{3}{2} |c_1|^2 + \tfrac{7}{12} |c_1|^4 \end{aligned}$$

and

$$\begin{aligned}{}[0,1]\ni t\rightarrow 1 - \tfrac{3}{2} t + \tfrac{7}{12} t^2 \end{aligned}$$

is a decreasing function, we conclude that

$$\begin{aligned} \left| c_2^2+\tfrac{1}{2} c_1^2c_2+\tfrac{1}{12} c_1^4 \right| \le 1\ . \end{aligned}$$

This and Lemma 3 result in the declared bound of $H_{2,2}$.

Furthermore, $|H_{2,2}| = \tfrac{1}{4}$ if $c_2=1$ and $c_k=0$ for $k\ne 2$, i.e. if $\omega (z)=z^2$. $\square$

Theorem 5

If $f\in {\mathcal {S}}_S^*(e^z)$ is of the form (1.1), the following sharp bound holds

$$\begin{aligned}|H_{2,3}|\le \tfrac{1}{8}\ .\end{aligned}$$

Proof

Assume that $f\in {\mathcal {S}}_S^*(e^z)$ is of the form (1.1). Then

$$\begin{aligned} |H_{2,3}| & = {} \left| a_3a_5-a_4^2\right| \\ & = {} \tfrac{1}{8}\left| \left( c_2+\tfrac{1}{2} c_1^2\right) c_4 - \tfrac{1}{2} c_3^2 - \tfrac{1}{2}\left( c_2-\tfrac{1}{6} c_1^2\right) c_1c_3 + c_2^3 + \tfrac{3}{8} c_1^2c_2^2 + \tfrac{1}{24} c_1^4c_2 - \tfrac{1}{288} c_1^6\right| \ . \end{aligned}$$

By the triangle inequality and Lemma 2,

$$\begin{aligned} |H_{2,3}|&\le {} \tfrac{1}{8}\left[ \left( |c_2|+\tfrac{1}{2} |c_1|^2\right) \left( 1-|c_1|^2-|c_2|^2\right) + \tfrac{1}{2} \left( 1-|c_1|^2-\frac{|c_2|^2}{1+|c_1|}\right) ^2 \right. \\&\left. \quad + \tfrac{1}{2}\left( |c_2|+\tfrac{1}{6} |c_1|^2\right) |c_1|\left( 1-|c_1|^2-\frac{|c_2|^2}{1+|c_1|}\right) \right. \\&\left. \quad + |c_2|^3 + \tfrac{3}{8} |c_1|^2|c_2|^2 + \tfrac{1}{24} |c_1|^4|c_2| + \tfrac{1}{288} |c_1|^6\right] \ . \end{aligned}$$

Let $h(|c_1|,|c_2|)$ denote the right hand side of the above inequality and let $c=|c_1|$, $d=|c_2|$. Since

$$\begin{aligned} \frac{\partial h}{\partial d} & = {} \frac{1}{24(1+c)^2} \left[ 48d^3 -36c(1+c)d^2-2(1+c)\left( 24-21c^2+5c^3\right) d\right. \\&\quad+\left. \left( 24+12c-24c^2-12c^3+c^4\right) (1+c)^2\right] \ , \end{aligned}$$

replacing $d^2$ by d and omitting $c^4$ in the last component, we get

$$\begin{aligned} \frac{\partial h}{\partial d} \ge \frac{g(c,d)}{24(1+c)^2} \end{aligned}$$

with

$$\begin{aligned} g(c,d) = 48d^3-2(1+c)\left( 24+18c-21c^2+5c^3\right) d+12(2+c)(1-c)(1+c)^3\ . \end{aligned}$$

A straightforward algebraic computation shows that the critical points of g satisfy

$$\begin{aligned}{\left\{ \begin{array}{ll} \left( 21-3c-24c^2+10c^3\right) d-3\left( 5-6c-5c^2\right) (1+c)^2 = 0 \\ 72d^2-(1+c)\left( 24+18c-21c^2+5c^3\right) = 0, \end{array}\right. }\end{aligned}$$

so in $\varOmega$ given by (2.7), there is only one critical point $(c_0,d_0)$ where $c_0=0.345\ldots$ and $d_0=0.722\ldots$. For this point,

$$\begin{aligned} g(c_0,d_0) = 8.695\ldots \ . \end{aligned}$$

On the boundary of $\varOmega$, we have

$$\begin{aligned} g(c,0) & = {} 12(2+c)(1-c)(1+c)^3 \ge g(1,0) = 0\ , \\ g(0,d) & = {} 24\left( 1-2d+2d^3\right) \ge g\left( 0,\tfrac{\sqrt{3}}{3}\right) = \tfrac{8}{3} \left( 9-4\sqrt{3} \right) ,\\ g\left( c,1-c^2\right) & = {} 2(1-c)(1+c)^2\left( 12-24c+3c^2+19c^3\right) \ . \end{aligned}$$

It is a simple task to show that $12-24c+3c^2+19c^3>0$ in [0, 1], so

$$\begin{aligned} g(c,d)\ge 0\quad \text {for }\ (c,d)\in \varOmega \ . \end{aligned}$$

This means that

$$\begin{aligned} \frac{\partial h}{\partial d} \ge 0\quad \text {for }\ (c,d)\in \varOmega \ . \end{aligned}$$

Consequently,

$$\begin{aligned} h(c,d) \le h\left( c,1-c^2\right) = 1-\tfrac{5}{8} c^2-\tfrac{9}{8} c^4+\tfrac{217}{288} c^6\ . \end{aligned}$$

The function $h(c,1-c^2)$ is decreasing for $c\in [0,1]$, so

$$\begin{aligned} h(c,d) \le h\left( c,1-c^2\right) \le 1\ . \end{aligned}$$

Similarly as in Theorem 4, the equality $|H_{2,3}| = \tfrac{1}{8}$ holds if $c_2=1$ and $c_k=0$ for $k\ne 2$, i.e. if $\omega (z)=z^2$. $\square$

Theorem 6

If $f\in {\mathcal {S}}_S^*(e^z)$ is of the form (1.1), then

$$\begin{aligned} |H_{3,1}|\le \tfrac{13}{128}\ . \end{aligned}$$

Proof

If $f\in {\mathcal {S}}_S^*(e^z)$, then

$$\begin{aligned} |H_{3,1}| = \tfrac{1}{16}\left| 2c_2c_4 - c_3^2 + \left( c_2+\tfrac{1}{6} c_1^2\right) c_1c_3 - \tfrac{1}{4} c_1^2c_2^2 - \tfrac{1}{12} c_1^4c_2 - \tfrac{1}{144} c_1^6\right| \ . \end{aligned}$$

(3.1)

At the beginning, it should be noted that

$$\begin{aligned} \left| c_3\left( c_3-c_1c_2-\tfrac{1}{6} c_1^3\right) \right| \le |c_3| \end{aligned}$$

by Lemma 1 with $\mu =-1$ and $\nu =-\tfrac{1}{6}$. Under Lemma 2,

$$\begin{aligned} |c_3|\le 1-|c_1|^2 -\tfrac{|c_2|^2}{1+|c_1|} \le 1-|c_1|^2 -\tfrac{|c_2|^2}{2}\ . \end{aligned}$$

Applying it and $|c_4|\le 1-|c_1|^2 -|c_2|^2$, we get

$$\begin{aligned} |H_{3,1}| \le \tfrac{1}{16} h\left( |c_1|,|c_2|\right) \ , \end{aligned}$$

where

$$\begin{aligned} h(c,d) = 1-c^2-\frac{d^2}{2} + 2d\left( 1-c^2-d^2\right) + \tfrac{1}{4} c^2d^2 + \tfrac{1}{12} c^4d + \tfrac{1}{144} c^6\quad ,\quad c=|c_1|, d=|c_2|\ . \end{aligned}$$

But h is a decreasing function of the variable c; consequently,

$$\begin{aligned} h(c,d) \le h(0,d) = 1+2d-\tfrac{1}{2} d^2-2d^3 . \end{aligned}$$

The function h(0, d) achieves its greatest value in [0, 1] if $d=1/2$, so $h(0,d)\le \tfrac{13}{8}$, which completes the proof. $\square$

This result is not sharp. Based on Formula (3.1), it is expected that the sharp bound of $|H_{3,1}|$ is equal to $\tfrac{1}{16}$.

The method used in the proof of Theorem 2 for the bound $|\gamma _5|\le \tfrac{1}{12}$ may be adopted to prove that if $f\in {\mathcal {S}}_S^*(e^z)$ is of the form (1.1), then $|a_6-a_2a_5|\le \tfrac{1}{6}$. This result is also sharp. Unfortunately, the sharp bound of $a_6$ has not been obtained. From (2.2), we can only obtain that

$$\begin{aligned}|a_6|\le \tfrac{1}{6} + \varepsilon ,\quad \varepsilon =0.016\ldots \ \end{aligned}$$

with an obvious conjecture that the exact value of the bound is equal to 1/6.