On the third logarithmic coefficient in some subclasses of close-to-convex functions

Abstract

For analytic functions f in the unit disk ${\mathbb {D}}$ normalized by $f(0)=0$ and $f'(0)=1$ satisfying in ${\mathbb {D}}$ respectively the conditions ${{\,\mathrm{Re}\,}}\{ (1-z)f'(z) \}> 0,\ {{\,\mathrm{Re}\,}}\{ (1-z^2)f'(z) \}> 0,\ {{\,\mathrm{Re}\,}}\{ (1-z+z^2)f'(z) \}> 0,\ {{\,\mathrm{Re}\,}}\{ (1-z)^2f'(z) \} > 0,$ the sharp upper bound of the third logarithmic coefficient in case when $f''(0)$ is real was computed.

1 Introduction

Let ${\mathbb {D}}:= \left\{ z \in {\mathbb {C}} : |z|<1 \right\} ,$$\overline{{\mathbb {D}}}:=\{z\in {\mathbb {C}}: |z|\le 1\}$ and ${\mathbb {T}}:=\partial {\mathbb {D}}.$ Let ${\mathcal H}$ be the class of all analytic functions in ${\mathbb {D}},$${\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)

and $\mathcal S$ be the subclass of $\mathcal A$ of all univalent functions.

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)

The numbers $\gamma _n$ are called logarithmic coefficients of f. As is well known, the logarithmic coefficients play a crucial role in Milin conjecture ([23], see also [10, p. 155]), namely that for $f\in \mathcal S,$

$$\begin{aligned} \sum _{m=1}^{n}\sum _{k=1}^n\left( k|\gamma _k|^2-\frac{1}{k}\right) \le 0. \end{aligned}$$

De Branges [8] showing Milin conjecture confirmed the famous Bieberbach conjecture (e.g., [10, p. 37]). 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,

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

and are unknown for $n\ge 3.$

As usual, instead of the whole class $\mathcal S$ one can take into account their subclasses for which the problem of finding sharp estimates of logarithmic coefficients can be studied. When $f \in {\mathcal S}^*,$ the class of starlike functions, the inequality $|\gamma _n| \le 1/n$ holds for $n \in {\mathbb {N}}$ (see e.g. [30, p. 42]). Moreover, for $f \in {\mathcal {SS}}^*(\beta ),$ the class of strongly starlike function of order $\beta$ ($0<\beta \le 1$), it holds that $|\gamma _n| \le \beta /n$ ($n\in {\mathbb {N}}$) (see [28]). Also, the bounds of $\gamma _n$ for functions in the class of gamma-starlike functions, close-to-convex functions and Bazilevič functions were examined in [30, p. 116], [9, 27, 29], respectively. In two recent papers, namely, in [15] the bounds of early logarithmic coefficients of the subclasses $\mathcal F_1,\mathcal F_2,\mathcal F_3$ of $\mathcal S$ and in [1] of the subclass $\mathcal F_4$ of $\mathcal S$ of functions f satisfying respectively the condition

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

(3)

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

(4)

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

(5)

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

(6)

were computed. Let us note that each class defined above is the subclass of the well known class of close-to-convex functions, so therefore families $\mathcal F_i,\ i=1,\dots ,4,$ contain only univalent functions (e.g., [12, Vol. II, p. 2]). Both cited paper contains sharp bounds of $\gamma _1$ and $\gamma _2$ and partial results for $\gamma _3$ only. The first three results in theorem below were shown in [15], and the last one in [1].

Theorem 1

Let $f\in \mathcal A$ be of the form (1). Then

1. if $f\in \mathcal F_1$ and $1\le a_2\le 3/2,$ then

$$\begin{aligned} |\gamma _3|\le \frac{1}{288}(11+15\sqrt{30}) = 0.323466\dots ; \end{aligned}$$

2. if $f\in \mathcal F_2$ and $0\le a_2\le 1,$ then

$$\begin{aligned} |\gamma _3|\le \frac{1}{972}(95+23\sqrt{46}) = 0.258223\dots ; \end{aligned}$$

3. if $f\in \mathcal F_3$ and $1/2\le a_2\le 3/2,$ then

$$\begin{aligned} |\gamma _3|\le \frac{1}{7776}(743+131\sqrt{262}) = 0.368238\dots ; \end{aligned}$$

4. if $f\in \mathcal F_4$ and $1\le a_2\le 2,$ then

$$\begin{aligned} |\gamma _3|\le \frac{1}{243}(28+19\sqrt{19}) = 0.456045\dots \end{aligned}$$

In this paper we improve all results in Theorem 1 for $\gamma _3$ for the general case when $a_2$ is real. Differentiating (2) and using (1) we get

$$\begin{aligned} \gamma _3=\frac{1}{2}\left( a_4-a_2a_3+\frac{1}{3}a_2^3\right) . \end{aligned}$$

(7)

Since each class $\mathcal F_i,\ i=1,\dots ,4,$ has a representation by using the Carathéodory class $\mathcal P$, i.e., the class of functions $p \in {\mathcal H}$ of the form

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

(8)

having a positive real part in ${\mathbb {D}},$ the coefficients of functions in $\mathcal F_i,$ so $\gamma _3$ has a suitable representation expressed by the coefficients of functions in $\mathcal P.$ Therefore to get the upper bound of $\gamma _3$ our computing is based on parametric formulas for the second and third coefficients in $\mathcal P.$ The proof of results of Theorem 1 are based on the well known formula on $c_2$ and on the formula $c_3$ due to Libera and Zlotkiewicz [21, 22] with the restriction that $c_1\ge 0.$ Since all classes $\mathcal F_i$ are not rotation invariant, to omit the assumption $c_1\ge 0.$ we will use a general formula for $c_3,$ which was found in [4]. However to be self contained we present a proof for $c_3$ here. Moreover in our computation of the sharp bound of $\gamma _3$ we use a lemma due to Ohno and Sugawa [24].

Let us mention that the conditions (3), (4) and (6) were discovered by Ozaki [25] as useful criteria of univalence. Recall also that the classes $\mathcal F_2$ and $\mathcal F_4$ have nice geometrical interpretations, and therefore they play an important role in the geometric function theory. Each function $f\in \mathcal F_2$ maps univalently ${\mathbb {D}}$ onto a domain $f({\mathbb {D}})$ convex in the direction of the imaginary axis, i.e., for every $w_1,w_2\in f({\mathbb {D}})$ such that ${{\,\mathrm{Re}\,}}w_1={{\,\mathrm{Re}\,}}w_2$ the line segment $[w_1,w_2]$ lies in $f({\mathbb {D}}),$ with the additional property that there exist two points $\omega _1,\omega _2\in \partial f({\mathbb {D}})$ for which $\{\omega _1+\mathrm {i}t: t> 0\}\subset {\mathbb {C}}{\setminus } f({\mathbb {D}})$ and $\{\omega _2-\mathrm {i}t: t> 0\}\subset {\mathbb {C}}{\setminus } f({\mathbb {D}})$ (see e.g., [12, p. 199]). Each function in the class $\mathcal F_4$ maps univalently ${\mathbb {D}}$ onto a domain $f({\mathbb {D}})$ called convex in the positive direction of the real axis, i.e., $\{w+it:t\ge 0\}\subset f({\mathbb {D}})$ for every $w\in f({\mathbb {D}})$ [2, 6, 7, 11, 18, 19].

At the end, let us say that the conditions (3)–(6) were generalized by replacing polynomials standing at $f'$ by any quadratic polynomial [16, 17], and by any polynomial of any degree having their roots in ${\mathbb {C}}{\setminus }{\mathbb {D}}$ [13, 14].

2 Lemmas

The formula (9) is due to Carathéodory [3] (see e.g., [10, p. 41]). The formula (10) can be found in [26, p. 166]. In a recent paper [4] the formula (11) was shown and the extremal functions (13) and (14) were computed also. When $c_1\ge 0$ the formula (11) was found by Libera and Zlotkiewicz [21, 22] (see also [20]).

Lemma 1

If $p \in {\mathcal P}$ is of the form (8), then

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

(9)

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

(10)

and

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

(11)

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 (9), namely,

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

(12)

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 (9)–(10), namely,

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

(13)

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 (9)–(11), namely,

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

(14)

The next lemma is a special case of more general results due to Choi, Kim and Sugawa [5] (see also [24]). Define

$$\begin{aligned} Y(a,b,c):=\max _{z\in \overline{{\mathbb {D}}}}\left( |a+bz+cz^2|+1-|z|^2\right) ,\quad a,b,c\in {\mathbb {R}}. \end{aligned}$$

Lemma 2

[5] If $ac\ge 0,$ then

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

If $ac<0,$ then

$$\begin{aligned}&Y(a,b,c) \\&\quad =\left\{ \begin{array}{lll} 1-|a|+\dfrac{b^2}{4(1-|c|)}, &{} \quad -4ac(c^{-2}-1)\le b^2 \wedge |b|<2(1-|c|), \\ 1+|a|+\dfrac{b^2}{4(1+|c|)}, &{}\quad b^2<\min \left\{ 4(1+|c|)^2,-4ac(c^{-2}-1)\right\} , \\ R(a,b,c), &{} {\mathrm{otherwise}} , \end{array} \right. \end{aligned}$$

where

$$\begin{aligned} R(a,b,c)=\left\{ \begin{array}{lll} |a|+|b|-|c|, &{}\quad |c|(|b|+4|a|)\le |ab|,\\ -|a|+|b|+|c|, &{}\quad |ab|\le |c|(|b|-4|a|), \\ (|c|+|a|)\sqrt{1-\dfrac{b^2}{4ac}}, &{}\quad {\mathrm{otherwise}}. \end{array} \right. \end{aligned}$$

3 Logarithmic coefficients

Now we will prove the main results of this paper.

3.1 The class ${\mathcal F}_1$

Recall that $f\in \mathcal F_1$ if $f\in \mathcal A$ and

$$\begin{aligned} {{\,\mathrm{Re}\,}}\{ (1-z)f'(z) \} > 0,\quad z\in {\mathbb {D}}. \end{aligned}$$

Theorem 2

If $f \in {\mathcal F}_1$ is of the form (1) with $a_2 \in {\mathbb {R}}$, then

$$\begin{aligned} |\gamma _3| \le \frac{1}{288}(11+15\sqrt{30})= 0.323466\dots \end{aligned}$$

(15)

The inequality is sharp with the extremal function

$$\begin{aligned} f(z)= \int _{0}^{z} \frac{p(t)}{1-t} \mathrm {d}t,\quad z\in {\mathbb {D}}, \end{aligned}$$

(16)

where

$$\begin{aligned} p(z) = \frac{ (1+z)(6+(7-2\sqrt{30})z+6z^2) }{ (1-z)( 6+(1+\sqrt{30})z + 6z^2 ) },\quad z\in {\mathbb {D}}. \end{aligned}$$

(17)

Proof

Let $f \in {\mathcal F}_1$ be of the form (1) with $a_2 \in {\mathbb {R}}$. Then there exists $p \in {\mathcal P}$ of the form (8) such that

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

(18)

Substituting the series (1) and (8) into (18) and equating the coefficients we get

$$\begin{aligned} a_2=\frac{1}{2}(1+c_1),\quad a_3=\frac{1}{3}(1+c_1+c_2),\quad a_4=\frac{1}{4}(1+c_1+c_2+c_3). \end{aligned}$$

(19)

Note first that since $a_2$ is real, so is $c_1,$ and (9) holds with some $\zeta _1\in [-1,1].$ Moreover, from (19) it follows that $a_2\in [-1/2,3/2].$

By (7) and (19) we get

$$\begin{aligned} 48\gamma _3=3+c_1-c_1^2+c_1^3-4c_1c_2+2c_2+6c_3. \end{aligned}$$

Using now (9)–(11) we have

$$\begin{aligned} \begin{aligned} 48\gamma _3 =&\, 3+2\zeta _1+4\zeta _1^3 +4(1-\zeta _1^2)\zeta _2 +8(1-\zeta _1^2)\zeta _1\zeta _2 \\&-12(1-\zeta _1^2)\zeta _1\zeta _2^2 +12(1-\zeta _1^2)(1-|\zeta _2|^2)\zeta _3, \end{aligned} \end{aligned}$$

(20)

with $\zeta _1 \in [-1,1]$ and $\zeta _2$, $\zeta _3 \in \overline{{\mathbb {D}}}$.

Hence for $\zeta _1=1$ and $\zeta _1=-1$ we respectively have

$$\begin{aligned} \gamma _3=\frac{3}{16}=0.1875,\quad \gamma _3=-\frac{1}{16}=-0.0625. \end{aligned}$$

(21)

Let now $\zeta _1 \in (-1,1).$ Then from (20) we obtain

$$\begin{aligned} 48|\gamma _3| \le 12(1-\zeta _1^2) \varPsi (A,B,C), \end{aligned}$$

(22)

where

$$\begin{aligned} \varPsi (A,B,C) := |A+B\zeta _2+C\zeta _2^2| + 1 - |\zeta _2|^2, \end{aligned}$$

(23)

with

$$\begin{aligned} A:= \frac{3+2\zeta _1+4\zeta _1^3}{12(1-\zeta _1^2)}, \quad B:= \frac{1}{3}(1+2\zeta _1), \quad C:= -\zeta _1. \end{aligned}$$

Note that

$$\begin{aligned} AC<0,\quad \zeta _1 \in [-1,\zeta ') \cup (0,1], \end{aligned}$$

and

$$\begin{aligned} AC \ge 0, \quad \zeta _1 \in [\zeta ',0], \end{aligned}$$

where $\zeta '= -0.72808\dots$ is the zero of the equation $3+2x+4x^3=0,\ x\in (-1,0).$

A. Let $\zeta _1 \in [\zeta ',0]$. Then the inequality $|B|<2(1-|C|)$ holds, so by (22) and Lemma 2 we have

$$\begin{aligned} 48|\gamma _3|\le 12(1-\zeta _1^2)\left( 1 +|A| + \frac{B^2}{4(1-|C|)}\right) = \varphi (\zeta _1), \end{aligned}$$

(24)

where

$$\begin{aligned} \varphi (x) := \frac{1}{3}(46+9x-36x^2+8x^3),\quad x\in [-1,1]. \end{aligned}$$

(25)

Since $\varphi$ is increasing on $[\zeta ',0]$, so $\varphi (x) \le \varphi (0) = 46/3$ for all $x\in [\zeta ',0]$. Therefore from (24) we get

$$\begin{aligned} |\gamma _3| \le \frac{23}{72} = 0.319444\dots \end{aligned}$$

B. Let $\zeta _1 \in (0,1)$. Then the following inequalities hold:

$$\begin{aligned} B^2 +4AC(C^{-2}-1) = \frac{1}{9\zeta _1} (-9-5\zeta _1+4\zeta _1^2-8\zeta _1^3) <0 \end{aligned}$$

and

$$\begin{aligned} B^2-4(1+|C|)^2 = - \frac{1}{9}(35+68\zeta _1+32\zeta _1^2)<0. \end{aligned}$$

Therefore from (22) and Lemma 2 it follows that

$$\begin{aligned} 48|\gamma _3| \le 12(1-\zeta _1^2)\left( 1 +|A| + \frac{B^2}{4(1+|C|)}\right) = \varphi (\zeta _1), \end{aligned}$$

(26)

where $\varphi$ is the function defined by (25). Since $\varphi '(x)=0$ occurs only at $x=(6-\sqrt{30})/4=:x_0$ in (0, 1) and $\varphi ''(x_0)=-4\sqrt{30}<0$, it follows that

$$\begin{aligned} \varphi (x) \le \varphi (x_0) = \frac{1}{6}(11+15\sqrt{30}), \quad x\in (0,1). \end{aligned}$$

Thus by (26) we get

$$\begin{aligned} |\gamma _3| \le \frac{1}{288}(11+15\sqrt{30}) = 0.323466\dots \end{aligned}$$

C. Let $\zeta _1 \in (-1,\zeta ')$. Note that then $B^2<4(1+|C|)^2.$ Furthermore, $B^2+4AC(C^{-2}-1)<0$ holds if and only if $\zeta _1 \in [-1,\zeta '']$, where $\zeta ''=-0.73448\dots$ is the zero of the equation $9+5x-4x^2+8x^3=0,\ x\in (-1,1).$ Therefore, when $\zeta _1 \in (-1,\zeta '']$, by (22) and Lemma 2 we have

$$\begin{aligned} \begin{aligned} |\gamma _3|&\le \frac{1}{4}(1-\zeta _1^2)\left( 1+ |A|+ \frac{B^2}{4(1+|C|)} \right) \\&= \frac{1}{144}(28-\zeta _1-28\zeta _1^2-8\zeta _1^3) < \frac{1}{8}=0.125. \end{aligned} \end{aligned}$$

(27)

For $\zeta _1 \in (\zeta '',\zeta ']$ it holds $B^2+4AC(C^{-2}-1)>0$ and $|B|<2(1-|C|)$. Hence by (22) and Lemma 2 we get

$$\begin{aligned} \begin{aligned} |\gamma _3|&\le \frac{1}{4}(1-\zeta _1^2) \left( 1- |A|+ \frac{B^2}{4(1-|C|)} \right) \\&= \frac{1}{144}(46 +9\zeta _1 -36\zeta _1^2 +8\zeta _1^3) < \frac{7}{48}=0.145833\dots . \end{aligned} \end{aligned}$$

(28)

Summarizing, from (21) and parts A-C it follows that the inequality (15) is true.

By tracking back the above proof, we see that equality in (15) holds when it is satisfied that

$$\begin{aligned} \zeta _1 = \frac{1}{4}(6-\sqrt{30}), \quad \zeta _3=1 \end{aligned}$$

(29)

and

$$\begin{aligned} | A +B\zeta _2 +C\zeta _2^2| + 1 - |\zeta _2|^2 = 1 +|A| + \frac{B^2}{4(1+|C|)}, \end{aligned}$$

(30)

where

$$\begin{aligned} A = \frac{-2490+731\sqrt{30}}{5460}, \quad B = \frac{8-\sqrt{30}}{6}, \quad C =-\frac{1}{4}(6-\sqrt{30}). \end{aligned}$$

Indeed we can easily check that one of the solutions of Eq. (30) is

$$\begin{aligned} \zeta _2 = \frac{1}{105}(25-\sqrt{30}). \end{aligned}$$

(31)

By Lemma 1 a function p of the form (14) with $\zeta _i$ ($i\in \{1,2,3\}$) given by (29) and (31), i.e., the function (17) belongs to $\mathcal P.$ Thus the function (16) belongs to $\mathcal F_1.$ Substituting (29) and (31) into (20) we get equality in (15). This ends the proof of the theorem. $\square$

3.2 The class ${\mathcal F}_2$

Recall that $f\in \mathcal F_2$ if $f\in \mathcal A$ and

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

Theorem 3

If $f \in {\mathcal F}_2$ is of the form (1) with $a_2 \in {\mathbb {R}}$, then

$$\begin{aligned} |\gamma _3| \le \frac{1}{972}(95+23\sqrt{46})=0.258223\dots \end{aligned}$$

(32)

The inequality is sharp with the extremal function

$$\begin{aligned} f(z)= \int _{0}^{z} \frac{p(t)}{1-t^2} \mathrm {d}t,\quad z\in {\mathbb {D}}, \end{aligned}$$

(33)

where

$$\begin{aligned} p(z) = \frac{(1+z)(9+(7-2\sqrt{46})z+9z^2)}{(1-z)(9+(1+\sqrt{46})z+9z^2)},\quad z\in {\mathbb {D}}. \end{aligned}$$

(34)

Proof

Let $f \in {\mathcal F}_2$ be of the form (1). Then there exists $p \in {\mathcal P}$ of the form (8) such that

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

(35)

Substituting the series (1) and (8) into (35) by equating the coefficients we get

$$\begin{aligned} a_2=\frac{1}{2}c_1, \quad a_3 = \frac{1}{3}(1+c_2), \quad a_4 = \frac{1}{4}(c_1 + c_3). \end{aligned}$$

(36)

Note first that since $a_2$ is real, so is $c_1$ and (9) holds with some $\zeta _1\in [-1,1].$ Moreover, from (36) it follows that $a_2\in [-1,1].$

By (7) and (36) we get

$$\begin{aligned} 48\gamma _3 = 2c_1 + c_1^3 -4c_1 c_2 +6c_3. \end{aligned}$$

Using now (9)–(11) we have

$$\begin{aligned} \begin{aligned} 12\gamma _3 =&\, \zeta _1^3 + \zeta _1 + 2(1-\zeta _1^2)\zeta _1 \zeta _2 \\&\, -3(1-\zeta _1^2)\zeta _1 \zeta _2^2 +3(1-\zeta _1^2)(1-|\zeta _2|^2)\zeta _3. \end{aligned} \end{aligned}$$

(37)

with $\zeta _1 \in [-1,1]$ and $\zeta _2$, $\zeta _3 \in \overline{{\mathbb {D}}}.$

Hence for $\zeta _1=1,$$\zeta _1=-1$ and $\zeta _1=0$ we respectively have

$$\begin{aligned} \gamma _3 = \frac{1}{6},\quad \gamma _3 = -\frac{1}{6},\quad |\gamma _3| \le \frac{1}{4}(1-|\zeta _2|^2) \le \frac{1}{4}. \end{aligned}$$

(38)

Now let $\zeta _1 \in (-1,1) {\setminus } \{ 0 \}=:I$. Then from (37) we obtain

$$\begin{aligned} 12 |\gamma _3| \le 3(1-\zeta _1^2) \varPsi (A,B,C), \end{aligned}$$

(39)

where $\varPsi$ is defined by (23) with

$$\begin{aligned} A:= \frac{\zeta _1 (1+\zeta _1^2)}{3(1-\zeta _1^2)}, \quad B:= \frac{2}{3}\zeta _1, \quad C:= -\zeta _1. \end{aligned}$$

Note now that $AC<0$ for $\zeta _1 \in I$. Moreover,

$$\begin{aligned} B^2 \le -4AC(C^{-2}-1), \quad \zeta _1 \in I, \end{aligned}$$

since

$$\begin{aligned} B^2 + 4AC(C^{-2}-1) = -\frac{4}{9}(3+2\zeta _1^2) <0, \quad \zeta _1 \in I, \end{aligned}$$

and

$$\begin{aligned} B^2<4(1+|C|)^2,\quad \zeta _1 \in I, \end{aligned}$$

since

$$\begin{aligned} B^2 -4(1+|C|)^2 = -\frac{4}{9}(8\zeta _1^2 + 18|\zeta _1|+9) <0, \quad \zeta _1 \in I. \end{aligned}$$

Therefore by Lemma 2 we get

$$\begin{aligned} \begin{aligned} \varPsi (A,B,C)&\le 1+|A|+\frac{B^2}{4(1+|C|)} \\&= 1 + \frac{ |\zeta _1|(1+\zeta _1^2) }{ 3(1-\zeta _1^2) } + \frac{\zeta _1^2}{9(1+|\zeta _1|)}. \end{aligned} \end{aligned}$$

Hence and from (39) it follows that

$$\begin{aligned} 12|\gamma _3|\le \varphi (\zeta _1), \end{aligned}$$

(40)

where

$$\begin{aligned} \varphi (x):=\frac{1}{3}(9 +3|x| -8x^2 +2|x|^3),\quad x\in I. \end{aligned}$$

We note that the function $\varphi$ is even in I. As easy to verify

$$\begin{aligned} \varphi (x)\le \varphi (x_0)=\frac{1}{81}(95+23\sqrt{46}),\quad x\in I, \end{aligned}$$

where $x_0:=(8-\sqrt{46})/6=0.202945\dots$ Hence and by (40) we obtain

$$\begin{aligned} |\gamma _3| \le \frac{1}{12}\varphi (x_0) \le \frac{1}{972}(95+23\sqrt{46}). \end{aligned}$$

This and (38) show that the inequality (32) is true.

By tracking back the above proof, we see that equality in (32) holds when it is satisfied that

$$\begin{aligned} \zeta _1 = \frac{1}{6}(8-\sqrt{46}), \quad \zeta _3=1 \end{aligned}$$

(41)

and

$$\begin{aligned} | A +B\zeta _2 +C\zeta _2^2| + 1 - |\zeta _2|^2 = 1 +|A| + \frac{B^2}{4(1+|C|)}, \end{aligned}$$

(42)

where

$$\begin{aligned} A = \frac{ -1688 +283\sqrt{46} }{ 3150 }, \quad B = \frac{1}{9}(8-\sqrt{46}), \quad C =-\frac{1}{6}(8-\sqrt{46}). \end{aligned}$$

Indeed we can easily check that one of the solutions of the equation (42) is

$$\begin{aligned} \zeta _2 = \frac{1}{75}(11-\sqrt{46}). \end{aligned}$$

(43)

By Lemma 1 a function p of the form (14) with $\zeta _i$ ($i\in \{1,2,3\}$) given by (41) and (43), i.e., the function (34) belongs to $\mathcal P.$ Thus the function (33) belongs to $\mathcal F_2.$ Substituting (41) and (43) into (37) we get equality in (32). This ends the proof of the theorem. $\square$

3.3 The class ${\mathcal F}_3$

Recall that $f\in \mathcal F_3$ if $f\in \mathcal A$ and

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

Theorem 4

If $f \in {\mathcal F}_3$ is of the form (1) with $a_2 \in {\mathbb {R}}$, then

$$\begin{aligned} |\gamma _3| \le \frac{1}{7776}(743+131\sqrt{262})=0.368238\dots \end{aligned}$$

(44)

This result is sharp.

Proof

Let $f \in {\mathcal F}_3$ be of the form (1) with $a_2 \in {\mathbb {R}}$. Then there exists $p \in {\mathcal P}$ of the form (8) such that

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

(45)

Substituting the series (1) and (8) into (45) by equating the coefficients we get

$$\begin{aligned} a_2=\frac{1}{2}(1+c_1), \quad a_3 = \frac{1}{3}(c_1+c_2), \quad a_4 = \frac{1}{4}(-1+c_2 + c_3). \end{aligned}$$

(46)

Note first that since $a_2$ is real, so is $c_1$ and (9) holds with some $\zeta _1\in [-1,1].$ Moreover, from (46) it follows that $a_2\in [-1/2,3/2].$

By (7) and (46) we get

$$\begin{aligned} 48\gamma _3=-5-c_1-c_1^2+c_1^3-4c_1c_2+2c_2+6c_3. \end{aligned}$$

Using now (9)–(11) we have

$$\begin{aligned} \begin{aligned} 48\gamma _3 =&\, -5 -2\zeta _1 +4\zeta _1^3 +4(1-\zeta _1^2)\zeta _2 +8(1-\zeta _1^2)\zeta _1\zeta _2 \\&\, -12(1-\zeta _1^2)\zeta _1\zeta _2^2+12(1-\zeta _1^2)(1-|\zeta _2|^2)\zeta _3, \end{aligned} \end{aligned}$$

(47)

with $\zeta _1 \in [-1,1]$ and $\zeta _2$, $\zeta _3 \in \overline{{\mathbb {D}}}$.

Hence for $\zeta _1=1$ and $\zeta _1=-1$ we respectively have

$$\begin{aligned} \gamma _3 = -\frac{1}{16},\quad \gamma _3 = -\frac{7}{48}. \end{aligned}$$

(48)

Now let $\zeta _1 \in (-1,1).$ Then from (47) we get

$$\begin{aligned} 48 |\gamma _3| \le 12(1-\zeta _1^2) \varPsi (A,B,C), \end{aligned}$$

(49)

where $\varPsi$ is defined by (23) with

$$\begin{aligned} A:= \frac{-5-2\zeta _1+4\zeta _1^3}{12(1-\zeta _1^2)}, \quad B:= \frac{1}{3}(1+2\zeta _1), \quad C:= -\zeta _1. \end{aligned}$$

Note that $A<0$ for $\zeta _1 \in (-1,1)$.

Let $\zeta _1\in (-1,0)$. Then $AC<0$ and it can be easily checked that the following inequalities are true:

$$\begin{aligned} B^2 +4AC(C^{-2}-1) = \frac{1}{9\zeta _1} (15+7\zeta _1+4\zeta _1^2-8\zeta _1^3) <0 \end{aligned}$$

and

$$\begin{aligned} B^2-4(1+|C|)^2 = - \frac{1}{9}(35-76\zeta _1+32\zeta _1^2)<0. \end{aligned}$$

Hence from (49) and Lemma 2 and it follows that

$$\begin{aligned} \begin{aligned} |\gamma _3|&\le \frac{1}{4}(1-\zeta _1^2) \left( 1+ |A|+ \frac{B^2}{4(1+|C|)} \right) \\&= \frac{1}{144}(52+11\zeta _1-28\zeta _1^2-8\zeta _1^3) \le \frac{13}{36} = 0.361111\dots \end{aligned} \end{aligned}$$

(50)

Let now $\zeta _1\in [0,1)$. Then $AC\ge 0$ and we consider two subcases, i.e., $\zeta _1 \in [5/8,1)$ and $\zeta _1 \in [0,5/8).$ For $\zeta _1 \in [5/8,1)$, it holds $|B| \ge 2(1-|C|).$ Thus by (49) and Lemma 2 we have

$$\begin{aligned} \begin{aligned} |\gamma _3|&\le \frac{1}{4}(1-\zeta _1^2) ( |A|+|B|+|C| ) \\&=\frac{1}{48}\left( 9+22\zeta _1-4\zeta _1^2-24\zeta _1^3\right) \le \frac{327}{1024} = 0.319335\dots \end{aligned} \end{aligned}$$

(51)

For $\zeta _1 \in [0,5/8)$ it holds $|B| < 2(1-|C|).$ Thus (49) and Lemma 2 lead to

$$\begin{aligned} 48 |\gamma _3|\le 12(1-\zeta _1^2)\left( 1+ |A|+ \frac{B^2}{4(1-|C|)}\right) =\varphi (\zeta _1), \end{aligned}$$

(52)

where

$$\begin{aligned} \varphi (x):= \frac{1}{3}(52+11x-28x^2-8x^3),\quad x\in [0,5/8). \end{aligned}$$

As easy to verify, for $x\in [0,5/8],$

$$\begin{aligned} \varphi (x) \le \varphi (x_0) = \frac{1}{162}(743+131\sqrt{262}) = 17.675433\dots , \end{aligned}$$

where $x_0 = (-14+\sqrt{262})/12 = 0.182201\ldots \in [0,5/8)$. Hence and by (52) it follows that

$$\begin{aligned} |\gamma _3| \le \frac{1}{7776}(743+131\sqrt{262}). \end{aligned}$$

(53)

Summarizing (48), (50), (51) and (53) show that the inequality (44) is true.

By tracking back the above proof, we see that equality in (44) holds when it is satisfied that

$$\begin{aligned} \zeta _1 = \frac{1}{12}(-14+\sqrt{262}), \quad \zeta _3=1 \end{aligned}$$

(54)

and

$$\begin{aligned} | A +B\zeta _2 +C\zeta _2^2| + 1 - |\zeta _2|^2 = 1 +|A| + \frac{B^2}{4(1-|C|)}, \end{aligned}$$

(55)

where

$$\begin{aligned} A = \frac{ 9526 -1601\sqrt{262} }{ 35604 }, \quad B = -\frac{1}{18}(8-\sqrt{262}), \quad C =\frac{1}{12}(14-\sqrt{262}). \end{aligned}$$

Indeed we can check that $\zeta _2$ defined by

$$\begin{aligned} \zeta _2 = \frac{ 4924-269\sqrt{262} \pm 2\mathrm {i}\sqrt{ 57399872-3438382\sqrt{262} } }{ 11946-555\sqrt{262} } \end{aligned}$$

(56)

satisfies the equation (55).

By Lemma 1 a function p of the form (14) with $\zeta _i$ ($i\in \{1,2,3\}$) given by (54) and (56) belongs to $\mathcal P.$ Note that $|\zeta _2|=0.912\dots$ Thus the corresponding function function

$$\begin{aligned} f(z) = \int _{0}^{z} \frac{p(t)}{1-t+t^2} \mathrm {d}t,\quad z\in {\mathbb {D}}, \end{aligned}$$

belongs to $\mathcal F_3.$ Substituting such chosen $\zeta _1,\zeta _2$ and $\zeta _3$ into (47) we get equality in (44). This ends the proof of the theorem. $\square$

3.4 The class ${\mathcal F}_4$

Recall that $f\in \mathcal F_4$ if $f\in \mathcal A$ and

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

Theorem 5

If $f \in {\mathcal F}_4$ is of the form (1) with $a_2 \in {\mathbb {R}}$, then

$$\begin{aligned} |\gamma _3| \le \frac{1}{243}(28+19\sqrt{19})=0.456045\dots \end{aligned}$$

(57)

The inequality is sharp with the extremal function

$$\begin{aligned} f(z)= \int _{0}^{z} \frac{p(t)}{(1-t)^2} \mathrm {d}t,\quad z\in {\mathbb {D}}, \end{aligned}$$

(58)

where

$$\begin{aligned} p(z) = \frac{ (1+z)\left( 9+(14-4\sqrt{19})z+9z^2\right) }{ (1-z)\left( 9+2(1+\sqrt{19})z + 9z^2 \right) },\quad z\in {\mathbb {D}}. \end{aligned}$$

(59)

Proof

Let $f \in {\mathcal F}_4$ be of the form (1) with $a_2 \in {\mathbb {R}}$. Then there exists $p \in {\mathcal P}$ of the form (8) such that

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

(60)

Putting the series (1) and (8) into (60) by equating the coefficients we get

$$\begin{aligned} \begin{aligned}&a_2=\frac{1}{2}(2+c_1), \quad a_3 = \frac{1}{3}(3+2c_1+c_2),\\&a_4 = \frac{1}{4}(4+3c_1+2c_2+c_3). \end{aligned} \end{aligned}$$

(61)

As in earlier consideration, $\zeta _1\in [-1,1]$ and from (61) it follows that $a_2\in [0,1].$

By (7) and (61) we get

$$\begin{aligned} 12\gamma _3=\frac{1}{4}(8+2c_1-2c_1^2+c_1^3-4c_1c_2+4c_2+6c_3). \end{aligned}$$

Using now (9)–(11) we have

$$\begin{aligned} \begin{aligned} 12\gamma _3=&\, \zeta _1^3 +\zeta _1 +2 +2(1-\zeta _1^2)\zeta _2 +2(1-\zeta _1^2)\zeta _1\zeta _2 \\&\, -3(1-\zeta _1^2)\zeta _1\zeta _2^2+3(1-\zeta _1^2)(1-|\zeta _2|^2)\zeta _3, \end{aligned} \end{aligned}$$

(62)

with $\zeta _1 \in [-1,1]$ and $\zeta _2$, $\zeta _3 \in \overline{{\mathbb {D}}}$.

Hence for $\zeta _1=1$ and $\zeta _1=-1$ we respectively have

$$\begin{aligned} \gamma _3 = 1/3,\quad \gamma _3 = 0. \end{aligned}$$

(63)

Let now $\zeta _1 \in (-1,1)$. Then

$$\begin{aligned} 12 |\gamma _3| \le 3(1-\zeta _1^2) \varPsi (A,B,C), \end{aligned}$$

(64)

where $\varPsi$ is defined by (23) with

$$\begin{aligned} A:= \frac{2-\zeta _1+\zeta _1^2}{3(1-\zeta _1)}, \quad B:= \frac{2}{3}(1+\zeta _1), \quad C:= -\zeta _1. \end{aligned}$$

For $\zeta _1 \in (-1,0]$ it holds $AC\ge 0$ and $|B|<2(1-|C|).$ Thus by (64) and Lemma 2 we have

$$\begin{aligned} \begin{aligned} |\gamma _3|&\le \frac{1}{4}(1-\zeta _1^2)\left( 1+ |A|+ \frac{B^2}{4(1-|C|)} \right) \\&= \frac{1}{18}(8+2\zeta _1-5\zeta _1^2+\zeta _1^3) \le \frac{4}{9}. \end{aligned} \end{aligned}$$

(65)

For $\zeta _1\in (0,1)$ it can be easily checked that

$$\begin{aligned} AC<0, \quad B^2< -4AC(C^{-2}-1), \quad B^2 < 4(1+|C|)^2. \end{aligned}$$

Therefore by (64) and Lemma 2 we get

$$\begin{aligned} 12 |\gamma _3| \le 3(1-\zeta _1^2) \left( 1+ |A|+ \frac{B^2}{4(1+|C|)} \right) = \varphi (\zeta _1), \end{aligned}$$

(66)

where

$$\begin{aligned} \varphi (x) := \frac{2}{3}(8+2x-5x^2+x^3),\quad x\in (0,1). \end{aligned}$$

As easy to verify

$$\begin{aligned} \varphi (x) \le \varphi (x_0) = \frac{4}{81}(28+19\sqrt{19}),\quad x\in (0,1), \end{aligned}$$

where $x_0:=(5-\sqrt{19})/3$. Hence and by (66) it follows that

$$\begin{aligned} |\gamma _3| \le \frac{1}{243}(28+19\sqrt{19}). \end{aligned}$$

(67)

Summarizing, (63), (65) and (67) show that the inequality (57) is true.

A simlar method used for the proof of Theorem 2, the equality in (57) when it is satisfied that

$$\begin{aligned} \zeta _1=\frac{1}{3}(5-\sqrt{19}), \quad \zeta _2=\frac{1}{3}, \quad \zeta _3=1. \end{aligned}$$

(68)

By Lemma 1 a function p of the form (14) with $\zeta _i$ ($i\in \{1,2,3\}$) given by (68), i.e., the function (59) belongs to $\mathcal P.$ Thus the function (58) belongs to $\mathcal F_4.$ Substituting (68) into (62) we get equality in (57). This ends the proof of the theorem. $\square$