On the Fekete–Szegö Type Functionals for Close-to-Convex Functions

Abstract

In this paper, we consider two functionals of the Fekete–Szegö type ${\mathsf{\Theta }}_f\left(\mu \right)=a_4-\mu a_2{a}_3$ and ${\mathsf{\Phi }}_f\left(\mu \right)=a_2{a}_4-\mu {a_3}^2$ for a real number $\mu$ and for an analytic function $f\left(z\right)=z+a_2{z}^2+a_3{z}^3+\dots$, $\left|z\right|<1$. This type of research was initiated by Hayami and Owa in 2010. They obtained results for functions satisfying one of the conditions Re $\left\{f\left(z\right)/z\right\}>\alpha$ or Re $\left\{f^{\prime }\left(z\right)\right\}>\alpha$, $\alpha \in [0,1)$. Similar estimates were also derived for univalent starlike functions and for univalent convex functions. We discuss ${\mathsf{\Theta }}_f\left(\mu \right)$ and ${\mathsf{\Phi }}_f\left(\mu \right)$ for close-to-convex functions such that $f^{\prime }\left(z\right)=h\left(z\right)/{(1-z)}^2$, where h is an analytic function with a positive real part. Many coefficient problems, among others estimating of ${\mathsf{\Theta }}_f\left(\mu \right)$, ${\mathsf{\Phi }}_f\left(\mu \right)$ or the Hankel determinants for close-to-convex functions or univalent functions, are not solved yet. Our results broaden the scope of theoretical results connected with these functionals defined for different subclasses of analytic univalent functions.

1. Introduction

Let $\mathcal{A}$ be the family of all functions analytic in $\Delta =\{z\in \mathbb{C}:|z|<1|\}$ having the power series expansion:

$$f\left(z\right)=z+a_2{z}^2+a_3{z}^3+\dots ,$$

(1)

and let ${\mathcal{S}}^{*}$ denote the class of univalent starlike functions in $\mathcal{A}$ (for the definitions and properties of ${\mathcal{S}}^{*}$ and other classes, see [1]). For a given real argument $\beta \in (-\pi /2,\pi /2)$ and a given function $g\in {\mathcal{S}}^{*}$, a function $f\in \mathcal{A}$ is called close-to-convex with argument $\beta$ with respect to g if:

$$\mathrm{Re}\left\{\frac{e^{i\beta }z{f}^{\prime }\left(z\right)}{g\left(z\right)}\right\}>0,z\in \Delta .$$

Let ${\mathcal{C}}_{\beta }\left(g\right)$ be the class of all such functions. Moreover, let:

$${\mathcal{C}}_{\beta }=\bigcup _{g\in {\mathcal{S}}^{*}}{\mathcal{C}}_{\beta }\left(g\right).$$

Let $\mathcal{C}$ denote the family of all close-to-convex functions (see [2,3]). It is obvious that:

$$\mathcal{C}=\bigcup _{\beta \in (-\pi /2,\pi /2)}{\mathcal{C}}_{\beta }.$$

All functions in $\mathcal{C}$ are univalent.

In this paper, we consider the class ${\mathcal{C}}_0\left(k\right)$, where k is the Koebe function:

$$k\left(z\right)=\frac{z}{{(1-z)}^2}=z+2z^2+3z^3+\dots .$$

(2)

The class ${\mathcal{C}}_0\left(k\right)$ is sometimes denoted by ${\mathcal{CR}}^{+}$. Such functions have a well known geometrical meaning. Namely, for each function f in this class, the set $f(\Delta )$ is a domain such that $\{w+t:t\ge 0\}\subset f(\Delta )$ for every $w\in f(\Delta )$. Such functions f are convex in the positive direction of the real axis.

For a function f analytic in $\Delta$ of the form (1), we define two functionals for a fixed real $\mu$:

$${\mathsf{\Theta }}_f\left(\mu \right)=a_4-\mu a_2{a}_3$$

(3)

and:

$${\mathsf{\Phi }}_f\left(\mu \right)=a_2{a}_4-\mu {a_3}^2.$$

(4)

The functionals ${\mathsf{\Theta }}_f\left(\mu \right)$ and ${\mathsf{\Phi }}_f\left(\mu \right)$ are the generalizations of two well known expressions: $a_4-a_2{a}_3$ and $a_2{a}_4-{a_3}^2$. Both functionals are symmetric, or invariant, under rotations. The first one is a particular case of the generalized Zalcman functional. It was investigated, among others, by Ma [4] and Efraimidis and Vukotić [5]. The second functional is known as the second Hankel determinant, and it was studied in many papers. The investigation of Hankel determinants for analytic functions was started by Pommerenke (see [6,7]) and continued by many mathematicians in various classes of univalent functions (see, for example [8,9,10,11,12,13,14,15,16]). The functional ${\mathsf{\Phi }}_f\left(\mu \right)$ was first studied by Hayami and Owa [17]. They discussed an even more general functional $a_n{a}_{n+2}-\mu {a_{n+1}}^2$ for the classes $\mathcal{Q}\left(\alpha \right)$ and $\mathcal{R}\left(\alpha \right)$, $\alpha \in [0,1)$, of functions $f\in \mathcal{A}$ such that Re $\left\{f\left(z\right)/z\right\}>\alpha$ and Re $\left\{f^{\prime }\left(z\right)\right\}>\alpha$, respectively. The functionals ${\mathsf{\Phi }}_f\left(\mu \right)$ and ${\mathsf{\Theta }}_f\left(\mu \right)$ for the classes ${\mathcal{S}}^{*}$ and $\mathcal{K}$ of starlike and convex functions, respectively, were discussed in [18].

It is worth pointing out a particularly interesting property of ${\mathsf{\Phi }}_f\left(\mu \right)$. The sharp estimates of this functional are often symmetric with respect to a certain point. It was shown in [18] that such points for ${\mathcal{S}}^{*}$ and $\mathcal{K}$ are 8/9 and one, respectively. We have:

$$|{\mathsf{\Phi }}_f\left(\mu \right)|\le max\{|9\mu -8|,1\}\mathrm{for}{\mathcal{S}}^{*}$$

(5)

and:

$$|{\mathsf{\Phi }}_f\left(\mu \right)|\le max\{|\mu -1|,1/8\}\mathrm{for}\mathcal{K}.$$

A similar situation occurs for $\mathcal{Q}(1/2)$ and for the class ${\mathcal{C}}_0\left(h\right)$, where $h\left(z\right)=z/(1-z^2)$; this point is 1/2 (see [17,19]). This situation appears even in the class $\mathcal{T}$ of typically real functions, which do not necessarily have to be univalent (see [19]).

In this work, we derive bounds of ${\mathsf{\Theta }}_f\left(\mu \right)$ and ${\mathsf{\Phi }}_f\left(\mu \right)$ for functions in ${\mathcal{C}}_0\left(k\right)$.

2. Preliminary Results

Let $\mathcal{P}$ denote the class of all analytic functions h with a positive real part in $\Delta$ satisfying the normalization condition $h\left(0\right)=1$. Let $h\in \mathcal{P}$ have the Taylor series expansion:

$$h\left(z\right)=1+p_{1}z+p_2{z}^2+p_3{z}^3+\dots .$$

(6)

We shall need here three results. The first one is known as Caratheodory’s lemma (see, for example, ref. [1]). The second one is due to Libera and Złotkiewicz ([20,21]), and the third one is the result of Hayami and Owa.

Lemma 1

([1]). If $h\in \mathcal{P}$ is given by (6), then the sharp inequality $|p_n|\le 2$ holds for $n\ge 1$.

Lemma 2

([20,21]). Let h be given by (6) and $p_1$ be a given real number, $p_1\in [-2,2]$. Then, $h\in \mathcal{P}$ if and only if:

$$2p_2={p_1}^2+(4-{p_1}^2)x$$

and:

$$4p_3={p_1}^3+2(4-{p_1}^2)p_{1}x-(4-{p_1}^2)p_1{x}^2+2(4-{p_1}^2){(1-|x|}^2)y$$

for some complex numbers x, y such that $\left|x\right|\le 1$, $\left|y\right|\le 1$.

Lemma 3

([17]). If $h\in \mathcal{P}$ is given by (6), then:

$$|p_3-\mu p_1{p}_2|\le max\{2,|2-4\mu |\}.$$

The next lemma is an improvement of Lemma 3 for $\mu \in [1/2,1]$.

Lemma 4

([22]). If $h\in \mathcal{P}$ is given by (6) and $\mu \in [1/2,1]$, then:

$$\begin{array}{c}\hfill |p_3-\mu p_1{p}_2|\le \left\{\begin{array}{cc}\frac{1}{4}{\mu }^2{p}^3-\frac{1}{2}\mu (2-\mu )p^2+2,\hfill & p\in \left[0,2/(2-\mu )\right],\hfill \\ (3-2\mu )p-(1-\mu )p^3,\hfill & p\in \left[2/(2-\mu ),2\right],\hfill \end{array}\right.\end{array}$$

(7)

where $p=|p_1|$. The inequality is sharp.

The following lemma was proven by Lecko (see Corollary 2.3 in [23]).

Lemma 5

([23]). If $h\in \mathcal{P}$ is given by (6), then:

$$|p_{n+1}+2p_n+p_{n-1}|\le 2(2+\mathrm{Re}\left\{p_1\right\}).$$

(8)

We have proven the next lemma.

Lemma 6.

If $h\in \mathcal{P}$ is given by (6), then:

$$|p_1{p}_3-{p_2}^2|\le 4-|p_1{|}^2.$$

(9)

The inequality is sharp.

Proof. 

By Lemma 2,

$$4(p_1{p}_3-{p_2}^2)=(4-p_1{}^2)\left[2p_1{(1-|x|}^2)y-4x^2\right].$$

Applying the invariance of $|p_1{p}_3-{p_2}^2|$ under rotation, we can assume that $p_1$ is a non-negative real number. Writing $r=\left|x\right|\in [0,1]$ and $p=p_1\in [0,2]$, we get by the triangle inequality and the assumption $\left|y\right|\le 1$:

$$4|p_1{p}_3-{p_2}^2|\le (4-p^2)[2p(1-r^2)+4r^2]=(4-p^2)[2p+(4-2p)r^2]\le 4(4-p^2),$$

which gives the desired bound. The equality (9) holds for:

$$h\left(z\right)=\left(1-\frac{p}{2}\right)\frac{1+z^2}{1-z^2}+\frac{p}{2}\frac{1+z}{1-z}=1+pz+2z^2+p{z}^3+\dots ,$$

(10)

which means that there is equality in (9) for rotations of (10). □

The next lemma is a special case of more general results due to Choi et al. [24] (see also [9]). Let $\overline{\Delta }=\{z\in \mathbb{C}:|z|\le 1\}$. Define:

$$Y(a,b,c)=\underset{z\in \overline{\Delta }}{max}\left(|a+bz+c{z}^2{|+1-|z|}^2\right),a,b,c\in \mathbb{R}.$$

Lemma 7.

If $ac<0$, then:

$$\begin{array}{l}Y(a,b,c)=\left\{\begin{array}{cc}{\displaystyle 1+\left|a\right|+\frac{b^2}{4(1+|c\left|\right)},}\hfill & \left|b\right|<2(1+\left|c\right|)\mathrm{and}{b}^2<-4a(1-c^2)/c,\hfill \\ {\displaystyle 1-\left|a\right|+\frac{b^2}{4(1-|c\left|\right)},}\hfill & \left|b\right|<2(1-\left|c\right|)\mathrm{and}{b}^2\ge -4a(1-c^2)/c,\hfill \\ R(a,b,c),\hfill & \mathrm{otherwise},\hfill \end{array}\right.\end{array}$$

where:

$$\begin{array}{l}R(a,b,c)=\left\{\begin{array}{cc}\left|a\right|+\left|b\right|-\left|c\right|,\hfill & \left|ab\right|\ge \left|c\right|\left(\left|b\right|+4\left|a\right|\right),\hfill \\ -\left|a\right|+\left|b\right|+\left|c\right|,\hfill & \left|ab\right|\le \left|c\right|\left(\left|b\right|-4\left|a\right|\right),\hfill \\ {\displaystyle \left(\left|c\right|+\left|a\right|\right)\sqrt{1-\frac{b^2}{4ac}},}\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\end{array}$$

If $ac\ge 0$, then:

$$\begin{array}{l}Y(a,b,c)=\left\{\begin{array}{cc}\left|a\right|+\left|b\right|+\left|c\right|,\hfill & \left|b\right|\ge 2(1-|c\left|\right),\hfill \\ {\displaystyle 1+\left|a\right|+\frac{b^2}{4(1-|c\left|\right)},}\hfill & \left|b\right|<2(1-|c\left|\right).\hfill \end{array}\right.\end{array}$$

Applying the correspondence between the functions in ${\mathcal{C}}_0\left(k\right)$ and $\mathcal{P}$:

$${(1-z)}^2{f}^{\prime }\left(z\right)=h\left(z\right),f\in {\mathcal{C}}_0\left(k\right),h\in \mathcal{P}$$

(11)

and Expansions (1) and (6) we get:

$$2a_2=2+p_1,3a_3=3+2p_1+p_2,4a_4=4+3p_1+2p_2+p_3.$$

(12)

Moreover, by Lemma 1, $Re\left\{a_2\right\}\ge 0$ with equality if and only if $p_1=-2$. The equality is possible only for the function $h\left(z\right)=\frac{1-z}{1+z}\in \mathcal{P}$, and then, $f\left(z\right)=\frac{1}{2}log\frac{1+z}{1-z}\in {\mathcal{C}}_0\left(k\right)$.

Hence, we can express ${\mathsf{\Theta }}_f\left(\mu \right)$ and ${\mathsf{\Phi }}_f\left(\mu \right)$ for $f\in {\mathcal{C}}_0\left(k\right)$ as coefficients of a corresponding function $h\in \mathcal{P}$ in the following way:

$${\mathsf{\Theta }}_f\left(\mu \right)=\frac{1}{4}{p}_3+(\frac{1}{2}-\frac{1}{3}\mu )p_2+(\frac{3}{4}-\frac{7}{6}\mu )p_1-\frac{1}{6}\mu p_1{p}_2-\frac{1}{3}\mu {p_1}^2+1-\mu$$

(13)

and:

$$\begin{array}{ll}{\mathsf{\Phi }}_f\left(\mu \right)& =\frac{1}{8}{p}_1{p}_3-\frac{1}{9}\mu p_2{}^2+\frac{1}{4}{p}_3+\left(\frac{1}{4}-\frac{4}{9}\mu \right)p_1{p}_2+\left(\frac{1}{2}-\frac{2}{3}\mu \right)p_2\hfill \\ & +\left(\frac{3}{8}-\frac{4}{9}\mu \right)p_1{}^2+\left(\frac{5}{4}-\frac{4}{3}\mu \right)p_1+1-\mu .\hfill \end{array}$$

(14)

3. Example

Let us consider the function:

$$F\left(z\right)=\frac{1}{2}(1-\alpha )log\frac{1+z}{1-z}+\alpha \frac{z}{{(1-z)}^2},\alpha \in [0,1],$$

(15)

which has the following Taylor series expansion:

$$\begin{array}{ll}F\left(z\right)& =(1-\alpha )\left(z+\frac{1}{3}{z}^3+\dots \right)+\alpha \left(z+2z^2+3z^3+4z^4+\dots \right)\hfill \\ & =z+2\alpha z^2+\frac{1}{3}(1+8\alpha )z^3+4\alpha z^4+\dots .\hfill \end{array}$$

Since:

$${(1-z)}^2{F}^{\prime }\left(z\right)=(1-\alpha )\frac{1-z}{1+z}+\alpha \frac{1+z}{1-z}\in \mathcal{P},$$

so $F\in {\mathcal{C}}_0\left(k\right)$. Moreover,

$$F(\Delta )=\mathbb{C}\backslash \left\{x\pm i(1-\alpha )\frac{\pi }{4}:x\le \frac{1}{4}[(1-\alpha )ln\frac{1-\alpha }{\alpha }-1]\right\}.$$

For F, we have:

$${\mathsf{\Theta }}_F\left(\mu \right)=\frac{2}{3}\left[-8\mu {\alpha }^2+(6-\mu )\alpha \right]$$

and:

$${\mathsf{\Phi }}_F\left(\mu \right)=\frac{1}{9}\left[8(9-8\mu ){\alpha }^2-16\mu \alpha -\mu \right].$$

For $\mu <0$, we have: ${\mathsf{\Theta }}_F\left(\mu \right)\le 4-6\mu$ and ${\mathsf{\Phi }}_F\left(\mu \right)\le 8-9\mu$. We find the estimation of ${\mathsf{\Theta }}_F\left(\mu \right)$ and ${\mathsf{\Phi }}_F\left(\mu \right)$ for $\mu \ge 0$.

Let us denote:

$$f\left(\alpha \right)=\frac{2}{3}\left[-8\mu {\alpha }^2+(6-\mu )\alpha \right]\mathrm{and}g\left(\alpha \right)=\frac{1}{9}\left[8(9-8\mu ){\alpha }^2-16\mu \alpha -\mu \right].$$

The critical point ${\alpha }_0=(6-\mu )/\left(16\mu \right)$ of $f\left(\alpha \right)$ is in $(0,1)$ if $\mu \in (6/17,6)$. Hence,

$$|{\mathsf{\Theta }}_F\left(\mu \right)|\le max\{|f\left({\alpha }_0\right)|,|f\left(1\right)|,|f\left(0\right)\left|\right\}=max\left\{\left|\frac{{(6-\mu )}^2}{48\mu }\right|,|4-6\mu |,0,\right\}$$

for $\mu \in (6/17,6)$ and:

$$|{\mathsf{\Theta }}_F\left(\mu \right)|\le max\{\left|f\left(1\right)\right|,\left|f\left(0\right)\right|\}=|4-6\mu |\mathrm{for}\mu \in [0,6/17]\cup [6,\infty ).$$

Similarly, the critical point ${\alpha }_1=\mu /(9-8\mu )$ of $g\left(\alpha \right)$ is in $(0,1)$ if $\mu \in (0,1)$. Hence,

$$|{\mathsf{\Phi }}_F\left(\mu \right)|\le max\{|g\left({\alpha }_1\right)|,|g\left(1\right)|,|g\left(0\right)\left|\right\}=max\left\{\left|\frac{\mu }{8\mu -9}\right|,|8-9\mu |,\left|-\frac{\mu }{9}\right|\right\}$$

for $\mu \in (0,1)$ and:

$$|{\mathsf{\Phi }}_F\left(\mu \right)|\le max\{\left|g\left(1\right)\right|,\left|g\left(0\right)\right|\}=|9\mu -8|\mathrm{for}\mu \in \left\{0\right\}\cup [1,\infty ).$$

Finally, for a function F given by (15), we obtain:

$${\mathsf{\Theta }}_F\left(\mu \right)\le \left\{\begin{array}{cc}4-6\mu ,\hfill & \mu \le 6/17=0.352\dots \hfill \\ {\displaystyle \frac{{(6-\mu )}^2}{48\mu },}\hfill & 6/17\le \mu \le 6(15+16\sqrt{2})/287=0.786\dots \hfill \\ 6\mu -4,\hfill & \mu \ge 6(15+16\sqrt{2})/287\hfill \end{array}\right.$$

and:

$${\mathsf{\Phi }}_F\left(\mu \right)\le \left\{\begin{array}{cc}8-9\mu ,\hfill & \mu \le (73-\sqrt{145})/72=0.846\dots \hfill \\ {\displaystyle \frac{\mu }{9-8\mu },}\hfill & (73-\sqrt{145})/72\le \mu \le 1\hfill \\ 9\mu -8,\hfill & \mu \ge 1.\hfill \end{array}\right.$$

4. Bounds of $\left|\mathsf{\Theta }\right(\mathit{\mu }\left)\right|$ for the Class ${\mathcal{C}}_{\mathbf{0}}\left(\mathit{k}\right)$

In the main theorem of this section, we establish the sharp bounds of $\left|\mathsf{\Theta }\right(\mu \left)\right|$ for the class ${\mathcal{C}}_0\left(k\right)$. The proof is divided into six lemmas. The first one is a particular case of the result obtained in [22] (Theorem 3.1 or Theorem 3.3 in [22]), and the second one is obvious.

Lemma 8.

Let $f\in {\mathcal{C}}_0\left(k\right)$. Then, $|{\mathsf{\Theta }}_f\left(1\right)|=|a_4-a_2{a}_3|\le 2$. The result is sharp.

Lemma 9.

Let $f\in {\mathcal{C}}_0\left(k\right)$ and $\mu \le 0$. Then, $|{\mathsf{\Theta }}_f\left(\mu \right)|\le 4-6\mu$. The result is sharp.

Lemma 10.

Let $f\in {\mathcal{C}}_0\left(k\right)$ and $\mu >1$. Then, $|{\mathsf{\Theta }}_f\left(\mu \right)|\le 6\mu -4$. The result is sharp.

Proof. 

From (13), we can write ${\mathsf{\Theta }}_f\left(\mu \right)$ as follows:

$${\mathsf{\Theta }}_f\left(\mu \right)=\frac{1}{4}\left(p_3-\frac{2}{3}\mu p_1{p}_2\right)+\left(\frac{1}{2}-\frac{1}{3}\mu \right)p_2-\frac{1}{3}\mu {p_1}^2+\left(\frac{3}{4}-\frac{7}{6}\mu \right)p_1+1-\mu .$$

If $\mu \ge 3/2$, then, taking into account Lemmas 1 and 3, we get:

$$|{\mathsf{\Theta }}_f\left(\mu \right)|\le \frac{1}{4}\left(\frac{8}{3}\mu -2\right)+2\left(\frac{1}{3}\mu -\frac{1}{2}\right)+\frac{4}{3}\mu +2\left(\frac{7}{6}\mu -\frac{3}{4}\right)+\mu -1=6\mu -4.$$

If $\mu \in (1,3/2)$, then we have:

$${\mathsf{\Theta }}_f\left(\mu \right)=(3-2\mu )(a_4-a_2{a}_3)+(2\mu -2)(a_4-\frac{3}{2}{a}_2{a}_3).$$

Now, from Lemma 8 and the first part of this proof (i.e., $|a_4-\frac{3}{2}{a}_2{a}_3|\le 5$), we obtain:

$$|{\mathsf{\Theta }}_f\left(\mu \right)|\le 2(3-2\mu )+5(2\mu -2)=6\mu -4.$$

It is clear that ${\mathsf{\Theta }}_f\left(\mu \right)=4-6\mu$ only when $p_1=p_2=p_3=2$, which means that this equality holds only for the Koebe function (2). In other words, the Koebe function is the extremal function for $\mu >1$. □

Taking into account (13) and Lemma 2, we can write ${\mathsf{\Theta }}_f\left(\mu \right)$ as follows:

$$\begin{array}{ll}{\mathsf{\Theta }}_f\left(\mu \right)& =1+\frac{1}{16}{p_1}^3+\frac{1}{4}{p_1}^2+\frac{3}{4}{p}_1-\frac{1}{12}\mu {p_1}^3-\frac{1}{2}\mu {p_1}^2-\frac{7}{6}\mu p_1-\mu \hfill \\ & +\left[\frac{1}{8}{p}_1+\frac{1}{4}-\frac{1}{12}\mu (2+p_1)\right](4-{p_1}^2)x-\frac{1}{16}(4-{p_1}^2)p_1{x}^2+\frac{1}{8}(4-{p_1}^2){(1-|x|}^2)y.\hfill \end{array}$$

From the above formula, we can obtain bounds of $|{\mathsf{\Theta }}_f\left(\mu \right)|$, while $\mu \in (0,1)$ and $f\in {\mathcal{C}}_0\left(k\right)$, but only with an additional assumption that $a_2$ is a positive real number. The assumption of Lemma 2 enforces that $p_1\in [-2,2]$. Notice that if $p_1=2$, then $f\left(z\right)=k\left(z\right)$ given by (2), and we have:

$${\mathsf{\Theta }}_f\left(\mu \right)=4-6\mu .$$

(16)

If $p_1=-2$, then $f\left(z\right)=\frac{1}{2}log\frac{1+z}{1-z}=z+\frac{1}{3}{z}^3+\frac{1}{5}{z}^5+\dots$ is in ${\mathcal{C}}_0\left(k\right)$, and so:

$${\mathsf{\Theta }}_f\left(\mu \right)=0.$$

(17)

To shorten notation, we write p instead of $p_1$. One can observe that ${\mathsf{\Theta }}_f\left(\mu \right)$ can be written as:

$$8{\mathsf{\Theta }}_f\left(\mu \right)=(4-p^2)\left[a+bx+c{x}^2{+(1-|x|}^2)y\right],$$

(18)

where:

$$\begin{array}{ll}a& =\frac{48(1-\mu )+4(9-14\mu )p+12(1-2\mu )p^2+(3-4\mu )p^3}{6(4-p^2)},\\ \hfill b& =(2+p)(1-\frac{2}{3}\mu ),\hfill \\ \hfill c& =-\frac{1}{2}p.\hfill \end{array}$$

(19)

From (18), the triangle inequality, $\left|y\right|\le 1$, and Lemma 2, we get:

$$8|{\mathsf{\Theta }}_f\left(\mu \right)|\le (4-p^2)\left[|a+bx+c{x}^2{|+1-|x|}^2\right],$$

(20)

where a, b, and c are given by (19).

Lemma 11.

Let $f\in {\mathcal{C}}_0\left(k\right)$, $a_2$ be a real number, $a_2\in [0,2]$ and $\mu \in (0,1/3]$. Then, $|{\mathsf{\Theta }}_f\left(\mu \right)|\le 4-6\mu$. The result is sharp.

Proof. 

For $\mu =1/3$, we have (20) with:

$$a=\frac{5p^2+2p+48}{18(2-p)},b=\frac{7(2+p)}{9},c=-\frac{p}{2}.$$

We use Lemma 7. Clearly, $ac<0$ for $p\in (0,2)$. Note that the inequality $\left|b\right|<2(1+|c\left|\right)$ from the first case of Lemma 7 is equivalent to the obviously true inequality:

$$7(2+p)<18(1+p/2).$$

(21)

The inequality $b^2<-4a(1-c^2)/c$, which can be written as:

$$\frac{4(2+p)(p^2+20p-108)}{81p}<0,$$

holds for all $p\in (0,2)$. Hence, for $p\in (0,2)$, we have:

$$Y(a,b,c)=1+\left|a\right|+\frac{b^2}{4(1+|c\left|\right)}=\frac{2(238-36p-p^2)}{81(2-p)}.$$

(22)

For $p\in (-2,0]$, we have $ac\ge 0$, and the inequality $\left|b\right|<2(1-|c\left|\right)$ from the last case of Lemma 7 is equivalent to (21). Therefore, $Y(a,b,c)$ is also given by (22).

Thus, from (20) for $\mu =1/3$, Lemma 7, (16) and (17), we obtain:

$$|{\mathsf{\Theta }}_f(1/3)|\le g\left(p\right),$$

(23)

where $g\left(p\right)=(2+p)(238-36p-p^2)/324$ and $p\in [-2,2]$ according to the assumption. The function g is increasing for $p\in [-2,2]$; therefore:

$$|{\mathsf{\Theta }}_f(1/3)|\le g\left(2\right)=2.$$

(24)

Moreover, we have by the triangle inequality:

$$|{\mathsf{\Theta }}_f\left(\mu \right)|=(1-3\mu )a_4+3\mu (a_4-\frac{1}{3}{a}_2{a}_3),\mu \in \left(0,1/3\right).$$

From Lemma 9 and from (24), we get:

$$|{\mathsf{\Theta }}_f\left(\mu \right)|\le 4(1-3\mu )+2·3\mu =4-6\mu ,$$

and the proof is complete. Equality holds for the Koebe function (2). □

Let us denote:

$$\begin{array}{cc}\hfill p_0& =2(\sqrt{103}-10)/3=0.099\dots ,\hfill \\ \hfill K& =16(103\sqrt{103}-910)/2187=0.9901\dots .\hfill \end{array}$$

(25)

Lemma 12.

Let $f\in {\mathcal{C}}_0\left(k\right)$, $a_2$ be a real number, $a_2\in [0,2]$, and K be given by (25). Then, $|{\mathsf{\Theta }}_f(2/3)|\le K$. The result is sharp.

Proof. 

For $\mu =2/3$, we have (20) with:

$$a=\frac{12-p}{18},b=\frac{5(2+p)}{9},c=-\frac{p}{2}.$$

We use Lemma 7. Clearly, $ac<0$ for $p\in (0,2]$. First, note that the inequality $\left|b\right|<2(1+|c\left|\right)$ is equivalent to the obviously true inequality:

$$5(2+p)<18(1+p/2).$$

(26)

The inequality $b^2<-4a(1-c^2)/c$, which is equivalent to:

$$\frac{8(2+p)(2p^2+22p-27)}{81p}<0,$$

holds for $p\in \left(0,(5\sqrt{7}-11)/2\right]$. For $p\in \left(0,(5\sqrt{7}-11)/2\right]$, we have:

$$Y(a,b,c)=1+\left|a\right|+\frac{b^2}{4(1+|c\left|\right)}=\frac{8(p+20)}{81},$$

(27)

so from (20) for $\mu =2/3$ and Lemma 7, we obtain:

$$|{\mathsf{\Theta }}_f(2/3)|\le (4-p^2)(p+20)/81.$$

(28)

From Lemma 7, the inequality system consists of $\left|b\right|<2(1-|c\left|\right)$, and $b^2\ge -4a(1-c^2)/c$ is contradictory, because the first inequality gives $p<4/7$, while the second one yields $p\ge (5\sqrt{7}-11)/2$.

Now, consider the third case of Lemma 7. Let $p\in \left[(5\sqrt{7}-11)/2,2\right]$. The inequality $\left|ab\right|\ge \left|c\right|\left(\right|b|+4|a\left|\right)$ is equivalent to $60-128p-16p^2\ge 0$, and it is not satisfied for any $p\in \left[(5\sqrt{7}-11)/2,2\right]$. The inequality $\left|ab\right|\le \left|c\right|\left(\right|b|-4|a\left|\right)$, which can be written as $30+44p-17p^2\le 0$, is also not satisfied for any $p\in \left[(5\sqrt{7}-11)/2,2\right]$. Thus, for $p\in \left[(5\sqrt{7}-11)/2,2\right]$, we have:

$$Y(a,b,c)=\left(\left|c\right|+\left|a\right|\right)\sqrt{1-\frac{b^2}{4ac}}=\frac{4(2p+3)}{27}\sqrt{\frac{(2p+25)(2p+1)}{(12-p)p}}.$$

(29)

From (20) for $\mu =2/3$ and Lemma 7, we obtain:

$$|{\mathsf{\Theta }}_f(2/3)|\le \frac{(4-p^2)(2p+3)}{54}\sqrt{\frac{(2p+25)(2p+1)}{(12-p)p}}.$$

(30)

For $p\in [-2,0]$, we have $ac\ge 0$, and the inequality $\left|b\right|<2(1-|c\left|\right)$ from the last case of Lemma 7 is equivalent to the inequality in (26).

Thus, $Y(a,b,c)$ is given by (27). Finally, from (16), (28) and (30), we obtain:

$$|{\mathsf{\Theta }}_f(2/3)|\le g\left(p\right),$$

where:

$$\begin{array}{c}\hfill g\left(p\right)=\left\{\begin{array}{cc}{\displaystyle \frac{1}{81}(4-p^2)(p+20),}\hfill & p\in \left[-2,(5\sqrt{7}-11)/2\right)\hfill \\ {\displaystyle \frac{1}{54}(4-p^2)(2p+3)\sqrt{\frac{(2p+25)(2p+1)}{(12-p)p}},}\hfill & p\in \left[(5\sqrt{7}-11)/2,2\right].\hfill \end{array}\right.\end{array}$$

Now, let us consider the function g for $p\in \left[(5\sqrt{7}-11)/2,2\right]$. We have:

$$g^{\prime }\left(p\right)=\frac{M\left(p\right)}{54{(12-p)}^2{p}^2}\sqrt{\frac{(12-p)p}{(2p+25)(2p+1)}},$$

where $M\left(p\right)=24p^6-52p^5-3802p^4-4801p^3+4242p^2+1500p-1800$ and:

$$M\left(p\right)=24p^5(p-13/6)+900(p-2)+3802p^2(1-p^2)+p(-4801p^2+440p+600)<0$$

for $p\in (1,2]$. Hence, $g^{\prime }\left(p\right)<0$ for $p\in \left[(5\sqrt{7}-11)/2,2\right]$.

Taking the above into account, one can check that the function g is increasing for $p\in \left[-2,p_0\right)$ and is decreasing for $p\in \left(p_0,2\right]$, where $p_0$ is given by (25). Therefore,

$$|{\mathsf{\Theta }}_f(2/3)|\le g\left(p_0\right)=16(103\sqrt{103}-910)/2187=0.9901\dots ,$$

so we have the desired result. □

Lemma 13.

Let $f\in {\mathcal{C}}_0\left(k\right)$, $a_2$ be a real number, and $a_2\in [0,2]$.

1. 

If $\mu \in \left(1/3,2/3\right)$, then $|{\mathsf{\Theta }}_f\left(\mu \right)|<3-3\mu$.

2. 

If $\mu \in \left(2/3,1\right)$, then $|{\mathsf{\Theta }}_f\left(\mu \right)|<3\mu -1$.

Proof. 

We have:

$$|{\mathsf{\Theta }}_f\left(\mu \right)|=|(2-3\mu )(a_4-\frac{1}{3}{a}_2{a}_3)+(3\mu -1)(a_4-\frac{2}{3}{a}_2{a}_3)|,\mu \in \left(1/3,2/3\right).$$

From Lemmas 11 and 12, and the triangle inequality, we get the first part of Lemma 13, i.e.,:

$$|{\mathsf{\Theta }}_f\left(\mu \right)|\le 2(2-3\mu )+K·(3\mu -1)<2(2-3\mu )+1·(3\mu -1)=3-3\mu .$$

Since:

$$|{\mathsf{\Theta }}_f\left(\mu \right)|=|(3-3\mu )(a_4-\frac{2}{3}{a}_2{a}_3)+(3\mu -2)(a_4-a_2{a}_3)|,\mu \in \left(2/3,1\right),$$

from Lemma 12, Lemma 8, and the triangle inequality, we get the second part of Lemma 13, i.e.,:

$$|{\mathsf{\Theta }}_f\left(\mu \right)|\le 1·(3-3\mu )+2(3\mu -2)=3\mu -1<1·(3-3\mu )+2(3\mu -2)=3\mu -1.$$

 □

The results presented in Lemmas 8–13 can be collected as follows.

Theorem 1.

Let $f\in {\mathcal{C}}_0\left(k\right)$, $a_2$ be a real number, and $a_2\in [0,2]$. Then:

$$\begin{array}{c}\hfill |{\mathsf{\Theta }}_f\left(\mu \right)|\le \left\{\begin{array}{cc}4-6\mu ,\hfill & \mu \le 1/3,\hfill \\ 3-3\mu ,\hfill & \mu \in \left(1/3,2/3\right),\hfill \\ K,\hfill & \mu =2/3,\hfill \\ 3\mu -1,\hfill & \mu \in \left(2/3,1\right),\hfill \\ 6\mu -4,\hfill & \mu \ge 1,\hfill \end{array}\right.\end{array}$$

where K is given by (25). The results are sharp for $\mu \le 1/3$, $\mu =2/3$, and $\mu \ge 1$. The equality holds for the Koebe function (2) in the first and the last case. The assumption $a_2\in [0,2]$ is not necessary for $\mu \le 0$ and $\mu \ge 1$.

5. Bounds of $\left|\mathsf{\Phi }\right(\mathbf{\mu }\left)\right|$ for the Class ${\mathcal{C}}_{\mathbf{0}}\left(\mathit{k}\right)$

At the beginning of this section, we will quote the well known theorem of Marjono and Thomas [14].

Theorem 2

([14]). If $f\in {\mathcal{C}}_0\left(k\right)$, then:

$$|{\mathsf{\Phi }}_f\left(1\right)|=|a_2{a}_4-{a_3}^2|\le 1.$$

Now, we shall prove the bound for $\mu \ge 1$.

Theorem 3.

Let $f\in {\mathcal{C}}_0\left(k\right)$ and $\mu \ge 1$. Then, $|{\mathsf{\Phi }}_f\left(\mu \right)|\le 9\mu -8$. The result is sharp.

Proof. 

Rearranging the components in (14):

$$\begin{array}{ll}{\mathsf{\Phi }}_f\left(\mu \right)& =\frac{1}{8}(p_1{p}_3-p_2{}^2)-(\frac{1}{9}\mu -\frac{1}{8})p_2{}^2+\frac{1}{4}(p_3-p_1{p}_2)-(\frac{4}{9}\mu -\frac{1}{2})p_1{p}_2\hfill \\ & -(\frac{2}{3}\mu -\frac{1}{2})p_2-(\frac{4}{9}\mu -\frac{3}{8})p_1{}^2-(\frac{4}{3}\mu -\frac{5}{4})p_1-(\mu -1),\hfill \end{array}$$

and writing p instead of $|p_1|$, by Lemmas 1, 3, and 6, for $\mu \ge 9/8$, we obtain:

$$\begin{array}{ll}|{\mathsf{\Phi }}_f\left(\mu \right)|& \le \frac{1}{8}(4-p^2)+(\frac{4}{9}\mu -\frac{1}{2})+\frac{1}{2}+(\frac{8}{9}\mu -1)p+(\frac{4}{3}\mu -1)\hfill \\ & +(\frac{4}{9}\mu -\frac{3}{8})p^2+(\frac{4}{3}\mu -\frac{5}{4})p+(\mu -1)\hfill \\ & =(\frac{4}{9}\mu -\frac{1}{2})p^2+(\frac{20}{9}\mu -\frac{9}{4})p+\frac{25}{9}\mu -\frac{3}{2}\hfill \\ & \le 9\mu -8.\hfill \end{array}$$

If $\mu \in (1,9/8)$, then:

$${\mathsf{\Phi }}_f\left(\mu \right)=(9-8\mu )\left(a_2{a}_4-a_3{}^2\right)+(8\mu -8)\left(a_2{a}_4-\frac{9}{8}{a}_3{}^2\right).$$

From the previous part of this proof $|a_2{a}_4-\frac{9}{8}{a_3}^2|\le \frac{17}{8}$ and from Theorem 2, after using the triangle inequality, we get:

$$|{\mathsf{\Phi }}_f\left(\mu \right)|\le (9-8\mu )·1+(8\mu -8)·\frac{17}{8}=9\mu -8.$$

It is easy to verify that for the Koebe function (2), we have ${\mathsf{\Phi }}_k\left(\mu \right)=8-9\mu$, so the derived estimate is sharp. □

In the next step, we shall prove that the Koebe function (2) is the extremal function for $\mu \le 63/92$.

Theorem 4.

Let $f\in {\mathcal{C}}_0\left(k\right)$ and $\mu \le 63/92$. Then, $|{\mathsf{\Phi }}_f\left(\mu \right)|\le 8-9\mu$. The result is sharp.

Proof. 

At the beginning, let us discuss the case $\mu =63/92$. From (14), it follows that:

$$\begin{array}{ll}184{\mathsf{\Phi }}_f\left(\frac{63}{92}\right)& =14(p_1{p}_3-p_2{}^2)+9p_1{p}_3+20(p_3-\frac{1}{2}{p}_1{p}_2)\hfill \\ & +4(p_3+2p_2+p_1)+22p_3+58p_1+13p_1{}^2+58.\hfill \end{array}$$

Now, applying Lemmas 1 and 4 for $\mu =1/2$, Lemma 5 (remembering that $2(2+\mathrm{Re}{p}_1)\le 2(2+|p_1\left|\right)$), Lemma 6, and the triangle inequality and writing p instead of $|p_1|$, we obtain:

$$184|{\mathsf{\Phi }}_f\left(\frac{63}{92}\right)|\le 14(4-p^2)+18p+20h\left(p\right)+8(2+p)+44+58p+13p^2+58,$$

where:

$$h\left(p\right)=\left\{\begin{array}{cc}\frac{1}{16}{p}^3-\frac{3}{8}{p}^2+2,\hfill & p\in [0,4/3],\hfill \\ 2p-\frac{1}{2}{p}^3,\hfill & p\in [4/3,2].\hfill \end{array}\right.$$

Hence,

$$184|{\mathsf{\Phi }}_f\left(\frac{63}{92}\right)|\le H\left(p\right),$$

where:

$$H\left(p\right)=\left\{\begin{array}{cc}\frac{5}{4}{p}^3-\frac{17}{2}{p}^2+84p+214,\hfill & p\in [0,4/3],\hfill \\ -10p^3-p^2+124p+174,\hfill & p\in [4/3,2].\hfill \end{array}\right.$$

(31)

Is it clear that H is an increasing function for $p\in [0,2]$, so:

$$|{\mathsf{\Phi }}_f\left(\frac{63}{92}\right)|\le H\left(2\right)=\frac{338}{184}=8-9·\frac{63}{92}.$$

If $\mu \in (0,63/92)$, then:

$${\mathsf{\Phi }}_f\left(\mu \right)=(1-\frac{92}{63}\mu )a_2{a}_4+\frac{92}{63}\mu \left(a_2{a}_4-\frac{63}{92}{a}_3{}^2\right).$$

From the previous part of this proof and the bound $|a_n|\le n$ valid for all functions in ${\mathcal{C}}_0\left(k\right)$,

$$|{\mathsf{\Phi }}_f\left(\mu \right)|\le (1-\frac{92}{63}\mu )·8+\frac{92}{63}\mu ·\frac{338}{184}=8-9\mu .$$

Equality holds for the Koebe function. □

It is worth adding that the function H given by (31) is decreasing for $p>2$, so the choice $\mu =63/92$ is important.

Now, we will find the exact bound of ${\mathsf{\Phi }}_f\left(\mu \right)$ for $\mu$ close to one. Namely, we will discuss the case $\mu \in [{\mu }_0,1]$, where:

$${\mu }_0=18/19=0.947\dots .$$

(32)

In this result, we need in addition that the coefficient $a_2$ should be real and $a_2\in [0,2]$. From (12), we get $p=p_1\in [-2,2]$. In the proof, we are going to apply Lemma 7.

Taking into account (14) and Lemma 2, we can write ${\mathsf{\Phi }}_f\left(\mu \right)$ as follows:

$$144{\mathsf{\Phi }}_f\left(\mu \right)=A_0+A_{1}x+A_2{x}^2{+B(1-|x|}^2)y,$$

where:

$$\begin{array}{cc}\hfill A_0& =\frac{1}{2}\left(9-8\mu \right)p^4+\left(27-32\mu \right)p^3+2\left(45-56\mu \right)p^2+12\left(15-16\mu \right)p+144(1-\mu ),\hfill \\ \hfill A_1& =(4-p^2)\left[12(3-4\mu )+4(9-8\mu )p+(9-8\mu )p^2\right],\hfill \\ \hfill A_2& =-\frac{1}{2}(4-p^2)(2+p)\left[(9-8\mu )p+16\mu \right],\hfill \\ \hfill B& =9(4-p^2)(2+p).\hfill \end{array}$$

If $p=-2$ and $p=2$, then $f\left(z\right)=\frac{1}{2}log\frac{1+z}{1-z}$ and $f\left(z\right)=\frac{z}{{(1-z)}^2}$, respectively, so:

$${\mathsf{\Phi }}_f\left(\mu \right)=-\mu /9\mathrm{and}{\mathsf{\Phi }}_f\left(\mu \right)=8-9\mu .$$

(33)

We will show that these values are less than or equal to the real bound of $|{\mathsf{\Phi }}_f\left(\mu \right)|$ for all $f\in {\mathcal{C}}_0\left(k\right)$. Now and on, we assume that $p\in (-2,2)$. Taking into account (14) and Lemma 2, by the triangle inequality and the assumption $\left|y\right|\le 1$, we get:

$$|{\mathsf{\Phi }}_f\left(\mu \right)|\le \frac{1}{16}(4-p^2)(2+p)\left[\left|a+bx+c{x}^2\right|+1-{\left|x\right|}^2\right],$$

(34)

where:

$$\begin{array}{cc}\hfill a& =\frac{1}{9(4-p^2)(2+p)}\left[\frac{1}{2}(9-8\mu )p^4+(27-32\mu )p^3+2(45-56\mu )p^2\right.\hfill \\ \hfill & \left.+12(15-16\mu )p+144(1-\mu )\right],\hfill \\ \hfill b& =\frac{1}{9(2+p)}\left[(9-8\mu )p^2+4(9-8\mu )p+12(3-4\mu )\right],\hfill \\ \hfill c& =-\frac{1}{18}\left[(9-8\mu )p+16\mu \right].\hfill \end{array}$$

(35)

Now, we are ready to establish the main theorem of this section.

Theorem 5.

Let $f\in {\mathcal{C}}_0\left(k\right)$, $a_2$ be a real number, $a_2\in [0,2]$, and $\mu \in [{\mu }_0,1]$, where ${\mu }_0=18/19$. Then:

$$|{\mathsf{\Phi }}_f\left(\mu \right)|\le \frac{\mu }{9-8\mu }.$$

(36)

Equality holds for the function F given by (15).

In the proof of this theorem, we will need the two lemmas that follow. We assume that a, b, and c are given by (35).

Lemma 14.

If $(p,\mu )\in (-2,2)\times [{\mu }_0,1]$ are such that $a\le 0$, then (36) holds.

Lemma 15.

If $(p,\mu )\in (-2,2)\times [{\mu }_0,1]$ are such that $a>0$, then the following inequalities hold:

$$b<0,\left|b\right|\ge 2(1-|c\left|\right),b^2\ge -4a(1-c^2)/c,\left|ab\right|\le \left|c\right|\left(\right|b|-4\left|a\right|).$$

Proof of Lemma 14. 

At the beginning, observe that if $(p,\mu )\in (-2,2)\times [{\mu }_0,1]$, then:

$$c=-\frac{1}{18}\left[9p+8(2-p)\mu \right]\le -\frac{1}{18}\left[9p+8(2-p)·\frac{18}{19}\right]=-\frac{1}{38}(3p+32)<0.$$

(37)

According to Lemma 7 from (34), we obtain:

$$|{\mathsf{\Phi }}_f\left(\mu \right)|\le \frac{1}{16}(4-p^2)(2+p)·Y(a,b,c),$$

where:

$$Y(a,b,c)=\left\{\begin{array}{cc}{\displaystyle -a+\left|b\right|-c}\hfill & ,\left|b\right|\ge 2(1+c),\hfill \\ {\displaystyle 1-a+\frac{b^2}{4(1+c)}}\hfill & ,\left|b\right|<2(1+c).\hfill \end{array}\right.$$

If $\left|b\right|<2(1+c)$, then from (34), we get:

$$\begin{array}{ll}144|{\mathsf{\Phi }}_f\left(\mu \right)|& \le 9(4-p^2)(2+p)\hfill \\ & -\left[\frac{1}{2}(9-8\mu )p^4+(27-32\mu )p^3+2(45-56\mu )p^2+12(15-16\mu )p+144(1-\mu )\right]\hfill \\ & +\frac{{\left[(9-8\mu )p^2+4(9-8\mu )p+12(3-4\mu )\right]}^2}{2(9-8\mu )}.\hfill \end{array}$$

Because the right hand side of this inequality is constant and equal to $144\mu /(9-8\mu )$; hence, $|{\mathsf{\Phi }}_f\left(\mu \right)|\le \mu /(9-8\mu )$.

If $\left|b\right|\ge 2(1+c)$, then:

$$|{\mathsf{\Phi }}_f\left(\mu \right)|\le \left\{\begin{array}{cc}\frac{1}{16}(4-p^2)(2+p)(-a+b-c),\hfill & b\ge 0,\hfill \\ \frac{1}{16}(4-p^2)(2+p)(-a-b-c),\hfill & b\le 0.\hfill \end{array}\right.$$

(38)

The first expression in (38) is equal to:

$$\begin{array}{cc}\hfill & \frac{1}{144}\left[-2(9-8\mu )p^4-8(9-8\mu )p^3-24(3-4\mu )p^2+64\mu p+16\mu \right]\hfill \\ \hfill & =-\frac{1}{72}\left[(9-8\mu )p^2{(p+2)}^2-16\mu {(p+1)}^2+8\mu \right].\hfill \end{array}$$

Substituting $q=p+1$, $q\in (-1,3)$, we obtain:

$$W_1\left(q\right)=-\frac{1}{72}\left[(9-8\mu )q^4-18q^2+9\right]=-\frac{1}{72}\left[(3-2\sqrt{2\mu })q^2-3\right]·\left[(3+2\sqrt{2\mu })q^2-3\right].$$

Hence, the maximum value of $W_1\left(q\right)$ is achieved for:

$$q_{*}{}^2=\frac{1}{2}\left(\frac{3}{3-2\sqrt{2\mu }}+\frac{3}{3+2\sqrt{2\mu }}\right)=\frac{9}{9-8\mu }.$$

This value is equal to $W_1\left(q_{*}\right)=\mu /(9-8\mu )$.

The second expression in (38) is equal to:

$$W_2\left(p\right)=\frac{1}{18}\left[-(9-8\mu )p^2-4(9-10\mu )p-2(18-25\mu )\right],$$

so:

$$W_2\left(p\right)\le W_2\left(\frac{2(10\mu -9)}{9-8\mu }\right)=\frac{\mu }{9-8\mu }.$$

It is easy to check that for $p_{*}=q_{*}-1=3/\sqrt{9-8\mu }-1$ and $p_{**}=2(10\mu -9)/(9-8\mu )$, we have $b=2(1+c)$ and $b=-2(1+c)$, respectively. This means that the maximum value of $|{\mathsf{\Phi }}_f\left(\mu \right)|$ for $\left|b\right|\ge 2(1+c)$ is obtained if $\left|b\right|=2(1+c)$. □

Proof of Lemma 15. 

Let $(p,\mu )\in (-2,2)\times [{\mu }_0,1]$. At the beginning, we want to constrain the range of variability of p to some subset of $(-2,2)$ for which $a>0$.

From (35) for $a=0$, we have:

$$\frac{1}{2}(9-8\mu )p^4+(27-32\mu )p^3+2(45-56\mu )p^2+12(15-16\mu )p+144(1-\mu )=0,$$

which is equivalent to:

$$9(p^2+2p+8){(2+p)}^2-8{(p^2+4p+6)}^2\mu =0.$$

If $p=0$, $\mu =0$, then from (35), $a=2$. Hence, points for which $a>0$ lie below the curve $a=0$. For the function $M\left(p\right)=9(p^2+2p+8){(2+p)}^2/8{(p^2+4p+6)}^2$, $p\in (-2,2)$, there is:

$$M^{\prime }\left(p\right)=\frac{9(2+p)}{4{(p^2+4p+6)}^3}·(p^3+2p^2-10p-4).$$

Consequently, $M\left(p\right)$ is an increasing function if $p\in (-2,p_0)$ and a decreasing function if $p\in (p_0,2)$ for $p_0=-0,376\dots$, where $p_0$ is the only solution of $M^{\prime }\left(p\right)=0$ in $(-2,2)$. Since $M(-1)<{\mu }_0$ and $M(2/3)<{\mu }_0$, then $M\left(p\right)<{\mu }_0$ for $p\in (-2,-1]\cup [2/3,2)$. This means that $a>0$ and $\mu \in [{\mu }_0,1]$ hold for $p\in I$, $I\subset (-1,2/3)$ (in other words, if $a>0$ and $\mu \in [{\mu }_0,1]$, then $-1<p<2/3$).

I.

Since $\mu \in [{\mu }_0,1]$ and:

$$b=\frac{1}{9}\left(9-8\mu \right)(2+p)-\frac{16\mu }{9(2+p)}$$

as a function of $p\in (-1,2/3)$, is increasing, it is enough to estimate this expression taking $p=2/3$ as a limit value. Therefore,

$$b<\frac{2}{27}(36-41\mu )<0.$$

II.

The inequality $-b\ge 2(1+c)$ can be written as $(8\mu -9)p+20\mu -18\ge 0$. For $\mu \in [{\mu }_0,1]$ and $p\in (-1,2/3)$,

$$(8\mu -9)p+20\mu -18>\frac{76}{3}\left(\mu -\frac{18}{19}\right)\ge 0.$$

III.

With the notation $W=b^2+4a(1-c^2)/c$ and:

$$g(p,\mu )=(9-8\mu )\left[(16\mu -9)p^3+18(4\mu -3)p^2+4(25\mu -27)p\right]-8(32{\mu }^2-117\mu +81),$$

we can write:

$$W=\frac{8g(p,\mu )}{9{(2+p)}^2[(9-8\mu )p+16\mu ]}.$$

We shall prove that $g(p,\mu )\ge 0$ for $\mu \in [{\mu }_0,1]$ and $p\in (-1,2/3)$. We have:

$$\frac{\partial g}{\partial p}(p,\mu )=(9-8\mu )\left[3(16\mu -9)p^2+36(4\mu -3)p+4(25\mu -27)\right].$$

For $\mu \in [{\mu }_0,1]$, we obtain:

$$\frac{\partial g}{\partial p}\left(-1,\mu \right)=(4\mu -27)(9-8\mu )<0,$$

and:

$$\frac{\partial g}{\partial p}\left(2/5,\mu \right)=4\left(1033\mu -972\right)\left(9-8\mu \right)/25>0.$$

This means that:

$$min\{g(p,\mu ):p\in (-1,2/3),\mu \in [{\mu }_0,1]\}=min\{g(p,\mu ):p\in (-1,2/5],\mu \in [{\mu }_0,1]\}.$$

Since $(16\mu -9)p+18(4\mu -3)\ge 0$ for $\mu \in [{\mu }_0,1]$ and $p\in (-1,2/3)$, we have:

$$\begin{array}{l}min\{g(p,\mu ):p\in (-1,2/3),\mu \in [{\mu }_0,1]\}\hfill \\ >min\left\{4(25\mu -27)(9-8\mu )p-8(32{\mu }^2-117\mu +81):p\in (-1,2/5],\mu \in [{\mu }_0,1]\right\}\hfill \\ \ge 4(25\mu -27)(9-8\mu )·2/5-8(32{\mu }^2-117\mu +81)\hfill \\ =144(-20{\mu }^2+57\mu -36)/5>0.\hfill \end{array}$$

In this way, we have proven that $b^2+4a(1-c^2)/c\ge 0$.

IV.

Let us denote $V=c(b+4a)+ab$ and:

$$h(p,\mu )=32(p^2+4p+6){(2p+5)}^2{\mu }^2-36(2+p)(16p^2+71p+82)\mu +81(2-p){(2+p)}^3.$$

We have

$$V=\frac{4h(p,\mu )}{81{(2+p)}^2(4-p^2)}.$$

The function h of a variable $\mu$ increases for $\mu \in [{\mu }_0,1]$. Indeed, for a fixed $p\in (-1,2/3)$,

$$\begin{array}{ll}\frac{\partial h}{\partial \mu }(p,\mu )& =64(p^2+4p+6){(2p+5)}^2\mu -36(2+p)(16p^2+71p+82)\hfill \\ & \ge 64(p^2+4p+6){(2p+5)}^2·\frac{18}{19}-36(2+p)(16p^2+71p+82)\hfill \\ & =\frac{36}{19}\left[32(p^2+4p+6){(2p+5)}^2-19(2+p)(16p^2+71p+82)\right]\hfill \\ & =\frac{36}{19}\left[351+474(p+1)+395{(p+1)}^2+336{(p+1)}^3+128{(p+1)}^4\right]\hfill \end{array}$$

and is greater than zero. Finally,

$$\frac{361}{81}h(p,\frac{18}{19})=151p^4+732(p+1)p^2+\frac{176}{3}{p}^2+\frac{4}{3}{(6-7p)}^2>0,$$

so h, as well as V are positive for $\mu \in [{\mu }_0,1]$ and $p\in (-1,2/3)$. □

Proof of Theorem 4. 

From Lemma 14, we know that if $a\le 0$ and $\mu \in [{\mu }_0,1]$, then (36) holds. Assume now that $a>0$ and $\mu \in [{\mu }_0,1]$. By Lemmas 7 and 15, and Formula (37),

$$|{\mathsf{\Phi }}_f\left(\mu \right)|\le \frac{1}{16}(4-p^2)(2+p)(-|a|+|b|+|c\left|\right)=\frac{1}{16}(4-p^2)(2+p)(-a-b-c).$$

This expression is the same as in the second line in (38), and it takes the maximum value $\mu /(9-8\mu )$ for $p=p_{**}=2(10\mu -9)/(9-8\mu )$. Observe that the function $[{\mu }_0,1]\ni \mu \mapsto 2(10\mu -9)/(9-8\mu )$ increases. Hence, $2/3\le p_{**}\le 2$, so $p_{**}$ is not less than $2/3$. For this reason, the maximum value of $|{\mathsf{\Phi }}_f\left(\mu \right)|$ is equal to $\mu /(9-8\mu )$, but this value is obtained if $a\le 0$.

It is easy to check that both values of ${\mathsf{\Phi }}_f\left(\mu \right)$ for $f\left(z\right)=\frac{1}{2}log\frac{1+z}{1-z}$ and $f\left(z\right)=k\left(z\right)$, which are given in (33), are less than or equal to $\mu /(9-8\mu )$. This completes the proof. □

Theorem 6.

Let $f\in {\mathcal{C}}_0\left(k\right)$, $\mu \in [63/92,18/19]$, and $a_2$ be a real number, $a_2\in [0,2]$. Then:

$$|{\mathsf{\Phi }}_f\left(\mu \right)|\le (396-361\mu )/81.$$

Proof. 

By Theorems 4 and 5, $|{\mathsf{\Phi }}_f(63/92)|\le 169/92$ and $|{\mathsf{\Phi }}_f(18/19)|\le 2/3$. Putting $\alpha =4(414-437\mu )/459$, we can write:

$${\mathsf{\Phi }}_f\left(\mu \right)=\alpha (a_2{a}_4-\frac{63}{92}{a}_3{}^2)+(1-\alpha )(a_2{a}_4-\frac{18}{19}{a}_3{}^2).$$

Applying the triangle inequality, we obtain our claim. □

The results presented in Theorems 2–6 can be collected as follows.

Corollary 1.

Let $f\in {\mathcal{C}}_0\left(k\right)$ be given by (1), $a_2$ be a real number, and $a_2\in [0,2]$. Then:

$$\begin{array}{c}\hfill |{\mathsf{\Phi }}_f\left(\mu \right)|\le \left\{\begin{array}{cc}8-9\mu ,\hfill & \mu \le 63/92,\hfill \\ (396-361\mu )/81,\hfill & \mu \in \left[63/92,18/19\right],\hfill \\ {\displaystyle \frac{\mu }{9-8\mu },}\hfill & \mu \in \left[18/19,1\right],\hfill \\ 9\mu -8,\hfill & \mu \ge 1,\hfill \end{array}\right.\end{array}$$

The results are sharp for $\mu \le 63/92$ and $\mu \ge 18/19$. The equality holds for the Koebe function (2) in the first and the last case. The function F given by (15) is an extremal function when $\mu \in \left[18/19,1\right]$. The assumption $a_2\in [0,2]$ is not necessary for $\mu \le 63/92$ and $\mu \ge 1$.

Observe that for $\mu \in \left(18/19,1\right)$, we have $\mu /(9-8\mu )<1$, so the sharp bound for ${\mathcal{C}}_0\left(k\right)$ is less than the sharp bound for ${\mathcal{S}}^{*}$ given by (5).

6. Concluding Remarks

In this paper, we estimated two functionals ${\mathsf{\Theta }}_f\left(\mu \right)=a_4-\mu a_2{a}_3$ and ${\mathsf{\Phi }}_f\left(\mu \right)=a_2{a}_4-\mu {a_3}^2$ for the family ${\mathcal{C}}_0\left(k\right)$, where $\mu$ is a real number. This family is a subset of the class $\mathcal{C}$ of all close-to-convex functions.

The results presented above broaden our knowledge about the behavior of the coefficient functionals defined for functions not only in $\mathcal{C}$, but also generally in the class $\mathcal{S}$ of univalent functions. Unfortunately, there are no good estimates of the discussed functionals in the whole classes $\mathcal{C}$ and $\mathcal{S}$. It seems that further research on the classes of the type ${\mathcal{C}}_0\left(f\right)$, where f is different from k, may result in obtaining some conclusions about $\mathcal{S}$.

In our opinion, the most important problem to be solved now is the estimating of the second Hankel determinant, or in other words ${\mathsf{\Phi }}_f\left(1\right)$ for $f\in \mathcal{S}$. Even in the class ${\mathcal{C}}_0$, the exact bound is unknown. It is only known that for ${\mathcal{C}}_0$, there is $|a_2{a}_4-{a_3}^2|<1.242\dots$ (see [25]). On the other hand, the conjecture posed by Thomas [26] about 30 years ago that $|a_n{a}_{n+2}-{a_{n+1}}^2|\le 1$ for $\mathcal{S}$ and $n\ge 2$ was disproven. This means that there are functions in $\mathcal{S}$ for which $|a_n{a}_{n+2}-{a_{n+1}}^2|>1$. Finding (even non-sharp) estimates of ${\mathsf{\Phi }}_f\left(1\right)$ for $f\in \mathcal{S}$ remains an interesting open problem.