Coefficient Problems in the Subclasses of Close-to-Star Functions

Abstract

For two subclasses of close-to-star functions we estimate early logarithmic coefficients, coefficients of inverse functions, Hankel determinant $H_{2,2}$ and Zalcman functional $J_{2,3}$. All results are sharp.

1 Introduction

Given $r>0,$ let ${\mathbb {D}}_r:=\{ z \in {\mathbb {C}}{:}\,|z|<r \},$ and let ${\mathbb {D}}:={\mathbb {D}}_1.$ Let $\overline{{\mathbb {D}}}:=\{z\in \mathbb C{:}\,|z|\le 1\}$ and $\mathbb T:=\partial {\mathbb {D}}.$ Let ${{\mathcal {H}}}$ be the class of all analytic functions in ${\mathbb {D}}$ and ${{\mathcal {A}}}$ be its subclass of f normalized by $f(0):=0$ and $f'(0):=1,$ i.e., of the form

$$\begin{aligned} f(z)= \sum _{n=1}^{\infty }a_nz^n, \quad a_1:=1,\ z\in {\mathbb {D}}. \end{aligned}$$

(1.1)

Let ${\mathcal {S}}$ be the subclass of ${\mathcal {A}}$ of all univalent functions and ${\mathcal {S}}^*$ be the subclass of ${\mathcal {S}}$ of all starlike functions, namely, $f\in {\mathcal {S}}^*$ if $f\in {\mathcal {A}}$ and

$$\begin{aligned} {{\,\mathrm{Re}\,}}\frac{zf'(z)}{f(z)}>0,\quad z\in {\mathbb {D}}. \end{aligned}$$

A function $f\in {\mathcal {A}}$ is called close-to-star if there exist $g\in {\mathcal {S}}^*$ and $\beta \in \mathbb R$ such that

$$\begin{aligned} {{\,\mathrm{Re}\,}}\frac{\mathrm {e}^{\mathrm {i}\beta }f(z)}{g(z)}>0,\quad z\in {\mathbb {D}}. \end{aligned}$$

(1.2)

Denote by ${\mathcal {C}}{\mathcal {S}}{\mathcal {T}}$ the class of all close-to-star functions introduced by Reade [30]. Note that $f\in {\mathcal {C}}{\mathcal {S}}{\mathcal {T}}$ if and only if a function

$$\begin{aligned} F(z):=\int _0^z\frac{f(t)}{t}dt,\quad z\in {\mathbb {D}}, \end{aligned}$$

(1.3)

is close-to-convex [15, 12, Vol. II, p. 3]. The class of close-to-star functions and its subclasses were intensively studied by various authors (e.g., MacGregor [25], Sakaguchi [32], Causey and Merkes [4]; for further references, see [12, Vol. II, pp. 97–104]). Given $g\in {\mathcal {S}}^*$ and $\beta \in \mathbb R,$ let ${\mathcal {C}}{\mathcal {S}}{\mathcal {T}}_\beta (g)$ be the subclass of ${\mathcal {C}}{\mathcal {S}}{\mathcal {T}}$ of all f satisfying (1.2). The classes ${\mathcal {C}}{\mathcal {S}}{\mathcal {T}}_0(g_i),\ i=1,2,3,$ where

$$\begin{aligned} g_1(z):=\frac{z}{1-z^2},\quad g_2(z):=\frac{z}{(1-z)^2},\quad g_3(z):=z,\quad z\in {\mathbb {D}}, \end{aligned}$$

are particularly interesting and were separately studied by authors. In this paper we deal with the classes ${\mathcal {C}}{\mathcal {S}}{\mathcal {T}}_0(g_1)=:{\mathcal {S}\mathcal {T}}(\mathrm {i})$ and ${\mathcal {C}}{\mathcal {S}}{\mathcal {T}}_0(g_2)=:{\mathcal {S}\mathcal {T}}(1)$ which elements f in view of (1.2) satisfy the condition

$$\begin{aligned} {{\,\mathrm{Re}\,}}\left\{ (1-z^2)\frac{f(z)}{z}\right\} >0,\quad z\in {\mathbb {D}}, \end{aligned}$$

(1.4)

and

$$\begin{aligned} {{\,\mathrm{Re}\,}}\left\{ (1-z)^2\frac{f(z)}{z}\right\} >0,\quad z\in {\mathbb {D}}, \end{aligned}$$

(1.5)

respectively. Let us add the inequality (1.4) defines the subclass of the class of functions starlike in the direction of the real axis introduced by Robertson [31]. Moreover, each function F given by (1.3) over the class ${\mathcal {S}\mathcal {T}}(\mathrm {i})$ maps univalently ${\mathbb {D}}$ onto a domain $F({\mathbb {D}})$ convex in the direction of the imaginary axis. The concept of convexity in one direction belongs to Roberston [31] (see e.g., [12, p. 199]). Each function F given by (1.3) over the class ${\mathcal {S}\mathcal {T}}(1)$ maps univalently ${\mathbb {D}}$ onto a domain $F({\mathbb {D}})$ called convex in the positive the direction of the real axis, i.e., $\{w+it{:}\,t\ge 0\}\subset f({\mathbb {D}})$ for every $w\in f({\mathbb {D}})$ [2, 8, 9, 11, 20, 21]. Let us remark that the condition (1.4) was generalized by replacing the expression $1-z^2$ by the expression $1-\alpha ^2z^2$ with $\alpha \in [0,1]$ in [13].

In this paper we find the sharp estimates of early logarithmic coefficients (Sect. 2), of the Hankel determinant $H_{2,2}$ and of Zalcman functional $J_{2,3}$ (Sect. 3) and of the early inverse coefficients (Sect. 4) of functions in the classes ${\mathcal {S}\mathcal {T}}(\mathrm {i})$ and ${\mathcal {S}\mathcal {T}}(1).$ Since both classes ${\mathcal {S}\mathcal {T}}(\mathrm {i})$ and ${\mathcal {S}\mathcal {T}}(1)$ have a representation using the Carathéodory class ${\mathcal {P}}$, i.e., the class of functions $p \in {{\mathcal {H}}}$ of the form

$$\begin{aligned} p(z) = 1 + \sum _{n=1}^{\infty }c_{n}z^{n}, \quad z\in {\mathbb {D}}, \end{aligned}$$

(1.6)

having a positive real part in ${\mathbb {D}},$ the coefficients of functions in ${\mathcal {S}\mathcal {T}}(\mathrm {i})$ and ${\mathcal {S}\mathcal {T}}(1)$ have a suitable representation expressed by the coefficients of functions in ${\mathcal {P}}.$ Therefore to get the upper bounds of considered functionals our computing is based on parametric formulas for the second and third coefficients in ${\mathcal {P}}.$ However both classes are rotation non-invariant. Thus to solve discussed problems we will apply a general formula for $c_3$ recently found in [7]. The formula (1.7) was proved by Carathéodory [3] (see e.g., [10, p. 41]). The formula (1.8) can be found in [28, p. 166]. The formula (1.9) was shown in a recent paper [7], where the extremal functions (1.11) and (1.12) were computed also. For $c_1\ge 0$ the formula (1.9) is due to by Libera and Zlotkiewicz [22, 23].

Lemma 1.1

If $p \in {{\mathcal {P}}}$ is of the form (1.6), then

$$\begin{aligned} c_1= & {} 2\zeta _1, \end{aligned}$$

(1.7)

$$\begin{aligned} c_2= & {} 2\zeta _1^2+2(1-|\zeta _1|^2)\zeta _2 \end{aligned}$$

(1.8)

and

$$\begin{aligned} \begin{aligned} c_3&=2\zeta _1^3+4(1-|\zeta _1|^2)\zeta _1\zeta _2\\&\quad -\,2(1-|\zeta _1|^2)\overline{\zeta _1}\zeta _2^2+2(1-|\zeta _1|^2)(1-|\zeta _2|^2)\zeta _3 \end{aligned} \end{aligned}$$

(1.9)

for some $\zeta _i\in \overline{{\mathbb {D}}},$ $i\in \{1,2,3\}.$

For $\zeta _1\in \mathbb T,$ there is a unique function $p\in {\mathcal {P}}$ with $c_1$ as in (1.7), namely,

$$\begin{aligned} p(z)=\frac{1+\zeta _1z}{1-\zeta _1z},\quad z\in {\mathbb {D}}. \end{aligned}$$

(1.10)

For $\zeta _1\in {\mathbb {D}}$ and $\zeta _2\in \mathbb T,$ there is a unique function $p\in {\mathcal {P}}$ with $c_1$ and $c_2$ as in (1.7)–(1.8), namely,

$$\begin{aligned} p(z)=\frac{1+\left( \overline{\zeta _1}\zeta _2+\zeta _1\right) z+\zeta _2z^2}{1+\left( \overline{\zeta _1}\zeta _2-\zeta _1\right) z-\zeta _2z^2},\quad z\in {\mathbb {D}}. \end{aligned}$$

(1.11)

For $\zeta _1,\zeta _2\in {\mathbb {D}}$ and $\zeta _3\in \mathbb T,$ there is a unique function $p\in {\mathcal {P}}$ with $c_1,$ $c_2$ and $c_3$ as in (1.7)–(1.9), namely,

$$\begin{aligned}&p(z) \nonumber \\&\quad =\frac{1+\left( \overline{\zeta _2}\zeta _3+\overline{\zeta _1}\zeta _2+\zeta _1\right) z+\left( \overline{\zeta _1}\zeta _3+\zeta _1\overline{\zeta _2}\zeta _3+\zeta _2\right) z^2+\zeta _3z^3}{1+\left( \overline{\zeta _2}\zeta _3+\overline{\zeta _1}\zeta _2-\zeta _1\right) z+\left( \overline{\zeta _1}\zeta _3-\zeta _1\overline{\zeta _2}\zeta _3-\zeta _2\right) z^2-\zeta _3z^3},\quad z\in {\mathbb {D}}.\nonumber \\ \end{aligned}$$

(1.12)

2 Logarithmic Coefficients

Given $f\in {\mathcal {S}}$ let

$$\begin{aligned} \log \frac{f(z)}{z}=2\sum _{n=1}^\infty \gamma _n z^n,\quad z\in {\mathbb {D}}{\setminus }\{0\},\ \log 1:=0. \end{aligned}$$

(2.1)

The numbers $\gamma _n$ are called logarithmic coefficients of f. Differentiating (2.1) and using (1.1) we get

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

(2.2)

As it well known, the logarithmic coefficients play a crucial role in Milin conjecture ([26], see also [10, p. 155]). It is surprising that for the class ${\mathcal {S}}$ the sharp estimates of single logarithmic coefficients ${\mathcal {S}}$ are known only for $\gamma _1$ and $\gamma _2,$ namely,

$$\begin{aligned} |\gamma _1|\le 1,\quad |\gamma _2|\le \frac{1}{2}+\frac{1}{\mathrm {e}}=0.635\dots \end{aligned}$$

and are unknown for $n\ge 3.$ Logarithmic coefficients is one of the topic recently being of interest by various authors (e.g., [1, 18, 33]).

Logarithmic coefficients can be considered for functions f from the class ${\mathcal {A}}$ however under the assumption that the branch of logarithm ${\mathbb {D}}\ni z\mapsto \log f(z)/z$ exists. From (1.4) and (1.5) it follows that $g(z):=f(z)/z\not =0$ in ${\mathbb {D}}{\setminus }\{0\}$ for $f\in {\mathcal {S}\mathcal {T}}(\mathrm {i})$ and $f\in {\mathcal {S}\mathcal {T}}(1)$. However $g({\mathbb {D}})$ needs not be necessarily a simply connected domain. Therefore, let ${\mathcal {S}\mathcal {T}}_0(\mathrm {i})$ and ${\mathcal {S}\mathcal {T}}_0(1)$ be the subclasses of ${\mathcal {S}\mathcal {T}}(\mathrm {i})$ and ${\mathcal {S}\mathcal {T}}(1)$ respectively, of all functions f for which the branch ${\mathbb {D}}\ni z\mapsto \log f(z)/z$ with $\log 1:=0$ exists.

Theorem 2.1

If $f \in {\mathcal {S}\mathcal {T}}_0(\mathrm {i})$ is of the form (1.1), then

$$\begin{aligned} |\gamma _1|\le 1,\quad |\gamma _2|\le \frac{3}{2},\quad |\gamma _3| \le 1. \end{aligned}$$

All inequalities are sharp.

Proof

By (1.4) there exists $p \in {{\mathcal {P}}}$ of the form (1.6) such that

$$\begin{aligned} (1-z^2)\frac{f(z)}{z} = p(z). \end{aligned}$$

(2.3)

Substituting the series (1.1) and (1.6) into (2.3) by equating the coefficients we get

$$\begin{aligned} a_2=c_1, \quad a_3 = c_2+1, \quad a_4 = c_1 + c_3. \end{aligned}$$

(2.4)

The inequality $|\gamma _1|\le 1$ follows directly from (2.2), (2.4) and (1.7) with sharpness for the function f given by (2.3), where p is as in (1.10).

Substituting (1.7) and (1.8) into (2.4) from (2.2) it follows that

$$\begin{aligned} \begin{aligned} |\gamma _{2}|&= \frac{1}{2}\left| a_{3}-\frac{1}{2}a_{2}^{2}\right| =\frac{1}{2}\left| c_2-\frac{1}{2}c_1^2+1\right| \\&=\frac{1}{2}\left| 2(1-|\zeta _1|^2)\zeta _2+1\right| \le \frac{1}{2}+(1-|\zeta _1|^2)|\zeta _2|\le \frac{3}{2} \end{aligned} \end{aligned}$$

with sharpness for the function f given by (2.3), where p is as in (1.11) with $\zeta _1=0$ and any $\zeta _2\in \mathbb T.$

By (2.2) and (2.4) we have

$$\begin{aligned} 6\gamma _3 = c_1^3 -3c_1c_2 +3c_3. \end{aligned}$$

Hence and by (1.7)–(1.9) we get

$$\begin{aligned} 3\gamma _3= 2\zeta _1^3 -3(1-|\zeta _1|^2){\overline{\zeta }}_1 \zeta _2^2 +3(1-|\zeta _1|^2)(1-|\zeta _2|^2)\zeta _3, \end{aligned}$$

where $\zeta _i \in \overline{{\mathbb {D}}},$ $i=1,2,3$. Thus by setting $x:=|\zeta _1|\in [0,1]$ and $y:=|\zeta _2|\in [0,1]$ we obtain

$$\begin{aligned} \begin{aligned} 3|\gamma _3|&\le 2x^3+3(1-x^2)xy^2+3(1-x^2)(1-y^2)\\&= 2x^3-3x^2+3 - 3(1-x^2)(1-x)y^2\\&\le 2x^3-3x^2+3\le 3,\quad (x,y)\in [0,1]\times [0,1]. \end{aligned} \end{aligned}$$

Thus $|\gamma _3|\le 1$ with sharpness for the function f given by (2.3), where p is as in (1.12) with $\zeta _1=\zeta _2=0$ and any $\zeta _3\in \mathbb T.$ $\square$

Theorem 2.2

If $f \in {\mathcal {S}\mathcal {T}}_0(1)$ is of the form (1.1), then

$$\begin{aligned} |\gamma _1|\le 2,\quad |\gamma _2|\le \frac{3}{2},\quad |\gamma _3| \le \frac{1}{3}(1+\sqrt{2}). \end{aligned}$$

All inequalities are sharp.

Proof

By (1.5) there exists $p \in {{\mathcal {P}}}$ of the form (1.6) such that

$$\begin{aligned} (1-z)^2\frac{f(z)}{z} = p(z). \end{aligned}$$

(2.5)

Substituting the series (1.1) and (1.6) into (2.5) by equating the coefficients we get

$$\begin{aligned} a_2=c_1+2, \quad a_3 = 3+2c_1+c_2, \quad a_4 = 4+3c_1+2c_2+c_3. \end{aligned}$$

(2.6)

The inequality $|\gamma _1|\le 2$ follows directly from (2.2), (2.6) and (1.7) with sharpness for the function f given by (2.5), where p is as in (1.10).

Substituting (1.7) and (1.8) into (2.6) from (2.2) it follows that

$$\begin{aligned} \begin{aligned} |\gamma _{2}|&=\frac{1}{2}\left| a_{3}-\frac{1}{2}a_{2}^{2}\right| =\frac{1}{2}\left| c_2-\frac{1}{2}c_1^2+1\right| \\&=\frac{1}{2}\left| 2(1-|\zeta _1|^2)\zeta _2+1\right| \le \frac{1}{2}+(1-|\zeta _1|^2)|\zeta _2|\le \frac{3}{2} \end{aligned} \end{aligned}$$

with sharpness for the function f given by (2.3), where p is as in (1.11) with $\zeta _1=0$ and any $\zeta _2\in \mathbb T.$

By (2.2) and (2.6) we have

$$\begin{aligned} 6\gamma _3 = 2+c_1^3+3c_3-3c_1c_2. \end{aligned}$$

Hence and by (1.7)–(1.9) we get

$$\begin{aligned} 3\gamma _3= 1 +\zeta _1^3 -3(1-|\zeta _1|^2){\overline{\zeta }}_1 \zeta _2^2 +(1-|\zeta _1|^2)(1-|\zeta _2|^2)\zeta _3, \end{aligned}$$

where $\zeta _i \in \overline{{\mathbb {D}}},$ $i=1,2,3$. Thus by setting $x:=|\zeta _1|\in [0,1]$ and $y:=|\zeta _2|\in [0,1]$ we obtain

$$\begin{aligned} \begin{aligned} 3|\gamma _3|&\le 1+x^3 -3(1-x^2)xy^2 +(1-x^2)(1-y^2)\\&= 2-x^2+x^3+(1-x^2)(3x-1)y^2=:F(x,y). \end{aligned} \end{aligned}$$

(2.7)

We have $F(1/3,y)=52/27.$ Moreover for $x\in (1/3,1]$ and $x\in [0,1/3)$ we get

$$\begin{aligned} F(x,y)\le F(x,1)\le 1+3x-2x^3 \le 1+\sqrt{2},\quad y\in [0,1], \end{aligned}$$

and

$$\begin{aligned} F(x,y)\le F(x,0)\le 2-x^2+x^3 \le 2,\quad y\in [0,1], \end{aligned}$$

respectively. Thus by (2.7), $|\gamma _3| \le (1+\sqrt{2})/3$ with sharpness for the function f given by (2.3), where p is as in (1.12) with $\zeta _1=1/\sqrt{2},$ $\zeta _2=\mathrm {i}$ and any $\zeta _3\in \mathbb T.$ $\square$

3 Zalcman Functional and Hankel Determinant

Now we compute the sharp upper bound of the Zalcman functional $J_{2,3}(f):=a_2 a_3 - a_4$ being a special case of the generalized Zalcman functional $J_{n,m}(f):=a_n a_m - a_{n+m-1},\ n,m \in {\mathbb {N}}{\setminus }\{1\},$ which was investigated by Ma [24] for $f \in {{\mathcal {S}}}$ (see also [29] for relevant results on this functional). We will find also the sharp bound of the second Hankel determinant $H_{2,2}(f)=a_2a_4-a_3^2.$ Both functionals $J_{2,3}$ and $H_{2,2}$ have been studied recently by various authors (see e.g., [5, 6, 14, 16, 17, 19, 27]).

Theorem 3.1

If $f \in {\mathcal {S}\mathcal {T}}(\mathrm {i})$ is of the form (1.1), then

$$\begin{aligned} |a_2 a_3-a_4| \le 2. \end{aligned}$$

The inequality is sharp with the extremal function

$$\begin{aligned} f(z)=\frac{z}{(1-z)^2},\quad z\in {\mathbb {D}}. \end{aligned}$$

(3.1)

Proof

From (2.4) by using (1.7)–(1.9) it follows that

$$\begin{aligned} \begin{aligned} |a_2 a_3 -a_4|&= | c_1 c_2 -c_3|\\&= 2\left| \zeta _1^3+2(1-|\zeta _1|^2)\overline{\zeta _1}\zeta _2^2-(1-|\zeta _1|^2)(1-|\zeta _2|^2)\zeta _3\right| \\&\le 2\left[ |\zeta _1|^3+2(1-|\zeta _1|^2)|\zeta _1||\zeta _2|^2-(1-|\zeta _1|^2)(1-|\zeta _2|^2\right] \\&= 2\left[ 1-|\zeta _1|^2+|\zeta _1|^3-2(1-|\zeta _1|^2)(1-|\zeta _1|)|\zeta _2|^2\right] \\&\le 2\left( 1-|\zeta _1|^2+|\zeta _1|^3\right) \le 2, \end{aligned} \end{aligned}$$

(3.2)

with sharpness for the function (3.1).

To find sharp estimate for $H_{2,2}$ over ${\mathcal {S}\mathcal {T}}(\mathrm {i})$ we use the following lemma.

$\square$

Proposition 3.2

$$\begin{aligned} | 4z^2-4z-1| \le {\left\{ \begin{array}{ll} 1+4|z|-4|z|^2, \quad |z| \le (-1+\sqrt{2})/2, \\ \sqrt{2}(1+4|z|^2), \quad (-1+\sqrt{2})/2 \le |z| \le 1. \end{array}\right. } \end{aligned}$$

(3.3)

Proof

Since the inequality (3.3) clearly holds for $z=0,$ assume that $z=r \mathrm {e}^{\mathrm {i}\theta }$ with $0 < r\le 1$ and $0\le \theta <2\pi$. A simple computation gives

$$\begin{aligned} |4z^2-4z-1|^2 = \varphi (\cos \theta ), \end{aligned}$$

(3.4)

where $\varphi {:}\,[-1,1]\rightarrow {\mathbb {R}}$ is a function defined by

$$\begin{aligned} \varphi (x) := -16r^2x^2 -8r(4r^2-1)x +16r^4 +24r^2 +1. \end{aligned}$$

Note that $\varphi '(x)=0$ occurs only when $x=(1-4r^2)/(4r)=:x_0$.

When $r \le (-1+\sqrt{2})/2,$ we have $x_0>1$ or $1-4r-4r^2>0$. Therefore

$$\begin{aligned} \varphi '(x) \ge 8r(1-4r-4r^2) >0, \quad x\in [-1,1]. \end{aligned}$$

Hence we get

$$\begin{aligned} \varphi (x) \le \varphi (1) = (1+4r-4r^2)^2. \end{aligned}$$

(3.5)

Thus from (3.4) and (3.5) it follows that the inequality (3.3) holds for $|z|\le (-1+\sqrt{2})/2$.

When $(-1+\sqrt{2})/2 \le r \le 1$, we have $x_0 \in [-1,1].$ Then

$$\begin{aligned} \varphi (x) \le \varphi (x_0) = 2(1+4r^2)^2, \quad x\in [-1,1]. \end{aligned}$$

(3.6)

Combining (3.4) and (3.6) we see that the inequality (3.3) holds for $(-1+\sqrt{2})/2 \le |z| \le 1$. $\square$

Theorem 3.3

If $f \in {\mathcal {S}\mathcal {T}}(\mathrm {i})$ is of the form (1.1), then

$$\begin{aligned} |a_2 a_4 -a_3^2| \le \frac{28}{3}. \end{aligned}$$

(3.7)

The inequality is sharp with the extremal function

$$\begin{aligned} f(z) = \frac{z(3+z+3z^2)}{3(1-z^2)^2},\quad z\in {\mathbb {D}}. \end{aligned}$$

(3.8)

Proof

From (2.4) by using (1.7)–(1.9)we have

$$\begin{aligned} \begin{aligned} a_2a_4-a_3^2&=c_1^2+c_1c_3-c_2^2-2c_2-1\\&= 4\zeta _1^2 -4\zeta _1-1 -4(1-|\zeta _1|^2)\zeta _2 -4(1-|\zeta _1|^2)\zeta _2^2 \\&\quad +\,4\zeta _1(1-|\zeta _1|^2)(1-|\zeta _2|^2)\zeta _3, \end{aligned} \end{aligned}$$

(3.9)

where $\zeta _i \in \overline{{\mathbb {D}}},\ i=1,2,3$. Let $x:=|\zeta _1|\in [0,1]$ and $y=|\zeta _2|\in [0,1].$

Assume first that $x\in [0,x_0],$ where $x_0:=(-1+\sqrt{2})/2.$ Then by (3.9) and Proposition 3.2 for $y\in [0,1]$ we get

$$\begin{aligned} \begin{aligned} |a_2a_4-a_3^2|&\le 1+8x-4x^2-4x^3 +4(1-x^2)y\\&\quad +\,4x(1-x^2)y^2=:F(x,y). \end{aligned} \end{aligned}$$

Clearly, for each $x\in [0,x_0],$ the function $[0,1]\ni y\mapsto F(\cdot ,y)$ is increasing and therefore for $y\in [0,1],$

$$\begin{aligned} F(x,y) \le F(x,1) = 9+4x-12x^2\le \frac{28}{3}=9.333\dots . \end{aligned}$$

(3.10)

Assume now that $x\in [x_0,1].$ Then by (3.9) and Proposition 3.2 for $y\in [0,1]$ we get

$$\begin{aligned} \begin{aligned} |a_2a_4-a_3^2|&\le \sqrt{2}+4x+4\sqrt{2}x^2-4x^3 +4(1-x^2)y\\&\quad +\,4(1-x^2)(1-x)y^2=:G(x,y). \end{aligned} \end{aligned}$$

Note first that

$$\begin{aligned} G(1,y)=5\sqrt{2}=7.071\ldots ,\quad y\in [0,1]. \end{aligned}$$

(3.11)

Clearly, for each $x\in [x_0,1],$ the function $[0,1]\ni y\mapsto G(\cdot ,y)$ is increasing and therefore for $y\in [0,1],$

$$\begin{aligned} \begin{aligned} G(x,y)&\le G(x,1) = 8+\sqrt{2} -4(2-\sqrt{2})x^2\\&\le -2+8\sqrt{2}=9.133\dots . \end{aligned} \end{aligned}$$

Hence, from (3.10) and (3.11) it follows that the inequality (3.7) is true. Equality in (3.7) holds for the function f given by (2.3), where p is given by (1.12) with $\zeta _1:=1/6$ and $\zeta _2=\zeta _3:=1$, i.e., for the function (3.8). $\square$

Theorem 3.4

If $f \in {\mathcal {S}\mathcal {T}}(1)$ is of the form (1.1), then

$$\begin{aligned} |a_2a_3-a_4| \le 20. \end{aligned}$$

The inequality is sharp with the extremal function

$$\begin{aligned} f(z) = \frac{z(1+z)}{(1-z)^3},\quad z\in {\mathbb {D}}. \end{aligned}$$

(3.12)

Proof

From (2.6), by using (1.7) and the inequality $|c_1c_2-c_3|\le 2$ which was proved in (3.2), we obtain

$$\begin{aligned} \begin{aligned} |a_2a_3-a_4|&= |2+4c_1+2c_1^2+c_1c_2-c_3|\\&\le 2+4|c_1|+2|c_1|^2+|c_1c_2-c_3|\le 20. \end{aligned} \end{aligned}$$

with sharpness for the function (3.12). $\square$

Theorem 3.5

If $f \in {\mathcal {S}\mathcal {T}}(1)$ is of the form (1.1), then

$$\begin{aligned} |a_2a_4-a_3^2| \le 17. \end{aligned}$$

(3.13)

The inequality is sharp with the extremal function (3.12).

Proof

From (2.6) by using (1.7)–(1.9)we have

$$\begin{aligned} \begin{aligned} a_2a_4-a_3^2&=-1-2c_1-c_1^2 -2c_1c_2 -2c_2 -c_2^2 +c_1c_3 +2c_3\\&= -1-4\zeta _1-8\zeta _1^2-4\zeta _1^3 -4(1-|\zeta _1|^2)\zeta _2 \\&\quad -\,4(1+\zeta _1)(1-|\zeta _1|^2)\overline{\zeta _1} \zeta _2^2\\&\quad +\,4(1+\zeta _1)(1-|\zeta _1|^2)(1-|\zeta _2|^2)\zeta _3, \end{aligned} \end{aligned}$$

(3.14)

where $\zeta _i\in \overline{{\mathbb {D}}},\ i\in \{1,2,3\}$. Set $x:=|\zeta _1| \in [0,1]$ and $y=:|\zeta _2| \in [0,1]$. By (3.14) we have

$$\begin{aligned} \begin{aligned} |a_2a_4-a_3^2|&\le 5+8x+4x^2 +4(1-x^2)y\\&\quad -\,4(1-x^2)^2 y^2=:F(x,y),\quad x,y\in [0,1]. \end{aligned} \end{aligned}$$

Note first that

$$\begin{aligned} F(1,y)=17,\quad y\in [0,1]. \end{aligned}$$

(3.15)

Let now $x\in [0,1).$ Then for $y\in [0,1]$ we have

$$\begin{aligned} \frac{\partial F}{\partial y}= 4(1-x^2)[1-2(1-x^2)y]=0 \end{aligned}$$

iff $y=1/2(1-x^2)=:y_0.$ Since $y_0\ge 1$ for each $x\in [1/\sqrt{2},1),$ so then the function $[0,1]\ni y\mapsto F(\cdot ,y)$ is increasing and therefore

$$\begin{aligned} F(x,y) \le F(x,1) =5+8x+8x^2-4x^4 \le 17,\quad y\in [0,1]. \end{aligned}$$

(3.16)

For $x\in [0,1/\sqrt{2})$ we have

$$\begin{aligned} \begin{aligned} F(x,y)&\le F(x,y_0) = F\left( x,\frac{1}{2(1-x^2)}\right) \\&= 6+8x+4x^2\le 8+4\sqrt{2}=13.656\dots ,\quad y\in [0,1]. \end{aligned} \end{aligned}$$

Hence by (3.15) and (3.16) it follows that the inequality (3.13) is true. Equality in (3.13) holds for the function f defined by (3.12). $\square$

4 Inverse Coefficients

Since ${\mathcal {S}\mathcal {T}}(\mathrm {i})$ is a compact class and $f'(0)=1$ for every $f\in {\mathcal {S}\mathcal {T}}(\mathrm {i}),$ there exists $r_0\in (0,1)$ such that every $f\in {\mathcal {S}\mathcal {T}}(\mathrm {i})$ is invertible in the disk ${\mathbb {D}}_{r_0}.$ Thus there exists $\delta >0$ such that the inverse function ${\hat{f}}$ of $f_{|{\mathbb {D}}_{r_0}}$ has a series expansion in the disk ${\mathbb {D}}_{\delta }$ of the form

$$\begin{aligned} {\hat{f}}(w) = w+\sum _{n=2}^{\infty }\beta _n w^n,\quad w\in {\mathbb {D}}_\delta . \end{aligned}$$

(4.1)

Thus for $f\in {\mathcal {S}\mathcal {T}}(\mathrm {i})$ of the form (1.1) the following relations hold (see e.g., [12, Vol. I, p. 57])

$$\begin{aligned} \beta _2=-a_2, \quad \beta _3=2a_2^2-a_3,\quad \beta _4=-5a_2^3 +5a_2a_3 -a_4. \end{aligned}$$

(4.2)

Similar situation holds for the class ${\mathcal {S}\mathcal {T}}(1).$

Theorem 4.1

If ${\hat{f}}$ is the inverse function of $f \in {\mathcal {S}\mathcal {T}}(\mathrm {i})$ of the form (4.1), then

• (i) $|\beta _2| \le 2;$

• (ii) $|\beta _3| \le 7;$

• (iii) $|\beta _4| \le 30$.

All inequalities are sharp with the extremal function

$$\begin{aligned} f(z) = \frac{z(1+\mathrm {i}z)}{ (1-z^2)(1-\mathrm {i}z) },\quad z\in {\mathbb {D}}. \end{aligned}$$

(4.3)

Proof

Substituting (2.4) into (4.2) we get

$$\begin{aligned} \beta _2 = -c_1, \quad \beta _3 = 2c_1^2-c_2-1 \end{aligned}$$

(4.4)

and

$$\begin{aligned} \beta _4 = -5c_1^3 +5c_1c_2 +4c_1 -c_3. \end{aligned}$$

(4.5)

By (4.4) and (1.7) the inequality (i) follows immediately. From (4.4) with (1.7) and (1.8) we have

$$\begin{aligned} \begin{aligned} |\beta _3|&=|2c_1^2-c_2-1|=\left| 6\zeta _1^2-2(1-|\zeta _1|^2)\zeta _2-1\right| \\&\le 6|\zeta _1|^2+2(1-|\zeta _1|^2)||\zeta _2|+1\le 4|\zeta _1|^2+3\le 7. \end{aligned} \end{aligned}$$

Now we prove (iii). By (4.5) and (1.7)–(1.9) we have

$$\begin{aligned} \begin{aligned} |\beta _4|&= \left| -22\zeta _1^3 +8\zeta _1 +16(1-|\zeta _1|^2)\zeta _1\zeta _2\right. \\&\quad \left. +\,2(1-|\zeta _1|^2){\overline{\zeta }}_1 \zeta _2^2 -2(1-|\zeta _1|^2)(1-|\zeta _2|^2)\zeta _3\right| \\&\quad \le 2+8x-2x^2+22x^3 + 16x(1-x^2)y -2(1-x)^2(1+x)y^2\\&=:F(x,y), \end{aligned} \end{aligned}$$

where $\zeta _i \in \overline{{\mathbb {D}}},\ i=1,2,3,$ $x:=|\zeta _1|\in [0,1]$ and $y:=|\zeta _2|\in [0,1]$.

Note first that

$$\begin{aligned} F(1,y)=30,\quad y\in [0,1]. \end{aligned}$$

(4.6)

Let now $x\in [0,1).$ Then for $y\in [0,1]$ we have

$$\begin{aligned} \frac{\partial F}{\partial y}= 4(1-x^2)[4x-(1-x)y]=0 \end{aligned}$$

iff $y=4x/(1-x)=:y_0.$ Since $y_0\ge 1$ for each $x\in [1/5,1),$ so then the function $[0,1]\ni y\mapsto F(\cdot ,y)$ is increasing and therefore

$$\begin{aligned} F(x,y) \le F(x,1) = 26x+4x^3 \le 30,\quad y\in [0,1]. \end{aligned}$$

(4.7)

For $x\in [0,1/5)$ we have

$$\begin{aligned} \begin{aligned} F(x,y)&\le F(x,y_0) = F\left( x,\frac{4x}{1-x}\right) \\&= 2+72x+30x^2-10x^3\le \frac{438}{25}=15.52,\quad y\in [0,1]. \end{aligned} \end{aligned}$$

(4.8)

Hence by (4.6)–(4.8) it follows that the inequality in (iii) is true.

All inequalities are sharp with the extremal function (4.3). $\square$

Theorem 4.2

If ${\hat{f}}$ is the inverse function of $f \in {\mathcal {S}\mathcal {T}}(1)$ of the form (4.1), then

• (i) $|\beta _2| \le 4;$

• (ii) $|\beta _3| \le 23;$

• (iii) $|\beta _4| \le 156$.

All inequalities are sharp with the extremal function

$$\begin{aligned} f(z) = \frac{z(1+z)}{(1-z)^3},\quad z\in {\mathbb {D}}. \end{aligned}$$

(4.9)

Proof

Substituting (2.6) into (4.2) we get

$$\begin{aligned} \beta _2 = -c_1-2, \quad \beta _3 = 2c_1^2+6c_1-c_2+5 \end{aligned}$$

(4.10)

and

$$\begin{aligned} \beta _4 = -5c_1^3-20c_1^2+2c_1+5c_1c_2+8c_2-c_3-14. \end{aligned}$$

(4.11)

By (4.10) and (1.7) the inequality (i) follows immediately. From (4.10) with (1.7) and (1.8) we have

$$\begin{aligned} \begin{aligned} |\beta _3|&=|2c_1^2+6c_1-c_2+5|=\left| 6\zeta _1^2+12\zeta _1-2(1-|\zeta _1|^2)\zeta _2+5 \right| \\&\le 6|\zeta _1|^2+12|\zeta _1|+2(1-|\zeta _1|^2)+5=4|\zeta |^2+12|\zeta |+7\le 23. \end{aligned} \end{aligned}$$

Now we prove (iii). By (4.11) and (1.7)–(1.9) we have

$$\begin{aligned} \begin{aligned} |\beta _4|&= \left| -22\zeta _1^3 -64\zeta _1^2 -56\zeta _1 -14 + 16(1-|\zeta _1|^2)\zeta _2\right. \\&\quad +\,16(1-|\zeta _1|^2)\zeta _1\zeta _2 +2(1-|\zeta _1|^2){\overline{\zeta }}_1 \zeta _2^2\\&\quad \left. -\,2(1-|\zeta _1|^2)(1-|\zeta _2|^2)\zeta _3\right| \\&\le 22x^3 +62x^2 +56x +16+16(1-x^2)(1+x)y\\&\quad -\,2(1-x^2)(1-x)y^2=:F(x,y), \end{aligned} \end{aligned}$$

where $\zeta _i \in \overline{{\mathbb {D}}},\ i=1,2,3,$ $x:=|\zeta _1|\in [0,1]$ and $y:=|\zeta _2|\in [0,1]$.

Note first that

$$\begin{aligned} F(1,y)=156,\quad y\in [0,1]. \end{aligned}$$

(4.12)

Let now $x\in [0,1).$ Then for $y\in [0,1]$ we have

$$\begin{aligned} \frac{\partial F}{\partial y}= 4(1-x^2)[4(1+x)-(1-x)y=0 \end{aligned}$$

iff $y=4(1+x)/(1-x)=:y_0.$ Since $y_0\ge 1$ for each $x\in (0,1),$ so the function $[0,1]\ni y\mapsto F(\cdot ,y)$ is increasing and therefore

$$\begin{aligned} F(x,y) \le F(x,1) = 4x^3+48x^2+74x+30 \le 156,\quad y\in [0,1]. \end{aligned}$$

Hence and from (4.12) it follows that the inequality in (iii) is true.

All inequalities are sharp with the extremal function (4.9). $\square$